---
title: "Cantilever strain energy: bending, shear, torsion"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:cantilever-strain-energy-circular-4m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cantilever-strain-energy-circular-4m
site: BCT Engineering
---

# Cantilever strain energy: bending, shear, torsion

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

- Asked in: 2073 Bhadra, 8 marks
- Video: [Analysis of cantilever beam using strain energy method](https://www.youtube.com/watch?v=lgMiA8P7sho) — Mastering Civil Engineering
- 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-4m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

A cantilever beam of length 4 m and having circular cross section of diameter 15 cm is subjected to a concentrated load of 10 kN and a twisting moment 5 kNm at its end. Calculate the strain energies due to bending, shear and torsion. $E = 200\ \text{kN/mm}^2$, $G = 80\ \text{kN/mm}^2$.

## Short answer

For a cantilever with end load 10kN and torque 5kNm, length 4m, diameter 0.15m, E=200GPa, G=80GPa: bending strain energy U_b = P²L³/(6EI) = 0.2146 kNm; shear strain energy U_s = αP²L/(2GA) = 1.57×10⁻⁴ kNm; torsional strain energy U_t = T²L/(2GJ) = 0.01258 kNm. Shear energy is negligible.

## Answer

#### Given data
$L = 4$ m, $d = 15$ cm $= 0.15$ m, $P = 10$ kN, $T = 5$ kNm, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $G = 80$ kN/mm$^2 = 8\times10^7$ kN/m$^2$.

#### Section properties
$$
\begin{aligned}
A &= \frac{\pi d^2}{4} = 0.017671\ \text{m}^2\\
I &= \frac{\pi d^4}{64} = 2.4850\times10^{-5}\ \text{m}^4\\
J &= \frac{\pi d^4}{32} = 4.9701\times10^{-5}\ \text{m}^4
\end{aligned}
$$

#### Bending energy
$M = Px$, so

$$
U_b = \frac{P^2L^3}{6EI} = \frac{10^2\times 4^3}{6\times2\times10^8\times2.4850\times10^{-5}} = \frac{6400}{29820.6} = 0.2146\ \text{kNm}
$$

#### Shear energy
$V = P$ constant. Shape factor for a solid circle $\alpha = 10/9$:

$$
U_s = \frac{\alpha P^2 L}{2GA} = \frac{(10/9)\times 100\times 4}{2\times 8\times10^{7}\times 0.017671} = \frac{444.44}{2827433} = 1.572\times10^{-4}\ \text{kNm}
$$

#### Torsion energy
$$
U_t = \frac{T^2L}{2GJ} = \frac{5^2\times4}{2\times8\times10^7\times4.9701\times10^{-5}} = \frac{100}{7952.2} = 0.01258\ \text{kNm}
$$

**Answer:**

| Energy | Value |
|---|---|
| Bending $U_b$ | 0.2146 kNm (214.6 J) |
| Shear $U_s$ | $1.57\times10^{-4}$ kNm (0.157 J) |
| Torsion $U_t$ | 0.01258 kNm (12.58 J) |

Shear energy is negligible compared with bending energy.

## Check yourself

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

1. The shear form factor for a solid circle is:
   A. 6/5
   B. 10/9
   C. 1
   D. 4/3
   Answer: B. Standard value. (Key point: Shape factor)

2. Shear energy for constant V = P is:
   A. P^2L^3/(6EI)
   B. α P^2L/(2EA)
   C. α P^2 L/(2GA)
   D. P^2L/(GJ)
   Answer: C. V is constant. (Key point: Shear energy)

3. Bending energy is about:
   A. 0.01258 kNm
   B. 1.572e-4 kNm
   C. 2.146 kNm
   D. 0.2146 kNm
   Answer: D. 6400/29820.6. (Key point: Bending result)

4. Compared with bending, shear energy is:
   A. Negligible
   B. Larger
   C. About equal
   D. About half
   Answer: A. 1.57e-4 vs 0.2146 kNm. (Key point: Comparison)

## Close variants set in other exams

- [Cantilever strain energy: bending and torsion](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acantilever-strain-energy-circular-8m) — Theory of Structures I, chapter 2, asked 2×

## Related in Theory of Structures I

- [Strain energy due to shear in rectangular beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Astrain-energy-shear-rect-beam) — Theory of Structures I, chapter 2, asked 3×
- [Strain energy in rectangular beam, central load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Ass-beam-strain-energy-rect) — Theory of Structures I, chapter 2
- [Strain energy due to bending, rectangular beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Astrain-energy-bending-derivation) — Theory of Structures I, chapter 2, asked 2×
- [Circular arch 50 m: rolling UDL, 15 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-50m-rolling-50knm-10m) — Theory of Structures I, chapter 6, asked 2×
- [Strain energy and real work deflection, overhang beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Areal-work-strain-energy-overhang-5m) — Theory of Structures I, chapter 3
- [Strain energy, gradual vs sudden load, shear](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Astrain-energy-gradual-sudden-shear) — Theory of Structures I, chapter 2

## Related in other subjects

- [Design 200x400 beam for bending, shear, torsion](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-200x400-beam-for-bending-shear) — Design of RCC Structure, chapter 5
- [Beam design for bending, torsion, shear, 425x550](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-design-for-bending-torsion-shear) — Design of RCC Structure, chapter 5
- [Castigliano: fixed-end moment, 15 kN load](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-15kn-2m-3m) — Theory of Structures II, chapter 2, asked 2×
- [T-beam: bending stress 15 cm above base](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-udl-5m-bending-stress) — Strength of Materials, chapter 7