Buckling of Deep Beams. 217 



satisfies the condition that t = where x = 0, which is one o£ 

 the conditions of the problem. 



Another condition is that the torque is zero at the middle. 

 This follows from symmetry, for clearly neither half can be 

 exerting a torque on the other. Then 



dp + dp =0 when ^ „ ^ ^ ^ (166) 

 ax ax 



that is, writing y for (^ml)*, 



!u u 2 u 3 ) 



1 -i + 4T5T»- 4.o.8.9.12 + -j 



rht (,, 10k 14w a 



+ a ^576r _ 479 + 4.8.9.13 



- 4.8.9 1 lfl3.1 7 + }=°- • • ( 167 > 

 Now let M w 2 



/( " ) = 1 -4 + 4. 5 .8— • 



4. 9^4. 8. y. 13 •-' 



then our equation is 



/(u)+^F.(u)=0 (168) 



Let us put w=5 f «, 



where s is the solution when W = 0; that is, 



/W = 0, (169) 



and, by equation (33) in the first paper, 



s=4-482 (170) 



Then, retaining only the first powers of v and r and no 

 products, equation (168) becomes 



or .» , rl 



•/(•)+ go P(«)-0. 



Therefore ,-l F(s) ,„„„ 



