Principal Stresses Calculator (Mohr’s Circle)

Enter the normal stresses σx, σy and shear stress τxy to get the principal stresses and maximum shear.

How it is calculated

For a plane (2D) stress state the tool finds the centre and radius of Mohr's circle: C = (σx + σy) / 2 and R = √[((σx − σy) / 2)² + τxy²]. The principal stresses are σ1 = C + R and σ2 = C − R, and the maximum in-plane shear stress is τmax = R. Inputs are the normal stresses σx and σy (tension positive, compression negative) and the shear stress τxy. Results are labelled in MPa, so enter MPa. Reversing the sign of τxy does not change the values, only the plane orientation, which the tool does not report. It covers the in-plane case only: out-of-plane stress is taken as zero and the principal directions are not computed.

Worked example

Take σx = 100 MPa, σy = −20 MPa and τxy = 40 MPa. The centre is C = (100 + (−20)) / 2 = 40. The radius is R = √[(60)² + 40²] = √(3600 + 1600) = √5200 = 72.11. So σ1 = 40 + 72.11 = 112.11 MPa, σ2 = 40 − 72.11 = −32.11 MPa (compressive), and the maximum in-plane shear stress is 72.11 MPa.

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Common mistakes

  • Entering the stresses in Pa, kPa or psi. The results are labelled MPa, so a value of 100000 kPa would have to be entered as 100.
  • Dropping the sign of compressive stress. A compressive σy is negative; entering it as positive changes the radius and both principal values.
  • Treating τmax as the largest shear stress in the part. It is only the in-plane maximum; when both principal stresses have the same sign, a three-dimensional check with the zero out-of-plane stress gives a larger shear.
  • Expecting the angle of the principal planes. The tool returns magnitudes only, so the orientation must be found separately.
  • Applying the plane-stress result to a point where the out-of-plane stress is not negligible, such as a thick-walled part.

About this calculator

This calculator finds the principal stresses of a 2D stress state (σx, σy, τxy) using the closed-form equivalent of Mohr's circle: σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²], and the maximum in-plane shear stress equals the radius of that circle, √[((σx − σy)/2)² + τxy²].

The principal stresses are the normal stresses acting on the planes where shear stress drops to zero — the orientation an engineer actually cares about, since materials typically fail along planes of maximum normal or shear stress. Structural, mechanical and civil engineers use this transformation constantly when checking whether a loaded beam, pressure vessel or machine part will yield or fracture.

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