28/02/2012STRESS AND STRAIN 2 1 28/02/2012 Topics: • Introduction • Main Principles of Statics Stress • Normal Stress • Shear Stress • Bearing Stress • Thermal Stre Mechanics : The study of how bodies react to forces acting on them RIGID BODIES (Things that do not change shape) DEFORMABLE BODIES (Things that do change shape) FLUIDS Statics : The study of bodies in an equilibrium Incompressible Compressible Dynamics : 1. Kinematics – concerned with the geometric aspects of the motion 2. Kinetics – concerned with the forces causing the motion. Mechanics of Materials : The study of the relationships between the external loads applied to a deformable body and the intensity of internal forces acting within the body. 1.1 Introduction 4 2 28/02/2012 External Loads Body Force - developed when one body exerts a force on another body without direct physical contact between the bodies. - e.g earth’s gravitation (weight) Surface Forces - caused by direct contact of one body with the surface of another. concentrated force linear distributed load, w(s) 1.2 Main Principles of Statics 5 ` ` ` ` ` ` ` ` Axial Load Normal Stress Shear Stress Bearing Stress Allowable Stress Deformation of Structural under Axial Load Statically indeterminate problem h l Stress Thermal 3 Strain = describe the deformation by changes in length of line segments and the changes in the angles between them ` ` • Normal Stress : stress which acts perpendicular. cross section of the load-carrying : can be either compressive or tensile. • Shear Stress : stress which acts tangent to the cross section of (τ) the load-carrying member. the ( σ) y g member.1 Introduction 8 4 . b d Stress = the intensity of the internal force on a specific plane (area) passing through a point. or normal to. 1.28/02/2012 ` Mechanics of material is a study of the relationship between the external loads applied to a deformable body and the intensity of i t internal lf forces acting ti within ithi the th body. : refers to a cutting-like action. acting normal to ΔA σ = P / A A positive sign will be used to indicate a tensile stress (member in tension) A negative sign will be used to indicate a compressive stress (member in compression) (a) (b) •Unit: Nm -² •N/mm2 or MPa Stress ( σ ) = Force (P) N/m2 or Pa Cross Section (A) 5 .28/02/2012 ` Normal Stress. σ the intensity of force. or force per unit area. de o at o . and cross section remains flat or plane during deformation 2.4 Axial Loading – Normal Stress + ↑ FRz = ΣFz . force o ce P be applied along centroidal axis of cross section C 11 1.28/02/2012 Assumptions : 1. ∫ dF = ∫ σ dA A P = σA σ= P A σ = average normal stress at any point on cross sectional area P = internal resultant normal force A = cross-sectional area of the bar 12 1. In order for uniform deformation. Uniform deformation: Bar remains straight before and after load is applied.4 Axial Loading – Normal Stress 6 . Knowing that d1=30mm and d2=20mm.4 Axial Loading – Normal Stress Example 1. 7 . find average normal stress at the midsection of (a) rod AB.28/02/2012 • Use equation of σ = P/A for cross-sectional area of a member when section subjected to internal resultant force P Internal Loading • Section member perpendicular to its longitudinal axis at pt where normal stress is to be determined • Draw free-body diagram • Use equation of force equilibrium to obtain internal axial force P at the section Average Normal Stress • Determine member’s x-sectional area at the section • Compute average normal stress σ = P/A 13 1. (b) rod BC BC.1: Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. 28/02/2012 Example 1. Knowing that d1 = 30 mm and d2 = 50 mm. (b) rod BC.2 Two solid cylindrical roads AB and BC are welded together at B and loaded as shown. find the average normal stress in the mid section of (a) rod AB. 8 . ε is the elongation or contraction of a line segment per unit of length ε = ΔL / Lo ΔL = elongation Lo = length ε = δ L = normal strain * ΔL= δ 9 .28/02/2012 ` Normal strain. 37mm.4 A cable and strut assembly ABC supports a vertical load P=12kN.3: Determine the corresponding strain for a bar of length L=0. (b) If the cable elongates 1.600m and uniform cross section which undergoes a deformation δ=150×10-6m.28/02/2012 Example 1.1mm. what is the strain? (c) If the strut shortens 0. what is the strain? 10 .600m = 250×10−6 @250μ 1. The cable has an effective cross sectional area of 160mm². and the strut has an area of 340mm². (a) Calculate the normal stresses in the cable and strut. δ 150×10−6 m ε= = = 250×10−6m/ m L 0. Data from tensile test can be plot into stress and strain diagram.8m ` ` ` Tensile test is an experiment to determine the load-deformation behavior of the material. L=2.5 The bar shown has a square cross section (20mm x 40mm) and length. Example of test specimen .8m. determine the stress and strain if the bar end up with 4m length. If an axial force of 70kN is applied along the centroidal axis of the bar cross sectional area.note the dog-bone geometry 28 11 .28/02/2012 1. length 70kN 70kN 2. etc. loads and measure its response 29 Stress-Strain Diagrams A number of important mechanical properties of materials that can be deduced from the stress-strain diagram are illustrated in figure above. bending.28/02/2012 ` Universal Testing Machine .equipment used to subject a specimen to tension. compression. 30 12 . Unit: MPa Point B = yield point Point B-C = specimen continues to elongate without any increase in stress. 31 13 . True stress is calculated using the actual cross sectional area at the instant the load is measured.28/02/2012 ` ` Point O-A = linear relationship between stress and strain Point A = proportional limit (σPL) The ratio of stress to strain in this linear region of stress-strain diagram is called Young Modulus or the Modulus of Elasticity given Ε ` ` ` ` ` ` = At point A-B. specimen b i t begins to neck-down k d Point E = fracture stress Δ σ Δ ε σ < σPL 31 Point O to A Point C to D Point D to E At point E Normal or engineering stress can be determined by dividing the applied load by the specimen original cross sectional area area. specimen begins yielding. Its refer as perfectly plastic zone Point C = stress begins to increase Point C-D = refer as the zone of strain hardening Point D = ultimate stress/strength . 2% 32 Brittle material such as ceramic and glass have low tensile stress value but high in compressive stress.002/0. Therefore. Stress-strain diagram for brittle material. Knowing that the maximum stress in the cable must not exceed 190MPa and that the elongation of the cable must not exceed 6mm.28/02/2012 Some of the materials like aluminum (ductile). ` Drawing a straight line that best fits the data in initial (linear) portion of the stress stress-strain strain diagram ` Second line is then drawn parallel to the original line but offset by specified amount of strain ` The intersection of this second line with the stress-strain curve determine the offset yield stress. find the maximum load P that can be applied as shown 33 14 . stress value called the offset yield stress. Example 1. σYL is used in line of a yield point stress. the offset yield stress is determine by.6 The 4 mm diameter cable BC is made of a steel with E=200GPa. As illustrated. does not have clear yield point likes structural steel. ` Commonly used offset value is 0. the material is said to be linear elastic. Beyond the elastic limit. some residual strain or permanent strains will remain in the material upon unloading . Any material which deforms when subjected to load and returns to its original dimensions when unloaded i said is id to b be elastic. otherwise it is non-linear elastic. Elasticity refers to the property of a material such that 34 • The amount of strain which is recovered upon unloading is called the elastic recovery. l i If the stress is proportional to the strain. 35 15 . The residual elongation corresponding to the permanent strain is called the permanent set .28/02/2012 ` ` ` ` ` it returns to its original dimensions after unloading . Calculate the elongation and the decrease in diameter Δd. laterally (sideways). The ratio of lateral strain to axial strain is a constant known as the Poisson's ratio. if the material is subjected to axial compression it shortens axially but bulges out compression. It is made from aluminum with modulus of elasticity of 70 GPa and poisson's ratio ν= 1/3. The bar has length of 3 m and diameter of 30 mm.28/02/2012 ` ` ` When an elastic. Similarly.7 A prismatic bar of circular cross-section is loaded by tensile forces P = 85 kN. homogenous and isotropic material is subjected to uniform tension. 37 16 . it stretches axially but contracts laterally along its entire length. v= where the strains are caused by uniaxial stress only ε ε lateral axial δL L δb δd εsisi @ ε y = − =− b d ε paksi @ ε x = 36 Example 1. 28/02/2012 Example 1.8 A 10 cm diameter steel rod is loaded with 862 kN by tensile forces. 00053 ε l = = ∴ −ν (ε Δ d a ) = − o . 00154 cm 38 Exercises 1 1. 1 ) − 0 . 7 MPa 207 x 10 3 MPa 862 x 10 1 π ( 0 .The material has modulus of elasticity E= 200GPa and Poisson’s Ratio v = 0.0125mm) d) the increase ∆t in the wall thickness (ans: 0.1 ) 4 3 2 N m 2 = 109 . 00053 ) )( 0 .Determine : a) the shortening. 7 MPa = 0 . ε = σ = 109 . 000154 = ε l ( D ) = ( − 0 .9x10-6) c) the increase ∆d2 in the outer diameter and the increase ∆d1 in the inner diameter (ans: 0.30. δ ( ans :-0. i ε lateral (ans: ( 113 9 10 6) 113. outside diameter d2=150mm and inside diameter d1=110mm is compressed by an axial force P= 620kN. 29 ( 0 . determine the deformation of rod diameter after being loaded.0171 mm and 0. σ = a p = A E Lateral strain. 000154 = − 0 .2 m. Knowing that the E=207 GPa and ν= 0.455 mm) b) the h l lateral l strain. Solution σ in rod rod.29. A steel pipe of length L=1.00228 mm) 39 17 . dBC 57 mm and tBC=9mm. A hollow circular post ABC as shown in Figure 2 supports a load P1=7. σAB in the upper part of the post post. (ans: 9. tAB= 12mm. what should be the magnitude of the load P2? (ans : P2=6kN) 40 3.95 9 95 MPa) b) If it is desired that the lower part of the post have the same compressive stress as the upper part. 49 18 . l i the h modulus d l of f rigidity. a) Calculate the normal stress. The diameters and thicknesses of the upper and lower parts of the post are dAB=32 mm.62 mm are observed in a 120 mm gage length. A standard tension test is used to determine the properties of an experimental plastic. The test specimen is a 15 mm diameter rod and it is subjected to a 3. and Poisson’s ratio of the material. respectively. A second load P2 is uniformly distributed around the cap plate at B. Knowing that an elongation of 11 mm and a decrease in diameter of 0.5 kN acting at the top.5 kN tensile force.28/02/2012 2. D Determine i the h modulus d l of f elasticy. `Shear 41 ` Depending on the type of connection. in the plane of the material) is called a The shear force intensity. i. a connecting element (bolt. . it is known as direct shear force stresses can also arise indirectly as a result of tension. pin) may be subjected to single shear or double shear as shown.e.τ over which it acts.28/02/2012 ` A force acting parallel or tangential to a section taken through a material (i. rivet. shear force divided by the area .e. Rivet in Single Shear τ= V P = d2 A π 4 42 19 . torsion or bending of a member. τ = shear stress V = τ V = shear force A A = cross-sectional area ` ` shear force Shear stress arises as a result of the direct action of forces trying to cut through a material. is called the average g shear stress. 9 For the 12 mm diameter bolt shown in the bolted joint below.28/02/2012 Rivet in Double Shear τ= V = A P 2P = 2 2 d πd 2(π ) 4 Example 1. determine the average shearing stress in the bolt. bolt 43 Single Shear Double Shear τ ave = P F = A A τave= = P F A 2A 20 . γ is a measure of the angular distortion of the body. x L V γ γ = x L (units: degrees. radians) 44 ` Bearing stress is also known as a contact stress Bearing stress in shaft key. σb = P M r 2M = = Ab (h 2) L rhL Bearing stress in rivet and plat. σb = P td 45 21 .28/02/2012 ` ` The effect of shear stress is to distort the shape of a body by inducing shear strains The shear strain. what is the average shear stress in the plate and the average compressive stress in the punch if the required force to create the hole is P = 110kN. ` τ ` = G γ Unit : Pa The modulus young (E). . Assume that a punch having diameter d=20 mm is used to punch a hole in an 8 mm plates. ` Value of shear modulus can be obtained from the linear region of shear stress-strain diagram. P 20 mm 8 mm 46 It also known as Shear Modulus of Elasticity or the Modulus of Rigidity.0 A punch for making holes in steel plates is shown in the figure. poisson’s ratio(ν) and the modulus of rigidity (G) can be related as G = E 2 (1 + ν ) 48 22 .28/02/2012 Example 2. and z directions respectively. Vf = (a + aε)(b .bνε)(c .νε)2 59 23 . Initial b d body Initial volume of body body. y and z directions. the volume of the body also changes within the elastic limit. 58 ` The tensile force P causes an axial elongation of aε and lateral contractions of bνε and cνε in the x. ` Consider a rectangular g parallel p piped pp having g sides a. respectively. Hence. . Vo = abc Final volume. b and c in the x. y.28/02/2012 ` Because of the change in the dimensions of a body as a result of tension or compression.cνε) = abc(1 + ε)(1 . 1) = abc(ε .2 νε .2 νε) . Vf = abc(1 + ε . Since all strain satisfy ε << 1.2νε) Change in volume.28/02/2012 Expanding and neglecting higher orders of ε (since ε is very small). ΔV = Final Volume .Initial Volume = abc(1 + ε . σy and σz.2 νε) = Vo ε(1 . Final volume.abc = abc(1 + ε . Δ V V o = ε (1 (1 − − 2 ν 2 ν ) ) ε = σ E 60 ` ` Isotropic I i material i li is subjected bj d to general l triaxial i i l stress σx. so εv = εx + εy + εz εx = εy = εz = ε v 1 σ x −ν (σ y + σ z ) E [ ] 1 σ y −ν (σ x + σ z ) E [ ] ] x 1 σ z − ν (σ x + σ y ) E [ = 1 − E 2ν (σ + σ y + σ z ) 61 24 .2 ν) Hence. εv. ` FS Allowable-Stress Design = F F fail allow FS > 1 σ allow = σ yield y FS or τ allow = τ yield y FS 63 25 . and z = 2cm. as indicated in figure below. (maximum load) One method of specifying the allowable load for the design or analysis of a member is use a number called the Factor of Safety (FS).33 for the titanium alloy.6 N. The remaining stresses (σz. Let E = 16 kN and ν = 0. τxz and τyz) are all zero.28/02/2012 Example 2. (a)Determine the changes in the length for Δx. y 6N 14 N 14 N x z 6N 62 ` Applied load that is less than the load the member can fully support. (b) Determine the dilatation. τxy. y = 4cm.1 A titanium alloy bar has the following original dimensions: x = 10cm. Δy and Δz. The bar is subjected to stresses σx = 14 N and σy = . the bar is called statically indeterminate. Therefore. FA L AC FBL CB − =0 AE AE FA L AC FBL CB = AE AE FBL CB AE FA = × AE L AC F L FA = B CB L AC F L P − FB = B CB L AC F L P = B CB + FB L AC ⎛ L CB ⎞ + 1⎟ P = FB ⎜ ⎟ ⎜L AC ⎝ ⎠ ⎛ LCB L AC ⎞ ⎟ + P = FB ⎜ ⎜L ⎟ ⎝ AC L AC ⎠ ⎛ LCB + L AC ⎞ ⎟ P = FB ⎜ ⎜ ⎟ L AC ⎝ ⎠ ⎛ L ⎞ ⎟ P = FB ⎜ ⎜L ⎟ ⎝ AC ⎠ ⎛ L AC ⎞ FB = P⎜ ⎜ L ⎟ ⎟ ⎝ ⎠ 26 . two unknown axial reactions occurs. • the relative displacement of one end of the bar with respect to the other end is equal to zero since the ends supports are fixed. FA = P − FB • Realizing that the internal force in segment AC is +FA. since the equilibrium equation are not sufficient to determine the reactions. δ= PL AE δA + δB = 0 FB + FA − P = 0. as shown in fig. the equation can be written as. δA / B = 0 • the relationship between the forces acting on the bar and its changes in length are known as force-displacement relations δ A / B = 0. and the force equilibrium equation becomes. . CB the internal force is –FB. ΣF = 0. and in segment CB.28/02/2012 ` If a bar is fixed at both ends. Hence. +↑ y FB + FA − P = 0 • In this case. (a). ..(1) FB = 20(103 ) − FA δB / A = 0.......8m ) = 3927.39kN 27 ... − FA − FB + 20(103 )N = 0..4m ) FB(0..4m ) − FB (0..0025m )2 ⎤ ⎡ 200 × 109 Nm −2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ or FA (0.0025m )2 ⎤ ⎡ 200 × 109 Nm −2 ⎤ ⎡ π ( 0.0N...8m ) − = 0.0N FA = 16.4m ) − ( 20.001m AE AE FA (0.001m ⎡ π ( 0..2: + → ΣFX = 0.001m FA L AC FB L CB − = 0.....001m δA − δB = 0...28/02/2012 Example 2..( 2) Substitute eq (1)int o eq ( 2) FA (0...8m ) = 3927... 000N − FA )(0....6kN FB = 3. 5 ⎢ ⎥ + 0 . 3FA + 0 . .. .5 ⎢ ⎥ + 0 . 2 ) + F E ( 0 .4 δC − δE δ − δE = A 0 . . CCW + ∑ M C F A + F C + F E − 1 5 ( 1 0 3 ) N = 0 . .. . ( 2 ) The applied load will cause the horizontal line ACE move to inclined line A’C’E’ δ − δE δA − δE = C 0 .4 0.. (1 ) = 0 − F A ( 0 . .8 δ − δE δC − δE = A × 0 . . . ⎢ ⎢ 2 .. .. .. 5 × 10 − 5 E st 1 .28/02/2012 Example 2.4 0. 33 × 10 3 FC = 10 × 10 3 FA + 10 × 10 3 FE FC = 33 . 4 ) + 1 5 (1 0 3 ) ( 0 .8 0 .8 0 . 5 × 10 E st ⎥ ⎢ ⎦ ⎣ 2 . .eq ( 3 ) 28 ..5δ A + 0 .. 5 × 10 E st ⎣ 10 × 10 3 FA + 10 × 10 3 FE ⎤ ⎥ ⎥ ⎦ ⎤ ⎥ ⎥ ⎦ 33 . 4 ) = 0 ... . 5 × 10 − 5 E st FC ( 0 .5 ⎢ −5 −5 ⎢ 2 . . . .4 δ A − 0 . . 5 ) ⎡ FA ( 0 ... . ...4 δ E δC = + δE 0 . 3FE . .5δ E 1 . . . .8 δ C = 0 . 5 ) FC L CD ⎡ FA L AB ⎤ ⎡ FE L EF = 0 . 5 × 10 E st st ⎥ ⎣ ⎦ ⎣ ⎡ FE ( 0 . . 5 ) ⎤ = 0 . ..3: Solution: + ↑ Σ Fy = 0..5 ⎢ −5 −5 2 5 × 10 E . 33 × 10 3 FC = 0 . . ... generally a material expands.52(103 ) = 2.4) + 15(103 )(0..3FE ...3(9.3FE = 15(103 ) FE = 15(103 ) − 1.52kN int o eq ( 4) = 9.28/02/2012 + ↑ ΣFy = 0.3( 2....... and the material is homogenous and isotropic.4) = 0 .4) + 3(103 ) + 4....52kN FE =11.. 3 Substitute eq ( 4) int o eq( 2) − FA (0..... If the temperature increases..3FE )FE − 15(103 ) = 0 1..eq( 4) Re place FA = 9.....3FA + 1......(1) FA + (0..4) + 3(103 ) + (0..... the material will contract.615(103 ) − 0....615103 −0..3FA + 0.4FA = 0 FA = −7.2) + FE (0..02 kN Re place FE = 2.538(103 ) − FA ...8 = 9.538(103 ) − FA = 11.3FA + 0...4) + 15(103 )(0.4) ⎡11..3FA 1.. whereas if the temperature decreases. Where 1/C°) δT=αΔTL α=linear coefficient of thermal expansion (unit: ΔT=change in temperature L=original length of the member δT=change in length of the member 29 .2) + FE (0.538(103 ) − 9...02 kN int o eq(3) FC = 0. If this is the case.462 kN ` ` ` A change in temperature can cause material to change its dimensions..eq(3) Substitute eq (3) int o eq(1) FA + FC + FE − 15(103 )N = 0...........3FA + 0.02 × 103 ) = 3..538(103 ) − FA ⎤ = 0 ⎣ ⎦ − FA (0..4) = 0 − FA (0..(1) − FA (0.. it has been found from experiment that the deformation of a member having a length L can be calculated using the formula.......3FE = 0... FA + FC + FE − 15(103 )N = 0.( 2) CCW + ∑ M C = 0 FC = 0......519(103 ) = 9.519(103 ) + 0..52kN FE =11. 4: Given: α=12x10-6/C° Solution: + ↑ ΣFY = 0 FA = FB = F δAB = 0 ( + ↑) The change in length of the bar is zero (because the supports do not move) To determine the change in length.28/02/2012 Example 2. δAB = δT − δF So the bar will elongate by an amount δT when only temperature change is acting And the bar shortens by an amount δF when only the reaction is acting 30 . remove the upper support of the bar and obtain a bar is fixed at the base and free to displace at the upper end. 012 ( 200 × 109 ) = 7.6 × 10−4 × 0.012 ( 200 × 109 ) =0 12 × 10−6 ( 60° − 30°)(1) − 3.28/02/2012 ( + ↑) δAB = δT − δF δT − δ F = 0 αΔTL − FL =0 AE F(1) 0. = 72 MPa A 0.2kN Average normal thermal stress: σ= F 7.012 Example 2.1 × 109 Pa 31 .2kN .6 × 10−4 = F(1) 0.5 Given: αst = 12 × 10−6 / °C α al = 23 × 10−6 / °C E st = 200 × 109 Pa E al = 73.012 ( 200 × 109 ) F = 3. .........1×109 ) Fal (0..02)2(200 ×109 ) = Fal (0....432Fal = 421.eq (2) (+ ↑) δst = (δst )T −(δst )F δal = (δal )T −(δal )F (δst )T −(δst )F = (δal )T −(δal )F αΔTL − Fst L FL =αΔTL − al Ast E AalE Fst (0.45 ×10−4 − −4 π(0..21×10−9 Fal − 9..8 ×10−4 Fst = 1.89 × 103 ) = − 16.25) π(0....45 ×10 − 9.947 ×10 Fst = 3..25) (206.28/02/2012 + ↑ ΣFy = 0.76 × 103 Fal = 122. THE STEEL POSTS ARE IN TENSION and ALUMINIUM POSTS IS IN COMPRESSION 32 ...216Fal = − 165.03)2(73.45 ×10−4 −1.....88×103 +1..216(122.8 ×10 −4 Fst (0...25) 12 ×10−6(80° − 20°)(0..445 kN The negative value for F steel indicates that the force acts opposite to arrow shown.......216Fal ..76 × 103 + 2.eq (3) Substitute eq (3)int o eq(1) 2Fst + Fal − 90(103 )N = 0 2( −165.21×10−9 Fal −1...8 ×10−4 − 1..25) − 1..88 × 103 + 1...88 × 103 + 1.eq(1) δst = δal.327 ×106 ) −10 = 3.947 ×10−10 Fst = 3.88 × 103 + 1..21×10−9 Fal − 9....216Fal ) + Fal − 90(103 )N = 0 − 331..25) 251.947 ×10−10 = −165.25) − 23×10−6(80° − 20°)(0.685 ×106 ) −1..65 ×10−4 −1.89 kN Substitute Fal = 122.432Fal + Fal − 90(103 )N = 0 3.89 kN int o eq (3) Fst = − 165..... 2Fst + Fal − 90(103 )N = 0. assuming a close fit at both supports before the loads are applied. are joined at C and restrained by rigid supports at A and D. RA= 323 kN.9kN. CD made of steel (E=200 GPa) and AC made of aluminum (E=72 GPa).1 kN) (b) The deflection of point C (0. RD= 87.086 (0 086 mm) 67 33 .28/02/2012 TUTORIAL 1 Determine the reactions at A and B for the steel bar and loading shown. Determine (a) the reactions at A and D (RA=52. Rb= 577kN 66 TUTORIAL 2 Two cylindrical rods. Answer. determine (a) ( )The normal stress in the aluminum rod (σa =-150. At a later time when the temperature has reached 1600C.28/02/2012 TUTORIAL 3 At room temperature (21oC) a 0.5 mm gap exists between the ends of the rods shown.369 mm) 69 34 .6 MPa) (b)The change in length of the aluminum rod (δa= 0.