---
title: "Suspension 120 m: UDL on left half, SF and BM at 30 m"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-120m-live-load-left-half"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-120m-live-load-left-half
site: BCT Engineering
---

# Suspension 120 m: UDL on left half, SF and BM at 30 m

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2069 Bhadra, 12 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-live-load-left-half
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A suspension bridge of 120 m span has two three hinged stiffening girder supported by two cables having a central dip of 12 m. The road way has a width of 6 m. The dead load on the bridge is 5 kN/m² while the live load is 10 kN/m² which act on the left half of span. Determine the shear force and bending moment in the girder at 30 m from left end. Also find maximum tension in the cable for this position of live load.

## Short answer

Per girder, H = 4500 kN. At 30 m from the left end, M = 6750.0 kN m sagging and S = 0.0 kN. Maximum cable tension for this loading is 4846.65 kN in each cable.

## Answer

**Loads per girder.** The roadway is 6 m wide and is carried by two girders and two cables, each taking half:

- Dead load: $5\times 6 = 30$ kN/m on the bridge, so $w_d$ = 15 kN/m per girder
- Live load on the left half (0 to 60 m): $10\times 6 = 60$ kN/m, so $w_l$ = 30 kN/m per girder

Span $L$ = 120 m, dip $d$ = 12 m.

#### Cable pull $H$
- Dead load: $H_d = \dfrac{w_d L^2}{8d} = \dfrac{15(120)^2}{8(12)} = 2250.0$ kN
- Live load on the left half: $R_A = \dfrac{30(60)(90)}{120} = 1350.0$ kN, then $M_{0C} = 1350.0(60) - 30(60)(30) = 27000.0$ kN m
- $H_l = \dfrac{M_{0C}}{d} = \dfrac{27000}{12} = 2250.0$ kN

$$
H = H_d + H_l = 2250.0 + 2250.0 = 4500.0\ \text{kN}
$$

#### SF and BM in the girder at 30 m from the left end
The dead load causes no girder BM or SF (the cable carries it). For the live load:

$$
y_{30} = \frac{4(12)(30)(90)}{120^2} = 9.00\ \text{m},\qquad \frac{dy}{dx} = \frac{4(12)(120-60)}{120^2} = 0.20
$$

$$
\begin{aligned} M_{0,30} &= 1350(30) - 30(30)(15) = 27000.0\ \text{kN m} \\ M_{30} &= M_{0,30} - H_l\,y_{30} = 27000.0 - 2250.0(9.00) = 6750.0\ \text{kN m} \\ S_{30} &= \left[R_A - 30(30)\right] - H_l\,\frac{dy}{dx} = 450.0 - 2250.0(0.20) = 0.0\ \text{kN} \end{aligned}
$$

#### Maximum tension in the cable for this loading
$\tan\theta = 4d/L = 0.40$ at the supports, so $\sec\theta = 1.0770$:
$$
T_{max} = H\sec\theta = 4500.0(1.0770) = 4846.65\ \text{kN}
$$

**Answer:** At 30 m from the left end, $M = 6750.0$ kN m (sagging) and $S = 0.0$ kN (the shear is zero, so this is the section of maximum BM in the loaded half); maximum cable tension = 4846.65 kN in each cable.

## Check yourself

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

1. Roadway 6 m wide, dead 5 and live 10 kN/m^2, two girders. What are wd and wl per girder?
   A. 30 kN/m and 60 kN/m
   B. 15 kN/m and 30 kN/m
   C. 5 kN/m and 10 kN/m
   D. 15 kN/m and 60 kN/m
   Answer: B. Area loads times 6 m give 30 and 60 kN/m, and each girder takes half: 15 and 30 kN/m. (Key point: Load per girder)

2. What is the total H?
   A. 2250 kN
   B. 4500 kN
   C. 6750 kN
   D. 1125 kN
   Answer: B. Hd = 15 x 120^2/96 = 2250 kN and Hl = 27000/12 = 2250 kN, so H = 4500 kN. (Key point: Horizontal pull)

3. At 30 m, S = 450 - 2250 x 0.20 = 0. What does this indicate?
   A. The BM is zero there
   B. The BM is a maximum in the loaded half
   C. The cable is slack
   D. The load ends there
   Answer: B. Zero shear means dM/dx = 0, so the BM of 6750 kN m is a maximum in the loaded half. (Key point: Meaning of zero shear)

4. How is M at 30 m calculated?
   A. M = M0 - Hl y = 27000 - 2250 x 9 = 6750 kN m
   B. M = M0 - H y = 27000 - 4500 x 9
   C. M = M0 + Hl y
   D. M = Hl y = 20250
   Answer: A. Only the live load acts on the girder, so the live-load pull Hl is used, not the total H (dead load is carried by the cable). (Key point: Girder BM)

## Close variants set in other exams

- [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×

## 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: 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×
- [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

## 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 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
- [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