This is a free interactive Mohr's circle calculator for plane stress and plane strain. Enter the stress state (σx, σy, τxy) and the tool draws the circle, computes the principal stresses, maximum shear stress, and principal angles, and shows the transformed stresses on an element rotated to any angle θ. Drag directly on the circle to rotate the element, or type an exact angle. A strain mode handles εx, εy, γxy the same way.
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Mohr's circle is a graphical representation of the plane stress transformation equations. Every point on the circle is the pair (σ, τ) acting on one cut plane through the material point. The circle is centered on the σ axis at the average normal stress and its radius equals the in-plane maximum shear stress.
Center: C = (σx + σy)/2. Radius: R = √(((σx − σy)/2)² + τxy²). Principal stresses: σ1,2 = C ± R. In-plane max shear: τmax = R, acting on planes 45° from the principal planes.
A rotation of the physical element by θ corresponds to a rotation of 2θ around the circle. That factor of two is the most common source of mistakes when reading angles off the diagram, and it is why the principal planes (180° apart on the circle) are 90° apart on the element.
The default plot puts positive shear downward, so a counterclockwise rotation of the element is a counterclockwise sweep on the circle. This matches Hibbeler's and Beer's Mechanics of Materials. If your course plots τ upward, flip the axis with the τ Axis Positive toggle; the numbers do not change, only the picture.
For plane stress, the third principal stress is σ3 = 0 (the free surface). When σ1 and σ2 have the same sign, the largest of the three circles is the one between σmax and 0, so the absolute maximum shear τabs = (σmax − σmin)/2 is larger than the in-plane value. The 3D mode draws all three circles and lets you set σz to a nonzero principal value if the out-of-plane direction is loaded.
Problem: σx = 80 MPa, σy = −40 MPa, τxy = 25 MPa.
Center. C = (80 − 40)/2 = 20 MPa.
Radius. R = √(((80 − (−40))/2)² + 25²) = √(60² + 25²) = 65 MPa.
Principal stresses. σ1 = 20 + 65 = 85 MPa, σ2 = 20 − 65 = −45 MPa.
Principal angle. θp = ½ atan(25/60) = 11.3°. In-plane τmax = 65 MPa at θs = −33.7°.
| State | σ1 | σ2 | In-plane τmax |
|---|---|---|---|
| Uniaxial tension σ | σ | 0 | σ/2 |
| Pure shear τ | +τ | −τ | τ |
| Equal biaxial tension σ | σ | σ | 0 (in-plane); σ/2 absolute |
| Thin-walled pressure vessel | pr/t (hoop) | pr/2t (axial) | pr/4t in-plane |
The transformation equations contain cos 2θ and sin 2θ, so one full trip around the circle (360°) covers only 180° of physical rotation. Normal stresses repeat every 180° of element rotation, which is exactly one revolution of the circle.
Yes. Switch the mode to Strain and enter εx, εy, and the engineering shear strain γxy. The circle plots γ/2 on the vertical axis, so the radius equals γmax/2 and the reported γ values are already converted back to engineering shear strain.
When both in-plane principal stresses have the same sign, the largest Mohr's circle is the one between the bigger principal stress and σ3 = 0. Switch to 3D circles to see it. This matters for failure criteria like Tresca, which uses the absolute maximum shear.
Positive τxy acts in the +y direction on the +x face, the standard convention in US undergraduate texts. Point X plots at (σx, τxy) and point Y at (σy, −τxy).