---
title: "Suspension 120 m: BMD and BM at 25, 40, 80 m"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:suspension-120m-240kn-300kn-bmd"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-120m-240kn-300kn-bmd
site: BCT Engineering
---

# Suspension 120 m: BMD and BM at 25, 40, 80 m

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2066 Kartik, 10 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-240kn-300kn-bmd
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A symmetrical suspension bridge with a three hinged stiffening girder of span 120 m and having a central dip of 12 m is loaded with two point loads of magnitude 240 kN and 300 kN at a distance 25 m and 80 m respectively from the left end. Draw bending moment diagram for the girder and also calculate bending moments at the distances 25 m, 40 m and 80 m from the left support.

## Short answer

H = 750.00 kN. Bending moments are M25 = 1312.50 kN m sagging, M40 = 0.00 kN m, and M80 = 2000.00 kN m sagging. The BMD is zero at supports and at the central hinge.

## Answer

Single three-hinged stiffening girder: $L$ = 120 m, $d$ = 12 m, loads 240 kN at 25 m and 300 kN at 80 m from the left end A.

#### Cable pull $H$
$$
\begin{aligned} R_A &= \frac{240(95)+300(40)}{120} = 290.00\ \text{kN} \\ M_{0C} &= 290.00(60) - 240(35) = 9000.0\ \text{kN m} \\ H &= \frac{M_{0C}}{d} = \frac{9000}{12} = 750.00\ \text{kN} \end{aligned}
$$

#### Bending moment in the girder
$M = M_0 - H\,y$ with $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{4(12)x(120-x)}{14400}$.

| x (m) | $M_0$ (kN m) | y (m) | $Hy$ (kN m) | $M$ (kN m) |
|---|---|---|---|---|
| 25 | 7250.00 | 7.917 | 5937.50 | 1312.50 |
| 40 | 8000.00 | 10.667 | 8000.00 | 0.00 |
| 60 | 9000.00 | 12.000 | 9000.00 | 0.00 |
| 80 | 10000.00 | 10.667 | 8000.00 | 2000.00 |

$M = 0$ at both supports and at the central hinge C (60 m).

In each region, $M$ is a quadratic in $x$:

- $0 \le x \le 25$: $M = 2.5x^2 - 10x$
- $25 \le x \le 80$: $M = 2.5x^2 - 250x + 6000$ (zero at 40 m and 60 m, minimum $-250$ kN m at 50 m)
- $80 \le x \le 120$: $M = 2.5x^2 - 550x + 30000$ (zero at 100 m and 120 m, minimum $-250$ kN m at 110 m)

```text
BMD (kN m): sagging above the line, hogging below
                                *

                               * *
          *                   *   *
         * *                 *     *
        *
       *    *               *       *
      *      *             *         *
    **        **         **           **
****------------**-----**---------------**-----**
                  *****                   *****
(top = 2000.0, bottom = -250.0; '-' is zero line)
```

The BMD is sagging from A to 40 m, with a peak of 1312.5 kN m under the 240 kN load (apart from a negligible hogging of about -10 kN m within the first 4 m). It is hogging from 40 m to 60 m (minimum -250 kN m at 50 m, just left of hinge C), sagging from 60 m to 100 m with a peak of 2000 kN m under the 300 kN load, and hogging again from 100 m to B (minimum -250 kN m at 110 m).

**Answer:** $H = 750.00$ kN; $M_{25} = 1312.50$ kN m (sagging), $M_{40} = 0.00$ kN m, $M_{80} = 2000.00$ kN m (sagging, positive).

## Check yourself

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

1. How is H found for the single girder?
   A. H = M0C/d = 9000/12 = 750 kN
   B. H = RA = 290 kN
   C. H = M0C x d
   D. H = (240 + 300)/2
   Answer: A. The moment at the hinge is zero, M0C - H d = 0, so H = 750 kN. (Key point: Cable pull)

2. What is the formula for the girder BM at a section?
   A. M = M0 + H y
   B. M = M0 - H y
   C. M = H y - RA x
   D. M = M0 / y
   Answer: B. The cable's horizontal pull acts at an offset y, reducing the simple-beam moment: M = M0 - H y. (Key point: Girder BM expression)

3. Why is the BM at 40 m exactly zero?
   A. M0 = 8000 kN m equals H y = 750 x 10.667 = 8000 kN m
   B. It is a support
   C. The load is there
   D. The shear is zero
   Answer: A. At 40 m the simple-beam moment equals the cable moment, so M = 0 there. (Key point: Zero BM at 40 m)

4. What is the BM at 25 m?
   A. 7250 kN m
   B. 5937.5 kN m
   C. 1312.5 kN m
   D. 13187.5 kN m
   Answer: C. M = 7250 - 750 x 7.917 = 1312.5 kN m. (Key point: BM at 25 m)

## Related in Theory of Structures I

- [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 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 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 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 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 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 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 parabolic arch: horizontal reaction, symmetric loads](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-secant-symmetric-loads-80m) — Theory of Structures II, chapter 2
- [Suspension bridge triangle ABC; span AB](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Asuspension-bridge-triangle-abc-span-ab) — Surveying I, chapter 7
- [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