---
title: "Castigliano: reaction at C of L-frame"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-frame-reaction-c-wkn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-frame-reaction-c-wkn
site: BCT Engineering
---

# Castigliano: reaction at C of L-frame

Theory of Structures II · Chapter 2: Force method

- Asked in: 2076 Chaitra, 6 marks
- Video: [EX 07 | ENERGY PRINCIPLE | PORTAL FRAME | REACTIONS USING MINIMUM ENERGY CONCEPT | LOADED WITH UDL](https://www.youtube.com/watch?v=2FTPLATSjyo) — 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-frame-reaction-c-wkn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Find the reaction at support 'C' using Castigliano's theorem. [Figure: L-shaped frame; horizontal member BC of L m carrying W kN/m UDL, C on a roller; vertical member AB of L m, A fixed; EI constant.]

## Short answer

Set the vertical displacement at the roller C to zero using Castigliano's theorem. This gives RC = 15WL/32 upward.

## Answer

Reading of the figure: A (0, 0) is fixed at the bottom of the vertical member AB (length $L$), B is the corner, BC is the horizontal member (length $L$) carrying $W$ kN/m, and C is on a roller that gives a vertical reaction $R_C$. EI is constant. The frame has one redundant, $R_C$.

```text
        W kN/m
  B ================= C  (roller)
  |                   o
  |
  | L
 A///   |<---- L ---->|
```

#### Bending moments (take $R_C$ upward)
- Member CB ($x$ from C): $M = R_C\,x - \dfrac{Wx^2}{2}$, $\ \dfrac{\partial M}{\partial R_C} = x$
- Member BA ($y$ from B downward): the moment is constant along the column, $M = R_C L - \dfrac{WL^2}{2}$, $\ \dfrac{\partial M}{\partial R_C} = L$

#### Castigliano's theorem
The roller does not move vertically, so $\dfrac{\partial U}{\partial R_C} = 0$:

$$
\frac{1}{EI}\left[\int_0^L \left(R_C x - \frac{Wx^2}{2}\right)x\,dx + \int_0^L \left(R_C L - \frac{WL^2}{2}\right)L\,dy\right] = 0
$$

$$
\frac{R_C L^3}{3} - \frac{WL^4}{8} + R_C L^3 - \frac{WL^4}{2} = 0
$$

$$
\frac{4}{3}R_C L^3 = \frac{5}{8}WL^4 \ \Rightarrow\ R_C = \frac{15\,WL}{32}
$$

#### Other reactions
$$
V_A = WL - \frac{15WL}{32} = \frac{17\,WL}{32}\ (\uparrow),\qquad H_A = 0,\qquad M_A = \left|\frac{15WL^2}{32} - \frac{WL^2}{2}\right| = \frac{WL^2}{32}
$$

The moment in the column is $-WL^2/32$, constant over its height, with tension on the outer face. The maximum sagging moment in the beam occurs at $x = 15L/32$ from C.

**Answer: $R_C = \dfrac{15\,W L}{32}$ (upward).**

## Check yourself

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

1. Why can the Castigliano condition dU/dR_C = 0 be applied?
   A. The roller C has no vertical deflection, and dU/dR_C is the deflection of C in the direction of R_C
   B. Because R_C equals zero
   C. Because the frame is determinate
   D. Because the column has no strain energy
   Answer: A. A support that does not move gives zero displacement in the direction of its reaction. (Key point: Condition)

2. Why is the moment in column BA constant, equal to R_C L - WL^2/2?
   A. Because the column carries the UDL
   B. The forces from the beam are vertical, so they act at a constant lever arm L about every section of the column
   C. Because the column has no axial force
   D. Because the lever arm increases along the column
   Answer: B. Vertical forces on the beam have the same horizontal distance L from every point of the vertical column, so the moment does not vary and dM/dR_C = L. (Key point: Column moment)

3. What does (4/3) R_C L^3 = (5/8) W L^4 give?
   A. R_C = 3WL/8
   B. R_C = WL/2
   C. R_C = 15WL/32 upward
   D. R_C = 5WL/32
   Answer: C. R_C = (5/8)(3/4)WL = 15WL/32. Then V_A = 17WL/32 and the column moment is WL^2/32. (Key point: Result)

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