For a two-dimensional stress state, Mohr's circle reads the principal stresses, the maximum in-plane shear stress, and the stress on a rotated plane straight off a single sketch. Start with the plane-stress components σx\sigma_x, σy\sigma_y, and τxy\tau_{xy}. The circle has center and radius

C=(σx+σy2,0)R=(σxσy2)2+τxy2C = \left(\frac{\sigma_x + \sigma_y}{2}, 0\right) \qquad R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}

where CC lies on the normal-stress axis at the average normal stress and RR is the radius. If all you need is principal stress and maximum in-plane shear stress, these two expressions already give almost everything.

Why The Circle Works

The horizontal axis is normal stress σ\sigma and the vertical axis is shear stress τ\tau. Every point on the circle represents the stress state on some plane through the same material point. The center is the average normal stress; moving left or right changes the normal stress on the plane, and moving up or down changes the shear stress.

This is not an arbitrary picture. It comes directly from the stress-transformation equations: as the physical element rotates by an angle θ\theta, the plotted point sweeps around the circle by 2θ2\theta. The two horizontal intercepts are the principal stresses because shear is zero there, and the top and bottom points give the maximum in-plane shear stress, whose magnitude is exactly the radius. That doubled angle is why the geometry collapses into a clean circle in the first place.

The Formulas You Read Off

Once you know CC and RR, the main results are

σ1=σx+σy2+Rσ2=σx+σy2Rτmax, in-plane=R\sigma_1 = \frac{\sigma_x + \sigma_y}{2} + R \qquad \sigma_2 = \frac{\sigma_x + \sigma_y}{2} - R \qquad \tau_{max,\ in\text{-}plane} = R

These results are for the plane-stress picture shown here. If the stress state is fully three-dimensional, the overall maximum shear stress depends on the full set of principal stresses, not just this single circle. For the principal-plane angle, use

tan(2θp)=2τxyσxσy\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}

when σxσy\sigma_x \ne \sigma_y and your sign convention matches the circle you drew.

Worked Example: Find The Principal Stresses

Suppose a point in a plate has

σx=80 MPa,σy=20 MPa,τxy=30 MPa\sigma_x = 80\ \mathrm{MPa}, \quad \sigma_y = 20\ \mathrm{MPa}, \quad \tau_{xy} = 30\ \mathrm{MPa}

Center:

C=(80+202,0)=(50,0)C = \left(\frac{80 + 20}{2}, 0\right) = (50, 0)

Radius:

R=(80202)2+302=302+302=180042.4 MPaR = \sqrt{\left(\frac{80 - 20}{2}\right)^2 + 30^2} = \sqrt{30^2 + 30^2} = \sqrt{1800} \approx 42.4\ \mathrm{MPa}

Read off the results:

σ1=50+42.4=92.4 MPaσ2=5042.4=7.6 MPaτmax, in-plane=42.4 MPa\sigma_1 = 50 + 42.4 = 92.4\ \mathrm{MPa} \qquad \sigma_2 = 50 - 42.4 = 7.6\ \mathrm{MPa} \qquad \tau_{max,\ in\text{-}plane} = 42.4\ \mathrm{MPa}

So the stress state has one large tensile principal stress, one much smaller tensile principal stress, and a maximum in-plane shear stress of 42.4 MPa42.4\ \mathrm{MPa}. One sketch shows the important extremes immediately.

Check It Yourself

Set τxy=0\tau_{xy} = 0 in the formulas: the radius collapses to half the difference of the normal stresses, and the principal stresses become σx\sigma_x and σy\sigma_y themselves. That is the degenerate case where the original axes are already the principal axes, and it is a fast way to confirm your center and radius logic. Try the same numbers with σx=σy\sigma_x = \sigma_y and watch the center sit exactly between equal normal stresses while R=τxyR = \tau_{xy}.

Calculation Traps To Avoid

Do not use the standard classroom circle blindly. The construction here assumes plane stress, so it is not the complete answer for a general 3D stress state.

Do not confuse the center with one of the original stresses. The center is the average normal stress, so it only matches σx\sigma_x or σy\sigma_y in special cases.

Do not mix up the radius and the intercepts. The radius gives the maximum in-plane shear stress, while the principal stresses are C+RC + R and CRC - R.

Keep the same sign convention for shear in both the transformation formulas and the sketch. Different textbooks place positive shear in opposite vertical directions, so the circle may appear reflected; if the convention is consistent, the principal stress values still agree, but a mismatch makes the angle or the plotted points come out mirrored.

Where Mohr's Circle Is Used

Mohr's circle appears in mechanics of materials, machine design, structural analysis, and failure analysis. It is especially useful when a part carries combined loading, such as bending plus torsion or tension plus shear. Even when software does the arithmetic, the circle still shows what is large, what drops to zero, and which planes are most critical.

Frequently Asked Questions

Frequently Asked Questions

What is Mohr's circle used for?
Mohr's circle is a graph for a two-dimensional stress state. In plane stress, it lets you read the principal stresses, the maximum in-plane shear stress, and the stress on a rotated plane without recomputing the full transformation each time. Every point on the circle represents the stress on some plane through the same material point.
How do you find the center and radius of Mohr's circle?
Start with the plane-stress components sigma x, sigma y, and tau xy. The center sits on the normal-stress axis at the average normal stress, half of sigma x plus sigma y. The radius is the square root of the squared half-difference of the normal stresses plus the squared shear stress tau xy.
How do you find principal stresses from Mohr's circle?
The principal stresses are the two horizontal intercepts of the circle, where shear stress is zero. Numerically, sigma 1 equals the average normal stress plus the radius, and sigma 2 equals the average normal stress minus the radius. For example, with sigma x of 80 MPa, sigma y of 20 MPa, the center is at 50 MPa.
What is the maximum in-plane shear stress on Mohr's circle?
The maximum in-plane shear stress equals the radius of the circle, and it occurs at the top and bottom points. Note this is the in-plane result only: if the stress state is fully three-dimensional, the overall maximum shear stress depends on the full set of principal stresses, not just this single circle.
How do you find the principal-plane angle?
Use the stress-transformation relation: tangent of twice the principal angle equals two tau xy divided by sigma x minus sigma y. This applies when sigma x is not equal to sigma y and your sign convention matches the circle you drew. The angle locates the plane on which the principal stresses act.

Need help with a problem?

Upload your question and get a verified, step-by-step solution in seconds.

Open GPAI Solver →