---
title: "Max BM at 10 m: UDL and two point loads"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:beam-udl-4m-two-10kn-25m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-udl-4m-two-10kn-25m
site: BCT Engineering
---

# Max BM at 10 m: UDL and two point loads

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

- Asked in: 2081 Chaitra, 8 marks
- Video: [Influence line diagram moving udl in beam](https://www.youtube.com/watch?v=t2GHc6ovDUQ) — Civil learning online
- 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/beam-udl-4m-two-10kn-25m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine maximum bending moment at section 10 m from left support of a simply supported beam of span 25 m when the given load traverses across the span from left to right. [Figure: load system: 20 kN/m UDL of length 4 m, followed by two 10 kN point loads, the dimensions along the train being 4 m, 2 m, 2 m.]

## Short answer

Place the load system with a 10 kN load at 16 m from left support and the UDL covering 8–12 m. Using influence line ordinates, the maximum bending moment at 10 m is 520 kN m.

## Answer

**Data:** span $L = 25$ m, section $C$ at $a = 10$ m from the left support, $b = 15$ m. The train is a 20 kN/m UDL over 4 m (= 80 kN), then a 10 kN load 2 m beyond the UDL and a second 10 kN load another 2 m further. The train moves left to right, so the 10 kN loads lead.

#### ILD for $M_C$
Peak ordinate under $C$: $y_C = \dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. Ordinate at $x$ from $A$: $0.6x$ (left of $C$), $0.4(25-x)$ (right of $C$).

```text
   A o----------C---------------o B
                ^ 6.0
   ILD: 0 -> 6.0 at C -> 0
```

#### Critical position
Try positions of the leading 10 kN load (distance $p$ from $A$):

| Lead load at p (m) | M_C (kN m) |
|---|---|
| 14.0 | 480.0 |
| 15.0 | 510.0 |
| 16.0 | 520.0 |
| 17.0 | 510.0 |
| 18.0 | 480.0 |

The maximum is near $p = 16$ m. Then the loads are: 10 kN at 16 m, 10 kN at 14 m, and the UDL spans 8 m to 12 m (covering $C$).

Ordinates: at 16 m: $0.4\times 9 = 3.6$; at 14 m: $0.4\times 11 = 4.4$; under the UDL: $4.8$ at 8 m, $6.0$ at 10 m, $5.2$ at 12 m.

$$
\begin{aligned} M_{max} &= 10(3.6) + 10(4.4) + 20\left[\tfrac{4.8+6.0}{2}\times 2 + \tfrac{6.0+5.2}{2}\times 2\right]\\ &= 36 + 44 + 20\,(10.8 + 11.2) = 36+44+440 = 520\ \text{kN m} \end{aligned}
$$

**Answer: maximum bending moment at the section = 520 kN m.**

## Check yourself

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

1. What is the peak ILD ordinate for M at 10 m from the left of a 25 m span?
   A. 6 m
   B. 10 m
   C. 2.5 m
   D. 15 m
   Answer: A. y_C = ab/L = 10 x 15/25 = 6 m. (Key point: ILD peak)

2. What is the ILD ordinate at x from A to the right of the section C (x > 10)?
   A. 0.6(25 - x)
   B. 0.4(25 - x)
   C. 0.6x
   D. 0.4x
   Answer: B. Right of C the ordinate is a(L - x)/L = 10(25 - x)/25 = 0.4(25 - x). (Key point: Ordinate slope)

3. Where does the maximum moment occur from the trial table?
   A. Lead load at 14 m
   B. Lead load at 18 m
   C. Lead 10 kN load at about 16 m, with the UDL covering the section
   D. UDL entirely beyond the section
   Answer: C. Trials give 480, 510, 520, 510 and 480 kN m, with the maximum 520 at p = 16 m. (Key point: Critical position)

4. What is the maximum bending moment at the section?
   A. 480 kN m
   B. 440 kN m
   C. 600 kN m
   D. 520 kN m
   Answer: D. M = 36 + 44 + 440 = 520 kN m. (Key point: Result)

## Related in Theory of Structures I

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