Civil Engineering Reference
In-Depth Information
In some situations the diagonal members are unsuitable for compressive forces so that
counterbracing
is required. This consists of diagonals inclined in the opposite direction
to the original diagonals as shown in Fig. 20.18(a) for the two centre panels. The
original diagonals are then assumed to be carrying zero force while the counterbracing
is in tension.
It is clear from Ex. 20.7 that the shear force in all the panels, except the two outer
ones, of a Pratt truss can be positive or negative so that all the diagonals in these panels
could experience compression. Therefore it would appear that all the interior panels
of a Pratt truss require counterbracing. However, as we saw in Section 20.3, the dead
load acting on a beam has a beneficial effect in that it reduces the length of the beam
subjected to shear reversal. This, in turn, will reduce the number of panels requiring
counterbracing.
E
XAMPLE
20.8
The Pratt truss shown in Fig. 20.19(a) carries a dead load of
1
.
0 kN/m applied at its upper chord joints. A uniformly distributed live load, which
exceeds 9m in length, has an intensity of 1
.
5 kN/m and is also carried at the upper
chord joints. If the diagonal members are designed to resist tension only, find which
panels require counterbracing.
1
2
3
4
5
6
7
8
9
10
1.2 m
A
B
10 0.9 m
(a)
l
m
n
o
p
q
r
s
t
n
2
n
6
n
7
n
8
n
9
a
b
n
3
n
4
n
5
k
j
i
h
F
IGURE
20.19
Counterbracing in a
Pratt truss
g
f
e
d
c
(b)
A family of influence lines may be drawn as shown in Fig. 20.19(b) for the shear force
in each of the 10 panels. We begin the analysis at the centre of the truss where the DLS
force has its least effect; initially, therefore, we consider panel 5. The shear force,
S
5
,
in panel 5 with the head of the live load at n
5
is given by
S
5
=
1
.
0 (area n
5
qa
−
area n
5
gb)
+
1
.
5 (area n
5
qa)
i.e.
S
5
=−
1
.
0
×
area n
5
gb
+
2
.
5
×
area n
5
qa
(i)
