(^336) „„„„„ A Textbook of Engineering Mechanics
or (^2)
45
15 m
3
l ==
and l 1 = 2l 2 = 2 × 15 = 30 m
∴ Length of the cable,
2 2
12
2( ) 2
33
Ll ydccy
ll

• =+ +
  2(2 6)^22 2(2)
  45 46.6 m
  330 315


• 
  

  =+ + =
  ××
  Ans.
  (b) Horizontal component of tension in the cable
  We know that horizontal component of tension in the cable,
  (^22)
  1 20 (30) 1125 kN
  2(c ) 2(2 6)
  wl
  H
  yd

  
  

  ++
  Ans.
  (c) Magnitude and position of maximum tension occurring in the cable
  We also know that maximum tension will occur at the support B. Vertical reaction B,
  20 45 1125 6
  600 kN
  B 2245
  wl H d
  V
  l
  ××
  =+ = + =
  ∴ Maximum tension,
  22 2 2
  TVHmax=+=B (600) +(1125) =1275 kN^ Ans.
  15.10. THE CATENARY
  The shape, which a string (cable or rope) takes up under its own weight (without any external
  load) is called catenary. It will be interesting to know that catenary, as a curve, has many technical
  and scientific properties.
  (a) Space diagram (b) Portion CP
  Fig. 15.14.
  Consider a string or cable suspended at two points A and B at different levels (or same level)
  hanging freely under its own weight with C as the lowest point as shown in Fig. 15.14.
  Let w = Weight per unit length of the cable.
  Now consider any point (P) on the cable such that length of the cable CP be s. Now draw the
  tangent at P. Let ψ be the inclination of the tangent with the horizontal as shown in the figure. We
  know that part CP of the string is in equilibrium under the action of the following forces.

  
  

1. Load (w.s) of the cable acting vertically downwards.


2. Horizontal pull (H) acting horizontally at C.


3. Tension (T) acting at P along the tangent.