300 Dr. J. Prescott on the 



when the ends are constrained in no way except that the end 

 sections are maintained upright. 



Case 2. — The problem just worked out is similar to the 

 strut problem with pin joints at the end so that the direction 

 o£ the elastic central line is not fixed at the ends. In the 

 buckling question, just as in the strut question, we have also 



the case where -r is zero at the ends. In order to maintain 

 ax 



these end conditions there must be applied at the ends of the 

 beam another pair of couples, which we shall denote by M, 

 these couples acting in a horizontal plane. Then the equa- 

 tions for the equilibrium of the portion All become 



E0g=-Gr + M, (7) 



Kk^I = N + G^ (8) 



ax ax 



From these we get M 



»■--*(*-?)■ • • • • < 9 > 



the solution of which is , r 



r = A sin mx 4- x> cos mx + p- . 



The condition t=0 when x=*0 gives 



M 



B== -G- 



Therefore # ]y[ 



T = Asinm#-f p (1— cosm.t'). . . . (10) 



From (8) and (10) 



G~ = ~Knm I A cos mx 4- f1 sin mx \— N. . (11) 



We have still to satisfy the three conditions 



t = and ~ =0 when x = l 

 ax 



and -f- =0 when x = 0. 

 ax 



The last of these gives 



N = KnmA (12) 



Therefore 



G ~=Knm I -^smmx—A(l—cosmx) >. 



