What principal stresses are and why you need them
The stresses you calculate from a load — bending stress from My/I, torsional shear from Tr/J, direct stress from P/A — are tied to the axes you happened to choose. Rotate the element and every one of those numbers changes. The principal stresses are the values you find on the one orientation where the shear stress vanishes entirely, leaving pure tension and pure compression. They are properties of the stress state itself, not of your coordinate choice.
That matters for three practical reasons. Brittle materials crack on the plane of maximum tensile stress, so σ1 tells you where the crack will start and which way it will run. Ductile materials yield on planes of maximum shear, so τmax and the von Mises stress predict yielding. And fatigue life is driven by the principal stress range, not by the component stresses.
Christian Otto Mohr's 1882 construction turns the algebra into a circle. Plot each face of the element as a point (σ, τ) and every possible orientation of that element lies on one circle whose centre is the average normal stress and whose radius is the maximum shear stress. Rotating the physical element by θ moves you 2θ around the circle. Once you see that, most transformation problems become geometry rather than trigonometry.
Reading the formula off the circle
Start with the two quantities that define the circle. The centre sits at (σx+σy)/2 on the σ axis, always — the sum of the normal stresses is invariant, so no rotation moves it. The radius is R = √( ((σx−σy)/2)² + τxy² ), which is just Pythagoras applied to the point representing the x face: it sits (σx−σy)/2 to the right of the centre and τxy above it.
Everything else follows. The circle crosses the σ axis at centre ± R, and those two crossings are σ1 and σ2. They are the extreme normal stresses, and because they lie on the horizontal axis, the shear there is zero — which is the definition of a principal plane. The top and bottom of the circle are at ±R, so the maximum in-plane shear stress equals the radius. Those planes are a quarter turn around the circle from the principal planes, which is 45° in the real element.
The angle is the part students most often get wrong. tan2θp = 2τxy / (σx − σy), and a plain arctangent loses the quadrant. This calculator uses atan2 with the numerator and denominator kept separate, which resolves it correctly and always returns the angle to the σ1 plane rather than to whichever principal plane happens to be nearer. Halving the result converts circle angle to element angle.
The von Mises stress is not a Mohr's circle quantity but is calculated from the principal stresses: σvM = √(σ1² − σ1σ2 + σ2²) for plane stress. Compare it directly to the yield strength — if it exceeds it, the material yields somewhere in the element.
Worked example: σx = −20, σy = 90, τxy = 60 MPa
This is a classic mechanics-of-materials state: compression in one direction, tension in the other, with substantial shear.
- Centre. (−20 + 90) ÷ 2 = 70 ÷ 2 = 35 MPa.
- Half-difference. (−20 − 90) ÷ 2 = −110 ÷ 2 = −55 MPa. This is the horizontal offset of the x-face point from the centre; it is negative, so that point sits to the left.
- Radius. R = √((−55)² + 60²) = √(3,025 + 3,600) = √6,625 = 81.394 MPa.
- Principal stresses. σ1 = 35 + 81.394 = 116.394 MPa. σ2 = 35 − 81.394 = −46.394 MPa. Check the invariant: 116.394 + (−46.394) = 70 = −20 + 90. ✓
- Maximum in-plane shear. τmax = R = 81.394 MPa, acting on planes where the normal stress is the average, 35 MPa.
- Principal angle. 2θp = atan2(2 × 60, −20 − 90) = atan2(120, −110). The reference angle is arctan(120/110) = 47.490°, and because the point is in the second quadrant, 2θp = 180 − 47.490 = 132.510°. So θp = 66.255° counter-clockwise.
- Maximum-shear planes. 66.255 − 45 = 21.255°.
- Von Mises. √(116.394² − 116.394 × (−46.394) + (−46.394)²) = √(13,547.6 + 5,399.9 + 2,152.4) = √21,099.9 = 145.26 MPa. Against a 250 MPa mild-steel yield, the element has a factor of 1.72 against yielding.
Notice how much larger σ1 is than either input normal stress. Shear stress always pushes the principal values further apart than the normal stresses you started with, which is why a shaft in combined bending and torsion is more highly stressed than either loading alone suggests.
How to interpret the result
Compare the right quantity to the right property. For a ductile metal, compare the von Mises stress to the yield strength. If you prefer the more conservative Tresca criterion, compare twice the absolute maximum shear stress to the yield strength instead — Tresca and von Mises agree in uniaxial tension and differ by at most 15.5%, with Tresca always the safer of the two. For a brittle material such as grey cast iron or concrete, compare σ1 to the tensile strength, because it fails on the maximum-tension plane.
Do not forget the third principal stress. Plane stress means σ3 = 0, and zero is still a principal stress. When both in-plane principals are tensile, σ3 = 0 is the smallest of the three, so the true maximum shear stress is σ1/2 on an out-of-plane plane, which is larger than the in-plane radius R. The same applies with the signs reversed when both are compressive. This calculator reports both the in-plane value and the absolute value in the steps, and warns you when they differ.
The sign of θp follows the sign convention you used for τxy. A positive τxy acts on the +x face in the +y direction. Many textbooks plot Mohr's circle with the shear axis inverted so that rotations on the circle and in the element go the same way; the numerical answers are identical either way, only the picture flips.
A circle of zero radius is a real answer, not an error. Equal normal stresses with no shear is the hydrostatic state, where every plane is principal and no plane carries shear. That is why a sphere under uniform pressure has no preferred failure plane.
Principal stresses for common loading states
| Stress state | σ₁ | σ₂ | τmax (in-plane) | θp | von Mises |
|---|---|---|---|---|---|
| Uniaxial tension σ | σ | 0 | 0.5σ | 0° | σ |
| Uniaxial compression −σ | 0 | −σ | 0.5σ | 0° | σ |
| Pure shear τ | τ | −τ | τ | 45° | 1.7321τ |
| Equal biaxial tension σ, σ | σ | σ | 0 | any | σ |
| Equal tension and compression σ, −σ | σ | −σ | σ | 0° | 1.7321σ |
| Thin cylinder under pressure (2σ, σ) | 2σ | σ | 0.5σ | 0° | 1.7321σ |
The pure-shear row is the origin of the shear yield strength τy = σy/√3 = 0.577σy used in the von Mises criterion; Tresca gives 0.5σy for the same quantity.
Pitfalls that produce a wrong principal stress
- Using arctan instead of atan2 for the angle. arctan(2τ/(σx−σy)) returns a value between −90° and +90° for 2θp, which silently maps some states onto the σ₂ plane instead of the σ₁ plane. The error is exactly 90° in the element and easy to miss.
- Halving the wrong thing. The circle angle is 2θ. Compute 2θp first, then divide by two. Halving the input stresses is a different and much more visible mistake; halving at the wrong step is not.
- Forgetting that σ₃ = 0 is a principal stress. For a biaxial tensile state, the in-plane radius understates the true maximum shear stress, so a Tresca check based on R alone is unconservative.
- Mixing sign conventions for shear. Some texts define positive shear as acting counter-clockwise on the element. The principal stresses are unaffected because τxy is squared, but the sign of θp flips.
- Applying plane stress where the body is thick. Deep inside a thick component under multiaxial load, σ₃ is not zero and you need the full three-dimensional principal stress solution, which is a cubic equation rather than a circle.
- Comparing von Mises stress to ultimate tensile strength. Von Mises predicts the onset of yielding, so it belongs against the yield strength. Fracture in ductile materials involves large plastic strain that the criterion does not model.
Where this fits in a strength check
Mohr's circle sits between the load analysis and the failure check. You compute component stresses first — P/A for axial, Mc/I for bending, Tr/J for torsion, VQ/It for transverse shear — superpose them at the point of interest, then transform. In a shaft under combined bending and torsion, the critical point is the outer surface where bending stress is maximum and transverse shear is zero, giving σx = Mc/I, σy = 0, τxy = Tr/J. Feed those three numbers in and the von Mises output is the number you compare to the allowable.
The same transformation mathematics applies to strain, with strain gauge rosettes as the everyday application: three measured normal strains give you the full plane-strain state, and the identical circle construction returns the principal strains and their direction. It also applies to the second moment of area, which is why a Mohr's circle of inertia gives the principal axes of an angle section — the axes about which the buckling check and the section modulus must be evaluated.
For beams in ordinary bending, the principal stress at the extreme fibre is simply the bending stress and no transformation is needed; the transformation earns its keep at the neutral axis of a thin web, and in any member where torsion is present. If your problem is a deflection limit rather than a strength limit, the deflection check is the governing one instead, and stress transformation does not enter it.
Key terms
- Plane stress
- A state in which one principal stress is zero. It holds at any free surface, in thin plates loaded in their own plane, and at the outer fibre of beams and shafts — which covers most points where failure actually starts.
- Principal plane
- An orientation on which the shear stress is zero. There are two in plane stress, always 90° apart in the element and 180° apart on the circle.
- Stress invariant
- A combination of stress components that does not change with rotation. In plane stress, σx + σy and σxσy − τxy² are both invariant, which gives you two independent checks on any hand transformation.
- Von Mises stress
- A single equivalent uniaxial stress that produces the same distortion energy as the actual multiaxial state. Comparing it to the yield strength is the standard ductile yield criterion.
- Tresca criterion
- Yield occurs when the maximum shear stress reaches half the yield strength. Simpler than von Mises and never less conservative, with the largest gap — 15.5% — occurring in pure shear.