---
title: "Shear stress in rectangular and circular beams"
subject: "Strength of Materials"
chapter: 5
key: "semester_4/strength_of_materials_mechanical:shear-distribution-rectangle-circle-ratio"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shear-distribution-rectangle-circle-ratio
site: BCT Engineering
---

# Shear stress in rectangular and circular beams

Strength of Materials · Chapter 5: Transverse loading

- Asked in: year not given, 5 marks
- Video: [Lecture 22 Horizontal Shear Stress for Rectangular & Circular Cross Section](https://www.youtube.com/watch?v=k6azcf0JWj0) — HN Sharma
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shear-distribution-rectangle-circle-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-5

## Question

Show that the maximum shear stress in a rectangular beam section is 1.5 times the average shear stress, and that in a solid circular section it is 4/3 times the average. Draw the shear stress distributions.

## Short answer

Rectangle: τ=(6V/bh³)(h²/4 – y²), maximum at neutral axis is 3V/(2bh)=1.5τ_avg. Solid circle: τ=(4V/3πr⁴)(r² – y²), maximum is 4V/(3πr²)=4/3 τ_avg. Both distributions are parabolic, zero at edges.

## Answer

#### Rectangular section ($b\times h$)
At a level $y$ above the NA, the area above is $b\left(\frac{h}{2}-y\right)$, with its centroid at $\frac12\left(\frac{h}{2}+y\right)$ from the NA:

$$
Q = b\left(\frac h2 - y\right)\cdot\frac12\left(\frac h2+y\right) = \frac b2\left(\frac{h^2}{4}-y^2\right)
$$

With $I = bh^3/12$ and width $b$:

$$
\tau = \frac{VQ}{Ib} = \frac{6V}{bh^3}\left(\frac{h^2}{4} - y^2\right)
$$

This is a **parabola**: $\tau = 0$ at $y = \pm h/2$, and it is maximum at $y = 0$:

$$
\tau_{max} = \frac{6V}{bh^3}\cdot\frac{h^2}{4} = \frac{3V}{2bh} = 1.5\,\tau_{avg}, \qquad \tau_{avg} = \frac{V}{bh}
$$

#### Solid circular section (diameter $d$, radius $r$)
At a level $y$ above the NA, the chord has width $b = 2\sqrt{r^2-y^2}$. The first moment of the segment above $y$ is

$$
Q = \int_y^r 2\sqrt{r^2-y_1^2}\;y_1\,dy_1 = \frac23\,(r^2-y^2)^{3/2}
$$

With $I = \pi r^4/4$:

$$
\tau = \frac{VQ}{Ib} = \frac{V\cdot\frac23(r^2-y^2)^{3/2}}{\frac{\pi r^4}{4}\cdot2\sqrt{r^2-y^2}} = \frac{4V}{3\pi r^4}(r^2-y^2)
$$

Maximum at $y = 0$:

$$
\tau_{max} = \frac{4V}{3\pi r^2} = \frac43\cdot\frac{V}{\pi r^2} = \frac43\,\tau_{avg}
$$

#### Distributions
```text
   Rectangle                    Circle
  +--------+  0                 .--.    0
  |        |\                  /    \   |\
  |        | \                |      |  | \
  |   NA   |  > 1.5 avg       |  NA  |  |  > 1.33 avg
  |        | /                |      |  | /
  |        |/                  \    /   |/
  +--------+  0                 '--'    0
```

Both distributions are parabolic, zero at the top and bottom edges and maximum at the neutral axis.

## Check yourself

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

1. How does shear stress vary over a rectangular section?
   A. Uniform
   B. Parabolic, zero at top and bottom and maximum at the NA
   C. Linear from zero at the NA
   D. Maximum at the top
   Answer: B. tau = (6V/bh^3)(h^2/4 - y^2) is parabolic in y. (Key point: Rectangular distribution)

2. What is tau_max for a rectangle in terms of the average?
   A. 1.33 x average
   B. 1.5 x average, tau_max = 3V/(2bh)
   C. 2 x average
   D. Equal to average
   Answer: B. At y = 0, tau_max = 3V/(2bh) = 1.5 V/(bh). (Key point: Max to average ratio, rectangle)

3. Which is the result for a solid circular section?
   A. tau_max = 1.5 tau_avg
   B. tau_max = (4/3) tau_avg = 4V/(3 pi r^2)
   C. tau_max = 2 tau_avg
   D. tau_max = tau_avg
   Answer: B. With Q = (2/3)(r^2-y^2)^(3/2), I = pi r^4/4, b = 2 sqrt(r^2-y^2), tau_max = 4V/(3 pi r^2). (Key point: Circular section)

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