---
title: "Castigliano: reaction at B, 150 kN overhang"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-reaction-b-150kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-reaction-b-150kn
site: BCT Engineering
---

# Castigliano: reaction at B, 150 kN overhang

Theory of Structures II · Chapter 2: Force method

- Asked in: 2075 Asoj, 6 marks
- Video: [EX 01 | ENERGY PRINCIPLE | PROPPED CANTILEVER BEAM | REACTIONS USING MINIMUM ENERGY CONCEPT](https://www.youtube.com/watch?v=D0GEsufQ6a4) — Structural Funda by:- KheNiSiR
- 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-reaction-b-150kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine reaction at support B of the beam shown in figure below by Castigliano's method. [Figure: beam A-B-C, A fixed, B roller at 4 m from A, overhang BC = 2 m with 150 kN downward at C.]

## Short answer

Treat RB as redundant; set ∂U/∂RB = 0, integrate M(∂M/∂RB) ds over the beam. RB = 262.5 kN upward.

## Answer

Beam ABC: A fixed, B a roller 4 m from A, overhang BC = 2 m with 150 kN downward at C. EI is constant. The beam has one redundant; take the roller reaction $R_B$ (upward).

```text
 ||==============o=======+ 150 kN
 A (fixed)       B       C
 |<----- 4 ----->|<- 2 ->|
```

#### Bending moments (x measured from C towards A, sagging positive)
$$
\begin{aligned}
CB\ (0 \le x \le 2):\quad & M = -150x, & \frac{\partial M}{\partial R_B} &= 0 \\
BA\ (2 \le x \le 6):\quad & M = -150x + R_B(x-2), & \frac{\partial M}{\partial R_B} &= x-2
\end{aligned}
$$

#### Castigliano's theorem
The deflection at B is zero, so $\dfrac{\partial U}{\partial R_B} = 0$. Let $s = x-2$ (from B):

$$
\int_0^4 \left[R_B\,s - 150(s+2)\right]s\,ds = 0
$$

$$
\frac{R_B (4)^3}{3} - 150\left(\frac{4^3}{3} + 2\cdot\frac{4^2}{2}\right) = 0
$$

$$
21.333\,R_B - 150(21.333 + 16) = 0 \ \Rightarrow\ R_B = \frac{5600}{21.333} = 262.5\ \text{kN}\ (\uparrow)
$$

#### Other reactions (check)
$$
V_A = 150 - 262.5 = -112.5\ \text{kN (i.e. 112.5 kN downward)}
$$

$$
M_A = R_B(4) - 150(6) = 1050 - 900 = 150\ \text{kNm}
$$

#### Bending moment
In the overhang $M = -150x$, so $M_B = -300$ kNm (hogging). In AB, with $s$ from B, $M = R_B s - 150(s+2) = -300 + 112.5\,s$, which is zero at $s = 2.667$ m from B and equals $+150$ kNm (sagging) at A ($s = 4$).

**Answer: $R_B = 262.5$ kN (upward); also $V_A = 112.5$ kN (downward) and $M_A = 150$ kNm.**

## Check yourself

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

1. What is the redundant in this beam?
   A. Roller reaction R_B
   B. The 150 kN load
   C. Moment at C
   D. Slope at A
   Answer: A. Fixed A plus roller B gives one degree of indeterminacy; R_B is taken as redundant. (Key point: Redundant)

2. Which equation is solved?
   A. Sum of V = 0 only
   B. dU/dR_B = 0 since deflection at B is zero
   C. dU/dR_B = 150
   D. U = 150 R_B
   Answer: B. B is a rigid roller, so its deflection dU/dR_B = ∫M(∂M/∂R_B)dx/EI = 0. (Key point: Castigliano condition)

3. Why does the overhang BC not contribute to the integral?
   A. The moment there is zero
   B. EI is zero
   C. ∂M/∂R_B = 0 there
   D. The load is a point load
   Answer: C. Sections in BC lie beyond B, so M = -150x contains no R_B and its derivative is zero. (Key point: Why overhang drops out)

4. What is R_B?
   A. 150 kN
   B. 112.5 kN
   C. 300 kN
   D. 262.5 kN upward
   Answer: D. 21.333 R_B = 150(21.333 + 16) gives R_B = 262.5 kN; then VA = 112.5 kN down and MA = 150 kNm. (Key point: Result)

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