(^330) A Textbook of Engineering Mechanics
∴ l 1 = 1.58 l 2
and l 1 + l 2 = 120 m
∴ 1.58l 2 + l 2 = 120 m ...(Q l 1 = 1.58 l 2 )
or 2
120
46.5 m
2.58
l ==
and l 1 = 120 – 46.5 m = 73.5 m
∴ Horizontal thrust in the cable,
2 2
1 5(73.5) 2701kN
2(c ) 2(2 3)
wl
H
yd
×
== =
++
Ans.
Maximum tension in the cable
We know that the maximum tension will take place at the higher end A. We also know that
*vertical reaction at A,
5 120 2701 3
367.5 kN
A 2 2 120
wl H d
V
l
××
=+ = + =
∴ Maximum tension in the cable,
22 2 2
TVHmax=+= + =( A) ( ) (2701) (367.5) 2726 kN Ans.
EXERCISE 15.1
- A string supported at A and B, at the same level over a span of 30 m is loaded as shown in
the figure given below :
Fig. 15.8.
If the depth of the point D is 8 m from the supports, find the tensions in AC, CD, DE and
EB of the string. Also find horizontal thrusts in the strings at A and B and draw shape of
the loaded string. [Ans. 21.8 kN ; 17.8 kN ; 21.2 kN ; 24.4 kN]- A suspension bridge of 40 m span, made of a cable of uniform thickness, has a central dip
of 6.25 m. The cable is loaded with a uniformly distributed load of 12.5 kN/m throughout
the span. What is the maximum tension in the cable? [Ans. 471.7 kN] - A light foot bridge is made up of cable of uniform thickness over a span of 75 m. The
supports are 8 m and 2 m higher than the lowest point of the cable. Find the horizontal
thrust and maximum tension in the cable, when it is loaded with a uniformly distributed
load of 2 kN/m. [Ans. 312.5 kN ; 328.1 kN]
* The vertical reaction at A may also be found out by considering the load from A to C. i.e.,
VA = 5 × 73.5 = 367.5 kN.