---
title: "Castigliano: deflection at 50 kN load, cantilever"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:castigliano-deflection-50kn-cantilever-100kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-deflection-50kn-cantilever-100kn
site: BCT Engineering
---

# Castigliano: deflection at 50 kN load, cantilever

Theory of Structures II · Chapter 2: Force method

- Asked in: 2068 Chaitra, 8 marks
- Video: [Castigliano's Theorem - Cantilever Beam Distributed Load - Slope and Deflection - Choosing Origin](https://www.youtube.com/watch?v=TF9lngl48kA) — AF Math & Engineering
- 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-50kn-cantilever-100kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Using Castigliano's second theorem, determine the vertical deflection at the 50 kN load in the beam shown in figure below. [Figure: cantilever beam, fixed at the left end; 100 kN at 3 m from the fixed end; 50 kN at the free end, 3 m further; $E = 211000\ \text{N/mm}^2$, $I = 46875\times10^{4}\ \text{mm}^4$.]

## Short answer

Apply Castigliano's theorem by differentiating bending moment with respect to the 50 kN load and integrating over the beam length. The vertical deflection under the 50 kN load is 59.15 mm downward.

## Answer

**Castigliano's second theorem:** $\delta=\partial U/\partial P$. Call the 50 kN load $P_2$ at the free end ($x=6$ m from the fixed end) and the 100 kN load $P_1$ at $x=3$ m. Treat both as symbols, differentiate with respect to $P_2$, then substitute the numbers.

#### Bending moments (x from the fixed end A, hogging taken as negative)
- $0\le x\le3$: $M=-[P_1(3-x)+P_2(6-x)]$, $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$
- $3\le x\le6$: $M=-P_2(6-x)$, $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$

#### Deflection under the 50 kN load
$$
\delta_2=\frac{1}{EI}\left[\int_0^3[100(3-x)+50(6-x)](6-x)\,dx+\int_3^6 50(6-x)^2dx\right]
$$
$$
\int_0^3[100(3-x)+50(6-x)](6-x)dx=5400\ \text{kN m}^3,\qquad \int_3^650(6-x)^2dx=450\ \text{kN m}^3
$$
$$
\delta_2=\frac{5850}{EI}\quad(\text{numerator in kN m}^3,\ EI\ \text{in kN m}^2)
$$

#### Flexural rigidity
$$
EI=211000\ \text{N/mm}^2\times46875\times10^4\ \text{mm}^4=9.8906\times10^{13}\ \text{N mm}^2=98906.25\ \text{kN m}^2
$$
$$
\delta_2=\frac{5850}{98906.25}=59.15\ \text{mm}\ \text{(downward)}
$$

Check with standard results: tip deflection due to 100 kN at 3 m is $Pa^2(3L-a)/6EI=2250/EI$ and due to 50 kN at the tip is $PL^3/3EI=3600/EI$; total $5850/EI$ ✓.

**Answer: vertical deflection under the 50 kN load = 59.15 mm downward.**

## Check yourself

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

1. How is deflection under the 50 kN load found?
   A. Differentiate U with respect to the 50 kN load, keeping loads as symbols until after differentiating
   B. Differentiate U with respect to the 100 kN load
   C. Set both loads equal to 1 before differentiating
   D. Integrate the shear force
   Answer: A. delta = dU/dP at that load, so the 50 kN load is kept as P2 until after dM/dP2 is found. (Key point: Castigliano's second theorem)

2. What is dM/dP2 along the cantilever?
   A. -(6 - x) for the whole length
   B. -(3 - x) only in the first 3 m
   C. -1 everywhere
   D. 0 for x > 3
   Answer: A. The 50 kN load is at x = 6 m so its moment arm is (6 - x) over the whole span. (Key point: Moment derivative)

3. What is the deflection under the 50 kN load?
   A. 5850/EI = 59.15 mm downward
   B. 2250/EI = 22.75 mm
   C. 3600/EI = 36.4 mm
   D. 5850 mm
   Answer: A. The sum 5400 + 450 = 5850 kN m^3 divided by EI = 98906.25 kN m^2 gives 59.15 mm. (Key point: Final deflection)

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