---
title: "Wheel load train: BM at 8 m and absolute max"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:train-120-80-160-wheel-load"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/train-120-80-160-wheel-load
site: BCT Engineering
---

# Wheel load train: BM at 8 m and absolute max

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

- Asked in: 2079 Chaitra, 10 marks
- Video: [Lect-90 Absolute max. bending moment any where in the span due to train wheel loads by using ILD](https://www.youtube.com/watch?v=dgJtsJ0PHOE) — Dr. Ashish Agrawal
- 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-120-80-160-wheel-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

A train of wheel load as shown below crosses a girder of 25 m span with 120 kN load leading. Determine the value of (i) maximum bending moment at the section 8 m from the left end of the girder, (ii) absolute maximum bending moment of the girder. [Figure: wheel loads 80 kN, 160 kN, 160 kN and 120 kN with spacings 2 m, 2 m and 3 m; 120 kN load leading; girder AB of 25 m, section C at 8 m from A.]

## Short answer

With rear 160 kN load placed at the section, maximum bending moment at 8 m is 2425.6 kN m. Absolute maximum bending moment is 2751.5 kN m under the 160 kN load placed 0.269 m from mid-span.

## Answer

**Data:** span $L = 25$ m. Wheel loads (from the rear): 80, 160, 160 and 120 kN at 2 m, 2 m and 3 m spacing; the 120 kN load leads (the train moves left to right). Total load $W = 520$ kN.

#### (i) Maximum BM at the section C, 8 m from the left end
ILD for $M_C$: peak ordinate $\dfrac{ab}{L} = \dfrac{8\times 17}{25} = 5.44$ m. Condition for maximum: the average load on the left of $C$, $\dfrac{W_L}{a}$, must be approximately equal to $\dfrac{W}{L}$ (it should exceed $W/L$ with the load at $C$ counted on the left, and fall below it without that load).

Try the rear 160 kN load (the second load from the rear) at $C$: the load on the left of $C$, including it, is $80 + 160 = 240$ kN.

$$
\frac{240}{8} = 30 > \frac{520}{25} = 20.8, \qquad \frac{80}{8} = 10 < 20.8
$$

So the condition is satisfied with the rear 160 kN load at $C$. Positions from the left support: 80 kN at 6 m, 160 kN at 8 m (at $C$), 160 kN at 10 m, 120 kN at 13 m.

| Load | Position | Ordinate (m) | Load x ordinate (kN m) |
|---|---|---|---|
| 80 | 6 m | 4.080 | 326.4 |
| 160 (rear) | 8 m | 5.440 | 870.4 |
| 160 (front) | 10 m | 4.800 | 768.0 |
| 120 | 13 m | 3.840 | 460.8 |

$$
M_{C,max} = 80(4.080) + 160(5.440) + 160(4.800) + 120(3.840) = 2425.6\ \text{kN m}
$$

(Check by reaction: $R_A = 323.2$ kN, $M_C = R_A\times 8 - 80\times 2 = 2425.6$ kN m.)

#### (ii) Absolute maximum bending moment
Resultant $R = 520$ kN. Its distance from the 120 kN (leading) load $= \dfrac{80\times 7 + 160\times 5 + 160\times 3}{520} = \dfrac{1840}{520} = 3.538$ m behind it. The load nearest to the resultant is the 160 kN load that is 3 m behind the 120 kN load, at 0.538 m from the resultant. Place this load and the resultant symmetrically about mid-span: this 160 kN load is 0.269 m on one side of the centre and the resultant 0.269 m on the other.

- 160 kN (critical) at 12.769 m from the left support; resultant at 12.231 m; 120 kN at 15.769 m; other 160 kN at 10.769 m; 80 kN at 8.769 m (all on the girder).
- $R_A = 520\times\dfrac{25-12.231}{25} = 265.60$ kN.

$$
M_{max} = R_A\times 12.769 - 80\times(12.769-8.769) - 160\times(12.769-10.769) = 2751.5\ \text{kN m}
$$

**Answer: (i) $M_C = 2425.6$ kN m; (ii) absolute maximum $BM = 2751.5$ kN m under the 160 kN load.**

## Check yourself

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

1. The condition for maximum BM at C is
   A. W_L/a on the left is about equal to W/L
   B. W_L = W_R
   C. The heaviest load is at mid-span
   D. The lightest load is at C
   Answer: A. The average load to the left of C should match the average over the span. (Key point: Trial condition)

2. For the section at 8 m, which load placed at C satisfies the test?
   A. The 120 kN load
   B. The rear 160 kN load
   C. The 80 kN load
   D. None of the loads
   Answer: B. 240/8 = 30 > 20.8 > 80/8 = 10. (Key point: Critical load)

3. What is the maximum BM at the section 8 m from the left?
   A. 2751.5 kN m
   B. 2000 kN m
   C. 2425.6 kN m
   D. 3000 kN m
   Answer: C. The sum of load times ordinate is 326.4 + 870.4 + 768 + 460.8 = 2425.6 kN m. (Key point: Result (i))

4. How is absolute maximum BM located?
   A. Place the heaviest load at mid-span
   B. Place the resultant at the support
   C. Place the first load at mid-span
   D. Place the load nearest the resultant and the resultant symmetrically about mid-span
   Answer: D. Absolute maximum BM occurs under a load when it and the resultant are equidistant from the centre. Its value here is 2751.5 kN m. (Key point: Absolute maximum)

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