---
title: "Stress transformation equations and Mohr's circle"
subject: "Mechanics of Solids"
chapter: 2
key: "semester_5/mechanics_of_solids:derive-stress-transformation-mohr"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/derive-stress-transformation-mohr
site: BCT Engineering
---

# Stress transformation equations and Mohr's circle

Mechanics of Solids · Chapter 2: Stress Tensor

- Asked in: year not given, 8 marks
- Video: [Understanding Stress Transformation and Mohr's Circle](https://www.youtube.com/watch?v=_DH3546mSCM) — The Efficient Engineer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/derive-stress-transformation-mohr
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-2

## Question

Derive the equations for the normal and shear stress on an inclined plane in plane stress. Hence explain the construction of Mohr's circle and show how the principal stresses and maximum shear stress are read from it.

## Short answer

From wedge equilibrium, σ_n = (σ_x+σ_y)/2 + (σ_x-σ_y)/2 cos2θ + τ_xy sin2θ, τ_nt = -(σ_x-σ_y)/2 sin2θ + τ_xy cos2θ. These describe Mohr's circle with center (σ_avg,0) and radius R. Principal stresses are the σ-axis intercepts; maximum shear stress τ_max=R at normal stress σ_avg.

## Answer

#### Derivation

Consider a wedge cut from a plane-stress element. Let the inclined plane have its normal $n$ at angle $\theta$ (anticlockwise) from the x-axis and area $dA$. The faces on the x and y sides then have areas $dA\cos\theta$ and $dA\sin\theta$.

```text
        y
        |   /  n (normal at angle theta)
   sigma_y  /
  +-----+/
  |     /|
  |    / |-- tau_xy
  +---+--+---- x
```

Resolve forces along the normal $n$:

$$
\sigma_n\, dA = \sigma_x (dA\cos\theta)\cos\theta + \sigma_y (dA\sin\theta)\sin\theta + \tau_{xy}(dA\cos\theta)\sin\theta + \tau_{yx}(dA\sin\theta)\cos\theta
$$

Using $\tau_{yx}=\tau_{xy}$ and double-angle identities:

$$
\sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta
$$

Resolving along the plane gives

$$
\tau_{nt} = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta
$$

**Principal planes**: $\tau_{nt}=0$ gives $\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$, and
$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$.

#### Mohr's circle

Squaring and adding the two equations eliminates $\theta$:

$$
\left(\sigma_n - \sigma_{avg}\right)^2 + \tau_{nt}^2 = R^2
$$

This is a circle with centre $(\sigma_{avg}, 0)$ and radius $R$ in the $\sigma$-$\tau$ plane, where $\sigma_{avg} = \dfrac{\sigma_x+\sigma_y}{2}$ and $R = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$.

**Construction**

1. Draw $\sigma$ (tension positive) to the right and $\tau$ downward, so that the angle $2\theta$ is measured anticlockwise on the circle. Plot the point for the x-face as $X(\sigma_x, \tau_{xy})$. (In the other common convention, $\tau$ is drawn upward and clockwise shear is positive. State which one you use.)
2. Plot the point for the y-face as $Y(\sigma_y, -\tau_{xy})$.
3. Join $XY$; it cuts the $\sigma$-axis at the centre $C$.
4. Draw the circle with diameter $XY$.

```text
   tau
    |        . S (tau max)
    |     .     .
    |   .         .
 ---+--+----C------+----- sigma
    | sigma2       sigma1
    |   .         .
    |     .     .
    |        . S'
```

**Reading results**

- The points where the circle cuts the $\sigma$-axis give $\sigma_1 = OC + R$ and $\sigma_2 = OC - R$.
- The highest and lowest points give $\tau_{max} = \pm R$, with normal stress $\sigma_{avg}$.
- The angle between $CX$ and the $\sigma$-axis (towards $\sigma_1$) is $2\theta_p$. Angles on the circle are **twice** the physical angles, and they turn in the same sense as the plane turns.
- The point on the circle at angle $2\theta$ from $X$ gives $(\sigma_n, \tau_{nt})$ for the plane at $\theta$.

## Check yourself

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

1. How is sigma_n on the inclined plane derived?
   A. By resolving forces on a wedge element along the plane's normal using areas dA cos(theta) and dA sin(theta)
   B. By integrating the strain energy
   C. By using Hooke's law only
   D. By taking moments about the plane
   Answer: A. A wedge is cut, face areas are dA cos(theta) and dA sin(theta), and forces are resolved along the normal, then double-angle identities are used. (Key point: Derivation basis)

2. What eliminates theta to give the Mohr's circle?
   A. Dividing the two equations
   B. Differentiating sigma_n with respect to theta
   C. Squaring and adding the sigma_n and tau_nt equations
   D. Setting tau_nt = 0
   Answer: C. Squaring and adding gives (sigma_n - sigma_avg)^2 + tau_nt^2 = R^2, a circle with centre (sigma_avg, 0). (Key point: Mohr's circle equation)

3. How do angles on Mohr's circle compare with physical angles?
   A. They are equal
   B. They are half the physical angles
   C. They are twice the physical angles
   D. They are 90 degrees apart always
   Answer: C. A plane rotated by theta corresponds to a point rotated by 2theta on the circle. (Key point: Angles on Mohr's circle)

4. Where on the circle are sigma_1, sigma_2 and tau_max read?
   A. sigma_1, sigma_2 at highest and lowest points; tau_max on the sigma-axis
   B. All three at the centre
   C. sigma_1, sigma_2 where the circle cuts the sigma-axis; tau_max at the highest and lowest points
   D. tau_max equals the diameter
   Answer: C. Intercepts on the sigma-axis are OC +/- R (principal stresses); the top and bottom points give tau_max = R with normal stress sigma_avg. (Key point: Reading principal and max shear)

## Related in Mechanics of Solids

- [Plane stress: principal stresses and oblique plane](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aplane-stress-principal-numerical) — Mechanics of Solids, chapter 2
- [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
- [Strain rosette: principal strains and stresses](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Astrain-rosette-principal-numerical) — Mechanics of Solids, chapter 3

## Related in other subjects

- [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×
- [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
- [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×
- [Principal stresses and Mohr's circle (30, 18, 75)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-stresses-mohr-30-18-75) — Strength of Materials, chapter 4