---
title: "Stresses on 30 degree plane; Mohr's circle"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:inclined-plane-30deg-50-30-mohr"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-30deg-50-30-mohr
site: BCT Engineering
---

# Stresses on 30 degree plane; Mohr's circle

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2074 Asoj, 12 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
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-30deg-50-30-mohr
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4

## Question

For an infinitesimal element normal and shearing stress in the two mutually perpendicular planes are shown in figure below. Determine the normal and shearing stress on the inclined plane at an angle of $30^\circ$ with vertical. Also calculate principal stresses their planes, maximum shear stress and their planes. Verify your result using Mohr's circle. [Figure: element with $50\ \text{MPa}$ on the left and right faces and $30\ \text{MPa}$ on the top and bottom faces; inclined plane at $30^\circ$ to the vertical.]

## Short answer

On the 30° inclined plane: normal stress 63.48 MPa, shear stress -6.65 MPa. Principal stresses: 64.05 MPa and -14.05 MPa at 25.10° and 115.10°; maximum shear 39.05 MPa, normal stress 25.00 MPa.

## Answer

**Reading of the figure:** $\sigma_x = 50$ MPa on the left/right faces, $\sigma_y = 0$ and a shear stress $\tau_{xy} = 30$ MPa on all four faces. The plane is at $30^\circ$ to the vertical, so its normal is at $30^\circ$ to the $x$-axis.

#### Stresses on the inclined plane ($\theta = 30^\circ$)

$$
\begin{aligned}
\sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\
&= 25.00 + (25.00)\cos 60^\circ + (30.00)\sin 60^\circ = 63.48\ \text{MPa} \\
\tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\
&= -(25.00)\sin 60^\circ + (30.00)\cos 60^\circ = -6.65\ \text{MPa}
\end{aligned}
$$

Resultant stress on the plane $= \sqrt{63.48^2+-6.65^2} = 63.83\ \text{MPa}$, inclined at $5.98^\circ$ to the normal of the plane.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 25.00 \pm \sqrt{(25.00)^2+(30.00)^2} = 25.00 \pm 39.05
$$

$\sigma_1 = 64.05\ \text{MPa}$ and $\sigma_2 = -14.05\ \text{MPa}$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{50.00} \Rightarrow 2\theta_p = 50.19^\circ
$$

So $\theta_p = 25.10^\circ$ (normal of the plane carrying $\sigma_1$) and $115.10^\circ$ (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

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 39.05\ \text{MPa}
$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 70.10^\circ$ and $160.10^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 25.00\ \text{MPa}$.

#### Verification by Mohr's circle

- Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
- Plot $X(50, 30)$ and $Y(0, -30)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (25.00, 0)$.
- Draw the circle with centre C and radius $CX = 39.05$.
- It cuts the $\sigma$-axis at $\sigma_1 = 64.05$ and $\sigma_2 = -14.05$; $2\theta_p = \angle XCP_1 = 50.19^\circ$ (scaled by protractor).
- The highest and lowest points of the circle give $\tau_{max} = \pm39.05$ with $\sigma = 25.00$.
- Turn CX through $2\theta = 60^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 63.48$ and $\tau \approx -6.65$, which match the analytical values.

**Answer:** $\sigma_n = 63.48$, $\tau = -6.65$ on the inclined plane; $\sigma_1 = 64.05$, $\sigma_2 = -14.05$ at $\theta_p = 25.10^\circ$ and $115.10^\circ$; $\tau_{max} = 39.05$ with normal stress $25.00$ (MPa).

## Check yourself

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

1. What are the inputs?
   A. sigma_x = 50, sigma_y = 30, tau = 0
   B. sigma_x = 50, sigma_y = 0, tau = 30 MPa, theta = 30 deg
   C. sigma_x = 30, sigma_y = 50, tau = 30
   D. sigma_x = 50, sigma_y = 30, tau = 30
   Answer: B. Only 50 MPa normal acts on left/right faces; 30 MPa is shear on all faces. (Key point: Stress inputs)

2. What are sigma_n and tau?
   A. 25 and 30
   B. 64.05 and -14.05
   C. 50 and 30
   D. 63.48 and -6.65 MPa
   Answer: D. 25 + 25 cos60 + 30 sin60 = 63.48. (Key point: Normal and shear on plane)

3. What are sigma_1 and sigma_2?
   A. 50 and 0
   B. 64.05 and -14.05 MPa
   C. 75 and -25
   D. 63.48 and -6.65
   Answer: B. 25 +/- 39.05. (Key point: Principal stresses)

4. Where are the centre and radius?
   A. Centre (50,0), radius 30
   B. Centre (0,30), radius 25
   C. Centre (25,0), radius 30
   D. Centre (25, 0), radius 39.05
   Answer: D. Plot X(50,30), Y(0,-30). (Key point: Mohr's circle)

## Close variants set in other exams

- [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×
- [Stresses on 20 degree plane; principal and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-20deg-50-40-80) — Strength of Materials, chapter 4, asked 4×

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- [Principal stresses, plane BE, Mohr's circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-40-50-30-plane) — Strength of Materials, chapter 4
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- [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×
- [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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