---
title: "Suspension 150 m: UDL right half, SF and BM at 50 m"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-150m-two-cables-udl-right-half"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-150m-two-cables-udl-right-half
site: BCT Engineering
---

# Suspension 150 m: UDL right half, SF and BM at 50 m

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2078 Chaitra, 10 marks
- Video: [Cables - Problem No 1 (supported on the same level & subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=vEWoLBnX1FQ) — 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-150m-two-cables-udl-right-half
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension bridge has a span of 150 m attached with two 3 hinged stiffening girders supported by two cables. The central sag in the cable is 15 m. The UDL of 15 kN/m acts on the right-half of the span. Determine the shear force and bending moment at a distance of 50 m from the right end. Also, determine the maximum tension in the cable. Assume, the road has a width of 5 m.

## Short answer

Per girder: at 50 m from right end, shear force = 46.88 kN and bending moment = 2343.75 kN m (sagging). Maximum cable tension is 757.3 kN.

## Answer

**Data and assumptions:** span $L = 150$ m, two cables with two three-hinged stiffening girders, central sag $d = 15$ m. The UDL of 15 kN/m on the right half is the total load on the bridge, shared equally by the two girders and cables, so each girder-cable system carries $w = 7.5$ kN/m on the right half ($x = 75$ to 150 m). (The road width is not needed for the force calculation.) If 15 kN/m acted on each girder, all answers below would be doubled. Section at 50 m from the right end, i.e. $x = 100$ m from the left. Cable: $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{4\times 15\,x(150-x)}{22500}$.

#### Forces for one cable and girder
Total load on the right half $= 7.5\times 75 = 562.5$ kN acting at $x = 112.5$ m.

$$
R_A = \frac{562.5\times 37.5}{150} = 140.625\ \text{kN}, \qquad M_C = R_A\times 75 = 10546.875\ \text{kN m}
$$

The mid-span hinge gives:

$$
H = \frac{M_C}{d} = \frac{10546.875}{15} = 703.125\ \text{kN}
$$

#### SF and BM at x = 100 m
Cable ordinate: $y = \dfrac{4\times 15\times 100\times 50}{22500} = 13.333$ m; slope: $\tan\alpha = \dfrac{4\times 15\,(150-200)}{22500} = -0.1333$.

Simple-beam values: $M_b = R_A\times 100 - 7.5\times 25\times 12.5 = 11718.75$ kN m; $V_b = R_A - 7.5\times 25 = -46.875$ kN.

$$
\begin{aligned} M &= M_b - H\,y = 11718.75 - 703.125\times 13.333 = 2343.75\ \text{kN m}\\ V &= V_b - H\tan\alpha = -46.875 - 703.125\times(-0.1333) = 46.88\ \text{kN} \end{aligned}
$$

For the whole bridge (two girders) these values are doubled: $M = 4687.50$ kN m, $V = 93.75$ kN.

#### Maximum tension in the cable
The cable carries a uniform load $p = \dfrac{8Hd}{L^2} = 3.750$ kN/m (total $pL = 562.5$ kN), so it is a parabola and the vertical component at each support is $V_s = \dfrac{pL}{2} = 281.250$ kN.

$$
T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{703.125^2 + 281.250^2} = 757.3\ \text{kN (in each cable)}
$$

**Answer (per girder): $V = 46.88$ kN, $M = 2343.75$ kN m (sagging) at 50 m from the right end; maximum cable tension $= 757.3$ kN.**

## Check yourself

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

1. The 15 kN/m UDL on the right half acts on the whole bridge with two girders and two cables. What load does each girder-cable system carry?
   A. 15 kN/m
   B. 7.5 kN/m
   C. 3.75 kN/m
   D. 15 kN/m divided by the road width, 3 kN/m
   Answer: B. The total load is shared equally by the two girder-cable systems, so each carries 7.5 kN/m. Road width is not needed for the force calculation. (Key point: Load sharing between girders)

2. Which condition gives the horizontal cable pull H of a three-hinged stiffening girder?
   A. The shear force at the support is zero
   B. The bending moment in the girder at the mid-span hinge is zero, so H = M0C/d
   C. The cable tension at mid-span equals the support reaction
   D. The vertical reactions of the cable are equal
   Answer: B. The hinge cannot carry moment, so M0C - H d = 0 and H = M0C/d. (Key point: Horizontal pull from the hinge)

3. At x = 100 m, the simple-beam moment is 11718.75 kN m, H = 703.125 kN and the cable ordinate y = 13.333 m. What is the girder BM per girder?
   A. 11718.75 + 9375 = 21093.75 kN m
   B. 2343.75 kN m
   C. 9375 kN m
   D. 4687.5 kN m
   Answer: B. M = Mb - H y = 11718.75 - 703.125 x 13.333 = 2343.75 kN m. The 4687.5 value is the doubled figure for both girders. (Key point: Girder BM at a section)

4. For this cable, H = 703.125 kN and the load is uniform on the cable of total 562.5 kN. How is Tmax found?
   A. Tmax = H, since the cable is horizontal at the support
   B. Tmax = sqrt(H^2 + Vs^2) with Vs = 281.25 kN at each support
   C. Tmax = H + Vs = 984.4 kN
   D. Tmax = Vs with Vs = 562.5 kN
   Answer: B. Tension is the resultant of H and the vertical support component Vs = pL/2 = 281.25 kN, giving 757.3 kN. (Key point: Maximum cable tension)

## 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: 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 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 150 m: BM and SF at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-120-dead-5) — Theory of Structures I, chapter 7, asked 2×

## Related in other subjects

- [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
- [Two-hinged arch: horizontal reaction, BM at C](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-horizontal-reaction-bm-c-50knm) — Theory of Structures II, chapter 2
- [Two-hinged parabolic arch: BM at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-point-d-30-50) — Theory of Structures II, chapter 2