THE EQUILIBRIUM OF AN ISOTROPIC ELASTIC PLATE. 149 
‘Calculating the stresses by means of (5), we find 
oe a ae Be o = = (2«°h sinh «xh cosh xz — 2x?z cosh kh sinh xz)F 
2 2 
= a e 22 : "3 = 2x" (cosh kh + xh sinh kh) sinh xz-F — 2«°z cosh kh cosh «z-F 
Z a 
both of which vanish when z= = h. 
We have further in (20) a solution of the type 
(ii) b= —2F + 1ay2F . | 
F a funct f x, as Se) 
6= 2F—day?F - 2h22y2F a function of 2, y with 7 
From this, by (4), (5), 
da 
| —(a+1)(zF —42y7F) + 2(42 - h*z)y? HL j @3) 
“a; \ 
w= (a+ 1)(F — 422y?F) + 2(2 — A?) v?F 
- | 
za = 4y(2? — i vF 
zy = 4plz 12) — sro 
2z=0 ) 
The solution for unit normal traction on z=/ contains a strain of this type, with 
F = x(R) - 3/327uh’. 
Lastly, in (20) there is a solution of the type 
(ii1) d= -2F F a function of a, 
j= alt with y?F=0, 
giving 
=—-(at+ eee 
dx 
Fes he fat eg ROMO ; . (24) 
w= (a+1)F J 
Obviously this is merely a degenerate case of (ii). 
Again in (21) we have a series of solutions of the type 
(iv) p= —sinh kh cosh xz F(x, y) | where (vy?+«2)F=0 he 
6 = (sinh xh + 2h cosh kh) cosh xz F(x, y) J sinh 2xh + 2xh=0 } he 
In this, as in G), w=z=%=0onz=+h. 
(v) The persistent part of (21) is of the type 
o= F(a, y); O= — 3F (a, 4 i where y7F =0 l 
=~ 
. 26 
giving “u= (a = = v= (a - ee ; a = ip =) j ( ) 
