Mohr's circle, explained: principal stresses and maximum shear
Why the stress at a point changes with the direction you look, how Otto Mohr turned the transformation equations into one circle, and how to read principal stresses, maximum shear and the angles they act at — including the out-of-plane shear that catches people out.
Stress depends on which way you look
Picture a tiny square cut from a loaded part. Its faces carry a normal stress pushing or pulling straight on them, and a shear stress sliding along them. Rotate the square and those numbers change, even though nothing about the load has changed. The stress state is one physical thing; the numbers depend on the axes you choose.
A bar in simple tension shows it. On a cut straight across the bar there is only the tensile stress σ. On a cut at 45° there is a shear stress of σ/2 as well — which is why a ductile metal bar pulled to yield shows slip lines at about 45°, and why a stick of chalk twisted until it snaps breaks along a 45° spiral.
The transformation equations
For plane stress — stresses σx, σy and τxy in one plane, nothing through the thickness — the stresses on a square rotated θ counter-clockwise are: σx' = (σx + σy)/2 + (σx − σy)/2 · cos 2θ + τxy · sin 2θ, and τx'y' = −(σx − σy)/2 · sin 2θ + τxy · cos 2θ. The stress on the other face, σy', follows from the fact that σx' + σy' always equals σx + σy.
The angles appear doubled, as 2θ. That is not an accident of algebra: two faces 90° apart are opposite faces of the same kind, so every behaviour repeats twice as the square turns through a full circle.
Mohr's drawing
In 1882 the German engineer Otto Mohr noticed that those two equations are the equation of a circle. Plot normal stress along one axis and shear stress along the other, and every orientation of the square lands on the same circle. Its centre sits at the average normal stress (σx + σy)/2, and its radius is R = √(((σx − σy)/2)² + τxy²).
Each face of the square is one point. The x face sits at (σx, τxy) and the y face at (σy, −τxy), at opposite ends of a diameter. Rotate the square by θ and the diameter turns by 2θ around the circle. That is the whole method: draw the circle once, and you can read the stresses for any orientation straight off it.
Reading the circle
Where the circle crosses the horizontal axis the shear is zero. Those two points are the principal stresses, σ1 = centre + R and σ2 = centre − R: the largest and smallest normal stresses the point sees in any direction. Half the angle round the circle from the x point to σ1 is θp, the angle you must turn the square to reach it, given by tan 2θp = 2τxy/(σx − σy).
The top and bottom of the circle are the planes of maximum in-plane shear, equal to the radius R and always 45° from the principal planes. The normal stress on those planes is not zero but the centre value, (σx + σy)/2 — a detail that matters for anything whose strength depends on whether a crack is being pulled open.
Sign conventions trip everyone up
Textbooks disagree on which way positive shear points and which way to plot it, and the circle's rotation sense flips with the choice. A common convention, used here, takes tension as positive, τxy as positive when it pushes in +y on the +x face, and plots shear positive downward. Then the point for a face turns the same way as the square itself: rotate the square counter-clockwise and its point moves counter-clockwise around the circle.
Whatever convention you use, write it down before you start, and check your answer against one easy case — pure tension has σ1 = σ and no shear on the axes — before trusting the rest.
The third stress hiding in plane stress
Plane stress still has a third principal stress: the zero stress through the thickness. When σ1 and σ2 have opposite signs, the in-plane circle is the biggest one and R really is the largest shear. When they have the same sign, a circle drawn between σ1 (or σ2) and zero is bigger, and the true maximum shear is half the larger of |σ1| and |σ2|.
A thin cylinder under pressure is the classic case. With a hoop stress of 100 MPa and an axial stress of 50 MPa, the in-plane shear peaks at 25 MPa, but the absolute maximum is 50 MPa, acting on planes tilted through the wall. The Tresca yield criterion uses that absolute value; the von Mises criterion combines the principals as √(σ1² − σ1σ2 + σ2²). The calculator reports both, so neither one is missed.
Turn the element, watch its point go twice as far
Every orientation of the same stress state is a point on one circle. Turn the element by θ and the point for its face moves 2θ, the same way round. At the crossings of the σ axis the shear vanishes: those are the principal stresses. Real numbers for your own stress state come from the calculator.
Check yourself
1. A bar is pulled in simple tension with stress σ. What is the largest shear stress in it, and on which planes?
2. An element is rotated by 30°. How far around Mohr's circle does the point for one of its faces move?
3. A thin cylinder wall has a hoop stress of 100 MPa, an axial stress of 50 MPa and no shear. What is the largest shear stress, counting the stress-free direction through the wall?
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