Budding of Deep Beams. 



313 



Now putting 

 we get 



d 2 r m 2 / 6 2 



where 





= - c V-i; 



r_ 16 2 E^CK 



(62) 

 (63) 



We want a solution of this last equation which will make t 

 a maximum at the middle of the beam, where 5 = 0, and t zero 

 at the ends, where s= ±1 or where * 2 =1. 



Assuming 



T = <2 + , 



a 4 ** f a 6 s 6 + 



(64) 



and substituting in the differential equation we get 

 2a 2 -f 4.3a 4 s 2 + 0.5tf 6 * 4 + 



= - Q ^ s ^2s 2 +l)\a + a 2 s 2 + a^ + a 6 s Q + ...,}. 

 Equating coefficients of like powers of s we find 

 2a 2 = — c 2 a , ~] 



4 . 3a 4 = — c 2 (a 2 — 2a Q ), j 



6 . 5a 6 = — c 2 (« 4 — 2a 2 + a ), ^ . . (65) 

 8 . la 8 = —c 2 (aQ — 2a± + a 2 ), I 

 etc. J 



By means of these equations each of the coefficients can 

 be expressed as the product of a and a function of c. 

 Thus 



- «o| 



V 6-V + 4) A 

 2 |4 



_....}. 



(66) 



Since this involves only even powers of s it follows 

 that t must be either a maximum or a minimum when 

 8 = 0, and if we choose the proper value of c then t will be 

 a maximum. 



To make t = when s=+l we have to satisfy the 

 equation 



o = i-| + 



c-V + 4) 



or 



= 1 + ^+-* + ^+.... 



a a a Q 



(67) 



(68) 



a 9 a< 



The terms — , — , etc. are functions of c 2 only, the 



a a 



Phil. Mag. S. 6. Vol. 36. No. 214. Oct. 1918. Y 



