Mohr's circle for plane stress and plane strain

Why it matters

Real parts are loaded in several ways at once: a motor shaft carries bending and torsion, a robot base plate carries bending in two directions plus shear. The stresses you calculate on x and y faces are not the largest ones; the critical stress acts on some inclined plane. Mohr's circle finds those planes and values in one picture, and the same circle for strain is how a three-gauge strain rosette, the standard experimental tool, is turned into principal stresses.

Key ideas

Plane stress. All stresses lie in one plane: σ_x, σ_y and τ_xy are non-zero, and σ_z = τ_xz = τ_yz = 0. It applies to thin plates, sheet-metal parts and the free surface of any component, which is exactly where strain gauges sit and where cracks usually start. The out-of-plane strain ε_z is not zero (Poisson effect).

Plane strain. All strains lie in one plane: ε_z = γ_xz = γ_yz = 0. It applies to long bodies restrained along their length, such as a long dam, a retaining wall or a long pressurised pipe section away from its ends. The out-of-plane stress is not zero: σ_z = ν·(σ_x + σ_y).

Stress transformation. The state of stress at a point is fixed, but the normal and shear stresses on a plane depend on its orientation θ. The transformation formulas come from equilibrium of a small wedge only, so they hold for any material.

Sign convention used here. Tension positive. τ_xy positive when it acts in the +y direction on the +x face. θ measured anticlockwise from the x axis to the normal of the plane. On Mohr's circle, plot the x-face point as (σ_x, −τ_xy) and the y-face as (σ_y, +τ_xy) with τ positive upward, or equivalently plot (σ_x, τ_xy) with τ positive downward; either way a rotation of θ in the element is a rotation of 2θ on the circle in the same sense.

Mohr's circle for stress. Centre C = (σ_x + σ_y)/2 on the σ axis; radius R = √[((σ_x − σ_y)/2)² + τ_xy²]. Points on the circle are the (σ, τ) pairs on all planes through the point. Its properties:

• Principal stresses σ₁ = C + R and σ₂ = C − R, where the circle cuts the σ axis; shear stress on principal planes is zero.
• Principal planes are 90° apart in the body (180° on the circle).
• Maximum in-plane shear stress = R, on planes at 45° to the principal planes, with normal stress C on them.
• σ_x + σ_y = σ₁ + σ₂ is invariant (the same for every orientation).
• Special cases: pure shear gives a circle centred at the origin with σ₁ = +τ, σ₂ = −τ at 45°; equal biaxial stress with no shear gives a point (every plane is principal).

Absolute maximum shear. In plane stress the third principal stress is σ₃ = 0. If σ₁ and σ₂ have the same sign, the largest shear stress is not R but max(|σ₁|, |σ₂|)/2, on planes inclined out of the x-y plane. Draw all three circles (σ₁-σ₂, σ₁-0, σ₂-0) to see it.

Mohr's circle for strain. Identical with ε in place of σ and γ/2 (not γ) in place of τ: centre (ε_x + ε_y)/2, radius √[((ε_x − ε_y)/2)² + (γ_xy/2)²]. The maximum in-plane shear strain is γ_max = 2R = ε₁ − ε₂. Principal strain directions coincide with principal stress directions in an isotropic material.

Strain rosettes. A gauge measures normal strain in one direction only, so three gauges are needed to find ε_x, ε_y and γ_xy. For a 0°-45°-90° (rectangular) rosette: ε_x = ε₀, ε_y = ε₉₀, γ_xy = 2ε₄₅ − ε₀ − ε₉₀. Principal stresses then follow from the plane-stress Hooke's law.

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_θ, τ_θ: normal and shear stress on a plane whose normal is at θ to x (Pa); σ_x, σ_y, τ_xy: given stresses (Pa).

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

• Principal stresses (Pa).

tan 2θ_p = 2τ_xy / (σ_x − σ_y)

• θ_p: angle from x to a principal plane; check which of θ_p and θ_p + 90° gives σ₁ by substituting into σ_θ.

τ_max,in-plane = √[((σ_x − σ_y)/2)² + τ_xy²] = (σ₁ − σ₂)/2

• On planes at θ_p ± 45°.

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)

• Principal strains; γ_xy: engineering shear strain (rad).

γ_xy = 2ε₄₅ − ε₀ − ε₉₀

• Rectangular rosette.

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

• Plane-stress Hooke's law; E (Pa), ν (dimensionless).

Worked examples

Example 1 (standard): stresses on an inclined plane and principal stresses. Given: at a point on a bracket, σ_x = 80 MPa, σ_y = −40 MPa, τ_xy = 30 MPa. Find the principal stresses and planes, the maximum shear stress, and the stresses on the plane at θ = 30°.

1. Centre: C = (80 − 40)/2 = 20 MPa. Radius: R = √(60² + 30²) = 67.08 MPa.
2. σ₁ = C + R = 87.08 MPa; σ₂ = C − R = −47.08 MPa.
3. tan 2θ_p = 2 × 30/120 = 0.5, so 2θ_p = 26.57° and θ_p = 13.28°. Substituting θ = 13.28° into σ_θ gives 87.08 MPa, so σ₁ acts on the plane at 13.28° anticlockwise from x.
4. In-plane τ_max = R = 67.08 MPa, at θ = 13.28° ± 45°, with normal stress 20 MPa. Since σ₁ and σ₂ have opposite signs, this is also the absolute maximum (σ₃ = 0 lies inside the main circle).
5. Plane at 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, σ₂ ≈ −47.1 MPa at θ_p ≈ 13.3°; τ_max ≈ 67.1 MPa; on the 30° plane σ ≈ 76.0 MPa, τ ≈ −37.0 MPa.

Example 2 (GATE level): strain rosette on a steel arm. Given: a rectangular rosette on the surface of a steel robot arm reads ε₀ = 600 με, ε₄₅ = 500 με, ε₉₀ = −200 με. E = 200 GPa, ν = 0.3. Find principal strains, principal stresses, their direction and the maximum in-plane shear stress.

1. γ_xy = 2ε₄₅ − ε₀ − ε₉₀ = 1000 − 600 + 200 = 600 με.
2. Strain circle: centre (600 − 200)/2 = 200 με; radius √(400² + 300²) = 500 με.
3. ε₁ = 700 με, ε₂ = −300 με.
4. tan 2θ_p = γ_xy/(ε_x − ε_y) = 600/800 = 0.75, so θ_p = 18.43° from the 0° gauge.
5. E/(1 − ν²) = 200 000/0.91 = 219 780 MPa.
6. σ₁ = 219 780 × (700 − 0.3 × 300) × 10⁻⁶ = 219 780 × 610 × 10⁻⁶ = 134.1 MPa.
7. σ₂ = 219 780 × (−300 + 0.3 × 700) × 10⁻⁶ = 219 780 × (−90) × 10⁻⁶ = −19.8 MPa.
8. τ_max = (σ₁ − σ₂)/2 = 76.9 MPa. Check: G = E/[2(1 + ν)] = 76 923 MPa and maximum shear strain = 2R = 1000 με, so τ = 76 923 × 0.001 = 76.9 MPa. Answer: ε₁ = 700 με, ε₂ = −300 με; σ₁ ≈ 134 MPa, σ₂ ≈ −19.8 MPa at 18.4° from gauge 0; τ_max ≈ 76.9 MPa.

Common mistakes

• Forgetting that angles double on the circle: 2θ on the circle is θ in the body.
• Using γ_xy instead of γ_xy/2 on the strain circle.
• Calling R the absolute maximum shear when σ₁ and σ₂ have the same sign.
• Getting σ₁'s plane wrong: tan 2θ_p has two solutions 90° apart; substitute to see which gives σ₁.
• Treating the principal strains as principal stresses; use plane-stress Hooke's law with ν.
• Saying the formulas need linear elasticity; stress transformation is pure equilibrium.

For GATE ME

Expect principal stresses and maximum shear from given σ_x, σ_y, τ_xy, stresses on an inclined plane, special cases (pure shear, equal biaxial, uniaxial), absolute maximum shear with σ₃ = 0, and rosette or principal-strain problems leading to principal stresses. Practise drawing the circle quickly to check signs and angles.

Quick check

1. σ_x = σ_y = 50 MPa, τ_xy = 0. What does Mohr's circle look like?
2. In pure shear τ = 40 MPa, what are the principal stresses and at what angle?
3. σ_x = 80, σ_y = 40, τ_xy = 0 MPa (plane stress). What are the in-plane and absolute maximum shear stresses?
4. What is the normal stress on the planes of maximum in-plane shear?
5. Rosette: ε₀ = 400, ε₄₅ = 300, ε₉₀ = 100 με. Find γ_xy.

Answers: 1. A point at σ = 50 MPa. 2. +40 and −40 MPa on planes at 45° to the shear planes. 3. 20 MPa in-plane; 40 MPa absolute (σ₃ = 0). 4. The average stress (σ_x + σ_y)/2. 5. 2 × 300 − 400 − 100 = 100 με.