Mohr's Circle Calculator

Compute principal stresses, maximum shear stress, and stress transformations for any plane stress state.

📐 Mohr's Circle Calculator
Normal Stress σx80 MPa
MPa
-200200
Normal Stress σy30 MPa
MPa
-200200
Shear Stress τxy25 MPa
MPa
-100100
Rotation Angle θ45°
deg
0°90°
Major Principal Stress (σ1)
—
Minor Principal Stress (σ2)
—
Maximum Shear Stress (τmax)
—
Average Normal Stress (σavg)
—
Principal Plane Angle (θp)
—
Max Shear Plane Angle (θs)
—
Rotated Normal Stress σx'
—
Rotated Normal Stress σy'
—
Rotated Shear Stress τx'y'
—
Major Principal Stress (σ1)
—
Minor Principal Stress (σ2)
—
Maximum Shear Stress (τmax)
—

📐 What is Mohr's Circle?

Mohr's circle is a graphical construction that represents the complete state of plane stress at a point in a loaded body. Developed by German civil engineer Christian Otto Mohr in 1882, it transforms the mathematical stress transformation equations into a circle drawn on a coordinate system where the horizontal axis represents normal stress and the vertical axis represents shear stress. Every point on the circle corresponds to a specific plane orientation through the stress element, with the x-coordinate giving the normal stress on that plane and the y-coordinate giving the shear stress.

Engineers use Mohr's circle across a wide range of applications. In mechanical engineering, it helps verify that shaft fillets, welded joints, and press-fit hubs do not exceed the shear yield strength under combined bending and torsion loads. In structural engineering, designers apply it to beam-columns under eccentric axial loads, to reinforced concrete sections under combined flexure and shear, and to steel connections where both direct and shear forces act together. In geotechnical engineering, Mohr's circle is the foundation of the Mohr-Coulomb failure criterion, which governs slope stability, bearing capacity, and retaining wall design for cohesive and frictional soils alike.

A common misconception is that a 45-degree rotation of the stress element produces a 45-degree rotation on the circle. In fact, the angle on Mohr's circle is double the physical angle: a 45-degree physical rotation corresponds to a 90-degree arc on the circle. This doubling arises because the transformation equations use sin(2θ) and cos(2θ). As a result, the maximum shear stress planes, which are 90 degrees from the principal planes on the circle, are only 45 degrees away in the physical material.

This calculator computes all key Mohr's circle outputs in two modes. Principal Stresses mode returns σ1, σ2, τmax, σavg, and the orientation of the principal planes directly. Stress at Angle mode applies the full transformation equations to find the exact normal and shear stresses acting on any arbitrarily rotated plane, making it easy to check specific cut planes in design or verification work.

📐 Formula

σ1,2  =  σavg  ±  R
σavg = (σx + σy) / 2 — center of Mohr's circle (average normal stress)
R = √[((σx − σy) / 2)² + τxy²] — radius of Mohr's circle = τmax
σ1 = σavg + R — major principal stress (maximum normal stress)
σ2 = σavg − R — minor principal stress (minimum normal stress)
τmax = R — maximum in-plane shear stress
θp = (1/2) × atan2(2τxy, σx − σy) — principal plane angle
θs = θp + 45° — maximum shear stress plane angle
Stress Transformation
σx' = σavg + ((σx − σy)/2) cos(2θ) + τxy sin(2θ)
σy' = σavg − ((σx − σy)/2) cos(2θ) − τxy sin(2θ)
τx'y' = −((σx − σy)/2) sin(2θ) + τxy cos(2θ)
θ = rotation angle of the new x'-axis from the original x-axis
Example: For σx = 80 MPa, σy = 30 MPa, τxy = 25 MPa: σavg = 55 MPa, R = 35.36 MPa, σ1 = 90.36 MPa, σ2 = 19.64 MPa, τmax = 35.36 MPa, θp = 22.50°

📖 How to Use This Calculator

Steps

1
Enter the stress components: Input σx and σy (normal stresses in MPa, positive for tension) and τxy (shear stress in MPa). Drag the sliders or type values directly. Negative values indicate compression or reversed shear direction.
2
Choose the calculation mode: Click Principal Stresses to find σ1, σ2, τmax, σavg, and the principal plane angle θp. Click Stress at Angle to find the transformed stresses on a rotated plane.
3
Enter the rotation angle for Stress at Angle mode: Type or drag θ from 0 to 90 degrees. The calculator returns σx', σy', and τx'y' on the plane rotated θ degrees from the x-axis. Setting θ = θp produces zero shear stress on the rotated plane.
4
Read the results: All outputs update instantly. Use the Copy link button to save the URL with your exact inputs for sharing or revisiting later.

💡 Example Calculations

Example 1: Uniaxial Tension

A steel bar under uniaxial tension: σx = 100 MPa, σy = 0 MPa, τxy = 0 MPa

1
Compute center: σavg = (100 + 0) / 2 = 50 MPa
2
Compute radius: R = sqrt((50)^2 + (0)^2) = 50 MPa
3
Principal stresses: σ1 = 50 + 50 = 100 MPa, σ2 = 50 - 50 = 0 MPa. τmax = 50 MPa at θs = 45°
σ1 = 100 MPa, σ2 = 0 MPa, τmax = 50 MPa, θp = 0.00°
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Example 2: Pure Shear State

A shaft under pure torsion: σx = 0 MPa, σy = 0 MPa, τxy = 50 MPa

1
Center: σavg = 0 MPa. Radius: R = sqrt(0 + 50^2) = 50 MPa
2
Principal stresses: σ1 = 0 + 50 = 50 MPa (tension), σ2 = 0 - 50 = -50 MPa (compression)
3
Principal plane angle: θp = 0.5 x atan2(100, 0) = 0.5 x 90° = 45°. This explains why brittle shafts fail in tension at 45 degrees under torsion.
σ1 = 50 MPa, σ2 = -50 MPa, τmax = 50 MPa, θp = 45.00°
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Example 3: Combined Biaxial and Shear

A pressure vessel lug: σx = 80 MPa, σy = 30 MPa, τxy = 25 MPa

1
Center: σavg = (80 + 30) / 2 = 55 MPa. Half-difference: (80 - 30) / 2 = 25 MPa
2
Radius: R = sqrt(25^2 + 25^2) = sqrt(1250) = 35.36 MPa
3
σ1 = 55 + 35.36 = 90.36 MPa, σ2 = 55 - 35.36 = 19.64 MPa, τmax = 35.36 MPa, θp = 0.5 x atan2(50, 50) = 22.50°
σ1 = 90.36 MPa, σ2 = 19.64 MPa, τmax = 35.36 MPa, θp = 22.50°
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Example 4: Stress Transformation at 30 Degrees

Find stresses on a plane at 30° from the x-axis: σx = 80 MPa, σy = 20 MPa, τxy = 40 MPa

1
σavg = (80 + 20) / 2 = 50 MPa. Half-difference = (80 - 20) / 2 = 30 MPa. Use 2θ = 60°: cos(60°) = 0.5, sin(60°) = 0.866
2
σx' = 50 + 30 x 0.5 + 40 x 0.866 = 50 + 15 + 34.64 = 99.64 MPa
3
σy' = 50 - 30 x 0.5 - 40 x 0.866 = 50 - 15 - 34.64 = 0.36 MPa. τx'y' = -30 x 0.866 + 40 x 0.5 = -25.98 + 20 = -5.98 MPa
σx' = 99.64 MPa, σy' = 0.36 MPa, τx'y' = -5.98 MPa
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❓ Frequently Asked Questions

What is Mohr's circle and what is it used for?+
Mohr's circle is a graphical method for visualizing the state of plane stress at a point. Developed by Otto Mohr in 1882, it maps every possible plane orientation to a point on a circle where x-coordinates equal normal stress and y-coordinates equal shear stress. Engineers use it to find principal stresses, maximum shear stress, and the angles of critical planes in mechanical design, structural analysis, and geotechnical engineering.
How do you find principal stresses using Mohr's circle formulas?+
Calculate the center σavg = (σx + σy) / 2 and the radius R = sqrt(((σx - σy)/2)^2 + τxy^2). The major principal stress σ1 = σavg + R and the minor principal stress σ2 = σavg - R. Both principal stresses act on planes where the shear stress is exactly zero, which are the two points where Mohr's circle crosses the horizontal axis.
What is the maximum shear stress formula from Mohr's circle?+
The maximum in-plane shear stress equals the radius of Mohr's circle: τmax = sqrt(((σx - σy)/2)^2 + τxy^2). This is also equal to (σ1 - σ2) / 2. The maximum shear stress acts on planes oriented 45 degrees from the principal planes in the physical element, which appear as 90 degrees apart on the Mohr's circle diagram.
How do you calculate the principal plane angle θp?+
The principal plane angle is θp = 0.5 times atan2(2τxy, σx - σy). If τxy is positive and σx > σy, θp is positive (counterclockwise from the x-axis). At angle θp, the shear stress on the face reaches zero and the normal stress reaches its maximum value σ1. The perpendicular plane at θp + 90 degrees carries the minimum normal stress σ2.
What are the stress transformation equations in Mohr's circle?+
For a plane rotated by θ from the x-axis: σx' = σavg + ((σx - σy)/2)cos(2θ) + τxy sin(2θ), σy' = σavg - ((σx - σy)/2)cos(2θ) - τxy sin(2θ), and τx'y' = -((σx - σy)/2)sin(2θ) + τxy cos(2θ). These equations are exact for any plane orientation and trace the circumference of Mohr's circle as θ varies from 0 to 180 degrees.
Why does Mohr's circle double the angles?+
Angles are doubled because the stress transformation equations contain sin(2θ) and cos(2θ) rather than sin(θ) and cos(θ). As a result, a physical rotation of θ degrees moves the point around the circle by 2θ degrees. This is why the maximum shear stress planes at 45 degrees from the principal planes physically correspond to 90 degrees on the circle, bringing the point to the top or bottom.
Can Mohr's circle be used for 3D stress analysis?+
Yes. In 3D, three Mohr's circles are constructed for the three principal stress pairs: (σ1, σ2), (σ2, σ3), and (σ1, σ3). The largest circle spans σ1 to σ3 and its radius gives the absolute maximum shear stress = (σ1 - σ3) / 2. The interior region between the three circles represents all reachable stress states for oblique planes. This representation is used in Tresca and von Mises yield criteria.
What does it mean when Mohr's circle is a point (zero radius)?+
A zero-radius Mohr's circle means the stress state is hydrostatic or isotropic: σx = σy and τxy = 0. Every plane through the point carries the same normal stress (equal to σavg) and zero shear stress. This occurs inside a fluid under uniform pressure or in a body submerged in a pressurized vessel where the stress is equal in all directions.
How does Mohr's circle connect to the Tresca and von Mises failure criteria?+
The Tresca criterion predicts yielding when the maximum shear stress τmax = (σ1 - σ2)/2 reaches the shear yield strength Ssy = Sy/2. Since τmax is the radius of Mohr's circle, yield is predicted when the circle's radius reaches Sy/2. The von Mises criterion uses the distortional energy equivalent stress σe = sqrt(σ1^2 - σ1σ2 + σ2^2) and predicts yielding when σe reaches Sy.
How is Mohr's circle used in soil mechanics and geotechnical engineering?+
In geotechnical engineering, Mohr's circle represents the stress state on a soil element. The Mohr-Coulomb failure envelope (a line tangent to failure circles) defines the shear strength as τ = c + σ tan(φ), where c is cohesion and φ is friction angle. If Mohr's circle touches or crosses this envelope, the soil fails by shearing. This criterion drives slope stability analysis, bearing capacity calculations, and lateral earth pressure design.
What sign convention should I use for Mohr's circle inputs?+
This calculator uses the standard solid mechanics convention: tensile normal stresses are positive, compressive normal stresses are negative. For shear stress, τxy is positive when it acts in the positive y-direction on the positive x-face (and simultaneously in the negative y-direction on the negative x-face). Apply this convention consistently to get correct principal stress magnitudes and orientations.
How do principal stresses relate to von Mises stress for yield checking?+
For plane stress (σ3 = 0), the von Mises equivalent stress is σe = sqrt(σ1^2 - σ1 x σ2 + σ2^2). In terms of component stresses: σe = sqrt(σx^2 - σx σy + σy^2 + 3τxy^2). Both formulas give the same result. Yielding is predicted when σe reaches the uniaxial yield strength Sy. This calculator supplies σ1 and σ2 directly for substitution into the von Mises formula.
Tags: Mohr's Circle Calculator Principal Stress Calculator Maximum Shear Stress Calculator Stress Transformation Calculator Plane Stress Analysis Calculator Formula Free Online

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What is Mohr's circle used for in engineering?

Mohr's circle is a graphical method for determining stresses on any plane through a point in a loaded body. Engineers use it to find principal stresses, maximum shear stress, and the orientation of critical planes. It is widely applied in structural, mechanical, and geotechnical engineering for failure analysis, material selection, and design verification.

How do you calculate principal stresses using Mohr's circle formulas?

Compute the center C = (σx + σy) / 2 and radius R = sqrt(((σx - σy)/2)^2 + τxy^2). Then σ1 = C + R and σ2 = C - R. σ1 is the maximum (most tensile) and σ2 is the minimum (most compressive) normal stress. Both act on planes of zero shear stress.

What is the center and radius of Mohr's circle?

The center lies on the horizontal axis at σavg = (σx + σy) / 2. The radius R = sqrt(((σx - σy)/2)^2 + τxy^2) equals the maximum in-plane shear stress. A large radius indicates high stress variation with plane orientation, which is critical for failure assessment.

How do you find the principal plane angle from stress components?

The principal plane angle is θp = 0.5 times atan2(2τxy, σx - σy). The 0.5 factor appears because angles on Mohr's circle are double the physical angles. If τxy = 0 and σx > σy, then θp = 0, meaning the x-face is already a principal plane.

What is the maximum shear stress and how is it calculated?

The maximum in-plane shear stress equals the radius of Mohr's circle: τmax = sqrt(((σx - σy)/2)^2 + τxy^2). It acts on planes oriented 45 degrees from the principal planes. On the circle diagram, maximum shear stress corresponds to the topmost and bottommost points.

What are stress transformation equations?

Stress transformation equations give stresses on a plane rotated by θ: σx' = σavg + ((σx - σy)/2)cos(2θ) + τxy sin(2θ) and τx'y' = -((σx - σy)/2)sin(2θ) + τxy cos(2θ). Each angle θ from 0 to 180 degrees traces a complete revolution around Mohr's circle.

Can Mohr's circle be used for three-dimensional stress analysis?

Yes. In 3D, three Mohr's circles are drawn for the three principal stress pairs (σ1, σ2), (σ2, σ3), and (σ1, σ3). The absolute maximum shear stress is half the range between the largest and smallest principal stresses. This approach underpins the Tresca and von Mises failure criteria.

What does it mean when Mohr's circle degenerates to a point?

A Mohr's circle of zero radius means the stress state is hydrostatic: σx = σy and τxy = 0. Every plane through that point carries the same normal stress and zero shear stress. This occurs inside a fluid under uniform pressure, where pressure acts equally in all directions.

How are principal stresses used in failure analysis?

Principal stresses are the key inputs to failure criteria. The Rankine criterion predicts failure when σ1 reaches tensile strength. The Tresca criterion predicts yielding when τmax = (σ1 - σ2)/2 reaches shear yield strength. Von Mises uses distortional energy derived from all three principal stresses.

What is the sign convention used in this Mohr's circle calculator?

Positive normal stress is tensile; negative is compressive. Positive shear stress τxy acts in the positive y-direction on a positive x-face (and in the negative y-direction on a negative x-face). Apply the same convention consistently for correct principal stress and angle results.

Why are angles doubled on Mohr's circle?

Angles are doubled because the stress transformation equations contain sin(2θ) and cos(2θ) terms rather than sin(θ) and cos(θ). A physical rotation of θ corresponds to a 2θ rotation on the diagram. Therefore, a 45-degree physical rotation to reach the maximum shear stress plane appears as 90 degrees on the circle.

📌 Quick Tips

💡Principal stresses act on planes where shear stress is exactly zero. These are the planes where Mohr's circle intersects the horizontal normal-stress axis.
💡The radius of Mohr's circle equals the maximum in-plane shear stress. A larger radius means higher stress variation across all plane orientations at that point.
💡When σx equals σy and τxy equals zero, the stress state is hydrostatic: Mohr's circle shrinks to a point and every plane carries the same normal stress with zero shear.
💡The principal plane angle θp is half the angle you measure on Mohr's circle. A 90-degree rotation on the diagram equals a 45-degree physical rotation of the element.
💡Maximum shear stress planes are always oriented 45 degrees from the principal planes in the physical element, and 90 degrees from them on the Mohr's circle diagram.