---
title: "Normal/shear stress on 25 degree plane, principal stresses"
subject: "Strength of Materials"
chapter: 4
key: "semester_3/strength_of_materials:inclined-plane-30-25deg-mohr-verify"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-30-25deg-mohr-verify
site: BCT Engineering
---

# Normal/shear stress on 25 degree plane, principal stresses

Strength of Materials · Chapter 4: Stress and Strain Analysis

- Asked in: 2080 Baisakh, 10 marks
- Video: [Understanding Stress Transformation and Mohr's Circle](https://www.youtube.com/watch?v=_DH3546mSCM) — The Efficient 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/inclined-plane-30-25deg-mohr-verify
- 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 AB at an angle of $25^\circ$ with vertical. Also calculate principal stresses, their planes, maximum shear stresses and their planes. Verify your result using Mohr's circle. [Figure: element with $\sigma_x = 60\ \text{MPa}$ tensile on the left and right faces, $\sigma_y = 30\ \text{MPa}$ tensile on the top and bottom faces, shear stress 50 MPa on all faces; inclined plane AB at $25^\circ$ to the vertical.]

## Short answer

σn = 92.94 MPa, τθ = 20.65 MPa. Principal stresses = 97.20 and −7.20 MPa; planes at 36.65° from x-axis. τmax = 52.20 MPa, acting on planes at 81.65°. Mohr’s circle: centre (45,0), radius 52.20, rotating CX by 50° gives matching coordinates.

## Answer

**Reading of the figure:** $\sigma_x = 60$ MPa (tensile) on the vertical faces, $\sigma_y = 30$ MPa (tensile) on the horizontal faces, $\tau_{xy} = 50$ MPa. Tensile stress is taken as positive and compressive as negative. The plane AB is at $25^\circ$ to the vertical, so its normal is at $\theta = 25^\circ$ from the $x$-axis, measured anticlockwise.

#### Stresses on the inclined plane ($\theta = 25^\circ$)

$$
\begin{aligned}
\sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\
&= 45.00 + (15.00)\cos 50^\circ + (50.00)\sin 50^\circ = 92.94\ \text{MPa} \\
\tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\
&= -(15.00)\sin 50^\circ + (50.00)\cos 50^\circ = 20.65\ \text{MPa}
\end{aligned}
$$

Resultant stress on the plane $= \sqrt{92.94^2+20.65^2} = 95.21\ \text{MPa}$, inclined at $12.53^\circ$ to the normal of the plane.

#### Principal stresses

$$
\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 45.00 \pm \sqrt{(15.00)^2+(50.00)^2} = 45.00 \pm 52.20
$$

$\sigma_1 = 97.20\ \text{MPa}$ and $\sigma_2 = -7.20\ \text{MPa}$.

#### Principal planes

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{30.00} \Rightarrow 2\theta_p = 73.30^\circ
$$

So $\theta_p = 36.65^\circ$ (normal of the plane carrying $\sigma_1$) and $126.65^\circ$ (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

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 52.20\ \text{MPa}
$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 81.65^\circ$ and $171.65^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 45.00\ \text{MPa}$.

#### Verification by Mohr's circle

- Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
- Plot $X(60, 50)$ and $Y(30, -50)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (45.00, 0)$.
- Draw the circle with centre C and radius $CX = 52.20$.
- It cuts the $\sigma$-axis at $\sigma_1 = 97.20$ and $\sigma_2 = -7.20$; $2\theta_p = \angle XCP_1 = 73.30^\circ$ (measured with a protractor).
- The highest and lowest points of the circle give $\tau_{max} = \pm52.20$ with $\sigma = 45.00$.
- Turn CX through $2\theta = 50^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 92.94$ and $\tau \approx 20.65$, which match the analytical values.

**Answer:** $\sigma_n = 92.94$ MPa, $\tau = 20.65$ MPa on AB; $\sigma_1 = 97.20$ MPa, $\sigma_2 = -7.20$ MPa; $\tau_{max} = 52.20$ MPa.

## Check yourself

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

1. Which stresses are used?
   A. sigma_x = 30, sigma_y = 60, tau_xy = 25
   B. sigma_x = 60, sigma_y = 30, tau_xy = 50 MPa, theta = 25 deg
   C. sigma_x = 60, sigma_y = 30, tau_xy = 25
   D. sigma_x = 50, sigma_y = 60, tau_xy = 30
   Answer: B. The figure gives 60 and 30 MPa normal and 50 MPa shear; the plane's normal is 25 deg from x. (Key point: Stress inputs)

2. What does the formula give for sigma_n?
   A. 45 MPa
   B. 20.65 MPa
   C. 97.20 MPa
   D. 92.94 MPa
   Answer: D. 45 + 15 cos50 + 50 sin50 = 92.94 MPa. (Key point: Normal stress on plane)

3. What are the principal stresses?
   A. 110 and 10 MPa
   B. 97.20 and -7.20 MPa
   C. 92.94 and 20.65 MPa
   D. 60 and 30 MPa
   Answer: B. 45 +/- sqrt(15^2 + 50^2) = 45 +/- 52.20. (Key point: Principal stresses)

4. What is tan 2 theta_p and the result?
   A. (sigma_x - sigma_y)/2tau, theta_p = 16.7 deg
   B. 2 tau/(sigma_x + sigma_y), theta_p = 29 deg
   C. tau/sigma_x, theta_p = 39.8 deg
   D. 2 tau/(sigma_x - sigma_y), 2 theta_p = 73.30 deg, theta_p = 36.65 deg
   Answer: D. tan2theta_p = 100/30. (Key point: Principal planes)

## Close variants set in other exams

- [Stresses on 30 degree plane; Mohr's circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30deg-50-30-mohr) — 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×

## Related in Strength of Materials

- [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×
- [Stresses on 40 degree plane; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-40deg-50-30) — Strength of Materials, chapter 4, asked 2×
- [Principal planes, stresses on plane AC](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-planes-max-shear-120-180) — Strength of Materials, chapter 4
- [Resultant stress on planes EF and GE, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aresultant-stress-plane-ef-ge) — Strength of Materials, chapter 4
- [Normal and resultant stress on 60 degree plane](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Anormal-resultant-stress-60deg-plane-100) — Strength of Materials, chapter 4

## Related in other subjects

- [Stresses on inclined plane and principal stresses](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ainclined-plane-stresses-principal-stresses) — Strength of Materials, chapter 6
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- [Mohr's circle for a plane stress element](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Amohr-circle-plane-stress-construction) — Strength of Materials, chapter 6
- [Principal stresses from a 3D stress tensor](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Athree-d-stress-invariants-numerical) — Mechanics of Solids, chapter 2