---
title: "Shear deflection negligible vs bending"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:shear-deflection-negligible"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/shear-deflection-negligible
site: BCT Engineering
---

# Shear deflection negligible vs bending

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

- Asked in: 2069 Poush, 8 marks
- 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/shear-deflection-negligible
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

Define and explain strain energy. Use the strain energy method to show that the deflection due to shear in an ordinary beam can be neglected in comparison to the deflection due to bending. Assume the ratio of Young's modulus to modulus of rigidity to be 2.4 and the shape factor for shear 1.2.

## Short answer

Strain energy is work done by gradually applied loads, U = ½PΔ. For a simply supported beam with central load, bending deflection δ_b = PL³/(48EI), shear deflection δ_s = αPL/(4GA). Ratio δ_s/δ_b = α(E/G)(d/L)² = 2.88(d/L)². For ordinary beams (L/d=10-20), shear deflection is only 0.7-2.9% of bending deflection, hence negligible.

## Answer

**Strain energy** is the energy stored in an elastic body due to deformation produced by loads. It equals the work done by the gradually applied loads: $U = \tfrac12 P\Delta$. For a beam,

$$
U = \int \frac{M^2}{2EI}\,dx + \int \frac{\alpha V^2}{2GA}\,dx
$$

with the first term due to bending and the second due to shear.

#### Comparison of bending and shear deflections
Take a simply supported beam of span $L$, depth $d$ and rectangular section with a central load $P$. Here $M = Px/2$ (for $x \le L/2$) and $V = P/2$.

**Bending energy:**

$$
U_b = \frac{P^2L^3}{96EI}\ \Rightarrow\ \delta_b = \frac{\partial U_b}{\partial P} = \frac{PL^3}{48EI}
$$

**Shear energy:**

$$
U_s = \frac{\alpha (P/2)^2 L}{2GA} = \frac{\alpha P^2L}{8GA}\ \Rightarrow\ \delta_s = \frac{\partial U_s}{\partial P} = \frac{\alpha PL}{4GA}
$$

**Ratio:**

$$
\frac{\delta_s}{\delta_b} = \frac{\alpha PL}{4GA}\times\frac{48EI}{PL^3} = 12\,\alpha\,\frac{E}{G}\,\frac{I}{AL^2}
$$

For a rectangular section $I/A = d^2/12$, so

$$
\frac{\delta_s}{\delta_b} = \alpha\,\frac{E}{G}\left(\frac{d}{L}\right)^2
$$

Substituting $\alpha = 1.2$ and $E/G = 2.4$:

$$
\frac{\delta_s}{\delta_b} = 1.2\times 2.4\times\left(\frac{d}{L}\right)^2 = 2.88\left(\frac{d}{L}\right)^2
$$

For an ordinary beam, $L/d$ is 10 to 20, so

| $L/d$ | $\delta_s/\delta_b$ |
|---|---|
| 10 | 0.0288 (2.9%) |
| 15 | 0.0128 (1.3%) |
| 20 | 0.0072 (0.7%) |

Hence the shear deflection is only about 1 to 3% of the bending deflection and can be neglected in an ordinary beam. It becomes important only for deep beams ($L/d < 5$ or so).

## Check yourself

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

1. Central deflection of a simply supported beam due to bending is:
   A. PL^3/(3EI)
   B. PL^3/(48EI)
   C. 5PL^4/(384EI)
   D. PL^3/(96EI)
   Answer: B. δb = dUb/dP. (Key point: Bending deflection)

2. Shear deflection under the central load is:
   A. PL/(GA)
   B. αPL^3/(4GA)
   C. αPL/(4GA)
   D. αPL/(2EA)
   Answer: C. From Us = αP^2L/(8GA). (Key point: Shear deflection)

3. For a rectangular section δs/δb equals:
   A. α (G/E)(d/L)
   B. α (E/G)(L/d)^2
   C. (E/G)(d/L)
   D. α (E/G)(d/L)^2
   Answer: D. Using I/A = d^2/12. (Key point: Ratio)

4. With α=1.2, E/G=2.4, L/d=10, the ratio is about:
   A. 0.0288 (2.9%)
   B. 0.288
   C. 0.0072
   D. 0.12
   Answer: A. 2.88 x 0.01 = 0.0288. (Key point: Conclusion)

## Related in Theory of Structures I

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