---
title: "Castigliano: deflection at B of propped cantilever"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-deflection-b-propped-p-midspan"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-deflection-b-propped-p-midspan
site: BCT Engineering
---

# Castigliano: deflection at B of propped cantilever

Theory of Structures II · Chapter 2: Force method

- Asked in: 2072 Chaitra, 12 marks
- Video: [Castigliano's theorem to find deflections in a beam](https://www.youtube.com/watch?v=RcZjw0YUmqM) — Abhijit Pawar
- 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-deflection-b-propped-p-midspan
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Using Castigliano's theorem, find the deflection at point B of the beam shown in figure below. Take constant EI through the length. [Figure: beam AC, A fixed, C roller; span L/2 + L/2 with point load P at B (mid-span).]

## Short answer

Castigliano's theorem used. Redundant R_C = 5P/16 found from zero deflection at roller C. Deflection at midspan B: δ_B = 7PL³/(768EI) downward.

## Answer

Beam AC of span $L$: A fixed, C a roller, point load $P$ at B (mid-span). EI is constant. The beam is indeterminate to the 1st degree. First find the redundant $R_C$ and then the deflection at B.

```text
          P
          v
 ||=======B=======o
 A (fixed)        C (roller, R_C)
 |<-- L/2 -->|<-- L/2 -->|
```

#### Step 1: Bending moments (x measured from C)
$$
\begin{aligned}
CB\ (0 \le x \le L/2):\quad & M = R_C\,x \\
BA\ (L/2 \le x \le L):\quad & M = R_C\,x - P\left(x - \frac{L}{2}\right)
\end{aligned}
$$

#### Step 2: Redundant reaction
The deflection at C is zero, so $\dfrac{\partial U}{\partial R_C} = 0$:

$$
\int_0^{L/2} R_C\,x^2\,dx + \int_{L/2}^{L}\left[R_C\,x - P\left(x-\frac L2\right)\right]x\,dx = 0
$$

$$
R_C\frac{L^3}{3} - P\left[\frac{x^3}{3} - \frac{Lx^2}{4}\right]_{L/2}^{L} = 0
$$

$$
R_C\frac{L^3}{3} - P\,\frac{5L^3}{48} = 0 \ \Rightarrow\ R_C = \frac{5P}{16}
$$

(The bracket equals $\left(\dfrac{L^3}{3} - \dfrac{L^3}{4}\right) - \left(\dfrac{L^3}{24} - \dfrac{L^3}{16}\right) = \dfrac{L^3}{12} + \dfrac{L^3}{48} = \dfrac{5L^3}{48}$.)

#### Step 3: Deflection at B by Castigliano's theorem
The load $P$ acts at B, so

$$
\delta_B = \frac{\partial U}{\partial P} = \frac{1}{EI}\int M\,\frac{\partial M}{\partial P}\,dx
$$

Because $\partial U/\partial R_C = 0$, the dependence of $R_C$ on $P$ does not add anything, so only the explicit $P$ in $M$ is differentiated: $\dfrac{\partial M}{\partial P} = -\left(x - \dfrac L2\right)$ for $L/2 \le x \le L$ and $0$ for the part CB.

With $R_C = 5P/16$:

$$
\delta_B = \frac{1}{EI}\int_{L/2}^{L}\left[\frac{5P}{16}x - P\left(x - \frac L2\right)\right]\left[-\left(x-\frac L2\right)\right]dx
$$

Let $u = x - L/2$ (from 0 to $L/2$), so $x = u + L/2$:

$$
\delta_B = \frac{P}{EI}\int_0^{L/2}\left[u - \frac{5}{16}\left(u + \frac L2\right)\right]u\,du
= \frac{P}{EI}\int_0^{L/2}\left[\frac{11}{16}u^2 - \frac{5L}{32}u\right]du
$$

$$
\delta_B = \frac{P}{EI}\left[\frac{11}{48}\cdot\frac{L^3}{8} - \frac{5L}{64}\cdot\frac{L^2}{4}\right]
= \frac{P L^3}{EI}\left(\frac{11}{384} - \frac{5}{256}\right)
$$

$$
\delta_B = \frac{PL^3}{EI}\cdot\frac{22 - 15}{768} = \frac{7PL^3}{768\,EI}
$$

**Answer: the deflection at B is $\delta_B = \dfrac{7PL^3}{768\,EI}$ downward; $R_C = \dfrac{5P}{16}$.**

## Check yourself

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

1. How is the roller reaction R_C found by Castigliano's theorem?
   A. Setting dU/dP = 0
   B. Setting dU/dR_C = 0, since the deflection at C is zero
   C. Setting U = 0
   D. Setting M = 0 at the fixed end
   Answer: B. The partial derivative of strain energy with respect to R_C is the deflection at C, which is zero for a roller support. (Key point: Condition for the redundant)

2. With x measured from C, what is the moment in the part BA (L/2 <= x <= L)?
   A. M = R_C x
   B. M = R_C x + P(x - L/2)
   C. M = R_C x - P(x - L/2)
   D. M = P(x - L/2)
   Answer: C. Beyond B the load P adds a hogging term -P(x - L/2) to the roller-reaction moment R_C x. (Key point: Bending moment expressions)

3. What is the value of R_C?
   A. P/2
   B. 3P/16
   C. 11P/16
   D. 5P/16
   Answer: D. R_C L^3/3 = P(5L^3/48) gives R_C = 5P/16. (Key point: Redundant reaction value)

4. Why can dM/dP be taken with R_C held as it is, and what is the final deflection at B?
   A. Because dU/dR_C = 0, so only the explicit P in M is differentiated; delta_B = 7PL^3/(768EI)
   B. Because R_C does not depend on P; delta_B = PL^3/(48EI)
   C. Because dU/dP = 0; delta_B = 0
   D. Because EI is constant; delta_B = 5PL^3/(48EI)
   Answer: A. Since dU/dR_C = 0, the implicit dependence of R_C on P adds nothing; integrating gives delta_B = 7PL^3/(768EI). (Key point: Deflection under the load)

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