Mohr’s circle is a picture of the stress-transformation equations. For a point in plane stress, plot the x-face stresses (σx, τxy) and the y-face stresses (σy, −τxy), join them with a diameter, and draw the circle. Its centre is the average normal stress C, its radius R is the maximum in-plane shear stress, and the ends of the horizontal diameter are the principal stresses C ± R.
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Why stress depends on the plane
The stress at a point is not one number. Cut the material at a different angle and the same internal forces resolve into different normal and shear stresses. A part can fail on a plane that is not aligned with x or y: brittle shafts in torsion crack at 45°, and ductile metals yield along planes of maximum shear.
The stress-transformation equations give σ and τ on a plane rotated by θ. Square them and add, and θ drops out, leaving the equation of a circle in σ–τ coordinates. That is Mohr’s circle. Every point on it is the stress on one plane through the point.
How to draw it
This lesson uses the convention of Hibbeler and Gere, also used by the Mohr’s circle calculator: tension positive, τxy positive when it acts in +y on the +x face, and the τ axis plotted pointing downward. With that choice, rotating the element θ counterclockwise is the same as rotating 2θ counterclockwise on the circle.
Draw the σ axis to the right and the τ axis downward.
Plot X = (σx, τxy) and Y = (σy, −τxy). The line XY is a diameter and crosses the σ axis at the centre C = (σx + σy)/2.
The radius is R = √[((σx − σy)/2)² + τxy²].
Principal stresses σ1 = C + R and σ2 = C − R are where the circle crosses the σ axis; shear is zero there.
The top and bottom of the circle are the planes of maximum in-plane shear, τ = ±R, with normal stress C.
To find the stress on a plane at θ, rotate from X by 2θ in the same direction and read the new point.
Absolute maximum shear
In plane stress the third principal stress is σ3 = 0. If σ1 and σ2 have opposite signs, the absolute maximum shear stress equals R. If they have the same sign, it is half the larger of |σ1| and |σ2|, on a plane out of the x–y plane. Ductile failure criteria use these values, so always check it. Torsion of a circular shaft is a useful special case: pure shear gives a circle centred on the origin.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Centre and radius
C = (σx + σy) / 2, R = √[((σx − σy) / 2)² + τxy²]
σx, σy = normal stresses on the x and y faces (MPa), tension positive; τxy = shear stress on the +x face in the +y direction (MPa)
Hibbeler, Mechanics of Materials
Principal stresses and in-plane maximum shear
σ1,2 = C ± R, τmax,in-plane = R
σ1 = algebraically largest principal stress (MPa)
Gere & Goodno, Mechanics of Materials
Principal angle
tan 2θp = 2τxy / (σx − σy)
θp = angle from the x axis to a principal plane, counterclockwise positive (degrees); the two solutions are 90° apart
θ = rotation of the element from the x axis, counterclockwise positive (degrees); σx′, τx′y′ in MPa
Beer & Johnston, Mechanics of Materials
Absolute maximum shear (plane stress)
τabs = max(|σ1 − σ2|, |σ1|, |σ2|) / 2
uses σ3 = 0
Hibbeler, Mechanics of Materials
Worked examples
Example: Principal stresses and maximum shear at a point
Given
At a point in plane stress, σx = 80 MPa, σy = −20 MPa and τxy = 30 MPa.
Find
The principal stresses and the orientation of σ1.
The maximum in-plane shear stress, its plane, and the absolute maximum shear stress.
Solution
Plot X = (80, 30) and Y = (−20, −30). The centre is midway between them:C = (80 + (−20)) / 2 = 30 MPa
The radius is the distance from C to X:R = √[(80 − 30)² + 30²] = √3400 = 58.31 MPa
Principal stresses where the circle meets the σ axis:σ1 = 30 + 58.31 = 88.31 MPa, σ2 = 30 − 58.31 = −28.31 MPa
The angle from X to σ1 on the circle is 2θp. Both 2τxy and σx − σy are positive:tan 2θp = 2(30) / (80 − (−20)) = 0.6 → 2θp = 30.96°, θp = 15.48° counterclockwise
Confirm that θp gives σ1, with the transformation equation:σx′ = 30 + 50·cos 30.96° + 30·sin 30.96° = 30 + 42.87 + 15.43 = 88.31 MPa ✓
Maximum in-plane shear is the radius, on planes 45° from the principal planes, with normal stress C:τmax = 58.31 MPa at θs = 15.48° − 45° = −29.52°, σ = 30 MPa
σ1 and σ2 have opposite signs, so the absolute maximum shear is the in-plane value:τabs = (88.31 − (−28.31)) / 2 = 58.31 MPa
Answer: σ1 = 88.3 MPa on a plane 15.5° counterclockwise from x, σ2 = −28.3 MPa 90° from it. τmax = 58.3 MPa (in-plane and absolute) on planes at −29.5° and 60.5°, each with a normal stress of 30 MPa.
Example: Stresses along an inclined weld seam
Given
At a point on a welded plate, σx = 50 MPa, σy = 10 MPa and τxy = −25 MPa.
A straight weld seam runs at 35° counterclockwise from the x axis.
Find
The normal stress across the seam and the shear stress along it.
Solution
The seam lies on a plane whose normal is at 35° + 90° = 125°. Rotating the element by θ = 35° puts the y′ axis on that normal, so the seam is the y′ face.
Rotate the element by θ = 35°, so 2θ = 70°. Normal stress on the x′ face:σx′ = 30 + 20·cos 70° + (−25)·sin 70° = 30 + 6.84 − 23.49 = 13.35 MPa
The y′ face (the seam plane) is 90° further on the element, 180° on the circle, so it is the opposite end of the diameter:σy′ = 2C − σx′ = 60 − 13.35 = 46.65 MPa
Shear on the rotated element:τx′y′ = −20·sin 70° + (−25)·cos 70° = −18.79 − 8.55 = −27.34 MPa
Check that the rotated point lies on the circle:(13.35 − 30)² + 27.34² = 277.2 + 747.5 = 1024.7 ≈ R² = 1025 ✓
Answer: Across the seam (on the y′ face) the normal stress is 46.7 MPa tension, and the shear along it is 27.3 MPa in magnitude (τx′y′ = −27.34 MPa in the sign convention used). Both values, not just σx, are what the weld must be checked against.
Common mistakes
Plotting Y at (σy, τxy) instead of (σy, −τxy). The shear on the y face has the opposite sign on the circle. X and Y must be at opposite ends of a diameter.
Rotating θ on the circle instead of 2θ. Planes 90° apart on the element are 180° apart on the circle. Double every element angle.
Rotating the wrong way. With the τ axis pointing down, counterclockwise on the element is counterclockwise on the circle. If your course plots τ up, the sense reverses; state your convention.
Taking the in-plane maximum shear as the absolute maximum. If σ1 and σ2 have the same sign, compare with |σ1|/2 and |σ2|/2 using σ3 = 0.
Confusing the plane with its normal. θ is the angle of the face normal (the x′ axis). A seam or crack lies on a plane; its normal is 90° from its line.
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Frequently asked questions
Is Mohr’s circle needed if I have the equations?
No, the transformation equations give the same results. The circle makes the whole stress state visible at once, which helps you spot the governing plane and check signs.
Why do textbooks plot τ in different directions?
There are two common conventions. Plotting positive τ downward makes rotations on the circle match the element; plotting it upward is also used, with the rotation sense reversed. The stresses you compute are the same.
What is the normal stress on the maximum-shear planes?
The average normal stress C = (σx + σy)/2. It is zero only when σx = −σy, as in pure shear.
Can Mohr’s circle handle 3D stress?
Yes, as three circles drawn between the three principal stresses. In plane stress, one principal stress is zero, which is why the absolute maximum shear check matters.