Engineering Mechanics

Chapter 15 : Equilibrium of Strings „„„„„ 335

We know that ratio of the horizontal lengths,

1

2

39

2

3

c

c

l yd

ly

+ +

===

∴ l 1 = 2l 2

and l 1 + l 2 = 60 m

∴ 2 l 2 + l 2 = 60 or 3 l 2 = 60

2

60

20 m

3

l == and l 1 = 2 × 20 = 40 m

We also know that length of cable,

2 2

12

2( ) 2

33

Ll ydccy

ll

+

=+ +

2(3 9)^22 2(3)

60

340 320

+

=+ +

××

m

=62.7 m Ans.

Example 15.8. A Cable is suspended and loaded as showns in Fig 15.13 below :

Fig. 15.13.
(a)Compute the length of the cable;
(b)Compute horizontal component of tension in the cable, and
(c)Determine the magnitude and position of the maximum tension occuring in the cable.
Solution. Given : Span (l) = 45 m ; Depth of the lowest point from it support A (yc) = 2 m or
difference between the levels of the supports (d) = 8 – 2 = 6 m and uniformly distributed load over
the span (w) = 20 kN/m

(a) Length of the cable

Let l 1 = Horizontal length of CB, and

l 2 = Horizontal length of AC.

We know that ratio of horizontal lengthe,

1

2

26

2

2

c

c

l yd

ly

+ +

===

∴ l 1 = 2l 2

and l 1 + l 2 = 45 m

∴ 2 l 2 + l 2 = 45 or 3 l 2 = 45