---
title: "Stresses on 20 degree plane; principal and max shear"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:inclined-plane-20deg-50-40-80"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-20deg-50-40-80
site: BCT Engineering
---

# Stresses on 20 degree plane; principal and max shear

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2078 Bhadra, 8 marks
- Video: [HOW TO PLOT THE STRESSES ON AN IN PLANE STRESS ELEMENT ON THE MOHR'S CIRCLE | STRENGTH OF MATERIALS](https://www.youtube.com/watch?v=4koyXom4PlE) — Engr. Jobert S. De La Cruz
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-20deg-50-40-80
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4

## Question

For an infinitesimal element normal and shearing stresses in the two mutually perpendicular planes are given below. Determine the normal and shearing stresses on the inclined plane at an angle of $20^\circ$ with vertical. Also calculate principal stresses, their planes, maximum shear stresses and their planes. [Figure: element with $50\ \text{MPa}$ acting vertically on the top and bottom faces, $80\ \text{MPa}$ acting horizontally on the left and right faces, and shear stress $40\ \text{MPa}$ on all faces.]

## Short answer

σn = 102.20 MPa, τ = 21.00 MPa. Principal stresses: 107.72 and 22.28 MPa, planes at 34.72° and 124.72° from x-axis. τmax = 42.72 MPa at 79.72°.

## Answer

**Reading of the figure:** $\sigma_x = 80$ MPa on the vertical faces (horizontal direction), $\sigma_y = 50$ MPa on the horizontal faces (vertical direction), $\tau_{xy} = 40$ MPa; the plane is at $20^\circ$ to the vertical, so its normal is $20^\circ$ from the $x$-axis.

#### Stresses on the inclined plane ($\theta = 20^\circ$)

$$
\begin{aligned}
\sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\
&= 65.00 + (15.00)\cos 40^\circ + (40.00)\sin 40^\circ = 102.20\ \text{MPa} \\
\tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\
&= -(15.00)\sin 40^\circ + (40.00)\cos 40^\circ = 21.00\ \text{MPa}
\end{aligned}
$$

Resultant stress on the plane $= \sqrt{102.20^2+21.00^2} = 104.34\ \text{MPa}$, inclined at $11.61^\circ$ to the normal of the plane.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 65.00 \pm \sqrt{(15.00)^2+(40.00)^2} = 65.00 \pm 42.72
$$

$\sigma_1 = 107.72\ \text{MPa}$ and $\sigma_2 = 22.28\ \text{MPa}$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(40.00)}{30.00} \Rightarrow 2\theta_p = 69.44^\circ
$$

So $\theta_p = 34.72^\circ$ (normal of the plane carrying $\sigma_1$) and $124.72^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

#### Maximum shear stress

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 42.72\ \text{MPa}
$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 79.72^\circ$ and $169.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 65.00\ \text{MPa}$.

#### Check with Mohr's circle

- Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
- Plot $X(80, 40)$ and $Y(50, -40)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (65.00, 0)$.
- Draw the circle with centre C and radius $CX = 42.72$.
- It cuts the $\sigma$-axis at $\sigma_1 = 107.72$ and $\sigma_2 = 22.28$; $2\theta_p = \angle XCP_1 = 69.44^\circ$ (scaled by protractor).
- The highest and lowest points of the circle give $\tau_{max} = \pm42.72$ with $\sigma = 65.00$.
- Turn CX through $2\theta = 40^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 102.20$ and $\tau \approx 21.00$, which match the analytical values.

**Answer:** $\sigma_n = 102.20$ MPa, $\tau = 21.00$ MPa; $\sigma_1 = 107.72$ MPa, $\sigma_2 = 22.28$ MPa at $\theta_p = 34.72^\circ$ and $124.72^\circ$; $\tau_{max} = 42.72$ MPa.

## Check yourself

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

1. Which values are taken?
   A. sigma_x = 50, sigma_y = 80, tau = 40, theta = 70
   B. sigma_x = 80, sigma_y = 40, tau = 50
   C. sigma_x = 65, sigma_y = 15, tau = 40
   D. sigma_x = 80, sigma_y = 50, tau_xy = 40 MPa, theta = 20 deg
   Answer: D. The 80 MPa acts on vertical faces. (Key point: Reading figure)

2. What are sigma_n and tau?
   A. 65 MPa and 40 MPa
   B. 102.20 MPa and 21.00 MPa
   C. 107.72 MPa and 0
   D. 94.5 MPa and 42.72 MPa
   Answer: B. 65 + 15 cos40 + 40 sin40 = 102.20 and -15 sin40 + 40 cos40 = 21.00. (Key point: Normal and shear on plane)

3. What are sigma_1, sigma_2?
   A. 130 and 15
   B. 65 and 42.72
   C. 122.28 and 7.72
   D. 107.72 and 22.28 MPa
   Answer: D. 65 +/- 42.72. (Key point: Principal stresses)

4. What is tau_max and its normal stress?
   A. 65 MPa with 42.72 normal
   B. 42.72 MPa with 65 MPa normal stress
   C. 21.00 with 102.20
   D. 85.44 with 65
   Answer: B. tau_max = R = sqrt(15^2 + 40^2), normal stress is the average. (Key point: Max shear)

## Close variants set in other exams

- [Stresses on 40 degree plane; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-40deg-50-30) — Strength of Materials, chapter 4, asked 2×
- [Stresses on 30 degree plane; Mohr's circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30deg-50-30-mohr) — Strength of Materials, chapter 4, asked 3×
- [Normal/shear stress on 25 degree plane, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30-25deg-mohr-verify) — Strength of Materials, chapter 4, asked 3×

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- [Plane stress: principal, max shear, normal](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-80-60-50-mpa) — Strength of Materials, chapter 4, asked 2×
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