---
title: "Castigliano: fixed moment, propped cantilever UDL"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-fixed-moment-udl-5knm-6m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-fixed-moment-udl-5knm-6m
site: BCT Engineering
---

# Castigliano: fixed moment, propped cantilever UDL

Theory of Structures II · Chapter 2: Force method

- Asked in: 2072 Kartik, 6 marks
- Video: [Strain Energy Method - Analysis of Propped Cantilever Beam ( With Uniformly Distributed Load )](https://www.youtube.com/watch?v=ZdP5LJ-H_Gk) — Stan Academy
- 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-fixed-moment-udl-5knm-6m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Using Castigliano's theorem, determine the moment at the fixed support A of the propped cantilever beam loaded as shown in figure below. [Figure: beam AB, A fixed, B roller, span 6 m, UDL 5 kN/m; $E = 232\ \text{kN/mm}^2$, $I = 112.5\times10^{6}\ \text{mm}^4$.]

## Short answer

Take roller reaction as redundant, apply Castigliano’s theorem to set deflection at B zero. Solving the integral gives M_A = 22.5 kNm (hogging).

## Answer

Propped cantilever AB: A fixed, B a roller, span $L = 6$ m, UDL $w = 5$ kN/m. $E = 232$ kN/mm$^2$ and $I = 112.5\times10^6$ mm$^4$ give $EI = 232\times10^6\times1.125\times10^{-4} = 26100$ kNm$^2$, but the reactions of a single-span propped cantilever do not depend on $EI$.

Take the roller reaction $R_B$ as the redundant. With $x$ measured from B:

$$
M = R_B\,x - \frac{wx^2}{2},\qquad \frac{\partial M}{\partial R_B} = x
$$

The deflection at B is zero:

$$
\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^6\left(R_Bx - 2.5x^2\right)x\,dx = 0
$$

$$
\frac{R_B(6)^3}{3} - \frac{2.5(6)^4}{4} = 0 \ \Rightarrow\ 72R_B = 810 \ \Rightarrow\ R_B = 11.25\ \text{kN}
$$

(This is $3wL/8 = 3(5)(6)/8 = 11.25$ kN.)

#### Moment at the fixed support A
$$
M_A = R_B(6) - \frac{5(6)^2}{2} = 67.5 - 90 = -22.5\ \text{kNm}
$$

The fixed-end moment is **22.5 kNm** (hogging, $= wL^2/8 = 5(36)/8$).

The vertical reaction at A is $30 - 11.25 = 18.75$ kN. The maximum sagging moment is $\dfrac{9wL^2}{128} = 12.66$ kNm at $x = \dfrac{3L}{8} = 2.25$ m from B.

**Answer: $M_A = 22.5$ kNm (hogging); $R_B = 11.25$ kN, $R_A = 18.75$ kN.**

## Check yourself

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

1. Which condition determines the roller reaction R_B?
   A. Slope at A is zero
   B. dU/dR_B = 0, since the deflection at B is zero
   C. Moment at B equals wL^2/8
   D. Shear at A is zero
   Answer: B. The roller does not allow deflection, so the partial derivative of strain energy with respect to R_B is zero. (Key point: Choice of redundant)

2. What is R_B for UDL 5 kN/m over 6 m?
   A. 15 kN
   B. 18.75 kN
   C. 11.25 kN
   D. 7.5 kN
   Answer: C. R_B L^3/3 = w L^4/8 gives R_B = 3wL/8 = 11.25 kN. (Key point: Redundant value)

3. What is the moment at A and does EI affect it?
   A. 22.5 kNm hogging; no, the reactions do not depend on EI
   B. 22.5 kNm sagging; yes, depends on EI
   C. 45 kNm hogging; no
   D. 11.25 kNm hogging; yes
   Answer: A. M_A = R_B(6) - 5(36)/2 = 67.5 - 90 = -22.5 kNm (= wL^2/8); EI cancels for a single span. (Key point: Fixed-end moment)

## Close variants set in other exams

- [Castigliano: fixed-end moment, 15 kN/m, L = 5 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-propped-udl-15knm-5m) — Theory of Structures II, chapter 2, asked 2×

## Related in Theory of Structures II

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