Stress transformation and Mohr's circle

By the end of this topic you will be able to

Calculate the stresses on an inclined plane from σx, σy and τxy.
Find the principal stresses, the principal directions and the maximum shear stress.
Construct and read Mohr's circle.

One point, many stresses

The stress at a point in a loaded body depends on the plane you look at. A bar in tension has pure normal stress on a cross-section, but on a plane tilted at 45° it also has shear. Failure may begin on whichever plane is most critical, so engineers need the stresses on every plane.

In plane stress we know the stresses on the x and y faces of a small element: $\sigma_x$ , $\sigma_y$ and $\tau_{xy}$ . The stresses on a plane whose normal is rotated by $\theta$ counter-clockwise from the x axis are

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

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

Principal stresses

The maximum and minimum normal stresses, the principal stresses, occur on planes with no shear stress:

\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}, \qquad \tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}

The largest in-plane shear stress is the radius of the circle below, and it acts on planes at $45^\circ$ to the principal planes.

Mohr's circle

The formulas above are the equation of a circle. Plot the points $X(\sigma_x, \tau_{xy})$ and $Y(\sigma_y, -\tau_{xy})$ and join them: the line is a diameter. Its centre is at $\sigma_{\text{avg}} = (\sigma_x + \sigma_y)/2$ , and the radius is $R = \sqrt{[(\sigma_x - \sigma_y)/2]^2 + \tau_{xy}^2}$ .

The circle's rightmost and leftmost points are $\sigma_1$ and $\sigma_2$ .
Its top and bottom points give $\pm\tau_{\text{max}} = \pm R$ .
Turning the plane by $\theta$ moves the point by $2\theta$ round the circle, in the same direction.

Interactive modelChange the stresses and the plane angle. The point on the circle shows the stress on that plane.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

At a point in a shaft, $\sigma_x = 80$ MPa, $\sigma_y = 20$ MPa and $\tau_{xy} = 30$ MPa. Find the principal stresses, their direction and $\tau_{\text{max}}$ .

Centre. $(80 + 20)/2 = 50$ MPa.

Radius. $R = \sqrt{30^2 + 30^2} = 42.4$ MPa.

Principal stresses. $\sigma_1 = 50 + 42.4 = 92.4$ MPa and $\sigma_2 = 50 - 42.4 = 7.6$ MPa.

Direction. $\tan2\theta_p = 2\times30/60 = 1$ , so $2\theta_p = 45^\circ$ and $\theta_p = 22.5^\circ$ (from the x axis to the $\sigma_1$ plane).

Maximum shear. $\tau_{\text{max}} = R = 42.4$ MPa, on planes at $22.5^\circ - 45^\circ = -22.5^\circ$ (and $67.5^\circ$ ).

Check with the model, setting the plane angle to 22.5° to see $\tau_\theta = 0$ and $\sigma_\theta = \sigma_1$ .

Practice problems

At a point $\sigma_x = 100$ MPa, $\sigma_y = 0$ , $\tau_{xy} = 0$ (a bar in tension). What is $\tau_{\text{max}}$ and on what planes?

Show answer

$R = 50$ MPa, so $\tau_{\max} = 50$ MPa on planes at 45° to the axis. (This is why ductile bars slip along 45° lines.)
For $\sigma_x = 60$ MPa, $\sigma_y = -20$ MPa, $\tau_{xy} = 30$ MPa, find $\sigma_1$ , $\sigma_2$ .

Show answer

Centre $= 20$ MPa; $R = \sqrt{40^2 + 30^2} = 50$ MPa. $\sigma_1 = 70$ MPa, $\sigma_2 = -30$ MPa.
Find the normal and shear stress on a plane at $\theta = 30^\circ$ for the worked-example stresses.

Show answer

$\sigma_\theta = 50 + 30\cos60^\circ + 30\sin60^\circ = 91.0$ MPa; $\tau_\theta = -30\sin60^\circ + 30\cos60^\circ = -11.0$ MPa.
What is the sum $\sigma_x + \sigma_y$ equal to?

Show answer

It equals $\sigma_1 + \sigma_2$ : the sum of normal stresses on perpendicular planes is invariant.

Key formulae

Stress on a plane at angle θ: $\sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta$
Shear stress on that plane: $\tau_\theta = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta$
Principal stresses: $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$
Principal direction: $\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}$
Maximum shear stress: $\tau_{\text{max}} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (50 min). 10 min: tilt a chalk stick or draw the bar-in-tension planes; 15 min: transformation equations briefly (give them, derive on request); 15 min: build the circle by hand with a compass for the worked example; 10 min: model.

Activity. Students draw the circle on graph paper, then use the model slider to check any plane angle chosen by a neighbour.

Discussion. Which plane matters for brittle materials (principal tension) and for ductile materials (maximum shear)?

Assessment idea. Ask for the stresses on an inclined weld seam.