Mohr's Circle Calculator (stress on a plane rotated by any angle θ)

Understanding the Mohr's circle calculator — concept and principle

This calculator applies plane-stress transformation and Mohr's circle. From the normal stresses σx, σy and shear stress τxy it gives the stresses on a plane rotated by any angle, the circle's centre and radius, the principal stresses, the maximum in-plane shear and the angles where they occur. Structural and mechanical engineering students use it for strength-of-materials problems.

The two face stresses sit on opposite ends of a diameter; the circle crosses the σ axis at the principal stresses.

How it is calculated

Centre C = (σx + σy) ÷ 2; radius R = √(((σx − σy)/2)² + τxy²)

σx′ = C + ((σx − σy)/2) cos 2θ + τxy sin 2θ

σy′ = C − ((σx − σy)/2) cos 2θ − τxy sin 2θ

τx′y′ = −((σx − σy)/2) sin 2θ + τxy cos 2θ

Principal stresses σ₁,₂ = C ± R at θp = ½ atan2(2τxy, σx − σy); maximum shear = R at θs = θp − 45°

Tension is positive; θ is counter-clockwise; stresses in MPa.

The principle

The stress at a point looks different depending on the orientation of the plane you cut through it. Plotting normal stress against shear stress for every orientation traces a circle, Mohr's circle. Rotating the cutting plane by θ moves the point around the circle by 2θ. The circle's rightmost and leftmost points are the principal stresses, where shear is zero, and its top and bottom are the maximum shear stresses. The sum σx′ + σy′ stays constant, a handy check.

Worked example

Default inputs: σx = 50 MPa, σy = −10 MPa, τxy = 40 MPa, θ = 30°.

C = (50 − 10) ÷ 2 = 20 MPa; R = √(30² + 40²) = 50 MPa.
2θ = 60°: σx′ = 20 + 30 × 0.5 + 40 × 0.866 = 69.64 MPa; σy′ = −29.64 MPa.
τx′y′ = −30 × 0.866 + 40 × 0.5 = −5.98 MPa.
σ₁ = 20 + 50 = 70 MPa; σ₂ = 20 − 50 = −30 MPa; maximum shear = 50 MPa.
θp = ½ atan2(80, 60) = 26.57°; θs = −18.43°.

Each point is the calculator re-run with only “Normal stress σx (MPa, tension +)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Normal stress σx (MPa, tension +)	σx′ on the rotated plane (MPa)
12.5	41.52
25	50.89
37.5	60.27
50	69.64
62.5	79.02
75	88.39
100	107.14

How to read the result

Compare the principal stresses and maximum shear with the material's allowable values using the appropriate failure theory: maximum shear (Tresca) or von Mises for ductile metals, maximum normal stress for brittle ones. At 30° the stresses are already close to principal, as the small shear shows.

Common mistakes

Rotating by θ on the circle instead of 2θ.
Using a different sign convention for shear than the formulas assume.
Forgetting that the absolute maximum shear in 3D may exceed R when both principal stresses have the same sign.

Keep in mind: This is a plane-stress analysis. Real designs need full load cases, stress concentrations, applicable codes and a qualified engineer.

Frequently asked questions

Why does the circle use 2θ?

The transformation equations depend on sin 2θ and cos 2θ, so a 90° physical rotation is 180° on the circle.

Where is shear zero?

At the principal planes, angle θp.

Is the maximum in-plane shear the absolute maximum?

Not always; with σ₃ = 0 out of plane, the absolute maximum can be larger.

How do I check my answer?

σx′ + σy′ must equal σx + σy.

Sources: Standard plane-stress transformation equations and Mohr's circle construction (strength-of-materials texts)

Checked 9 Nov 2026. Standard plane-stress transformation equations and Mohr’s circle construction (as in strength-of-materials texts); no external constants.

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