Plane Strain in a Wedge. 
O 
295 
1912-13.] 
(7) For pressure Pr on 0 = a and tension Pr on 0 = — a put 
X = - Pr 3 (cos a sin 30-3 cos 3a sin 0)/6(cos a sin 3a - 3 cos 3a sin a) 
= - Pr 
giving stresses 
>s a sin 30-3 cos 3a sin 0)/6n , say , . 
■ (15) 
fr = Pr(cos a sin 30 + cos 3a sin 6)/m , ^ 
00 = - Pr(cos a sin 30-3 cos 3a sin 6)/m , \ 
• (16) 
rO = Pr(cos a cos 30 - cos 3a cos 0)/m . J 
(17) 
(18) 
For pressure Pr on both sides of the wedge put 
X= - P/ ,3 (sin a cos 30-3 sin 3a cos 0)/6(sin a cos 3a - 3 sin 3a cos a) 
= - Pr 3 (sin a cos 30 - 3 sin 3a cos 0)/6m , .... 
giving stresses 
fr — Pr(sin a cos 30 + sin 3a cos 6)/m , 'i 
00 = - Pr(sin a cos 30 - 3 sin 3a cos 6)/m , V 
rO = - Pr(sin a sin 30 - sin 3a sin 0)/m . J 
To obtain the case of pressure Pr on 0 = a, the face 6= — a being free 
from stress, take half the sum of x as given by equations (15) and (17). 
After some reduction there is obtained 
y = - Pr 3 [(3 sin 2 a - sin 2 0) sin 0 + tan 3 a(3 cos 2 a - cos 2 0) cos 0]/24 sin 3 a, . (19) 
giving stresses 
fr = Pr[(cos 2 a — sin 2 0) sin 0 - tan 3 a(cos 2 0 — sin 2 a) cos 0]/4 sin 3 a , 'j 
00 = - Pr[(3 sin 2 a — sin 2 0) sin 0 + tan 3 a(3 cos 2 a - cos 2 0) cos 0]/4 sin 3 a. 
rO = Pr[(sin 2 a - sin 2 0) cos 0 + tan 3 a(cos 2 0 - cos 2 a) sin 0]/4 sin 3 a . 
(8) Weight of Wedge. 
If the weight of the wedge is taken into consideration, its central plane 
being placed at an angle /3 with the vertical, we have for the stress 
equations 
i 
( 20 ) 
\ !•(»•") + 1 ? + M cos iP ~ 6 ) 
W 
and 
l 0 , ~ i,i ^ 
r6) + rde eo 
+ prj sin (/3 - 0) = 0 . 
If we assume 
- 7 ' 1 d\ 1 d 2 x a ' 
r r = v _i A + A , 
rde r*d0 2 
Te aA'- 1 - 
and rO 
d(ldx 
0r\ r 80 
( 21 ) 
( 22 ) 
and substitute these values in equation (21) and differentiate out the values 
of A' and IT can be obtained. A solution is 
A' = IV = - pgr cos (ft - 0) . 
Again, if we assume 
X = pgr 3 ( A sin 30 + B cos 30 + C sin 0 + 1) cos 0) 
(23) 
