Engineering Mechanics

Chapter 15 : Equilibrium of Strings „„„„„ 333

Solution. Given : Central dip ()

c 12

l

y = (where l is the span); Maximum stress in steel wire

(f) = 120 N/mm^2 = 120 × 10^6 N/m^2 = 120 × 10^3 kN/m^2 and mass density of steel (ρ) = 7800 kg/m^3

= 7800 × 9.81 = 76520 N/m^3 = 76.52 kN/m^3

Let A = Cross-sectional area of the wire.

We know that length of the wire,

8 22 8 ( /12) 55

3354

yc l l

Ll l

ll

=+ =+ =

and total weight of the wire,

55

76.52 77.94 kN

54

l

WAL A=ρ=× × = Al

∴ Horizontal thrust in the wire,

1.5

88(/12)c

Wl Wl

HW

yl

== =

and vertical reaction at the support,
V = W/2 = 0.5 W
∴ Maximum tension in the wire,

(^) TVHmax=+=^22 (0.5W W W)^2 +(1.5 )^2 =1.58 kN
= 1.58 × 77.94 Al = 123.15 Al kN ...(Q W = 77.94 Al)
We know that maximum stress in the wire
= f A = 120 × 10^3 A
∴ Equating maximum tension in the wire to the maximum strees in it,
123.15 Al = 120 × 10^3 A
120 103
974.4 m
123.15
l
×
== Ans.
15.9.LENGTH OF A STRING WHEN THE SUPPORTS ARE AT DIFFERENT LEVELS
Fig. 15.11.
Consider a string ACB, supported at different levels A and B as shown in Fig 15.11. Let C be
the lowest point of the cable.
Let l = Span of the cable,
yc= Depth of the lowest point of the cable C from the support B,
d= Difference between the levels of the two supports,