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Principal Stresses and Mohr's Circle

Stress at a point depends on the plane you ask about. One circle answers every plane at once — and then two failure theories disagree by fifteen per cent.

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Stress at a point depends on the plane you ask about, and the transformation equations trace a circle — so principal stresses, maximum shear and the planes they act on can all be read from one drawing, which is what lets a combined stress state be compared against a uniaxial test result.

Stress depends on the plane

Cut through a point on one plane and you get one pair of stresses; cut on another and you get a different pair. Nothing about the loading changed — only the question. So "the stress here is 80 MPa" is an incomplete statement.

A two-dimensional stress state takes three numbers: σx, σy and τxy. Everything on any other plane follows from those three.

Principal planes

The transformation equations are sinusoidal in , so a full cycle takes 180° of physical rotation. Where the direct stress reaches its maximum, the shear stress is exactly zero — those are the principal planes, and the values there are the principal stresses σ₁ and σ₂.

The circle

Squaring and adding the transformation equations eliminates the angle and leaves the equation of a circle. Every plane through the point maps to one point on it, and rotating the physical plane by θ moves you 2θ around the circle — which is why it closes after 180°.

Feature of the circleWhat it means
Centre, on the σ axisThe average direct stress, (σx + σy)/2
RadiusThe maximum shear stress
Where it crosses the σ axisσ₁ and σ₂, the principal stresses
Top and bottom of the circlePlanes of maximum shear, 45° from the principal planes
Angle around the circleTwice the physical rotation

Because the top of the circle is 90° around from the axis crossings, τ_max acts on planes 45° from the principal planes. That single geometric fact explains slip bands, torsional fractures and the 45° shear failure of a ductile tensile specimen.

Pure shear

Pure shear has zero direct stress on the reference planes, so the circle is centred on the origin and the principal stresses are ±τ. A twisted shaft therefore carries equal tension and compression on planes at 45°.

Brittle materials fail on the tension plane and crack in a 45° helix; ductile ones fail in shear and break flat across the section. Twisting a stick of chalk demonstrates the first in about a second.

Why any of this is needed

Material data comes from a uniaxial test: one stress, one direction, one number. A real component is under a combined state produced by bending, torsion and axial load together. The bridging question — which combination is as severe as the tensile yield stress — is what a failure theory answers, and Mohr's circle supplies its inputs.

Failure theories

TheoryCriterionSuitsAllows in pure shear
Maximum shear stress (Tresca)τ_max = σy/2Ductile metals; simple and conservative0.500 σy
Distortion energy (von Mises)Energy of shape changeDuctile metals; matches test data best0.577 σy
Maximum principal stress (Rankine)σ₁ = σutBrittle materials — failure is tensile1.000 σut

The 15% gap between Tresca and von Mises is not academic when the factor of safety is 1.5. And using a ductile theory on a brittle material — or the reverse — is a genuine design error, not a matter of preference.

The three-dimensional caution

Two-dimensional analysis assumes the third principal stress is zero. That is fine for a free surface, which is where the maximum stress usually is — but in a thick pressure vessel or a rolling contact, all three are significant and the two-dimensional circle understates the shear.

The numbers you will be asked for

Principal stresses

σ₁,₂ = (σx+σy)/2 ± √[((σx−σy)/2)² + τxy²]

Maximum shear

τ_max = (σ₁ − σ₂) / 2

Principal angle

tan 2θ = 2τxy / (σx − σy)

Circle centre and radius

C = (σx+σy)/2 · R = τ_max

Von Mises stress

σ_v = √(σ₁² − σ₁σ₂ + σ₂²)

Watch it work

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Check yourself

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One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Why is 'the stress at this point is 80 MPa' an incomplete statement?
What is special about the principal planes?
Why does a ductile tensile specimen shear at 45° rather than pulling apart square?
Tresca and von Mises disagree by 15% in pure shear. Does it matter?

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4 still unanswered — the dots above jump straight to them.

 

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