Statics index frame bottomSTATICS & STRENGTH OF MATERIAL [UW-Stout - Physics 372-321] Welcome Email Instructor Statics Web-Syllabus Student Grades Class Room Schedule Spring 08 Class Room Syllabus Spring 08 This page has been accessed Hit Counter times since 1/20/98 TABLE OF CONTENTS Topic 1: Statics I - Principals Topic 2: Statics II - Applications Topic 3: Stress, Strain & Hooke's Law Topic 4: Beams Topic 5: Torsion, Rivets & Welds Topic 6: Columns & Pressure Vessels Topic 7: Friction (may be omitted due to time considerations) Topic 8: Special Topics I Reference: Tables and Links Reference: Search file:///Z|/www/Statstr/indexfbt.htm (1 of 5) [5/8/2009 4:39:30 PM] Statics index frame bottom Physics Home Page Top Of Page Expanded Table of Contents Topic 1: Statics I - Principals 1.1 Algebra/Trigonometry/Vectors Basic Trigonometric Review Problem Assignment - Trigonometry Basic Vector Review Problem Assignment -Vectors 1.2 Translational Equilibrium Concurrent Forces - Example 1 Concurrent Forces - Example 2 Problem Assignment - Coplanar 1.3 Rotational Equilibrium Subtopic 1.31 Torque Torque - Example 1 Torque - Example 2 Torque - Example 3 Problem Assignment - Torque Problem Assignment 2 - simply supported beams 1.4 Statics I - Topic Sample Exam Topic 2: Statics II - Applications 2.1 Frames (non-truss, rigid body structures) Frames - Example 1 Frames - Example 2 Frames - Example 3 Frames - Example 4 Additional Examples: #5, #6, #7, #8, #9, #10 [Previous test problems] Problem Assignment - Frames 1 (Required) Problem Assignment - Frames 2 (Supplemental - should attempt) Problem Assignment - Frames 3 (Supplemental - more difficult) 2.2 Trusses Trusses - Example 1 Trusses - Example 2 Trusses - Example 3 Additional Examples: #5, #6, #7, #8, #9, #10 [Previous test problems] Problem Assignment - Trusses 1 (Required) Problem Assignment - Trusses 2 (Supplemental) Problem Assignment - Trusses 3 (Supplemental) 2.3 Statics II - Sample Exam Topic 3: Stress, Strain & Hooke's Law 3.1 Stress, Strain, Hooke's Law - I file:///Z|/www/Statstr/indexfbt.htm (2 of 5) [5/8/2009 4:39:30 PM] Statics index frame bottom 3.2 Stress, Strain, Hooke's Law - II 3.2a Statically Determinate - Example 1 3.2b Statically Determinate - Example 2 3.2c Statically Determinate - Example 3 Additional Examples: #4, #5, #6, #7, #8, #9 [Previous test problems] 3.3 Statically Indeterminate Structures 3.3a Statically Indeterminate - Example 1 3.3b Statically Indeterminate - Example 2 3.4 Shear Stress & Strain 3.4a Shear Stress & Strain - Example 1 3.4b Shear Stress & Strain - Example 2 3.5 Problem Assignments - Stress/Strain Determinate 3.5a Problem Assignment 1 - Determinate [required] 3.5b Problem Assignment 2 - Determinate [supplemental] 3.5c Problem Assignment 3 - Determinate [required] 3.6 Problem Assignment -Stress/Strain Indeterminate [required] 3.7 Stress, Strain & Hooke's Law - Topic Examination 3.8 Thermal Stress, Strain & Deformation I 3.81 Thermal Stress, Strain & Deformation II 3.82 Mixed Mechanical/Thermal Examples 3.82a Mixed Mechanical/Thermal - Example 1 3.82b Mixed Mechanical/Thermal - Example 2 3.82c Mixed Mechanical/Thermal - Example 3 Additional Examples: #4, #5, #6, #7, #8, #9 [Previous test problems] 3.83 Thermal Stress,Strain & Deformation - Assignment Problems [required] 3.84 Thermal Stress, Strain & Deformation - Topic Examination Topic 4: Beams 4.1 Shear Forces and Bending Moments I 4.2 Shear Forces and Bending Moments II 4.3 Shear Forces and Bending Moments Examples 4.3a Simply Supported Beam - Example 1 4.3b Simply Supported Beam - Example 2 4.3c Cantilever Beam - Example 3 Additional examples: (simply supported) #4, #5, #6, #7, #8, #9 [Previous test problems] Additional examples: (cantilever) #4, #5, #6, #7, #8, #9 [Previous test problems] 4.4a1 Shear Force/Bending Moment Problems - Simply Supported Beams 1 [supplementary] 4.4a2 Shear Force/Bending Moment Problems - Simply Supported Beams 2 [required] 4.4b1 Shear Force/Bending Moment Problems - Cantilevered Beams 1 [supplementary] 4.4b2 Shear Force/Bending Moment Problems - Cantilevered Beams 2 [required] 4.5 Shear Forces and Bending Moments - Topic Examination 4.6 Calculus Review (Brief) 4.7 Beams - Bending Stress 4.7a Centroids and the Moment of Inertia 4.7b Bending Stress - Example 1 file:///Z|/www/Statstr/indexfbt.htm (3 of 5) [5/8/2009 4:39:30 PM] Statics index frame bottom 4.8 Beams - Bending Stress (cont) 4.8a Bending Stress - Example 2 4.8b Bending Stress - Example 3 4.9 Beams - Horizontal Shear Stress 4.9a Horizontal Shear Stress - Example 1 4.9b Horizontal Shear Stress - Example 2 Additional Bending & Shear Stress Examples: #1, #2, #3, #4, #5, #6 [Previous test problems] 4.10 Beams - Beam Selection 4.10a Beam Selection - Example 1 Additional Beam Selection Examples: #2, #3, #4, #5, #6, #7 [Previous test problems] 4.11a Beams - Problem Assignment 1 - Inertia, Moment, Stress [required] 4.11b Beams - Problem Assignment 2 - Bending Stress [supplementary] 4.11c Beams - Problem Assignment 3 - Horizontal Shear Stress [supplementary] 4.11d Beams - Problem Assignment 4 - Bending & Shear Stress [required] 4.12a Beams - Problem Assignment 5 - Beam Selection [required] 4.12b Beams - Problem Assignment 6 - Beam Deflection [supplementary] 4.13 Beams -Topic Examination Topic 5: Torsion, Rivets & Welds 5.1 Torsion: Transverse Shear Stress 5.1a Shear Stress - Example 1 5.1b Shear Stress - Example 2 5.2 Torsion: Deformation - Angle of Twist 5.3 Torsion: Power Transmission 5.3a Power Transmission - Example 1 5.3b Power Transmission - Example 2 5.3c Power Transmission - Example 3 5.3d Compound Shaft - Example 4 Additional General Torsion Examples: #1, #2, #3, #4, #5, #6 [Previous test problems] 5.4a Torsion - Problem Assignment 1 (required) 5.4b Torsion - Problem Assignment 2 (supplementary) 5.4c Torsion - Problem Assignment 3 (supplementary) 5.4d Torsion - Problem Assignment 4 (required) 5.5 Rivets & Welds - Riveted Joints 5.5a Riveted Joints - Example 1 5.5b Riveted Joints - Example 2 5.6 Rivets & Welds - Riveted Joint Selection 5.6a Riveted Joint Selection - Example1 Additional Rivet Examples: #2, #3, #4, #5, #6, #7 [Previous test problems] 5.7 Rivets & Welds - Welded Joints 5.7a Welded Joints - Example 1 Additional Weld Examples: #2, #3, #4, #5 [Previous test problems] 5.8a Rivets & Welds - Problem Assignment 1 (required) 5.8b Rivets & Welds - Problem Assignment 2 (supplementary) file:///Z|/www/Statstr/indexfbt.htm (4 of 5) [5/8/2009 4:39:30 PM] Statics index frame bottom 5.9 Torsion, Rivets & Welds - Topic Examination Topic 6: Columns & Pressure Vessels 6.1 Columns & Buckling - I 6.1a Euler Buckling - Example 1 6.1b Euler Buckling - Example 2 6.2 Columns & Buckling - II 6.2a Secant Formula - Example 1 6.2b Structural Steel - Example 2 6.2c Structural Steel Column Selection - Example 3 6.2d Aluminum Columns - Example 4 6.2e Wood Columns - Example 5 6.3 Columns & Buckling - Problem Assignment (required) 6.4 Columns & Buckling - Topic Examination 6.5 Pressure Vessels - Thin Wall Pressure Vessels 6.6a Pressure Vessels - Problem Assignment 1 (required) 6.6b Pressure Vessels - Problem Assignment 2 (supplementary) Topic 7: Friction Friction Problems 1 Friction Problems 2 Friction Problems 3 Friction Problems 4 Friction Problems 5 Friction Problems 6 Topic 8: Special Topics I 8.1 Special Topics I: Combined Stress 8.1a Combined Stress - Example 1 8.1b Combined Stress - Example 2 8.2 Special Topics I: Stresses on Inclined Planes 8.3 Special Topics I: Non-Axial Loads 8.4 Special Topics I: Principal Stresses 8.4a Principal Stresses - Example 1 8.5 Special Topics I: Mohr's Circle 8.6 Special Topics I: Problem Assignment 1 (required) 8.7 Special Topics I: Problem Assignment 2 (required) 8.8 Special Topics I: Topic Examination Top Of Page file:///Z|/www/Statstr/indexfbt.htm (5 of 5) [5/8/2009 4:39:30 PM] Topic 1.1: Statics I - Principles Topic 1: Statics I - Principles ● ● Topic 1: Statics I - Principles 1.1 Algebra/Trigonometry/Vectors r Basic Trigonometric Review r Problem Assignment - Trigonometry r Basic Vector Review r Problem Assignment -Vectors 1.2 Translational Equilibrium r Concurrent Forces - Example 1 r Concurrent Forces - Example 2 r Problem Assignment - Coplanar 1.3 Rotational Equilibrium r Subtopic 1.31 Torque r Torque - Example 1 r Torque - Example 2 r Torque - Example 3 r Problem Assignment - Torque r Problem Assignment 2 - simply supported beams 1.3a Statics Summary Sheet 1.3b Additional Important Examples 1.4 Statics I - Topic Sample Exam ● ● ● ● ● Topic 1.1: Algebra/Trigonometry/Vectors 1.1a. Algebra/Trigonometry: It is anticipated that the student will have introductory level college algebra skills, including the ability to solve simple algebraic equations, quadratics, and simultaneous equations. An appropriate level of skill would be one which a student would be expected to acquire in a one-year College Algebra sequence. Minimal trigonometric skills should include being able to find sides and angles in both right and nonright triangle. See Basic Trigonometric Review. 1.1b. Vectors: file:///Z|/www/Statstr/statics/StatI/stat1.htm (1 of 2) [5/8/2009 4:39:31 PM] Topic 1.1: Statics I - Principles The student should have a working knowledge of vectors (as, of course, forces are vectors and we will be summing forces in many of our problems.) A working knowledge of vectors would include vector addition, subtraction and resolving vectors into perpendicular components. We will be using the component method for vector addition. For a short review of this method see: Basic Vector Review. Also, Vector Review Problems are available. Return to: Topic 1: Statics I - Principles Table of contents or select: Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/stat1.htm (2 of 2) [5/8/2009 4:39:31 PM] Topic 1.1a Triangles Basic Trigonometric Review For Right Triangles: 1. Pythagorean Law: C2 = A2 + B2 2. Sine Ø = opposite side/hypotenuse = B/C 3. Cosine Ø = adjacent side/hypotenuse = A/C 4. Tangent Ø = opposite side/adjacent side = B/A For Non-Right Triangles: 1. Law of Cosines: C2 = A2 + B2 - 2AB cos c 2. Law of Sines: (A/sin a) = (B/sin b) = (C/sin c) Where A,B,C are length of the sides, and a, b, c are the corresponding angles opposite the sides. file:///Z|/www/Statstr/statics/StatI/triangle.htm (1 of 2) [5/8/2009 4:39:31 PM] Topic 1.1a Triangles Some general Trigonometric Identities are shown below. Return to: Topic 1.1: Algebra/Trigonometry/Vectors continue to: Problem Assignment - Trigonometry or select Topic 1: Statics I - Principles Table of Contents Statics & Strength of Materials - Course Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/triangle.htm (2 of 2) [5/8/2009 4:39:31 PM] Trig I Problems Statics & Strength of Materials Problem Assignment - Trigonometry 1. A 40 ft long ladder leaning against a wall makes an angle of 60° with the ground. Determine the vertical height to which the ladder will reach. (Answers at bottom of page.) 2. In the roof truss shown, the bottom chord members AD and DC have lengths of 18ft. and 36ft respectively. The height BD is 14 ft. Determine the lengths of the top chords AB and BC and find the angles at A and C. file:///Z|/www/Statstr/statics/StatI/Trig%20I.htm (1 of 3) [5/8/2009 4:39:32 PM] Trig I Problems 3. One side of a triangular lot is 150 ft. and the angle opposite this side is 55°. Another angle is 63°. Sketch the shape of the lot and determine how much fencing is needed to enclose it. 4. An Egyptian pyramid has a square base and symmetrical sloping faces. The inclination of a sloping face is 42° 08'. At a distance of 500 ft. from the base, on level ground the angle of inclination to the apex is 25° 15'. Find the vertical height(h), the slant height of the pyramid, and the width of the pyramid at its base. #5. Triangle ABC shown is a triangular tract of land. The one acre tract DEFG is to be subdivided. AE is 500ft and DC is 300 ft. Determine the lengths of DG and CB. [This is a harder problem.] file:///Z|/www/Statstr/statics/StatI/Trig%20I.htm (2 of 3) [5/8/2009 4:39:32 PM] Trig I Problems Some Answers 1. H =34.64'; 2. AB=22.8', BC=38.6', A = 37.9o, C = 21.25o; 3. L=474.8'; 4. h=411.3', slant = 554.6', width=744.2' 5. ED=68.4', DG=677.4', CB=1035' Return to: Topic 1.1a: Algebra/Trigonometry/Vectors or select Topic 1: Statics I - Principles Table of Contents Statics & Strength of Materials - Course Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/Trig%20I.htm (3 of 3) [5/8/2009 4:39:32 PM] Topic 1.1b-Vectors Basic Vector Review 1. Definitions: ● ● Scalar: Any quantity possessing magnitude (size) only, such as mass, volume, temperature Vector: Any quantity possessing both magnitude and direction, such as force, velocity, momentum 2. Vector Addition: Vector addition may be done several ways including, Graphical Method, Trigonometric Method, and Component Method. We will be reviewing only the Component Method, as that is the method which will be used in the course. Other methods are detailed in your textbook. 3. Vector Addition - Component Method: (2-dimensional) The component method will follow the procedure shown below: ● 1. Choose an origin, sketch a coordinate system, and draw the vectors to be added (or summed). 2. Break (resolve) each vector into it's "x" and "y" components, using the following relationships: r A = A cosine Ø, and, A = A sine Ø, where A is the vector, and Ø is the vector's x y angle. ● ● 3. Sum all the x-components and all the y-components obtaining a net resultant Rx, and Ry vectors. r R = A + B + C + . . ., &, R = A + B + C + . . . x x x x y y y y ● 4. Recombine Rx and Ry to obtain the final resultant vector (magnitude and direction) using See Example below Example - Vector Addition file:///Z|/www/Statstr/statics/StatI/vectors.htm (1 of 4) [5/8/2009 4:39:32 PM] Topic 1.1b-Vectors Three ropes are tied to a small metal ring. At the end of each rope three students are pulling, each trying to move the ring in their direction. If we look down from above the students, the forces and directions they are applying the forces are as follows: (See diagram to the right) Find the net (resultant) force (magnitude and direction) on the ring due to the three applied forces. Choose origin, sketch coordinate system and vectors (done above) Resolve vectors into x & y components (See Diagram) Ax = 30 lb cos 37o = + 24.0 lbs ; Ay = 30 lb sin 37o = + 18.1 lb Bx = 50 lb cos135o = - 35.4 lbs ; By = 50 lb sin135o = + 35.4 lb Cx = 80 lb cos240o = - 40.0 lbs ; Cy = 80 lb sin240o = - 69.3 lb file:///Z|/www/Statstr/statics/StatI/vectors.htm (2 of 4) [5/8/2009 4:39:32 PM] 8 lbs 'Recombine' (add) Rx and Ry to determine final resultant vector.4 lbs .Topic 1.htm (3 of 4) [5/8/2009 4:39:32 PM] . Rx = 24.4 lbs .1b-Vectors Sum x & y components to find resultant Rx and Ry forces.4 lbs Ry =18.1 lbs + 35.0 lbs = -51.0 lbs .40.35. file:///Z|/www/Statstr/statics/StatI/vectors.3 lbs = -15.69. Return to: Topic 1.Course Table of Contents Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/vectors.8 pounds acting at an angle of 197.Vectors continue to: Vector Review Problems or select: Topic 1: Statics I .Topic 1.Principles Table of Contents Strength of Materials .1b-Vectors Thus the resultant force on the ring is 53.1b: Statics I .1 degrees.htm (4 of 4) [5/8/2009 4:39:32 PM] . Find the x and y components of the following vectors. at 0o b) 27 lb. a) 15 lb. d)x = -13ft = -24ft 3. a) x = 5 feet y = 17 feet. c)x = 3 lb.Vector review prob Statics & Strength of Materials Problems Assignment .b. y = 3 lb..Vector Review Problems 1. Solve the following vector problems by adding vectors A and B to find the resultant vector for problems a. and c. at 63o c) 15 ft at 237o d) 18 ft at -39o 2.. Find the vectors (magnitude and direction) that have the following components. file:///Z|/www/Statstr/statics/StatI/vectpb11c. specifically stating each component. b) x = -8 lb. y = -8 lb.htm (1 of 3) [5/8/2009 4:39:32 PM] . b. (c). B.Vector review prob (a). 5. and C and find the resultant vector in diagrams a. 4. (b). and d. (d). Using diagram (a) above. (b).htm (2 of 3) [5/8/2009 4:39:32 PM] . find the following resultants: a) R = 2 * A .B + 3 * C b) R = -3 * A + 2 * B .c.C file:///Z|/www/Statstr/statics/StatI/vectpb11c. (a). Add vectors A. (c). .3 lb at 103.Course Table of Contents Statcis & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/vectpb11c. R = 15. Rx = -4. R = 54.4o 4c.Principles Table of Contents Statics & Strength of Materials .11 m. Rx = -10. at 131o.99 m..3o 5a..Vectors or select: Topic 1: Statics I ..57 lb.Vector review prob Some Answers: 3a. Ry = . R = 4..3o 4a.16 ft/s.htm (3 of 3) [5/8/2009 4:39:32 PM] . 5b.95 lb. R = 164..83 lb.45 m. at 354o.62 m...7 n at 280..82 m. Rx = 12. Ry = 52. R = 16..61 m.31 ft/s. Rx = 16.5o 4b.51 ft/s.57 n at 263o 4d.54 m. Rx = -12. Ry = 12.4o 3b.16 n.11n. Return to: Topic 1. Ry = -7. Ry = -9.07 m.1b: Statics I . R = 9. R = 16.. at 84. Ry = 1.1.11 n.. Rx = -15.5 ft/s.39 lb. Rx = 14. at 160.13 lb. Rx = -1..74 n at 205o 3c. Ry = -35. Ry = 164.38 ft/s at 287. Rx = 6. R = 35. R = 41..46 n. Ry = -39.12 m... This is important as most problems in statics and strength of materials are not the kind of problem in which we can easily see the answer. We will be normally looking at structures which are at rest.Concurrent Forces. . forces in the x-direction must sum to zero.Equilb Topic 1. Mathematically this may be expressed as: or. Select: Example 2. The Translational Equilibrium condition states that for an object or a structure to be in translational equilibrium (which means that the structure as a whole will not experience linear acceleration) the vector sum of all the external forces acting on the structure must be zero. we will be able to apply the Conditions of Equilibrium to help us solve for the force in and on the structures. and. the forces in the y-direction must sum to zero. but rather we must relay on our problem solving techniques. To see the application of the first condition of equilibrium and also the application of a standard problem solving technique. for translational equilibrium in the z-direction. That is.Principles Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/stat12. for translational equilibrium in the y-direction. and.htm [5/8/2009 4:39:33 PM] . using a well defined methodology. There are two general equilibrium conditions: Translational Equilibrium. that is structures that are at rest or in uniform.Topic 1. (non-accelerated) motion. and Rotational Equilibrium. the forces in the z-direction must sum to zero. To determine forces on and in structures we will proceed carefully. For static equilibrium problems.2 Trans. in 3-dimensions: .2 .Concurrent Forces Select: Topic 1: Statics I . Select: Example 1. for translational equilibrium in the x-direction. For these structures we will be interested in determining the forces (loads and support reactions) acting on the structure and forces acting within members of the structure (internal forces).Translational Equilibrium The topic of statics deal with objects or structures which are in equilibrium. let's look carefully at introductory examples. the forces are said to be Concurrent. (The effect of the pulleys is just to change the direction of the force. In later problems this will not necessarily be the case.Topic 1. When the problem involves forces in two dimensions only. (Notice in this problem. the directions the support forces act are easy to determine. as follows: 1. and the vector representing the load force will all intersect at one point.) To "Solve" this problem. When the force vectors all intersect at one point. Draw a Free Body Diagram (FBD) of the structure or a portion of the structure. that forces lie in the x-y plane only. The two cords each make an angle of 50o with the vertical.htm (1 of 3) [5/8/2009 4:39:33 PM] . Coplanar Forces In this relatively simple structure. This Free Body Diagram should include a coordinate system and vectors representing all the external forces (which include support forces and load forces) acting on the structure. and will be discussed later.Concurrent. that is to determine the weight supported by forces (tensions) in cable 1 and cable 2.) If we examine the first diagram for a moment we observe this problem may be classified as a problem involving Concurrent. The second diagram 2 is the Free file:///Z|/www/Statstr/statics/StatI/state12a.2a example1-coplanar Example 1 . Coplanar Forces. the forces are said to be Coplanar. just above the body. the vectors representing the two support forces in Cable 1 and Cable 2. Additionally. These forces should be labeled either with actual known values or symbols representing unknown forces. Determine the weight of the body. we will now follow a very specific procedure or technique. and cables can only be in tension. That is. that since the two supporting members are cables. we note that this is a two-dimensional problem. which run over pulleys (which we will assume are very low friction) and are attached to 100 lb. we have a weight supported by two cables. weights as shown in the diagram. it may be considered to not effect the value of the tensions in the ropes. remembering to put the correct sign with the force. -100 cos 40o + 100 cos 40o = 0 (Just as we would expect. 2.Topic 1. Notice for Cable 1. In the third diagram. We now place our forces into these equations.2a example1-coplanar Body Diagram of point just above the weight where with all forces come together. we must be replaced with their horizontal and vertical components. the x-forces balance each file:///Z|/www/Statstr/statics/StatI/state12a. it is negative in the equation. the components of Cable 1 and Cable 2 are shown. In this step we will now apply the actual equilibrium equations. Thus the forces Cable 1. and Cable 2. Since the problem is in two dimensions only (coplanar) we have the following two equilibrium conditions: The sum of the forces in the x direction.htm (2 of 3) [5/8/2009 4:39:33 PM] . that is. 3. and the sum of the forces in the y direction must be zero. Resolve (break) forces not in x or y direction into their x and y components. Apply the Equilibrium Conditions and solve for unknowns. that is if the force acts in the positive direction it is positive and if the force acts in the negative direction. the vectors representing the tensions in the cables were acting at angles with respect to the xaxis. Cable 2. they are not simply in the x or y direction. or. weight of body = 0 In this instance.Principles Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/state12a. 100 sin 40o + 100 sin 40o .htm (3 of 3) [5/8/2009 4:39:33 PM] .56 lb.Topic 1. and find: Weight of body = 128. Return to: Topic 1.2a example1-coplanar other.Translation Equilibrium or select: Topic 1: Statics I . it is very easy to solve for the weight of the body from the y-equation.) or.2: . we will now follow a very specific procedure or technique. which in turn are attached to walls. we note that this is a two-dimensional problem. Additionally. that forces lie in the x-y plane only. that is to determine the forces (tensions) in cable 1 and cable 2. When the problem involves forces in two dimensions only. Coplanar Forces. See Diagram 2). In later problems this will not necessarily be the case. the forces are said to be Concurrent. Draw a Free Body Diagram (FBD) of the structure or a portion of the structure.Concurrent.Topic 1. the vectors representing the two support forces in Cable 1 and Cable 2. and cables can only be in tension. When the force vectors all intersect at one point. This Free Body Diagram should include a coordinate system and vectors representing all the external forces (which include support forces and load forces) acting on the structure.) To "Solve" this problem. the directions the support forces act are easy to determine. Coplanar Forces In this relatively simple structure. These forces should be labeled either with actual known values or symbols representing unknown forces. That is. file:///Z|/www/Statstr/statics/StatI/state12b. as follows: 1. If we examine Diagram-1 for a moment we observe this problem may be classified as a problem involving Concurrent.2b example2-coplanar Example 2 . the forces are said to be Coplanar. Diagram 2 is the Free Body Diagram of point C with all forces acting on point C shown and labeled. and the vector representing the load force will all intersect at one point (Point C. we have a 500 lb. load supported by two cables. and will be discussed later. Let's say that we would like to determine the forces (tensions) in each cable. that since the two supporting members are cables. (Notice in this problem.htm (1 of 3) [5/8/2009 4:39:33 PM] . they are not simply in the x or y direction. Notice that we do not have to do this for the load force of 500 lb. Notice that T1. since it is already acting in the ydirection only.Topic 1. we must replace them with their horizontal and vertical components. In Diagram 3. that is. file:///Z|/www/Statstr/statics/StatI/state12b. Thus for the forces T1. T2 cos 30o) are equivalent to T1 and T2. T2 sin 30o. we remove T1 and T2 which are now represented by their components. and T2.2b example2-coplanar 2. the vectors representing the tensions in the cables are acting at angles with respect to the x-axis. T1 cos 53o. Since the components of T1 and T2 (T1 sin 53o. in the final diagram 1d. Resolve (break) forces not in x or y direction into their x and y components.htm (2 of 3) [5/8/2009 4:39:33 PM] . T2. the components of T1 and T2 are shown.. and the sum of the forces in the y direction must be zero.Translation Equilibrium or select: Topic 1: Statics I . T2 = 302 lb.2: . if the structure is to be in equilibrium. (T1 = T2 cos 30o/cos 53o). T1 sin53o + T2 sin30o . then the forces in the cable must be as found above.T2 cos30o = 0 or. T1 cos53o .500 lb= 0 Notice we have two equations and two unknowns (T1 and T2). if the cables. giving us only one equation and one unknown.. We now place our forces into these equations. we must be sure they will support loads at least equal to the tensions we found.2b example2-coplanar 3. it is negative in the equation. are to support the 500 lb. respectively. load.. and substitute the expression for T1 into the second equation [(T2 cos30o/cos53o)sin53o + T2 sin30o .Topic 1.500 lb= 0]. and therefore can solve for the unknowns. that is if the force acts in the positive direction it is positive and if the force acts in the negative direction. In this step we will now apply the actual equilibrium equations. remembering to put the correct sign with the force. and 302 lb.htm (3 of 3) [5/8/2009 4:39:33 PM] . or. There are several ways to solve these two 'simultaneous' equations. Return to: Topic 1.Principles Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/state12b. On solving the equations for T1 and T2 we obtain: T1 = 436 lb. Since the problem is in two dimensions only (coplanar) we have the following two equilibrium conditions: The sum of the forces in the x direction. acting at the angles given. Apply the Equilibrium Conditions and solve for unknowns. Thus. So when we go to purchase cables for our structure. 436 lb. We could solve the first equation for T1 in terms of T2. Calculate the horizontal force F that should be applied to the 200lb weight shown in order that the cable AB be inclined at an angle of 30° with the vertical. 1.Coplanar probs Statics & Strength of Materials Problem Assignment .4o) 2. Calculate the force in cable AB and the angle q for the support system shown.Coplanar Draw a free body diagram and a force diagram as a part of the solution for each problem.) file:///Z|/www/Statstr/statics/StatI/coplnp.htm (1 of 4) [5/8/2009 4:39:33 PM] .2 lb. 63. (115 lb. (447. All problems are coplanar with concurrent forces. Calculate the force in each cable for the suspended weight shown. (439 lb. Two forces of 100 lb.Coplanar probs 3.htm (2 of 4) [5/8/2009 4:39:33 PM] . each act on a body at an angle of 120o with each other.) file:///Z|/www/Statstr/statics/StatI/coplnp..) 4. What is the weight of the body the two forces are supporting? (100 lb. 538 lb. Coplanar probs 5. What weight suspended from the middle of the wire will break it? (96.) file:///Z|/www/Statstr/statics/StatI/coplnp.htm (3 of 4) [5/8/2009 4:39:33 PM] .8 lb. The ends are fastened to two points 21 inches apart on the same level. A wire 24 inches long will stand a straight pull of 100 lb. Principles Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/coplnp. The cylinder weighs 150 lbs. 102.htm (4 of 4) [5/8/2009 4:39:33 PM] .8 lb. (79.) Select: Topic 1: Statics I ..Coplanar probs 6. Calculate the reactions of the two smooth inclined planes against the cylinder shown.6 lb. The second condition for equilibrium states that if we are to have rotational equilibrium. We can see the need for this second condition if we look at the diagram 1. it rotates about the center of mass of the bar.Rotational Equilibrium The second condition for equilibrium is rotational equilibrium. in two dimensions.3 . even though it rotates. and the center of mass of the bar will not move.htm (1 of 2) [5/8/2009 4:39:34 PM] . . if we apply the 1st condition of equilibrium and sum the forces in the y-direction. the sum of the torque acting on a structure with equilibrium condition will be: respect to any point selected must equal zero.3a.Topic 1. . the sum of the Torque acting on the structure must be zero.Equil Topic 1. That is.Principles Table of Contents or select: file:///Z|/www/Statstr/statics/StatI/stat13. Select Torque Example 1. Now let's take a moment to go very slowly through an introductory problem involving both conditions of equilibrium. (For an overview and review of Torque. Where. In this diagram. we obtain zero. That is. Select Torque Example 3 Return to: Topic 1: Statics I . please select: Subtopic 1. one can only have counterclockwise(+) or clockwise(-) torque.31 Torque) The 2nd condition for equilibrium may be written: or in three dimensions: . our normal rotational . and will experience rotational motion (and rotational acceleration). However. and applying our statics problem solving techniques. Since. Select Torque Example 2. Torque (or Moments) is normally covered in the first semester of a General College Physics course. (+100 lb. = 0).3 Rot. most of our problems will be dealing with structures in two dimensions. Please notice that the object actually is in translational equilibrium. This would indicate that the object is in translational equilibrium.100 lb. we almost instinctively recognize that the object certain will not remain at rest. Equil Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/stat13.3 Rot.htm (2 of 2) [5/8/2009 4:39:34 PM] .Topic 1. where theta is the angle between the distance vector and force vector. In this case it could be expressed this way.) If this all seems confusing at this point.in this case. is shown with 100 lb. Example 1. out of the screen. don't panic. The direction of the torque vector is perpendicular to both the distance and force vectors. as shown in the following example(s). There are a variety of ways of expressing the right hand rule. given a distance vector R (from origin. This cross product is a vector and has a resultant magnitude (size) of . what this cross product effectively does is to take the component of the force vector perpendicular to the distance vector and then multiply it by the distance vector. the torque is defined as: . and is given by the right hand rule. (+ z direction. We will use a somewhat less formal approach to determining torque acting on structures.Topic 1.31 . As diagram 2 shows. In Diagram 3. force acting upward on the beam. force. pinned at point P.htm (1 of 4) [5/8/2009 4:39:34 PM] . Torque is always file:///Z|/www/Statstr/statics/StatI/stat131. pinned at the origin. a torque (also know as the Moment of Force) is a vector cross product. our thumb will point the direction or the torque vector . If vector R were a rod. a 10 foot long beam.Torque Formally. If we now curl the fingers of our right hand in the same way as the rotation. as shown in diagram 1) and a force vector F. The very first thing we must do when determining torque is to PICK A POINT.31 Torque Subtopic 1. that is. notice that force component would cause R to start to rotate counterclockwise about the origin. We wish to determine the torque on the beam due to the 100 lb. force to halfway down the beam toward point P. we mean.5 ft. at the left end of the beam.). force with respect to point P now becomes: Torque = F x d = 100 lb. In this example we will use point P.Topic 1.. We next calculate the torque caused by the force by Torque(about P) = Force x perpendicular distance. force three quarters length down the beam toward point P. In diagram 3c. x 5 ft. force with respect to point P now becomes: Torque = F x d = 100 lb. that there is only one line we can draw which begins at point P and intersects the line of action (direction) of the force at 90 degrees. so the torque due to the 100 lb.31 Torque calculated with respect to a point (or axis). In diagram 3a. we have moved the 100 lb. That is. In diagram 3b. and the torque due to the 100 lb. and so before we can proceed we first choose a point (sometimes called the pivot point) to calculate torque about. file:///Z|/www/Statstr/statics/StatI/stat131. = 250 ft-lb. and the torque due to the 100 lb. and so the perpendicular distance is now 2. = 500 ft-lb. and the length of this line is the perpendicular distance. force with respect to point P will be the 100 lb.htm (2 of 4) [5/8/2009 4:39:34 PM] .. force with respect to point P will be: Torque = F x d = 100 lb. x 10 ft = 1000 ft-lb. the torque caused by the 100 lb. force times the perpendicular distance from point P to the line of action of force (d = 10 ft. it is the thin dark line with 'd' below it. and its value is 10 ft. and so the perpendicular distance is changed to 5 ft. we have moved the 100 lb.. By perpendicular distance. x 2.5 ft. it does. force does not push on point P. x 6 ft = 600 ft-lb. force acts directly below point P. This is the perpendicular distance 'd' from the pivot point to the line of the force. force with respect to point P. We can find the value of the perpendicular distance d by noting that the force. To see what happens if we do not have a force acting a 'nice' direction as in example 1 above. (Positive torque. We will first do this directly from our less formal definition of torque: Torque = Force (times) perpendicular distance. force would start the beam rotating in the counterclockwise direction.htm (3 of 4) [5/8/2009 4:39:34 PM] .Topic 1. however it produces no torque (no rotation) with respect to point P. and beam length form a right triangle with the beam length as the hypotenuse. and then we have started at point P and drawn a line from point P which intersects the force line at 90o. Once again we wish to calculate the torque produced by the 100 lb.) file:///Z|/www/Statstr/statics/StatI/stat131. perpendicular distance d. and therefore there is zero torque due to the 100 lb. Example 2: In this example we use the same beam as above but now the 100 lb.31 Torque Finally in diagram 3d. Therefore from trigonometry. force acts at an angle of 37o with respect to the horizontal (below) as shown in diagram 4. since if the beam were actually pinned at point P. d = 10 sin 37o = 6 ft. the 100 lb. we have extended the line of action of the force. we will do a second example. and the torque will be: Torque = 100 lb. the 100 lb. and so the perpendicular distance is 0 ft. force with respect to point P. Notice in Diagram 4a. This does not mean the 100 lb. Summing the two torque we get a total of 600 ft-lb. While we have determined torque here in only 2-dimensions.htm (4 of 4) [5/8/2009 4:39:34 PM] . the procedure is the same 3-dimensions. Then the resultant torque about point P will be the sum of the torque produced by each component force. while the x-component force results in Torque = 100 lb. certainly in problems with a number of forces. and we have shown this process in Diagram 4b. = 600 ft-lb. just as we found in the first method. it actually is more effective in problems. cos 37o x 0 ft. resulting in zero torque. and thus its perpendicular distance is zero. Return to Topic 1. with the exception that there are 3-axis of rotation possible.. we will not go into calculating torque in 3dimensions at this point. This is the process we will normally use.. = 0 ft-lb since its line of action passes through point P. We first found the x and y-components of the 100 lb. force as shown in the diagram. sin 37o x 10 ft. Notice the ycomponent force results in Torque = 100 lb. As we continue with examples of statics problems.Topic 1.3 Statics . to first break the force into it's equivalent x and y-components before calculating torque. Since the type of problems we will consider are mainly 2-dimensional.Rotational Equilibrium or select: Topic 1: Statics I -Principles Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/stat131.31 Torque While this method of calculating the torque is fine and does work. the applications of torque in solving problems will be shown in complete detail. Topic 1. We proceed with this problem with a well establish statics technique. and labeled these forces TA and TB. One painter weighs 160 lb.3a Torque Example 1 Example 1 . what is the force (tension) in each cable when the painters are standing in the positions shown. Notice that for a uniform beam or bar. The question is. file:///Z|/www/Statstr/statics/StatI/state13a. all the forces are already acting in the y-direction only. Draw a Free Body Diagram (FBD) showing and labeling all external forces acting on the structure. the painters begin wondering what force (tension) is in each cable. 1. long. two painters are standing on a 300 lb. The scaffolding is supported by two cables. the beam weight may be considered to act at the center (of mass) of the bar. and the second painter weighs 140 lb. one at each end. as far as equilibrium conditions are concerned. we have shown the forces in the cables supporting the beam as arrows upward. Notice in the diagram to the right. As they paint.htm (1 of 3) [5/8/2009 4:39:35 PM] . 2. In this example.Torque review problems In this example. Resolve all forces into x and y-components. and including a coordinate system. so nothing more needs to be done. scaffolding (beam) which is 12 ft. as we always calculate torque with respect to a point (or axis). Apply the (2-dimensional) equilibrium conditions: (No external x-forces acting on the structure. if we look at the 160 lb. as there are two unknowns (TA and TE) and only one equation so far.Topic 1.160 lb . Before we can sum torque we must first PICK A POINT. select Torque. that force does not produce a torque with respect to the point (since the perpendicular distance is zero). as shown or -TA(12) +300 lb(6ft) + 160 lb(3 ft) + 140 lb(1 ft) = 0 Where we determined each term by looking at the forces acting on the beam one by one. the 160 lb.3a Torque Example 1 3. if a force acts through a point. weight of painter one (and ignore the other forces). weight would cause the beam to start rotating counterclockwise (+).140 lb + TE = 0 Here we have summed the external y-forces. Thus summing torque about point E will result in an equation with only one unknown.sum of torque must equal zero. The sign of the torque is determined by considering which way the torque would cause the beam to rotate. however some points result in an easier equation(s) to solve. For a review of torque. if the beam were pinned at point E.) TA . and calculating the torque produced by each force with respect to the chosen point E from: Torque = Force x perpendicular distance (from the point E to the line of action of the force). We can't solve this equation yet. if we sum torque with respect to point E. and. As an example. Any point on (or off) the structure will work. if the beam were actually pinned at the chosen point. we notice that unknown force TE acts through point E.300 lb . That is. We obtain our second equation from the 2nd condition of equilibrium . file:///Z|/www/Statstr/statics/StatI/state13a. so this equation gives no information.htm (2 of 3) [5/8/2009 4:39:35 PM] . As an additional thought problem.67 lb (round off to 202 lb).3a Torque Example 1 It is important to note that the sign of the torque depends on the direction of the rotation it would produce with respect to the chosen pivot point.) Return to: Topic 1.3 Rotational Equilibrium or select: Topic 1: Statics I .Principles Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/state13a.htm (3 of 3) [5/8/2009 4:39:35 PM] .300 lb . We have now found the forces in the cables when the painters are in the positions shown in the problem. We then place the value for TA back into the y-force equation and find the value of TE = 398 lb. That is. not on the direction of the force. one might consider the question of what is the minimum cable strength required so that the painters could move anywhere on the scaffolding safely. but it produces a positive torque with respect to point E in the sum of torque equation. finding TA = 201.Topic 1. TA .160 lb -140 lb + TE = 0 or. force is a negative force (downward) in the sum of forces equation. See equations below: or. (Answer = 450 lb. -TA(12) +300 lb(6ft) + 160 lb(3 ft) + 140 lb(1 ft) = 0 We now solve the torque equation for TA. the 160 lb. 3 Rotational Equilibrium or select: Topic 1: Statics I . Eq. The weight of the bridge may be considered to act at its center. shown above.000 lb . 2. Fb = 4500 lb Return to: Topic 1. Choose an appropriate coordinate system. Resolve all forces into their x and y components. Statics Problems: Techniques 1.000 lb = 0 S ta = -10. Determine needed dimensions and angles. 3. has a 5 ton truck parked 10 feet from the left end of the bridge.000 lb x 20 ft + Fb x 40 ft = 0 Solving for unknowns: Fa = 9500 lb. 4000 lb bridge. All forces are in x or y direction 3.000 lb x 10 ft .4. Cond: S Fx = 0 (no external x forces acting on structure. Draw Free Body Diagram of entire structure.Principles Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/state13b. 1.4.Torque Review Problems A simply supported 40 foot. We would like to determine the compressional force in each support. Free body diagram is shown above. Apply the equilibrium conditions and solve for unknown external forces and torques as completely as possible.) S Fy = + Fa + Fb .10. showing and labeling all external forces.Torque example 2 Statics & Strength of Materials Example 2 . 2.htm [5/8/2009 4:39:35 PM] . including support forces and loads. + 250 lb. and then direction from Tan φ = Ry/Rx = -150/400 = -. Determine where the resultant intersects the bottom of the body with respect to point O.6 or 339.. =Ry We next find the resultant vector from it's components: R = square root (Rx2+Ry2) = Sqrt (4002 + (-150) 2) R = 427 lb. = -150 lb.375 The negative y-component and the positive x-component tell us that the resultant must be in the fourth quadrant. = 400 lb.100 lb. Sum Fx = +500 lb.Torque example 3 Statics & Strength of Materials Example 3 . For the first part of this problem we sum the forces in both the x and y direction to determine the resultant components our final force vector will have.4 degrees.Torque Review Problems Determine the magnitude. Solving for φ we find φ = -20. direction. and sense of the resultant forces shown. we realize file:///Z|/www/Statstr/statics/StatI/state13c. =Rx Sum Fy = -400 lb. .htm (1 of 2) [5/8/2009 4:39:35 PM] . To determine where the resultant intersects the bottom of the body with respect to point O. This actually simplifies the solution.htm (2 of 2) [5/8/2009 4:39:35 PM] .3" Return to: Topic 1.Torque example 3 that our resultant vector must produce the same torque with respect to point O as the orginal forces produce..500 lb. and therefore d = 22.Principles Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/state13c. x 17" . x 4" -400 lb. and so we may write: -150 lb. (clockwise direction) Our resultant must produce the same torque when it is applied or acts at the bottom of the body. Only the y-component (-150 lb..3 Rotational Equilibrium or select: Topic 1: Statics I .) produces no torque as its line of action passes through point O. Thus we must determine the resultant torque of our original forces. * 12" = -3350 in-lb.downward) produces a torque. the x-component of our vector (400 lb. x 5" + 250 lb. x (d") = -3350 in-lb. Torque = + 100 lb. At the bottom. ) 2. Net = -49 ft-lb. Compute the moment of the resultant with respect to point O and compare with the results of problem 1. (F1: 104 ft-lb.89 lb @ 240o).. Four coplanar concurrent forces act as shown.htm (1 of 4) [5/8/2009 4:39:35 PM] .Torque probs1 Statics & Strength of Materials Problem Assignment .. a) Calculate the moment of each force about point O that lies in the line of action of the F4 force.) file:///Z|/www/Statstr/statics/StatI/torquep1. 3. Moment of resultant should be same as in previous problem. b) Calculate the algebraic summation of the four moments and determine the direction of rotation.Torque 1. a) Calculate the algebraic summation of the moments of the forces shown about point A. Determine the resultant of the four forces in problem #1 (magnitude and angle of inclination with respect to the X axis)..) b) Calculate the algebraic summation of the moments of the forces about point B. (-330 ftlb. F4 = 0. (13.F3: -40 ft-lb. by 15 ft. F2: -113 ft-lb. The rectangular body shown measures 6 ft. (-90 ft-lb. .Torque probs1 4.. (260 lb. Each painter weighs 180 lbs. 8. The beam weights 100 pounds. Determine the magnitude a resultant force which would be equivalent to the load system shown.] 5. Determine the tension in each rope (AB & FE). (8820 lb. A beam is supporting two painters as shown below. Determine where it would act. 200 lb.htm (2 of 4) [5/8/2009 4:39:35 PM] .91 ft right of point A) [Do not calculate support forces in this problem. if it were to produce the same torque with respect to the center of mass of the beam as the load system does.) file:///Z|/www/Statstr/statics/StatI/torquep1. htm (3 of 4) [5/8/2009 4:39:35 PM] . A & B. (Include the directions of the forces) (-390 lb. (500 ft-lb. ccw) file:///Z|/www/Statstr/statics/StatI/torquep1.Torque probs1 6. A 160 lb person is standing at the end of a diving board as shown below. 690 lb.) 7. The diving board weighs 140 lbs. and this weight may be considered to act at the center of the board.. Compute the magnitude and direction of the resultant couples action on the bodies shown. Calculate the vertical forces acting at each support. htm (4 of 4) [5/8/2009 4:39:35 PM] .Torque probs1 Select: Topic 1: Statics I .Principles Table of contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/torquep1. A weightless horizontal plank is ten feet long.htm (1 of 6) [5/8/2009 4:39:36 PM] . It is simply supported at its two ends.Rigid Body Probs 1 Statics & Strength of Materials Problems 1. 180 lb. Determine the support forces at each of the two ends.) file:///Z|/www/Statstr/statics/StatI/Rbodyp1. (300 lb. Determine the support forces at each of the two ends.. (120 lb. A weightless horizontal plank is 16 feet long and is simply supported at its two ends.. A single load of 300 pounds rests at 4 feet from the left end.) 2. A load of 200 pounds is 4 feet from the left end and a load of 400 pounds is 6 feet from the right end. 300 lb. Rigid Body Probs 1 3. Determine the forces acting at each end.) The beam is simply supported at its two ends. A 12 foot long uniform steel beam weighs 500 pounds and has no additional loads. (250 lb.) file:///Z|/www/Statstr/statics/StatI/Rbodyp1.. 250 lb. (For static equilibrium calculations the entire weight of a uniform beam can be considered to act at its midpoint.htm (2 of 6) [5/8/2009 4:39:36 PM] . (1400 lb. 2400 lb.Rigid Body Probs 1 4.htm (3 of 6) [5/8/2009 4:39:36 PM] . The beam is simply supported 4 feet from the ends..) file:///Z|/www/Statstr/statics/StatI/Rbodyp1. A 20 foot long uniform steel beam weighs 800 pounds. A load of 1200 pounds is at one end of the beam and 1800 pounds at the other end. Determine the support forces acting on the beam. Rigid Body Probs 1 5.htm (4 of 6) [5/8/2009 4:39:36 PM] . 150 lb. There is a load of 150 pounds placed at the free end and a load of 600 pounds placed midway between the supports.) file:///Z|/www/Statstr/statics/StatI/Rbodyp1.. An 8 foot beam (weightless) is simply supported at its left end and its midpoint. Determine the support forces acting on the beam.(600 lb. A 140 lb. Determine the support forces acting on the diving board. .Rigid Body Probs 1 6.) file:///Z|/www/Statstr/statics/StatI/Rbodyp1.420 lb. diver stands at the right end of the board.. (560 lb. A diving board is 16 feet long and is supported at its left end and a point 4 feet from the left end.htm (5 of 6) [5/8/2009 4:39:36 PM] . htm (6 of 6) [5/8/2009 4:39:36 PM] .Rigid Body Probs 1 file:///Z|/www/Statstr/statics/StatI/Rbodyp1. Σ t = 0 (right hand rule +).000 lb = 0 Σ τa = -10.10.htm (1 of 3) [5/8/2009 4:39:36 PM] .000 lb x 10 ft .4. Σ Fz = 0 Rotational Σ t = 0 or Σ t = 0 (right hand rule +). Σ tz = 0 (right hand rule x y +) where t = R x F or (τ = force x perpendicular distance to pivot point) II Equilibrium Conditions 2-dimensions Translational: Σ F = 0 or Σ Fx = 0.Statics-Review/Procedure/Problems Statics . Cond: Σ Fx = 0 (no external x forces acting on structure. All forces are in x or y direction 3. Σ Fy = 0. Eq.) Σ Fy = + Fa + Fb .4. Σ Fy = 0 Rotational Σ t = 0 or Σ tp = 0 (ccw = +). Fb = 4500 lb file:///Z|/www/Statstr/statics/StatI/stprb1rv.000 lb x 20 ft + Fb x 40 ft = 0 Solving: Fa = 9500 lb. 2. 1. has a 5 ton truck parked 10 feet from the left end of the bridge.Review/Summary/Problem Sheet I Equilibrium Conditions 3-dimensions Translational: Σ F = 0 or Σ Fx = 0. where t = R x F or (τ = force x perpendicular distance to pivot point) Example 1 A simply supported 40 foot. Free body diagram is shown above. We would like to determine the compressional force in each support.000 lb . 4000 lb bridge. shown above. The weight of the bridge may be considered to act at its center. Problem #1. Determine the tension in each rope (AB & FE). 200 lb. 3. To do this (with non-truss problems) we continue the procedure above. Often we wish to know the internal forces (tension & compression) in each member of the structure in addition to the external support forces.. Draw Free Body Diagram of a member (s) of the structure of interest. Apply the equilibrium conditions and solve for unknown external forces and torques acting on the member as completely as possible. including support forces and loads. Show and label all external forces and loads acting on the selected member. but with members of the structure. A beam is supporting two painters as shown below. Choose an appropriate coordinate system. not the entire structure. Each painter weighs 180 lbs. Resolve all forces into their x and y components. Draw Free Body Diagram of entire structure. A 160 lb person is standing at the end of a diving board as shown below. 2. Determine needed dimensions and angles. the equilibrium equations for both the entire structure and for its members may have to be written and solved simultaneously before all the forces on the structure and in its members can be determined. 4. In certain problems. Determine needed dimensions and angles. 150 lb. Apply the equilibrium conditions and solve for unknown external forces and torques as completely as possible. Rework problem #1.(260 lb.Statics-Review/Procedure/Problems III Statics Problems: Techniques 1.) (210 lb. 5. (Neglect the weight of the beam. Choose an appropriate coordinate system. showing and labeling all external forces. Assume that the plank weighs 100 pounds and that this weight may be considered to act at the center of the span.) Probem #2. The diving board file:///Z|/www/Statstr/statics/StatI/stprb1rv.) Problem #3..htm (2 of 3) [5/8/2009 4:39:36 PM] . Resolve all forces into their x and y components. 6. Determine the tension in the cable and the forces acting at the hinge. 690 lb.) Problem #4. (BC=917 lb.htm (3 of 3) [5/8/2009 4:39:36 PM] ..) file:///Z|/www/Statstr/statics/StatI/stprb1rv. and this weight may be considered to act at the center of the board. Ax = 733 lb. and has a uniformly distributed weight of 100 lbs.. (Include the directions of the forces) (-390 lb.. The beam is attached to a wall by a hinge. A & B.Statics-Review/Procedure/Problems weighs 140 lbs. Ay = 50lb. A 500 pound sign is supported by a beam and cable as shown below. Calculate the vertical forces acting at each support. Three vectors are shown in the diagram below.Statics-Review/Procedure/Problems Additional Important Examples Some additional examples with brief solutions 1.htm (1 of 4) [5/8/2009 4:39:36 PM] . Find the resultant vector (magnitude and direction): R = -2 A .2 B -3 C file:///Z|/www/Statstr/statics/StatI/stprb. Determine the resultant force (magnitude & direction) and where along the base it would act to produce the same torque about point C file:///Z|/www/Statstr/statics/StatI/stprb.htm (2 of 4) [5/8/2009 4:39:36 PM] .Statics-Review/Procedure/Problems 2. A block has four forces acting on it as shown in the diagram. Determine the resultant (net) torque (magnitude & direction) about the center of mass (point O). so x = .4^2) = 204 lb.2 deg 123.3 lb + 126.6 lb * 3 ft -128.2 lb + 200 lb = 123.953 ft-lb Sum Force x = -64.6 lb – 153.4 lb (x ft) – 162.Statics-Review/Procedure/Problems Sum Torque-c: +64.3 lb Sum Force y = 76.6 lb * 2 ft – 153 lb * 3 ft + 100 lb * 2ft – 200 lb * 2 ft = .3 ^2 + 123.78 ft (to left of center of bottom) file:///Z|/www/Statstr/statics/StatI/stprb.3 .htm (3 of 4) [5/8/2009 4:39:36 PM] . 4 lb R = sqrt( 162.3 lb * 3 ft – 76.4/162. Tan (Theta) = 123. Theta = 37.3 lb * (3 ft) = -953 ft-lb.6 lb + 100 lb = 162.3. 5 ft) – (720 lb * 5 ft) – (5000 lb * 12. Ey = + 3907 lb.htm (4 of 4) [5/8/2009 4:39:36 PM] . The beam shown below is supported by a roller at point B and by a pinned joint at point E. By = + 4813 lb Return to: Topic 1: Statics I .Principles Table of contents or select: Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/stprb.5 ft) + Ey * 15 ft = 0 Ex = 1860 lb (negative x direction). Determine the support forces at points B and E Solution: Sum Fx: 900 lb + 960 lb – Ex = 0 Sum Fy: .Statics-Review/Procedure/Problems 3.(600 lb/ft * 5ft) + By – 720 lb – (1000 lb/ft * 5 ft) + Ey = 0 Sum Torque B: +(3000 lb * 2. ) file:///Z|/www/Statstr/statics/StatI/statx1. (T1 = 500 lb. B. Find one vector (magnitude and direction) that will have the same effect as the three vectors shown below. T2 = 71 lb. T3) in each cable... @ 140o) 2.htm (1 of 3) [5/8/2009 4:39:36 PM] .Statics 1 . Determine the tensions (T1. T2.) (71. & C) are shown in the diagram below.Exam 1. T3 = 505 lb.Examination-1 Topic 1: Statics I .7 lb. Three vectors (A. (Remember to show your work. x = 4.Statics 1 .. Dy = 1712 lb. Determine the support forces acting on the beam at points A & D. a) Determine the resultant torque due to all the forces with respect to Point O.htm (2 of 3) [5/8/2009 4:39:36 PM] .33 ft) 4.Examination-1 3.. (142 lb @ 132 deg. A square block has four forces acting on it as shown in the diagram. (Ax = -707 lb. direction and location (acting at the base) which would be equivalent to the four forces shown below. Each force has strength of 100 lb.) Select: file:///Z|/www/Statstr/statics/StatI/statx1. Ay = 995 lb. (+ 407.2 ft-lb) b) Determine a resultant force – magnitude. htm (3 of 3) [5/8/2009 4:39:36 PM] .Examination-1 Topic 1: Sample Test Solution Topic 1: Statics I .Statics 1 .Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/statx1. Three vectors (A. T2.) T1 = 500 lb (pulley just changes direction of forces. Determine the tensions (T1. T3 = 505 lb. B. & C) are shown in the diagram below..T3 sin 37 = 0 (solve equations for T2 & T3) file:///Z|/www/Statstr/statics/StatI/statx1ans.Statics 1 .Exam 1.T2 sin 45 .) (71. T2 = 71 lb.7 lb. (Remember to show your work.htm (1 of 3) [5/8/2009 4:39:37 PM] . @ 140o) Rx = 100 lb cos 37 +100 lb cos 150 + 80 lb cos 233: Ry = 100 lb sin 37 + 100 lb sin 150 + 80 lb sin 233 R = square root [ Rx2 + Ry2] = 71..) -500 cos 45 . (T1 = 500 lb. Find one vector (magnitude and direction) that will have the same effect as the three vectors shown below.7 lb Tan(angle) = Ry/Ry => angle = 140 degrees 2.Examination-1 Topic 1: Statics I . T3) in each cable.T2 cos 45 + T3 cos 37 = 0 +500 sin 45 . htm (2 of 3) [5/8/2009 4:39:37 PM] . A square block has four forces acting on it as shown in the diagram.2 ft-lb) b) Determine a resultant force – magnitude. (+ 407. direction and location (acting at the base) which would be equivalent to the four forces shown below.Statics 1 .Examination-1 3.100 lb sin 50 + 100 lb R = square root [ Rx2 + Ry2] = 142 lb Tan(angle) = Ry/Ry => angle = 132 degrees file:///Z|/www/Statstr/statics/StatI/statx1ans.(100 lb sin 50) x 6 ft + 100 lb x 3 ft + 100 lb x 4 ft = 407.2 ft-lb b) Rx = -100 lb cos 45 + 100 lb cos 50 .100 lb Ry = +100 lb sin 45 .(100 lb cos 50) x 4 ft . Each force has strength of 100 lb. x = 4.33 ft) a)(100 lb cos 45)x 6 ft . a) Determine the resultant torque due to all the forces with respect to Point O. (142 lb @ 132 deg. .5 ft) + (Dy * 10 ft) = 0 Solve for Ax.Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/statx1ans.htm (3 of 3) [5/8/2009 4:39:37 PM] .(2000 lb * 7. (Ax = -707 lb. Ay and Dy Select: Topic 1: Statics I . Dy = 1712 lb.Statics 1 .Examination-1 4..(400 lb/ft * 5 ft)+ Ay + Dy = 0 Sum Torque about point A: (-1000 lb sin 45 * 3 ft) .) Sum Fx: 1000 lb cos 45 + Ax = 0 Sum Fy: -1000 lb sin 45 . Determine the support forces acting on the beam at points A & D. Ay = 995 lb. Three vectors (A. Find the resultant of : 2A + B + 2C. & C) are shown in the diagram below. and remember to show your work.) 2. (Use the component method.Exam 1. file:///Z|/www/Statstr/statics/StatI/statx2. Find the magnitude. Determine the tensions in each cable.Statics 1 Examination-2 Topic 1: Statics I . B.htm (1 of 2) [5/8/2009 4:39:37 PM] . direction and location of a force which will place the bar (shown below) in translational equilibrium. 3. Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatI/statx2. supporting the beam ABCDE.Statics 1 Examination-2 4.htm (2 of 2) [5/8/2009 4:39:37 PM] . Determine the vertical forces in Rod1 and Rod 2. Select: Topic 1: Statics I . T3) in each cable. Find the resultant vector (magnitude and direction): R=2A+B 2C 2. 3. file:///Z|/www/Statstr/statics/StatI/statx4. Three vectors are shown in the diagram below.htm (1 of 2) [5/8/2009 4:39:37 PM] . Determine the tensions (T1.Sample Examination Statics I Exam 1. A square block has four forces acting on it as shown in the diagram. Each force has strength of 100 lb. T2. The beam shown below is supported by a roller at point B and by a pinned joint at point F. 4.htm (2 of 2) [5/8/2009 4:39:37 PM] . b) Determine the resultant torque due to all the forces with respect to Point O. Determine the support forces at points B and F.Sample Examination a) Determine the resultant force (magnitude and direction) due to all the forces acting on the block. file:///Z|/www/Statstr/statics/StatI/statx4. Sample Examination Statics I Exam 1. The angular forces act at an angle of 45 degrees from the horizontal.htm (1 of 2) [5/8/2009 4:39:37 PM] . Three vectors are shown in the diagram below. file:///Z|/www/Statstr/statics/StatI/statx5. T3) in each cable. A block has four forces acting on it as shown in the diagram. T2. Each force has strength of 100 lb. 3. Determine the tensions (T1. a) Determine the resultant torque due to all the forces with respect to Point O. Find the resultant vector (magnitude and direction): R=3A-B 2C 2. The beam shown below is supported by a roller at point B and by a pinned joint at point E. Determine the values of support forces at points B and E. file:///Z|/www/Statstr/statics/StatI/statx5. 4.Sample Examination b) Determine a resultant force (magnitude and direction) which would be equivalent to the four forces.htm (2 of 2) [5/8/2009 4:39:37 PM] . file:///Z|/www/Statstr/statics/StatI/statx6. direction which would be equivalent to the four forces shown below.Sample Examination Statics I Exam 1. a) Determine the resultant torque due to all the forces with respect to Point O. Each force has strength of 100 lb.htm (1 of 2) [5/8/2009 4:39:37 PM] . The forces at angles all act at 45 degrees. A square block has four forces acting on it as shown in the diagram. 3. T2. T3) in each cable. Determine the tensions (T1. Find the resultant vector (magnitude and direction): R = -2 A .2 B -3 C 2. Three vectors are shown in the diagram below. b) Determine a resultant force – magnitude. htm (2 of 2) [5/8/2009 4:39:37 PM] . The beam shown below is supported by a roller at point B and by a pinned joint at point E. file:///Z|/www/Statstr/statics/StatI/statx6.Sample Examination 4. Determine the support forces at points B and E . Example 2 r Trusses . #6. Bending and deformation effects will be considered in later materials.Example 2 r Frames . #7.1 Frames (non-truss. We will begin by looking at problems involving rigid bodies. rigid body structures) r Frames .3 Statics II .Example 1 r Frames .Example 1 r Trusses . a roller or bearing can only be placed in compression. especially as we draw our free body diagrams.Example 3 r Additional Examples: #5.Frames 2 (Supplemental) r Problem Assignment .Topic 2 Statics II . while a hinged or pinned support point may exert both a horizontal file:///Z|/www/Statstr/statics/StatII/stat21. Different types of supports will result in different types of support forces (reactions) acting on the structure. #6. #8. #8. #9. For example. rigid body structures) We are now ready to begin looking at somewhat more involved problems in statics.Trusses 1 (Required) r Problem Assignment . which are often small compared to the loads applied to the structures. As we do so there will be a number of concepts we need to keep in mind (and apply) as we approach these problems.Trusses 3 (Supplemental) 2.more difficult) 2.2 Trusses r Trusses . thus the force it will exert on a structure will be a normal or perpendicular force only.htm (1 of 4) [5/8/2009 4:39:38 PM] . #10 [Previous test problems] r Problem Assignment . As we begin to analyze structures there are two important considerations to keep in mind.1: Frames (non-truss.Trusses 2 (Supplemental) r Problem Assignment .Applications Topic 2: Statics II . #10 [Previous test problems] r Problem Assignment . At this point in the course we will also ignore the weight of the members.Frames 1 (Required) r Problem Assignment . This simply means that at this point we will not be concerned with the fact that a body (or member of a structure) may actually bend or deform (change length) length under the applied loads.Sample Exam ● Topic 2. The first concern will be Structure Supports.Applications ● ● 2. #9.Example 3 r Frames .Frames 3 (Supplemental .Example 4 r Additional Examples: #5. #7. The net effect is that a hinged or pinned support may be replaced by x and y support forces.] See Diagram 1 below. Diagram 1a is the Free Body Diagram of the beam with the roller and the pinned joint now replaced by the support forces which they apply on the beam. If we knew the value of the load. file:///Z|/www/Statstr/statics/StatII/stat21.Topic 2 Statics II .Applications and vertical force. In Diagram 1 we have shown a horizontal beam supported at point A by a roller and at point C by a pinned support. we could apply statics principles to find the actual value of the support forces. [Or to be more specific. Diagram 2 below shows examples of supports and the types of forces and/or torque which they may exert on a structure.htm (2 of 4) [5/8/2009 4:39:38 PM] . We shall do this process in great detail later for a somewhat more complex example. a hinged or pinned support (or joint) will exert a support force acting at a particular angle. this force may then always be broke into an x and y-component. The roller applies the vertical force Ay and the pinned support applies the forces Cx and Cy. A Non-Axial Member is a member which is not simply in tension or compression. That is.it is an axial member. (The importance of this will be seen in more detail when we look at our first extended example.Axial or Non-Axial Members. A simple way to tell if a member is an axial member is by the number of Points at which forces act on a member. the resultant of the forces must be two single equal forces acting in opposite directions along axis of the member. It may have shear forces acting perpendicular to the member and/or there may be different values of tension and compression forces in different parts of the member.htm (3 of 4) [5/8/2009 4:39:38 PM] . A member with forces acting at More Than Two Points (locations) on the member is a non-axial member. The internal force in the member is constant and acts only along the axis of the member. file:///Z|/www/Statstr/statics/StatII/stat21. If forces (no matter how many) act at only Two Points on the member .) In equilibrium or statics problems. See Diagram 4. an Axial Member is a member which is only in simple tension or compression.Topic 2 Statics II . See Diagram 3.Applications The second important concept to keep in mind as we begin looking at our structure is the type of members the structure is composed of . or frame problems.2 .Applications EXAMPLES: To show the application of the concepts discussed above and of our general statics problem-solving technique. Select Example 3 In Example 4.Trusses Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/stat21. we will examine a relatively straight forward problem which points out several features concerning torques and beam loading. we will look at a problem which seems to be a statically indeterminate problem. Select Example 2 In Example 3. Select Example 1 In Example 2. In Example 1 we will concentrate on finding the values of the external support forces acting on the structure.Topic 2 Statics II . Return to: Topic 2: Statics II .htm (4 of 4) [5/8/2009 4:39:38 PM] .Topic Table of Contents or Select: Topic 2. we will now look in careful detail at several statics problems. we will see how both the external support reactions and also the internal forces in a member of the structure may be found. Select Example 4 Additional Examples: The following additional examples all demonstrate different aspects and features of a variety of non-truss. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces. 1. Making the FBD is probably the most important part of the problem.Example 1 Example 1: Frames (non-truss. the load forces are already shown by the downward arrows.Topic 2. A correct FBD usually leads to a quick solution. and are pinned to the wall at points A and D. It is important to get the method and concepts we need to keep in mind firmly established. Resolve (break) all forces into their x and y-components. We next look at the forces file:///Z|/www/Statstr/statics/StatII/stat21e1. In our example. while an inaccurate FBD can leave a student investing frustrating unsuccessful hours on a problem. Step 1: Free Body Diagram (FBD).1 Frames . 2. we will proceed very carefully and methodically. 3. include any needed dimensions and angles. Apply the Equilibrium Equations ( ) and solve for the unknown forces. Loads of 4000 lb. Once again our procedure consists basically of three steps. We note that the structure is composed of members ABC. are applied to member ABC as shown in Diagram 1. In this problem we wish to determine all the external support forces (reactions) acting on the structure shown in Diagram 1 below.htm (1 of 5) [5/8/2009 4:39:38 PM] . and CD. and 2000 lb. rigid body) problems In this first example. These two members are pinned together at point C. With this in mind we will discuss in near excruciating detail the process of making a good FBD. Example 1 exerted on the structure by the supports. D and C). This is an accurate FBD. Since each support is a pinned joint. in our first FBD on the right (Diagram 2). We can draw a better FBD by reflecting on the concept of axial and non-axial members. as shown in Diagram 3). A negative value when solving for a force does not mean the force necessarily acts in the negative direction. rather than having two unknown forces. In other problems the directions the support forces act is not clear at all. If our guess is wrong. This means that at point D. Ay.1 Frames . Dx. The difficulty is that for our problem. rather it means that the force acts in the direction OPPOSITE to the one we initially chose. This is important. However. Dy) in this FBD. we have shown unknown x and y support forces acting on the structure at each support point. we can not solve for more unknowns than we have independent equations. We simple make our best guess for the directions of the support reactions. In Diagram 3. Ay. Thus. when we solve for the value of the support forces. this is not really a problem. but we have four unknowns (Ax. Since CD is an axial member the force acting on it from the wall (and in it) must act along the direction of the member. We also must choose directions for the x and y support forces. we have three equilibrium conditions ( ). This means we have only three unknowns. the worst case we could have is an unknown x and y-force acting on the structure at each support point. breaking any forces not in the x or y-direction into x and yfile:///Z|/www/Statstr/statics/StatII/stat21e1. and D. Ax.Topic 2. This is important. but it is not the best.htm (2 of 5) [5/8/2009 4:39:38 PM] . that value will be negative. In some problems the directions of the support forces are clear from the nature of the problem. And as we are well aware. Notice in our structure that member ABC is a non-axial member (since forces act on it at more than two points). we can draw one unknown force acting at a known angle (force D acting at angle of 37o. while member CD is an axial member (since if we drew a FBD of member CD we would see forces act on it at only two points. we have also completed Step II. D cos 37o and D sin 37o. file:///Z|/www/Statstr/statics/StatII/stat21e1. D cos 37o being the x-component.htm (3 of 5) [5/8/2009 4:39:38 PM] . [Please notice that there are not three forces at point D. can't we just drop the Ay force? The answer is NO. that is put in one force acting at a known angle. A second question is often.Example 1 components. not of the wall. because member ABC is not an axial member. why can't we do at point A what we did at point D. so we want to consider forces which act on the structure due to the wall. Thus.1 Frames .Topic 2. YES. First.] Now before we proceed with the final step and determine the values of the support reactions. Member ABC is a horizontal member. doesn't the wall just push horizontally on member ABC. what about the wall. we have shown the two components of D (which act at 37o). because we are making a FBD of the STRUCTURE. not forces on the wall due to the structure. there is either D acting at 37o or its two equivalent components. which has been replaced by its components. yes and no. we should deal with several conceptual questions which often arise at this point. aren't there forces acting on the wall that we should consider? Well. in Diagram 3. Apply the Equilibrium conditions. and the wall does not just push horizontally on member ABC (as we will see in our solution). In Diagram 3 at this point we really should cross out the D force. in compliance with Newton's Third Law). Thus the best we can do at point A is unknown forces Ax and Ay. and D sin 37o being the y-component. But NO we should not consider them. there are forces acting on the wall (as a matter of fact they are exactly equal and opposite to the forces acting on the members. Now Step III. it is not simply in compression or tension. Notice that with respect to point A. and able to support 7500 lb. We have now solved our problem. Thus our torque equation will have less unknowns in it. and D sin 37o do not produce torque since their lines of action pass through point A. and Ay = 1500 lb.Topic 2. at A. keeping track of their direction signs. Thus the force at point A is 6185 lb. The support force at point D is 7500 lb.. and then use it in the two force equations to find the other unknowns. forces to right.htm (4 of 5) [5/8/2009 4:39:38 PM] . which means the directions we choose for them initially are the actual directions they act. Ax = +6000 lb. as shown. Ax and Ay. and if a force acts through a point.1 Frames . and will be easier to solve. it does not produce a torque with respect to that point. This information would help us purchase the correct size hinge (able to support 6185 lb. or estimate if the file:///Z|/www/Statstr/statics/StatII/stat21e1. forces Ax. +. Again keeping track of direction signs. We choose point A. D. as shown in Diagram 4. The support forces at A can be left as the two components. Ay = +1500 lb. Ax = 6000 lb. at D). Ay. (Completing the calculations. Note that all the support forces we solved for are positive.) D = +7500 lb. Sum of Torque about a point. including load forces. Sum of y-forces. We can solve for force D. acting at 37o. to left. we arrive at the following answers. or may be added (as vectors) obtaining one force at a known angle. Thus in this problem the torque equation has only one unknown. Point D is also a good point to sum torque about since unknowns act through both points A and D. acting at 14o.Example 1 Here we sum the x-forces. Frames Continue to Example 2 or select: Topic 2: Statics II .Topic Table of Contents file:///Z|/www/Statstr/statics/StatII/stat21e1. All very useful and interesting information.Topic 2.1 Frames .htm (5 of 5) [5/8/2009 4:39:38 PM] . Return to Topic 2.1 .Example 1 wall is strong enough to support the structure. one side due to the load.htm (1 of 3) [5/8/2009 4:39:39 PM] . These members are pinned together at point B. We now proceed normally. Point B must support both the load and the pull of the person which results in a total force of 400 lb. which may be considered to act at the center of member BCD. but points out some aspects of axial/nonaxial forces. (This results since both sides of the rope have 200 lb. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces. Member BCD has a weight of 160 lb. below) we will want to determine the value of the external support reactions. rigid body) Problems Again in this second example. and the other side due to the pull of the person. III. and torque. This second example is reasonably straight forward. Additionally. showing and labeling all loads and support reactions acting on the structure. ) and We observe. force in them . load. ● ● ● I. that is we first draw our FBD (Diagram 2). In this second example (Diagram 1.Example 2 Example 2: Frame (non-truss. and are pinned to the floor at points A and D. acting on point B) file:///Z|/www/Statstr/statics/StatII/stat21e2. II. Apply the Equilibrium Equations ( solve for the unknown forces. As we do so we will note several items: that member AB is an axial member (only in tension or compression).Topic 2.. include any needed dimensions and angles. Step 1: Free Body Diagram (FBD). point B supports a pulley with which a person is hoisting a 200 lb. Resolve (break) all forces into their x and y-components. downward load at point B. we will continue to proceed very carefully and methodically. The general procedure to find the external support reactions consists of three basic steps. as shown in Diagram 1. and that the person / load / pulley combination at point B produces a net 400 lb.1 Frames . that the structure is composed of members AB and BCD. when we solve for the value of the support forces. and will be easier to solve.htm (2 of 3) [5/8/2009 4:39:39 PM] . Again keeping track of direction signs.A cos 37o being the x-component. thus the floor must exert a force on the member also along the direction of the member (due to equal and opposite forces principle). If our guess is wrong. Apply the Equilibrium conditions. We chose Point D to calculate torque. it does not produce a torque with respect to that point. at point A we have shown one unknown support force 'A' acting at a known angle (37o). However. keeping track of their direction signs. In an axial member the force is along the direction of the member. -) (Sum of y-forces. We now proceed through the structure. +.Topic 2.1 Frames . at point D.indicating our original direction was incorrect]. and if a force acts through a point. since member D is a non-axial member. looking a each force and calculating the torque due to that force with file:///Z|/www/Statstr/statics/StatII/stat21e2. . that value will be negative . Since two unknown forces (Dx. forces to right. and A sin 37o being the y-component. Step III. we have shown the two components of A (which act at 37o) .) Sum of Torque about a point. We can do this at point A since we know member AB is an axial member. thus our torque equation will have fewer unknowns in it. to left. Thus. we have also included Step II: Resolve (break) any forces not in the x or y-direction into x and y-components.Example 2 In the FBD (Diagram 2). In Diagram 2. the best we can do is to show an unknown Dx and Dy support forces acting on the structure at point D [We simply make our best guess for the directions of the support reactions. including load forces. Dy) are acting at Point D. (Here we sum the x-forces. finding the support reactions (forces) acting on the structure.Example 2 respect to the chosen Point D.8o (Please note that the force at D does not act along the direction of member BCD.) Finally. and Dy acts in the positive y . which means the directions we chose for them initially are the actual directions they act.7 lb. In this example. @ 127.Topic Table of Contents file:///Z|/www/Statstr/statics/StatII/stat21e2. Tangent (angle) = Dy/Dx = 354lb/-274 lb = -1.direction. We observe that all the support forces we solved for are positive. We could add (as vectors) Dx & Dy to find one resultant force acting a some angle on point D. and entering it in our torque equation (above) with the correct sign (+ for counterclockwise acting torque. solving for our unknowns we obtain: A = +343 lb.for clockwise acting torque). . as follows: D = Square Root [Dx2 + Dy2] = Square Root [(-274 lb)2 + (354 lb)2] = 447. Dx act in the negative x-direction. which it would do if BCD were an axial member.29) = 127.1 .8o (from x axis) so support force at Point D could also be expressed as: D = 447. we must be careful to use the correct distance in the torque relationship .7 lb.1 Frames . so Angle = ArcTangent (-1.29.Torque = Force x Perpendicular Distance. Dx = +274 lb. Return to Topic 2.Frames Continue to Example 3 or select: Topic 2: Statics II . Dy = +354 lb.htm (3 of 3) [5/8/2009 4:39:39 PM] . (See Torque Review if needed. (Notice that means that A acts at 37o as shown.Topic 2.) We have now solved our problem . Since each support is a pinned joint. ) and solve for the We note that the structure is composed of members ABC. We also must choose directions for the x and y support forces. III. but we will also determine the internal force in member CD of the structure shown in Diagram 1. However. Apply the Equilibrium Equations ( unknown forces. and a load of 8000 lb. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces. is applied at point C. A negative value when solving for a force does not mean the force necessarily acts in the negative direction. the worst case we could have is an unknown x and y-force acting on the structure at each support point. AD.000 lb. this is not really a problem. A load of 12. CD. We first calculate the support forces. II.Topic 2. These forces are already shown by the downward arrows. In some problems the directions of the support forces are clear from the nature of the problem. file:///Z|/www/Statstr/statics/StatII/stat21e3. is acting on member ABC at point B. rather it means that the force acts in the direction OPPOSITE to the one we initially chose. and cable DE.Example 3 Example 3: Frame (non-truss.. rigid body) Problems This third example is somewhat similar to example one. This is important. when we solve for the value of the support forces. If our guess is wrong. These members are pinned together at several points as shown in Diagram 1. The procedure to find the external support reactions consists of our basic statics procedure. In other problems the directions the support forces act is not clear at all. We simply make our best guess for the directions of the support reactions. that value will be negative.htm (1 of 7) [5/8/2009 4:39:40 PM] .1 Frames . Resolve (break) all forces into their x and y-components. but we will extend the problem by not only determining the external support reactions acting on the structure. I. include any needed dimensions and angles. We next look at the forces exerted on the structure by the supports. we have shown unknown x and y support forces acting on the structure at point A. file:///Z|/www/Statstr/statics/StatII/stat21e3.htm (2 of 7) [5/8/2009 4:39:40 PM] . at point E we have shown one unknown force 'E' acting at a known angle (37o). however.1 Frames .Topic 2. In the FBD (Diagram 2).Example 3 Step 1: Draw a Free Body Diagram of the entire structure. axial member AD.Example 3 We can do this at point E since we know that ED is a cable. forces to right. and E sin 37o being the y-component. and we have three independent equations from our equilibrium conditions.E cos 37o being the x-component. Since the cable pulls axially on the wall. Because of these two members (as opposed to a single axial member. to left. and a cable is an axial member which can only be in tension. we have shown the two components of E (which act at 37o) . -) file:///Z|/www/Statstr/statics/StatII/stat21e3. Apply the Equilibrium conditions. We have also used given angles and dimensions to calculate some distance. However. we have a good FBD since we have only three unknown forces. (Here we sum the x-forces. In Diagram 2. the best we can do at point A is to replace the hinged joint by an unknown Ax and Ay support forces acting on the structure as shown in Diagram 2. We also note that at point A we have two members pinned together to the wall. and non-axial ABC.htm (3 of 7) [5/8/2009 4:39:40 PM] . Resolve any forces not in the x or y-direction into x and y-components. we have also included Step II. as shown. Thus. Step III. as shown in Diagram 2. +. the wall pulls equally and in the opposite direction on the structure. such as at point E)..1 Frames . keeping track of their direction signs. which may be needed when we apply the equilibrium equations.Topic 2. Ax = +8620 lb. We can solve for E.) II. That is. (The member selected should be one acted on by the member in which we wish to find the force. and if a force acts through a point. Thus. (choosing not the member we wish to find the force in. such as member ABC. and then use it in the two force equations to find the other unknowns. in this problem the torque equation has only one unknown .) Sum of Torque about a point. We see that all the support forces we solved for are positive. which means the directions we chose for them initially are the actual directions they act. Again keeping track of direction signs. Draw a Free Body Diagram of a member of the structure showing and labeling all external load forces and support forces acting on that member. but of a member of the structure. and Ay = 13. We have now solved part one of our problem. including load forces. The support force(s) at A can be left as the two components. we draw a FBD . Ay = +13500 lb. Ay. Thus. In this example we will use member ABC to find the force in member CD. To find the force in a member of the structure we will use the following steps: First. acting at 37o. Resolve (break) all forces into their x and y-components.500 lb. and will be easier to solve.) To determine the force in an internal member of a structure we use a procedure similar to that used to find the external support reactions. Point E is also a good point to sum torque about since unknowns act through both points A and D. We choose point A. (We know member CD is axial as there are only two points at which forces acts on CD. or may be added (as vectors) obtaining one force at a known angle.E.) E = +10. file:///Z|/www/Statstr/statics/StatII/stat21e3. include any needed dimensions and angles. and E sin 37o do not produce torque since their lines of action pass through point A. forces Ax.1 Frames . The second part of the problem is to determine the force in axial member CD. The support force at point E is 10. not of the entire structure..800 lb. Then continue with steps below ● ● ● I. Ax = 8.800 lb. point C and point D. it does not produce a torque with respect to that point .620 lb. III.not of member CD. if we wish to find the force in member CD. but a member CD acts on. our torque equation will have less unknowns in it. determine the external support reactions acting on the structure (as we did in the first part of this example). Apply the Equilibrium Equations ( ) and solve for the unknown forces. Notice that with respect to point A. or member AD. Ax and Ay. but a member it acts on). we draw a FBD.htm (4 of 7) [5/8/2009 4:39:40 PM] . Doing the mathematics we arrive at the following answers.Topic 2.Example 3 (Sum of y-forces.thus. rather than show both the wall forces and the force due to AD on ABC. This is fine to do. This second FBD is slightly easier than the first in that it will result in one less force (AD) in the equilibrium equations.Example 3 Step 1: FBD of Member ABC. and that is still present in our FBD. There are actually two good FBD for member ABC. Step 2: Resolve forces into their x and y-components (This is done in Diagram 5. and also the force from axial member AD which acts on member ABC.Topic 2. We will use the second FBD in the rest of the problem. Thus. What we have done in this diagram is to look more closely at the left end of member ABC and observe that the effect of the wall forces and the effect of member AD. In Diagram 3 we have shown the first of these. as we are looking for force CD. That is. At point C we have shown the force from axial member CD which acts on member ABC. we simple show an ACx and an ACy force which is the net horizontal and vertical force acting on ABC at the left end.htm (5 of 7) [5/8/2009 4:39:40 PM] .1 Frames . The second good FBD of member ABC is shown in Diagram 4. Notice at the left end we have shown both the wall support reactions at A. we have isolated member ABD and indicated the forces on it due to the other members (and the wall) attached to it. is to give some net x and y-force acting on member ABC.) Notice we have file:///Z|/www/Statstr/statics/StatII/stat21e3. file:///Z|/www/Statstr/statics/StatII/stat21e3..) are not the same as the forces Ax and Ay acting on the entire structure at joint A.1 Frames . but are distributed to both members ABC and AD.440 lb. Note in the diagram that if the forces ACx and ADx are summed (13010 lb. This results since the forces of the wall at point A are not just acting on member ABC.sign shows force acts opposite direction chosen).htm (6 of 7) [5/8/2009 4:39:40 PM] .8o (8 ft ) =0 Solving: CD = 7. but our solution will tell us if we have the right or wrong directions for the unknown forces.440 lb. that their vector sums equal the external forces (Ax and Ay) acting on point A. ACy = 6000 lb.Topic 2. as we expect they should. + CD sin 53. = 13500 lb. this indicates that we have chosen the correct direction for the force CD (which indicates it is in tension). Since the force is positive. This solves our problem.).8o = 0 Sum Fy: ACy -12.1 . Step 3: Apply the equilibrium conditions and solve for unknown forces. . (For a little further analysis of forces at point A.Topic Table of Contents Point A: As an aside. (. We have determined the force in CD to be 7.) Return to Topic 2. (4 ft) + CD sin 53. Sum Fx: ACx + CD cos 53. ACx = . as shown in Diagram 6.4390 lb.000 lb.4390 lb. + 7500 lb.Frames Continue to Example 4 or select: Topic 2: Statics II .000 lb. and if forces ACy and ADy are summed (6000 lb.8o = 0 Sum TA: -12.). = 8620 lb. These may not be the correct directions.Example 3 chosen directions earlier for the forces. select MORE. notice that the forces ACx and ACy (The horizontal and vertical forces acting on member ABC at end A. Topic 2.Topic Table of Contents file:///Z|/www/Statstr/statics/StatII/stat21e3.1 .htm (7 of 7) [5/8/2009 4:39:40 PM] .Example 3 Return to Topic 2.1 Frames .Frames Continue to Example 4 or select: Topic 2: Statics II . Topic 2.1 Frames - Example 4 Example 4: Frame (non-truss, rigid body) Problems In our fourth example, we will examine a problem in which it will initially seem that there are too many unknowns to allow us to determine the external support forces acting on the structure, however, by a slight variation of our approach we will find that we can determine all the unknowns. The problem then is this, for the structure shown in Diagram 1 below, determine the external support forces acting on the structure, and additionally, determine the force in member CF. Our usual procedure to find the external support reactions consists of three basic steps. ● ● I. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces, include any needed dimensions and angles. II. Resolve (break) all forces into their x and y-components. III. Apply the Equilibrium Equations ( solve for the unknown forces. ) and ● We note that the structure is composed of members ABC, DEF, DE, and CF. These members are pinned together at several points as shown in Diagram 1. A load of 6000 lb. is acting on member DEF at point E, and a load of 3000 lb. is applied at point F. These forces are already shown by the downward arrows. We next look at the forces exerted on the structure by the supports. Since each support is a pinned joint, the worst case we could have is an unknown x and y-force acting on the structure at each support point. We also must choose directions for the x and y support forces. In some problems the directions of the support forces are clear from the nature of the problem. In other problems the directions the support forces act is not clear at all. However, this is not really a problem. We simple make our best guess for the directions of the support reactions. If our guess is wrong, when we solve for the value of the support forces, that value will be negative. file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (1 of 6) [5/8/2009 4:39:40 PM] Topic 2.1 Frames - Example 4 This is important. A negative value when solving for a force does not mean the force necessarily acts in the negative direction, rather it means that the force acts in the direction OPPOSITE to the one we initially chose. Step 1: Free Body Diagram (FBD). In the FBD (Diagram 2), we have shown unknown x and y support forces acting on the structure at pinned support points A and D, (Ax, Ay, Dx, Dy). If we think ahead somewhat, we realize that there could be a problem. We have four unknown forces supporting the structure, but there are only three equations in our Equilibrium Conditions, . file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (2 of 6) [5/8/2009 4:39:40 PM] Topic 2.1 Frames - Example 4 Normally, one can not solve for more unknowns then there are independent equations. Our first reaction should be to see if we can draw a better FBD. Perhaps we can replace the two unknowns at either point A or D by one unknown acting at a known angle (which is possible if we have a single axial member acting at the support point). However in this case both member ABC and member DEF are non-axial members, and the forces in them (and on them from the wall) do not act along the axis of the member. Thus, we already have the best FBD possible - we can not reduce the number of external unknowns acting on the structure. At this point we will simply continue with our normal analysis procedure and see what results. Step II: Resolve any forces not in the x or y-direction into x and y-components. (All forces are already in either the x or y-direction. Step III. Apply the Equilibrium conditions. (Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left, -) (Sum of y-forces, including load forces. Again keeping track of direction signs.) Sum of Torque about a point. We choose point D. Forces Ay, Dx and Dy do not produce torque since their lines of action pass through point A. Thus, the torque equation has only one unknown, Ax. We solve for Ax, and then use it in the sum of forces in the x-direction equation to find the unknown, Dx . And if we do so, we find: Ax = +18000 lb. Dx = +18000 lb. (The positive signs indicate we initially chose the correct direction for the forces.) However, file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (3 of 6) [5/8/2009 4:39:40 PM] Topic 2.1 Frames - Example 4 please notice that while we found Ax and Dx, we can not find Ay and Dy. There are still two unknowns in the y-equation and not enough information to determine then at this point. Thus, analysis of the structure as a whole has enabled us to determine several of the external support forces, but not all of them. What now? First, an overview. There are problems for which the static equilibrium conditions are not enough to enable one to solve the problem. They are called Statically Indeterminate Problems, and we will be considering these a bit later. Then there are problems which, on first glance, appear to be statically indeterminate, but are not. That is the case here. To find the remaining unknown support forces (and at the same time, determining the force in member CF), we will now take out a member of the structure and apply our statics analysis procedure to the selected member of the structure (rather than the entire structure). We will select member ABC to analyze. (See Diagram 3) Step 1: Free Body Diagram (FBD) In Diagram 3, we have drawn a FBD of member ABC, showing and labeling all forces external to member ABC which act on it. That is at point A we have the forces of the wall acting on ABC, at point B we have an axial force on ABC due to member BE, and at point C we have a axial force on ABC due to the member CF. Both member BE and CF are axial members and so we know the directions their forces act - along their axis. We do have to guess if they push or pull on member ABC, and we have chosen those directions as shown, (if the direction chosen is wrong, the force's value will be negative when we solve). We also are happy with the FBD as it has only three unknowns acting on the member, which indicates that we should be able to solve completely for the unknowns. Step 2: Resolve forces into x/y components. Here we have resolved force CF into its horizontal and vertical components as shown in Diagram 4. All other forces are already in x or y-direction. file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (4 of 6) [5/8/2009 4:39:40 PM] Topic 2.1 Frames - Example 4 Step 3. Apply the Equilibrium conditions. We now solve the first equation for CF, then use that value in the last equation to find BE, and use both values in the middle equation to find Ay, giving us: CF = 30,000 lb., BE = 36,000 lb., Ay = -12, 000 lb. Please note that the value of Ay is negative, which indicates the direction we chose was incorrect. Ay acts downward rather than in the positive y-direction we initial chose. Additionally, we now can return to the equations for the entire structure (see below), and knowing the value for Ay, we can use it in the yforces equation to solve for the value of Dy. Equilibrium Conditions for entire structure (from first part of problem) ● ● ● From the y-forces equation: (-12,000 lb.) - Dy -6000 lb. -3000 lb = 0; Solving Dy = -21,000 lb. file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (5 of 6) [5/8/2009 4:39:40 PM] Topic 2.1 Frames - Example 4 Once again, the negative sign indicates we selected the wrong direction for Dy, rather than acting downward it actually acts upward. See Diagram 5 for final force values and directions. Ax = 18000 lb. Ay = 12,000 lb. Dx = 18000 lb. Dy = 21,000 lb. CF = 30,000 lb. BE = 36,000 lb. Return to: Topic 2.1 - Frames or select: Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/stat21e4.htm (6 of 6) [5/8/2009 4:39:40 PM] solution111 STATICS / STRENGTH OF MATERIALS - Example In the structure shown below members AD, DC, and ABC are assumed to be solid rigid members. Member ED is a cable. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the values of all the support forces acting on the structure. C. Determine the force (tension or compression) in member DC. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B: file:///Z|/www/Statstr/statics/Stests/statics/sol111.htm (1 of 3) [5/8/2009 4:39:40 PM] solution111 STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = -E cos (37o) + Ax = 0 Sum Fy = Ay + E sin (37o) - 10,000 lbs - 8,000 lbs = 0 Sum TA = E cos (37o)(12ft) - (10,000 lbs)(4 ft) - (8,000 lbs)(12ft) = 0 Solving for the unknowns: E = 14,200 lbs; Ay = 9,480 lbs; Ax = 11,400 lbs PART C: Now find internal force in member DC file:///Z|/www/Statstr/statics/Stests/statics/sol111.htm (2 of 3) [5/8/2009 4:39:40 PM] solution111 STEP 1: Draw a free body diagram of a member that DC acts on - member ABC. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply the equilibrium conditions. Sum Fx = Acx - DC cos (56.3o) =0 Sum Fy = Acy - 10,000 lbs 8,000 lbs + DC sin (56.3o) = 0 Sum TA = (-10,000 lbs)(4 ft) - (8,000 lbs)(12 ft) +DC sin (56.3o)(12 ft) = 0 Solving for the unknowns: DC = 13,600 lbs; Acx = 7,560 lbs; Acy = 6,670 lbs These are external forces acting on member ABC. The force in DC is 13,600 lbs (c). Select Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol111.htm (3 of 3) [5/8/2009 4:39:40 PM] solution112 STATICS / STRENGTH OF MATERIALS - Example In the structure shown below member ABC is assumed to be a solid rigid member. Member CD is a cable. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force (tension ) in member CD. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, file:///Z|/www/Statstr/statics/Stests/statics/sol112.htm (1 of 2) [5/8/2009 4:39:41 PM] solution112 as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = -D cos (37o) + Ax = 0 Sum Fy = D sin (37o) + Ay - 10,000 lbs - 8,000 lbs = 0 Sum TA = (-10,000 lbs)(6.93 ft) - (8,000 lbs)(10.4 ft) + D sin (37o)(5.61 ft) + D cos (37o)(9.61 ft) = 0 Solving for the unknowns: D = 13,800 lbs; Ax = 11,100 lbs; Ay = 9,710 lbs PART C - Now find internal force in member DC. In this problem member DC is a single axial member connected to the wall at point D. Therefore, the force in member DC is equal and opposite the force exerted on DC by the wall. From parts A and B, the force on DC due to the wall is 13,800 lbs. Therefore, force in DC is also 13,800 lbs (in tension since DC is a cable). Select Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol112.htm (2 of 2) [5/8/2009 4:39:41 PM] solution113 STATICS / STRENGTH OF MATERIALS - Example In the structure shown below members ABC, ADE, and DB are assumed to be solid rigid members. Members ABC and ADE are pinned to the wall at point A. Member ADE is supported by a roller at point E. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force (tension or compression) in member DB. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, file:///Z|/www/Statstr/statics/Stests/statics/sol113.htm (1 of 3) [5/8/2009 4:39:41 PM] solution113 as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = Ex + Ax = 0 Sum Fy = Ay - 12,000 lbs = 0 Sum TA = Ex(8 ft) - (12,000 lbs) (12 ft) = 0 Solving for the unknowns: Ex= 18,000 lbs; Ax= -18,000 lbs; Ay = 12,000 lbs PART C - Now find internal force in member DB STEP 1: Draw a free body diagram of a member that DB acts on - member ABC. STEP 2: Resolve all forces into x and y components (see diagram) STEP 3: Apply the equilibrium conditions. Sum Fx = Acx + DB cos (33.7o) = 0 Sum Fy = Acy + DB sin (33.7o) - 12,000 lbs = 0 Sum TA = DB sin (33.7o)(6 ft) - (12,000 lbs)(12 ft) = 0 Solving for the unknowns: DB = 43,300 lbs; Acx = -36,000 lbs; Acy = -12,000 lbs file:///Z|/www/Statstr/statics/Stests/statics/sol113.htm (2 of 3) [5/8/2009 4:39:41 PM] solution113 These are external forces acting on member ABC. The force in DB is 43,300 lbs (c). Select Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol113.htm (3 of 3) [5/8/2009 4:39:41 PM] Determine the force (tension or compression) in member CD. CD. Unless otherwise indicated.Example In the structure shown below members ABC. C.solution114 STATICS / STRENGTH OF MATERIALS . B. Draw a Free Body Diagram showing all support forces and loads. Determine the value of all the support forces acting on the structure.htm (1 of 3) [5/8/2009 4:39:41 PM] . For this structure: A. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. and AD are assumed to be solid rigid members. all joints and support points are assumed to be pinned or hinged joints. Member DE is a cable. file:///Z|/www/Statstr/statics/Stests/statics/sol114. 400 lbs PART C . STEP 2: Break any forces not already in x and y direction into their x and y components.800 lbs.Now find internal force in member DC.000 lbs)(13.86 ft) = 0 Solving for the unknowns: E = 10. STEP 3: Apply the equilibrium conditions.000 lbs)(4 ft) . Sum Fx = -E cos (37o) + Ax = 0 Sum Fy = E sin (37o) + Ay -12. STEP 1: Draw a free body diagram of a member that DC acts on . file:///Z|/www/Statstr/statics/Stests/statics/sol114.600 lbs.solution114 as well as any needed angles and dimensions.000 lbs = 0 Sum TA = E cos (37o)(18.htm (2 of 3) [5/8/2009 4:39:41 PM] .(8.8.44 ft) (12. Ax = 8.000 lbs . Ay = 13.member ABC. 440 lbs (t).solution114 STEP 2: Resolve all forces into x and y components (see diagram).000 lbs + DC sin (53. Acx = -4. STEP 3: Apply the equilibrium conditions. Acy = 6. Sum Fx = Acx + DC cos (53.440 lbs.390 lbs.8o)(8 ft) = 0 Solving for the unknowns: DC = 7.12.8o) = 0 Sum TA = (-12.Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol114.000 lbs)(4 ft) + DC sin(53.8o) = 0 Sum Fy = Acy . Select Topic 2: Statics II .000 lbs These are the external forces acting on member ABC. The force in DC is 7.htm (3 of 3) [5/8/2009 4:39:41 PM] . Determine the value of all the support forces acting on the structure. C. Determine the force (tension or compression) in member CD. all joints and support points are assumed to be pinned or hinged joints. file:///Z|/www/Statstr/statics/Stests/statics/sol115. Draw a Free Body Diagram showing all support forces and loads. Members AE and DE are cables. For this structure: A. and CD are assumed to be solid rigid members.solution115 STATICS / STRENGTH OF MATERIALS . Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. B.htm (1 of 3) [5/8/2009 4:39:41 PM] .Example In the structure shown below members BCE. Unless otherwise indicated. Bx = 4.htm (2 of 3) [5/8/2009 4:39:41 PM] . STEP 3: Apply the equilibrium conditions.(12.3o) + By 12.member BCE. STEP 1: Draw a free body diagram of a member that CD acts on .000 lbs) (24 ft) = 0 Solving for the unknowns: A = 7210 lbs.000 lbs PART C .A sin(56.000 lbs = 0 Sum TA = By(16 ft) . file:///Z|/www/Statstr/statics/Stests/statics/sol115.solution115 as well as any needed angles and dimensions.Now find internal force in member CD. By = 18.000 lbs. Sum Fx = -A cos(56. STEP 2: Break any forces not already in x and y direction into their x and y components.3o) + Bx = 0 Sum Fy = . 000 lbs)(24 ft) = 0 Solving for the unknowns: CD = 10.000 lbs (c).htm (3 of 3) [5/8/2009 4:39:41 PM] .000 lbs = 0 Sum Fy = Edy . STEP 3: Apply the equilibrium conditions: Sum Fx = Edx . Edx = 4.000 lbs = 0 Sum TE = -CD cos (37o)(12 ft) + (4.CD sin (37o) + 18.000 lbs These are the external forces acting on member BCE.CD cos (37o) + 4.000 lbs. The force in CD is 10.Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol115.solution115 STEP 2: Resolve all forces into x and y components (see diagram). Select Topic 2: Statics II . Edy = -12.000 lbs. The structure is pinned at A and supported by a roller at E. Unless otherwise indicated. BDE and CD are assumed to be solid rigid members.solution116 STATICS / STRENGTH OF MATERIALS .htm (1 of 3) [5/8/2009 4:39:42 PM] . C. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support(reaction) forces. Determine the force (tension or compression) in member CD. For this structure: A. as file:///Z|/www/Statstr/statics/Stests/statics/sol116.Example In the structure shown below members ABC . Determine the value of all the support forces acting on the structure. B. Draw a Free Body Diagram showing all support forces and loads. all joints and support points are assumed to be pinned or hinged joints. Ay = 6.member BDE.000 lbs)(3 ft) (4.000 lbs = 0 Sum TA = (8.000 lbs 4.solution116 well as any needed angles and dimensions.Now find internal force in member CD. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = Ax = 0 Sum Fy = Ay + Ey .880 lbs.5 ft) = 0 Solving for the unknowns: Ey = 5.htm (2 of 3) [5/8/2009 4:39:42 PM] . STEP 1: Draw a free body diagram of a member that CD acts on .000 lbs)(6.120 lbs PART C .8. file:///Z|/www/Statstr/statics/Stests/statics/sol116.5 ft) + Ey(8. STEP 3: Apply the equilibrium conditions: Sum Fx = -Bx + CD cos (60o) =0 Sum Fy = By .140 lbs (c).880 lbs = 0 Sum TB = -CD sin (60o)(3 ft) (4. Bx = 4. file:///Z|/www/Statstr/statics/Stests/statics/sol116. The force in CD is 8.140 lbs.000 lbs)(5 ft) + (5.CD sin (60o) 4.880 lbs)(7 ft) = 0 Solving for the unknowns: CD = 8.htm (3 of 3) [5/8/2009 4:39:42 PM] . By = 5.070 lbs.solution116 STEP 2: Resolve all forces into x and y components (see diagram).170 lbs These are the external forces acting on member BDE.000 lbs + 5. The structure shown is composed of members ACD and BC pinned together at point C. By = 5600 lb.. and the force in member CB. (Ax = -78. Determine the values of the external support reactions. Determine the values of the external support reactions. and the force in member BC. Ay = +178 lb.. (Ax = 0 lb.. 1. BC = 5600 lb. (T)) 2. and member CB is a cable pinned at the wall. The structure is pinned to ceiling at points A and B..Frames 1 Draw a complete free body diagram as a part of the solution for each problem. In the structure shown member AB is pinned at the floor. Ay = -1600 lb...Frames Statics & Strength of Materials Problems Assignment ..Assignment Problems . CB = 455 lb.T) file:///Z|/www/Statstr/statics/StatII/statp21f. C = 455 lb.htm (1 of 4) [5/8/2009 4:39:42 PM] .3 lb. 000 lb. C. The structure shown is composed of solid rigid members ABC.670 lb. and BDE pinned together at points B. and the force in member DC.. determine the values of the external support reactions.8.600 (T)) file:///Z|/www/Statstr/statics/StatII/statp21f.. (Ax = 8. DC = 15.htm (2 of 4) [5/8/2009 4:39:42 PM] .. For the structure shown.Assignment Problems .Frames 3. and pinned to the wall at point A. CD.670 lb. Ay = 10. The structure is supported by a roller at point E. and D. Ex =. . BD.. The structure is pinned to the floor at points A and C. and DE pinned together at points B.5419 lb. The structure is pinned to the floor at points A and E. (Ax = 6375 lb.htm (3 of 4) [5/8/2009 4:39:42 PM] .830 lb. Ay = 4170 lb. Cy = 15. Cx = . CD = 9000 lb. E = 10. Ay = 10500 lb.) file:///Z|/www/Statstr/statics/StatII/statp21f. The structure shown is composed of solid rigid members AB and BC pinned together at point B. and D. The structure shown is composed of solid rigid members ABC. Determine the values of the external support reactions (Ax = +419 lb. C) 5. and the force in member CD..625 lb. For the structure shown. CD..Assignment Problems . C. determine the values of the external support reactions.Frames 4. Assignment Problems .htm (4 of 4) [5/8/2009 4:39:42 PM] .1 .Frames or select: Topic 2: Statics II -Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp21f.Frames Return to Topic 2. (Ax = 404 lb. Ay = 296 lb. Determine the force in the cable and the reactions at point A.Frames 2 Draw a complete free body diagram as a part of the solution for each problem.) 2. B = 814 lb. (Ax = 938 lb. A 700 lb... as shown in the diagram. Neglect the weight of the beam.htm (1 of 6) [5/8/2009 4:39:42 PM] .) file:///Z|/www/Statstr/statics/StatII/statp21f2.. C = 572 lb. 1. Determine the reaction at supports A and B of the beam in the diagram shown.frame problems 2 Statics & Strength of Materials Problems Assignment . Ay = 728 lb. weight is carried by a boom-and-cable arrangement.. Ay = 45 lb. and what will be the values of the vertical and horizontal components of the reaction at hinge A? (Ax = 596 lb.) file:///Z|/www/Statstr/statics/StatII/statp21f2.. B = 994 lb. The beam weighs 250 lb. A brace is hinged at one end to a vertical wall and at the other end to a beam 14ft long.frame problems 2 3.. What will be the compressive force in the brace. and is also hinged to a vertical wall as shown.htm (2 of 6) [5/8/2009 4:39:42 PM] . at the free end. The beam carries load of 500 lb. climbs up on the gate at the point B.) file:///Z|/www/Statstr/statics/StatII/statp21f2. A gate has a weight of 200 lb. A small boy weighing 95 lb. which may be considered as uniformly distributed. What will be the reactions on the hinges? (Upper hinge horizontal force only = 318 lb.frame problems 2 4..htm (3 of 6) [5/8/2009 4:39:42 PM] . lower hinge horizontal force = 318 lb. see the diagram shown... vertical force = 295 lb. . and it may be skipped. These sockets may be considered as hinges. What will be the vertical and horizontal components of the reactions at A and B? The weight of the water in the flume supported by each frame is estimated as 18. Bx = 6389 lb.. 107 lb. it was found necessary to cross low ground or else swing the canal to the left by cutting into the solid rock. In an irrigation project.) file:///Z|/www/Statstr/statics/StatII/statp21f2. Ay = 5093 lb. (This is a somewhat more complicated problem then the others in this problem set. Or contact your instructor for hints.. It was decided to run the canal as a flume and support it on a number of frames as shown in the diagram.) (Ax = 6389 lb.200 lb.frame problems 2 5. The two members rest in sockets in solid rock at points A and B.htm (4 of 6) [5/8/2009 4:39:42 PM] . By = 13. Ax = 1250 lb. If the boat and its load weigh 2..frame problems 2 6.000 lb. what are the reactions at A and B? Two identical davits support each lifeboat. (Ay = 1000 lb. Bx = 1250 lb.htm (5 of 6) [5/8/2009 4:39:42 PM] ... There is a socket at A and a smooth hole through the deck rail at B.) file:///Z|/www/Statstr/statics/StatII/statp21f2. An Ocean liner has an arrangement for supporting lifeboats and for lowering them over the side as shown in the diagram. frame problems 2 Return to Topic 2.1 .Frames or select: Topic 2: Statics II -Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp21f2.htm (6 of 6) [5/8/2009 4:39:42 PM] . Dx = 1250 lb.. directions not indicated) file:///Z|/www/Statstr/statics/StatII/statp21f3.. Ey=1000 lb. Cx=1443 lb. Cy=7500 lb... and C and is subjected to loads as shown.Frames 3 Draw a complete free body diagram as a part of the solution for each problem. By=1750 lb.100 lb. and C. Dy=750 lb. directions not indicated) 2. A pin-connected A-frame supports a load as shown. Compute the pin reactions at all of the pins. 1..100 lb. Neglect the weight of the members. Bx=1250 lb.... Bx=10.htm (1 of 6) [5/8/2009 4:39:42 PM] .....frame problems 3 Statics & Strength of Materials Problems Assignment . Neglect the weight of the members.. B. Ay=2500 lb. Cy=250 lb. Compute the pin reactions at A. A simple frame is pin connected at points A. By=2500 lb. (Ax=10. B. (Ay=1500 lb.. Cx=1250 lb. . Cx=7768 lb.. 600 lbs. boom.1o . By=0. Calculate the pin reactions at pins A. ED = 12. directions not indicated) file:///Z|/www/Statstr/statics/StatII/statp21f3.frame problems 3 3. 650 lb @ 52. Cy=3279 lb. A pin-connected crane framework is loaded and supported as shown... B. 800lbs. Bx=3824 lb. Ay=8100 lb. D.. 700 lbs. and E. (Ax=3824 lb.htm (2 of 6) [5/8/2009 4:39:42 PM] . and brace. C. These weights may be considered to be acting at the midpoint of the respective members. The member weights are: post... By=750 lb.frame problems 3 4... Dy=750 lb. Sketch the structure and draw a free body diagram for the relevant member or the entire structure. directions not indicated) file:///Z|/www/Statstr/statics/StatII/statp21f3. Write the appropriate moment or force equations and solve them for the unknown forces.htm (3 of 6) [5/8/2009 4:39:42 PM] .. (Ax=500 lb. Bx=500 lb.. Dx=100 lb... Ay=250 lb. Bx=300 lb.. Cx=300 lb. Calculate the pin reactions at each point of the pins in the frame shown below.frame problems 3 5. Ey=150 lb. directions not indicated) file:///Z|/www/Statstr/statics/StatII/statp21f3.. Cy=0.. Dx =0. Ay=150 lb.. By=150 lb...htm (4 of 6) [5/8/2009 4:39:42 PM] .. (Ax=300 lb. Dy=150 lb. For an input force of 15 lb. on each handle.) Return to Topic 2.htm (5 of 6) [5/8/2009 4:39:42 PM] . (F = 48 lb. on object from each jaw.. The tongs shown are used to grip an object.Frames file:///Z|/www/Statstr/statics/StatII/statp21f3. Ay=63 lb. Ax = 0.1 . determine the forces exerted on the object and the forces exerted on the pin at A.frame problems 3 6. htm (6 of 6) [5/8/2009 4:39:42 PM] .frame problems 3 or select: Topic 2: Statics II -Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp21f3. riveted. forming triangular substructures (within the main structure). BC. For this structure we wish to determine the values of the support reactions. In real trusses.Topic 2.htm (1 of 6) [5/8/2009 4:39:43 PM] . determining the external support reactions. and the force (tension/ compression) in members BE. but it is often much less than the applied load and may be neglected with little error. and EF. however the method for determining the internal forces in members of a truss is not the same. the members have weight. and with the external loads applied only at the joints. and a horizontal load of 800 lb. and in fact. A vertical load of 500 lb. However as long as the centerline of the member coincide at the joint. these may be transmitted to a joint by use of a support system composed of stringers and cross beams. acts at point F. the assumption of a pinned joint maybe used. For the first part. In cases where there are distributed loads on a truss. The procedure for determining the external support reactions acting on a truss is exactly the same as the procedure for determining the support forces in non-truss problems. The procedures for finding internal forces in truss members are Method of Sections and Method of Joints (either of which may be used). Also in actual trusses the joints may be welded. or bolted to a gusset plate at the joint.2: Rigid Body Structures .Trusses A very common structure used in construction is a truss. Or the weight maybe included by dividing the weight in half and allowing half the weight to act at each end of the member. acts at Point C. An ideal truss is a structure which is composed completely of (weightless) Axial Members that lie in a plane. one must be very careful not to use these methods with non-truss problems as they will not give correct results. which is supported at the joints and transmits the load to the joints. Perhaps the best way to clarify these concepts is to work very slowly and carefully through a truss example. connected by pinned (hinged) joints. Example 1: In Diagram 1 we have a truss supported by a pinned joint at Point A and supported by a roller at point D.2 Trusses Topic 2. we apply the normal static equilibrium procedure: file:///Z|/www/Statstr/statics/StatII/stat22. of course. See Diagram 1. and we may apply the static equilibrium conditions (and procedure) to solve for the forces on the joint(s) due to the members. Ay = -33 lb. so the pin or joint will be in static equilibrium. Note that at the pinned support point A. we notice file:///Z|/www/Statstr/statics/StatII/stat22. rather then analyze the entire structure. This means that to solve completely for the forces acting on a joints we must have a joint which has. we may proceed to determining the values of the forces in the members themselves.2 Trusses I. we rather examine the joint (pin or hinge) where members come together. As the structure is in static equilibrium. (Sum of x-forces) (Sum of y-forces) (Sum of Torque about A. which will also equal the forces in the axial members . they have no perpendicular distance to that point and so produce no torque. the best we can do is to put both an unknown x and y support force. at most.) Part 2 Once we have determined the values of the external support reactions. There are several points to keep in mind as we use method of joints. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces. Ay acts downward. III. since all the forces pass through the same point. II. however at point D we only need a unknown y support force since a roller can only be in compression and so must support vertically in this problem.Topic 2. Resolve (break) all forces into their x and y-components. our sum of torque equation will be of no help. One is that since we are analyzing a point (joint) rather then an extended body. we will use Method of Joints to determine the force in the selected members. two unknown forces acting. That is. In Method of Joints.) Solving we obtain: Ax = 800 lb. In our example (Diagram 2). include any needed dimensions and angles.due to Newton's third law of equal and opposite reactions (forces). Dy = 533 lb (The negative sign for force Ay means that we initially chose it in the incorrect direction.htm (2 of 6) [5/8/2009 4:39:43 PM] ) and solve for . or even a member of the structure. Apply the Equilibrium Equations ( the unknown forces. In this first example. the internal forces. not upward as shown in FBD. pushing into the joint. if indeed. that the solution will tell us that AB is in the wrong direction. AB is in Tension. the y-support reaction.. However. 33 lb. and that member AE is in tension.2 Trusses that there are only two joints which initially have only two unknowns acting . which clearly will be in the . That is. And likewise.htm (3 of 6) [5/8/2009 4:39:43 PM] . We will leave AB as chosen to see. act along the directions of the members (since the members are axial).Joint A and Joint D. We will begin with Joint A. Step 3: Apply the Equilibrium Conditions: (Sum of x-forces) (Sum of y-forces) Solving: AE = 756 lb. In Diagram 3 we have shown Joint A with the all the forces which act on the joint. we start our process at one of these.Topic 2. In Diagram 3.y direction. if a member of a truss is in compression. The forces on Joint A. it will pull outward on the joint. Step 1. There are no other y-forces on the joint. due to members AB and AE. [A little consideration will show that we have actually chosen AB in an incorrect direction. Include needed angles. (Tension). not Compression. showing it acting out of the joint. Thus. if these values are negative it means that our chosen directions were incorrect and the forces act in the opposite direction. we have assumed member AB is in compression. is also in the -y direction. it will push outward on it ends . showing and labeling all forces acting on the joint. A force acting into the joint due to a member means that member is in compression. When we solve for the forces AB and AE. showing its direction into the joint.) file:///Z|/www/Statstr/statics/StatII/stat22. (The negative sign indicates we selected an incorrect initial direction for AB. FBD of the Joint A. if a truss member is in tension. We can see this if we consider the y-component of AB.] Step 2: Resolve (Break) all forces into their x and y components. We choose directions for AB and AE (into or out of the joint). AB = -55 lb. so it can not be in equilibrium is both forces act in the same direction. We now apply the Equilibrium Conditions (for joints). we are trying to find the forces in members BE. On the right side of Diagram 5 is the FBD with all forces resolved into x and y-components. we have the value of the force in member AB. tension) and AB (55 lb. since there were three unknowns (initially) at joint B (AB. Include needed angles.Method of Joints Step 1. tension). Step 2: Resolve (Break) all forces into their x and y components. (Compression) file:///Z|/www/Statstr/statics/StatII/stat22. (Sum of x-forces) (Sum of y-forces) Solving : BC = 44 lb. We could not have solved for the forces acting at joint B initially. and can proceed to joint B where we will now have only two unknowns to determine (BE & BC). Step 3: Apply the Equilibrium Conditions: In Diagram 5.Topic 2. Now that we have the forces in members AE (756 lb. If the directions we chose for the unknowns are correct.htm (4 of 6) [5/8/2009 4:39:43 PM] . If a value is negative it means the force acts in the opposite direction.2 Trusses Remember. showing and labeling all forces acting on the joint. Joint B: Procedure . we have on the left the FBD of joint B with all external forces acting on the joint shown. BC. and EF. BC. However now that we have analyzed joint A. (Tension) BE = 33 lb. their values will be positive in the solution. and our initial direction for the forces. and BE). FBD of the Joint B. we can move unto a second joint (joint B) and find two of the unknowns we are looking for. finding both the external forces and forces in members BE. their values will be positive in the solution. If a value is negative it means the force acts in the opposite direction. the forces in several other members . If the directions we chose for the unknowns are correct. we can now proceed to analyze joint E and determine the force in member EF.Topic 2.htm (5 of 6) [5/8/2009 4:39:43 PM] . The right hand drawing in Diagram 6 is the FBD of joint E with all forces resolved into x and y-components. Joint E: Procedure .Method of Joints Step 1. FBD of the Joint E.See Diagram 7 ) file:///Z|/www/Statstr/statics/StatII/stat22. (Tension) EF = 712 lb. Step 3: Apply the Equilibrium Conditions: In Diagram 6. (Tension) Since both force values came out positive in our solution. Step 2: Resolve (Break) all forces into their x and y components. and EF (and along the way. We now apply the Equilibrium Conditions (for joints). this means that the initial directions selected for the forces were correct. the FBD of joint E with all external forces acting on the joint shown. We have now solved our problem. showing and labeling all forces acting on the joint. we have on the left. Include needed angles. BC. and our initial direction for the forces.2 Trusses Finally. (Sum of x-forces) (Sum of y-forces) Solving : EC = 55 lb. select Example 1 For additional Examples of Truss using Method of Joints.Topic 2.2 Trusses Additional Examples To see an example of finding internal forces in a truss using Method of Sections. select Example 2 For additional Examples of Truss using Method of Sections. select Example 3 Select: Topic 2: Statics II .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/stat22.htm (6 of 6) [5/8/2009 4:39:43 PM] . we can only solve for three unknowns). For this structure we wish to determine the values of the internal force (tension/compression) in members BC. or even curved depending on the problem. The internal forces in members BC. Example 1: In Diagram 1 we have a truss supported by a pinned joint at Point A and supported by a roller at point D. but not to cut through more than three unknowns (since we will have three equilibrium conditions equations. EC. The forces must act along the direction of the cut member (since all members in a truss are axial members). A vertical load of 500 lb. This vertical line cuts through members BC. we have cut through the original truss with a vertical line just to the right of member BE.Example 1 Topic 2. The criteria for this line is that we would like to cut through the unknown members (whose internal force value we wish to determine). cutting the truss once will enable us to find our selected unknowns. and a horizontal load of 800 lb. at some angle. and we have selected an initial direction either into or away from the section for each of the forces.Example 1 In the Statics . we will not repeat that portion. The support reactions acting on the structure at points A and D are shown. If we have selected an incorrect initial direction for a force. we ended with a sample truss example in which we determined the external forces acting on the truss and the internal forces in several members by Method of Joints.2 Trusses . EC. acts at point F. one may have to repeat Method of Sections several times to determine all the unknowns.Topic 2. we would like to use the same truss and solve for several internal forces by Method of Sections. the value will be negative indicating the force acts in the opposite direction of the one file:///Z|/www/Statstr/statics/StatII/stat22e1. or for finding more internal forces. We have represented these forces with the arrows shown. when we solve for the value of the force.htm (1 of 3) [5/8/2009 4:39:43 PM] . EC. In Diagram 2. and EF now become external forces with respect to this section. We have shown the section of the truss to the left of the cut. acts at Point C. For example 2. and EF (the selected members whose internal forces we wish to determine). but begin with the external support forces given and move to determine the internal forces in the selected members. In this example. and EF.Truss Page. horizontal. we will 'cut' the truss into two sections by drawing a line through the truss. in some trusses. This line may be vertical. however. And since we previously solved for the external support reactions.2 Trusses . In Method of Sections. We now treat this section of the truss as if it were a completely new structure. EC. I. showing and labeling all external forces.) (Sum of Torque with respect to point E. (T). Resolve (break) all forces into their x and y-components. EC acts out of the section and so is in tension. (Here we sum the x-forces) (Sum of y-forces. We may now proceed with the analysis of this structure using standard Static's techniques.2 .htm (2 of 3) [5/8/2009 4:39:43 PM] .Trusses file:///Z|/www/Statstr/statics/StatII/stat22e1. and indicating needed dimensions and angles. Apply the Equilibrium Equations : and solve for the unknown forces. including load forces. and EF by method of sections. (T). we have solved for the internal forces in the members BC. (T) (The negative sign for force EC means that we initially chose it in the incorrect direction. EF = 712 lb. (Diagram 3) II. Draw a Free Body Diagram of the structure (section).) Solving we obtain: BC = 44 lb. Thus.Example 1 chosen initially.2 Trusses . not into the section as shown in the FBD). III. EC = -50 lb.Topic 2. Return to Topic 2. Topic 2.2 Trusses - Example 1 Continue to Example 2 or select: Topic 2: Statics II - Applications-Topic Table of Contents file:///Z|/www/Statstr/statics/StatII/stat22e1.htm (3 of 3) [5/8/2009 4:39:43 PM] Topic 2.2 Trusses - Example 2 Topic 2.2 Trusses - Example 2 The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and supported by a roller at point D. For this structure we wish to determine the value of all the support forces acting on the structure, and to determine the force in member FC by method of joints. For the first part of the problem we proceed using our normal static's procedure. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces as well as any needed angles and dimensions. (Note in Diagram 2, we have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced the roller support by the vertical unknown force Dy. file:///Z|/www/Statstr/statics/StatII/stat22e2.htm (1 of 4) [5/8/2009 4:39:44 PM] Topic 2.2 Trusses - Example 2 STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions : Ax = 0 : Ay + Dy - 12,000 lbs - 20,000 lbs = 0 : (-12,000 lbs)(4 ft) - (20,000 lbs)(12 ft) + Dy(24 ft) = 0 Solving for the unknowns: Dy = 12,000 lbs; Ay = 20,000 lbs. These are the external support reactions acting on the structure. PART 2: Determine the internal force in member FC by method of joints. We begin at a joint with only two unknowns acting, joint D. JOINT D: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and load, and including any needed angles.(Diagram 3) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected. file:///Z|/www/Statstr/statics/StatII/stat22e2.htm (2 of 4) [5/8/2009 4:39:44 PM] Topic 2.2 Trusses - Example 2 STEP 2: Resolve all forces into x and y components. (Diagram 3). STEP 3: Apply equilibrium conditions: : -CD + ED cos (66.4o) = 0 : 12,000 lbs - ED sin (66.4o)= 0 Solving for the unknowns: ED = 13,100 lbs (C); CD = 5,240 lbs (T) Now that we have calculated the values for ED and CD we can move to joint E. We could not solve joint E initially as it had too many unknowns forces acting on it. JOINT E: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any needed angles. (Diagram 4) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected. STEP 2: Resolve all forces into x and y components. (Diagram 4). STEP 3: Apply equilibrium conditions: FE -(13,100 lbs) cos (66.4o) - CE cos (66.4o) = 0 :(13,100 lbs) sin (66.4o) - CE sin (66.4o) = 0 Solving for the unknowns: FE = 10,500 lbs (C); CE = 13,100 lbs (T) (Since force values were positive, the initial direction chosen for the forces was correct.) Now that we have calculated the values for FE and CE we move to joint C. We could not solve joint C file:///Z|/www/Statstr/statics/StatII/stat22e2.htm (3 of 4) [5/8/2009 4:39:44 PM] Topic 2.2 Trusses - Example 2 initially as it had too many unknowns forces acting on it. JOINT C: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any need angles. (Diagram 5) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected. STEP 2: Resolve all forces into x and y components. (Diagram 5). STEP 3: Apply equilibrium conditions: : 5,450 + (13,100 lbs) cos (66.4o) + FC cos (66.4o) - BC = 0 : 13,100 lbs sin (66.4o) - FC sin (66.4o) = 0 Solving for the unknowns: FC = 13,100 lbs (c); BC = 15,950 lbs (t) Thus, member FC is in compression with a force of 13, 100 lbs. Return to Topic 2.2 - Trusses or select: Topic 2: Statics II - Applications-Topic Table of Contents Statics & Strength of Materials - Course Table of Contents file:///Z|/www/Statstr/statics/StatII/stat22e2.htm (4 of 4) [5/8/2009 4:39:44 PM] Topic 2.2 Trusses - Example 3 Topic 2.2 Trusses -Example 3 The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and supported by a roller at point H. For this structure we wish to determine the value of all the support forces acting on the structure, and to determine the force in member DG by method of sections. We begin by determining the external support reactions acting on the structure. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. (Note in Diagram 1, we have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced the roller support by the vertical unknown force Hy STEP 2: Break any forces not already in x and y direction into their x and y components. (All forces in x/y directions.) STEP 3: Apply the equilibrium conditions. : Ax = 0 : Ay + Hy -12,000 lbs - 20,000 lbs - 10,000 lbs = 0 : (-12,000 lbs)(20 ft) - (20,000 lbs)(40 ft) - (10,000 lbs)(60 ft) + Hy(80 ft) = 0 Solving for the unknowns: Ay = 21,500 lbs; Hy = 20,500 lbs. These are the external support reactions acting on the structure. Part 2: Now we will find internal force in member DG by method of sections. Cut the truss vertically with a line passing through members DF, DG, and EG. We have shown the section of the truss to the right of the cut. We now treat this section of the truss as if it were a completely new structure. The internal forces in members DF, DG, and EG now become external forces with respect to this section. We file:///Z|/www/Statstr/statics/StatII/stat22e3.htm (1 of 2) [5/8/2009 4:39:44 PM] Topic 2.2 Trusses - Example 3 have represented these forces with the arrows shown. The forces must act along the direction of the cut member (since all members in a truss are axial members), and we have selected an initial direction either into or away from the section for each of the forces. If we have selected an incorrect initial direction for a force, when we solve for the value of the force, the value will be negative indicating the force acts in the opposite direction of the one chosen initially. We may now proceed with the analysis of this structure using standard Static's techniques. I. Draw a Free Body Diagram of the structure (section), showing and labeling all external forces, and indicating needed dimensions and angles. (Diagram 2) II. Resolve (break) all forces into their x and y-components. (Diagram 2) III. Apply the Equilibrium Equations ( :EG + DG cos (51.3o) + DF cos (22.6o) = 0 : -10,000 lbs + 20,500 lbs - DG sin (51.3o) - DF sin (26.6o) = 0 : -DF cos (26.6o)(15 ft) + (20,500 lbs)(20 ft) = 0 ) Solving for the unknowns: DF = 30,600 lbs (C); DG = -4,090 (opposite direction)= 4,090 lbs (T); EG = -24,800(opposite direction)= 24,800 lbs (T) Return to Topic 2.2 - Trusses or select: Topic 2: Statics II - Applications-Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/stat22e3.htm (2 of 2) [5/8/2009 4:39:44 PM] solution121 STATICS / STRENGTH OF MATERIALS - Example The structure shown below is a truss which is pinned to the wall at point F, and supported by a roller at point A. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force (tension or compression) in member EB by method of joints. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components STEP 3: Apply the equilibrium conditions. Sum Fx = Ax + Fx = 0 Sum Fy = Fy - 8,000 lbs - 6,000 lbs = 0 Sum TA = (-8,000 lbs)(8 ft) - (6,000 lbs)(16 ft) - Fx(10 ft) = 0 Solving for the unknowns: Ax = 16,000 lbs; Fx = -16,000 lbs; Fy = 14,000 lbs file:///Z|/www/Statstr/statics/Stests/statics/sol121.htm (1 of 3) [5/8/2009 4:39:44 PM] solution121 PART C - Now find internal force in member EB by method of joints. JOINT F: STEP 1: Draw a free body diagram of the joint. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply equilibrium conditions: Sum Fx = FE - 16,000 lbs = 0 Sum Fy = 14,000 lbs - FA = 0 Solving for the unknowns: FE = 16,000 lbs (t); FA = 14,000 lbs (t) JOINT A: STEP 1: Draw a free body diagram of the joint. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply equilibrium conditions: Sum Fx = 16,000 lbs - AB -AE cos (68.2o) = 0 Sum Fy = 14,000 lbs - AE sin (68.2o) = 0 Solving for the unknowns: AE = 15,080 lbs (c); AB = 10,400 lbs (c) JOINT E: STEP 1: Draw a free body diagram of the joint. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply equilibrium conditions: Sum Fx = -16,000 lbs + ED + 5,600 lbs + EB cos (68.2o) = 0 Sum Fy = 14,000 lbs - EB sin (68.2o) = 0 Solving for the unknowns: ED = 4,800 lbs (t); EB = 15,080 lbs (t) file:///Z|/www/Statstr/statics/Stests/statics/sol121.htm (2 of 3) [5/8/2009 4:39:44 PM] solution121 Select Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol121.htm (3 of 3) [5/8/2009 4:39:44 PM] solution122 STATICS / STRENGTH OF MATERIALS - Example The structure shown below is a truss which is pinned to the wall at point E, and supported by a roller at point A. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force in member GC by method of sections. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B: file:///Z|/www/Statstr/statics/Stests/statics/sol122.htm (1 of 3) [5/8/2009 4:39:45 PM] solution122 STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = Ex = 0 Sum Fy = Ay + Ey - 12,000 lbs - 6,000 lbs = 0 Sum TE = (12,000 lbs)(4 ft) - Ay(12 ft) = 0 Solving for the unknowns: Ey = 14,000 lbs; Ay = 4,000 lbs PART C: - Now find the internal force in member GC by method of section STEP 1: Cut the structure into "2 sections" with a vertical line which cuts through members GE, GC and BC, and is just to the right of point G (see diagram). The internal forces in members GE, GC and BC now become external forces acting on the left hand section as shown. (We chose directions file:///Z|/www/Statstr/statics/Stests/statics/sol122.htm (2 of 3) [5/8/2009 4:39:45 PM] solution122 for these forces which may or may not be correct, but which will become clear when we solve for their values.) STEP 2: Now treat the section shown as a new structure and apply statics procedure - Draw a free body diagram of the left hand section. - Resolve all forces into x an y components (see diagram). - Apply equilibrium conditions: Sum Fx = - GE cos (37o) + GC cos (37o) + BC = 0 Sum Fy = 4,000 lbs - GE sin (37o) - GC sin (37o) = 0 Sum TG = (-4,000 lbs)(4 ft) + BC(3 ft) = 0 Solving for the unknowns: BC = 5,330 lbs; GE = 6,670 lbs; GC = 0 lbs Select Topic 2: Statics II - Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol122.htm (3 of 3) [5/8/2009 4:39:45 PM] solution123 STATICS / STRENGTH OF MATERIALS - Example The structure shown below is a truss which is pinned to the floor at point D, and supported by a roller at point A. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force in member GC by method of joints. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, file:///Z|/www/Statstr/statics/Stests/statics/sol123.htm (1 of 4) [5/8/2009 4:39:45 PM] AB = 7.Now find internal force in member GC by method of joints.000 lbs)(8 ft) .700 lbs PART C .3o) = 0 Solving for the unknowns: AG = 12.3o) = 0 Sum Fy = 10.140 lbs (t) JOINT B: file:///Z|/www/Statstr/statics/Stests/statics/sol123. STEP 3: Apply equilibrium conditions: Sum Fx = AB .000 lbs .300 lbs. Sum Fx = Dx = 0 Sum Fy = Ay + Dy . Ay = 10.12.700 lbs .000 lbs)(4 ft) = 0 Solving for the unknowns: Dy = 11. STEP 2: Break any forces not already in x and y direction into their x and y components.AG cos (56.solution123 as well as any needed angles and dimensions.10. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply the equilibrium conditions.(10. JOINT A: STEP 1: Draw a free body diagram of the joint.AG sin (56.000 lbs = 0 Sum TA = Dy(12 ft) (12.htm (2 of 4) [5/8/2009 4:39:45 PM] .900 lbs (c). ED sin (56.400 lbs (t).12.7.htm (3 of 4) [5/8/2009 4:39:45 PM] .3o) = 0 Solving for the unknowns: CD = 7. GC = 730 lbs (t) Select Topic 2: Statics II .140 lbs .600 lbs (c) JOINT C: STEP 1: Draw a free body diagram of the joint.3o) =0 Solving for the unknowns: CE = 11.300 lbs .560 lbs . STEP 3: Apply equilibrium conditions: Sum Fx = -CD +ED cos (56.140 lbs (t) JOINT D: STEP 1: Draw a free body diagram of the joint.GC sin (56.560 lbs (t).Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol123. ED = 13. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply equilibrium conditions: Sum Fx = 7. STEP 2: Resolve all forces into x and y components (see diagram). BC = 7.140lbs + BC = 0 Sum Fy = BG = 0 Solving for the unknowns: BG = 0 lbs .3o) = 0 Sum Fy = CE .GC cos (56.solution123 STEP 1: Draw a free body diagram of the joint.000 lbs .3o) = 0 Sum Fy = 11. STEP 2: Resolve all forces into x and y components (see diagram). STEP 3: Apply equilibrium conditions: Sum Fx = -7. solution123 file:///Z|/www/Statstr/statics/Stests/statics/sol123.htm (4 of 4) [5/8/2009 4:39:45 PM] . For this structure: A. Draw a Free Body Diagram showing all support forces and loads.solution124 STATICS / STRENGTH OF MATERIALS . B. Determine the force in member FC by method of joints. all joints and support points are assumed to be pinned or hinged joints. as file:///Z|/www/Statstr/statics/Stests/statics/sol124. and supported by a roller at point D. Determine the value of all the support forces acting on the structure.Example The structure shown below is a truss which is pinned to the floor at point A. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.htm (1 of 4) [5/8/2009 4:39:45 PM] . C. Unless otherwise indicated. Ay = 20.000 lbs. Sum Fx = Ax = 0 Sum Fy = Ay + Dy .12.solution124 well as any needed angles and dimensions.000 lbs)(12 ft) + Dy(24 ft) = 0 Solving for the unknowns: Dy = 12.(20.000 lbs)(4 ft) .000 lbs . STEP 3: Apply the equilibrium conditions.000 lbs = 0 Sum TA = (-12.Now find internal force in member FC by method of joints.htm (2 of 4) [5/8/2009 4:39:45 PM] .000 lbs PART C . STEP 2: Break any forces not already in x and y direction into their x and y components.20. JOINT D: file:///Z|/www/Statstr/statics/Stests/statics/sol124. 4o) = 0 Solving for the unknowns: FE = 10.Topic Table of Contents file:///Z|/www/Statstr/statics/Stests/statics/sol124.100 lbs) sin (66.FC sin (66. STEP 3: Apply equilibrium conditions: Sum Fx = (-13. CD = 5.4o) +FE . STEP 2: Resolve all forces into x and y components (see diagram).000 lbs .100 lbs (c).100 lbs (c).100 lbs (t) JOINT C: STEP 1: Draw a free body diagram of the joint.4o) = 0 Solving for the unknowns: FC = 13.BC = 0 Sum Fy = (13.100 lbs) cos (66. STEP 2: Resolve all forces into x and y components (see diagram).CE sin (66. STEP 3: Apply equilibrium conditions: Sum Fx = -CD + ED cos (66.4o) + FC cos (66. BC = 15.500 lbs (c).4o)= 0 Solving for the unknowns: ED = 13.4o) = 0 Sum Fy = 12.4o) .100 lbs) cos (66.4o) .450 lbs (t) JOINT E: STEP 1: Draw a free body diagram of the joint.ED sin (66.4o) .100 lbs) sin (66. CE = 13.htm (3 of 4) [5/8/2009 4:39:45 PM] .solution124 STEP 1: Draw a free body diagram of the joint.4o) = 0 Sum Fy = (13. STEP 3: Apply equilibrium conditions: Sum Fx = 5. STEP 2: Resolve all forces into x and y components (see diagram).450 + (13.950 lbs (t) Select Topic 2: Statics II .CE cos (66. solution124 Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol124.htm (4 of 4) [5/8/2009 4:39:45 PM] . Draw a Free Body Diagram showing all support forces and loads. Unless otherwise indicated. Determine the force in member FB by any method. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. all joints and support points are assumed to be pinned or hinged joints. B. For this structure: A. C. Determine the value of all the support forces acting on the structure.solution125 STATICS / STRENGTH OF MATERIALS . file:///Z|/www/Statstr/statics/Stests/statics/sol125.htm (1 of 3) [5/8/2009 4:39:46 PM] .Example The structure shown below is a truss which is pinned to the floor at point A and also at point H. STEP 1: Draw a free body diagram of the lower section (see diagram).htm (2 of 3) [5/8/2009 4:39:46 PM] . Ay = -30.000 lbs)(15 ft) = 0 Solving for the unknowns: Hy = 40. Cut horizontally through members BC. Sum Fx = Ax + 8. and FG.Now find internal force in member FB by section method.000 lbs PART C . STEP 3: Apply the equilibrium conditions.solution125 as well as any needed angles and dimensions.000 lbs + FB sin (51.000 lbs = 0 file:///Z|/www/Statstr/statics/Stests/statics/sol125. FB. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 2: Resolve all forces into x and y components (see diagram).8. Analyze lower section.000 lbs = 0 Sum Fy = BC 30.3o) .000 lbs)(4 ft) . STEP 3: Apply equilibrium conditions: Sum Fx = FB cos (51. Ax = -8.000 lbs.(10.(8.GF + 40.000 lbs = 0 Sum Fy = Ay + Hy .000 lbs.000 lbs = 0 Sum TA= Hy(4 ft) .3o) .10. htm (3 of 3) [5/8/2009 4:39:46 PM] .GF(4 ft) + (40.000 lbs (t) Select Topic 2: Statics II .solution125 Sum TB = (-8.000 lbs)(4 ft) = 0 Solving for the unknowns: FB = 12. BC = 10.Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol125.000 lbs (c).8000 lbs (t). GF = 30.000 lbs)(5 ft) . all joints and support points are assumed to be pinned or hinged joints. Determine the force in member CD by any method. Draw a Free Body Diagram showing all support forces and loads.Example The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point F. For this structure: A. file:///Z|/www/Statstr/statics/Stests/statics/sol126. Unless otherwise indicated. Solution: PARTS A & B: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.solution126 STATICS / STRENGTH OF MATERIALS . Determine the value of all the support forces acting on the structure. B. C.htm (1 of 3) [5/8/2009 4:39:46 PM] . CD sin (45o) = 0 Sum TC = (-3.solution126 as well as any needed angles and dimensions. STEP 3: Apply the equilibrium conditions: Sum Fx = BD cos (14o) . CD. Analyze left section as shown in diagram. Sum Fx = Ax = 0 Sum Fy = -4.360 lbs PART C .CE = 0 Sum Fy = -4. and CE near to points B and C.htm (2 of 3) [5/8/2009 4:39:46 PM] .000 lbs + 3. STEP 3: Apply the equilibrium conditions.360 lbs . Ay = 3.3.000 lbs + Ay + Fy = 0 Sum TA = (-4. STEP 2: Resolve all forces into x and y components. STEP 1: Draw a free body diagram of the left section.Now find internal force in member CD by sections.CD cos (45o) .000 lbs . Cut vertically through members BD.BD cos (14o)(15 ft) = 0 file:///Z|/www/Statstr/statics/Stests/statics/sol126.640 lbs.BD sin (14o) .360 lbs)(12 ft) .000 lbs)(24 ft) + Fy(33 ft) = 0 Solving for the unknowns: Fy = 3.000 lbs)(12 ft) (3. showing and labeling all external loads and forces. STEP 2: Break any forces not already in x and y direction into their x and y components. CE =2.770 lbs (c).770 lbs (opposite direction).htm (3 of 3) [5/8/2009 4:39:46 PM] .solution126 Solving for the unknowns: BD = -2.Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/statics/sol126. CD = 43 lbs (c).720 lbs (t) Select Topic 2: Statics II . 2. Fx = . Fy = 9. Determine the external force support reactions. The structure shown is a truss composed of axial members pinned together at the joints. and the force in member CD by method of joints.000 lb. (Ay = -9. Dy = 4800 lb.000 lb.000 lb.Trusses 1 1. The structure shown is a truss composed of axial members pinned together at the joints.htm (1 of 4) [5/8/2009 4:39:46 PM] . The stucture is pinned to the floor at points A and F. and the force in member BE by method of joints. BE = 10. (Ax = -4000 lb. CD = 970 lb (C)) file:///Z|/www/Statstr/statics/StatII/statp22t.000 lb (T)) 2.Assignment Problems .10. The stucture is pinned to the floor at point A and supported by a roller at point D.Trusses Statics & Strength of Materials Problem Assignment . Ay = 200 lb. Determine the external force support reactions. 22. and the force in members DE.140 (C). HC and DC by method of sections. DH = 14. (Fx = -28. Determine the external force support reactions and the force in members HI. HI = 12. IH = 35. HC = 3000 lb (T).000 lb. The stucture is pinned to the floor at point E and supported by a roller at point F.000 lb.htm (2 of 4) [5/8/2009 4:39:46 PM] .650 lb (T). CD = 10. The structure shown is a truss composed of axial members pinned together at the joints. Determine the external force support reactions.000 lb.500 lb. The structure shown is a truss composed of axial members pinned together at the joints.Trusses 3. Gy = 35. Fy = .500 lb (C)) file:///Z|/www/Statstr/statics/StatII/statp22t. DH and IH by method of sections (Ay = 35. The stucture is supported by a roller at point A and pinned to the floor at point G. DE = 25.500 lb.000 lb (T)) 4.000 lb (C).Assignment Problems .Ey = 22. Ly = 18. The stucture is pinned to the floor at point A and supported by a roller at point L.Assignment Problems .600 (C). DG = 10. EG = 23.Trusses 5. 420 (T).330 (T)) Select: Topic 2: Statics II .Determine the external force support reactions and the force in members DF. DG and EG by any method. (Ay = 14.000 lb. DF = 30.000 lb.htm (3 of 4) [5/8/2009 4:39:46 PM] . The structure shown is a truss composed of axial members pinned together at the joints.Applications-Topic Table of Contents file:///Z|/www/Statstr/statics/StatII/statp22t. Trusses Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp22t.Assignment Problems .htm (4 of 4) [5/8/2009 4:39:46 PM] . [Ay = 6.Truss probs 2 Statics & Strength of Materials Problem Assignment . AB = 8.230 lb. Calculate the forces in all members of the truss shown in the following diagram using the method of joints.Trusses 2 Note the typical designation of pin and roller supports in the diagrams shown.000 lb.htm (1 of 5) [5/8/2009 4:39:46 PM] . AC = 15. Cy = 8460 lb. BC=17.600 lb (C). Ax = -10.600 lb (C)] 2.375 lb.540 lb.420 lb) file:///Z|/www/Statstr/statics/StatII/statp22t2. Calculate the forces in all member of the truss shown in the following diagram using the method of joints. 1. [AB =10. CB = 10... Truss probs 2 3. Ax = -500 lb. BC = 950 lb.. Cy = 475 lb.htm (2 of 5) [5/8/2009 4:39:46 PM] .. AB = 450 lb.. Calculate the forces in all members of the truss shown in the following diagram using the method of joints. AD = 890 lb..) file:///Z|/www/Statstr/statics/StatII/statp22t2. (Ay = 225 lb. BF=0..(T)] file:///Z|/www/Statstr/statics/StatII/statp22t2. CD = 25.000 lb.000 lb. (C).. BC = 21. Fy=-20. AB = 0. right side same by symmetry] 5.000 lb.000 lb. CE = 90.. [Ey =90. (C). CG = 50. (C).360 (T).000 lb. BC = CD =65. (C).000 lb (T).. CF = 56.220 lb.htm (3 of 5) [5/8/2009 4:39:46 PM] . AF = FG = 40. Calculate the forces in all members of the truss shown in the following diagram using the method of joints. BD=20. (T).000 lb. [Ay=Ey=40. EF = 0. AB =56.000 lb.000 lb. DF = 20. BG = 35. (C).580 lb.580 lb.000 lb. (C).000 lb. Calculate the forces in all members of the trusses shown in the following diagram using the method of joints.Truss probs 2 4. Fx=40. (C). AC = 35.000 lb. [Ax=-26. Ex=26. BC = 8.. Ay=15. (T).366 lb.250 lb. (T)..000 lb. for the cantilever truss shown below.500 lb (C)] file:///Z|/www/Statstr/statics/StatII/statp22t2.Truss probs 2 6.250 lb.750 lb. FG=17. BG. by method of joints. BG=17.. and FG. Calculate the forces in members BC.htm (4 of 5) [5/8/2009 4:39:46 PM] . htm (5 of 5) [5/8/2009 4:39:46 PM] .Applications-Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp22t2.Truss probs 2 Select: Topic 2: Statics II . htm (1 of 4) [5/8/2009 4:39:47 PM] . and IJ. and AC. Determine the truss reactions and the force in members BD. Determine the truss reactions and the forces in members CD. ID. 2. file:///Z|/www/Statstr/statics/StatII/statp22t3.Trusses 3 1. Solve the following truss by the method of sections. Solve the following truss by the method of sections.Truss probs 3 Statics & Strength of Materials Problem Assignment . BC. Truss probs 3 3.htm (2 of 4) [5/8/2009 4:39:47 PM] . and MN. Determine the truss reactions and the forces in members EF. FK. KL. Solve the following truss by the method of sections. file:///Z|/www/Statstr/statics/StatII/statp22t3. Determine the truss reactions and the forces in members DE. and 4. JE. Solve the following truss by the method of sections. 6.Truss probs 3 5. DH. Solve the following truss by the method of sections. Determine the truss reactions and the force in members CD. JE. AND JI. Solve the following truss by the method of sections. Determine the truss reactions and the force in members DE. and HI. file:///Z|/www/Statstr/statics/StatII/statp22t3.htm (3 of 4) [5/8/2009 4:39:47 PM] . CI. and IJ by the method of sections. Select: Topic 2: Statics II .Truss probs 3 7. determine the support reactions and the forces in members BC. For the Howe roof truss shown below.Applications-Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statp22t3.htm (4 of 4) [5/8/2009 4:39:47 PM] . C. Draw a Free Body Diagram showing all support forces and loads. B. Unless otherwise indicated.) In the structure shown below. BDE and CD are assumed to be solid rigid members. For answers to Problem 1 select Solution-Exam Problem 1 2.Applications SAMPLE EXAMINATION 1. For this structure: file:///Z|/www/Statstr/statics/StatII/statx23. all joints and support points are assumed to be pinned or hinged joints. members ABC . The structure is pinned at A and supported by a roller at E. Determine the force (tension or compression) in member CD.) The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point H. Determine the value of all the support forces acting on the structure. For this structure: A.Sample Examination Topic 2: Statics II .htm (1 of 2) [5/8/2009 4:39:47 PM] . Determine the force in member DG by method of sections. Draw a Free Body Diagram showing all support forces and loads. all joints and support points are assumed to be pinned or hinged joints. B. C.Applications . For answers to Problem 2 select Solution-Exam Problem 2 Select: Topic 2: Statics II . Determine the value of all the support forces acting on the structure. Unless otherwise indicated.Sample Examination A.htm (2 of 2) [5/8/2009 4:39:47 PM] .Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/statx23. Sum Fx = Ax = 0 Sum Fy = Ay + Ey . B. STEP 2: Break any forces not already in x and y direction into their x and y components. Solution: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. C.4. The structure is pinned at A and supported by a roller at E. Draw a Free Body Diagram showing all support forces and loads. BDE and CD are assumed to be solid rigid members. as well as any needed angles and dimensions.000 lbs . Unless otherwise indicated. Determine the value of all the support forces acting on the structure.000 lbs = 0 file:///Z|/www/Statstr/statics/StatII/xsol1.htm (1 of 2) [5/8/2009 4:39:47 PM] . all joints and support points are assumed to be pinned or hinged joints. For this structure: A.Exam Problem 1 In the structure shown below members ABC . STEP 3: Apply the equilibrium conditions.Sample Exam solution 1 Solution .8. Determine the force (tension or compression) in member CD. Bx = 4.000 lbs + 5.Sample Exam solution 1 Sum TA = (-8.4. STEP 1: Draw a free body diagram of a member that CD acts on .140 lbs.880 lbs.880 lbs = 0 Sum TB = -CD sin (60o)(3 ft) . STEP 2: Resolve all forces into x and y components (see diagram).Now find internal force in member CD.5 ft) + Ey(8.120 lbs PART C .000 lbs)(6. Return to Statics II .880 lbs)(7 ft) = 0 Solving for the unknowns: CD = 8.170 lbs These are the external forces acting on member BDE.Sample Exam or Select: Statics Applications .(4.(4. The force in CD is 8.htm (2 of 2) [5/8/2009 4:39:47 PM] . Ay = 6.070 lbs. STEP 3: Apply the equilibrium conditions: Sum Fx = -Bx + CD cos (60o) = 0 Sum Fy = By . By = 5.CD sin (60o) .140 lbs (c).5 ft) = 0 Solving for the support reactions: Ey = 5.Topic Table of Contents Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/xsol1.000 lbs)(3 ft) .000 lbs)(5 ft) + (5.member BDE. all joints and support points are assumed to be pinned or hinged joints. as well as any needed angles and dimensions. Hy = 20.Exam Problem 2 The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point H. file:///Z|/www/Statstr/statics/StatII/xsol2. DG.500 lbs PART C .Sample Exam Solution 2 Solution . For this structure: A.000 lbs)(40 ft) . C. and EG.Now find the internal force in member DB by section method. Analyze right hand section. B. STEP 2: Break any forces not already in x and y direction into their x and y components. Cut vertically through members DF.500 lbs.000 lbs .000 lbs)(60 ft) + Hy(80 ft) = 0 Solving for the unknowns: Ay = 21. Solution: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Unless otherwise indicated.20.000 lbs)(20 ft) .(20. Determine the value of all the support forces acting on the structure.000 lbs = 0 Sum TA = (-12.(10.10. Sum Fx = Ax = 0 Sum Fy = Ay + Hy -12. Determine the force in member DG by method of sections.000 lbs . Draw a Free Body Diagram showing all support forces and loads.htm (1 of 2) [5/8/2009 4:39:47 PM] . STEP 3: Apply the equilibrium conditions. 3o) .800 lbs (T) Return to Statics II .090 lbs (T).500 lbs)(20 ft) = 0 Solving for the unknowns: DF = 30.090 (opposite direction) or DG= 4.DG sin (51.htm (2 of 2) [5/8/2009 4:39:47 PM] .Topic Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/StatII/xsol2. STEP 2: Resolve all forces into x and y components (see diagram).000 lbs + 20.Sample Exam Solution 2 STEP 1: Draw a free body diagram of the right hand section (see diagram).6o) = 0 Sum Fy = -10.800(opposite direction) or EG=24.6o)(15 ft) + (20. STEP 3: Apply equilibrium conditions: Sum Fx = EG + DG cos (51. DG = -4.3o) + DF cos (22.600 lbs (C).DF sin (26.Sample Exam or select: Topic 2: Statics II .500 lbs . EG = 24.6o) = 0 Sum TG = -DF cos (26.Applications . #8.7 Stress.Example 2 r 3.4a Shear Stress & Strain .Example 3 r Additional Examples: #4. Strain & Deformation II 3.1: Stress. knowing the force values is not enough.83 Thermal Stress.1 .Determinate [supplemental] r 3.Example 2 r 3. file:///Z|/www/Statstr/statics/Stress/strs31. Strain & Deformation I 3.I In the first general topic (Statics) we examined the process of determining both the external support forces acting on a structure and the internal forces acting in members of a structure (particularly axial members).2c Statically Determinate .Topic 3.Example 2 3.82c Mixed Mechanical/Thermal .4 Shear Stress & Strain r 3.Example 1 r 3.Example 1 r 3. Strain.3b Statically Indeterminate . #7.Strain & Deformation .Example 3 r Additional Examples: #4. somewhat harder] 3. Strain & Deformation . #6.Determinate [required] 3. Strain & Hooke's Law ● ● ● ● ● ● ● ● ● ● ● ● 3. Strain & Hooke's Law .6 Problem Assignment -Stress/Strain Indeterminate [required. While this is important and in fact indispensable when designing structures or determining the safety of a loaded structure.81 Thermal Stress.Stress/Strain Determinate r 3.Determinate [required] r 3. Hooke's Law .4b Shear Stress & Strain .Topic Examination 3.3 Statically Indeterminate Structures r 3.1 Stress.84 Thermal Stress. #9 [Previous test problems] 3.82 Mixed Mechanical/Thermal Examples r 3. Strain & Hooke's Law .5c Problem Assignment 3 .I Topic 3: Stress. We can see that from a simple example. #5. #6.II r 3.5a Problem Assignment 1 .Example 1 r 3.5b Problem Assignment 2 .Stress. Strain.2 Stress.2b Statically Determinate .Topic Examination Topic 3.Example 1 r 3. #7. #5. #8.82a Mixed Mechanical/Thermal .2a Statically Determinate .I 3. Hooke's Law .Assignment Problems [required] 3.htm (1 of 4) [5/8/2009 4:39:48 PM] . #9 [Previous test problems] 3.3a Statically Indeterminate . Strain & Hooke's Law .5 Problem Assignments .82b Mixed Mechanical/Thermal .Example 2 3.8 Thermal Stress. is under simple tension due to force F. So axial stress is really the 'pressure' in a solid member. acting at an angle of 150o (from +x-axis) This is also the value of the internal tension in the cable. Or In diagram 2. As you remember. is defined as the force perpendicular to the cross sectional area of the member divided by the cross sectional area. AXIAL STRESS What is known as Axial (or Normal) Stress. If we divide that axial force.. Now the question becomes.000 lb is attached to the structure at point C. this quotient would be the axial stress in the member. acting at 37o (along the direction of the member). we could ask the question. Is this structure safe? Are members BC and AC strong enough to support the load? We recognize right away that knowing the force in the cable BC is not enough to tell us if the cable is safe or if it will break. To address the first consideration. and cable BC which is pinned to the wall at point B. 13. Now. a load of 15. which is also the internal force in member AC.Stress.1 .090 lb.I In Diagram 1.090 lb. F. It depends on the size of the cable. A steel cable will clearly support more than an aluminum cable. Clearly it depends on several other factors in addition to the force in the cable.180 lb. Axial stress is the equivalent of pressure in a gas or liquid. pressure is the force/unit area. At point B. the external support force of the wall on the cable has a value of 13. often symbolized by the Greek letter sigma. we will turn to the concept of STRESS.htm (2 of 4) [5/8/2009 4:39:48 PM] . A 1" diameter steel cable will carry more load than a ¼ " diameter steel cable.Topic 3. by the cross sectional area of the rod (A). If we solve for the forces acting on and in the structure we will find that at point A there is a support force of 14. Strain & Hooke's Law . file:///Z|/www/Statstr/statics/Stress/strs31. It also depends on what the cable is made of. In addition. the structure shown is composed of axial member AC which is pinned to the floor at point A. how much 'pressure' can a material bear before it fails. a solid rod of length L.180 lb (compression). as shown. 14. file:///Z|/www/Statstr/statics/Stress/strs31.180 lb.htm (3 of 4) [5/8/2009 4:39:48 PM] .000 lb/in2.700 lb/in2 For the member AC: Axial Stress = F/A = 14.I Well.090 lb. and solve for the cable area needed. put in the axial force in the cable. Now let's return to our example shown in Diagram 1 (repeated in Diagram 3). however. what size cable should we use? Another interesting question whose answer we find by simply reversing our process. We see from the calculations that the stress in member AC (11. we will examine that question in some detail in a bit. using the stress equation to find the minimum size cable for the allowable stress of 30.700 lb/in2) is over twice the allowable stress of 30.000 lb/in2.25"2) = 66.820 lb/ in2) is well within the allowable stress of 30. we set the stress value to the allowable stress of 30. This means that the ½ inch diameter cable is much too small to support the load.000 lb/in2.000 lb/in2? For the cable BC: Axial Stress = F/A = 13.2 in2. a normal operating stress for carbon steel might be 30. but to give an example. and if the cross sectional area of the member is 1.Stress. are the stresses in the cable BC and in member AC within the 'allowable' stress for steel of 30./ (1. if we assume both the member and the cable are made of steel./ (p * .000 lb/in2.5 inches. That is. Well.000 lb/in2.2 in2) = 11.Topic 3. we also see clearly that the stress in the cable AC (66. In our structure .1 . and if the diameter of the cable is .820 lb/in2 These are interesting results. Strain & Hooke's Law . Our next step is to examine an associated property of stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs31. We checked the members in the structure. strain.436 in2.14 (r2).000 lb/in2 = 13./A. So the minimum diameter steel cable which would safely support the load is d = . found one was not safe according to the allowable stress for the material. solving for A = . Select Topic 3. and to examine the stress and strain in materials in somewhat more detail.1 .htm (4 of 4) [5/8/2009 4:39:48 PM] .14) = . Since the area of cable = 3. This is an important process. we can solve for the radius r = square root (.Stress.Topic 3.436 in2/3.2: Stress.II or Select: Topic 3: Stress. Strain & Hooke's Law . Strain & Hooke's Law .I Axial Stress = F/A : 30. and then calculated the size member needed so that the structure would be safe. Strain & Hooke's Law .746 inches ( or ¾ inch diameter cable).090 lb.373 inches. the amount of deformation is doubled. if the amount of force is doubled.II Topic 3.2: Stress. we see that. That is.Topic 3. we have deformation. the length (and other dimensions) change due to applied loads and forces. if our metal rod is tested by increasing the tension in the rod. In real members. If we then graph the tension (force) verses the deformation we obtain a result as shown in diagram 2.Stress. This is a form of Hooke's Law and could be file:///Z|/www/Statstr/statics/Stress/strs32. In fact. the deformation increases.II In our first topic. Strain & Hooke's Law . we examined structures in which we assumed the members were rigid . of course. if we look at a metal rod in simple tension as shown in diagram 1. In the first region the deformation increases in proportion to the force.htm (1 of 3) [5/8/2009 4:39:48 PM] .rigid in the sense that we assumed that the member did not deform due to the applied loads and resulting forces. That is. Strain & Hooke's Law . Static Equilibrium.2 . In diagram 2. we see that there will be an elongation (or deformation) due to the tension. these two points are not quite the same.2 .) file:///Z|/www/Statstr/statics/Stress/strs32. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). so we may write: Strain (where Lo is the original length of the member) Strain has no units . that is. it will not return to its original size and shape. and is usually symbolized by an E or Y.' in some texts. In the elastic (linear) region. we will treat them as the same in this course. Finally the rod actually breaks. Although these two points are slightly different. we will instead graph the axial stress verses the axial strain (diagram 3). rather than examining the applied force and resulting deformation. permanent deformation has occurred. or the proportional limit. the diameter becomes smaller as the rod is about to fail./in. but rather a small amount of increase in force produces a large amount of deformation. if the force is taken off the sample. Next.Stress. Strain & Hooke's Law . In this region. or: . As we see from diagram 3. the rod often begins to 'neck down'. The Elastic Limit is the point at which permanent deformation occurs. We have defined the axial stress earlier. the ratio of stress/strain will be a constant (and actually equal to the slope of the linear portion of the graph). This constant is known as Young's Modulus. The point at which the Elastic Region ends is called the elastic limit. since stress is directly proportional to strain. The axial strain is defined as the fractional change in length or Strain = (deformation of member) divided by the (original length of member) . the Stress verses Strain graph has the same shape and regions as the force verses deformation graph in diagram 2.htm (2 of 3) [5/8/2009 4:39:48 PM] . (This is another form of Hooke's Law. after the elastic limit.since its length divided by length. We will use E for Young's modulus. however it is sometimes expressed as 'in. In actuality. After enough force has been applied the material enters the plastic region where the force and the deformation are not proportional.II written this way: F = k (deformation). that is. and the deformation is often represented by the Greek symbol delta(δ). We may now write Young's Modulus = Stress/Strain.Topic 3. where k is a constant depending on the material (and is sometimes called the spring constant). Strain is often represented by the Greek symbol epsilon(ε). II The value of Young's modulus . and says simply that the amount of deformation which occurs in a member is equal to the product of the force in the member and the length of the member (usually in inches) divided by Young's Modulus for the material. Example 3 or Select: Topic 3: Stress.depends on the material.which is a measure of the amount of force needed to produce a unit deformation .Topic 3. for Aluminum E = 10 x 106 lb/in2. Strain & Hooke's Law . For more values. and divided by the cross sectional area of the member. To see applications of these relationships. select: Young's Modulus .htm (3 of 3) [5/8/2009 4:39:48 PM] . To summarize our stress/strain/Hooke's Law relationships up to this point. Strain & Hooke's Law . we have: The last relationship is just a combination of the first three. Young's Modulus for Steel is 30 x 106 lb/in2.Stress. we now will look at several examples. Example 2 .Table.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs32. Continue to: Example 1 .2 . and for Brass E = 15 x 106 lb/in2. Example 1 Topic 3. hopefully familiar. (Young's Modulus for steel = 30 x 106 lb/in2.) Step I: As the first step in our solution. the strain in member CE. For this structure we would like to determine the axial stress in cable CE.2a:Statically Determinate . and supported by steel cable CE.2a: Statically Determinate . procedure is as follows: file:///Z|/www/Statstr/statics/Stress/strse32a. Our.Example 1 In the structure shown in Diagram 1. member ABCD is a solid rigid member pinned to the wall at A.Statics).Topic 3.htm (1 of 3) [5/8/2009 4:39:48 PM] . we apply static equilibrium conditions to determine the value of the external support reactions (the process we studied in Topic 1 . Cable CE is pinned to the wall at E and has a diameter of 1 inch. and finally to determine the movement of point D due to the applied loads. the deformation of member CE. 46 ft) = 0 Solving for the unknowns: E = 25. 2: Resolve all forces into their x and y components.2a:Statically Determinate . there is only one axial member (CE) attached to the wall.10.. we determine the value of the force in the internal member in which we would like to find the stress.000 lbs . Step II. Sum Fx = Ax .htm (2 of 3) [5/8/2009 4:39:48 PM] . In this particular problem. Therefore the force of the wall acting on member CE is equal to the internal force in the member itself: Force in CE = 25. Ax = 22.600 lb. this is quite easy once we recognize that at point E.200 lb.Example 1 1: Draw a free body diagram showing and labeling all load forces and support forces.Topic 3.E sin (30o)(23. file:///Z|/www/Statstr/statics/Stress/strse32a. Now that we have the values of the external support forces. 3: Apply the static equilibrium conditions. Ay = 17.000 lbs = 0 Sum TA = (20. as well as any needed angles and dimensions (Diagram 2).600 lb..8 ft) + (10.E cos (30o) = 0 Sum Fy = Ay + E sin (30o) .000 lbs)(16 ft) + E cos (30o)(2 ft) .170 lb.000 lbs)(4.20. The amount that point D moves is related to the amount point C moves.) Movement of point D.Example 1 Once we have the force in member CE.5 in)2= 32. As is shown in diagram 3.600 lbs/(3. we can apply the appropriate stress/strain relationships and solve for the quantities of interest.which is pinned at point A .209 in C. Strain & Hooke's Law .Strain & Hooke's Law .2a:Statically Determinate .Stress.5 in)2 ) = . B.600 psi.600 lbs)(16 ft * 12 in/ft) / (30*106 lbs/in2 )(3.2 . and solving gives us Movement of D = .) Stress in CE = F/A = 25. Return to: Topic 3.) Deformation of CE = (FL/ EA)CE = (25. We can relate the movement of the two points from geometry by: [Movement of C / 12 ft = Movement of D / 20 ft] or [.00109 D. This part of the problem requires a bit of reflection on the geometry of the problem. member ACD .Topic 3.rotates downward.209 in / 12 ft = Movement of D / 20 ft].II or Select: Topic 3: Stress.348 in.) Strain in CE = (Deformation of CE)/(Length of CE) = (.14*(.14*(.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse32a. and point C moves the same amount that cable CE deforms.htm (3 of 3) [5/8/2009 4:39:48 PM] . A.209"/192") = . when member CE elongates. and is pinned to member ABC at point B.2b: Statically Determinate . (Young's Modulus for Aluminum is 10 x 106 lb/in2) For this structure we would like to determine the axial stress in member ABC both in section AB and section BC.Example 2 Topic 3. (See Diagram 2) file:///Z|/www/Statstr/statics/Stress/strse32b.Topic 3.2b: Statically Determinate .Statics) 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.htm (1 of 4) [5/8/2009 4:39:49 PM] . We proceed using our static equilibrium procedure (from Topic 1 . To solve the problem we first need to determine the external support forces acting on the structure. as well as any needed angles and dimensions. Member ABC is pinned to the wall at point A. and to determine the movement of point C due to the applied loads. Part I. members ABC and BDE are assumed to be solid rigid members. Member BDE is supported by a roller at point E. Member ABC is an aluminum rod with a diameter of 1 inch.Example 2 In the structure shown in Diagram 1. we are fortunate in that it is not a complex nonaxial member. but rather simply is in different amounts of tension above and below point B.htm (2 of 4) [5/8/2009 4:39:49 PM] .000 lbs)(8 ft) .000 lbs)(12 ft) + (12. Therefore to determine the amount of stress in each part of ABC. and so it is not in simple uniform tension or compression. However.Example 2 2: Resolve forces into x and y components. An interesting aspect to this problem is that member ABC is not an axial member. Ey = 4000 lbs Part II. It is not in shear.000 lbs. (Diagram 3) file:///Z|/www/Statstr/statics/Stress/strse32b. 1: FBD of member ABC. Sum Fx = Ax = 0 Sum Fy = Ay + Ey -16.000 lbs = 0 Sum TE = (16.Ay(12 ft) = 0 Solving for the unknowns: Ay = 24.2b: Statically Determinate . 3: Apply the equilibrium conditions. we first make a free body diagram of member ABC and apply static equilibrium principles.12.000 lbs .Topic 3. Now to find the force in section AB of member ABC.htm (3 of 4) [5/8/2009 4:39:49 PM] .000 lb. we see that for the section of AB shown to be in equilibrium.000 lb.By .000 lb force of the wall on the member at point A. We can do this since if a member is in static equilibrium.Topic 3. Cut the member between points A and B. Resolve all forces into x and y components. Sum Fx = Bx = 0 Sum Fy = 24. Apply equilibrium conditions.16. Looking at Diagram 4 (which is the free body diagram of the upper section of member ABC).000 lbs = 0 Solving for the unknowns:By = 8. the internal force (which becomes external when we cut the member) must be equal and opposite to the 24.2b: Statically Determinate . file:///Z|/www/Statstr/statics/Stress/strse32b. 3.and analyze the top section. then any portion of the member is also in static equilibrium.Example 2 2. . for equilibrium.52) Stress (BC) = 20.000 lb/(3. and by simply summing forces in the y-direction. and apply static equilibrium principles to the top section. Part III.Example 2 Once we know the tension in section AB of member ABD. That is.Topic 3. We cut member ABC between point B and point C. Strain & Hooke's Law .220 in) + (.14*(. we see that the internal force BC (which becomes an external force when we cut the member) must be 16. C = [ (FL / EA)AB + (FL / EA)BC ] Movement of C = [ (24.II or Select: Topic 3: Stress.400 psi.14 x .2b: Statically Determinate .Strain & Hooke's Law .318 in Return to: Topic 3.14*(. we find the stress from our relationship Stress = F/A = 24. We then use the same approach with section BC of member ABC.14 x .000 lbs)(72 in) / (10*106 psi)(3.000 lbs)(48 in) / (10*106 psi)(3. The stress in section BC is then given by Stress = F/A = 16.htm (4 of 4) [5/8/2009 4:39:49 PM] .5 in)2)]AB + [ (16.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse32b.000 lb/(3.0978 in) = .2: Stress.52) = 30.000 lb. Diagram 5 is the free body diagram of that section. the Movement of C = DefAB + DefBC Movement of.600 psi. To determine the movement of point C is a relatively simple problem when we realize that the movement of point C will be equal to the deformation (elongation) of section AB plus the deformation (elongation) of section BC.5 in)2)]BC Movement of C = (. We proceed using our static equilibrium procedure (from Topic 1 .] Part I.Example 3 In the structure shown in Diagram 1.Example 3 Topic 3. It is supported by a two cables.Statics) 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. To solve the problem we first need to determine the external support forces acting on the structure. [Young's Modulus for steel : Est = 30 x 106 psi. For this structure we would like to determine the axial stress in cable AB and DE.htm (1 of 3) [5/8/2009 4:39:49 PM] . Both cables have a cross sectional area of . Cable AB is steel and cable DE is aluminum. member BCDFG is assumed to be a solid rigid member. We would also like to determine the movement of point F due to the applied loads. as well as any needed angles and dimensions.2c: Statically Determinate .Topic 3.5 in2. file:///Z|/www/Statstr/statics/Stress/strse32c.2c: Statically Determinate . Young's Modulus for Aluminum: Eal = 10 x 106 psi. AB & DE. as is shown in the FBD. FDE = 24. Sum Fy = Ay + Ey .000 lb.000 lbs/. (tension).000 psi.000 lb. We apply the stress equation (from the Stress / Strain / Hooke's Law relationships shown to the right).000 lb. Part II.000 lbs/..000 lb.)(4 ft) + Ey(10 ft) .000 lb.(10.2c: Statically Determinate . All the forces are acting in the y-direction.5 in2 = 48. the external forces exerted by the ceiling on the members is also equal to the internal forces in the members. To find the stress in each cable is now straight forward. Stress DE = F/A = 24. So.htm (2 of 3) [5/8/2009 4:39:49 PM] .. Ay = 16. Part III To find the movement of point F requires us to use a bit of geometry. (tension). Point F moves since both file:///Z|/www/Statstr/statics/Stress/strse32c. Thus FAB = 16.Topic 3.30.)(12ft) = 0 Solving for the unknowns: Ey = 24. Since both the steel member AB and the aluminum member DE are single axial members connected to the supporting ceiling.5 in2 = 32. .000 psi.000 lb.000 lb. 3: Apply the equilibrium conditions.Example 3 2: Resolve forces into x and y components.10.000 lb. = 0 Sum TB = (-30. Stress AB = F/A = 16. In Diagram 3 we have shown the initial and final position (exaggerated) of member BCDFG. Solving for x we find: x = . point D moves down .II or Select: Topic 3: Stress.128 inches + "x" (where x is the distance below the horizontal line as shown in the diagram).2c: Statically Determinate .576 in 2: Movement of point F.2: Stress.htm (3 of 3) [5/8/2009 4:39:49 PM] .128 in DefED = (FL / AE)ED = (24.576 inches (the deformation of member DE).000 lbs)(120 in) / (10*106 lbs/in2)(.5 in2) = . Point B moves down .5376 inches. 1: Calculate deformation of members AB and ED.448 in / 10 ft). Point F moves down an intermediate amount.Example 3 member AB and ED deform and member BCDFG moves downward according to these deformations.Topic 3. and write: (x / 12 ft) = (. Strain & Hooke's Law . To determine this we have drawn a horizontal line from the final position of point B across to the right side of the beam as shown.5376 inches = . Strain & Hooke's Law . From this we see that the distance point F moves down is .5 in2) = .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse32c.128 inches (the deformation of member AB).128 inches + . DefAB = (FL / AE)AB = (16. We can determine the value of x from proportionality (since similar triangles are involved). So the Movement of F = .000 lbs)(120 in) / (30*106 lbs/in2)(.666 inches Return to: Topic 3. and is pinned to member ABC at point B. file:///Z|/www/Statstr/statics/Stests/stress/sol211.htm (1 of 4) [5/8/2009 4:39:49 PM] . Member BDE is supported by a roller at point E. all joints and support points are assumed to be pinned or hinged joints.Example In the structure shown below members ABC and BDE are assumed to be solid rigid members. Eal = 10 x 106 psi Unless otherwise indicated. Draw a Free Body Diagram showing all support forces and loads. Determine the movement of point C due to the two applied loads. For this structure: A. Member ABC is a aluminum rod with a diameter of 1 inch. Determine the axial stress in member ABC both in section AB and section BC. C. Solution: PART A: External support reaction . B. Member ABC is pinned to the wall at point A.Statics: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.solution211 STATICS / STRENGTH OF MATERIALS . Ey = 4.000 lbs PART B: STEP 1: Take out member ABC.Ay(12 ft) =0 Solving for the unknowns: Ay = 24. STEP 3: Apply the equilibrium conditions.000 lbs)(12 ft) + (12.000 lbs .12.htm (2 of 4) [5/8/2009 4:39:49 PM] . Draw a free body diagram of ABC file:///Z|/www/Statstr/statics/Stests/stress/sol211.000 lbs = 0 Sum TE = (16.000 lbs.solution211 as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. Analyze force acting on it.000 lbs)(8 ft) . Sum Fx = Ax = 0 Sum Fy = Ay + Ey 16. 600 psi Now to find the force in section BC of member ABC.000 lbs .785 in2 = 30.htm (3 of 4) [5/8/2009 4:39:49 PM] . Look at top section (see diagram).16.C=[ (FL / EA)AB + (FL / EA)BC ] file:///Z|/www/Statstr/statics/Stests/stress/sol211.By .000 lbs for equilibrium. Look at top section (see diagram). Cut member between points B and C. Def = Deformation To find the total movement of C = DefAB + DefBC move. StressAB = F/A = 24. The internal force in section BC must be 16.000 lbs = 0 Solving for the unknowns: By = 8.000 lbs Now to find the force in section AB of member ABC.000 lbs/.785 in2 = 20.000 lbs/. Sum Fx = Bx = 0 Sum Fy = 24.000 lbs (equal to A) for equilibrium.solution211 STEP 2: Resolve all forces into x and y components STEP 3: Apply equilibrium conditions. The internal force in section AB must be 24.400 psi PART C: C. StressBC = F/A = 16. Cut member between points A and B. 22 in) + (.5 in)2)]BC move.C=(. Strain & Hooke's Law .C=[(24.14*(.0978 in) = .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol211.318 in Select: Topic 3: Stress.htm (4 of 4) [5/8/2009 4:39:49 PM] .000 lbs)(48 in)/(10*106 psi)(3.solution211 move.5 in)2)]AB+[(16.14*(.000 lbs)(72 in)/(10*106 psi)(3. as well as any needed angles and dimensions. B. External support reaction . Sum Fx = Ax . Unless otherwise indicated.Example In the structure shown below member ABCD a solid rigid member pinned to the wall at A.E cos (30o) =0 Sum Fy = Ay + E sin (30o) file:///Z|/www/Statstr/statics/Stests/stress/sol212. For this structure: A. all joints and support points are assumed to be pinned or hinged joints. STEP 3: Apply the equilibrium conditions.Statics: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. STEP 2: Break any forces not already in x and y direction into their x and y components. and supported by steel cable CE. Determine the movement of point D due to the applied load. Cable CE is pinned to the wall at E and has a diameter of 1 inch. Solution: Part A. Determine the axial stress in cable CE. C. Draw a Free Body Diagram showing all support forces and loads.solution212 STATICS / STRENGTH OF MATERIALS .htm (1 of 2) [5/8/2009 4:39:50 PM] . 5 in)2 ) = .E sin (30o)(23. C / 12 ft = move. A = 17.348 in Select: Topic 3: Stress.600 lbs)(16 ft * 12 in/ft) / (30*106 lbs/in2 )(3.600 psi 2 Part C.10.htm (2 of 2) [5/8/2009 4:39:50 PM] . Stress CE = F/A = 25. D / 16 ft .000 lbs)(16 ft) + E cos (30o)(2 ft) .solution212 .46 ft) Solving for the unknowns: E = 25.785 in = 32. Def = Deformation Movement of D: First find deformation of cable CE STEP 1: DefCE = (FL / EA)CE = (25. D / 16 ft Solving for movement of D = .14*(.000 lbs)(4.000 lbs .20.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol212. A = 22. Strain & Hooke's Law .170 lbs.000 lbs = 0 Sum TA = (20.600 lbs.209 in / 12 ft = move. and point D moves a proportional amount compared to point C (see diagram).200 lbs x y Part B.600 lbs/.8 ft) + (10. move.209 in STEP 2: Point C moves the same amount that cable CE deforms. Cable AB is brass.5 square inches. Determine the movement of point F due to the applied loads. all joints and support points are assumed to be pinned or hinged joints. Solution: file:///Z|/www/Statstr/statics/Stests/stress/sol213.htm (1 of 2) [5/8/2009 4:39:50 PM] . Both cables have a cross sectional area of . AB. For this structure: A.Example In the structure shown below member BCDFG is assumed to be a solid rigid member.solution213 STATICS / STRENGTH OF MATERIALS . C. B. and DE. and cable DE is steel. Est = 30 x 106 psi Unless otherwise indicated. It is supported by a two cables. Determine the axial stress in cable AB. Draw a Free Body Diagram showing all support forces and loads. Strain & Hooke's Law . Def = Deformation Movement of point F.16 in). STEP 2: Break any forces not already in x and y direction into their x and y components.5 in2) = .(10. So. A = 10. DefAB = (FL / AE)AB = (10. movement of F = . Point F moves since both member AB and ED deform and member BCDFG moves downward according to these deformations.000 lbs)(4 ft) + Ey (10 ft) . then cross member BCDFG simply moves downward that amount.000 lbs)(120 in) / (30*106 lbs/in2)(.000 lbs)(120 in) / (15*106 lbs/in2)(.000 lbs y y Part B.000 lbs = 0 Sum TB = (-20.5 in = 20.000 lbs 10.solution213 Part A. STEP 1: Calculate deformation of members AB and ED. Stress AB = F/A = 10.5 in2) = . STEP 3: Apply the equilibrium conditions.000 lbs/.htm (2 of 2) [5/8/2009 4:39:50 PM] . External support reaction .16 in DefED = (FL / AE)ED = (20.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol213.000 psi 2 Part C.Statics: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.000 lbs)(12ft) = 0 Solving for the unknowns: E = 20.000 lbs. as does point F. as well as any needed angles and dimensions.16 in STEP 2: Movement of point F. Sum Fy = Ay + Ey .20. In this problem since both AE and ED deform the same amount (.16 in Select: Topic 3: Stress. αal = 23 x 10-6 /oC Unless otherwise indicated.Example In the structure shown below member ABCD is assumed to be a solid rigid member. Draw a Free Body Diagram showing all support forces and loads. Determine the movement of point D due to the applied loads. Determine the axial stress in cable CE. For this structure: A. Solution: file:///Z|/www/Statstr/statics/Stests/stress/sol214. all joints and support points are assumed to be pinned or hinged joints. αbr = 20 x 10-6 /oC. Cable CE is made of steel and has a diameter of 1 inch.htm (1 of 2) [5/8/2009 4:39:50 PM] . and is supported by cable CE. Ebr = 15 x 10 6 psi. B. It is pinned to the floor at point A. Eal = 10 x 10 6 psi αst = 12 x 10-6 /oC.solution214 STATICS / STRENGTH OF MATERIALS . Est = 30 x 10 6 psi. Strain & Hooke's Law . So we first determine the deformation of EC. STEP 3: Apply the equilibrium conditions. External support reaction . Def = Deformation STEP 1: Point D moves due to the deformation of cable EC.(10. Sum Fx = Ax .solution214 Part A.790 lbs. D / 16 ft and so Mov.000 lbs)(4.Statics: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.htm (2 of 2) [5/8/2009 4:39:50 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol214. as well as any needed angles and dimensions. Stress CE = F/A = 13. DefEC = (FL / EA)EC = (13.14 * .6 ft)= 0 Solving for the unknowns: E = 13.100 lbs x y Part B. D / 16 ft .790 lbs)(12.0912 in STEP 2: Point D moves in proportion to how much point C moves.000 lbs . A = 15.122 in Select: Topic 3: Stress.790 lbs/ . STEP 2: Break any forces not already in x and y direction into their x and y components. So we can write: Mov.000 lbs)(9.12. A = 11940 lbs.E cos (30o) = 0 Sum Fy = Ay + E sin (30o) 10.0912 in / 12 ft = Mov.785 in = 17.9 ft)(12 in/ft) / (30*106 lbs/in2)(3. C / 12 ft = Mov.000 lbs = 0 Sum TA = -E sin (30o)(4 ft) + E cos (30o)(16 ft) .(12.600 psi 2 Part C. D = .5 in2) = . And point C moves the amount cable EC stretches.8 ft). 75 inch. C. as well as any needed angles and dimensions. STEP 3: If we were to go about this problem in the normal fashion. B. Member ABC is pinned to the wall at A and is supported by a roller at point C. Member CDE is pinned to the wall at point E. Determine the movement of point B due to the applied load. Determine the axial stress in cable DF. we would have five unknowns and only three equations.Example In the structure shown below members ABC and CDE are assumed to be solid rigid members. we file:///Z|/www/Statstr/statics/Stests/stress/sol215. STEP 2: Break any forces not already in x and y direction into their x and y components. Therefore. Unless otherwise indicated. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Solution: Part A. all joints and support points are assumed to be pinned or hinged joints. Draw a Free Body Diagram showing all support forces and loads. and is supported by steel cable DF.htm (1 of 3) [5/8/2009 4:39:50 PM] .solution215 STATICS / STRENGTH OF MATERIALS . There would be no way to solve this problem using that approach. Cable DF has a diameter of . For this structure: A. Def = Deformation file:///Z|/www/Statstr/statics/Stests/stress/sol215.4418 in = 40.htm (2 of 3) [5/8/2009 4:39:50 PM] . Apply the equilibrium conditions: Sum Fx = Ax = 0 Sum Fy = Ay + Cy .000 lbs.000 lbs)(6 ft) + Cy (8 ft) = 0 Solving for the unknowns: C = 9.000 lbs y y Part B.000 lbs.9. A = 3.solution215 must take a different perspective.000 lbs)(4 ft) = 0 Solving for the unknowns: F = 18.000 lbs/ .12. By taking the member apart and analyzing member ABC there will only be three unknowns. Apply the equilibrium conditions: Sum Fx = Ex = 0 Sum Fy = Fy + Ey .000 lbs = 0 Sum TE = -Fy (2 ft) + (9.000 lbs = 0 Sum TA = (-12.740 psi 2 Part C.000 lbs y y Now we can analyze member CDE in a similar fashion. E = -9. Stress DF = F/A = 18. 0978 in/ 2 ft Mov. D / 2 ft Mov.4417 in2) = . Strain & Hooke's Law . DefFD = (FL / EA) = (18.1956 in STEP 3: Member ABC is resting on member CDE therefore.solution215 STEP 1: Point C moves due to the deformation of cable FD. at point C the movement is the same for members ABC and CDE (see diagram).1956 in/ 8 ft Mov.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol215. C / 8 ft Mov.000 lbs)(6 ft)(12 in/ft) / (30 *106)(. Mov. B = .htm (3 of 3) [5/8/2009 4:39:50 PM] . C / 4 ft = . B / 6 ft = Mov. B / 6 ft = . C = .0978 in STEP 2: Point C moves in proportion to how much point D moves (see diagram). Mov. C / 4 ft = Mov.1467 in Select: Topic 3: Stress. So we first determine the deformation of FD. solution216 STATICS / STRENGTH OF MATERIALS . and connected to member EFG by steel cable DE. (Cables BC and DE each have a cross sectional area of . Determine the axial stress in cable BC. supported by steel cable BC.5 square inches. as well as any needed angles and dimensions.htm (1 of 3) [5/8/2009 4:39:51 PM] . Draw a Free Body Diagram showing all support forces and loads. we would have four unknowns and only three equations. STEP 3: If we were to go about this problem in the normal fashion. C. all joints and support points are assumed to be pinned or hinged joints. Unless otherwise indicated. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. There would be no way to file:///Z|/www/Statstr/statics/Stests/stress/sol216. STEP 2: Break any forces not already in x and y direction into their x and y components.Example In the structure shown below member ABD is a solid rigid member pinned to the wall at A. B. Determine the movement of point G due to the applied load. For this structure: A.) Member EFG is supported by a roller at F and is loaded with 12000 lbs at G. Therefore. Apply the equilibrium conditions: Sum Fx = Ax = 0 Sum Fy = Ay .000 lbs)(3 ft) = 0 Solving for the unknowns: A = 24.000 lbs.5 in2) = .solution216 solve this problem using that approach.5 in = 96.000 lbs y y Now we can analyze member ABD in a similar fashion.Ey . Stress psi BC = F/A = 48. By taking the member apart and analyzing member EFG there will only be two unknowns.000 lbs)(6 ft) + Fy (2 ft) = 0 Solving for the unknowns: F = 36.Cy + 24.000 lbs = 0 Sum TE = (-12.12.000 lbs/ .000 lbs = 0 Sum TB = -Ay (3 ft) + (24.000 2 Part C. we must take a different perspective. Apply the equilibrium conditions: Sum Fy = Fy . STEP 1: Mov. E = 24.000 lbs)(24 in) / (30*106 lbs/in2)(. B = Deformation cable BC Mov.000 lbs y y Part B.0768 in STEP 2: Movement of point D is proportional to movement of point B. and we can write: file:///Z|/www/Statstr/statics/Stests/stress/sol216. Def = Deformation Point G moves due to the deformation of cable BC and Cable DE.htm (2 of 3) [5/8/2009 4:39:51 PM] .000 lbs. B = (FL / EA) = (48. C = 48. 0768 in / 3 ft Mov. E / 2 ft Mov. G / 4 ft = Mov. E = . G / 4 ft = . G = . D + (FL / EA)DE = . E = Mov.1536 in STEP 3: Movement of point E is equal to movement of point D plus the elongation of cable DE. B / 3 ft Mov.5 in2) Mov.192 in STEP 4: Finally the movement of point G is proportional to the movement of point E and we may write: Mov. D / 6 ft = Mov. Mov. D = .192 in / 2 ft Mov. D / 6 ft = .1536 in + .0384 in = . Strain & Hooke's Law .solution216 Mov.1536 in + (24.384 in Select: Topic 3: Stress.htm (3 of 3) [5/8/2009 4:39:51 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol216.000 lbs)(24 in) / (30*106 lbs/in2)(. This often effects the result only slightly. An external load of 10. known as Statically Indeterminate Structures. what is the stress in the steel and aluminum members. For this structure some of the things we might wish to know might be: after the 10. what are the external support forces acting on the structure.000 lb. Ignoring the bending in this case effects the result only to a small degree. is applied at point F. both of which are pinned to and support member ABDF. and aluminum member DE. and which we will deal with when the topic of beams is discussed).htm (1 of 5) [5/8/2009 4:39:51 PM] .3: Statically Indeterminate Structures The fact that real members deform under external forces and loads seems to add a complication when analyzing structures. we are going to ignore the fact that member ABDF will experience some bending (which in a sense is a deformation.Static Equilibrium Analysis 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. As we analyze this structure. As is also the case when the weight of the structural member is ignored in analyzing the structure.Topic 3. but for certain types of structures. as file:///Z|/www/Statstr/statics/Stress/strs33.000 lb. and both of which are pinned to the ceiling as shown. especially when the external force and loads are much larger than the weight of the members. what is the movement of point F due to the load.3 . steel member BC. To understand what we mean by a statically indeterminate structure. let us first try to analyze the structure using our standard static equilibrium procedure. Example 1: The structure shown in Diagram 1 is formed by member ABDF (which is pinned to the wall at point A).Statically Indeterminate Structures Topic 3. load is applied. Perhaps the best way to illustrate this is to examine a relatively simple statically indeterminate structure. it is the effect of the deformation that allows use to solve for the external forces on the structure and in the members of the structure. Part I . 10. We may get this from the way the problem is stated. Another way to state this difficulty is that we need another independent equation to solve for the unknowns.. (All forces are either in x or y direction. in most cases.) We might try to take the structure apart in some way. The steel and aluminum members are pinned to the ceiling. (We do see that Ax must be zero.Deformation Equation Step 1 is to find some general relationship between the deformations of the members of the structure. we observe that we have three unknowns and only two independent equations .Statically Indeterminate Structures well as any needed angles and dimensions. a FBD of the structure is shown. Therefore.in this problem. we replace the hinge by support forces Ax and Ay.000 lb. Part II.000 lb. where member ABDF is hinged to the wall. or redraw the FBD. (18 ft.Topic 3. but none of this will help. FSt and FAl. however in this case we can do better (meaning less unknowns). be able to solve for the external support reactions.and can not solve. Since both the steel and aluminum members have forces acting on them at only two points (each end). At point A. FAl.) 3: Apply the equilibrium conditions. FSt. but that is no help with the other two equations. they are axial members and are in simple tension or compression . The deformations of the steel and aluminum members will give us this additional equation.3 . from the first equation. However.10. = 0 : + FSt (6 ft. simple tension. the ceiling simply pulls vertically upward on each member with force. : Ax = 0 : -Ay + FSt + FAl . in finding Ay.htm (2 of 5) [5/8/2009 4:39:51 PM] .it is statically indeterminate. . as shown in Diagram 2.) .as in file:///Z|/www/Statstr/statics/Stress/strs33. Normally we would replace the pins by horizontal and vertical support forces. In Diagram 2.) + FAl (12 ft. 2: Resolve forces into x and y components.) = 0 At this point in a statically determinate problem. or often from the geometry of the structure . in this case. Static equilibrium conditions alone are not enough to solve this problem . we would. However. or def = FL/EA. we see we can write the deformation of a member as: deformation = [ force in member * length of member] / [ young's modulus of member * area of member]. since it involves deformations. At first it may not be clear how we can use this equation. but with deformation effects shown. This is somewhat like a Free Body Diagram. After substituting in the values for the members we have: file:///Z|/www/Statstr/statics/Stress/strs33. showing and labeling the deformations involved. not forces as in the equilibrium equations. if we recall our stress/strain/deformation relationships.Topic 3. We diagram this. load will be to elongate both the steel and aluminum members which will cause member ABDF to rotate downward about hinged point A. See Diagram 3. as it rotates downward we see we can write (from similar triangles): Deformation of Steel / 6 ft = Deformation of Aluminum / 12 ft or we can rewrite as: 2 * Deformation of Steel = Deformation of Aluminum or symbolically: This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure.000 lb. Substituting this relationship into our deformation equation we have: 2 * [FL/EA]St = [FL/EA]Al We now have our additional relationship between the forces.3 . The effect of the 10. Since member ABDF is pinned at point A.Statically Indeterminate Structures this case. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem.htm (3 of 5) [5/8/2009 4:39:51 PM] . 3 . from Def.htm (4 of 5) [5/8/2009 4:39:51 PM] . We have now solved our statically indeterminate problem and determined the values of the external support reactions acting on the structure.900 lb * 72 in. Continue to: Example 1 . we now will look at several examples. Stess Aluminum = 10. the stresses. F/18 ft.5 in2 = 16. we should be able to solve for all the external support forces acting on the structure.Statically Indeterminate Structures (2 * FSt *72")/(30 x 106 lb/in2 * . Summary: As a brief review.900 lb. 350 lb/in2.10.)/(10 x 106 lb.in2 * 1 in2) = .900 lb/1 in2 = 10.75 FAl We now substitute this into our torque equation from static equilibrium equations (shown below) : Ax = 0 : -Ay + FSt + FAl . = 0 : + FSt (6 ft.. Step 2 is to find a general relationship between the deformations of the members in the structure. and FSt = 8175 lb. finding Ay = 10. and we may write: . Point F moves in proportion to the elongation of member DE. We can now also solve for Ay.900 lb/in2.000 lb.075 lb We find the stress from: Stress Steel = 8175 lb/ .10.) + FAl (12 ft. and the movement of point F.118 in. = FL/EA = (10.) + FAl (12) . To see applications of these relationships.) .10.75 FAl)(6 ft. And finally we can find the movement of point F by first finding the elongation of member DE./12 ft = Move.) = 0 and obtain: (. Finally using the equations from Steps 1 and 2.0785 in.000 lb. Example 2 or Select: file:///Z|/www/Statstr/statics/Stress/strs33. F = .0785 in. (18 ft. and solving we have: FAl = 10. and then to rewrite this relationship using deformation = FL/EA to obtain an additional equation between the forces acting on the structure.000 lb. the aluminum member. our method for solving statically indeterminate problems consists of two basic steps: Step 1 is to apply static equilibrium conditions and write the static equilibrium equations.) = 0.Topic 3. solving Move.5 in2) = ( FAl * 72")/(10 x 106 lb/in2 * 1 in2 ) If we simplify this equation we obtain: FSt = . (18 ft. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs33.htm (5 of 5) [5/8/2009 4:39:51 PM] .Statically Indeterminate Structures Topic 3: Stress.3 .Topic 3. Strain & Hooke's Law . and the movement of point D due to the load. Since both the steel and brass members have forces acting on them at only two points (each end). is applied at point D. The 20. A downward external load of 20. The brass and steel members are pinned to the ceiling.Static Equilibrium Analysis 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. At point B. both of which are pinned to and support member ABCD. As we analyze this structure. and both of which are pinned to the ceiling as shown. Brass member AF.000 lb. file:///Z|/www/Statstr/statics/Stress/strse33a. as well as any needed angles and dimensions. Our first step is to apply our static equilibrium procedure to our structure Part I .htm (1 of 4) [5/8/2009 4:39:52 PM] .Example 1 The structure shown in Diagram 1 is formed by member ABCD (which is pinned to the floor at point B). we replace the pin by support forces Bx and By. the ceiling acts on the members as shown in the free body diagram. Normally we would replace the pins by horizontal and vertical support forces. As a result. they are axial members and are in simple tension or compression. however in this case we can do better (meaning less unknowns).Example 1 Topic 3. and which we will deal with when the topic of beams is discussed). load acting at end D tries to rotate the bar ABCD downward about point B. and Steel member CE.Topic 3. For this structure we would like to determine the stress in the steel and brass members. a FBD of the structure is shown. Thus putting the brass member (AF) in compression and the steel member (CE) in tension.3a: Statically Indeterminate . In Diagram 2.3a: Statically Indeterminate Structures . we ignore the fact that member ABCD will experience some bending (which in a sense is a deformation. where member ABCD is pinned to the floor. Ignoring the bending in this case effects the result only to a small degree.000 lb. and can not solve. or redraw the FBD. = 0 : + FBr (10 ft. However. from the first equation. in finding By.) 3: Apply the equilibrium conditions.Example 1 2: Resolve forces into x and y components.) . be able to solve for the external support reactions..) We might try to take the structure apart in some way. but that is no help with the other two equations.3a: Statically Indeterminate . but none of this will help.000 lb.000 lb. we observe that we have three unknowns and only two independent equations . : Bx = 0 : By -FBr + FSt . load will be to compress the brass member and to elongate the steel member which will cause member ABCD to rotate downward about hinged point B. (12 ft. (All forces are either in x or y direction.it is statically indeterminate.Topic 3. Another way to state the problem is that we need another independent equation to solve for the unknowns.000 lb. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem.) + FSt (6 ft. or often from the geometry of the structure .20. in this case.htm (2 of 4) [5/8/2009 4:39:52 PM] . Since member ABDF is pinned at point A. FSt. We may get this from the way the problem is stated.as in this case. See Diagram 3. in most cases. FBr. Part II. showing and labeling the deformations involved.Deformation Equation Step 1 is to find some general relationship between the deformations of the members of the structure. . We diagram this. we would.20. Static equilibrium conditions alone are not enough to solve this problem . (We do see that Bx must be zero. The deformations of the brass and steel members will give us this additional equation.) = 0 At this point in a statically determinate problem. as it rotates downward we see we can write (from similar triangles): file:///Z|/www/Statstr/statics/Stress/strse33a. The effect of the 20. since it involves deformations.20. not forces as in the equilibrium equations.Example 1 Deformation of Steel / 6 ft = Deformation of Brass / 10 ft or we can rewrite as: Deformation of Steel = . Substituting this relationship into our deformation equation we obtain: [FL/ EA]St = . = 0 file:///Z|/www/Statstr/statics/Stress/strse33a.5 in2 )] If we simplify this equation we obtain: FSt = .Topic 3. or def = FL/EA. if we recall our stress/strain/deformation relationships.75 in2) = .6 [FL/EA]Br We now have our additional relationship between the forces. we see we can write the deformation of a member as: deformation = [ force in member * length of member] / [ young's modulus of member * area of member].3a: Statically Indeterminate .000 lb.htm (3 of 4) [5/8/2009 4:39:52 PM] . After substituting in the values for the members we get: ( FSt *96")/(30 x 106 lb/in2 * . At first it may not be clear how we can use this equation.6 * Deformation of Brass or symbolically: This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure.6[( FBr * 96")/(15 x 106 lb/in2 * 1.6 FBr We now substitute this into our torque equation from static equilibrium equations (shown on right) : Bx = 0 : By -FBr + FSt . However. 600 lb/ . D/12 ft.5 in2 = 11.)/(30 x 106 lb. and FSt = 10.htm (4 of 4) [5/8/2009 4:39:52 PM] ..) + (.770 lb/in2.Example 1 : + FBr (10 ft. Stress Brass = 17.3a: Statically Indeterminate .000 lb.650 lb.6 FBr) (6 ft.) = 0 and obtain: (FBr)(10 ft. D = . the stresses.0452 in.600 lb * 96 in.650 lb/1.0904 in. solving Move. and we may write: .20.in2 * . and solving we have: FBr = 17. Return to: Topic 3. finding By = 27.) .3: Statically Indeterminate Structures or Select: Topic 3: Stress. We have now solved our statically indeterminate problem and determined the values of the external support reactions acting on the structure.) . the steel member.20. Point D moves in proportion to the elongation of member CE.000 lb. from Def. And finally we can find the movement of point D by first finding the elongation of member CE. and the movement of point D. (12 ft. = FL/EA = (10. (12 ft.) + FSt (6 ft.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse33a.130 lb/in2.Topic 3.050 lb We find the stress from: Stress Steel = 10.) = 0. We can now also solve for By.0452 in.75 in2 = 14./6ft = Move.600 lb. Strain & Hooke's Law .75 in2) = . file:///Z|/www/Statstr/statics/Stress/strse33b. Ignoring the bending in this case effects the result only to a small degree. Fst1 and Fst2. In this structure the load simply puts all the cables in tension. which is supported by two steel cables.Example 2 Topic 3. we ignore the fact that member BCE will experience some bending (which in a sense is a deformation.000 lb.3b: Statically Indeterminate Structures . and which we will deal with when the topic of beams is discussed). and a Brass cable. For this structure we would like to determine the stress in the steel and brass cables. A downward external load of 50. and the movement of point C due to the load. Our first step is to apply our static equilibrium procedure to our structure Part I . As we analyze this structure. In Diagram 2.3b: Statically Indeterminate . Notice that at this point we do not assume the force the ceiling exerts on each steel cable is the same (although the symmetry of problem indicates that should be true). as well as any needed angles and dimensions. and the ceiling acts on the structure as shown in Diagram 2. but rather we indicate two different forces on the steel cables.Example 2 The structure shown in Diagram 1 consists of horizontal member BCE.Topic 3. AB and EF.htm (1 of 4) [5/8/2009 4:39:52 PM] . Our static equilibrium equations will make clear the relationship between the forces in the two steel cables. BD pinned to the ceiling as shown. a FBD of the structure is shown.Static Equilibrium Analysis 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. is applied at point C. ) + FSt2 (4 ft.htm (2 of 4) [5/8/2009 4:39:52 PM] . load.Example 2 2: Resolve forces into x and y components. we still have too many unknowns. will be to elongate both the steel and the brass members equally. and only one independent equations left .000 lb. because of the symmetry of the problem. we see that Fst1 = Fst2. Part II. file:///Z|/www/Statstr/statics/Stress/strse33b.as in this case. or 2 FSt + FBr -50.) : FSt1 + FSt2 + FBr . = 0 : . or often from the geometry of the structure .000 lb. Static equilibrium conditions alone are not enough to solve this problem .and we cannot solve. and cause horizontal member BCE to move downward as shown in Diagram 3.000 lb. (All forces are either in x or y direction. We have two unknowns at this point. Another way to state the problem is that we need another independent equation to solve for the unknowns. and so we will now call the force in the steel cables just FSt. = 0.FSt1 (4 ft..3b: Statically Indeterminate .it is statically indeterminate. And we can rewrite the sum of y. The effect of the 50.Deformation Equation Step 1 is to find some general relationship between the deformations of the members of the structure. We may get this from the way the problem is stated. . : (no external x forces.) = 0 From the torque equation.50.50. = 0 However. The deformations of the brass and steel members will give us this additional equation.) 3: Apply the equilibrium conditions.000 lb.forces equation as: FSt + FSt + FBr .Topic 3. if we recall our stress/strain/deformation relationships.75 in2 ) If we simplify this equation we obtain: FSt = . since it involves deformations.89 FBr We now substitute this into our sum of y-forces equation from static equilibrium equations. Substituting this relationship into our deformation equation we have: [FL/ EA]St = [FL/EA]B We now have our additional relationship between the forces.Example 2 We now write a general relationship between the deformations of the members which we can get from the geometry of the problem./in2 * .) file:///Z|/www/Statstr/statics/Stress/strse33b. At first it may not be clear how we can use this equation. However. or def = FL/EA./in2 * .Topic 3.htm (3 of 4) [5/8/2009 4:39:52 PM] . : (no external x forces. After substituting in r the values for the members we have: ( FSt *72")/(30 x 106 lb.5 in2) = ( FBr * 48")/(15 x 106 lb. we see we can write the deformation of a member as: deformation = [ force in member * length of member] / [ young's modulus of member * area of member]. not forces as in the equilibrium equations.3b: Statically Indeterminate . For our structure: Deformation of Steel = Deformation of Brass or symbolically: This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure. = 0 : ./in2. Stress Steel = 16./ .) + FSt2 (4 ft. and solving we have: FBr = 18.75 in2 = 24. Strain & Hooke's Law .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse33b.. = 0.3: Statically Indeterminate Structures or Select: Topic 3: Stress./. = FL/EA = (18./in2. the Brass member.000 lb.3b: Statically Indeterminate .FSt1 (4 ft.000 lb. Return to: Topic 3.)/(15 x 106 lb. and FSt = 16. the stresses.000 lb.000 lb.000 lb.000 lb.000 lb.0768 in.htm (4 of 4) [5/8/2009 4:39:52 PM] . from Def.) = 0 and obtain: 2 (.000 lb.89 FBr) + FBr . And finally we can find the movement of point C simply by finding the elongation of member CD. * 48 in.50.in2 * .Topic 3.Example 2 : 2 FSt + FBr -50. We have now solved our statically indeterminate problem and determined the values of the external support reactions acting on the structure.000 lb.5 in2 = 32. We find the stress from: Stress Brass = 18.75 in2) = . and the movement of point C. we may also have what is known as Shear Stress and Shear Strain.In Diagram 1 we have shown a metal rod which is solidly attached to the floor. The component of the Force perpendicular to the surface area will produce an Axial Stress on the rod given by Force perpendicular to an area divided by the area. The units of both Axial and Shear Stress will normally be lb/in2 or N/m2. which is the change in the length divided by the original length of the member. or: where the Greek letter. This displacement (or horizontal deformation) divided by the length of the rod L is equal to the Shear Strain. Examining the small triangle made by . defined as Force parallel to an area divided by the area.4: Shear Stress & Strain SHEAR STRESS In additional to Axial (or normal) Stress and Strain (discussed in topic 2.htm (1 of 3) [5/8/2009 4:39:52 PM] . from its initial position. Both Axial Strain and Shear Strain are shown in Diagram 2.2). is also equal to the tangent of the angle gamma. L and the side of the rod.Topic 3. Tau. The edge of the rod is displaced a horizontal distance. F.4 . Or we may write: Shear Strain = file:///Z|/www/Statstr/statics/Stress/strs34. we see that the Shear Strain.Shear Stress & Strain Topic 3. acting at angle theta with respect to the horizontal.1and 2. is used to represent Shear Stress. or: The component of the Force parallel to the area will also effect the rod by producing a Shear Stress. on the rod. Shear Strain: Just as an axial stress results in an axial strain. We then exert a force. /L . so does shear stress produce a shear strain. and since the amount of displacement is quite small the tangent of the angle is approximately equal to the angle itself. The shear stress produces a displacement of the rod as indicated in the right drawing in Diagram 2. If a graph is made of Shear Stress versus Shear Strain. these two points are not quite the same. that is. it will normally exhibit the same characteristics as the graph of Axial Stress versus Axial Strain. There is an Elastic Region in which the Stress is directly proportional to the Strain.Topic 3. file:///Z|/www/Statstr/statics/Stress/strs34. and finally there is a Failure Point where the sample fails in shear. after the elastic limit.htm (2 of 3) [5/8/2009 4:39:52 PM] . it will not return to its original size and shape.4 . permanent deformation has occurred. the Shear Modulus) and has units of lb/ in2. or the proportional limit. if the force is taken off the sample. Although these two points are slightly different. The Modulus of Rigidity for Steel is approximately 12 x 106 lb/in2. The Elastic Limit is the point at which permanent deformation occurs. This relationship may be expressed as G = Shear Stress/Shear Strain. we will treat them as the same in this course. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). Shear Stress and Strain are proportional in the elastic region of the material. where a small increase in the Shear Stress results in a larger increase in Shear Strain. where G is a property of the material and is called the Modulus of Rigidity (or at times. In actuality. There is a Plastic Region. The point at which the Elastic Region ends is called the elastic limit.Shear Stress & Strain As with Axial Stress and Strain. with respect to Shear Stress and Strain.Shear Stress & Strain Example 2 . we have: While we will not go in any great depth. or Select: Topic 3: Stress.4 .htm (3 of 3) [5/8/2009 4:39:52 PM] . we will look at several relative easy examples.Topic 3. at this point.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs34. Strain & Hooke's Law .Shear Stress & Strain To summarize our shear stress/strain/Hooke's Law relationships up to this point.Shear Stress & Strain. Please Select: Example 1 . What is the shearing stress that develops in the steel bolts? If the Modulus of Rigidity for Steel in 12 x 106 lb/in2.) The surface area between the top and bottom interface is in shear. if we examine one bolt. are bolted together with two 3/4" inch diameter steel bolts. 650 lb/in2.Example 1 Topic 3.Example 1 Example 1 Two metal plates. That is.4a: Shear Stress & Strain .Topic 3.000 lb.14 * .4a: Shear Stress & Strain . From this we can calculate the shear stress in a straight forward manner from one of our Shear Stress/Strain Relationships: Stress = Force parallel to area/ area Stress = F / (p * r2) = 10. and so each bolt "carries" 10. what shear strain develops in the steel bolts? As we examine the structure we see the area of the bolt where the two plates surfaces come together are in shear. as shown.000 lb.375"2) = 22.. the top of the bolt experiences a force to the left while the bottom of the bolt experiences and an equal force to the right. The plates are loaded in tension with a force of 20.htm (1 of 2) [5/8/2009 4:39:53 PM] . We assume the bolts carry the load equally.000 lb/ (3. (See Diagram 2. as shown in Diagram 1. In similar manner the Shear Strain can be found from the appropriate form of Hooke's Law: file:///Z|/www/Statstr/statics/Stress/strse34a. 00189 Return to: Topic 3. 650 lb/in2)/(12 x 106 lb/in2) = .Topic 3.4: Shear Stress & Strain or Select: Topic 3: Stress. Strain & Hooke's Law . or (Shear Strain) = (Shear Stress)/ G = (22.Example 1 G = (Shear Stress) / (Shear Strain).4a: Shear Stress & Strain .htm (2 of 2) [5/8/2009 4:39:53 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse34a. That is.Topic 3.htm (1 of 2) [5/8/2009 4:39:53 PM] .4b: Shear Stress & Strain Example 2 Topic 3. that the driving force is really producing a torque about the center of the shaft. We are interested in determining the shear stress on the shear key. the wheel begins to rotate. In our problem the two 500 lb. the wheel. [The purpose of a shear key is to protect machinery connected to a shaft or gear. after a little thought. and that the torque produced by the driving force(s) must equal the torque produced by the force (of the inner ring) acting on the shear key.5 ft).Example 2 Example 2 In our second example we have a inner shaft (which perhaps drives a piece of machinery) and an outer driving wheel connected by spokes to an inner ring which is connected by means of a shear key to the inner shaft.) When force is applied to the wheel (through a driving belt perhaps).] To determine the shear stress we first need to determine the force trying to shear the key.4b: Shear Stress & Strain . (See Diagram 3. We do this by realizing. This is the force acing on the top half of the shear key. So we may write: 500 lb. and inner ring are one structure which is connected to the inner shaft by a shear key.(2 ft) + 500 lb. We will also determine the compressive stress (or bearing stress) acting on the shear key. Calculating torque about the center of the shaft we have: 500 lb. The key is 1/2 inch wide. Solving for F = 4000 lb. If the driving forces and/or torque become too large the shear key will "shear off " (fail in shear) and thus disconnect the driving force/torque before it can damage the connected machinery. spokes.(2 ft) = 2000 ft-lb. There is a equal force in the opposite direction acting on the bottom half of the shear key. driving forces are acting a distance of 2 ft from the center of the 1 ft diameter shaft. = F (. file:///Z|/www/Statstr/statics/Stress/strse34b. by 3/4 inch high. These two forces place the horizontal cross section of the key in shear.(2 ft) + 500 lb. and through the shear key causes the inner shaft to rotate. This must also be the torque produced by the force acting on the upper half of the shear key (shown in Diagram 4).(2 ft) = 2000 ft-lb. / (1/2" * 1") = 8000 lb/in2.4b: Shear Stress & Strain Example 2 by 1 inch deep as shown in Diagram 4.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse34b. / (3/8" * 1") = 10.4: Shear Stress & Strain or Select: Topic 3: Stress. There is an equal compressive stress on the bottom of the shear key. The compressive stress (also called the bearing stress) on the top half of the shear key will be given by: Compression (Bearing) Stress = Force normal to the area / area = 4000 lb. 700 lb/ in2. Strain & Hooke's Law . We calculate the shear stress by: Shear Stress = Force parallel to area / area = 4000 lb.htm (2 of 2) [5/8/2009 4:39:53 PM] . Return to: Topic 3. In addition to the shear stress on the horizontal cross sectional area. the forces acting on the key also place the key itself into compression.Topic 3. 4 tons.000 pounds. A short fir 4x4 has an allowable compressive stress of 1200 psi. (53. The entire house weighs 65. (1700 psi) 2. A cast iron pipe with a 6 inch outer diameter and a 1/4 inch wall thickness carries a load of 3200 pounds. parallel to the grain. A concrete post has a 20 inch diameter. It supports a load of 72. A 3/4 inch diameter steel cable hangs vertically and supports a 750 pound ball.htm [5/8/2009 4:39:53 PM] . Determine the stress in the cable. Find the stress in a 1. (708 psi) 6. (8840 lb) 7. Determine the bearing stress of the post in pounds per square inch. A 24 x 36 house with full basement rests on 18" wide concrete footings. It supports a compressive load of 8. Determine the bearing stress of the post.000 pounds. (133 psi) 5.000 psi. (19.5 psi) 4. Determine the maximum load that the 4x4 can carry. The allowable working stress for the cable is 20.5 psi) Select: Topic 3: Stress. Determine the maximum load that the crane can lift.5 feet. A rectangular concrete column is 1.5 feet by 2. (6667 psi) 3.200 lb) 8.5a Stress probs 1 determinate Statics & Strength of Materials Problem Assignment (determinate) Stress/Strain 1 1.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsp35a.Topic 3. (~ 2. Strain & Hooke's Law . Determine the compressive stress in the pipe. A construction crane has a 3/4 inch diameter cable. Determine the bearing stress on the soil beneath the footings.5 inch square cast iron rod that has a tensile force of 15.000 pounds applied to it. The beam is 80 feet long with a rectangular cross section 8 inches wide and 24 inches deep. A hammer strikes a 12 penny common nail with a maximum force of 500 pounds. Find the size cable required for a crane.? (. The ends of the bridge beams in problem 1 rest on 8 inch wide steel plates.064 psi. The boom on the crane is 100 feet long. The allowable tensile stress in the cable is 25. (150 lb) 3. That is. Assuming that the axle is solid steel. Strain & Hooke's Law . Each crane lifts an equal share of the weight. their booms are at an angle of 53 degrees. The density of concrete is 150 pounds per cubic foot. What is the compressive stress in the nail? (29.Table of Contents file:///Z|/www/Statstr/statics/Stress/strsp35b.000 psi. How long should the plates be if the allowable bearing stress is 45. Determine the weight of cable hanging from the boom when it is fully extended.98") 5.. The allowable working stress for the cable is 20. The density of steel is 490 pounds per cubic foot. Determine the maximum load that the crane can lift. how much does 100 feet of cable weigh. 20.5b Stress probs 2 (determinate) Statics & Strength of Materials Problem Assignment (determinate) Stress/Strain 2 1. The cable at the top of the boom runs over a pulley held in place with an axle. (3/4") 2.(8830 lb) Select: Topic 3: Stress. The tension in the cable is constant.000 psi. determine the minimum diameter of axle required so that the shear stress in the axle material does not exceed the allowable shear stress for hardened steel. Two cranes lift a concrete bridge beam. using nominal diameter) 6. It requires a drawing of the crane and then a free body diagram of the pulley assembly. Cable size is by 16ths to 1/2 inch.0223") 4. (d = . At one point while the cranes in problem 1 are lifting the bridge beams in problem 1. A construction crane has a 3/4 inch diameter cable.Topic 3.htm (1 of 2) [5/8/2009 4:39:53 PM] . by 1/8ths to 2 inches and by 1/4 thereafter.000 psi. Determine the shear force acting on the axle from static equilibrium.000 psi. htm (2 of 2) [5/8/2009 4:39:53 PM] .Topic 3.5b Stress probs 2 (determinate) Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsp35b. Dx = 5.725 lb) (answers: Stress AB = 6.480 psi.Stress Prob.084 lb. If we assume that member AB does not bend. member AB is a Brass rod with a diameter of 1.. determine the stress in member AB and the deformation of member AB.5 c. 280 psi..5 inches. Member BCD is also a Brass member. determine the stress in member BD and the deformation of member BD. member BD is a steel rod with a diameter of 1 inch. In the structure shown to the right.Determinate Statics & Strength of Materials Problem Assignment (determinate) Stress/Strain 3 1. In the structure shown to the right.. Both members are pinned to the wall.0081") file:///Z|/www/Statstr/statics/Stress/strsp35c..Topic 3. 570 lb) (answers: Stress BD = 17. (answers: AB = 11. Deform BD = . Deform AB = .104") 2. Both members are attached to the wall by pinned joints. (answers: Ax = 8723 lb.htm (1 of 4) [5/8/2009 4:39:54 PM] . D = 13. 450 lb. If we assume that member BCD does not bend. Dy = 5. Member ABC is also a steel member. Ay = 1605 lb. (answers: A = 40.Stress Prob.000 lb) (answers: Stress AB = 50. and the movement of point C. Member BCD is pinned at point E..Topic 3. member AB is a cable and BCD is a solid rigid member (that is.5 c. determine the stress in cable AB. Movement of C = .htm (2 of 4) [5/8/2009 4:39:54 PM] . Dy = 4. For this structure.930 psi.000 lb.. Dx = 32.109") file:///Z|/www/Statstr/statics/Stress/strsp35c. Member AB is pinned to the wall at point A.000 lb. You may assume that point B moves "down" the amount cable AB deforms.Determinate 3 In the structure shown to the right. and is supported by cable AB at point B. Member AB is a steel cable with a diameter of 1 inch. it does not bend). .5 square inches. BC = 48. Determine the stress which develops in cable AB. Movement of F = .000 psi. and the movement of point F. (answers: Fy = 36. Member EFG is pinned to the wall at G. Assume members BCE and EFG do not bend..Determinate 4. ED = 24. Ay = 24. (Cables BC and DE each have a cross sectional area of . Movement of G = . Cy = 14.000 lb. and connected to member EFG by steel cable DE. Cx = 10. and the movement of point G..000 lb. AB..000 lb.000 lb. In the structure shown to the right L-shaped member BCDE is connect by a steel cable.000 lb) (answers: Stress AB = 20.000 lb. For this structure determine the stress in cable BC.000 psi.) (answers: Stress BC = 96.) Member EFG is supported by a roller at F and is loaded with 12000 lb. In the structure shown to the right member ABD is a solid rigid member pinned to the wall at A.5 in2. (answers: A = 10.384") file:///Z|/www/Statstr/statics/Stress/strsp35c. at G. supported by steel cable BC.008") 5.000 lb.5 c. to the wall. and is resting on top of member BCDE. The steel cable has an area of .. The members are loaded as shown.Stress Prob. supported by a roller.htm (3 of 4) [5/8/2009 4:39:54 PM] .Topic 3. htm (4 of 4) [5/8/2009 4:39:54 PM] .Topic 3.5 c.Determinate Select: Topic 3: Stress.Stress Prob.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsp35c. Strain & Hooke's Law . Young's modulus for aluminum is 10 x 106 lb/in2.800 psi.000 lb acting to the left is applied at junction C.900 psi. Stress-Al = 4700 psi.5 in2. (answers: Faluminum = 4.6: Assignment Problems . 1 in2. Young's modulus for aluminum is 10 x 106 lb/in2. 6 ft.htm (1 of 4) [5/8/2009 4:39:54 PM] .75 in2. The cross sectional areas and lengths of the steel and aluminum rods are respectively 1.. . Stressal = 4. def.0565") 2. This force will compress the steel member while stretching the aluminum member. 8 ft as shown in the diagram. Fsteel = 35. and brass rods are respectively 1. Fal = 4.700 lb.000 psi. steel rod AB and aluminum rod BC are joined together and then held between two rigid walls as shown.6 Assignment Problems-Indeterminate Topic 3. aluminum rod BC. aluminum.. comp. Fbr = 15.000 lb. Young's modulus for brass is 15 x 106 lb/in2. Stressbr = 20.Statically Indeterminate 1. Young's modulus for steel is 30 x 106 lb/in2. 1 in2.. Determine the stress that develops in each member. tens.0576") file:///Z|/www/Statstr/statics/Stress/strsp36a.5 in2. and determine the deformation of the aluminum rod. ) (answers: Stressst = 29. and determine the deformation of the aluminum rod.800 lb.300. Determine the stress that develops in each member. 500 psi) (answers: Deformation-Al = .-al = -. (answers: Fst = 44. 10 ft as shown in the diagram. Young's modulus for steel is 30 x 106 lb/in2.000 lb acting to the left is applied at junction B. 6 ft. The cross sectional areas and lengths of the steel. A force of 40. Stress-St = 23. 10 ft. steel rod AB.comp.000 lb acting to the left is applied at junction B. and a force of 20.. In the structure shown.Topic 3. In the structure shown. A force of 40.800 lb.. and brass rod CD are joined together and then held between two rigid walls as shown. Young's modulus for Steel is 30 x 106 lb/in2. is applied at point D. Determine the stress in each rod. In the structure shown. and is supported by two steel rods. (answers: StressFB = StressEC = 8150 psi.0448") 4.htm (2 of 4) [5/8/2009 4:39:54 PM] . and the movement of point D. member ABCD is pinned to the ceiling at point A. Md=. two aluminum rods and a threaded steel rod connect to vertical plates. as shown in the Diagram.Topic 3. as file:///Z|/www/Statstr/statics/Stress/strsp36a. A load of 8000 lb. FB and EC.6 Assignment Problems-Indeterminate 3 In the structure shown. A load of 20. StressAl = 80. with lengths and dimensions as shown.123") Select: Topic 3: Stress. Strain & Hooke's Law .htm (3 of 4) [5/8/2009 4:39:54 PM] . is applied a point C. If the structure is initially unstressed and the nut on the threaded steel rod in screwed in . StressAl = 8570 psi.860 psi. what is the stress that develops in the aluminum and steel members? (answers: StressST = 71.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsp36a.25 inches.300 psi.. Determine the stress that develops in the steel and aluminum members. (answers: StressST = 12.) (Movement of E = .. and the movement of point E. In the structure shown. and supported by steel member BG and aluminum member DF.200 psi.6 Assignment Problems-Indeterminate shown in the diagram.Topic 3.) 5. solid rigid member ABCDE is pinned to the wall at point A.000 lb. htm (4 of 4) [5/8/2009 4:39:54 PM] .Topic 3.6 Assignment Problems-Indeterminate file:///Z|/www/Statstr/statics/Stress/strsp36a. htm (1 of 2) [5/8/2009 4:39:54 PM] . all joints and support points are assumed to be pinned or hinged joints. Eal = 10 x 10 6 psi Unless otherwise indicated. Ebr = 15 x 10 6 psi. The structure is initially unstressed and then a load of 24. and by a roller at point C. MemberABC is pinned to the wall at A and is supported by a roller at point C.) In the structure shown horizontal member BCDE is supported by vertical brass member. Draw a Free Body Diagram showing all support forces and loads. Strain & Hooke's Law . B. C.) In the structure shown members ABC and CDE are assumed to be solid rigid members.7 Stress. an aluminum member EF.75 inch.Topic Examination 1.5 in2.7 . For this structure: file:///Z|/www/Statstr/statics/Stress/strsx37. For this structure: A. is applied at point D. Determine the axial stress in cable DF. AB. Cable DF has a diameter of . and is supported by steel cable DF.000 lb. Determine the movement of point B due to the applied load.Topic 3. Both AB and EF have cross sectional areas of . Est = 30 x 10 6 psi. Member CDE is pinned to the wall at point E. Select: Solution Problem 1) 2.Stress.Strain Topic Exam Topic 3. (For Solution. (For Solution.htm (2 of 2) [5/8/2009 4:39:54 PM] . Draw a Free Body Diagram showing all support forces and loads. Determine the axial stress that develops in aluminum member EF.Strain Topic Exam A.Topic 3.7 . B.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsx37.Stress. Est = 30 x 10 6 psi. Eal = 10 x 10 6 psi Unless otherwise indicated. Determine the resulting movement of point E. all joints and support points are assumed to be pinned or hinged joints. Ebr = 15 x 10 6 psi. Select: Solution Problem 2) Select: Topic 3: Stress. C. Strain & Hooke's Law . STEP 3: If we were to go about this problem in the normal fashion. There would be no way to solve this problem using that approach.Solution Exam Question 1 The structure shown members ABC and CDE are assumed to be solid rigid members. as well as any needed angles and dimensions.htm (1 of 3) [5/8/2009 4:39:54 PM] . STEP 2: Break any forces not already in x and y direction into their x and y components. Member ABC is pinned to the wall at A and is supported by a roller at point C. Strain & Hooke's Law .(See Diagram) Then we apply the equilibrium conditions: file:///Z|/www/Statstr/statics/Stress/xsol37a. we would have up to five unknowns and only three equations.Exam Solution 1 Topic 3.7: Stress. If we first take the structure apart and analyze member ABC there will only be three unknowns. C. Determine the axial stress in cable DF. Solution: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Therefore.75 inch. Draw a Free Body Diagram showing all support forces and loads. Cable DF has a diameter of . Determine the movement of point B due to the applied load. we must take a different perspective. Ebr = 15 x 10 6 psi.Topic 3. Est = 30 x 10 6 psi. For this structure: A.7a . B. drawing a free body diagram of the entire structure. Member CDE is pinned to the wall at point E. all joints and support points are assumed to be pinned or hinged joints. and is supported by steel cable DF. Eal = 10 x 10 6 psi Unless otherwise indicated. Drawing a free body diagram and applying the equilibrium conditions: (See Diagram) Sum Fx = Ex = 0 Sum Fy = Fy + Ey .7a .000 lbs.000 lbs = 0 Sum TE = -Fy (2 ft) + (9.740 psi C.htm (2 of 3) [5/8/2009 4:39:54 PM] .9.4418 in2 = 40. Ay = 3.Topic 3.000 lbs We next analyze member CDE in a similar fashion.Exam Solution 1 Sum Fx = Ax = 0 Sum Fy = Ay + Cy .000 lbs/ .000 lbs. Point B moves due to the deformation of member DF. file:///Z|/www/Statstr/statics/Stress/xsol37a.000 lbs = 0 Sum TA = (-12. StressDF = F/A = 18.000 lbs Next we find the remaining quantities B.000 lbs)(6 ft) + Cy (8 ft) = 0 Solving for the unknowns: Cy = 9. Ey = -9.12.000 lbs)(4 ft) = 0 Solving for the unknowns: Fy = 18. Finally.Topic 3. C/4' = .14 *.196". Point C moves down a proportional amount: Move. Strain & Hooke's Law .000 lb * 72")/(30 x 106 lb/in2 * (3.B/6' = . Solving Move.147" Return to: Topic 3.7a .Topic Examination or Select: Topic 3: Stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/xsol37a. Strain & Hooke's Law .098"/2 ft.196"/8'.Exam Solution 1 Deformation of DF = FL/EA = (18. B =. Point B moves down a proportional amount to the movement of point C. solving Move.098".375"2)) = .htm (3 of 3) [5/8/2009 4:39:54 PM] . Move.C = .7: Stress. Exam Solution 2 Topic 3. load acting at end D tries to rotate the bar BCDE downward about point C.000 lb. : Cy .Static Equilibrium Analysis Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.Topic 3.Solution Exam Question 2 In the structure shown horizontal member BCDE is supported by vertical brass member.htm (1 of 3) [5/8/2009 4:39:55 PM] . Both AB and EF have cross sectional areas of . Determine the resulting movement of point E. as well as any needed angles and dimensions. Draw a Free Body Diagram showing all support forces and loads.7b . Eal = 10 x 10 6 psi Unless otherwise indicated.FBr + FAl .000 lb. and by a roller at point C. Step 2: Resolve forces into x and y components.000 lb. C. The structure is initially unstressed and then a load of 24. This puts the brass member. = 0 file:///Z|/www/Statstr/statics/Stress/xsol37b. Determine the axial stress that develops in aluminum member EF.7: Stress. Strain & Hooke's Law . The 24. AB. EF. Ebr = 15 x 10 6 psi. Est = 30 x 10 6 psi. B.24. in tension.) Step 3: Apply the equilibrium conditions. an aluminum member EF. all joints and support points are assumed to be pinned or hinged joints. (All forces are either in x or y direction. Solution: Our first step is to apply our static equilibrium procedure to our structure Part I . For this structure: A. As a result the ceiling acts on the members as shown in the free body diagram. is applied at point D.5 in2. AB. in compression and the aluminum member. or redraw the FBD.) = 0 At this point in a statically determinate problem. we would. The effect of the 24. We diagram this. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem.Exam Solution 2 : + FBr (4 ft.) .000 lb. Since member BCDE rotates downward about point C. or symbolically: This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure. Another way to state the problem is that we need another independent equation to solve for the unknowns. be able to solve for the external support reactions. we can substitute into the deformation relationship and obtain:. We may get this from the way the problem is stated. However.. (8 ft. Since deformation of a member = [ force in member * length of member] / [ young's modulus of member * area of member]. but none of this will help. After substituting in the values for the file:///Z|/www/Statstr/statics/Stress/xsol37b. . or often from the geometry of the structure . The deformations of the brass and steel members will give us this additional equation. [FL/EA]Al = 3 [FL/EA]Br We now have our additional relationship between the forces.htm (2 of 3) [5/8/2009 4:39:55 PM] .and can not solve.Deformation Equation Our first step is to find some general relationship between the deformations of the members of the structure. or def = FL/EA. showing and labeling the deformations involved.) + FAl (12 ft.7b .Topic 3. we see we can write (from similar triangles): Deformation of Br / 4 ft = Deformation of Aluminum / 12 ft or we can rewrite as: Deformation of Aluminum = 3* Deformation of Brass. Part II.000 lb. We might try to take the structure apart in some way.24.as in this case. load will be to compress the brass member and to elongate the aluminum member which will cause member BCDE to rotate downward about point C. in this case. we observe that we have three unknowns and only two independent equations . in most cases. And finally we can find the movement of point E by finding the elongation of member EF.5 in2 = 29.770 lb.htm (3 of 3) [5/8/2009 4:39:55 PM] . finding Cy = 12.Topic Examination or Select: Topic 3: Stress. and solving we have: FBr = 3. (8 ft.Topic 3. Stress Aluminum = 14.000 lb.142 in.920 lb We find the stress from: Stress Brass = 3.) + FAl (12 ft.7b .5 in2 )] If we simplify this equation we obtain: FAl = 4 FBr We now substitute this into our torque equation from static equilibrium equations : + FBr (4 ft.690 lb/ .5 in2) = . Return to: Topic 3.5 in2 = 7.Exam Solution 2 members we have: ( FAl *48")/(10 x 106 lb/in2 * . We can now also solve for Cy.) . and FAl = 14.540 lb/in2.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/xsol37b.000 lb. from Deformation of EF = FL/EA = (14..in2 * .690 lb.770 lb * 48 in.) = 0 and obtain: (FBr)(4 ft.24.380 lb/in2.24. Strain & Hooke's Law .)/(10 x 106 lb. Strain & Hooke's Law . (8 ft.5 in2) = 3[( FBr * 96")/(15 x 106 lb/in2 *.7: Stress. the aluminum member.) = 0.) + (4 FBr) (12ft.) .770 lb/. 83 Thermal Stress.8: Thermal Stress/Strain & Deformation Topic 3. Strain & Deformation I 3. Strain & Deformation . However. and if negative.Topic Examination Thermal Stress. and is given by: where (alpha) is the linear coefficient of expansion for the material.htm (1 of 3) [5/8/2009 4:39:55 PM] .which causes the material to expand.8: Thermal Stress. Thermal deformation simply means that as the "thermal" energy (and temperature) of a material increases. For a long rod the main thermal deformation occurs along the length of the rod. and this increased vibration results in what can be considered a stretching of the molecular bonds . Brass = 20 x 10-6/oC = 11 x 10-6/oF. The first effect we will consider is thermal deformation.5 x 10-6/oF (80oF. Aluminum = 23 x 10-6/oC = 13 x 10-6/oF. Example 1 A twelve foot steel rod is initially at a temperature of 0oF and experiences a temperature increase to a final temperature of 80oF.Topic 3. [Some values of the linear coefficient of expansion are: Steel = 12 x 10-6/oC = 6. so does the vibration of its atoms/molecules. Strain & Deformation II 3. the material will shrink or contract. Some of these thermal effects include thermal stress. Strain & Deformation I ● ● ● ● ● 3. which represents (Tf To).075 inches (The length of the rod was converted into inches in the equation since the deformations are normally quite small.] The term is the temperature change the material experiences. What is the resultant change in length of the steel? Solution: Deformation = = 6.0oF) (144 inches) = . and deformation. the final temperature minus the original temperature.82 Mixed Mechanical/Thermal Examples r Example 1 r Example 2 r Example 3 3.8 Thermal Stress.5 x 10-6/oF.I Changes in temperatures causes thermal effects on materials.Strain & Deformation . Of course.84 Thermal Stress.) We see the deformation is indeed quite small.81 Thermal Stress. and is the fractional change in length per degree change in temperature. Strain & Deformation . The term 'L' represents the initial length of the rod.Assignment Problems 3. if the structure or members of the structure are file:///Z|/www/Statstr/statics/Stress/strs38. strain. thermal contraction. if the thermal energy (and temperature) of a material decreases. If the change in temperature is positive we have thermal expansion. and in many cases the thermal deformation has no significant effect on the structure. or shrinks? When I ask this question in my classroom it is not unusual for the majority of the answers to be incorrect. stays the same. (or contract)? When this happens a force and resulting stress develop in the structure. This is also true of volume expansion. A flat round copper plate has a hole in the center.htm (2 of 3) [5/8/2009 4:39:55 PM] . While unconstrained thermal expansion is relatively straight forward effect. and These formulas. and then compress it back to its original length (a mechanical deformation) . what happens when we constrain a structure or member so it can not expand.that is. this is not the case. it still requires a bit of thought. and (L + L)3 for Volume] there are cross terms involving the linear coefficient of expansion squared in the area formula. resulting in the two equations above. These terms are very small and can be ignored. However. See diagram below.expands. A simple way to determine the amount of stress is to let the material expand freely due to thermal expansion.Topic 3. The resulting changes in area and volume are given by: . such as in the following question. then a significant thermal stress may arise which can effect the structure substantially . both the area and volume of a material will change with a corresponding change in temperature. The plate is heated and expands. file:///Z|/www/Statstr/statics/Stress/strs38. and the coefficient of expansion squared and cubed in the volume formula. In the derivations [using (L + L)2 for area.and which we will address shortly. The inside volume of a glass bottle expands as if it were made of glass. Our first thought often is that since the plate is expanding. A somewhat more interesting aspect of thermal expansion is when it "can't" . What happens to the hole in the center of the plate . are not exact.8: Thermal Stress/Strain & Deformation constrained such that the thermal expansion can not occur. The atoms/molecules all move away from each other with the result that the hole expands just as if it were made of the same material as the plate. as written. In addition to the length. the hole is the center must be getting smaller. Topic 3. and then cross multiply by E. The thermal stress which develops if a structure or member is completely constrained (not allowed to move at all) is the product of the coefficient of linear expansion and the temperature change and Young's modulus for the material.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs38. Strain & Deformation II Or Topic 3: Stress.htm (3 of 3) [5/8/2009 4:39:55 PM] . F/A is stress and we can finally write: . Select:Thermal Stress. Strain & Hooke's Law . arriving at: however. length L from each side of the equation.note that we can cancel the = F/A .8: Thermal Stress/Strain & Deformation If we equate these two effects (deformations) we have: = FL/EA . 0oF)(30 x 106 lb/in2) = 15.5 x 10-6/oF) (80oF . Strain & Deformation II 1. if a thermal stress develops. of course. what is the stress in the steel at 80oF? [Young's modulus for steel is 30 x 106 lb/in2. In this case there was no initial stress. However. is the reason bridges are built with expansion joints which allow the structure to expand and contract freely and thus avoid thermal stresses. allowing expansions to avoid thermal stresses. Partially Constrained Thermal Deformation: A more normal situation in a structure. The rod is initially at temperature of 0oF and experiences a temperature increase to a final temperature of 80oF.] Solution: In a completely constrained problem. this is why concrete sidewalks are built with spaces separating adjacent slabs. Notice that this is quite a sizable stress. the thermal stress which develops is given by: = (6. rather than completely constrained or completely free thermal deformation.81 . This. and the coefficient of linear expansion of steel is 6. but occasionally long hot periods will allow stresses to built up until the highway actually exploded in a area. there are many cases where structures and materials are near or at their allowable stresses. This means a member may expand (or contract) but not as much as it would if unconstrained.600 lb/in2. Normally they do withstand these stresses. In that case. Additionally. or Example: A twelve foot horizontal steel rod is fixed between two concrete walls. is a partially constrained thermal deformation. the total stress may well exceed the allowable stress and cause the structure to fail.5 x 10-6/oF. producing a large hole in the concrete.81: Thermal Stress. Perhaps the best way to demonstrate this situation file:///Z|/www/Statstr/statics/Stress/strs381. however modern concrete highways are designed without expansion spaces to withstand thermal stresses which develop. 2.Strain & Deformation -II Topic 3.Thermal Stress. so the stress which developed is well within the range of allowable stresses for steel. If the steel rod was initially unstressed. where the member can not move at all. Completely Constrained Thermal Deformation The thermal stress which develops if a structure or member is completely constrained (not allowed to move at all) is the product of the coefficient of linear expansion and the temperature change and Young's modulus for the material.htm (1 of 2) [5/8/2009 4:39:55 PM] .Topic 3. Concrete highways used to also have expansion spaces built-in. Example 2 .81 .htm (2 of 2) [5/8/2009 4:39:55 PM] . Please select the following examples: Example 1 . Strain & Hooke's . Stress.Topic 3.Thermal Stress.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strs381. Example 3 or Select: Topic 3.Strain & Deformation -II is to work slowly through an example(s). In Diagram 2 we have shown the amount each file:///Z|/www/Statstr/statics/Stress/strse382a. members more often are partially constrained. We would also like to determine the deformation of the brass member.Example 1 Partially Constrained Thermal Deformation: In a structure.htm (1 of 4) [5/8/2009 4:39:56 PM] . We will consider beam bending at a later point. BE. due to the temperature increase. This means a member may expand (or contract) but not as much as it would if unconstrained. However to do this effectively we first need to consider the physical effects of the temperature change to determine the directions of the forces acting on the structure. CD. and supported by a Aluminum member.) The linear coefficient of expansion. al = 23 x 10-6 /oC. and Young's modulus for brass and aluminum are : br = 20 x 10-6 /oC. rather than being completely constrained or completely free. For this structure we wish to determine the stress which develops in the aluminum and brass members. We do this by first considering how much the brass and aluminum members would expand if they were free to expand. (At this point we will assume that the horizontal member ABC does not bend due to forces acting on it. The structure is initially unstressed and then experiences a temperature increase of 40 degrees Celsius.Example 1 Topic 3.82a: Mixed Mechanical/Thermal .82a: Mixed Mechanical/Thermal . and by a Brass member.5 in2 cross sectional area. Eal = 10 x 10 6 psi Solution: The first part of the solution will be to consider the static equilibrium conditions for the structure. Ebr = 15 x 10 6 psi. Member BE has a .Topic 3. Example 1 In the structure shown in Diagram 1 horizontal member ABC is pinned to the wall at point A. Both the brass and aluminum members try to expand.75 in2 cross sectional area and Member CD has . We will obtain the additional equation from the deformation is the problem.82a: Mixed Mechanical/Thermal . This is due to the fact that the brass and aluminum are axial members. as well as any needed angles and dimensions. but not as much as it would if unconstrained . We are now ready to apply static equilibrium conditions. However both can not expand . We have shown in Diagram 2 the direction the external support forces will act on the structure.one will "win" causing the other to contract. At point A where the structure is pinned to the wall. PART II: DEFORMATION file:///Z|/www/Statstr/statics/Stress/strse382a. ABC.Topic 3.htm (2 of 4) [5/8/2009 4:39:56 PM] .Fal + Ay = 0 Sum TorqueA = + Fbr (14 ft) . PART I: STATICS STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Thus both the brass and aluminum members are in compression. We will assume the brass "wins" forcing the aluminum to compress.Example 1 member would expand if free to do so. (Already done in Diagram 2) Apply equilibrium conditions: Sum Fx = Ax = 0 (There is no external horizontal force acting at point A) Sum Fy = Fbr . We need an additional equation to solve for the unknowns. Notice the forces at E and D are along the direction of the members. Fal. will rotate upward about point A as shown in Diagram 2. The horizontal member. and thus simply in tension or compression. That is. Ay) and not enough equations to solve for our forces. putting it into compression also.since it is compressing the aluminum which is in turn pushing back on the brass. we have put in horizontal and vertical support forces Ax and Ay. the brass will expand.Fal (8 ft) = 0 At this point we see we have too many unknowns (Fbr. 650 lbs.200 lbs.75 Fbr ) = 0. (Notice that we need not be concerned that the temperature is in Celsius rather than Fahrenheit.18 x 10-5 (in. the aluminum gets shorter.82a: Mixed Mechanical/Thermal .75 in2)]. with elongation being positive.FL / EA ]al (Notice that we use a negative sign for each + mechanical term since both members are in compression.1.sign) if the member is in compression.28x10-5 Fbr ) = (-0. We see (from Diagram 2) that: br / 14 ft = .75 Fbr We now substitute this into our deformation equation above obtaining: 1.Fal (120 in) / (10x106 lbs./in2 )(.al / 8 ft or simplifying: br = -1./in2 PART III: Deformation of Brass Member The deformation of the brass member may now be found from the general expression for the . in our assumption.8x10-5 Fal ) or 1. and the mechanical deformation (FL / EA) due to deformation ( the forces which develop in the members. So the total deformation of a member may be written as total FL / EA ] where the mechanical deformation term is positive (use + sign) if the member is =[ in tension. The stress in brass = F/A = 4.5 in2)] = -1. the net deformation is the sum of the thermal ) due to the temperature change. which we find from the geometry of the problem. the units cancel.740 lbs.75 in2 = 10.htm (3 of 4) [5/8/2009 4:39:56 PM] ./.Example 1 We first write a general relationship between the deformations.5 in2 = 8. and in problems involving mixed thermal and mechanical deformations (as in this problem) we need to keep signs associated with deformations.. Fal = 7. Now from our torque equation from static equilibrium in Part I we have Fal = 1. If we substitute this expression for the deformations into our general relationship ( br = -1.28x10-5 Fbr + 2.FL / EA ]br or putting in values deformation: [ file:///Z|/www/Statstr/statics/Stress/strse382a./lb.27 This is our additional equation. Once we have this general deformation relationship. solving: Fbr = 4.370 lbs.) Then after combining terms we have: (0.8x10-5 Fal = 0.193 + 2. and the mechanical deformation term is negative (use .75 al The negative sign in front of the deformation of aluminum term comes from the fact that.370 lbs.75[ .8x10-5 (1./in2 The stress in aluminum = F/A = 7.FL / EA ]br = -1.28x10-5 Fbr + 2./in2 )(.27 in.75[(23x10-6 / oC)(40oC)(120 in) .650 lbs. That is.) Fbr = ./.Fbr (96 in) / (15x106 lbs. we now substitute in our deformation expressions for mechanical plus thermal deformation.75 a) we obtain: [ . and contractions being negative.27 or 6.0768 . We now substitute in numeric values given in our problem and get: [(20x10-6 / oC)(40oC)(96 in) . since the linear coefficient of expansion is also per o Celsius.Topic 3. /in2 )(.) Return to: Topic 3.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse382a. (Since this deformation is positive. Strain & Hooke's . Stress. it means brass member CD does expand as we assumed.0210 in.5 in2)] = 0.htm (4 of 4) [5/8/2009 4:39:56 PM] . If the value were negative it would have meant that the brass member was actually compressed by the aluminum member. Strain & Deformation II or Select: Topic 3. which then would have expanded.Example 1 Brass= [(20x10-6 / oC)(40oC)(96 in) -(4. However the values found for the forces and the stress would have been correct in either case.370 lbs.)(96 in) / (15x106 lbs.81: Thermal Stress.82a: Mixed Mechanical/Thermal .Topic 3. ) The linear coefficient of expansion. the brass would expand more if free to do so. and Young's modulus for brass and steel are : br = 20 x 10-6 /oC.5 in2. Notice that the brass and steel members have forces acting on them at only two points. Example 2 In the structure shown below horizontal member BDF is supported by two brass members. st = 12 x 10-6 /oC.Example 2 Partially Constrained Thermal Deformation: In a structure. Both AB and EF have cross sectional areas of . reflecting these forces. members more often are partially constrained. determine the axial stress that develops in steel member CD.Topic 3. Member CD has a cross sectional area of .82b: Mixed Mechanical/Thermal . (At this point we will assume that the horizontal member BDF does not bend due to forces acting on it. The structure is initially unstressed and then experiences a temperature increase of 40 degrees celsius. The steel pulls back on the brass placing the brass in compression. if free to do so. As the coefficient of expansion of the brass is larger than the coefficient of expansion of the steel. and a steel member CD.htm (1 of 3) [5/8/2009 4:39:56 PM] . due to the temperature increase. Est = 30 x 10 6 psi Solution: PART I: STATICS In this problem we first consider how much the brass and steel member would expand. (See Diagram 2) As the brass expands it pulls on the steel placing the steel in tension. rather than being completely constrained or completely free. This also means that the external forces on the members due to the floor are equal to the forces in the members.Example 2 Topic 3. and the resulting movement of point D. The support reactions. Ebr = 15 x 10 6 psi. so they are axial members. This means a member may expand (or contract) but not as much as it would if unconstrained. file:///Z|/www/Statstr/statics/Stress/strse382b.5 in2. are shown in diagram 2. AB and EF.82b: Mixed Mechanical/Thermal . We will consider beam bending at a later point. For this structure. we see that the final net deformation of the brass and steel members will be equal. We need an additional equation. we substitute into the general deformation relationship and obtain: [ .82b: Mixed Mechanical/Thermal .Example 2 Additionally we note that the forces in the brass member are equal from symmetry of the structure.) The result of the brass and steel members working against each other is that the horizontal member BDF moves to an intermediate position. as shown in Diagram 2. as well as any needed angles and dimensions. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.Topic 3.Fst + Fbr = 0 Sum TorqueB = -Fst (6 ft) + Fbr (12 ft) = 0 The torque equation gives us: Fst = 2 Fb and the sum of y forces also gives us: Fst = 2 Fbr r We do not have enough equations at this point to solve the problem. (or if we mentally sum torque about point D.htm (2 of 3) [5/8/2009 4:39:56 PM] .FL / EA ]br = [ + FL / EA ]st + file:///Z|/www/Statstr/statics/Stress/strse382b. we see that forces in the brass members would produce opposing torque which would need to be equal and opposite for equilibrium. Since the distances are equal. We are now ready to proceed with the Statics. which we will obtain from the deformation relationships. or br = st Notice that both deformations are positive since the members expand. the forces in the brass members need to be equal to produce equal amounts of torque. PART II: DEFORMATION From the geometry of the problem. the center of the member BDF. Next using our expression for combined thermal and mechanical deformations: δ total = [α ∆TL FL / EA ]. (Diagram 2) We now write the equilibrium equations: Sum Fx = 0 Sum Fy = Fbr . 64x10-5 (2 Fbr ) = 0. or Movement of D = [ + FL / EA ]st .28x10-5 Fbr + 0.1.Topic 3./lb) Fbr = .28x10-5 Fbr ) = (0. Stress.400 lbs/.0307 . Point D is connected to steel member CD and so moves the amount CD deforms.200 lb.06144 in. We now substitute values from our problem into the equation and obtain: [(20x10-6 / oC)(40oC)(96 in) .0461 + 0.400 lbs. which gave us : Fst = 2 Fbr Substituting.0307 This is our additional equation.56 x 10-5 (in.Example 2 The sign of the mechanical deformation term for the brass is negative (-) since the brass is in compression. Return to: Topic 3.htm (3 of 3) [5/8/2009 4:39:56 PM] . Strain & Deformation II or Select: Topic 3. or Movement of D = [(12x10-6 / oC)(40oC)(96 in) + (2.Fbr (96 in) / (15x106 lbs/in2 )(.0307 in.5 in2)] or (0.0768 .5 in2 = 4. and Fst = 2 Fbr = 2. We can now substitute our relationship between the brass and steel force from our static equilibrium equations in part I.28x10-5 (Fbr ) + 0.5 in2)] = [(12x10-6 / oC)(40oC)(96 in) + Fst (96 in) / (30x106 lbs/in2 )(.400 lbs)(96 in) / (30x106 lbs/in2 )(. or 2. and so the Stress in steel = F/A = 2.64x10-5 Fst = 0.82b: Mixed Mechanical/Thermal . The sign of the mechanical deformation term for the steel is positive (+) since the steel is in tension. Then solving Fbr = 1.5 in2)] = 0.800 lbs/in2 PART III: MOVEMENT Movement of point D.81: Thermal Stress. we obtain: 1.64x10-5 Fst ) or 1..Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse382b. Strain & Hooke's . Topic 3. rather than being completely constrained or completely free. The steel.82c: Mixed Mechanical/Thermal . brass) are attached to each other and constrained between two rigid walls. the members may expand or contract against each other.5 x 10-6/oF. Example 3 In the structure shown three metal rods (steel.75 in2. aluminum and brass rod have areas and lengths respectively of 1. We would like to determine the stress which develops in each rod. Brass = 11 x 10-6/oF. The linear coefficient of expansion for the materials are as follows: Steel = 6. The rods are initially unstressed and then experience a temperature increase of 80o F. Ebr = 15 x 10 6 psi. Eal = 10 x 10 6 psi. In response an external force due to the wall acts inward on the members at each end as shown in Diagram 2. file:///Z|/www/Statstr/statics/Stress/strse382c. aluminum. 10 ft. as shown in the diagram.Example 3 Topic 3. and the amount of deformation of the aluminum rod. And Young's modulus for the materials are Est = 30 x 10 6 psi.Example 3 Partially Constrained Thermal Deformation: In a structure.82c: Mixed Mechanical/Thermal . 8 ft. Since all three members are trying to expand they put each other into compression. . and also push outward on the walls. Even though the total length of the three members is fixed.. . Solution: PART I: STATICS In this problem we first consider the effect of the change in temperature. members more often are partially constrained. 1 in2.htm (1 of 3) [5/8/2009 4:39:57 PM] . Aluminum = 13 x 10-6/oF.5 in2. This means a member may expand (or contract) but not as much as it would if unconstrained. 6 ft. this same argument would hold if we cut the structure through the aluminum member. And. where we have cut the structure through the steel member.htm (2 of 3) [5/8/2009 4:39:57 PM] . Since the structure is in equilibrium. This may be seen in Diagram 3.82c: Mixed Mechanical/Thermal .Example 3 With a little thought we realize that the forces in each member are the same and equal to the force exerted by the wall on the structure. of course. this section of the structure must also be in equilibrium. But we can see that this is only possible if there is an internal force in the steel member which is equal and opposite to the force exerted by the wall. or through the brass member (as shown in Diagram 4) So we can write (from static equilibrium) that Fst = Fal = Fbr = F file:///Z|/www/Statstr/statics/Stress/strse382c.Topic 3. Finding: Steel Stress = 11.1248 in. its deformation may be calculated from .mechanical) we obtain: [ .75 in2 = 14. This is the force in each rod. After a little consideration we see that following deformation relationship must be true: st + al + st = 0 That is.. + .870 lb/in2. Strain & Hooke's .430 lb/in2.5 in2)] + [(13 x 10-6/oF)(80oF)(120 in) . and we then calculate the stress in each rod from Stress = Force/Area. the sum of the deformations (some of which may be positive expansions and some of which may be negative contractions) must be zero.F (72 in) / (30x106 lbs/in2 )(1.FL / EA ]. We now substitute values from our problem into the equation and obtain: [(6.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strse382c.Example 3 PART II: DEFORMATION We now consider the geometry of the problem to obtain an additional equation to use in determining the force in the member.150 lb. Strain & Deformation II or Select: Topic 3.FL / EA ]al + [ . Brass Stress = 11.150 lb (120 in) / (10x10 * 1 in ] = -.13 x 10-6 in/lb]F Now solving for F = 11.Topic 3. + .6 x 10-6 in/lb (F) + 12 x 10-6 in/lb (F) + 8.150 lb/1 in2 = 11.81: Thermal Stress.5 x 10-6/oF)(80oF)(72 in) .FL / EA ]br = 0 The sign of the mechanical deformation term is negative (-) for all the rods since they are all in compression. Return to: Topic 3.0374 in.FL / EA ]st + [ . The negative deformation means that the aluminum is forced to shrink by that amount due to the effects of the steel and brass acting on it.150 lb/. Stress.82c: Mixed Mechanical/Thermal . 150 lb/in2.009 in.75 in2)] =0 Rewriting the equation and combining terms we can write: .53 x 10-6 in/lb (F)]. since the rods are fixed between two walls.F (120 in) / (10x106 lbs/in2 )(1 in2)] + [(11 x 10-6/oF)(80oF)(96 in) .htm (3 of 3) [5/8/2009 4:39:57 PM] . PART III: DEFORMATION Now that we have the amount of force in the aluminum member.150 lb/1.2467 in = [22. or [ -6 o o -6 2 al = [(13 x 10 / F)(80 F)(120 in) .5 in2 = 7.11. = [1.F (96 in) / (15x106 lbs/in2 )(. On substituting our expression for the total deformation (thermal +/. or .0845 in. Aluminum Stress = 11. 83:Thermal Stress. Total def. A concrete sidewalk slab is 3 ft.384") file:///Z|/www/Statstr/statics/Stress/strsp383. Determine the final stress that develops in each rod.000 psi. & Deformation . (Al. the rods experience a temperature increase of 80 o F.Problems The following values will be need for the Problems below Linear coefficient of Expansion: -6 o -6 o -6 o Steel = 6. 000 lb horizontal force is applied to the end of the brass rod as shown. Strain Assignment Problems Topic 3. and the aluminum rod is attached to a wall as shown in the diagram.Topic 3..htm (1 of 4) [5/8/2009 4:39:57 PM] . The rods are initially unstressed and then a 20.) 2. long. wide by 4 ft. = . EAluminum = 10 x 106 lb/in2. concrete = 6 x 10-6/oF Young's Modulus: ESteel = 30 x 106 lb/in2. EBrass = 15 x 106 lb/in2. stress = 10.. stress = 20.000 psi. Aluminum = 13 x 10 / F . EConcrete = 5 x 106 lb/ in2 1.83: Thermal Stress. If the concrete slab is constrained so it can not expand. An aluminum rod and a brass rod are attached to each other. Br.5 x 10 / F. Additionally. what stress would develop in it due to thermal effects if it experienced a temperature increase of 60o F? (Stress = 1800 psi. and the total movement of point C. Brass = 11 x 10 / F. Strain. determine the stress which develops in the steel and aluminum members. ( St. stress = 16000 psi..074") file:///Z|/www/Statstr/statics/Stress/strsp383. of F = -. Br. Def. A horizontal bar ABDF is pinned to the wall at point A. determine the stress which develops both in the steel and in the brass rod. Al stress = 2951 psi. Strain Assignment Problems 3. A steel rod and a brass rod are attached to each other and mounted between two walls as shown in the diagram.stress =9444 psi. stress = 8000 psi.(St. If the structure is initially unstressed and then experiences a temperature decrease of 70o F... and the amount of deformation of the brass rod... and the movement of point F.=. of Br. Move.0182") 4. If the structure is initially unstressed and then experiences a temperature increase of 80o F.htm (2 of 4) [5/8/2009 4:39:57 PM] ..83:Thermal Stress. as shown in the diagram. and supported by steel member BC and aluminum member DE.Topic 3. and the movement of Point B. an aluminum member DE. (Al. In the structure shown below horizontal member BCD is supported by vertical brass member. If the structure is initially unstressed and then experiences a temperature increase of 90 degrees. AB.Topic 3. Both AB and ED have cross sectional areas of ..5 in2. Strain & Hooke's . Stress.htm (3 of 4) [5/8/2009 4:39:57 PM] .83:Thermal Stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsp383.01") Select: Topic 3.= -. and by a roller at point C. determine the stress that develops in aluminum member DE. Strain Assignment Problems 5. Move B. stress = 5470 psi. Topic 3. Strain Assignment Problems file:///Z|/www/Statstr/statics/Stress/strsp383.htm (4 of 4) [5/8/2009 4:39:57 PM] .83:Thermal Stress. Determine the resulting movement of point D.84 . The structure is initially unstressed and then experiences a temperature decrease of 60 degrees Celsius. Aluminum = 13 x 10 / F . In the structure shown the L-shaped member BCD is supported by Steel rod AB and Aluminum member DE. EAluminum = 10 x 10 6 lb/in2. EBrass = 15 x 10 6 lb/in2. and supported by a brass member.htm (1 of 2) [5/8/2009 4:39:57 PM] . For this structure: A.Strain -Topic Examination Topic 3. Brass = 11 x 10 / F. Both BE and CF have cross sectional areas of . C. For this structure: file:///Z|/www/Statstr/statics/Stress/strsx3844. B. Determine the axial stress that develops in brass member BE.5 in2. (Select Solution Problem 1 for solution) 2. BE.Topic Examination The following values will be needed for the problems below Linear coefficient of Expansion: -6 o -6 o -6 o Steel = 6. as shown.84: Thermal Stress. The structure is initially unstressed and then experiences a temperature increase of 50 degrees Celsius. and by a steel member.5 x 10 / F. CF. concrete = 11 x 10-6/oF Young's Modulus: ESteel = 30 x 10 6 lb/in2. Draw a Free Body Diagram showing all support forces and loads. Strain & Deformation . In the structure shown member ABCD is pinned to the wall at point A.Topic 3. Member DE has a cross sectional area of 1 in2 and member AB has a cross sectional area of . EConcrete = 5 x 10 6 lb/in2 1.5 in2.Thermal Stress. and pinned at point C. Topic 3. (Select Solution Problem 2 for solution) or Select: Topic 3.84 . Strain & Hooke's . Draw a Free Body Diagram showing all support forces and loads B.htm (2 of 2) [5/8/2009 4:39:57 PM] . C.Strain -Topic Examination A.Thermal Stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/strsx3844. Determine the axial stress that develops in steel rod AB. Stress. Determine the resulting movement of point D. CF.Thermal Stress Exam Solution 1 Topic 3.5 x 10-6/oF. and supported by a brass member. . as well as any needed angles and dimensions. EAluminum = 10 x 10 6 lb/in2.5 in2. BE. and by a steel member. Aluminum = 13 x 10-6/oF . For this structure: A. Both BE and CF have cross sectional areas of .Solution Problem 1 The following values may be needed for the problem below Linear coefficient of Expansion: Steel = 6. Draw a Free Body Diagram showing all support forces and loads. The structure is initially unstressed and then experiences a temperature increase of 50 degrees Celsius. Solution: PART I: STATICS Our first step is to draw a free body diagram showing and labeling all load forces and support (reaction) forces. In order to do this accurately.Topic 3. EConcrete = 5 x 10 6 lb/in2 1.84. C. B. file:///Z|/www/Statstr/statics/Stress/xsol384a. Determine the resulting movement of point D. EBrass = 15 x 10 6 lb/in2.84: Exam . Determine the axial stress that develops in brass member BE. concrete = 11 x 10-6/oF Young's Modulus: ESteel = 30 x 10 6 lb/in2.htm (1 of 4) [5/8/2009 4:39:57 PM] . we first determine the direction of support forces by examining what thermal deformation are trying to occur and how the structure will respond. In the structure shown member ABCD is pinned to the wall at point A. Brass = 11 x 10-6/oF. 84.Thermal Stress Exam Solution 1 As shown in Diagram 2. Thus. and stopping it from expanding as much as it would like to. (Since the thermal coefficient of expansion of brass is larger than the thermal coefficient of expansion of steel. it will rotate about point A to a middle position as shown in Diagram 3. . and the free body diagram may now be drawn as shown in Diagram 4. pulling on the steel member causing it to expand more than it would like (putting the steel member in tension).Topic 3.) Although we assume member ABCD will not bend. file:///Z|/www/Statstr/statics/Stress/xsol384a. the brass member is in compression and the steel member is in tension. From the steel member's point of view.htm (2 of 4) [5/8/2009 4:39:57 PM] . the steel member want to expand less and pulls back on the brass member putting it into compression. as the brass member expands. but the brass member continues to expand. it expands to a point (thermal expansion) and wants to stop. the brass member (BE) would like to expand more than the steel member (CF) due to thermal effects. That is. 84.6 [ (12x10-6/oC) (+50oC) (72 in) + FST (72 in) / (30x106 lbs/in2 ) (.Thermal Stress Exam Solution 1 Apply equilibrium conditions: Sum Fx = Ax = 0. and Solving for FBR = 4.6 ST The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as: total =( FL/EA).6( Substituting in values.6x10-6 FBR = 0.072 in . or 2. we have: [ (20x10-6/oC) (+50oC) (72 in) . Sum TA = FST (10 ft) .060 lbs FST = 2.FBR (6 ft) = 0 We have too many unknowns at this point.6x10-6 FBR = 0. PART II: DEFORMATION We now find a relationship between the deformations to develop an additional equation. .046 From our static torque equation we have: FST (10 ft) .FBR (6 ft) = 0. or FST = . then substituting this expression into our deformation + FL/EA)ST relationship gives us: ( .5 in2 )] .FBR (72 in) / (15x106 lbs/in2 ) (.046.026 in + 2. Sum Fy = Ay .88x10-6 (.6 FBR) + 9. From the geometry of the problem. or 0. so we now examine the deformations in the problem.FBR + FST = 0. we have: + BR / 6 ft = + ST / 10 ft or + BR = .Topic 3.5 in2 )] =.6FBR We now substitute into our deformation expression 2.88x10-6 FST + 9.6x10-6 FBR = 0.88x10-6 FST .440 lbs file:///Z|/www/Statstr/statics/Stress/xsol384a.htm (3 of 4) [5/8/2009 4:39:57 PM] .FL/EA)BR = .9. Stress.Thermal Stress Exam Solution 1 Then the stress in brass member BE is s = F/A = 4.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/xsol384a. Strain & Hooke's .Topic Examination or Select: Topic 3. Strain & Deformation . .84: Thermal Stress.htm (4 of 4) [5/8/2009 4:39:57 PM] .060 lbs/ .440 lbs) (72 in) / (30x106 lbs/in2) (.84.0659 in Return to: Topic 3. and we can write: Movement of D / 12 ft = δCF / 10 ft.5 in2) ] / 10 ft or Movement of D = (12 ft) [ 0.5 in2 = 8.120 lbs/in2 PART III: MOVEMENT Finally.Topic 3.0549 in / 10 ft ] = 0. point D moves in proportion to the movement of point C (which is equal to the deformation of member CF). Movement of D / 12 ft = [ (12x10-6/oC) ( +50oC) (72 in) + (2. htm (1 of 3) [5/8/2009 4:39:58 PM] . For this structure: A. as well as any needed angles and dimensions. B. C. Determine the resulting movement of point D. Aluminum = 13 x 10-6/oF . we first determine the direction of support forces by examining what thermal deformations are trying to occur and how the structure will respond. Solution: PART I: STATICS Our first step is to draw a free body diagram showing and labeling all load forces and support (reaction) forces. The structure is initially unstressed and then experiences a temperature decrease of 60 degrees Celsius.84 . Member DE has a cross sectional area of 1 in2 and member AB have a cross sectional area of .Solution Problem 2 The following values may be needed for the problem below Linear coefficient of Expansion: Steel = 6.84: Exam . EBrass = 15 x 10 6 lb/in2.Topic 3. In the structure shown the L-shaped member BCD is supported by Steel rod AB and Aluminum member DE. Brass = 11 x 10-6/oF. and pinned at point C.5 in2. concrete = 11 x 10-6/oF Young's Modulus: ESteel = 30 x 10 6 lb/in2. Draw a Free Body Diagram showing all support forces and loads. Determine the axial stress that develops in steel rod AB. We will assume that the aluminum "wins" and does contract. but not file:///Z|/www/Statstr/statics/Stress/xsol384b. In order to do this accurately.Thermal Stress Exam Solution 2 Topic 3. however both can not win. as shown. As shown in Diagram 2. EConcrete = 5 x 10 6 lb/in2 2. the steel member (AB) and the aluminum member (DE) would both like to contract. EAluminum = 10 x 10 6 lb/in2.5 x 10-6/oF. of course.Fal = 0.htm (2 of 3) [5/8/2009 4:39:58 PM] . putting the steel into tension .and in our assumption. and we now apply static equilibrium conditions: Sum Fx = -Fst + Cx = 0. with the external forces. so we now examine the deformations in the problem. Sum TC = Fst (4 ft) .Thermal Stress Exam Solution 2 as much as it would like to .due to the steel resisting. also pulling on the steel. The aluminum is. file:///Z|/www/Statstr/statics/Stress/xsol384b. Sum Fy = Cy .84 .Topic 3. Diagram 3 is now our free body diagram of the structure with all forces in either the x or y-direction. acting on the structure. The result is that the bar moves to position as shown in Diagram 3.Fal (6 ft) = 0 We have too many unknowns at this point. forcing the steel to actually expand. as shown. putting the aluminum in tension. substituting in values.5 [(12x10-6/oC) (-60oC) (24 in) + Fst (24 in) / (30x106 lbs/in2 ) (.) Next. Stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stress/xsol384b.059 in Solving for Fal = 9.Thermal Stress Exam Solution 2 PART II: DEFORMATION We now find a relationship between the deformations to develop an additional equation. point D movement is equal to the deformation of member DE. we have: + st / 4 ft = - al / 6 ft or 1. or 6 x 10-6 (in/lb.5 in2 )] = .4x10-6 Fal = 0.0259 + 2.Topic 3.830 lbs Fst = 14.059. Strain & Hooke's .[(23x10-6/oC) (-60oC) (24 in) + Fal (24 in) / (10x106 lbs/in2 ) (1 in2 )] .) Fal = .Fal (6 ft) = 0 OR Fst = 1. Return to: Topic 3. D = [-. From the geometry of the problem.4x10-6 Fal = 0.4x10-6 Fst + 2. we have: 1. The negative sign indicates we made a correct assumption. and the aluminum member actually does contract a small amount.5 st = - al (Where the + sign indicates the steel member expands.2.0095 in.5 ( (Where the + sign in front of each the mechanical deformation terms is used because both member are in tension.0236 in ] = .750 lbs Then the stress in steel member AB is s = F/A = 14.5 Fal We now substitute into our deformation expression 2. and the .84 .htm (3 of 3) [5/8/2009 4:39:58 PM] . D = [(23x10-6/oC) (-60oC) (24 in) + 9830 lb (24 in) / (10x106 lbs/in2 ) (1 in2 )] Movement.5 Fal ) + 2.5 in2 = 29.sign indicates the aluminum contract) The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as: total =( + FL/EA). Strain & Deformation .750 lbs/ . and we can write: Movement.4x10-6 Fal.059 From our static torque equation we have: Fst (4 ft) .4x10-6 (1. substituting this expression into our deformation relationship gives us: + FL/EA)st = .( + FL/EA)al 1.84: Thermal Stress.500 lbs/in2 PART III: MOVEMENT Finally.4x10-6 Fst = 0. or -.0331 in + .0331 ..Topic Examination or Select: Topic 3. or 2. For this structure: A. B. Both BE and CF have cross sectional areas of . and by a steel member. The structure is initially unstressed and then experiences a temperature increase of 50 degrees Celsius. Determine the axial stress that develops in brass member BE. Determine the resulting movement of point D . -6 o -6 o -6 o 6 6 6 Solution: PART A: STATICS STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. abr = 20 x 10 / C.Example In the structure shown on right member ABCD is pinned to the wall at point A. all joints and support points are assumed to be pinned or hinged joints.5 in2. Est = 30 x 10 psi. Eal = 10 x 10 psi ast = 12 x 10 / C. Draw a Free Body Diagram showing all support forces and loads. Ebr = 15 x 10 psi.solution221 STATICS / STRENGTH OF MATERIALS . file:///Z|/www/Statstr/statics/Stests/stress/sol221. C. CF. First determine the direction of support forces by examining what thermal deformation are trying to occur and how the structure will respond.htm (1 of 4) [5/8/2009 4:39:58 PM] . aal = 23 x 10 / C Unless otherwise indicated. and supported by a brass member. as well as any needed angles and dimensions. BE. the steel member want to expand less and pulls back on the brass member putting it into compression. it will rotate about point A to a middle position as shown in Diagram 3. the brass member (BE) would like to expand more than the steel member (CF) due to thermal effects. pulling on the steel member causing it to expand more than it would like (putting the steel member in tension). and stopping it from expanding as much as it would like to.solution221 As shown in Diagram 2. but the brass member continues to expand. as the brass member expands. From the steel member's point of view.htm (2 of 4) [5/8/2009 4:39:58 PM] . file:///Z|/www/Statstr/statics/Stests/stress/sol221. it expands to a point (thermal expansion) and wants to stop. That is. (Since the thermal coefficient of expansion of brass is larger than the thermal coefficient of expansion of steel.) Since we assume member ABCD will not bend. 046 From our statics torque equation we have: F ST (10 ft) .6x10 F -6 ST = 0.6 d BR 6ft / = ST 10ft / BR total ST total The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as: d total = (a DTL FL/EA).6F BR We now substitute into our deformation expression file:///Z|/www/Statstr/statics/Stests/stress/sol221.F (72 in) + F F ST 6 ST -6 o o BR (72 in) / (15x10 lbs/in ) (. we have: +d +d or d = .88x10 -6 OR 2.5 in )] = . From the geometry of the problem.072 in .6(a DTL + FL/EA) ST Substituting in values.FBR + FST = 0 Sum TA = FST (10 ft) .026 in + 2.5 in )] OR 0. we have: [ (20x10 / C) (+50 C) (72 in) . + substituting this expression into our deformation relationship gives us: (a DTL .FL/EA) BR = .6x10 F -6 BR = 0.solution221 Thus the brass member is in compression and the steel member is in tension.88x10 F + 9.6 [ (12x10 / C) (+50 C) 2 2 -6 BR 6 2 2 -6 o o (72 in) / (30x10 lbs/in ) (.FBR (6 ft) = 0 PART B: DEFORMATION Too many unknowns. and the free body diagram may now be drawn as shown in Diagram 4.htm (3 of 4) [5/8/2009 4:39:58 PM] . we now find a relation ship between the deformations to develop an additional equation. Apply equilibrium conditions: Sum Fx = Ax = 0 Sum Fy = Ay .9.F BR (6 ft) = 0 OR F ST = . 6 FBR) + 9.solution221 2.5 in2) ] / 10 ft or Mov.htm (4 of 4) [5/8/2009 4:39:58 PM] . D = (12 ft) [ 0. point D moves in proportion to the movement of point C (which is equal to the deformation of member CF).0659 in Select: Topic 3: Stress.060 lbs/ .5 in = 8. D / 12 ft = [ (12x10-6/oC) ( +50oC) (72 in) + (2. Strain & Hooke's Law .440 lbs 2 2 Then the stress in brass member BE is s = F/A = 4.440 lbs) (72 in) / (30x106 lbs/in2) (.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol221. D / 12 ft = dCF / 10 ft Mov.0549 in / 10 ft ] = 0.88x10-6 (.060 lbs F ST = 2.120 lbs/in PART C: MOVEMENT Finally.6x10-6 FBR = 0.046 Solving for F BR = 4. and we can write: Mov. solution222 STATICS / STRENGTH OF MATERIALS - Example In the structure shown on right horizontal member BCF is supported by vertical brass members, AB, and EF, and by steel member, CD. Both AB and EF have cross sectional areas of .5 in2. Member CD has a cross sectional area of .75 in2. The structure is initially unstressed and then experiences a temperature decrease of 50 degrees celsius. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the axial stress that develops in steel member CD. C. Determine the resulting movement of point E . Est = 30 x 10 psi; Ebr = 15 x 10 psi; Eal = 10 x 10 psi ast = 12 x 10 / C; abr = 20 x 10 / C; aal = 23 x 10 / C Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. -6 o -6 o -6 o 6 6 6 Solution: PART A: STATICS file:///Z|/www/Statstr/statics/Stests/stress/sol222.htm (1 of 4) [5/8/2009 4:39:59 PM] solution222 STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces into x and y components STEP 3: Apply equilibrium conditions: Sum Fx = 0 Sum Fy = Dy - Ay - Fy =0 Sum Tc = Ay (6 ft) - Fy (6 ft) = 0 From torque equation we get Ay = Fy; If we put this into the force equation we get: Dy - 2Ay = 0 or Ay = Dy / 2 To solve for forces Ay, Dy and Fy, we need a third equation - which we obtain from the relationship between the deformations. PART B: DEFORMATION Since the forces in members AB and FE are equal, since they are made of the same material - brass, and since they are the same length and area, they will deform the same amount (from δ = FL / EA). Both the bottom brass member (AB and FE) and the top steel member (CD) try to contract (because the change in temperature is negative). However clearly both top and bottom members can not contract, one will "win," contracting while forcing the other member (s) to expand. We will assume the steel member "wins" actually shrinking a certain amount. The brass members will elongate the same amount. The steel contracts, as shown in diagram 2. file:///Z|/www/Statstr/statics/Stests/stress/sol222.htm (2 of 4) [5/8/2009 4:39:59 PM] solution222 We write the deformation relationship as: - δ (steel) = + δ (brass) (negative indicates that the deformation of the steel is a contraction) The deformation of the members depend on both the thermal deformation and the mechanical deformation due to the forces that develop in the members. This may be written as follows: δ = [α∆TL (FL/EA) ]material ("+" sign if member is in tension, "-" sign if member is in compression) where: α = thermal coefficient of expansion ∆T = change in temperature L = length of member F = force in member E = Young's modulus for material A = cross sectional area of member We now substitute the expression into our deformation relationship so: - δ (steel) = + δ (brass) becomes -[α∆TL + (FL/EA) ]steel = [α∆TL + (FL/EA) ]brass We now substitute in values and get the expression: -[(12x10-6/oC) (-50oC) (72 in) + Dy (72 in) / ( 30x106 lbs/in2) (.75 in2)] = +[(20x10-6/oC) (-50oC) (96 in) + Ay (96 in) / ( 15x106 lbs/in2) (.5 in2)] Combining terms, the equation becomes: -[-0.0432 + (3.2x10-6 ) Dy ] = [-0.096 + ( 12.8x10-6 ) Ay ] OR 3.2x10-6 Dy + 12.8x10-6 Ay = 0.139 This is our third equation, we can now substitute our relationship from statics: Ay =Dy / 2, into the above expression and get: 3.2x10-6 Dy + 12.8x10-6 (Dy / 2) = 0.139 Solving for D = 14,500 lbs (force in steel member); A = 7,250 lbs (force in brass member(s)) y y + Then to find stress in member CD file:///Z|/www/Statstr/statics/Stests/stress/sol222.htm (3 of 4) [5/8/2009 4:39:59 PM] solution222 s = D / A = 14,500 lbs / 0.75 in = 19,300 lbs/in y 2 2 PART C: MOVEMENT Point E moves since it is attached to member FE, and its movement is equal to the deformation of member FE. d FE = [ (20x10 / C) (-50 C) (96 in) + (7,250 lbs) (96 in) / (15x10 lbs/in ) (.5 in )] = -0.0032 in -6 o o 6 2 2 Select: Topic 3: Stress, Strain & Hooke's Law - Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol222.htm (4 of 4) [5/8/2009 4:39:59 PM] solution223 STATICS / STRENGTH OF MATERIALS -Example In the structure shown on right horizontal member BCD is supported by vertical brass member, AB, an aluminum member DE, and by a roller at point C. Both AB and ED have cross sectional areas of .5 in2. The structure is initially unstressed and then experiences a temperature increase of 60 degrees Celsius. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the axial stress that develops in aluminum member DE. C. Determine the resulting movement of point B . Est = 30 x 10 psi; Ebr = 15 x 10 psi; Eal = 10 x 10 psi ast = 12 x 10 / C; abr = 20 x 10 / C; aal = 23 x 10 / C Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. -6 o -6 o -6 o 6 6 6 Solution: PART A: STATICS STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. file:///Z|/www/Statstr/statics/Stests/stress/sol223.htm (1 of 3) [5/8/2009 4:39:59 PM] solution223 We first examine how much the members would expand freely, due to temperature increase, if they were free to expand (Diagram 2). Since both members can not expand as shown (assuming member BCD does not bend), one member will "win," expanding and compression the other member. We will assume the brass member actually expands, compression the aluminum member (Diagram 3). Since the two members are pushing against each other, both members are put into compression, and the support reaction are as shown in diagram 3. Now we write the equilibrium equations: Sum Fx = 0 (None) Sum Fy = -FBR + FC FAL = 0 Sum TC = FBR (4 ft) - FAL (12 ft) = 0 Not enough independent equations to solve for our three unknowns, so we need another equation - which we will obtain from the deformation relationship. file:///Z|/www/Statstr/statics/Stests/stress/sol223.htm (2 of 3) [5/8/2009 4:39:59 PM] solution223 PART B: DEFORMAION General relationship from geometry of structure: δBR / 4 ft = - δAL / 12 ft ; or 3 δBR = - δAL (negative sign is due to our assumption that the aluminum got shorter (is compressed).) We now expand the expression for our deformations using the fact that the total deformation is given by: δtotal = [α ∆TL FL/EA]material so, 3 [α ∆TL - FL/EA]brass = - [α ∆TL - FL/EA]aluminum or, using known values: 3[ (20x10-6/oC) (+60oC) (96 in) - FBR (96 in) / (15x106 lbs/in2) (.5 in2) ] = [ (23x10-6/oC) (+60oC) (48 in) - FAL (48 in) / (10x106 lbs/in2) (.5 in2) ] or + (0.346 - 3.84x10-5 FBR) = - (0.0662 - 0.96x10-5 FAL) or 3.84x10-5 FBR + 0.96x10-5 FAL = 0.412 in From our statics, we had FBR (4 ft) - FAL (12 ft) = 0 or FBR = 3 FAL, which we now substitute into our deformation relationship 3.84x10-5 (3 FAL) + 0.96x10-5 FAL = 0.412 in Solving for F AL + = 3,330 lbs ; F BR = 9,900 lbs ; s AL = F/A = 3,300 lbs/.5 in = 6,600 lbs/in 2 2 PART C: MOVEMENT Mov. B = the deformation of member AB So, Mov. B = [(20x10-6 / oC)(60oC)(96 in) - (9,900 lbs)(96 in) / (15x106 lbs/in2 )(.5 in2 )] Mov. B = -.01152" (This indicates that the brass member is compressed this amount.) Select: Topic 3: Stress, Strain & Hooke's Law - Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol223.htm (3 of 3) [5/8/2009 4:39:59 PM] solution224 STATICS / STRENGTH OF MATERIALS - Example In the structure shown on right horizontal member BCD is supported by vertical brass member, AB, a steel member DE, and is pinned to the floor at point C. Both AB and DE have cross sectional areas of .5 in2. The structure is initially unstressed and then experiences a temperature decrease of 60 degrees celsius. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the axial stress that develops in steel member DE. C. Determine the resulting movement of point B . Est = 30 x 10 psi; Ebr = 15 x 10 psi; Eal = 10 x 10 psi ast = 12 x 10 / C; abr = 20 x 10 / C; aal = 23 x 10 / C Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. -6 o -6 o -6 o 6 6 6 Solution: PART A : STATICS file:///Z|/www/Statstr/statics/Stests/stress/sol224.htm (1 of 4) [5/8/2009 4:39:59 PM] solution224 STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. In this problem we first examine how much the brass and steel members would shrink if they were free to do so due to the temperature decrease. The brass member, AB, would shrink much more than the steel member since it's coefficient of linear expansion is nearly twice that of steel. (see diagram 2) As the brass shrinks, it causes the horizontal member BCD to rotate about point C. As it does so, point D moves downward - which is all right to a certain point, as the steel is also shrinking. At a certain pint the steel is done shrinking due to the temperature decrease and "wants" to stop; however file:///Z|/www/Statstr/statics/Stests/stress/sol224.htm (2 of 4) [5/8/2009 4:39:59 PM] solution224 the brass "wants" to shrink more and continues to pull upward on point B causing more rotation of member BCD about point C, and causing point D to move downward an additional amount, putting steel member DE into compression. The steel member DE pushes back on point D, stopping the brass from shrinking any more and putting the brass into tension. This thought analysis process is how the directions of the support reactions direction in diagram 1 were arrives at. The horizontal member BCD ends up in an intermediate position as shown in diagram 2. We now write the equilibrium equations: Sum Fx = Cx = 0 Sum Fy = Fbr - Cy + Fst = 0 Sum TC = -Fbr(4 ft) + Fst(8 ft) = 0 There are not enough independent equations to solve for our three unknowns, so we need another independent equation - which we will obtain from the deformation relationship. PART B: DEFORMATION From the geometry of the problem we see that - δ brass / 4 ft = - δ steel / 8 ft (negative signs since deformation and contractions - members get shorter.) 2d =d brass steel We now expand our expression for our deformations using: δtotal = [α ∆TL FL / EA]material ; so, we obtain 2[α ∆TL + FL / EA]brass = [α ∆TL - FL / EA]steel and putting values of materials and the structure into this equation we obtain: 2[(20x10-6 / oC)(-60oC)(96 in) + Fbr (96 in) / (15x106 lbs/in2 )(.5 in2)] = [(12x10-6 / oC)(-60oC)(48 in) + Fst (48 in) / (30x106 lbs/in2 )(.5 in2)] or (-0.2304 + 2.56x10-5 Fbr ) = (-0.0346 - 0.32x10-5 Fst ) or 2.56x10-5 Fbr + 0.32x10-5 Fst = 0.196 from our static equilibrium torque equilibrium we had: -Fbr(4) + Fst(8) = 0 or Fbr = 2Fst , we now substitute this into our deformation equation. file:///Z|/www/Statstr/statics/Stests/stress/sol224.htm (3 of 4) [5/8/2009 4:39:59 PM] + solution224 and obtain: 2.56x10-5 (2 Fst ) + 0.32x10-5 Fst = 0.196 solving: F = 3,600 lbs ; F st br = 7,200 lbs 2 2 and stress in steel = F/A = 3,600 lbs / .5 in = 7,200 lbs/in PART C: MOVEMENT Point B is attached to member AB and so movement of point B is equal to the deformation of brass member AB. Mov. B = [(20x10-6 / oC)(60oC)(96 in) +(7,200 lbs)(96 in) / (15x106 lbs/in2 )(.5 in2)] Mov. B = -0.023 in Select: Topic 3: Stress, Strain & Hooke's Law - Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol224.htm (4 of 4) [5/8/2009 4:39:59 PM] solution225 STATICS / STRENGTH OF MATERIALS - Example In the structure shown on the right, horizontal member BDF is supported by two brass members, AB and EF, and a steel member CD. Both AB and EF have cross sectional areas of .5 in2. Member CD has a cross sectional area of .75 in2. The structure is initially unstressed and then experiences a temperature increase of 40 degrees celsius. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the axial stress that develops in steel member CD. C. Determine the resulting movement of point D . Est = 30 x 10 psi; Ebr = 15 x 10 psi; Eal = 10 x 10 psi ast = 12 x 10 / C; abr = 20 x 10 / C; aal = 23 x 10 / C Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. -6 o -6 o -6 o 6 6 6 Solution: PART A: STATICS STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. In this problem we first examine how much the brass and steel member would expand, if free to do so, due to the temperature increase. As the coefficient of expansion of the brass in larger than the coefficient of expansion of the steel, the brass would expand more if free to do so. (see diagram 2) file:///Z|/www/Statstr/statics/Stests/stress/sol225.htm (1 of 3) [5/8/2009 4:40:00 PM] solution225 As the brass expands it pulls on the steel placing the steel in tension. The steel pulls back on the brass placing the brass in compression. The support reactions, reflecting these forces, are shown in diagram 2. The forces in the brass member are equal from symmetry of the structure. The horizontal member BDF moves to a middle position, as shown in diagram 2. We now write the equilibrium equations: Sum Fx = 0 Sum Fy = Fbr - Fst + Fbr = 0 Sum TD = -Fbr (6 ft) + Fbr (6 ft) = 0 The torque equation simply gives us: Fbr = Fbr and the sum of y forces gives us: Fst = 2 Fbr We do not have enough equation to solve the problem. We obtain an additional independent equation for the formation relationship. PART B: DEFORMATION δ br = δst from the geometry of the structureexpanding using: δ total = [α ∆TL FL / EA ], we obtain: + [(20x10-6 / oC)(40oC)(96 in) - Fbr (96 in) / (15x106 lbs/in2 )(.5 in2)] = [(12x10-6 / oC)(40oC)(96 in) + Fst (96 in) / (30x106 lbs/in2 )(.75 in2)] or (-0.0768 - 1.28x10-5 Fbr ) = (0.0461 + 0.427x10-5 Fst ) file:///Z|/www/Statstr/statics/Stests/stress/sol225.htm (2 of 3) [5/8/2009 4:40:00 PM] solution225 or 1.28x10-5 Fbr + 0.427x10-5 Fst = 0.0307 From our static equilibrium equation in part I, we have : Fst = 2 Fbr Substituting, we obtain: 0.427x10-5 (2 Fbr ) + 1.28x10-5 (Fbr ) = 0.0307 solving: Fbr = 1,440 lbs ; Fst = 2,880 lbs Stress in steel = F/A = 2,880 lbs/.75 in2 = 3,840 lbs/in2 PART C: MOVEMENT Movement of point D. Point D is connected to steel member CD and so moves the amount CD deforms or Mov. D = [(12x10-6 / oC)(40oC)(96 in) + (2,880 lbs)(96 in) / (30x106 lbs/in2 )(.75 in2)] Mov. D = 0.0584 in Select: Topic 3: Stress, Strain & Hooke's Law - Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol225.htm (3 of 3) [5/8/2009 4:40:00 PM] aal = 23 x 10 / C Unless otherwise indicated. The structure is initially unstressed and then experiences a temperature decrease of 50 degrees celsius. abr = 20 x 10 / C. Eal = 10 x 10 psi ast = 12 x 10 / C. all joints and support points are assumed to be pinned or hinged joints. B. Member CD has a .5 in2 cross sectional area. Est = 30 x 10 psi.solution226 STATICS / STRENGTH OF MATERIALS .Example In the structure shown on the right horizontal member ABC is supported by steel member CD. and brass member CE. For this structure: A.htm (1 of 2) [5/8/2009 4:40:00 PM] . -6 o -6 o -6 o 6 6 6 Solution: Part A: STATICS file:///Z|/www/Statstr/statics/Stests/stress/sol226. C. Draw a Free Body Diagram showing all support forces and loads. Determine the axial stress that develops in member CE. Determine the resulting movement of point B . Member CE has a cross sectional area of .75 in2. Ebr = 15 x 10 psi. htm (2 of 2) [5/8/2009 4:40:00 PM] . as well as any needed angles and dimensions.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/stress/sol226. Select: Topic 3: Stress. Strain & Hooke's Law .solution226 STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Bending & Shear Stress [required.4a2 Shear Force/Bending Moment Problems . except#5] 4.9a Horizontal Shear Stress .6 Calculus Review (Brief) 4. #8.7a Centroids and the Moment of Inertia r 4. #6. #7 [Previous test problems] 4. #9 [Previous test problems] r Additional examples: (cantilever) #4.Horizontal Shear Stress r 4.Example 1 r 4.Example 2 r 4.11d Beams .1 .Topic Examination 4.Example 1 4.Bending Stress r 4.Inertia.Cantilevered Beams 1 [supplementary] 4. #5.Beam Selection r 4.Example 3 4.11c Beams .Example 2 r 4.1 Shear Forces and Bending Moments I 4. #3.Example 1 r 4.3c Cantilever Beam .9b Horizontal Shear Stress .Beam Selection [required.Example 3 r Additional examples: (simply supported) #4. #2.10a Beam Selection . #5. #6. #6 [Previous test problems] 4.8 Beams .10 Beams . #4.Horizontal Shear Stress [supplementary] 4.5 Shear Forces and Bending Moments .4b1 Shear Force/Bending Moment Problems . Moment.Problem Assignment 4 .3a Simply Supported Beam .Topic 4.4a1 Shear Force/Bending Moment Problems . #7.Problem Assignment 3 .7 Beams .3 Shear Forces and Bending Moments Examples r 4.Cantilevered Beams 2 [required.8a Bending Stress .11b Beams .Bending Stress [supplementary] 4.Example 1 r Additional Beam Selection Examples: #2. #3.Beams Shear Forces/Bending Moments Topic 4: Beams ● ● ● 4.Problem Assignment 2 . except#5] 4. #8.Simply Supported Beams 1 [supplementary] ● ● ● ● ● ● ● ● ● ● ● ● ● ● 4.9 Beams .Simply Supported Beams 2 [required] 4.2 Shear Forces and Bending Moments II 4. #5.11a Beams . #5. #4.htm (1 of 5) [5/8/2009 4:40:00 PM] . except#5] file:///Z|/www/Statstr/statics/Beams/beam41.4b2 Shear Force/Bending Moment Problems .Problem Assignment 5 .Bending Stress (cont) r 4.7b Bending Stress . #7.Problem Assignment 1 .12a Beams .8b Bending Stress .Example 2 r Additional Bending & Shear Stress Examples: r #1. #9 [Previous test problems] 4. #6.3b Simply Supported Beam . Stress [required] 4. acting downward right at the center of the beam..1 .htm (2 of 5) [5/8/2009 4:40:00 PM] . We will cut the beam a arbitrary distance (x) between 0 and 10 feet. and have considered the forces. We will now look at a particular type of non-axial member .loaded horizontal beams. In this first topic. and deformations which occur in these beams.Beams Shear Forces/Bending Moments ● ● 4. since the axial and shear stresses which will develop in the beam depend on the values of the shear forces and bending moments in the beam. we will focus only on the SHEAR FORCES and BENDING MOMENTS (internal torque) which occur in loaded beams. And as we proceed on we will also consider the problem of beam design and/or beam selection.Problem Assignment 6 .Beam Deflection [supplementary] 4.13 Beams -Topic Examination Topic 4. Up to this point we have generally looked at only axial members .12b Beams . toque. and deformations which occur in such members.members in simple tension or compression. and apply static equilibrium conditions to the left end section as shown in Diagram 2. each. as we shall see.1: Shear Forces & Bending Moments I We will now turn our attention to the forces and torque which develop in a loaded beam. In Diagram 1. stresses. then a section of the beam must also be in equilibrium. beam with a load of 10. stresses. file:///Z|/www/Statstr/statics/Beams/beam41. This is the static equilibrium condition for the whole beam. These quantities are very important. Due to symmetry the two support forces will be equal. Next let's examine a section of the beam. with a value of 5000 lb. and will begin the process of determining the forces. we have shown a simply supported 20 ft.000 lb. To understand the shear forces and bending moments in a beam.Topic 4. we will look at a simple example. We can do this since as the entire beam is in static equilibrium. load which we originally applied to the beam. (We have symbolically shown these in Diagram 2c. we have shown the missing force and torque.) When we cut the beam. and the support force cause internal "shearing forces" and internal torque called "bending moments" to develop. In diagram 2b. long .5000 lb. It does not appear in the sum of forces equation when we apply static equilibrium to the section .V = 0 . we next substitute the value of V from the force equation into the torque equation: . external support force.htm (3 of 5) [5/8/2009 4:40:00 PM] . Please note that M is a moment or torque . The 10.Beams Shear Forces/Bending Moments In Diagram 2a.not a force. since we know the beam section is in equilibrium. . Notice if we just include the 5000 lb. file:///Z|/www/Statstr/statics/Beams/beam41.) These are the equations for the shear force and bending moments for the section of the beam from 0 to 10 feet. Equilibrium Conditions: Sum of Forces in y-direction: + 5000 lb. x feet. * x + M = 0 . to a value of 50. nor the sum of torque (rotational equilibrium) will sum to zero . at x = 0. solving V = 5000 lb.where x is an arbitrary distance greater than 0 ft. we have shown left section of the beam. Notice that the internal shear force is a constant value of 5000 lb. at x = 10 ft. Therefore.as required for equilibrium. but that the value of the internal torque (bending moment) varies from 0 ft-lb. Neither the sum of forces (translational equilibrium).000 ft-lb.which will be our next step.Topic 4. there must be some forces and/or torque not accounted for. the internal shear force and bending moment at that point then become an external force and moment (torque) acting on the section. We have shown these in Diagram 2b. for the section.1 . Sum of Toque about left end: -V * x + M = 0 . then solving for M = 5000x (ft-lb. the section of the beam is clearly not in equilibrium. and less than 10 ft.000 lb. and labeled them V (shear force) and M (bending moment). we have our forces acting at point .000 lb * 10 (ft) -V * x (ft) + M = 0 . The two expressions above give the value of the internal shear force and bending moment in the beam. That is. we have an internal shear force and bending moment . Sum of Toque about left end: -10.V = 0 . there are 'point loads/forces' acting. Still when we put values into our expressions we put in values such as x = 9. and 20 ft. It all may sound confusing. we actually skip around these points. However. then solving for M2 = -[5000x (ft-lb. however we will finish analyzing our simple beam.000 lb. a stress can not be infinite. So far we have found expressions for the shear force and bending moments (V1 = 5000 lb. -10. A useful way to visualize this information is to make Shear Force and Bending Moment Diagrams .99999999 ft. We cut the beam at distance x (ft) from the left end. in effect. Of course. To deal with this difficulty. Where the beam was cut.(-5000 lb) * x (ft) + M = 0 . between 0 and 10 ft.Topic 4. This is.1 . (We add the '2'.it is actually applied over some area (even if the area if small). (See Diagram 4. but only because the number we actually put in was rounded off to 10 ft. and 10 ft..) . to indicate we are looking at section two of the beam. so the stress (F/A) at these points would in theory be infinite. cheating a bit. and will become clear as we do several examples.which are really the graphs of the shear force and bending moment expressions over the length of the beam..) We next apply static equilibrium conditions to the beam section.and a point has zero area.000 lb * 10 ft. solving V2 = -5000 lb. and then look at entire section to the left of where we cut the beam (See Diagram 3). and round it off (numerically) to 10 ft. and less then 20 ft.] First. we next substitute the value of V from the force equation into the torque equation : -10. into our equation for the bending moment. and we can not apply a force at a point .htm (4 of 5) [5/8/2009 4:40:00 PM] .) file:///Z|/www/Statstr/statics/Beams/beam41.100. ..000] ft-lb. between the distances of the 10 ft. We cut our section at 0' < x < 6'. and obtain: Equilibrium Conditions: Sum of Forces in y-direction: + 5000 lb. where x is now greater than 10 ft. but it works. . M1 = 5000x ft-lb) for section 1 of the beam.which now become external. The reason is that at 0 and 10 ft. Now we will look at the next section of the beam. in 'book' problems we normally apply forces at a point.Beams Shear Forces/Bending Moments [We really should not put exactly 0 ft. These are shown in Diagram 3 as V2 and M2. We are putting in the value x = 10 ft. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beam41. such as choosing the direction of the shear forces and bending moments.2: Beams .1 .Shear Force and Bending Moments II or Select: Topic 4: Beams .htm (5 of 5) [5/8/2009 4:40:00 PM] . and have made accompanying shear force and bending moment diagrams. we want to step back for a moment and become a little more specific concerning some details. Now that we have the general concepts concerning shear forces and bending moments. We have completed our first Shear Force/Bending Moment Problem.Beams Shear Forces/Bending Moments These are a quite useful way of visualizing how the shear force and bending moments vary through out the beam.Topic 4. Continue to: Topic 4. We have determined the expressions for the shear forces and bending moments in the beam. 2 .htm (1 of 2) [5/8/2009 4:40:01 PM] . and always choosing V & M in a positive direction according to the shear force and bending moments conventions defined above. When summing forces. (Or if it tends to put the top of the beam into compression and the bottom of the beam into tension. we need to take a moment to discuss the sign associated with the shear force and bending moment.Topic 4. we have shown the Shear Force V and Bending Moment M acting in positive directions according to the definitions above. This approach will simplify the sign conventions. If we then cut the beam and look at a left end section. The Bending Moment is positive if it tends to bend the beam section concave facing upward. We will deal with possible confusion by always working from the left for our beam sections. The signs associated with the shear force and bending moment are defined in a different manner than the signs associated with forces and moments in static equilibrium. That is.) In the beam section shown in Diagram 1. as we will see in the next example.Shear Force & Bending Moments II Topic 4. Notice that there is a possibility for a degree of confusion with sign notation. However before the next example. In Diagram 2a. ● ● The Shear Force is positive if it tends to rotate the beam section clockwise with respect to a point inside the beam section. and have indicated in an exaggerated way the bending caused by the load. we have the Diagram 2b. we have shown a simply supported loaded beam.2: Shear Forces and Bending Moments II Before continuing with a second example of determining the shear forces and bending moments in a loaded beam. file:///Z|/www/Statstr/statics/Beams/beam42. the direction of V shown in the diagram is in the negative y-direction. we will look at the causes of the internal bending moment in a little greater detail. yet it is a positive shear force. This can lead to some confusion unless we are careful. we will always select the V & M directions as shown in Diagram 1. which we call the internal bending moment. Looking at Diagram 2c and mentally summing torque about the center of the beam. These forces develop since.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beam42. As a result there are internal horizontal (x-direction) forces acting in the beam. This is the cause of the internal bending moment (torque) inside a loaded beam. Thus the net horizontal (x-direction) internal force in the beam section is zero. We will also examine an alternate method for determining the bending moments in a beam. M. Example 2 . the top region of the beam is put into compression and the bottom region of the beam is put into tension. the torque produced by these x-forces is not zero. Example 3 Or: Topic 4: Beams . We now continue by proceeding very slowly and carefully through a somewhat extended example(s). for the sake of clarity. left out the vertical shear force which develops.2 . however for every positive x-force.Shear Force & Bending Moments II In this diagram we have. as the beam bends. However.htm (2 of 2) [5/8/2009 4:40:01 PM] . Please select: Example 1 . there is an equal and opposite negative x-force.Topic 4. we see that the horizontal x-forces cause a net toque . even though the actual x-forces cancel each other. but have shown horizontal forces (-Fx and + Fx). Topic 4.3a: Simply Supported Beam - Example 1 Topic 4.3a: Simply Supported Beam - Example 1 Example 1 A loaded, simply supported beam is shown. For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam, and to make shear force and bending moment diagrams for the beam. We will work very slowly and carefully, step by step, through the solution for this example. Solution: Part A. We first find the support forces acting on the structure. We do this in the normal way, by applying static equilibrium conditions for the beam. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fy = -4,000 lb. - (1,000 lb./ft)(8 ft) - 6,000 lb. + By + Dy = 0 Sum TB = (Dy)(8 ft) - (6,000 lb.)(4 ft) + (1,000 lb./ft)(8 ft)(4 ft) + (4,000 lb.)(8 ft) = 0 Solving for the unknowns: By =23,000 lb.; Dy = -5,000 lb. (The negative sign indicates that Dy acts the opposite of the initial direction we chose.) file:///Z|/www/Statstr/statics/Beams/beame43a.htm (1 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 Part B: Now we will determine the Shear Force and Bending Moment expressions for each section of the loaded beam. For this process we will cut the beam into sections, and then use the translational equilibrium condition for the beam section (Sum of Forces = zero) to determine the Shear Force expressions in each section. Determining the Bending Moment expression for each section of the beam may be done in two ways. ● 1) By applying the rotational equilibrium condition for the beam section (Sum of Torque = zero), and solving for the bending moment. 2) By Integration. The value of the bending moment in the beam may be found from That is, the bending moment expression is the integral of the shear force expression for the beam section. . ● We now continue with the example. We begin by starting at the left end of the beam, and cutting the beam a distance "x" from the left end - where x is a distance greater than zero and less the position where the loading of the beam changes in some way. In this problem we see that from zero to eight feet there is a uniformly distributed load of 1000 lb./ft. However this ends at eight feet (the loading changes). Thus for section 1, we will cut the beam at distance x from the left end, where x is greater than zero and less then eight feet. Section 1: Cut the beam at x, where 0 < x < 8 ft., and analyze left hand section. 1. Draw a FBD of the beam section shown and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V1 and M1, as this is section 1 of the beam. 2. We check that we have all forces in x & y components (yes) 3. Apply translational equilibrium conditions to determine the shear force expression. Sum Fx = 0 (no net external x- forces) Sum Fy = -4,000 lb. - 1,000 lb./ft *(x) ft - V1 = 0 ; and solving: V1 = [-4,000 - 1,000x] lb. This expression gives us the values of the internal shear force in the beam between 0 and 8 ft. Notice as x nears zero, the shear force value in the beam goes to - 4000 lb., and as x approaches 8 ft., the shear force value becomes -12,000 lb., and that is negative everywhere between 0 and 8 ft. Let's think for a moment what this negative sign tells us. Since we found the shear force (V) by static equilibrium conditions, the negative sign tells us that we choose the incorrect direction for the shear force - that the shear force acts in the opposite direction. file:///Z|/www/Statstr/statics/Beams/beame43a.htm (2 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 However, we choose the positive direction of the shear force (by its definition) and so the negative sign also tells us we have a negative shear force. To try to simplify a somewhat confusing sign situation we may say this: As long as we work from the left end of the beam, and choose the initial direction of the shear force and bending moment in the positive direction (by their definition), then when we solve for the shear force and bending moment, the sign which results is the correct sign as applies to the shear force and bending moment values. If we graph the shear force expression above, we obtain the graph shown of the internal shearing force in the beam for the first eight feet. We next will determine the bending moment expression for this first beam section. 4. We can find the bending moment from static equilibrium principles; summing torque about the left end of the beam. Referring to the free body diagram for beam section 1, we can write: Sum Torque left end = -1000 lb/ft * (x) * (x/2) - V1 (x) + M1 = 0 To make sure we understand this equation, let's examine each term. The first term is the torque due to the uniformly distributed load - 1000 lb./ft * (x) ft (this is the load) times (x/2) which is the perpendicular distance, file:///Z|/www/Statstr/statics/Beams/beame43a.htm (3 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 since the uniform load may be considered to act in the center, which is x/2 from the left end. Then we have the shear force V1 times x feet to the left end, and finally we have the bending moment M1 (which needs no distance since it is already a torque). Next we substitute the expression for V1 (V1 = [-4,000 - 1,000x] lb.) from our sum of forces result above into the torque equation to get: Sum Torque left end = -1000 lb/ft * (x) * (x/2) -[-4,000 - 1,000x] (x) + M1 = 0 ; and solving for M1 = [-500x2 - 4,000x] ft-lb. This is our expression for the internal torque inside the load beam for section 1, the first eight feet, which is graphed in the diagram below. 5. Finally, we may also obtain the expression for the bending moment by integration of the shear force expression. The integrals we will be using are basic types. For a simple, brief review and/or introduction to basic calculus concepts, Please Select: Simple Derivatives/Integrals. Continuing with our example: Integration: 1000(1/2 x2) - 4000 (x) + C1; so M1 = -500x2 - 4,000x + C1 As we the results above show, when we do an indefinite integral, the result include an arbitrary constant, in this case called C1. To determine the correct value for C1 for our problem we must apply a boundary condition: That is, we must know the value of the bending moment at some point on our interval into to find the constant. For simply supported beams (with no external torque applied to the beam) the value of the bending moment will be zero at the ends of the beam. (There are many ways to explain why this must be so. One of the easiest explanations is to remember that the bending moment value at a point in a simply supported beam is equal to the total area under the shear force diagram up to that point. However, at the left end, as x goes to zero, the area under the shear force diagram would file:///Z|/www/Statstr/statics/Beams/beame43a.htm (4 of 8) [5/8/2009 4:40:02 PM] = Topic 4.3a: Simply Supported Beam - Example 1 also go to zero, and thus so would the bending moment value.) So we have for our "boundary condition" that at x = 0, M1 = 0. We put these values into our expression for the bending moment (M1 = -500x2 - 4,000x + C1), and solve for the value of the integration constant, C1; that is: 0 = -500(0)2 - 4,000(0) + C1, and solving: C1 = 0 Therefore: M1 = [-500x2 - 4,000x] ft-lb. for 0 < x < 8 ft., is our final expression for the bending moment over the first section. (Note, it is the same as found above by summing torque for the beam section.) We now continue with the next section of the beam. Referring to the beam diagram, we see that at a location just greater than 8 ft., there is no loading, and that this continues until 12 ft. where there is a point load of 6,000 lb. So for our second section, we cut the beam at a location "x", where x is greater than 8 ft., and less than 12 ft and then analyze the entire left hand section of the beam. Section 2: We cut the beam at x, where 8 < x <12 ft., and analyze the entire section left of where we cut the beam. 1. Draw a FBD of the beam section shown and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 2) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V2 and M2, as this is section 2 of the beam. file:///Z|/www/Statstr/statics/Beams/beame43a.htm (5 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 2. We check that we have all forces in x & y components (yes) 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x- forces) Sum Fy = -4,000 lb. - 1,000 lb./ft(8 ft) + 23,000 lb. - V2 = 0, Solving: V2 = 11,000 lb. 4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section. Rotational Equilibrium: Sum of Toque left end = - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - V2 * x + M2 = 0; then we substitute the value for V2 (V2 = 11,000 lb) from above and obtain: - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - (11,000 lb.) * x + M2 = 0; and then solving for M2 we find: M2 = [11,000x - 152,000] ft-lb. From integration of the shear force, we find: Integration: , or M2 = 11,000x + C2 We get our boundary condition from another characteristic of the bending moment expression - which is that the bending moment must be continuous. That is, the value of the bending moment at the end of the first beam section, and the value of the bending moment at the beginning of the second beam section must agree - they must be equal. We determine the value of the bending moment from our M1 equation as x approaches 8 ft. (M1 = [-500 (8)2 - 4,000(8)] = -64,000 ft-lb.) Then our boundary condition to find C2 is: at x=8 ft M=-64,000 ft-lb. We apply our boundary condition to find C2. Apply BC: -64,000 ft-lb. = 11,000 lb. (8) + C2, Solving: C2 = -152,000 ft-lb. Therefore: M2 = [11,000x - 152,000] ft-lb. for 8 < x < 12 In like manner we proceed with section 3 of the beam, cutting the beam at a location greater than 12 ft. and less 16 ft., and then analyzing the entire section left of where we cut the beam. Section 3: Cut the beam at x, where 12 < x < 16 ft. Analyze left hand section. 1. FBD. (See Diagram Section 3) file:///Z|/www/Statstr/statics/Beams/beame43a.htm (6 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 2. All forces in x & y components (yes) 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x- forces)Sum Fy = -4,000 lb. - 1,000 lb./ft(8 ft) + 23,000 lb. -6,000 lb. - V3 = 0, and Solving: V3 = 5,000 lb. 4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section. Rotational Equilibrium: Sum of Toque left end = - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - 6,000 lb. * 12 ft -V3 * x + M3 = 0; then we substitute the value for V3 (V3 = 5,000 lb) from above and obtain: - (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. - 6,000 lb. * 12 ft - (5,000 lb.) * x + M3 = 0; and then solving for M3 we find: M3 = [5,000x - 80,000] ft-lb. We will find the bending moment expression for this section using integration only. Integration , or M3 = 5,000x + C3 We obtain a boundary condition for section 3 by remembering that at a free end or simply supported (no external torque) end, the bending moment must go to zero, thus we have the boundary condition to find C3: at x = 16 ft., M = 0 ft-lb. Apply BC: 0 = 5,000(16) + C3, and Solving: C3 = -80,000 ft-lb. Therefore: M3 = [5,000x - 80,000] ft-lb. for 12 < x < 16 We now have our expressions for the shear forces and bending moments in each section of the loaded beam (summarized below). Additional, we have shown the shear force and bending moment diagrams for the entire beam - which is a visual representation of the internal shear forces and internal torque in the beam due to the loading. Part C: Shear Force and Bending Moment Diagrams: Using the expressions found above, we can draw the shear force and bending moment diagrams for our loaded beam. V1 = -1,000x+4,000 lb.; V2 = 11,000 lb.; V3 = 5,000 lb M1 =-500x2+4,000x ft-lb.; M2 = 11,000x -152,000 ft-lb.; M3 = 5,000x-80,000 ft-lb file:///Z|/www/Statstr/statics/Beams/beame43a.htm (7 of 8) [5/8/2009 4:40:02 PM] Topic 4.3a: Simply Supported Beam - Example 1 Return to: Topic 4.2: Shear Forces and Bending Moments II or Select: Topic 4: Beams - Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beame43a.htm (8 of 8) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 Topic 4.3b: Simply Supported Beam - Example 2 Example 2 For the loaded, simply supported beam shown, determine expressions for the internal shear forces and bending moments in each section of the beam, and draw shear force and bending moment diagrams for the beam. Solution: We first need to determine the external support reaction by applying our standard static equilibrium conditions and procedure.. PART A STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Resolve any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions. Sum Fx = 0 (no external x-forces acting on structure.) Sum Fy = (-800 lbs/ft)(8 ft) - (1,200 lbs/ft)(8 ft) + Ay + Cy = 0 file:///Z|/www/Statstr/statics/Beams/beame43b.htm (1 of 7) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 Sum TA = (Cy)(12 ft) - (800 lbs/ft)(8 ft)(4 ft) - (1,200 lbs/ft)(8 ft)(12 ft) = 0 Solving for the unknowns: Cy = 11,700 lbs; Ay = 4,270 lbs Part B: Determine the Shear Force and Bending Moment expressions for each section of the loaded beam. For this process we will cut the beam into sections, and then use Statics - Sum of Forces to determine the Shear Force expressions, and Integration to determine the Bending Moment expressions in each section of the beam. Section 1: We note that the loading of the beam (800 lb./ft) remains uniform until 8 feet, where it changes to 1200 lb./ft. As a result, for our first beam section, we cut the beam at an arbitary position x, where 0 < x < 8 ft. Then we analyze the left hand beam section. 1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V1 and M1, as this is section 1 of the beam. 2. Resolve all forces into x & y components (yes) 3. Apply translational equilibrium conditions (forces only) to the section 1 of the beam: Sum Fx = 0 (no net external x- forces) Sum Fy = 4,300 lbs - 800 lbs/ft(x)ft - V1 = 0; Solving for the shear force: V1 = [4,300 - 800x] lbs 4. We can find the bending moment from static equilibrium principles; summing torque about the left end of the beam. Referring to the free body diagram for beam section 1, we can write: Sum Torque left end = -800 lb/ft * (x) * (x/2) - V1 (x) + M1 = 0 To make sure we understand this equation, let's examine each term. The first term is the torque due to the uniformly distributed load - 800 lb./ft * (x) ft (this is the load) times (x/2) which is the perpendicular distance, since the uniform load may be considered to act in the center, which is x/2 from the left end. Then we have the shear force V1 times x feet to the left end, and finally we have the bending moment M1 (which needs no distance since it is already a torque). file:///Z|/www/Statstr/statics/Beams/beame43b.htm (2 of 7) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 Next we substitute the expression for V1 (V1 = [-4,300 - 800x] lb.) from our sum of forces result above into the torque equation to get: Sum Torque left end = -1000 lb/ft * (x) * (x/2) -[-4,300 - 800x] (x) + M1 = 0 ; and solving for M1 = [-400x2 + 4,300x] ft-lbs for 0 < x < 8 ft. We will now also find the bending moment expression by integration of the shear force equation. Integration , solving M1 = -400x2 + 4,300x + C1 Our boundary condition to find the integration constant, C1, is at x = 0, M = 0 (since this is a simply support beam end.) Applying the boundary condition: 0 = -400(0)2 + 4,300(0) + C1, and solving gives us: C1 = 0. Therefore the bending moment expression for section 1 of the beam is: M1 = [-400x2 + 4,300x] ft-lbs for 0 < x < 8 ft. (The shear force and bending moment diagrams are shown at the bottom of this example page.) Section 2: We continue in the same manner with beam section 2. We note that the loading changes once more at 12 ft, due to the upward support force acting at that point. So for beam section 2, we cut the beam at location x, where 8 < x <12 ft., and then analyze left hand beam section from x to the end of the beam. 1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We have labeled them as V2 and M2, as this is section 2 of the beam. 2. Resolve all forces into x & y components (yes). 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x- forces) Sum Fy = 4,300 lbs - 800 lbs/ft*(8 ft) - 1,200lbs/ft *(x - 8)ft - V2 = 0; and solving for the shear force expression: V2 = [7,500 - 1,200x] lbs file:///Z|/www/Statstr/statics/Beams/beame43b.htm (3 of 7) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. We will do it both ways for this section. Rotational Equilibrium: Sum of Toque left end = - (800 lb./ft * 8 ft) * 4 ft. -1200 lb/ft * (x - 8')*[(x-8')/2+8'] - V2 * x + M2 = 0; Please notice the second term. In that term the quantity (1200 * (x-8')) is load due to the 1200 lb/ft acting over the distance (x-8). However we still need to multiply the force expression times the distance to obtain the torque. The uniform load of 1200 lb/ft acts at the center of its distance (x-8'), so the lever arm to point A would be [(x-8')/2 + 8'] (See diagram.) We next substitute the value for V2 (V2 = [7,500 - 1,200x] lb.) from above and obtain: - (800 lb./ft * 8 ft) * 4 ft. -1200 * (x - 8')*[(x-8')/2+8'] - [7,500 - 1,200x]* x + M2 = 0; and then solving for M2 we find: M2 = [-600x2 + 7,500x - 12,800] ft-lbs for 8 < x < 12 Next we find the bending moment, M2, from integration of shear force expression,V2. Integration: , and solving, M2 = -600x2 + 7,500x + C2 We obtain our boundary condition for beam section 2 by remembering that the bending moment must be continuous along the beam. This means that value of the bending moment at the end of section 1 (at x = 8 ft.) must also be the value of the bending moment at the beginning of section 2 (at x = 8 ft.). Thus our boundary condition to find C2 is: at x = 8 ft M = 8,560 ft-lbs (from equation M1). Now applying the boundary condition and solving for the integration constant, C2, we have: 8560 ft-lbs = -600(8)2 +7500(8) + C2, and solving: C2 = -13,000 ft-lbs Therefore our bending moment expression is: M2 = [-600x2 + 7,500x - 12,800] ft-lbs for 8 < x < 12 (The shear force and bending moment diagrams are shown at the bottom of this example page.) Section 3: Finally, we continue with the last section of the beam, cutting the beam at location x, where 12 < x < 16 ft., and analyzing the left hand beam section from x to the left end of the beams. file:///Z|/www/Statstr/statics/Beams/beame43b.htm (4 of 7) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 1. Draw a FBD of the beam section showing and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We have labeled them as V3 and M3, as this is section 3 of the beam. 2. All forces in x & y components (yes) 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x- forces) Sum Fy = 4,300 lbs - 800 lbs/ft(8 ft) + 11,700 lbs - 1,200lbs/ft (x - 8)ft - V3 = 0 Solving for the shear force expressing: V3 = [-1,200x + 19,200] lbs 4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section. Rotational Equilibrium: Sum of Toque left end = - (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] + 11,700 lb. * 12 ft -V3 * x + M3 = 0; Once again the distance from end A at which the effective load due to the uniform load of 1200 lb/ft [1200 lb/ft file:///Z|/www/Statstr/statics/Beams/beame43b.htm (5 of 7) [5/8/2009 4:40:02 PM] Topic 4.3b: Simply Supported Beam - Example 2 * (x-8')]may be considered to act must be determined carefully. That distance [(x-8')/2 + 8')] is shown at the top of the adjacent diagram. We next substitute the value for V3 (V3 = [-1,200x + 19,200] lb.) from above and obtain: - (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] + 11,700 lb. * 12 ft - [-1,200x + 19,200] lb* x + M3 = 0; and then solving for M3 we find: M3 = [-600x2 + 19,200x - 153,000] ft-lbs for 12 < x < 16. Determine the bending moment by Integration: or, M3 = -600x2 + 19,200x + C3 We find our boundary condition for beam section 3, by realizing at the end of the beam (a free end) the bending moment must go to zero, so our boundary condition to find C3 is: at x = 16 ft M = 0 ft-lb. Appling the boundary condition, and solving for the integration constant C3, we have: BC: 0 = -600(16)2 + 19,200(16) + C3; and then C3 = -153,000 ft-lbs So the final expressionn for the bending moment on section 3 will be: M3 = [-600x2 + 19,200x - 153,000] ft-lbs for 12 < x < 16 PART C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above, we can draw the shear force and bending moment diagrams for our loaded beam. V1 = -800x + 4,300 lb.; V2 = -1,200x + 7,500 lb.; V3 = -1,200x + 19,200 lb M1 = -400x2 + 4,300x ft-lb.; M2 = -600x2 + 7,500 - 12,800 ft-lb.; M3 = -600x2 + 19,200x - 153,000ft-lb , Return to: Topic 4.2: Shear Forces and Bending Moments II or Select: file:///Z|/www/Statstr/statics/Beams/beame43b.htm (6 of 7) [5/8/2009 4:40:02 PM] htm (7 of 7) [5/8/2009 4:40:02 PM] .3b: Simply Supported Beam .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beame43b.Topic 4.Example 2 Topic 4: Beams . Solution: Part A: Our first step will be to determine the support reactions and external torque acting on the loaded beam. For a cantilevered beam (with one end embedded or rigidly fixed at the wall).htm (1 of 6) [5/8/2009 4:40:03 PM] . For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam. file:///Z|/www/Statstr/statics/Beams/beame43c.Topic 4. STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.Example 3 Topic 4. and label Mext) acting on the beam .as we have shown in the free body diagram of the beam.Example 3 Example 3 In this example we have a loaded. the wall may exert horizontal and vertical forces and an external torque (which we will call an external moment. as well as any needed angles and dimensions.3c: Cantilever Beam . and to draw the shear force and bending moment diagrams for the beam.3c: Cantilever Beam . as shown . cantilever beam. Draw a FBD of the beam section showing and labeling all forces and toque acting .000 lbs .000 ft-lbs Part B: Determine the Shear Force and Bending Moment expressions for each section of the loaded beam.000 lbs)(4 ft) . Mext = 200.Topic 4. We now find the bending moment expression by summing torque about the left end. Solving for shear force: V1 = 21.000 lbs + Ay = 0 Sum TA = (-4. 2.000 lbs)(14 ft) +Mext = 0 Solving for the unknowns: Ay = 21. That equation would be as follows: file:///Z|/www/Statstr/statics/Beams/beame43c. as this is section 1 of the beam.000 lbs)(8 ft) -(2.htm (2 of 6) [5/8/2009 4:40:03 PM] . and have labeled them as V1 and M1. and analyze left hand beam section. Apply translational equilibrium conditions (forces only) to determine the shear force expression: Sum Fx = 0 (no net external x.2.(2. STEP 3: Apply the static equilibrium conditions. Sum Fx = Ax = 0 Sum Fy = -4. Resolve all forces into x & y components (yes) 3.000 lbs . and then use static equilibrium condition . For this process we will cut the beam into sections.Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention for the shear force and bending moment discussed earlier.) (See Diagram . Section 1: Cut the beam at an arbitrary location x. where 0 < x < 4 ft.000 lbs/ft)(6 ft)(11 ft) .forces) Sum Fy = 21.sum of forces to determine the shear force expressions.including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam. from x to the left end of the beam.V1 = 0 .000 lbs/ft)(6 ft) . the beam loading changes.3c: Cantilever Beam .000 lbs.3. and Integration to determine the bending moment expressions in each section of the beam. 1.000 lbs 4.(2.000 lbs .). (since at 4 ft.(3.Example 3 STEP 2: Break any forces not already in x and y direction into their x and y components. Since we have an external moment acting at the end of the cantilevered beam. as this is section 2 of the beam.htm (3 of 6) [5/8/2009 4:40:03 PM] . Resolve all forces into x & y components (yes). the value of the bending moment must become equal to the negative of the external moment at the cantilevered end of the beam. [This is the same result we will find by integration.000 lb * x + 200.000 ft-lb + M1 = 0.000] ft-lb for 0 < x < 4 ft. when we substitute in the value for V1 = 21. which we will use to determine the integration constant C1. file:///Z|/www/Statstr/statics/Beams/beame43c.000x . And solving for M1 = 21. 1. particularly for non-point loads. we obtain the equation: Sum of Torque left end = -21. for section 2 we cut the beam at location x.3c: Cantilever Beam . Thus our boundary condition to determine C1 is: at x = 0. and so C1 = -200. for a cantilever beam.000 x -200. and analyze the left hand beam section. In general. the value of the bending moment at the wall is equal to the negative of the external moment.000 ft-lb (That is.Topic 4.200.000 ft-lb + M1 = 0.) We label them as V2 and M2.000 = 21000(0) + C1. Integration . due to the point load and the beginning of the uniformly distributed load.. 2. Draw a FBD of the beam section showing and labeling all forces and toque acting . integration is the faster method.000 Then the bending moment expression is: M1 = [21. Section 2: Since the loading changes at 8 ft.000 lb. is different for the cantilevered end of a beam as compared with the end of a simply supported beam (or a free end).000x + C1 The boundary condition.including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.] Now we find the bending moment equation by integration of the shear force expression. in order for the beam to be in rotational equilibrium (sum of torque = 0).) Applying the boundary condition and solving for the integration constant: -200. and then M1 = 21. M = -200.Example 3 Sum of Torque left end = -V1 * x + 200. where 4 < x < 8 ft.000 ft-lb. forces) Sum Fy = 21.4. Draw a FBD of the beam section showing and labeling all forces and toque acting ..000 lb. when we substitute in the value for V2 = 17..000 ft-lbs = 17.htm (4 of 6) [5/8/2009 4:40:03 PM] .000x . and analyze left hand beam section.including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam. Thus our boundary condition to find the integration constant C2 is: at x = 4 ft.* 4 ft.000] ft-lb for 4 < x < 8 ft.000 lbs .184. (from equation M1).000 ft-lb. and our expression for the bending moment on beam section 2 is: M2 = [ 17. as this is section 3 of the beam.Topic 4.000 lbs .* 4 ft. Next we also determine the bending moment expression by integration of the shear force equation from above.000 ft-lb + M2 = 0. Now applying the boundary condition and solving for the integration constant. -17. and so C2 = -184.) must also be the value of the bending moment at the beginning of section 2 (at x = 4 ft.000] ft-lb for 4 < x < 8 ft. from x to the left end of the beam. This means that value of the bending moment at the end of section 1 (at x = 4 ft. Integration: . where 8 < x < 14 ft.3c: Cantilever Beam . * x + 200. file:///Z|/www/Statstr/statics/Beams/beame43c.184. we have: -116.000x . We continue in like manner for the last section of the beam Section 3: Cut the beam at x. We now find the bending moment expression by summing torque about the left end.000 ft-lb. and so M2 = 17.).000 ft-lb + M2 = 0 And solving for M2 = [ 17.000 lb 4. -V2 * x + 200. we obtain the equation: Sum of Torque left end = -4000 lb. 1. That equation would be as follows: Sum of Torque left end = -4000 lb.000x + C2 We obtain our boundary condition for beam section 2 by remembering that the bending moment must be continuous along the beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000(4) + C2.000 lb. C2.V2 = 0 . M = -116.) We label them as V3 and M3. and solving: V2 = 17.Example 3 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.2000 lb/ft *(x-8')*[(x-8)/2 + 8'] .000x+30. -3000 lb. .[-2.3.224.} When we substitute in the value for V3 = [-2.000x2 +30. We now find the bending moment expression by summing torque about the left end.224.2000 lb/ft *(x-8')*[(x-8)/2 + 8'] -V3 * x + 200.. we obtain the equation: Sum of Torque= -4000 lb. and then M3 =-1..000 lb. is: at x = 14 ft M = 0 ftlb Applying the boundary condition: 0 = -1. M1 =[21.000 lbs .000x + 30.000 ft-lbs So M3 = [-1.000x2 + 30.000] ft-lbs for 8 < x < 14 ft.* 8 ft. V1 = 21. [ If you are unsure how the third term in the equation was obtained. we can draw the shear force and bending moment diagrams for our loaded beam.000x . file:///Z|/www/Statstr/statics/Beams/beame43c. Resolve all forces into x & y components (yes) 3. -3000 lb. .000] ft-lb.3c: Cantilever Beam .* 4 ft.000] ft-lb.000 ft-lb + M3 = 0.000 lbs .000 lbs . That equation would be as follows: Sum of Torque left end = -4000 lb.000x .4.000] lb.000 lbs/ft)(x-8)ft .000x2+30. Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. So our boundary condition to find the integration constant. M2 = [17.000x-184. please see example 2.forces) Sum Fy = 21.htm (5 of 6) [5/8/2009 4:40:03 PM] .Example 3 2..000x .000] ft-lb.000] lb* x + 200.000] ft-lb. and solving C3 = -224.. V2 = 17.000x + C3 Since the right hand end of the beam is a free end.000 ft-lb + M2 = 0 And solving for M3 = [-1.000x + 30.224.000] lb 4. the bending moment must go to zero as x goes approaches 14 ft.000x2+30.000 lb.* 8 ft.000(14)2 + 30.000x + 30.000x-200.000(14) + C3 . V3 = [-2. M3 = [-1.Topic 4.* 4 ft. C3.(2. By Integration: .V3 = 0 Solving for the bending moment: V3 = [-2.000] lb. 2: Shear Forces and Bending Moments II or Select: Topic 4: Beams .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beame43c.Example 3 Return to: Topic 4.htm (6 of 6) [5/8/2009 4:40:03 PM] .3c: Cantilever Beam .Topic 4. Solution311 STATICS & STRENGTH OF MATERIALS . simply supported beam is shown below.(1. B. Unless otherwise indicated. all joints and support points are assumed to be pinned or hinged joints. A = 4. and then use Statics . as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components.000 lbs/ft)(6 ft)(13 ft) = 0 Solving for the unknowns: C = 9.(2. Determine expressions for the internal shear forces and bending moments in each sections of the beam.000 lbs/ft)(4 ft) .(1. showing all external loads and support forces (reactions).600 lbs y y Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. STEP 3: Apply the equilibrium conditions. where 0 < x < 4 ft. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. and Integration to determine the Bending Moment expressions in each section of the beam.Example A loaded. Section 1: Cut the beam at x.000 lbs/ft)(4 ft)(2 ft) . Sum Fy = (-2. (Shown in Diagram) file:///Z|/www/Statstr/statics/Stests/beams1/sol311. For this beam: A. For this process we will cut the beam into sections. 1.Sum of Forces to determine the Shear Force expressions.htm (1 of 4) [5/8/2009 4:40:03 PM] . Make shear force and bending moment diagrams for the beam. C. Draw a Free Body Diagram of the beam. FBD.400 lbs.000 lbs/ft)(6 ft) + Ay + Cy = 0 Sum TA = (Cy)(10 ft) . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.2.htm (2 of 4) [5/8/2009 4:40:03 PM] . All forces in x & y components (yes) 3. FBD.V1 = 0 Solving: V1 = (4. where 10 < x < 16 ft.2.600 lb . M = 0 Apply BC: 0 = -1000(0)2 + 4600(0) + C1.V2 = 0 Solving: V2 = -3400 lbs 4.000 lb/ft *x . Integration 2 M1 = -1000 x + 4600x + C1 a)Boundary condition to find C1: at x = 0.000] ft-lbs for 4 < x < 10 Section 3: Cut the beam at x.000 ft-lbs Therefore M2 = [-3400x + 16.Solution311 2.forces) Sum Fy = 4600 lbs . 1. (Shown in Diagram) 2. 1. Integration M2 = -3400x + C2 a)Boundary condition to find C2: at x=4 ft M=2400 ft-lbs (from equation M1) Apply BC: 2400 ft lbs = -3400(4) + C2 Solving: C2 = 16. Solving: C1 = 0 Therefore M1 = [-1000x + 4600x] ft-lbs for 0 < x < 4 ft.000x) lbs 4.forces) Sum Fy = 4. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. All forces in x & y components (yes) 3.600 .000 lbs/ft(4 ft) . All forces in x & y components (yes) 3. where 4 < x < 10 ft. Section 2: Cut the beam at x.2. FBD. (Shown in Diagram) 2. Apply translational equilibrium conditions (forces only): 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol311. 000 ft-lb 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol311. we can draw the shear force and bending moment diagrams for our loaded beam.600 ft-lb. V1 = 4.forces) Sum Fy = 4.Solution311 Sum Fx = 0 (no net external x.000 lb M1 = -1. V2 = -3.000 ft-lbs Therefore M3 = [-500x + 16. M3 = -500x +16. M2 = -3.128.400x+16.000 lb/ft(4 ft) + 9.V3 = 0 Solving: V3 = [-1000 x + 16.000(16) + C3 Solving: C3 = -128.000x+16.000] ft-lbs for 10 < x < 16 2 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000x .000 lb/ft(x-10)ft .2.400 lbs 1.000x +4. no external torque so M3=0) Apply BC: 0 = -500(16)2 + 16.000x-128.000] lbs 4. V3 = -1.000x lb.000x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (end of beam.htm (3 of 4) [5/8/2009 4:40:03 PM] .000 ft-lb. Integration 2 M3 = -500x + 16.400 lb.600-2.600 lbs . Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol311.htm (4 of 4) [5/8/2009 4:40:03 PM] .Solution311 Select: Topic 4: Beams . For this beam: A.000 lb = 0 STA = -(5. all joints and support points are assumed to be pinned or hinged joints. file:///Z|/www/Statstr/statics/Stests/beams1/sol312. and Integration to determine the Bending Moment expressions in each section of the beam.67 ft) + (By)(10 ft) . SFx = 0 SFy = Ay + By .000 lbs)(14 ft) = 0 Solving for the unknowns: By = 11. STEP 2: Break any forces not already in x and y direction into their x and y components. Unless otherwise indicated.735 lbs.000 lb/ft) 6. STEP 3: Apply the equilibrium conditions. C.000 lbs)(6.htm (1 of 3) [5/8/2009 4:40:04 PM] . Ay = -735 lbs (acts downward) Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. For this process we will cut the beam into sections. Construct the shear force and bending moment diagrams for the beam. showing all external loads and support forces (reactions).(6. Draw a Free Body Diagram of the beam. simply supported beam is shown below. B.Example A loaded.1/2(10 ft)(1.solution312 STATICS & STRENGTH OF MATERIALS .Sum of Forces to determine the Shear Force expressions. Solution: Part A: Statics STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Determine expressions for the internal shear forces and bending moments in each sections of the beam. and then use Statics . as well as any needed angles and dimensions. V1 =0 Solving: V1 = [-50x2 . (Shown in Diagram) 2. 1.forces) SFy = -735 lb . FBD. where (0 < x < 10) ft. (simply supported beam) Apply BC: 0 = -16.735] lbs 4.htm (2 of 3) [5/8/2009 4:40:04 PM] .forces) SFy = -735 lb .solution312 Section 1: Cut the beam at x.000 lb/ft) + file:///Z|/www/Statstr/statics/Stests/beams1/sol312. (Shown in Diagram) 2. All forces in x & y components (yes) 3.67x 3 . M = 0. All forces in x & y components (yes) 3. where (10 < x < 14) ft.735x] ft-lbs for (0 < x < 10) ft. Apply translational equilibrium conditions (forces only): SFx = 0 (no net external x.1/2(10 ft)(1.67x3 . FBD.67(0)3 . Apply translational equilibrium conditions (forces only): SFx = 0 (no net external x. Integration M1 = -16.000(lb/ft)/10ft)](x ft) .735(0) + C1. Solving: C1 = 0 Therefore M1 = [-16.735x + C1 Boundary condition to find C1: at x = 0. 1.1/2(x ft)[(1. Section 2: Cut the beam at x. htm (3 of 3) [5/8/2009 4:40:04 PM] . M = 0. Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000 lbs M1 = [-16. we can draw the shear force and bending moment diagrams for our loaded beam. M2 = [6.84.000 ft-lbs Therefore M2 = [6.67x3 .000] ft-lbs.000x + C2 Boundary condition from Section 1 at (x = 14 ft) . Select: Topic 4: Beams .000(14 ft) + C2 Solving: C2 = -84.735x] ft-lbs. V1 = [-50x2 . V2 = +6.000x .735] lbs.735 lb .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol312.solution312 11. (end of simply supported beam): Apply BC: 0 = 6.000] ft-lbs.000x .V2 = 0 Solving: V2 = +6.84. Integration M2 = 6.000 lbs 4. (1. Draw a Free Body Diagram of the beam.Example A loaded.Sum of Forces to determine the Shear Force expressions.(1. simply supported beam is shown below. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. B. C.500 lbs/ft)(4 ft) + By + Cy = 0 Sum TB = (Cy)(6 ft) + (1. as well as any needed angles and dimensions.500 lbs/ft)(4 ft)(8 ft) = 0 Solving for the unknowns: C = 6.000 lbs/ft)(4 ft) . and then use Statics . showing all external loads and support forces (reactions).Solution313 STATICS & STRENGTH OF MATERIALS . Determine expressions for the internal shear forces and bending moments in each sections of the beam. Make shear force and bending moment diagrams for the beam. STEP 3: Apply the equilibrium conditions.670 lbs. B = 3. and Integration to determine the Bending Moment expressions in file:///Z|/www/Statstr/statics/Stests/beams1/sol313. Sum Fy = (-1.000 lbs/ft)(4 ft)(2 ft) . STEP 2: Break any forces not already in x and y direction into their x and y components. Unless otherwise indicated.htm (1 of 4) [5/8/2009 4:40:04 PM] . For this process we will cut the beam into sections. For this beam: A. all joints and support points are assumed to be pinned or hinged joints.330 lbs y y Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. Analyze left hand section. Section 1: Cut the beam at x. Analyze left hand section. Section 2: Cut the beam at x. where 4 < x < 10 ft. FBD. (Shown in Diagram) 2. FBD. 1.V1 = 0 Solving: V1 = -1.000 lbs/ft (4 ft) + 3.000x lbs 4.htm (2 of 4) [5/8/2009 4:40:04 PM] 2 .V2 = 0 Solving: V2 = -667 lbs 4. Integration M2 = -667x + C2 a)Boundary condition to find C2: at x=4 ft M=-8000 ft-lbs (from equation M1) file:///Z|/www/Statstr/statics/Stests/beams1/sol313. where 0 < x < 4 ft. Integration 2 M1 = -500 x + C1 a)Boundary condition to find C1: at x=0 M=0 Apply BC: 0 = -500(0)2 + C1 Solving: C1 = 0 Therefore M1 = [-500x ] ft-lbs for 0 < x < 4 ft. All forces in x & y components (yes) 3. 1. (Shown in Diagram) 2.forces) Sum Fy = -1.Solution313 each section of the beam.forces) Sum Fy = -1. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 lbs/ft(x) .330 lbs . All forces in x & y components (yes) 3. FBD. Integration 2 M3 = -750x + 21. no external torque so M3=0) Apply BC: 0 = -750(14)2 + 21.V3 = 0 Solving: V3 = [-1.forces) Sum Fy = -1. V1 = -1. M3 = -750x +21.330] ft-lbs for 4 < x < 10 Section 3: Cut the beam at x.147. M2 = -667x-5.330 ft-lbs Therefore M2 = [-667x .500x+21. Analyze left hand section. All forces in x & y components (yes) 3. we can draw the shear force and bending moment diagrams for our loaded beam.000x lb.670 lbs 1500lbs/ft(x-10)ft . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.htm (3 of 4) [5/8/2009 4:40:04 PM] . where 10 < x < 14 ft. V2 = -667 lb.330 ft-lb.000 ft-lbs Therefore M3 = [-750x + 21.500x + 21.000 lb M1 = -500x ft-lb.5.000x + C3 a)Boundary condition to find C3: at x=14 ft M=0 ft-lbs (end of beam.000] ft-lbs for 10 < x < 14 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000] lbs 4. (Shown in Diagram) 2.Solution313 Apply BC: 8000 ft-lbs = -667(4) + C2 Solving: C2 = -5.000 lbs/ft(4 ft) + 3. 1.000x-147.000x .000(14) + C3 Solving: C3 = -147. V3 = -1.330 lbs + 6.000 ft-lb 2 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol313. Solution313 Select: Topic 4: Beams .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol313.htm (4 of 4) [5/8/2009 4:40:04 PM] . Example A loaded. Make shear force and bending moment diagrams for the beam. Unless otherwise indicated. B. Section 1: Cut the beam at x. showing all external loads and support forces (reactions).500 lbs. A = -1. and Integration to determine the Bending Moment expressions in each section of the beam.000 lbs/ft)(8 ft)(12 ft) = 0 Solving for the unknowns: C = 14.000 lbs)(4 ft) .Solution314 STATICS & STRENGTH OF MATERIALS . Sum Fy = (-5. For this beam: A. and then use Statics .htm (1 of 4) [5/8/2009 4:40:05 PM] . as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. Analyze left hand section.(5.000 lbs/ft)(8 ft) + Ay + Cy = 0 Sum TA = (Cy)(8 ft) .000 lbs) . where 0 < x < 4 ft. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. STEP 3: Apply the equilibrium conditions.(1.Sum of Forces to determine the Shear Force expressions. file:///Z|/www/Statstr/statics/Stests/beams1/sol314. Draw a Free Body Diagram of the beam. C. all joints and support points are assumed to be pinned or hinged joints.500 lbs y y Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. Determine expressions for the internal shear forces and bending moments in each sections of the beam. For this process we will cut the beam into sections.(1. simply supported beam is shown below. 000 ft-lbs Therefore M2 = [-6.000 lbs . Analyze left hand section. (Shown in Diagram) 2. file:///Z|/www/Statstr/statics/Stests/beams1/sol314. where 4 < x < 8 ft. All forces in x & y components (yes) 3.500x + C2 a)Boundary condition to find C2: at x=4 ft M=-8.500 lbs 4. where 8 < x < 16 ft.500 lbs . (Shown in Diagram) 2. Analyze left hand section.000 ft-lbs (from equation M1) Apply BC: 8000 ft-lbs = -6. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. Integration M2 = -6. 1.forces) Sum Fy = -1.V1 = 0 Solving: V1 = -1. FBD.forces) Sum Fy = -1. FBD.500x + 20.500 lbs 4. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.500(0) + C1 Solving: C1 = 0 Therefore M1 = [-1.htm (2 of 4) [5/8/2009 4:40:05 PM] .500x + C1 a)Boundary condition to find C1: at x=0 M=0 Apply BC: 0 = -1.500 lbs .000] ft-lbs for 4 < x < 8 Section 3: Cut the beam at x. All forces in x & y components (yes) 3.Solution314 1. Integration M1 = -1.5. Section 2: Cut the beam at x.V2 = 0 Solving: V2 = -6.500x] ft-lbs for 0 < x < 4 ft.500(4) + C2 Solving: C2 = 20. 000x .5.forces) Sum Fy = -1.000 lb M1 = -1. Integration 2 M3 = -500x + 16.V3 = 0 Solving: V3 = [-1. FBD.htm (3 of 4) [5/8/2009 4:40:05 PM] . V3 = -1.500 lbs .000] lbs 4.500 lb.000 ft-lbs Therefore M3 = [-500x +16. M2 = -6.500 lb.000x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (end of beam. V2 = -6.500x ft-lb.Solution314 1.000(16) + C3 Solving: C3 = -128.000] ft-lbs for 8 < x < 16 2 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. (Shown in Diagram) 2.000x-128. V1 = -1.000x + 16.500 lbs 1.000x+16. M3 = -500x +16.500x+20. All forces in x & y components (yes) 3. we can draw the shear force and bending moment diagrams for our loaded beam.000 lbs + 14.000 ft-lb 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol314.128.000ft-lb.000lbs/ft(x-8)ft . no external torque so M3=0) Apply BC: 0 = -500(16)2 + 16. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. htm (4 of 4) [5/8/2009 4:40:05 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol314.Solution314 Select: Topic 4: Beams . htm (1 of 4) [5/8/2009 4:40:05 PM] . Solution: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Determine expressions for the internal shear forces and bending moments in each section of the beam.200 lbs/ft)(8 ft)(12 ft) = 0 Solving for the unknowns: C = 11.Example A loaded.200 lbs/ft) (8 ft) + Ay + Cy = 0 Sum TA = (Cy)(12 ft) .Solution315 STATICS & STRENGTH OF MATERIALS . A = 4. file:///Z|/www/Statstr/statics/Stests/beams1/sol315. and then use Statics . For this process we will cut the beam into sections.300 lbs y y Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. and Integration to determine the Bending Moment expressions in each section of the beam. Make shear force and bending moment diagrams for the beam. For this beam: A.(1. Draw a Free Body Diagram of the beam. all joints and support points are assumed to be pinned or hinged joints. showing all external loads and support forces (reactions).(800 lbs/ft)(8 ft)(4 ft) .(1. as well as any needed angles and dimensions. B. Sum Fy = (-800 lbs/ft)(8 ft) . STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions.Sum of Forces to determine the Shear Force expressions. Unless otherwise indicated.700 lbs. simply supported beam is shown below. C. where 0 < x < 8 ft.1. (Shown in Diagram) 2. where 8 < x <12 ft. All forces in x & y components (yes) 3.270x] ft-lbs for 0 < x < 8 ft. Integration 2 M1 = -400x + 4. All forces in x & y components (yes) 3. 1. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.Solution315 Section 1: Cut the beam at x.htm (2 of 4) [5/8/2009 4:40:05 PM] .560 ft-lbs (from equation M1) Apply BC: -8560 ft-lbs = -600(8)2 +7470(8) + C2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol315. FBD.800 lbs/ft(x)ft .V1 = 0 Solving: V1 = [4. FBD.200lbs/ft (x-8)ft V2 = 0 Solving: V2 = [7. Integration 2 M2 = -600x + 7. Analyze left hand section.270(0) + C1 Solving: C1 = 0 Therefore M1 = [-400x + 4.270 . 1.270 lbs . (Shown in Diagram) 2.270x + C1 a)Boundary condition to find C1: at x=0 M=0 Apply BC: 0 = -400(0)2 + 4.200x] lbs 4.1. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. Analyze left hand section.800x] lbs 4.forces) Sum Fy = 4.270 lbs .470x + C2 a)Boundary condition to find C2: at x=8 ft M=-8.470 .800 lbs/ft(8 ft) .forces) Sum Fy = 4. Section 2: Cut the beam at x. V1 = -800x+4.forces) Sum Fy = 4. FBD.000 ft-lbs Therefore M3 = [-600x + 19. M3 = -600x +19.V3 = 0 Solving: V3 = [-1.000] ft-lbs for 8 < x < 12 Section 3: Cut the beam at x. All forces in x & y components (yes) 3.200 lb M1 =-400x +4. we can draw the shear force and bending moment diagrams for our loaded beam. 1.000] ft-lbs for 12 < x < 16 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.200(16) + C3 Solving: C3 = -153.200x + 19.000ft-lb 2 2 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol315.470-12.200x . M2 = -600x +7.470x . V2 = -1.200x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (end of beam.000 ft-lbs Therefore M2 = [-600x + 7. no external torque so M3=0) Apply BC: 0 = -600(16)2 + 19.470 lb. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.800 lbs/ft(8 ft) + 11.270 lbs .200] lbs 4.730 lbs .Solution315 Solving: C2 = -128.200x+7.200lbs/ft (x-8)ft .htm (3 of 4) [5/8/2009 4:40:05 PM] . Analyze left hand section.128.1. where 12 < x < 16 ft.200x-153.270x ft-lb. (Shown in Diagram) 2.200x+19.153. V3 = -1.800 ft-lb.270 lb. Integration 2 2 M3 = -600x + 19. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol315.htm (4 of 4) [5/8/2009 4:40:05 PM] .Solution315 Select: Topic 4: Beams . and Integration to determine the Bending Moment expressions in each section of the beam. For this beam: A.(6.000 lbs y y Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.000 lbs. simply supported beam is shown below. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.(1. D = -5. STEP 2: Break any forces not already in x and y direction into their x and y components.000 lbs/ft)(8 ft) . Unless otherwise indicated. B.000 lbs/ft)(8 ft)(4 ft) + (4. as well as any needed angles and dimensions.000 lbs)(4 ft) + (1.000 lbs)(8 ft) = 0 Solving for the unknowns: B =23. and then use Statics .Solution316 STATICS & STRENGTH OF MATERIALS . Determine expressions for the internal shear forces and bending moments in each section of the beam. Sum Fy = -4. C. all joints and support points are assumed to be pinned or hinged joints. STEP 3: Apply the equilibrium conditions.000 lbs . For this process we will cut the beam into sections. showing all external loads and support forces (reactions). file:///Z|/www/Statstr/statics/Stests/beams1/sol316.htm (1 of 4) [5/8/2009 4:40:05 PM] . Draw a Free Body Diagram of the beam.6.Example A loaded. Make shear force and bending moment diagrams for the beam.000 lbs + By + Dy = 0 Sum TB = (Dy)(8 ft) .Sum of Forces to determine the Shear Force expressions. 000 lbs .000(0) + C1 Solving: C1 = 0 Therefore M1 = [-500x . Integration 2 M1 = -500x . Section 2: Cut the beam at x.4.000x] lbs 4.000 lbs .000 lbs 4.1. FBD.000x] ft-lbs for 0 < x < 8 ft. Analyze left hand section. Integration M2 = 11.1.4. All forces in x & y components (yes) 3. Analyze left hand section.000 ft-lbs (from equation M1) file:///Z|/www/Statstr/statics/Stests/beams1/sol316.V1 = 0 Solving: V1 = [-4. where 8 < x <12 ft. 1. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external xforces) Sum Fy = -4.1. (Shown in Diagram) 2.V2 = 0 Solving: V2 = 11.htm (2 of 4) [5/8/2009 4:40:05 PM] 2 . All forces in x & y components (yes) 3.000x + C2 a)Boundary condition to find C2: at x=8 ft M=-64.000 . where 0 < x < 8 ft.000 lbs/ft(8 ft) + 23.Solution316 Section 1: Cut the beam at x.000x + C1 a)Boundary condition to find C1: at x=0 M=0 Apply BC: 0 = -500(0)2 . 1. (Shown in Diagram) 2. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 lbs . FBD.4.forces) Sum Fy = -4.000 lbs/ft(x)ft . 000 lb. All forces in x & y components (yes) 3.000x+4. where 12 < x < 16 ft.000x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (end of beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000] ft-lbs for 8 < x < 12 Section 3: Cut the beam at x.000 ft-lbs Therefore M2 = [11.000 lbs/ft(8 ft) + 23.Solution316 Apply BC: -64. Integration M3 = 5.000x-80.000x -152. M2 = 11.000 lb M1 =-500x +4.000 lbs 4.000 lb. V1 = -1. V2 = 11.000 lbs (8) + C2 Solving: C2 = -152.forces) Sum Fy = -4.80. we can draw the shear force and bending moment diagrams for our loaded beam.000 lbs -6.000 ft-lbs = 11. no external torque so M3=0) Apply BC: 0 = 5.000(16) + C3 Solving: C3 = -80. FBD.000 lbs .000 lbs .htm (3 of 4) [5/8/2009 4:40:05 PM] . V3 = 5.1.000x .000x .000] ft-lbs for 12 < x < 16 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. 1.V3 = 0 Solving: V3 = 5. M3 = 5.000 ft-lb 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol316.000 ft-lbs Therefore M3 = [5. (Shown in Diagram) 2.000 ft-lb.000x ft-lb.152. Analyze left hand section. Solution316 Select: Topic 4: Beams .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol316.htm (4 of 4) [5/8/2009 4:40:05 PM] . showing all external loads and support forces (reactions). Make shear force and bending moment diagrams for the beam. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. B. Unless otherwise indicated. C. Determine expressions for the internal shear forces and bending moments in each sections of the beam. as well as any needed angles and dimensions.(8.Example A loaded.000 lbs/ft)(8 ft) + Ay = 0 Sum TA = . M = 156. Sum Fx = Ax = 0 Sum Fy = -5.(1.Solution321 STATICS & STRENGTH OF MATERIALS .000 lbs)(12 ft) + Mext = 0 Solving for the unknowns: A = 13.000 lbs . cantilever beam is shown below. For this beam: A.(5. Draw a Free Body Diagram of the beam.000 lbs)(12 ft) .htm (1 of 4) [5/8/2009 4:40:06 PM] . STEP 3: Apply the equilibrium conditions.000 ft-lbs y ext file:///Z|/www/Statstr/statics/Stests/beams1/sol321.000 lbs. all joints and support points are assumed to be pinned or hinged joints. STEP 2: Break any forces not already in x and y direction into their x and y components. 8)ft) . and then use Statics . the value of the bending moment at the wall is equal to the negative of the external moment.000x .000 lbs .forces) Sum Fy = 13.000x + 21. for a cantilever beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. All forces in x & y components (yes) 3. (Shown in Diagram) 2. where 8 < x < 12 ft.000 ft-lbs (That is. All forces in x & y components (yes) 3. FBD.htm (2 of 4) [5/8/2009 4:40:06 PM] . For this process we will cut the beam into sections.Solution321 Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.forces) Sum Fy = 13. Section 2: Cut the beam at x.000x + C1 a)Boundary condition to find C1: at x=0 M=-156. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 lbs 4. Analyze left hand section.000 Therefore M1 = [13. Section 1: Cut the beam at x.156. Integration M1 = 13.000] ft-lbs for 0 < x < 8 ft.000 lbs/ft)((x .) Apply BC: -156. where 0 < x < 8 ft. (Shown in Diagram) 2.V2 = 0 Solving: V2 = [-1.Sum of Forces to determine the Shear Force expressions. Analyze left hand section. 1.000] lbs 4. FBD.000 lbs .000 = 13000(0) + C1 Solving: C1 = -156.(1. Integration file:///Z|/www/Statstr/statics/Stests/beams1/sol321. 1. and Integration to determine the Bending Moment expressions in each section of the beam.V1 = 0 Solving: V1 = 13. All forces in x & y components (yes) 3.000(16) + C3 Solving: C3 = -128. V2 = [-1.000x + C2 a)Boundary condition to find C2: at x=8 ft M=-52.000x+21.forces) Sum Fy = 13. 1. (Shown in Diagram) 2.000 ft-lbs Therefore M3 = [-500x + 16.000 lbs .Solution321 M2 = -500x + 21. no external torque so M3=0) Apply BC: 0 = -500(16)2 + 16.000] ft-lbs for 8 < x < 12 Section 3: Cut the beam at x.000] lbs 4.000x .000 ft-lbs (from equation M1) Apply BC: -52.000x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (free end of beam. V3 = [-1.000x .000] ft-lb. M3 = [-500x + 16.000 lbs/ft)((x . M2 = [-500x +21.8)ft) 5. V1 = 13.128.000] ftlb 2 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol321.000x-156.000(8) + C2 Solving: C2 = -188.000 ft-lbs Therefore M2 = [ -500x + 21.000x + 16. FBD.000 lbs . Analyze left hand section.128.000x .000 ft-lbs = -500(8)2 + 21. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000] ft-lb.000x + 16.000x-188.000] lb M1 =[13.188.htm (3 of 4) [5/8/2009 4:40:06 PM] .(1.000] ft-lbs for 12 < x < 16 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. Integration 2 2 2 M3 = -500x + 16.000] lb.V3 = 0 Solving: V3 = [-1.000 lb. where 12 < x < 16 ft. we can draw the shear force and bending moment diagrams for our loaded beam. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol321.Solution321 Select: Topic 4: Beams .htm (4 of 4) [5/8/2009 4:40:06 PM] . B. C. all joints and support points are assumed to be pinned or hinged joints.(1. Sum Fx = Ax = 0 Sum Fy =( -2.Solution322 STATICS & STRENGTH OF MATERIALS . cantilever beam is shown below. Make shear force and bending moment diagrams for the beam. STEP 2: Break any forces not already in x and y direction into their x and y components.000 lbs/ft)(2 ft) + Ay = 0 Sum TA = (-2. showing all external loads and support forces (reactions). Unless otherwise indicated.000 lbs/ft)(6 ft) .000 lbs/ft)(6 ft)(3 ft) .Example A loaded. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. as well as any needed angles and dimensions. STEP 3: Apply the equilibrium conditions.000 lbs/ft)(2 ft)(11 ft) + Mext = 0 Solving for the unknowns: file:///Z|/www/Statstr/statics/Stests/beams1/sol322.(1. For this beam: A. Draw a Free Body Diagram of the beam.htm (1 of 4) [5/8/2009 4:40:06 PM] . Determine expressions for the internal shear forces and bending moments in each sections of the beam. For this process we will cut the beam into sections.000x + 14. the value of the bending moment at the wall is equal to the negative of the external moment. 1.000x .) Apply BC: -58. FBD. All forces in x & y components (yes) 3. and then use Statics . M y ext = 58. where 0 < x < 6 ft. (Shown in Diagram) 2.000 lbs .htm (2 of 4) [5/8/2009 4:40:06 PM] 2 .(2.58.(2.000 Therefore M1 = [-1. where 6 < x < 10 ft.V2 = 0 file:///Z|/www/Statstr/statics/Stests/beams1/sol322.000 ft-lbs (That is.000x + 14.forces) Sum Fy = 14.000(0)2 + 14000(0) + C1 Solving: C1 = -58. Section 2: Cut the beam at x.000 lbs/ft)x ft . All forces in x & y components (yes) 3. FBD. Analyze left hand section.Solution322 A = 14.V1 = 0 Solving: V1 = [-2.000] lbs 4. Analyze left hand section.000 = -1.000 ft-lbs Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.Sum of Forces to determine the Shear Force expressions.000] ft-lbs for 0 < x < 6 ft.000x + 14. Integration 2 M1 = -1. Section 1: Cut the beam at x.000 lbs . for a cantilever beam. 1.forces) Sum Fy = 14.000 lbs/ft)(6 ft) . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. (Shown in Diagram) 2. and Integration to determine the Bending Moment expressions in each section of the beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 lbs.000x + C1 a)Boundary condition to find C1: at x=0 M=-58. All forces in x & y components (yes) 3.000 ft-lbs (from equation M1) Apply BC: -10.(1. Analyze left hand section.000x .000x + C3 a)Boundary condition to find C3: at x=12 ft M=0 ft-lbs (free end of beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 ft-lbs Therefore M3 = [-500x + 12.V3 = 0 Solving: V3 = [-1.Solution322 Solving: V2 = 2.000 lbs/ft)(x-10)ft .000x + C2 a)Boundary condition to find C2: at x=6 ft M=-10. 1. no external torque so M3=0) Apply BC: 0 = -500(12)2 + 12.000x + 12.forces) Sum Fy = 14.000(6) + C2 Solving: C2 = -22.htm (3 of 4) [5/8/2009 4:40:06 PM] .000x .000 lbs 4.000] ft-lbs for 10 < x < 12 2 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000 lbs/ft)(6 ft) . file:///Z|/www/Statstr/statics/Stests/beams1/sol322.22.000(12) + C3 Solving: C3 = -72. Integration M2 = 2.000] ft-lbs for 6 < x < 10 Section 3: Cut the beam at x. we can draw the shear force and bending moment diagrams for our loaded beam. (Shown in Diagram) 2.000] lbs 4.000 lbs .72. where 10 < x < 12 ft. FBD.000 ft-lbs = 2. Integration 2 M3 = -500x + 12.000 ft-lbs Therefore M2 = [ 2.(2. 000x + 12.htm (4 of 4) [5/8/2009 4:40:06 PM] .000x +14.72.000x-22.000] ftlb 2 2 Select: Topic 4: Beams .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol322.000] lb V2 = 2.000 lb V3 = [-1.000x .000] ft-lb M2 = [2.Solution322 V1 = [-2.000x-58.000x+14.000] ft-lb M3 = [-500x + 12.000] lb M1 =[-1. 000 lbs + Ay = 0 Sum TA = (-1.Solution323 STATICS & STRENGTH OF MATERIALS . Unless otherwise indicated.Example A loaded.(1.000 lbs/ft)(4 ft)(8 ft) . showing all external loads and support forces (reactions).500 lbs/ft)(6 ft)(3 ft) . Draw a Free Body Diagram of the beam. STEP 2: Break any forces not already in x and y direction into their x and y components. C.htm (1 of 4) [5/8/2009 4:40:07 PM] . Determine expressions for the internal shear forces and bending moments in each section of the beam.(1.000 ft-lbs y ext Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.000 lbs. For this beam: A.000 lbs/ft)(4 ft) 5. all joints and support points are assumed to be pinned or hinged joints. as well as any needed angles and dimensions.500 lbs/ft)(6 ft) . M = 139.(5. B. file:///Z|/www/Statstr/statics/Stests/beams1/sol323. STEP 3: Apply the equilibrium conditions. Make shear force and bending moment diagrams for the beam. Sum Fy = Ax = 0 Sum Fy =( -1.000 lbs)(16 ft) + Mext = 0 Solving for the unknowns: A = 18. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. cantilever beam is shown below. 000 lbs/ft)(x . and Integration to determine the Bending Moment expressions in each section of the beam. Analyze left hand section.000x + C2 file:///Z|/www/Statstr/statics/Stests/beams1/sol323. the value of the bending moment at the wall is equal to the negative of the external moment. Integration 2 M1 = -750x + 18. Section 1: Cut the beam at x.1.000 Therefore M1 = [-750x + 18. All forces in x & y components (yes) 3.htm (2 of 4) [5/8/2009 4:40:07 PM] .500 lbs/ft)(6 ft) -(1.139. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 ft-lbs (That is. (Shown in Diagram) 2.Solution323 For this process we will cut the beam into sections. All forces in x & y components (yes) 3. for a cantilever beam.V1 = 0 Solving: V1 = [-1.forces) Sum Fy = 18.000] ft-lbs for 0 < x < 6 ft.500 lbs/ft(x) . Integration 2 2 M2 = -500x + 15. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. (Shown in Diagram) 2. Section 2: Cut the beam at x.6) ft .(1. where 0 < x < 6 ft.) Apply BC: -139.Sum of Forces to determine the Shear Force expressions. FBD.forces) Sum Fy = 18.500x + 18. 1.000x . FBD.000] lbs 4. Analyze left hand section.000] lbs 4.V2 = 0 Solving: V2 = [-1. 1.000 = -750(0)2 + 18000(0) + C1 Solving: C1 = -139.000 lbs .000 lbs .000x + 15. where 6 < x < 10 ft.000x + C1 a)Boundary condition to find C1: at x=0 M=-139. and then use Statics . 130.000] ft-lbs for 10 < x < 16 2 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000x . we can draw the shear force and bending moment diagrams for our loaded beam. Integration M3 =5. where 10 < x < 16 ft.000] lb V2 = [-1.000 ft-lbs (from equation M1) Apply BC: -58.V3 = 0 Solving: V3 = 5.Solution323 a)Boundary condition to find C2: at x=6 ft M=-58.000] ft-lbs for 6 < x < 10 Section 3: Cut the beam at x.htm (3 of 4) [5/8/2009 4:40:07 PM] . 1.80.500x+18.000x .000x+15.000] ft-lb M2 = [-500x +15.000] ft-lb M3 = [5. FBD.000(16) + C3 Solving: C3 = -80. Analyze left hand section. (Shown in Diagram) 2.000(6) + C2 Solving: C2 = -130.000] lb V3 = 5.500 lbs/ft)(6 ft) .000x + C3 a)Boundary condition to find C3: at x=16 ft M=0 ft-lbs (free end of beam.000 lbs .000 ft-lbs Therefore M3 = [5.80.000 lbs 4.000 lb M1 =[-750x +18.forces) Sum Fy = 18. All forces in x & y components (yes) 3. no external torque so M3=0) Apply BC: 0 = 5.(1.000x-130.(1.000 ft-lbs Therefore M2 = [ -500x + 15.000] ft-lb 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol323. V1 = [-1.000x . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 lbs/ft)(4 ft) .000x-139.000 ft-lbs = -500(6)2 + 15. htm (4 of 4) [5/8/2009 4:40:07 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol323.Solution323 Select: Topic 4: Beams . Solution324 STATICS & STRENGTH OF MATERIALS .500 lbs/ft)(6 ft)(3 ft) .(5. cantilever beam is shown below.(1.(1.htm (1 of 4) [5/8/2009 4:40:07 PM] . Make shear force and bending moment diagrams for the beam. Draw a Free Body Diagram of the beam. Sum Fx = + Ax = 0 Sum Fy =( -1. STEP 3: Apply the equilibrium conditions.000 lbs/ft)(4 ft) 5. STEP 2: Break any forces not already in x and y direction into their x and y components. For this beam: A.000 lbs)(8 ft) + Mext = 0 Solving for the unknowns: file:///Z|/www/Statstr/statics/Stests/beams1/sol324. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. showing all external loads and support forces (reactions).Example A loaded. all joints and support points are assumed to be pinned or hinged joints. as well as any needed angles and dimensions. Unless otherwise indicated.500 lbs/ft)(6 ft) . B. Determine expressions for the internal shear forces and bending moments in each section of the beam.000 lbs + Ay = 0 Sum TA = (-1. C.000 lbs/ft)(4 ft)(10 ft) . 000 ft-lbs Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. (Shown in Diagram) 2.000 Therefore M1 = [-750x + 184. FBD. M y ext = 107. and then use Statics . 1.forces) Sum Fy = 18. For this process we will cut the beam into sections.000 lbs . Integration 2 M1 = -750x + 18. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.htm (2 of 4) [5/8/2009 4:40:07 PM] 2 .Sum of Forces to determine the Shear Force expressions.V2 = 0 file:///Z|/www/Statstr/statics/Stests/beams1/sol324. (Shown in Diagram) 2. Section 2: Cut the beam at x.000] ft-lbs for 0 < x < 6 ft.(1.) Apply BC: -107.000 = -750(0)2 + 18000(0) + C1 Solving: C1 = -107.Solution324 A = 18.000x . for a cantilever beam. Section 1: Cut the beam at x. Analyze left hand section.000 lbs.500 lbs/ft(x) . All forces in x & y components (yes) 3.000 lbs . the value of the bending moment at the wall is equal to the negative of the external moment.107.000x + C1 a)Boundary condition to find C1: at x=0 M=-107.000] lbs 4.500 lbs/ft)(6 ft) .000 ft-lbs (That is. 1. and Integration to determine the Bending Moment expressions in each section of the beam.forces) Sum Fy = 18. All forces in x & y components (yes) 3. Analyze left hand section. where 0 < x < 6 ft.1.500x + 18. FBD.V1 = 0 Solving: V1 = [-1. where 6 < x < 8 ft. where 8 < x < 12 ft.htm (3 of 4) [5/8/2009 4:40:07 PM] 2 2 2 2 .000 ft-lbs (from equation M1) Apply BC: -26.Solution324 Solving: V2 = 9.000x-80.000x .000 lbs 4.000 ft-lbs Therefore M3 = [500x + 12.000(6) + C2 Solving: C2 = -80.000x+12.V3 = 0 Solving: V3 = [-1.000x + C3 a)Boundary condition to find C3: at x=12 ft M=0 ft-lbs (free end of beam. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.72.000 ft-lbs Therefore M2 = [9.000 ft-lbs = 9.000 lbs .000x + 12. we can draw the shear force and bending moment diagrams for our loaded beam. FBD. Integration M2 = 9.000] ft-lbs for 8 < x < 12 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.80.forces) Sum Fy = 18. (Shown in Diagram) 2.000 lbs/ft)(x .500 lbs/ft)(6 ft) 5.000(12) + C3 Solving: C3 = -72.000x-72.(1.500x+18.000 lb V3 = [-1.000] ft-lbs for 6 < x < 8 Section 3: Cut the beam at x.000] ft-lb M2 = [9.000 lbs . 1.000x + C2 a)Boundary condition to find C2: at x=6 ft M=-26.000x .000x-107.8)ft . Integration M3 =-500x + 12.000] ft-lb file:///Z|/www/Statstr/statics/Stests/beams1/sol324. no external torque so M3=0) Apply BC: 0 = -500(12)2 + 12. V1 = [-1. Analyze left hand section.000] ft-lb M3 = [-500x +12.000] lb M1 =[-750x +18.000] lb V2 = 9. All forces in x & y components (yes) 3.000] lbs 4.(1. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol324.htm (4 of 4) [5/8/2009 4:40:07 PM] .Solution324 Select: Topic 4: Beams . STEP 2: Break any forces not already in x and y direction into their x and y components. For this beam: A.(1.Solution325 STATICS & STRENGTH OF MATERIALS . as well as any needed angles and dimensions. Sum Fx = Ax = 0 Sum Fy = ( -1.500 lbs/ft)(6 ft)(3 ft)-(1. C.500 lbs/ft)(6 ft) .Example A loaded. B. showing all external loads and support forces (reactions).000 lbs/ft)(2 ft) (800 lbs/ft)(4 ft) + Ay = 0 Sum TA = (-1. cantilever beam is shown below.000 lbs/ft)(2 ft)(7 ft)-(800 lbs/ft)(4 ft)(10 ft) + Mext = 0 Solving for the unknowns: file:///Z|/www/Statstr/statics/Stests/beams1/sol325.htm (1 of 4) [5/8/2009 4:40:08 PM] . all joints and support points are assumed to be pinned or hinged joints. Unless otherwise indicated. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Make shear force and bending moment diagrams for the beam. Draw a Free Body Diagram of the beam. STEP 3: Apply the equilibrium conditions. Determine expressions for the internal shear forces and bending moments in each section of the beam. the value of the bending moment at the wall is equal to the negative of the external moment.200x .200] lbs 4. (Shown in Diagram) 2.forces) Sum Fy = 14.(1.200x + C1 a)Boundary condition to find C1: at x=0 M=-73.) Apply BC: -73. (Shown in Diagram) 2. Analyze left hand section.1. Integration 2 M1 = -750x + 14. where 6 < x < 8 ft.Solution325 A = 14. for a cantilever beam. All forces in x & y components (yes) 3.htm (2 of 4) [5/8/2009 4:40:08 PM] 2 .200 lbs . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.500 lbs/ft(x) .000 ft-lbs (That is.500x + 14. and then use Statics .000 lbs/ft) (x .V2 = 0 file:///Z|/www/Statstr/statics/Stests/beams1/sol325. where 0 < x < 6 ft.(1.200 lbs. 1. FBD.6)ft . Analyze left hand section. M y ext = 73. Section 1: Cut the beam at x.V1 = 0 Solving: V1 = [-1. FBD. and Integration to determine the Bending Moment expressions in each section of the beam. For this process we will cut the beam into sections.73. All forces in x & y components (yes) 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.500 lbs/ft)(6 ft) .000 = -750(0)2 + 14200(0) + C1 Solving: C1 = -73. 1.200 lbs . Section 2: Cut the beam at x.forces) Sum Fy = 14.000] ft-lbs for 0 < x < 6 ft.Sum of Forces to determine the Shear Force expressions.000 ft-lbs Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.000 Therefore M1 = [-750x + 14. 600] ft-lbs for 8 < x < 12 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.200] lb V2 = [-1. no external torque so M3=0) Apply BC: 0 = -400(12)2 + 9.(1.64.500 lbs/ft)(6 ft) .600(12) + C3 Solving: C3 = -57.200x .000x+11.000] ft-lbs for 6 < x < 8 Section 3: Cut the beam at x.200] lbs 4.600] lbs 4.500x+14. Integration 2 M2 = -500x + 11. we can draw the shear force and bending moment diagrams for our loaded beam.600x + C3 a)Boundary condition to find C3: at x=12 ft M=0 ft-lbs (free end of beam.200(6) + C2 Solving: C2 = -64.000] ft-lb M2 = [-500x +11. Integration 2 2 M3 =-400x + 9.600x .000] ft-lb M3 = [-400x +9.600] lb M1 =[-750x +14.200x-64.000 ft-lbs Therefore M2 = [ -500x + 11. (Shown in Diagram) 2.000x + 11. V1 = [-1.htm (3 of 4) [5/8/2009 4:40:08 PM] 2 2 2 2 .Solution325 Solving: V2 = [-1.600x file:///Z|/www/Statstr/statics/Stests/beams1/sol325.V3 = 0 Solving: V3 = [-800x + 9.800 ft-lbs (from equation M1) Apply BC: -14.200] lb V3 = [-800x+9. 1.(800 lbs/ft)(x-8)ft .800 ft-lbs = -500(6)2 + 11.200x-73.200x + C2 a)Boundary condition to find C2: at x=6 ft M=-14.600 ft-lbs Therefore M3 = [-400x +9.200 lbs .57. Analyze left hand section. where 8 < x < 12 ft. All forces in x & y components (yes) 3.forces) Sum Fy = 14. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.(1.000 lbs/ft)(2 ft) . FBD. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol325.Solution325 57.htm (4 of 4) [5/8/2009 4:40:08 PM] .600] ft-lb Select: Topic 4: Beams . Unless otherwise indicated. showing all external loads and support forces (reactions).000 lbs . Make shear force and bending moment diagrams for the beam. cantilever beam is shown below.000 lbs + Ay = 0 Sum TA = (-4.(2. C.000 lbs/ft)(6 ft) .000 lbs . Sum Fx = Ax = 0 Sum Fy = -4. M = 200. STEP 2: Break any forces not already in x and y direction into their x and y components.2. STEP 3: Apply the equilibrium conditions. B. Draw a Free Body Diagram of the beam.3.000 lbs/ft)(6 ft)(11 ft) .000 ft-lbs y ext file:///Z|/www/Statstr/statics/Stests/beams1/sol326.(2. as well as any needed angles and dimensions. Solution: Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.000 lbs)(14 ft) + Mext = 0 Solving for the unknowns: A = 21.Example A loaded. For this beam: A.000 lbs)(8 ft) -(2.htm (1 of 4) [5/8/2009 4:40:08 PM] .(3.000 lbs)(4 ft) . Determine expressions for the internal shear forces and bending moments in each section of the beam.000 lbs.Solution326 STATICS & STRENGTH OF MATERIALS . all joints and support points are assumed to be pinned or hinged joints. Section 2: Cut the beam at x.Sum of Forces to determine the Shear Force expressions. Integration file:///Z|/www/Statstr/statics/Stests/beams1/sol326. FBD.000 = 21000(0) + C1 Solving: C1 = -200. (Shown in Diagram) 2. (Shown in Diagram) 2. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000 ft-lbs (That is. For this process we will cut the beam into sections. Analyze left hand section. Section 1: Cut the beam at x.000] ft-lbs for 0 < x < 4 ft.000 lbs 4.000x + C1 a)Boundary condition to find C1: at x=0 M=-200. All forces in x & y components (yes) 3. Integration M1 = 21.000 lbs . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.forces) Sum Fy = 21.000 lbs . Analyze left hand section.000 lbs . where 0 < x < 4 ft. FBD. where 4 < x < 8 ft.htm (2 of 4) [5/8/2009 4:40:08 PM] .Solution326 Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam.) Apply BC: -200.forces) Sum Fy = 21. and Integration to determine the Bending Moment expressions in each section of the beam.V2 = 0 Solving: V2 = 17.000 lbs 4. 1.4. 1. and then use Statics .000x . the value of the bending moment at the wall is equal to the negative of the external moment.000 Therefore M1 = [21.200. All forces in x & y components (yes) 3. for a cantilever beam.V1 = 0 Solving: V1 = 21. 000x-184. (Shown in Diagram) 2. V1 = 21.000(14) + C3 Solving: C3 = -224.000x + 30.000x . FBD.224.184.Solution326 M2 = 17.forces) Sum Fy = 21. All forces in x & y components (yes) 3.000x + C3 a)Boundary condition to find C3: at x=14 ft M=0 ft-lbs (free end of beam.htm (3 of 4) [5/8/2009 4:40:08 PM] .000x + C2 a)Boundary condition to find C2: at x=4 ft M=-116.000x +30.000x + 30.000x-200.000 lb V2 = 17.000] ft-lbs for 8 < x < 14 Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above.000x .000 lbs . where 8 < x < 14 ft. Integration 2 M3 =-1.000] ft-lb M2 = [17.3.000] lbs 4. no external torque so M3=0) Apply BC: 0 = -1.000(4) + C2 Solving: C2 = -184.000 ft-lbs (from equation M1) Apply BC: -116.000] ft-lbs for 4 < x < 8 Section 3: Cut the beam at x.000x +30. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.000(14)2 + 30. 1.000x .4.000x+30.000] ft-lb M3 = [-1.V3 = 0 Solving: V3 = [-2.000 lbs .000] ft-lb 2 2 file:///Z|/www/Statstr/statics/Stests/beams1/sol326.000 ft-lbs Therefore M3 = [-1. we can draw the shear force and bending moment diagrams for our loaded beam.000 lbs/ft)(x-8)ft .224.000 lb V3 = [-2.000 lbs (2.000] lb M1 =[21. Analyze left hand section.000 ft-lbs Therefore M2 = [ 17.000 ft-lbs = 17. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams1/sol326.htm (4 of 4) [5/8/2009 4:40:08 PM] .Solution326 Select: Topic 4: Beams . V2=-200 lb.500 lb. V1=300 lb. V3 = -1200 lb. V2 = 2000 lb. M2 = 500 x -7800 ft-lb. V4 = 800 lb. A 24 foot beam is simply supported at the quarter points (6 feet and 18 feet). M1 = 800x -100 x2 ft-lb. M3 = .200x2 ft-lb. V2 = 0 lb.800x ft-lb. M4 = 800x 19. A load of 500 pounds is placed 4 feet from the left end.200 ft-lb 4. M2 = 7200 ft-lb. V2 = 160 lb.4a1 . M1 = -250x2 ft-lb. Draw the Shear Force and Bending Moment Diagrams for each beam. M1= .200 x lb. A 16 foot beam is simply supported at its ends. A diving board is 16 feet long and pinned at the left end. M2 = 2000 x . V1 = 800 . B. M2 = 160 x -2560 ft-lb 6. C.Shear/Bending Diag Prob 1 Statics & Strength of Materials Topic 4.Shear Forces & Bending Moments 1 For the loaded. Draw a Free Body Diagram of the beam. A 160 pound person stands at the right end of the board. M1 = -267 x ft-lb.4a1 Problem Assignment .htm [5/8/2009 4:40:08 PM] . 1. 3. M3 = 1200 x + 19200 ft-lb 5. M2 = -200x + 2000 ft-lb 2. It has a uniformly distributed load of 400 pounds per foot over the first 6 feet and a point load of 1200 pounds at the 10 foot mark.500x +4200 ft-lb.000 ft-lb.Topic 4. M1 = 2400 x . Beams in these problems are considered weightless.800 lb. It has a uniformly distributed load of 200 pounds per foot along its entire length.32. V1 = . V2 = 500 lb. Find expressions for the Shear Forces and Bending Moments in each section of the beams. Use a sheet of graph paper. simply supported beams described below: A.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beamp44a1. It has a roller support at the 6 foot mark. It has a distributed load of 500 pounds per foot between the supports and a point load of 2000 pounds at the right end. V1 = -267 lb. M1= 300x ft-lb. It has point loads of 800 pounds at each end and a point load of 1000 points at its mid point. V1 = 2400 . and determine the external support reactions. A 10 foot beam is simply supported at its ends. V3 = . A 16 foot beam is simply supported at its left end and its midpoint. V1= . use topic half for shear diagram and the bottom half for the bending moment diagram.400x lb. Select: Topic 4: Beams . An 8 foot beam is simply supported at its ends.500 x lb. Ay = 4. Find expressions for the Shear Forces and Bending Moments in each section of the beams. simply supported beams shown in the diagrams below: A. file:///Z|/www/Statstr/statics/Beams/beamp44a2.400 lb. 1..000 x V2 = 0 lb.000x + 96.-lb.8.600 lb. 2. Dy = 8. and determine the external support reactions.000 lb. use topic half for shear diagram and the bottom half for the bending moment diagram. V3 = 10.99. M3 = 10.600 lb.400x + 0 V2 = -5. Prob 2 Statics & Strength of Materials Topic 4. C.4a2 Problem Assignment . M2 = 24.400 lb.000 lb..000 ft-lb V3 = -8..-lb.000 ft.000 ft. M1 = -1. Cy = 15. Draw a Free Body Diagram of the beam.htm (1 of 3) [5/8/2009 4:40:09 PM] .Shear/Bending Diag.000x + 4.-lb. M2 = . M1 = 8.800 ft. Use a sheet of graph paper... Ay = 8. V1 = 8.000 x2 + 4.000 lb. V1 = -2.5.Shear Forces & Bending Moments 2 For the loaded.000 lb. M3 = ..4a2.Topic 4.000 lb. Draw the Shear Force and Bending Moment Diagrams for each beam.000x .600x + 25. B. Shear/Bending Diag.000 ft.000 ft.500 x2 ft.4a2...5.lb.000x lb.500 x2 + 24.8.000 lb.-lb.000x + 160.M1 = -1. V1 = -3.000x .000lb.000x lb.-lb.000 lb.000 ft.000 lb.000 lb. Ey = 8.000x .-lb. V2 = -2.M3 = -8.000x2 + 10. Ay = 2.96.M4 = . V3 = 0.-lb. Cy = 6. file:///Z|/www/Statstr/statics/Beams/beamp44a2.000x ft.000x + 24.-lb. M3 = 24. Prob 2 3.. By = 24.-lb. 4.M1 = -1.000 lb. Dy = 8.000x . V1 = 2.72.000 x2 + 24..000 x2 ft.Topic 4. V2 = -3. M2 = -1.htm (2 of 3) [5/8/2009 4:40:09 PM] . M1 = 2.000 lb. By = 24. V2 = -4.000 ft.000x + 24..000x + 10..000 lb. M2 = -1. V1 = -2..000 lb.000 lb.000 ft.8.000 lb. V3 = 8.000 lb. M2 = -2.-lb.8. V4 = .-lb.000 ft.000x + 120.. Prob 2 Select: Topic 4: Beams .4a2.htm (3 of 3) [5/8/2009 4:40:09 PM] .Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beamp44a2.Shear/Bending Diag.Topic 4. 25.. C.Topic 4. An 8 foot cantilever was designed to carry a point load of 800 pounds at its free end.4b1 Problem Assignment . M2 = 600x . 1. A 12 foot cantilever carries a load of 1200 pounds 4 feet from the wall and a load of 600 pounds at its end. [V= 800 lb. and determine the external support reactions.35. [V1 = 1800 lb.. [V = 4000 . A 16 foot cantilever has a uniformly distributed load of 400 pounds per foot from the wall to the 10 foot mark and a load of 200 pounds per foot from that point to the end of the beam.] 5.Shear/Bending Diag Prob 1 Statics & Strength of Materials Topic 4. V2=800 lb.400x lb.20. M = 800x -8000 ft-lb.. M1 = 400x -4800 ft-lb.] Select: Topic 4: Beams . M = -2000x2 +4000x . Find expressions for the Shear Forces and Bending Moments in each section of the beams.] 3... M2 = 800x . 12 feet from the wall.htm (1 of 2) [5/8/2009 4:40:09 PM] . M1 = 1800x .7200 ft-lb.1800 ft-lb. B.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beamp44b1.3150 ft-lb.] 4. M1 = -200x2 +5200x . The beam joint of the beam with the wall was found to be failing and a prop was placed 4 feet from the wall that exerted a 400 pound upward point force on the beam. A 6 foot cantilever has a point load of 600 pounds at its mid point and a uniformly distributed load of 100 pounds per foot from the midpoint to the free end.. V2=600100x lb.000 ft-lb.. V2 = 3200-200x lb. V2 = 600 lb. Draw the Shear Force and Bending Moment Diagrams for each beam. Beams in these problems are considered weightless. A 10 foot cantilever carries a point load of 800 pounds at its free end... M2 = -100x2+3200x .6400 ft-lb. M1 = 900x .Shear Forces & Bending Moments 1 For the loaded. M2 = -50 x2 + 600x ..4b1 . [V1 = 5200 . [V1 = 400 lb.000 ft-lb..] 6.12... simply supported beams shown in the diagrams below: A.] 2.600 ft-lb. [V1 = 900 lb. use topic half for shear diagram and the bottom half for the bending moment diagram. Draw a Free Body Diagram of the beam. Use a sheet of graph paper..600 ft-lb.400x lb. A 10 foot cantilever has a uniformly distributed load of 400 pounds per foot over its entire length. htm (2 of 2) [5/8/2009 4:40:09 PM] .Shear/Bending Diag Prob 1 file:///Z|/www/Statstr/statics/Beams/beamp44b1.4b1 .Topic 4. -lb.000-3000x lb.1.000 lb. V2 = 28.-lb.-lb.000 ft.M1 = 20.4b .000x2 . Mext= 168.000 ft-lb.Shear Forces & Bending Moments 2 For the loaded.M3 = 36.000 .htm (1 of 3) [5/8/2009 4:40:09 PM] .00 lb. B.000 ft..000x -168.000 lb. 2 Ay = 18. Draw the Shear Force and Bending Moment Diagrams for each beam.184. and determine the external support reactions. V3 = 36.Problem Assignment 2 Statics & Strength of Materials Topic 4.000 ft.. Ay = 20. file:///Z|/www/Statstr/statics/Beams/beamp44b2. M2 = 28..Topic 4. Find expressions for the Shear Forces and Bending Moments in each section of the beams.-lb.000x lb. 1.000x . Use a sheet of graph paper. cantelever beams shown in the diagrams below: A.V1 = 20. Mext = 168.000 ft. use topic half for shear diagram and the bottom half for the bending moment diagram.2. Draw a Free Body Diagram of the beam. C..4b2 Problem Assignment .000 x -1500x2-216. 4b . V3 = 10.M3 = 10.000x .000 ft.000 ft/lb.-lb.000 lb.M2 = 4.M1 = .000 lb.M3 = 2.M2 = 12.000 ft..000x ..143.-lb.000 . V3 = -1.. M1 = 18..168.000 lb.000 ft.000x + 9.-lb.500 ft.000x ..M1 = -500x2 + 9.000x .48.500 + 24.000 lb.192.-lb.htm (2 of 3) [5/8/2009 4:40:09 PM] ..-lb. V2 = 12. Mext = 60.-lb. 5. V1 = . M4 = 0 3.500 ft/lb.184. V2 = 4.000 .000 ft/lb.000x + 17. Ay = 17.500 ft/lb.156.000 lb.000x .000x . ft. V2 = 26.M3 = -750x2 + 24.000 ft.M2 = 26. Mext = 156. V4 = 0.000x .000 x2 .500x2 + 17.-lb.-lb.60.1.32. 4.1.Problem Assignment 2 V1 = 18.000 lb.000x lb. V3 = 2. file:///Z|/www/Statstr/statics/Beams/beamp44b2. V1 = 1.000 lb...000 lb. Ay = 9.2.000 ft.000x -120.Topic 4.000 lb.500 ft.000 lb... 600 .Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beamp44b2.000] ft..600x . Select: Topic 4: Beams .-lb.42.36.100(x-6)2] or V3 = [-100x2 + 1..000] ft.htm (3 of 3) [5/8/2009 4:40:09 PM] . V3 = [3. V2 = 3. M3 = [-33.800] ft.600 lb.4b .600x .600. M2 = [3.600 lb.Topic 4. V1 = 5.3x3 + 600x2 .Problem Assignment 2 Ay = 5.200x]lb. Mext = 42.-lb.M1 = [5.28.-lb.000 ft.-lb. showing all external loads and support forces. Draw the Shear Force and Bending Moment Diagrams for the beam. For this beam: A. cantilever beam is shown below.) A loaded. simply supported beam is shown. Draw a Free Body Diagram of the beam. showing all external loads and support forces and moments.Topic 4. C.5: Shear Force & Bending Moment Topic Exam Topic 4.5a) 2. Determine expressions for the internal Shear Forces and Bending Moments in each section of the file:///Z|/www/Statstr/statics/Beams/beamx45. B. Draw a Free Body Diagram of the beam. (For Solution Select: Solution Beams 4. Determine expressions for the internal Shear Forces and Bending Moments in each section of the beam.Topic Exam (312 & 326) 1. For this beam: A.) A loaded. B.5: Shear Forces & Bending Moments .htm (1 of 2) [5/8/2009 4:40:09 PM] . (For Solution Select: Solution Beams 4. Make Shear Force and Bending Moment Diagrams for the beam.htm (2 of 2) [5/8/2009 4:40:09 PM] . C.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beamx45.5b) Select: Topic 4: Beams .5: Shear Force & Bending Moment Topic Exam beam.Topic 4. 5: Topic Exam Solution 1 A loaded.000 lbs + By + Dy = 0 Sum TB = (Dy)(8 ft) . C.000 lbs/ft)(4 ft)(6 ft) + (5.Sum of Forces to file:///Z|/www/Statstr/statics/Beams/xsol45a. Determine expressions for the internal shear forces and bending moments in each section of the beam. For this process we will cut the beam into sections. Draw the shear force and bending moment diagrams for the beam. Sum Fy = (-2.000 lbs)(4 ft) = 0 Solving for the unknowns: By = 9. For this beam: A. B.5. SOLUTION: Part A STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.Shear Force Exam Sol.500 lbs.(2. STEP 2: Resolve any forces not already in x and y direction into their x and y components.500 lbs Part B: Determine the Shear Force and Bending Moment expressions for each section of the loaded beam.000 lbs/ft)(4 ft) . Dy = 3. and then use Statics . as well as any needed angles and dimensions.htm (1 of 5) [5/8/2009 4:40:10 PM] . Draw a Free Body Diagram of the beam. showing all external loads and support forces. STEP 3: Apply the equilibrium conditions.1 Topic 4.5 . simply supported beam is shown.Topic 4. This enables us to find the value of C1. This results since the shear force and bending moment (torque) are internal. due to beginning of the 2000 lb/ft distributed load. Apply Eq. we must find that M1 =0 at that point.Shear Force Exam Sol. . therefore C1 = 0. where the beam is cut. In this case because it is a simply supported beam. where the beam is cut. for our second section because the loading changes at that point. we obtain: M1 = -5000 x + C1.Topic 4.) We now continue with our usual Static Equilibrium procedure. for the first section because the loading changes at that point. FBD. Notice in the Free Body Diagram. knowing the value of M1 somewhere on our interval. Now apply the relationship between the shear force and bending moment.) We now continue with our usual static equilibrium procedure. except when we cut the beam. All Forces in x & y components (yes) 3.5 . due to the 9500 lb. Notice in the Free Body Diagram. Conditions (forces only): Sum Fx = 0 (no net external x. where 0< x < 4 ft. . we again show the Shear Force (V2) and Bending Moment (M2) (torque) acting on the end of the section. We now apply the boundary condition: 0 = -5000 (0) + C1. (Shown in Diagram) 2.V1 = 0 (Solving: V1 = -5000 lb) 4. (We must stop just before 4 ft. (We must stop just before 8 ft. we show the Shear Force (V1) and Bending Moment (M1) (torque) acting on the end. support reaction.forces) Sum Fy = -5000 lb . to Integrating. we know that M1 = 0 at x = 0. then they become external to our section and act on it at the end. and our final expression for the bending moment on Section 1 is M1 = -5000 x Section 2: Cut the beam at x.1 determine the Shear Force expressions. (This is true for all beams which do not have an external moment applied . that is find the expression for M in section 1.the bending moment at the ends of the beams must be zero) Thus if we put x =0 in our expression. where 4< x < 8 ft. file:///Z|/www/Statstr/statics/Beams/xsol45a. 1.htm (2 of 5) [5/8/2009 4:40:10 PM] . and Integration to determine the Bending Moment expressions in each section of the beam. We find the value of this constant by using a boundary condition . Section 1: Cut the beam at x. Thus. where C1 is a constant of integration for section 1.that is. Apply Eq. (Shown in Diagram) file:///Z|/www/Statstr/statics/Beams/xsol45a. . (Shown in Diagram) 2. where C2 is a constant of integration for section 2. We now apply the boundary condition: -20000 = 4500 (4) + C2. M1 = -20. Thus we use our expression for section 1 (M1=-5000x) to find that at x = 4 ft. it must start with that value at the beginning of our second section . Now apply the relationship between the shear force and bending moment. whatever value the bending moment has at the end of our first section. due to 3500 lb. .Shear Force Exam Sol. Thus. we can find our boundary condition by use the fact that the bending moment must be continuous.1 1.knowing the value of M2 somewhere on our interval.Topic 4. This enables us to find the value of C2.) We now continue with our usual Static Equilibrium procedure. 1. support reaction. (We must stop just before 12 ft. that is find the expression for M in section 2. We find the value of this constant by using a boundary condition . FBD.htm (3 of 5) [5/8/2009 4:40:10 PM] .000 ft-lb. FBD.000 ft-lb. For section 2. for our third section because the loading changes at that point.5 . the statics are a little more interesting on section 3. and our final expression for the bending moment on Section 2 is M2 = 4500 x -38. where 8< x < 12 ft. where the beam is cut. All Forces in x & y components (yes) 3.forces) Sum Fy = -5000 lb + 9500 lb .V2 = 0 (Solving: V2 = 4500 lb) 4. therefore C2 = -38. we must find that M1 = -20000 ft-lb at that point. Now if we put x =4 ft. to Integrating we obtain: M2 = 4500 x + C2. Conditions (forces only): Sum Fx = 0 (no net external x. Section 3: Cut the beam at x. in our expression for section 2. we again show the Shear Force (V3) and Bending Moment (M3) (torque) acting on the end of the section. In the Free Body Diagram. (M2 = 4500x + C2).000 ft-lb. Due to the uniformly distributed load. That is. 000 ft-lb Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. in our expression for section 3.000 ft-lb file:///Z|/www/Statstr/statics/Beams/xsol45a. Conditions (forces only): Sum Fx = 0 (no net external x. That is. We find the value of this constant by using a boundary condition . For section 3. Thus. (: M3 = 1000 x2 + 20500x + C3).htm (4 of 5) [5/8/2009 4:40:10 PM] . the bending moment at the beginning of the third section must have that same value.Shear Force Exam Sol. and our final expression for the bending moment on Section 3 is : M3 = 1000 x2 + 20500x .500) 4. we obtain: M3 = 1000 x2 + 20500x + C3. to Integrating. Now if we put x =8 ft.forces) Sum Fy = -5000 lb + 9500 lb . that is find the expression for M in section 2. All Forces in x & y components (yes) 3. M2 = -2000 ft-lb. V2 = 4500 lb V3 = -2000 x +20. we must find that M2 = -2000 ft-lb at that point.500 lb M1 = -5000 x ft-lb M2 = 4500 x -38. where C3 is a constant of integration for section 3. .000) to find that at x = 8 ft. Apply Eq.Topic 4. whatever value the bending moment has at the end of our second section. This enables us to find the value of C3.knowing the value of M3 somewhere on our interval. V1 = -5000 lb.V3 = 0 (Solving: V3 = -2000 x +20. we can draw the shear force and bending moment diagrams for our loaded beam. we can find our boundary condition by use the fact that the bending moment must be continuous. Now apply the relationship between the shear force and bending moment. So we use our bending moment expression for section 2 (M2 = 4500 x -38. therefore C3 = -102. We now apply the boundary condition: -2000 = 1000 (8)2 +20500 (8) + C3.000 ft-lb M3 = -1000 x2 + 20500x . .102.8 ft) .102.1 2.2000 lb/ft (x .5 .000 ftlb. Topic 4.htm (5 of 5) [5/8/2009 4:40:10 PM] .Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/xsol45a.Topic Examination or Select: Topic 4: Beams .Shear Force Exam Sol.5 .1 Return to: Shear Force & Bending Moments . C. file:///Z|/www/Statstr/statics/Beams/xsol45b. support forces and moments.5 . STEP 3: Apply the equilibrium conditions. STEP 2: Resolve any forces not already in x and y direction into their x and y components. cantilever beam is shown.htm (1 of 5) [5/8/2009 4:40:10 PM] . Make shear force and bending moment diagrams for the beam. For this beam: A. as well as any needed angles and dimensions. Determine expressions for the internal shear forces and bending moments in each section of the beam. B. SOLUTION: Part A STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.Topic 4. showing all external loads. Draw a Free Body Diagram of the beam.5: Topic Exam Solution 2 A loaded.Shear Force Topic Exam Sol 2 Topic 4. 000 lbs + Ay = 0 Sum TA = (-4. For this process we will cut the beam into sections.000 = 21000(0) + C1. Mext = 200.000 lbs .) Applying the boundary condition: -200. 1. for a cantilever beam.200.000 lbs/ft)(6 ft) . 000 ft-lb.(2.000 lbs . where 0 < x < 4 ft. file:///Z|/www/Statstr/statics/Beams/xsol45b.Sum of Forces to determine the Shear Force expressions.(3. and then use Statics .Topic 4. Analyze left hand section.000 lbs 4. and Integration to determine the Bending Moment expressions in each section of the beam.000 lbs)(8 ft) -(2.V1 = 0. FBD.000 ft-lbs (That is. Resolve All forces into x & y components.000 lbs . and solving for C1 = .000 lbs.000 lbs)(14 ft) + Mext = 0 Solving for the unknowns: Ay = 21. the value of the bending moment at the wall is equal to the negative of the external moment.000] ft-lbs for 0 < x < 4 ft. Therefore M1 = [21.000 lbs)(4 ft) .200.5 . 3.000x + C1 Where C1 is an integration constant which we determine from a boundary condition.000 ft-lbs Part B: Determine the Shear Force and Bending Moment expressions for each section of the loaded beam.(2.000 lbs/ft)(6 ft)(11 ft) . Our boundary condition on section 1 is that at x=0 M=-200.000x . Section 1: Cut the beam at x.3. Solving: V1 = 21.htm (2 of 5) [5/8/2009 4:40:10 PM] . Integration . (Shown in Diagram) 2. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x. and solving M1 = 21.forces) Sum Fy = 21.Shear Force Topic Exam Sol 2 Sum Fx = Ax = 0 Sum Fy = -4.2. Analyze left hand section.000(4) + C2.000 lbs . and solving: C2 = 184. FBD.000] ft-lb. we again apply a boundary condition.Shear Force Topic Exam Sol 2 Section 2: Cut the beam at x. thus our boundary condition is: at x=4 ft M=-116. 3.htm (3 of 5) [5/8/2009 4:40:10 PM] .Topic 4.000x . Analyze left hand section.000 ft-lbs Therefore M2 = [ 17. In this case our boundary condition can be found by remembering that the bending moment must be continuous.5 . where 8 < x < 14 ft.000 lb 4. (Shown in Diagram) 2. Integration . That is. the value of the bending at the end of section1 must equal the value of the bending moment at the beginning of section 2. Resolve all forces into x & y components.V2 = 0 Solving: V2 = 17.184. for 4 < x < 8 ft Section 3: Cut the beam at x.000x + C2 To determine our integration constant C2.000 ft-lb (from equation M1) Applying the boundary condition using M2: -116.000 lbs . Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.forces) Sum Fy = 21. and solving M2 = 17. file:///Z|/www/Statstr/statics/Beams/xsol45b.4.000 ft-lbs = 17. where 4 < x < 8 ft. 1. Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above. For this section of the beam a boundary condition is: at x = 14 ft M = 0 ft-lb ( since the right end of the beam is a free end with no external torque. V1 = 21.000 lbs .000x2 +30. the bending moment must go to zero there) Applying the boundary condition: 0 = -1.000 lb V2 = 17.000 lbs/ft)(x-8)ft .000x-184. Integration: .000x + 30.5 . and solving: V3 = [-2.(2.000x .Shear Force Topic Exam Sol 2 1.000 lbs .224.Topic 4. and solving M3 =-1. for 8 < x < 14 ft.htm (4 of 5) [5/8/2009 4:40:10 PM] .000(14) + C3. we can draw the shear force and bending moment diagrams for our loaded beam. FBD.224.000x+30.000(14)2 + 30. (Shown in Diagram) 2.V3 = 0.000] lb 4.000] lb M1 =[21.000 lb V3 = [-2.000] ft-lb file:///Z|/www/Statstr/statics/Beams/xsol45b.000x2+30.000x-200.000x2 + 30.000] ft-lb M2 = [17.000] ft-lb M3 = [-1. and solving: C3 = -224. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x.4.000x .000 lbs .000] ft-lb.forces) Sum Fy = 21.000 ft-lb Therefore: M3 = [-1. Resolve all forces into x & y components.000x + C3 Where C3 is the integration constant which we find by applying a boundary condition. 3.3. Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/xsol45b.5 .htm (5 of 5) [5/8/2009 4:40:10 PM] .Topic 4.Topic Examination or Select: Topic 4: Beams .Shear Force Topic Exam Sol 2 Return to: Shear Force & Bending Moments . This function has a slope at every point. which is graphed in Diagram 2. The 'derivative' function gives us the value of the slope of our quadratic function at every point.50 = -14) file:///Z|/www/Statstr/statics/Beams/beam46.htm (1 of 3) [5/8/2009 4:40:11 PM] .50). the slope of the quadratic function is 18 * 2 .Topic 4.Integral Introduction Review . above Diagram 2. If we take the "derivative" of our quadratic function. we obtain a new function (y' = 18 x .6 .Basic Calculus Concepts This is a very basic review of introductory Calculus concepts. which is graphed in Diagram 1. (Thus at x = 2.50 x + 50. We first look at a quadratic function: y = 9 x2 . 25 x2 + 50 t).Topic 4. That is. With a little reflection we see that a possible function is: ½ x2 .what function must we take the derivative of to obtain what is inside the integral sign. Before we look at an example of this. Some basic integrals. we obtain a new function (y* = 3 x3 . Thus.000.Integral Introduction If on the other hand we "integrate" our quadratic function. the area under the quadratic function curve between zero and 2 is: A = 3 (2)3 -25 (2)2 + 50 (2) = 24. We can also do an "indefinite" integral which results in 'integrated' function involving a constant which is evaluated by applying a boundary condition. the answer is a function plus a constant which must be determined by an appropriate boundary condition. the initial x value is zero and the ending x-value is what ever value we choose. However. A second way to think of derivatives and integrals is as inverse functions of each other. or ½ x2 + C (where C is any arbitrary constant). the integral asks the question .what function Now we look at the indefinite integral must we take the derivative of to obtain 'x' (what is inside the integral sign). For the integrated function above. let's review some basic derivatives (which are normally covered in the first semester of Calculus). This integral asks the question . which is graphed in Diagram 3 directly below Diagram 1. It actually tells us the area between some beginning x-value and ending x-value (when we do what is called a definite integral). we do obtain x. That is for x = 2.000. if we take the derivative of ½ x2. the function ½ x2 + 5 is also a solution.htm (2 of 3) [5/8/2009 4:40:11 PM] . as is the function ½ x2 + 1. Thus every y value on the curve in Diagram 3 is equal to the sum of the area under the curve in Diagram 2 up to that point. The 'integrated' function tells us the net area under the quadratic function curve (between the function curve and the x axis). ● ● ● ● ● d/dx (any constant) = 0 d/dx (x) = 1 d/dx (x2) = 2x d/dx (x3) = 3x2 d/dx (xn) = n xn-1 . That is.6 . (which we will use in determine bending moment expressions in loaded beams) are: ● ● ● file:///Z|/www/Statstr/statics/Beams/beam46. we see that when one does an indefinite integral. htm (3 of 3) [5/8/2009 4:40:11 PM] .Topic 4. Return to: Example 1 or Select: Topic 4: Beams . where A is any constant that is. For an example of the application of integration and use of a boundary condition in solving a problem.Table of Contents Statics & Srength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/beam46. These simple integrals will be enough to solve beam problems involving standard type loads. the integral of a sum (or difference) is the sum (or difference) of the integrals.Integral Introduction ● ● ● .6 . which is often known as the Bending Stress in a beam. and return to this page. Before we continue with determining the Bending Stresses and the Horizontal Shear Stresses in a loaded beam we must examine two related topics which are used in the determination of the bending and shear stresses . review the material. This relationship is known as the Flexure Formula. In Diagram 2. First. Both of these are very important processes for the safety and efficiency of a beam. In Diagram 1 we have shown a simply supported beam loaded at the center. Second. but the forces act across the whole cross section as shown in the side view in Diagram 2b. Please select Topic 4.7a : Centroids and The Moment of Inertia .htm (1 of 3) [5/8/2009 4:40:11 PM] . when examining bending moments. it will enable us to select the best beam (from a table of beams) for a particular loading. As discussed previously.Shear Forces and Bending Moments) we examined how to determine the Shear Forces and Bending Moments in a loaded beam. Determining the axial stress . The horizontal forces decrease from maximum at the outer edges to zero at the neutral axis (an axis running through the centroid of beam cross section). horizontal forces act on the cross sectional face of the beam section. It deflects (or bends) under the load. We have shown only the horizontal forces along the top and bottom in Diagram 2a.often called the Horizontal Shear Stress (for reasons we will discuss) is important in two ways.7: Beams .Bending Stress BEAMS: BENDING STRESS In the proceeding topic (Beams . we have shown the left end section of the beam. This is particularly valuable since the Axial Stress (tension and compression) and the Shear Stress (vertical and horizontal) which develop in a loaded beam depend on the values of the Bending Moments and the Shear Forces in the beam. and determining the shear stress . file:///Z|/www/Statstr/statics/Beams/bdsn47.Bending Stress Topic 4. The Bending Stress: We will first develop a relationship for the bending stress which develops in a loaded beam.Centroids and The Moment of Inertia.7: Beams . it will enable us to determine if a particular loaded beam is safe under the applied loading.Topic 4. and finally we recognize that the summation term remaining is simply the Moment of Inertia (I) about the centroid of the beam cross section. arranging terms and isolating the stress term by itself. from the neutral axis to the outer edge of the beam file:///Z|/www/Statstr/statics/Beams/bdsn47. We first write an expression for the bending moment produced by the horizontal forces with respect to the neutral axis (which is a line passing through the centroid of the beam cross sectional area . the maximum "Bending Stress" at some location along the beam is equal to the bending moment. We can then write: dA. The expression for the bending moment is simply the sum of the forces times the perpendicular distance to the neutral axis. We can then write the stress at an arbitrary y as: substitute this expression into our relationship for the bending moment and obtain . the force acting on any small horizontal strip of area (dA) is the product of axial stress at that point and the amount of area (dA).Topic 4. We now term . We can now rewrite the expression for the bending moment as: We now will rewrite this expression one more time by noting that the horizontal forces and accompanying stresses increase linearly from zero at the neutral axis to a maximum value at an outer . or: We now note that we can express the force Fx as . We now rewrite one last time.htm (2 of 3) [5/8/2009 4:40:11 PM] . and ymax is the distance from the neutral axis to the outer edge of the beam cross-section. where y is the distance from the neutral axis to area edge. we finally obtain: M ymax / I That is.shown in Diagram 2a). we find: .Bending Stress We will now go through a relatively brief derivation to arrive at the flexure formula. (This simply comes from the definition of axial stress = Force/Area).. as shown in Diagram 2a. M. at that location "times" the distance. then rewriting slightly and factoring out the from the summation sign (which we may since it is a constant). that is.7: Beams . y. htm (3 of 3) [5/8/2009 4:40:11 PM] .8: Beams .Bending Stress (cont) or Select: Topic 4: Beams . While the formula above was derived for the maximum stress.Topic 4. If this seems somewhat confusing. Select: Topic 4. it actually holds for the stress at any point in the beam cross section and is known as the Flexure Formula.Example 1.7: Beams . Flexure Formula: My/I We now will look at a simple example to see the application of the flexure formula. Continue to: Topic 4.Bending Stress Example 1 When finished with Bending Stress . I.7b: Beams . it will become clearer as we work through several examples.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsn47.Bending Stress "divided" by the moment of inertia. of the beam cross sectional area. as in a very thin plate. summing over the entire object.7a: Centroids and The Moment of Inertia Centroids: The concept of the centroid is nearly the same as the center of mass of an object in two dimensions.7a . The center of mass is obtained by breaking the object into very small bits of mass dM.Topic 4. and Yct location for the Centroid (center of area) of the object. and finally dividing by the total mass of the object to obtain the Center of Mass which may be considered to be the point at which the entire mass of the object may be considered to "act". This results in an Xct.htm (1 of 5) [5/8/2009 4:40:11 PM] . See Diagram 2 file:///Z|/www/Statstr/statics/Beams/bdsn47a. we instead divided the object into small bits of areas dA. See Diagram 1. The only difference between the center of mass and the centroid is that rather than summing the product of each bit of mass dM and the distance xi (and yi) to an axis then dividing by the total mass.Centroids/Inertia Topic 4. multiplying these bits of mass by the distance to the x (and y) axis. and then take the sum of the product of each bit of area dA and the distance xi (and yi) to an axis then divide by the total area of the object. Centroids of Composite Areas: Some objects or beams may be formed from several simple areas.7a .Centroids/Inertia Several points to mention. (See Diagram 3) In this case the centroid of the compose area may be found by taking the sum of the produce of each simple area and the distance it's centroid is from the axis.Topic 4. and x4 are the distances from the centroid of each simple area to the y-axis as shown in the Diagram 3. such as rectangles. We will assume all our beams have uniform density and will not consider the case of non-uniform density beams. The location of the y . the location of it's x . the centroid will simply be at the geometric center of the cross section. We will also point out that for any beam cross section (or object) which is symmetry. etc. although the y distances are not shown in Diagram 3: file:///Z|/www/Statstr/statics/Beams/bdsn47a. For the composite area shown in Diagram 3.htm (2 of 5) [5/8/2009 4:40:11 PM] .centroid would be given in like manner. Thus for rectangular beam and I-beams. the centroid is located at the exact center of the beam.centroid would be given by: X ct = (A1 * x1 + A2 * x2 + A3 * x3 + A4 * x4)/(A1 +A2 +A3 + A4) where x1. This is not the case for T-beams. divided by the sum of the areas. triangles. x3. x2. we will once again use an Area Moment of Inertia. Once again.Centroids/Inertia Y ct = (A1 * y1 + A2 * y2 + A3 * y3 + A4 * y4)/(A1 +A2 +A3 + A4) Moment of Inertia A second quantity which is of importance when considering beam stresses is the Moment of Inertia. For use with beam stresses.htm (3 of 5) [5/8/2009 4:40:11 PM] . rather than using the Moment of Inertia as discussed above. The Moment of Inertia is obtained by breaking the object into very small bits of mass dM.Topic 4. the Moment of Inertia as used in Physics involves the mass of the object. See Diagram 5. This Area Moment of Inertia is obtained by breaking the object into very small bits of area dA. multiplying these bits of area by the square of the distance to the x (and y) axis and summing over the entire object.7a . See Diagram 4. multiplying these bits of mass by the square of the distance to the x (and y) axis and summing over the entire object. file:///Z|/www/Statstr/statics/Beams/bdsn47a. Note that the moment of inertia of any object has its smallest value when calculated with respect to an axis passing through the centroid of the object. That is. Radius of Gyration: rxx = (Ixx/A)1/2 The radius of gyration is the distance from an axis which. while the moment of inertia with respect to an axis through the base of the rectangle is: I = 1/3 (base * depth3) in4. See Diagram 6. it would result in the same moment of inertia about the axis that the object has.7a . the polar moment of inertia in related to the x & y moments of inertia by: J = Ixx + Iyy.Centroids/Inertia The actual value of the moment of inertia depends on the axis chosen to calculate the moment of the inertia with respect to. the moment of inertia about an axis passing through the centroid of the rectangle is: I = 1/12 (base * depth3) with units of inches4. In a case as shown in Diagram 7. file:///Z|/www/Statstr/statics/Beams/bdsn47a. if the entire area of the object were located at that distance.Topic 4. r2 dA The polar moment of inertia is the sum of the produce of each Polar Moment of Inertia J = bit of area dA and the radial distance to an origin squared. for a rectangular object. We lastly take a moment to define several other concepts related to the Moment of Inertia. which may be written: Ixx = Icc + Adc-x2 This says that the moment of inertia about any axis (Ixx) parallel to an axis through the centroid of the object is equal to the moment of inertia about the axis passing through the centroid (Icc) plus the product of the area of the object and the distance between the two parallel axis (Adc-x2). Parallel Axis Theorem: Moments of inertia about different axis may calculated using the Parallel Axis Theorem.htm (4 of 5) [5/8/2009 4:40:11 PM] .. 7: Beams .all the summations shown above become integrations as we let the dM's and dA's approach zero. And.Centroids/Inertia One final comment .Bending Stress or Select: Topic 4: Beams .7a . the summation method is just as useful for understanding the concepts involved Return to: Topic 4.htm (5 of 5) [5/8/2009 4:40:11 PM] .Topic 4. while this is important and useful when calculating Centroids and Moments of Inertia.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsn47a. 000 lb. we have shown a simply supported 20 ft. We really don't need the shear force diagram at this point.Topic 4. the shear force remains constant at 5000 lb. the 10. Then as there are no additional loads for the next 10 file:///Z|/www/Statstr/statics/Beams/bdsne47b. We first draw the shear force diagram. each as shown in Diagram 2.5000 lb.7b: Bending Stress .Example 1 In Diagram 1. beam with a load of 10. Due to the 5000 lb. Step 1: Out first step in solving this problem is. We would like to determine the maximum bending (axial) stress which develops in the beam due to the loading. between 0 and 10 feet. of course. to apply static equilibrium conditions to determine the external support reactions. but point out that the two support forces will support the load at the center equally with forces of 5000 lb. If necessary we will determine the shear force and bending moment expressions and make the shear force and bending moment diagrams from these expressions.htm (1 of 2) [5/8/2009 4:40:12 PM] . because of the symmetry of the problem. In this particular example. We will normally be able to draw the shear force diagram by simply looking at the load forces and the support reactions. Step 2: The second step is to draw the shear force and bending moment diagrams for the beam. acting downward at the center of the beam. support force. downward load drives the shear force down by that amount.000 lb.. At 10 feet.7b: Bending Stress . except we will use it to make the bending moment diagram.Example 1 Topic 4. to a value of . the shear force value begins at +5000 lb. we will not go through the statics in detail. as we did in the proceeding topic. and since there is no additional loads for the next 10 feet. from + 5000 lb. The beam used is a rectangular 2" by 4" steel beam. we develop the bending moment graph shown in Diagram 3b. the value of y would be: y = 3". Notice that the maximum bending moment does not depend on the type of beam.8: Beams . if we had used a rectangular 2"x6" beam (instead of a 2" x 4" beam). Then. (from bending moment diagram) y = distance from the neutral axis of the cross section to outer edge of beam = 2 inches I = moment of inertia of cross section. Maximum Bending Stress = M y / I = (600.Topic 4. Thus if we tried to use a rectangular 2"x4" steel beam.000 in-lb)*(3 in)/(36in4) = 50. When finished with this example. it would fail under the load. This value is in a more reasonable range for acceptable axial stresses in steel. Graphing the shear force values produces the result in Diagram 3a. 500 lb/in2 This is the correct value. which occurs at the outer edge of the beam so: M = maximum bending moment = 50.67 in4) = 112. and the value of I would be: I = (1/12) (2" * 6"3) = 36 in4. Then using the fact that for non-cantilevered beams the bending moment values are equal to the area under the shear force diagram.000 ft-lb.000 lb/in2 . Then the maximum bending stress for this beam would be: Maximum Bending Stress = M y / I = (600. The values of "y" and "I" in the flexure formula do depend on the beam used. Continue to: Topic 4. We now apply the flexure formula: Bending Stress = M y / I We wish to find the maximum bending stress. Step 3. the shear force will remain constant over the remainder of the beam. for rectangle I = (1/12) bd3 = 1/12 (2" * 4"3) = 10. Thus.000 in-lb)*(2 in)/(10.000 in-lb. We will have to use a stronger beam. = 600.htm (2 of 2) [5/8/2009 4:40:12 PM] .Example 1 feet.67 in4.7b: Bending Stress .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne47b.Bending Stress (cont) or Select: Topic 4: Beams . but it is clearly excessive for normal steel. The Beam Designation. and the connecting region is known as the web. gives us the following information. the moment of inertia. and the lower cost of wood rectangular beams offsets their lower efficiency.htm (1 of 4) [5/8/2009 4:40:13 PM] .8: Beams . I-Beams We will now consider an I-beam example. with a weight of 20 lb/ft.Bending Stress (cont) Topic 4. section modulus.. beam information is shown. Rectangular beams are used in wood construction where the cost of using a metal I or T beam is excessive. Rectangular metal beams are not normally used in large building construction. as they are not very efficient.Topic 4. In the sample Beam Data Table shown below. In Bending Stress Example 1.8:Beams. eight inch deep I-beam.).14"). which we will discuss shortly. The "20" indicates the weight per foot of the beam in a standard type of steel. we applied the flexure formula to a loaded rectangular beam.(You may wish to print file:///Z|/www/Statstr/statics/Beams/bdsn48. with respect to the x-x neutral axis. The horizontal line passing through the center (centroid) of the beam is the neutral axis. shown in its general form. such as W 8 x 20. The "W" indicates a Wide Flange beam. there is also a moment of inertia. something called the section modulus. we have shown an I-beam cross section. M y / I. The "8" gives the approximate depth of the beam in inches. S. In Diagram 1. r. W 8 x 20.Bending Stress (cont) We continue our discussion of the Flexure Formula. and T beams are used instead. about the x-x neutral axis of the beam (shown in Diagram 1). (This would apply if we flipped the beam on its side and loaded it in that orientation. the web thickness. I. rather metal I beams. Additional information given in the table includes the flange width and depth. This is not usually done and so we will not expect to use the y-y axis information. (Notice the actual depth of the beam is 8. Additionally. and the radius of gyration. The top and bottom sections are known was the flanges. Thus the first beam in the table. is a wide flange. and radius of gyration about the yy neutral axis. 54 1.12 10.525 6.2 35.50 4.60 23.0 We now look at an example of a loaded I-beam.htm (2 of 4) [5/8/2009 4:40:13 PM] .60 70.76 21.30 37.0 25.68 6.20 44.00 1.776 6.99 23.073 7.618 0.349 0.61 7.030 4.985 8.60 15.964 4.00 23.70 16. The horizontal line passing through a portion of the T is the neutral axis.350 0.628 0.29 22.313 0.424 0.59 9.260 0.13 d in 8.000 6. and the vertical section is known as the stem.6 80.5 54.82 1.416 1820.69 9. we have shown a T-beam cross section.27 4.98 1.5 221.20 14.292 0.50 9. In Diagram 2.40 15.94 9.92 1.86 9.60 26.09 thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I in4 69. Cross Section Info.250 8.528 0.52 1.12 5.1 42.7 48.6 249.0 171.88 6.307 0.83 6.14 8.299 13.1 49.465 0.36 4.90 36.6 56.83 13.31 12.0 239.85 27. Please select Topic 4.54 1.428 0.90 18.3 39.86 6.98 1.0 0.71 0.50 8.0 447. In the diagram. The top horizontal section is known as the flange.00 S in3 3.40 23.Bending Stress (cont) out this topic and the beam tables to study them more effectively.89 5.25 1.00 82.60 r in 1.16 12.8a: Bending Stress .0 340.75 12.40 108.453 0.0 0.30 11.682 0.50 9.0 14.85 0.80 24.44 53.308 0.287 0.47 5.Example 2 T-Beams We will next consider a T-beam example.94 0.380 0.0 210.20 9.268 5.990 7.88 5.80 16.007 6.59 1.022 7.22 2.230 0.22 7.31 1.582 0.20 4.0 517.) I-Beams Flange Flange Web bf in 5.25 11.Topic 4.8:Beams.64 3.061 8.318 0.00 14.503 0.0 S in3 17.56 1.10 28.3 21.23 6.87 0. file:///Z|/www/Statstr/statics/Beams/bdsn48.20 10.00 9.89 10.516 2690.16 2.52 1.62 6.872 0.70 24.378 0.400 2100. y is the distance from the top of the T to the neutral axis of the beam.4 56.433 0.01 11.0 153.992 9.0 176.0 r in 3.0 657.91 20.33 4.0 386.71 6.248 0.0 156.0 130. Designation Area Depth Width thick W 8x20 W 8x17 W 10x45 W 10x39 W 10x33 W 12x22 W 12x19 W 12x31 W 14x38 W 14x34 W 16x50 W 16x40 W 16x36 W 24x94 W 24x76 W 24x68 A in2 5.50 2.00 10.00 14.8 64.750 7.53 I in4 9.43 3.237 0.82 5.961 tf in tw in Cross Section Info.513 0.91 4.265 0.12 11. 36 5. The "11" indicates the weight per foot of the beam in a standard type of steel. the moment of inertia.90 42.40 26.890 2.630 1.89 5. about the x-x neutral axis of the beam (shown in Diagram 2).5 WT 7x26.339 0.20 27.590 2.91 8.80 Info.031 7.70 25.93 10.57 Width bf in 4.13 8.380 0. The "WT" indicates a Wide Flange Tee beam.299 0. Additional information given in the table includes the flange width and depth. six inch deep T-beam.410 3.80 20. T-Beams Designation WT 6x11 WT 6x9.350 1./ft. is a wide flange. with respect to the x-x neutral axis.08 6.900 file:///Z|/www/Statstr/statics/Beams/bdsn48.349 0.590 WT10. the radius of gyration.062 8.00 12.490 6.20 30. the stem thickness.Topic 4.900 1.628 0.00 7.503 0.960 4. beam information is given.260 0. - x-x axis x-x axis x-x axis S in3 2.16"). gives us the following information.30 Depth Flange Flange of T d in 6.030 4.424 0. Thus the first beam in the table.593 0.40 21.650 1.685 Stem thick tw in 0. and y.00 103.007 8.97 6. the section modulus. the distance from the top of the tee to the neutral axis of the beam.5 WT 7 x24 WT 8x25 WT 8x20 WT 8x18 Area A in2 3.307 0.770 5.200 y in 1.8:Beams.380 5. The "6" gives the approximate depth of the beam in inches.880 2.20 33.70 24. I.79 7.370 0.Bending Stress (cont) In the Beam Data Table shown below.06 7.430 d/tw 23.992 8. The Beam Designation.380 1.70 10.400 2.110 r in 1.24 2.073 7.890 1.270 thick tf in 0.880 1. r.820 1.60 Cross Section x-x axis I in4 11.300 4.16 6.000 6.428 0.658 0. such as WT 6 x 11. with a weight of 11 lb.50 24. S. (Notice the actual depth of the beam is 6.80 7.910 1.htm (3 of 4) [5/8/2009 4:40:13 PM] .370 2.10 26.237 0. W 6 x 11.5 x34 10.70 18. 551 10.615 0.05 11.15 12.061 9.00 20.htm (4 of 4) [5/8/2009 4:40:13 PM] . Continue to: Topic 4.985 10.50 25.760 1.690 5.00 18.850 0.400 0.810 4.900 421.300 166.Bending Stress (cont) WT 10x31 WT 12x47 WT 12x42 WT 12x38 WT 15x66 WT 15x58 WT 15x54 WT18x97 WT 18x75 9.500 10.670 3.630 5.00 37.8b: Bending Stress .680 4.600 350.210 3.60 26.10 10.00 32.000 0.20 23.90 28.770 0.60 27.20 24.970 2.00 16.117 11.60 22.900 3.92 8.60 27.470 0.872 0.40 17.400 372.40 11.70 28.900 3.00 67.10 15.80 12.625 26.8:Beams.670 4.13 13.400 698.100 905.972 0.100 We now look at an example of a loaded T-beam.650 4.440 0.660 3.91 18.484 12.020 4.20 19.564 0.15 15.940 0.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsn48.260 0.000 2.80 11.682 1.015 8.580 3.620 2.516 0.9: Beams .300 151.Horizontal Shear Stress or Select: Topic 4: Beams .96 15.930 4.70 93.990 3.24 17.00 33.20 23.00 53.548 0.Topic 4.780 186.772 0.Example 3 When finished with Bending Stress Example 3. Please Select: Topic 4.615 0.00 14.50 12.240 9. 500 lbs STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.000 lbs = 0 Sum TB = 5.2.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .2. at x= 0'. and also the bending stress at that location along the beam and 8 inches from the bottom of the beam cross section.Example 2 A loaded. Then we will determine the maximum bending stress at that location..Topic 4. the shear force begins with a value of -5000 lb. Dy = 3. Next nothing happens (no loading) for the next 4 ft. so the file:///Z|/www/Statstr/statics/Beams/bdsne48a. That is.htm (1 of 3) [5/8/2009 4:40:13 PM] .8a: Bending Stress .) Resolve all forces into x & y components 3. (at x = 0') due to the downward acting load of 5000 lb. reasonably accurately.000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0 Solving: By = 9.8a: Bending Stress Example 2 Topic 4.000 lbs (4 ft) . For this beam we will first determine the maximum bending moment (and where it occurs in the beam).) Draw Free Body Diagram of structure (See Diagram 2) 2. by simple looking at the loading and the support reactions.500 lbs. STEP 1: Apply Static Equilibrium Principles and determine the external support reactions: 1. for this example.000 lbs/ft (4 ft) .5. simply supported W 10 x 45 beam is shown in Diagram 1. We can normally do this for the Shear Force Diagram. That is. the value of shear force is -3. At that point. where the section modulus is equal to the moment of inertia of the beam cross section divided by the maximum distance from the neutral axis of the beam to an outer edge of the beam. Applying this.500 lb. If we rewrite the standard flexure formula several times as follows for the maximum stress: M (ymax / I) = M / (I / ymax) = M / s . this special form of the flexure formula can only be used to find the maximum bending stress. Also at x = 12 ft. driving the shear force downward at a rate of 2000 lb.000 lb.8a: Bending Stress Example 2 shear force remains constant. Then at x = 4 ft there is an upward support force of 9.8000 lb. is applied. from a value of -5000 lb. a uniform load of 2.500 lb.htm (2 of 3) [5/8/2009 4:40:13 PM] ./ft.). Again there is no change in loading for the next 4 ft. which can be considered to bring the shear force back to zero. we then see that the section modulus is defined as s = I / ymax. where s is what is known as the section modulus. is equal to the area under the shear force diagram up to that point. We may use Flexure Formula: M y / I. or a special form of the Flexure Formula: M / s. The Bending Moment Diagram may be drawn using the Shear Force Diagram. is the upward support reaction of +3. and uses the section modulus. STEP 3: We will now Apply the Flexure Formula to determine the maximum bending stress for the beam.500 lb.Topic 4. so the shear force remains constant until x = 8 ft.(See Diagram 3). until at x = 12 ft.. per each foot for the 4 ft (at total of . file:///Z|/www/Statstr/statics/Beams/bdsne48a.. we obtain the Bending Moment Diagram shown in Diagram 4. to + 4. which drives the shear force value up this amount.500 lb. by remembering that (for non-cantilever beams) the value of the bending moment at a given location. x feet from the left end. -lb.000 ft-lb.06 inches above the bottom of the beam (at the beam center).) We also then find the x-x axis section modulus of the beam as listed in the beam table below.) or Select: Topic 4: Beams .5.0 Section Info. and since the neutral axis is 5. and occurs at x = 4 ft.30 r in 2.) with the top in tension and the bottom in compression. the stress at the top of the beam and at the bottom of the beam are equal in value (4.8: Beams . Since we wish to find the bending stress 8 inches above the bottom of the beam./ft.-lb..1 r in 4.000 in-lb.000 in-lb. For this we need to use the flexure formula in the form My/ 4..06 = 2. this is a tensile stress.)/ 49.000 ft.94 inches.8a: Bending Stress Example 2 As an example we apply this form to determine the maximum bending stress in our beam.1 in3 Flange Flange Web thick tf in 0. And since the location is above the beam centroid (and the bending 2. and 8 inches above the bottom of the beam.Topic 4.022 x-x axis x-x axis S in3 49.618 thick tw in 0. where M = 20.890 psi.)(12 in.94 in. we will determine the bending stress at 4 ft (where the maximum bending moment occurs). * 2. Designation Area Depth Width W 10x45 A in2 d in bf in 8. This means the top of the beam is in tension and the bottom of the beam is in compression.Bending Stress (cont. Then 240.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne48a. = 2830 lb.000 ft-lb./in moment is positive).20 10./in2. to in. for the units to be consistent) We have thus determined the maximum bending stress (axial stress) in the beam.33 y-y axis y-y axis S in3 13. First we determine the maximum bending moment from our bending moment diagram .12 Now M / s = (20.00 13. and y = distance I. Cross y-y axis I in4 53. (We will drop the negative sign which simply tells us that the beam is bent concave facing downward at this point. then y = 8 . = 240. / 249 in4. s = 49. I = moment of inertia of beam = 249 in from the neutral axis to point at which we wish to find the bending stress. (Notice that we had to convert the bending moment in ft.350 Cross x-x axis I in4 249.1 in3 = 4.htm (3 of 3) [5/8/2009 4:40:13 PM] .-lb. Finally. 890 lb. Since the beam is symmetric about the neutral axis. Return to: Topic 4.20 Section Info.which we observe from Diagram 4 is: Mmax = 20. We will also determine the bending stress 4 ft from the left end of the beam and 2 inches above the bottom of the beam.Example 3 A simply supported WT 8 x 25 T-beam is loaded is shown in Diagram 1. Cy = 6. That is.Example 3 Topic 4.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Cy . For this beam we will determine the maximum bending stress in the beam.330 lbs. also at x = 4 ft.8b: Bending Stress . at x = 4 ft.) FBD of structure (See Diagram 2) 2.1.000 lbs/ft (4 ft) . per foot for the first 4 feet due to the uniform load.) Resolve all forces into x/y components 3. STEP 1: Apply Static Equilibrium Principles and determine the external support reactions: 1.1. resulting in a shear force value of -4000 lb. by simple looking at the loading and the support reactions.Topic 4. the shear force begins with a value of zero (since there is not point load or reaction at x = 0). We can normally do this for the Shear Force Diagram.8b: Bending Stress . and then decreases at a rate of 1000 lb..500 lbs/ft (4 ft) (8 ft) + Cy(6 ft) = 0 Solving: By = 3.500 lbs/ft (4 ft) = 0 Sum TB = 1.670 lbs STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam. there is an file:///Z|/www/Statstr/statics/Beams/bdsne48b.htm (1 of 3) [5/8/2009 4:40:14 PM] . reasonably accurately.1. for this example. Then.000 lbs/ft (4 ft) (2 ft) . We may use Flexure Formula: M y / I.. The Bending Moment Diagram may be drawn using the Shear Force Diagram. which drives the value of the shear force upward (from -4000 lb.-lb. from a value of -670 lb. we obtain the Bending Moment Diagram shown in Diagram 4.) by that amount to a value of -670 lb. Applying this. driving the shear force downward at a rate of 1500 lb. x feet from the left end. by remembering that (for non-cantilever beams) the value of the bending moment at a given location. (We will drop the negative sign which simply tells us that the beam is bent concave facing downward at this point. where s is the section modulus.Topic 4. Which brings the shear force value down to zero at x = 14 ft.(See Diagram 3).) We also then find the x-x axis section modulus of the beam as listed in the beam table./ft.77 in3 file:///Z|/www/Statstr/statics/Beams/bdsne48b.500 lb. and occurs at x = 10 ft.Example 3 upward support reaction of 3. or a special form of the Flexure Formula: M / s.which we observe from Diagram 4 is: Mmax = 12. This special form of the flexure formula can only be used to find the maximum bending stress.. where the section modulus is equal to the moment of inertia of the beam cross section divided by the maximum distance from the neutral axis of the beam to an outer edge of the beam.670 lb.htm (2 of 3) [5/8/2009 4:40:14 PM] . Next nothing happens (no loading) for the next 6 ft. First we determine the maximum bending moment from our bending moment diagram .6000 lb). so the shear force remains constant.8b: Bending Stress . As an example we apply this form to determine the maximum bending stress in our beam.330 lb. to 14 ft. Then at x = 10 ft there is an upward support force of 6. per each foot for the last 4 ft (for total of .000 ft. and uses the section modulus. which drives the shear force value up this amount. Then from 10 ft. to +6000 lb. STEP 3: We will now Apply the Flexure Formula to determine the maximum bending stress for the beam. This means the top of the beam is in tension and the bottom of the beam is in compression. a uniform load of 1. s = 6. is equal to the area under the shear force diagram up to that point. is applied. I = moment of inertia of beam = 42.-lb./ft.2" = 4. Return to: Topic 4. Thus 96. for the units to be consistent) We have thus determined the maximum bending stress (axial stress) in the beam.13 Width bf in 7.Example 3 Designation WT8x25 Area A in2 7.890 Now M / s = (12. Since we wish to find bending stress 2 inches above the bottom of the beam.8b: Bending Stress .htm (3 of 3) [5/8/2009 4:40:14 PM] .24 in. Since the bending moment is negative.) or Select: Topic 4: Beams .8: Beams .Bending Stress (cont.073 thick tf in 0. then the neutral axis is 8.-lb.) Finally.770 r in 2. since that is the outer edge of the beam which is farthest from the neutral axis.89 inches below the top of the beam (from the beam data table).89" = 6. we will determine the bending stress at 4 ft from the left end of the beam and 2 inches above the bottom of the beam.270 lb/in2.20 S in3 6.40 x-x axis x-x axis x-x axis x-x axis I in4 42.000 in-lb * 4.000 in-lb (which we determine from the bending moment graph shown in Diagram 4).24" from the bottom of the beam.2 in4.24" from the neutral axis to where we wish to determine the bending stress. / 42.13"-1. meaning the beam is bent concave facing in4. (The stress at the top of the tee is less than that at the bottom of the stem.628 thick tw in 0.77 in3 = 21. to in. = 9.450 lb/in downward. then y = 6..)/ 6.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne48b.000 ft-lb)(12 in.Topic 4.400 y in 1. The neutral axis is 1.380 d/tw 21. then this stress is compressive. and y = distance from the neutral axis to point at which we wish to find the bending stress. and since the location is below the beam centroid.2 2.36 of T d in 8. (Notice that we had to convert the bending moment in ft. For this we need to use the flexure formula in the form M y / I. Please note that this maximum stress is compressive and occurs at the bottom of the stem. where M = 8.000 ft-lb = -96.24" . since the top of the tee is closer to the neutral axis than the bottom of the stem. We will take a moment to derive the formula for the Horizontal Shear Stress. we have shown a side view of section dx. for static equilibrium. we have cut a section dx long out of the left end of the beam.9: Horizontal Shear Stress Topic 4. and a Horizontal (longitudinal) Shear Stress. In Diagram 2b. in which we see the bending moment increases from a value of zero at the left end to a maximum at the center of the beam.htm (1 of 3) [5/8/2009 4:40:14 PM] . Notice that the bending moment is larger on the right hand face of the section by an amount dM.Topic 4. there is also a shear stress which develops. (This is clear if we make the bending moment diagram for the beam. including both a Vertical Shear Stress. It can be shown that at any given point in the beam. we have shown a simply supported loaded beam. As a result it is usual to discuss and calculate the horizontal shear stress in a beam (and simply remember that the vertical shearing stress is equal in value to the horizontal shear stress at any given point). and have shown the internal horizontal forces acting on the section.) file:///Z|/www/Statstr/statics/Beams/bdsn49. at that point. In Diagram 1.9: Beams -Horizontal Shear Stress In addition to the bending (axial) stress which develops in a loaded beam. In Diagram 2a. the values of vertical shear stress and the horizontal shear stress must be equal. where V = Shear force at location along the beam where we wish to find from the horizontal shear stress A = cross sectional area. we have shown a top slice of section dx. and in terms of the shear stress. we have: F= * b dx = (dM y/I)dA. however dM/dx is equal to the shear force V (as discussed in the previous = [(dM/dx)/Ib] topic). and y dA is the first moment of the area of the section. from point where we wish to find the shear stress at. that is F1 = (My/I) dA. horizontally.Topic 4. the differential shearing force. and may be written as A y'. and F2 = (M+dM)y/I dA. file:///Z|/www/Statstr/statics/Beams/bdsn49. Rewriting in a final form we have: Horizontal Shear Stress: = VAy'/Ib. If we now combine the two F = expressions. to an outer edge of the beam cross section (top or bottom) y' = distance from neutral axis to the centroid of the area A. F. and then rewriting to solve for the shear stress: y dA. This means there is a shear stress on the section.htm (2 of 3) [5/8/2009 4:40:14 PM] . shown in Diagram 2c. can be written as F = times the longitudinal area of the section (b dx). A second way of expressing the shear force is by expressing the forces in terms of the bending stress. where A is the area of the section and y' is the distance from the centroid of the area A to the neutral axis of the beam cross section. Since the forces are different between the top of the section and the bottom of the section (less at the bottom) there is a differential (shearing) force which tries to shear the section.9: Horizontal Shear Stress In Diagram 2c. then the differential force is (dM y/I)dA. and zero at the neutral axis. The Horizontal Shear Stress is (normally) a maximum at the neutral axis of the beam. Continue to: Topic 4.htm (3 of 3) [5/8/2009 4:40:14 PM] .9: Horizontal Shear Stress I = moment of inertia for the beam cross section. (In some texts.9b: Horizontal Shear Stress .Topic 4. To help clarify the Horizontal Shear Stress equation we will now look at at several example of calculating the Horizontal Shear Stress.since the area A is zero there. Please select: Topic 4.10: Beams .9a: Horizontal Shear Stress . This is the opposite of the behavior of the Bending Stress which is maximum at the other edge of the beam.Example 2 When finished with the above examples. b = width of the beam at the point we wish to determine the shear stress.Example 1 Topic 4.Beam Selection or Select: Topic 4: Beams . we observe that the Horizontal Shear Stress is zero at the outer edge of the beam . the product Ay' is given the symbol Q and used in the shear stress equation) If we consider our shear relationship a little.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsn49. Horizontal Shear Stress .9a: Horizontal Shear Stress .Example 1 Topic 4.7b:Bending Stress . for a more complete explanation of how the shear force and bending moment diagrams were made. acting downward at the center of the beam. In this particular example.Topic 4. but will include it for completeness. file:///Z|/www/Statstr/statics/Beams/bdsne49a. but point out that the two support forces will support the load at the center equally with forces of 5000 lb. We will also determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum.9a . Step 2: The second step is to draw the shear force and bending moment diagrams for the beam. we have shown a simply supported 20 ft. This beam is the same beam used in Topic 4.000 lb.Example 1 In Diagram 1. We would like to determine the maximum Horizontal Shear Stress which develops in the beam due to the loading. because of the symmetry of the problem. beam with a load of 10. if needed.htm (1 of 3) [5/8/2009 4:40:14 PM] . Step 1: Out first step in solving this problem is to apply static equilibrium conditions to determine the external support reactions. we will not go through the statics in detail. We have shown the shear force and bending moment graphs in Diagram 3a and 3b. We really don't need the bending moment diagram. each as shown in Diagram 2. The beam used is a rectangular 2" by 4" steel beam.Example 1 Please see that example. which occurs at the neutral axis of the beam: V = maximum shear force = 5. (from the shear force diagram) I = moment of inertia of cross section.htm (2 of 3) [5/8/2009 4:40:14 PM] . y' = distance from neutral axis to the centroid of the area "a" which we used. b = width of beam section where we wish to find shear stress at. we find: Maximum Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(4 in2)*(1 in)/ (10.9a . (See Diagram 4) Placing the values into the equation. b= 2 in.000 ft-lb. y'= 1 in.Horizontal Shear Stress .Example 1 Step 3. We will now apply the Horizontal Shear Stress formula: Shear Stress = Vay'/Ib We wish to find the maximum shear stress.67 in4)(2 in)= 937 lb/in2 This is the correct value. we notice it is not very large.Topic 4.67 in4. The beam is clearly able to carry the load without failing in shear. a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam a= (2" x 2")= 4 in2. for rectangle I = (1/12) bd3 = 1/12 (2" * 4"3) = 10. file:///Z|/www/Statstr/statics/Beams/bdsne49a. 5000 lb.Topic 4. We will go to the top edge of the beam.y'= 1.Horizontal Shear Stress . b = width of beam section where we wish to find shear stress at.5 in)/ (10.9: Beams .000 ft-lb. then a = (2" x 1")= 2 in2. Return to: Topic 4. for rectangle I = (1/12) bd3 = 1/12 (2" * 4"3) = 10.Horizontal Shear Stress or Select: Topic 4: Beams .) We again apply the Horizontal Shear Stress formula: Horizontal Shear Stress = Vay'/Ib We wish to find the shear stress 3 inches above the bottom of the beam cross section.5 in. (See Diagram 5) V = shear force = 5. y' = distance from neutral axis to the centroid of the area "a" which we used.67 in4)(2 in)= 703 lb/in2 Notice. since the value of shear force is either +5000 lb.67 in4.. (from the shear force diagram) I = moment of inertia of cross section.Example 1 Part II We now would also like to determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum (which is actually through out the beam. or . as we expect. the horizontal shear stress value becomes smaller as we move toward the outer edge of the beam cross section. through out the beam. b = 2 in. (See Diagram 5) Then the horizontal shear stress 3 inches above the bottom of the beam is: Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(2 in2)*(1.9a .htm (3 of 3) [5/8/2009 4:40:14 PM] . a = area from point we wish to find shear stress at (3" above bottom of the beam) to an outer edge of beam.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne49a. We have shown the shear force and bending moment graphs in Diagram 3 and 4.Example 2 A loaded.htm (1 of 7) [5/8/2009 4:40:15 PM] .1.9b: Horizontal Shear Stress .000 lbs/ft (4 ft) (2 ft) . We will also determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum. STEP 1: Apply Static Equilibrium Principles and determine the external support reactions: 1. This beam is the same beam used in Beams - file:///Z|/www/Statstr/statics/Beams/bdsne49b. simply supported beam is shown in Diagram 1.500 lbs/ft (4 ft) (8 ft) + Cy(6 ft) = 0 Solving: By = 3.1.Example 2 Topic 4.1. and a W 10 x 45 beam) we will determine the maximum Horizontal Shear Stress which would develop in the beam due to the loading. For two different beam cross sections (a WT 8 x 25 T-beam. but will include it for completeness.. Cy = 6. Step 2: The second step is to draw the shear force and bending moment diagrams for the beam.) FBD of structure (See Diagram 2) 2.000 lbs/ft (4 ft) .670 lb.) Resolve all forces into x/y components 3.500 lbs/ft (4 ft) = 0 Sum TB = 1.9b: Horizontal Shear Stress .Topic 4. We really don't need the bending moment diagram.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Cy .330 lb. 9b: Horizontal Shear Stress . Please see that example.13 Width bf in 7.073 thick tf in 0.628 thick tw in 0.20 S in3 6.36 of T d in 8. For the WT 8 x 25 T-beam (table above) we will now apply the Horizontal Shear Stress formula: file:///Z|/www/Statstr/statics/Beams/bdsne49b. Beam Table for WT 8 x 25 Designation WT 8x25 Area A in2 7.380 d/tw 21. if needed.htm (2 of 7) [5/8/2009 4:40:15 PM] .770 r in 2.400 y in 1.40 x-x axis x-x axis x-x axis x-x axis I in4 42.Example 2 Bending Stress Example III. for a more complete explanation of how the shear force and bending moment diagrams were made.890 Step 3.Topic 4. (from the shear force diagram) I = moment of inertia of cross section.24" )= 2.20 in4. which occurs at the neutral axis of the beam: V = maximum shear force = 6. where the shear force is a maximum.12 in.20 in4)(.37 in2.000 lb. b = width of beam where we wish to find shear stress (neutral axis for maximum) from table.Example 2 Shear Stress = Vay'/Ib.38" * 6. to find the maximum shear stress.37 in2)*(3. y' = 3. Then a = (.38 in.9b: Horizontal Shear Stress . In this case we will go to bottom of beam.htm (3 of 7) [5/8/2009 4:40:15 PM] . (See Diagram 5) Then placing values into our expression we find: Maximum Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(2. a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam. from beam table.12 in)/ (42.Topic 4. I = 42. b = .38 in) = 2770 lb/in2 We now would also like to determine the Horizontal Shear Stress 3 inches above the bottom of the beam at the position in the beam where the shear force is a maximum We again apply the Horizontal Shear Stress formula: Shear Stress = Vay'/Ib We wish to find the shear stress 3 inches above the bottom of the beam cross section. y' = distance from neutral axis to the centroid of the area "a" . (See Diagram 6)\ file:///Z|/www/Statstr/statics/Beams/bdsne49b. b = width of beam where we wish to find shear stress (3" above bottom of beam) from table. b = . and the shear stress 3 inches above the bottom of the beam (at the point where the shear force is a maximum).00 13. Then a = (.20 10. y' = 4.9b: Horizontal Shear Stress .33 y-y axis y-y axis S in3 13.Topic 4.350 Cross x-x axis I in4 249. and the shear force and bending moment diagrams are shown in the first file:///Z|/www/Statstr/statics/Beams/bdsne49b.000 lb. a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam.20 Section Info. the horizontal shear stress value becomes smaller as we move toward an outer edge of the beam cross section. (from the shear force diagram) I = moment of inertia of cross section.38 in.2 in4)(. as we expect.38 in)= 2020 lb/in2 Notice.Example 2 V = maximum shear force = 6.618 thick tw in 0.74 in)/ (42.12 We have already done the statics.74 in.htm (4 of 7) [5/8/2009 4:40:15 PM] .022 x-x axis x-x axis S in3 49. but now find these values for a W 10 x 45 I-beam. from beam table. In this case we will go to bottom of beam. Designation Area Depth Width W 10 x 45 A in2 d in bf in 8. Cross y-y axis I in4 53.20 in4. Beam Table for W 10 x 45 I-Beam Flange Flange Web thick tf in 0. Part 2: W 10 x 45 beam: We would like to again determine the maximum horizontal shear stress.14 in2.38" * 3" )= 1.30 r in 2. (See Diagram 6) Then the horizontal shear stress 3 inches above the bottom of the beam is: Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(1.1 r in 4. I = 42.14 in2)*(4. (See Diagram 6) y' = distance from neutral axis to the centroid of the area "a" .0 Section Info. 35" * 8. We now wish to find the shear stress 3 inches above the bottom of the beam cross section. however since we are looking for the maximum shear stress in the I-Beam. To do so. we can use the approximate formula for I-beam. we must use the standard formula as we will do in the next part. (See Diagram 8). (See Diagram 7) Then Maximum Horizontal Shear Stress = (6000 lb)/(3. the approximate formula is only for the maximum horizontal shear stress (which occurs are the neutral axis) in an I-Beam. Applying this we have: Vmax = maximum shear force = 6. (from the shear force diagram) Amax = area of web: A = (. where the shear force is a maximum.11 in2) = 1930 lb/in2 As long as this approximate value is reasonably below the allowable shear stress for the beam material there is no need to use the exact formula for the maximum shear stress.88" )= 3. If we need to know the shear stress at any other location. however.Beam is equal to the maximum shear force divided by the area of the web of the I-Beam. Please remember. so we continue at the point where we apply the horizontal shear stress formula to find the values we desire.htm (5 of 7) [5/8/2009 4:40:15 PM] .11 in2.Topic 4.Example 2 part of this example above.9b: Horizontal Shear Stress .000 lb. Maximum Horizontal Shear Stress = Vmax/Aweb. This says the approximate maximum shear stress in an I . For the WT 8 x 25 T-beam we apply the Horizontal Shear Stress formula: Shear Stress = Vay'/Ib. we apply the standard horizontal shear stress formula: Shear Stress = Vay'/Ib file:///Z|/www/Statstr/statics/Beams/bdsne49b. I = 249.) Using the values shown in Diagram 9.35in2) = 4. (Which is simply the distance from the neutral axis to the center of each of the respective areas.35 in. where A1 and A2 are the two areas.0 in4.794 in2 (See Diagram 9) y' = distance from neutral axis to the centroid of the area "a" Notice that in this case.000 lb.022") + (2.htm (6 of 7) [5/8/2009 4:40:15 PM] . we have: file:///Z|/www/Statstr/statics/Beams/bdsne49b.96 in2 + . To find the centroid of this compound area we use: y' = (A1 y1 + A2 y2)/(A1 +A2).9b: Horizontal Shear Stress . b = width of beam where we wish to find shear stress (3" above bottom of beam) from table. since for a rectangle the centroid is at the center. Notice that the area is composed of the area of the flange (A1) and part of the area of the web (A2). (from the shear force diagram) I = moment of inertia of cross section.834 in2= 5. but rather two rectangles. for an I-Beam. a = area from point we wish to find shear stress at (3" from the bottom) to an outer edge of beam. (See Diagrams 8 and 9. and y1 and y2 are the distances from the neutral axis of the beam to the centroid of each of the respective areas.618" x 8. The area we wish to find the centroid of is not a simple rectangle. this is not a entirely simple matter. In this case we will go to bottom of beam.383 in2 x .) Then a = (A1 + A2)= (. from beam table. b = .Example 2 V = maximum shear force = 6.Topic 4. Return to: Topic 4.794 in2)*(4. the horizontal shear stress value becomes smaller as we move toward an outer edge of the beam cross section.834 in2 x 2.834 in2) = 4.35 in)= 1790 lb/in2 Notice.96 in2 + . as we expect.94 in)/(4.49 in.Topic 4. Then the horizontal shear stress 3 inches above the bottom of the beam is: Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(5.htm (7 of 7) [5/8/2009 4:40:15 PM] .0 in4)(.Horizontal Shear Stress or Select: Topic 4: Beams .Example 2 y' = (A1 y1 + A2 y2)/(A1 +A2) = (4.9b: Horizontal Shear Stress .9: Beams .49 in)/ (249.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne49b.96 in2 x 4.75 in + . Determine the maximum bending stress 12 feet from the left end of the beam. 2.Solution 411 STATICS & STRENGTH OF MATERIALS .) Cut Beam at 12 ft.) Apply Equilibrium Conditions: Sum Fx: none Sum Fy: Ay + Cy . 1. Solution: Part A: STEP 1: Determine the external support reactions: 1.2000 lb/ft(4 ft) + 4600 . simply supported W T 12 x 50 beam is shown below. Determine the horizontal shear stress at a point 7 inches above the bottom of the beam cross section and 12 feet from the left end of the beam.(6000 lb)(13 ft) = 0 Solving: Cy = 9400 lb.2000 lb/ft(4 ft) .1000 lb/ft(6 ft) = 0 Sum Torque(A) = TA = -(8000 lb)(2 ft) + Cy (10 ft) . all joints and support points are assumed to be pinned or hinged joints.) Resolve all forces into x/y components 3.) Resolve all forces into x/y directions 3.Example A loaded. Unless otherwise indicated. Ay = 4600 lb STEP 2: Determine the shear force and bending moment at x = 12 ft.) Apply Equilibrium Conditions: Sum Fx: none Sum Fy: 9400 lb . B. Draw the FBD of left end of beam.htm (1 of 2) [5/8/2009 4:40:16 PM] .V12 = 0 Sum Torque(A) = TA = -(8000 lb)(2 ft) + 9400 lb (10 file:///Z|/www/Statstr/statics/Stests/beams2/sol411.) FBD of structure (See Diagram) 2. For this beam: A.1000 lb/ft(2 ft) . showing and labeling all external forces. The section modulus is available from the Beam Tables.97")]/[(177 in )(.)(7" x .V(12) + M12= 0 Solving: V 12 = 4000 lb.47)] = 944 lb/in Select: Topic 4: Beams . The Maximum bending stress occurs at the bottom of the Tbeam (since it is the outer edge most distance from the neutral axis) The negative sign means that the beam is bent concave downward at this point. M 12 = -8000 ft-lb STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress at 12 ft.Table of Contents Statics & Strength of Materials Home Page 4 2 3 2 file:///Z|/www/Statstr/statics/Stests/beams2/sol411. The form we will use is: HSS = Vay'/lb Where: V = Shear force 12 ft from the end of the beam a = cross sectional area from 7 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (177 in4 for W 12 x 50 beam) b = width of beam a 7 in above the bottom of the beam HSS = [(4000 lb.. Maximum Bending Stress (MBS) = M12 /S (Where M12' is the bending moment at 12 ft.Solution 411 ft) . The section modulus for this beam from the beam tables is 18.) MBS = -8000 ft-lb. and S is the ' section modulus for the beam.htm (2 of 2) [5/8/2009 4:40:16 PM] .(2000 lb)(11 ft) . so the bottom of the beam is in compression. Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 12 ft from the left end of the beam and 7 inches above the bottom of the beam./ft)/18.7 in3. The WT 12 X 50 beam is shown in the diagram to the right. apply the horizontal shear stress formula..7 in = -5130 lb/in . (12in.47")(5. 000 lbs = 0 Sum TB = 5.htm (1 of 2) [5/8/2009 4:40:16 PM] .2.) Resolve all forces into x/y directions.) FBD of structure (See Diagram) 2. Solution: Part A: STEP 1: Determine the external support reactions: 1.) Resolve all forces into x/y components 3.000 lbs (4 ft) . D = 3.000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0 Solving: B = 9. simply supported W 10 x 45 beam is shown below. Determine the maximum bending stress 6 feet from the left end of the beam.5.500 lbs. file:///Z|/www/Statstr/statics/Stests/beams2/sol412. all joints and support points are assumed to be pinned or hinged joints. B.) Cut beam at 6 ft.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy . For this beam: A. Unless otherwise indicated. Determine the horizontal shear stress at a point 4 inches above the bottom of the beam cross section and 6 feet from the left end of the beam.000 lbs/ft (4 ft) .2. showing and labeling all external forces.500 lbs y y STEP 2: Determine the shear force and bending moment at x=6 ft.Solution412 STATICS & STRENGTH OF MATERIALS . 2. Draw the FBD of left end of beam.Example A loaded. 1. 388 psi Select: Topic 4: Beams .37 in)]/[(249 in )(.htm (2 of 2) [5/8/2009 4:40:16 PM] .500 lbs)(6.1 in3.35 in)] = 1. apply the horizontal shear stress formula.Solution412 3.4.500 lbs (4 ft) .500 lbs .500 lbs (6 ft) + M6 = 0 Solving: V = 4. The 6' section modulus is available from the Beam Tables. The form we will use is: HSS = Vay'/Ib Where: V = Shear force 6 ft from the end of the beam a = cross sectional area from 4 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (249 in4 for W 10 x 45 beam) b = width of beam a 4 in above the bottom of the beam HSS = [(4. The W 10 x 45 beam has a section modulus for the beam from the beam tables is 49.1 in = -2.500 lbs. MBS = M /S (Where M6' is the bending moment at 6 ft.688 lbs/in 3 2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 6 ft from the end of the beam and 4 inches above the bottom of the beam.V6 = 0 Sum TA = 9.Table of Contents Statics & Strength of Materials Home Page 2 4 file:///Z|/www/Statstr/statics/Stests/beams2/sol412.153 in )(4.000 ft-lbs(12 in/ft)/49. and S is the section modulus for the beam.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = -5. M = -11.) MBS = -11.000 ft-lbs 6 6 STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 6'.000 lbs + 9. ) Resolve all forces into x/y components 3.000 lbs/ft (4 ft) .1. Determine the horizontal shear stress at a point 3 inches above the bottom of the beam cross section and 8 feet from the left end of the beam. showing and labeling all external forces.1. C = 6. Solution: Part A: STEP 1: Determine the external support reactions: 1.htm (1 of 2) [5/8/2009 4:40:16 PM] . all joints and support points are assumed to be pinned or hinged joints.) Cut beam at 8 ft. 1.500 lbs/ft (4 ft) (8 ft) + Cy(6 ft) = 0 Solving: B = 3.Solution413 STATICS & STRENGTH OF MATERIALS .000 lbs/ft (4 ft) (2 ft) . B.1.330 lbs.670 lbs y y STEP 2: Determine the shear force and bending moment at x=8 ft. For this beam: A.) FBD of structure (See Diagram) 2. Determine the maximum bending stress 8 feet from the left end of the beam.500 lbs/ft (4 ft) = 0 Sum TB = 1.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Cy . Draw the FBD of left end of beam.Example A simply supported rectangular 2 x 12 beam is loaded as shown below. file:///Z|/www/Statstr/statics/Stests/beams2/sol413. Unless otherwise indicated. 000 lbs/ft (4 ft) + 3.3 psi Select: Topic 4: Beams . The section modulus is available from the Beam Tables. 3. This is a 2 x 12 beam.330 lbs (4 ft) + M8 = 0 Solving: V = 667 lbs.Table of Contents Statics & Strength of Materials Home Page 2 4 file:///Z|/www/Statstr/statics/Stests/beams2/sol413.) Resolve all forces into x/y directions.667 lbs/in 3 2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 8 ft from the end of the beam and 3 inches above the bottom of the beam.5 in)]/[(288 in )(2 in)] = 31. MBS = My/I (Where M8' is the bending moment at 8 ft.Solution413 2.670 ft-lbs 8 8 STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 8'.330 lbs .htm (2 of 2) [5/8/2009 4:40:16 PM] .) MBS = -10. apply the horizontal shear stress formula. M = 10.670 ft-lbs(12 in/ft)(6 in) /[(1/12)(2 in)(12 in) ] = 2. and S is the section modulus for the beam. The form we will use is: HSS = Vay'/Ib Where: V = Shear force 8 ft from the end of the beam a = cross sectional area from 3 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (288 in4 for 2 x 12 beam) b = width of beam a 3 in above the bottom of the beam HSS = [(667 lbs)(6 in )(4.V8 = 0 Sum TA = -1.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = -1.000 lbs/ft (4 ft) (2 ft) + 3. 000 lbs = 0 Sum TA = -5.5.Solution414 STATICS & STRENGTH OF MATERIALS . Determine the horizontal shear stress at a point 6 inches above the bottom of the beam cross section and 4 feet from the left end of the beam.Example A loaded.000 lbs (12 ft) .000 lbs/ft (8 ft) . file:///Z|/www/Statstr/statics/Stests/beams2/sol414. cantilever 2 x 10 rectangular beam is shown below. M y ext = 156. 1. Determine the maximum bending stress 4 feet from the left end of the beam.000 ft-lbs STEP 2: Determine the shear force and bending moment at x=4 ft. all joints and support points are assumed to be pinned or hinged joints. showing and labeling all external forces. Draw the FBD of left end of beam.) FBD of structure (See Diagram) 2.000 lbs/ft (8 ft) (12 ft) + Mext = 0 Solving: A = 13.1.) Resolve all forces into x/y components 3.htm (1 of 2) [5/8/2009 4:40:16 PM] . B.000 lbs. Solution: Part A: STEP 1: Determine the external support reactions: 1. For this beam: A.1.) Apply equilibrium conditions: Sum Fx = Ax = 0 Sum Fy = Ay .) Cut beam at 4 ft. Unless otherwise indicated. ) Resolve all forces into x/y directions.000 ft-lbs 4 4 STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 4'. This is a 2 x 10 beam.000 ft-lbs(12 in/ft)(5 in) /[(1/12)(2 in)(10 in) ] = 37.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = 13.400 lbs/in 3 2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 4 ft from the end of the beam and 6 inches above the bottom of the beam. The section modulus is available from the Beam Tables.Table of Contents Statics & Strength of Materials Home Page 2 4 file:///Z|/www/Statstr/statics/Stests/beams2/sol414.htm (2 of 2) [5/8/2009 4:40:16 PM] . and S is the section modulus for the beam.000 lbs)(12 in )(2 in)]/[(167 in )(2 in)] = 934 psi Select: Topic 4: Beams . MBS = My/I (Where M4' is the bending moment at 4 ft.V4 = 0 Sum TA = -V4(4 ft) +156. apply the horizontal shear stress formula.000 lbs . The form we will use is: HSS = Vay'/Ib Where: V = Shear force 4 ft from the end of the beam a = cross sectional area from 6 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (167 in4 for 2 x 10 beam) b = width of beam a 6 in above the bottom of the beam HSS = [(13.) MBS = 104. M = 104.Solution414 2.000 lbs. 3.000 ft-lbs + M4 = 0 Solving: V = 13. ) Apply equilibrium conditions: Sum Fx = Ax = 0 Sum Fy = Ay . M y ext = 58.1. file:///Z|/www/Statstr/statics/Stests/beams2/sol415. cantilever WT 8 x 29 beam is shown below.000 lbs.000 lbs/ft (6 ft) .000 lbs/ft (2 ft) = 0 Sum TA = 2.000 lbs/ft (6 ft) (3 ft) .Solution415 STATICS & STRENGTH OF MATERIALS .) Resolve all forces into x/y components 3.htm (1 of 2) [5/8/2009 4:40:17 PM] . For this beam: A.) Cut beam at 5 ft. Determine the maximum bending stress 5 feet from the left end of the beam. Draw the FBD of left end of beam. 1.2. Solution: Part A: STEP 1: Determine the external support reactions: 1.Example A loaded.) FBD of structure (See Diagram) 2. Determine the horizontal shear stress at a point 6 inches above the bottom of the beam cross section and 5 feet from the left end of the beam. Unless otherwise indicated. all joints and support points are assumed to be pinned or hinged joints.000 ft-lbs STEP 2: Determine the shear force and bending moment at x=5 ft. B. showing and labeling all external forces.000 lbs/ft (2 ft) (11 ft) + Mext = 0 Solving: A = 14.1. Table of Contents Statics & Strength of Materials Home Page 2 4 file:///Z|/www/Statstr/statics/Stests/beams2/sol415.000 ft-lbs -V5(5 ft) + M5 = 0 Solving: V = 4.Solution415 2.) MBS = 13.19 in)]/[(44 in )(. The form we will use is: HSS = Vay'/Ib Where: V = Shear force 5 ft from the end of the beam a = cross sectional area from 6 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (44 in4 for WT 8 x 29 beam) b = width of beam a 6 in above the bottom of the beam HSS = [(4.900 lbs/in 3 2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 5 ft from the end of the beam and 6 inches above the bottom of the beam.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = 14.000 lbs)(2.000 lbs.5 ft)+ 58. The WT 8 x 29 beam has a section modulus for the beam from the beam tables is 7.000 lbs/ft (5 ft) . MBS = M /S (Where M5' is the bending moment at 5 ft. and S is the section modulus for the beam.000 lbs . M = -13.46 in )(3.000 ft-lbs(12 in/ft)/7. The 5' section modulus is available from the Beam Tables. apply the horizontal shear stress formula.000 ft-lbs 5 5 STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 5'.41 in)] = 7.2.) Resolve all forces into x/y directions.000 lbs/ft (5 ft) (2.V5 = 0 Sum TA = -2.740 psi Select: Topic 4: Beams .0 in3.0 in = 21. 3.htm (2 of 2) [5/8/2009 4:40:17 PM] . B. Determine the horizontal shear stress at a point 6 inches above the bottom of the beam cross section and 10 feet from the left end of the beam.htm (1 of 2) [5/8/2009 4:40:17 PM] .000 lbs = 0 Sum TB = 1.000 lbs STEP 2: Determine the shear force and bending moment at x=10 ft.) Resolve all forces into x/y components 3.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .6.) FBD of structure (See Diagram) 2.Example A loaded.000 lbs/ft (8 ft) . Unless otherwise indicated. simply supported W 8 x 28 beam is shown below.4. For this beam: A.000 lbs.6.000 lbs (4 ft) + Dy (8 ft) = 0 Solving: By = 23.Solution416 STATICS & STRENGTH OF MATERIALS . all joints and support points are assumed to be pinned or hinged joints.000 lbs .000 lbs/ft (8 ft) (4 ft) + 4. Solution: Part A: STEP 1: Determine the external support reactions: 1. file:///Z|/www/Statstr/statics/Stests/beams2/sol416.000 lbs (8 ft) . Determine the maximum bending stress 10 feet from the left end of the beam. Dy = -5.1. apply the horizontal shear stress formula. The section modulus is available from the Beam Tables. MBS = M10'/S (Where M10' is the bending moment at 10 ft.3 in3 = 20.29 in)] = 4. 2. 3.) Resolve all forces into x/y directions.htm (2 of 2) [5/8/2009 4:40:17 PM] .) Cut beam at 10 ft.000 ft-lbs(12 in/ft)/24.000 ft-lbs STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 10'.Solution416 1.000 lbs .8 in4)(.000 lbs (4 ft) + 23.3 in3.8 in4 for W 8 x 28 beam) b = width of beam a 6 in above the bottom of the beam HSS = [(11. Draw the FBD of left end of beam. M10 = -42.740 lbs/in2 Part B: STEP 4: To determine the Horizontal Shear Stress (HSS) at 10 ft from the end of the beam and 6 inches above the bottom of the beam.8. The form we will use is: HSS = Vay'/Ib Where: V = Shear force 10 ft from the end of the beam a = cross sectional area from 6 in above the bottom of the beam to bottom of beam y' = distance from neutral axis to the centroid of area a I = moment of inertia of the beam (97.V10 = 0 Sum TA = -8.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams2/sol416. The W 8 x 28 beam has a section modulus for the beam from the beam tables is 24.950 psi Select: Topic 4: Beams .000 lbs . showing and labeling all external forces.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = -4.000 lbs)(4.62 in2)(2.000 lbs (8 ft) -V10(10 ft) + M10 = 0 Solving: V10 = 11. and S is the section modulus for the beam.000 lbs.) MBS = 42.76 in)]/[(97.000 lbs + 23. 000 lbs (4 ft) . 1.500 lbs. For this beam and loading we would like to select the best I-Beam to use (from a selection of IBeams.10: Beams -Beam Selection Another very important use of the flexure formula is in Beam Selection.htm (1 of 4) [5/8/2009 4:40:18 PM] .2.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .Bending Stress Example II).2. Dy = 3.000 lbs/ft (4 ft) .000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0 Solving: By = 9.) Draw Free Body Diagram of structure (See Diagram 2) 2.Topic 4.000 lbs = 0 Sum TB = 5. how does one decide on the best (safe and least expensive) beam to use with a particular loading. Perhaps the best way to explain this process is to work carefully through an example of the procedure.10: Beams .5. In Diagram 1 we have shown a loaded beam (same loading as in Beams II .Beam Selection Topic 4.) Resolve all forces into x & y components 3. which we will discuss shortly) STEP 1: Apply Static Equilibrium Principles and determine the external support reactions.500 lbs file:///Z|/www/Statstr/statics/Beams/bdsn410. That is. and solve for the value of the section modulus. = Mmax / S. and determine the values of the maximum bending moment and maximum shear force. compressive. and Vmax = -5000 lb. For this example we will use the following allowable stresses for the beam material: . and then S = 240. If you need more detail please see that example. and shear) for the beam material. This value for the section modulus is the smallest value possible if the maximum bending stress is not to file:///Z|/www/Statstr/statics/Beams/bdsn410.Example 2. Placing values into the equation we have: 20.)(12 in.)/ S.000 lb/in2 = 12 in3 .000 in-lb.10: Beams ..8a: Bending Stress . and use the lowest allowable axial stress for the maximum bending stress.Beam Selection STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.000 lb/in2 = (20.000 ft-lb.) From the Diagrams we observe that Mmax = -20. We now use the flexure formula form: . By material specification we mean the allowable stresses (tensile.htm (2 of 4) [5/8/2009 4:40:18 PM] . Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load.000 ft-lb. The graphs are shown in Diagram 3 and Diagram 4. This information is normally furnished by the beam supplier with their selection of beams./ft. (We determine these diagrams in some detail in Topic 4.Topic 4./ 20. 40 12.0 17.027 8.Table of Contents file:///Z|/www/Statstr/statics/Beams/bdsn410. Designation Area Depth Width W 6x25 W 6x20 W 6x15. Please Select Topic 4.01 Step 4.12 d in 6.933 0.00 9.37 6.00 42.83 21.10: Beams .00 8.56 5. Select: Topic 4: Beams .558 0.493 0.367 0.71 3.53 3.230 0.47 y-y axis y-y axis S in3 5.00 Section Info.0 110.315 0.43 3.22 7.250 8.575 0.0 Section Info.88 4. Flange Flange Web thick tf in 0.258 0.1 69. We now look at an example selecting the best T-Beam to use for a particular loading.25 8.53 1. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.67 9. that the maximum horizontal shear stress for the selected beam is within the allowable shear stress for the beam material.30 9.433 thick tw in 0. We see that this value is well within the allowable shear stress of 16.66 2.0 146.80 10.00 bf in 6.10 13.7 13.077 8. Thus we have selected the best beam to use from the given list of possible beam.69 2. The last step is to now check that the beam we have selected is also safe with respect to the horizontal shear stress that is.Beam Selection exceed the allowable axial stress for the beam material.6 272.20 6.Topic 4.080 6. After examining the selections.25 1.320 0.43 3.248 0.288 Cross x-x axis I in4 53.3 41.12 8.5 30.50 3.89 5.230 x 7.4 56.1 60.4 r in 2.htm (3 of 4) [5/8/2009 4:40:18 PM] .269 0. Shown below is a selection of beams.30 9.4 10. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam).03 2.57 3.365 0.287 8.235 0.60 9. We therefore now apply our formula for the maximum 2 horizontal shear stress in an I-Beam: max = Vmax/Aweb = 5000 lb/ (. Cross y-y axis I in4 17.378 0.24 r in 1.0 14.384) = 2944 lb/in .01 19.51 1. We would like to now selected the best beam based on the minimum value of the section modulus determined above.60 49.018 5.04 2.308 0.23 3.50 37. we determine W 8 x 17 is the best beam from the selection listed.1 27.995 5.4 35.50 2.14 8.000 lb/in2 given above.10 10.0 126.268 5.456 0.70 11.5 31.000 x-x axis x-x axis S in3 16. It has a section modulus of 14.36 3.62 4. and a weight of 17 lb/ ft.00 8.5 W 8x20 W 8x17 W 8x67 W 8x40 W 8x35 W 8x31 A in2 7.22 2.12 2.44 88.35 5.46 1.1 in3 (greater than the minimum section modulus of 12 in3).10a: Beam Selection Example 1 When finished with Example of Beam Selection. htm (4 of 4) [5/8/2009 4:40:18 PM] .10: Beams .Topic 4.Beam Selection Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsn410. ) FBD of structure (See Diagram 1) 2.500 lb.000 ft-lb./ft.htm (1 of 3) [5/8/2009 4:40:18 PM] .Topic 4./ft. The maximum allowable bending stress = 35.) Apply equilibrium conditions: Sum Fx =Ax = 0 Sum Fy = Ay . and determine the values of the maximum bending moment and maximum shear force. cantilever beam is shown in Diagram 1. STEP 1: Apply Static Equilibrium Principles and determine the external support reactions: 1.000 lb.10a: Beam Selection .200 lb.)(4 ft) = 0 Sum TB = -(1.) + Mext = 0 Solving: Ay = 14.)(6 ft) .)-(800 lb./ft.(800 lb. (both in tension and compression).) Resolve all forces into x/y components 3./ft.)(4 ft)(10 ft.)(6 ft)(3 ft.)(2 ft) .000 lb.Example 1 A loaded./in2.)-(1.(1.)(2 ft)(7 ft. and the maximum allowable shear stress = 15. Mext = 73. We would like to choose the best T-beam to use from the T-Beam selection shown at the lower part of this page. (See Diagrams 2 and 3.500 lb.000 lb./ft.. STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.) file:///Z|/www/Statstr/statics/Beams/bdsne410a.(1.000 lb/in2 for the beam material used./ft.10a: Beam Selection Example 1 Topic 4. 000 in-lb./ 35.000 r in 1.)/ S. and solve for the value of the section modulus.283 Stem thick tw in 1.200 lb.10a: Beam Selection Example 1 From the Diagrams we observe that Mmax = -73. - x-x axis x-x axis x-x axis x-x axis I in4 179. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. and then S = 876. Designation WT 7 x157 Area A in2 46. It has a section modulus of 30.07 Cross Section Info. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam). and shear) for the beam material./ft.htm (2 of 3) [5/8/2009 4:40:18 PM] . Placing values into the equation we have: 35./ft.1 in3 (greater than the minimum section modulus of 25 in3).000 ft-lb.235 Flange thick tf in 2. We would like to now selected the best beam based on the minimum value of the section modulus determined above.000 psi. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load. compressive. and Vmax = 14.60 Flange Width bf in 16.970 y in 1. and use the lowest allowable axial stress for the maximum bending stress.5 is the best beam from the selection listed. This information is normally furnished by the beam supplier with their selection of beams. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.980 file:///Z|/www/Statstr/statics/Beams/bdsne410a.20 Depth of T d in 8. After examining the selections. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 35. Shown below is a selection of beams.415 d/tw 6. and Maximum Allowable Shear Stress = 15. we determine WT 15 x 49. and a weight of 49.000 lb/in2 = 25 in3 .000 lb/in2 = (73.000 ft-lb. We now use the flexure formula form: =Mmax / S.000 psi.00 S in3 27.)(12 in. By material specification we mean the allowable stresses (tensile.Topic 4..5 lb.. 00 372.35 15. a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam.10 15.40 19.40 18.458 2.00 139.200 lb.500 19.093 1.400 35.5x51 WT 13.36 in)/ (323 in4)(.5 WT 7x132 WT 7x123 WT 13.100 30.10 27.150 4.82 16.551 10.990 9.54 13. (from the shear force diagram) I = moment of inertia of cross section.930 0.Topic 4.80 36.15 15.025 15.410 3.020 4.00 395.50 28.463 0. Maximum Horizontal Shear Stress = Vay'/Ib = (14.130 16.710 1.300 33.200 lb)*(5.650 4.000 37.08 15.518 0.20 16.490 0.42 6.300 25.41 8.100 1.484 10.410 3.6 in2. y' = 5.20 38.90 14.100 21.380 3.670 4.36 in.522 in.10a: Beam Selection Example 1 WT 7x143.522 in) = 2530 lb/in2 We see that this value is well within the allowable shear stress of 15.900 3.5x47 WT 13.00 289.5 42.150 4.890 3.125 0.650 4.636 1.690 4.670 1.890 1.930 4.522" * 10. y' = distance from neutral axis to the centroid of the area "a" .600 32.6 in2)*(5. b = width of beam where we wish to find shear stress (neutral axis for maximum) from table.548 0.180 4. I = 323 in4.00 258.000 0.500 10.60 27.310 1.140 4.945 10.00 239.60 8.585 0.00 323.000 lb/in2 given above.564 0.522 6. from beam table.90 26.00 126. Return to: Topic 4.13 13.813 0.60 25. Thus we have selected the best beam to use from the given list of possible beam.205 1.100 Step 4.46 13. b = .72" )= 5.25 8.521 10.850 0. We therefore now apply our formula for the horizontal shear stress for the WT 15 x 49.930 1.5 x42 WT 15x66 WT 15x62 WT 15x58 WT 15x54 WT 15x49.00 216.80 12. that the maximum horizontal shear stress for the selected beam is within the allowable shear stress for the beam material.5x57 WT 13.85 7.760 0.5 T-beam: Shear Stress = Vay'/Ib The maximum shear stress occurs at the neutral axis of the beam: V = maximum shear force = 14.860 4.00 14.htm (3 of 3) [5/8/2009 4:40:18 PM] .80 24.800 22.780 1.20 28.827 0.80 26.00 13.00 350.615 0.018 9.64 13.00 24.Beam Selection or Select: Topic 4: Beams .500 3.870 1.710 3.400 23.22 23. The last step is to now check that the beam we have selected is also safe with respect to the horizontal shear stress that is.932 0.00 421.20 17.91 14.938 1. Then a = (.80 15.963 10.570 0.600 28.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsne410a.10: Beams . In this case we will go to bottom of beam.747 0.070 10.40 157. 000 lbs (4 ft) + Cy (8 ft) . as well as any needed angles and dimensions.htm (1 of 3) [5/8/2009 4:40:19 PM] . all joints and support points are assumed to be pinned or hinged joints. 1.500 lbs y y Step 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.000 lbs/ft (8 ft)(12 ft)= 0 Solving: A = -1500 lbs. and determine the values of the maximum bending moment and maximum shear force. Solution: Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.000 lb/in2. The maximum allowable shear stress = 12.000 lbs .1. simply supported beam is shown below. (See Diagrams 2 and 3.) file:///Z|/www/Statstr/statics/Stests/beams2/sol421.1. Unless otherwise indicated.) FBD of structure (See Diagram) 2.Solution421 STATICS & STRENGTH OF MATERIALS .) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = Ay + Cy .Example A loaded.000 lb/in2.000 lbs/ft (8 ft)= 0 Sum TA = -5. The maximum allowable bending stress = 30. For this beam: Select the best T-beam to use if: 1). C = 14.) Resolve all forces into x/y components 3. 2).5. After examining the selections. We now use the flexure formula form: M / S.000 in-lb. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.8 in3).000 lb/in = 12.. and a weight of 34 lb. Shown below is a selection of beams. It has a section modulus of 15.6 in3 (greater than the minimum section modulus of 12.htm (2 of 3) [5/8/2009 4:40:19 PM] . and then S = 384. We would like to now selected the best beam based on the minimum value of the section modulus determined above. and solve for the value of the section modulus.000 ft-lb. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. and use the lowest allowable axial stress for max the maximum bending stress.8 in . Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 30.000 ft-lb. and Maximum Allowable Shear Stress = 12.Solution421 From the Diagrams we observe that M max = -32. Placing values into the equation we have: 30./ 30. compressive.000 psi.000 lb./ft. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam). we determine WT 12 x 34 is the best beam from the selection listed. By material specification we mean the allowable stresses (tensile.)/ S.000 lb/in = (32.000 psi.)(12 in. and V max = 8./ft. 2 2 3 T-Beam Data file:///Z|/www/Statstr/statics/Stests/beams2/sol421. This information is normally furnished by the beam supplier with their selection of beams. Step 4.. and shear) for the beam material. )(8. So the beam is safe and is our best choice of the beams given to choose from.86 8.600 r in 3.416 28.42") / (137 in * .961 0.00 11.700 y in 3. t = Vay' / Ib = (8000 lb. Select: Topic 4: Beams .42") = 2280 lb/in which is lower than the allowable shear stress of 12.htm (3 of 3) [5/8/2009 4:40:19 PM] .50 137.Table of Contents Statics & Strength of Materials Home Page 4 2 file:///Z|/www/Statstr/statics/Stests/beams2/sol421.00 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress. - Designation Area WT 12x34 A in2 x-x axis x-x axis x-x axis x-x axis I in4 S in3 15.000 lb/in2.Solution421 - - Depth Flange Flange Stem of T d in Width bf in thick tf in thick tw in d/tw - Cross Section Info.070 10.83" * .582 0.42")(4. 200 lbs/ft (8 ft)= 0 Sum TA = -800 lbs (8 ft)(4 ft) + Cy (12 ft) . Unless otherwise indicated.) FBD of structure (See Diagram) 2.1. as well as any needed angles and dimensions.1. The maximum allowable bending stress = 35.) file:///Z|/www/Statstr/statics/Stests/beams2/sol422.Solution422 STATICS & STRENGTH OF MATERIALS .800 lbs/ft (8 ft) . 1. all joints and support points are assumed to be pinned or hinged joints.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = Ay + Cy . simply supported beam is shown below.200 lbs/ft (8 ft)(12 ft)= 0 Solving: A = 4270 lbs. (See Diagrams 2 and 3. The maximum allowable shear stress = 10.000 lb/in2. C = 11.000 lb/in2.htm (1 of 3) [5/8/2009 4:40:19 PM] . and determine the values of the maximum bending moment and maximum shear force.) Resolve all forces into x/y components 3. Solution: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. For this beam: Select the best I-beam to use if: 1).730 lbs y y STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam. 2).Example A loaded. )(12 in.400 ft-lb.Solution422 From the Diagrams we observe that M max = 11. We now use the flexure formula form: M / S.000 lb/in = 3. and solve for the value of the section modulus.800 in-lb. and use the lowest allowable axial stress for max the maximum bending stress. compressive.000 psi. and then S = 136./ 35. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive file:///Z|/www/Statstr/statics/Stests/beams2/sol422.)/ S.000 lb/in = (11. Shown below is a selection of beams. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load./ft. and V max = 6. and shear) for the beam material.000 psi. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 35. Placing values into the equation we have: 35.htm (2 of 3) [5/8/2009 4:40:19 PM] 2 2 3 .400 ft-lb..91 in . This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. By material specification we mean the allowable stresses (tensile. and Maximum Allowable Shear Stress = 10. We would like to now selected the best beam based on the minimum value of the section modulus determined above. This information is normally furnished by the beam supplier with their selection of beams..930 lb. 170 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress.)= 7.83 thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I in4 14. Step 4.5 A in2 2.1 r in 2.89 0.940 tw in Cross Section Info.5 is the best beam from the selection listed. * .442 in. Cross Section Info.Solution422 beam).8 S in3 5.1 in3 (greater than the minimum section modulus of 3. Designation Area Depth Width W 6x8. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below./ft.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams2/sol422.01 r in 0.5 lb. we determine W 6 x 8.000 lb/in2.98 S in3 1.194 0. and a weight of 8.51 d in 5. I-Beam Data Flange Flange Web thick tf in bf in 3.91 in3). So the beam is safe and is our best choice of the beams given to choose from. t max =V max /a web = 6930 lb / (5.43 I in4 1.htm (3 of 3) [5/8/2009 4:40:19 PM] . It has a section modulus of 5. After examining the selections.490 lb/in which is lower than the allowable 2 shear stress of 12.170 in. Select: Topic 4: Beams . The maximum allowable shear stress = 10.Solution423 STATICS & STRENGTH OF MATERIALS . Solution: Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.000 lbs .000 lbs/ft (4 ft)= 0 Solving: B = 23. simply supported beam is shown below.000 lbs y y Step 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.) file:///Z|/www/Statstr/statics/Stests/beams2/sol423. (See Diagrams 2 and 3. 1.) FBD of structure (See Diagram) 2. 2).000 lbs (16 ft) + 1.000 lbs.000 lb/in2.4. as well as any needed angles and dimensions.) Resolve all forces into x/y components 3.000 lbs/ft (8 ft) .000 lbs/ft (8 ft)(12 ft) . Unless otherwise indicated.6.Example A loaded.5. and determine the values of the maximum bending moment and maximum shear force.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .By (8 ft) + 6. D = . For this beam: Select the best T-beam to use if: 1).1.000 lb/in2.htm (1 of 3) [5/8/2009 4:40:19 PM] . all joints and support points are assumed to be pinned or hinged joints. The maximum allowable bending stress = 20.000 lbs = 0 Sum TD = 4. 5 x 65 is the best beam from the selection listed.000 in-lb./ft. We would like to now selected the best beam based on the minimum value of the section modulus determined above.4 in3)./ 20. It has a section modulus of 42. and Maximum Allowable Shear Stress = 10.4 in . Shown below is a selection of beams. 2 2 3 T-Beam Data file:///Z|/www/Statstr/statics/Stests/beams2/sol423.000 lb. and shear) for the beam material.Solution423 From the Diagrams we observe that M max = -64.2 in3 (greater than the minimum section modulus of 38. and then S = 768. By material specification we mean the allowable stresses (tensile. and solve for the value of the section modulus. and use the lowest allowable axial stress for max the maximum bending stress. compressive. This information is normally furnished by the beam supplier with their selection of beams.. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.)/ S.000 psi. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load. we determine W T 16.)(12 in.000 lb/in = (64. We now use the flexure formula form: M / S.000 lb/in = 38.000 ft-lb. and V max = 12. Step 4. After examining the selections.000 ft-lb. Placing values into the equation we have: 20. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam)./ft..htm (2 of 3) [5/8/2009 4:40:19 PM] .000 psi. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 20. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. and a weight of 65 lb. 180 y in 4.Table of Contents Statics & Strength of Materials Home Page 4 2 file:///Z|/www/Statstr/statics/Stests/beams2/sol423.370 WT 16.Solution423 - - Depth Flange Flange Stem of T d in Width bf in thick tf in thick tw in d/tw - Cross Section Info.580 28. t = Vay' / Ib = (12.58")(6.50 514.000 lb/in2.5x65 19.09") / ( 514 in * .200 r in 5.510 0.732 lb/in which is lower than the allowable shear stress of 10.58") = 1.855 0.18" * .)(12. Select: Topic 4: Beams . So the beam is safe and is our best choice of the beams given to choose from.00 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress. - Designation Area A in2 x-x axis x-x axis x-x axis x-x axis I in4 S in3 42.000 lb.htm (3 of 3) [5/8/2009 4:40:19 PM] .20 16.55 11. and determine the file:///Z|/www/Statstr/statics/Stests/beams2/sol424.000 ft-lbs Step 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = Ay .000 lbs (3 ft) .000 lb/in2. cantilever beam is shown below. 1. all joints and support points are assumed to be pinned or hinged joints.000 lbs .000 lbs (8 ft) 4.Example A loaded. M y ext = 107.9.5.000 lb/ft (4 ft) = 0 Sum TA = Mext . Unless otherwise indicated.1. The maximum allowable bending stress = 25.000 lb/in2.htm (1 of 3) [5/8/2009 4:40:20 PM] .1. 2).5.000 lbs (10 ft)= 0 Solving: A = 18.) Resolve all forces into x/y components 3. The maximum allowable shear stress = 12.) FBD of structure (See Diagram) 2.000 lbs.500 lb/ft (6 ft) . Solution: Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. For this beam: Select the best I-beam to use if: 1).Solution424 STATICS & STRENGTH OF MATERIALS . as well as any needed angles and dimensions. / 25.000 lb/in = 51. compressive.) From the Diagrams we observe that M max = -107. Shown below is a selection of beams.4 in .Solution424 values of the maximum bending moment and maximum shear force. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below. Step 4. and a weight of 36 lb. and use the lowest allowable axial stress for max the maximum bending stress.4 in3). and Maximum Allowable Shear Stress = 12. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 25.000 lb.284.htm (2 of 3) [5/8/2009 4:40:20 PM] .5 in3 (greater than the minimum section modulus of 51./ft. Placing values into the equation we have: 25. and V max = 18. After examining the selections./ft. By material specification we mean the allowable stresses (tensile. We would like to now selected the best beam based on the minimum value of the section modulus determined above. 2 2 3 I-Beam Data file:///Z|/www/Statstr/statics/Stests/beams2/sol424.000 lb/in = (107.000 in-lb. and solve for the value of the section modulus.000 ft-lb.)(12 in. (See Diagrams 2 and 3.000 psi. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load.)/ S. This information is normally furnished by the beam supplier with their selection of beams.000 ft-lb. We now use the flexure formula form: M / S. and shear) for the beam material. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam)..000 psi. we determine W 14 x 36 is the best beam from the selection listed. It has a section modulus of 56. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. and then S = 1.. * . Select: Topic 4: Beams .99 in.0 S in3 56. Cross Section Info.000 lb/in2. So the beam is safe and is our best choice of the beams given to choose from.40 S in3 6.000 lb / (14.60 15.299 in.htm (3 of 3) [5/8/2009 4:40:20 PM] . t max =V max /a web = 18. Designation Area Depth Width W 14x36 A in2 d in thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I in4 447.) = 4.299 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress.52 10.50 I in4 24.Solution424 - - - Flange Flange Web thick tf in bf in tw in Cross Section Info.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams2/sol424.85 6.99 r in 1.020 lb/in which is lower than the allowable 2 shear stress of 12.5 r in 6.428 0.992 0. M y ext = 73.Solution425 STATICS & STRENGTH OF MATERIALS .) file:///Z|/www/Statstr/statics/Stests/beams2/sol425.) Resolve all forces into x/y components 3.9. 1.500 lb/ft (6 ft) .3.000 lb/in2. and determine the values of the maximum bending moment and maximum shear force. all joints and support points are assumed to be pinned or hinged joints. For this beam: Select the best T-beam to use if: 1). The maximum allowable shear stress = 15.2.000 lbs/ft (2 ft) 800 lb/ft (4 ft) = 0 Sum TA = Mext .1.000 lb/in2. as well as any needed angles and dimensions.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = Ay . (See Diagrams 2 and 3. 2). Solution: Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces. Unless otherwise indicated. The maximum allowable bending stress = 35. cantilever beam is shown below.Example A loaded.200 lbs (10 ft)= 0 Solving: A = 14.1.000 lbs (7 ft) .000 lbs (3 ft) .htm (1 of 3) [5/8/2009 4:40:20 PM] .000 ft-lbs Step 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.200 lbs.) FBD of structure (See Diagram) 2. compressive.000 in-lb. and then S = 876. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 35.)(12 in. and Maximum Allowable Shear Stress = 15. We would like to now selected the best beam based on the minimum value of the section modulus determined above. and solve for the value of the section modulus.000 lb/in = 25 in . By material specification we mean the allowable stresses (tensile. and use the lowest allowable axial stress for max the maximum bending stress. Shown below is a selection of beams.)/ S. This information is normally furnished by the beam supplier with their selection of beams.000 ft-lb..Solution425 From the Diagrams we observe that M max = -73. and V max = 14. and shear) for the beam material.000 psi.htm (2 of 3) [5/8/2009 4:40:20 PM] 2 2 3 .000 psi. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. Placing values into the equation we have: 35. We now use the flexure formula form: M / S. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load.200 lb. We select the beam but find the one with a section modulus equal or greater than the minimum file:///Z|/www/Statstr/statics/Stests/beams2/sol425./ 35./ft.000 lb/in = (73.000 ft-lb.. 140 y in 3. So the beam is safe and is our best choice of the beams given to choose from. It has a section modulus of 25. Select: Topic 4: Beams .518 26. After examining the selections.10 258.000 lb/in2.518")(5.16" * .4 in3 (greater than the minimum section modulus of 25 in3).)(10.08") / ( 258 in * . Step 4.827 0.518") = 2./ft.00 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress.54 10.5x51 15.380 WT 13. we determine WT 13. t = Vay' / Ib = (14. - Designation Area A in2 x-x axis x-x axis x-x axis x-x axis I in4 S in3 25.Solution425 section modulus and with the least pounds per foot weight (which normally means the least expensive beam).480 lb/in which is lower than the allowable shear stress of 15. and a weight of 51 lb.200 lb.018 0.00 13.htm (3 of 3) [5/8/2009 4:40:20 PM] .Table of Contents Statics & Strength of Materials Home Page 4 2 file:///Z|/www/Statstr/statics/Stests/beams2/sol425. T-Beam Data Depth Flange Flange Stem of T d in Width bf in thick tf in thick tw in d/tw Cross Section Info.400 r in 4.5 x 51 is the best beam from the selection listed. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below. 3.000 lbs (11 ft) .000 lb/in2. all joints and support points are assumed to be pinned or hinged joints.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = Ay .3.4.) file:///Z|/www/Statstr/statics/Stests/beams2/sol426.000 lbs .Solution426 STATICS & STRENGTH OF MATERIALS .000 lbs (4 ft) . and determine the values of the maximum bending moment and maximum shear force.000 lbs (8 ft) .Example A loaded. 2).000 lbs/ft (6 ft) .htm (1 of 3) [5/8/2009 4:40:21 PM] .000 ft-lbs y ext Step 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam. The maximum allowable shear stress = 15. The maximum allowable bending stress = 40.4. (See Diagrams 2 and 3.000 lbs .000 lb/in2. 1.12. cantilever beam is shown below.2. Unless otherwise indicated.) Resolve all forces into x/y components 3. as well as any needed angles and dimensions.2000 lb (14 ft) = 0 Solving: A = 21. For this beam: Select the best I-beam to use if: 1).2000 lbs = 0 Sum TA = Mext . Solution: Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces.) FBD of structure (See Diagram) 2. M = 200.000 lbs. This information is normally furnished by the beam supplier with their selection of beams. This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material.htm (2 of 3) [5/8/2009 4:40:21 PM] .6 in3 (greater than the minimum section modulus of 60 in3). Placing values into the equation we have: 40.400.000 lb/in = 60 in . we determine W 14 x 40 is the best beam from the selection listed.000 psi.)(12 in.000 ft-lb.000 psi. and shear) for the beam material.)/ S. and use the lowest allowable axial stress for max the maximum bending stress. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 40.000 in-lb. Step 4.000 ft-lb. We select the beam but find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam).. and V max = 21.Solution426 From the Diagrams we observe that M max = -200./ft. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.. By material specification we mean the allowable stresses (tensile. Shown below is a selection of beams./ 40.000 lb. and then S = 2. It has a section modulus of 64. We now use the flexure formula form: M / S.000 lb/in = (200./ft. and a weight of 40 lb. 2 2 3 I-Beam Data file:///Z|/www/Statstr/statics/Stests/beams2/sol426. compressive. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load. We would like to now selected the best beam based on the minimum value of the section modulus determined above. and solve for the value of the section modulus. and Maximum Allowable Shear Stress = 15. After examining the selections. 307 in.0 S in3 64.6 r in 6.htm (3 of 3) [5/8/2009 4:40:21 PM] .000 0.62 I in4 28.80 S in3 8. So the beam is safe and is our best choice of the beams given to choose from.94 in.000 lb / (14.56 11.Solution426 - - - Flange Flange Web thick tf in bf in tw in Cross Section Info.580 lb/in which is lower than the allowable 2 shear stress of 15. Cross Section Info.503 0.307 Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress.23 r in 1.80 16.000 lb/in2. * . Designation Area Depth Width W 14x40 A in2 d in thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I in4 517. t max =V max /a web = 21.00 7.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/beams2/sol426.) = 4. Select: Topic 4: Beams . Bending Moment. Determine the maximum bending moment that can be applied to a rectangular 6"x10" beam if the allowable stress in the member is 1200 psi. is to support the load. Also determine the moment of inertia and section modulus of a beam made of a 3 inch diameter pipe with 1/4 inch walls. Moment. (1386 in4.Topic 4. 12" high.833 in4..Problem Assignment 1 .9") 6. If the allowable axial stress in 2000 psi. Stress 1. Determine the moment of inertia and section modulus of a beam made of a solid rod 3 inches in diameter. A wooden cantilever beam has a maximum bending moment of 3600 foot pounds. 1. A rectangular wood beam. what is the minimum width of the beam needed to support the load? (.98 in4.65 in3.Beam Inertia. The maximum bending moment in a rectangular 2"x 8".Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp411a. Stress Statics & Strength of Materials Topic 4. (See Image) Calculate the moment of inertia and section modulus of the beam. 2.Inertia.000 in-lb = 10. 1. (120. Determine the moment of inertia and section modulus of the 1"x10".11a: Probs .000 ft-lb) 5. 2. (647 psi) 4.06 in4. Determine the maximum bending stress in the beam.66 in3) 3.htm (1 of 2) [5/8/2009 4:40:21 PM] . Asquare box beam is made of four full size 2"x10"s. 12 foot long beam is 1150 foot pounds.11a . (.38in3) Select: Topic 4: Beams . (3. 231 in3) 2. A book shelf is to be made of a clear pine board 1"x10". Stress file:///Z|/www/Statstr/statics/Beams/bdsnp411a.htm (2 of 2) [5/8/2009 4:40:21 PM] .11a: Probs .Topic 4. Moment.Inertia. (5630 lb. The beams are designed to carry a uniformly distributed load of 5000 pounds per foot and are 72 feet long.) 4.(1620 in3) 6. Select the appropriate wood beam to carry this load.260 lb. (6480 in3) Select: Topic 4: Beams .11b: Probs . The horizontal concrete beams are supported at the quarter points. Find the maximum point load that a 14 foot long steel W24x84 can carry at its midpoint without exceeding the allowable bending stress. Determine the minimum section modulus for the beams.4 in3. A parking ramp floor is supported by horizontal concrete beams. The beam is simply supported at its ends.000 psi.11b . for concrete. The beam is simply supported at its ends. for wood. (83. The beam is simply supported at its ends. Find the maximum point load that a 30 foot long steel W30x108 can carry at its midpoint without exceeding the allowable bending stress.Bending Stresses For the following problems use a maximum allowable bending stress of 1500 psi.Problem Assignment 2 . 25.) 3. that is 1/4 of the beam length in from each end. for steel and 6000 psi. Find the minimum section modulus for the parking ramp in problem 5 if the beams are supported at their ends.) 2.Bending Stress Statics & Strength of Materials Topic 4. Determine the maximum point load that a the beam can carry at its end without exceeding the allowable bending stress. (117.330 lb. Determine the minimum section modulus needed for a 14 foot beam to carry a uniformly distributed load of 160 pounds per foot.(31.htm [5/8/2009 4:40:21 PM] .Topic 4. 1.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp411b. A W10x21 steel cantilever beam is 8 feet long. 2" x 10" selection may depend on table used) 5. (26. What is the maximum horizontal shear stress in the beam? (65 lb/ft.11c . Determine the maximum horizontal shear stress in the beam. The beam is simply supported at its ends. Find the maximum horizontal shear stress in the beam. (5.htm [5/8/2009 4:40:21 PM] . Determine the depth of beam if it is 2" wide and has maximum horizontal shear stress of 95 psi. It has a point load of 200 pounds at its midpoint.11c: Probs . Determine the maximum uniformly distributed load that can be carried by an 8 long 2"x 10" beam which is simply supported at its ends. What is the maximum horizontal shear stress in the beam? (12.25 psi) 3. The allowable bending stress is 1500 psi.Topic 4. The maximum allowable horizontal shear stress for construction fir is 95 psi. 19.5 psi) Select: Topic 4: Beams .Horizontal Shear Stresses 1.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp411c. A 10 long cantilever 4"x 4" beam has a 150 pound load at its extreme end.Horizontal Shear Stress Statics & Strength of Materials Topic 4.5 psi) 2. 4 and 6 marks. (14. (1013 lb) 6.1 psi) 4. A 8 long rectangular beam has point loads of 500 pounds at the 2. A 14 long rectangular 2"x 10" beam has a distributed load of 50 pounds per foot.Problem Assignment 3 .92") 5. It is simply supported at its ends. Determine the maximum point load that can be suspended from the end of a 4 long 4"x 4" cantilever beam. A 6 long rectangular 2"x 6" beam is simply supported at its ends. and then using the beam designated by the problem: )at the location stated. determine the support reactions. and at x = 10 ft.Bending & Shear Stress Statics & Strength of Materials Topic 4..) 2. For a rectangular 2 in x 8 in beam. file:///Z|/www/Statstr/statics/Beams/bdsnp411d. 3..htm (1 of 3) [5/8/2009 4:40:21 PM] . For W 8 x 40 beam. 2 inches above bottom 1. (2250 psi. 1. For WT 8 x 32 beam. at x = 3 ft. 0 psi.) Determine the maximum bending stress ( 2.11d . find (53. find (8113 psi.) Determine the shear stress ( ) at the location stated..Problem Assignment 4 .630 psi. and at x = 3 ft. find of beam. and at x = 9 ft.) at x = 10 ft.) at x = 9 ft. 4 inches above bottom of beam.Topic 4. 2160 psi.Beam Stresses For the loaded beams shown below.11d: Probs . 3 inches above bottom of beam. 141 psi. . and at x = 9 ft. find psi. 3092 psi.Topic 4. 4 inches above bottom of beam.. 5. For WT 12 x 60 beam. find (23.Bending & Shear Stress 4.(1000 file:///Z|/www/Statstr/statics/Beams/bdsnp411d. and at x = 4 ft.) at x = 4 ft.467 psi. 870 psi. For W 10 x 60 beam.) at x = 9 ft.11d: Probs . 8 inches above bottom of beam.htm (2 of 3) [5/8/2009 4:40:21 PM] . Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp411d.Topic 4.11d: Probs .Bending & Shear Stress Select: Topic 4: Beams .htm (3 of 3) [5/8/2009 4:40:21 PM] . Make Shear Force and Bending Moment diagrams for each beam. For all the beams use: Allowable Tensile Stress = 20.. best I-beam. Determine the support reactions. (Minimum Section Modulus = 36 in3.12a .7 in3. T-beam depends on beam table used) 2.000 psi.000 psi 1 (Minimum Section Modulus = 2. Allowable Compressive Stress = 25. determine the Maximum Shear Stress and see if it is with in the allowable shear stress. C. For the selected Beam.Beam Selection Statics & Strength of Materials Topic 4. B.Topic 4. Allowable Shear Stress = 15. Select the best I-Beam & Best T-Beam to use D.. bottom of beam in tension. T-beam depends on beam table used) file:///Z|/www/Statstr/statics/Beams/bdsnp412a. best I-beam.Beam Selection/Design For the loaded beams shown: A. bottom of beam in tension.htm (1 of 3) [5/8/2009 4:40:22 PM] .000 psi.12a :Probs .Problem Assignment 5 . best I-beam. bottom of beam in tension. = 58. (Minimum Section Modulus = 14. (Minimum Section Modulus = 72 in3 for I-Beam. T-beam depends on beam table used) 5. bottom of beam in compression.Topic 4.12a :Probs . bottom of beam in compression.6 in3 for T-Beam.htm (2 of 3) [5/8/2009 4:40:22 PM] . T-beam depends on beam table used) 4. best I-beam. best I-beam.4 in3. T-beam depends on beam table used) file:///Z|/www/Statstr/statics/Beams/bdsnp412a. (Minimum Section Modulus = 45 in3 for I-Beam. = 36 in3 for T-Beam.Beam Selection 3. htm (3 of 3) [5/8/2009 4:40:22 PM] .Topic 4.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp412a.Beam Selection Select: Topic 4: Beams .12a :Probs . Maximum deflection occurs at the mid point and is: ymax = FL3/(48EI) Beam simply supported at its end points with a uniformly distributed load. values are as shown below: E= I= F= y= L= Modulus of Elasticity Moment of Inertia Total Load Deflection Length of Beam psi in4 lbs inches inches (N/m2) (m4) (N) (m) (m) For steel E = 30x106 psi. The following formulae work only for the specific stated beam configurations.12b: Probs . Maximum deflection occurs at the free end and is: ymax = FL3/(3EI) Cantilever beam with a uniformly distributed load.Problem Assignment 6 . Total load is W. Maximum deflection occurs at the mid point and is: ymax = 5FL3/(384EI) Cantilever beam with a point load at the extreme end. Determine the deflection of the midpoint of a 14 foot long fir 2"x 10" beam which carries a uniformly file:///Z|/www/Statstr/statics/Beams/bdsnp412b.Topic 4. Maximum deflection occurs at the free end and is: ymax = FL3/(8EI) In all of the above formulae.Beam Deflection The maximum deflection of various beams can be given by specific formulae. 1.12b . Beam simply supported at its end points with a point load at its midpoint.76x106 psi.Beam Deflection Statics & Strength of Materials Topic 4.htm (1 of 2) [5/8/2009 4:40:22 PM] . Total load is W. and for Douglas fir or yellow pine E = 1. Beam Deflection distributed load of 75 pounds per foot.552") 3. (768 in4) Select: Topic 4: Beams .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnp412b. (.221") 2. steel beam to deflect no more than 1/2 inch when a load of 20 tons is applied at the center of the beam. Find the maximum deflection of a 4"x 6" pine cantilever beam 6 feet long that carries a load of 250 pounds per foot.12b: Probs . Find the maximum point load (acting at the end) that can be carried by a 16 foot long steel W14x38 cantilever beam if the beam end is limited to a maximum deflection of 1/4 inch. Find the minimum moment of inertia necessary for a 20 foot long. (3.htm (2 of 2) [5/8/2009 4:40:22 PM] . simply supported.270 lb) 4. (.Topic 4. x-x axis x-x axis x-x axis S ./in2 for the beam material. simply supported WT 6 x 36 T-Beam is shown.in 1.000 lb. For this beam select the best I-beam to use from the beam table shown below.000 lb.in 0.in4 23.040 Flange thick tf ./in2 (for tension and compression). For this beam: A. file:///Z|/www/Statstr/statics/Beams/bdsnx413. (For solution select: Topic Exam Solution .in3 4.Topic Exam Topic 4. B.Topic Examination 1. A loaded.430 d/tw 14. simply supported beam is shown. and the maximum allowable shear stress = 10.13: Beams .02 2.20 Section Info. Determine the maximum bending stress 10 feet from the left end of the beam.in 1.13 Flange Width bf -in 12.htm (1 of 3) [5/8/2009 4:40:23 PM] . .540 r .in 6.60 Depth of T d . Determine the horizontal shear stress at a point 4 inches above the bottom of the beam cross section and 10 feet from the left end of the beam.671 Stem thick tw .Problem 1) Designation WT 6 x 36 Area A .13: Beams .20 Cross x-x axis I . The maximum allowable bending stress = 35. A loaded.in 0.Topic 4.480 y.in2 10. 0 133.027 8.83 3.in 0.028 10.075 4.50 9.0 146.0 42.799 7.00 13.022 10.990 8.007 4.230 0.367 0.237 0.64 Section y-y axis S .365 0.12 10.76 4.000 6.88 2.70 23.50 13.424 Web thick tw .in3 3.528 0.456 0.60 12.61 9.308 0.3 Cross y-y axis I .70 16.in3 8.500 0.3 25.Problem 2) Designation W 5x16 W 6x20 W 6x25 W 8x17 W 8x20 W 8x31 W 8x35 W 8x40 W 10x25 W 10x29 W 10x33 W 10x39 W 10x45 W 10x54 W 10x60 W 12x19 W 12x22 Area A-in2 4.5 53.37 8.71 11.94 10.in4 21.50 49.0 130.5 30.43 5.22 9.289 0.in 0.3 41.0 210.315 0.4 16.318 0.51 13.1 35.0 158.4 67.0 Section x-x axis S .000 8.378 0.6 69.30 11.62 2.00 42.0 156.080 5.in 5.12 8.5 26.35 5.30 20.16 12.4 110.433 0.618 0.50 44.25 10.54 9.0 126.0 27.Topic 4.430 0.349 0.250 5.350 0.260 Cross x-x axis I .10 1.59 6.1 60.268 8.288 0.558 0.75 9.0 344.3 56.4 31.31 Flange Width bf .36 8.htm (2 of 3) [5/8/2009 4:40:23 PM] .433 0.10 4.415 0.44 9.030 Flange thick tf .248 0.292 0.00 3.47 Depth d .018 6.762 5.0 306.16 11.258 0.80 7.Topic Exam (For solution select: Topic Exam Solution .20 104.20 6.00 4.88 7.00 6.70 5.10 7.70 5.30 17.12 10.618 0.320 0.20 13.0 171.1 21.76 5.360 0.90 17.00 8.12 10.in 5.14 8.in4 7.00 116.30 36.493 0.5 13.24 10.20 15.90 53.077 5.08 10.31 file:///Z|/www/Statstr/statics/Beams/bdsnx413.22 37.683 0.2 49.00 8.1 17.7 14.13: Beams .89 9.01 5.964 7.240 0.252 0.0 249.368 0.25 12.8 35. 29 35.10 34.htm (3 of 3) [5/8/2009 4:40:23 PM] .60 23.695 0.576 0.497 6.0 447.841 204.30 15.06 13.00 5.13 10.0 952.000 10.516 0.86 19.0 641.345 0.24 6.0 34.688 0.32 13.0 108.0 153.7 70.89 7.260 1.582 0.20 21.420 0.95 9.313 0.80 50.70 76.91 14.0 18.5 138.061 16.558 8.750 6.60 17.Topic Exam W 12x27 W 12x31 W 14x34 W 14x38 W 12x53 W 14x61 W 14x87 W 16x36 W 16x50 W 18x45 W 18x60 W 24x68 W 24x94 W 36x230 W 36 x 260 7.465 0.30 21.00 350.00 14.500 6.10 107.287 0.073 7.Topic 4.00 1090.0 15000.0 967.60 10.88 36.513 0.90 25.85 16.0 386.96 12.50 48.000 14.40 37.0 2690.6 54.0 986.70 13.70 67.551 0.0 706.237 0.0 1820.453 0.265 0.00 940.8 79.0 221.400 0.60 96.00 108.525 6.20 17.00 24.0 426.2 39.416 0.0 56.416 0.0 17300.63 6.992 7.378 0.12 12.09 14.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/bdsnx413.00 132.961 9.499 0.60 14.7 92.428 0.25 23.00 27.30 26.00 11.13: Beams .00 Select: Topic 4: Beams .86 18.0 657.50 11.0 837.61 6.299 0.872 1.71 24.477 7.99 10.10 70.440 0.380 0.761 0.50 9.20 15.628 0.335 0.70 20.00 15.25 17.0 239.643 0.90 114.0 340.471 16.5 48.60 23.5 80.20 6.776 10. Determine the maximum bending stress 10 feet from the left end of the beam. simply supported W 10 x 29 beam is shown below.000 lbs . file:///Z|/www/Statstr/statics/Beams/xsol413a.Problem 1 A loaded.000 lbs (4 ft) .4.Beams Exam Solution .) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = By + Dy .Problem 1 Topic Exam Solution .000 lbs .6.000 lbs (4 ft) + Dy (8 ft) = 0 Solving: By = 21.) FBD of structure (See Diagram) 2. For this beam: A.htm (1 of 2) [5/8/2009 4:40:23 PM] . B.000 lbs STEP 2: Determine the shear force and bending moment at x=10 ft.000 lbs/ft (8 ft) = 0 Sum TB = +4. Determine the horizontal shear stress at a point 6 inches above the bottom of the beam cross section and 10 feet from the left end of the beam.1.8. Dy = -3.000 lbs.) Resolve all forces into x/y components 3. Solution: STEP 1: Determine the external support reactions: 1.000 lbs (8 ft) + 6. Beams Exam Solution .000 ft-lbs STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress (MBS) at 10'.Topic Examination or Select: Topic 4: Beams . showing and labeling all external forces.V10(10 ft) + M10 = 0 Solving: V10 = 9.030 lbs/in2 STEP 4: To determine the Horizontal Shear Stress (HSS) at 10 ft from the end of the beam and 6 inches above the bottom of the beam. a = cross sectional area from 6 in above the bottom of the beam to bottom of beam =(2. M10 = -36.000 lbs (9 ft) . The W 10 x 29 beam has a section modulus for the beam from the beam tables is 30.000 lbs (4 ft) + 21. and S is the section modulus for the beam.000 lbs.8 in3.htm (2 of 2) [5/8/2009 4:40:23 PM] . MBS = M10'/S (Where M10' is the bending moment at 10 ft.000 lbs)(2.9 in2 + 1.000 lb.) Apply equilibrium conditions: Sum Fx = 0 none Sum Fy = -4. The form we will use is: HSS = Vay'/Ib: Where: V = Shear force 10 ft from the end of the beam = 9.000 lbs . apply the horizontal shear stress formula. Draw the FBD of left end of beam.)Therefore: MBS = 36. 2.8 in)]/[(158 in4)(.29 in) Then putting in values: HSS = [(9.8 in3 = 14. 3.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/xsol413a.1.9 in2 + 1.69 in2)(3.8 in) I = moment of inertia of the beam (158 in4 for W 10 x 29 beam) b = width of beam a 6 in above the bottom of the beam =(.000 lbs .29 in)] = 3360 psi Return to: Beams .000 lbs/ft (2 ft) + 21.000 ft-lbs(12 in/ft)/30.69 in2) y' = distance from neutral axis to the centroid of area a = (3.6.000 lbs .Problem 1 1.000 lbs (8 ft) -2.) Resolve all forces into x/y directions. The section modulus is available from the Beam Tables.) Cut beam at 10 ft.V10 = 0 Sum TA = -6. Solution: STEP 1: Apply Static Equilibrium Principles and determine the external support reactions: 1.) Resolve all forces into x/y components 3./in2.)(8 ft) . file:///Z|/www/Statstr/statics/Beams/xsol413b.(800 lb.htm (1 of 3) [5/8/2009 4:40:23 PM] .000 lb.) FBD of structure (See Diagram 1) 2./ft.)(8 ft) + = 0 Sum TA = -(800 lb.)(8 ft)(12 ft.(1. For this beam select the best I-beam to use if: the maximum allowable bending stress = 35.730 ft-lb. and the maximum allowable shear stress = 10. Cy = 11.)-(1.Problem 2 Topic Exam Solution ./in2(tension & compression).Problem 2 A loaded.) = 0 Solving: Ay = 4.200 lb.200 lb.Beams: Exam Solution .) +Cy(12 ft./ft../ft. simply supported beam is shown below.) Apply equilibrium conditions: Sum Fx =Ax = 0 Sum Fy = Ay .000 lb./ft.270 lb.)(8 ft)(4 ft. Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load.)/ S. and shear) for the beam material. By material specification we mean the allowable stresses (tensile. Placing values into the equation we have: 35. and Vmax = -6.000 lb.)(12 in.htm (2 of 3) [5/8/2009 4:40:23 PM] . We now use the flexure formula form: Maximum Bending Stress = Mmax / S.91 in3 . This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material.Beams: Exam Solution . compressive.and find the one with a section modulus equal or greater than the minimum section file:///Z|/www/Statstr/statics/Beams/xsol413b. and determine the values of the maximum bending moment and maximum shear force. and then S = 136. (See Diagrams 2 and 3.930 lb. and Maximum Allowable Shear Stress = 10. This information is normally furnished by the beam supplier with their selection of beams./ft./ 35.400 ft-lb. and use the lowest allowable axial stress for the maximum bending stress.000 lb. We would like to now selected the best beam based on the minimum value of the section modulus determined above. and solve for the value of the section modulus. We select the beam from the beam table ./in2 = 3..000 psi.400 ft-lb.. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 35.000 psi./in2 = (11.Problem 2 STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam.) From the Diagrams we observe that Mmax = 11.800 in-lb. 03 in2 (from the beam table for the W 5 x 16 beam.24" * 4.930 lb.000 lb./in2 As long as this approximate value is reasonably below the allowable shear stress for the beam material there is no need to use the exact formula for the maximum shear stress. Return to: Beams .Topic Examination or Select: Topic 4: Beams .Problem 2 modulus and with the least pounds per foot weight (which normally means the least expensive beam). We now must check the maximum shear stress in the I-Beam to see if it is below the allowable shear stress.28" )= 1. we determine W 5 x 16 is the best beam from the selection listed./in2 given above. which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below. It has a section modulus of 8.Beams: Exam Solution ./ft. We can use the approximate formula for I-beam. Thus we have selected the best beam to use from the given list of possible beam.930 lb.)/(1. and a weight of 16 lb.03 in2) = 6730 lb. (from the shear force diagram) Amax = area of web = (. We see that this value is well within the allowable shear stress of 10. Applying this we have: Vmax = maximum shear force = 6.Beam is equal to the maximum shear force divided by the area of the web of the I-Beam. After examining the selections. however.htm (3 of 3) [5/8/2009 4:40:23 PM] .5 in3 (greater than the minimum section modulus of 3.91 in3). If we need to know the shear stress at any other location.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Beams/xsol413b.) Then Maximum Horizontal Shear Stress = (6. This says the approximate maximum shear stress in an I . Step 4. the approximate formula is only for the maximum horizontal shear stress (which occurs are the neutral axis) in an IBeam. Maximum Horizontal Shear Stress = Vmax/Aweb. Please remember. we must use the standard formula. 4c Torsion .6 Rivets & Welds . and an internal torque. to which an external torque is applied (such as in power transmission).Riveted Joint Selection r 5.4a Torsion .2 Torsion: Deformation .Example 3 r 5.Example 2 5. #4. #4. #7 [Previous test problems] 5.Problem Assignment 2 (supplementary) 5. #2.Riveted Joints r 5.1 Torsion: Transverse Shear Stress r 5. We will consider solid and hollow circular shafts. shear stress.1a Shear Stress .Example 4 r Additional General Torsion Examples: #1.Topic Examination Topic 5.Problem Assignment 1 (required) 5.Example 1 r 5.8a Rivets & Welds . #5.Angle of Twist 5.1b Shear Stress . #4.Problem Assignment 3 (supplementary) 5.3c Power Transmission . Rivets & Welds .Example 1 r Additional Weld Examples: #2.5b Riveted Joints .3 Torsion: Power Transmission r 5. Rivets & Welds ● ● ● ● ● ● ● ● ● ● ● ● ● 5. Rivets & Welds Topic 5: Torsion.6a Riveted Joint Selection .4b Torsion .1: Torsion.7a Welded Joints .3b Power Transmission .5 Rivets & Welds .htm (1 of 4) [5/8/2009 4:40:24 PM] . #3.Topic 5.Problem Assignment 4 (required) 5. #5.1: Torsion: Transverse Shear Stress In this topic we wish to consider shafts.3a Power Transmission .Example 1 r 5.Welded Joints r 5.9 Torsion. and that plane sections of the shaft remain plane under the applied torque.3d Compound Shaft .4d Torsion .Example 2 5. #5 [Previous test problems] 5. in which we assume the material is homogeneous and isotropic (that is properties of the materials are the same in all directions in the material).Problem Assignment 1 (required) 5.Example1 r Additional Rivet Examples: #2.Problem Assignment 2 (supplementary) 5. #6 [Previous test problems] 5.8b Rivets & Welds . that the stress which develop remain within the elastic limits. #3.5a Riveted Joints . #3.Example 1 r 5. and deformation (twist) develops in response to the externally applied torque. file:///Z|/www/Statstr/statics/Torsion/tors51.7 Rivets & Welds .Example 2 r 5. #6. The result will be a shearing of the chalk into two pieces. Rivets & Welds In Diagram 1 we have shown a section of a solid shaft. shown in Diagram 1. we have shown the cross sectional area of the shaft of radius R. grasp each end and twist strongly and quickly in opposite directions at each end. We will now go through a relative brief derivation to develop a torsion formula to determine transverse shear stress. This may be thought of as being due to the adjacent cross sectional areas of the shaft trying to twist passed each other. Additionally there is a corresponding deformation (angle of twist) which results from the applied torque and the resisting internal torque causing the shaft to twist through an angle. take a long stick of chalk. and indicated a small area A. (As a practical example of this. In Diagram 2.htm (2 of 4) [5/8/2009 4:40:24 PM] . a radial distance r from the center. file:///Z|/www/Statstr/statics/Torsion/tors51. There is also an internal shear stress which develops inside the shaft. and an equal internal torque T develops inside the shaft. theta. An external torque T is applied to the left end of the shaft. If it is really a torsion failure the sheared end of the chalk will have a conical twisting fracture at the end.1: Torsion.) This shearing stress on the cross sectional area varies from zero at the center of the shaft linearly to a maximum at the outer edge as also shown in Diagram 1.Topic 5. A's. select Centroids/Moment of Inertia: ) And now we solve for = T R/J. = T r / J. And then we can write: T= The summation of r2 A. we now mentally divide the cross sectional area into a large number of small areas.): T= . and J = polar moment of inertia. and J = polar moment of inertia. J. and J = ( /32)(do4 -di4) for a hollow shaft. where T is the internal torque in that section of the shaft. Then the shear force acting on the small area may be written as: We next may write the small amount of torque due to the force as: .1: Torsion. where as above T is the internal torque in that section of the shaft. We may rewrite the last summation by taking radius R and the maximum shear stress to the outside of the summation sign. is the quantity known as the polar moment of inertia. [ J = ( /32) d4 for a solid shaft.htm (3 of 4) [5/8/2009 4:40:24 PM] . . is the relationship for the transverse shearing stress in a shaft under torsion. We may now write the total internal torque as the sum of the small torque acting on the areas A's. where do = outer diameter and di = inner diameter. since these values are constant (they are the same for each A). r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. Rivets & Welds The shear stress at that location may be written in terms of the maximum shear stress as: = (r/R) . We may also extend this formula for the shear stress at any arbitrary radius as follows: = T r / J. (that is.] The formula.Topic 5. (For a brief review of the moment of inertia. file:///Z|/www/Statstr/statics/Torsion/tors51. and sum the torque on these areas. R = maximum radius of the shaft. Please select : Topic 5. the shaft will fail in longitudinal shear stress. along the shaft.1a: Shear Stress .Example 2 When finished with Torsion Shear Stress examples. We will now look at two examples of determining the shear stress in a shaft. Rivets & Welds There is also a longitudinal shear stress in the shaft and.htm (4 of 4) [5/8/2009 4:40:24 PM] . Rivets & Welds . And often in wood shafts.Topic 5.Angle of Twist or Select: Topic 5: Torsion. without going through the derivation.2: Torsion: Deformation .1: Torsion.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/tors51. where the failure stress in lower than across the grain (transverse).1b: Shear Stress . along the grain of the wood. the value of the transverse shear stress and the longitudinal shear stress are equal. Continue to: Topic 5.Example 1 Topic 5. we will simply state that at any point in the shaft. 57 in4. at end B. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.1a: Shear Stress . In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum. * 1 in. where: T is the internal torque in that section of the shaft = 1000 ft-lb = 12./1. We would like to determine the maximum transverse shear stress in the shaft due to the applied torque.000 in-lb. we first need to determine the internal torque in the shaft. r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. We cut the shaft a distance x from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 1b. The shaft is in equilibrium. which in this case must be equal and opposite to the torque at end A for equilibrium. = 7. This is the Maximum Transverse (and longitudinal) Shear Stress in the shaft.57 in4. at end A.Example 1 Topic 5.htm (1 of 2) [5/8/2009 4:40:24 PM] . but a hollow shaft with an outer file:///Z|/www/Statstr/statics/Torsion/torse51a. = T r / J = 12.1a: Shear Stress . We next simply apply the torsion formula for the shear stress: = T r / J. To solve.Topic 5. r = 1 in. and a load torque of 1000 ft-lb. J = polar moment of inertia = ( /32) d4 for a solid shaft = (3.Example 1 Part I.1416/32) (24in4) = 1.640 lb/in2. Where we cut the shaft there is an internal torque. In Diagram 1a we have shown a solid shaft with what we will call a driving external torque of 1000 ft-lb.000 in-lb. Part II We now would like to consider the case where the shaft is not solid. So. the polar moment of inertia. We next apply the torsion formula for the shear stress for the hollow shaft: = T r / J. * 1 in.1: Torsion: Transverse Shear Stress. as shown in Diagram 2a. J = polar moment of inertia = (3. We still apply the same driving and load torque. and still have the same value of the internal torque.1416/32) [do4 . except for the value of J. This then is the Maximum Transverse (and longitudinal) Shear Stress in the hollow shaft.000 in-lb.1416/32) [(24in4) (1. Continue to: Topic 5. T is the internal torque in that section of the shaft = 1000 ft-lb = 12.di4] for a hollow shaft = (3.Example 1 diameter of 2" and an inner diameter of 1". In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum. Rivets & Welds . So. r = the radial distance from the center of the shaft to the point where we wish to find the shear stress.04in4)] = 1. as is shown in Diagram 2b. where we observe that all the values are the same as in part one.Topic 5. Return to: Topic 5.Example 2 or Select: Topic 5: Torsion./1.1a: Shear Stress . r = 1 in.47 in4.htm (2 of 2) [5/8/2009 4:40:24 PM] .47 in4.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse51a. = T r / J = 12.150 lb/in2.1b: Shear Stress . = 8.000 in-lb. 900 ft-lb. Notice that the shaft is in rotational equilibrium.Topic 5. where: T is the internal torque in that section of the shaft = 400 ft-lb. and it falls in a reasonable file:///Z|/www/Statstr/statics/Torsion/torse51b. = T r / J = 4. at end D. This value is the Maximum Transverse Shear Stress in the shaft section AB. and 300 ft-lb.1b: Shear Stres . r = . and load torque of 400 ft-lb. We next apply the torsion formula for the shear stress: = T r / J.500 lb. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.800 in-lb. Where we cut the shaft there is an internal torque./in2./. = 24. * . We cut the shaft a distance 0' < x < 1' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 2. We would like to determine the maximum transverse shear stress in each section of the shaft due to the applied torque. To solve.1b: Shear Stress .Example 2 In Diagram 1 we have shown a solid compound shaft with what we will call the driving external torque of 1600 ft-lb. So. at end A. which in this case must be equal and opposite to the torque at end A for equilibrium.800 in-lb.098 in4. at point C. acting at point B.1416/32) (14in4) = .Example 2 Topic 5.5 in. r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem r is to the outer edge of the shaft since that is where the transverse shear stress is a maximum. we first need to determine the internal torque in each section of the shaft.098 in4.htm (1 of 3) [5/8/2009 4:40:25 PM] . = 4.5 in. J = polar moment of inertia = ( /32) d4 for a solid shaft = (3. Where we cut the shaft there is an internal torque. = T r / J = 14. We apply the torsion formula for the shear stress once again: = T r / J. Where we cut the shaft there is an internal torque. J = polar moment of inertia = ( /32) d4 for a solid shaft = (3. From the Free Body Diagram we see that in order to have rotational equilibrium we must have an internal torque in section CD of 300 ft-lb.htm (2 of 3) [5/8/2009 4:40:25 PM] .170 lb. r = 1 in. We cut the shaft a distance 3' < x < 4' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 4. acting in the direction shown. file:///Z|/www/Statstr/statics/Torsion/torse51b.400 in-lb. where: T is the internal torque in that section of the shaft = 1. the shear stress is much lower in BC./in2.57 in4. This then is the Maximum Shear Stress in shaft section BC.1b: Shear Stres . We note that even though the internal torque is much larger in section BC as compared to section AB./ 1. acting in the direction shown.200 ft-lb.Example 2 range for allowable shear stresses for metals. = 9.Topic 5. because of the size of the shaft in section BC. r = the radial distance to the outer edge of the shaft since that is where the transverse shear stress is a maximum. We now determine the internal torque in the next section of the shaft. we see that in order to have rotational equilibrium we must have an internal torque in section BC of 1200 ft-lb.57 in4. So. and by mentally summing torque. We now determine the internal torque in the next section of the shaft. = 14.1416/32) (24in4) = 1. We cut the shaft a distance 1' < x < 3' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 3. * 1 in.400 in-lb. * . This is the Maximum Shear Stress in shaft section CD. thus this section of the shaft would fail .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse51b. We note that this is much larger than the ultimate shear stress most metals. Rivets & Welds .1b: Shear Stres .htm (3 of 3) [5/8/2009 4:40:25 PM] . = 147.1: Torsion: Transverse Shear Stress or Select: Topic 5: Torsion.Example 2 We apply the torsion formula for the shear stress: = T r / J. r = the radial distance from the center of the shaft to the outer edge of the shaft since that is where the transverse shear stress is a maximum.a larger diameter is needed to carry the torque.54in4) = . = T r / J = 3. J = polar moment of inertia = ( /32) d4 for a solid shaft = (3. where: T is the internal torque in that section of the shaft = 300 ft-lb.600 in-lb. So.0061 in4. = 3.1416/32) (.25 in.600 in-lb.Topic 5./ .500 lb.0061 in4.25 in. r = ./in2. Return to: Topic 5. htm (1 of 3) [5/8/2009 4:40:25 PM] . G brass = 6 x 106 lb/in2.2: Torsion: Deformation . An external torque T is applied to the left end of the shaft. where T = the internal torque in the shaft L = the length of shaft being "twisted" J = the polar moment of inertia of the shaft G = the Modulus of Rigidity (Shear Modulus) for the material.Angle of Twist Topic 5. for example for steel and brass we have. G steel = 12 x 106 lb/in2.Angle of Twist Another effect of applying an external torque to a shaft is a resulting deformation or twist as the material is stressed. Additionally there is a corresponding deformation (angle of twist) which results from the applied torque and the resisting internal torque causing the shaft to twist through an angle.2: Torsion: Deformation . In Diagram 2a we have shown a solid steel circular shaft with an external torque of 1000 ft-lb. being applied at each end of the shaft.Topic 5. file:///Z|/www/Statstr/statics/Torsion/tors52. phi. In Diagram 1 we have shown a section of a solid shaft. The angle of twist may be calculated from: = T L / J G . We will now look at an example of determining the angle of twist in a shaft. in opposite directions. shown in Diagram 1. The resulting shaft deformation is expressed as an Angle of Twist of one end of the shaft with respect to the other. and an equal internal torque T develops inside the shaft. 2: Torsion: Deformation .000 lb/in2. We next apply the Angle of Twist formula: = T L / J G .000 in-lb. and in this problem is clockwise with respect to end A as shown in Diagram 3. We would like to determine the angle of twist of end B with respect to end A.htm (2 of 3) [5/8/2009 4:40:25 PM] .Topic 5.5 in4. and make a free body diagram of the left section of the shaft . = 24 inches J = polar moment of inertia = ( /32) d4 for a solid shaft = (3. From the free body diagram. The angle of twist will have units of radians.Angle of Twist The shaft has a diameter of 1. file:///Z|/www/Statstr/statics/Torsion/tors52. where T = 1000 ft-lb./. to satisfy rotational equilibrium.shown in Diagram 2b. * .54in4) = . just out of interest in it's value.5 in4 * 12 x 106 lb/in2) = . = T r / J = 12.000 in-lb.5 inch. = 12.000 in-lb. we see that the internal torque must be 1000 ft-lb.5 in4.1416/32) (1. We also take a moment to calculate the maximum shear stress in the shaft.048 radians = 2. G steel = 12 x 106 lb/in2 Then. We cut the shaft a distance x feet from the left end.75o. = 18.75 in. L = 2 ft. = T L / J G = (12. To find the angle of twist we first determine the internal torque in the shaft.* 24 in) / (. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/tors52.Power Transmission or Select: Topic 5: Torsion.Topic 5. Rivets & Welds .htm (3 of 3) [5/8/2009 4:40:25 PM] .2: Torsion: Deformation .3: Torsion .Angle of Twist Continue to: Topic 5. for each revolution of the shaft.. usually at the outer edge of the shaft.3: Torsion . This force may be applied through use of a belt or gear. n = # rev/sec This is the formula for power transmitted in foot-pounds/second. rotation rate. Example In Diagram 2. or we may write: Work = F x d = F * (2 p r) * (# revolutions) The Power sent down the shaft is then the Work per unit time. and Power transmitted./sec) where T = Torque in ft-lb. as below: Power = 2 p ( F * r) * (# rev/sec) Then we recognize the (F * r) term is the torque in the shaft. at end A and an equal and opposite load torque at end B. That is.which is the circumference. or if we divide the equation for Work above by the time we can write: Power = Work/Time = F * (2 p r) * (# rev/time) If we now rewrite the above equation slightly. the force act through a distance of one circumference. F. In Diagram 1a we have shown a shaft with a drive torque of 1000 ft-lb. The product of the force (F) and the radius (r) is the applied (or load) torque.Power Transmission One important process which involves applying torque to a shaft is power transmission.Power Transmission Topic 5. we have shown a solid shaft with an applied driving torque of 1000 ft-lb. A short derivation will gives us a relationship between Torque.Topic 5./sec) as shown below. In Diagram 1b we have shown the shaft from end on.htm (1 of 2) [5/8/2009 4:40:25 PM] . Notice that to apply a torque to a shaft we must exert a force. and an equal and file:///Z|/www/Statstr/statics/Torsion/tors53. and we can rewrite as: Power = 2 p T n (ft-lb. Power hp = [2 p T n / 550 ft-lb/sec/hp] Next we look at a simple example of determining transmitted horsepower. The work done as we rotate the shaft will be the product of the force and the distance the force acts through .3:Torsion . To transmit power a drive torque is applied to a rotating shaft. It is often more convenient to express it in horsepower (1 hp = 550 ft-lb. Next we apply the Horsepower equation: Power hp = [2 p T n / 550 ft-lb/sec/hp] . rotating at a speed of 1800 rpm (30 rev/sec).htm (2 of 2) [5/8/2009 4:40:25 PM] .Power Transmission opposite load torque.1416* 1000 ft-lb.* 30 rev/sec / 550 ft-lb/sec/hp] = 343 hp.Topic 5. n = the number of revolution per second = 30 rev/sec. Rivets & Welds . then the internal torque will be equal in value to the external torque of 1000 ft-lb.Example 1 or Select: Topic 5: Torsion.3a: Power Transmission . Solution: First we mentally note that since this is a simple shaft with equal external driving and load torque.3:Torsion . Continue to: Topic 5.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/tors53. So we have: Power hp = [2* 3. where T = internal torque in shaft (in foot-pounds) = 1000 ft-lb. We would like to determine the horsepower being transmitted down the shaft. 098 in4. we can determine the largest amount of power we can transmitted from the horsepower equation.Example 1 Topic 5. The formula for the shear stress in a shaft is: = T r / J .098 in4 / . (We use the allowable shear stress as the maximum stress in the shaft. (40 rev/s)/ (550 ft-lb/s/hp) = 90 hp Continue to: Topic 5.Topic 5. r = . using the maximum allowable shear stress.Example 1 Example 1: A solid steel shaft.1416) d4 / 32 = .3a: Power Transmission . then solving for the torque: T = J / r .1416) 196 ft-lb. and = 12.htm [5/8/2009 4:40:25 PM] .3b: Power Transmission . Step 2: Now that we have the maximum torque we can safely apply.000 lb/in2.) Putting values into the equation and solving: T = 12.3a: Power Transmission .Example 2 or Select: Topic 5: Torsion.5 in = 2356 in-lb = 196 ft-lb. we determine the largest torque which may be applied to the shaft.000 lb/ in2. where J = (3.000 lb/in2 * . What is the largest amount of power which could safely be transmitted down the shaft if it is to rotate at 2400 rpm? Solution: Step 1: First. Php = 2 pi T n/ 550 = 2 (3. has a 1 inch diameter and an allowable shear stress of 12. shown in Diagram 1.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse53a. Rivets & Welds .5 inch. We first use the horsepower equation to determine the amount of torque which is being applied to the shaft. = T r / J . What is the maximum shear stress in the shaft? Solution: Step 1.65 in4 = 4080 lb/in2.3b: Power Transmission . where T = 2650 in-lb.75"4) = . we use the transverse shear stress equation for circular shafts to determine the maximum shear stress in the shaft. Now that we have the torque being applied to the shaft.di4) = (3.Example 3 or Select: Topic 5: Torsion.65 in4. J = (pi/32)(do4 . = T r / J = 2650 in-lb.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse53b.1416/32)(2"4 .Example 2 Example 2: A car has a hollow drive shaft with a two inch outer diameter and a 1/8 inch wall thickness as shown in Diagram 2.. Continue to: Topic 5. Power hp = [2 pi T n / 550 ft-lb/s/hp] Solving for T = (Php 550 ft-lb/s/hp) / 2 pi n.1416) 4400/60 rev/sec] = 221 ft-lb = 2650 in-lb Step 2.Topic 5.* 1 in / . then putting in the values. The maximum power transmitted down the shaft is 185 hp at a shaft speed of 4400 rpm.3c: Power Transmission . r = 1 in.htm [5/8/2009 4:40:26 PM] .3b: Power Transmission . Rivets & Welds . we have: T = 185 hp (550 ft-lb/s/hp)/[ 2 (3.1..Example 2 Topic 5. TBC = 400hp (550 ft-lb/sec/hp)/[2 (3. file:///Z|/www/Statstr/statics/Torsion/torse53c. and shaft CD had a diameter of 2". and with power being taken off the shaft . We first determine the amount of horsepower being transmitted through each section of the compound shaft. = 8412 in-lb. BC = 21024 in-lb * 1. = T r / J.Example 3 Topic 5.3c: Power Transmission . we use the horsepower equation to determine the amount of torque which is being applied to each shaft. and 160 hp at point D.5 in / [pi(3")4 / 32] = 3966 lb/in2. Now that we know the horsepower in each section of the shaft.1416) (1200/60 rev/s)] = 701 ft-lb. In section AB.1416) (1200/60 rev/s)] = 1752 ft-lb. we realize that the amount of power being transmitted through BC must be 400 hp . we have: TAB = 100hp (550 ft-lb/sec/hp)/[2 (3. Solution: Step 1. continues through shaft CD. shaft BC had a diameter of 3". then putting in the values. 240 hp at point C. and is taken off at point D.770 lb/in . we can see that 100 hp must being transmitted internally through the shaft from B to A. Step 2.of which 240 hp is taken off at point C. Step 3.5 in / [pi(1") / 32] = 26. Finally we use the shear stress formula.htm (1 of 2) [5/8/2009 4:40:26 PM] . = 21024 in-lb.Topic 5. to find maximum shear stress in each section 4 2 AB = 5256 in-lb * . If the shaft is rotating at 1200 rpm.Example 3 Example 3: We have shown a compound shaft with 500 horsepower being applied at point B.3c: Power Transmission . determine the maximum shear stress in each section of the shaft.1416) (1200/60 rev/s)] = 438 ft-lb. Shaft AB has a diameter of 1". Power hp = [2 pi T n / 550 ft-lb/s/hp] Solving for T = (Php 550 ft-lb/s/hp) / 2 p n. and the remaining 160 hp. with a little thought.100 hp at point A. = 5256 in-lb. In section BC. TCD = 160hp (550 ft-lb/sec/hp)/[2 (3. 3d: Compound Shaft .htm (2 of 2) [5/8/2009 4:40:26 PM] .Example 3 CD = 8412 in-lb * 1 in / [pi(2")4 / 32] = 5355 lb/in2.Topic 5.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse53c.3c: Power Transmission . Continue to: Topic 5. Rivets & Welds .Example 4 or Select: Topic 5: Torsion. Modulus of Rigidity: Steel = 12 x 106 lb/in2. Determine the resultant angle of twist of a point on end D with respect to a point on end A.htm (1 of 3) [5/8/2009 4:40:26 PM] . which in this case must be equal and opposite to the torque at end A for equilibrium. A solid compound shaft with a driving torque of 1800 ft-lb.3d: Compound Shaft . The shaft is rotating at 2400 rpm. The length and diameters of the shafts are shown in the Diagram 1. Section CD is made of steel. 800 ft-lb..Topic 5.Example 4 Topic 5.Example 4 This last example looks at determining a number of quantities in a compound shaft. we first need to determine the internal torque in each section of the shaft. Section AB is made of steel. Section BC is made of brass... Step 1. To solve. and 400 ft-lb. So for this shaft the value of the internal torque is equal to the value of the externally applied torque. TAB = 600 ft-lb. file:///Z|/www/Statstr/statics/Torsion/torse53d. Determine the maximum shear stress in each of the sections of the shaft. and load toque of 600 lf-lb. Brass = 6 x 106 lb/in2. We cut the shaft a distance 0' < x < 1' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 2. Aluminum = 4 x 106 lb/in2. 2. Where we cut the shaft there is an internal torque. Determine the horsepower being transmitted through each section of the shaft.. 3. is shown in Diagram 1. For this compound shaft : 1.3d: Compound Shaft . 400 in-lb * 1 in / [pi(2")4 / 32] = 9. From the Free Body Diagram we see that in order to have rotational equilibrium we must have an internal torque in section CD of TCD = 400 ft-lb. 4.200 in-lb * ..170 lb/in2.800 in-lb * . Where we cut the shaft there is an internal torque.htm (2 of 3) [5/8/2009 4:40:26 PM] .400 in-lb. we see that in order to have rotational equilibrium we must have an internal torque in section BC of TBC = 1200 ft-lb. Step 2.Topic 5.800 in-lb. we now apply the torsion formula for the shear stress: =Tr/J 4 2 AB = 7.. = 4. We now determine the internal torque in the last section of the shaft. Step 3. acting in the direction shown. TBC = 1200 ft-lb.200 in-lb. We cut the shaft a distance 1' < x < 3' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 3. Using the internal torque in each section (TAB = 600 ft-lb = 7.700 lb/in .75")4 / 32] = 58.Example 4 We now determine the internal torque in the next section of the shaft. = 14. and by mentally summing torque.5 in / [pi(1") / 32] = 36.375 in / [pi(. We cut the shaft a distance 3' < x < 4' from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 4. acting in the direction shown. BC = CD = 14.000 lb/in2. Again using the internal torque in each section. we determine the horsepower transmitted file:///Z|/www/Statstr/statics/Torsion/torse53d.3d: Compound Shaft . Where we cut the shaft there is an internal torque.). TCD = 400 ft-lb. resulting in the (+.037 radians = 2. =T L/J G= (14.155 radians = 8.800 in-lb. where our signs are taken from the direction of total the internal torque in each section.*(40 rev/s) /(550 ft-lb/s/hp)= 274 hp PBC=2 pi Tn/(550 ft-lb/s/hp)=2*3. AB .htm (3 of 3) [5/8/2009 4:40:26 PM] .12o.counter clockwise.1416*1200 ft-lb.200 in-lb. =T L/J G= (4. -.Problem Assignment or Select: Topic 5: Torsion.57 in4 * 6 x 106 lb/in2)=. and CD clockwise.098 in4 * 12 x 106 lb/in2)=.073 radians = 4.* 24 in)/(1. Then the total twist of end D with respect to end A will be: = + 4.clockwise.*(40 rev/s) /(550 ft-lb/s/hp)= 183 hp Step 4.*(40 rev/s) /(550 ft-lb/s/hp)=548 hp PCD=2 pi Tn/(550 ft-lb/s/hp)=2*3. AB BC CD =T L/J G= (7.Example 4 through each shaft section.18o.18 o . In this example. Rivets & Welds .* 12 in)/(.1416*600 ft-lb.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torse53d.4: Torsion . BC .3d: Compound Shaft .Topic 5.6.1416*400 ft-lb. Apply the Angle of Twist relationship to each section of the shaft.82o (clockwise). PAB=2 pi Tn/(550 ft-lb/s/hp)=2*3.12o .031 in4 * 12 x 106 lb/in2)=.88o = .2.88o. -) signs of the angles of twist.* 12 in)/(.400 in-lb. Continue to: Topic 5.8. Section AB is made of steel.Solution511 STATICS & STRENGTH OF MATERIALS .1416 * (2 in)4 / 32] = 10. Determine the resultant angle of twist of a point on end D with respect to a point on end A. For this shaft: A. 2 2 Part A: Section I: From equilibrium condtions: Sum of Torque = 600 ft-lb .5 in) /[3. Section BC is made of brass. Determine the maximum shear stress in each of the sections of the shaft.Example A compound shaft with applied torques and dimensions is shown below.TAB = 0.htm (1 of 2) [5/8/2009 4:40:27 PM] . The modulus of rigidity for brass = 6 x 106 lb/in .700 psi Section II: From equilibrium condtions: Sum of Torque = 600 ft-lb .2000 ft-lb + TBC = 0.700 psi file:///Z|/www/Statstr/statics/Stests/torsion/sol511. Section CD is made of steel. So TBC = 1400 ft-lb Then τBC = Tr / J = (1400 ft-lb)(12 in/ft)(1 in) /[3.1416 * (1 in)4 / 32] = 36. B. So TAB = 600 ft-lb Then τAB = Tr / J = (600 ft-lb)(12 in/ft)(. The modulus of rigidity for steel = 12 x 106 lb/in . 75 in)4 / 32)(12x106 lb/in2)] = .1416 * (2 in)4 / 32)(6x106 lb/in2)] = .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol511./ft)(2 ft)(12 in/ft)/[(3.Solution511 Section III: From equilibrium condtions: Sum of Torque = 600 ft-lb .155 radians (ccw) f =-f +f Total AB BC + fCD = .1416 * (.400 ft-lb)(12 in./ft)(1 ft)(12 in/ft)/[(3.2000 ft-lb + 1000 ft-lb + TCD = 0.1416 * (.0733 radians (cw) fBC = TL/JG = (1./ft)(1 ft)(12 in/ft)/[(3.900 psi Part B: Resultant angle of twist: fAB = TL/JG = (600 ft-lb)(12 in.75 in)4 / 32] = 57.1416 * (1 in)4 / 32)(12x106 lb/in2)] = . So TCD = 400 ft-lb Then τCD = Tr / J = (400 ft-lb) (12 in/ft)(.375 in) /[3.htm (2 of 2) [5/8/2009 4:40:27 PM] .1245 radians (ccw) Topic 5: Torsion.0428 radians (ccw) fCD = TL/JG = (400 ft-lb)(12 in. Rivets & Welds . 75 in) /[3. Determine the maximum shear stress in each of the sections of the shaft. So TAB = 400 ft-lb Then τAB = Tr / J = (400 ft-lb)(12 in/ft)(. The modulus of rigidity for brass = 6 x 10 lb/sq.5 in)4 / 32] = 18.900 psi Section II: From equilibrium condtions: Sum of Torque = 400 ft-lb .000 ft-lb Then τBC= Tr / J = (1.solution512 STATICS & STRENGTH OF MATERIALS . in.400 ft-lb + TBC = 0. For this shaft: A. Section BC is made of brass. The modulus of rigidity for steel = 12 x 10 lb/sq. Determine the resultant angle of twist of a point on end D with respect to a point on end A.1. Section AB is made of brass.75 in)4 / 32] = 57.375in) /[3. 6 6 Part A: Section I: From equilibrium condtions: Sum of Torque = 400 ft-lb .htm (1 of 2) [5/8/2009 4:40:27 PM] .1416 * (1.TAB = 0.100 psi file:///Z|/www/Statstr/statics/Stests/torsion/sol512. B. in.Example A compound shaft with applied torques and dimensions is shown below.1416 * (. So TBC = 1. Section CD is made of steel.000 ft-lb)(12 in/ft)(. htm (2 of 2) [5/8/2009 4:40:27 PM] .0966 radians (ccw) fCD = TL/JG = (200 ft-lb)(12 in.309 radians (cw) fBC = TL/JG = (1.587 radians (ccw) f =-f +f Total AB BC + fCD = .1416 * (.400 ft-lb + 800 ft-lb + TCD = 0.375 radians (ccw) Topic 5: Torsion. Rivets & Welds .1416 * (.solution512 Section III: From equilibrium condtions: Sum of Torque = 400 ft-lb .5 in)4 / 32)(12x106 lb/in2)] = .5 in)4 / 32)(6x106 lb/in2)] = .75 in)4 / 32)(6x106 lb/in2)] = ./ft)(2 ft)(12 in/ft)/[(3.5 ft)(12 in/ft)/[(3.1416 * (.1.800 psi Part B: Resultant angle of twist: fAB = TL/JG = (400 ft-lb)(12 in.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol512./ft)(1. So TCD = 200 ft-lb Then τCD = Tr / J = (200 ft-lb) (12 in/ft)(.1416 * (1./ft)(1 ft)(12 in/ft)/[(3.000 ft-lb)(12 in.25 in) /[3.5 in)4 / 32] = 97. 375in) /[3. in. & B (TD.000 lb/in2 = TCD(. determine the resultant angle of twist of a point on end D with respect to a point on end A. The modulus of rigidity for steel = 12 x 10 lb/sq.000 psi A.1416 * (. Section AB is made of brass.Example A compound shaft is attached to a fixed wall as shown below.TD = 0. Part A: Section I: From equilibrium condtions: Sum of Torque = TCD . & TB) without exceeding the allowable shear stress in any section of the shaft.000 psi t brass = 12.htm (1 of 3) [5/8/2009 4:40:27 PM] .Solution513 STATICS & STRENGTH OF MATERIALS . Sections BC and CD are made of steel. in.75 in)4 / 32] Solving: T CD 6 6 = T = 1491 in-lb = 124 ft-lb D file:///Z|/www/Statstr/statics/Stests/torsion/sol513. So TD = TCD Then: τsteel = TCD r / J 18. Determine the maximum torques which could be applied at points D. If the allowable shear stress for steel and brass are: t steel = 18. Using the torques found in part A. The modulus of rigidity for brass = 6 x 10 lb/sq.TC. B. C. TCD .124 ft-lb = 0.Solution513 Section II: From equilibrium condtions: Sum of Torque = TBC . T = (1570 ft-lb + 171 ft-lb + 124 ft-lb) = 1865 ft-lb file:///Z|/www/Statstr/statics/Stests/torsion/sol513.htm (2 of 3) [5/8/2009 4:40:27 PM] . So TC = (TBC .000 lb/in2 = TBC(.1416 * (2 in)4 / 32] Solving: T B AB = 18.000 lb/in2 = TAB(1 in) /[3.124 ft-lb = 0.850 in-lb = 1570 ft-lb.124 ft-lb) = 171 ft-lb Section III: From equilibrium condtions: Sum of Torque = -TAB + TB . So TB = (TAB + 171 ft-lb + 124 ft-lb) Then τbrass = TABr / J 12.171 ft-lb .124 ft-lb) Then τsteel = TBCr / J 18.1416 * (1 in)4 / 32] Solving: T C BC = 3534 in-lb = 295 ft-lb T = (295 ft-lb .5in) /[3. /ft)(1 ft)(12 in/ft)/[(3.1416 * (.06 radians (cw) Topic 5: Torsion./ft)(1 ft)(12 in/ft)/[(3. Rivets & Welds ./ft)(1 ft)(12 in/ft)/[(3.1416 * (1 in)4 / 32)(12x106 lb/in2)] = .75 in)4 / 32)(12x106 lb/in2)] = .570 ft-lb)(12 in.0361 radians (cw) fCD = TL/JG = (124 ft-lb)(12 in.1416 * (2 in)4 / 32)(6x106 lb/in2)] = .htm (3 of 3) [5/8/2009 4:40:27 PM] .0479 radians (cw) f =+f -f Total AB BC -fCD = .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol513.024 radians (ccw) fBC = TL/JG = (245 ft-lb)(12 in.Solution513 Part B: Resultant angle of twist: fAB = TL/JG = (1. A driving torque of 2000 ft-lb is applied as shown. Section AB is made of steel. The modulus of rigidity for brass = 6 x 10 lb/sq.000 lb/sq. and 400 ft-lbs are also shown. So TAB = 600 ft-lb Then: HP = 2πηT / 550 ft-lb/s / hp = 2π (40 Rev/s )( 600 ft-lb) / 550 ft-lb/s / hp = 274 hp Section II: From equilibrium condtions: Sum of Torque = 600 ft-lb .in.2000 ft-lb + TBC = 0. Section CD is made of steel. Section BC is made of brass. Load torques of 600 ft-lbs.. The shaft is rotating at 2400 rev/minute. Determine the horsepower being transfer through each shaft. So TBC = 1400 ft-lb file:///Z|/www/Statstr/statics/Stests/torsion/sol514.TAB = 0. 6 6 Part A: Section I: From equilibrium condtions: Sum of Torque = 600 ft-lb . The modulus of rigidity for steel = 12 x 10 lb/sq. B. The allowable shear stress of steel is 20. such that it can safely carry the transmitted power.htm (1 of 3) [5/8/2009 4:40:27 PM] .Example A compound shaft with applied torques is shown below. Determine the minimum diameter for each shaft.000 lb/sq. For this shaft: A. in.in. and the allowable shear stress for brass in 18. 1000 ft-lb. in.Solution514 STATICS & STRENGTH OF MATERIALS . 400 ft-lb) / 550 ft-lb/s / hp = 640 hp Section III: From equilibrium condtions: Sum of Torque = 600 ft-lb .400 ft-lb)(12 in/ft)( 1/2 ) /[(3.8335 in3 so: dAB = 1.htm (2 of 3) [5/8/2009 4:40:27 PM] .1416 * (d)4 / 32] Solve for d3BC = (1.22 inches Section II: τBC = Tr / J => 18.7534 in3 so: dBC = 1.000 1b/in2)] = 1.Solution514 Then: HP = 2πηT / 550 ft-lb/s / hp = 2π (40 Rev/s)( 1.68 inches file:///Z|/www/Statstr/statics/Stests/torsion/sol514.1416 * (d)4 / 32] Solve for d3AB = (600 ft-lb)(12 in/ft)( 1/2 ) /[(3.000 1b/in2)] = 4. So TCD = 400 ft-lb Then: HP = 2πηT / 550 ft-lb/s / hp = 2π (40 Rev/s)( 400 ft-lb) / 550 ft-lb/s / hp = 183 hp Part B: Section I: τAB = Tr / J => 20.1416 / 32) (20.000 = (600 ft-lb)(12 in/ft)( d/2 ) /[3.000 = (1.400 ft-lb)(12 in/ft)( d/2 ) /[3.1416 / 32) (18.2000 ft-lb + 1000 ft-lb + TCD = 0. 000 = (400 ft-lb)(12 in/ft)( d/2 ) /[3.07 inches Topic 5: Torsion.Solution514 Section III: τCD = Tr / J => 20.000 1b/in2)] = 1.2223 in3 so: dCD = 1.1416 / 32) (20. Rivets & Welds .1416 * (d)4 / 32] Solve for d3CD = (400 ft-lb)(12 in/ft)( 1/2 ) /[(3.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol514.htm (3 of 3) [5/8/2009 4:40:27 PM] . Solution515 STATICS & STRENGTH OF MATERIALS . Determine the angle of twist of end C with respect to end A. For this shaft: A. and the torque acting at end C is 180 ft-lbs.Example The diagram below represents two hollow brass shafts attached to a solid wall at end A. The modulus of rigidity for brass = 6 x 10 lb/sq.5 in) /[3.5 in. The torque acting at point B is 800 ft-lbs.8 in)4] / 32] = 18. Determine the maximum shear stress in each shaft.8 in 6 6 Part A: Section I: From equilibrium condtions: Sum of Torque = TBC .180 ft-lb = 0. in. inner diameter of AB = 2 in outer diameter of BC = 1 in . So TBC = 180 ft-lb Then τAB = Tr / J = (180 ft-lb)(12 in/ft)(. in. inner diameter of BC = .1416 * [(1 in)4 .htm (1 of 2) [5/8/2009 4:40:28 PM] . The modulus of rigidity for steel = 12 x 10 lb/sq.600 psi file:///Z|/www/Statstr/statics/Stests/torsion/sol515. The diameters of each shaft are as follows: outer diameter of AB = 2. B.(. 1416 * [(2.0312 radians (cw) fBC = TL/JG = (180 ft-lb)(12 in. So TAB = 980 ft-lb Then τAB = Tr / J = (980 ft-lb)(12 in/ft)(1./ft)(2 ft)(12 in/ft)/[(3.0149 radians (cw) = .Solution515 Section II: From equilibrium condtions: Sum of Torque = TAB .f = -.1416 * [(2.0461 radians (cw) f Total AB BC Topic 5: Torsion.180 ft-lb = 0.800 ft-lb .5 in)4 ./ft)(3 ft)(12 in/ft)/[(3.htm (2 of 2) [5/8/2009 4:40:28 PM] .(2 in)4] / 32] = 6.8 in)4] / 32)(6x106 lb/in2)] = .f .490 psi Part B: Resultant angle of twist: fAB = TL/JG = (980 ft-lb)(12 in.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol515. Rivets & Welds .(.1416 * [(1 in)4 .5 in)4 .(2 in)4] / 32)(6x106 lb/in2)] = .25 in) /[3. The torque acting at point B is 600 ft-lbs. For this shaft: A.htm (1 of 2) [5/8/2009 4:40:28 PM] . The modulus of rigidity for brass = 6 x 10 lb/sq.5 in. inner diameter of AB = 2.Example The diagram below represents two hollow brass shafts attached to a solid wall at end A.6 in)4] / 32] = 2.(1. B. The modulus of rigidity for steel = 12 x 10 lb/sq. Determine the angle of twist of end C with respect to end A.1416 * [(2 in)4 . in. in. inner diameter of BC = 1.8 in outer diameter of BC = 2 in .600 psi file:///Z|/www/Statstr/statics/Stests/torsion/sol516. Determine the maximum shear stress in each shaft.Solution516 STATICS & STRENGTH OF MATERIALS .200 ft-lb = 0. The diameters of each shaft are as follows: outer diameter of AB = 3.6 in 6 6 Part A: Section I: From equilibrium condtions: Sum of Torque = TBC . and the torque acting at end C is 200 ft-lbs. So TBC = 200 ft-lb Then τAB = Tr / J = (200 ft-lb)(12 in/ft)(1 in) /[3. 1416 * [(3.f = -./ft)(2 ft)(12 in/ft)/[(3. So TAB = 800 ft-lb Then τAB = Tr / J = (800 ft-lb)(12 in/ft)(1.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/torsion/sol516. Rivets & Welds .5 in)4 .0155 radians (cw) = .(2.(2.600 ft-lb .1416 * [(3.8 in)4] / 32] = 1.(1.75 in) /[3.5 in)4 .0199 radians (cw) f Total AB BC Topic 5: Torsion.Solution516 Section II: From equilibrium condtions: Sum of Torque = TAB .1416 * [(2 in)4 .8 in)4] / 32)(6x106 lb/in2)] = .f .200 ft-lb = 0.htm (2 of 2) [5/8/2009 4:40:28 PM] .150 psi Part B: Resultant angle of twist: fAB = TL/JG = (800 ft-lb)(12 in./ft)(3 ft)(12 in/ft)/[(3.0044 radians (cw) fBC = TL/JG = (200 ft-lb)(12 in.6 in)4] / 32)(6x106 lb/in2)] = . outer radius) 2.Torsion 1. The shaft is solid 3/8 inch diameter steel. A 3 inch diameter solid steel shaft is 15 inches long. Determine the maximum shear stress in the shaft.Problem Assignment 1 . A vice is clamped by applying a force of 50 pounds at each end of a 4 inch arm on a shaft. (142 psi.Topic 54a . A torque of 750 inch pounds is applied to the shaft. (19.htm (1 of 3) [5/8/2009 4:40:28 PM] .Torsion probs 1 Statics & Strength of Materials Topic 5.320 psi) file:///Z|/www/Statstr/statics/Torsion/torsp54a. Where does the maximum shear stress occur. Determine the maximum shear stress in the shaft..4a . The maximum allowable shear stress in the shaft is 8000 psi. The clamp screw is effectively a 5/16 inch diameter steel shaft.Topic 54a . (2.34 in4. 1410 psi) file:///Z|/www/Statstr/statics/Torsion/torsp54a. (16 lb) 4. Determine the polar section modulus of the shaft. The drive shaft of a car is a hollow steel shaft with an outside diameter of 3 inches and a wall thickness of 1/8 inch.Torsion probs 1 3. Find the maximum force that can be applied in the couple.htm (2 of 3) [5/8/2009 4:40:28 PM] . A C-clamp is tightened by applying a couple (equal forces in opposite directions on opposite sides) to a 3" diameter lever at the end of the clamp screw. What is the maximum shear stress in the shaft when it has a torque applied equal to 2200 inch pounds. Determine the angle of twist of drill. G = 1. A hollow steel shaft 2 inches in diameter and 3 feet long with 1/16 inch walls is used in an auger.178 rad) 7.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torsp54a. The drive shaft is 4 feet long and 2. Determine the maximum shear stress in the shaft and the angle of twist.042 rad. What is the angle of twist for the shaft? G= 12x106 psi.Topic 54a . (16. A 3/4 inch solid steel shaft is 8 inches long.0213 rad) 6.04x106 psi. 13. A three foot wooden dowel 1/4 inch in diameter has a torque of 5 inch pounds applied.5 inches in diameter with a 1/16 diameter wall thickness. 990 psi) 9. . . Determine the maximum torque that can be transmitted by the shaft.01733 rad) 8. . Determine the angle of twist of the shaft when a torque of 5000 inch pounds is applied to one end of the shaft.4513 rad) Select: Topic 5: Torsion.000 psi.htm (3 of 3) [5/8/2009 4:40:28 PM] . The steel drive shaft of a car has a torque applied of 1600 inch pounds.7 x 103 psi. Determine the angle of twist of the dowel and the maximum shear stress in the wood. The free length of the drill is 2 inches long. (82. The maximum allowable stress for the shaft is 12. Find the maximum torsional stress in the drill. (1630 psi. . Rivets & Welds . (5415 psi.Torsion probs 1 5. A 1/32 inch twist steel drill has a torque of 0.1 inch pounds applied. What is the maximum shear stress in the shaft? (.84 ft-lb. 000 psi. A quarter horsepower electric drill turns at 750 rpm. The shaft on the propeller transmits 3000 horse power.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torsp54b. It develops a torque of 75 inch pounds at 6000 rpm.(38.64 in. Determine the maximum power that can be delivered by the shaft if it is rotating at 6000 rpm. Determine the power delivered by the shaft when the angular speed of the shaft is 1500 rpm. A motorcycle is shaft driven with a 1 inch solid steel shaft.14 hp) 2. An outboard motor has a steel shaft 1/2 inch in diameter and 18 inches long.4b .) Select: Topic 5: Torsion.htm [5/8/2009 4:40:28 PM] . The drive shaft of a car has a torque applied of 1600 inch pounds. G = 12x106 psi.1 hp) 4.000 psi. Determine the horse power that is developed.5 in4.Topic 54b . What is the required diameter of the shaft if it is solid? (334. (150 hp. What is the torque that is developed in the drill bits? (1.5 inches in diameter with a 1/16 diameter wall thickness. Rivets & Welds . The propeller on an ship has a 72 inch pitch and turns at 180 rpm. The drive shaft is 4 feet long and 2. Determine the minimum polar moment of inertia of the shaft if the shaft steel has a maximum allowable shear stress of 12. The maximum allowable transverse shear stress in the shaft is 8.Torsion 1.Torsion probs 2 Statics & Strength of Materials Topic 5. (7. What is the angle of twist of the shaft if the shaft is 24 inches long. 7.Problem Assignment 2 .75 ft-lb) 3.032 rad) 5. . The chain ring on a bicycle has 42 teeth and the free wheel has 28 teeth. The pin is has an ultimate shear strength of 23. Determine the rotational speed of the second shaft. If this force is 120 pounds. Determine the torque delivered to the free wheel. A force of 150 pounds is applied to the 7 inch peddle arm. These forces are equal and opposite. Determine the maximum torque that can be transmitted if the shaft is 3/4 inch in diameter and the pin is 1/4 inch in diameter. Determine the torque delivered to the chain ring. b. c. c. The belt on the pulley has a tensile force of 65 pounds on one side and 35 pounds on the other side? Determine the torque applied to the shaft. The teeth of the gear on the first shaft push on the teeth of the gear on the second shaft. file:///Z|/www/Statstr/statics/Torsion/torsp54c. 8. a. b.htm (1 of 2) [5/8/2009 4:40:29 PM] . Determine the torque delivered to the free wheel. 6. Determine the power delivered if the free wheel rotates at 90 rpms. A V-belt is attached to a 6 inch diameter pulley. a. A propeller is attached to the drive shaft of an outboard by means of a brass shear pin. The chain ring has 3 teeth per inch. The first shaft is rotating at 300 rpms. The chain ring on a bicycle has 42 teeth and the free wheel has 15 teeth. Determine the power transmitted by the shaft if it is rotating at 600 rpms.4c . A force of 150 pounds is applied to the 7 inch peddle arm. d. Determine the power transmitted by the pulley if the shaft is turning at 1200 rpms.Torsion probs 3 Statics & Strength of Materials Topic 5. The belt tension on the two sides of the pulley are 60 pounds and 100 pounds. 3.Topic 54c . Determine the power transmitted by the torques in each of the two shafts in the problems above. The first shaft has an 8 inch gear with 40 teeth and the second shaft has a 5 inch gear with 25 teeth. determine the torques acting on each of the two shafts. d. The chain ring has 3 teeth per inch.Torsion 1.Problem Assignment 3 . A 3/4 inch diameter steel shaft has a 6 inch diameter pulley attached. Determine the torque delivered to the chain ring. Two rotating shafts are connected with gears. 5. Determine the force in the chain. 2. Determine the power delivered if the free wheel rotates at 90 rpms.000 psi. Determine the force in the chain. 4. Determine the torque applied to the pulley by the V-belt. Consider the shafts and gears in problem 4. 7. Torsion probs 3 Select: Topic 5: Torsion.Topic 54c .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torsp54c. Rivets & Welds .htm (2 of 2) [5/8/2009 4:40:29 PM] . Topic 5. A hollow steel shaft with an outer diameter of 2 inches and an inner diameter of 1.) If the allowable shear stress in the shaft is 20.000 lb/in2. B.400 lb/in2.htm (1 of 3) [5/8/2009 4:40:29 PM] . 1. If the maximum allowable shear stress in the shaft is 20. as shown in Diagram 1. What is the angle of twist which develops in the shaft? C.000 lb/in2) file:///Z|/www/Statstr/statics/Torsion/torsp54d.15. how large a force could be applied to the outer edge of the disk? (Answers: A.000 lb/in2.Torsion probs 4 Statics & Strength of Materials Topic 5.) If we were not given the outer diameter of the shaft.75 inches is attached to circular disk with a radius of 5 inches.031 radians.4c . A 2 foot long hollow steel shaft with an outer diameter of 1 inches is to transmit 100 horsepower while being driven a 1800 rev/min.000 lb.. .600 lb..) 2.Torsion Modulus of Rigidity for several materials: Steel = 12 x 106 lb/in2. what would be the minimum outer diameter of the shaft which could safely transmit the horse power ? (The allowable shear stress in the shaft is 20. A. A force of 2. B. Aluminum = 4 x 106 lb/in2. Brass = 6 x 106 lb/in2. but were told that the inner diameter was to be nine-tenths of the outer diameter. what is the maximum possible value of the inner diameter of the shaft. is applied to the outer edge of the disk.4d . A.Problem Assignment 3 . What is the maximum shear stress which develops in the hollow shaft? B. C. 2. The allowable shear stress in the aluminum is 14. C. Determine the maximum values of the applied torque.htm (2 of 3) [5/8/2009 4:40:29 PM] .Topic 5. file:///Z|/www/Statstr/statics/Torsion/torsp54d. do = 1.5 inches and an inner diameter of 1 inches is to transmit power while being driven a 3600 rev/min.4d .) If we were not given the outer diameter of the shaft. ) (Answers: A 548 hp.46") 4.37") 3. B. what is the maximum horsepower which can be transmitted down the shaft ? B. Torque of T1 and T2 are applied to the shaft in the directions shown in Diagram 4. A. what would be the minimum outer diameter of the shaft which could safely transmit the horsepower found in part A? (The allowable shear stress in the shaft is 18.. T1 and T2. . Shaft section BC is aluminum and has a diameter of 3 inches. Shaft section AB is steel and has a diameter of 2 inches..Torsion probs 4 (Answers: A.000 lb/ in2.) If we were to transmit the maximum power.000 lb/in2. A 2 foot long hollow brass shaft with an outer diameter of 1.000 lb/in2.000 lb/in2. A compound shaft is attached to the wall at point C. as shown in Diagram 4. such that the allowable shear stress is not exceeded in either shaft section.096 radians.573" . what would be the resulting angle of twist of the shaft due to the applied torque. B. but were told that the inner diameter was to be six-tenths of the outer diameter. C. and the allowable stress in the steel is 18. di = .) If the allowable shear stress in the shaft is 18. 1. C AB=. Load torque of 500 ft-lb. CD=1. and the allowable shear stress in the aluminum is 14. CD=137 hp. Determine the minimum diameter for each shaft.4d .Topic 5.71". Total = .. such that it can carry the transmitted power within the allowable stress limit.htm (3 of 3) [5/8/2009 4:40:29 PM] . The shaft is rotating at 2. A driving torque of 1. T2 = 8545 ft-lb.BC=056 rad-ccw.Torsion probs 4 (Answers: T1=2360 ft-lb..000 lb/in2. calculate the angle of twist of end D with respect to end A. and 400 ft-lb. C and D. For this shaft: A. the allowable shear stress in the brass is 16.) 5. Determine the horsepower being transferred through each shaft. act at points A. A compound shaft with applied torque is shown in Diagram 5.000 lb/in2.099 rad-ccw) Select: Topic 5: Torsion.800 ft-lb.09". 1000 ft-lb. AB=228 hp. Section AB is made of steel.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torsp54d. Section CD is made of aluminum. .. CD=078 rad-ccw. Rivets & Welds . B. The allowable shear stress in the steel is 20. is applied at point B. Using the diameters determined in part B. BC=594 hp. (Answers: A.15".400 rev/minute. Section BC is made of brass.035 rad-cw. AB=1. BC=1. C. B.000 lb/in2. Generally for bolts the hole is slightly larger than the bolt. in a riveted joint a heated rivet is forced into a hole connecting two "plates" (or beams).Riveted Joints In this topic we will discuss riveted and bolted joints. In our discussion. we will list some assumptions used in our discussion. We will also ignore the effect of friction in carrying the load. This is a conservative approach in the sense that by ignoring the effect of friction. Basically.5: Rivets & Welds . Because the friction effect can vary substantially.Topic 5.htm (1 of 5) [5/8/2009 4:40:30 PM] . 3. A lap joint is shown in Diagram 1.Lap Joints and Butt Joints. for both rivets and bolts. This friction may play a significant part is the amount of load a joint can carry. In a Lap Joint. 4.5: Rivets & Welds: Riveted Joints Topic 5. 1. That is. a tension develops in the rivet and the plates are forced together. file:///Z|/www/Statstr/statics/Torsion/rivw55. We will generally treat riveted and bolt joints in the same way and use the same method of analysis. the joint in actuality will normally carry more than the calculated strength of the joint. That the rivets and bolts completely filled the connecting holes. There are two basic types of riveted (bolted) joints . we will assume the hole diameter is the same as the rivet or bolt diameter. That the applied loads are carried equally by the rivets (bolts). there is significant friction between plates riveted or bolted together. Before we examine the specific modes of failure. two plates are overlapped and rivets or bolts penetrate the two plates connecting them together. In bolted joints. 2. That the rivet (bolt) shear stress is distributed uniformly over the cross sectional rivet (bolt) area. we will not try to include the contribution of friction to supporting the load. and this is taken into account in a careful analysis. That the tensile load carried by the plate is also distributed equally across the plate material. and a butt joint is shown in Diagram 2. As the rivet cools. high strength bolts are inserted into connecting holes between plates (or beams) and then tightened to a percentage (initially approximately 70 percent) of the allowable bolt tensile strength. 1. Rivet Shear: As shown in Diagram 3. There is a symmetric rivet pattern above each main plate as shown in Diagram 2. the two main plates are butted up against each other and then covered with one or two cover plates. The distance between rivets in a row in riveted (bolted) joint pattern is known as the Pitch. The first row (row 1) of a pattern is the row which is closest to the applied load. the minimum pitch is three times the diameter of the rivet (bolt). Thus if we take the allowable shear stress for the rivet material times the cross sectional area of the rivet this gives us the load one rivet area could carry in shear before it failed (By failed. a side view of a lap joint. the rivet area between the two main plates is in shear.5 times the rivet (bolt) diameter.which is the force parallel to the area in shear divided by area. we mean having exceeded the allowable stress. As a general guideline for steel or aluminum plates.) So we can write: file:///Z|/www/Statstr/statics/Torsion/rivw55. The distance between rows in a riveted (bolted) joint pattern is known as the Back Pitch (or Transverse Pitch. and the edge pitch (distance from the nearest rivet to the edge) is 1.Topic 5.htm (2 of 5) [5/8/2009 4:40:30 PM] .5: Rivets & Welds: Riveted Joints In a Butt Joint. or Gauge). We obtain the formula for the Strength of the Joint in Rivet Shear by simply using the definition of the shear stress . Then rivets or bolts penetrate the cover plates and the two main plates connecting them together. The load is transferred through one main plate to the cover plates by the rivets and then transferred back to the second main plate. Joint Failure: There are a number of ways in which a riveted (bolted) joint may fail. 2. This puts the plate material behind the rivet into compression. of course. In this case. on the maximum allowable compressive stress for the rivet and plate material .Topic 5. and this puts the rivet into compression. Again if the load is large enough. the rivet material may fail in compression. Where: N = Number of areas in shear. From the rivet's perspective. the plate is being pulled into it. This equals the number of rivets in a lap joint or in a butt joint with one cover plate. it is common practice to take the area in compression as the vertical cross sectional area of the rivet (Diagram 5). To determine the load which will cause failure. Which will fail first in compression depends.5: Rivets & Welds: Riveted Joints Privet shear = N (pi * d2/4) all. and twice the number of rivets in a butt joint with double cover plates. for both the area of the rivet in compression and the area of plate in compression. the top main plate is being pulled into the fixed rivet. As shown in middle diagram in Diagram 4. file:///Z|/www/Statstr/statics/Torsion/rivw55. A = pi * d2/4 (or pi * r2) is the cross sectional area of the rivet in shear. and if the load is large enough the plate material may fail in compression. Rivet/Plate Bearing Failure: This is compression failure of either the rivet or the plate material behind the rivet. all = The allowable shear stress for the rivet material. when we consider the top main plate. we again multiply the stress by the area.the lowest allowable compressive stress material will fail first.htm (3 of 5) [5/8/2009 4:40:30 PM] . n d)t all. that is.5: Rivets & Welds: Riveted Joints So we can write: Pbearing = N (d*t) all . we see that for equilibrium the plate material is in tension.Topic 5.. and we have to subtract the diameter of the rivet from the width of the plate (since the area of the plate is reduce due to the rivet hole). P. the plate will tear first where the holes are in the plate. we multiply the allowable tensile stress by the area in tension. where w = Width of the main plate file:///Z|/www/Statstr/statics/Torsion/rivw55. As is shown in Diagram 6. However. the plate can carry before it would fail in tension. we have cut the plate at rivet row 1. To determine the applied load. just as paper towels tear where the perforations are located. Then we can write: Prow1 = (w . if solid. where N = Number of rivets in compression d = Diameter of rivet t = Thickness of the main plate all = Allowable compressive stress of the rivet or plate material 3. This area is the cross sectional area of the plate which. Plate Tearing: This is a tensile failure of the plate material normally at the rivet row positions. would be the width of the plate times the thickness of the plate (A = w*t).htm (4 of 5) [5/8/2009 4:40:30 PM] . if we cut the plate material at rivet row 1 and look at the left end section. including plate shear .5b: Riveted Joints .htm (5 of 5) [5/8/2009 4:40:30 PM] .which may occur if a rivet is placed to close to the end of the plate. row 1.Topic 5.Example 2 When finished with Riveted Joints Examples. We will go into this in more detail in a later example.5: Rivets & Welds: Riveted Joints n = number of rivets in row (in this example. this mode is not normally a problem.6:Rivets & Welds .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivw55. Please select : Topic 5. If proper placing of rivets is maintained. since some of the load has already be transferred to the second plate. We will consider only the three main modes of failure discussed above. There are several additional ways the joint may fail. Continue to: Topic 5. due to the fact that rows beyond row 1 are no longer carrying the entire load.Riveted Joint Selection or Select: Topic 5: Torsion. We will now look at several examples of Riveted Joints.Example 1 Topic 5.5a: Riveted Joints . and the plate material behind the rivet fails in shear. Rivets & Welds . P. 1 rivet) d = Diameter of rivet t = Thickness of the main plate all = Maximum allowable tensile stress for the plate material The formula for plate tearing are rivet rows beyond row 1 have to be modified somewhat. we calculate the load. Plate Tearing (Row 1): We now determine the load the joint (plate) can carry before failing in file:///Z|/www/Statstr/statics/Torsion/rivwe55a..1416 * (5/8)2/4]*16. depending on how much the allowable stress is exceeded.700 lb. 1. 000 lb/in . Part 1: To determine the Strength of the Joint. 2 c = 24. Bearing (compression) Failure: We next determine the load the rivet or plate can carry before failing in compression.Example 1 Topic 5. The allowable stresses are as follows: Rivets: Plate: = 16. 3. which will cause the joint to fail in each of the main modes of failure (Rivet Shear. Notice we use the smaller of the allowable compressive stress between the rivet and plate.Topic 5. Please note we used the allowable shear stress for the rivet material in our equation. Rivet Shear: The load the joint can carry before failing in rivet shear is given by: P = N (pi * d2/4) = ( 9 rivet areas)* [3. and Plate Tearing). = 14. and the Efficiency of the Joint. The lowest load which will cause the joint to fail is known as the Strength of the Joint.5a: Riveted Joints . the joint will fail in shear. P. At this point the joint may or may not actually fail depending on the safety factor used in the determining the allowable stress. 000 lb/in2.Example 1 A riveted lap joint is shown in Diagram 1. as it is the rivet (not the plate) which fails in shear.000 lb/in2. Bearing. and of course. Please remember that the term failure of the joint in these problems does not mean when the joint actually breaks. 000 lb/in We would like to determine the Strength of the Joint.700 lb.200 lb. 2. 000 lb/in2 = 44.htm (1 of 3) [5/8/2009 4:40:30 PM] . Thus at a load of 44. 200 lb. will cause the joint (plate) to fail in compression.000 lb/in2) = 64. 000 lb/in .5a: Riveted Joints . 000 lb/in 2 c = 23. The width of the plates is 6 inches. and the thickness of the plates is 1/2 inch. Thus a load of 64. 2 t = 20. The diameter of the rivets is 5/8 inch. Pbearing = N (d*t) all = ( 9 rivets) * (5/8" *1/2") * (23. 2 t = 22. but rather we consider the joint to have failed when the stress in the joint exceed the allowable stress. P.1* 5/8") *(1/2") * (20. This is due to the fact that part of the load has been transferred to the bottom plate by the rivet in row 1.500 lb.3* 5/8") *(1/2") * (20. Then we can write: 2 (6/9)Prow3 = (w . Up to this point. Of all the loads which will cause the joint to fail (either in shear. and thus the plate material at row 3 carries only 6/9 of the load. and then: Prow3 = (9/6) * 41. We see that it is lower than the load that will cause row 1 to fail (that is. Prow1 = (w . The way we determine how good or efficient a joint we have is to compare the strength of the joint with the strength of the plate if it were solid (no riveted or bolted joint). By itself the strength of the joint does not really tell us how 'good' the joint is. We see that of the three main modes of failure above. = 53. however we still need to check plate tearing failure at rivet row 2 (and perhaps row 3) to see if the joints fails at a lower load there.250 lb.htm (2 of 3) [5/8/2009 4:40:30 PM] . See Diagram 2.500 lb. = 61.250 lb.at row 1. Part II. the joint would fail at row 2 before row 1). 2 (8/9)Prow2 = (w ..200 lb.200 lb. however the joint will still fail first in rivet shear (44.. Since there are 9 rivets in the pattern and we assume that the rivets share the load equally. The strength of the plate we determine by realizing that if it is solid the joint will fail in tension (plate tearing). This is the load which will cause shear row 2 to fail in tension. and this is the final Strength of the Joint.n d)t all = (6". the largest load we can safely apply..400 lb. we need now to continue and determine the load at which row 3 would fail.000 lb/in ) = 47.n d)t all = (6". 3a.000 lb/in ) = 41. P. found above.2* 5/8") *(1/2") * (20. this means that 3/9 of the load has been transferred to the bottom plate. We do this as follows.Example 1 tension . Here we see that load which causes row 3 to fail in tension is much larger than the loads which would cause row 1 and row 2 to fail. or tension) the lowest is still the rivet shear . and we may write. The strength of the joint is 44. compression.000 lb/in2 ) = 53. Plate Tearing (Row 3): Since the joint would fail at row 2 before row 1. This is the load which will cause the plate to fail in tension at row 1.200 lb. 3b.900 lb.5a: Riveted Joints .) . this is the Strength of the Joint. and then: Prow2 = (9/8) * 47. Thus row 2 carries of 8/9 of the load P.since that is the lowest applied load which will cause the joint to exceed the allowable stress. And we can write: file:///Z|/www/Statstr/statics/Torsion/rivwe55a. then one ninth (1/9) of the load has been transferred to the bottom plate. As there are a total of 3 rivets in row 1 and row 2.200 lb.750 lb. the joint fails first in rivet shear at the lowest load of 44. This means we can stop here.44.n d)t all = (6"..Topic 5. Plate Tearing (Row 2): The main difference between Plate Tearing at row 2 and at row 1 (other than the fact that there are different numbers of rivet in the rows) is that plate material at row 2 does not carry the entire load. 000 lb. This is the plate strength.htm (3 of 3) [5/8/2009 4:40:30 PM] .5a: Riveted Joints . and we then define the joint efficiency as the ratio of the Joint Strength to the Plate Strength.7 percent of what the solid plate could carry. Rivets & Welds .Riveted Joints or Select: Topic 5: Torsion.Example 1 Plate Strength = area of plate cross section times allowable tensile stress for plate material. or Efficiency = Joint Strength / Plate Strength = 44.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwe55a.5: Rivets & Welds . or Pplate = (w * t) * all = (6" * 1/2") * 20. = .000 lb.000 lb/in2 = 60.737 = 73. This tells us that the joint can carry 73.7 %. / 60.200 lb. Return to: Topic 5.Topic 5. The width of the plates is 6 inches. 000 lb/in2 = 95. 000 lb/in . The allowable stresses are as follows: Rivets: Plate: = 18. For a butt joint we use only one half of the joint pattern. Rivet Shear: The load the joint can carry before failing in rivet shear is given by: P = N (pi * d2/4) = ( 12 rivet areas)* [3. The lowest load which will cause the joint to fail is known as the Strength of the Joint. file:///Z|/www/Statstr/statics/Torsion/rivwe55b.000 lb/in2.400 lb. since if one side fails.Topic 5. 000 lb/in We would like to determine the Strength of the Joint.1416 * (3/4)2/4]*18. P.Example 2 Topic 5. the main plate on that side has failed and the joint has failed. and the thickness of the plates is 1/2 inch. Bearing. 1. the joint will fail in shear. 2 t = 21. Part 1.htm (1 of 3) [5/8/2009 4:40:30 PM] . We will consider the left half of the joint pattern in this example. The diameter of the rivets is 3/4 inch. Notice that we use 12 rivet areas since each of the six rivets in the left half of the pattern has two areas in shear in the double cover plate butt joint..Example 2 A riveted butt joint is shown in diagram 1.400 lb. 2 c = 24.5b: Riveted Joints . and Plate Tearing). (See Diagram 2) Thus at a load of 95. which will cause the joint to fail in each of the main modes of failure (Rivet Shear. we calculate the load. = 16. 2 t = 22. 000 lb/in2. 000 lb/in 2 c = 22. and the Efficiency of the Joint. To determine the Strength of the Joint.5b: Riveted Joints . 000 lb/in . We see that of the three main modes of failure above.n d)t all = (6". 3b. This is the load which will cause the plate to fail in tension at row 1. the joint would fail at row 1 before row 2). . and we may write. We do this as follows.at row 1. Pbearing = N (d*t) all = ( 6 rivets) * (3/4" *1/2") * (22. Note that we use the smaller of the allowable compressive stress between the rivet and plate.1* 3/4") *(1/2") * (21.100 lb. 3a.500 lb.Example 2 2. since that should fail at a even larger load. we normally do not have to check row 3. Plate Tearing (Row 2): The main difference between Plate Tearing at row 2 and at row 1 (other than the fact that there are different numbers of rivet in the rows) is that plate material at row 2 does not carry the entire load. Plate Tearing (Row 1): We now determine the load the joint (plate) can carry before failing in tension .500 lb.500 lb. We see that it is larger than the load that will cause row 1 to fail (that is. Bearing (compression) Failure: We next determine the load the rivet or plate can carry before failing in compression.htm (2 of 3) [5/8/2009 4:40:30 PM] . and then: Prow2 = (6/5) * 47. the joint fails first in bearing (compression) failure at the lowest load of 49. this is the Strength of the Joint.Topic 5.000 lb/in2 ) = 55. Thus a load of 49..500 lb.2* 3/4") *(1/2") * (21. This is due to the fact that part of the load has been transferred to the bottom plate by the rivet in row 1.000 lb/in2 ) = 47. P. This is the load which will cause shear row 2 to fail in tension. (5/6)Prow2 = (w . Up to this point.250 lb. 3.700 lb. Since there are 6 rivets in left side pattern and we assume that the rivets share the load equally.000 lb/in2) = 49.5b: Riveted Joints .since that is the lowest applied load which will cause the joint to exceed the allowable stress. Thus row 2 carries of 5/6 of the load P. then one sixth (1/6) of the load has been transferred to the cover plates and on the second main plate. Plate Tearing (Row 3): Since the load which causes row 2 to fail is larger than the load which causes row 1 to fail.n d)t all = (6". = 56. Prow1 = (w .500 lb. will cause the joint (plate) to fail in compression. file:///Z|/www/Statstr/statics/Torsion/rivwe55b. however we still need to check plate tearing failure at rivet row 2 (and perhaps row 3) to see if the joints fails at a lower load there. however the joint will still fail first in bearing (compression) failure at 49. or Pplate = (w * t) * all = (6" * 1/2") * 21.5: Rivets & Welds . and this is the final Strength of the Joint. and thus the plate material at row 3 carries only 3/6 of the load.000 lb. Here we see that load which causes row 3 to fail in tension is much larger than the loads which would cause row 1 and row 2 to fail. Rivets & Welds . and we then define the joint efficiency as the ratio of the Joint Strength to the Plate Strength. = 78.Riveted Joints Continue to: Topic 5. Of all the loads which will cause the joint to fail (either in shear.000 lb. The way we determine how good or efficient a joint we have is to compare the strength of the joint with the strength of the plate is it were solid (no riveted or bolted joint). P. As there are a total of 3 rivets in row 1 and row 2. This tells us that the butt joint can carry 78. or Efficiency = Joint Strength / Plate Strength = 49.000 lb/in2 = 63. Return to: Topic 5.Topic 5.786 = 78.800 lb. this means that 3/6 of the load has been transferred to the bottom plate.Example 2 However.. we will check row 3..500 lb.6 percent of what the solid plate could carry. See Diagram 3. and then: Prow2 = (6/3) * 39.000 lb/in2 ) = 39. = . simply as an exercise.500 lb. or tension) the lowest is still the bearing failure.Riveted Joint Selection or Select: Topic 5: Torsion.400 lb. compression.n d)t all = (6". The strength of the plate we determine by realizing that if it is solid the joint will fail in tension (plate tearing).Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwe55b.400 lb..6: Rivets & Welds . Part II.5b: Riveted Joints . This is the plate strength. Then we can write: (3/6)Prow2 = (w . By itself the strength of the joint does not really tell us how 'good' the joint is. found above. 49. / 63.6 %.3* 3/4") *(1/2") * (21. the largest load we can safely apply.500 lb. And we can write: Plate Strength = area of plate cross section times allowable tensile stress for plate material. The strength of the joint is 49.htm (3 of 3) [5/8/2009 4:40:30 PM] . Use the number of rivets found in step 4 to design a joint pattern. Calculate the load which would cause the joint to fail in plate tearing at row 1 . (See Diagram 1 for example of possible rivet patterns. We will apply the following procedure to this question: 1.6: Rivets & Welds . Calculate the maximum load which one rivet (or the plate material behind one rivet) could carry in bearing (compression) 4. use the number of rivets to select the best joint pattern from a given set of patterns. calculated in step 1.Topic 5.6: Rivets & Welds: Riveted Joint Selection Topic 5. and allowable stresses).htm (1 of 4) [5/8/2009 4:40:31 PM] .) Example .assuming one rivet in row 1.Joint Selection file:///Z|/www/Statstr/statics/Torsion/rivw56. A related and interesting question is how do we go about determining (or selecting) the most efficient joint pattern for given main plates. 2. 3. Calculate the maximum load which one rivet could carry in rivet shear. and round up to determine the number of rivets which will result in the most efficient joint (for the given main plate dimension. by the smallest of the loads one rivet could carry from steps 2 and 3. 5. Divide the allowable load in plate tearing. or more usually. rivet dimensions.Riveted Joint Selection We have so far examined how we determine the strength and efficiency of a given riveted (bolted) joint pattern. . N = 1 d = 3/4 inch all = 16. 2 c = 25. Step 2. 000 lb./in2. Calculate the strength and efficiency of the joint./in2. The maximum allowable stresses for the rivet and weld materials are as follows: Rivets: Plate: = 16./in 2 c = 24./in . 000 lb.1 * 3/4")*(1/2") * 20. B.000 lb. 000 lb.6: Rivets & Welds: Riveted Joint Selection A lap joint (See Diagram 2) is to connect two steel plates both with a width of 6 inches and a thickness of 1/2 inch./in A. 000 lb. C. 000 lb.assuming one rivet in row 1. The rivets to be used have a diameter of 3/4 inch./in2. 2 t = 20. 1 rivet) d = diameter of rivet = 3/4 inch t = thickness of the main plate = 1/2 inch all = Maximum allowable tensile stress for the plate material = 20. Select the best pattern from Diagram 1 above.000 lb. 2 t = 22. We use the plate tearing relationship: Prow1 = (w . Solution: Part A Step 1: Calculate the load which would cause the joint to fail in plate tearing at row 1 ./in2 = 52. Determine the number of rivets for the most efficient joint. We use the rivet shear relationship: Privet shear = N (pi * d2/4) all. where w = width of the main plate = 6 inches n = number of rivets in row (in row 1. Where: N = Number of areas in shear For 1 rivet in a lap joint this will be 1 area. Prow1 = (6" .n d)t all . = 17.500 lb. Calculate the maximum load which one rivet could carry in rivet shear./in . 000 lb./in2 file:///Z|/www/Statstr/statics/Torsion/rivw56.htm (2 of 4) [5/8/2009 4:40:31 PM] . 000 lb.Topic 5. The total load the joint can carry in bearing (compression) will be the product of the number of rivets and the allowable load per rivet: Pbearing = 8 * 9000 lb. Calculate the maximum load which one rivet (or the plate material behind one rivet) could carry in bearing (compression) Pbearing = N (d*t) all. Prow1 = (6" . (See Diagram 3.) Rivet Shear: We have already determined that one rivet can carry 7070 lb. ./in2 = 7070 lb.Topic 5.) Plate Tearing (row 1): We have already calculated plate tearing at row 1.000 lb. 000 lb. 1. we calculate the strength and efficiency of the joint. calculated in step 1.500 lb. However. where N = Number of rivets in compression = 1 d = Diameter of rivet = 3/4 inch t = Thickness of the main plate = 1/2 inch all = Lowest allowable compressive stress of the rivet or plate material = 24./in2 Pbearing = 1 * [(3/4") * (1/2")] * 24.htm (3 of 4) [5/8/2009 4:40:31 PM] ./in2 = 9000 lb. 3.1416 * (3/4")2/4] * 16. so the total load the joint can carry in rivet shear will be the product of the number of rivets and the allowable load per rivet: Pshear = 8 * 7070 lb. From Diagram 1 above we select the a lap joint pattern with correct number of rivets.43 .) Solution: Part C Finally./rivet Step 4. rounding up # rivets = 8 Solution: Part B We use the number of rivets found in Part A above to select the best joint pattern from the given set of patterns. / 7070 lb./ rivet. in step 1 above.) Bearing Failure: Again. 2. by the small of the loads one rivet could carry from steps 2 and 3. and round up to determine the number of rivets which will result in the most efficient joint. now that we have selected the rivet file:///Z|/www/Statstr/statics/Torsion/rivw56. using the selected joint pattern from part B. 000 lb. Divide the allowable load in plate tearing.6: Rivets & Welds: Riveted Joint Selection Privet shear = 1 [3./rivet = 72./in2 = 52.000 lb./rivet. 000 lb./rivet Step 3.1 * 3/4")*(1/2") * 20. we have already calculated the allowable load per rivet in bearing = 9000 lb./rivet = 56. 500 lb./rivet = 7. # Rivets = 52.560 lb. Rivets & Welds .2 * 3/4")*(1/2") * 20. (plate tearing row 2).400 lb. However. so we write: (7/8)Prow2 = (6" ..000 lb. Considering all the modes of failure. up to this point. so we can stop here. = 72.000 lb.000 lb. 000 lb.000 lb. we will continue and check the load which causes row 3 to fail.6a: Riveted Joint Selection ./in2 = 45.000 lb. we also need to calculate plate tearing at row 2 (and perhaps row 3) 4.400 lb. and then Prow2 = (8/7) 45. to cause failure. so we write: (5/8)Prow3 = (6" . 000 lb. Notice.Topic 5. This loading is much larger than the loading which will cause row 2 to fail. we see: Strength of the Joint is 51. that this loading is the lowest./in2 = 45.6: Rivets & Welds: Riveted Joint Selection pattern.. This type of problem is easily done using current computer spreadsheet methods.) Plate Tearing (row 3): Row 2 carries 5/8 of the load. By changing the size of the rivets (and the material and associated allowable stresses) we could arrive at a joint of greater efficiency. = 51./(6"*1/2")*20. Continue to: Topic 5. 5.) Plate Tearing (row 2): Row 2 carries 7/8 of the load.86 =86% This is the most efficient joint for the dimensions and allowable stresses of both the plates and rivets.htm (4 of 4) [5/8/2009 4:40:31 PM] .2 * 3/4")*(1/2") * 20. and then Prow3 = (8/5) 45. 400 lb.Example 1 or Select: Topic 5: Torsion./in2 = . 000 lb. and Efficiency = Joint Strength / Plate Strength = 51. and so it is currently the strength of the joint.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivw56. 6a: Riveted Joint Selection . 000 lb/in2. The rivets to be used have a diameter of 5/8 inch.n d)t all .Example 1 Topic 5.htm (1 of 4) [5/8/2009 4:40:31 PM] . B. = 16.Example 1 Riveted Joint Selection . 2 c = 26.000 lb/in2. We use the plate tearing relationship: Prow1 = (w . 000 lb/in . Solution: Part A Step 1: Calculate the load which would cause the joint to fail in plate tearing at row 1 .6a: Riveted Joint Selection . 000 lb/in 2 c = 24. where w = width of the main plate = 7 inches file:///Z|/www/Statstr/statics/Torsion/rivwe56a. C.assuming one rivet in row 1. 2 t = 22. Determine the number of rivets for the most efficient joint. Select the best pattern from the patterns in Diagram 2 . 000 lb/in .Example 1 A butt joint (Diagram 1) is to connect two steel plates both with a width of 7 inches and a thickness of 3/4 inch. 2 t = 24. 000 lb/in A. The maximum allowable stresses for the rivet and weld materials are as follows: Rivets: Plate: = 15. Calculate the strength and efficiency of the joint.Topic 5. . where N = Number of rivets in compression = 1 d = Diameter of rivet = 5/8 inch t = Thickness of the main plate = 3/4 inch all = Lowest allowable compressive stress of the rivet or plate material = 24./rivet Step 4. 1 rivet) d = diameter of rivet = 5/8 inch t = thickness of the main plate = 3/4 inch all = Maximum allowable tensile stress for the plate material = 22. Prow1 = (7" ./rivet Step 3. From Diagram 1 above we select the a lap joint pattern with correct number of rivets. Pbearing = N (d*t) all. Divide the allowable load in plate tearing.htm (2 of 4) [5/8/2009 4:40:31 PM] . so N = 2 d = 5/8 inch all = 15.6a: Riveted Joint Selection . We use the rivet shear relationship: Privet shear = N (pi * d2/4) all. Calculate the maximum load which one rivet could carry in rivet shear. Step 2. 000 lb/in2 Pbearing = 1 * [(5/8") * (3/4")] * 24.Topic 5.250 lb. 200 lb. . by the smallest of the loads one rivet could carry from steps 2 and 3.000 lb/in2 Privet shear = 2 [3.43 .1 * 5/8")*(3/4") * 22. 000 lb/in2 = 11. and round up to determine the number of rivets which will result in the most efficient joint. (See Diagram 3.) file:///Z|/www/Statstr/statics/Torsion/rivwe56a. 000 lb/in2 = 105. / 9200 lb. rounding up # rivets = 12 Solution: Part B We use the number of rivets found in Part A above to select the best joint pattern from the given set of patterns. calculated in step 1.200 lb. For 1 rivet in a double cover plate butt joint this will be 2 area.Example 1 n = number of rivets in row (in row 1. Where: N = Number of areas in shear. Calculate the maximum load which one rivet (or the plate material behind one rivet) could carry in bearing (compression).000 lb/in2 = 9200 lb. # Rivets = 105.1416 * (5/8")2/4] * 15./rivet = 11. 000 lb/in2. we see:Strength of the Joint is 103. and then Prow2 = (12/11) 94.6a: Riveted Joint Selection .875 lb. that this loading is the lowest.500 lb. 3. we calculate the strength and efficiency of the joint. Considering all the modes of failure.500 lb.1 * 5/8")*(3/4") * 22. (plate tearing row 2). and so it is currently the strength of the joint.875 lb.400 lb.Example 1 Solution: Part C Finally. now that we have selected the rivet pattern. By changing the size of the rivets (and the material and associated allowable stresses) we could arrive at a joint of greater efficiency./rivet = 135. so we write: (11/12)Prow2 = (7" .875 lb.500 lb. = 103. This type of problem is easily done using current computer spreadsheet methods.. 2. Notice. However.000 lb. This loading is much larger than the loading which will cause row 2 to fail. we have already calculated the allowable load per rivet in bearing = 9000 lb.2 * 5/8")*(3/4") * 22. 000 lb/in2 = 105./(7"*3/4")*22.) Bearing Failure: Again. in step 1 above.Topic 5.2 * 5/8")*(3/4") * 22.896 = 89..) Plate Tearing (row 3): Row 3 carries 9/12 of the load. The total load the joint can carry in bearing (compression) will be the product of the number of rivets and the allowable load per rivet: Pbearing = 12 * 11.) Plate Tearing (row 2): Row 2 carries 11/12 of the load. using the selected joint pattern from part B.875 lb. we also need to calculate plate tearing at row 2 (and perhaps row 3) 4. However./rivet.) Plate Tearing (row 1): We have already calculated plate tearing at row 1.250 lb. so we can stop here. up to this point. and Efficiency = Joint Strength / Plate Strength = 103.200 lb. 5. 000 lb/in2 = 94. 000 lb/in2 = 94. 1./rivet. file:///Z|/www/Statstr/statics/Torsion/rivwe56a.200 lb.000 lb/in2 = . = 126.500 lb. to cause failure. It is usually possible to arrive at a joint efficiency of 90% or greater.6% This is the most efficient joint for the dimensions and allowable stresses of both the plates and rivets.) Rivet Shear: We have already determined that one rivet can carry 9. so we write: (9/12)Prow3 = (7" . Prow1 = (7" . we will continue and check the load which causes row 3 to fail. so the total load the joint can carry in rivet shear will be the product of the number of rivets and the allowable load per rivet: Pshear = 12 * 9200 lb/rivet = 110.htm (3 of 4) [5/8/2009 4:40:31 PM] . and then Prow3 = (12/9) 94. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwe56a.htm (4 of 4) [5/8/2009 4:40:31 PM] . Rivets & Welds .6a: Riveted Joint Selection .Riveted Joint Selection Continue to:Topic 5.7: Rivets & Welds .Example 1 Return to: Topic 5.Welded Joints or Select: Topic 5: Torsion.6: Rivets & Welds .Topic 5. plate = 62.500lb/ (6in*0. The diameter of the rivets is 1 inch.Solution601 STATICS & STRENGTH OF MATERIALS .000 psi) = 117. Part A: 1).(2* 1 in))(0.nd)t*σ = (6 in . Pbearing = n(d*t)σ = 5(1 in* 0.htm [5/8/2009 4:40:32 PM] .row 1 = (w . Rivets & Welds .500 lb Then strength of joint equals lowest load causing failure = 62.(1* 1 in))(0. 000 psi A.000 lb T Pplate.5 in)(30.000 psi σt = 25. 000 psi σc = 35.) Calculate the strength of the joint.000 psi) = 75. 000 psi Plate: τ = 12.row 2 = (5/4)(50. The width of the plates is 6 inches.Examples A riveted butt joint is shown below.nd)t*σ = (6 in .5in) (25.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol601.) Calculate the efficiency of the joint. and the thickness of the plates is 1/2 inch.500 lb T (4/5) Pplate.row 2 = (w .000 psi) = 50. 000 psi σc = 30.000 lb C 3).000 lb) = 62. Pshear = n(πd2/4)τrivets = 10(π*(1 in)2/4)(15.5 in)(25.000 psi) = 62. 000 psi σt = 30. The allowable stresses are as follows: Rivets: τ = 15.833 Topic 5: Torsion.5 in)(25.000 psi) = . Pplate.800 lb 2).500 lb Part B: efficiency = strength of joint/strength of solid plate = strength/(w*t)sT. B. The allowable stresses are as follows: Rivets: τ = 15. The diameter of the rivets is 3/4 inch.75 in))(0.(1* 0.75in)(28.000 psi) = 132.500 lb) = 128.000 psi) = 168.75 in)(30.350 lb/ . Pshear = n(πd2/4)τrivets = 20(π*(0.550 lb 2).Table of Contents file:///Z|/www/Statstr/statics/Stests/rivweld/sol602.75 in))(0.75 in)(28.750 lb C 3). plate = 128.75 in* 0.nd)t*σ = (7 in .Solution602 STATICS & STRENGTH OF MATERIALS . 000 psi σt = 32. Pbearing = n(d*t)σ = 10(0. and the thickness of the plates is 3/4 inch.) Calculate the efficiency of the joint.Examples A riveted butt joint is shown below. 000 psi A.(2* 0.75 in)2/4)(15. 000 psi Plate: τ = 14.htm (1 of 2) [5/8/2009 4:40:32 PM] T. The width of the plates is 7 inches.350 lb Part B: efficiency = strength of joint/strength of solid plate = strength/(w*t)s (7in*0. 000 psi σc = 34.row 2 = (w .nd)t*σ = (7 in .row 1 = (w .) Calculate the strength of the joint.500 lb T Pplate.250 lb T (9/10) Pplate.75 in)(28.350 lb Then strength of joint equals lowest load causing failure = 128.000 psi) = .000 psi) = 131.000 psi σt = 28. 000 psi σc = 30.873 Topic 5: Torsion. B. Part A: 1). Rivets & Welds .row 2 = (10/9)(115. Pplate.000 psi) = 115. Solution602 Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol602.htm (2 of 2) [5/8/2009 4:40:32 PM] . row 1/ (smallest) load/rivet = 191.htm (1 of 2) [5/8/2009 4:40:32 PM] . B.row 1 = (w . The main plates have a width of 8 inches and a thickness of 1 inch.Examples Two steel plates are to be connected using a double cover plate butt joint as shown below.365 lb/rivet = 26. The rivets to be used have a diameter of 5/8 inch.000 psi) = 191.365 lb/rivet 3). Part A: 1).000 psi σc = 28. In this case no shown pattern is large enough. and three rivets in row three. using the selected pattern.625 in)2/4)(12. Pshear.04 @ 27 Rivets. and select the best pattern to use. two rivets in row 2. # of Rivets = Plate strength. Pbearing = n(d*t)σ = 1(0.000 psi Rivet: τ = 12.000 psi A.000 psi) = 17.625 in* 1 in)(28.(1* 0. we will use one rivet in row 1.000 psi σt = 24.) Determine the strength and efficiency of the joint of maximum efficiency.nd)t*σ = (8 in . In checking strength of joint in plate tearing.500 lb C 4).000 psi σc = 30.750 lb T 2). 1 rivet = n(πd2/4)τshear = 2(π*(0.) Determine the number of rivets which will result in a joint of maximum efficiency.750 lb/ 7.625 in))(1 in)(26.000 psi σt = 26.000 psi) = 7. file:///Z|/www/Statstr/statics/Stests/rivweld/sol603.Solution603 STATICS & STRENGTH OF MATERIALS . Pplate. The allowable stress are: Plate: τ =15. So check Row 2.160 lbs.000 psi) = 175.250 lb (Since row 2 is lower than row 1.000 psi) = .htm (2 of 2) [5/8/2009 4:40:32 PM] . Rivets & Welds .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol603.861 Topic 5: Torsion.) Row 3: (24/27) Pplate.160 lb So Strength of joint is plate Tearing Row 3 = 179.nd)t*σ = (8 in . Unless Row 2 is lower.Solution603 Part B: Strength of joint will be Plate Tearing at Row 1 = 191. Row 2: (26/27) Pplate.row 2 = (27/26)(175.500 lb T Pplate. plate = 179.250 lb) = 179.625 in))(1 in)(26.625 in))(1 in)(26. row 2 = (w . efficiency = strength of joint/strength of solid plate = strength/(w*t)sT.nd)t*σ = (8 in .row 3 = (27/24)(159.000 psi) = 159.750 lbs. row 2 = (w .160 lb/ (8in*1in) (26.(2* 0.(3* 0.250 lb T Pplate. also check row 3.500 lb) = 182. 25 in* 0.row 1 = (w .000 psi A.375 lb/rivet C 4).000 psi) = 17.Solution604 STATICS & STRENGTH OF MATERIALS .375 lb/ 17. 1 rivet = n(πd2/4)τshear = 1(π*(1. The rivets to be used have a diameter of 1 1/4 inch. Allowable stress are: Rivet : τ = 14. Pbearing = n(d*t)σ = 1(1.32 @ 11 Rivets file:///Z|/www/Statstr/statics/Stests/rivweld/sol604.75 in)(22.25 in)2/4)(14.000 psi σt = 22.180 lb/rivet 3).000 psi) = 177.000 psi σt = 26.000 psi σc = 26.htm (1 of 2) [5/8/2009 4:40:32 PM] .000 psi σc = 30.Examples Two steel plates are to be connected by a lap joint as shown below.) Determine the strength and efficiency of the joint of maximum efficiency using the selected pattern.75 in)(26.000 psi) = 24.000 psi Plate: τ =12. B. Pplate.) Determine the number of rivets which will result in a joint of maximum efficiency and select the best pattern to use. Pshear. # of Rivets = Plate strength. Part A: 1).25 in))(0.375 lb T 2). row 1/ (smallest) load/rivet = 177. The main plates have a width of 12 inches and a thickness of 3/4 inch.nd)t*σ = (12 in .(1* 1.180 lb/rivet = 10. (2* 1.750 lb T P plate.75 in)(22.750 lb) = 171.000 psi) = 156. row 2 = (12/11)(156.375 lbs. (11/12) Pplate. efficiency = strength of joint/strength of solid plate = strength/(w*t)s (12in*0. Unless Row 2 is lower.Solution604 Part B: Strength of joint will be Plate Tearing at Row 1 = 177. plate file:///Z|/www/Statstr/statics/Stests/rivweld/sol604.nd)t*σ = (12 in . So check Row 2.864 Topic 5: Torsion.000 lbs.htm (2 of 2) [5/8/2009 4:40:32 PM] . Rivets & Welds .000 lb/ T.75in)(22.Table of Contents Statics & Strength of Materials Home Page = 171.000 lb So Strength of joint is plate Tearing Row 2 = 171. row 2 = (w .000 psi) = .25 in))(0. 050 lb/rivet = 8. Pshear. # of Rivets = Plate strength.htm (1 of 2) [5/8/2009 4:40:33 PM] .000 psi σc = 30. and select the best pattern to use.000 psi σt = 20.03 @ 9 Rivets. Pplate. The main plates have a width of 10 inches and a thickness of 3/4 inch. Part A: 1).(1* 0.) Determine the number of rivets which will result in a joint of maximum efficiency.000 psi 2). The rivets to be used have a diameter of 7/8 inch.250 lb/rivet 3). Pbearing = n(d*t)σ = 1(0.000 psi) = 19.Examples Two steel plates are to be connected using a double cover plate butt joint as shown below. row 1/ (smallest) load/rivet = 136.Solution605 STATICS & STRENGTH OF MATERIALS .) Determine the strength and efficiency of the joint of maximum efficiency using the selected pattern . file:///Z|/www/Statstr/statics/Stests/rivweld/sol605.000 psi A.875 lb/ 17. The allowable stress are: Plate: τ =14.875 in)2/4)(16.75 in)(20.row 1 = (w . B.000 psi) = 17.875 lb T Rivet: τ = 16.875 in))(0.000 psi σc = 26.000 psi) = 136.000 psi σt = 24.75 in)(26.nd)t*σ = (10 in . 1 rivet = n(πd2/4)τshear = 2(π*(0.875 in* 0.050 lb C 4). row 2 = (9/8)(123.000 psi) = 123.875 in))(0.Solution605 Part B: Strength of joint will be Plate Tearing at Row 1 = 136.913 Topic 5: Torsion.75in)(20.(2* 0.875 lb/ T. Rivets & Welds .875 lbs. row 2 = (w .000 psi) = . (8/9) Pplate. So check Row 2.220 lb So Strength of joint is plate Tearing Row 1 = 136.875 lbs.Table of Contents Statics & Strength of Materials Home Page = 136.750 lb) = 139.750 lb T P plate.nd)t*σ = (10 in . plate file:///Z|/www/Statstr/statics/Stests/rivweld/sol605. Unless Row 2 is lower.htm (2 of 2) [5/8/2009 4:40:33 PM] . efficiency = strength of joint/strength of solid plate = strength/(w*t)s (10in*0.75 in)(20. nd)t*σ = (9 in .300 lb Part B: efficiency = strength of joint/strength of solid plate = strength/(w*t)s (9in*0.000 psi) = 152. Pshear = n(πd2/4)τrivets = 16(π*(0. 000 psi σc = 34.875 in))(0.htm (1 of 2) [5/8/2009 4:40:33 PM] T.875 lb) = 178.75 in)(28.75 in)(28.75 in)(30.500 lb Then strength of joint equals lowest load causing failure = 144. Pplate. 000 psi σt = 32.000 psi) = 133.Examples A riveted butt joint is shown below.763 Topic 5: Torsion.Table of Contents file:///Z|/www/Statstr/statics/Stests/rivweld/sol606. The width of the plates is 9 inches. Pbearing = n(d*t)σ = 8(0.row 1 = (w . Part A: 1). The diameter of the rivets is 7/8 inch. 000 psi A.300 lb 2).875 in* 0.250 lb T (6/8) Pplate.(2* 0.(3* 0. plate = 144.Solution606 STATICS & STRENGTH OF MATERIALS . B.875 in))(0.row 2 = (8/6)(133.500 lb C 3).000 psi) = 144.row 2 = (w .) Calculate the strength of the joint.000 psi) = .75in)(28.000 psi) = 157.000 psi σt = 28.nd)t*σ = (9 in . 000 psi Plate: τ = 14. 000 psi σc = 30.300 lb/ .875 in)2/4)(15.875 lb T Pplate. Rivets & Welds .) Calculate the efficiency of the joint. The allowable stresses are as follows: Rivets: τ = 15. and the thickness of the plates is 3/4 inch. htm (2 of 2) [5/8/2009 4:40:33 PM] .Solution606 Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol606. depending on the weld material used.Topic 5. Since the two sides of the right triangle formed by the weld are equal forming a 45o right triangle. the end edges to be welded together may by sloped or vee-ed to generate a better weld. 3. the weld is known as a 45o fillet weld. or: Pweld = (w * t) * tension. In theory. In some situations the weld may be x-rayed to determine its quality. shown in Diagrams 2. and 4. P. Depending on the size and thickness of the plates. In practice. whether or not there are voids or cracks in the weld. and the plates are welded together.Welded Joints There are a number of types of welded joints.7: Rivets & Welds: Welded Joints Topic 5. it is assumed to fail in shear across the throat region. shown in Diagram 2a. a butt weld can be made as strong or stronger than the plate. where w and t are the width and thickness of the weld (plate). The fillet weld runs from the top edge of the top plate to an equal distance horizontally outward on the top of the bottom plate. In an example of this type of weld a top plate is welded to a bottom plate. To determine the load a weld can carry before failing we take the product of the area of the weld which fails in shear ( the throat distance times the file:///Z|/www/Statstr/statics/Torsion/rivw57. One type of welded joint is the Butt Joint.htm (1 of 5) [5/8/2009 4:40:33 PM] . When the weld fails. A second type of weld which we will consider is the Lap Joint using a 45 degree fillet weld.7: Rivets & Welds . The perpendicular line bisecting the 90o angle and intersecting the hypotenuse of the 45o fillet weld is know as the throat of the weld. We will consider several common types in this discussion. forming a triangular weld as shown in the cross sectional view in Diagram 2a. the weld can carry in tension is simply the product of the cross sectional area of the weld and the allowable tensile stress for the weld material. the strength of a weld depends on how well the weld is made. The gap is then filled with weld material as from an arc welder. where two plates are brought together with a small gap separating them. as shown in Diagram 1a. the load. For a welded butt joint loaded in tension as shown in Diagram 1b. 7: Rivets & Welds: Welded Joints length of weld. where t = thickness of plate (and height & base of weld) L = length of weld. all = allowable shear stress for the weld material file:///Z|/www/Statstr/statics/Torsion/rivw57. Diagram 2b) and the allowable shear stress for the weld material. and finally Pweld = (.707 t * L) * . We can write: Pweld = Area* all = (throat * length) * all all = (t sin 45o * L) * all.Topic 5.htm (2 of 5) [5/8/2009 4:40:33 PM] . t = 30..000 lb/in2 = (7./(7. it is apply closer to one edge of the top plate than the other. load and then to decide how the weld should be distributed along sides AB and FG. load by using the weld formula and setting the load the weld can carry before failing to 80.000 pound load is file:///Z|/www/Statstr/statics/Torsion/rivw57.7: Rivets & Welds: Welded Joints Welded Joint Example Two steel plates are shown in Diagram 5.000 lb/in2 Plate Material: Dimensions: AB = GF = 20 inches./in.000 lb.424 lb.000 lb/in2 . Weld Material: = 14. We first determine the minimum inches of weld needed to carry the 80. Following that. however since the 80. The top plate is 3/4 inch thick and 8 inches wide. and is to be welded to the bottom plate with a 45o fillet weld.000 lb = (. DE = 5 inches Solution: Part 1.707 * 3/4" * L) * 14.000 lb.htm (3 of 5) [5/8/2009 4:40:33 PM] .000 lb/in2 .(There is only one way for the minimum inches of weld. or 80.78 inches.)L. This is the minimum inches of weld needed to carry the load./in. CD = 3 inches. we will then determine the minimum number of inches of weld needed to make the weld strength as great as the plate strength. We would first like to determine the minimum inches of weld need to carry the 80.Topic 5.000 lb. The allowable stresses are as follows.424 lb.707 t * L) * all .) Notice also that the load is not applied symmetrically. t = 24. The top plate is to be welded along sides AB and FG.000 lb. that is. Pweld = (. then solving for L L = 80.000 lb/in2 = 15.) = 10. or Pplate = (w * t ) solve for the length of weld needed.000 lb. and FGF = (7.424 lb./in.(7. Part 3.) Solving the torque equation for LAB = 6.78 inches.424 lb.6. In Diagram 6 we have shown the weld distributed with an amount LAB on side AB and an amount LGF on side GF. or = Pplate =(w * t ) file:///Z|/www/Statstr/statics/Torsion/rivw57. We then set this equal to the strength of the riveted joint and t .707 t * L) * all t ./in. Pweld = (. as shown in Diagram 6. To determine how the weld should be distributed. if we try to put equal amounts of weld on sides AB and GF. The lengths LAB and LGF must.73" = 4. The other question of interest in welded joints is exactly how much weld is needed to make the welded joint as strong as the plate itself.78" .)* LGF ./in. The maximum force these weld can resist with is given by FAB = (7.78 inches ( the minimum inches of weld needed).htm (4 of 5) [5/8/2009 4:40:33 PM] .424 lb.000 lb. . we can then solve for LGF since LAB and LGF sum to 10.) * LAB . the weld will fail in this case. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material. We now apply static equilibrium conditions to the top plate: Sum of Forces: 80.424 lb. Part 2.000 lb. In this case we assume the plate is loaded in tension.Topic 5.)* LGF = 0 Sum of Torque about G: -80. we once again apply static equilibrium conditions. Therefore LGF = 10. That is.)* LAB .73 inches. since its line of action passes through point G.* (5") + (7./in. we can not apply the weld symmetrically./in. of course sum to 10.7: Rivets & Welds: Welded Joints not applied symmetrically to the plate.)* LAB * (8") = 0 (Notice that the force FGF does not produce a torque about point G.(7.424 lb.05" This is how the weld must be distributed in order to carry the 80. load (and satisfy static equilibrium conditions). 7: Rivets & Welds: Welded Joints (. and then (.Example 1 or Select: Topic 5: Torsion.Topic 5.707 t * L) * all = (w * t ) t . since we are not dealing with a specific loading.000 lb/in2 .000 lb/in2 = (8" * 3/4" ) 30.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivw57. Continue to: Topic 5. then solving for L L = 24.htm (5 of 5) [5/8/2009 4:40:33 PM] .707 *3/4" * L) * 14.7a: Welded Joints . We can not state how the weld should be distributed. Rivets & Welds .25 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension). 707 t * L) * all .Example 1 Topic 5.000 lb = (.)*L. load and specify how many inches of weld should be placed along each side AB and FG. then solving for L L = 90. t = 24.000 lb.5 inch thick and 10 inches wide.Example 1 Two steel plates are shown in Diagram 1./in./(5.) Determine the minimum inches of weld need to carry the 90.000 lb. We first determine the minimum inches of weld needed to carry the 90.. file:///Z|/www/Statstr/statics/Torsion/rivwe57a. DE = 6 inches Solution: Part 1.98 inches. Pweld = (. The top plate is to be welded completely across end AG and partially along sides AB and FG. B.000 lb/in2 .000 lb. CD = 4 inches. The top plate is .5" * L) * 15.000 lb/in2 Plate Material: Dimensions: AB = GF = 20 inches.300 lb.7a: Welded Joints ./in. Weld Material: = 15.) = 16.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength. The allowable stresses are as follows.Topic 5.000 lb.707 * . t = 28. and is to be welded to the bottom plate with a 45o fillet weld. load by using the weld formula and setting the load the weld can carry before failing to 90.000 lb/in2 = (5. or 90.000 lb/in2 .300 lb. 000 lb/in2 = 14.htm (1 of 3) [5/8/2009 4:40:34 PM] .7a: Welded Joints . A. /in.300 lb.* (6") + (5.000 lb.300 lb.)* LAB * (10") + 53. We then set this equal to the strength of the riveted joint and solve for the length of weld needed. or Pplate = (w * t ) t .300 lb/ in * 10" = 53.Topic 5. and then t . since its line of action passes through point G.) Solving the torque equation for LAB = 5. we once again apply static equilibrium conditions.)* LGF .10" = 1.000 lb. Therefore LGF = 16.000 lb. load (and satisfy static equilibrium conditions).(5. we can then solve for LGF since LAB + LGF + 10" must sum to 16. The maximum force these weld can resist with is given by FAB = (5.707 t * L) * (. The lengths LAB . We assume the plate is loaded in tension.53. LGF on side GF. .300 lb. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material.htm (2 of 3) [5/8/2009 4:40:34 PM] .98 inches ( the minimum inches of weld needed).000 lb.000 lb. FGF = (5.)* LAB ./in. In Diagram 2 we have shown the weld distributed with an amount LAB on side AB. however since the 90. Part 3. That is. = 0 Sum of Torque about G: -90. We now apply static equilibrium conditions to the top plate: Sum of Forces: 90. * 5" = 0 (Notice that the force FGF does not produce a torque about point G.98" .98 inches..300 lb.79" This is how the weld must be distributed in order to carry the 80.5. LGF .000 pound load is not applied symmetrically to the plate. and FAG = 5.300 lb. we can not apply the weld symmetrically.000 lb.707 t * L) * all all = Pplate =(w * t ) t .(5./in. Pweld = (. Part 2. the weld will fail in this case. if we try to put equal amounts of weld on sides AB and GF.7a: Welded Joints . To determine how the weld should be distributed. and the 10 inches of weld must sum to 16./in./in.19" .Example 1 This is the minimum inches of weld needed to carry the load.)* LGF . as shown in Diagram 2.)* LAB .19 inches. and 10 inches of weld completely across the end AG. or = (w * t ) file:///Z|/www/Statstr/statics/Torsion/rivwe57a. Return to: Topic 5.Problem Assignment or Select: Topic 5: Torsion.4 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension). since we are not dealing with a specific loading.htm (3 of 3) [5/8/2009 4:40:34 PM] .5" ) 28. Rivets & Welds .Example 1 (.Topic 5.Welded Joints Continue to: Topic 5.7a: Welded Joints . then solving for L L = 26.5" * L) * 15.000 lb/in2 = (10" * .707 *.7: Rivets & Welds . We can not state how the weld should be distributed.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwe57a.000 lb/in2 .8: Rivets & Welds . Example Two steel plates are shown below. The dimension and stresses are as follows. The top plate is to be weld completely across end AG and partially along sides AB and FG.) = 18.and the 8 inches of weld must Weld Material: τ=12.000 lb = (.000 psi Dimensions: AB = GF = 20 inches: CD = 3 inches: DE = 5 inches Solution: Part A.. load by using the weld formula and setting the load the weld can carry before failing to 80. A.000 lb.000 lb/in2 = (4.The lengths LAB ./in.707 * .000 lb load and specify how many inches of weld should be placed along sides AB and FG.242 lb. To determine how the weld should be distributed. This is the minimum inches of weld needed to carry the load. if we try to put equal amounts of weld on sides AB and GF.000 lb.000 psi file:///Z|/www/Statstr/statics/Stests/rivweld/sol711.5" * L) * 12. then solving for L L = 80.000 pound load is not applied symmetrically to the plate.86 inches. Part II. and 8 inches of weld completely across the end AG. In Diagram 2 we have shown the weld distributed with an amount LAB on side AB. σt= 24. LGF on side GF./in. we once again apply static equilibrium conditions./(4. σt= 30. or 80. Allowable stresses : Plate Material: τ = 15. Part I: We first determine the minimum inches of weld needed to carry the 80.242 lb.707 t * L) * all . Pweld = (. and is to be welded to the bottom plate with a 45o fillet weld. however since the 80. The top plate is .000 psi.)*L.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength.000psi.htm (1 of 2) [5/8/2009 4:40:34 PM] . That is.000 lb. we can not apply the weld symmetrically.Solution711 STATICS & STRENGTH OF MATERIALS . LGF . the weld will fail in this case. B.) Determine the minimum inches of weld need to carry the 80.5 inches thick and 8 inches wide. Therefore LGF = 18.)* LGF .936 lb.242 lb. We = Pplate =(w * t ) t . and FAG = 4.707 t * L) * (. or = (w * t ) (.936 lb.79" .707 t * L) * all all t . = 0 Sum of Torque about G: -80.936 lb.242 lb. Part B: We assume the plate is loaded in tension.242 lb/in * 8" = 33. The maximum force these weld can resist with is given by FAB = (4.242 lb.(4.(4. then solving for L L = 28.)* LAB * (8") + 33.)* LAB .8" = 3.5" * L) * 12. We now apply static equilibrium conditions to the top plate: Sum of Forces: 80. load (and satisfy static equilibrium conditions). Rivets & Welds .707 *./in. Topic 5: Torsion.)* LAB .) Solving the torque equation for LAB = 7.242 lb.)* LGF ./in.86 inches. * 4" = 0 (Notice that the force FGF does not produce a torque about point G.07" This is how the weld must be distributed in order to carry the 80. or Pplate = (w * t ) then set this equal to the strength of the riveted joint and solve for the length of weld needed.000 lb/in2 . We can not state how the weld should be distributed.5" ) 30.79 inches.33.000 lb/in2 = (8" * .000 lb.7. Pweld = (./in./in.242 lb.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol711. and then t . FGF = (4. we can then solve for LGF since LAB + LGF + 8" must sum to 18. as shown in Diagram 2./in. since we are not dealing with a specific loading. since its line of action passes through point G.* (5") + (4. .000 lb.29 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension).Solution711 sum to 18.000 lb.htm (2 of 2) [5/8/2009 4:40:34 PM] . and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material.86" .86 inches ( the minimum inches of weld needed). the weld will fail in this case. however since the 80. A.000 lb.707 t * L) * all . The top plate is to be weld completely across end AG and partially along sides AB and FG. σt= 28.000 lb load and specify how many inches of weld should be placed along each side AB and FG. load by using the weld formula and setting the load the weld can carry before failing to 80.and the 6 inches of weld must 2 file:///Z|/www/Statstr/statics/Stests/rivweld/sol712. That is. B.000 psi.Solution712 STATICS & STRENGTH OF MATERIALS .) Determine the number of inches of weld needed to make the weld strength as great as the plate strength.000 psi Dimensions: AB = GF = 30 inches: CD = 2 inches: DE = 4 inches Weld Material: τ = 16. Part II..7" * L) * 16.The lengths LAB . P = (. and is to be welded to the bottom plate with a 45o fillet weld.000 lb/in = (7. The dimension and stresses are as follows.918 lb.1 inches. or weld 80.918 lb.) = 10. σt= 30.Example Two steel plates are shown below.707 * ./in./in. we can not apply the weld symmetrically. To determine how the weld should be distributed.)*L.000 lb = (. LGF . LGF on side GF. Allowable stresses : Plate Material: τ = 14. This is the minimum inches of weld needed to carry the load. and 6 inches of weld completely across the end AG. 000 psi Solution: Part A.) Determine the minimum inches of weld need to carry the 80. The top plate in . we once again apply static equilibrium conditions.000 lb.000 pound load is not applied symmetrically to the plate.000 lb. then solving for L L = 80. In Diagram 2 we have shown the weld distributed with an amount LAB on side AB.htm (1 of 3) [5/8/2009 4:40:34 PM] . Part I: We first determine the minimum inches of weld needed to carry the 80. if we try to put equal amounts of weld on sides AB and GF./(7.7 inches thick and 6 inches wide. 000 psi. Solution712 sum to 10.000 lb/in .1" .000 lb.(7. since its line of action passes through point G. Therefore L = 10.1 inches.7" ) 28. P weld t .) Solving the torque equation for L AB = 3.3.36" This is how the weld must be distributed in order to carry the 80.918 lb/in * 6" = 47.74" .000 lb/in = (6" * .918 lb.918 lb.707 *.7" * L) * 16. or (. FGF = (7.918 lb. . Rivets & Welds . Topic 5: Torsion.000 lb.1 inches ( the minimum inches of weld needed). as shown in Diagram 2. or Pplate = (w * t ) then set this equal to the strength of the riveted joint and solve for the length of weld needed.* (4") + (7.74 inches. then solving for L L = 14.918 lb.6" = ./in. load (and satisfy static equilibrium conditions).)* LGF ./in. We now apply static equilibrium conditions to the top plate: Sum of Forces: 80.htm (2 of 3) [5/8/2009 4:40:34 PM] .918 lb.707 t * L) * = (w * t ) 2 t .707 t * L) * all all =P plate =(w * t ) t .)* L Sum of Torque about G AB .510 lb.85 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension).)* LAB .)* L * (6") + 47. The maximum force these weld can resist with is given by FAB = (7.)* L AB GF . and then 2 (./in. = 0 : -80. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material./in. * 3" = 0 (Notice that the force FGF does not produce a torque about point G. we can then solve for LGF since LAB + LGF + 6" GF must sum to 10.000 lb.510 lb. We = (./in.(7. since we are not dealing with a specific loading. Part B: We assume the plate is loaded in tension. We can not state how the weld should be distributed.510 lb.47. and FAG = 7.Table of Contents file:///Z|/www/Statstr/statics/Stests/rivweld/sol712. htm (3 of 3) [5/8/2009 4:40:34 PM] .Solution712 Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol712. Part II. The top plate in . however since the 60. 000 psi = (. of course sum to 11. and is to be welded to the bottom plate with a 45o fillet weld. The lengths LAB and LGF must./(5. B./in. σt= 28.5 inches thick and 8 inches wide.) = 11.31 inches ( the minimum file:///Z|/www/Statstr/statics/Stests/rivweld/sol713.Example Two steel plates are shown below. σt= 26. In Diagram 2 we have shown the weld distributed with an amount LAB on side AB and an amount LGF on side GF./in. This is the minimum inches of weld needed to carry the load. we can not apply the weld symmetrically. DE = 5 inches Solution: Part A: Part I.000 psi Dimensions: AB = GF = 30 inches: . 000 psi.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength.5" * L) * 15.htm (1 of 3) [5/8/2009 4:40:35 PM] .707 * . A.000 psi. CD = 3 inches: . if we try to put equal amounts of weld on sides AB and GF..) Determine the minimum inches of weld need to carry the 60. Allowable stresses : Plate Material: τ = 14.000 pound load is not applied symmetrically to the plate.000 lb = (. the weld will fail in this case. The top plate is to be weld along sides AB and FG only. we once again apply static equilibrium conditions. or 2 60. That is. P weld Weld Material: τ = 15. To determine how the weld should be distributed.000 lb. load by using the weld formula and setting the load the weld can carry before failing to 60.303 lb.Solution713 STATICS & STRENGTH OF MATERIALS . The dimension and stresses are as follows.707 t * L) * all .000 lb/in = (5.000 lb load and specify how many inches of weld should be placed along each side AB and FG.303 lb. then solving for L L = 60.)L.000 lb.000 lb.31 inches. We first determine the minimum inches of weld needed to carry the 60. Part B.000 lb.24" This is how the weld must be distributed in order to carry the 60. We then set this equal to the strength of the riveted joint and solve for =(w * t ) t ./in./in. The maximum force these weld can resist with is given by FAB = (5.707 t * L) * = (w * t ) 2 t .) Solving the torque equation for L 11.303 lb.07 inches. as shown in Diagram 2.5" ) 28.htm (2 of 3) [5/8/2009 4:40:35 PM] . The other question of interest in welded joints is exactly how much weld is needed to make the welded joint as strong as the plate itself.)* L Sum of Torque =0 : -60. we can then solve for LGF since LAB and LGF sum to = 11. (5. We can not state how the weld should be distributed./in.)* L AB about G * (8") = 0 (Notice that the force FGF does not produce a torque about point G.31" . then solving for L L = 21. In this case we assume the plate is loaded in tension.000 lb/in .12 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension).7./in.)* LGF .707 * ./in. since its line of action passes through point G. and then 2 (. or = (. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material. or Pplate = (w * t ) the length of weld needed.303 lb.303 lb. since we are not dealing with a specific loading file:///Z|/www/Statstr/statics/Stests/rivweld/sol713.07" = 4.Solution713 inches of weld needed).)* LAB .303 lb.707 t * L) * all all =P plate (. We now apply static equilibrium conditions to the top plate: Sum of Forces: 60.303 lb. P weld t .000 lb.5" * L) * 15.000 lb.31 inches. Therefore L GF AB = 7.000 lb/in = (8" * . load (and satisfy static equilibrium conditions).)* L AB GF (5.* (5") + (5. and FGF = (5. Solution713 Topic 5: Torsion. Rivets & Welds .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol713.htm (3 of 3) [5/8/2009 4:40:35 PM] . This is the minimum inches of weld needed to carry the load.32 inches ( the minimum file:///Z|/www/Statstr/statics/Stests/rivweld/sol714. σt= 22.929 lb. In Diagram 2 we have shown the weld distributed with an amount LAB on side AB and an amount LGF on side GF. of course sum to 17.000 pound load is not applied symmetrically to the plate.htm (1 of 2) [5/8/2009 4:40:35 PM] . The top plate is to be welded along sides AB and FG. The dimension and stresses are as follows./in.000 lb load and specify how many inches of weld should be placed along each side AB and FG. CD = 3 inches. 000 psi Plate Material: τ = 15. Pweld = (. then solving for L L = 120.Example Two steel plates are shown below. we can not apply the weld symmetrically.The lengths LAB and LGF must.000 psi.000 lb/in2 = (6.7 inches thick and 9 inches wide. or 120.) Determine the minimum inches of weld need to carry a Load P of 120. and is to be welded to the bottom plate with a 45o fillet weld. however since the 120.Solution714 STATICS & STRENGTH OF MATERIALS .707 * . 000 psi.. DE = 6 inches Solution: Part A: Part I. We first determine the minimum inches of weld needed to carry the 120. Allowable stresses : Weld Material: τ = 14.31 inches.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength./(6. B.000 lb. load by using the weld formula and setting the load the weld can carry before failing to 120./in.000 psi Dimensions: AB = GF = 30 inches.) = 17. The top plate in . if we try to put equal amounts of weld on sides AB and GF.7" * L) * 14. A.)L.000 lb. the weld will fail in this case. we once again apply static equilibrium conditions. Part II. σt= 35.000 lb = (.929 lb. That is.000 lb. To determine how the weld should be distributed.707 t * L) * all . )* LGF .) Solving the torque equation for LAB = 11.929 lb.32" .000 lb.htm (2 of 2) [5/8/2009 4:40:35 PM] . since we are not dealing with a specific loading. We then set this equal to the strength of the riveted joint and solve for t .000 lb/in2 . or Pplate = (w * t ) the length of weld needed. or = Pplate =(w * t ) t .)* LAB . Therefore LGF = 17.)* LAB . In this case we assume the plate is loaded in tension./in.11.Solution714 inches of weld needed)./in.707 t * L) * all all t . as shown in Diagram 2.707 t * L) * (.7" ) 35.000 lb.929 lb. Pweld = (.(6. We now apply static equilibrium conditions to the top plate: Sum of Forces: 120.32 inches./ in. Part B. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material. Topic 5: Torsion. Rivets & Welds .)* LAB * (9") = 0 (Notice that the force FGF does not produce a torque about point G.929 lb.929 lb. The other question of interest in welded joints is exactly how much weld is needed to make the welded joint as strong as the plate itself.55 inches. ./in. The maximum force these weld can resist with is given by FAB = (6./in.* (6") + (6. and then = (w * t ) (.707 * .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Stests/rivweld/sol714. then solving for L L = 31.)* LGF = 0 Sum of Torque about G: -120. we can then solve for LGF since LAB and LGF sum to 17.(6.7" * L) * 14.000 lb/in2 = (9" * . We can not state how the weld should be distributed.82 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension).000 lb. since its line of action passes through point G. load (and satisfy static equilibrium conditions). and FGF = (6.55" = 5.929 lb.77" This is how the weld must be distributed in order to carry the 120. Pr1 = 72. 2 t = 22. 000 lb/in2. 2 t = 20.8a . and the thickness of the plates is 1/2 inch. The width of the plates is 8 inches. file:///Z|/www/Statstr/statics/Torsion/rivwp58a. 000 lb/in Determine the Strength of the Joint. and the Efficiency of the Joint. 000 lb/in . 000 lb/in2. 2 c = 25. The allowable stresses are as follows: Rivets: Plate: = 18.000 lb/in2.000 lb.700 lb) [Joint Strength = 70. 2 c = 24. A riveted butt joint is shown in Diagram 2. and the thickness of the plates is 3/4 inch. 000 lb/in Determine the Strength of the Joint. (Ps = 79. The width of the plates is 6 inches. 000 lb/in .886] 2. eff. The diameter of the rivets is 1/2 inch.8a: Rivets & Welds . and the Efficiency of the Joint.520 lb..Problem Assignment 1 . 2 t = 20. 2 t = 22. = 14.500 lb. 000 lb/in 2 c = 21. The allowable stresses are as follows: Rivets: Plate: = 15. 000 lb/in 2 c = 24. = . 000 lb/in . = 17. Pb = 108. Pr2 = 70.910 (lowest of failure loads above). A riveted lap joint is shown in Diagram 1.Probs 1 Statics & Strength of Materials Topic 5.910 lb.Rivets/Welds 1. 000 lb/in .htm (1 of 5) [5/8/2009 4:40:36 PM] . Pr3 = 76. The diameter of the rivets is 3/4 inch.Topic 5.000 lb/in2. 000 lb/in 2 c = 25. 000 lb/in2.12 pattern) C.11.7 = 10 rivets) B. Pr1 = 82. = . Pr2 = 84. 000 lb/in . Pr3 = 101. 10.000 lb/in2. Pr2 = 82.Topic 5. The rivets to be used have a diameter of 3/4 inch. Pb = 70. file:///Z|/www/Statstr/statics/Torsion/rivwp58a. 000 lb/in A.. 2 t = 22. A lap joint (Diagram 3) is to connect two steel plates both with a width of 7 inches and a thickness of 5/8 inch. (ans.htm (2 of 5) [5/8/2009 4:40:36 PM] . (ans. Determine the number of rivets for the most efficient joint. 000 lb/in . Select the best pattern from those shown in Diagram 4. (ans.500.620 lb.707] 3. 9. = 17.Probs 1 (Ps = 63.880 lb.500 lb. eff. eff = . 2 t = 21.8a: Rivets & Welds .375 lb. A double cover plate butt joint (Diagram 5) is to connect two steel plates both with a width of 10 inches and a thickness of 5/8 inch.250 lb) [Joint Strength = 63.857) 4. The maximum allowable stresses for the rivet and plate materials are as follows: Rivets: Plate: = 20.620 (lowest of failure loads above). 2 c = 23. Calculate the strength and efficiency of the joint. 2 t = 20. Select the best pattern from the patterns in Diagram 6. 000 lb/in 2 c = 21. (ans. = 14.8a: Rivets & Welds . Calculate the strength and efficiency of the joint. Determine the number of rivets for the most efficient joint.75 = 12 rivets) B. A lap joint is to connect two steel plates each with thickness of 3/4 inch (Diagram 7). (ans.Topic 5.Probs 1 The rivets to be used have a diameter of 3/4 inch.htm (3 of 5) [5/8/2009 4:40:36 PM] . 000 lb/in2. 000 lb/in A. 000 lb/in . The maximum allowable stresses for the rivet and plate materials are as follows: file:///Z|/www/Statstr/statics/Torsion/rivwp58a.000 lb/in2. The maximum allowable stresses for the rivet and plate materials are as follows: Rivets: Plate: = 16. 000 lb/in . 11. 625 lb. Pr1 = 115. eff = . (11-12 pattern) C. The rivets to be used have a diameter of 1/2 inch. 2 t = 22. 2 c = 23.925) 5. The top plate is 1/2 inch thick and 10 inches wide. 2 t = 24. file:///Z|/www/Statstr/statics/Torsion/rivwp58a. (ans. 2 t = 25. = 15. not a very good joint] 6. = . Using the width found in part A. Pr2 = 78.750 lb.Topic 5. 000 lb/in .8a: Rivets & Welds . The maximum allowable stresses for the rivet and plate materials are as follows: Rivets: Plate: = 16. Calculate the width needed for the steel plate such that the joint will fail in bearing and plate tearing at row 1at the same load. 2 c = 27. Pb = 58. The top plate is to be weld completely across end AG and partially along sides AB and FG.Probs 1 Rivets: Plate: = 15.790 lb) [Joint Strength = 23. = .000 lb/in2..750 (lowest of failure loads above).) B. The rivets to be used have a diameter of 1/2 inch.27.000 lb/in2.500 lb. (Ps = 37. determine the strength and efficiency of the joint.000 lb. (ans. and is to be welded to the bottom plate with a 45o fillet weld.700 lb. 000 lb/in 2 c = 26.562 lb. Two steel plates are shown below.940 lb. 000 lb/in A. 2 t = 24. = 14. 000 lb/in . 000 lb/in A. very small width. Calculate the width needed for the steel plate such that the joint will fail in rivet shear and plate tearing at row 1 at the same load. 2 t = 22.83") B.72] 7. Pr1 = 37. 000 lb/in . Pr1 = 77. 2. 2 c = 28.500 lb) [Joint Strength = 33.500 lb.562 (lowest of failure loads above). w = 4.5". eff. Pr3 = 37. eff. not very realistic. Using the width found in part A.htm (4 of 5) [5/8/2009 4:40:36 PM] . Pb = 78. 000 lb/in2. Pr2 = 33. 000 lb/in 2 c = 26. 000 lb/in . 000 lb/in2.. (Ps = 23. A a double cover plate butt joint is to connect two steel plates each with thickness of 3/4 inch ( See Diagram 8). determine the strength and efficiency of the joint. Part A.000 lb/in2 Plate Material: Dimensions: AB = GF = 20 inches.000 lb/in2 .) Determine the number of inches of weld needed to make the weld strength as great as the plate strength. Ltotal = 22.htm (5 of 5) [5/8/2009 4:40:36 PM] . Rivets & Welds . 000 lb/in2 = 15. B.35") (ans. LAG = 10" (given).Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwp58a. LAB = 9. t = 26.8a: Rivets & Welds . CD = 3 inches. Ltotal =21. Weld Material: = 16.000 lb. LGF = 1.Topic 5. The allowable stresses are as follows. DE = 7 inches (ans.2".85".98") Select: Topic 5: Torsion.000 lb/in2 .Part B.Probs 1 A. load and specify how many inches of weld should be placed along each side AB and FG.) Determine the minimum inches of weld need to carry the 120. t = 25. 5)St(ult) 87 (600) Sy 36 (250) Allowable Tensile (Gross) (0.0(207) 28.threads excludes from shear plane A502 Grade 1 hot-driven rivets A502 Grade 2 hot-driven rivets 17. Plate Tearing= 77. The plates are ASTM A36 steel and the bolts are A325.0(68. Plate Bearing = 391.Rivets & Welds .5(121) 22..Probs 2 Statics & Strength of Materials Topic 5.0(193) 40.Rivets/Welds Allowable stresses [ksi(MPa)] Allowable Bearing Stress Structural steels A36 carbon St(ult) 58 (400) Sp(all) = (1. Determine the allowable tensile load for the single shear lap joint shown.htm (1 of 7) [5/8/2009 4:40:36 PM] .0(145) 21. Assume that the threads are excluded from the shear plane.8b .0(145) 30.Problem Assignment .0(117) 21.000 lb. The Diameter of the Rivets is 1 inch.50)St(ult) 29*(200) Stress St(all) = Allowable shear stress on steel fasteners.Topic 58b .000 lb) file:///Z|/www/Statstr/statics/Torsion/rivwp58b.. [Use gross allowable tensile stress for ASTM A36] [ For Joint use Bearing Type connection] (Rivet Shear = 212.0(152) 1.0(145) Bearing-Type Connection 10.threads excluded from shear plane A490 bolts --.500 lb.0(117) 17.threads in shear plane A490 bolts --.0(276) 17.60)Sy 22 (152) St(all) = (Net) (0.9) 21.threads in shear plane A325 bolts --. Allowable Shear Stress (Ss(all))[ksi MPa)] Description of Fastener Slip-Critical Connection A307 low-carbon bolts A325 bolts --. Probs 2 2. The plates are ASTM A36 steel and the bolts are A325.500 lb.Topic 58b . The Diameter of Rivets is 1 inch.750 lb) file:///Z|/www/Statstr/statics/Torsion/rivwp58b.. Determine the allowable tensile load for the double shear butt joint shown. Plate Bearing = 174.Rivets & Welds . Plate Tearing= 68. [Use gross allowable tensile stress for ASTM A36] [ For Joint use Bearing Type connection] (Rivet Shear = 188.000 lb..htm (2 of 7) [5/8/2009 4:40:36 PM] . Assume that the threads are excluded from the shear plane. The plates are ASTM A36 steel and the bolts are A325. and allowable shear stress of 30% of tensile stress] [Use gross allowable tensile stress for ASTM A36] (Weld strength (shear) = 178. ] file:///Z|/www/Statstr/statics/Torsion/rivwp58b. Plate strength (tension) = 99.Rivets & Welds . fillet weld. and A325 bolts --.160 lb. Determine the number of 3/4 inch bolts needed for a maximum efficiency lap joint of two 8" x 1/2" plates.Probs 2 3.. The cover plates are 8" x 9/16".htm (3 of 7) [5/8/2009 4:40:36 PM] . which is made using an E70 electrode. The plates are ASTM A36 steel and the bolts are A325. [E70 electrode has allowable tensile stress of 70.threads excluded from shear plane] (3 rivets) 5. [Use gross allowable tensile stress for ASTM A36] [ For Joint use Bearing Type connection. Determine the number of 3/4 inch bolts needed for a maximum efficiency butt joint of two 8" x 1/2" plates. and A325 bolts --.000 psi. Calculate the allowable tensile load for the connection in the diagram shown.threads excluded from shear plane] (6 rivets) 4.000 lb. [Use gross allowable tensile stress for ASTM A36] [ For Joint use Bearing Type connection.. The plates are ASTM A36 steel and the weld is a 3/4 in.Topic 58b . side and end fillet welds are used to connect the 2.Rivets & Welds . 1/4 in. Find the required dimension L.000 psi. and allowable shear stress of 30% of tensile stress] [Use gross allowable tensile stress for ASTM A36] (L = 7.0 in tension member to the plate. In the connection in the diagram shown. however joint fails in plate tearing at 44. The steel is ASTM A36 and the electrode used is an E70.htm (4 of 7) [5/8/2009 4:40:36 PM] . The applied load is 65. [E70 electrode has allowable tensile stress of 70. by 1.000 lb..000 lb) file:///Z|/www/Statstr/statics/Torsion/rivwp58b.Topic 58b .76".Probs 2 6.0 in. by 3/8 in. ASTM A36 steel plate in the diagram shown. and allowable shear stress of 30% of tensile stress] [Use gross allowable tensile stress for ASTM A36] (L = 1. [E70 electrode has allowable tensile stress of 70. The electrode is an E70. Design the fillet welds parallel to the applied load to develop the full allowable tensile load of the 6 in.Rivets & Welds .Topic 58b .. The Plate is welded along the far end in addition to the sides.000 psi.45") file:///Z|/www/Statstr/statics/Torsion/rivwp58b.Probs 2 7.htm (5 of 7) [5/8/2009 4:40:36 PM] . Probs 2 8.Rivets & Welds .57kips/in. Determine the strength of the weld in kips/in.A fillet weld between two steel plates intersecting at right angles was made with one 3/8 in.) file:///Z|/www/Statstr/statics/Torsion/rivwp58b.htm (6 of 7) [5/8/2009 4:40:36 PM] . [E70 electrode has allowable tensile stress of 70. leg and one 1/2 in leg using an E70 electrode. and allowable shear stress of 30% of tensile stress] (5570 lb/in = 5.000 psi..Topic 58b . Topic 58b .Rivets & Welds .htm (7 of 7) [5/8/2009 4:40:36 PM] .Probs 2 Select: Topic 5: Torsion. Rivets & Welds .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/rivwp58b. Topic Exam Modulus of Rigidity for several materials: Steel = 12 x 106 lb/in2.. determine the resultant angle of twist of end D with respect to end A. & TB) without exceeding the allowable shear stress in any section of the shaft.000 lb/in2.Topic Exam Topic 5.htm (1 of 4) [5/8/2009 4:40:37 PM] . C. and section CD is brass. file:///Z|/www/Statstr/statics/Torsion/torsx59. and for brass is 16. A compound shaft is attached to a fixed wall as shown in Diagram 1. B. A. section BC is steel. & B (TD.TC. Aluminum = 4 x 106 lb/in2. (For Solution Select: Torsion . Determine the maximum torque which could be applied at points D. Brass = 6 x 106 lb/in2.5 inches is to transmit power while being driven a 3000 rpm.9: Torsion.9: Torsion. Rivets & Welds . If the allowable shear stress for steel is 18. Rivets & Welds .. 1.000 lb/in2.Topic Examination Torsion . Using the torque found in part A.Topic 5. Section AB brass. A 2 foot long hollow steel shaft with an outer diameter of 2 inches and an inner diameter of 1.Solution Problem 1) 2. what would be the minimum outer diameter of the shaft which could safely transmit the horse power found in part A? (The allowable shear stress in the shaft is 15. A riveted lap joint is shown in Diagram 1.9: Torsion. The width of the plates is 12 inches.Topic Exam A. A double cover plate butt joint (Diagram 2) is to connect two steel plates both with a width of 9 inches and a thickness of 7/8 inch. = 17.Topic Exam 1. 2 c = 28. The allowable stresses are as follows: Rivets: Plate: = 20. Rivets & Welds . what is the maximum horsepower which can be transmitted down the shaft ? B.000 lb/in2.000 lb/in2 ) (For Solution Select: Torsion . file:///Z|/www/Statstr/statics/Torsion/torsx59. 000 lb/in 2 c = 24.Solution Problem 1) 2.. but were told that the inner diameter was to be four-tenths of the outer diameter. The rivets to be used have a diameter of 3/4 inch. and the Efficiency of the Joint. 000 lb/in Determine the Strength of the Joint.Solution Problem 2) Rivets & Welds .Topic 5. 000 lb/in . 000 lb/in2. 2 t = 26. (For Solution Select: Rivets & Welds . The diameter of the rivets is 1 inch.htm (2 of 4) [5/8/2009 4:40:37 PM] . 2 t = 22. and the thickness of the plates is 5/8 inch.000 lb/in2.) If we were not given the outer diameter of the shaft. 000 lb/in .) If the allowable shear stress in the shaft is 15. file:///Z|/www/Statstr/statics/Torsion/torsx59. = 18.9: Torsion. Calculate the strength and efficiency of the joint. The top plate is to be weld completely across end AG and partially along sides AB and FG. 000 lb/in .Solution Problem 2) 3. and is to be welded to the bottom plate with a 45o fillet weld. Rivets & Welds .000 lb/in2.Topic Exam The maximum allowable stresses for the rivet and plate materials are as follows: Rivets: Plate: = 17. 2 c = 23.htm (3 of 4) [5/8/2009 4:40:37 PM] . (For Solution Select: Rivets & Welds . 2 t = 22. Two metal plates are shown below. C. The top plate is 3/4 inch thick and 9 inches wide.Topic 5. 000 lb/in A. 000 lb/in2. Select the best pattern from the patterns in Diagram 3 . 000 lb/in 2 c = 26. Determine the number of rivets for the most efficient joint. 2 t = 24. B. 000 lb/in . Rivets & Welds .Topic 5. Weld Material: = 12. load and specify how many inches of weld should be placed along each side AB and FG. 000 lb/in2 = 13. B. CD = 3 inches.000 lb/in2 .htm (4 of 4) [5/8/2009 4:40:37 PM] .9: Torsion.Solution Problem 3) Select: Topic 5: Torsion.000 lb/in2 . t = 22. t = 24.000 lb/in2 Plate Material: Dimensions: AB = GF = 20 inches.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/torsx59. Rivets & Welds .) Determine the minimum inches of weld need to carry the 100.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength. The allowable stresses are as follows. DE = 6 inches (For Solution Select: Rivets & Welds .000 lb.Topic Exam A. . A.htm (1 of 4) [5/8/2009 4:40:37 PM] . Brass = 6 x 106 lb/in2.TC.. & B (TD. 1. C. Using the torque found in part A. and section CD is brass. cutting shaft CD and drawing the section shown in Diagram 2.000 lb/in2. Rivets & Welds . & TB) without exceeding the allowable shear stress in any section of the shaft.Topic 5. Determine the maximum torque which could be applied at points D. B. Solution: We start by determining the maximum allowable internal torque in each section. If the allowable shear stress for steel is 18. We begin from the right end and work to the left.9a: Torsion Exam .9 Torsion. Aluminum = 4 x 106 lb/in2. Section AB brass. A compound shaft is attached to a fixed wall as shown in Diagram 1.000 lb/in2. and for brass is 16. section BC is steel.Solution Topic 5.Topic Examination Torsion Exam Solutions Modulus of Rigidity for several materials: Steel = 12 x 106 lb/in2. determine the resultant angle of twist of end D with respect to end A. file:///Z|/www/Statstr/statics/Torsion/xsol59a. and is equal to the externally applied TD. finding TC = 185 ft-lb file:///Z|/www/Statstr/statics/Torsion/xsol59a.9a: Torsion Exam . as shown in Diagram 3.) Putting values into the equation and solving: T = 16. r = .5 in = 3534 in-lb = 295 ft-lb.031 in4 / . and look at section to the right.000 lb/in2. This is the internal torque in section CD. as is clear from rotational equilibrium conditions and Diagram 2.) Putting values into the equation and solving: T = 18.5 inch.Topic 5. This is the internal torque in section BC.1416)(. (We use the allowable shear stress as the maximum stress in the shaft.375 inch.htm (2 of 4) [5/8/2009 4:40:37 PM] . and = 16. then solving for the torque: T = J / r .031 in4.000 lb/in2. where J = (3. and = 18.098 in4. = T r / J . r = . We next applied the torsion formula and solve for the internal torque in section BC. We next cut the shaft in section BC. From Diagram 3 we see that the internal toque is equal to the sum of the external torque TD(110 ft-lb) and TC.098 in4 / . We can solve for the value of TC.1416)(1")4 / 32 = .000 lb/in2 * .Solution We next applied the torsion formula and solve for the internal torque.375 in = 1325 in-lb = 110 ft-lb. where J = (3. (We use the allowable shear stress as the maximum stress in the shaft. = T r / J . then solving for the torque: T = J / r .000 lb/in2 * .75)4 / 32 = . 2.06o .5.1o (clockwise). Apply the Angle of Twist relationship to each section of the shaft.Topic 5.000 lb/in2. 2.9a: Torsion Exam .* 12 in)/(1.132 in-lb = 2094 ft-lb.036 radians = 2. r = 1 inch.Solution We next applied the torsion formula and solve for the internal torque in section AB.* 12 in)/(. (We use the allowable shear stress as the maximum stress in the shaft. -) signs of the angles of twist. A 2 foot long hollow steel shaft with an outer diameter of 2 inches and an inner diameter of 1. and = 16. In this example.clockwise.06o(cw). finding TB = 2389 ft-lb Part B: The angle of twist of end D with respect to end A can be found by determining the angle of twist in each section and then finding the net angle of twist.085 radians = 4. From Diagram 4 we see that TB is equal to the sum of the internal toque (2094 ft-lb) + TD(110 ft-lb) + TC(185 ft-lb). -.098 in4 * 12 x 106 lb/in2)=.57 in4.1416)(2")4 / 32 = 1. then solving for the torque: T = J / r .87o = . = + 1.000 lb/in2 * 1.* 12 in)/(.534 in-lb. =T L/J G= (3. and CD clockwise. AB . AB BC CD =T L/J G= (25.5 inches is to transmit power while being driven a 3000 rpm. file:///Z|/www/Statstr/statics/Torsion/xsol59a.132 in-lb. BC .032 radians =1.031 in4 * 6 x 106 lb/in2)=.4.) Putting values into the equation and solving: T = 16. where J = (3. This is the internal torque in section AB. = T r / J . where our signs are taken from the direction of the Then the total twist of end D with respect to end A will be: total internal torque in each section.57 in4 / 1 in = 25.counter clockwise.325 in-lb. We can solve for the value of TB. =T L/J G= (1.83o(ccw).57 in4 * 6 x 106 lb/in2)=. resulting in the (+.87o(cw).83 o .htm (3 of 4) [5/8/2009 4:40:37 PM] . Rivets & Welds .9: Torsion. and = 15.07 in4 / 1 in = 16. Inserting values and solve for Do we find: 15.000 lb/in2 * 1. PAB=2 pi Tn/(550 ft-lb/s/hp)=2*3. Rivets & Welds .1416)(2"4 . what is the maximum horsepower which can be transmitted down the shaft ? B. r = Do /2.000 lb/in2.and T = 16107 in-lb.(.1416/32)[Do4 .000 lb/in2 = (16107 in-lb * Do /2)/(3.. (We use the allowable shear stress as the maximum stress in the shaft.4Do)4]/ 32. where J = (3.000 lb/in2.1416*1342 ft-lb.1.000 lb/in2 ) Solution: Part A We use the shear stress relationship to first find the maximum torque which may be applied to the shaft: = T r / J .Topic Examination or Select: Topic 5: Torsion.107 in-lb = 1342 ft-lb.5"4)/ 32 = 1. r = 1 inch. where J = (3.78 inches Return to: Topic 5.*(50 rev/s) /(550 ft-lb/s/hp)= 767 hp Part B: Using the torque found in part A. Then we apply the power equation to determine the maximum power which may be transmitted. then solving for the torque: T = J / r .) If the allowable shear stress in the shaft is 15.htm (4 of 4) [5/8/2009 4:40:37 PM] . but were told that the inner diameter was to be four-tenths of the outer diameter.07 in4.Topic 5. what would be the minimum outer diameter of the shaft which could safely transmit the horse power found in part A? (The allowable shear stress in the shaft is 15.1416)[Do4 .) Putting values into the equation and solving: T = 15.4Do)4] so. we can now solve of the outer diameter. and the maximum allowable shear stress.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/xsol59a.9a: Torsion Exam . Do= 1.) If we were not given the outer diameter of the shaft.000 lb/in2.Solution A.(. = 15. = T r / J . Bearing (compression) Failure: We next determine the load the rivet or plate can carry before failing in compression.800 lb. A riveted lap joint is shown in Diagram 1. = 17.1* 1") *(5/8") * (22. 000 lb/in2 = 251.Topic 5. 2 t = 22.000 lb/in2. Rivets & Welds . Solution: Part 1 1. 3.Topic Exam Solutions Rivets & Welds .000 lb. 000 lb/in . will cause the joint (plate) to fail in compression.Solution Problem 1 1. and the Efficiency of the Joint.Exam Solution 1 Topic 5. file:///Z|/www/Statstr/statics/Torsion/xsol59b.1416 * (1)2/4]*20.. Prow1 = (w . Notice we use the smaller of the allowable compressive stress between the rivet and plate material. Rivet Shear: The load the joint can carry before failing in rivet shear is given by: P = N (pi * d2/4) = ( 16 rivet areas)* [3.250 lb. as it is the rivet (not the plate) which fails in shear. Thus at a load of 238.000 lb.n d)t all = (12".000 lb/in2 ) = 151. and the thickness of the plates is 5/8 inch. 000 lb/in Determine the Strength of the Joint.300 lb. The width of the plates is 12 inches. Thus a load of 240. 2. 2 c = 28. 000 lb/in 2 c = 24. The diameter of the rivets is 1 inch.000 lb/in2) = 240. The allowable stresses are as follows: Rivets: Plate: = 20. 2 t = 26.9b: Rivets & Welds . Please note we used the allowable shear stress for the rivet material in our equation. 000 lb/in . 000 lb/in2. Pbearing = N (d*t) all = ( 16 rivets) * (1" * 5/8") * (24. Plate Tearing (Row 1): We now determine the load the joint (plate) can carry before failing in tension . This is the load which will cause the plate to fail in tension at row 1.htm (1 of 2) [5/8/2009 4:40:37 PM] .9: Torsion.at row 1. the joint will fail in shear. 000 lb. the joint fails first in plate tearing at row 1 with the lowest load of 151.9: Torsion.700 lb. = 152. Then we can write: (13/16)Prow3 = (w . Plate Tearing (Row 2): . and then: Prow3 = (16/13) * 123.n d)t all = (12". As there are a total of 3 rivets in row 1 and row 2. We do this as follows. This says that joint can carry 89 percent of what the solid plate could carry. then one sixteenth (1/16) of the load has been transferred to the bottom plate by the rivet in row 1. however we still need to check plate tearing failure at rivet row 2 (and perhaps row 3) to see if the joints fails at a lower load there. compression. or Efficiency = Joint Strength / Plate Strength = 146.250 lb.500 lb. Rivets & Welds . or Pplate = (w * t) * all = (6" * 1/2") * 20. and we may write.n d)t all = (12".000 lb/in2 ) = 123. and this is the final Strength of the Joint.. Rivets & Welds . and we then define the joint efficiency as the ratio of the Joint Strength to the Plate Strength.700 lb. Thus row 2 carries of 15/16 of the load P.000 = .3* 1") *(5/8") * (22. the largest load we can safely apply. and then: Prow2 = (16/15) * 137. Return to: Topic 5. Plate Tearing (Row 3): Since the joint would fail at row 2 before row 1. This means we can stop here. (15/16)Prow2 = (w .2* 1") *(5/8") * (22.Table of Contents Statics & Strength of Materials . Since there are 16 rivets in the pattern and we assume that the rivets share the load equally.9b: Rivets & Welds .Course Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/xsol59b.700 lb.300 lb. and thus the plate material at row 3 carries only 13/16 of the load.Exam Solution 1 We see that of the three main modes of failure above. and thus it is the strength of the joint to this point. This is the load which will cause shear row 2 to fail in tension.146.000 lb/in2 = 60. P.000 lb/in2 = 146.89 = 89%. = 146. This is the plate strength. Part 2 Plate Strength = area of plate cross section times allowable tensile stress for plate material. /(12" * 5/8") * 22.Topic Examination or Select: Topic 5: Torsion.500 lb.htm (2 of 2) [5/8/2009 4:40:37 PM] . We see that it is lower than the lowest load so far that will cause the joint to fail. 3a. / 165. Up to this point.000 lb/in2 ) = 137. Of all the loads which will cause the joint to fail (either in shear. we need now to continue and determine the load at which row 3 would fail. or tension)? The lowest is plate tearing at row 2 .750 lb.700 lb.750 lb. Here we see that load which causes row 3 to fail in tension is larger than the loads which would cause row 1 and row 2 to fail. 3b.. this means that 3/16 of the load has been transferred to the bottom plate.Topic 5. this is the Strength of the Joint.. /in A. B. Select the best pattern from the patterns in Diagram 3 ./in2. C. 000 lb. The rivets to be used have a diameter of 3/4 inch. = 18./in . We use the plate tearing relationship: Prow1 = (w .Topic Exam Solutions Rivets & Welds .Solution Problem 2 2. Solution: Part A Step 1: Calculate the load which would cause the joint to fail in plate tearing at row 1 . 2 t = 24. 000 lb. 000 lb. 2 c = 23.9c: Rivets & Welds Exam Solution 2 Topic 5.000 lb.htm (1 of 4) [5/8/2009 4:40:38 PM] . Rivets & Welds . 2 t = 22. The maximum allowable stresses for the rivet and plate materials are as follows: Rivets: Plate: = 17./in ./in 2 c = 26. .assuming one rivet in row 1. A double cover plate butt joint (Diagram 2) is to connect two steel plates both with a width of 9 inches and a thickness of 7/8 inch./in2. 000 lb.9: Torsion. Determine the number of rivets for the most efficient joint.Topic 5. Calculate the strength and efficiency of the joint. 000 lb.n d)t all . where w = width of the main plate = 9 inches file:///Z|/www/Statstr/statics/Torsion/xsol59c. Calculate the maximum load which one rivet (or the plate material behind one rivet) could carry in bearing (compression) Pbearing = N (d*t) all.9c: Rivets & Welds Exam Solution 2 n = number of rivets in row (in row 1.020 lb./in2.250 lb. (See Diagram below) file:///Z|/www/Statstr/statics/Torsion/xsol59c. # Rivets = 173. Where: N = Number of areas in shear For 1 rivet in a double cover plate butt joint this will be 2 area. Prow1 = (9" ./rivet = 11. Divide the allowable load in plate tearing./rivet Step 4./rivet Step 3./in2 = 15. Step 2. 000 lb. 000 lb.090 lb.000 lb.Topic 5. calculated in step 1.000 lb. rounding up # rivets = 12 Solution: Part B We use the number of rivets found in Part A above to select the best joint pattern from the given set of patterns.htm (2 of 4) [5/8/2009 4:40:38 PM] . N = 2 d = 3/4 inch all = 17.1 * 3/4")*(7/8") * 24.53 . From Diagram 3 above we select the a lap joint pattern with correct number of rivets. . 000 lb./in2 = 15.020 lb. We use the rivet shear relationship: Privet shear = N (pi * d2/4) all./in2 Privet shear = 2 [3.1416 * (3/4")2/4] * 17. by the small of the loads one rivet could carry from steps 2 and 3. and round up to determine the number of rivets which will result in the most efficient joint. Calculate the maximum load which one rivet could carry in rivet shear.250 lb. 000 lb. where N = Number of rivets in compression = 1 d = Diameter of rivet = 3/4 inch t = Thickness of the main plate = 7/8 inch all = Lowest allowable compressive stress of the rivet or plate material = 23. / 15. 1 rivet) d = diameter of rivet = 3/4 inch t = thickness of the main plate = 7/8 inch all = Maximum allowable tensile stress for the plate material = 24./in2 Pbearing = 1 * [(3/4") * (7/8")] * 23./in2 = 173. Rivets & Welds .820 lb.) Rivet Shear: We have already determined that one rivet can carry 15.500 lb. This loading is much larger than the loading which will cause row 2 to fail. Considering all the modes of failure. 2.Table of Contents file:///Z|/www/Statstr/statics/Torsion/xsol59c./rivet. using the selected joint pattern from part B. = 210.090 lb. in step 1 above. now that we have selected the rivet pattern. Notice.2 * 3/4")*(7/8") * 24. we have already calculated the allowable load per rivet in bearing = 15. 000 lb.9% Return to: Topic 5. 1. The total load the joint can carry in bearing (compression) will be the product of the number of rivets and the allowable load per rivet: Pbearing = 12 * 15.100 lb.) Plate Tearing (row 3): Row 2 carries 9/12 of the load. we also need to calculate plate tearing at row 2 (and perhaps row 3) 4. we calculate the strength and efficiency of the joint.020 lb. However. so we can stop here./in2 = 173. 3.2 * 3/4")*(7/8") * 24. so we write: (11/12)Prow2 = (9" .) Plate Tearing (row 1): We have already calculated plate tearing at row 1. and then Prow3 = (12/9) 157. we see:Strength of the Joint is 171. and then Prow2 = (12/11) 157.1 * 3/4")*(7/8") * 24.htm (3 of 4) [5/8/2009 4:40:38 PM] .9c: Rivets & Welds Exam Solution 2 Solution: Part C Finally. 000 lb./rivet = 180.250 lb.000 lb.Topic Examination or Select: Topic 5: Torsion. to cause failure.090 lb.909 = 90. so we write: (9/12)Prow2 = (9" .) Bearing Failure: Again. Prow1 = (9" .500 lb. that this loading is the lowest./rivet. 5.020 lb.500 lb.820 lb.240 lb.000 lb./in2 = 157./(9"*7/8")*24..500 lb. Rivets & Welds . and so it is currently the strength of the joint. up to this point./in2 = 157. = 171. However.9: Torsion./rivet = 181.) Plate Tearing (row 2): Row 2 carries 11/12 of the load. so the total load the joint can carry in rivet shear will be the product of the number of rivets and the allowable load per rivet: Pshear = 12 * 15./in2 = . (plate tearing row 2) and Efficiency = Joint Strength / Plate Strength = 171.Topic 5.820 lb. we will continue and check the load which causes row 3 to fail.. 000 lb. htm (4 of 4) [5/8/2009 4:40:38 PM] .Topic 5.Course Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/xsol59c.9c: Rivets & Welds Exam Solution 2 Statics & Strength of Materials . load by using the weld formula and setting the load the weld can carry before failing to 85.000 lb/in2 . Two metal plates are shown below. load and specify how many inches of weld should be placed along each side AB and FG.Topic Exam Solutions Rivets & Welds . file:///Z|/www/Statstr/statics/Torsion/xsol59d.000 lb. t = 22.9: Torsion.000 lb/in2 Plate Material: Dimensions: AB = GF = 20 inches. Rivets & Welds . B. DE = 6 inches Solution: Part 1.000 lb/in2 .Solution Problem 3 3.Topic 5. CD = 3 inches.. The top plate is 3/4 inch thick and 9 inches wide.htm (1 of 3) [5/8/2009 4:40:38 PM] . A.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength. and is to be welded to the bottom plate with a 45o fillet weld.000 lb. Weld Material: = 12. The allowable stresses are as follows.9d: Rivets & Welds Exam Solution 3 Topic 5.) Determine the minimum inches of weld need to carry the 100. 000 lb/in2 = 13.000 lb. t = 24. The top plate is to be weld completely across end AG and partially along sides AB and FG. We first determine the minimum inches of weld needed to carry the 85. . We assume the plate is loaded in tension.363 lb./in. we can not apply the weld symmetrically.32 inches ( the minimum inches of weld needed).)* LGF ./in.98 inches.57. Part 3. Part 2.270 lb.000 lb./in. To determine how the weld should be distributed. LGF .)* LGF .000 lb/in2 .270 lb.) Solving the torque equation for LAB = 5.000 lb. and FAG = 6.707 t * L) * all .5. and then t . and 9 inches of weld completely across the end AG The lengths LAB .)* LAB * (9") + 57.363 lb.)* LAB .000 lb. or = (w * t ) (./in * 9" = 57. This is the minimum inches of weld needed to carry the load.Topic 5. The weld will be distributed with an amount LAB on side AB. We now apply static equilibrium conditions to the top plate: Sum of Forces: 100.000 lb/in2 = (9" * 3/4" ) 22.(6.363 lb. if we try to put equal amounts of weld on sides AB and GF.363 lb. however since the 85. we once again apply static equilibrium conditions. Rivets & Welds .363 lb.98" . = 0 Sum of Torque about G: -100. since we are not dealing with a specific loading. we can then solve for LGF since LAB + LGF + 9" must sum to 15.000 pound load is not applied symmetrically to the plate.707 t * L) * (.707 t * L) * all all = Pplate =(w * t ) t .(6. Rivets & Welds ./in. The maximum force these weld can resist with is given by FAB = (6. the weld will fail in this case./in. .Topic Examination or Select: Topic 5: Torsion.74" This is how the weld must be distributed in order to carry the 100.363 lb.5" = 0 (Notice that the force FGF does not produce a torque about point G.) = 15. Therefore LGF = 15. or 100.707 *3/4" * L) * 12. since its line of action passes through point G.000 lb. We then set this equal to the strength of the riveted joint and solve for the length of weld needed.72 inches. LGF on side GF. Return to: Topic 5.000 lb = (.)*L. then solving for L L = 23.* (6") + (6.270 lb. Pweld = (./in.htm (2 of 3) [5/8/2009 4:40:38 PM] .Table of Contents file:///Z|/www/Statstr/statics/Torsion/xsol59d. or Pplate = (w * t ) t .000 lb/in2 = (6.33 inches This is the minimum inches of weld needed to make the weld as strong as the plate (in tension).363 lb.9" = . load (and satisfy static equilibrium conditions).9d: Rivets & Welds Exam Solution 3 Pweld = (. * 4.72" . FGF = (6. and the 9 inches of weld must sum to 12.707 * 3/4" * L) * 12.72 inches. and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material. We can not state how the weld should be distributed.363 lb./in./( 6. then solving for L L = 100.)* LAB .9: Torsion. That is. 9d: Rivets & Welds Exam Solution 3 Statics & Strength of Materials .Course Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Torsion/xsol59d.Topic 5.htm (3 of 3) [5/8/2009 4:40:38 PM] . Or.Topic 6. we do not use the actual length but an effective length.6a Pressure Vessels .Thin Wall Pressure Vessels 6.2a Secant Formula .2 Columns & Buckling .Example 1 r 6.Example 2 6.Problem Assignment 2 (supplementary) 6.Example 5 6. The values for K depend on how the column is supported.1b Euler Buckling .2c Structural Steel Column Selection .Problem Assignment (required) 6.6b Pressure Vessels .2b Structural Steel . if the entire cross sectional area of the object (beam) were located at that distance. The distinction between types of columns is not well defined. where K could be called an effective length constant. Notice that we have put a subscript "e" by the length of the column. the formulas we will utilize also apply to horizontal compression members. Fixed-Free: K = 2 (See Diagram.Example 1 r 6.3 Columns & Buckling .2d Aluminum Columns . often axially loaded. and Long Columns.5 Pressure Vessels .1a Euler Buckling .6c Columns & Buckling .Example 2 r 6.I When we speak of columns (and buckling) we are talking about members loaded in compression. Slenderness Ratio = Le / r. Intermediate Columns. Fixed-Pinned: K= .) file:///Z|/www/Statstr/statics/Columns/cols61.2e Wood Columns . depending on how the column is supported. Fixed-Fixed: K = .I r 6. For instance. The effective length is given by: Le = K L.5.Topic Examination (with thin wall pressure vessel problem) Columns & Buckling . Pinned-Pinned: K= 1. however.Topic Examination 6. it may be expressed as: Radius of Gyration: rxx = (Ixx/A)1/2 (radius of gyration about xx-axis) So. or to compression members in general. The radius of gyration is the distance from an axis which.II r 6.7. The Slenderness Ratio is the (effective) length of the column divided by its radius of gyration.Problem Assignment 1 (required) 6.0: Columns & Pressure Vessels Topic 6: Columns & Pressure Vessels ● ● ● ● ● ● ● ● 6. although columns may be loaded eccentrically. compression members of a truss may be considered to be columns pinned at each end point. Columns may be divided into three general types: Short Columns.htm (1 of 5) [5/8/2009 4:50:09 PM] .1 Columns & Buckling .Example 3 r 6. but a generally accepted measure is based on the Slenderness Ratio. it would result in the same moment of inertia that the object (beam) possesses.Example 4 r 6. This is to indicate that.4 Columns & Buckling . We also tend to think of columns as vertical members. as the ratio of the column length to the effective cross sectional area is too small. Leonard Euler (pronounced Oiler) developed a relationship for the critical column load which would produce buckling.htm (2 of 5) [5/8/2009 4:50:09 PM] .Topic 6. II. Short Columns: When the slenderness ratio is less than 60. A very brief derivation of Euler's equation goes as follows: file:///Z|/www/Statstr/statics/Columns/cols61.0: Columns & Pressure Vessels A generally accepted relationship between the slenderness ratio and the type of Column is as follows: Short Column: 0< Le / r < 60 Intermediate Column: 60< Le / r <120 Long Column: 120< Le / r < 300 I. axially loaded. P/A > s allow. Long Columns & Euler's Equation: In 1757. Idealized buckling in long columns was first treated by the famous mathematician Leonard Euler. which we will look at later in this section. 'thick' column. a column will not fail due to buckling. We also will put off discussion of intermediate length columns until we have discussed long columns. will fail in simple compressive failure: that is when the load/area of the column exceeds the allowable stress. Eccentrically loaded short columns have a slightly more complicated result for compression failure. Rather a short. and in terms of the load P. we can write: 1.P y. Then substituting from EQ. and y = 0 at x = L. That is. which. the value of B must be zero (since cos (0) = 1). (M /EI) = (d2y /dx2) (See Strength of Material text chapters on beams and beam deformations. 2. 1 to EQ. which has a general solution form of 4. then either A must be zero (which leaves us with no equation at all. While for y to be zero at x = L. We also can write that for beams/columns the bending moment is proportional to the curvature of the beam.0: Columns & Pressure Vessels A loaded pinned-pinned column is shown in the diagram. (d2y /dx2) = -(P/EI)y or (d2y /dx2) + (P/EI)y = 0 This is a second order differential equation. or . we obtain: 3.Topic 6. for small deflection can be expressed as: 2. M = . if A and B are both zero). and I = moment of Inertial. Applying these conditions (putting these values into the equation) gives us the following results: For y to be zero at x =0.) Where E = Young's modulus. Which results in the fact that And we can now solve for P and find: file:///Z|/www/Statstr/statics/Columns/cols61. and the deflection distance y. : We next apply boundary conditions: y = 0 at x = 0.htm (3 of 5) [5/8/2009 4:50:09 PM] . the deflection of the column must be zero at each end since it is pinned at each end. A top section of the diagram is shown with the bending moment indicated. as compared to I = (1/12) b d3 = (1/12) (2")(4")3 = 10. but not with eccentrically loaded columns. and Fixed-Free columns.htm (4 of 5) [5/8/2009 4:50:09 PM] . so that when given a column cross section. As an example: For structural steel with a proportional limit stress of 35. we can arrive at an equation for the minimum slenderness ratio such that Euler's equation will be valid. (I = (1/12) b d3 = (1/12) (4")(2")3 = 2. Le. By replacing L with the effective length.000 psi) = 92 This is the minimum slenderness ratio for which we could use Euler's buckling equation with this column material. Another form may be obtained by solving for the critical stress: and then remembering that the Radius of Gyration: rxx = (Ixx/A)1/2 . Fixed-Pinned. and a Young's Modulus of E = 30 x 106 psi. the column will buckle in the 2" direction rather than the 4" direction. the student must be sure to determine the minimum moment of inertia.. Fixed-Fixed. as buckling will occur about that axis. as the moment of inertia for the 2" direction is considerable smaller than the moment of inertia in the 4" direction. That is. Euler's Equation for columns while useful. and substituting we can obtain: which gives the critical stress in terms of Young's Modulus of the column material and the slenderness ratio.142 x 30 x 106 psi / 35. the critical stress must not exceed the proportional limit stress.67 in4 ) It is important to point this out. which then applies to Pinned-Pinned. we can generalize the formula to: . Some additional points need to be mentioned. file:///Z|/www/Statstr/statics/Columns/cols61. If we now substitute the proportional limit stress for the critical stress. That is. is only reasonably accurate for long columns. 6. and in addition will work for axially loaded members with stress in the elastic region. where Pcr stands for the critical load which can be applied before buckling is initiated. we can obtain (Le/r) =sqrt( 3. Let us at this time also point out that Euler's formula applies only while the material is in the elastic/proportional region. or slenderness ratios of general range: 120 < Le / r < 300.0: Columns & Pressure Vessels 5. (rxx = (Ixx/A)1/2 & ryy = (Iyy/A)1/2 ). which in turns depends on the moment of inertia of the column cross section.000 psi. .Topic 6. The slenderness ratio (Le/r) depends on the radius of gyration.. The value of the moment of inertia depends on the axis about which it is calculated. This equation is a form of Euler's Equation. for a rectangular 2" x 4" cross section.67 in4. which was defined above. variations in straightness of the member. and defects in the material. it should not be assumed the values are completely accurate. Several examples of buckling calculations follow.1b: Euler Buckling .htm (5 of 5) [5/8/2009 4:50:09 PM] .Topic 6. Continue to: Topic 6.1a: Euler Buckling . Buckling is a complicated phenomena. presence of initial unknown stresses in the column.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/cols61. and the buckling in any individual column may be influenced by misalignment in loading.Example 1 Topic 6.Example 2 or Select: Topic 6: Columns & Pressure Vessels .0: Columns & Pressure Vessels It should also be pointed out that while these formulas give reasonable values for critical loads causing buckling. with respect to the y-y axis. The Euler Buckling Load is then give by: .Example 1 Topic 6. For this column.289 158.14)2*29x106 lb/in2 * 16. Or.1a: Euler Buckling .in 4.in I . We also notice that the slenderness ratio is large enough to apply Eulers buckling formula to this beam. as that is the axis about which buckling will occur.1a: Euler Buckling .61 A-in2 d . and are also listed below.428 lb/8.Example 1 An 16 ft.0 S -in3 30. and yield stress = 36.in 1.38 in = 139 Notice that we must use the smallest radius of gyration.22 5. after finding for ASTM-A36 Steel.1: Columns & Buckling .in4 0.in wf .30 in4/(16x12"/ft)2] = 126. x-x axis r .30 S -in3 5.in tf .54 10. the load that will result in Euler buckling.I Continue to:Topic 6./ 1. W10x29 I-Beam is to be used as a column with pinned ends. determine the slenderness ratio. * 12 in. and the associated Euler buckling stress. long ASTM-A36 steel. y-y axis r .htm (1 of 2) [5/8/2009 4:50:09 PM] .54 in2 = 14. This means that the column will fail in buckling before axial compressive failure.500 The slenderness ratio = Le / r = 16 ft. Notice that this stress which will produce buckling is much less than the yield stress of the material. The beam characteristics may be found in the I-Beam Table.38 Designation Area Depth Width W 10x29 thick x-x axis x-x axis tw .Topic 6. we obtain: Pcr = [(3.1b: Euler Buckling .800 lb/in2.000 lb/in2. To verify this we use the relationship for the minimum slenderness ratio for Eulers equation to be valid.8 y-y axis y-y axis I . Flange Flange Web thick Cross Section Info.in4 16.in 8.428 lb c) The Euler Stress is then easily found by Stress = Force/Area = 126. E = 29 x 106 lb/in2./ft.Example 2 file:///Z|/www/Statstr/statics/Columns/colse61a. Return to:Topic 6.799 0.30 Cross Section Info. we can solve and determine that Le/r = 89. and after substituting values. Example 1 or Select: Topic 6: Columns & Pressure Vessels .htm (2 of 2) [5/8/2009 4:50:09 PM] .1a: Euler Buckling .Topic 6.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse61a. htm (1 of 2) [5/8/2009 4:50:09 PM] . To determine the slenderness ratio in this problem. the load that will produce Euler buckling. we obtain: Pcr = [(3.420 lb/(2"*4") in = 678 lb/in . The yield stress for the wood is 6. determine the slenderness ratio. where this is the radius of gyration about an x-x axis. Then Slenderness ratio is given by: Le / r = ( 8 ft x 12"/ft)/.14)2*1. 2 2 file:///Z|/www/Statstr/statics/Columns/colse61b. so we use d =2". We now simplify and obtain: rxx = [(1/12)d2]1/2 = . and after substituting values.Topic 6.1b: Euler Buckling .5774" = 166 which puts the beam in the long slender category.500 lb/in2.420 lb. buckling will first occur about the x-x axis shown is the diagram.Example 2 A 8 ft.5774 in. and where I = (1/12)bd3 for a rectangular cross section. Or rxx = [(1/12)bd3/bd]1/2 . we first have to find the radius of gryration (smallest). The Euler Buckling Load is then give by: . which we may do from the relationship: Radius of Gyration: rxx = (Ixx/A)1/2 . For this column. The slenderness ratio = Le / r . That is. c) The Euler Stress is then easily found by Stress = Force/Area = 5. where we have substituted A = bd. Notice that this stress which will produce buckling is much less than the yield stress of the material.9x106 lb/in2 * (4*23/12) in4/(8x12"/ft)2] = 5.1b: Euler Buckling .5774(d/2) We want the smallest radius of gyration.Example 2 Topic 6. and Youngs modulus is 1. long southern pine 2" x 4" is to be used as a column.9 x 106 lb/in2. and the associated Euler buckling stress. and r = . Example 2 Return to:Topic 6.1b: Euler Buckling .II or Select: Topic 6: Columns & Pressure Vessels .I Continue to:Topic 6.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse61b.Topic 6.2: Columns & Buckling .htm (2 of 2) [5/8/2009 4:50:09 PM] .1: Columns & Buckling . B.II III. We first note that at the point where the Johnson formula and Euler's formula are tangent.htm (1 of 7) [5/8/2009 4:50:10 PM] .Topic 6. We will not derive this formula. such that the corresponding stress is one-half of the yield stress.II Topic 6.000 psi. Additionally.2: Columns & Buckling .2: Columns & Buckling . and a Young's modulus of 30 x file:///Z|/www/Statstr/statics/Columns/cols62. In the diagram below. but make several comments. For a graph of stress versus slenderness ratio. The J. One of these is the J. Johnson Formula. Johnson formula is the equation of a parabola with the following characteristics.B. the parabola has its vertex at the value of the yield stress on the y-axis. Notice the parabolic curve beginning at the yield stress and arriving tangent to the Euler curve at 1/2 the yield stress. Intermediate Columns There are a number of semi-empirical formulas for buckling in columns in the intermediate length range. we can relate the stress to Euler's formula as follows (where C represents the slenderness ratio when the stress is 1/2 the yield stress): From this we find an expression for C (critical slenderness ratio) of: For our particular case. the parabola is tangent to the Euler curve at a value of the slenderness ratio. where we have a steel member with a yield stress of 40.000 psi. we have a steel member with a yield stress of 40. 0 S -in3 78. (Beam information and Johnson formula shown below.641 J.720 psi. Designation Area Depth Width W 12x58 thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis tw . Cross Section Info. If our actual beam has a slenderness ratio greater than the critical slenderness ratio we may use Eulers formula.in 2. This is the critical stress that would produce buckling.51 A-in2 d . The Secant Formula: Another useful formula is known as the Secant formula.19 10. we find C = sqrt(2 * 3. VI.6 (where 2.B. Note we did not have a safety factor in this problem.) Flange Flange Web thick Cross Section Info.in4 107. As a result we really would not want to load the column to near the critical stress.in tf .(95. If on the other hand our actual slenderness ratio is smaller than the critical slenderness ratio. we may use the J. but not with other end conditions.in 5. The Critical Load will equal the product of the critical stress and the area. but focus on its application.in I . about y-y axis).htm (2 of 7) [5/8/2009 4:50:10 PM] .B. and with fixed-free (Le = 2L) end columns. Example: As an example let us now take a 20 foot long W12 x 58 steel column (made of same steel as above).00 S -in3 21. It may be used with pinnedpinned (Le = L).II 106 psi..51 in. and calculate the critical stress using the J.1 r .28 I .40 r .2: Columns & Buckling . Inserting values we find: Critical Stress = [ 1 .Topic 6. or Pcr = 27.B. Johnson formula.10 in2 = 474.000 psi.51in = 95.142 * 30 x 106 / 40.62/2* 1222)]*40.720 psi.000 psi) = 122. is the smallest radius of gyration.in 17. * 17. but at a lower 'allowable' stress. We will not go through the derivation of this relationship. = 27.359 476.10 12. Johnson Formula. Johnson's formula: For our beam the slenderness ratio = (20 ft * 12 in/ft)/2. file:///Z|/www/Statstr/statics/Columns/cols62.012 lb.in4 0.014 0. The Secant formula may be used for both axially loaded and eccentrically loaded columns.in wf . the easiest method of solution is to simply try different values of P. P. etc. Once we have the maximum compressive stress due to the load. the eccentricity ratio (ec/r2). etc. until we find a satisfactory fit. See following example.II The Secant formula gives the maximum compressive stress in the column as a function of the average axial stress (P/A). we may use the Secant formula to determine the maximum load we can safely apply to the column. for a given column. we can compare this stress with the allowable stress for the material and decide if the column will be able to carry the load. On the other hand a common practice with axially loaded structural steel columns is to use an eccentricity ratio of . When the eccentricity value is zero (corresponding to an axial loading) we have the special case that the maximum load is the critical load: and the corresponding stress is the critical stress or This is one way to look at axial loads.2a: Secant Formula .25 to account for the effects of imperfections. file:///Z|/www/Statstr/statics/Columns/cols62.. the load. the slenderness ratio (L/r). Then the allowable stress does not have to be reduced to account for column imperfections.Example 1 The eccentricity ratio has a normal range from 0 to 3.Topic 6.2: Columns & Buckling . Thus. with most values being less than 1. Please Select 6. are known. On the other hand. If. and eccentricity of the load. as this is taken into account in eccentricity ratio. then the maximum compressive stress can be calculated. In this case we will be solving for P. and we take note that the equation is a transcendental equation when solved for P. and Youngs Modulus for the material. if we know the allowable compressive stress for the column. e.htm (3 of 7) [5/8/2009 4:50:10 PM] . Mechanic.Example 2 for structural steel example. aluminum and wood. and National Design Specifications for Wood Constuction.Topic 6.2c: Structural Steel Column Selection . Aluminum Construction Manuel. 2. Structural Steel: Please Select 6. 1. Timber Construction Manual.2b: Structural Steel . Empirical Design Formulas for Columns: A number of empirical design formulas have been developed for materials such as structural steel. and may be found in such publications as the Manual of Steel Construction.2: Columns & Buckling .II V.htm (4 of 7) [5/8/2009 4:50:10 PM] . Aluminum file:///Z|/www/Statstr/statics/Columns/cols62. Specifications for Aluminum Structures.Example 3 for structural steel example. Please Select 6. file:///Z|/www/Statstr/statics/Columns/cols62.2d: Aluminum Column .Topic 6.htm (5 of 7) [5/8/2009 4:50:10 PM] .II Please Select 6.2: Columns & Buckling .Example 4 for Aluminum example. we add the two stress and obtain Total Maximum Compressive Stress = (P/A)(1 + A e c1/I) file:///Z|/www/Statstr/statics/Columns/cols62. with the force.Example 5 for wood column example. Finally. P.II 3. P. Wood Columns Please Select 6. plus a moment whose value will be M = P x e. P. acting a distance. which will simply be P/ A. Next we calculate the compressive stress due to the axial force.Topic 6.2: Columns & Buckling . which gives (Pe)c/I where the bending stress will be a compressive maximum on the right side of the column and a tensile maximum on the left side of the column (and zero at the centroid). with an axial force. P .2e: Wood Column . VI Short Eccentrically Loaded Columns An eccentrically loaded short column is shown in the diagram. We may replace the eccentrically acting force. e. Then we calculate the bending stress due to the moment P x e.htm (6 of 7) [5/8/2009 4:50:10 PM] . from the centroid of the column cross sectional area. the left side on the column may be in tension.Problem Assignment or Select: Topic 6: Columns & Pressure Vessels .Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/cols62.II (right side of column).A e c1/I) (left side of column).2: Columns & Buckling .htm (7 of 7) [5/8/2009 4:50:10 PM] . and the Total Minimum Compressive Stress = (P/A)(1 . if the bending stress is large enough. And in fact.3: Columns & Buckling . Continue to: Topic 6.Topic 6. and buckling will occur first with respect to that axis. Since the slenderness ratio of the column is less than the critical slenderness ratio.650 y .in tf . Determine the maximum allowable axial load. The beam is 12 ft.Example 2 A structural steel T-beam (WT 15 x 66) is used as a column.Example 2 Topic 6.00 We will first do this problem assuming that radius of gyration is lowest about the X-X Axis.in 3.2b: Structural Steel .94 Next we calculate the critical slenderness ratio Cc2 = ( 2 * 3.400 r .in wf .000 lb/in2) = 15568. and C = 124. Calculating the slenderness ratio = Le/r = (2*12ft * 12"/ ft)/4.000 tw .900 A-in2 d .60 421.in 19.551 1.in d/tw 0.htm (1 of 3) [5/8/2009 4:50:10 PM] .8.2b: Structural Steel .40 15. file:///Z|/www/Statstr/statics/Columns/colse62b. - Designation Area WT15x66 x-x axis x-x axis x-x axis x-x axis I .142 * 30 x 106 lb/in2 /38.in4 S -in3 37. Repeat the problem if the column is 20 ft long. Youngs Modulus for the steel is 30 x 106 psi. long and is fixed on one end and free on the other.615 24. (Note that since this is a fixed-free column. we use the intermediate formula to find the allowable stress.in 4.).15 10. the effective length is equal to 2 * L.Topic 6.000 psi. Depth Flange Flange Stem of T Width thick thick Cross Section Info.65 = 61. and the yield stress for the steel is 38.. 110 lb/in2.94/124. Since the slenderness ratio of the column is less than the critical slenderness ratio. before we can file:///Z|/www/Statstr/statics/Columns/colse62b. FS = (5/3) + (3/8)(61. However.(1/8)(61.142 * 30 x 106 lb/in2 /38.Topic 6.8. we first calculate the factor of safety.65 in = 103. Calculating the slenderness ratio = Le/r = (2*20ft * 12"/ft)/4.Example 2 Before we can determine the allowable stress. Then Allowable Stress = (38. We again assume that radius of gyration is lowest about the X-X Axis. PART 2: Repeat if the column is 20 ft long. we use the formula for intermediate columns to find the allowable stress.4 in2 = 351.htm (2 of 3) [5/8/2009 4:50:10 PM] . and C = 124.2 Next we calculate the critical slenderness ratio Cc2 = ( 2 * 3.94/124. Finally the allowable load = Stress * Area = 18.(1/2)(61.94/124.8) .8)2] = 18.000 lb/in2) = 15568.000/1. and buckling will occur first with respect to that axis.8)3 = 1.84.110 lb/in2 * 19.84)[1 .334 lb.2b: Structural Steel . 8)3 = 1.090 lb/in2 Finally the Allowable Load = Stress * Area = 13. FS = (5/3) + (3/8)(103. Return to:Topic 6.(1/8)(103.2/124. we must first calculate the factor of safety.II Continue to:Topic 6.090 lb/in2 * 19.2: Columns & Buckling .2b: Structural Steel .8) .Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse62b.91)[1 -(1/2)(103.000/1.htm (3 of 3) [5/8/2009 4:50:10 PM] .Topic 6.Example 2 determine the allowable stress.2c: Structural Steel Column Selection .2/124.91 Then Allowable Stress = (38.8)2] = 13.4 in2 = 253.2/124.Example 3 or Select: Topic 6: Columns & Pressure Vessels .946 lb. Notice how much the change in length has reduced the allowable load. and which is to carry an axial load of 120.2 in2 and a minimum radius of gyration of 1. FS = 5/3 = 1.) Use the Table of I-Beams to select from.000 lb/in2/1. From our table we will try a W10 x 21.Example 3 Topic 6.400 lb/in2 = 5. which has an area of 6.000 psi.7 * L.400 lb/in2.Example 3 Select the best (safe & lightest) I-beam to be used as a 16 foot vertical column.Topic 6.htm (1 of 3) [5/8/2009 4:50:10 PM] . Next we choose (guess) a beam with an area greater than the minimum area found in step 1. One method of determining the best I-beam to use is an iterative type process. We first find the minimum cross sectional area of the beam by assuming the slenderness ratio.2c: Structural Steel Column Selection .88 in2.000 lb/20. (Note that since this is a fixed-pinned column.667 = 20.2c: Structural Steel Column Selection . and the yield stress for the steel is 34. we can solve for the minimum area: Area = 120. and the allowable stress = (yield stress/safety factor) = 34.. with one end fixed and the other end pinned. the effective length is equal to .667.000 lb. Le/r =0. Since the allowable stress is also Force/Area. Then the factor of safety. Youngs Modulus for the steel is 30 x 106 psi.32 file:///Z|/www/Statstr/statics/Columns/colse62c. that this I-beam would work.3/132) .71 in2 and a minimum radius of gyration of 1.54 in.000 lb/in2) = 17417. = 69.(1/8)(69. we use formula for intermediate columns to find the allowable stress.9)[1 . The Slenderness ratio for this column = Le/r = (.3/132)3 = 1.13 in2 and a minimum radius of gyration of 1. This means. = 101.000 lb/in2) = 17417. The critical slenderness ratio remains the same [Cc2 = ( 2 * 3.(1/2)(69.8/132)3 = 1.142 * 30 x 106 lb/in2 /34.(1/8)(101. and C = 132] Since the slenderness ratio of the column is less than the critical slenderness ratio.855 lb. which has an area of 9. We notice that the allowable load is somewhat higher than the load we would like to apply.2c: Structural Steel Column Selection . FS = (5/3) + (3/8) (69. We notice that the allowable load is about half of the load we would like to apply.142 * 30 x 106 lb/in2 /34.88. The critical slenderness ratio remains the same [Cc2 = ( 2 * 3. Before we can determine the allowable stress.85. The Slenderness ratio for this column = Le/r = (.142 * 30 x 106 lb/in2 /34. Then the Allowable Stress = (34. FS = (5/3) + (3/8) (87.54 in. 3.3/132)2] = 15.(1/8)(87.3.(1/2)(87.000 lb/in2) = 17417.Example 3 in.85)[1 . So we now select a larger cross section beam.845 lb/in2.3. Before we can determine the allowable stress. Then the Allowable Stress = (34.94 in. and 4.3/132)3 = 1. The Slenderness ratio for this column = Le/r = (. FS = (5/3) + (3/8) (101.000/1.88)[1 . = 87. we first calculate the factor of safety.000/1.000/1. Then the Allowable Stress = (34. We need a beam with a slightly lower area and/or radius of gyration. and C = 132] Since the slenderness ratio of the column is less than the critical slenderness ratio.3/132) . Finally the Allowable Load = Stress * Area = 15. So we make one more guess/estimate.htm (2 of 3) [5/8/2009 4:50:10 PM] .7*16' * 12"/ft)/ 1. we first calculate the factor of safety. and repeat steps 2.(1/2)(101.8/132)2] = 10. We also calculate the critical slenderness ratio Cc2 = ( 2 * 3. which has area of 9.845 lb/in2 * 9. and C = 132 Since the slenderness ratio of the column is less than the critical slenderness ratio.9.7*16' * 12"/ft)/ 1.2 in2 = 68.994 lb/in2 * 6. we use the formula for intermediate columns to find the allowable stress.165 lb.8.994 lb/in2 Finally the Allowable Load = Stress * Area = 10. From our table we will try a W10 x 33. but it is probably not the best (lightest).7*16' * 12"/ft)/ 1.Topic 6.32 in. Before we can determine the allowable stress. of course.94 in.8/132) .3/132)2] = 14. we first calculate the factor of safety.71 in2 = 153.130 lb/in2 file:///Z|/www/Statstr/statics/Columns/colse62c. From our table we will try a W12 x 31. This means we need an I-beam with a larger area and/or a larger radius of gyration. we use the formula for intermediate columns to find the allowable stress. htm (3 of 3) [5/8/2009 4:50:10 PM] .2d: Aluminum Columns . Here we see our maximum allowable load is just slightly above the load we would like to apply.2: Columns & Buckling . There are other design procedures for selecting columns than the one above. but we will end here.000 lb.13 in2 = 129.130 lb/in2 * 9. with computers and spreadsheets today. (120. This beam is a very good candidate for the best beam. It might be possible to find a slightly better beam.Example 3 Finally the Allowable Load = Stress * Area = 14. it is not a hard process to develop a program to try a number of different columns and arrive at the best one in a short time.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse62c.2c: Structural Steel Column Selection . Return to:Topic 6.II Continue to:Topic 6.000 lb. Most of these also utilize a trial and error type of procedure.Topic 6.).Example 4 or Select: Topic 6: Columns & Pressure Vessels . However. as shown in the diagram. and inner dimensions of 3" x 7". The rectangular cross section has outer dimension of 4" x 8". fixed-fixed.Example 4 Topic 6.2d: Aluminum Columns . however we need the minimum radius of gyration.Example 4 A hollow rectangular tube is to be used as a 10-foot long vertical. Determine the maximum axial load that may be applied to the column before the allowable stress for buckling will be exceeded. The tube material is the aluminum alloy 2014-Ty (Alclad).Topic 6. which may be calculated from r = (I/A)1/2. where I is taken about the axis which gives the smallest value.2d: Aluminum Columns . column. We first wish to determine the slenderness ratio.htm (1 of 2) [5/8/2009 4:50:11 PM] . file:///Z|/www/Statstr/statics/Columns/colse62d. 2: Columns & Buckling .Example 5 or Select: Topic 6: Columns & Pressure Vessels .4 Since the slenderness ratio is greater than 12 and less then 55. Then r = (I/A)1/2 = 1.870 lb/in2. we use the appropriate formula (for Alclad) for intermediate columns. Return to:Topic 6. resulting in the following value: I = (1/12)[bodo3 bidi3] = (1/12)[8"*4"3 7"*3"3] = 26. Additionally.23(Le/r)] ksi.9 in4. The Allowable Stress = [30.56in. = 21.htm (2 of 2) [5/8/2009 4:50:11 PM] . And slenderness ratio = Le/r = (.870 lb/in2 * 11 in2 = 240. = 21. the area = bodo bidi = 32 in2 21 in2 = 11 in2.Topic 6. = 38.7.87 ksi.2e: Wood Columns .5*10' * 12"/ft)/1.56 in..570 lb.Example 4 In this case that would be about a vertical y-y axis through the center of the cross section.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse62d. Finally the Allowable Load = Stress * Area = 21.II Continue to:Topic 6.2d: Aluminum Columns . The Allowable Stress = [.3E/(Le/d)2] = .5*25' * 12"/ft)/ 5 in. We next need to calculate the transition slenderness ratio.88 Since the column's slenderness ratio is greater than k. = 30.Example 5 Topic 6. The columns are rough cut and exactly 5" by 5".htm (1 of 2) [5/8/2009 4:50:11 PM] .Topic 6.671 sqrt(1. file:///Z|/www/Statstr/statics/Columns/colse62e.671 sqrt(E/allowable compressive stress along grain) = . How heavy can the observation platform be before buckling will occur? (Assume the wood columns are fixed on both ends. k = .3 x 106 lb/in2 /(30)2 = 433 lb/in2.3 x 106 lb/in2 / 6 x 103 lb/in2) = 9.2e: Wood Columns .2e: Wood Columns .) We first wish to determine the slenderness ratio we use slenderness ratio = Le/d = (.3*1.Example 5 Four twenty five foot long Douglas Fir columns are to support a observation tower. we use the appropriate formula (for wood) for long slender columns. 3: Columns & Buckling ./per column.2e: Wood Columns .Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colse62e. And since there are four columns.Topic 6.300 lb. Return to:Topic 6. = 44.825 lb. the platform weight could be 4 * 10.Problem Assignment or Select: Topic 6: Columns & Pressure Vessels .Example 5 Finally the Allowable Load = Stress * Area = 433 lb/in2 * 25 in2 = 10.825 lb.htm (2 of 2) [5/8/2009 4:50:11 PM] .II Continue to:Topic 6.2: Columns & Buckling . free column.Topic 6. 67100 lb. pinned-pinned. and the allowable yield stress is 35 x 10 3 lb/in2 .360 lb. c) fixed . Young's modulus is 30 x 10 6 lb/in2. [fixed-free: 34 lb. 1220 lb. A 16 foot long WT 6 x 46 steel T-beam is used as a fixed .810 lb. d) fixed .pinned columns and compare your answers. nominally 4" by 8" (actual 3. 127. Young's modulus is 1. the Euler Buckling load.Problem Assignment 1. (133. thickness = 3/8". 8200 lb. (277.3: Columns & Buckling .Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/cols63. [ a. 28. and the allowable yield stress is 36 x 10 3 lb/in2 . 32. what force will cause the yard stick to fail in Euler buckling. and the allowable yield stress is 6000 psi.5" by 3.).000 lb. b. Young's modulus is 30 x 10 6 lb/in2 .Problem Assignment Topic 6. A 16 foot long wood beam. b) pinned-fixed. d.free. If one end of the year stick is held fixed and a person pushes on the other end.780 psi)] 6. (45psi. length = 36". Determine the slenderness ratio.pinned column.) 3. (254. and the Euler Buckling load for this column. and the allowable yield stress is 6600 psi. and the Euler Buckling load for this column. Select the best (safe & lightest) structural steel wide flange I-beam (from web table) to be used as a 16 foot long. (W8x31) Select: Topic 6: Columns & Pressure Vessels . nominally 2" by 4" (actual 1.3: Columns & Buckling .950 lb (2790 psi). A 10 foot long wood beam. width = 2".htm [5/8/2009 4:50:11 PM] . and fixed . 4580 psi.fixed conditions for ends of the beam.) 4.)] 5. fixed-pinned: 274 lb.5").200 lb. is used as a fixed . 61. A wood yard stick has dimensions. A 40 foot long W8 x 40 structural steel I-beam is used as a column. Determine the slenderness ratio. and the axial stress when the Euler Buckling load is applied. (365 psi. (10. Assume both fixed-free. Determine the maximum safe load assuming: a) pinned-pinned. Determine the slenderness ratio. is used as a pinned-pinned column.) 2. and the allowable yield stress is 38 x 10 3 lb/in2.25"). Young's modulus is 30 x 10 6 lb/ in2 .8 x 10 6 lb/in2. c.. column to support an axial load of 120. (5680 psi).5" by 7. and the allowable yield stress of 6000 psi. (695 psi). Young's modulus is 2 x 10 6 lb/in2. Use a Young's modulus of 2 x 10 6 lb/in2. 000 psi.Topic 6. Determine the maximum allowable length permittted if a load of a) 50. (See tables below) [More than one beam may work well.365 146.660 lb. Young's modulus is 30 x 10 6 lb/in2. Determine a) the slenderness ratio.25 4.] file:///Z|/www/Statstr/statics/Columns/cols64.420 psi for the 20 ft beam with a allow steel stress of 36.free column. A W 8x40 aluminum 6061 -T6 alloy I-beam is used as a pinned-pinned column.000 lb is applied.4 in2] [a) 246. moment of inertia about y-y axis = 47. The yield stress of the steel is 36. Determine the slenderness ratio.in tf . b) 209.] 3.2 in4.000 lb is applied.5 3..04 W8x40 11.in 8.890 psi. and if a load of b) 100.077 0. [ a) 18. Young's modulus is 30 x 10 6 lb/in2.in4 Syy -in3 ryy . c) 4.di4)] / 64 [a) 111.0 35. A large steel pipe is used as a pinned-pinned column.4: Columns & Buckling .000 lb.80 8. The W12 x 36 gives a maximum load of 68. b) 60.4: Columns & Buckling .030 lb. the Euler Buckling load. The radius of gyration of hollow circular area: r = [sqrt (do2 + di2)] / 4 . Moment of Inertia: I = [pi * (do4 . b) 13.Topic Examination Topic 6.000 lb. with an allowable buckling stress of 6.558 0.in tw . [radius of gryration about y-y axis = 1.200 psi..53 49. area = 12.htm (1 of 3) [5/8/2009 4:50:11 PM] .10 2.00 12.in Ixx ..] 2. c) the axial stress in the column when the Euler load is applied. The pipe is 18 feet long with an outer diameter of 6 inches and an inner diameter of 5 inches. and the axial stress when the Euler Buckling load is applied. b) the Euler buckling load. See table below.in wf .in4 Sxx -in3 rxx. Select the best (safe and lightest) structural streel I-beam to be used as a pinned-pinned column of length 20 feet which is to carry a load of 60.95".65 ft. c) 24.Topic Examination 1.2 ft. A 20 foot long WT 12 x 42 steel T-beam is used as a fixed .in Iyy .000 psi.] A-in2 d . 75 14.0 r in 2.80 14.24 6.29 9.575 0.70 16.00 23.558 21.81 1.47 8.2 16.535 1360.10 224.38 2.48 3.00 14.305 10.695 0.3 272.618 0.0 707.776 0.in4 386.500 0.7 53.287 tw in Cross Section Info.20 16.025 4.0 I .72 7.0 151.60 16.0 S -in3 54.628 0.00 thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I in4 25.43 4.9 10.00 0.380 0.0 0.in4 26.80 13.10 88.12 6.513 0.30 53. Cross Section Info.0 797.799 12.456 0.0 80.in wf .260 0.19 10.4 31.420 0.0 221.30 17.416 657.61 13.565 A-in2 d .516 2690.97 1.00 50.in tf .10 70.91 5.31 7.450 40.0 108.77 4.86 I in4 8.3 46.1 25.77 S -in3 7.50 38.71 4.32 11.60 12.31 7.8 166.37 9.022 10.30 4.289 0.0 2360.660 1590.85 1.783 0.7 112.740 1.93 7.0 249.35 19.50 21.15 r .19 5.47 d in 5.455 1600.89 4.0 W 14x426 125.00 568.in tw .88 6.00 37.350 0.in 11.55 r .54 2.64 25.70 24.54 2.25 6.030 13.69 16.70 18.4: Columns & Buckling .Topic Examination - - - Flange Flange Web thick tf in wf in 5.59 2.260 0.40 5.4 30.072 0.320 0.0 0.0 216.20 14.05 6.0 14.872 0.86 26.00 file:///Z|/www/Statstr/statics/Columns/cols64.in 1.0 S in3 9.00 S in3 3.875 6610.265 0.in 5.12 6.62 21.31 4.00 283.60 108.42 17.82 1.073 17.5 W 6x16 W 6x25 W 8x67 W 10x29 W 10x45 W 12x22 W 12x36 W 14x38 W 14x74 W 14x136 W 16x50 W 16x96 W 18x60 W 21x73 W 24x94 A in2 5.68 6.53 2.933 0.htm (2 of 3) [5/8/2009 4:50:11 PM] .533 0.Topic 6.540 0.404 0.12 8.22 5.080 8.24 8.12 1.25 7.0 156.34 1.50 I .033 1.740 0.70 8.030 6.875 0.0 28.54 6.64 9.0 158.50 77.69 3.60 133.25 7.90 r in 1.424 0.00 10.00 18.16 2.98 0.061 0.7 60.8 49.313 21.20 10.33 4.295 27.0 986.063 0.695 3.30 2.59 2.20 4.26 6.68 1. Designation Area Depth Width W 5x18.0 281.28 0. 655 1.932 0.00 2.00 Select: Topic 6: Columns & Pressure Vessels .00 933.60 33.4: Columns & Buckling .20 159.680 0.60 118.60 27.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/cols64.945 20300.00 13.830 13600.90 15.28 10.0 300.00 31.18 3.Topic Examination W 27x114 W 33x240 W 36x300 33.83 88.400 0.570 4090.0 813.0 1110.72 16.50 15.0 70.0 1300.30 36.00 156.htm (3 of 3) [5/8/2009 4:50:11 PM] .070 0.64 3.865 1.Topic 6.0 11. ) The longitudinal stress may be found by equating the force due to internal gas/fluid pressure with the force due to the longitudinal stress as follows: P(A) = (A'). By thin wall pressure vessel we will mean a container whose wall thickness is less than 1/10 of the radius of the container. the stress in the wall may be considered uniform. then canceling terms and solving for the longitudinal stress. t = wall thickness To determine the relationship for the transverse stress.Thin Wall Pressure Vessels Thin wall pressure vessels are in fairly common use.htm (1 of 4) [5/8/2009 4:50:12 PM] . We first look at a cylindrical pressure vessel shown in Diagram 1. file:///Z|/www/Statstr/statics/Columns/cols65.1416 * R2) = (2 * 3. and spherical pressure vessels. where P = internal pressure in cylinder. but first cut the cylinder lengthwise as shown in Diagram 2. we have: = P R / 2 t . and the balancing forces due to the longitudinal stress which develops in the vessel walls. and have shown the forces due to the internal pressure.Topic 6. Under this condition. R = radius of cylinder. and which we will consider next. (There is also a transverse or circumferential stress which develops.5:Pressure Vessels . we use the same approach. Cylindrical pressure vessels.Pressure Vessels Topic 6. or P(3. where we have cut a cross section of the vessel. We would like to consider two specific types.1416 * R * t). often called the hoop stress.5 . htm (2 of 4) [5/8/2009 4:50:12 PM] . effective area the internal pressure acts on may be consider to be the flat cross section given by (2*R*L). So we may write: P(A) = have: (A'). or P(2*R*L) = (2*t*L). file:///Z|/www/Statstr/statics/Columns/cols65.Pressure Vessels We once again equate the force on the cylinder section wall due to the internal pressure with the . and is normally the limiting factor. and a wall thickness of the 1".5 . t = wall thickness We note that the hoop stress is twice the value of the longitudinal stress. The resistive force which develops in walls and may be expressed in terms of the hoop stress. Determine the longitudinal and hoop stresses in the cylindrical region. R = radius of cylinder. It's cylindrical section has a radius of 2 feet. we = P R / t . The internal pressure is 500 lb/in2. then canceling terms and solving for the hoop stress. Example A: A thin wall pressure vessel is shown in Diagram 3. the stress in the cylindrical section is given by the relationships above. In any thin wall pressure vessel in which the pressure is uniform and which has a cylindrical section.Topic 6. where P = internal pressure in cylinder. The vessel does not have to be a perfect cylinder. t = wall thickness Note that we have not called this a longitudinal or hoop stress. Using the approach as in cylindrical vessels. 000 lb/in2. or P(3. where P = internal pressure in sphere.1416 * R2) = (2 * 3.Table of Content file:///Z|/www/Statstr/statics/Columns/cols65.000 lb/in2. If we once again equal the force due to the internal pressure with the resistive force expressed in term of the stress.Topic 6. Select: Topic 6: Columns & Pressure Vessels .5 . determine the stress in the spherical region. above. Solution: We apply our spherical relationship: = P R / 2 t = 500 lb/in2 * 24"/2 * 1" = 6000 lb/in2. Thus in the container in Example A. = P R / 2 t = 500 lb/in2 * 24"/2 * 1" = 6000 lb/in2. In Example A. We do not do so since the symmetry of the sphere means that the stress in equal in what we could consider a longitudinal and/or transverse direction. the limiting (maximum) stress occurs in the in the cylindrical region and has a value of 12. we have: P(A) = (A'). then canceling terms and solving for the stress. if the radius of the spherical section of the container is also 2 feet.htm (3 of 4) [5/8/2009 4:50:12 PM] . in Diagram 4 we have shown a half section of a spherical vessel. R = radius of sphere. = P R / t = 500 lb/in2 * 24"/ 1" = 12. we have: = P R / 2 t .1416 * R * t). Next we consider the stress in thin wall spherical pressure vessels.Pressure Vessels Solution:We apply the relationships developed for stress in cylindrical vessels. Example B. 5 .Pressure Vessels Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/cols65.Topic 6.htm (4 of 4) [5/8/2009 4:50:12 PM] . Problem Assignment 1 1. Calculate the circumferential stress developed in the walls of the pipe.Topic 66a . A welded water pipe has a diameter of 8 ft. (9300 psi. After fabrication. The boiler is subjected to an internal gage pressure of 155 psi.700 psi) 2. thick. Calculate the tensile stresses developed (circumferential and longitudinal) in the walls of a cylindrical boiler 5 ft. 4650 psi) file:///Z|/www/Statstr/statics/Columns/colsp66a.Pressure Vessel problems Statics & Strength of Materials Topic 6. in diameter with a wall thickness of 1/2 in. (14. and a wall of steel plate 3/4 in.htm (1 of 5) [5/8/2009 4:50:12 PM] . this pipe was tested under an internal pressure of 230 psi.6a :Pressure Vessels . htm (2 of 5) [5/8/2009 4:50:12 PM] . for the pipe.500 psi. (1. Calculate the internal water pressure that will burst a 15 in. Calculate the wall thickness required for a 5 ft.500 psi.Pressure Vessel problems 3. (8333 psi) 4. Use an ultimate tensile strength of 62. diameter cylindrical steel tank containing gas at an internal gage pressure of 600 psi.Topic 66a . diameter cast-iron water pipe if the wall thickness is 1/2 in. The allowable tensile stress for the steel is 17.03") file:///Z|/www/Statstr/statics/Columns/colsp66a. Topic 66a .Pressure Vessel problems 5.500 psi. The inside diameter is 15 in.000 lb.htm (3 of 5) [5/8/2009 4:50:12 PM] . in diameter. and a simultaneous external axial tensile load of 45.570 psi) file:///Z|/www/Statstr/statics/Columns/colsp66a.000 psi. The allowable tensile stress is 20. A spherical gas container.3") 6. Calculate the tensile stresses (circumferential and longitudinal) developed in the walls of a cylindrical pressure vessel. 50 ft. (13. The wall thickness is 1/4 in. The vessel is subjected to an internal gage pressure of 450 psi. Calculate the thickness of the steel wall required. 10. is to hold gas at a pressure of 40 psi. (. Determine the axial and hoop stress in the cylindrical region. The larger spherical region has a radius of 3 ft and a wall thickness of 1. 9600 psi. 2: 19.Topic 66a .htm (4 of 5) [5/8/2009 4:50:12 PM] .5 ft and a wall thickness of 3/4". 3: 8.000 psi) file:///Z|/www/Statstr/statics/Columns/colsp66a.760 psi. The smaller spherical region has a radius of 2. Which stress is the largest? (1: 5. A thin wall pressure vessel is composed of two spherical regions and a cylindrical region is shown in Diagram 5.Pressure Vessel problems 7.25". and the wall stresses in the two spherical regions.200 psi. The cylindrical region has a radius of 2 ft and a wall thickness of 1/2". A thin wall pressure vessel is composed of two spherical regions and is shown in Diagram 4.htm (5 of 5) [5/8/2009 4:50:12 PM] . The larger spherical region has a radius of 3 ft . .45". Determine wall thickness for each spherical region.Topic 66a .Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colsp66a. and the smaller spherical region has a radius of 2.(. If we wish the maximum stress in both spherical sections to at the allowable stress.5 ft.Pressure Vessel problems 8.375") Select: Topic 6: Columns & Pressure Vessels .000 lb/in2. The vessel is made of steel with an allowable tensile stress of 24. Determine the maximum pressure that the pipes can withstand. (667 psi) 2. Copper water pipe in a house is 1.Problem Assignment 2 1.htm (1 of 5) [5/8/2009 4:50:13 PM] .Topic 66b .) file:///Z|/www/Statstr/statics/Columns/colsp66b.5 inches in diameter with 1/16 inch walls. Determine the stress in the plastic when the hose is filled with water at 50 psi. (200 psi.6b :Pressure Vessels . The allowable tensile stress in the copper is 8000 psi.Pressure Vessel Problems Statics & Strength of Materials Topic 6. A plastic garden hose has an inside diameter of 3/4 inch and a wall thickness of 3/16 inch. 49 mi/hr) 4.000 psi. (3. The steam is transferred in an 18 inch SA-202 chrome-magnesium silicon steel pipe having an allowable tensile stress of 15. Determine the speed of the water in the hose in problem 2 if it is used to fill a five gallon pail in one minute.Topic 66b .)] (If t = 1".8 psi) file:///Z|/www/Statstr/statics/Columns/colsp66b. Process steam is generated at a central plant and distributed to various buildings in a city. (3/4 in.) 5.000 psi. P = 45. The steam is at a temperature of 300 Celsius which has a pressure of 1246 psi.8 lb/in3 * (t in. Determine the minimum thickness of the pipe where wall thickness is given in sixteenths of an inch.htm (2 of 5) [5/8/2009 4:50:13 PM] . (one gallon is 231 cubic inches) Express the speed in feet per second and convert to miles per hour if you can.65 ft/sec = 2.Pressure Vessel Problems 3. Determine the maximum pressure of gas that is stored in a spherical container 80 feet in diameter made of SA-515-55 carbon-silicon steel having an allowable tensile stress of 11.[P = 45. and that the allowable shear stress in the rivets is 12.05 x 106 lb.000 psi.000 psi.000 psi.htm (3 of 5) [5/8/2009 4:50:13 PM] .4 pounds per cubic foot) and the pressure in a liquid is equal to the weight density of the liquid times the depth of the liquid. F = Stress * A = 4. # rivet/vertical seam = 764 x 2 =1528] file:///Z|/www/Statstr/statics/Columns/colsp66b. (1/4 in.Topic 66b . The silo is 16 feet in diameter. Determine the thickness necessary in a steel silo (Harvester or similar) that is 80 feet tall. Give the thickness to the closest sixteenth of an inch. It is made of 1/2 inch steel plate with single cover plate riveted joints. A water tank is 60 feet in diameter and 45 feet high.Pressure Vessel Problems 6. [Hoop stress in steel ~ 15. The maximum allowable pressure in the steel is 15. Silage has the same density as water (62.) 7. Determine the number of 3/4 inch diameter rivets needed to secure the joint. Assume that the allowable tensile strength of the steel is 15. f/rivet = 5300 lb/rivet.000 psi (depends a little on assumptions). Fend = 12. The tank is made of welded 1/8 inch steel.Topic 66b . The allowable tensile stress in the steel file:///Z|/www/Statstr/statics/Columns/colsp66b.htm (4 of 5) [5/8/2009 4:50:13 PM] . A propane tank used for home heating has a diameter of 4 feet. Determine the force acting on the tank ends that the weld must withstand. Determine the tensile stress in the tank.Pressure Vessel Problems 8. An air compressor used for a nail gun has a diameter of 8 inches and is charged to a pressure of 250 psi. [ hoop = 8000 psi. The ends of the tank are welded to the cylindrical body. 570 lb] 9. [ t 1atm = . 80 F 20 atm. Propane has the following vapor pressures. 34.137 F 40 atm.htm (5 of 5) [5/8/2009 4:50:13 PM] .5 F 10 atm.0294 in. -40 F 2 atm. Determine the required thickness of the tank.Pressure Vessel Problems of the tank is 12.176 in] P T 1 atm. -14 F 5 atm.Table of Content Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/Columns/colsp66b.000 psi.Topic 66b . t 40atm = 1.203 F Select: Topic 6: Columns & Pressure Vessels . 8596 lb. Determine the maximum pressure that the pipes can withstand. Young's modulus for the wood is 2 x 10 6 lb/in2.in4 0. 18 ft long.30 S -in3 5. and the allowable axial stress for the steel is 45.289 158. Copper water pipe in a house is 1.799 0.3.38 A-in2 d . A 16 foot long wood 2" by 6" is used as a fixed-fixed column.in I . c) the axial stress in the column when the Euler load is applied.in tf . 103. (long column. file:///Z|/www/Statstr/statics/Columns/cols64a.30 I .in4 16. A W10x29 steel I-Beam. Cross Section Info. Designation Area Depth Width W 10x29 thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis tw .8 r .htm (1 of 3) [5/8/2009 4:50:13 PM] . is to be used as a column with pinned-pinned ends.Test page Columns & Buckling . 716 psi) 2. Johnson's formula: C (critical slenderness ratio) : 3.54 10.in 8. The allowable tensile stress in the copper is 8000 psi. b) the Euler buckling load.Topic Examination 1. Determine: a) the slenderness ratio.440 lb) - - - Flange Flange Web thick Cross Section Info. [Hint: Determine the slenderness ratio. Then use the Euler formula if it is a long column or the J.in 1.61 r . Determine the load which will cause the beam to buckle.in 4. (166.B.5 inches in diameter with 1/16 inch walls.in wf . Young's modulus for the steel is 30 x 10 6 lb/in2.22 5.500 J.0 S -in3 30.B. Johnson formula (see below) is it is an intermediate column.000 psi.] The beam characteristics are listed below. 86 W 18x114 33. 504.in Ixx . The yield stress of the steel is 42. (See tables below) (W 14 x 136) - - - Flange Flange Web thick tf wf tw Cross Section Info.00 46. A W 18 x 114 STRUCTURAL STEEL I-beam.in Iyy .in4 Sxx -in3 rxx.833 0.0 7.in tf . (See structural steel relationships after next problem. 24 feet long.000 psi. The yield stress of the steel is 32. Determine the maximum load it can carry before buckling will occur.in 2040.in tw .0 220.) (15.in wf . Young's modulus for the steel is 30 x 10 6 lb/in2. Cross Section Info. Select the best (safe and lightest) structural steel I-beam to be used as a fixed-free column of length 16 feet which is to carry a load of 300.000 lb. Designation Area Depth Width A d thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis I S r I S r file:///Z|/www/Statstr/statics/Columns/cols64a.000 psi.500 lb) Beam A-in2 d .79 274. is used as a fixed-pinned column.30 2.Test page 4.060 psi.in4 Syy -in3 ryy .48 11.991 0.595 5.htm (2 of 3) [5/8/2009 4:50:13 PM] .50 18. 0 in3 9.450 40.70 16.1 25.695 3.655 1.2 16.47 in 5.00 14.30 4.31 7.061 0.400 0.00 23.64 25.60 12.72 16.54 6.88 6.18 3.in tw .59 2.022 10.30 17.865 1.10 70.8 49.783 0.5 W 6x16 W 6x25 W 8x67 W 10x29 W 10x45 W 12x22 W 12x36 W 14x38 W 14x74 W 14x136 W 16x50 W 16x96 W 18x60 W 21x73 W 24x94 W 27x114 W 33x240 W 36x300 in2 5.3 46.50 77.456 0.7 53.20 14.565 A-in2 d .404 0.25 7.80 13.64 9.68 1.0 707.533 0.072 0.0 221.32 11.3 272.575 0.54 2.00 in 1.50 21.0 0.0 28.628 0.71 4.0 S -in3 54.61 13.540 0.289 0.15 r .24 8.12 6.063 0.htm (3 of 3) [5/8/2009 4:50:13 PM] .416 657.080 8.8 166.030 13.945 20300.53 2.93 7. 2005 file:///Z|/www/Statstr/statics/Columns/cols64a.260 0.680 0.in tf .54 2.7 112.830 13600.0 151.77 4.933 0.25 6.70 18.0 158.287 in in in4 25.24 6.420 0.12 1.0 1110.43 4.in 5.20 4.25 7.59 2.695 0.in 1.455 1600.00 Last revised: April 28.380 0.320 0.0 156.00 0.073 17.872 0.618 0.26 6.033 1.295 27.10 88.0 813.260 0.0 300.30 53.42 17.Test page W 5x18.64 3.80 14.33 4.932 0.31 4.81 1.60 27.4 31.60 33.00 159.558 21.37 9.0 1300.070 0.20 10.875 0.0 281.0 I .00 156.82 1.55 r .60 133.875 6610.90 15.77 S -in3 7.89 4.60 108.16 2.00 in3 3.48 3.47 8.0 70.69 3.799 12.0 2360.0 249.00 283.90 31.12 8.0 88.0 80.40 5.313 21.00 in 5.516 2690.9 10.28 10.00 568.97 1.00 37.0 0.35 19.350 0.34 1.513 0.740 0.86 11.535 1360.0 14.7 60.62 21.00 33.86 26.025 4.740 1.60 16.030 6.31 7.19 5.10 224.70 8.570 4090.00 10.424 0.22 5.500 0.776 0.265 0.0 108.4 30.69 16.05 6.0 in 2.30 2.70 24.83 0.00 50.660 1590.50 15.20 in4 8.0 W 14x426 125.50 38.in wf .20 16.72 7.75 14.0 216.12 6.30 36.00 18.in 11.85 1.68 6.50 I .0 797.in4 26.19 10.98 2.60 118.28 0.0 986.38 2.91 5.305 10.00 13.29 9.00 933.in4 386. Topic 7.htm [5/8/2009 4:50:13 PM] .0: Friction Topic 7: Friction (This topic may be omitted due to time considerations) Friction Problems 1 Friction Problems 2 Friction Problems 3 Friction Problems 4 Friction Problems 5 Friction Problems 6 file:///Z|/www/Statstr/statics/Friction/fric70. htm (1 of 3) [5/8/2009 4:50:14 PM] .Friction probs 1 Statics & Strength of Materials Problems Sketch the structure and draw a free body diagram for the relevant member or the entire structure. 1. A 160 pound painter stands 4 feet from the top of the ladder. Determine the normal force acting on the ladder by the building and the frictional force acting on the ladder at the point it makes contact with the ground.51.62 and the coefficient of sliding friction for wood on concrete is 0. 2. The ladder rests against the side of a building with the base of the ladder 6 feet from the building. file:///Z|/www/Statstr/statics/Friction/fricp1. Write the appropriate moment or force equations and solve them for the unknown forces. The coefficient of static friction for wood on concrete is 0. A wooden crate weighing 250 pounds rests on a concrete floor. A 16 foot uniform ladder has a weight of 40 pounds. Friction probs 1 a.) b. A Bobcat is parked on a 15 degree incline. c. Determine the force required to start the crate sliding on the floor. (The force is applied parallel to the floor.htm (2 of 3) [5/8/2009 4:50:14 PM] . file:///Z|/www/Statstr/statics/Friction/fricp1. How long can you continue to apply the force indicated in part a? Hint: for the constant force found in part a. Determine the force required to keep the crate sliding once it has started. Determine the frictional force acting on the Bobcat if it weighs 1500 pounds. determine the acceleration of the crate 3. file:///Z|/www/Statstr/statics/Friction/fricp1. Determine the minimum angle required for a steel chute so that sand will slide down the chute.htm (3 of 3) [5/8/2009 4:50:14 PM] . The coefficient of static friction for sand on steel is 0.25.Friction probs 1 4. Friction probs 2 Statics & Strength of Materials Problems Sketch the structure and draw a free body diagram for the relevant member or the entire structure.htm (1 of 3) [5/8/2009 4:50:14 PM] . Write the appropriate moment or force equations and solve them for the unknown forces. The coefficient of sliding friction for the crate static on a concrete floor is 0. Find the force required to start the crate sliding across the floor. file:///Z|/www/Statstr/statics/Friction/fricp2. 1.35. A uniform wood crate (center of gravity is at the center of the crate) is 3 feet wide and 6 feet high and weighs 500 pounds. 45. Find the space required for a sand pile containing 50 yards of sand.Friction probs 2 2. Determine the angle of a sand pile. The coefficient of friction for sand sliding on sand is 0.) file:///Z|/www/Statstr/statics/Friction/fricp2. Will the crate slide on the floor or tip over if the force found above is applied at a point 4 feet above the floor? 3. (The pile will be conical and the volume of a cone is 1/3pr2h.htm (2 of 3) [5/8/2009 4:50:14 PM] . file:///Z|/www/Statstr/statics/Friction/fricp2.htm (3 of 3) [5/8/2009 4:50:14 PM] . Pav = .4 pounds per cubic foot.Friction probs 2 4. Assume that dry earth has a weight density of 200 pounds per cubic foot. Find the force exerted on the dam due to the pressure of the water.4.5dwy The density of water is 62. The lake behind the dam comes to within 3 feet of the top of the dam. Is the force exerted on the dam sufficient to make the dam slide down the valley? The coefficient of friction for dry earth sliding on dry earth is 0. A dam is 75 feet wide and 25 feet high with a base of 150 feet. F = Pavyw where y is the depth of the water on the dam and Pav is the average pressure due to the water. A solid steel shaft is fabricated as shown. A cylindrical cast-iron casting has an axial hole extending partway through the casting.htm (1 of 2) [5/8/2009 4:50:14 PM] . Locate the center of gravity of the casting.Friction probs 3 Statics & Strength of Materials Problems 1. Locate the center of gravity with respect to the left end. as shown. file:///Z|/www/Statstr/statics/Friction/fricp3. 2. Awood mallet has a cylindrical head 5 in. Locate the X-X centroidal axis.5 in. plate welded to its top flange. 6. long and 2. by 1/2 in. Locate the centroid of the area. horizontal plate to a 10 in. file:///Z|/www/Statstr/statics/Friction/fricp3. 5. If the cylindrical handle is 1.5 in. in diameter.) 4. A lintel in the form of an inverted tee is made by welding a 12 in. by 1 in. as shown. in diameter.Friction probs 3 3. Locate the X-X and Y-Y centroidal axis for the areas shown.htm (2 of 2) [5/8/2009 4:50:14 PM] . by 1 in. vertical plate as shown. from where the handle enters the mallet head? (Assume the unit weight of the wood to be 40 lb. how long must it be for the mallet to balance at a point 6 in. A built-up steel member is composed of a W20x50 wide flange section with a 10 in./ft3. as shown. Abuilt-up steel member is composed of a W20x60 wide flange section with a C10x22 American Standard channel welded to its top flange.Friction probs 4 Statics & Strength of Materials Problems 1. For the area shown. the X-X centroidal axis is required to be located as shown. Locate the X-X centroidal axis. file:///Z|/www/Statstr/statics/Friction/fricp4. 2.htm (1 of 3) [5/8/2009 4:50:14 PM] . Calculate the required dimension h. Locate the X-X and Y-Y centroidal axis for the built-up structural steel area shown below.Friction probs 4 3.htm (2 of 3) [5/8/2009 4:50:14 PM] . file:///Z|/www/Statstr/statics/Friction/fricp4. Friction probs 4 file:///Z|/www/Statstr/statics/Friction/fricp4.htm (3 of 3) [5/8/2009 4:50:14 PM] . htm (1 of 4) [5/8/2009 4:50:14 PM] . Calculate the moment of inertia with respect to the X-X centroidal axis in the diagram shown. Calculate the moment of inertia of the triangular area in the diagram shown with respect to the X-X centroidal axis.Friction probs 5 Statics & Strength of Materials Problems 1. 2. file:///Z|/www/Statstr/statics/Friction/fricp5. and a height of 14 in. Using the approximate method.Friction probs 5 3. A rectangle has a base of 7 in. determine the moment of inertia with respect to the base.htm (2 of 4) [5/8/2009 4:50:14 PM] . file:///Z|/www/Statstr/statics/Friction/fricp5. 4. Calculate the moment of inertia of the built-up member with respect to the X-X centroidal axis. A structural steel wide-flange section is reinforced with two steel plates attached to the web of the member as shown. Dividethe area into four equal horizontal strips. file:///Z|/www/Statstr/statics/Friction/fricp5.Friction probs 5 a. b.htm (3 of 4) [5/8/2009 4:50:14 PM] . Divide the area into six equal horizontal strips. Friction probs 5 file:///Z|/www/Statstr/statics/Friction/fricp5.htm (4 of 4) [5/8/2009 4:50:14 PM] . Friction probs 6 Statics & Strength of Materials Problems 1.) file:///Z|/www/Statstr/statics/Friction/fricp6. (Use dressed sizes of lumber. (There are at least three ways you can make this beam.) Sketch the beam and calculate the moment of inertia of the box beam about its X-X centroidal axis.htm (1 of 4) [5/8/2009 4:50:15 PM] . A box beam is made from four 2x8s. Friction probs 6 2. file:///Z|/www/Statstr/statics/Friction/fricp6. A wooden I beam is made by nailing 2x6s on the edges of a 2x10.htm (2 of 4) [5/8/2009 4:50:15 PM] . Calculate the moment of inertia of this I-beam. file:///Z|/www/Statstr/statics/Friction/fricp6. Calculate the moment of inertia of this I-beam. 4. A steel I beam is made of a steel plate 20 inches high and 1 inch thick with 4 L6x6 angles attached. Find the moment of inertia made by 2 2x4s attached to the edges of a 3/4 inch plywood web 10 inches tall.Friction probs 6 3.htm (3 of 4) [5/8/2009 4:50:15 PM] . htm (4 of 4) [5/8/2009 4:50:15 PM] . a.Friction probs 6 Calculate the radius of gyration of the following timbers about their x axes. 4x4 c. 6x8 Calculate the radius of gyration of the beams in problems 1-4 about their x-axes. 2x10 b. file:///Z|/www/Statstr/statics/Friction/fricp6. 6 Special Topics I: Problem Assignment 1 (required) 8.1: Combined Stress Topic 8 . We will look at several examples of relatively simple combined stress problems. such as the axial stress due to an axial load and a bending stress.1 Special Topics I: Combined Stress r 8.at least in the case of similar stresses acting along the same line. Many situations involve more than one type of stress occurring simultaneously in a structure. One of the principles we will apply in these examples is the principle of superposition.Example 1 r 8. rivet. These problems can become relatively complicated.5 Special Topics I: Mohr's Circle 8.Example 2 8. We look at this type of problem in the example below. We would like to determine the maximum axial stress acting in the beam cross section.Example 1 8.8 Special Topics I: Topic Examination ● ● ● ● ● ● ● Topic 8.3 Special Topics I: Non-Axial Loads 8.7 Special Topics I: Problem Assignment 2 (required) 8.Combined Stress Up to this point we have considered only or mainly one type of applied stress acting on a structure.4a Principal Stresses . Example. beam. or weld.1a Combined Stress .4 Special Topics I: Principal Stresses r 8. file:///Z|/www/Statstr/statics/spec1/spec81.Special Topics I ● 8. and has a load of 10. that is. shaft. acting (at the centroid of the beam) at angle of 20o below the horizontal. the resultant stress will be the algebraic sum of the individual stresses .Topic 8.2 Special Topics I: Stresses on Inclined Planes 8. A four foot long cantilever beam (shown in Diagram 1) is attached to the wall at point A.htm (1 of 3) [5/8/2009 4:50:15 PM] .1b Combined Stress . member of a structure.1: Special Topics I . 000 lb. 67 in4) = 30. The maximum bending stress occurs at the outer edge of the beam.700 ft-lb.820 lb/in2.995 lb.400 lb. and so add to a total tensile stress of 30. and is the same everywhere in the beam. We note that this stress will be tensile and constant through out the length of the beam.400 lb. We can now combine (sum) the axial stress at the very top and bottom of the beam to determine the maximum axial stress./in2 (compression). the bending stress is compressive and the normal axial stress in tensile so the resultant bottom stress is . At the bottom of the beam. / (2" x 4") = 1175 lb/in2.1: Combined Stress Solution: We first apply static equilibrium conditions to the beam and determine the external support reactions. We see in the beam section in Diagram 2. and the bottom on the beam is in compression.30. Notice we have resolved the 10. file:///Z|/www/Statstr/statics/spec1/spec81.) We can then calculate the maximum bending moment by: Maximum Bending Stress = M y/I = (164. load into its perpendicular x and y components. The horizontal component of the load (9. and the external moment acting on the beam at point A.400 in-lb./in2 + 1.700 ft-lb) to act on the beam (balanced by the external moment).000 lb.) causes a torque about point A (13. Since the bending moment was negative.) produces a normal horizontal axial stress in the beam. that the stresses at the top of the beam are both tensile./in2 = -29. this means that the top of the beam (above the centroid) is in tension.420 lb.htm (2 of 3) [5/8/2009 4:50:15 PM] . and at the point in the beam where the bending moment is a maximum.Topic 8./in2 = 31.)(2")/(10. the maximum bending moment occurs at the wall and is equal to the (negative of) external bending moment.175 lb.400 in-lb./in2. The total axial stress at a point in the beam will be the sum of the normal axial stress and the axial bending stress.820 lb. (M = -13.645 lb.175 lb. So the maximum normal Axial Stress is 1175 lb/in2.820 lb./in2 + 1. = -164. The Normal Axial Stress = Force/Area = 9. The vertical component of the load (-3. The resulting internal bending moment(s) in the beam produces an axial bending stress. In the cantilever beam. Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spec81.Topic 8.Example 1 Topic 8.htm (3 of 3) [5/8/2009 4:50:15 PM] . Select : Topic 8.1b: Combined Stress .1: Combined Stress We will now look at several additional examples of combined stresses.1a: Combined Stress .Example 2 or select: Topic 8: Special Topics I . Example 1 Topic 8. Ax = 21.1a: Combined Stress .Example 1 A loaded beam (shown in Diagram 1) is pinned to the wall at point A.8 ft.. and use the Bending Moment Diagram to determine the Maximum Bending Stress in the beam. Sum of Forcex = Ax . and Ay = -3990 lb. = 0 Sum of TorqueA = -6. with the characteristics shown in Diagram 1. We would like to determine the maximum axial stress acting in the beam cross section.htm (1 of 3) [5/8/2009 4:50:15 PM] .000 lb. = distance from A to D) Solving: T = 23.8 ft. acting downward at point C. (See Diagram 2. The supporting rod makes an angle of 25o with respect to the beam.Topic 8. The beam cross section is a W8 x 24 I-Beam.1a . and is supported by a rod DB.6.000 lb.) = 0 (where 2.430 lb.000 lb (10ft) + T cos 25o (2. (Ay acts downward) We next draw the Shear Force and Bending Moment Diagrams.T cos 25o = 0 Sum of Forcey = Ay + T sin 25o .Combined Stress .640 lb..) file:///Z|/www/Statstr/statics/spec1/spece81a. attached to the wall at point D and to the beam at point B. Solution: We first apply static equilibrium to the beam and determine the external support reactions acting on the beam at point A. The beam has a load of 6. Since the bending moment was negative.780 lb/in2. there is no horizontal axial force due to an external horizontal force. Then: Maximum Bending Stress = (288.000 in-lb.) For beam section BC. In section AB the beam is in compression with horizontal axial force of 21. and so add to a total compressive stress of 13. (We have considered the force to act along the centroid of the beam./ in2 + 3.)/20.) The compressive horizontal axial stress in section AB is given simply by: F/A = 21.1a . (However.htm (2 of 3) [5/8/2009 4:50:15 PM] . This will be considered in a moment.430 lb.08 in2 = 3. and the bottom of the beam will be in compression. = -288.) There is a bending stress also acting in the beam. From our bending moment diagram. We see in the beam section (at 6 ft from left end) in Diagram 3. the bending stress is tensile and the normal file:///Z|/www/Statstr/statics/spec1/spece81a. (Due to the force Ax and the horizontal component of the force in rod DB. In this example S = 20.000 ft-lb. since it is to the right of where the support rod is attached. that the stresses at the bottom of the beam are both compressive.000 in-lb.Combined Stress .030 lb. / 7. the top of the I-Beam will be in tension. (See Diagram 3) We can now combine (sum) the axial stresses at the very top and bottom of the beam to determine the maximum axial stress./in2. The maximum bending stress occurs at the outer edge of the beam. and at the point in the beam where the bending moment is a maximum.780 lb. we see that the maximum bending moment occurs at 6 feet from the left end.9 in3 = 13.030 lb/in2.) We can then calculate the maximum bending moment by: Maximum Bending Stress = M / S Where S is the section modulus for the beam. That is.Example 1 We next will consider the axial stress due to the horizontal force acting on the beam.430 lb. and has a value of -24. ( The negative sign indicating that the top of the beam is in tension and the bottom of the beam is in compression. At the top of the beam./in2 = 16. but section BC is not experiencing normal horizontal stress. which is in turn due to the vertical loads being applied. section AB is in compression. The total axial stress at a point in the beam will be the sum of the normal axial stress and the axial bending stress. there is a horizontal bending stress due to the bending moment.9 in3. 810 lb.Topic 8. Return to: Topic 8.1a .Combined Stress .030 lb.780 lb.Example 1 axial stress in compressive so the resultant bottom stress is: +13.Topic 8./in2 = 10.3.htm (3 of 3) [5/8/2009 4:50:15 PM] ./ in2 (tension)./in2 .1: Combined Stress or select: Topic 8: Special Topics I .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spece81a.750 lb. a torque and perpendicular force develops in the supporting wall. The two stresses may be summed to find the maximum shear stress in the shaft. a solid 1 foot long shaft is attached to a wall at point A. In this example.1416 * 2"4/32) = 1. (Notice also that an external moment develops at the wall. since the shaft is also acting as a cantilevered beam.Example 2 A second example of a structure where stress may be added using the principle of superposition is shown in Diagram 1.1b: Combined Stress . Maximum Transverse Shear Stress = Tr/J = (2000 in-lb.Combined Stress .)*(1") / (3. (The student is referred to a complete Strength of file:///Z|/www/Statstr/statics/spec1/spece81b. so the result will simply be stated here. The 1000 lb.Example 2 Topic 8.) The internal torque in the shaft produces transverse shearing stress. such that the shaft is in translational and rotational equilibrium. force acting on the outside edge of the disk produces a torque of 2000 in-lb. force perpendicular to the shaft. but we will not consider this at this point.Topic 8. and has a disk attached at end B. An axial bending stress will also develop in the beam. In response to this.270 lb/in2. as shown. is applied to the outer edge of the disk. with respect to the center (centroid) of the shaft.1b . Maximum Horizontal Shear Stress = Vay'/Ib = (4/3) V/A This last is the expression for the maximum horizontal (and vertical) shear stress in a circular cross section beam. The derivation for this is a bit messy. We would like to determine the maximum shear stress in the shaft. Additionally it also effectively exerts a 1000 lb. since we are concerned with the total shear stress in the beam.) * (1") / (pi * d4/32) = (2000 in-lb. We first consider the static equilibrium conditions.htm (1 of 3) [5/8/2009 4:50:16 PM] . The perpendicular forces produce an internal shearing force in the shaft. as shown in the box in Diagram 1. It can be considered to produce a torque and a force acting at the centroid. A force of 1000 lb. which also produces a shear stress. and cross sectional views showing the directions of the two shear stresses. We can now sum the two shear stresses to determine the maximum shear stress in the shaft.htm (2 of 3) [5/8/2009 4:50:16 PM] . In Diagram 2 and Diagram 3.694 lb/in2.1b .270 lb/in2 + 424 lb/in2 = 1.1: Combined Stress Continue to: Topic 8.Example 2 Materials Text for the derivation. As we can see from the diagram. at the top the cross sections the two stresses add giving the result of 1.1416 * 2"2/4) = 424 lb/in2. This is the maximum shear stress in the shaft.Stresses on Inclined Planes or select: file:///Z|/www/Statstr/statics/spec1/spece81b.Topic 8.Combined Stress .) Maximum Horizontal Shear Stress = (4/3)V/A = (4/3) (1000 lb.)/3. Return to: Topic 8.2: Special Topics I . we have shown a section of the shaft. htm (3 of 3) [5/8/2009 4:50:16 PM] .1b .Combined Stress .Topic 8.Example 2 Topic 8: Special Topics I .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spece81b. and where these relationships allow us to calculate the axial and shear stress on an incline plane section at an angle theta.htm (1 of 3) [5/8/2009 4:50:16 PM] . However if we cut the section at an angle theta as shown in Diagram 1b.2: Special Topics I . or A' = A/cos (theta).) The area of incline plane can be written as the cross sectional area of the rectangular section divided by the cosine of angle theta. using a trigonometric identity. file:///Z|/www/Statstr/statics/spec1/spec82.2 . To determine the stresses on the inclined plane area we break the force into components perpendicular and parallel to the incline plane. We can then write the axial and shear stress on the inclined area as follows: Axial Stress = F cos (theta) / A /cos (theta) = (F/A) cos2(theta) Shear Stress = F sin (theta) / A /cos (theta) = (F/A) sin(theta) * cos(theta) or. The normal stress on the end on the rectangular section is simply given by Normal Stress = F/A. (Again as shown in Diagram 1b. we have: .Topic 8.Incline Plane Stress Topic 8. the force F is no longer perpendicular to the inclined plane area.Stresses on Incline Planes In Diagram 1a we have shown a rectangular section in simple tension with an axial force F apply as shown to each end. we can rewrite the shear stress as Shear Stress = (F/2A)Sin (2*theta). and in terms of the normal stress on the rectangular area. and finally writing symbolically. which is half of the maximum axial stress. as shown in Diagram 2). We now look at a simple example of this application. As shown in Diagram 2. The maximum shear stress occurs at theta = 45o. Up to this point we have considered a section with an axial stress only in one direction.Non-Axial Loads or select: file:///Z|/www/Statstr/statics/spec1/spec82. we have shown a square. we first calculate the normal stress on the square cross section. acting on each end.2 . We will wish to determine what are called the principal stresses and the principal planes.Topic 8. 2" by 2". and find that the normal axial stress on the 30o incline plane is 375 lb/in2. Continue to: Topic 8. and the shear stress on the 30o incline plane is 217 lb/in2.htm (2 of 3) [5/8/2009 4:50:16 PM] . and is equal to F/(2A). and find Normal Stress (sigma) = 500 lb/in2. section in tension with a normal force of 2000 lb. Example In Diagram 2.3: Special Topics I . The maximum axial stress is just the initial normal stress on the rectangular cross section = F/A. We would like to know the axial and shear stress on a 30o incline plane cut through the section. We will shortly look at the general case where we have several axial and shear stresses acting.Incline Plane Stress From our relationships we can determine the maximum stresses. We next apply the our formula for axial and shear stress on an incline plane (again. htm (3 of 3) [5/8/2009 4:50:16 PM] .Topic 8.Incline Plane Stress Topic 8: Special Topics I .2 .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spec82. F. file:///Z|/www/Statstr/statics/spec1/spec83. we have replaced the 5000 lb. it is not acting along the axis of the element.Topic 8. The load force F is non-axial. we have shown a structural element in tension with a load F acting at the edge of the element. we have shown a 2" by 2" structural element with a tensile force of 5000 lb.3 .Non-Axial Loads We next look at the case where there is a non-axial loading on a structure or structural element.htm (1 of 3) [5/8/2009 4:50:16 PM] . equal to F and acting at the centroid. In Diagram 1a. When we represent the structural element in this manner. as shown in the following example. and a moment (torque) about the centroid. however we can represent the effect of this edge force. edge force with a 5000 lb. applied at the edge of the element as shown.3: Special Topics I . * 1" = 5000 in-lb. by an axial force. as shown in Diagram 1b. Example: In Diagram 2a.Non-Axial Loads Topic 8. we can determine the axial stress acting on the element by superposition of stresses. In Diagram 2b. force acting at the centroid of the element and a moment about the centroid of 5000 lb. In Diagram 3a we have shown the normal stress distribution on the area. And the bending stress is given by: = My/I = 5000 in-lb. Continue to: Topic 8.4: Special Topics I . and subtract along the back side the element.Principle Stresses file:///Z|/www/Statstr/statics/spec1/spec83./(2"x 2") = 250 lb/in2. Where for y we have used the maximum distance from the neutral axis to the outer edge of the area.Topic 8.* 1" /[(1/12)2"*2"3] = 3750 lb/in2.htm (2 of 3) [5/8/2009 4:50:16 PM] . The normal axial stress is given by: = F/A = 5000 lb. We can then find the maximum axial stress from: Maximum Axial Stress = 250 lb/in2 + 3750 lb/in2 = 4000 lb/in2. giving us the maximum bending stress. From the direction of stresses we see that they add along the front side of the element face. In Diagram 3b we have shown the bending stress distribution along one edge (it is the same across rest of the area).3 .Non-Axial Loads We next determine the normal axial stress and the bending stress on the top cross sectional area. Non-Axial Loads or select: Topic 8: Special Topics I .Topic 8.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spec83.3 .htm (3 of 3) [5/8/2009 4:50:16 PM] . In Diagram 2 we have shown the structure element with a plane cut through it at angle theta.Principal Stresses We next look at the general case where we have several axial stresses and shear stresses acting. we have the forces on each surface. we need to establish a sign convention as follows: file:///Z|/www/Statstr/statics/spec1/spec84. We may then apply static equilibrium conditions and write the equilibrium equations.Principle Stress Topic 8. We will look at the plane stresses on an inclined plane section to determine what are called the principal stresses and the principal planes.htm (1 of 3) [5/8/2009 4:50:17 PM] . we have shown a triangular element with axial and shear stresses shown. We would like this plane will be both an axial stress to write relationships which allow us to calculate the value of these two stress for any arbitrary plane section. We remember at this point that for static equilibrium the shear stresses Tauxy and Tauyx must be equal in magnitude.Topic 8. If we multiply these stresses by the appropriate areas. In Diagram 2b. Acting on and a shear stress . as shown in Diagram 2a.4: Special Topics I . Before we do so. In Diagram 1 we have shown a structure element with both normal axial stresses and shear stresses acting on the element.4 . where the axial stress is a maximum or minimum. The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element produce a counterclockwise torque (couple). the stresses acting on the incline plane shown in Diagram 2. In Diagram 3 a structural element is shown with axial and shear stress given by: normal xstress = 4000 lb/in2. The maximum and minimum normal axial stresses are known as the Principal Stresses. it is clear that we will may have both maximum and minimum stress values. In the above equations. This will be the positive direction for the angle.Principle Stress 1. We would like to find the file:///Z|/www/Statstr/statics/spec1/spec84. counterclockwise to the plane.Topic 8. We now write the following equilibrium equations: Sum of Forcesx Sum of Forcesy where we have used that face that magnitude of = magnitude of .htm (2 of 3) [5/8/2009 4:50:17 PM] . and . The incline plane angle will be measure from the vertical. (Some text use the opposite direction for the positive shear stress. The results of this process are as follows: The equations may be referred to as transformation equations. At the principal planes. and will not be presented. In this case we solve for . and Compressive Stresses will be considered negative. The two equation above may be solved for two "unknowns".4 . The value of the principal(maximum/minimum) stresses are given by: and for the principal plane.) 3. The planes for maximum shear stress vary by 45o from the principal planes. The details of solving these two simultaneous equations involve a number of trigonometric identities and some extended algebraic manipulations. shear stress = 1000 lb/in2. the shear stress will be zero. normal y-stress = 3000 lb/in2. The principal stresses may also be related as follows: Example. Tensile Stress will be considered positive. so we must be somewhat careful of signs when working problems and examples. and the planes at which they occur are known as the Principal Planes. This changes a sign in several equations. 2. We will calculate the principal stress two ways. and the maximum shear stress. and theta = 31.3000 lb/in2)/2}2 + (1000 lb/in2)2] = 3500 lb/in2 + 1118 lb/in2 = 4618 lb/in2 = 3500 lb/in2 . and 243.894 lb/in2 = 2382 lb/in2 The second method is from specific formula for the maximum/minimum stresses: = (4000 lb/in2 + 3000 lb/in2)/2 +/Sqrt[{(4000 lb/in2 . Now please select: Topic 8.7o.1118 lb/in2 = 2382 lb/in2 Note that we arrive at the same result by both methods.3000 lb/in2)/2}2 + (1000 lb/in2)2] = +/. These are the angles of the principal planes.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spec84.4 . then 2(theta) = 63.4o) = 3500 lb/in2 + 224 lb/in2 + 894 lb/in2 = 4618 lb/in2 and for 2nd principal plane = (4000 lb/in2 + 3000 lb/in2)/2 + [(4000 lb/in2 . First from the general formula for plane stresses = (4000 lb/in2 + 3000 lb/in2)/2 + [(4000 lb/in2 .4a: Principal Stresses .4o.Principle Stress principal planes. And finally we calculate the maximum shear stress from: = Sqrt[{(4000 lb/in2 .4o.4o) = 3500 lb/in2 . the principal stresses. and 121.4o) + 1000 lb/in2*sin(63.3000 lb/in2)/2] = 2. We first apply the formula for determining the angle of the principal planes: = 1000 lb/in2/[(4000 lb/in2 .1118 lb/in2.htm (3 of 3) [5/8/2009 4:50:17 PM] .3000 lb/in2)/2]cos(243.7o.Topic 8.3000 lb/in2)/2]cos(63.224 lb/in2 .Example 1 or: Topic 8: Special Topics I .4o) + 1000 lb/in2*sin(243. also applied at end B is balance by an equal 20.4a: Principal Stresses . stress will occur at an element on the outer edge of the shaft as that is where the shear stress is a maximum. 2. 1.htm (1 of 2) [5/8/2009 4:50:17 PM] . shown below. Thus the shaft is in equilibrium. We next show a shaft element with these values indicated in Diagram 2.Topic 8. principal stresses and the maximum shear stress.000 lb.000 lb.1416 * 2"4/32) = 15. The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element will produce a counterclockwise torque (couple).300 lb/in2. We wish to determine the principal planes. Also remembering our sign conventions. force by: = F/A = 20. The applied horizontal force of 20. acting at end B.000 lb. maximum. (Some text use the opposite direction for the positive shear stress.1416 * 1"2) in2 = 6.Example 1 Topic 8. and Compressive Stresses will be considered negative.) We then determine the maximum transverse shear stress by: Maximum Transverse Shear Stress = Tr/J = (2000 ft-lb * 12 in/ft) * (1") / (pi * d4/32) = (24. (We have ignored any weight of the shaft.370 lb/in2. (We assume the stress in uniform across the circular cross section. The applied torque of 2000 lb/in2 at end B is balance by an equal torque which develops in the wall acting on the shaft. force which the wall exerts on the shaft. (The principal. Tensile Stress will be considered positive.) Solution: We first review briefly the static equilibrium conditions for the shaft.4a: Principle Stress .) We next determine the axial (normal) stress due to the applied 20. and an axial force of 20. / (3. This changes a sign in several equations. while the normal axial stress we will assume is uniform across the area of the shaft.)*(1") / (3.000 inlb. so we must be file:///Z|/www/Statstr/statics/spec1/spece84a.Example 1 In Diagram 1 we have shown a shaft attached to a wall at end A with a torque of 2000 ft-lb.000 lb.000 lb. We calculate the principal stress from specific formula for the maximum/minimum stresses: = (6370 lb/in2 + 0)/2 +/.htm (2 of 2) [5/8/2009 4:50:17 PM] .4o.2o.4a: Principle Stress . and 258.815 lb/in2 = 3185 lb/in2 .630 lb/in2 = -12445 lb/in2 And finally we calculate the maximum shear stress from: = Sqrt[{(6370 lb/in2 .Mohr's Circle Or select: Topic 8: Special Topics I .300 lb/in2)2] = +/15.Sqrt[{(6370 lb/in2 .300 lb/in2/[(6370 lb/in2 .Example 1 somewhat careful of signs when working problems and examples. counterclockwise to the plane.2o.Topic 8. And finally we apply our plane stress and transformation equations.) 3. The incline plane angle will be measure from the vertical. and 129. and theta = 39.300 lb/in2)2] = 3185 lb/in2 + 15. then 2(theta) = 78. This will be the positive direction for the angle.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spece84a.4o.15.630 lb/in2.0)/2] = 4. Return to: Topic 8.8.0)/2}2 + (15.0)/2}2 + (15.5: Special Topics I . We first apply the formula for determining the angle of the principal planes: = 15.4 Principal Stress Or continue to:Topic 8. These are the angles of the principal planes.630 lb/in2 = 18. Thus Point A is above the x-axis as shown. we graph two initial points. however they are not easily remembered. This is the center of Mohr's Circle. 2. Point A with coordinates ( ). The distance from the origin to Points D and E are the values of the maximum and minimum principal stresses. as shown in Diagram 2.5: Special Topics I . the principal plane is represented by the line ED. developed by a German engineer. We draw a coordinate system with the x-axis representing the normal stresses. We now connect points A and B. In Mohr's Circle.Mohr's Circle Topic 8.Mohr's Circle The equations for the axial and shear stress at any plane in a structural element. and the y-axis representing the shear stresses.htm (1 of 4) [5/8/2009 4:50:17 PM] . A very useful way of expressing and visualizing the plane stresses in a loaded structural element is method known as Mohr's Circle. so Point B is below the x-axis in the Mohr Circle Diagram.5 . Otto Mohr (1835 . and Point B with coordinates ( ) as shown in Diagram 2. the normal x-stress is positive (tension) . The method for drawing Mohr's Circle is as follows: 1.1918). The normal y-stress in Diagram 1 is also positive. The line connecting points A and B intersects the x-axis.}] 3. and the equations for the principal stress present in Topic 8. which has an angle of zero and zero shear stress. but the shear stress on that face is negative. file:///Z|/www/Statstr/statics/spec1/spec85.as is the shear stress (counterclockwise) on the associated face. [According to our sign convention.Topic 8. Using the values from a given structural element (Diagram 1).3 Plane Stresses are accurate and useful. principal stresses. We will now work a Mohr's Circle example. The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element will produce a counterclockwise torque (couple). R= which is also equal to the maximum shear stress. (Some text use the opposite direction for the positive shear stress. This will be the positive direction for the angle. The incline plane angle will be measure from the vertical. normal y-stress = 3000 lb/in2. so the angle the plane of maximum shear stress makes is actually 45o different from the angle of the principal plane. This changes a sign in several equations. Example.) 3.Mohr's Circle The radius of the circle is given by . and the maximum shear stress.5 . In Diagram 3 a structural element is shown with axial and shear stress of normal x-stress = 4000 lb/in2. We would like to find the principal planes.Topic 8. the maximum shear stress is represented by line CF. In Diagram 2. This Mohr's Circle angle however is twice the angle in real space. so we must be somewhat careful of signs when working problems and examples. and Compressive Stresses will be considered negative. Tensile Stress will be considered positive. file:///Z|/www/Statstr/statics/spec1/spec85. We also note that the location of the center of Mohr's Circle is from the origin. 2. Structural Element Sign Conventions 1. We note that the plane represented by line FCG make an angle of 90o with respect to the principal plane (ACB). counterclockwise to the plane.htm (2 of 4) [5/8/2009 4:50:17 PM] . shear stress = 1000 lb/in2. +1000 lb/in2). Point A (+4000 lb/in2. This value is also the value of the maximum shear stress as we see from Mohr's Circle.5 . We also calculate the radius of the circle from = {[(4000 lb/in2 -3000 lb/in2)/2]2 + (1000 lb/in2)2}1/2 R= R = 1118 lb/in2.Mohr's Circle We begin by drawing Mohr's Circle for this problem.Topic 8. and Point B (3000 lb/in2 . -1000 lb/in2) are drawn and connected.Radius = 3500 lb/in2 .1118 lb/in2 = 2382 lb/in2 file:///Z|/www/Statstr/statics/spec1/spec85.htm (3 of 4) [5/8/2009 4:50:17 PM] . From Mohr's Circle we see that: Maximum Stress = Location of Center + Radius = 3500 lb/in2 + 1118 lb/in2 = 4618 lb/in2 and Minimum Stress = Location of Center . We next calculate the location of the center of Mohr's Circle = (4000 lb/in2 +3000 lb/in2)/2 = 3500 lb/ in2. We then use the value of the radius and the location of the center of the circle to find the principal stresses. and 121. and due to the sign conventions used.Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/spec85. counterclockwise from the vertical. In real space they will be in the opposite direction.htm (4 of 4) [5/8/2009 4:50:17 PM] .7o.Problem Assignment 1 or select: Topic 8: Special Topics I . we find (2 theta) = 63. and 243.7o.4o. and 243.4o. The Mohr Circle angles. and solving for(2 theta). are double the angles in real space.5 .4o.Mohr's Circle And from the geometry of the circle we can determine the angle the principal axis makes with respect to the element from Tan (2*theta) = (1000 lb/in2)/(4000 lb/in2 .3500 lb/in2). the angles in Mohr's circle are clockwise from the structural element. so the actual angles the principal planes make are 31. 63. The other important point with respect to the angles is that due to the way the initial points were chosen for Mohr's Circle.6 Special Topics I .) Continue to: Topic 8. and 121. (See Diagram 5.7o.Topic 8. then theta = 31.4o .7o. attached to the wall at point D and to the beam at point B. compression] 2.topic 8.Special Topics Problems 1 Topic 8.htm (1 of 4) [5/8/2009 4:50:18 PM] . The beam cross section is a W8 x 24 I-Beam. and is supported by a rod DB. with the characteristics shown in Diagram 1. A loaded beam (shown in Diagram 2) is pinned to the wall at point A.7a . and is supported a vertical support at point C.000 lb. file:///Z|/www/Statstr/statics/spec1/specp86. The beam has a load of 8.Problem Assignment 1 1. Determine the maximum axial stress acting in the beam cross section and state where it occurs.000 lb. The supporting rod makes an angle of 25o with respect to the beam.Special Topics . acting downward at point C and a second load of 10. [25.7a .400 psi @ x = 6' at bottom of beam. A loaded beam (shown in Diagram 1) is pinned to the wall at point A. acting downward at point B. file:///Z|/www/Statstr/statics/spec1/specp86.000 lb. at an angle of 30o with respect to the horizontal as shown in the Diagram. is applied at the centroid of the beam.192 psi @ x = 6' at bottom of beam. Determine the maximum axial stress acting in the beam cross section and state where it occurs.Special Topics Problems 1 A force of 20.topic 8. The beam cross section is a WT12 x 34 T-Beam. [20. with the characteristics shown in Diagram 2. and has a disk with a radius of 2" attached at end B. tension] 3. a solid 1 foot long shaft with a radius of 1" is attached to a wall at point A.7a . As shown in Diagram 3.htm (2 of 4) [5/8/2009 4:50:18 PM] . and state where it acts.. is applied to the outer edge of the disk.topic 8. [849 psi + 2. as shown. Determine the maximum shear stress in the shaft. we have shown a rectangular.395 psi at top of beam] 4.[axial 1. acting on each end. shear 940 psi.342 psi. In Diagram 4.5".Special Topics Problems 1 A force of 2000 lb. 2" by 1. section in tension with a normal force of 6000 lb.546 psi = 3.7a .] file:///Z|/www/Statstr/statics/spec1/specp86.htm (3 of 4) [5/8/2009 4:50:18 PM] . Determine the axial and shear stress on a 35o incline plane cut through the section. 7a . A cantilever beam is shown in Diagram 5. = 41. and state where is acts.[ = 36.Table of Contents Strength of Materials -Course Table of Contents Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/specp86.7b Special Topics . and a 5000 lb. The beam cross section is a 2" by 2" square.htm (4 of 4) [5/8/2009 4:50:18 PM] .Special Topics Problems 1 5.topic 8.000 psi. At end B at 2000 lb. Determine the maximum axial stress in the beam. + 3750 psi.000 psi @ wall and at top of beam on far edge] Continue to Topic 8. horizontal force is applied in the vertical center. Or Select Topic 8: Stress Analysis .Problem Assignment 2. + 1250 psi. downward load is applied at the centroid of the beam. but at the far outside edge as shown in Diagram 5. .axial min -1.7b . the principal stresses.8o. A structural element with given axial and shear stresses is shown in Diagram 1.4o. and the maximum shear stress. [axial 5750 psi.7b .] file:///Z|/www/Statstr/statics/spec1/specp87..htm (1 of 3) [5/8/2009 4:50:18 PM] .] Also determine the axial and shear stresses on a plane which makes an angle of 37 degrees counterclockwise from the negative vertical axis in the given element.. and the maximum shear stress.Special Topics .Topic 8. shear -2433 psi. Determine principal planes.8o. shear = 3606 psi.] Also determine the axial and shear stresses on a plane which makes an angle of 60 degrees counterclockwise from the negative vertical axis in the given element. [41. Determine principal planes. 106.Problem Assignment 2 1.606 psi. axial max 9.. the principal stresses.axial min -606 psi.Special Topics Problems 2 Topic 8.4o. [16. A structural element with given axial and shear stresses is shown in Diagram 2.] 2. 131. axial max 6.530 psi.470 psi.. shear = 4030 psi.. shear -2235 psi.[axial 8714 psi. axial max 6.414 psi. and a horizontal axial force of 24.Special Topics Problems 2 3. (We will assume that the normal axial stress is uniform across the area of the shaft. principal stresses and the maximum shear stress for a structural element on the outer edge of the shaft. In Diagram 3 we have shown a shaft attached to a wall at end A with a torque of 1800 ft-lb.000 lb. file:///Z|/www/Statstr/statics/spec1/specp87.) [56.] 4. A loaded cantilever beam is shown in Diagram 4. The beam cross section is a 2" by 2" square. shear = 4. 146.716 psi.112 psi.7b . acting at end B..3o.htm (2 of 3) [5/8/2009 4:50:18 PM] .axial min -2. Determine the principal planes.3o..Topic 8. Topic Examination. as shown in Diagram 4. shear = 10.Topic 8.4o.7b .Table of Contents Strength of Materials -Course Table of Contents Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/specp87.8 Special Topics . principal stresses.axial min -35 psi. axial max 20.] Continue to Topic 8.160 psi.4o.htm (3 of 3) [5/8/2009 4:50:18 PM] .Special Topics Problems 2 Determine the principal planes.. [2.285 psi.. 92. and maximum shear stress on a structural element 1/2 foot from the wall. Or Select Topic 8: Stress Analysis . and at point P in the cross sectional area. The beam cross section is a rectangular 2" by 4". A force of 18. at an angle of 37o with respect to the horizontal as shown in the Diagram.000 psi @ 4. the principle axis stress in slightly greater.Topic 8. is applied at the centroid of the beam. ~ 41. and is supported a vertical support at point C.8: Special Topics I .Topic Exam Topic 8.htm (1 of 3) [5/8/2009 4:50:18 PM] . compression.000 lb. [(38880 psi + 1800 psi = 40680 psi.8: Special Topics I .Topic Examination 1.6o angle off vertical plane. at top of beam 8 ft from left end) This is considering only sum of axial stress acting. A cantilever beam is shown in Diagram 2. file:///Z|/www/Statstr/statics/spec1/specx88. Determine the maximum axial stress acting in the beam cross section and state where it occurs.] 2. A loaded beam (shown in Diagram 1) is pinned to the wall at point A. If one takes a small section at point indicated and finds principle stresses. [(2333 psi + 3000 psi + 4000 psi = 9333 psi. The beam cross section is a 2" by 3" rectangle. tension at top right corner of cross section. A structural element with applied axial and shear stresses is shown in Diagram 3. horizontal force is applied at the horizontal center of the beam. and the other on the vertical axis 1.000 lb. horizontal force is applied in the vertical center.8: Special Topics I .000 lb. and state where it acts. One would be on horizontal neutral axis . Determine the maximum axial stress in the beam.htm (2 of 3) [5/8/2009 4:50:18 PM] . A second 6.583" to left of center point (2333 psi (t) + 0 + 2333 psi (c) = 0).167" below the center point of cross section (2333 psi (t) + 2333 psi (c) + 0 = 0)] 3. but at the far outside edge as shown in Diagram 5. file:///Z|/www/Statstr/statics/spec1/specx88. Again considering only combined axial stresses. There are two points of zero axial stress in cross section. but at the top edge of the area as shown in Diagram 5.Topic Exam At end B a 8.Topic 8. 8485 psi) Also determine the axial and shear stresses on a plane which makes an angle of 40 degrees counterclockwise from the negative vertical axis in the given element.Topic Exam Determine principal planes. (-22. and the maximum shear stress. +/. -6485 psi.Topic 8. 10.5o.485 psi. the principal stresses. 6951 psi) Select: Topic 8: Special Topics I .htm (3 of 3) [5/8/2009 4:50:18 PM] .Table of Contents Statics & Strength of Materials Home Page file:///Z|/www/Statstr/statics/spec1/specx88.8: Special Topics I . (6867 psi. +67.5o. 2. Pa Fundamental Terminology of Structural Behavior by Kirk Martini. University of Oklahoma Statics Tutorial by Jack Kayser and Michael Massetti. Hermann Gruenwald.Tables TABLES 1. Easton. Poisson's Ratio Yield Strength and Ultimate Strength Fundamental Constants from NATIONAL INSTITUTE OF STANDARDS AND TECHNOLOGY LINKS 1.htm [5/8/2009 4:50:18 PM] . University of Virginia Engineering Technical Data Reference Tables file:///Z|/www/Statstr/statics/tables/tables. 4. Beam Tables: T-Beams Beam Tables: I-Beams Weight Density & Coefficient of Linear Expansion Young's Modulus. 6. Lafayette College. 3. 5. Arch 4333 . 3. Shear Modulus. 4. 2.Steel Structures I by Professor Dr. in tw .204 0.16 I .5 x8.910 0.96 7.74 4.in 0.269 0.269 0.in tf .5x9.258 12.80 A-in2 d .in tf .in tw .964 0.10 3.25 x8 WT3x12.94 4.27 1.62 5.601 0.40 A-in2 d .995 0.in 1.250 1.394 0.70 0.080 6.67 2.280 1.456 0.13 5.in d/tw 2.60 0.190 1.000 3.5 x5 WT5x9.in wf .in 1.559 y .72 2.75 1.97 4.06 5.48 4.in tw .030 0.in 0.in 1.00 4.5 x19.015 4.370 1.00 0.94 2.292 16.19 3.in wf .in 1.70 5.771 0.in 1.230 17.329 0.35 Depth Flange Flange Stem of T d in 2.260 23.235 12.40 0.070 0.250 20.010 4.240 10.5 x10 x7.609 0.510 1.630 0.70 6.591 S -in3 1.270 r .30 8.5 WT6x11 Area A in2 2.16 4.5 x5.570 r .20 1.210 r .020 4.367 0.480 2.56 2.618 0.550 1.60 0.457 y .957 y .250 1.in wf .10 0.95 0.in d/tw 3.000 thick tf in thick tw in d/tw Cross Section Info.340 y .66 0.5 x16.318 15. x-x axis x-x axis x-x axis x-x axis I in4 0.570 1.in d/tw 3.20 A-in2 d .75 WT4x7.00 3.85 5.70 A-in2 d .06 4.in4 11.in d/tw 6.in tf .000 3.75 WT5x22.320 1.80 I .28 3.000 1.620 1.950 S -in3 2.07 5.314 0.875 y .44 I .90 2.254 0.180 27.in wf .T-Beam T-Beam Data Table Designation WT2.719 S -in3 1.in 1.350 14.in wf .50 0.787 0.420 0.70 S in3 0.49 2.265 9.92 1.in tw .481 0.230 1.in4 3.46 4.98 0.in tf .560 1.5 x6.in4 6.245 16.976 0.025 5.in tw .883 0.22 1.220 1.230 21.81 2.883 0.170 23.in4 10.740 1.528 0.688 0.160 S -in3 2.360 0.018 5.030 0.70 file:///Z|/www/Statstr/statics/tables/tbd.40 A-in2 d .424 0.00 6.htm (1 of 5) [5/8/2009 4:50:19 PM] .320 9.990 7.84 I .900 y in 0.5 x7.557 0.in 0.590 r in 0.29 2.06 4.in 1.in tf .795 r .in d/tw 2.15 I .950 0.204 0.240 21.411 S -in3 0.433 0.95 4.50 Width wf in 5.598 r .471 0.in4 2.50 0.24 6.940 0.022 7.88 8. 5 x39.920 r .in 1.67 19.10 0.031 8.44 12.06 12.20 10.60 12.160 1.61 I .730 1.40 I .50 6.5 2.371 16.105 0.90 7.063 0.19 12.150 A-in2 d .873 0.T-Beam x9.000 - 0.540 13.480 5.560 r .294 20.000 0.03 5.in wf .500 9.in tw .70 33.736 0.650 0.365 1.610 12.5 x66.in d/tw 20.349 0.5 x24 x21.20 A-in 2 d .70 6.5 x59.198 30.040 0.40 14.921 0.10 10.60 6.70 24.in tf .080 0.43 2.000 0.060 1.690 1.269 0.000 3.50 7.91 6.470 1.25 14.06 6.360 7.890 9.930 1.03 7.330 y .5 x55.40 2.062 8.30 7.310 1.230 26.94 12.80 25.050 8.5 x60 x53 x49.658 0.5 x51.570 12.870 r .220 7.545 11.606 0.5 x46 x42.856 0.19 12.10 7.755 8.230 0.60 27.70 0.00 5.470 1.380 1.510 1.998 0.38 12.70 7.97 8.in tw .620 10.24 15.31 12.820 7.736 1.in 1.339 20.60 6.020 0.in d/tw 27.70 6.495 12.300 2.19 14.750 y .60 6.in 1.56 12.80 0.490 1.20 I .170 1.600 1.in4 79.10 54.650 1.in wf .320 1.80 2.010 6.490 4.40 0.70 16.78 23.60 11.5 x20 WT6x95 x80.5 WT7x68 x63.480 1.020 S -in3 9.90 42.55 7.in tw .576 0.5 x36 x32.796 0.5 WT7x26.79 7.641 0.570 1.570 1.130 1.690 0.540 4.5 x8.in 1.100 1.htm (2 of 5) [5/8/2009 4:50:19 PM] .84 8.70 50.in d/tw 7.in4 18.00 7.593 0.620 0.20 18.910 1.90 0.08 6.486 0.528 0.800 3.190 1.938 0.155 0.060 4.380 5.192 0.13 14.985 1.236 0.38 14.in 1.260 1.50 6.530 1.200 11.660 11.336 17.96 4.106 0.671 0.40 36.520 1.880 1.940 S -in3 14.042 8.97 6.00 27.00 13.670 1.224 0.490 1.10 6.86 17.89 6.710 9.700 1.470 13.968 0.90 22.680 1.990 5.50 6.200 1.70 16.60 14.620 1.370 18.710 1.280 1.810 S -in3 3.080 y .60 6.330 1.80 23.40 0.986 0.36 6.740 1.in tf .060 6.516 0.60 6.30 A-in2 d .00 17.077 8.813 0.06 6.30 15.in4 60.20 9.25 x7 WT6x25 x22.60 48.308 22.007 4.40 46.32 6.515 1.630 1.130 1.575 0.20 20.20 9.350 1.030 4.760 1.in wf .670 1.00 62.40 file:///Z|/www/Statstr/statics/tables/tbd.220 1.880 1.390 15.040 8.13 12.237 25.40 43.582 11.905 7.69 12.62 5.495 14.400 2.130 1.25 12.590 1.31 14.80 31.960 4.in tf .430 14.in 1.080 1. 60 9.00 I .890 y .in tw .60 0.in tw .50 A-in2 d .50 8.843 1.10 7.280 2.5 x25 14.00 179.307 26.550 y .20 12.98 58.283 1.110 S -in3 9.70 59.270 2.83 8.in 2.100 5.in tw .795 0.860 r .545 5.380 5.30 0.00 258.00 7.60 16.80 7.600 S -in3 6.945 1.390 23.00 6.205 6.80 I .370 2.970 1.16 16.in tf .20 0.16 10.130 2.033 1.140 r .090 1.00 230.41 16.160 2.07 42.20 8.50 8.36 9.930 1.260 r .10 48.365 2.00 7.in 2.13 8.10 8.320 8.5 WT7x213 x199 x185 x171 x157 x143.570 7.860 1.870 1.80 33.97 16.in d/tw 11.25 16.90 59.08 11.570 0.563 0.400 2.30 8.50 9.658 1.025 1.in wf .13 15.820 1.710 2.06 9.730 1.50 I .010 S -in3 9.in wf .795 0.880 S -in3 41.in tf .in tf .60 54.40 8.90 I .35 16.770 6.10 0.125 7.in4 288.5 x32 x29 WT9x30 x27.5 x20 x18 WT8x48 x44 WT8x39 x35.695 0.415 6.346 23.580 5.50 39.770 5.670 1.in 1.in wf .860 1.000 30.992 0.468 1.630 0.in 2.020 1.in d/tw 7.20 37.in4 64.450 8.535 15.140 2.060 2.in tf .400 2.42 38.in4 42.93 7.16 11.780 1.78 16.875 4.53 8.500 27.443 18.70 A-in2 d .890 1.20 8.00 204.416 21.770 1.22 A-in2 d .420 16.100 2.89 5.688 0.00 139.710 2.110 S -in3 9.300 2.628 0.503 0.68 46.533 0.80 8.235 2.715 0.140 2.645 0.810 1.875 0.500 19.00 157.in wf .590 2.08 7.in 2.in 2.htm (3 of 5) [5/8/2009 4:50:19 PM] .in tf .10 file:///Z|/www/Statstr/statics/tables/tbd.T-Beam x47.000 24.85 36.20 8.100 21.in wf .42 50.in4 60.40 0.532 7.938 1.90 8.710 1.36 6.650 r .545 0.in 1.695 3.820 9.06 8.120 1.in tw .in4 64.299 26.358 25.486 16.63 5.504 16.30 43.40 0.in d/tw 8.90 0.13 9.20 30.130 A-in2 d .500 0.00 54.41 8.558 7.543 8.in d/tw 14.410 r .30 8.390 2.06 14.428 0.464 0.17 54.30 12.710 y .529 15.640 7.475 2.500 8.10 34.60 I .875 0.720 7.in 2.160 2.073 7.00 7.400 37.in 1.586 8.655 5.270 2.5 x132 x123 WT8x25 x22.700 34.10 0.5 x43.465 15.in 2.980 1.407 19.500 0.080 y .in d/tw 62.in tw .890 1.08 9.93 8.039 7.00 14.00 A-in2 d .380 21.710 y .813 1.748 0.310 6.00 126.502 0.093 1.190 2.570 1.000 6. 93 0.in 3.00 11.930 4.in tf .410 3.80 395.5x57 16.in4 WT10.660 3.690 4.80 9.900 10.in wf .700 S -in3 13.463 28.700 r .in tf .05 9.10 258.780 y .00 0.500 0.09 0.150 4.00 0.62 3.300 18.499 0.in -3.46 9.20 151.in tf .in d/tw I .615 0.747 0.200 3.380 3.455 23.961 0.970 2.64 10.00 17.160 y .932 0.000 0.990 3.60 166.50 137.00 0.827 0.40 12.20 0.00 I .5 20.458 0.833 0.in d/tw 17.70 49.210 3.5 10.600 S -in3 28.50 239.10 10.00 15.470 2.60 14.57 8.00 0.20 350.in 2.40 I .in d/tw 19.00 85.564 26.00 I .477 0.400 26.900 11.070 0.40 9.50 8.582 0.15 10.93 7.800 11.in tf .40 8.40 I .240 8.70 10.00 15.5 WT12x60 x47 x42 x38 x34 10.300 33.580 2.900 S -in3 22.30 110.300 25.615 24.40 15.470 25.in tf .100 S -in3 2.in tw .in 3.50 103.in d/tw 16.15 9.963 0.480 2.16 11.685 0.in wf .516 23.850 1.80 13.640 y .in tw .750 0.820 1.670 3.00 0.850 0.in 4.295 x34 x31 x27.518 26.650 4.in wf .in tw .930 0.772 0.416 28.in wf .5 WT9x57 x52.90 289.460 r .522 0.900 3.522 28.600 32.35 9.792 0.760 0.60 372.in d/tw 94.710 r .50 186.230 r .in 4.80 12.215 0.00 18.00 14.991 0.10 9.720 r .800 12.in wf .020 4.5 x48 - 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x314 x287 x264 x246 W 16x50 x45 x40 x36 W 16x96 x88 W 18x60 x55 x50 x45 W 18x114 x105 x96 W 21x73 x68 x62 x55 W 24x94 109.039 11.12 7.in tf .in4 37.68 6.32 11.80 I .82 2.in wf .in 2.24 6.in4 274.0 92.77 1.0 72.in wf .in 1.00 225.30 38.25 15.00 241.563 0.0 file:///Z|/www/Statstr/statics/tables/ibd.00 7.0 30.32 11.in tw .50 48.0 S -in3 220.in wf .5 64.80 35.00 4.42 7.307 0.00 S -in3 10.70 16.945 1.400 1330.in 6.87 r .0 101.in4 70.558 16.375 1140.0 84.833 0.86 7.0 25.00 40.0 559.80 1.658 1.70 r .62 6.80 24.554 1850.64 6.84 2.502 0.0 127.20 6.in4 224.80 8.90 1630.in4 108.in tw .0 512.30 I .93 6.in 6.532 14.0 0.061 0.0 77.54 8.0 A-in2 d .0 A-in2 d .358 0.00 4.655 5450.20 18.740 0.in 2.80 16.522 0.0 584.70 13.in wf .30 S -in3 17.90 11.in tw .00 221.in4 33.24 8.in4 50.80 S -in3 23.00 154.00 17.0 706.0 S -in3 108.00 17.in tw .62 r .0 465.0 0.70 9.992 0.50 r .57 1.68 1230.0 A-in2 d .795 0.30 11.in I .00 166.25 7.60 16.20 18.430 1480.47 7.30 r .in I .365 2.13 8.00 182.17 6.in I .630 0.27 6.0 447.240 16.295 20.00 249.64 8.380 0.516 2690.0 891.10 32.750 0.99 8.in tw .in tf .00 202.000 10.00 4.8 72.215 0.025 1.813 1.125 3230.283 1.1 79.468 1.695 0.67 1.00 I .416 0.00 15.79 r .in 1.415 4400.346 0.628 0.0 517.85 6.0 397.in 7.12 r .615 0.19 16.14 6.10 45.74 1330.in wf .0 427.335 986.82 r .390 0.in wf .50 18.40 r .86 2.16 11.595 2040.10 S -in3 13.5 S -in3 166.29 9.093 1.0 110.30 16.792 0.0 A-in2 d .32 S -in3 46.00 4.70 24.685 0.0 S -in3 221.0 S -in3 80.831 0.12 7.30 16.56 16.in tf .in4 28.0 98.htm (4 of 5) [5/8/2009 4:50:21 PM] .0 28.00 4.270 18.in 9.in4 27.in tf .504 1220.130 2.60 8.99 S -in3 38.20 16.60 15.299 657.81 16.59 1.30 20.in tw .0 608.00 21.65 1.25 7.90 r .68 1.in 1.48 11.0 0.90 18.20 17. 53 r .0 671.0 A-in2 d .0 15.1a: Euler Buckling .0 15.0 0.90 124.in4 70.40 23.150 0.00 3.932 0.760 0.963 0.761 15000.458 0.000 0.750 1.90 146.19 2.in 88.827 0.570 0.670 0.60 27.15 2.in wf .92 1.I-Beam x84 x76 x68 W 27x114 x102 x94 x84 W 30x132 x124 x116 x108 x99 W 33x240 x220 x200 W 36x300 x280 x260 x245 x230 24.00 13.0 1030.Example 1 Topic 6.615 5760.070 0.80 33.75 14.885 18900.0 153.81 76.990 24.70 750.0 34.0 952.40 36.00 3.00 27.585 5360.40 31.70 26.0 S -in3 300.79 9.865 1.73 Last Update:05/20/03 Return to: Topic 6.0 27.09 9.in I .90 30.0 67.in tf .25 15.70 24.20 S -in3 3.23 2.0 837.15 2.91 8.512 1.00 10.00 18.0 S -in3 380.0 A-in2 d .90 933.90 33.470 2370.0 212.00 13.0 58.0 29.70 24.30 36.945 20300.00 r .in4 118.463 2830.0 15.522 4000.0 11.00 114.in4 33.00 164.680 0.in I .400 2100.802 16100.50 82.00 15.30 10.775 12300.00 r .0 329.10 1200.90 24.0 64.00 123.518 3610.00 3.00 132.57 r .0 270.in I .87 r .06 16.416 1820.0 36.in I .00 156.20 196.582 0.850 0.015 22.30 27.551 1.0 894.50 30.2c: Structural Steel Column Selection .00 11.841 17300.82 10.00 10.747 0.00 12.in4 13.00 3.00 10.24 16.00 3.88 16.00 3.in 94.in tw .90 21.484 0.in tw .961 0.810 1.69 9.0 30.69 9.0 197.490 3270.60 70.80 29.25 2.77 15.0 243.in tw .in wf .in wf .00 I .50 36.64 10.in4 38.50 S -in3 1.0 S -in3 813.018 0.00 1090.in tf .0 72.in I .564 4930.548 4470.40 15.in4 A-in2 d .70 35.in tf .00 r .00 3.50 16.0 S -in3 9.0 267.71 8.00 12.00 23.00 159.10 r .0 1110.400 0.930 0.985 20.18 2.00 1010.70 105.0 300.60 95.60 27.521 0.28 10.551 1.772 0.in4 12.in I .00 11.50 15.95 1.64 106.in 2.10 181.16 10.in I .in wf .595 1.20 1300.0 31.655 1.471 1.275 0.80 26.500 0.90 940.Example 3 file:///Z|/www/Statstr/statics/tables/ibd.570 4090.in 2.in A-in2 d .636 0.80 841.in4 21.0 0.830 13600.00 144.0 176.682 0.20 34.00 139.12 2.10 36.10 S -in3 37.0 742.10 29.in tw .20 30.0 355.72 16.83 82.00 11.715 11100.440 0.60 S -in3 31.350 0.0 0.60 33.htm (5 of 5) [5/8/2009 4:50:21 PM] .70 128.07 10.in tf .91 9.06 r .260 0. 5 17. sandstone Rubber Sand.5 15.htm (1 of 2) [5/8/2009 4:50:21 PM] .5 83 70 2014-T6 7075-T6 6061-T6 Brass Bronze Cast iron Concrete Plain Reinforced Lightweight Copper glass Magnesium alloys Monel (67% Ni.5 4.15 19. gravel Steel 145 150 92.5 70 105 490 27 24 11 16.75 11.8 11.5 556 165 112 550 550 23 24 16.5 27 28 26 83.25 7. quartz Limestone.Constants Combined Page 1 Material Constants Weight Density 'ρ' and Coefficient of Thermal Expansion 'α' Material lb/ft3 Aluminum alloys r kN/m3 Material Aluminum alloys Brass Bronze Cast iron Concrete Copper & copper alloys Glass Magnesium alloys Monel (67% Ni.5 77 file:///Z|/www/Statstr/statics/tables/cdensity. 30% Cu) Nickel a 10-6/ oF 13 11.5 152.1 8 27. marble.5 10.2 10-6/ oC 23 20.7 7.2 10.7 172.6 6 9.95 10. soil.5 87 87 62.45 14 13 170 175 170 530 530 447. 30% Cu) Nickel Plastics Nylon Polyethylene Rock Granite.5 87 26 17.5 75 9. 7 5.5 37.4 63.Constants Combined Page 1 Titanium Tungsten Water.81 10 32.5 5.9 file:///Z|/www/Statstr/statics/tables/cdensity.5 42.1 6.200 62.8 44 190 9. fresh sea Wood (air dry) Douglas fir Oak Southern pine 280 1.htm (2 of 2) [5/8/2009 4:50:21 PM] . 3 86.32 .33 0.500 6.32 .600 10.5 4.000 3.500 27.000 207 207 207 207 207 200 200 11.32 .000 24.600 10.000 10.3 .000 26.27 11.900 3.000 30.300 43.500 12.4 0.000 15.000 25.000 4.500 G Gpa 27.htm (1 of 2) [5/8/2009 4:50:22 PM] .Poisson's Ratios 'v' Material Aluminum alloys 2014-T4 2014-T6 2024-T4 6061-T6 7075-T6 Brass (Red.600 10.15 .600 11.6 26.000 29.28 0.Shear Modulus'G' .35 .000 400 150 .4 0.32 .750 5.2% C hot-rolled .34 0.600 12.400 . cold rolled) Brass (Red.76 1 ksi 4.8% C hot-rolled Stainless (cold-rolled) Stainless (heat-treated) E ksi 10.000 25.2 27 27 26 38 Poisson's ratio n 0.6 138 65.600 11.3 6.000 9.400 10.32 0.000 30.31 0.5 - .000 30.000 30.6 11.500 5.500 17.2% C cold-rolled .48 - 38 44.000 15.9 44.4 GPa 73 73 73 72 69 104 104 104 104 173 173 31 117 69 166 179 207 2.500 12.500 12.900 3.600 80 80 80 80 80 86.000 20. hot rolled) Nickel alloy Plastics Nylon Polyethylene Rubber (average) Steel .000 15.33 0.000 15.27 .Constants Combined Page 3 Material Constants Youngs' Modulus 'E' .3 86.00276 - 30.000 29.2 0.34 0.500 file:///Z|/www/Statstr/statics/tables/cyoungs.7 41.2% C hardened .9 86.4% C hot-rolled .28 0.600 11.000 30.33 0.33 0.0007 - 78.33 0.32 .34 0.500 6.35 0.34 0. annealed) Bronze (cold rolled) Bronze (annealed) Cast iron (tension) Cast iron (compression) Concrete (compression) Copper (cold-drawn) Plate glass Magnesium alloy Monel (wrough.800 3. 4 11.htm (2 of 2) [5/8/2009 4:50:22 PM] .32 .300 1.000 11.1 12.along grain) Douglas Fir Southern Pine Red Oak 29.000 11.900 1.32 .3 file:///Z|/www/Statstr/statics/tables/cyoungs.9 - .Constants Combined Page 3 Steel.29 .9 75.000 29.000 - 75.32 . structural ASTM-A36 ASTM-A572 ASTM-A514 Wood (comp.000 29.3 .000 1.9 75.800 200 200 200 9 13. 5 13. annealed) Bronze (cold rolled) Bronze (annealed) Cast iron (tension) Cast iron (compression) Concrete (compression) Copper (cold-drawn) Plate glass Magnesium alloy Monel (wrough.Constants Combined Page 2 Material Constants Yield Strength and Ultimate Strength Material Aluminum alloys 2014-T4 2014-T6 2024-T4 6061-T6 7075-T6 Brass (Red.5 870 35 310 70 280 621 552 2 40 13.5 2 60 17.htm (1 of 2) [5/8/2009 4:50:22 PM] .6 4 9 2. cold rolled) Brass (Red.8 280 22 50 60 150 345 414 0. hot rolled) Nickel alloy Plastics Nylon Polyethylene Rubber (average) Steel sy ksi MPa ksi su Mpa 41 60 48 40 70 60 15 75 20 29.5 - - - - file:///Z|/www/Statstr/statics/tables/cyield.5 283 410 331 276 483 414 104 772 138 205 62 70 68 45 80 75 40 100 50 40 125 5 45 10 40 90 80 428 480 470 310 552 518 276 515 345 274. 2% C hardened .htm (2 of 2) [5/8/2009 4:50:22 PM] .along grain) Douglas Fir Southern Pine Red Oak 62 60 53 53 76 165 132 428 414 366 366 524 1140 911 90 85 62 84 122 190 150 620 587 428 580 842 1310 1040 36 50 100 6 6.2% C cold-rolled .2% C hot-rolled .4 6.6 250 340 700 41 45 32 60 70 120 7.4 8.4% C hot-rolled .8% C hot-rolled Stainless (cold-rolled) Stainless (heat-treated) Steel.9 400 500 830 51 58 48 file:///Z|/www/Statstr/statics/tables/cyield. structural ASTM-A36 ASTM-A572 ASTM-A514 Wood (comp.Constants Combined Page 2 .5 4.