﻿36 Dr. Dorothy Wrinch on the 



giving a solution or* the form 



Vi^a^ + bjr-^r 2 



and b 1 = 0, since t] y is not infinite at the axis. The boundary 

 conditions determine the constants a : and b x ; for the radial 

 traction T x is given by 



T ] = (X +2 ,)(J-1) + X(^1) 



= q [{\ + 2 f L)d Vl /dr + \r ]l lr]\ 



and the radial traction must vanish over the two boundary 

 surfaces. Thus 



(X + 2yLt) dffifdr | r =a +Xrj 1 /'r '< r =a =0. 



The constant a x is therefore determined by the equation, 



(\ + /x)a 1 = (2X + 3,a)a-78, 

 giving 



T 1 = ? [2(X + /A )a 1 ~r 2 /4(2X+,>)] 



= 7(2X + 3/z)(a 2 -r 2 )/4r 2 

 and 



7?1 = ^(a 1 r — ?' 3 /8) 



= ^[(2X + 3 / ,)(a 2 + & 2 )/ l X+/.)->- 2 ]. 



77i£ Effect of a Circular Hole in the Cylinder. 



If the cylinder at rest has two boundary surfaces r = a, 

 r=b (b<a) the solution stands in the form, 



Vl = (ll r-^b l /r~ir\ 



and the conditions for zero radial traction on both the 

 bounding surfaces yield 



(X + /x)a l — fjbbi/a 2 + (2X + 3/a) a 2 /S, 



(\ + fi)a l = /ib 1 /b' 2 i- (2\ + 3yu &-/8. 

 These give 



a 1 = (2X-I- 3/x) {a 2 + & 2 )/M(X 1 /*), 



b\={2\ + 3fM)a 2 b 2 l${i. 

 Thus 



T 1 =g[2(X + /A )a 1 -2 / ,/Vr 2 -(2X + 3^> 2 /4] 



= 7(2x + 3/x)(a 2 — ?•'-') (r 2 — b 2 )/-kr 2 . 



