---
title: "Plane stress: principal stresses and oblique plane"
subject: "Mechanics of Solids"
chapter: 2
key: "semester_5/mechanics_of_solids:plane-stress-principal-numerical"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/plane-stress-principal-numerical
site: BCT Engineering
---

# Plane stress: principal stresses and oblique plane

Mechanics of Solids · Chapter 2: Stress Tensor

- Asked in: year not given, 8 marks
- Video: [Principal Stresses and MOHR'S CIRCLE in 12 Minutes!!](https://www.youtube.com/watch?v=aumN-Vl0OhQ) — Less Boring Lectures
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/plane-stress-principal-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-2

## Question

At a point in a loaded steel member the stresses on the x- and y-planes are $\sigma_x = 80$ MPa (tension), $\sigma_y = -40$ MPa (compression) and $\tau_{xy} = 30$ MPa (positive). Find (a) the principal stresses and the orientation of the principal planes, (b) the maximum shear stress, and (c) the normal and shear stress on a plane whose normal makes $30^\circ$ anticlockwise with the x-axis.

## Short answer

Principal stresses: σ1=87.1 MPa (tension), σ2=-47.1 MPa (compression), with principal plane at 13.3° anticlockwise from x-axis. Maximum shear stress τ_max=67.1 MPa. On plane at30°, normal stress σ_n=76.0 MPa, shear stress τ=-37.0 MPa.

## Answer

Use the plane-stress transformation equations (Rajput, Timoshenko).

#### Given

$\sigma_x = 80$, $\sigma_y = -40$, $\tau_{xy} = 30$ MPa.

#### (a) Principal stresses

$$
\begin{aligned}
\sigma_{avg} &= \frac{\sigma_x+\sigma_y}{2} = \frac{80-40}{2} = 20.0\ \text{MPa} \\
R &= \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{60^2+30^2} = 67.08\ \text{MPa} \\
\sigma_{1,2} &= \sigma_{avg} \pm R = 87.08,\ -47.08\ \text{MPa}
\end{aligned}
$$

Orientation:

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{60}{120} = 0.5
\quad\Rightarrow\quad 2\theta_p = 26.57^\circ,\ \theta_p = 13.28^\circ
$$

Substituting $\theta = 13.28^\circ$ in the $\sigma_n$ equation gives $87.08$ MPa, so this plane carries $\sigma_1$. The other principal plane is at $13.28 + 90 = 103.28^\circ$. Shear stress is zero on both.

#### (b) Maximum shear stress

$$
\tau_{max} = R = 67.08\ \text{MPa}
$$

It acts on planes at $\theta_s = \theta_p + 45^\circ = 58.28^\circ$, with a normal stress $\sigma_{avg} = 20$ MPa on those planes. (The out-of-plane principal stress is zero; $(\sigma_1 - 0)/2 = 43.5$ MPa is smaller, so the in-plane value governs.)

#### (c) Plane at $\theta = 30^\circ$

$$
\begin{aligned}
\sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\
&= 20 + 60\cos 60^\circ + 30 \sin 60^\circ = 20 + 30.00 + 25.98 = 75.98\ \text{MPa} \\[4pt]
\tau_{n} &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\
&= -60\sin 60^\circ + 30\cos 60^\circ = -51.96 + 15.00 = -36.96\ \text{MPa}
\end{aligned}
$$

Check: the normal stress on the perpendicular plane ($120^\circ$) is $-35.98$ MPa, and $75.98 + (-35.98) = 40 = \sigma_x + \sigma_y$ (first invariant preserved).

**Answer:** $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $\theta_p = 13.3^\circ$; $\tau_{max} = 67.1$ MPa; on the $30^\circ$ plane $\sigma_n = 76.0$ MPa and $\tau = -37.0$ MPa.

## Check yourself

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

1. For sigma_x = 80, sigma_y = -40, tau_xy = 30 MPa, what is the radius R used to find the principal stresses?
   A. sqrt(60^2 + 30^2) = 67.08 MPa
   B. sqrt(80^2 + 30^2) = 85.4 MPa
   C. (80 + 40)/2 = 60 MPa
   D. sqrt(20^2 + 30^2) = 36.06 MPa
   Answer: A. R = sqrt(((sx - sy)/2)^2 + tau^2) = sqrt(60^2 + 30^2) = 67.08 MPa, so sigma = 20 +/- 67.08. (Key point: Principal stress formula)

2. How is the principal plane angle found here?
   A. tan 2theta = (sx - sy)/(2 tau) = 2
   B. tan 2theta = 2 tau/(sx - sy) = 0.5, theta_p = 13.28 degrees
   C. tan theta = 0.5, theta = 26.57 degrees
   D. tan 2theta = tau/sx
   Answer: B. tan 2theta_p = 2(30)/(120) = 0.5 gives 2theta_p = 26.57 degrees, theta_p = 13.28 degrees. (Key point: Principal plane angle)

3. What are tau_max and the normal stress on its planes?
   A. 43.5 MPa with normal stress 0
   B. 87.1 MPa with normal stress 20 MPa
   C. 67.08 MPa with normal stress 80 MPa
   D. 67.08 MPa with normal stress 20 MPa
   Answer: D. tau_max = R = 67.08 MPa, with normal stress equal to sigma_avg = 20 MPa on planes 45 degrees from the principal planes. (Key point: Maximum shear stress)

4. What is sigma_n on the plane at 30 degrees?
   A. 75.98 MPa
   B. 20 MPa
   C. -35.98 MPa
   D. 87.08 MPa
   Answer: A. sigma_n = 20 + 60 cos60 + 30 sin60 = 20 + 30 + 25.98 = 75.98 MPa. (Key point: Stresses on inclined plane)

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