---
title: "Castigliano: central deflection of simply supported beam"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-central-deflection-ss-beam-udl"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-central-deflection-ss-beam-udl
site: BCT Engineering
---

# Castigliano: central deflection of simply supported beam

Theory of Structures II · Chapter 2: Force method

- Asked in: 2066 Bhadra, 10 marks
- Video: [Ex 6: Strain Energy and Central  Deflection of a simply supported beam carrying a UDL over the span](https://www.youtube.com/watch?v=-_rrgIe9r04) — G M Basha
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-central-deflection-ss-beam-udl
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Compute the maximum central vertical deflection for a simply supported beam of span L loaded with a uniformly distributed load of w/unit length, EI is constant. Use Castigliano's theorem.

## Short answer

Apply a fictitious central load P, differentiate the strain energy with respect to P, then set P=0. The central deflection is δ_C=5wL^4/(384EI) downward.

## Answer

**Castigliano's second theorem:** the deflection at a point equals $\partial U/\partial P$, where $P$ is a load at that point. There is no point load at mid-span, so apply a fictitious load $P$ downward at the centre and set $P=0$ after differentiating.

#### Reactions
$$
R_A=R_B=\frac{wL}{2}+\frac{P}{2}
$$

#### Bending moment (x from A; the beam is symmetrical, so work with half and double)
For $0\le x\le L/2$:
$$
M=\left(\frac{wL}{2}+\frac{P}{2}\right)x-\frac{wx^2}{2},\qquad \frac{\partial M}{\partial P}=\frac{x}{2}
$$

#### Deflection at the centre
$$
\delta_C=\frac{\partial U}{\partial P}=2\int_0^{L/2}\frac{M}{EI}\frac{\partial M}{\partial P}dx\Big|_{P=0}
=\frac{2}{EI}\int_0^{L/2}\left(\frac{wLx}{2}-\frac{wx^2}{2}\right)\frac{x}{2}\,dx
$$
$$
=\frac{w}{2EI}\int_0^{L/2}\left(Lx^2-x^3\right)dx=\frac{w}{2EI}\left[\frac{Lx^3}{3}-\frac{x^4}{4}\right]_0^{L/2}
=\frac{w}{2EI}\left[\frac{L^4}{24}-\frac{L^4}{64}\right]
$$
$$
=\frac{w}{2EI}\cdot\frac{(8-3)L^4}{192}=\frac{5wL^4}{384EI}
$$

**Answer: maximum central deflection $\delta_C=\dfrac{5wL^4}{384\,EI}$ (downward).**

## Check yourself

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

1. Why is a fictitious load P applied at mid-span?
   A. Castigliano needs a load at the point where deflection is wanted; P is set to zero after differentiating
   B. P represents the actual load
   C. P is added to increase the answer
   D. It is used to find the reactions
   Answer: A. No actual load is at mid-span, so a dummy P is applied, differentiated against, then put to zero. (Key point: Fictitious load)

2. What is dM/dP in 0 <= x <= L/2, and why double the integral?
   A. x/2; the beam is symmetrical so the second half gives the same
   B. x; the second half is zero
   C. 1; moments are constant
   D. L/2; the integral is not doubled
   Answer: A. Reactions increase by P/2, so dM/dP = x/2, and the two halves contribute equally. (Key point: Use of symmetry)

3. What is the central deflection?
   A. wL^4/(8EI)
   B. 5wL^4/(384EI)
   C. wL^4/(48EI)
   D. wL^3/(24EI)
   Answer: B. Integrating gives (w/2EI)[L^4/24 - L^4/64] = 5wL^4/384EI. (Key point: Result)

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