302 
[Sess. 
Proceedings of the Royal Society of Edinburgh, 
and the stresses are 
rr— Pr[(cos 2 a — sin 2 #) sin # — tui 3 a(cos 2 # - sin 2 a) cos #]/4m sin 3 a 
— p^r[s(cos 2 a - sin 2 #) sin # — c(cos 2 # - sin 2 a) cos 0 + 2 cos (y 3 - #)]/2m 
+ mVr sin (a + #)/m sin 2a , 
## = — Pr[(3 sin 2 a - sin 2 #) sin # + tan 3 a(3 cos 2 a - cos 2 #) cos 0\im sin 3 a , 
+ p^r[s(3 sin 2 a - sin 2 #).sin # + c(3 cos 2 a - cos 2 #) cos # - 2 cos (/3 - #)]/2m f 
+ mPr sin (a + #)/m sin 2a , 
r# = Pr[(sin 2 a - sin 2 #) cos # + tan 3 a(cos 2 # - cos 2 a) sin #]/4m sin 3 a 
- p|gs(sin 2 a - sin 2 #) cos # + c(cos 2 # — cos 2 a) sin #]/2m , / 
where 
8 = sin /?/sin 2 a, c = cos /3/cos 2 a . 
The stresses exerted across unit area of any section are 
rr = m rr - m p , 
## = m 60 - mp , 
and 
(36) 
r# = m r# • 
These give exactly the same resultant stresses across any plane as in 
the case of a non-porous dam. The interpretation of this result appears 
to be clear. 
(13) If it is assumed that the front face is vertical, or /3 = a, the 
conditions that rr should vanish at the front face when the dam is full 
of water are as follows : — 
Por a porous dam . . . cot 2 2a =p/p-m. 
For a non-porous dam . . cot 2 2a = p/p. 
It should perhaps be remarked that the value of p in the porous 
dam would in general be different from p for the non-porous dam. 
(14) Displacements. 
The movements for a wedge of isotropic material acted on by forces 
have been calculated for three cases : — 
Case I. Bending due to pressure Pr/2 on 0 = a and tension Pr/2 on0 = - a. 
Case II. Compression due to pressure Pr/2 on both faces. 
Case III. Displacements due to the weight of the wedge. 
The displacements are in each case calculated on the assumption that at 
x — h and y = 0 the wedge is held fixed, so that 
In Case I. 
u — v = du/dy = 0. 
9 iiu = — 
2 pu g 
2 m 
X + 2p , v 
c ' + 2(XT/I) Ca ~ Cl _ 
(h - x)y 
2pv = - 
4w 
/ V‘C“'/ i L (c - c-,)(x 2 + h 2 - y 2 - 2 hx) + cry 2 + cd?>h 2 - 2 hx - x 2 ) 
2(A + /a) 
h (37) 
