Mohr's Circle & Plane-Stress Transformation, Explained

The problem Mohr's circle solves

At a point in a loaded part, the stress you calculate depends on the orientation of the plane you look at. The same point can show pure shear on one plane and pure tension on another rotated 45°. The question a strength check really needs answered is: across all orientations, what are the largest normal stress and the largest shear stress? Mohr's circle is the graphical answer — and the plane-stress transformation equations are the algebra behind it.

From a stress state to a circle

A 2-D (plane) stress state is three numbers: σx, σy, and the shear τxy. Plot the two faces as points — the x-face at (σx, τxy) and the y-face at (σy, −τxy) — and they turn out to be the ends of a diameter of a circle on a normal-stress (horizontal) vs shear (vertical) plot. That circle is centred at the average normal stress and has a radius equal to the maximum shear:

• Centre: σ_avg = (σx + σy) / 2
• Radius: R = √[((σx − σy)/2)² + τxy²]

Everything else is read straight off the circle. Rotating the physical element by an angle θ corresponds to rotating twice that angle (2θ) around the circle — the factor-of-two relationship that makes the construction work.

What you read off it

• Principal stresses — the max and min normal stresses, where the circle crosses the horizontal axis: σ1,2 = σ_avg ± R. On these planes the shear is zero. Their orientation is θp = ½·atan2(2τxy, σx − σy), and the two principal planes are 90° apart.
• Maximum in-plane shear — equals the radius R, on planes 45° from the principal planes, where the normal stress is σ_avg.
• Stresses on any rotated plane — just another point on the circle.

The subtlety simpler calculators get wrong

Because this is a plane-stress state, the third principal stress is zero. The absolute maximum shear is taken over the full set {σ1, σ2, 0} — not just σ1 and σ2. When σ1 and σ2 have the same sign (e.g. both tensile), the governing shear plane is out-of-plane, involving that zero third stress, and the absolute max shear is larger than the in-plane value. Miss this and you underestimate the shear that actually drives yielding.