Buckling of Deep Beams. 309 



Writing, for the sake of shortness in the argument, 



X = 1 -4 + 4.5.8 



4.5.8.9.12 



Y = 3_ i _i_ 



6 "^ 6.7.10 6.7.10.11.14 



^,..(45) 



V = 



+ 



3 4. 5. 7' 4. 5. 8. 9. 11 



W = 6-^ + 



5 6.7.9 6.7.10.11.13 

 equations (43) and (44) become 



a x X=--6*% . . (43) 

 a x V= foc»W. . . (44) 



By division 



X 



V 



Y 

 W 



or 



XW+YV=0 (46) 



Equation (46) has to be solved for s, and this will give the 

 critical load. i 



The smallest value of s satisfying (46) is 



5 = 10*47 approximately. 

 Therefore 



P 2 / 4 



= 10-47 



or 



64EnCK 



PZ 2 =8x 3-236 VE^CK 



= 25-89 VEnCK. . . . (47) 



Case 6. — The beam carries a load W uniformly distributed 

 Fig. 5. 



Plan of Central Line AB. 



along its length, is quite free at one end, and held rigidly at 

 the other. 



