200 



Dr. J. Prescott on the 



from Case 2 only in having a thrust in addition to the 

 couples. 



Equation (79) is true in this case as in the last. In 

 equation (76) M is not zero for this case, and therefore the 

 equation corresponding to (80) is 



ec 3=-{£ +r > +m • • • «*> 



This is the same equation as for a strut with clamped ends 



(P 2 \ 

 jr — KR I . Also the rest of the conditions for 



the beam are the same as for the strut. Therefore, by 

 analogy with the strut, 



G 2 , ^ 4tt 2 E(J 

 tt +K= — y — 

 i\ n r 



(84) 



If a tension R' is applied instead of the thrust R the 

 equation becomes 



4tt 2 EC 





It should be observed that the two cases just worked out 

 can be regarded as solutions of the strut problem. Suppose 

 the two thrusts at the end are each applied at distance p 

 from the centre of the end section in the direction parallel 

 to the depth as in fig. 12. Then the couple G is Rp and the 



Fig. 12. 



~ T 

 P 



thrust at which instability begins for a pair of free ends is R 

 given by equation (81), that is, by the equation 



ivn 



EC 



7T 



(85) 



If Rp 2 is small compared with Kn we may use the approximate 

 equation 



R = EC^-£(EC5) 2 



= EC 



7T 2 (, TrVEC 



m 1 - 



VEC], 

 / 2 Kn J * 



