---
title: "Propped cantilever by strain energy method"
subject: "Mechanics of Solids"
chapter: 5
key: "semester_5/mechanics_of_solids:propped-cantilever-redundant-reaction"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/propped-cantilever-redundant-reaction
site: BCT Engineering
---

# Propped cantilever by strain energy method

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

- Asked in: year not given, 8 marks
- Video: [Strain Energy Method - Problem No 1 - Analysis of Propped Cantilever Beams](https://www.youtube.com/watch?v=HAHixWNyVLk) — Stan Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/propped-cantilever-redundant-reaction
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-5

## Question

A cantilever AB of length 4 m, fixed at A, carries a uniformly distributed load of 12 kN/m over its whole length and is propped at the free end B so that the level of B remains the same as that of A. Using Castigliano's theorem find the prop reaction, the fixed-end moment, the position of the point of contraflexure and the maximum sagging moment. $EI$ is constant.

## Short answer

Take prop reaction as redundant, set ∂U/∂R_B = 0. Prop reaction = 18 kN, fixed-end moment = 24 kN m (hogging). Point of contraflexure is 1 m from the fixed end; maximum sagging moment = 13.5 kN m at 1.5 m from the prop.

## Answer

#### Method

The beam is statically indeterminate to the first degree. Take the prop reaction $R_B$ as the redundant. Since the prop does not allow vertical movement at B:

$$
\delta_B = \frac{\partial U}{\partial R_B} = 0
$$

#### Bending moment

Take $x$ from B toward the fixed end A ($L = 4$ m, $w = 12$ kN/m):

$$
M(x) = R_B\,x - \frac{w x^2}{2},\qquad \frac{\partial M}{\partial R_B} = x
$$

```text
 A (fixed)                       B
 |#==============================o   <- prop R_B
 |#  v v v v v v v v v v v v v
 |#  w = 12 kN/m,  L = 4 m
```

#### Compatibility

$$
\begin{aligned}
\frac{1}{EI}\int_0^L \left(R_B x - \frac{wx^2}{2}\right)x\,dx &= 0 \\
\frac{R_B L^3}{3} - \frac{wL^4}{8} &= 0 \\
R_B &= \frac{3wL}{8} = \frac{3\times12\times4}{8} = 18.0\ \text{kN}
\end{aligned}
$$

#### Reactions at A

$$
\begin{aligned}
R_A &= wL - R_B = 48 - 18 = 30.0\ \text{kN} \\
M_A &= \frac{wL^2}{2} - R_B L = \frac{12\times16}{2} - 18\times4 = 96 - 72 = 24.0\ \text{kN m (hogging)}
\end{aligned}
$$

Check: $M_A = wL^2/8 = 24$ kN m. $\checkmark$

#### Contraflexure

$M(x) = 18x - 6x^2 = 0 \Rightarrow x = 3$ m from B (i.e. 1 m from the fixed end).

#### Maximum sagging moment

Shear force $= R_B - wx = 0 \Rightarrow x = 18/12 = 1.5$ m from B:

$$
M_{max} = 18(1.5) - 6(1.5)^2 = 27 - 13.5 = 13.5\ \text{kN m (sagging)}
$$

#### Bending moment diagram (shape)

```text
 -24 at A                  0 at B
  \                          .
   \____ 0 at x=3 m _____ +13.5 at x=1.5 m
```

**Answer:** $R_B = 18$ kN, $R_A = 30$ kN, $M_A = 24$ kN m (hogging), contraflexure 1 m from the fixed end, maximum sagging moment $13.5$ kN m at 1.5 m from the prop.

## Check yourself

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

1. What condition determines the prop reaction R_B?
   A. Slope at B is zero
   B. dU/dR_B = 0 since B cannot deflect
   C. Total moment is zero
   D. R_B = wL
   Answer: B. The prop prevents vertical movement at B, so the partial derivative of strain energy wrt R_B is zero. (Key point: Redundant and compatibility)

2. What is R_B?
   A. 48 kN
   B. 24 kN
   C. 18 kN
   D. 30 kN
   Answer: C. R_B L^3/3 = wL^4/8 gives R_B = 3wL/8 = 18 kN. (Key point: Prop reaction value)

3. What is the fixed-end moment M_A and its nature?
   A. 24 kN m hogging
   B. 72 kN m sagging
   C. 96 kN m hogging
   D. 13.5 kN m sagging
   Answer: A. M_A = wL^2/2 - R_B L = 96 - 72 = 24 kN m hogging. (Key point: Fixed-end moment)

4. Where is the point of contraflexure and what is the maximum sagging moment?
   A. 1.5 m from B, 24 kN m
   B. 3 m from B, 13.5 kN m at 1.5 m from B
   C. 2 m from B, 18 kN m
   D. 3 m from B, 27 kN m
   Answer: B. M = 18x - 6x^2 = 0 at x = 3 m; shear is zero at x = 1.5 m where M = 13.5 kN m. (Key point: Contraflexure and max sagging)

## Related in Mechanics of Solids

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