---
title: "Cable over 300 m gap: tension and length"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:cable-gap-300m-1-5m-dip"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-gap-300m-1-5m-dip
site: BCT Engineering
---

# Cable over 300 m gap: tension and length

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2065 Chaitra, 8 marks
- Video: [Cables - Problem No 1 (supported on the same level & subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=vEWoLBnX1FQ) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-gap-300m-1-5m-dip
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A cable is stretched over a gap of 300 m and carries uniformly distributed load of 300 kN/m horizontally. If the central dip is 1.5 m, calculate the maximum tension in the cable. Also find the length of cable. [Load value read as 300 kN/m from the scan, partly unclear.]

## Short answer

Horizontal tension H = wL^2/(8d) = 2.25×10^6 kN, vertical reaction V = wL/2 = 45,000 kN. Maximum tension at supports is Tmax = sqrt(H^2+V^2) ≈ 2,250,450 kN. Cable length S ≈ 300 + 8×1.5^2/(3×300) = 300.020 m.

## Answer

**Data:** span $L = 300$ m, central dip $d = 1.5$ m (supports at the same level), uniformly distributed load $w = 300$ kN/m on the horizontal span (the value is as read from the scan; the method is the same for any $w$). The cable is a parabola.

#### Horizontal tension
$$
H = \frac{wL^2}{8d} = \frac{300\times 300^2}{8\times 1.5} = 2250000\ \text{kN}
$$

#### Vertical reaction at each support
$$
V = \frac{wL}{2} = \frac{300\times 300}{2} = 45000\ \text{kN}
$$

#### Maximum tension (at the supports)
$$
T_{max} = \sqrt{H^2 + V^2} = \sqrt{(2250000)^2 + (45000)^2} = 2250450.0\ \text{kN}
$$

Slope at the supports: $\tan\alpha = \dfrac{4d}{L} = \dfrac{6}{300} = 0.02$, $\alpha = 1.15^\circ$.

#### Length of the cable
$$
S = L\left[1 + \frac{8}{3}\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 300\left[1 + \frac{8}{3}(0.005)^2 - \frac{32}{5}(0.005)^4\right] = 300.020\ \text{m}
$$

**Answer: $T_{max} = 2250450.0$ kN (at the supports); length of the cable $\approx 300.020$ m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives the horizontal tension of a cable with level supports?
   A. H = wL/(8d)
   B. H = wL^2/(8d)
   C. H = wL^2/(2d)
   D. H = wd^2/(8L)
   Answer: B. For a parabolic cable under UDL w on the span, H = wL^2/(8d) = 2,250,000 kN here. (Key point: Horizontal tension)

2. Where is the tension maximum and how is it computed?
   A. At mid-span; T = H
   B. At the supports; T = sqrt(H^2 + V^2) with V = wL/2
   C. At mid-span; T = V
   D. At the supports; T = H + V
   Answer: B. The slope is greatest at the supports, where the vertical component V = wL/2 combines with H as T = sqrt(H^2 + V^2) = 2,250,450 kN. (Key point: Where maximum tension occurs)

3. How is the length of the parabolic cable obtained?
   A. S = L + d
   B. S = L [1 + (8/3)(d/L)^2 - (32/5)(d/L)^4], about 300.020 m
   C. S = 2L
   D. S = L / (1 + 8d/3L)
   Answer: B. The binomial series gives S = L + 8d^2/(3L) - ..., and for d/L = 0.005 the cable is only about 0.02 m longer than the span. (Key point: Cable length)

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