How this calculation works
A stress state depends on the direction of the plane on which it is resolved. Principal stresses identify directions with zero shear, while Mohr’s circle organizes the transformed normal and shear components. This page evaluates a two-dimensional plane-stress state.
c = (σx+σy)/2; R = √[((σx−σy)/2)²+τxy²]; σ1,2 = c ± RA worked example
With sigma x = 80 MPa, sigma y = 20 MPa and tau xy = 40 MPa, the circle centre is 50 MPa and radius is 50 MPa. The in-plane principal stresses are 100 and zero MPa; maximum in-plane shear is 50 MPa.
Good to know
The third principal stress is assumed zero. This does not represent a general triaxial state or plane strain. Material strength, residual stresses, fatigue and acceptance criteria are not inferred from the transformed values.
Reference: Jönköping University — Mohr circle and strain transformation (opens in a new tab) ↗
What this model answers
A stress state depends on the direction of the plane on which it is resolved.
Principal stresses identify directions with zero shear, while Mohr’s circle organizes the transformed normal and shear components. This page evaluates a two-dimensional plane-stress state.
The Rectangular strain rosette principal strain calculator recovers a strain tensor from gauge readings. Converting strain to stress additionally requires a valid material model.
Keep the input conventions consistent
Tensile normal stress is positive.
The angle refers to the plane normal rotating counterclockwise from the x axis. The transformation uses twice that angle inside its trigonometric terms; a ninety-degree physical rotation therefore makes a full circle in the Mohr representation.
Read the result carefully
Maximum in-plane shear and absolute three-dimensional maximum shear are different when the zero third principal stress lies outside the two in-plane values.
The table includes the latter explicitly. Von Mises equivalent stress is reported as a scalar, without a material-yield pass or fail.
Rotating coordinate axes changes reported normal and shear components while leaving principal values invariant. This provides a useful interpretation check: the same physical stress state should not acquire different principal stresses merely because its components were described in another in-plane orientation. The out-of-plane plane-stress assumption remains separate.
Work through the numbers
With sigma x = 80 MPa, sigma y = 20 MPa and tau xy = 40 MPa, the circle centre is 50 MPa and radius is 50 MPa.
The in-plane principal stresses are 100 and zero MPa; maximum in-plane shear is 50 MPa.
Where the calculation stops
The third principal stress is assumed zero.
This does not represent a general triaxial state or plane strain. Material strength, residual stresses, fatigue and acceptance criteria are not inferred from the transformed values.
A few questions, answered
Is maximum in-plane shear always the absolute maximum?
No. The out-of-plane zero principal stress must also be considered for the absolute three-dimensional value.
What should I change in the comparison?
Rotate the evaluation plane without changing the stress state or its principal values.