---
title: "Max shear for 150 MPa stress limit; principal stresses"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:direct-stresses-100t-60c-shear-limit"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/direct-stresses-100t-60c-shear-limit
site: BCT Engineering
---

# Max shear for 150 MPa stress limit; principal stresses

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2076 Chaitra, 8 marks
- Video: [Mohr's Circle Construction | Calculation of Principal Stress | Strength of Materials](https://www.youtube.com/watch?v=4hgJmOxz2YA) — The Mechanical Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/direct-stresses-100t-60c-shear-limit
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4

## Question

Direct stresses of 100 MPa in tension and 60 MPa in compression are applied to an elastic material at a certain point on planes right angles to each other. If the maximum stress is not to exceed 150 MPa, to what shearing stress can the material be subjected at the point? What is then the maximum shearing stress in the material? Also find the magnitude of the principal stresses and its planes.

## Short answer

Limiting shear stress is 102.47 MPa, giving a maximum shear stress of 130 MPa. Principal stresses are 150 MPa and -110 MPa, oriented at 26.01° and 116.01° to the 100 MPa direction.

## Answer

**Reading:** the maximum (principal) stress must not exceed 150 MPa. Let $\sigma_x = 100$ MPa (tension), $\sigma_y = -60$ MPa (compression) and $\tau$ the unknown shear stress.

#### Shear stress that can be applied
$$
\frac{\sigma_x+\sigma_y}{2} = \frac{100-60}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 80\ \text{MPa}
$$

The maximum principal stress is

$$
\sigma_1 = \frac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 150
$$

$$
\sqrt{80^2 + \tau^2} = 150 - 20 = 130 \Rightarrow \tau^2 = 130^2 - 80^2 = 10500
$$

$$
\tau = 102.47\ \text{MPa}
$$

#### Maximum shear stress in the material
$$
\tau_{max} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 130\ \text{MPa}
$$

#### Principal stresses and planes
$$
\sigma_1 = 20 + 130 = 150\ \text{MPa}, \qquad \sigma_2 = 20 - 130 = -110\ \text{MPa}
$$

$$
\tan 2\theta_p = \frac{2\tau}{\sigma_x-\sigma_y} = \frac{2(102.47)}{160} = 1.2809 \Rightarrow 2\theta_p = 52.02^\circ,\ \theta_p = 26.01^\circ
$$

$\sigma_1$ acts on the plane whose normal is at $26.01^\circ$ to the direction of the 100 MPa stress, and $\sigma_2$ on the plane at $116.01^\circ$ (anticlockwise, with $\tau$ taken positive as in the standard sign convention).

**Answer:** shear stress $= 102.47$ MPa; $\tau_{max} = 130$ MPa; $\sigma_1 = 150$ MPa, $\sigma_2 = -110$ MPa at $26.01^\circ$ and $116.01^\circ$.

## Check yourself

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

1. What equation finds the shear stress?
   A. 100 + tau = 150
   B. sigma_1 = 80 + tau = 150
   C. sigma_1 = 20 + sqrt(80^2 + tau^2) = 150
   D. tau = 150 - 100
   Answer: C. The maximum principal stress is limited to 150 MPa. (Key point: Governing condition)

2. What shear stress can be applied?
   A. 102.47 MPa
   B. 50 MPa
   C. 130 MPa
   D. 90 MPa
   Answer: A. tau^2 = 130^2 - 80^2 = 10500. (Key point: Shear stress)

3. What is tau_max?
   A. 102.47 MPa
   B. 150 MPa
   C. 130 MPa
   D. 80 MPa
   Answer: C. tau_max = sqrt(80^2 + 102.47^2) = 130 MPa. (Key point: Max shear)

4. What is sigma_2?
   A. -110 MPa
   B. -60 MPa
   C. -130 MPa
   D. 20 MPa
   Answer: A. sigma_2 = 20 - 130. (Key point: Principal stresses)

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