---
title: "Cantilever strain energy: bending and torsion"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:cantilever-strain-energy-circular-8m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cantilever-strain-energy-circular-8m
site: BCT Engineering
---

# Cantilever strain energy: bending and torsion

Theory of Structures I · Chapter 2: Analysis by the Strain Energy Method

- Asked in: 2078 Chaitra, 6 marks
- Video: [Understanding Strain Energy In Torsion | GATE Strength Of Materials Explained](https://www.youtube.com/watch?v=jod8tu8zyEY) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cantilever-strain-energy-circular-8m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

A cantilever beam of length 8 m having circular cross section of diameter 20 cm is subjected to a load of 15 kN and a twisting moment of 8 kNm at its end. Calculate the strain energies due to bending and torsion. Assume necessary data.

## Short answer

For a cantilever with end load P=15kN and torque T=8kNm, length 8m, diameter 0.2m, E=200GPa, G=80GPa: bending strain energy U_b = P²L³/(6EI) = 1.222 kNm; torsional strain energy U_t = T²L/(2GJ) = 0.0204 kNm. Torsional energy is much smaller than bending energy.

## Answer

#### Given data
Length $L = 8$ m, diameter $d = 20$ cm $= 0.2$ m, end load $P = 15$ kN, twisting moment $T = 8$ kNm.

**Assumed:** $E = 200$ GPa $= 2\times10^8$ kN/m$^2$, $G = 80$ GPa $= 8\times10^7$ kN/m$^2$.

#### Section properties
$$
I = \frac{\pi d^4}{64} = \frac{\pi(0.2)^4}{64} = 7.854\times10^{-5}\ \text{m}^4
$$

$$
J = \frac{\pi d^4}{32} = 1.5708\times10^{-4}\ \text{m}^4
$$

#### Strain energy due to bending
At distance $x$ from the free end, $M = Px$:

$$
U_b = \int_0^L \frac{(Px)^2}{2EI}dx = \frac{P^2L^3}{6EI}
$$

$$
U_b = \frac{15^2\times 8^3}{6\times 2\times10^{8}\times 7.854\times10^{-5}} = \frac{115200}{94247.8} = 1.2223\ \text{kNm}
$$

#### Strain energy due to torsion
Torque is constant along the length:

$$
U_t = \frac{T^2L}{2GJ} = \frac{8^2\times 8}{2\times 8\times10^{7}\times 1.5708\times10^{-4}} = \frac{512}{25132.7} = 0.02037\ \text{kNm}
$$

**Answer:** $U_b = 1.222$ kNm ($\approx 1222$ J) and $U_t = 0.0204$ kNm ($\approx 20.4$ J).

The torsional energy is small compared to the bending energy.

## Check yourself

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

1. Bending energy of a cantilever with end load is:
   A. P^2L^3/(96EI)
   B. P^2L^3/(6EI)
   C. P^2L^3/(2EI)
   D. P^2L/(2GA)
   Answer: B. M = Px. (Key point: Bending formula)

2. Torsional energy under constant T is:
   A. T^2L/(2EI)
   B. T^2L^3/(6GJ)
   C. T^2L/(2GJ)
   D. T^2L/(GA)
   Answer: C. Torque is constant along length. (Key point: Torsion formula)

3. For a solid circle, J equals:
   A. πd^4/64
   B. πd^2/4
   C. πd^3/16
   D. πd^4/32
   Answer: D. Polar moment is twice I. (Key point: Section property)

4. The strain energy values are:
   A. Ub about 1.222 kNm and Ut about 0.0204 kNm
   B. Ub about 0.0204 kNm and Ut about 1.222 kNm
   C. Both about 1.22 kNm
   D. Ub about 12.2 kNm
   Answer: A. Bending energy dominates. (Key point: Result)

## Close variants set in other exams

- [Cantilever strain energy: bending, shear, torsion](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acantilever-strain-energy-circular-4m) — Theory of Structures I, chapter 2, asked 2×

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