Collapse of Tubes by External Pressure. 

 where ^ denotes the quantity 



(n x -n 2 )a 



695 



nr 



EA 



m — 1 



A solution of (22) is of the form 



u= U sin &(£ cos \.£, 



v = V cos kcf> sin \# , !> (23) 



w=W sin &(£ sin Xx, J 



where U, V, W, k and X are constants, k being necessarily 

 integral. 



Substituting in (22) from (23), we obtain the following 

 determinantal equation for *$? : — 



1¥ 



^ 2 + ! T- 



Zm 

 m+1 k 2 



a 2 ' V 



2m 



m 



a 



v 



*+=£>'. 



\ ??2 / 



2ra a' 5 



a 2 ' 



i, 



1 -¥(*■- 1) 



0. (24) 



This equation, when expanded and simplified, gives 



^~ x ^ (*-u - * [(*-!>(*+ ?)'+ a + 



m^ 



mf 



\ 4 = 0. (25) 



But it is easy to see from its expression in (22) that M* 

 must be very small in any case of practical importance, so 

 that the term in "VP 2 may be neglected in equation (25) ; and 

 from the latter equation that M* is of the fourth order in 

 terms of X, so that X also must be very small. Hence we 

 may write with sufficient accuracy 



^ m 2 — 1 a 4 \ 4 



m 2 k\k 2 -l) J 



or 



n 1 -n 2 =2E- 



h a 4 \ 4 



(26) 



ak\k 2 —l)' ' ' 



Considering the problem again in general terms, we see 

 that the resistance offered by the tube to distortion will be 

 due partly to its extension and partly to its bending. The 

 complete expression for (Ili — II 2 ) will be of the form 



uh + /3h z + vh 5 + ... , . . . . (27) 



where «, (3, <y, etc. are functions of the radius and elastic 

 properties of the tube, and of k and X. In this paper we are 



3 B 2 



