---
title: "UDL transferred to main cable from point load"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:udl-transferred-to-main-cable"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/udl-transferred-to-main-cable
site: BCT Engineering
---

# UDL transferred to main cable from point load

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2075 Bhadra, 6 marks
- Video: [Cables Problem No - 10 ( Point Load & Uniformly Distributed Load  )](https://www.youtube.com/watch?v=B3N4jQDWbHY) — 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/udl-transferred-to-main-cable
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

Calculate the uniform distributed load transferred to the main cable due to a point load applied to the three hinged stiffening girder. Assume suitable data.

## Short answer

For a three-hinged stiffening girder of span L, a point load W at distance x from the left end (x ≤ L/2) transfers a uniform load p = 4Wx/L^2 to the cable; for x ≥ L/2, p = 4W(L-x)/L^2. Example: W=100 kN, x=10 m, L=50 m gives p = 1.6 kN/m.

## Answer

In a suspension bridge with a **three-hinged stiffening girder** (hinges at the two ends and at mid-span $C$), the hangers carry a point load to the cable as a uniform load.

#### Assumptions
The hangers are closely spaced and inextensible, so the cable transmits an upward **uniform** load $p$ per unit length to the girder. The cable is a parabola of dip $d$.

#### Derivation
Let a point load $W$ act on the girder (span $L$) at a distance $x$ from the left end, $x\le L/2$.

```text
        W
   A    |        C(hinge)       B
   o----+---------o-------------o
   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^   p per unit length (from the cable)
```

The bending moment at the middle hinge is zero. Treating the girder as a simple beam under $W$ and the upward uniform load $p$:

$$
M_C = \frac{W x}{2} - \frac{p L^2}{8} = 0 \Rightarrow p = \frac{4Wx}{L^2}
$$

Here $\dfrac{Wx}{2}$ is the mid-span moment of the simple beam under $W$ (the reaction $R_B = Wx/L$ times the lever arm $L/2$). For a load on the other half, put $L - x$ in place of $x$:

$$
p = \frac{4W\,(L-x)}{L^2} \quad (x\ge L/2)
$$

#### Result
The load transferred to the main cable is a uniformly distributed load of intensity

$$
p = \frac{4Wx}{L^2}\ \ (x\le L/2), \qquad p = \frac{4W(L-x)}{L^2}\ \ (x\ge L/2)
$$

The total load on the cable is $pL = \dfrac{4Wx}{L}$, which is $4\times$ the simple-beam reaction $R_B = Wx/L$. It is largest ($p = \dfrac{2W}{L}$) when $W$ is at mid-span. The horizontal pull in the cable is

$$
H = \frac{pL^2}{8d} = \frac{Wx}{2d}
$$

**Example:** $W = 100$ kN at $x = 10$ m on a girder of span $L = 50$ m: $p = \dfrac{4\times 100\times 10}{2500} = 1.6$ kN/m.

## Check yourself

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

1. Which condition is used to derive the uniform load p carried by the cable from a point load on a three-hinged stiffening girder?
   A. Bending moment at the middle hinge C is zero
   B. Shear at the support is zero
   C. Total vertical load is zero
   D. Cable tension equals the load W
   Answer: A. The girder has a hinge at mid-span, so W x/2 - p L^2/8 = 0. (Key point: Condition used)

2. For a load W at distance x (x <= L/2), what is the load intensity on the cable?
   A. p = W/L
   B. p = 4Wx/L^2
   C. p = 2Wx/L
   D. p = Wx/(2L^2)
   Answer: B. M_C = Wx/2 - pL^2/8 = 0 gives p = 4Wx/L^2. (Key point: Resulting intensity)

3. When is p largest and what is its value?
   A. Load at the support, p = 4W/L
   B. Load at the quarter point, p = W/L
   C. Load at the support, p = 0
   D. Load at mid-span, p = 2W/L
   Answer: D. Putting x = L/2 gives p = 2W/L; the total load on the cable pL = 4Wx/L is twice the simple-beam reaction. (Key point: Maximum intensity)

4. For W = 100 kN at x = 10 m on L = 50 m, what is p?
   A. 0.8 kN/m
   B. 3.2 kN/m
   C. 1.6 kN/m
   D. 2.0 kN/m
   Answer: C. p = 4 x 100 x 10/2500 = 1.6 kN/m, and then H = pL^2/(8d). (Key point: Thrust and example)

## Related in Theory of Structures I

- [Cable area, BMD of girder, cable length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-girder-hinge-8m-cable-area) — Theory of Structures I, chapter 7
- [Conjugate beam: free end of overhang, UDL and point load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-10kn-4m-3m-2071b) — Theory of Structures I, chapter 4
- [Suspension 120 m: girder with different support levels](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-hinged-ends-middle-6m-diff) — Theory of Structures I, chapter 7
- [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
- [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2×
- [Suspension 120 m: UDL on left half, SF and BM at 30 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-live-load-left-half) — Theory of Structures I, chapter 7, asked 2×

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