---
title: "Castigliano: fixed-end moment, 15 kN/m, L = 5 m"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-propped-udl-15knm-5m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-propped-udl-15knm-5m
site: BCT Engineering
---

# Castigliano: fixed-end moment, 15 kN/m, L = 5 m

Theory of Structures II · Chapter 2: Force method

- Asked in: 2070 Asar, 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-propped-udl-15knm-5m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine the moment at the fixed end of the propped cantilever beam shown in figure below using Castigliano's theorem. [Figure: beam, left end fixed, right end roller, span L = 5 m with UDL 15 kN/m.]

## Short answer

Using Castigliano's theorem with roller reaction as redundant, fixed-end moment M_A = 46.875 kNm (hogging). Roller reaction R_B = 28.125 kN, fixed-end reaction R_A = 46.875 kN.

## Answer

Propped cantilever: fixed at the left end A, roller at B, span $L = 5$ m, UDL $w = 15$ kN/m. Take the roller reaction $R_B$ as the redundant and measure $x$ from B.

$$
M = R_Bx - \frac{wx^2}{2} = R_Bx - 7.5x^2,\qquad \frac{\partial M}{\partial R_B} = x
$$

The deflection at B is zero:

$$
\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^5\left(R_Bx^2 - 7.5x^3\right)dx = 0
$$

$$
R_B\frac{5^3}{3} - 7.5\frac{5^4}{4} = 0 \ \Rightarrow\ 41.667\,R_B = 1171.875 \ \Rightarrow\ R_B = 28.125\ \text{kN} \ \left(= \frac{3wL}{8}\right)
$$

#### Moment at the fixed end
$$
M_A = R_B(5) - 7.5(5)^2 = 140.625 - 187.5 = -46.875\ \text{kNm}
$$

The fixed-end moment is **46.875 kNm** (hogging), equal to $\dfrac{wL^2}{8} = \dfrac{15(25)}{8}$.

Check: $R_A = wL - R_B = 75 - 28.125 = 46.875$ kN.

**Answer: $M_A = 46.875$ kNm (hogging); $R_B = 28.125$ kN and $R_A = 46.875$ kN.**

## Check yourself

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

1. For the propped cantilever, which condition fixes R_B using Castigliano?
   A. ∂U/∂R_B = 0, since deflection at the roller is zero
   B. ∂U/∂R_B = wL
   C. Rotation at B is zero
   D. M at B is maximum
   Answer: A. The roller prevents deflection, so ∂U/∂R_B = 0. (Key point: Redundant and condition)

2. The result of the integral gives R_B equal to:
   A. wL/2 = 37.5 kN
   B. 5wL/8 = 46.875 kN
   C. 3wL/8 = 28.125 kN
   D. wL/8 = 9.375 kN
   Answer: C. R_B(125/3) = 7.5(625/4) gives 28.125 kN = 3wL/8. (Key point: Prop reaction)

3. M_A = R_B(5) - 7.5(5)^2 equals:
   A. +46.875 kNm sagging
   B. 46.875 kNm hogging
   C. 187.5 kNm
   D. 0
   Answer: B. 140.625 - 187.5 = -46.875 kNm, i.e. wL^2/8 hogging. (Key point: Fixed-end moment)

## Close variants set in other exams

- [Castigliano: fixed moment, propped cantilever UDL](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-udl-5knm-6m) — Theory of Structures II, chapter 2, asked 2×

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