---
title: "Castigliano: beam with yielding middle support"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-yielding-support-abc"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-yielding-support-abc
site: BCT Engineering
---

# Castigliano: beam with yielding middle support

Theory of Structures II · Chapter 2: Force method

- Asked in: 2079 Baishakh, 6 marks
- Video: [When & How to use Castigliano's Theorem ? in Deflection of Beams | Strength of Material](https://www.youtube.com/watch?v=dX8hvaFczY4) — GATE ACADEMY by Umesh Dhande
- 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-yielding-support-abc
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

In the beam ABC, support B settles by $\lambda R_B$ units. Find the reaction $R_B$ using Castigliano's theorem. [Figure: beam ABC, AB = BC = l, EI constant, hinge supports at A and C, support B on a spring, UDL of $P/2l$ over the full length 2l.]

## Short answer

By symmetry, R_A = R_C = (P – R_B)/2. With settlement λ R_B, Castigliano’s theorem gives ∂U/∂R_B = -λ R_B. Integrating M = ((P–R_B)/2)x – (P/4l)x² and ∂M/∂R_B = –x/2 over half span, solving yields R_B = 5Pl³/(8(l³ + 6EIλ)) upward.

## Answer

Beam ABC has $AB = BC = l$, hinged ends A and C, a spring-type support at B, and UDL $w = \dfrac{P}{2l}$ over $2l$ (total load $P$). Support B settles $\lambda R_B$, where $R_B$ is its upward reaction.

```text
 w = P/2l
 vvvvvvvvvvvvvvvvvvvvvvvv
 o---------o---------o
 A         B(R_B)    C
 |<-- l -->|<-- l -->|
```

By symmetry $R_A = R_C = \dfrac{P - R_B}{2}$.

For $0 \le x \le l$ from A (the other half is the same):

$$
M = \frac{P - R_B}{2}\,x - \frac{P}{4l}\,x^2, \qquad \frac{\partial M}{\partial R_B} = -\frac{x}{2}
$$

Castigliano's theorem for the upward redundant $R_B$ gives the upward displacement of B. B moves down by $\lambda R_B$, so

$$
\frac{\partial U}{\partial R_B} = -\lambda R_B
$$

$$
\frac{2}{EI}\int_0^l \left[\frac{P - R_B}{2}x - \frac{P}{4l}x^2\right]\left(-\frac{x}{2}\right)dx = -\lambda R_B
$$

Evaluate the integral:

$$
-\frac{1}{EI}\left[\frac{P - R_B}{2}\cdot\frac{l^3}{3} - \frac{P}{4l}\cdot\frac{l^4}{4}\right] = -\lambda R_B
$$

$$
\frac{(P - R_B)\,l^3}{6} - \frac{P l^3}{16} = EI\,\lambda R_B
$$

$$
\frac{5Pl^3}{48} - \frac{R_B l^3}{6} = EI\,\lambda\,R_B
$$

$$
R_B = \frac{5P}{8}\cdot\frac{1}{1 + \dfrac{6EI\lambda}{l^3}}
$$

**Answer: $R_B = \dfrac{5Pl^3}{8\,(l^3 + 6EI\lambda)}$ upward.** For $\lambda = 0$ (rigid support), $R_B = \dfrac{5P}{8}$ (the continuous-beam value).

## Check yourself

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

1. By symmetry the end reactions are:
   A. (P - R_B)/2 each
   B. P/2 each
   C. R_B each
   D. P - R_B
   Answer: A. Total load P is shared by R_B and two equal end reactions. (Key point: Symmetry)

2. Since B moves down by lambda R_B, dU/dR_B equals:
   A. 0
   B. -lambda R_B
   C. +lambda R_B
   D. P
   Answer: B. R_B is upward and B moves downward, so the upward displacement is negative. (Key point: Condition for yielding support)

3. dM/dR_B for 0 <= x <= l is:
   A. -x/2
   B. x
   C. x/2
   D. -x
   Answer: A. M = (P - R_B)x/2 - Px^2/4l. (Key point: Derivative)

4. Result for R_B is:
   A. 5Pl^3/[8(l^3 + 6EI lambda)]
   B. 5P/8 always
   C. P/2
   D. 3P/8
   Answer: A. For lambda = 0 it reduces to 5P/8. (Key point: Result)

## Related in Theory of Structures II

- [Castigliano: reaction at B, 20 kN/m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-reaction-b-20knm) — Theory of Structures II, chapter 2
- [Castigliano: reaction at C, beam with overhang](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-support-c-6knm) — Theory of Structures II, chapter 2
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- [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4
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- [Castigliano's second theorem: slope at B](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-slope-b-fixed-moment-m) — Theory of Structures II, chapter 2

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