202 Dr. J. Prescott on the 



The origin being at the free end of the beam the bending 

 moment at x is 



G = P#, 



and therefore equation (75) gives 



rf 2 T 



ECKn~=-PV 2 T, 



dx 

 or d 2 r 



da? 



where ± P 



= — m 4 x 2 r, . . . . (86) 





* 



EraCK 



(87) 



This is exactly the same equation as for Case 3 in the first 

 paper. The only difference is in the end conditions. These 

 conditions are now 



t=0 where x — l, (88) 



— Kn-r- = torque 



= F(9 T +P) where a? = 0. : . (89) 



The negative sign is necessary because r decreases as x 

 increases. 



The solution of equation (86) in series is 



f m*x h m 8 x 9 ~] 



T=g t/-0- + 4.5.8.9 -J 



+ H l -HT4+ 3TOT8 i">- • • (90) 



Now condition (89) gives 



Kna=-F(qb+p), ..... (91) 

 and condition (88) gives 



+»{'-S+n?n ! --} ; - • • « 



Substituting the value of a from (9.1) in equation (92), 

 and writing 5 for mH*, we get 



Fljqb+p) t t I 



Km X 4.5^4.5.8.9 ••"( 



- 6 { 1 - 3 74 + 3^7T8- }= -< 93 ) 



