THE EQUILIBRIUM OF AN ISOTROPIC ELASTIC PLATE. 221 
Further, with the values of F, , g, so determined, the equations (1), (2), (8) are 
all satisfied, the traction xz vanishes exactly, and the residual tractions nm 1s are of 
order 1. 
(ii) Tangential traction. 
S being of order 0, z¥ and at will again be of this order, but it will be seen that 9, 
is of order 1. For, making these suppositions, we have 
0 — — 29,F . . . . (1) | 
NS) ay 
on =— oto K + dz . “ . . (2) 
3 Z dy 
=HP-M)KE + pe + BieLZ(K2) i xe) 
From (1), S10 (0) : ; : (4) 
7 1 5 s2 2 d 
} From (2), ran Sdz= —4( -hW)3 B+ a : ; (5) 
Differentiate this with respect to s, and subtract from (3). Thus 
td ; se 
<5 | _Sde= M212) (IF +5, 99F) +39Ll02) (6) 
and, integrating the last from —/h to h, 
Chg Sd fh a 
JH ot 7s Dok = 4uh3 ae | pe : ; “ (7) 
(4) and (7) define F', and (5) integrated from 0 to z gives Y. With the values so 
determined, and with (6) satisfied by g, (and this is possible in virtue of (7)), the equa- 
tions (1), (2), (3) are all satisfied, the traction nz vanishes exactly, and the residual 
’ tractions nn, ns are of order 1. 
By combining this with the preceding case, we see that the results do in fact give a 
first approximation to the solution, since the residual stresses N , 8 are each of an order 
higher than their original given values. 
The additional equation required to define g, might be found by carrying out the 
process of (1) with the residual normal traction 4u = = . The analysis would be prac- 
tically the same as will be given in connection with the next case. 
(iii) Perpendicular traction. 
P 
Z being of order zero, we shall have to take zF and = of order —1, gy, of order 0. 
On this hypothesis, we shall write down the exact expression for xz, and the terms of 
order —1 and 0 in yn and xs. We are then to have 
f _d db 
0 = —23,k . +2 5 = + SgeN(«z) ; ee) | 
2 NS rat 
0 = =25P 5 oy | +E Oe 
ee z —h*)$,F a S92 | 
yn =h(?- V) Is La at Je Z( KZ) ° ° (3) 
