A cable is supported at both ends at the same level and is subjected to U.D.L. of intensity w/unit length over the entire span. If yc is the central dip and ‘l’ is the span of cable, then horizontal thrust developed at the support is
- Awl³/8yc
- Bwl²/3yc
- Cwl²/8ycCorrect
- Dwl²/4yc
Solution
Concept
Horizontal thrust in cable under UDL
Formula
H = wl2 / (8 × yc)
Step-by-step solution
1. For a cable supported at the same level subjected to a uniform distributed load w over span l, the bending moment at any section is zero. 2. The maximum sagging moment due to vertical loading at the center is wl2 / 8. 3. Equating this to horizontal thrust moment H × yc gives H = wl²/8yc. Answer: C
Explanation
For a cable subjected to a uniform distributed load w across a span l with central dip yc, taking moments about the central point gives H multiplied by yc equals w times l squared divided by 8. Solving for H yields the horizontal thrust formula wl²/8yc. This horizontal reaction remains constant along the entire length of the cable under uniform loading.
Why the other options are wrong
A) The span length l is squared in the formula, not cubed. B) The formula uses 8yc in the denominator rather than 3yc. D) The denominator requires a factor of 8yc instead of 4yc.
Shortcut
Horizontal thrust in a cable under UDL = wl²/8yc.
Common mistake
Confusing the span exponent as cubed (wl³/8y
Correct answer: C) wl²/8y_c
