297 
1912-13.] Plane Strain in a Wedge. 
water pressure ; taking the weight of a cubic foot of water as the unit, we 
readily obtain 
{xx + yy)/ 2= -0-9852x + 0*7570t/. 
xx-yy ==) - 0*7430x -f 3*8435?/ . 
2 xy = 0-9072# - 166 15?/ . 
If it is found what these become at the points, say, y — 0, y = ±^/10, 
2/ = =±=2^/10, 2/ = ±3^/10, and y = zhx tan a, we shall then have the stresses 
at nine points across the section of the wedge. The results are as follows : — 
y • 
x tan a. 
Bx/10. 
o 
o< 
x/10. 
0. 
— xj\ 0. 
-2^/10. 
- 3a?/ 10. 
- x tan a. 
pjx 
-1-0512 
- 1-0471 
-1*1218 
- 1 *3255 
-1*5712 
- 1*8389 
-2*1136 
-2-3923 
-2-4622 
p 2 /x 
-0-4272 
- 0-4691 
-0*5458 
- 0-4935 
-0*3992 
-0-2829 
-0-1596 
-0-0323 
-0*0002 
0 
OO 
o 
O 
O 
22° 28' 
43° 42' 
32° 14' 
25° 19' 
21° 48' 
19° 40' 
18° 16' 
OO 
o 
Those values of <p under the line are measured from the axis of x, the 
others from the axis of y, the axis being determined by the change of sign 
of 2 xy/(xx — yy). 
The maximum shear in the wedge occurs at the outer face, and is easily 
seen to be rr/2 at that face. 
Maximum shear = - J[P r cos 2a/sin 2 2a + pgr sin (a - /?)/ sin 2a] . . (29) 
This shear acts on those planes parallel to the z axis which cut the back 
face at 45°. 
(10) From the values of <p just obtained the curves along which the 
shear vanishes can be drawn. Each curve cuts one face normally, bends 
round and becomes practically parallel to the other face. If the stress 
functions are transformed to new axes having the axis of x in the face 
0 = a, the axis of y being drawn into the wedge, then the nature of the 
curve in the region of this face becomes amenable to algebra. The stress 
functions for pressure Pr on 6 = a and the weight respectively are 
aXp = - j^[# 3 - 3 XI/ 2 cot 2 2a 4- 2 y 3 cot 3 2a] , j 
a-Xg — COs(a - /3) + §xy 2 cB 2 2asin(a - /3) ^ ^ 
+ ?/ 3 {cos /3 sin a(3 - tan 2 a) - sin /3 cos a(3 - cot 2 a)}] . , 
Similarly, if the plane y = 0 is taken as lying in the face 0——a 
we have 
-aXr = - P[3 xy 2 sin 2a cos 2a - y 3 ( 3 cos 2 2a - l)]/6 sin 2 2a , ) 
-aXg = if [2^ 3 COS (a + J3) + 6xy 2 sin (a + f3) j- (31) 
+ y 3 {sin /3 cos a(3 - cot 2 a) + cos /3 sin a(3 - tan 2 a) }] . J 
