Civil Engineering Reference
In-Depth Information
Integrating Eq. (5.9) with respect to
x
we have
H
d
y
d
x
=+
wx
+
C
1
(5.10)
again integrating
w
x
2
Hy
=+
2
+
C
1
x
+
C
2
(5.11)
The boundary conditions are
y
=
0at
x
=
0 and
y
=
h
at
x
=
L
. The first of these gives
C
2
=
0 while from the second we have
w
L
2
H
(
+
h
)
=+
2
+
C
1
L
so that
H
h
L
Equations (5.10) and (5.11) then become, respectively
wL
2
C
1
=−
+
d
y
d
x
=+
w
H
x
wL
2
H
+
h
L
−
(5.12)
and
wL
2
H
−
x
w
2
H
x
2
h
L
y
=+
−
(5.13)
Thus the cable in this case takes up a parabolic shape.
Equations (5.12) and (5.13) are expressed in terms of the horizontal component,
H
,
of the tension in the cable, the applied load and the cable geometry. If, however, the
maximum sag,
D
, of the cable is known,
H
may be eliminated as follows.
The position of maximum sag coincides with the point of zero slope. Thus from
Eq. (5.12)
w
H
x
wL
2
H
+
h
L
0
=+
−
so that
L
2
−
Hh
wL
=
x
=
L
1
(see Fig. 5.7)
Then the horizontal distance,
L
2
, from the lowest point of the cable to the support at
B is given by
Hh
wL
Now considering the moment equilibrium of the length CB of the cable about B we
have, from Fig. 5.7
L
2
+
L
2
=
L
−
L
1
=
w
L
2
HD
−
2
=
0
