---
title: "ILDs for reactions, SF and BM, 20 m beam"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:beam-ild-reactions-sf-bm-5m-section"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-ild-reactions-sf-bm-5m-section
site: BCT Engineering
---

# ILDs for reactions, SF and BM, 20 m beam

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

- Asked in: 2075 Baisakh, 6 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/beam-ild-reactions-sf-bm-5m-section
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw the influence lines for support reactions, shear force and bending moment at a section 5 m from the left support of a simply supported beam of 20 m span.

## Short answer

Influence line for RA: 1 at A to 0 at B. For RB: 0 at A to 1 at B. Shear at C: 0 at A, -0.25 just left of C, +0.75 just right, 0 at B. Bending moment at C: triangle peaking at 3.75 m under C.

## Answer

**Data:** simply supported beam $AB$, $L = 20$ m. Section $C$ at $a = 5$ m from $A$, $b = 15$ m. Let a unit load be at a distance $x$ from $A$.

#### ILD for reaction $R_A$
$R_A = \dfrac{20-x}{20}$: a straight line from 1 at $A$ to 0 at $B$.

#### ILD for reaction $R_B$
$R_B = \dfrac{x}{20}$: a straight line from 0 at $A$ to 1 at $B$.

#### ILD for shear force at C
- Load to the left of $C$ ($x<5$): from the right part, $V_C = -R_B = -\dfrac{x}{20}$. At $C$ the ordinate is $-\dfrac{5}{20} = -0.25$.
- Load to the right of $C$ ($x>5$): $V_C = R_A = \dfrac{20-x}{20}$. At $C$ the ordinate is $+\dfrac{15}{20} = +0.75$.

So the ILD is two parallel lines: from 0 at $A$ to $-0.25$ just left of $C$, a jump of 1 at $C$ to $+0.75$, and then down to 0 at $B$.

#### ILD for bending moment at C
- Load left of $C$: $M_C = R_B\times 15 = \dfrac{15x}{20}$.
- Load right of $C$: $M_C = R_A\times 5 = \dfrac{5(20-x)}{20}$.
- Peak at $C$: $\dfrac{ab}{L} = \dfrac{5\times 15}{20} = 3.75$ m.

```text
 R_A:  1 \
          \_____ 0 at B
 SF_C: 0 \-0.25 | +0.75 \ 0
 BM_C: 0 /\ 3.75 (at C) \ 0
```

| Quantity | at A | just left of C | just right of C | at B |
|---|---|---|---|---|
| $R_A$ | 1 | 0.75 | 0.75 | 0 |
| $R_B$ | 0 | 0.25 | 0.25 | 1 |
| $V_C$ | 0 | −0.25 | +0.75 | 0 |
| $M_C$ (m) | 0 | 3.75 | 3.75 | 0 |

The positive area of the SF ILD is $\tfrac12\times 15\times 0.75 = 5.625$ m and the negative area is $\tfrac12\times 5\times 0.25 = 0.625$ m. The area of the BM ILD is $\tfrac12\times 20\times 3.75 = 37.5$ m².

## Check yourself

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

1. What are the ILDs of R_A and R_B for the 20 m beam?
   A. R_A falls from 1 at A to 0 at B; R_B rises from 0 at A to 1 at B
   B. Both are triangles with peak at the section
   C. R_A rises from 0 to 1; R_B falls from 1 to 0
   D. Both are constant 0.5
   Answer: A. R_A = (20-x)/20 and R_B = x/20 are straight lines with unit ordinates at their own supports. (Key point: ILD of reactions)

2. What are the shear ordinates just left and right of C (5 m from A)?
   A. +0.25 and -0.75
   B. -0.25 and +0.75 with a jump of 1
   C. -0.75 and +0.25
   D. -0.25 and +0.25
   Answer: B. Left of C, V = -R_B = -5/20 = -0.25; right of C, V = R_A = 15/20 = +0.75; the jump is 1. (Key point: Shear ILD at C)

3. What is the peak ordinate of the M_C influence line?
   A. 5 m
   B. 3.75 m
   C. 15 m
   D. 7.5 m
   Answer: B. ab/L = 5 x 15 / 20 = 3.75 m, at the section. (Key point: Peak of moment ILD)

4. What is the area of the BM influence line?
   A. 37.5 m^2
   B. 75 m^2
   C. 3.75 m^2
   D. 18.75 m^2
   Answer: A. It is a triangle: 1/2 x 20 x 3.75 = 37.5 m^2. (Key point: Area of BM ILD)

## Close variants set in other exams

- [ILDs for BM and SF at 10 m, 25 m beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-beam-bm-sf-10m-25m) — 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
- [ILD: reactions, BM and SF beneath loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-reactions-sf-bm-beneath-loads) — Theory of Structures I, chapter 5, asked 2×
- [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — Theory of Structures I, chapter 6
- [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 20 m: reactions, BM, thrust, shear, BMD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-20m-udl-2knm-left-8m) — Theory of Structures I, chapter 6
- [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2×

## Related in other subjects

- [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: 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
- [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