What a Mohr circle actually tells you

The Mohr circle is a drawing that replaces a calculation. Everything it shows can be obtained algebraically, so it earns its place only if it makes relationships visible that the algebra hides — which it does, provided you remember that every angle on it is doubled.

Every number in this article is recomputed from its inputs by a script that runs on each build of this site, and the diagram’s coordinates are checked against the same arithmetic.

The circlePermalink to “The circle”#

Take a two-dimensional stress state: normal stresses $\sigma_x$ and $\sigma_y$ on two perpendicular planes, with shear stress $\tau_{xy}$ on both. Rotating the reference planes by an angle $\theta$ traces out a circle in the $(\sigma, \tau)$ plane with

$$\text{centre} = \left(\frac{\sigma_x + \sigma_y}{2},\ 0\right), \qquad R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^{2} + \tau_{xy}^{2}}$$

The principal stresses are where the circle crosses the $\sigma$ axis, because that is where the shear stress vanishes:

$$\sigma_1 = \text{centre} + R, \qquad \sigma_3 = \text{centre} - R$$

and the radius is therefore the maximum shear stress in the plane, $R = (\sigma_1 - \sigma_3)/2$. Take $\sigma_x = 260$ kPa, $\sigma_y = 140$ kPa, $\tau_{xy} = 60$ kPa:

$$\text{centre} = 200\ \text{kPa}, \qquad R = \sqrt{60^{2} + 60^{2}} = 60\sqrt{2} = 84.85\ \text{kPa}$$

giving $\sigma_1 = 284.85$ kPa and $\sigma_3 = 115.15$ kPa.

The doubled anglePermalink to “The doubled angle”#

This is the fact that causes the errors. The angle subtended at the centre of the circle is twice the angle through which the physical plane has rotated. The principal planes are found from

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

For the numbers above, $2\theta_p = \arctan(120/120) = 45^\circ$, so the major principal plane is at $\theta_p = 22.5^\circ$ — not 45°.

The reason is structural, not a convention that could have been chosen differently. The stress transformation equations contain $\cos 2\theta$ and $\sin 2\theta$, so a rotation of the plane by $\theta$ moves the stress point by $2\theta$ around the circle. Two planes 90° apart in the ground sit on opposite sides of the circle, 180° apart — which is exactly why they have the same shear stress magnitude and why the circle closes after a physical rotation of only 180°.

If an angle taken off a Mohr circle looks twice what it should be, it is.

Adding a failure envelopePermalink to “Adding a failure envelope”#

Now give the soil a strength: a straight envelope

$$\tau = c + \sigma \tan\varphi$$

A stress state is safe while its circle stays clear of the line, and at failure the circle touches it. Tangency is a distance condition, and that is all the Mohr-Coulomb criterion is.

Write the envelope as $\sigma\tan\varphi - \tau + c = 0$. The perpendicular distance from the circle centre $(s, 0)$, where $s = (\sigma_1 + \sigma_3)/2$, to that line is

$$d = \frac{\left|s\tan\varphi + c\right|}{\sqrt{\tan^{2}\varphi + 1}} = \left(s\tan\varphi + c\right)\cos\varphi = s\sin\varphi + c\cos\varphi$$

Setting $d = R = (\sigma_1 - \sigma_3)/2$ gives

$$\frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_1 + \sigma_3}{2}\sin\varphi + c\cos\varphi$$

Collecting terms in $\sigma_1$,

$$\sigma_1\left(1 - \sin\varphi\right) = \sigma_3\left(1 + \sin\varphi\right) + 2c\cos\varphi$$

and using $\cos\varphi = \sqrt{(1-\sin\varphi)(1+\sin\varphi)}$, so that $\cos\varphi/(1-\sin\varphi) = \sqrt{K_p}$, this becomes the form everyone quotes:

$$\sigma_1 = K_p\,\sigma_3 + 2c\sqrt{K_p}, \qquad K_p = \frac{1 + \sin\varphi}{1 - \sin\varphi}$$

The triaxial formula and the tangency condition are the same statement. Nothing was assumed about how the soil fails beyond “the circle touches the line”.

A worked failure statePermalink to “A worked failure state”#

Take $c = 20$ kPa, $\varphi = 30^\circ$, $\sigma_3 = 100$ kPa. Then $K_p = 3$ exactly, and

$$\sigma_1 = 3 \times 100 + 2 \times 20 \times \sqrt{3} = 369.28\ \text{kPa}$$ $$s = \frac{369.28 + 100}{2} = 234.64\ \text{kPa}, \qquad R = \frac{369.28 - 100}{2} = 134.64\ \text{kPa}$$

Check the tangency independently, from the distance formula:

$$d = s\sin\varphi + c\cos\varphi = 234.64 \times 0.5 + 20 \times 0.8660 = 134.64\ \text{kPa}$$

which equals $R$. The circle touches the envelope, as it must.

Where the soil failsPermalink to “Where the soil fails”#

The radius to the tangent point is perpendicular to the envelope. The envelope makes an angle $\varphi$ with the $\sigma$ axis, so that radius makes an angle $90^\circ + \varphi$ with it. That angle is measured on the circle, so it is twice the physical one:

$$2\theta = 90^\circ + \varphi \qquad\Longrightarrow\qquad \theta = 45^\circ + \frac{\varphi}{2}$$

The failure plane sits at $45^\circ + \varphi/2$ from the major principal plane. For $\varphi = 30^\circ$ that is 60°, and the tangent point is at

$$\sigma = s + R\cos\left(90^\circ + \varphi\right) = 167.32\ \text{kPa}, \qquad \tau = R\sin\left(90^\circ + \varphi\right) = 116.60\ \text{kPa}$$

Substituting back into the envelope: $20 + 167.32 \times \tan 30^\circ = 116.60$ kPa. The point lies on the line, which it must, and computing it two ways is the cheapest check available.

Note what this angle is not. It is not 45°, which is the answer only for $\varphi = 0$ — the undrained case. And it is not measured from the horizontal unless the major principal plane happens to be horizontal.

Total and effective stressPermalink to “Total and effective stress”#

Pore pressure $u$ acts equally in all directions, so it subtracts equally from $\sigma_1$ and $\sigma_3$:

$$\sigma'_1 = \sigma_1 - u, \qquad \sigma'_3 = \sigma_3 - u$$

The difference $\sigma_1 - \sigma_3$ is unchanged. The effective stress circle is the same size as the total stress circle, shifted left by $u$. Nothing about its diameter changes; only where it sits.

Two consequences follow immediately from that picture. Strength is governed by the effective stress circle, so the total stress circle can sit anywhere and tell you nothing on its own. And an undrained test on a saturated soil, where the effective stresses are fixed by the initial state, produces total stress circles of constant diameter — which is drawn as a horizontal envelope, $\tau = s_u$, with $\varphi_u = 0$. That apparent zero friction angle is a consequence of the drainage condition, not a property of the soil.

What the circle does not tell youPermalink to “What the circle does not tell you”#

It is a two-dimensional construction, and $\sigma_2$ does not appear anywhere in it. The Mohr-Coulomb criterion inherits that blindness: strength is made to depend only on $\sigma_1$ and $\sigma_3$.

That is a real modelling decision with a visible consequence. Because the criterion ignores the intermediate principal stress, its surface in principal stress space is an irregular hexagonal pyramid rather than a cone — six flat faces, one for each ordering of the three principal stresses, meeting at edges where the gradient is undefined. Implementing the criterion means dealing with those edges, and it is why the strength measured in a triaxial compression test does not transfer unchanged to a plane strain problem. Whether that matters depends on the problem; whether it is true does not.

ReferencesPermalink to “References”#

• Mohr, O. (1900). Welche Umstände bedingen die Elastizitätsgrenze und den Bruch eines Materials? Zeitschrift des Vereines Deutscher Ingenieure, 44, 1524–1530 and 1572–1577. The origin of the graphical construction and of the failure hypothesis that bears his name.
• Coulomb, C.A. (1776). Essai sur une application des règles de maximis et minimis à quelques problèmes de statique relatifs à l’architecture. Mémoires de Mathématique et de Physique, Académie Royale des Sciences, 7, 343–382. The origin of the linear strength relation $\tau = c + \sigma\tan\varphi$.
• Terzaghi, K. (1943). Theoretical Soil Mechanics. John Wiley & Sons, New York. The effective stress principle applied throughout soil mechanics.

The derivations above are carried out in full rather than cited, so that they can be checked rather than trusted. The circle geometry, the tangency condition and the failure plane angle follow from the two definitions at the top of the article and nothing else.

Classification decides which strength parameters are even applicable, and the arithmetic there has its own trap — see soil classification from a grading curve. The implementation of the criterion derived above, edges and all, is what KATAI 2D has to get right.