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Steel Designers' Manual - 6th Edition (2003)
C
or
EId
P =
L
This load would cause a BM at A or B equal to
12
Therefore
3
d P L 2
=
2 3EI
(being the standard deflection formula)
3
( )
L 12EId
L 6EId
P ¥ = ¥ =
3 2
2 L 2 L
-
The solution in any given case consists of adding to the ordinary diagram of BM,
the BM diagram A1DCEB1.
Shear forces in fixed beams
In the case of fixed beams it is necessary to evaluate the BM before the SF can be
determined. This is the converse of the procedure for the case of simply-supported
beams.
The SF at the ends of a beam is found in the following manner:
M - M
SFA = the simple support reaction at A =+
L
A B
M - M
SFB = the simple support reaction at B = +
L
Fixed, built-in or encastré beams 331
Fig. 10.5 Bending moment diagram for fixed-end beam with supports at different levels
B A
where MA and MB are the numerical values of the moments at the ends of the beam.
These formulae must be followed exactly with respect to the signs shown since if
MA is smaller than MB the signs will adjust themselves.
E