Mohr’s Circle Explained: Principal Stresses & Maximum Shear
Construct Mohr’s circle from any plane-stress state (σx, σy, τxy), read off principal stresses, maximum shear, and plane orientations — with verified worked examples.
Key Engineering Takeaways
- The circle is defined by center C = (σ_avg, 0) and radius R = √[((σx − σy)/2)² + τxy²].
- Principal stresses σ₁,₂ = σ_avg ± R act on planes oriented at θp where tan(2θp) = 2τxy / (σx − σy).
- Maximum shear stress equals the circle radius, τ_max = R, and acts on planes 45° from the principal planes.
- Invariant self-check: I₁ = σx + σy = σ₁ + σ₂ — the sum never changes with rotation.
1. What is Mohr’s Circle?
Mohr’s circle is a graphical representation of the plane-stress transformation equations. Every stress state (σx, σy, τxy) maps to a circle whose geometry encodes the principal stresses, the maximum shear stress, and the orientation of the principal planes — without solving any transcendental equations.
2. Sign Conventions
- Tensile normal stress is positive (compression negative) on the σ-axis.
- Counter-clockwise shear is positive on the face, plotted positive on the τ-axis.
- Point A(σx, τxy) represents the x-face; point B(σy, −τxy) represents the y-face — the shear components are equal and opposite across perpendicular faces.
3. Construction: Center and Radius
Step 1 — Plot the two face points:
A(σx, τxy) and B(σy, −τxy)
Step 2 — Find the center C (midpoint of AB):
C = (σ_avg, 0) where σ_avg = (σx + σy) / 2
Step 3 — Radius R (distance C → A):
R = √[ ((σx − σy)/2)² + τxy² ]
4. Principal Stresses & Orientation
- Principal stresses: σ₁,₂ = σ_avg ± R, the two intersections of the circle with the σ-axis.
- Principal plane orientation: tan(2θp) = 2τxy / (σx − σy) — rotate the σ-axis 2θp from the radius to A.
- Maximum shear: τ_max = R, acting on planes rotated 45° from the principal planes (θp + 45°).
- Invariant: I₁ = σx + σy = σ₁ + σ₂ — a built-in arithmetic self-check.
5. Worked Example: Biaxial Tension
Take σx = 80 MPa, σy = 40 MPa, τxy = 20 MPa — the standard example in Mohr Master's suite.
σ_avg = (80 + 40)/2 = 60 MPa
R = √[ ((80 − 40)/2)² + 20² ] = √(400 + 400) = 28.28 MPa
σ₁ = 60 + 28.28 = 88.28 MPa · σ₂ = 60 − 28.28 = 31.72 MPa
τ_max = R = 28.28 MPa
tan(2θp) = 2·20 / (80 − 40) = 1 ⟹ θp = 22.5°
Invariant check:
I₁ = 80 + 40 = 120 = 88.28 + 31.72 = 120 ✓
6. Worked Example: Pure Shear
For a shaft in pure torsion, σx = σy = 0 with τxy = 50 MPa:
σ_avg = 0 ⟹ center at the origin (0, 0)
R = √[ 0² + 50² ] = 50 MPa
σ₁ = +50 MPa · σ₂ = −50 MPa
θp = 45° — principal planes bisect the shear faces
Pure shear is the classic result: equal and opposite principal stresses of magnitude τ_max at 45° — exactly why ductile shafts in torsion fail along 45° helices.
Verify your stress transformations in real time.
Build the circle from σx, σy and τxy, read principal stresses, max shear and plane orientations off the interactive diagram, and follow Formula → Substitution → Result steps instantly.
Textbook References & Verification
- 01
Hibbeler, R.C. — Mechanics of Materials
Plane-stress transformation equations and Mohr’s circle construction; worked numbers cross-checked in Mohr Master.
- 02
Beer & Johnston — Mechanics of Materials
Center/radius construction and sign conventions for τxy on the two face points.
- 03
Philpot, T.A. — Mechanics of Materials, Ch. 13
Stress transformation and principal-orientation derivations verified against Mohr Master modes.