---
title: "ILDs for BM and SF at 10 m, 25 m beam"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-beam-bm-sf-10m-25m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-beam-bm-sf-10m-25m
site: BCT Engineering
---

# ILDs for BM and SF at 10 m, 25 m beam

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

- Asked in: 2068 Bhadra, 8 marks
- Video: [Influence Line Diagram Rolling Load Part I  |Maximum Shear Force in Section](https://www.youtube.com/watch?v=yfDeuexd0Nc) — Parash Joshi - Civil Construction and Tutor
- 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/ild-beam-bm-sf-10m-25m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw the influence lines for bending moment and shear force at a section 10 m from the left support of a simply supported beam of 25 m span.

## Short answer

For a simply supported beam of 25 m span, the ILD for bending moment at 10 m is a triangle with peak ab/L = 6 m. The shear ILD has two parallel lines: −0.4 left of the section and +0.6 right, with a jump of 1 at the section.

## Answer

**Data:** simply supported beam $AB$, $L = 25$ m. Section $C$ at $a = 10$ m from $A$, $b = 15$ m. Unit load at $x$ from $A$.

#### Reactions
$R_A = \dfrac{25-x}{25}$, $R_B = \dfrac{x}{25}$.

#### ILD for bending moment at C
- Load left of $C$ ($x\le 10$): take the right part: $M_C = R_B\times 15 = 0.6x$
- Load right of $C$ ($x\ge 10$): take the left part: $M_C = R_A\times 10 = 0.4\,(25-x)$

Peak under $C$: $\dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. The ILD is a triangle: 0 at $A$, $+6$ at $C$, 0 at $B$.

```text
  A o----------C-------------------o B
               /\ 6.0
  ILD M_C:  0 /  \ 0
```

#### ILD for shear force at C
- Load left of $C$ ($x<10$): $V_C = -R_B = -\dfrac{x}{25}$; at $C^-$: $-0.4$
- Load right of $C$ ($x>10$): $V_C = R_A = \dfrac{25-x}{25}$; at $C^+$: $+0.6$

```text
  ILD V_C:   0 \ -0.4 | +0.6 \ 0     (two parallel lines, jump of 1 at C)
```

| Position | $A$ (0) | $C^-$ (10) | $C^+$ (10) | $B$ (25) |
|---|---|---|---|---|
| $M_C$ (m) | 0 | 6 | 6 | 0 |
| $V_C$ | 0 | −0.4 | +0.6 | 0 |

Areas: $M_C$ ILD $=\tfrac12\times 25\times 6 = 75$ m²; $V_C$ ILD: positive $\tfrac12\times 15\times 0.6 = 4.5$ m, negative $\tfrac12\times 10\times 0.4 = 2$ m.

For a UDL $w$ over the whole span, $M_C = 75w$ and $V_C = 4.5w - 2w = 2.5w$; for a single load $P$ at $x$, $M_C = P\,y$ and $V_C = P\,y$ with $y$ read from the ILD.

## Check yourself

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

1. For a section 10 m from the left support of a 25 m span, what is the peak ordinate of the M_C ILD?
   A. 10 m
   B. 15 m
   C. 6 m
   D. 12.5 m
   Answer: C. Peak = ab/L = 10 x 15/25 = 6 m, at the section, with zero at both supports. (Key point: Peak of BM ILD)

2. What are the ordinates of the shear force ILD just left and right of the section?
   A. -0.4 and +0.6
   B. -0.6 and +0.4
   C. +0.4 and -0.6
   D. -0.4 and +0.4
   Answer: A. At C- V = -R_B = -10/25 = -0.4; at C+ V = R_A = 15/25 = +0.6. The jump is 1. (Key point: Shear ILD ordinates)

3. What is the shape of the V_C influence line?
   A. A single triangle
   B. Two parallel straight lines with a jump of 1 at C
   C. A parabola
   D. Two lines meeting at a peak at C
   Answer: B. Both segments have slope 1/L, so they are parallel, and a unit load crossing C changes the shear by 1. (Key point: Shape of the shear ILD)

4. For a UDL w over the whole span, V_C = ?
   A. 4.5w
   B. 2.5w
   C. 6.5w
   D. 75w
   Answer: B. V_C = w x (positive area - negative area) = w(4.5 - 2) = 2.5w; 75w is the bending moment. (Key point: UDL over full span)

## Close variants set in other exams

- [ILDs for reactions, SF and BM, 20 m beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-ild-reactions-sf-bm-5m-section) — Theory of Structures I, chapter 5, asked 2×

## Related in Theory of Structures I

- [ILD at mid-span: BM and SF under given loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-mid-span-bm-sf-50-10-30) — Theory of Structures I, chapter 5
- [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: 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×
- [Arch 40 m: ILDs for thrust, BM at 8 m, shear](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-40m-rise10-ild-thrust-bm-shear) — Theory of Structures I, chapter 6
- [ILD for SF and BM at C of beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-sf-bm-c-40-60-5knm) — Theory of Structures I, chapter 5
- [Overhang beam: max BM and SF at D, either direction](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-max-bm-sf-d-either-direction) — Theory of Structures I, chapter 5

## Related in other subjects

- [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4
- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4
- [ILD: moment at section 5, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-section-5-two-span) — Theory of Structures II, chapter 4