A P P L I E D M E C H A N I C S I I 1. 1 I N T R O D U C T I O N

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Hilary Greene
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1 A PPLIED MECHANICS II 1. 1 INTRODUCTION T h e second aspect of the course on A pplied Mechanics deals with the internal stress and strain g e n e r a t e d by eternally applied forces. 1. STRESS AND STRAIN I n the study of stresses and strains, it is important that the s t udents have a good understanding of the concept of free body d i a g r a m s and equilibrium of a body under a set of applied loads which m u s t have been acquired in the first part of Applied Mechanics. This f a miliarity will be assumed in the following sections. I f a cutting plane X-X is passed through the member in Fig ( a ), it may be assumed that the internal forces generated is uniform a s shown in Fig.1.. The intensity of the force perpendicular or n o r m a l to the section is called a normal stress at a point. It is d e s i g n a t e d by the greek letter, σ ( s i g m a ) in this note. T h i s stress, if considered uniform over the cross- sectional area, A, a s is the case in most Engineering problems, is defined mathematically a s, N o r m a l stress Force( orload) Area P A [ N ] [ m ]...(1.1)

2 I t should be noted that stress is a vector quantity having both m a g n i t ude and direction.the sense of the stress is also important. It d e p e n d s on the action of stress on the body and may not be determined f r o m the sign of the force vector. If the stress is tending to stretch the b o d y or pull it apart, it is called a t e n s ile s t r e ss as in Fig.1.. If the b o d y stress is compressing or quashing the body, it is called a c o m p r e ssive s t r e ss. Using a sign convention, tension stress is c o n s i d e r e d positive, while compression stress carries a negative sign. T h e SI unit of force is the Newton and the unit of area is the meter s quare. The unit of stress is thus Newton per meter squared or Pascal ( P a ) as shown in Eq.1.1. O t h e r multiples of the Pascal, commonly used because of the fact t h a t the Pascal is a relatively small unit of stress are the KiloPascals ( K P a 10 3 P a ) 1kN/m M e g a P a s cals (1MPa 10 6 / P a ) 1 N / mm G i g a P a s cal(1gpa 10 9 / P a ), etc S H E AR STRESS I n general, the applied forced P, can be at an angle to the cross ection under consideration as shown in Fig.1.3. The force can then be r e s o l v e d into a normal component P y a n d a tangential component P. T h e normal component P y causes normalcauses normal stress as in F i g. 1.. The other component of the force acts parallel to the plane and t e n d s to shear the member. An average shear stress may then be e p r e ssed as a v P / A...(1.)

3 I t is to be noted that Eq.1. differs considerably from the true stress s i t u a t i o n, but is widely used in many engineering applications for m a n y practical reasons.the Greek letter, ( t a u ) is used to represent s h ear stress and the subscript A is used to indicate the average nature o f the stress being calculated. Again for practical purposes, the stress d i s t r i b u t i o n is assumed to be uniform BEARING STRESS T h i s is a special type of normal stress, occurring when two o b j ects come into contact or bear on each other. An eample of bearing s t r e ss is that eerted on the floor by a table leg. It isalways c o m p r e ssive in nature. E a m p l e 1.1 D e t e r mine the normal stress in a 100 mm square short column carrying a tensile force of 900 kn. S o l u t i o n : σ P A 3 N 900kN 10 kn 100mm 100mm N 90 mm 90MPa

4 E a m p l e 1. F i n d the bearing stress eerted on the floor by each leg of a t a b l e carrying a load of 140 kn. The table has four legs and weighs 0kN. The c r o ss-sectional dimension of each leg of the table is 0mm square. S o l u t i o n : T o t a l force kN F o r ce/leg 160kN 40kN 4 C r o ss-sectional area of each leg 0mm 0mm 400mm T h e r e f o r e bearing stress/leg P A 3 40X10 N 400mm 100MPa E a m p l e 1.3 A 00N load is carried by a 5mm diameter rivet as shown in fig Find the average shear stress in the rivet.

5 S o l u t i o n A free body diagram for the riveted bars is drawn as shown in Fig 1. 4 (b) S u mming all the horizontal forces, we have: T h e shear load,, is 00 N. Thus the average shear s t r e ss is F N A A D 4 (4)(00N) X10 (.05m) 3 4 D kn N 407.4KPa

6 E a m p l e 1.4: I f the plate were double riveted as shown in Fig. 1.5, find the a v e r a g e shear stress in the rivet. S o l u t i o n : F B T B N T h e free body diagram for the top plate is shown in Fig. 1.5b. F T T 10 0 N F r o m the free body diagram of the middle plate in Fig. 1.5c T h i s shows that the shear load on either side of the middle plate is 100 N. This condition is known as double shear and in it the rivet arries o n l y one-half of the total load. In single shear it carries the total load. T h u s average shear stress is

7 P L τ A A πd 4 3 ( 4 )( 100) 10 π ( 0. 5) kp a 4 ( ) πd T T T Deformation under aial load P 1. 5 NORMAL STRAIN T h e application of an aial tensile load P as in Fig 1.6 will normally result in the member stretching in the direction of the l o a d. This elongation per unit lengh of the member is referred t o as normal strain If is the total deformation in a given length, L, N o r m a l Strain, ( e p s ilon) / L I t is to be noted that strain is usually dimensionless and is g e n e r a lly very small ecept for a few materials like rubber.

8 E XAMPLE1.6 T h e unloaded length of a beam carrying an aial load of 60kN is 3m long. If the beam has a diameter of 50mm and was s t r e t c h e d by 0.00m, find the stress and strain in the beam. y y y (a) (b) y y L (c) S O L U T I O N F r o m equation 1.1 S t r e ss P/A A 3.14*(0.05) / m 60000/ S t r a i n, / L 0.000/ m/m 1. 6 SHEAR STRAIN C o n s i d e r a square element of a material in an undisturbed state as s h o w n in Fig. 1.7 (a). T o this element a shear stress i s applied to each face in the d i r ections shown, so that equilibrium conditions are met. The shear s t r e sses produce the shear deformation shown in Fig 1.7(b). Fig ( c ) shows the element rotated clockwise through the angle / s o t h a t the entire deformation can be easily seen as the angle ( Greek g a mma). This angle is defined as shear strain. It is measured in radians

9 w h i c h is a dimensionless term. T h e shear strain,, is calculated as δ tanγ γ......(1.4) L s i n ce for very small angles, tan. I t should be noted that whereas equation 1.4 looks very similar to e quation 1.3, a n d L are in the same direction in equation 1.3 but they a r e measured perpendicularly in equation1.4. However, in both cases t h e deformation,, is in the direction of the applied stress.

both an analytical approach and the pole method, determine: (a) the direction of the

Quantitative Problems Problem 4-3 Figure 4-45 shows the state of stress at a point within a soil deposit. Using both an analytical approach and the pole method, determine: (a) the direction of the principal