LIMITING STRENGTH OF ROCKS UNDER STRESS 125 
U'=A'r' +B’ /r’ 
5 / 
ry =2(\' tp!)\A’— = Bl 4 Ne! 
phe, 
7 
(4) 
66 =2(N’ +p!) A’ +7 BIEN! 
ga = 2d! A’ +(r’ + 2p’ Je! 
Let a be the radius of the cylindrical cavity in the rock specimen, 
6 the radius of the specimen, and c the radius of the nickel-steel 
jacket. The boundary conditions give six equations to determine 
the constants A, B, e, A’, B’, e’. 
At r=a, rr=o, giving 
2(A-+-p)A — 2¢B/a?+de=0 (5) 
At r=r' =), rr=rr'’, giving 
2(A+p)A — 24B/b?+A\e = 2(A’ +p’) A’ — 2p’ B’/b?+N'e! (6) 
Nt —7 0 —— Oe soya: 
Ab+B/b=A'b+B'/b (7) 
At r’=c, rr=o, giving 
2(A’+p’) A’ — 2p’ B'/c?+2X'e' =o (8) 
In addition to these we have in the rock specimen, 22=—P, 
where P is the pressure per unit area applied to the end of the test 
cylinder; this condition gives 
2\A+(A+2m)e= —P (9) 
Also since the nickel-steel jacket is free from longitudinal stress, 
23’ =o or 2d’A’+(A'+ 2p’ )e’=0 (10) 
We denote by S the quantity, 
I+o BS I—o’ ye 
ee ° ie ] (4255, C2 
We then find from the six equations (5) to (10), 
B Sia o I 
Gua tao. 1228 
and (12) 
awe P g B 
oe er wieed| at 
Me al ha 
pie oe : (Garg) 
