126 APPLIED MECHANICS 
133. Continuous Beams and Theorem of Three Moments.—A beam 
which rests on more than two supports is called a continuous beam. Let 
BCD (Fig. 167) be a portion of a continuous beam, BC =L, and CD=L, 
~ being two consecutive spans. Let b’ec’ and c’fd’ be the bending moment 
diagrams on the base ’c'd’ for BC and CD considered as separate beams 
supported at their ends. The separate heams BC and CD would have no 
bending moments at their supports, but the continuous beam BCD will 
have bending moments M,, M,, and M, at the supports B, C, and D 
Fie. 167. 
respectively, but at present these bending moments are unknown. Sup- 
pose, however, that M,, M,, and M, are known. Make 0‘=M,, ce=M,, 
and d’d=M,. Considering the portion BC, the bending moments M, and 
M, may be considered as arising from the loading to the left of B and to 
the right of C, and these bending moments will affect the whole of BC, as 
shown by the diagram b’bec’, where bc is a straight line. The resulting 
bending moment diagram for BC as a part of the continuous beam will 
be the shaded diagram b’ec’cb, the base of which is bc. 
Let a,=area of bending moment diagram 6’ec’, and a,=area of 
bending moment diagram c’fd’. Let z,=horizontal distance of the 
centre of gravity of the diagram 2 A cn 
‘ec’ 2 = hori ~ = RSs Ses WH 
Yee fom Band ce 
diagram ¢’fd’ from D. Also let 
x, = horizontal distance of the centre 
of gravity of the shaded diagram 
b’ec’cb from B, parts below be being 
reckoned as negative, and parts 
above be as positive. Lastly, let 8, denote the effective area of b’ec’ch, 
that is, the algebraical sum of the positive and negative parts. 
Now consider the shaded bending moment diagram to be a load 
diagram. Let TCT’ be the tangent to the bent continuous beam at C, 
and let it meet the verticals through B and D at T and T’ respectively. 
By Art. 126, BT =S,x,+EL. Dividing the figure b’bce’ into a rectangle 
and a triangle, as shown in Fig. 168, where G, is the centre of gravity of 
Fig, 168. 
