Civil Engineering Reference
In-Depth Information
in which
F
BA
,
F
AC
and
F
DC
have the same values as before. Then, taking moments
about, say, the mid-point of the top chord member AB, we have
0
.
5 sin 45
◦
=
0
.
5 sin 45
◦
=
M
=
F
DC
×
1
+
F
AC
×
1
.
0
W
×
1
+
0
.
707
W
×
1
.
25
W
the value of the applied moment.
From the discussion above it is clear that, in trusses, shear loads are resisted by inclined
members, while all members combine to resist bending moments. Furthermore, pos-
itive (sagging) bending moments induce compression in upper chord members and
tension in lower chord members.
Finally, note that in the truss in Fig. 4.10 the forces in the members GE, BC and HF
are all zero, as can be seen by considering the vertical equilibrium of joints E, B and F.
Forces would only be induced in these members if external loads were applied directly
at the joints E, B and F. Generally, if three coplanar members meet at a joint and
two of them are collinear, the force in the third member is zero if no external force is
applied at the joint.
4.6 M
ETHOD OF
J
OINTS
We have seen in Section 4.4 that the axial forces in the members of a simple pin-jointed
triangular structure may be found by examining the equilibrium of their connecting
pins or hinges in two directions at right angles (Eq. (2.10)). This approach may be
extended to plane trusses to determine the axial forces in all theirmembers; themethod
is known as the
method of joints
and will be illustrated by the following example.
E
XAMPLE
4.2
Determine the forces in the members of the Warren truss shown in
Fig. 4.14; all members are 1m long.
2kN
3kN
B
C
A
D
E
1kN
F
IGURE
4.14
Analysis of a Warren
truss
R
A
2.75 kN
R
D
3.25 kN
