Page 296 - Introduction to Continuum Mechanics

280 Plane Strain

(b) The stress components are

i.e., for the plane strain problem

(c) On the boundaries, KI - ±h/2, the tractions are

But, we wish the lateral surface fa - ±h/2) to be traction-free, therefore

On the boundary*! = 0,

This shearing traction can be made equipollent to an applied load P*2 by setting

where A = bh and / = b h?/l2. Substituting for ft, we have

3
Therefore, a = 2P/bh 'fi = -3P/2bh and the stresses are

In order that the state of plane strain is achieved, it is necessary to have normal tractions
acting on the side faces x-$ = ±6/2. The tractions are in fact t = ± ^363 = ±6 v a x.\ X2 ^

(d) Since Ty$ is not a linear function of the coordinates jcj and *2, from example 5.16.2, we
see that we cannot simply remove r 33 from the plane strain solution to arrive at a the stress