LECTURE 3
Consider a point ‘q' in some sort of structural member like as shown in figure below. Assuming that at point exist. ‘q' a plane state of stress exist. i.e. the state of state stress is to describe by a parameters s
This is a commen way of representing the stresses. It must be realize a that the material is unaware of what we have called the x and y axes. i.e. the material has to resist the loads irrespective less of how we wish to name them or whether they are horizontal, vertical or otherwise further more, the material will fail when the stresses exceed beyond a permissible value. Thus, a fundamental problem in engineering design is to determine the maximum normal stress or maximum shear stress at any particular point in a body. There is no reason to believe apriori that s
If the applied load P consists of two equal and opposite parallel forces not in the same line, than there is a tendency for one part of the body to slide over or shear from the other part across any section LM. If the cross section at LM measured parallel to the load is A, then the average value of shear stress t = P/A . The shear stress is tangential to the area over which it acts. If the shear stress varies then at a point then t may be defined as
Let ABCD be a small rectangular element of sides x, y and z perpendicular to the plane of paper let there be shear stress acting on planes AB and CD It is obvious that these stresses will from a couple ( t . xz )y which can only be balanced by tangential forces on planes AD and BC. These are known as complementary shear stresses. i.e. the existence of shear stresses on sides AB and CD of the element implies that there must also be complementary shear stresses on to maintain equilibrium. Let t' be the complementary shear stress induced on planes AD and BC. Then for the equilibrium ( t . xz )y = t' ( yz )x
Thus, every shear stress is accompanied by an equal complementary shear stress.
A plane stse of stress is a 2 dimensional stae of stress in a sense that the stress components in one direction are all zero i.e s examples of plane state of stress includes plates and shells. Consider the general case of a bar under direct load F giving rise to a stress s The stress acting at a point is represented by the stresses acting on the faces of the element enclosing the point. The stresses change with the inclination of the planes passing through that point i.e. the stress on the faces of the element vary as the angular position of the element changes. Let the block be of unit depth now considering the equilibrium of forces on the triangle portion ABC Resolving forces perpendicular to BC, gives s but AB/BC = sinq or AB = BCsinq Substituting this value in the above equation, we get s
t again AB = BCcosq t
If q = 90 By examining the equations (1) and (2), the following conclusions may be drawn (i) The value of direct stress s (ii) The shear stress t (iii) The stresses s
Consider the element shown to which shear stresses have been applied to the sides AB and DC
Complementary shear stresses of equal value but of opposite effect are then set up on the sides AD and BC in order to prevent the rotation of the element. Since the applied and complementary shear stresses are of equal value on the x and y planes. Therefore, they are both represented by the symbol t Now consider the equilibrium of portion of PBC
Assuming unit depth and resolving normal to PC or in the direction of s s = t
PB/PC = sinq BC/PC = cosq
s s (1) Now resolving forces parallel to PC or in the direction t -ve sign has been put because this component is in the same direction as that of t again converting the various quantities in terms of PC we have t
the negative sign means that the sense of t From equation (1) i.e, s The equation (1) represents that the maximum value of s Let us take into consideration the equation (2) which states that
It indicates that the maximum value of t From equation (1) it may be noticed that the normal component s Hence the system of pure shear stresses produces and equivalent direct stress system, one set compressive and one tensile each located at 45
Now consider a rectangular element of unit depth, subjected to a system of two direct stresses both tensile, s
for equilibrium of the portion ABC, resolving perpendicular to AC s converting AB and BC in terms of AC so that AC cancels out from the sides s Futher, recalling that cos Similarly (1 + cos2q)/2 = cos Hence by these transformations the expression for s = 1/2s On rearranging the various terms we get
Now resolving parallal to AC s The – ve sign appears because this component is in the same direction as that of AC. Again converting the various quantities in terms of AC so that the AC cancels out from the two sides. (4)
The following conclusions may be drawn from equation (3) and (4) (i) The maximum direct stress would be equal to s (ii) The maximum shear stress in the plane of the applied stresses occurs when q = 45 |