﻿Elastic Equilibrium under Surface Tractions. 505 

 We therefore find 

 2V = X(*+/+;/) 3 + X 44 (« 2 -4/;/) ■+ X 55 (& 2 -4^) + \ m (&-Aef) 



+ 2\ 56 (bc-2ae) + 2\+ 6 (ac-2bf) + 2\ ib {ab - 2cg) . (9) 

 and 



B B \ . „ . . B . B 



^1 = —2X561 -v. — 5^v — I + (^55 — "^66) W —^45 

 \OA.55 OA-66/ OA.5t5 



B^46 



T A-46 zrr . 

 0^45 



I shall now write, for convenience, 



^44 = v l-> ^b'o—V'2-) ^-66— ^35 ^<56 = — ^45 ^4d=— ^5) ^45= — ^6 



and 



2V=X(.+/+ i /) 2 + ^ 1 (a 3 -4/5/)4-v 2 (6 2 -%)+v 3 (c 2 -^/) 



+ 2v4(2ae-bc)+2v 5 {2bf-ae)+2v 6 (2cff-ab), . (10) 

 where 



n 9/B B \ , , \ B B . B 



\d^ 2 0^3/ 0^4 0^5 B^ 6 



The requirement that the solid should act as an elastic 

 solid under tractions has reduced the 21 eolotropic constants 

 in (3) to 7 constants of elastic equilibrium in (10). 



Accordingly, we find from (5), 



01 = ^(#2 + 03) — V.5^2 — "6^3, 

 </) 2 = ^ 2 (#2 + #l) — Vii? 1—^3, 



3 =^i + ^ a )-^i-v 5 ^ 2 . . . . (11) 

 The value of 2% x is deduced from X2 1 2 = 



2%i = ^(26*! -i- # 2 + 0s) — (>2 + v 3 )^i - ^^2 — ^3, with 



2%2 = V -5(^l + 20 2 + 3 ) — l/fifa — (>i + ^2)^2 — ^3, 



2%3= r e (0i + #2 + 2^3) -v 5 fi-^2- ("1 + ^2)^. (12) 



From the value of 2V given by (10), we rind the elements 

 of elastic stress, as usual, by differentiation, 



p = x e + (X-2v 3 )/+(X-2v 2 ^ + 2v 4 a, 



Q= (\-2vi)e + \f+(\-2v l )g + 2v b b, 



R=(\-2v 2 )e + (\-2v 1 )f+\g + 2v 6 c, 



S= 2v±e + Via — v 6 b — v b c, 

 T = 2v 5 f— v & a + v. 2 b — i/ 4 c, 

 U= 2v 6 (/ — v 5 a — V4& + 1/3C (13) 



