Civil Engineering Reference
In-Depth Information
As in the case of the catenary the maximum tension will occur, since
H
constant, at
the point where the vertical component of the tension is greatest. Thus, in the cable of
Fig. 5.7, the maximum tension occurs at B where, as
L
2
>
L
1
, the vertical component
of the tension (
=
=
wL
2
) is greatest. Hence
(
wL
2
)
2
T
max
=
+
H
2
(5.19)
in which
L
2
is obtained from Eq. (5.17) and
H
from one of Eqs (5.14), (5.16) or (5.18).
At B the slope of the cable is given by
tan
−
1
wL
H
α
=
(5.20)
or, alternatively, from Eq. (5.12)
d
y
d
x
w
H
L
wL
2
H
+
h
L
=+
wL
2
H
+
h
L
L
=+
−
(5.21)
z
=
For a cable in which the supports are on the same horizontal level, i.e.
h
=
0, Eqs
(5.12), (5.13), (5.14) and (5.19) reduce, respectively, to
x
d
y
d
x
w
H
L
2
=
−
(5.22)
w
2
H
(
x
2
y
=
−
Lx
)
(5.23)
wL
2
8
D
H
=
(5.24)
L
4
D
2
wL
2
T
max
=
1
+
(5.25)
We observe from the above that the analysis of a cable under its own weight, that is a
catenary, yields a more complex solution than that in which the load is assumed to be
uniformly distributed horizontally. However, if the sag in the cable is small relative to
its length, this assumption gives results that differ only slightly from the more accurate
but more complex catenary approach. Therefore, in practice, the loading is generally
assumed to be uniformly distributed horizontally.
E
XAMPLE
5.4
Determine the maximum tension and the maximum slope in the
cable shown in Fig. 5.8 if it carries a uniform horizontally distributed load of intensity
10 kN/m.
From Eq. (5.17)
200
18
−
6
18
L
2
=
1
=
110
.
1m
+
