---
title: "ILD: reactions, BM and SF beneath loads"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-reactions-sf-bm-beneath-loads"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-reactions-sf-bm-beneath-loads
site: BCT Engineering
---

# ILD: reactions, BM and SF beneath loads

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

- Asked in: 2070 Bhadra, 10 marks; 2066 Kartik, 10 marks (asked 2 times)
- Video: [Influence Line Diagrams for Simply Supported Beams - Problem No 7  (Absolute SF & BM)](https://www.youtube.com/watch?v=dwk9gBGj6HI) — Stan Academy
- 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-reactions-sf-bm-beneath-loads
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Use influence line diagrams to determine reactions at the supports, bending moments and shear forces beneath the applied forces in the beam shown in the earlier question (simply supported beam 0-1, 240 kN at 3 m from support 0 and 160 kN at 4.5 m from support 1, span 14 m).

## Short answer

Reactions: R0 = 240 kN, R1 = 160 kN. Bending moment under both loads is 720 kN m. Shear force: +240 kN from support 0 to 3 m, zero between loads, −160 kN from 9.5 m to support 1.

## Answer

**Data:** simply supported beam $0$–$1$, $L = 14$ m. $P_1 = 240$ kN at 3 m from support 0; $P_2 = 160$ kN at 4.5 m from support 1, i.e. 9.5 m from support 0.

#### Reaction at support 0
ILD for $R_0$: a straight line from 1 at support 0 to 0 at support 1; the ordinate at distance $s$ from 0 is $(L-s)/L$.

$$
R_0 = 240\left(\frac{11}{14}\right) + 160\left(\frac{4.5}{14}\right) = 240.0\ \text{kN}
$$

#### Reaction at support 1
Ordinate $s/L$:

$$
R_1 = 240\left(\frac{3}{14}\right) + 160\left(\frac{9.5}{14}\right) = 160.0\ \text{kN}
$$

(Check: $R_0 + R_1 = 400$ kN = total load.)

#### Bending moment beneath the loads
The ILD for $M$ at a section $x$ is a triangle with peak $x(L-x)/L$. For a load at $s$ the ordinate is $s(L-x)/L$ if $s\le x$, and $x(L-s)/L$ if $s\ge x$.

**Under 240 kN ($x = 3$ m):** peak ordinate $=3\times 11/14 = 2.357$; ordinate under the 160 kN load $=3\times 4.5/14 = 0.964$.

$$
M_{3} = 240(2.357) + 160(0.964) = 720.0\ \text{kN m}
$$

**Under 160 kN ($x = 9.5$ m):** peak ordinate $=9.5\times 4.5/14 = 3.054$; ordinate under 240 kN $=3\times 4.5/14 = 0.964$.

$$
M_{9.5} = 240(0.964) + 160(3.054) = 720.0\ \text{kN m}
$$

#### Shear force beneath the loads
The ILD for shear at $x$ has ordinate $-s/L$ for $s<x$ and $+(L-s)/L$ for $s>x$ (a jump of 1 at the section). A load standing exactly at the section counts as being on the right of the section for "just left" and on the left for "just right".

| Section | Just left | Just right |
|---|---|---|
| Under 240 kN (3 m) | $240\left(\tfrac{11}{14}\right) + 160\left(\tfrac{4.5}{14}\right) = +240.0$ kN | $-240\left(\tfrac{3}{14}\right) + 160\left(\tfrac{4.5}{14}\right) = 0.0$ kN |
| Under 160 kN (9.5 m) | $-240\left(\tfrac{3}{14}\right) + 160\left(\tfrac{4.5}{14}\right) = 0.0$ kN | $-240\left(\tfrac{3}{14}\right) - 160\left(\tfrac{9.5}{14}\right) = -160.0$ kN |

**Answer: $R_0 = 240$ kN, $R_1 = 160$ kN; $M = 720$ kN m under the 240 kN load and $720$ kN m under the 160 kN load; the shear force is +240 kN from support 0 to the 240 kN load, 0 between the loads, and −160 kN from the 160 kN load to support 1.**

## Check yourself

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

1. What is the ILD ordinate for R0 at distance s from support 0?
   A. (L - s)/L
   B. s/L
   C. s(L - s)/L
   D. 1
   Answer: A. R0 varies linearly from 1 at support 0 to 0 at support 1. (Key point: Reaction ILD)

2. What are R0 and R1 for 240 kN at 3 m and 160 kN at 9.5 m on a 14 m span?
   A. R0 = 160 kN, R1 = 240 kN
   B. R0 = 240 kN, R1 = 160 kN
   C. R0 = 200 kN, R1 = 200 kN
   D. R0 = 280 kN, R1 = 120 kN
   Answer: B. R0 = 240(11/14) + 160(4.5/14) = 240 kN and R1 = 400 - 240 = 160 kN. (Key point: Reaction values)

3. What is the peak ILD ordinate for bending moment at 3 m from support 0?
   A. 3 x 3/14 = 0.643 m
   B. 3 m
   C. 3 x 11/14 = 2.357 m
   D. 11/14 = 0.786 m
   Answer: C. Peak ordinate is x(L - x)/L. (Key point: BM ordinate)

4. What is the shear force between the two loads?
   A. +240 kN
   B. -160 kN
   C. +80 kN
   D. 0 kN
   Answer: D. The reaction 240 kN is cancelled by the 240 kN load, so the shear between loads is zero. (Key point: Shear between loads)

## 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
- [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
- [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×
- [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
- [Cantilever ILD for SF and BM at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-cantilever-sf-bm-c) — 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

- [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
- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Flexural rigidity; T-beam bending stresses (60 kN loads)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aflexural-rigidity-t-beam-60kn) — Strength of Materials, chapter 7
- [Beam thrust and bending moment diagrams with max moments](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-thrust-bmd-2kn-5kn-1knm) — Strength of Materials, chapter 1