Engineering Rock Mechanics

404 Appendix A: Stress and strain analysis

Cancelling the S2 term, and remembering that zq = ryx, then

ax! = a, cos2 e + a,, sin2 e + 2rv sin ecos e.

Notice that each term has a trigonometric identity of order 2 associated

with it: this is because in the transformation force is resolved once and area

is resolved once.

CFg = 0

• (q S~COS elsin e + (ay S2sin e)cos e +(zq S~COS e)cos e -
  (zy. S2sin e) sin e - rx3, a2 = 0.

Again, cancelling the i? term, and puttirtg zyx = T~, we find that

zxy = - ox cos 8 sin 8 + ay sin e cos e + zq cos2 e - rv sin2 e

or

Zx3' = rq (cos2 e - sin2 e) - (ox - cy) sin e COS e.

Determining ayt can be done either by cutting a prism parallel to the

x-axis, or simply by replacing 8 with (0 + d2) and ox* with ayt in the

expression for 0,. on the previous page (this is valid because we know cy*

is perpendicular to ox,):

sin [e + (n/2)] = COS e and COS [e + (n/2)] = - sin 8

so

ox. = a, COS' e + oy sin2 e + 2rV sin e cos e

becomes

oYr = 0, sinz e + cy COS^ e - 2rq sin e COS e

so the three equations are:

ox.= O,COS~ e+ sin^ e+ 2 rqsin ecos e

oYr = 0, sin2 e+ cos2 e- 2 rq sin COS e

zx3< = rq (COS' e - sin2 - (ox - oy> sin e cos e.

stress transformation equations