Buckling of Deep Beams, 

 The other two conditions give 



= A sin ml + ^- (1 — cos ml) , 



= ^-sin ml— A (1 — cos ml). 



From these last two equations we get 



AM 



301 



AM 



G 



or 



sin 2 ml = — ^-(1 — cos?w/) : 

 sin 2 ml+ (1 — cos ml) 2 — 0, 



which can only be satisfied when the two following equations 

 are simultaneously true : 



sin ml = 0, 



1— cos ml = 0. 

 Therefore ml=2ir^ 



or Gl = 27r^/EnCK. . . . (13) 



Thus the buckling couple is twice as great as when the 

 ends were not constrained in the y direction. 



There is one condition we have not used in arriving at the 

 last solution, namely, that y has the same value at both ends. 

 If we do make use of this condition it only tells us that the 

 couple N is zero. The same is true for the first case we 

 dealt with. 



Case 3.— A beam is built into a wall at one end and is 

 Fur. 3. 



|p 



Fi2. 4. 



Pian of Central Line. 



quite free at the other. A load P is applied to the middle 

 of the section at the free end. For the equilibrium of the 

 whole beam in the buckled state there must clearly be a 



