Chapter 15 : Equilibrium of Strings „„„„„ 329

The term l 1 / l 2 is known as ratio of the horizontal length of the string or cable. Now take C as
the origin. The co-ordinates of the support B are l 2 and yc, whreas co-ordinates of the support A are
(– l 1 ) and (yc + d). We have discussed in Art. 15.4 that the horizontal thrust,
2 2
2

(^22) c
wx wl
H
yy
== ...(iv)
2 2
(– ) 11
2(cc) 2( )
wl wl
yd yd

++
...(v)
Now we can find out the vertical reactions at A and B. Taking moments about B and equating
the same, i.e.
2
2
A
wl
Vl=+Hd
∴
2
A
wl H d
V
l
=+
Similarly, vertical reaction at B,

• 2
  B

wl Hd

V

l

=

Now tension in the string at A,

(^22)
TRHAA=+
Similarly TRHBB=+^22
Note. Since the value of RA (the support A being higher than B) is more than RB, therefore the
maximum tension in the string will be at A.
Example 15.4. A cable of uniform thickness hangs between two points 120 m apart, with
one end 3 m above the other. The cable loaded with a uniformly distributed load of 5 kN/m and the
sag of the cable, measured from the higher end, is 5 m.
Find the horizontal thrust and maximum tension in the cable.
Solution. Given : Span (l) = 120 m ; Difference between the levels of supports (d) = 3 m ;
Uniformly distributed load (w) = 5 kN/m and sag of the cable (yc ) = 5 – 3 = 2 m
Fig. 15.7.
Horizontal thrust in the cable
Let l 1 = Horizontal length of AC, and
l 2 = Horizontal length of CB.
We know that ratio of the horizontal langths,
1
2
23
1.58
2
c
c
l yd
ly

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