Engineering Mechanics

Chapter 15 : Equilibrium of Strings „„„„„ 339

We know that for the peg B,

y= 2.5 m and s = l m

Similarly for peg D, y= 2.0 m and s = (5.5 – l) m

Substituting above values in the general equation of the catenary,

y^2 =c^2 + s^2

∴ (2.5)^2 =c^2 + l^2

6.25 =c^2 + l^2 ...(i)

and (2.0)^2 =c^2 + (5.5 – l)^2

∴ 4=c^2 + 30.25 + l^2 – 11l ...(ii)

Subtracting equation (ii) from (i),

2.25 = – 30.25 + 11l

∴ 11 l= 30.25 + 2.25 = 32.5

or l=3 m Ans.

Example 15.10. A cable 20 metres long weighs 25 N/m. It hangs between two points A and
B at the same level. If the central dip of the cable is 5 m, find the distance between the two supports.
Also find the maximum tension in the cable.
Solution. Given : Length of the cable (L) = 20 m ; Weight of the cable (w) = 25 N/m and
central dip = 5 m
Distance between the two supports

Let 2x = Distance between the two supports,

ψ = Angle, which the tangent at B makes with the X - X axis, and

c = Parameter of the catenary.

We know that for the support B,

y = (5 + c) and s = 10 m

Substituting above values in the general equation of the catenary,

y^2 =c^2 + s^2

(5 + c)^2 =c^2 + 10^2

25 + c^2 + 10c=c^2 + 100

10 c= 100 – 25 = 75

or c= 7.5 m

∴ Horizontal pull at C,

H=wc = 25 × 7.5 = 187.5 N

and vertical tension at B =ws = 25 × 10 = 250 N

∴

250

tan 1.3333

187.5

ws

H

ψ= = = or ψ = 53.1°

We know that distance between the two supports

2 x = 2 [2.3 c log (sec ψ + tan ψ)]

= 2 [2.3 × 7.5 log (sec 53.1° + tan 53.1°)]

= 2 × 17.25 log (1.6667 + 1.333) = 34.5 log 3

= 34.5 × 0.4771 = 16.46 m Ans.

Fig. 15.16.