A CONTINUOUS BEAM.
417
Multiply the first équation throughout by - a, the second by b, and
re-arrange—
S2 a2 = - MB a + MAa +
Sjoi
Sj 62 = MB6 - Mc 6 -
Subtract
S2 a2 - Si 62 = - MB (a + b) + MA a+ Mcb+^ (a3 + bs).
Divide by 6, and substitute in equation (96) above—
Mb“J6 - MB^ b+ MA“ + McI+g(«3+ 63)-g(«8 + &s) = o, ■
which reduces to
MAa + Mc b + 2 MB(a + 6) + j(a3 + i3) = 0. . (97)
This equation is known as the Theorem of Three Moments, and its first
enunciation is attributed to Clapeyron. By means of the relationship thus
established, if the bending moments at two of the points of support of a
uniformly loaded beam are known, the third can be deduced, The bending
moments at the end supports are sufficiently obvious. If the beam project a
W C“
distance, c, beyond the outer support, C, the moment at C is —^. If the beam
do not project, the moment at the point of support is zero.
The shearing stresses can then be obtained from the formula already
eiven, viz. :—
Mg Mc wb
b ~ b 2
MB Ma wa
a 2
The shear at any point, X, is Sj - wx. Accordingly, at A and C it is
SA = - S2 + wa and SC = Sj - wb respectively.
From these, the reactions at the points of support are readily forthcoming,
for RA = SA and Rc = Sc, if there be no overhang. If there be an over-
hanging portion, as c at C, Rc = So. + wc.
Also
Assuming that there is no overhang this equation simplifies into
„ w, (a* + ?>ab + b-\
Rb = s(«+i)( -------------------b -----i
Equation (98) may be confirmed by an independent investigation which
is worthy of notice, for it gives an expression for the current moment in
terms of the moments at the points of support.