---
title: "Cable with four point loads, max tension in segments"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:cable-25m-different-levels-four-loads"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-25m-different-levels-four-loads
site: BCT Engineering
---

# Cable with four point loads, max tension in segments

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2075 Bhadra, 10 marks
- Video: [Cables - Problem No 8 ( With 4 concentrated loads and supports in different level )](https://www.youtube.com/watch?v=N7oFTlEZDbA) — 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-25m-different-levels-four-loads
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension cable is suspended between two supports at a distance 25 m and loaded as shown in figure below. Calculate maximum tension in each segment of cable, if level difference between A and B is 0.5 m and dip under second load is 2 m. [Figure: cable with five segments of 5 m each; vertical loads 30 kN, 40 kN, 25 kN and 30 kN at the four intermediate points 1, 2, 3 and 4; end B is 0.5 m higher than A; point 2 is 2 m below the chord line.]

## Short answer

Horizontal pull H = 245.0 kN. Segment tensions (A–1 to 4–B): 252.03, 246.72, 245.24, 247.62, 253.71 kN; maximum 253.71 kN in segment 4–B.

## Answer

The cable is a funicular polygon. Take A as the origin, the span as horizontal (25 m) and let $V_A$, $V_B$ be the vertical reactions and $H$ the (constant) horizontal pull. Loads: 30, 40, 25, 30 kN at $x$ = 5, 10, 15, 20 m. B is 0.5 m above A, and point 2 ($x$ = 10 m) is 2 m below the chord AB, i.e. $2 - 0.5\times\frac{10}{25} = 1.8$ m below the level of A.

#### Step 1: Equations for $H$ and $V_A$
Depth of the cable below A at a section $x$ is $\dfrac{V_A x - \sum P(x-x_i)}{H}$.

- At B (depth = $-0.5$ m, B is higher): $V_A(25) - [30(20)+40(15)+25(10)+30(5)] = -0.5H$, so $25V_A - 1600 = -0.5H$ ... (i)
- At point 2 (depth = 1.8 m): $V_A(10) - 30(5) = 1.8H$, so $10V_A - 150 = 1.8H$ ... (ii)

From (i), $V_A = 64 - 0.02H$. Put in (ii): $640 - 0.2H - 150 = 1.8H$, so $2H = 490$.

$$
\begin{aligned} H &= 245.0\ \text{kN} \\ V_A &= 64 - 0.02(245.0) = 59.1\ \text{kN} \\ V_B &= 125 - 59.1 = 65.9\ \text{kN} \end{aligned}
$$

Check: $V_A + V_B = 125$ kN = total load.

#### Step 2: Cable levels (depth below A)
Points 1, 2, 3, 4 are 1.206 m, 1.800 m, 1.578 m and 0.845 m below A; B is 0.5 m above A.

#### Step 3: Tension in each segment
Vertical component in a segment = $V_A$ minus the loads to its left; $T = \sqrt{H^2+V^2}$.

| Segment | Vertical component V (kN) | Tension T (kN) |
|---|---|---|
| 1 | 59.1 | 252.03 |
| 2 | 29.1 | 246.72 |
| 3 | -10.9 | 245.24 |
| 4 | -35.9 | 247.62 |
| 5 | -65.9 | 253.71 |

**Answer:** $H = 245.0$ kN; segment tensions are 252.03, 246.72, 245.24, 247.62 and 253.71 kN for segments A-1 to 4-B. The maximum tension is 253.71 kN in segment 4-B (next to the higher support B).

## Check yourself

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

1. In a cable with several point loads, which quantity is the same in every segment?
   A. The vertical component
   B. The horizontal component H
   C. The tension
   D. The slope
   Answer: B. With only vertical loads, the horizontal component H is constant; tension and vertical components vary segment by segment. (Key point: Funicular cable)

2. Point 2 is 2 m below the chord AB, and B is 0.5 m higher than A. How far below A is point 2?
   A. 2.5 m
   B. 2.0 m
   C. 1.8 m
   D. 1.5 m
   Answer: C. The chord is 0.5 x 10/25 = 0.2 m above A at x = 10 m, so the depth below A is 2 - 0.2 = 1.8 m. (Key point: Depth of point 2 below A)

3. How are H and VA obtained?
   A. From the depth equation at two points (B and point 2), solving two simultaneous equations
   B. By taking H = total load / 2
   C. By assuming VA = VB
   D. By taking H from the tension in the first segment
   Answer: A. The depth at a section is (VA x - sum P(x - xi))/H. Applying it at B and at point 2 gives two equations, solved for H = 245 kN and VA = 59.1 kN. (Key point: Equations for H and VA)

4. Why is the maximum tension in segment 4-B (253.71 kN)?
   A. It has the largest vertical component magnitude (65.9 kN) with constant H
   B. It has the largest horizontal force
   C. It is the longest segment
   D. Its vertical component is zero
   Answer: A. T = sqrt(H^2 + V^2) with constant H, so the segment with the largest V magnitude, 4-B, has the largest tension. (Key point: Maximum tension segment)

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