Chapter 15 : Equilibrium of Strings „„„„„ 327

∴

2

(^8) c
wl
H
y

We also know that the maximum tension in the cable will be at the supports A and B, and will
be given by the relation.
2 2
() ( )^22
max A A (^28) c
wl wl
TVH
y
⎛⎞⎛⎞
=+= +⎜⎟⎜⎟
⎝⎠⎝⎠
2
2 1 16 2
c
wl l
y
=+
and minimum tension in the cable will be at C and is equal to.
2
min
(^8) c
wl
TH
y

Example 15.2. A suspension cable, with supports at the same level, has a span of 30 m and
maximum dip of 3 m. The cable is loaded with a uniformly distributed load of 10 kN/m throughout its
length. Find, from first principles, the maximum tension in the cable.
Solution. Given : Span (l) = 30 m; Maximum dip (yc ) = 3 m and uniformaly distributed
load (w) = 10 kN/m
We know that vertical reaction at the supports,
10 30
150 kN
22
wl
V
×
== = ...(i)
and horizontal thrust in the cable,
(^2210) (30)
375 kN
883
wl
H
yc
×
== =
×
...(ii)
∴ Maximum tension in the cable,
(^) TVHmax=+=^22 (150)^2 + =(375)^2 404 kN Ans.
Example 15.3. A suspension bridge of 40 m span with 1.5 m wide platform is subjected to
an average load of 20 kN/m^2. The bridge is supported by a pair of cables having a central dip of 5 m.
Find the necessary cross sectional area of the cable, if the maximum permissible strees in the
cable material is not to exceed 1050 N/mm^2.
Solution. Given : Span (l) = 40 m ; Width of platform = 1.5 m ; Load on platform = 20 kN/m^2 ;
central dip (yc ) = 5 m and maximum permissible strees in the cable (f) = 1050 N/mm^2
We know that total load per metre length of the cable
= 1.5 × 20 = 30 kN/m
Since the bridge is supported by a pair of cables, therefroe load on each cable,
30
15 kN/m
2
w==
∴ Maximum tension in the cable,
2 2
max 22
15 40 (40)
11kN
(^2216165)
wl l
T
yc
×
=+ = +
×
= 670.1 kN = 670.1 × 10 3 N
and necessary cross-sectional area of the cable
3
max 670.1 10 638 mm 2
1050
T
A
f
×
== = Ans.