---
title: "Castigliano: reactions of two-span beam"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-two-span-udl-reactions"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-two-span-udl-reactions
site: BCT Engineering
---

# Castigliano: reactions of two-span beam

Theory of Structures II · Chapter 2: Force method

- Asked in: 2068 Baishakh, 10 marks
- Video: [EX 08 | ENERGY PRINCIPLE | TWO SPAN CONTINUOUS BEAM | REACTIONS USING MINIMUM ENERGY CONCEPT](https://www.youtube.com/watch?v=jT8XVk30OQw) — 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-two-span-udl-reactions
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Find support reactions of the given loaded beam using Castigliano's theorem. [Figure: two-span continuous beam, spans L and L, UDL w per metre on the left span; A hinged, B roller, C roller.]

## Short answer

Take the intermediate support reaction RB as redundant and impose ∂U/∂RB = 0. Reactions: RA = 7wL/16 upward, RB = 5wL/8 upward, RC = wL/16 downward.

## Answer

**Method:** the two-span beam has 3 vertical reactions and 2 equilibrium equations, so it is indeterminate to the 1st degree. Take $R_B$ (upward) as the redundant. Since B does not settle, Castigliano's theorem gives $\partial U/\partial R_B=0$.

#### Reactions in terms of $R_B$
$\sum M_C=0$ with the UDL of total load $wL$ acting at $L/2$ from A (that is $3L/2$ from C):
$$
R_A(2L)+R_B(L)=wL\cdot\frac{3L}{2}\ \Rightarrow\ R_A=\frac{3wL}{4}-\frac{R_B}{2}
$$

#### Bending moments (x from A)
- $0\le x\le L$: $M=R_Ax-\dfrac{wx^2}{2}$, $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}$
- $L\le x\le2L$: $M=R_Ax-wL\left(x-\dfrac{L}{2}\right)+R_B(x-L)$, $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}+(x-L)=\dfrac{x}{2}-L$

#### Condition $\partial U/\partial R_B=0$
$$
\int_0^L M_1\left(-\frac{x}{2}\right)dx+\int_L^{2L}M_2\left(\frac{x}{2}-L\right)dx=0
$$
Evaluating the integrals (with $R_A$ substituted) gives
$$
\frac{L^{3} \left(- 5 L w + 8 R_{B}\right)}{48}=0\ \Rightarrow\ R_B=\frac{5wL}{8}
$$

#### Other reactions
$$
R_A=\frac{3wL}{4}-\frac{5wL}{16}=\frac{7wL}{16}\ \uparrow,\qquad R_C=wL-R_A-R_B=-\frac{wL}{16}
$$
$R_C$ is negative, so the support at C pulls the beam **downward** with $wL/16$.

Check: $R_A+R_B+R_C=\frac{7+10-1}{16}wL=wL$ ✓.

**Answer: $R_A=\dfrac{7wL}{16}$ (up), $R_B=\dfrac{5wL}{8}$ (up), $R_C=\dfrac{wL}{16}$ (down).**

## Check yourself

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

1. Why is the two-span beam indeterminate to the 1st degree?
   A. 3 vertical reactions and 2 equilibrium equations
   B. 2 reactions and 2 equations
   C. 4 reactions and 2 equations
   D. 3 reactions and 3 equations
   Answer: A. Only vertical forces act, so there are 2 equations for 3 unknowns, leaving R_B as the single redundant. (Key point: Degree of indeterminacy)

2. What condition is used to find R_B?
   A. dU/dR_B = 0, since B does not settle
   B. dU/dR_B = wL
   C. dU/dw = 0
   D. R_B equals R_A
   Answer: A. The derivative of strain energy with respect to a reaction gives the deflection there, which is zero at an unyielding support. (Key point: Castigliano condition)

3. What are the reactions?
   A. R_A = 5wL/8, R_B = 7wL/16, R_C = wL/16 up
   B. R_A = 7wL/16, R_B = 5wL/8, R_C = wL/16 down
   C. R_A = wL/2, R_B = wL/2, R_C = 0
   D. R_A = 3wL/8, R_B = 5wL/8, R_C = wL/16 up
   Answer: B. R_B = 5wL/8, R_A = 7wL/16 and R_C = -wL/16, so C pulls downward; total is wL. (Key point: Results)

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