Mohr's Circle for Stress

Turn the transformation equations into a geometric construction you can sketch in seconds.

Mechanics of MaterialsMechanical Engineering Year 2Free preview
⏱️ About 22 min

If you had to find principal stresses for a hundred different points on a turbine blade, you'd never want to re-derive the transformation equations every time. Mohr's circle compresses the entire algebra into one geometric sketch: draw a diameter, find the centre, read the radius — and every quantity you need is visible at a glance.

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The big idea: Mohr's circle is not a new theory — it is the stress-transformation equations plotted parametrically. Every point on the circle represents the normal and shear stress on one particular plane through the point. Rotating the physical element by an angle θ corresponds to moving around the circle by 2θ, which is why the graphical method is so compact.
🎯 By the end, you'll be able to
  • Construct Mohr's circle from a given σx, σy, τxy state
  • Read principal stresses, principal angles, and maximum in-plane shear directly from the circle
  • Distinguish in-plane maximum shear from the absolute maximum shear in three dimensions
  • Apply the Hibbeler sign convention consistently on the circle
📎 Helpful to know first

From equations to geometry

Solve the transformation equations for sin(2θ) and cos(2θ), substitute back, and eliminate θ. The result is the equation of a circle in (σ, τ) space:

(σx′ − σavg)² + τx′y′² = R²

That means every possible (σ, τ) pair at the point lies on a circle centred at (σavg, 0) with radius R. The circle's intersection with the σ-axis gives the principal stresses. The top and bottom of the circle give the maximum in-plane shear. Everything you need is one sketch away.

🔑 Locked sign convention: Hibbeler (positive shear plots downward)

We adopt the convention used in Hibbeler's Mechanics of Materials — the standard undergraduate text — and we apply it everywhere in this module:

  • Positive shear stress acts on the positive x-face in the positive y-direction.
  • On Mohr's circle, positive shear on the reference (x) face plots downward.
  • A rotation of 2θ on the circle corresponds to a rotation of θ on the element, and the sense is the same: counterclockwise on the element means counterclockwise on the circle.

Using this convention, point X is plotted at (σx, τxy) with τxy measured downward if positive. Point Y — the y-face — is plotted at (σy, −τxy), directly across the diameter. The centre lies on the σ-axis at σavg = (σx + σy)/2.

Mohr's circle for a plane-stress state with positive shear plotted downward. The diameter joins point X (sigma-x, tau-xy) and point Y (sigma-y, minus tau-xy). Principal stresses sigma-1 and sigma-2 lie on the sigma-axis; the radius R equals the maximum in-plane shear stress. σ τ (+ down) C = σavg X (σx, τxy) Y (σy, −τxy) σ1 σ2 R 2θ
Mohr's circle construction with the Hibbeler convention: positive τ on the reference face plots below the σ-axis.

Reading the circle

  • Principal stresses: where the circle crosses the σ-axis. σ1 = σavg + R (rightmost), σ2 = σavg − R (leftmost).
  • Principal angle: the angle from point X to the σ1 intersection, measured on the circle, is 2θp. Half that angle gives θp on the element, same sense.
  • Maximum in-plane shear: the top and bottom of the circle, at τ = ±R. The normal stress on those planes is σavg — read directly from the centre's horizontal coordinate.
  • Any rotated plane: move around the circle by 2θ from X; the coordinates of that point are (σx′, τx′y′) on the transformed plane.
⚠️ Mohr's circle is at one point, not along a member

A common and serious misconception: students sometimes think Mohr's circle describes how stress varies along the length of a beam or shaft. It does not. Mohr's circle represents the stress state at a single point under rotation of the viewing plane. Every point on the circle is a different plane through the same material point. If you move to a different point in the body, you draw a completely new circle.

In-plane vs absolute maximum shear

The circle we've drawn lives in the x-y plane. The maximum shear it shows — R — is the largest shear stress acting on any plane perpendicular to the x-y plane. But in three dimensions there is also the out-of-plane principal stress σ3 = 0 for plane stress.

The absolute maximum shear stress at the point is the largest of:

  • |σ1 − σ2| / 2 = R (the in-plane value)
  • |σ1 − σ3| / 2 = |σ1| / 2
  • |σ2 − σ3| / 2 = |σ2| / 2

For most plane-stress states where σ1 and σ2 have the same sign, the absolute maximum shear is actually |σ1|/2 (or |σ2|/2), occurring on a plane that is not in the original x-y plane. This matters when you are designing against shear failure in ductile materials.

🎮 Interactive: build and read Mohr's circle LIVE
Predict first: Set σx = 60, σy = 20, τxy = 40. Read σ1, σ2, and R from the circle. Does the principal angle agree with tan(2θp) = τxy / ((σx−σy)/2)?

An interactive Mohr's circle simulator with sliders for sigma-x, sigma-y, and tau-xy, showing the circle, principal stresses, and a rotation slider that moves the transformed point around the circumference.

Adjust the stress state and watch the circle resize and shift. The principal-stress points, radius, and rotated point all update live.
📝 Worked example: A stress element has σx = 60 MPa, σy = 20 MPa, and τxy = 40 MPa. Construct Mohr's circle and read the principal stresses, principal angle, and maximum in-plane shear stress.
  1. Centre: σavg = (60 + 20)/2 = 40 MPa.
  2. Radius R = √[((60 − 20)/2)² + 40²] = √[20² + 40²] = √2000 ≈ 44.72 MPa.
  3. Principal stresses: σ1 = 40 + 44.72 = 84.72 MPa; σ2 = 40 − 44.72 = −4.72 MPa.
  4. Sanity check: σ1 + σ2 = 80 MPa = σx + σy — confirmed.
  5. On the circle, point X is at (60, +40) with τ plotted downward. The angle from X to σ1 is 2θp = arctan(40/20) = 63.43° CCW.
  6. Principal angle: θp = 31.7° CCW from the x-axis.
  7. Maximum in-plane shear: τmax = R = 44.72 MPa, on planes at θp + 45° ≈ 76.7°.
  8. Absolute maximum shear: since σ1 and σ2 have opposite signs, the absolute max is |σ1 − σ2|/2 = R = 44.72 MPa — same as in-plane here.
✓ σ1 ≈ 84.7 MPa, σ2 ≈ −4.7 MPa, θp ≈ 31.7° CCW, τmax(in-plane) ≈ 44.7 MPa.
✏️ Practice: A stress element has σx = 60 MPa, σy = 20 MPa, τxy = 40 MPa. What is the radius R of Mohr's circle, in MPa?
MPa
Solution
  1. R = √[((60−20)/2)² + 40²] = √[400 + 1600] = √2000 ≈ 44.72 MPa.
✏️ Practice: For the same state (σx = 60 MPa, σy = 20 MPa, τxy = 40 MPa), what is the principal angle θp, to the nearest degree, measured from the x-axis to the plane of σ1?
°
Solution
  1. tan(2θp) = τxy / ((σx−σy)/2) = 40 / 20 = 2.
  2. 2θp = arctan(2) = 63.43°.
  3. θp = 31.7° ≈ 32° CCW from the x-axis.

Check your understanding

1. Mohr's circle represents the stress state:
Every point on the circle is a different plane through the same material point — not a different point in the body.
2. With the Hibbeler convention, where is positive shear stress plotted on Mohr's circle?
In the Hibbeler convention, positive shear on the reference (x) face plots downward, below the σ-axis.
3. A rotation of θ on the stress element corresponds to what rotation on Mohr's circle?
The transformation equations contain 2θ, so a physical rotation of θ maps to a rotation of 2θ around the circle, and the sense is the same (CCW stays CCW).
🔎 Related lessons
Plane Stress Transformation Equations Combined Loading: Axial + Bending + Torsion + Shear · soon
✅ Key takeaways
  • Mohr's circle is the stress-transformation equations plotted in (σ, τ) space.
  • Centre = σavg; radius R = maximum in-plane shear; σ-axis intersections = principal stresses.
  • Hibbeler convention: positive shear on the reference face plots downward; 2θ on circle = θ on element, same sense.
  • Absolute maximum shear may exceed the in-plane value when σ1 and σ2 have the same sign.
➡️ With stress transformation mastered, the next step is to connect stress to strain for multi-axial loading — Generalized Hooke's Law.
Want to test yourself on this? Try the Mechanical Aptitude test →
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