Civil Engineering Reference
In-Depth Information
the bending moment has a turning value and is sagging if the supports at B and C are
spaced a sufficient distance apart. Suppose that B and C are positioned such that the
value of the hogging bending moment at B and C is numerically equal to the sagging
bending moment at the mid-span point. If now B and C are moved further apart the
mid-span moment will increase while the moment at B and C decreases. Conversely, if
B and C are brought closer together, the hogging moment at B and C increases while
the mid-span moment decreases. It follows that the maximum bending moment will
be as small as possible when the hogging moment at B and C is numerically equal to
the sagging moment at mid-span.
The solution will be simplified if use is made of the relationship in Eq. (3.7). Thus,
when the supports are in the optimumposition, the change in bending moment fromA
to B (negative) is equal to minus half the change in the bending moment from B to the
mid-span point (positive). It follows that the area of the shear force diagram between
A and B is equal to minus half of that between B and the mid-span point. Then
L
2
−
a
w
L
a
1
2
awa
1
2
1
2
+
=−
−
2
−
which reduces to
L
2
4
a
2
+
La
−
=
0
the solution of which gives
a
=
0
.
21
L
(the negative solution has no practical significance)
3.6 T
ORSION
The distribution of torque along a structural member may be obtained by considering
the equilibrium in free body diagrams of lengths of member in a similar manner to
that used for the determination of shear force distributions in Exs 3.4-3.9.
E
XAMPLE
3.12
Construct a torsion diagram for the beam shown in Fig. 3.21(a).
There is a loading discontinuity at B so that we must consider the torque at separate
sections X
1
and X
2
in AB and BC, respectively. Thus, in the free body diagrams shown
in Fig. 3.21(b) and (c) we insert positive internal torques.
From Fig. 3.21(b)
T
AB
−
10
+
8
=
0
so that
T
AB
=+
2kNm
