---
title: "Plane stress: principal planes and max shear"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:plane-stress-15-30-60-principal"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-stress-15-30-60-principal
site: BCT Engineering
---

# Plane stress: principal planes and max shear

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2074 Chaitra, 8 marks
- Video: [Plane Stress and Transformation of Plane Stress: An introduction](https://www.youtube.com/watch?v=s2vdUgnEQtk) — CIVIL Solved
- 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-15-30-60-principal
- 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) the principal planes (ii) principal stresses (iii) the maximum shearing stress and the corresponding normal stress. [Figure: element with $60\ \text{MPa}$ tensile on the left and right faces; shear stresses labelled $15\ \text{MPa}$ on the top and bottom faces and $30\ \text{MPa}$ on the right face, as printed.]

## Short answer

Principal stresses are 75.00 MPa and 0.00 MPa on planes at 26.57° and 116.57° from x. Maximum shear stress is 37.50 MPa, with a constant normal stress of 37.50 MPa.

## Answer

**Reading of the figure:** shear stresses on perpendicular faces must be equal, so the labels are read as $\sigma_x = 60$ MPa (left/right faces), $\sigma_y = 15$ MPa (top/bottom faces) and $\tau_{xy} = 30$ MPa.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50 \pm \sqrt{(22.50)^2+(30.00)^2} = 37.50 \pm 37.50
$$

$\sigma_1 = 75.00\ \text{MPa}$ and $\sigma_2 = 0.00\ \text{MPa}$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{45.00} \Rightarrow 2\theta_p = 53.13^\circ
$$

So $\theta_p = 26.57^\circ$ (normal of the plane carrying $\sigma_1$) and $116.57^\circ$ (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

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50\ \text{MPa}
$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 71.57^\circ$ and $161.57^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 37.50\ \text{MPa}$.

**Answer:** $\sigma_1 = 75.00$, $\sigma_2 = 0.00$ at $\theta_p = 26.57^\circ$ and $116.57^\circ$; $\tau_{max} = 37.50$ with normal stress $37.50$ (MPa).

## Check yourself

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

1. Why is tau_xy = 30 and sigma_y = 15?
   A. Shear can differ on perpendicular faces
   B. Shear stresses on perpendicular faces are equal, so the 15 is a normal stress
   C. All labels are shear
   D. 15 is subtracted from 30
   Answer: B. Complementary shears are equal. (Key point: Resolving the figure labels)

2. What are they?
   A. 60 and 15
   B. 37.5 and 37.5
   C. 90 and -15
   D. 75 and 0 MPa
   Answer: D. 37.5 +/- 37.5. (Key point: Principal stresses)

3. What is theta_p for sigma_1?
   A. 53.13 deg
   B. 26.57 deg
   C. 45 deg
   D. 36.87 deg
   Answer: B. tan 2 theta_p = 60/45. (Key point: Principal plane)

4. What are tau_max and the normal stress?
   A. 75 and 0
   B. 30 and 37.5
   C. 37.5 and 75
   D. 37.5 MPa and 37.5 MPa
   Answer: D. Radius and centre are both 37.5. (Key point: Max shear)

## Close variants set in other exams

- [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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