---
title: "Four loads on 100 m girder: BM at C, absolute"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:train-100-160-180-120-girder-100m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/train-100-160-180-120-girder-100m
site: BCT Engineering
---

# Four loads on 100 m girder: BM at C, absolute

Theory of Structures I · Chapter 5: Influence Lines for Simple Structures

- Asked in: 2074 Bhadra, 12 marks
- Video: [Absolute Maximum Shear and Moment Due To Series of Concentrated Loads in a Simply Supported Beam](https://www.youtube.com/watch?v=LKdXQyK1hgk) — Master Civil Engineering
- 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/train-100-160-180-120-girder-100m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine maximum bending moment at section C and also the absolute maximum bending moment when the set of concentrated loads moves from left to right of the girder shown in figure below. [Figure: loads 100 kN, 160 kN, 180 kN, 120 kN with spacings 4 m, 3 m, 2 m; girder AB of 100 m span; section C at 35 m from A.]

## Short answer

Maximum bending moment at C is obtained by placing the 160 kN load at C, giving 12081.0 kN m (or 12171.0 kN m if the train direction is reversed). Absolute maximum bending moment is 13293.9 kN m, occurring under the 180 kN load when placed at 50.84 m from A.

## Answer

**Data:** girder $AB$, $L = 100$ m; section $C$ at $a = 35$ m from $A$, $b = 65$ m. Wheel loads from the rear: 100, 160, 180, 120 kN at spacings 4 m, 3 m, 2 m; the train moves left to right (assumed as drawn, so the 120 kN load leads). Total $W = 560$ kN.

#### (i) Maximum BM at C
ILD for $M_C$: peak $\dfrac{ab}{L} = \dfrac{35\times 65}{100} = 22.75$ m; ordinate $0.65x$ for $x\le 35$ and $0.35(100-x)$ for $x\ge 35$.

Place the 160 kN load at $C$. Load on the left of $C$ including it: $100 + 160 = 260$ kN.

$$
\frac{260}{35} = 7.43 > \frac{560}{100} = 5.6 > \frac{100}{35} = 2.86
$$

The condition for maximum is satisfied. Positions from $A$: 100 kN at 31 m, 160 kN at 35 m (at $C$), 180 kN at 38 m, 120 kN at 40 m.

| Load (kN) | Position (m) | Ordinate (m) | P x y (kN m) |
|---|---|---|---|
| 100 | 31 | 20.150 | 2015.0 |
| 160 | 35 | 22.750 | 3640.0 |
| 180 | 38 | 21.700 | 3906.0 |
| 120 | 40 | 21.000 | 2520.0 |

$$
M_{C,max} = 12081.0\ \text{kN m}
$$

(If the train moves the other way, with the 100 kN load leading, the same method gives 12171.0 kN m.)

#### (ii) Absolute maximum BM
Resultant from the 120 kN (leading) load: $\bar x = \dfrac{100(9)+160(5)+180(2)}{560} = 3.679$ m behind it. For each load, place it and the resultant symmetrically about mid-span and find the moment under the load:

| Load at mid-span position | Position of load (m) | M under it (kN m) |
|---|---|---|
| 100 kN | 47.34 | 12549.6 |
| 160 kN | 49.34 | 13232.4 |
| 180 kN | 50.84 | 13293.9 |
| 120 kN | 51.84 | 12988.9 |

The largest is under the 180 kN load. With this load at 50.84 m from $A$, the loads are at 100 kN: 43.84 m, 160 kN: 47.84 m, 180 kN: 50.84 m, 120 kN: 52.84 m (all on the girder).

$$
R_A = 284.7\ \text{kN}, \qquad M_{max} = R_A\times 50.84 - (\text{moments of the loads on its left})= 13293.9\ \text{kN m}
$$

**Answer: (i) $M_C = 12081.0$ kN m; (ii) absolute maximum BM = 13293.9 kN m under the 180 kN load.**

## Check yourself

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

1. Which condition checks that a load at C gives the maximum BM at C?
   A. (W left incl. load)/a > W/L > (W left excl. load)/a
   B. W/L = P/a
   C. W left = W right
   D. Load at mid-span only
   Answer: A. 260/35 = 7.43 > 560/100 = 5.6 > 100/35 = 2.86 confirms the 160 kN load at C. (Key point: Criterion for maximum BM at C)

2. What is the maximum BM at C (35 m from A)?
   A. 12081.0 kN m
   B. 13293.9 kN m
   C. 22.75 kN m
   D. 12171.0 kN m
   Answer: A. Sum of P x y with ordinates 20.15, 22.75, 21.70, 21.00 gives 12081.0 kN m. (Key point: Maximum BM at C)

3. How is the absolute maximum BM position found?
   A. Place the heaviest load at mid-span
   B. Place a load and the resultant of the train symmetrically about mid-span and find the moment under that load
   C. Place the resultant at the support
   D. Place the leading load at C
   Answer: B. The moment under a load is greatest when it and the resultant are equidistant from mid-span. (Key point: Absolute maximum method)

4. Which load gives the absolute maximum BM and its value?
   A. 100 kN load, 12549.6 kN m
   B. 160 kN load, 13232.4 kN m
   C. 180 kN load, 13293.9 kN m
   D. 120 kN load, 12988.9 kN m
   Answer: C. The largest of the trial values is under the 180 kN load, 13293.9 kN m. (Key point: Absolute maximum result)

## Close variants set in other exams

- [Four loads on 60 m girder: BM at C, absolute](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atrain-170-215-150-120-60m) — Theory of Structures I, chapter 5, asked 2×

## Related in Theory of Structures I

- [Four loads on 50 m girder: BM at 20 m, absolute](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Afour-loads-100-150-200-150-50m) — Theory of Structures I, chapter 5
- [Four wheel loads on 20 m beam: absolute max BM](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Afour-wheel-30-80-60-30-20m) — Theory of Structures I, chapter 5
- [Wheel load train: BM at 8 m and absolute max](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atrain-120-80-160-wheel-load) — Theory of Structures I, chapter 5
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- [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×

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