PLANE STRAIN IN BIPOLAR CO-ORDINATES. 
289 
We must now choose the coefficients in (66) so as to satisfy (48) and (49), and 
there is no difficulty in finding the following values for the coefficients :— 
Ai = ^e -2a, sech 2a,. 
B 0 = sech 2a x 
» _ _ n 2 sinti 2 ^ — n sinh cl x cosh a 1 + e~ nai sinh ?iot x 
2 {sinh 2 na. x —n 2 sinb 2 ^} 
(67) 
( 68 ) 
E. = 
n sinh 2 aj 
(69) 
2 {sinh^aj — n 2 sinh 2 a T } ' 
Substituting in (66) we have for the complete stress-function, 
hx = «T a sech 2a x (cosh a —cos /3)+|- sinh a + sech 2aj cosh {2a x — a) sinh a cos /3 
\n sinh a x sinh (a — a x ) sinh na 
—sinh a sinh n (a — a x ) sinh naj cos n/3 
sinh 2 wa]— n 2 smlra. 
We may now calculate the stress /3/3 in the boundaries by means of (6). We find, 
on the circular boundary a = a l5 
/3/3 x = 2T (cosh aj — cos /3) i sinh a x sech 2a x + 2 M n cos n/3? . . . (70) 
+ 2 
71 = 2 
where 
n = 2 
iyr _ n (n — l) sinh (n+ 1) — n (n + l) sinh (n— l) ol x _ v 
2 {sinffinaj—n 2 sinti 2 ^} 
The stress in the straight boundary cannot be directly determined from (69), for it 
is found that the resulting series diverges for a = 0. We can, however, find without 
difficulty from (66) that when a = 0, 
/3/3 0 = T j 1 + (l — cos 0) 2 P n cos n/3 j.(72) 
where P n = 4nA re . 
The series in (70) converges only slowly, unless ot x is large, and for convenience in 
computation we may transform it by separating the more slowly converging part. 
Let 
M„ = 2 n (n sinh a! — cosh a x ) e~ na ' + N„,.(73) 
and we readily obtain 
2 sinh 2 a x sin 2 /3 
2 (cosh <x x — cos /3) 2 n (n sinh a. x — cosh a x ) e~ na ' cos n/3 = 1 — 
n = 1 
(cosh a. x — COS /3) 2 
Substituting in (70) we have 
/3/3 1 = 2T 1 — 
2 sinh 2 a x sin 2 /3 
(cosh a,—cos /3) 2 ' 
+ 2T (cosh a x —cos /3 )\sinh a x sech 2a x + 2e~ 3a ' cos /3+ 2 N n cosn/3}>. (74) 
2 s 
n = 2 
VOL. CCXXI.—A. 
