Buckling of Deep Beams. 199 



Integrating again we get 



Knr=-G//, (79) 



no constant being added in this case because both r and y 

 are zero at the ends. The negative sign on the right side is 

 due to the fact that t is negative by our convention. 

 Now equation (76) gives, since M is zero, 



eo S=-{£ +r >' • • • <*» 



which is the same equation as for a strut under a thrust 



/G 2 \ 



I p — \- R I . By exactly the same reasoning as in Euler's 



theory of struts instability occurs when 



£ +R )= EC S < w 



If the couple G is zero then we get Euler's value of the 

 thrust, namely 



77-2 

 R=EC-y2~, 



and if R is zero we get the value of the end couple G that 

 causes instability when no thrust acts, namely 



IT 



(: 



G= j VEnCK, 



which result has already been obtained in Case 1 in the first 

 paper. 



If the eud force is a tension of magnitude R/ we need 

 only put — R' for R, and then we find 



G 2 it' 1 



fc- R '= E( ^ ( 82 > 



From this it follows that a very big couple could be 

 applied at each end and the beam would remain stable 

 provided a suitable tension R' is also applied. That is, the 

 tension wholly or partially neutralizes the effect of the end 

 couples in producing instability, while the couple G weakens 

 the beam when used as a strut. 



Case 9. — Beam under the same forces as in the last case 

 with the addition of a pair of couples in horizontal planes 



applied at the ends to keep -j- zero at those points. 



This corresponds to a strut with clamped ends, and differs 



