A CONTINUOUS BEAM. 421
Next, let us consider the span, 6, as a whole, and take moments about
the points B and C respectively. In the first case, we have
So& - W^ + Mb = 0, .........................(115)
and in the second,
S^ _ W(6 - dj) - MB = 0.................................(116)
Substitute the values for Sx and Sc given by these equations in (113)
and (114), re-arranging as below—
EI[Ødj - a (6 - dj)J = Mb|(5 -3dj- W^(6 - dj) (6 - 2dj).
EI(/8+a)=-MBJ +W^(6 - dj). . (117)
Multiply the latter equation by (6 - dj) and eliminate a by addition—
EI/36=-MbJ +WJ(è-dj) (2&-dj). . (118)
Now, let us deal in a similar manner with the span, q, to the left of the
point, B. It is only necessary to re-write the previous equation, making the
requisite changes in sign—
- EI/3q= - MB^ +W^(q-d2) (2q-d2). (119)
Whence, eliminating ß between the two equations, we get
1 )Wjdj(6 - dj) (26 - dj) W2d2(a - d2) (2 q - d2) )
Ms = 2(^6)1 6 + *
an expression which furnishes us with the value of the bending moment at
the intermediate support, B. The bending moment at each end is, of
course, zero.
It is sufficiently obvious that, when there is a load on only one of the
spans (as Wj on the span, 6) the bending moment at the intermediate
support is given by
M - 1 ! W ^<& ~ ^^ .... (121)
B 2 (q + 6)l 1 6 J
and that, for any system of concentrated loads on a single span, the
general equation may be written—
Ma = o * -^{Wd(6 - d)(26- d)}...........................(122)
The reactions at the point of supports will be easily determined fiom a
considération of the conditions of equilibrium in each span. Assuming one
span to be loaded as above—
.....................................
MB = R06-2Wd.......................(124)