Buckling of Deep Beams. 211 



The condition to be satisfied at the free end of the beam is 

 that the torque is zero, that is, 



— =0, where # = 0.' 

 ax 



This condition is satistied by the value of t in (126) if p 

 has the value given by (133). 

 The other condition is that 



t=0 where x = L 

 Let 



p — — rbxF(m 2 x 2 ). 

 Then we have to solve the following equation for m 2 l 2 



bf(m 2 l 2 )-rblF(m 2 l 2 ) = 0. . . . (134) 

 Let m 2 l 2 = s-hv 



where /(*)=0, (135) 



that is, from equation (25), 



5 = 4-012 (136) 



Then f(s + v)-rlF{s + v)=0, 



or f(s) + vf(s)-rlF{s) = .... (137) 



on neglecting v 2 . rv, and small quantities of higher orders. 

 Equations (135) and (137) give 



f {*) 



After the necessary arithmetic we arrive at the results 

 F(«)= 0-1014x5, 



/ v ( 5 ) = -0-0869x5. 

 Therefore v= -1-167 rl 



— 1-1672? 



r o 



W 

 = -1-167™ , 



r o 



W being the total distributed load, and P the buckling 

 load when "W is zero. 



i (> 2 



