306 Dr. J. Prescott on the 



Now using the last value of s in the s 3 and s 4 terms we get 



4.-4^3 4-48 4 



<- 20 > , = M0 + yn3- ».li f 13.16 

 = 240*814, 



5 = 20- ^240-814 

 = 20-15-518 

 = 4-482 (33} 



This is probably correct to the Inst figure. It follows that 



P 2 / 4 = 64 x 4-482 EnCK, 



PZ 2 = 16-94VEnCK (34) 



Case 5. — The beam carries a concentrated load at the 

 middle as in the last case, but the ends are constrained as in 

 Case 2 ; that is, a pair of couples act on the ends in a hori- 

 zontal plane preventing the ends from bending sideways. 



The difference between this and the last case is that thero 



is an unknown horizontal couple M at each end, and ~ is 

 zero at the ends. 



Measuring x from one end the equations of equilibrium are 







Ecg = -*P OT + M, . . . 



• (35) 







^%^^t-yY • 



• (36) 



From these we 



get 



Kn^-i-Pr^ 

 ' dx 2 ~ 2ra/ d^ 



-P 2 „ PM 



~4E(T' lT ! 2JEO*' 





or 





^ T 4 2 ^7 



doc 2 



• (37) 



where 





4_ P 2 



. (38> 





4EnOK' ■••■•■ 







PM 



2EnCK' 





