Civil Engineering Reference
In-Depth Information
or
M
E
h
F
CD
=
(20.27)
Since
M
E
is positive for all load positions (Fig. 20.17(c)),
F
CD
is compressive.
The force in the member GE is obtained from the
M
C
influence line in Fig. 20.17(d).
Thus
F
GE
h
=
M
C
which gives
M
C
h
F
GE
=
(20.28)
F
GE
will be tensile since
M
C
is positive for all load positions.
It is clear from Eqs (20.25)-(20.28) that the influence lines for the forces in the mem-
bers could be constructed from the appropriate shear force and bending moment
influence lines. Thus, for example, the influence line for
F
CE
would be identical in
shape to the shear force influence line in Fig. 20.17(b) but would have the ordinates
factored by 1/sin
θ
and the signs reversed. The influence line for
F
DE
would also have
the
S
CD
influence line ordinates factored by 1/sin
θ
.
E
XAMPLE
20.7
Determine the maximum tensile and compressive forces in the
member EC in the Pratt truss shown in Fig. 20.18(a) when it is crossed by a uniformly
distributed load of intensity 2
.
5 kN/m and length 4m; the load is applied on the bottom
chord of the truss.
Counterbracing
E
1.4 m
A
B
H
D
C
M
8
1.4 m
(a)
f
j
c
a
b
F
IGURE
20.18
Determination of
the force in a
member of the Pratt
girder of Ex. 20.7
d
g
(b)
S
DC
IL
The vertical component of the force in the member EC resists the shear force in the
panel DC. Therefore we construct the shear force influence line for the panel DC as
shown in Fig. 20.18(b). From Eq. (20.19) the ordinate df
=
2
×
1
.
4
/
(8
×
1
.
4)
=
0
.
25
while from Eq.
(20.20)
the ordinate cg
=
(8
×
1
.
4
−
3
×
1
.
4)
/
(8
×
1
.
4)
=
0
.
625.
