Principal stresses and Mohr’s circle

Cut a stressed body along any plane through a point and you find a normal stress and a shear stress acting on that cut — and those two numbers change as you rotate the cutting plane. Somewhere in that rotation there is a special orientation where the shear vanishes entirely and the normal stress hits its extreme values. Those extreme normal stresses are the principal stresses, and the plane they act on is the principal plane. Finding that orientation by algebra is one route; Mohr's circle is the other — a geometric device that turns the same transformation equations into a picture you can read stresses off of directly, for any angle, at a glance. Both routes give identical numbers; the circle is simply faster once you trust it, and it makes the trickiest part of the topic — how a physical rotation maps onto the mathematics — visually obvious. This node builds directly on plane-stress transformation and feeds straight into combined-loading problems (axial plus torsion plus bending, superposed at a point) and thin-walled pressure vessels. It stays inside plane (2-D) stress; 3-D principal stresses and yield criteria such as Tresca or von Mises belong to a later Solid Mechanics course.

The core insight

The principal stresses are the maximum and minimum normal stresses at a point, found either from $\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$ or by reading the $\sigma$ -axis crossings of Mohr's circle, whose centre is the average stress and whose radius is the maximum in-plane shear stress.

The core idea

Start from the plane-stress transformation equations for a stress state $(\sigma_x,\sigma_y,\tau_{xy})$ rotated by an angle $\theta$ . Setting the shear term to zero and solving for the stationary normal stress gives the 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}

where: $\sigma_1$ is the larger (major) principal stress, $\sigma_2$ the smaller (minor) principal stress, and $\sigma_x,\sigma_y,\tau_{xy}$ are the stress components on the original element. The two principal planes are $90°$ apart, and their orientation follows from

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

where: $\theta_p$ is the angle, measured on the physical element, from the $x$ -axis to a principal plane. This equation always returns two candidate angles $90°$ apart; you must substitute back (or read the circle) to know which one carries $\sigma_1$ and which carries $\sigma_2$ . Halfway between the two principal planes, at $45°$ to them, lies the plane of maximum in-plane shear stress, with magnitude equal to the square-root term above:

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

On that plane the normal stress is not zero — it equals the average stress $(\sigma_x+\sigma_y)/2$ . This is one of the most useful facts in the whole topic: the average stress is an invariant, unchanged by rotation, and it reappears in three places at once — as the centre of Mohr's circle, as the normal stress at maximum shear, and as half of $\sigma_1+\sigma_2$ .

Mohr's circle packages all of this graphically. Set up $\sigma$ (horizontal) and $\tau$ (vertical) axes, plot the two points representing the two faces of the element — $(\sigma_x,-\tau_{xy})$ and $(\sigma_y,\tau_{xy})$ under the usual "clockwise shear plots downward" convention — and join them with a straight line. That line is a diameter of the circle. Its midpoint, at $\left(\frac{\sigma_x+\sigma_y}{2},0\right)$ , is the circle's centre, and half the diameter's length equals $\tau_{max}$ , exactly matching the radical above. Every point on the circumference represents the stress state on some plane through the point; the principal stresses are where the circle crosses the $\sigma$ -axis (zero shear, by definition of principal planes), and the top and bottom of the circle give $\tau_{max}$ together with the accompanying normal stress. The single geometric fact that makes the circle powerful — and the single fact that trips students up — is the angle doubling described below.

Figure

Mohr circle for the stress state eighty twenty thirty showing the centre at fifty the radius forty two point four and the two principal stress crossings on the sigma axis

The angle-doubling relationship between element and circle

A physical rotation of the cutting plane by an angle $\theta$ corresponds to a rotation of $2\theta$ around Mohr's circle, in the same rotational sense. This is why the two principal planes, $90°$ apart physically, appear $180°$ apart on the circle — diametrically opposite, exactly as the two $\sigma$ -axis crossings must be. It is also why the maximum-shear plane, at $45°$ to the principal planes physically, sits at $90°$ from them on the circle — at the top or bottom, where $\tau$ is extreme. Internalising this doubling is the single most valuable habit in the whole topic: every angle you read off a Mohr's-circle sketch must be halved before it means anything on the actual stressed element.

Derivation ADeriving the principal-stress formula from the transformation equations

The goal is to show, starting only from the plane-stress transformation equation for normal stress on a rotated plane, why setting its slope to zero and solving picks out exactly the two extreme normal stresses given above — and why the maximum shear stress that falls out along the way is numerically the same square-root quantity, confirming the two are not independent facts but two faces of one calculation.

Step 1. Write the transformation equation for normal stress on a plane rotated by $\theta$ from the $x$ -axis: $\sigma_{x'} = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta$ . This is a sinusoidal function of $2\theta$ sitting on top of a constant offset, the average stress.

Step 2. Find the stationary points by differentiating with respect to $\theta$ and setting the result to zero: $\dfrac{d\sigma_{x'}}{d\theta} = -(\sigma_x-\sigma_y)\sin2\theta + 2\tau_{xy}\cos2\theta = 0$ . Rearranging gives $\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$ , the principal-plane angle equation quoted above. Notice this is exactly the condition for the shear transformation equation, $\tau_{x'y'} = -\dfrac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta$ , to vanish — confirming that the stationary-normal-stress planes are indeed the zero-shear planes.

Step 3. Build a right triangle from $\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$ with legs $2\tau_{xy}$ and $\sigma_x-\sigma_y$ , hypotenuse $R=\sqrt{(\sigma_x-\sigma_y)^2+(2\tau_{xy})^2}$ . Then $\cos2\theta_p = \dfrac{\sigma_x-\sigma_y}{R}$ and $\sin2\theta_p=\dfrac{2\tau_{xy}}{R}$ .

Step 4. Substitute these back into the Step 1 expression for $\sigma_{x'}$ at $\theta=\theta_p$ : $\sigma_{x'} = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{\sigma_x-\sigma_y}{2}\cdot\dfrac{\sigma_x-\sigma_y}{R} + \tau_{xy}\cdot\dfrac{2\tau_{xy}}{R} = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{(\sigma_x-\sigma_y)^2/2+2\tau_{xy}^2}{R}$ . Writing $R = 2\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$ and simplifying the fraction reduces this to exactly $\dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$ , the two signs corresponding to the two roots $\theta_p$ and $\theta_p+90°$ . This recovers the principal-stress formula, and the same square-root term is, by inspection of the shear equation's own amplitude, identically $\tau_{max}$ — one calculation yields both results.

Worked example

Using $\sigma_x = 80\text{ MPa}$ , $\sigma_y = 20\text{ MPa}$ , $\tau_{xy} = 30\text{ MPa}$ , find the principal stresses, the principal-plane angle, $\tau_{max}$ , and confirm the result with Mohr's circle.

Step 1 — average and radius. $\text{avg} = \dfrac{80+20}{2} = 50\text{ MPa}$ , and $R = \sqrt{\left(\dfrac{80-20}{2}\right)^2+30^2} = \sqrt{30^2+30^2} = \sqrt{1800} = 42.43\text{ MPa}$ . This average is the Mohr's-circle centre; the radius $R$ is $\tau_{max}$ .

Step 2 — principal stresses. $\sigma_1 = 50+42.43 = 92.4\text{ MPa}$ , $\sigma_2 = 50-42.43 = 7.6\text{ MPa}$ . On the circle these are the two points where the circumference crosses the $\sigma$ -axis.

Step 3 — principal-plane angle. $\tan2\theta_p = \dfrac{2(30)}{80-20} = \dfrac{60}{60} = 1 \implies 2\theta_p = 45° \implies \theta_p = 22.5°$ . This is the physical rotation of the element; on the circle the corresponding rotation from the plotted point $(80,-30)$ to the $\sigma_1$ crossing is $45°$ , exactly double.

Step 4 — maximum shear stress. $\tau_{max} = R = 42.4\text{ MPa}$ , occurring at $\theta_p+45°=67.5°$ physically, where the normal stress equals the average, $50\text{ MPa}$ — not zero.

Step 5 — invariant check. $\sigma_1+\sigma_2 = 92.4+7.6 = 100 = \sigma_x+\sigma_y$ . This equality must always hold and is the fastest arithmetic check available. Answer: $\sigma_1 = 92.4\text{ MPa}$ , $\sigma_2 = 7.6\text{ MPa}$ , $\theta_p = 22.5°$ , $\tau_{max} = 42.4\text{ MPa}$ .

Competitive stretch

For $\sigma_x=-40\text{ MPa}$ , $\sigma_y=20\text{ MPa}$ , $\tau_{xy}=-25\text{ MPa}$ : centre $C=\dfrac{-40+20}{2}=-10\text{ MPa}$ , radius $R=\sqrt{30^2+25^2}=\sqrt{1525}=39.05\text{ MPa}$ , giving $\sigma_1=-10+39.05=29.1\text{ MPa}$ and $\sigma_2=-10-39.05=-49.1\text{ MPa}$ . Here $\tan2\theta_p=\dfrac{2(-25)}{-40-20}=\dfrac{-50}{-60}=0.833\implies2\theta_p=39.8°\implies\theta_p\approx19.9°$ , showing the method carries unchanged through negative stresses — only the sign bookkeeping needs care.

Common traps

Confusing the angle read on the circle with the angle on the physical element — a rotation of $\theta$ on the element is $2\theta$ on the circle, and forgetting this factor of two is the single most common Mohr's-circle mistake.
Mixing up the shear-plotting sign convention — different textbooks plot clockwise shear as positive or negative $\tau$ ; whichever you adopt, apply it consistently to both plotted points and to every angle you read afterward.
Assuming the normal stress is zero on the maximum-shear plane — it equals the average stress $(\sigma_x+\sigma_y)/2$ , not zero, except in the special case where that average happens to be zero.
Reporting the magnitude of $\theta_p$ without checking which principal stress it belongs to — the arctangent gives two candidates $90°$ apart, so substitute back into the transformation equation, or simply read the circle, to confirm which physical plane carries $\sigma_1$ versus $\sigma_2$ .

Ask the tutor: if you want to practice constructing Mohr's circle step by step for a fresh stress state, or need help reconciling different sign-convention choices between textbooks.

Why does Mohr's circle double every angle instead of just copying it directly

Because the transformation equations for $\sigma_{x'}$ and $\tau_{x'y'}$ are sinusoidal functions of $2\theta$ , not $\theta$ — a consequence of how stress components transform under rotation of the cutting plane. Plotting the circle in terms of $2\theta$ is what makes a full rotation of the physical element, $180°$ of physically distinct planes, map onto one complete $360°$ trip around the circle. It is a bookkeeping convenience baked into the mathematics, not an arbitrary rule, and it is exactly why the two principal planes, $90°$ apart physically, sit $180°$ apart — diametrically across the circle.

Do I need to memorise the sign convention for plotting tau on Mohr's circle

You need to fix one and use it consistently, but the specific convention matters less than most students fear. The two common choices — clockwise shear on the $x$ -face plotted as positive $\tau$ , or the reverse — both give the same principal stresses, the same $\tau_{max}$ , and the same circle shape; only the labelling of which plotted point is which flips. Pick your course's or textbook's convention, state it, and apply it identically to both the plotting step and to any angle you subsequently read off. Switching mid-problem is the real hazard, not the choice itself.

Why bother with the algebraic formula if Mohr's circle gives the same answer graphically

The formula and the circle are two representations of the same equations, and each has a use. The formula is faster when you only need the numerical principal stresses and don't need a picture, and it is what you differentiate or manipulate symbolically in later, more involved problems. The circle is faster and far less error-prone when you need the stress on an arbitrary plane, want to see several quantities — $\sigma_1$ , $\sigma_2$ , $\tau_{max}$ , and the stress at any angle — at once, or want a visual check against silly sign mistakes. Competent practice uses the circle to sketch and sanity-check, then the formula to nail down exact numbers.