---
title: "Derive structural quantities using ILD"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:ild-derive-structural-quantities"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ild-derive-structural-quantities
site: BCT Engineering
---

# Derive structural quantities using ILD

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

- Asked in: 2076 Baisakh, 6 marks; 2073 Bhadra, 6 marks; 2073 Magh, 6 marks; 2072 Asoj, 6 marks (asked 4 times)
- Video: [DERIVATION OF INFLUENCE LINE DIAGRAM | SIMPLY SUPPORTED BEAM | MOVING UNIT LOAD | ROLLING LOAD | ILD](https://www.youtube.com/watch?v=WxryiiLU7N4) — Structural Funda by:- KheNiSiR
- 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-derive-structural-quantities
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Derive the expression for calculation of structural quantities by using influence line diagram when the loads applied are concentrated force, uniformly distributed load and couple.

## Short answer

For a concentrated load P at ordinate y, F = P·y. For a UDL w over length, F = w × area under ILD. For a couple M0, F = M0 × (slope of ILD at that point).

## Answer

Let $F$ be a structural quantity (reaction, shear, moment or member force) at a fixed section. Let $y$ be its influence ordinate (value of $F$ for a unit load) at the load position. By the **principle of superposition** (linear elastic structure), the effect of several loads is the sum of their separate effects.

#### 1. Concentrated loads
A single load $P$ at a point where the ILD ordinate is $y$ gives $F = P\,y$, because a unit load gives $y$ and the effect is proportional to the load. For loads $P_1, P_2, \dots, P_n$ at ordinates $y_1, y_2, \dots, y_n$:

$$
F = P_1y_1 + P_2y_2 + \dots + P_ny_n = \sum P_i\,y_i
$$

#### 2. Uniformly distributed load
Consider a UDL of intensity $w$ per unit length between $x_1$ and $x_2$. An elementary strip $dx$ carries load $w\,dx$, acting as a concentrated load at ordinate $y$ (the ILD ordinate at $x$):

$$
dF = w\,dx\;y \Rightarrow F = w\int_{x_1}^{x_2} y\,dx = w\times(\text{area of ILD between }x_1\text{ and }x_2)
$$

Areas above the base line are positive and those below are negative. For the maximum positive effect, load only the positive portions.

#### 3. Couple
A couple $M_0$ is a pair of equal and opposite loads $P$ at a small distance $d$ apart, with $M_0 = P\,d$. If the ordinates at the two points are $y_1$ and $y_2$ (the second load acting opposite to the first):

$$
F = P\,(y_2 - y_1) = P\,d\;\frac{y_2-y_1}{d} = M_0\,\frac{dy}{dx}
$$

So the effect of a couple equals the **couple magnitude multiplied by the slope of the ILD** at the point of application (the sign follows the sense of the couple: positive when clockwise, for an ILD drawn with upward positive ordinates).

```text
 Summary:   P  ->  F = P . y
            w  ->  F = w . (area of ILD under load)
            M0 ->  F = M0 . (slope of ILD)
```

## Check yourself

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

1. On what principle is F = sum P y based?
   A. Superposition for a linear elastic structure
   B. Equilibrium of a rigid body only
   C. Maxwell's reciprocal theorem
   D. Castigliano's theorem
   Answer: A. Effects are proportional to the load and add up, so F = P y for each load. (Key point: Basis)

2. A load P acts where the ILD ordinate is y. The effect is
   A. P x y
   B. P / y
   C. P + y
   D. y / P
   Answer: A. A unit load gives y, so a load P gives P times y. (Key point: Concentrated load)

3. How is the effect of a UDL w over x1 to x2 found?
   A. w times the area of the ILD between x1 and x2
   B. w times the ILD ordinate at the centre only
   C. w times the slope of the ILD
   D. w times the length divided by the peak ordinate
   Answer: A. dF = w y dx integrates to w times the area. (Key point: UDL)

4. The effect of a couple M0 on a quantity is
   A. M0 times the slope of the ILD at the couple
   B. M0 times the ILD ordinate
   C. M0 times the ILD area
   D. M0 divided by the slope
   Answer: A. A couple is two equal opposite loads at distance d: F = P(y2 - y1) = M0 dy/dx. (Key point: Couple)

## Related in Theory of Structures I

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