---
title: "Four loads on 50 m girder: BM at 20 m, absolute"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:four-loads-100-150-200-150-50m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/four-loads-100-150-200-150-50m
site: BCT Engineering
---

# Four loads on 50 m girder: BM at 20 m, absolute

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

- Asked in: 2077 Chaitra, 10 marks
- Video: [Absolute maximum bending moment | structural Analysis| rolling load](https://www.youtube.com/watch?v=VMNJdTQoQeg) — KK  Sir ki classes
- 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/four-loads-100-150-200-150-50m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine maximum bending moment at a point at a distance 20 m from right support in a simply supported girder of span 50 m, when four concentrated loads of 100 kN, 150 kN, 200 kN and 150 kN each separated from adjacent load by 3 m move from right to left with 100 kN load leading. Also determine the absolute maximum bending moment in the beam.

## Short answer

Place the 200 kN load at the section 20 m from the right support. The ILD peak ordinate is 12 m; resulting BM = 6510.0 kN m. Absolute maximum BM occurs with the 200 kN load 0.5 m left of mid-span, giving 6753.0 kN m.

## Answer

**Data:** span $L = 50$ m. Four loads 100, 150, 200, 150 kN, 3 m apart, moving from right to left with the 100 kN load leading. Total $W = 600$ kN. Distances below are measured from the **right** support (where the train enters); the section $C$ is at $a = 20$ m from it, $b = 30$ m.

#### (i) Maximum BM at C
ILD for $M_C$: peak ordinate $\dfrac{ab}{L} = \dfrac{20\times 30}{50} = 12$ m. Trial: the 200 kN load at $C$. The loads between the entry support and $C$ are the 150 kN (17 m), so $W_L = 150$ kN without the 200 kN and $350$ kN with it.

$$
\frac{350}{20} = 17.5 \ge \frac{600}{50} = 12 \ge \frac{150}{20} = 7.5
$$

The condition is satisfied, so the 200 kN load at $C$ gives the maximum. Positions from the right support: 100 kN at 26 m, 150 kN at 23 m, 200 kN at 20 m (at $C$), 150 kN at 17 m.

| Load | Position from right support | Ordinate (m) | Load x ordinate (kN m) |
|---|---|---|---|
| 100 | 26 m | 9.600 | 960.0 |
| 150 | 23 m | 10.800 | 1620.0 |
| 200 | 20 m | 12.000 | 2400.0 |
| 150 | 17 m | 10.200 | 1530.0 |

$$
M_{C,max} = 6510.0\ \text{kN m}
$$

Check with other loads placed at $C$ (kN m): 100 kN at $C$: 5400.0; first 150 kN at $C$: 6180.0; last 150 kN at $C$: 6240.0; 200 kN at $C$: 6510.0. The largest is 6510.0 kN m.

#### (ii) Absolute maximum BM
Resultant from the leading load: $\bar x = \dfrac{150(3)+200(6)+150(9)}{600} = 5.0$ m behind the 100 kN load, i.e. 1.0 m from the 200 kN load. Place the 200 kN load and the resultant symmetrically about mid-span: 200 kN at 24.50 m, resultant at 25.50 m from the right support.

$$
R_{right} = 600\times\frac{50-25.50}{50} = 294.0\ \text{kN}
$$

Moment under the 200 kN load, taking the part of the girder between the right support and the 200 kN load (it carries one 150 kN load, 3 m from the 200 kN load):

$$
M_{max} = R_{right}\times 24.50 - 150\times 3 = 6753.0\ \text{kN m}
$$

All four loads lie on the girder (the train spans 9 m).

**Answer: (i) maximum BM at 20 m from the right support = 6510.0 kN m; (ii) absolute maximum BM = 6753.0 kN m (under the 200 kN load, 0.5 m from mid-span).**

## Check yourself

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

1. Which load at C gives the maximum BM at 20 m from the right support?
   A. The 200 kN load
   B. The 100 kN load
   C. The first 150 kN load
   D. The last 150 kN load
   Answer: A. 350/20 = 17.5 >= 12 >= 150/20 = 7.5 is satisfied. (Key point: Critical load)

2. What is the ILD peak ordinate at the section?
   A. 20 m
   B. 12 m
   C. 30 m
   D. 25 m
   Answer: B. ab/L = 20 x 30/50 = 12. (Key point: Peak ordinate)

3. What is the maximum BM at C?
   A. 6753 kN m
   B. 6240 kN m
   C. 6510 kN m
   D. 5400 kN m
   Answer: C. Sum of load x ordinate = 960 + 1620 + 2400 + 1530 = 6510 kN m. (Key point: Result (i))

4. Where is the 200 kN load placed for the absolute maximum?
   A. At mid-span
   B. At the support
   C. 1 m from mid-span, same side as the resultant
   D. 0.5 m from mid-span, on the opposite side to the resultant, which is 1.0 m from it
   Answer: D. The load and resultant are placed symmetrically about mid-span, giving 6753 kN m. (Key point: Absolute max)

## Related in Theory of Structures I

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