The shape of a cable under a load uniformly distributed along the horizontal span is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
A cable has no bending stiffness, so at every point the moment of the loads equals the horizontal tension H times the sag y. The cable therefore takes the shape of the bending moment diagram of a simple beam with the same loads.
\[ H\,y = M_{beam} = \frac{w x (L-x)}{2} \Rightarrow y = \frac{w x (L-x)}{2H},\qquad \text{central sag } h = \frac{wL^2}{8H} \]
Why “parabolic” Is Correct
The BMD of a UDL is a parabola, so the cable is parabolic — the case of a suspension bridge deck hung from the main cable.
Why the Other Options Are Wrong
- B. catenary: A catenary results from a load uniform along the cable's own length (self-weight).
- C. circular: No common loading gives a circular cable.
- D. triangular: A triangular shape results from a single point load.
Key Point
UDL per horizontal length → parabola; self-weight along the cable → catenary.
Key Concepts for Students
- H = wL²/(8h) for a parabolic cable of sag h.
- Max tension at supports = √(H² + V²).
- For small sag, parabola and catenary are almost the same.
Practice more: this is question 27 in Bending and Shear Stress MCQs – Part 2. Try the whole set and check every answer instantly.