Civil Engineering Reference
In-Depth Information
F
B
E
C
A
D
F
IGURE
4.9
Investigation into truss stability
Suppose that ACD is the initial triangle. The additional triangle ACB is formed by
adding the two members AB and BC and the single joint B. The triangle DCE follows
by adding the two members CE and DE and the joint E. Finally, the two members BF
and EF and the joint F are added to form the rectangular portion CBFE. We therefore
conclude that the truss in Fig. 4.9 is stable and statically determinate. Compare the
construction of this truss with that of the statically indeterminate truss in Fig. 4.7(c).
A condition, similar to Eq. (4.1), applies to space trusses; the result for a space truss
having
m
members and
j
pinned joints is
m
=
3
j
−
6
(4.2)
4.5 R
ESISTANCE OF A
T
RUSS TO
S
HEAR
F
ORCE AND
B
ENDING
M
OMENT
Although the members of a truss carry only axial loads, the truss itself acts as a beam
and is subjected to shear forces and bending moments. Therefore, before we consider
methods of analysis of trusses, it will be instructive to examine the manner in which a
truss resists shear forces and bending moments.
The Pratt truss shown in Fig. 4.10(a) carries a concentrated load
W
applied at a joint
on the bottom chord at mid-span. Using the methods described in Section 3.4, the
shear force and bending moment diagrams for the truss are constructed as shown in
Fig. 4.10(b) and (c), respectively.
First we shall consider the shear force. In the bay ABCD the shear force is
W
/
2 and
is negative. Thus at any section mm between A and B (Fig. 4.11) we see that the
internal shear force is
W
/
2. Since the horizontal members AB and DC are unable
to resist shear forces, the internal shear force can only be equilibrated by the vertical
component of the force
F
AC
in the member AC. Figure 4.11 shows the direction of
the internal shear force applied at the section mm so that
F
AC
is tensile. Then
−
W
2
F
AC
cos 45
◦
=
