Buckling of Deep Beams. 

 From equations (48) and (50) we get 



or 

 where 



dx' 



= — mW, 



4E/iCK 

 The solution of (52) in series is 



T = a | 1— 



+ a x x i 



311 



(51) 

 (52) 

 (53) 



5.6 



+ 



5.6.11.12 



6 m l V 



+ 



...} 



6.7 6.7. 12. 13 



f 



(54) 



At the free end, where x is zero, the twisting couple is 



LIT 



zero ; that is, j~ is zero. This makes aj=0. 



At the other end, where x = l, the twist t is zero. 

 Therefore 



mH 6 



m 12 Z 12 



m l8 l 



18718 



5.6 5.6.11.12 

 The smallest root of this is 



m«P = 41-30 : 

 that is, w 2 l 6 



5.6.11.12.17.18 



4E?iUK 



= 41-30, 



from which 



W/ 2 = 2v / 41-30i / EnCK 



= 12-86 VEnCK (55) 



Case 7. — The beam carries a total load W distributed as a 

 Fig. 6. 



'2\rA 



'' 2 w« 

 B 



Fig. 7. 



Plan of Central Line. 



uniform load w per unit length and is supported at the ends 

 as in Case 4. 



