---
title: "Cantilever deflection and slope by Castigliano"
subject: "Mechanics of Solids"
chapter: 5
key: "semester_5/mechanics_of_solids:castigliano-cantilever-slope-deflection"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/castigliano-cantilever-slope-deflection
site: BCT Engineering
---

# Cantilever deflection and slope by Castigliano

Mechanics of Solids · Chapter 5: Deflections and Slope of Statically Determinate and Indeterminate Structures

- Asked in: year not given, 6 marks
- Video: [3323 Castigliano 2   Shaft Slope Homework Problem](https://www.youtube.com/watch?v=5261jtOFV2A) — Ronald Delahoussaye
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/castigliano-cantilever-slope-deflection
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-5

## Question

A cantilever of length 2 m carries a uniformly distributed load of 10 kN/m over the whole span and a point load of 8 kN at its free end. Using Castigliano's theorem, find the vertical deflection and the slope at the free end. Take $E = 200$ GPa and $I = 8\times10^{7}$ mm$^4$.

## Short answer

Using Castigliano's theorem with a fictitious couple for slope, integrate M(∂M/∂P) and M(∂M/∂M0) over the beam length. Vertical deflection at free end = 2.58 mm downward; slope = 0.00183 rad.

## Answer

#### Data

$L = 2$ m, $w = 10$ kN/m, $P = 8$ kN, $EI = 200\times10^{9}\times 8\times10^{-5} = 16\times10^{6}$ N m$^2$ $= 16000$ kN m$^2$.

#### Bending moment

Take $x$ from the free end. Add a fictitious couple $M_0$ at the free end for the slope (to be set to zero after differentiation). Let $P$ be treated as a variable load for the deflection.

$$
M(x) = -\left(Px + \frac{w x^2}{2} + M_0\right)
$$

(hogging; the sign cancels in $M\,\partial M/\partial P$.)

```text
  P = 8 kN   w = 10 kN/m (all along)
   |  v v v v v v v v v v v v
   v
   o==============================# fixed end
   x = 0  <---------- 2 m ------>
```

#### Deflection at the free end

$$
\frac{\partial M}{\partial P} = -x
$$

$$
\begin{aligned}
\delta &= \frac{\partial U}{\partial P} = \frac{1}{EI}\int_0^L M\frac{\partial M}{\partial P}\,dx
= \frac{1}{EI}\int_0^L \left(Px + \frac{wx^2}{2}\right)x\,dx \\
&= \frac{1}{EI}\left[\frac{PL^3}{3} + \frac{wL^4}{8}\right]
\end{aligned}
$$

$$
\delta = \frac{1}{16000}\left[\frac{8\times2^3}{3} + \frac{10\times 2^4}{8}\right]
= \frac{21.333 + 20}{16000} = 2.583\times10^{-3}\ \text{m}
$$

Here $PL^3/3EI = 1.333$ mm (point load) and $wL^4/8EI = 1.250$ mm (UDL).

#### Slope at the free end

$$
\frac{\partial M}{\partial M_0} = -1
$$

$$
\theta = \frac{\partial U}{\partial M_0}\Big|_{M_0=0} = \frac{1}{EI}\int_0^L\left(Px + \frac{wx^2}{2}\right)dx = \frac{1}{EI}\left[\frac{PL^2}{2} + \frac{wL^3}{6}\right]
$$

$$
\theta = \frac{16 + 13.333}{16000} = 0.00183\ \text{rad}
$$

The positive signs mean that the free end moves in the direction of the load (downward) and rotates in the same sense as the fictitious couple $M_0$, which is the sense of the rotation caused by the loads.

**Answer:** Deflection at the free end $= 2.58$ mm downward; slope $= 0.00183$ rad ($0.105^\circ$).

## Check yourself

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

1. Which expression gives free-end deflection of the cantilever?
   A. PL^3/(3EI) + wL^4/(8EI)
   B. PL^2/(2EI) + wL^3/(6EI)
   C. PL^3/(3EI) + wL^4/(4EI)
   D. PL^3/(8EI) + wL^4/(3EI)
   Answer: A. Integrating M dM/dP over L gives PL^3/3EI + wL^4/8EI. (Key point: Deflection formula)

2. What is the free-end deflection with EI = 16000 kN m^2?
   A. 1.25 mm
   B. 1.33 mm
   C. 5.17 mm
   D. 2.58 mm
   Answer: D. (21.333 + 20)/16000 m = 2.583 mm; point load gives 1.333 mm and UDL 1.250 mm. (Key point: Deflection value)

3. How is the slope at the free end found?
   A. Differentiate U with respect to P
   B. Apply a fictitious couple M0, differentiate U wrt M0, then set M0 = 0
   C. Differentiate twice with respect to x
   D. Use dM/dP = -x
   Answer: B. theta = dU/dM0 with dM/dM0 = -1, giving integral (Px + wx^2/2)/EI dx. (Key point: Slope via fictitious couple)

4. What is the slope at the free end?
   A. 0.00183 rad
   B. 0.00258 rad
   C. 0.0183 rad
   D. 0.00100 rad
   Answer: A. theta = (PL^2/2 + wL^3/6)/EI = (16 + 13.333)/16000 = 0.00183 rad. (Key point: Slope value)

## Related in Mechanics of Solids

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- [Inclined load on a rectangular cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Ainclined-load-cantilever-rectangular) — Mechanics of Solids, chapter 7
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## Related in other subjects

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