---
title: "Castigliano: fixed-end moment with 50 kNm at overhang"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-fixed-moment-50knm-overhang"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-fixed-moment-50knm-overhang
site: BCT Engineering
---

# Castigliano: fixed-end moment with 50 kNm at overhang

Theory of Structures II · Chapter 2: Force method

- Asked in: 2074 Asoj, 6 marks
- Video: [Structural Theory | Castigliano's Second Theorem (Beam Deflection) Part 1 of 3](https://www.youtube.com/watch?v=7qHLMuyRonU) — the Engineer's Insight
- 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-50knm-overhang
- 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 following loaded beam using Castigliano's theorem. Take EI constant. [Figure: beam ABC, A fixed, B roller at 5 m from A, overhang BC = 2 m with a 50 kNm moment applied at C.]

## Short answer

Taking roller reaction RB as redundant, Castigliano’s theorem gives δB=0. Integrating virtual work yields RB = 15 kN upward. The fixed-end moment MA = RB*5 - 50 = 25 kNm clockwise. Reactions: RA = 15 kN downward.

## Answer

Reading of the figure: A fixed; the roller B is 5 m from A; the overhang BC (2 m) carries a couple of 50 kNm at C, taken as clockwise. EI is constant. The beam has one redundant; take the roller reaction $R_B$ (upward).

```text
 ||==================o=========  <- 50 kNm (cw) at C
 A (fixed)           B        C
 |<------ 5 m ------>|<- 2 ->|
```

#### Bending moments (x from C for the overhang, $s$ from B for AB)
- Overhang BC: $M = -50$ kNm (constant, hogging), independent of $R_B$.
- Span AB: $M = R_B\,s - 50$, $\ \dfrac{\partial M}{\partial R_B} = s$

#### Castigliano's theorem ($\delta_B = 0$)
$$
\int_0^5 (R_B\,s - 50)\,s\,ds = 0 \ \Rightarrow\ \frac{R_B (5)^3}{3} - 50\frac{(5)^2}{2} = 0
$$

$$
41.667\,R_B = 625 \ \Rightarrow\ R_B = 15\ \text{kN}\ (\uparrow)
$$

#### Moment at the fixed support
At A ($s = 5$):

$$
M_A = R_B(5) - 50 = 15(5) - 50 = +25\ \text{kNm}
$$

The reaction moment at A is **25 kNm**, which is half the applied couple (the carry-over factor of a propped cantilever is 1/2). It acts in the clockwise sense on the beam. The vertical reaction at A is $15$ kN downward.

Check: $\Sigma V = 0$: $R_A + R_B = 0$ so $R_A = -15$ kN.

**Answer: moment at the fixed support $M_A = 25$ kNm (clockwise); $R_B = 15$ kN upward, $R_A = 15$ kN downward.** For an anticlockwise couple the results have the same magnitudes with all senses reversed.

## Check yourself

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

1. What is the bending moment in the overhang BC carrying a 50 kNm couple at C?
   A. Constant 50 kNm independent of R_B
   B. Zero
   C. Varies linearly with R_B
   D. Depends on EI
   Answer: A. A couple produces a constant moment along the overhang, and it contains no R_B. (Key point: Overhang moment)

2. Which equation gives R_B?
   A. ∫(R_B s - 50) s ds = 0 over AB
   B. ∫(R_B s - 50) ds = 0 over BC
   C. R_B = 50/5 only
   D. ∫ 50 s ds = 0
   Answer: A. With δ_B = 0, dU/dR_B = ∫ M (∂M/∂R_B) ds = ∫(R_B s - 50)s ds = 0 over 0 to 5. (Key point: Equation)

3. What is R_B?
   A. 10 kN
   B. 25 kN
   C. 15 kN upward
   D. 50 kN
   Answer: C. 41.667 R_B = 625 gives R_B = 15 kN. (Key point: Result for R_B)

4. What is the moment at A, and how does it relate to the applied couple?
   A. 50 kNm, equal to it
   B. 25 kNm, half the couple (carry-over 1/2)
   C. 75 kNm
   D. 0
   Answer: B. M_A = 15 x 5 - 50 = 25 kNm, half the couple. (Key point: Fixed support moment)

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