---
title: "Suspension 400 m: max/min tension, cable length"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-400m-dip30-3x300kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-400m-dip30-3x300kn
site: BCT Engineering
---

# Suspension 400 m: max/min tension, cable length

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2068 Bhadra, 16 marks
- Video: [Suspension Bridges Unsymmetrical Cable Problem 1 | Tension in Cables Supported At Different Level](https://www.youtube.com/watch?v=peLiegSKNes) — Ganesh Shegar
- 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/suspension-400m-dip30-3x300kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension bridge, 400 m span, has two three-hinged stiffening girders supported by two cables with a central dip of 30 m. The dead load of the bridge is 30 kN/m run and in addition, it supports three point loads of 300 kN each placed along the centre line of the roadway and dividing the span in four equal parts. Calculate the maximum tension and minimum tension with their locations in the cable and the length of the cable.

## Short answer

Per cable, H = 11000 kN. Maximum tension is 11484.34 kN at the supports; minimum is 11000.0 kN at mid-span. Cable length is 405.92 m per cable.

## Answer

**Assumptions.** The data are for the whole bridge; two girders and two cables share the load equally, so per cable $w$ = 30/2 = 15 kN/m and $P$ = 300/2 = 150 kN. (If a single cable is intended, all forces below double: $H$ = 22000 kN, $T_{max}$ = 22968.7 kN. The length is unchanged.) Span 400 m, dip 30 m, point loads at 100, 200 and 300 m.

#### Horizontal pull $H$
- Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{15(400)^2}{8(30)} = 10000.0$ kN
- Point loads: $R_A = 1.5P = 225.0$ kN, $M_{0C} = 225.0(200) - 150(100) = 30000.0$ kN m, so $H_p = \dfrac{30000}{30} = 1000.0$ kN

$$
H = 10000.0 + 1000.0 = 11000.0\ \text{kN}
$$

#### Maximum and minimum tension
$\tan\theta = 4d/L = 0.300$ at the supports, $\sec\theta = 1.04403$.

- **Maximum** (at the two supports): $T_{max} = H\sec\theta = 11000.0(1.04403) = 11484.34$ kN
- **Minimum** (at mid-span, where the cable is horizontal): $T_{min} = H = 11000.0$ kN

#### Length of the cable
Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.075$:
$$
l = 400\left[1+\frac{8}{3}(0.075)^2-\frac{32}{5}(0.075)^4\right] = 405.92\ \text{m}
$$

**Answer:** $T_{max} = 11484.34$ kN at the supports; $T_{min} = 11000.0$ kN at mid-span; length of cable = 405.92 m (per cable, loads shared equally by the two cables).

## Check yourself

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

1. The 30 kN/m and 300 kN loads are for the whole bridge with two cables. What are w and P per cable?
   A. 30 kN/m and 300 kN
   B. 15 kN/m and 150 kN
   C. 60 kN/m and 600 kN
   D. 15 kN/m and 300 kN
   Answer: B. Two cables share equally, halving both. (Key point: Load per cable)

2. What is H?
   A. 10000 kN
   B. 1000 kN
   C. 11000 kN
   D. 12500 kN
   Answer: C. Hd = 15 x 400^2/240 = 10000 kN and Hp = 30000/30 = 1000 kN, so H = 11000 kN. (Key point: Horizontal pull)

3. Where are Tmax and Tmin, and why?
   A. Tmax at the supports (H sec theta), Tmin at mid-span (= H)
   B. Tmax at mid-span, Tmin at the supports
   C. Both equal H everywhere
   D. Tmax at the quarter points
   Answer: A. The slope is 0.3 at the supports so T = 11000 x 1.04403 = 11484.34 kN; at mid-span the cable is horizontal so T = H. (Key point: Maximum and minimum tension)

4. If the loads were for a single cable instead, what changes?
   A. All forces double but the cable length is unchanged
   B. Only H doubles
   C. The length doubles
   D. Nothing changes
   Answer: A. Forces scale with load (H = 22000 kN), while the length depends only on span and dip. (Key point: Effect of single-cable assumption)

## Close variants set in other exams

- [Suspension 200 m: max/min tension, cable length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-200m-dip20-3x800kn) — Theory of Structures I, chapter 7, asked 3×
- [Suspension 100 m: max/min tension, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-dip10-3x200kn-dead-40) — Theory of Structures I, chapter 7, asked 3×

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