---
title: "Castigliano's second theorem: slope at A"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-slope-a-50kn-10kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-slope-a-50kn-10kn
site: BCT Engineering
---

# Castigliano's second theorem: slope at A

Theory of Structures II · Chapter 2: Force method

- Asked in: 2069 Asar, 4 marks
- Video: [Castigliano's Second Theorem - Deflection at a Point - Continuous Beam with Cantilever](https://www.youtube.com/watch?v=pAhp20WsNNc) — AF Math & Engineering
- 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-a-50kn-10kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Using Castigliano's second theorem, determine the slope at A of the beam shown in figure below. EI is constant. [Figure: beam A-B-C, A hinged, B roller, AB = 5 m (2.5 m + 2.5 m) with 50 kN at mid-span; overhang BC = 2 m with 10 kN at C.]

## Short answer

Apply a fictitious clockwise couple M0 at A. Express bending moments including M0, differentiate, set M0 = 0, and integrate over the beam. Slope θA = 61.46/EI rad clockwise.

## Answer

**Castigliano's second theorem:** the rotation at a point equals the partial derivative of the strain energy with respect to a couple applied there:
$$
\theta_A=\frac{\partial U}{\partial M_0}=\int \frac{M}{EI}\frac{\partial M}{\partial M_0}\,dx
$$
Since no couple acts at A, apply a fictitious clockwise couple $M_0$ at A, write $M$ in terms of $M_0$, differentiate, then put $M_0=0$.

#### Reactions (with $M_0$)
Taking moments about B (AB = 5 m, BC = 2 m):
$$
5R_A + M_0 = 50(2.5) - 10(2) = 105 \;\Rightarrow\; R_A = 21-\frac{M_0}{5}\ \text{kN},\qquad R_B = 39+\frac{M_0}{5}\ \text{kN}
$$
(Check: $R_A+R_B=60=50+10$.)

#### Bending moments (x from A, sagging +)
- Portion A to mid-span ($0\le x\le2.5$): $M=R_Ax+M_0=\left(21-\frac{M_0}{5}\right)x+M_0$, so $\dfrac{\partial M}{\partial M_0}=1-\dfrac{x}{5}$
- Mid-span to B ($2.5\le x\le5$): $M=R_Ax+M_0-50(x-2.5)$, same derivative $1-\dfrac{x}{5}$
- Overhang BC: $M=-10x'$ (x' from C), independent of $M_0$, so it contributes nothing.

#### Slope at A (put $M_0=0$)
$$
EI\,\theta_A=\int_0^{2.5}21x\left(1-\frac{x}{5}\right)dx+\int_{2.5}^{5}\big[21x-50(x-2.5)\big]\left(1-\frac{x}{5}\right)dx
$$
$$
= 175/4 + 425/24 = 1475/24 = 61.46\ \text{kN m}^2
$$
$$
\theta_A=\frac{61.46}{EI}\ \text{rad (clockwise, same sense as }M_0)
$$
Check: $50\times5^2/16=78.13$ (central load) minus $20\times5/6=16.67$ (hogging moment from the overhang) = 61.46. ✓

**Answer: $\theta_A = 61.46/EI$ rad (with EI in kNm²), clockwise.**

## Check yourself

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

1. How is the slope at A obtained using Castigliano's second theorem?
   A. Differentiate U with respect to a force at A
   B. Differentiate U with respect to a couple at A (fictitious if none), then set it to zero
   C. Integrate M over the span
   D. Divide deflection by the length
   Answer: B. θ = ∂U/∂M0 with a fictitious couple M0 at A, set to zero afterward. (Key point: Castigliano for rotation)

2. Why is ∂M/∂M0 = 1 - x/5 for AB and 0 for overhang BC?
   A. Overhang moment depends only on the 10 kN load, not on the couple at A
   B. The overhang is rigid
   C. The couple affects only BC
   D. It has no meaning
   Answer: A. M = -10x' in BC is independent of M0, so it contributes nothing. (Key point: Derivative)

3. The slope at A comes out as:
   A. 0 rad
   B. 61.46/EI clockwise
   C. 78.13/EI anticlockwise
   D. 16.67/EI clockwise
   Answer: B. EIθ = 175/4 + 425/24 = 61.46, equal to 78.13 - 16.67. (Key point: Result)

## Related in Theory of Structures II

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