Buckling of Deep Beams. 213 



Case 13. — The same problem as Case 12 except that the 

 load P on the end is small compared with the uniformly 

 distributed load W. 



Here, as in the last case, 



G=Fx + iwx 2 . 



\ wxj 

 = \w 2 x x ( 1 + — I approximately. 

 Therefore 



dx 2 ~ 4EnCKV W T 



=- m6 *i 1+ Sh • • • • (142 > 



^=4eSk 0*0 



Now let t = t 1 H- / 3, (144) 



where (\ >» 6 ^ 6 m 12 c£ 12 ") ,-, , ^ 



Tl=a t^SJS+ S. 6.11.12 — •)' • < 145 > 



and therefore *H = _ wVti (U6) 



J. 



Then equation (142) becomes 



V * 



4P 



^ 2 Ti d 2 p - .A 4P\ , 



= -mV{T 1+ — Tl + pJ, 



neglecting the product Pp since both factors are small. 

 By means of equation (146) this last equation gives 



f£ + mVp= -^ xhl . . . . (U7) 



ax* r iv 



Now if we differentiate through (146) we get 



^- +m 6A/=-4mVT 1 , . . . (148) 



dr 

 t/ being written for — ^ . 



