CONCEPT OF STRESSSTRESS IN THE MEMBERS OF A STRUCTURE • The force per unit area, or intensity of the forces distributed over a given section, is called the stress on that section. • Denoted by Greek letter σ (Sigma) A P = o 2 9 9 2 6 6 2 3 3 2 2 / 10 10 1 / 10 10 1 / 10 10 1 ) ( 1 1 : . m N Pa GPa m N Pa MPa m N Pa kPa Pascal Pa m N m N Units SI = = = = = = = AXIAL LOADING : NORMAL STRESS AXIAL LOADING: NORMAL STRESS • The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy } } = = = A ave dA dF A P o o • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. A P A F ave A = A A = ÷ A o o 0 lim • The force intensity on that section is defined as the normal stress. • A uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading. • If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment. • A uniform distribution of stress in a section infers that the line of action for the resultant of the internal forces passes through the centroid of the section. • The stress distributions in eccentrically loaded members cannot be uniform or symmetric. Centric & Eccentric Loading • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • Forces P and P’ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. A P = ave t • The corresponding average shear stress is, • Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shear stress distribution cannot be assumed to be uniform. Shearing Stress A F A P = = ave t Single Shear A F A P 2 ave = = t Double Shear Shearing Stress Examples • Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect. d t P A P = = b o • Corresponding average force intensity is called the bearing stress, • The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin. Bearing Stress in Connections STRESS IN TWO FORCE MEMBERS • Axial forces on a two force member result in only normal stresses on a plane cut perpendicular to the member axis. • The forces acting on each end of member must have the same magnitude, same line of action, and opposite sense. STRESS ON AN OBLIQUE PLANE • Pass a section through the member forming an angle u with the normal plane. u u u u t u u u o u u cos sin cos sin cos cos cos 0 0 2 0 0 A P A P A V A P A P A F = = = = = = • The average normal and shear stresses on the oblique plane are u u sin cos P V P F = = • Resolve P into components normal and tangential to the oblique section, • From equilibrium conditions, the distributed forces (stresses) on the plane must be equivalent to the force P. MAXIMUM STRESSES • The maximum normal stress occurs when the reference plane is perpendicular to the member axis, 0 0 m = ' = t o A P • The maximum shear stress occurs for a plane at + 45 o with respect to the axis, o t ' = = = 0 0 2 45 cos 45 sin A P A P m u u t u o cos sin cos 0 2 0 A P A P = = • Normal and shearing stresses on an oblique plane Stress Under General Loadings • A member subjected to a general combination of loads is cut into two segments by a plane passing through Q • For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member. A V A V A F x z A xz x y A xy x A x A A = A A = A A = ÷ A ÷ A ÷ A lim lim lim 0 0 0 t t o • The distribution of internal stress components may be defined as, State of Stress • Stress components are defined for the planes cut parallel to the x, y and z axes. For equilibrium, equal and opposite stresses are exerted on the hidden planes. • It follows that only 6 components of stress are required to define the complete state of stress • The combination of forces generated by the stresses must satisfy the conditions for equilibrium: 0 0 = = = = = = ¿ ¿ ¿ ¿ ¿ ¿ z y x z y x M M M F F F ( ) ( ) yx xy yx xy z a A a A M t t t t = A ÷ A = = ¿ 0 zy yz zy yz t t t t = = and similarly, • Consider the moments about the z axis: Stress matrix in 3D= Quantity One dimension Two dimension Three dimension Scalar Vector Tensor 0 1 1= 1 1 1= 2 1 1= 1 3 3 = 2 3 9 = 2 2 4 = 1 2 2 = 0 3 1= 0 2 1= ( ( ( ¸ ( ¸ = zz zy zx yz yy yx xz xy xx o t t t o t t t o o Comparison of number of components FACTOR OF SAFETY stress allowable stress ultimate safety of Factor all u = = = o o FS FS Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material. Factor of safety considerations: • uncertainty in material properties • uncertainty of loadings • uncertainty of analyses • number of loading cycles • types of failure • maintenance requirements and deterioration effects • importance of member to structures integrity • risk to life and property • influence on machine function PROBLEM NO: 1 PROBLEM NO: 2 PROBLEM NO: 3 PROBLEM NO: 4