---
title: "Derivation of plane stress transformation equations"
subject: "Strength of Materials"
chapter: 6
key: "semester_4/strength_of_materials_mechanical:plane-stress-transformation-principal-derivation"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/plane-stress-transformation-principal-derivation
site: BCT Engineering
---

# Derivation of plane stress transformation equations

Strength of Materials · Chapter 6: Transformation of stress and strain

- Asked in: year not given, 8 marks
- Video: [DERIVATION OF THE TRANSFORMATION SHEAR AND NORMAL STRESS EQUATIONS](https://www.youtube.com/watch?v=qXjooqfjP0k) — Davis Hunter
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/plane-stress-transformation-principal-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-6

## Question

For a body in a state of plane stress with stresses sigma_x, sigma_y and tau_xy, derive expressions for the normal and shear stresses on an inclined plane at angle theta. Hence obtain the principal stresses, the position of the principal planes and the maximum shear stress.

## Short answer

Wedge equilibrium yields σθ=(σx+σy)/2+((σx-σy)/2)cos2θ+τxy sin2θ and τθ=-((σx-σy)/2)sin2θ+τxy cos2θ. Principal stresses σ1,2=(σx+σy)/2±√[((σx-σy)/2)²+τxy²] occur where τθ=0; tan2θp=2τxy/(σx-σy). Maximum shear stress equals the radius of Mohr’s circle, on planes at 45° to principal planes.

## Answer

#### Sign convention
Tensile normal stress positive. Shear stress $\tau_{xy}$ is positive when it acts in the $+y$ direction on the face whose outward normal is $+x$. Angle $\theta$ is measured anticlockwise from the $x$ axis to the normal of the inclined plane.

```text
        sy (top face)
      ^   ^
   +-----------+
   |           |  txy on the +x face points up (+y)
 <-|           |->  sx
   |           |
   +-----------+
      sy (bottom face)
```

#### Equilibrium of a wedge
Take a triangular wedge with the inclined face of area $dA$, normal at $\theta$. The areas of the vertical face and horizontal face are $dA\cos\theta$ and $dA\sin\theta$.

Forces along the normal $n$ ($\sum F_n = 0$):

$$
\sigma_\theta\,dA = \sigma_x\cos^2\theta\,dA + \sigma_y\sin^2\theta\,dA + 2\tau_{xy}\sin\theta\cos\theta\,dA
$$

Forces along the plane ($\sum F_t = 0$):

$$
\tau_\theta\,dA = -(\sigma_x-\sigma_y)\sin\theta\cos\theta\,dA + \tau_{xy}(\cos^2\theta-\sin^2\theta)\,dA
$$

Using double-angle identities:

$$
\boxed{\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta}
$$

$$
\boxed{\tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta}
$$

#### Principal stresses
Principal planes have $\tau_\theta = 0$:

$$
\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y}
$$

This gives two values of $2\theta_p$ that differ by $180^\circ$, so the two principal planes are $90^\circ$ apart. Substituting back:

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

(Equivalently $d\sigma_\theta/d\theta = 0$.)

#### Maximum shear stress
Setting $d\tau_\theta/d\theta = 0$:

$$
\tan2\theta_s = -\frac{\sigma_x-\sigma_y}{2\tau_{xy}}
$$

so $\theta_s = \theta_p \pm 45^\circ$ (planes of maximum shear are at $45^\circ$ to the principal planes).

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

On these planes the normal stress is $\sigma_{avg} = (\sigma_x+\sigma_y)/2$.

## Check yourself

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

1. What condition defines the principal planes, and what does it give for their orientation?
   A. Shear stress on the plane is zero, giving tan2θp = 2τxy/(σx - σy)
   B. Normal stress on the plane is zero, giving tan2θp = (σx - σy)/2τxy
   C. Shear stress is maximum, giving tan2θp = -(σx - σy)/2τxy
   D. Normal stress equals the average stress, giving tan2θp = 2τxy/(σx + σy)
   Answer: A. Principal planes carry no shear, so setting τθ = 0 gives tan2θp = 2τxy/(σx - σy). (Key point: Principal planes condition)

2. Why are the two principal planes 90° apart?
   A. Because σx and σy are always perpendicular
   B. tan2θp gives two values of 2θp differing by 180°
   C. Because the shear stress is zero on both planes only when τxy = 0
   D. Because tan2θp has a period of 90° in 2θp
   Answer: B. The two roots of 2θp differ by 180°, so θp values differ by 90°. (Key point: Two planes 90° apart)

3. What is the maximum shear stress in terms of the principal stresses?
   A. (σ1 + σ2)/2
   B. σ1 - σ2
   C. (σ1 - σ2)/2, equal to sqrt(((σx - σy)/2)^2 + τxy^2)
   D. sqrt((σx - σy)^2 + τxy^2)
   Answer: C. τmax is the radius of the stress circle, which equals half the difference of the principal stresses. (Key point: Maximum shear stress)

4. Where do the planes of maximum shear lie and what normal stress acts on them?
   A. On the principal planes, with normal stress σ1
   B. At 90° to the principal planes, with zero normal stress
   C. At 45° to the principal planes, with normal stress (σ1 - σ2)/2
   D. At 45° to the principal planes, with normal stress (σx + σy)/2
   Answer: D. Setting dτθ/dθ = 0 gives θs = θp ± 45°, and the normal stress there is the average (σx + σy)/2. (Key point: Planes of maximum shear)

## Related in Strength of Materials

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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
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