Mohr's circle for plane stress and plane strain

Why it matters

A drive shaft under bending and torsion, a pressure vessel wall, a weld under combined load: the stresses you calculate on x and y faces are rarely the largest ones. Failure happens on whichever plane carries the highest normal or shear stress. Mohr's circle finds those planes and values in one sketch, and the same circle turns strain-gauge readings into stresses.

Key ideas

Plane stress. A state where all stresses act in one plane: σ_z = τ_xz = τ_yz = 0. It applies to thin plates, sheet-metal panels and the free surface of any part (where strain gauges are fixed). The third principal stress is then zero, but the third strain ε_z is not.

Plane strain. A state where ε_z = γ_xz = γ_yz = 0, found in long bodies constrained along their length (a long thick cylinder away from its ends, a dam, a rolled strip). Here σ_z = ν·(σ_x + σ_y) is not zero.

Stress transformation. Rotating the element by θ (counter-clockwise from x) changes the stresses on its faces. Both the transformation equations and Mohr's circle give the same answers; the circle is just a picture of them.

Constructing Mohr's circle for stress.

1. Axes: σ horizontal (tension positive), τ vertical. Use one shear convention and stick to it. A common convention: plot the x-face as point X(σ_x, τ_xy) and the y-face as Y(σ_y, −τ_xy), with τ_xy positive when it acts in +y on the +x face, and τ plotted downward-positive so that rotations on the circle and on the element are in the same sense. Textbooks differ; what matters is consistency.
2. Centre C at (σ_avg, 0) with σ_avg = (σ_x + σ_y)/2. Radius R = CX.
3. A rotation of θ on the element corresponds to 2θ on the circle in the same sense.
4. The points where the circle cuts the σ-axis are the principal stresses σ₁ and σ₂ (zero shear on those planes). The top and bottom of the circle give the maximum in-plane shear τ = R, on planes at 45° to the principal planes, where the normal stress is σ_avg, not zero.

Special cases.

• Pure shear (σ_x = σ_y = 0, τ_xy = τ): centre at the origin, R = τ, principal stresses +τ and −τ on planes at 45°. This is why a brittle shaft in torsion cracks along a 45° helix.
• Equal biaxial stress (σ_x = σ_y, τ = 0): the circle shrinks to a point; every in-plane direction is principal.

Absolute maximum shear. In 3D there are three principal stresses and three circles. For plane stress the third principal stress is 0. If σ₁ and σ₂ have opposite signs, the in-plane circle is the largest and τ_abs max = (σ₁ − σ₂)/2. If both are tensile (or both compressive), the largest circle runs from 0 to σ₁, so τ_abs max = σ₁/2, larger than the in-plane value. This matters for thin pressure vessels and for Tresca's criterion.

Mohr's circle for strain. Identical construction with ε in place of σ and γ/2 in place of τ. Principal strain directions coincide with principal stress directions for isotropic materials. Strain rosettes (three gauges at 0°, 45°, 90° or 0°, 60°, 120°) give ε_x, ε_y and γ_xy; the circle then gives principal strains, and the plane-stress Hooke's law converts them to stresses.

Formulas

σ_θ = (σ_x + σ_y)/2 + [(σ_x − σ_y)/2]·cos 2θ + τ_xy·sin 2θ τ_θ = −[(σ_x − σ_y)/2]·sin 2θ + τ_xy·cos 2θ

• σ_x, σ_y: normal stresses (Pa); τ_xy: shear stress (Pa); θ: angle of the plane's normal from x, counter-clockwise.

σ₁,₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²]

• Principal stresses. σ₁ + σ₂ = σ_x + σ_y (the first invariant).

tan 2θ_p = 2·τ_xy / (σ_x − σ_y)

• θ_p: principal plane angle. Gives two planes 90° apart; substitute back to see which carries σ₁.

τ_max,in-plane = R = √[((σ_x − σ_y)/2)² + τ_xy²], on planes at θ_p ± 45°

• Normal stress on those planes is σ_avg.

τ_abs,max = max(|σ₁ − σ₂|, |σ₁|, |σ₂|) / 2

• Plane stress, with σ₃ = 0.

ε₁,₂ = (ε_x + ε_y)/2 ± √[((ε_x − ε_y)/2)² + (γ_xy/2)²], tan 2θ_p = γ_xy / (ε_x − ε_y)

• Strains (dimensionless); γ_xy: engineering shear strain (rad).

σ₁ = E·(ε₁ + ν·ε₂)/(1 − ν²), σ₂ = E·(ε₂ + ν·ε₁)/(1 − ν²)

• Plane-stress conversion from principal strains to principal stresses.

Worked examples

Example 1 (standard): principal stresses and a rotated plane. Given: σ_x = 80 MPa, σ_y = −40 MPa, τ_xy = 30 MPa. Find the principal stresses and their directions, the maximum in-plane shear, and the stresses on a plane at θ = 30°.

1. σ_avg = (80 − 40)/2 = 20 MPa. R = √(60² + 30²) = √4500 = 67.08 MPa.
2. σ₁ = 20 + 67.08 = 87.08 MPa; σ₂ = 20 − 67.08 = −47.08 MPa.
3. tan 2θ_p = 2 × 30 / 120 = 0.5, so 2θ_p = 26.57° and θ_p = 13.28° (this plane carries σ₁, since substituting gives 87.08 MPa).
4. τ_max,in-plane = 67.08 MPa, on planes at 13.28° + 45° = 58.28° and −31.72°, with normal stress 20 MPa.
5. At θ = 30°: σ_θ = 20 + 60·cos 60° + 30·sin 60° = 20 + 30 + 25.98 = 75.98 MPa; τ_θ = −60·sin 60° + 30·cos 60° = −51.96 + 15 = −36.96 MPa. Answer: σ₁ = 87.1 MPa at 13.3°, σ₂ = −47.1 MPa, τ_max = 67.1 MPa; on the 30° plane σ = 76.0 MPa and τ = −37.0 MPa.

Example 2 (GATE level): in-plane versus absolute maximum shear. Given: plane stress σ_x = 120 MPa, σ_y = 50 MPa, τ_xy = 30 MPa. Find the in-plane and absolute maximum shear stresses.

1. σ_avg = 85 MPa; R = √(35² + 30²) = √2125 = 46.10 MPa.
2. σ₁ = 131.10 MPa, σ₂ = 38.90 MPa, σ₃ = 0.
3. In-plane maximum shear = R = 46.10 MPa.
4. Both in-plane principal stresses are tensile, so the largest circle is the one between σ₃ = 0 and σ₁: τ_abs max = 131.10 / 2 = 65.55 MPa. Answer: in-plane 46.1 MPa; absolute 65.5 MPa. Using the in-plane value in Tresca's criterion here would be unsafe.

Example 3: from strain gauges to stresses. Given: on a steel surface (E = 200 GPa, ν = 0.3), ε_x = 600 × 10⁻⁶, ε_y = −200 × 10⁻⁶, γ_xy = 400 × 10⁻⁶.

1. Centre (600 − 200)/2 = 200 × 10⁻⁶; radius √(400² + 200²) × 10⁻⁶ = 447.2 × 10⁻⁶.
2. ε₁ = 647.2 × 10⁻⁶, ε₂ = −247.2 × 10⁻⁶; tan 2θ_p = 400/800, so θ_p = 13.3°.
3. E/(1 − ν²) = 219 780 MPa. σ₁ = 219 780 × (647.2 − 0.3 × 247.2) × 10⁻⁶ = 125.9 MPa; σ₂ = 219 780 × (−247.2 + 0.3 × 647.2) × 10⁻⁶ = −11.7 MPa. Answer: σ₁ ≈ 125.9 MPa, σ₂ ≈ −11.7 MPa, at 13.3° from x.

Common mistakes

• Forgetting that angles double on the circle (2θ), or rotating the wrong way.
• Plotting X and Y with the same shear sign; they must be on opposite sides of the σ-axis.
• Using γ instead of γ/2 on the strain circle.
• Assuming normal stress is zero on the maximum-shear planes; it equals σ_avg.
• Quoting the in-plane maximum shear as the absolute maximum when both principal stresses have the same sign.
• Picking the wrong root of tan 2θ_p: always substitute back to see which plane carries σ₁.

For GATE ME

Expect principal stresses and maximum shear for given σ_x, σ_y, τ_xy, the angle of the principal plane, pure shear and equal biaxial special cases, the radius or centre of the circle as a direct question, and conversion of strain-gauge readings. Practise sketching the circle quickly to check signs and to spot when the absolute maximum shear differs from the in-plane value.

Quick check

1. What are the principal stresses for pure shear of 50 MPa?
2. σ_x = σ_y = 60 MPa, τ_xy = 0. What is the radius of Mohr's circle?
3. On the strain circle, what is plotted on the vertical axis?
4. σ₁ = 100 MPa, σ₂ = 40 MPa in plane stress. Find the absolute maximum shear.

Answers: 1. +50 MPa and −50 MPa at 45°. 2. Zero (a point). 3. γ/2. 4. 50 MPa (σ₁/2, using σ₃ = 0).