---
title: "Suspension 120 m: SF, BM at 25 m, cable area"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-120m-road6m-5knm2-200-150"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-120m-road6m-5knm2-200-150
site: BCT Engineering
---

# Suspension 120 m: SF, BM at 25 m, cable area

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2071 Magh (old course), 16 marks
- Video: [Cables - Problem No 13 ( Three Hinged Stiffening Girder)](https://www.youtube.com/watch?v=cf-8qzK6pMo) — 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-120m-road6m-5knm2-200-150
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension cable bridge has a three hinged girder supported by two cables. The roadway is 6 m wide. The girder has its self weight 5 kN/m². The live load consists of two concentrated loads 200 kN at 20 m from left support and 150 kN at 10 m right from the 200 kN load. The span is 120 m and central dip is 12 m. The live loads are acting at the central of the girder. Determine shear force and bending moment at section 25 m from left support of the girder. Also determine required cross sectional area of the cable if the allowable tensile stress of cable material is 120 N/mm².

## Short answer

Per girder, H = 2427.08 kN. At 25 m from left support, M = 1587.67 kN m sagging and S = -1.74 kN. Maximum cable tension is 2614.05 kN, so required cable area is 21784 mm² per cable.

## Answer

**Load shares.** The bridge has two cables and two girders, so each takes half. Dead load: $5\times 6 = 30$ kN/m, so 15 kN/m per girder. The live loads (200 kN at 20 m and 150 kN at 30 m from the left support, both on the centre line) give 100 kN and 75 kN per girder. $L$ = 120 m, $d$ = 12 m.

#### Cable pull $H$
$$
\begin{aligned} R_A &= \frac{100(100)+75(90)}{120} = 139.58\ \text{kN} \\ M_{0C} &= 139.58(60) - 100(40) - 75(30) = 2125.0\ \text{kN m} \\ H_p &= \frac{2125}{12} = 177.08\ \text{kN}, \qquad H_d = \frac{15(120)^2}{8(12)} = 2250.0\ \text{kN} \\ H &= 2250.0 + 177.08 = 2427.08\ \text{kN} \end{aligned}
$$

#### SF and BM at 25 m (one of the two girders)
The dead load is carried entirely by the cable and produces no moment or shear in the girder.
$$
y_{25} = \frac{4(12)(25)(95)}{120^2} = 7.9167\ \text{m},\qquad \frac{dy}{dx} = \frac{4(12)(120-50)}{120^2} = 0.2333
$$

$$
\begin{aligned} M_{25} &= R_A(25) - 100(5) - H_p\,y_{25} = 2989.58 - 1401.91 = 1587.67\ \text{kN m} \\ S_{25} &= (R_A - 100) - H_p\,\frac{dy}{dx} = 39.58 - 41.32 = -1.74\ \text{kN} \end{aligned}
$$

(If the whole roadway is taken as carried by a single girder, the girder values double: $M_{25} = 3175.3$ kN m and $S_{25} = -3.47$ kN. The cable force per cable is unchanged.)

#### Cross-sectional area of the cable
Maximum tension (at the supports, $\tan\theta = 4d/L = 0.40$, $\sec\theta = 1.0770$):
$$
T_{max} = H\sec\theta = 2427.08(1.0770) = 2614.05\ \text{kN}
$$
$$
A = \frac{T_{max}}{\sigma_{allow}} = \frac{2614.05\times 10^3}{120} = 21784\ \text{mm}^2
$$

**Answer:** $M_{25} = 1587.67$ kN m (sagging); $S_{25} = -1.74$ kN (almost zero, so 25 m is close to the section of maximum BM); required cable area = 21784 mm² for each of the two cables.

## Check yourself

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

1. Dead load 5 kN/m^2 on a 6 m road, two girders, live loads 200 and 150 kN. What does each girder take?
   A. 30 kN/m, 200 kN and 150 kN
   B. 15 kN/m, 100 kN and 75 kN
   C. 5 kN/m, 100 kN and 75 kN
   D. 15 kN/m, 200 kN and 150 kN
   Answer: B. The loads are shared equally between the two cables and girders. (Key point: Load per girder)

2. Why is the dead load not included in M25 and S25?
   A. It is carried entirely by the cable
   B. It is small
   C. It acts at the supports
   D. It is balanced by the live load
   Answer: A. A uniform load on a three-hinged girder is carried by the parabolic cable, producing no girder BM or SF. It still adds to H. (Key point: Girder forces)

3. S25 = -1.74 kN, about zero. What does it indicate?
   A. The section is near the point of maximum BM
   B. The girder is unstable
   C. The BM is zero
   D. The cable is slack
   Answer: A. Shear near zero means the BM curve is near a maximum at 25 m. (Key point: Meaning of S25 near zero)

## Related in Theory of Structures I

- [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×
- [Suspension 120 m: BMD and BM at 25, 40, 80 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-120m-240kn-300kn-bmd) — Theory of Structures I, chapter 7
- [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
- [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×
- [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

## Related in other subjects

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- [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
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- [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