Engineering Calculators
● Mohr's Circle Calculator

Mohr's Circle Stress Calculator

Enter a 2-D (plane) stress state — normal stresses σx, σy and shear τxy — to get the principal stresses, maximum shear, and their orientations, with the transformed stresses on any rotated plane and a to-scale Mohr's circle showing the whole geometry.

1Stress stateThe three components of a 2-D (plane) stress state at a point. Tension is positive. τxy is the shear on the x-face acting along +y (the complementary shear on the y-face is equal). These are the stresses resolved on the x/y faces of your element — Mohr's circle then gives them on any other plane.
Tension +, compression −.
Tension +, compression −.
+x face acting in +y (right-hand convention).
2Plane rotationRotate the element (its axes) counter-clockwise by θ to read the normal and shear stresses on the rotated faces — the X′ point on the circle. Set θ = θp to land on the principal plane (τ → 0), or θ = θs for the maximum-shear plane. This does not change the circle, only which point on it you are reading.
Principal & shear results
σ1 (max principal)
124.5MPa
σ2 (min principal)
-14.5MPa
τmax (in-plane)
69.5MPa
θp (to σ1)
29.87°
Mohr's circle
σ (normal)τ (shear)Oσavg 55 MPaσ₁ 124.5 MPaσ₂ -14.5 MPaτmax 69.5 MPaX (σx, τxy)YX′ (θ=30°)2θσ–τ plane · true scale · X=(σx, τxy), Y=(σy, −τxy) · stresses in MPaσxσyτxyInput elementσx=90 MPa σy=20 MPa τxy=60 MPaσx′σy′τ′Rotated θ = 30°σx′=124.5 MPa σy′=-14.5 MPa τ′=-0.3 MPa
Principal & maximum shearPrincipal stresses are the extreme normal stresses (where shear vanishes); the maximum in-plane shear is the circle radius and acts 45° from the principal planes, carrying a normal stress equal to σavg. The absolute maximum shear also accounts for the zero out-of-plane principal (σ3 = 0) and governs when σ1 and σ2 have the same sign.
Max principal σ1124.46 MPa
Min principal σ2-14.46 MPa
Mean stress σavg (centre)55 MPa
Principal angle θp1 / θp229.87° / 119.87°
Max in-plane shear τmax69.46 MPa
Max-shear plane angle θs-15.13°
Normal stress at τmax55 MPa
Absolute max shear (3-D)69.46 MPa
Stresses on the rotated plane (θ = 30°)The normal and shear stresses on the element after rotating its axes counter-clockwise by θ — i.e. the X′/Y′ points on the circle. Note σx′ + σy′ = σx + σy always (the first stress invariant).
σx′ (rotated x-face)124.46 MPa
σy′ (rotated y-face)-14.46 MPa
τx′y′ (rotated shear)-0.31 MPa
Failure-criterion equivalentsSingle-number equivalent stresses for comparing against a uniaxial material allowable. von Mises (distortion-energy) suits ductile metals; Tresca (maximum-shear) is more conservative. Both use the full principal set including σ3 = 0.
von Mises equivalent132.29 MPa
Tresca equivalent138.92 MPa
Reference & assumptions

New to Mohr's Circle? Read the guide: Mohr's Circle & Plane-Stress Transformation, Explained — a plain-language explainer of the standard behind this calculator.

This tool applies the standard plane-stress transformation equations and their Mohr's-circle construction (Gere & Goodno, Mechanics of Materials; Hibbeler, Mechanics of Materials, Ch. 9). Sign convention: normal stress tension-positive; τxy positive on the +x face acting in the +y direction; the rotation angle θ positive counter-clockwise (the transformed x′ face is the x face rotated CCW by θ). The circle centre is σavg = (σx + σy)/2 and its radius is R = √[((σx − σy)/2)² + τxy²]; the principal stresses are σ1,2 = σavg ± R and the principal orientation is θp = ½·atan2(2τxy, σx − σy), with the σ1 and σ2 planes 90° apart. The maximum in-plane shear equals R and acts on planes 45° from the principal planes, where the normal stress is σavg. Because this is a plane-stress state the third principal stress is zero; the absolute maximum shear is taken over the full set {σ1, σ2, 0} and therefore exceeds the in-plane value whenever σ1 and σ2 share the same sign (the governing shear plane is then out-of-plane). The von Mises equivalent uses the plane-stress form √(σ1² − σ1σ2 + σ2²) and the Tresca equivalent is σmax − σmin over the same principal set. On the diagram the horizontal axis is normal stress (tension right) and the vertical axis is shear (positive up); the x-face is plotted at (σx, τxy) and the y-face at (σy, −τxy). Idealisations: a linear-elastic, homogeneous, isotropic material in true plane stress (σz = τxz = τyz = 0); no stress concentrations, plane-strain effects, or fully three-dimensional stress states beyond the zero third principal, and no yielding. Use the equivalent stresses together with the relevant material allowable and factor of safety for a strength check.

Validated: checked against two independent hand-worked examples using the equations above. (σx, σy, τxy) = (−20, 90, 60) MPa gives σ1 = 116.4 MPa, σ2 = −46.4 MPa, θp1 = 66.3° — matches this calculator exactly. (80, 50, −25) MPa gives σ1 = 94.2 MPa, σ2 = 35.8 MPa, and correctly shows the absolute maximum shear (94.2 MPa) exceeding the in-plane maximum (29.2 MPa) once both principal stresses share a sign — a subtlety simpler stress-transformation calculators often get wrong.

Calculation steps
1. Circle centre and radius
σavg = (σx + σy)/2, R = √[ ((σx − σy)/2)² + τxy² ]
σx = 90, σy = 20, τxy = 60 MPa
σavg = 55 MPa, R = 69.46 MPa
2. Principal stresses
σ1 = σavg + R, σ2 = σavg − R (σ3 = 0 for plane stress)
σavg = 55 MPa, R = 69.46 MPa
σ1 = 124.46 MPa, σ2 = -14.46 MPa
3. Principal orientation
θp = ½·atan2(2τxy, σx − σy) (CCW from x-axis to the σ1 plane)
2τxy = 120, σx − σy = 70 MPa
θp1 = 29.87° (σ1), θp2 = 119.87° (σ2)
4. Maximum shear stress
τmax(in-plane) = R; τabs,max = (σmax − σmin)/2 over {σ1, σ2, 0}
σ1 = 124.46, σ2 = -14.46, σ3 = 0 MPa
τmax(in-plane) = 69.46 MPa at θs = -15.13° (σ = σavg = 55 MPa); τabs,max = 69.46 MPa (= in-plane)
5. Stress transformation at θ = 30°
σx' = σavg + (σx−σy)/2·cos2θ + τxy·sin2θ; τx'y' = −(σx−σy)/2·sin2θ + τxy·cos2θ
θ = 30° (CCW)
σx' = 124.46, σy' = -14.46, τx'y' = -0.31 MPa
6. Equivalent stresses
von Mises = √(σ1² − σ1σ2 + σ2²); Tresca = 2·τabs,max = σmax − σmin
σ1 = 124.46, σ2 = -14.46 MPa
σ_vM = 132.29 MPa, σ_Tresca = 138.92 MPa