Mohr's Circle Stress Calculator

Mohr's circle is a graphical restatement of the plane-stress transformation equations.

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Normal stress σx (MPa)

Normal stress σy (MPa)

Shear stress τxy (MPa)

Enter your values and press Calculate to see results.

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How to use this calculator

Enter the three components of the 2D stress state at your point: the normal stress on the x-face (σx), the normal stress on the y-face (σy) and the shear stress on those faces (τxy). Use the ± toggle for tension-positive/compression-negative and the sign convention of your source.
Read off the major and minor principal stresses σ1 and σ2 (the largest and smallest normal stresses, which act on planes with zero shear), plus τmax, the average stress and the principal-plane angle θp.
Compare σ1, σ2 and τmax against your material's allowable/yield limits with an adequate factor of safety, and use θp to orient the principal planes on your element.

How it works

Mohr's circle is a graphical restatement of the plane-stress transformation equations. For a given stress state (σx, σy, τxy), all possible normal/shear stress pairs on planes through the point lie on a circle in σ–τ space centred at σavg = (σx + σy)/2 with radius R = √( ((σx − σy)/2)² + τxy² ). The far-right and far-left points of that circle are the principal stresses σ1 = σavg + R and σ2 = σavg − R, which act on planes carrying no shear; the top of the circle gives the maximum in-plane shear τmax = R.

The orientation of the principal planes follows from θp = ½·atan2(2τxy, σx − σy), measured from the x-axis to the σ1 direction; the planes of maximum shear lie 45° away at θp − 45°. This tool solves those closed-form equations exactly (using atan2 so the quadrant is always correct). Note it works with the two in-plane stresses only: for a plane-stress element the out-of-plane principal stress is zero, so if σ1 and σ2 have the same sign the true absolute-maximum shear can exceed the in-plane τmax reported here.

Worked example

Plane-stress element: σx = 100 MPa, σy = 40 MPa, τxy = 30 MPa. Centre σavg = (100 + 40)/2 = 70 MPa. Radius R = √(((100 − 40)/2)² + 30²) = √(30² + 30²) = 42.43 MPa. So the major principal stress σ1 = 70 + 42.43 = 112.43 MPa, the minor principal stress σ2 = 70 − 42.43 = 27.57 MPa, and the maximum in-plane shear τmax = R = 42.43 MPa. The principal planes are oriented at θp = ½·atan2(2·30, 100 − 40) = 22.5° to the x-axis, and the maximum-shear planes sit 45° from those at −22.5°.

Common mistakes

Sign of τxy and of compressive normal stresses. Compression must be entered as a negative value and the shear sign follows your chosen convention — flipping a sign moves the whole circle and changes σ1, σ2 and θp.
Assuming τmax (the in-plane radius) is always the governing shear. For plane stress the out-of-plane principal stress is 0, so when σ1 and σ2 are both positive (or both negative) the absolute-maximum shear is max(|σ1|, |σ2|)/2, which is larger than the in-plane R.
Reading θp as the angle to σ2 or doubling/halving it. θp here is the physical plane rotation to the σ1 direction (the transformation equations use 2θ internally, but the reported angle is the real, halved plane angle).

Frequently asked questions

What sign convention does this use?

Tension-positive: enter tensile normal stresses as positive and compressive as negative using the ± toggle. τxy is the shear on the x and y faces; a positive value follows the standard convention where a positive shear tends to rotate the element counter-clockwise. Principal stresses and angles come out consistent with whatever convention you feed in.

Is τmax the absolute maximum shear stress?

It is the maximum in-plane shear (the radius of the in-plane circle). Because this is a plane-stress state, the third principal stress is zero. If σ1 and σ2 have the same sign, the true absolute-maximum shear is max(|σ1|, |σ2|)/2 and is larger than the in-plane τmax shown here — check this for failure criteria such as Tresca.

How is the principal angle θp defined?

θp is the rotation (in degrees) from the x-axis to the plane on which σ1 acts, computed as ½·atan2(2τxy, σx − σy). The maximum-shear planes are 45° from the principal planes, reported as θs = θp − 45°.

Can I use it for principal strains too?

The same circle construction applies to strain if you substitute εx, εy and γxy/2 for σx, σy and τxy, but this tool is labelled and validated for stress. For strain analysis use the values as an analogy only and confirm the factor-of-two on the shear-strain term.

Tip: Enter any known values to calculate the remaining results.

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