---
title: "Principal planes, stresses on plane AC"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:principal-planes-max-shear-120-180"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-planes-max-shear-120-180
site: BCT Engineering
---

# Principal planes, stresses on plane AC

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2068 Baisakh (old course), 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/principal-planes-max-shear-120-180
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4

## Question

The state of the stress in a two dimensional stress system is as shown in the figure. Find the principal planes and maximum shear stress. Determine also the normal and tangential stress on plane AC. Verify the results by drawing Mohr's circle. [Figure: rectangular element with $120\ \text{N/mm}^2$ acting horizontally on the top face and $180\ \text{N/mm}^2$ tensile on the right face, shear stress $40\ \text{N/mm}^2$; plane AC runs diagonally from A at the bottom left to C at the top right, inclined at $60^\circ$ to the horizontal at C.]

## Short answer

With σx=180, σy=120, τxy=40 N/mm², on plane AC (normal -30°): σn=130.36, τ=45.98 N/mm². Principal stresses 200.00, 100.00 N/mm² at 26.57°, 116.57°. Maximum shear 50.00 N/mm², normal stress 150.00.

## Answer

**Reading of the figure:** $\sigma_x = 180$ N/mm² (right face, tensile), $\sigma_y = 120$ N/mm² (top face), $\tau_{xy} = 40$ N/mm². The plane AC rises at $60^\circ$ to the horizontal, so its normal is at $\theta = 60^\circ - 90^\circ = -30^\circ$ from the $x$-axis (anticlockwise positive).

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

$$
\begin{aligned}
\sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\
&= 150.00 + (30.00)\cos(-60^\circ) + (40.00)\sin(-60^\circ) = 130.36\ \text{N/mm}^2 \\
\tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\
&= -(30.00)\sin(-60^\circ) + (40.00)\cos(-60^\circ) = 45.98\ \text{N/mm}^2
\end{aligned}
$$

Resultant stress on the plane $= \sqrt{130.36^2+45.98^2} = 138.23\ \text{N/mm}^2$, inclined at $19.43^\circ$ to the normal of the plane.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 150.00 \pm \sqrt{(30.00)^2+(40.00)^2} = 150.00 \pm 50.00
$$

$\sigma_1 = 200.00\ \text{N/mm}^2$ and $\sigma_2 = 100.00\ \text{N/mm}^2$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(40.00)}{60.00} \Rightarrow 2\theta_p = 53.13^\circ
$$

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

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 50.00\ \text{N/mm}^2
$$

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 $\frac{\sigma_x+\sigma_y}{2} = 150.00\ \text{N/mm}^2$.

#### Verification by Mohr's circle

- Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
- Plot $X(180, 40)$ and $Y(120, -40)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (150.00, 0)$.
- Draw the circle with centre C and radius $CX = 50.00$.
- It cuts the $\sigma$-axis at $\sigma_1 = 200.00$ and $\sigma_2 = 100.00$; $2\theta_p = \angle XCP_1 = 53.13^\circ$ (measured with a protractor).
- The highest and lowest points of the circle give $\tau_{max} = \pm50.00$ with $\sigma = 150.00$.
- Turn CX through $2\theta = 60^\circ$ clockwise (the same sense as $\theta = -30^\circ$) to reach point Q; its coordinates read $\sigma_n \approx 130.36$ and $\tau \approx 45.98$, which match the analytical values.

**Answer:** $\sigma_n = 130.36$, $\tau = 45.98$ on plane AC; $\sigma_1 = 200.00$, $\sigma_2 = 100.00$ at $\theta_p = 26.57^\circ$ and $116.57^\circ$; $\tau_{max} = 50.00$ with normal stress $150.00$ (N/mm²).

## Check yourself

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

1. With sx = 180, sy = 120, txy = 40, what are the principal stresses?
   A. 180 and 120 N/mm2
   B. 190 and 110 N/mm2
   C. 200 and 100 N/mm2
   D. 150 and 50 N/mm2
   Answer: C. 150 +/- sqrt(30^2 + 40^2) = 150 +/- 50. (Key point: Principal stresses)

2. What is the maximum shear stress and the normal stress with it?
   A. 40 N/mm2 with normal stress 150 N/mm2
   B. 50 N/mm2 with normal stress 150 N/mm2
   C. 50 N/mm2 with normal stress 0
   D. 100 N/mm2 with normal stress 150 N/mm2
   Answer: B. tau_max = (200 - 100)/2 = 50 on planes with normal stress (sx+sy)/2 = 150. (Key point: Maximum shear stress)

3. How is Mohr's circle drawn to verify the results?
   A. Plot X(180, 40) and Y(120, -40); join XY to find the centre; radius CX = 50
   B. Plot X(180, 40) and Y(120, 40) on the same side
   C. Centre at (180, 0) with radius 40
   D. Centre at origin with radius 200
   Answer: A. The two shear stresses go on opposite sides of the sigma-axis; XY cuts the axis at the centre 150. (Key point: Mohr's circle construction)

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