---
title: "Plane stress: principal, max shear, normal"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:plane-stress-80-60-50-mpa"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-stress-80-60-50-mpa
site: BCT Engineering
---

# Plane stress: principal, max shear, normal

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2075 Asoj, 8 marks
- Video: [Mechanics of Materials Lecture 19: Principal stresses and maximum in-plane shear stress](https://www.youtube.com/watch?v=6mWlkqfAjz8) — Yiheng Wang
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-stress-80-60-50-mpa
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4

## Question

For the state of plane stress shown in figure below determine. i) principal stresses ii) orientation of principal planes iii) maximum shearing stress iv) normal stress on the plane of maximum shear stress [Figure: element with $60\ \text{MPa}$ on the top and bottom faces, $80\ \text{MPa}$ on the left and right faces, shear stress $50\ \text{MPa}$.]

## Short answer

Principal stresses are 120.99 MPa and 19.01 MPa at 39.35° and 129.35° anticlockwise from x. Maximum shear stress is 50.99 MPa, accompanied by 70.00 MPa normal stress.

## Answer

**Reading of the figure:** $\sigma_x = 80$ MPa (left/right faces), $\sigma_y = 60$ MPa (top/bottom faces), $\tau_{xy} = 50$ MPa.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(50.00)^2} = 70.00 \pm 50.99
$$

$\sigma_1 = 120.99\ \text{MPa}$ and $\sigma_2 = 19.01\ \text{MPa}$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{20.00} \Rightarrow 2\theta_p = 78.69^\circ
$$

So $\theta_p = 39.35^\circ$ (normal of the plane carrying $\sigma_1$) and $129.35^\circ$ (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

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 50.99\ \text{MPa}
$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 84.35^\circ$ and $174.35^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$.

**Answer:** $\sigma_1 = 120.99$, $\sigma_2 = 19.01$ at $\theta_p = 39.35^\circ$ and $129.35^\circ$; $\tau_{max} = 50.99$ with normal stress $70.00$ (MPa).

## Check yourself

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

1. What are sigma_1 and sigma_2?
   A. 130 and 10
   B. 80 and 60
   C. 70 and 50.99
   D. 120.99 and 19.01 MPa
   Answer: D. 70 +/- sqrt(10^2 + 50^2). (Key point: Principal stresses)

2. Which is theta_p?
   A. 78.69 deg
   B. 39.35 deg
   C. 45 deg
   D. 19.01 deg
   Answer: B. tan 2 theta_p = 100/20, 2 theta_p = 78.69 deg. (Key point: Principal plane)

3. What is tau_max?
   A. 50 MPa
   B. 70 MPa
   C. 101.98 MPa
   D. 50.99 MPa
   Answer: D. tau_max = (sigma_1 - sigma_2)/2. (Key point: Maximum shear)

4. What is it?
   A. 50.99 MPa
   B. 70 MPa
   C. 0
   D. 120.99 MPa
   Answer: B. The average of sigma_x and sigma_y. (Key point: Normal stress on max shear plane)

## Close variants set in other exams

- [Plane stress: principal planes and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-15-30-60-principal) — Strength of Materials, chapter 4, asked 2×

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- [Principal stresses, max shear: 80, 60, 20 MPa](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-stresses-60-20-80) — Strength of Materials, chapter 4
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