Buckling of Deep Beams. 299 



The couple G has also another component, of magnitude 

 G~, about the tangent at R, and this component twists 

 the beam just as it would twist a straight prism. Therefore 



*%5-*+ G J W 



The couple N is introduced as a twisting couple at the end 

 to maintain the upright position of the end section if any 

 couple is necessary for this. 



Eliminating y from (1) and (2) by differentiating (2) and 



d' 2 y 

 using thejvalue of -~ from (1) we get 



v cPt G 2 



whence di 2 



J? 



= -m?r, ...... (3) 



where 2 G 2 



"=Eril ..•■(*) 



The solution of this equation is 



T = Asinm# + Bcosm#. . . . . (5) 



The conditions to [be satisfied at the ends in the present 

 case, since we have assumed that the end sections are held 

 upright, are that 



r = when # = 



and when # = /, 



I being the length of the beam. 

 The first of these conditions gives 



B = 0, 

 and the second AsinmZ=0, 



which means that either A is zero, in which case the beam 

 is not buckled at all, or 



sin ml = 0, 



from which ml = 7r, 



that is, Gl = 7r\/En~CK (6) 



This gives the couple G for which the beam is unstable 

 6 X2 



