---
title: "Suspension 100 m: UDL on right half, SF and BM at 20 m"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-100m-5knm2-10knm2-right-half"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-100m-5knm2-10knm2-right-half
site: BCT Engineering
---

# Suspension 100 m: UDL on right half, SF and BM at 20 m

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2076 Bhadra, 12 marks
- Video: [Cables supported on the same level subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=GGBX9CC5Iw0) — 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/suspension-100m-5knm2-10knm2-right-half
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A cable suspension bridge of 100 m span has two three hinged stiffening girder supported by two cables having a central dip of 10 m. The roadway has a width of 5 m. The dead load on the bridge is 5 kN/m² while the live load is 10 kN/m² which act on the right half of the span. Determine the SF and BM in the girder at 20 m from left support. Also find the maximum tension in the cable for this position of live load.

## Short answer

Per girder: at 20 m from left, V = -62.5 kN, M = -3750.0 kN m (hogging). Max tension in each cable = 3365.7 kN.

## Answer

**Data and assumptions:** span $L = 100$ m, two cables with two three-hinged stiffening girders, dip $d = 10$ m, roadway width 5 m. Dead load $5$ kN/m² and live load $10$ kN/m² (right half only). Load per metre run of the bridge: dead $5\times 5 = 25$ kN/m, live $10\times 5 = 50$ kN/m. Each of the two cables and girders carries half: **per cable/girder**, dead load $w_d = 12.5$ kN/m (whole span, taken by the cable) and live load $w_l = 25$ kN/m on the right half ($x = 50$ to 100 m). Cable: $y = 0.004\,x(100-x)$.

#### Horizontal pull (per cable)
$$
H_d = \frac{w_dL^2}{8d} = \frac{12.5\times 10^4}{80} = 1562.5\ \text{kN}
$$

Live load: total $25\times 50 = 1250$ kN acting at $x = 75$ m.

$$
R_A = \frac{1250\times 25}{100} = 312.5\ \text{kN}, \qquad M_C = R_A\times 50 = 15625.0\ \text{kN m}, \qquad H_l = \frac{M_C}{d} = 1562.5\ \text{kN}
$$

$$
H = H_d + H_l = 3125.0\ \text{kN}
$$

#### SF and BM in the girder at 20 m from the left support
The dead load is carried by the cable, so only the live load acts on the girder. At $x = 20$ m: $y = 0.004\times 20\times 80 = 6.40$ m, $\tan\alpha = 0.004\,(100-40) = 0.24$. For the section (left of the load): $M_b = R_A\times 20 = 6250.0$ kN m, $V_b = R_A = 312.5$ kN.

$$
\begin{aligned} M &= M_b - H_l\,y = 6250.0 - 1562.5\times 6.40 = -3750.0\ \text{kN m}\\ V &= V_b - H_l\tan\alpha = 312.5 - 1562.5\times 0.24 = -62.5\ \text{kN} \end{aligned}
$$

(These values are per girder; the negative moment is hogging.)

#### Maximum tension in the cable
At the supports, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$, $V_s = H\tan\alpha_s = 1250.0$ kN:

$$
T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{3125.0^2 + 1250.0^2} = 3365.7\ \text{kN (each cable)}
$$

**Answer (per girder): $V = -62.5$ kN, $M = -3750.0$ kN m at 20 m from the left support; maximum tension in each cable $= 3365.7$ kN.**

## Check yourself

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

1. How is the load per cable/girder found from the area loads?
   A. Multiply by width 5 m only: dead 25 kN/m, live 50 kN/m on each girder
   B. Multiply by width and halve for two girders: dead 12.5 kN/m, live 25 kN/m
   C. Divide by width: 1 and 2 kN/m
   D. Multiply by width and by span: 2500 kN and 5000 kN
   Answer: B. Width gives 25 and 50 kN/m for the bridge; two cables and girders each take half. (Key point: Converting area loads)

2. Which load acts on the stiffening girder for SF and BM?
   A. Dead plus live
   B. Live load only, since the cable carries the dead load
   C. Dead load only
   D. Neither, because the hinge removes all load
   Answer: B. The full-span dead load is carried by the cable (parabola), so only the live load causes girder SF and BM, though both add to H. (Key point: Dead load on girder)

3. What is the total H?
   A. 1562.5 kN
   B. 3125.0 kN
   C. 4687.5 kN
   D. 1250 kN
   Answer: B. Hd = 12.5 x 100^2/80 = 1562.5 kN and Hl = 15625/10 = 1562.5 kN, so H = 3125 kN. (Key point: Total horizontal pull)

4. At 20 m, M = 6250 - 1562.5 x 6.4 = -3750 kN m. What does the negative sign mean?
   A. Hogging moment
   B. Sagging moment
   C. Zero shear
   D. The cable is in compression
   Answer: A. A negative girder moment means hogging. The cable pull H y exceeds the simple-beam moment there. (Key point: Sign of girder BM)

## Close variants set in other exams

- [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×

## Related in Theory of Structures I

- [Suspension 150 m: UDL right half, SF and BM at 50 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-two-cables-udl-right-half) — Theory of Structures I, chapter 7
- [Suspension 120 m: SF, BM at 25 m, cable area](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-road6m-5knm2-200-150) — Theory of Structures I, chapter 7
- [Suspension 100 m: SF and BM at 30 m, tension](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-200-300kn-30m) — Theory of Structures I, chapter 7
- [Suspension 150 m: four moving loads, BM at 40 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-four-moving-loads-200-150-300-100) — Theory of Structures I, chapter 7
- [Suspension 150 m: SF, BM at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-150-dead-7-5) — Theory of Structures I, chapter 7, asked 2×
- [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×

## Related in other subjects

- [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2
- [Two-hinged arch: horizontal reaction, UDL on half](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-12knm-horizontal-reaction-50m-8m) — Theory of Structures II, chapter 2
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