---
title: "Strain energy in rectangular beam, central load"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:ss-beam-strain-energy-rect"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ss-beam-strain-energy-rect
site: BCT Engineering
---

# Strain energy in rectangular beam, central load

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

- Asked in: 2072 Magh, 4 marks
- Video: [EX  2 :-Strain Energy and deflection of beam II simply supported beam II](https://www.youtube.com/watch?v=P70ssvZgu3M) — G M Basha
- 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/ss-beam-strain-energy-rect
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

A rectangular beam 25 cm × 50 cm (b × d) is simply supported on a span of 6 m and carries a central load of 100 kN. Calculate the strain energy due to shear and bending. Take $E = 2\times10^6\ \text{kg/cm}^2$ and $G = 0.85\times10^6\ \text{kg/cm}^2$. Neglect self weight of the beam.

## Short answer

For a simply supported rectangular beam (25cm×50cm) with central load 100kN, span 6m: bending strain energy U_b = P²L³/(96EI) = 448.9 kg-cm; shear strain energy U_s = αV²L/(2GA) = 8.80 kg-cm. Shear energy is only about 2% of bending energy.

## Answer

#### Given data
$b = 25$ cm, $d = 50$ cm, $L = 6$ m $= 600$ cm, $P = 100$ kN, $E = 2\times10^6$ kg/cm$^2$, $G = 0.85\times10^6$ kg/cm$^2$.

**Assumption:** $1\ \text{kgf} = 9.81$ N, so $P = 100\,000/9.81 = 10\,193.7$ kgf. Shear form factor for rectangle $\alpha = 1.2$.

$$
I = \frac{25\times50^3}{12} = 2.604\times10^5\ \text{cm}^4,\qquad A = 25\times50 = 1250\ \text{cm}^2
$$

#### Strain energy due to bending
For a central load, $M = Px/2$ for $x \le L/2$:

$$
U_b = 2\int_0^{L/2}\frac{(Px/2)^2}{2EI}dx = \frac{P^2L^3}{96EI}
$$

$$
U_b = \frac{(10193.7)^2\times(600)^3}{96\times2\times10^6\times2.604\times10^5} = 448.9\ \text{kg-cm}
$$

#### Strain energy due to shear
$V = P/2$ throughout the span:

$$
U_s = \frac{\alpha V^2 L}{2GA} = \frac{1.2\times(5096.8)^2\times600}{2\times0.85\times10^6\times1250} = 8.80\ \text{kg-cm}
$$

**Answer:**
- Bending: $U_b = 448.9$ kg-cm ($\approx 44.0$ N·m)
- Shear: $U_s = 8.80$ kg-cm ($\approx 0.863$ N·m)

Shear energy is only about 2% of the bending energy.

## Check yourself

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

1. Bending energy for central load on simply supported beam is:
   A. P^2L^3/(6EI)
   B. P^2L^3/(96EI)
   C. P^2L^3/(48EI)
   D. P^2L/(8GA)
   Answer: B. 2 times integral of P^2x^2/(8EI). (Key point: Bending energy)

2. Shear force in the span is:
   A. P
   B. P/4
   C. P/2 throughout
   D. Varies from 0 to P
   Answer: C. Reactions equal P/2. (Key point: Shear force)

3. Shear energy is:
   A. V^2 L/(2EI)
   B. α V^2 L/(2EA)
   C. α V L/(GA)
   D. α V^2 L/(2GA) with α = 1.2
   Answer: D. Uses G and A. (Key point: Shear energy)

4. Shear energy is about what fraction of bending energy?
   A. About 2%
   B. About 50%
   C. About 20%
   D. About 100%
   Answer: A. 8.80 / 448.9. (Key point: Comparison)

## Related in Theory of Structures I

- [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×
- [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×
- [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×
- [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
- [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
- [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 other subjects

- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4
- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7