---
title: "Castigliano: fixed moment, propped cantilever"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-fixed-moment-propped-100kn-50kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-fixed-moment-propped-100kn-50kn
site: BCT Engineering
---

# Castigliano: fixed moment, propped cantilever

Theory of Structures II · Chapter 2: Force method

- Asked in: 2071 Chaitra, 6 marks
- Video: [Strain Energy Method - Analysis of Propped Cantilever Beam ( Point Load in the Center )](https://www.youtube.com/watch?v=lLjjJT2QS6c) — 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-propped-100kn-50kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine the moment at the fixed support of the propped cantilever beam using Castigliano's method. [Figure: beam AB, A fixed, B roller, span 12 m with 100 kN at 4 m from A and 50 kN at 8 m from A.]

## Short answer

Use Castigliano’s theorem with roller reaction as redundant; set end deflection zero. The fixed-end moment is M_A = 311.11 kNm (hogging).

## Answer

Propped cantilever AB: A fixed, B a roller, span 12 m, loads 100 kN at 4 m and 50 kN at 8 m from A. EI is constant. Take the roller reaction $R_B$ as the redundant. Measure $x$ from B towards A.

```text
         50 kN      100 kN
           v          v
 B o-------+----------+---------|| A (fixed)
 |<- 4 ->|<--- 4 --->|<--- 4 --->|
```

(Positions from B: 50 kN at 4 m, 100 kN at 8 m.)

#### Bending moments
$$
\begin{aligned}
0 \le x \le 4:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\
4 \le x \le 8:\quad & M = R_Bx - 50(x-4), & \frac{\partial M}{\partial R_B} &= x \\
8 \le x \le 12:\quad & M = R_Bx - 50(x-4) - 100(x-8), & \frac{\partial M}{\partial R_B} &= x
\end{aligned}
$$

#### Castigliano's theorem ($\delta_B = 0$)
$$
\int_0^{12}R_Bx^2\,dx - 50\int_4^{12}(x-4)x\,dx - 100\int_8^{12}(x-8)x\,dx = 0
$$

Evaluate each term:

$$
\begin{aligned}
R_B\frac{12^3}{3} &= 576\,R_B \\
50\int_4^{12}(x^2 - 4x)\,dx &= 50\left[\frac{x^3}{3} - 2x^2\right]_4^{12} = 50\,(576 - 288 - 21.333 + 32) = 14933.3 \\
100\int_8^{12}(x^2 - 8x)\,dx &= 100\left[\frac{x^3}{3} - 4x^2\right]_8^{12} = 100\,(0 - (170.667 - 256)) = 8533.3
\end{aligned}
$$

$$
576R_B = 14933.3 + 8533.3 = 23466.7 \ \Rightarrow\ R_B = 40.741\ \text{kN}
$$

#### Moment at the fixed support
$$
M_A = R_B(12) - 50(8) - 100(4) = 40.741(12) - 800 = -311.11\ \text{kNm}
$$

The fixed-end moment is **311.11 kNm (hogging)**. The vertical reaction is $R_A = 150 - 40.741 = 109.259$ kN.

**Answer: $M_A = 311.11$ kNm (hogging); $R_B = 40.741$ kN.**

## Check yourself

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

1. With x from B, what is dM/dR_B in every part of the beam?
   A. x
   B. 1
   C. x - 4
   D. 0
   Answer: A. R_B acts at x = 0 so its moment is R_B x in each portion, and the derivative is x. (Key point: Redundant and derivative)

2. From 576 R_B = 23466.7, what is R_B?
   A. 60 kN
   B. 109.26 kN
   C. 25 kN
   D. 40.74 kN
   Answer: D. R_B = 23466.7/576 = 40.741 kN. (Key point: Redundant value)

3. What is M_A?
   A. 311.11 kNm hogging
   B. 800 kNm hogging
   C. 488.9 kNm sagging
   D. 150 kNm hogging
   Answer: A. M_A = 40.741(12) - 50(8) - 100(4) = -311.11 kNm. (Key point: Fixed-end moment)

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