Mohr's circle for plane stress and strain

Why it matters

Loads on shafts, pressure vessels, welds and brackets produce normal and shear stresses on the planes we happen to choose (along and across the member). Failure, however, happens on whichever plane is worst. Mohr's circle shows at a glance the stresses on every plane through a point, the principal stresses, and the maximum shear stress, and the same construction works for strains measured with strain gauges.

Key ideas

Plane stress. A thin plate or the free surface of a part: σz = τxz = τyz = 0. The state at a point is fully described by σx, σy and τxy. (Plane strain is the counterpart with εz = γxz = γyz = 0, as in long dams, tunnels and thick shafts.)

Stress transformation. On a plane whose normal makes angle θ (anticlockwise) with the x-axis, the normal stress σθ and shear stress τθ follow the transformation equations below. They depend on 2θ, which is why angles on Mohr's circle are doubled.

Sign convention used here. Tensile normal stress positive. τxy is positive when it acts in +y on the face whose outward normal is +x. When plotting, take σ to the right and τ positive downward; point X (the x-face) is at (σx, τxy) and point Y (the y-face) is at (σy, −τxy). With τ downward, a rotation of 2θ on the circle is in the same sense as the rotation θ of the plane in the element. Many textbooks plot τ upward instead and reverse the rotation rule; use one convention consistently and check angles with the formula.

Construction.

1. Plot X and Y. The line XY is a diameter.
2. Centre C = ((σx + σy)/2, 0); radius R = distance CX.
3. The circle meets the σ-axis at the principal stresses σ1 = C + R and σ2 = C − R; shear is zero there.
4. The top and bottom points give the maximum in-plane shear stress R, on planes at 45° to the principal planes, carrying normal stress equal to C.
5. To find stresses on a plane at θ, rotate the diameter XY by 2θ.

Invariants. σx + σy = σ1 + σ2 for every pair of perpendicular planes. Principal planes are 90° apart in the element (180° on the circle).

Special cases.

• Uniaxial tension σ: circle through 0 and σ; τmax = σ/2 at 45°.
• Pure shear τ: centre at origin, R = τ; principal stresses ±τ at 45°.
• Equal biaxial stress (σx = σy, τxy = 0): the circle shrinks to a point; every in-plane direction is principal.

Absolute maximum shear. In plane stress the third principal stress is σ3 = 0. If σ1 and σ2 have the same sign, the absolute maximum shear is max(|σ1|, |σ2|)/2, larger than the in-plane value R. If they have opposite signs, it equals R. This matters for thin cylinders and for the Tresca criterion.

Mohr's circle for strain. Replace σ by ε and τ by γ/2. Centre (εx + εy)/2, radius √[((εx − εy)/2)² + (γxy/2)²]. Principal strains and directions follow as for stress; for an isotropic material the principal strain and stress directions coincide. Strain-gauge rosettes give εx, εy and γxy.

Formulas

σθ = (σx + σy)/2 + (σx − σy)/2 · cos 2θ + τxy sin 2θ

τθ = −(σx − σy)/2 · sin 2θ + τxy cos 2θ

• σ, τ in Pa (MPa); θ measured anticlockwise from the x-axis to the plane's normal.

C = (σx + σy)/2, R = √[((σx − σy)/2)² + τxy²]

σ1,2 = C ± R, τmax,in-plane = R = (σ1 − σ2)/2

tan 2θp = 2τxy / (σx − σy) (principal planes; θs = θp ± 45° for maximum shear)

τabs,max = max(|σ1|, |σ2|, |σ1 − σ2|)/2 (plane stress, σ3 = 0)

ε1,2 = (εx + εy)/2 ± √[((εx − εy)/2)² + (γxy/2)²], γmax = ε1 − ε2 (in-plane)

σ1 = E (ε1 + ν ε2)/(1 − ν²), σ2 = E (ε2 + ν ε1)/(1 − ν²) (plane stress, isotropic)

Worked examples

Example 1 (standard). At a point, σx = 80 MPa, σy = −40 MPa, τxy = 30 MPa. Find the principal stresses and their planes, the maximum in-plane shear stress, and the stresses on a plane at θ = 30°.

1. Centre: C = (80 − 40)/2 = 20 MPa.
2. Radius: R = √(60² + 30²) = √4500 = 67.08 MPa.
3. Principal stresses: σ1 = 20 + 67.08 = 87.08 MPa, σ2 = 20 − 67.08 = −47.08 MPa.
4. Direction: tan 2θp = 2 × 30 / 120 = 0.5, so 2θp = 26.57°, θp = 13.28° (to the σ1 plane; check: σθ at 13.28° = 20 + 60 cos 26.57° + 30 sin 26.57° = 20 + 53.67 + 13.42 = 87.08 MPa ✓).
5. Maximum in-plane shear: τmax = R = 67.08 MPa, on planes at 13.28° ± 45°, with normal stress 20 MPa.
6. Plane at 30° (2θ = 60°): σθ = 20 + 60 × 0.5 + 30 × 0.866 = 75.98 MPa; τθ = −60 × 0.866 + 30 × 0.5 = −36.96 MPa.
7. Since σ1 > 0 > σ2, the absolute maximum shear equals R = 67.08 MPa.

Answer: σ1 = 87.1 MPa, σ2 = −47.1 MPa at θp = 13.3°; τmax = 67.1 MPa; on the 30° plane σ = 76.0 MPa, τ = −37.0 MPa.

Example 2 (GATE level, strain). A rosette on a steel plate gives εx = 500 × 10⁻⁶, εy = −100 × 10⁻⁶, γxy = 400 × 10⁻⁶. E = 200 GPa, ν = 0.3. Find the principal strains, the maximum in-plane shear strain and the principal stresses.

1. Centre: (500 − 100)/2 = 200 × 10⁻⁶.
2. Radius: √(300² + 200²) × 10⁻⁶ = 360.6 × 10⁻⁶.
3. ε1 = 560.6 × 10⁻⁶, ε2 = −160.6 × 10⁻⁶; γmax = ε1 − ε2 = 721.1 × 10⁻⁶.
4. Direction: tan 2θp = γxy/(εx − εy) = 400/600, so θp = 16.8°.
5. E/(1 − ν²) = 200 000 / 0.91 = 219 780 MPa.
6. σ1 = 219 780 × (560.6 − 0.3 × 160.6) × 10⁻⁶ = 219 780 × 512.4 × 10⁻⁶ = 112.6 MPa.
7. σ2 = 219 780 × (−160.6 + 0.3 × 560.6) × 10⁻⁶ = 219 780 × 7.6 × 10⁻⁶ = 1.67 MPa.

Answer: ε1 = 560.6 μ, ε2 = −160.6 μ, γmax = 721.1 μ, σ1 ≈ 112.6 MPa, σ2 ≈ 1.7 MPa.

Common mistakes

• Forgetting that circle angles are twice the physical angles.
• Plotting γ instead of γ/2 on a strain circle.
• Calling R the absolute maximum shear when σ1 and σ2 have the same sign.
• Mixing sign conventions for τ between the formula and the plot, which flips the principal direction.
• Assuming principal stresses are the largest normal stresses on x and y faces; they are always at least as large as σx and σy.
• Converting principal strains to stresses with σ = E ε alone; Poisson coupling must be included.

For GATE PI

Expect principal stresses and maximum shear from given σx, σy, τxy, the radius or centre of the circle, stresses on an inclined plane, pure shear and uniaxial special cases, and principal strains from a rosette. Questions often combine this with shafts under bending and torsion or with thin cylinders. Practise the formulas until you can do them without drawing.

Quick check

1. For σx = 80 MPa, σy = 40 MPa, τxy = 0, what are C and R?
2. In pure shear of 50 MPa, what are the principal stresses?
3. How far apart in the element are the principal planes and the maximum-shear planes?
4. In plane stress with σ1 = 100 MPa and σ2 = 40 MPa, what is the absolute maximum shear stress?

Answers: 1. C = 60 MPa, R = 20 MPa. 2. +50 MPa and −50 MPa. 3. 45°. 4. 50 MPa (σ1/2, because σ3 = 0).