Mohr’s Circle Calculator for Plane Stress

Mohr's circle for plane stress

UnitsSIUS customary

Normal stress σx (MPa)Tension positive.

Normal stress σy (MPa)Tension positive.

Shear stress τxy (MPa)Positive on the +x face in +y.

Rotation θ (°)Counterclockwise from x, for the rotated element.

Principal stress σ1

87.08 MPa

Principal stress σ2

-47.08 MPa

Principal angle θp (to σ1)

13.28°

Max in-plane shear

67.08 MPa

Max shear angle θs

-31.72°

Absolute max shear (σ3 = 0)

67.08 MPa

Rotated 30°: σx′ = 75.98 MPa, σy′ = -35.98 MPa, τx′y′ = -36.96 MPa.

Mohr's circle (τ positive downward). Rotated 30°: σx′ = 75.98 MPa, σy′ = -35.98 MPa, τx′y′ = -36.96 MPa.

Intermediate steps
Step	Formula	Value
Centre	C = (σx + σy) / 2 = (80 + (-40)) / 2	20 MPa
Radius	R = √[((σx − σy) / 2)² + τxy²] = √[60² + 30²]	67.08 MPa
Principal stress σ1	σ1 = C + R	87.08 MPa
Principal stress σ2	σ2 = C − R	-47.08 MPa
Principal angle	θp = ½·atan2(2τxy, σx − σy) = ½·atan2(60, 120)	13.28°
Max in-plane shear	τmax, in-plane = R	67.08 MPa
Max shear angle	θs = θp − 45°	-31.72°
Absolute max shear	τabs = max(\|σ1\|, \|σ2\|, \|σ1 − σ2\|) / 2, with σ3 = 0	67.08 MPa

Inputs

σx and σy (MPa or ksi) — Normal stresses on the x and y faces. Tension is positive, compression negative.
τxy (MPa or ksi) — Shear stress on the +x face acting in the +y direction. Its sign matters.
Rotation θ (°) — Angle of a rotated element, counterclockwise from the x axis, between −180° and 180°.

Key formulas

Formulas used on this page, with the textbook each is cited from
Formula	Expression	Where	Cited from
Centre and radius of the circle	`C = (σx + σy) / 2, R = √[((σx − σy) / 2)² + τxy²]`	C is the average normal stress	Hibbeler, Mechanics of Materials
Principal stresses	`σ1,2 = C ± R`	shear stress is zero on the principal planes	Gere & Goodno, Mechanics of Materials
Principal angle	`tan 2θp = 2τxy / (σx − σy)`	the calculator uses atan2 so θp points at σ1	Gere & Goodno, Mechanics of Materials
Maximum in-plane shear	`τmax = R, at θs = θp − 45°`	normal stress on those planes is C	Beer, Johnston, DeWolf & Mazurek, Mechanics of Materials
Stress transformation	`σx′ = C + ((σx − σy)/2)·cos 2θ + τxy·sin 2θ, τx′y′ = −((σx − σy)/2)·sin 2θ + τxy·cos 2θ`	θ counterclockwise from x	Hibbeler, Mechanics of Materials

Assumptions

Plane stress: σz = τxz = τyz = 0. The third principal stress is therefore zero, which the absolute maximum shear stress accounts for.
Stresses at a single point. The calculator transforms a known stress state; it does not find stresses from loads.
Signs follow Hibbeler and Gere: tension positive, τxy positive on the +x face in the +y direction, θ counterclockwise positive.
On the circle, point X is (σx, τxy) and the τ axis points down, so rotating 2θ counterclockwise on the circle matches rotating the element θ counterclockwise.

Worked example

Example: Principal stresses at a point on a loaded bracket

Given

At a point on a bracket, σx = 80 MPa (tension), σy = −40 MPa (compression) and τxy = 30 MPa.
Plane stress, so σz = 0.

Find

The principal stresses and the angle to σ1.
The maximum in-plane shear stress and the absolute maximum shear stress.

Solution

Find the circle’s centre.C = (80 + (−40)) / 2 = 20 MPa
Find the radius from the half-difference of the normal stresses and the shear stress.R = √(60² + 30²) = √4500 = 67.08 MPa
The principal stresses lie on the σ axis, one radius either side of the centre.σ1 = 20 + 67.08 = 87.08 MPa, σ2 = 20 − 67.08 = −47.08 MPa
Find the principal angle. Both 2τxy and σx − σy are positive, so 2θp is in the first quadrant.θp = ½·atan(60 / 120) = ½ × 26.57° = 13.28°
Maximum in-plane shear equals the radius, on planes 45° from the principal planes.τmax = 67.08 MPa at θs = 13.28° − 45° = −31.72°
σ1 and σ2 have opposite signs, so with σ3 = 0 the absolute maximum shear is still the in-plane value.τabs = (σ1 − σ2) / 2 = (87.08 + 47.08) / 2 = 67.08 MPa

Answer: σ1 = 87.08 MPa at θp = 13.28°, σ2 = −47.08 MPa, τmax = 67.08 MPa (in-plane and absolute) at θs = −31.72°.

Common mistakes

Getting the sign of τxy wrong. τxy is positive when it acts in +y on the +x face. Flipping it mirrors θp and changes every rotated stress.
Plotting the angle once instead of twice. A rotation of θ on the element is 2θ on Mohr’s circle.
Reporting the wrong principal plane. tan 2θp has two solutions 90° apart. Substitute θp back, or use atan2 as this calculator does, to see which gives σ1.
Forgetting σ3 = 0. When σ1 and σ2 are both tensile (or both compressive), the absolute maximum shear is |σ1|/2 or |σ2|/2, larger than the in-plane R.
Mixing up the τ axis direction between textbooks. Check which way your course plots positive τ. The numbers are the same; only the picture flips.

Frequently asked questions

What is Mohr’s circle used for?

It is a graphical form of the stress-transformation equations. It shows, on one diagram, the normal and shear stress on every plane through a point, which makes principal stresses and maximum shear easy to read off.

What is the difference between in-plane and absolute maximum shear stress?

In-plane maximum shear is the radius of the circle for the x–y plane. Absolute maximum shear also considers the out-of-plane direction, where σ3 = 0 for plane stress, and is half the largest difference between any two of σ1, σ2 and σ3.

Can I use it for strain?

The same construction works for plane strain if you enter εx, εy and γxy/2 in place of σx, σy and τxy. The units become strain, and the results are principal strains.

How does this relate to failure criteria?

The principal stresses feed directly into the Tresca (maximum shear) and von Mises criteria used to compare a stress state with the material’s yield strength.

More engineering calculators

Stress and strain lesson · All engineering calculators

Sources

R. C. Hibbeler, Mechanics of Materials, Pearson
J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage
F. P. Beer, E. R. Johnston, J. T. DeWolf, D. F. Mazurek, Mechanics of Materials, McGraw Hill

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