Common Mistakes When Using Mohr’s Circle 2D (and How to Avoid Them)

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.

1. 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

2. Principal stresses:

   • σ1 = 30 + 58.31 = 88.31 MPa
   • σ2 = 30 − 58.31 = −28.31 MPa

3. Maximum shear:

   • τmax = 58.31 MPa (occurs on planes at 45° from principal planes)

4. 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.

1. σavg = 0, (σx − σy)/2 = 0, R = sqrt(0 + 50^2) = 50 MPa.
2. Principal stresses:
   • σ1 = 0 + 50 = 50 MPa
   • σ2 = 0 − 50 = −50 MPa
3. τmax = 50 MPa (occurs at planes 45° from x-axis).
4. 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.

1. σavg = (40 + 40)/2 = 40 MPa; (σx − σy)/2 = 0.
2. R = sqrt(0 + 20^2) = 20 MPa.
3. Principal stresses:
   • σ1 = 40 + 20 = 60 MPa
   • σ2 = 40 − 20 = 20 MPa
4. τmax = 20 MPa.
5. 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)

1. Plot point A = (σx, τxy) and point B = (σy, −τxy) on the σ–τ plane.
2. The circle center is at σavg on the σ-axis; radius R as above.
3. Principal stresses are intersection points of circle with σ-axis (τ = 0): σ1 and σ2.
4. 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)

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.