1886.] = Myr Webb, The problem of three moments. 43 
figure correspond bending the rod over each support. Then we 
have 
A,. AB+M,4—Mz=0) 
—B,. AB+ M,—My=05’ 
so that A,+B,=0 as is otherwise obvious, and the equation of 
upward deflection between A and B is, 
where y is measured upwards, and w# from O. Now multiply by 
(«— OA)/E, integrate from A to B and we have 
0B (OB —«) (w — OA) dx 
ABtan B= | i 
eal me a AB E 
OB(x— OA) dx 
M, | Cama ee ag 
Tet Re 
Whence we have 
0B (OB —«x) (x— OA) da ( (c— OA) da... 
tan R@=M i é zy gl eg Cmalcaeee) agrees 
ena AB | Ta Be 
Similarly, if between B and C the deflection be measured down- 
wards, an exactly similar process leads to 
0¢ (OC—a)(a—OB) da (ee (OC — a)” w 
t —M, He : — M . 
an B= C Be BC E B aR BC? (iil ii), 
and on subtraction there results the equation 
0B (OB —«2) (a— OA) dx 0¢ (OC — x) (a — OB) da 
Ma Aa ABO pee i BC? E 
OB (@ — QA)? dx (eas ow) da 
titel [pap Et |p BOB [=o 
We have now to pass on to the case in which a weight W is 
