Calculator · Structural Mechanics
Principal Stresses Calculator
Every plane-stress state has one orientation on which the shear vanishes. The normal stresses on that orientation are the principal stresses — the algebraic extremes of normal stress at the point.
Stress state
σ is tensile positive. τxy acts on the +x face in the +y direction. θ is a counter-clockwise rotation of the axes in physical space.
Results
| Major principal stress σ₁ | 88.31 MPa |
|---|---|
| Minor principal stress σ₂ | -28.31 MPa |
| Orientation of σ₁ (CCW from +x) | 15.482° |
| Orientation of σ₂ | 105.48° |
| Maximum in-plane shear τmax | 58.31 MPa |
| Plane of τmax (CCW from +x) | -29.518° |
| Average normal stress σavg | 30 MPa |
| Third principal stress σ₃ (out of plane) | 0 MPa — plane stress |
Principal element
Open in the Lab
Technical reference
σavg = (σx + σy)/2
R = √( ((σx−σy)/2)² + τxy² )
σ₁ = σavg + R, σ₂ = σavg − R
θp = ½ · atan2( 2τxy , σx − σy )
τmax(in-plane) = R, on planes at θp − 45°
- σ₁, σ₂
- principal stresses, algebraically ordered σ₁ ≥ σ₂[Pa]
- R
- radius of Mohr's circle, equal to the maximum in-plane shear[Pa]
- θp
- counter-clockwise angle from +x to the σ₁ direction[rad]
The principal planes carry no shear; the planes of maximum in-plane shear sit 45° from them and do carry a normal stress equal to σavg.
τmax here is the maximum IN-PLANE shear. Because plane stress has a third principal stress of zero, the maximum ABSOLUTE shear is (σmax − σmin)/2 over all three principal values — which is larger whenever σ₁ and σ₂ have the same sign.
When R = 0 the state is equal-biaxial and every direction is principal, so no unique angle exists; the engine reports that rather than returning an arbitrary one.
Assumptions
- · Plane stress: the out-of-plane face is free, σz = 0.
- · Angles are counter-clockwise from the +x axis, matching the transformation convention.
- · Principal values are ordered algebraically, not by magnitude: σ₁ ≥ σ₂ including sign.