Mohr’s Circle 2D Worked Examples: From Stress Components to Principal Values
Mohr’s Circle is a graphical method that makes stress transformation intuitive and helps find principal stresses and maximum shear stress for a 2D stress state. Below are concise worked examples that take you from given stress components to principal stresses, orientations, and maximum shear.
Key formulas (2D plane stress)
- Normal and shear stresses on an arbitrary plane at angle θ:
σn = (σx + σy)/2 + (σx – σy)/2cos(2θ) + τxy * sin(2θ)
τn = −(σx – σy)/2 * sin(2θ) + τxy * cos(2θ) - Principal stresses (σ1, σ2):
σavg = (σx + σy)/2
R = sqrt(((σx – σy)/2)^2 + τxy^2)
σ1 = σavg + R
σ2 = σavg − R - Maximum in-plane shear stress:
τmax = R - Principal plane angle (θp) measured from x-axis to principal plane:
tan(2θp) = 2τxy / (σx − σy)
Example 1 — Tension/compression with shear
Given: σx = 80 MPa (tension), σy = −20 MPa (compression), τxy = 30 MPa.
-
Compute averages and radius:
- σavg = (80 + (−20))/2 = 30 MPa
- (σx − σy)/2 = (80 − (−20))/2 = 50 MPa
- R = sqrt(50^2 + 30^2) = sqrt(2500 + 900) = sqrt(3400) ≈ 58.31 MPa
-
Principal stresses:
- σ1 = 30 + 58.31 = 88.31 MPa
- σ2 = 30 − 58.31 = −28.31 MPa
-
Maximum shear:
- τmax = 58.31 MPa (occurs on planes at 45° from principal planes)
-
Principal angle:
- tan(2θp) = 2·30/(80 − (−20)) = ⁄100 = 0.6 → 2θp = arctan(0.6) ≈ 30.96° → θp ≈ 15.48°
- Principal plane orientations: θp ≈ 15.5° and θp + 90° ≈ 105.5°
Interpretation: The material experiences one tensile principal stress (~88.3 MPa) and one compressive (~−28.3 MPa). Maximum shear magnitude is ~58.3 MPa.
Example 2 — Pure shear
Given: σx = 0, σy = 0, τxy = 50 MPa.
- σavg = 0, (σx − σy)/2 = 0, R = sqrt(0 + 50^2) = 50 MPa.
- Principal stresses:
- σ1 = 0 + 50 = 50 MPa
- σ2 = 0 − 50 = −50 MPa
- τmax = 50 MPa (occurs at planes 45° from x-axis).
- Principal angle:
- tan(2θp) = 2·50/(0 − 0) → infinite → 2θp = 90° → θp = 45°.
Interpretation: Pure shear transforms to equal and opposite principal normal stresses ±50 MPa at 45°.
Example 3 — Equal biaxial tension with shear
Given: σx = 40 MPa, σy = 40 MPa, τxy = 20 MPa.
- σavg = (40 + 40)/2 = 40 MPa; (σx − σy)/2 = 0.
- R = sqrt(0 + 20^2) = 20 MPa.
- Principal stresses:
- σ1 = 40 + 20 = 60 MPa
- σ2 = 40 − 20 = 20 MPa
- τmax = 20 MPa.
- Principal angle:
- tan(2θp) = 2·20/(40 − 40) → infinite → θp = 45°.
Interpretation: With equal biaxial normal stress, shear alone splits the stresses into 60 and 20 MPa along ±45°.
How to draw Mohr’s Circle (quick steps)
- Plot point A = (σx, τxy) and point B = (σy, −τxy) on the σ–τ plane.
- The circle center is at σavg on the σ-axis; radius R as above.
- Principal stresses are intersection points of circle with σ-axis (τ = 0): σ1 and σ2.
- Points on circle give stresses on planes at various θ; angle on circle is 2θ (graphical relation).
Quick checklist for using formulas
- Use consistent sign convention for τxy (positive if shear on x-face causing counterclockwise positive rotation).
- Ensure units match (MPa, psi).
- For angles: arctan gives 2θ; divide by 2 and add 90° as needed for the second principal plane.
- Maximum shear plane orientations are ±45° from principal planes.
Summary table (results from examples)
| Example | σx (MPa) | σy (MPa) | τxy (MPa) | σ1 (MPa) | σ2 (MPa) | τmax (MPa) | θp (°) |
|---|---|---|---|---|---|---|---|
| 1 | 80 | −20 | 30 | 88.31 | −28.31 | 58.31 | 15.48 |
| 2 | 0 | 0 | 50 | 50 | −50 | 50 | 45.00 |
| 3 | 40 | 40 | 20 | 60 | 20 | 20 | 45.00 |
If you want, I can provide step-by-step plots of the Mohr’s circles for these examples or a short Python script to compute and draw them.