312 Dr. J. Prescott on the 



Here the origin is taken at one end A, and y is measured 

 towards the side to which the beam buckles. For the 

 point K on the central line Ar=x, Rr=y. 



A pair of couples N act at the ends in this case to keep 

 the twist zero there. 



The moment about Rr of the forces on the part AR is 



\wlx — \wx 2 . 



The component of this about the depth of the twisted section 

 at R is 



\wx(l — x)t. 

 Therefore 



E0g = -i«.^-*)T (56) 



The twisting couple at R due to the uniform load on AR 

 is expressed by the same integral as in the last case. The 

 twisting couple due to the force and couple at A is 



dr 

 In the present case r increases as x increases so that -j- is 



positive. The total twisting couple at R is 



Therefore 



KnJ=N->z(y-^g)+J^Q^^. . (57) 



^-i.j.^_i..jA 



Kn-j-s =iwlx~ — ^wx* j-^ 

 dx 2 2 dx 2 2 dx 1 



— 2 wx\i X) , 2 



=— i^ 2 (l-^ s T, (58) 



dx 



EC 



and consequently 



d 2 r 



^-mVQ-vVr, (59) 



where w 2 



m =mMK ( 60 > 



Putting X = x— \l, 



thus measuring X from the middle of the beam, our dif- 

 ferential equation becomes 



U = - m *(X*-iP)V (61) 



