---
title: "Cable with UDL: length, H and max tension"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:cable-45m-udl-20knm-different-levels"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-45m-udl-20knm-different-levels
site: BCT Engineering
---

# Cable with UDL: length, H and max tension

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2071 Magh, 10 marks
- Video: [Cables supported on different levels subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=R4ALc2U5zdQ) — 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-45m-udl-20knm-different-levels
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A cable is suspended and loaded as shown in figure. Calculate: (a) length of cable, (b) horizontal component of tension in cable, (c) magnitude and position of maximum tension occurring in cable. [Figure: cable between supports 45 m apart horizontally; 20 kN/m UDL over the horizontal span; left support A is 2 m and right support B is 8 m above the lowest point of the cable as read from the scan.]

## Short answer

Cable length is 46.54 m, horizontal tension component H = 1125.0 kN. Maximum tension is 1275.00 kN, occurring at the higher support B.

## Answer

**Set-up.** Take the lowest point O of the cable as origin. A is 2 m and B is 8 m above O. For a parabolic cable under a UDL on the horizontal span, $y = \dfrac{wx^2}{2H}$.

#### Position of the lowest point
$$
\frac{x_A^2}{x_B^2} = \frac{y_A}{y_B} = \frac{2}{8} \Rightarrow x_A = 0.5\,x_B,\qquad x_A + x_B = 45
$$
So $x_B = 30.0$ m and $x_A = 15.0$ m. The lowest point is 15 m from A horizontally (30 m from B).

#### (b) Horizontal component of tension
$$
H = \frac{w x_B^2}{2y_B} = \frac{20(30)^2}{2(8)} = 1125.0\ \text{kN}
$$
Check from A: $\dfrac{20(15)^2}{2(2)} = 1125.0$ kN.

#### (c) Maximum tension and its position
Tension is greatest at the higher support B (30 m from the lowest point):
$$
V_B = w x_B = 20(30) = 600\ \text{kN},\qquad T_{max} = \sqrt{H^2 + V_B^2} = \sqrt{1125.0^2 + 600^2} = 1275.00\ \text{kN}
$$
At A: $V_A = 300$ kN and $T_A = 1164.31$ kN.

#### (a) Length of the cable
For each side, with end slope $s = 2y/x$: $l = \dfrac{x}{2}\left[\sqrt{1+s^2} + \dfrac{\sinh^{-1}s}{s}\right]$

| Side | x (m) | y (m) | s | Length (m) |
|---|---|---|---|---|
| O to A | 15.0 | 2 | 0.2667 | 15.176 |
| O to B | 30.0 | 8 | 0.5333 | 31.367 |

Total length = 46.54 m.

**Answer:** (a) length = 46.54 m; (b) $H$ = 1125.0 kN; (c) $T_{max}$ = 1275.00 kN at support B (the higher end).

## Check yourself

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

1. Supports A and B are 2 m and 8 m above the lowest point O, with a 45 m horizontal span. How is the position of O found?
   A. Using x_A/x_B = y_A/y_B = 2/8, so the horizontal distances are in the ratio 1:2
   B. Using x_A/x_B = sqrt(y_A/y_B) = 1/2, so the distances are in the ratio 1:2
   C. Using x_A = x_B = 22.5 m since the span is symmetric
   D. Using x_A/x_B = y_B/y_A = 4
   Answer: B. Since y = wx^2/2H, x^2 is proportional to y, so x_A^2/x_B^2 = 2/8 and x_A = 0.5 x_B. With x_A + x_B = 45 this gives 15 m and 30 m. (Key point: Locating the lowest point)

2. Which expression gives the horizontal tension H for the cable (w = 20 kN/m, x_B = 30 m, y_B = 8 m)?
   A. H = w x_B / y_B = 75 kN
   B. H = w x_B^2 / (2 y_B) = 1125 kN
   C. H = w x_B^2 / y_B = 2250 kN
   D. H = w (45)^2 / (8 y_B) = 633 kN
   Answer: B. From the parabola y = wx^2/2H, H = w x^2/(2y). Using B gives 20(900)/16 = 1125 kN, which agrees with the check from A. (Key point: Horizontal tension H)

3. Where does the maximum tension occur and why?
   A. At the lowest point, because H acts there alone
   B. At the lower support A, because it is closer to the lowest point
   C. At the higher support B, because it is farthest from the lowest point and carries the largest vertical component
   D. At mid-span (22.5 m), because the load is symmetric
   Answer: C. H is constant, so T = sqrt(H^2 + V^2) is largest where V = wx is largest, at the end farthest horizontally from O, which is B (30 m). (Key point: Position of maximum tension)

4. What is T_max at B?
   A. sqrt(1125^2 + 600^2) = 1275 kN
   B. 1125 + 600 = 1725 kN
   C. sqrt(1125^2 + 300^2) = 1164 kN
   D. 600 kN
   Answer: A. V_B = w x_B = 20(30) = 600 kN and T is the resultant of H and V_B, giving 1275 kN. The 1164 kN value is the tension at A. (Key point: Computing T_max)

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