496 Profs. John Perry and W. E. Ayrtoii. [Dec. 11, 



If ct x is the inclination at the support Q, then 



* 1 =* + M cZ-S /+< 7l (9). 



Integrating (8) again we see that the deflection at any point P is 



y = X x + M n + ¥-S q (10). 



and, as the supports are on the same level, 



y=0 



when x=l } , 



so that — \a Q + + F x — S 2 l3 



and hence ^-^h-^i^Jh (H). 



Substituting the value of S from (6) we have therefore 



^l 2 



and using this in equation (9) we find 



Xl = ( M »+ »'i ~ M i)gi 7 W ~*Wi + Mpd, - S /\ + gl . (13), 



Now, let m 2 , J. 2 , / 2 , # 2 , ^ be ^ ne values of the summations 



made in the next span, QJB, Q, corresponding with 0, and jB with Q. 

 If lf 2 is the bending moment at the support B, then equation (12) 

 becomes, putting x 1 instead of a , M x instead of Ji~ , JLT 3 instead of 



(M T + m 2 -M 2 ) go - ?oFo - UjnjL n A , 



— - ~f — .... (14) 



M M, P M 2 



A A " x A 



O Q R 



< ....^ X h.... > 



Thus we have in equations (13) and (14) distinct values of the 

 angle of inclination at Q. Putting these values equal to one another 



* If $ is the small angle of discontinuity of the neutral line of the beam at Q, 

 that is if B is equal to 



inclination downwards of the I f inclination downwards of the 

 neutral hue of QR at Q I I neutral line of O Q at Q, 



then we see that a x in (14) minus a x in (13) equals 0, and hence we must put — 9 

 instead of on the right hand side of equation (15) . The angle 6 is not supposed 

 to vary with alteration of loading. 



