Civil Engineering Reference
In-Depth Information
In the arrangement shown in Fig. 5.10(b) the cable passes over a saddle which is
supported on rollers on the top of the tower. The saddle therefore cannot resist a
horizontal force and adjusts its position until
T
A
cos
β
=
T
C
cos
α
(5.29)
For a given value of
β
, Eq. (5.29) determines the necessary value of
T
A
. Clearly, since
there is no resultant horizontal force on the top of the tower, the bending moment
in the tower is everywhere zero. Finally, the vertical compressive load on the tower is
given by
V
T
=
T
C
sin
α
+
T
A
sin
β
(5.30)
E
XAMPLE
5.5
The cable of a suspension bridge, shown in Fig. 5.11, runs over a
frictionless pulley on the top of each of the towers at A and B and is fixed to anchor
blocks at D and E. If the cable carries a uniform horizontally distributed load of
120 kN/m determine the diameter required if the permissible working stress on the
gross area of the cable, including voids, is 600N/mm
2
. Also calculate the bending
moment and direct load at the base of a tower and the required weight of the anchor
blocks.
A
B
30 m
C
50 m
120 kN/m
D
E
45
°
45
°
F
IGURE
5.11
Suspension bridge
of Ex. 5.5
300 m
W
A
The tops of the towers are on the same horizontal level, so that the tension in the
cable at these points is the same and will be the maximum tension in the cable. The
maximum tension is found directly from Eq. (5.25) and is
300
4
2
120
×
300
T
max
=
1
+
=
48 466
.
5kN
2
×
30
The maximum direct stress,
σ
max
, is given by
T
max
π
d
2
/
4
σ
max
=
(see Section 7.1)
in which
d
is the cable diameter. Hence
10
3
48 466
.
5
×
600
=
π
d
2
/
4
