Civil Engineering Reference
In-Depth Information
positive directions for the shear force,
S
AB
, and bending moment,
M
AB
, in the free
body diagram of Fig. 3.12(b). Then, for vertical equilibrium of the length XB of
the beam
S
AB
+
w
(
L
−
x
)
=
0
so that
S
AB
=−
w
(
L
−
x
)
(i)
Therefore
S
AB
varies linearlywith
x
and varies fromzero at B to
−
wL
atA(Fig. 3.12(c)).
Now consider themoment equilibriumof the lengthABof the beamand takemoments
about X
w
2
(
L
x
)
2
M
AB
+
−
=
0
which gives
w
2
(
L
x
)
2
M
AB
=−
−
(ii)
Note that the total load on the length XB of the beam is
w
(
L
−
x
), which we may
consider acting as a concentrated load at a distance (
L
x
)/2 from X. From Eq. (ii)
we see that the bending moment,
M
AB
, is negative at all sections of the beam and
varies parabolically as shown in Fig. 3.12(d) where the bending moment diagram is
again drawn on the tension side of the beam. The actual shape of the bending moment
diagrammay be found by plotting values or, more conveniently, by examining Eq. (ii).
Differentiating with respect to
x
we obtain
−
d
M
AB
d
x
=
w
(
L
−
x
)
(iii)
so that when
x
0 and the bending moment diagram is tangential to
the datum line AB at B. Furthermore it can be seen from Eq. (iii) that the gradient
(d
M
AB
/d
x
) of the bending moment diagram decreases as
x
increases, so that its shape
is as shown in Fig. 3.12(d).
=
L
,d
M
AB
/d
x
=
E
XAMPLE
3.6
Simply supported beam carrying a central concentrated load.
In this example it is necessary to calculate the value of the support reactions, both
of which are seen, from symmetry, to be
W
/2 (Fig. 3.13(a)). Also, there is a loading
discontinuity at B, so that we must consider the shear force and bending moment first
at an arbitrary section X
1
say, between A and B and then at an arbitrary section X
2
between B and C.
From the free body diagram in Fig. 3.13(b) in which both
S
AB
and
M
AB
are in positive
directions we see, by considering the vertical equilibrium of the length AX
1
of the
