Hall A.S., Archer F.E., Gilbert R.I. Engineering Statics
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0.0~0
1.006
FLEXIBLE
CABLES
141
rts are at ~if~erent levels, the m~i~u~
sag
of
a
occur at the centre
of
the span.
ve the level
of
A.
The
loadin
1
found by setting
dy/&
=
O
and solving for
x.
Performing this operation yields the co-
ordinates of the lowest point,
(xo,
yo):
L
Hh
xo=---
2
WL
h
Hh2
wL2
Yo=2-~-~
WL2
If
H
the cable will not sag below
A.
The cable tension at any point can be obtained from
=
H
sec
ti!
The m~imum
tension will occur at the higher support, and is:
In a problem in which
W,
L,
h
and
yo
are specified,
H
may be found from Equation
9.17, and then
x.
from Equation 9.18,
The length of the cable
AB
can now be found
as
the
sum
of the lengths
AC
and
CB.
The length
AC
will be half the length of the symmetrical parabola of base 2xo and
sag
yo,
while the length
CB
will be half the length of the parabola of the base 2(L
-
XJ
and the sag
(h
+
yo).
A
uniform cable carrying its own weight hangs in the shape of a curve called a
catenary
which is a hyperbolic cosine curve. This problem is not of great engineering interest,
since cables are rarely used simply to carry their own weight. Moreover, unless the ratio
of sag
to
span
is
quite large, the catenary does not differ greatly from a parabola, since
for fairly
flat
curves the weight per metre along the cable is much the same as the weight
per horizontal metre.
(a)
Y
t
T
X
.c
ws
H(=
WC)
ris
0.00
1.0000
0.0000
0.05 1.000s 0.0125
0.10 1.0017
0.0250
0.15
1.0038 0.0376
0.20
1.0067 0.0502
0.25
1.0104 0,062
0.30 1.0151 0.0756
0.55 1.0512 0.1410
0.75 1.0964 0.1965
0.90 1.1406 0.2406
1.10
1.2142
0.303~
1.15
1.2355 0.320
1.20
1.2579 0.337
1.30 1.3064 0.3734
1.35 1.3327 O"3920
1.40 1.3602 0.4110
1.45 1.3892 0.430~
1.50 1.4195 0*4~08
1
1
1.90 1.7201 ~.63~
Unless the cable sag
is
large compared with the span, the shape
of
the catenary diEers
little from that
of
a parabola. For practical purposes, Example
9.3
could have been solved
on the basis
of
a parabolic shape.
FLEXIBLE CABLES
147
the
cable
is
weightless.
A
El
1.5
m
EAMS
AND
CABLES
The supports
A
and of the cable shown
m tension in the cable
is
7
kN
in Figure
P9.3
is shortened un
eterrnine the tension in
C
8
kN
A
uniform cable
AB
is
130
m
long and wei /m. It is suspended from
mo
points
120
m
apart and at rhe same l
surne
that the cable hangs in
the shape of
a
parabola, i.e. tbar it carrie
r
~o~~~o~t~~
metre.
(i)
Find the sag at the centre.
(ii Find the tension at the centre and at the end.
(iii) Find the sag at a point
40
m from
A.
Solve Problem
9.4
if the weight is correctly taken as
N/m
along the cable,
i.e. if the cable shape is
a
catenary.
The cable
of
Problem
9.4
is to be used to support a load suspended above a
ravine. If the load of
60
kN
is suspended at
C,
40
m
(horizont~~y) from
A.
Find:
(i)
the sag at
C
(neglect the cable weight)
(ii)
the increase in sag at
C
(compared with the freely hanging cable in
(iii) the maximum cable tension.
Problem
9.5)
G
has a total length of
440
m
and
nts
A
and
C
and passes over a fric
supports are at the same level. The distance
Find the m~i~um sag in each section
o
tension,
*
~i~cult problems, suitable
for
later study.