---
title: "Cable at different levels: length and thrust"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:cable-50m-different-levels-udl-15"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-50m-different-levels-udl-15
site: BCT Engineering
---

# Cable at different levels: length and thrust

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2080 Chaitra, 10 marks
- Video: [Cable Support at Different Level Problem 3 - Structural Analysis 1](https://www.youtube.com/watch?v=h11y8VBiO4k) — Ekeeda
- 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-50m-different-levels-udl-15
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A cable is suspended between supports at different levels with span of 50 m. The left support is 3 m below the right and the lowest point of the cable is 5 m below the lower support. The cable carries a udl of 15 kN/m on the entire span. Calculate the length of the cable, horizontal thrust and maximum tension in the cable.

## Short answer

Using parabolic cable with vertex at lowest point: horizontal thrust H = 731.0 kN, maximum tension T_max = 842.5 kN at the higher support, and cable length S = 52.19 m.

## Answer

**Data:** span $L = 50$ m; the left support is 3 m below the right one; the lowest point of the cable is 5 m below the lower (left) support. So the heights above the lowest point are $h_1 = 5$ m (left) and $h_2 = 5 + 3 = 8$ m (right). UDL $w = 15$ kN/m on the horizontal span. The cable is a parabola with its vertex at the lowest point.

#### Horizontal thrust
With $y = \dfrac{w x^2}{2H}$ from the lowest point and $x_1 + x_2 = L$:

$$
H = \frac{wL^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2} = \frac{15\times 50^2}{2\left(\sqrt5+\sqrt8\right)^2} = \frac{37500}{2\times(5.0645)^2} = 731.0\ \text{kN}
$$

#### Position of the lowest point
$$
x_1 = \frac{L\sqrt{h_1}}{\sqrt{h_1}+\sqrt{h_2}} = 22.08\ \text{m (from the left support)}, \qquad x_2 = 50 - x_1 = 27.92\ \text{m}
$$

Check: $\dfrac{w x_1^2}{2H} = 5.000$ m and $\dfrac{w x_2^2}{2H} = 8.000$ m.

#### Maximum tension
The tension is greatest at the higher (right) support:

$$
T_{2} = \sqrt{H^2 + (w x_2)^2} = \sqrt{731.0^2 + (15\times 27.92)^2} = 842.5\ \text{kN}
$$

(At the left support, $T_1 = \sqrt{H^2 + (w x_1)^2} = 802.5$ kN.)

#### Length of the cable
For each portion from the lowest point, with $u = \dfrac{w x}{H}$:

$$
s = \frac{H}{w}\cdot\frac{u\sqrt{1+u^2} + \sinh^{-1}u}{2}
$$

Left portion: $u_1 = 0.4530$, $s_1 = 22.81$ m. Right portion: $u_2 = 0.5730$, $s_2 = 29.38$ m.

$$
S = s_1 + s_2 = 52.19\ \text{m}
$$

(Approximation: $s \approx x\left[1 + \tfrac23\left(\tfrac{h}{x}\right)^2\right]$ gives 52.28 m.)

**Answer: $H = 731.0$ kN; maximum tension $= 842.5$ kN (at the higher support); length of the cable $= 52.19$ m.**

## Check yourself

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

1. The left support is 3 m below the right, and the lowest point is 5 m below the left support. What are h1 and h2?
   A. h1 = 5 m, h2 = 3 m
   B. h1 = 3 m, h2 = 5 m
   C. h1 = 5 m, h2 = 8 m
   D. h1 = 8 m, h2 = 5 m
   Answer: C. The left (lower) support is 5 m above the lowest point; the right is 3 m higher, so 8 m. (Key point: Heights above lowest point)

2. Which formula gives the horizontal thrust for supports at different levels?
   A. H = wL^2/(8 h2)
   B. H = wL^2/(2 (sqrt h1 + sqrt h2)^2), about 731 kN
   C. H = wL^2/(h1 + h2)
   D. H = wL/2
   Answer: B. From y = w x^2/(2H) and x1 + x2 = L, H = 15 x 2500/(2 x 5.0645^2) = 731 kN. (Key point: Horizontal thrust formula)

3. How far from the left support is the lowest point?
   A. x1 = L/2 = 25 m
   B. x1 = L sqrt(h2)/(sqrt h1 + sqrt h2) = 27.92 m
   C. x1 = L sqrt(h1)/(sqrt h1 + sqrt h2) = 22.08 m
   D. x1 = L h1/(h1 + h2) = 19.2 m
   Answer: C. x1 is proportional to sqrt(h1), giving 50 x 2.236/5.0645 = 22.08 m (the lowest point is nearer the lower support). (Key point: Position of the lowest point)

4. Where is the tension maximum and what is it?
   A. At the higher right support, sqrt(H^2 + (w x2)^2) = 842.5 kN
   B. At the lowest point, 731 kN
   C. At the left support, 802.5 kN
   D. At mid-span, 842.5 kN
   Answer: A. The vertical component w x is larger at the higher support (x2 = 27.92 m), so T is largest there. (Key point: Maximum tension)

## Related in Theory of Structures I

- [Cable with UDL: length, H and max tension](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-45m-udl-20knm-different-levels) — Theory of Structures I, chapter 7
- [Cable on piers: tension, length, pier thrust (180 m)](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-180m-5m-above-60-degree-backstay) — Theory of Structures I, chapter 7, asked 2×
- [Cable on piers: tension, length, pier thrust (200 m)](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-200m-5m-above-60-degree-backstay) — Theory of Structures I, chapter 7, asked 2×
- [Cable 80 m: lowest point, length, tensions](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-80m-different-levels-udl-1-5) — Theory of Structures I, chapter 7
- [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4×
- [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×

## Related in other subjects

- [Horizontal length of 30 m span, 50 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-50m-span-30m-26c) — Surveying I, chapter 2
- [Frame AFD, SFD, BMD (50 kN, 15 kN/m)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-50kn-15knm) — Strength of Materials, chapter 1
- [Left and right focal point ratios](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afocal-point-ratios-left-right) — Theory of Structures II, chapter 4
- [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2