---
title: "Castigliano's second theorem: slope at B"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-slope-b-fixed-moment-m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-slope-b-fixed-moment-m
site: BCT Engineering
---

# Castigliano's second theorem: slope at B

Theory of Structures II · Chapter 2: Force method

- Asked in: 2068 Chaitra, 5 marks
- Video: [CASTIGLIANO'S THEOREM in Just Over 10 Minutes!](https://www.youtube.com/watch?v=JH6iOhOe8h0) — Less Boring Lectures
- 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-slope-b-fixed-moment-m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

For the beam as shown, determine the slope at support B. Use Castigliano's second theorem. Take EI = constant. [Figure: beam AB of length l, A fixed, B hinged support with an applied moment M at B.]

## Short answer

Treat the reaction at B as redundant and apply Castigliano's second theorem. The slope at B is Ml/(4EI) in the same clockwise sense as the applied moment.

## Answer

**Castigliano's second theorem:** $\theta=\partial U/\partial M$. The beam is a propped cantilever (A fixed, B on a roller/hinge support) with a clockwise couple $M$ at B. The vertical reaction $R_B$ is the redundant, found from $\partial U/\partial R_B=0$ because the support does not settle.

Take $x$ measured from B towards A (B at $x=0$, A at $x=l$), $R_B$ upward.

#### Bending moment (sagging +)
$$
M_x = M + R_Bx,\qquad \frac{\partial M_x}{\partial R_B}=x,\qquad \frac{\partial M_x}{\partial M}=1
$$

#### Step 1: find $R_B$
$$
\frac{\partial U}{\partial R_B}=\frac{1}{EI}\int_0^l(M+R_Bx)\,x\,dx=0
\;\Rightarrow\; \frac{Ml^2}{2}+\frac{R_Bl^3}{3}=0
\;\Rightarrow\; R_B=-\frac{3M}{2l}
$$
(i.e. $3M/2l$ acting downward.)

#### Step 2: slope at B
$$
\theta_B=\frac{\partial U}{\partial M}=\frac{1}{EI}\int_0^l(M+R_Bx)(1)\,dx=\frac{1}{EI}\left[Ml+R_B\frac{l^2}{2}\right]
$$
$$
\theta_B=\frac{1}{EI}\left[Ml-\frac{3M}{2l}\cdot\frac{l^2}{2}\right]=\frac{1}{EI}\left[Ml-\frac{3Ml}{4}\right]=\frac{Ml}{4EI}
$$

**Answer: $\theta_B=\dfrac{Ml}{4EI}$, in the same (clockwise) sense as the applied moment.** The fixed far end stiffens the beam, so the slope is smaller than the simply-supported value $Ml/3EI$.

## Check yourself

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

1. How does Castigliano's second theorem give the slope at B?
   A. Differentiate strain energy U with respect to the moment M applied at B
   B. Differentiate U with respect to the load R_B
   C. Integrate M dx/EI without differentiating
   D. Differentiate U with respect to EI
   Answer: A. Rotation at a point equals dU/dM where M is the couple applied at that point. (Key point: Castigliano for slope)

2. How is the reaction R_B found?
   A. From dU/dR_B = 0 since B does not settle
   B. From dU/dM = 0
   C. From sum of moments at B equal to zero only
   D. It is taken as zero
   Answer: A. The support is rigid, so deflection at B, dU/dR_B, is zero; this gives R_B = -3M/2l (downward). (Key point: Finding the redundant)

3. What is the slope at B?
   A. Ml/3EI
   B. Ml/4EI
   C. Ml/EI
   D. Ml/2EI
   Answer: B. theta_B = (1/EI)[Ml - (3M/2l)(l^2/2)] = Ml/4EI, smaller than the simply supported Ml/3EI because the fixed end stiffens the beam. (Key point: Result)

## Related in Theory of Structures II

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## Related in other subjects

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