Stress transformation and Mohr's circle

Why it matters

A wing skin panel carries tension from bending and shear from torsion at the same time, and a pressurised fuselage has hoop and axial stresses plus shear. Cracks, yielding and buckling depend on the stresses on the worst plane, not on the x and y planes you happened to choose. Stress transformation and Mohr's circle find those worst planes and their stresses.

Key ideas

State of stress at a point. In plane stress (thin plates and skins, σz = τxz = τyz = 0), the stress at a point is fully described by σx, σy and τxy. On a plane whose outward normal makes an angle θ with the x-axis (counter-clockwise positive), the normal and shear stresses are different, but they are fixed by those three numbers through the transformation equations. These come from force equilibrium of a small wedge, so they hold for any material, elastic or not.

Sign convention used here. Tensile normal stress is positive. τxy is positive when it acts in the +y direction on the face whose outward normal is +x. θ is measured counter-clockwise from x to the plane's normal.

Principal stresses and planes. As θ varies, σθ reaches a maximum σ₁ and a minimum σ₂ on two perpendicular planes on which the shear stress is zero. These are the principal planes, at angle θp given by tan 2θp = 2τxy/(σx − σy). Of the two solutions 90° apart, check which gives σ₁ by substituting back.

Maximum shear. The maximum in-plane shear stress is τmax = (σ₁ − σ₂)/2, on planes at 45° to the principal planes. Those planes also carry a normal stress equal to the average (σx + σy)/2.

Absolute maximum shear in plane stress. The third principal stress is σ₃ = 0. If σ₁ and σ₂ have the same sign, the largest shear acts out of plane and equals max(|σ₁|, |σ₂|)/2, which exceeds the in-plane value. This is a frequent exam trap.

Invariants. σx + σy = σθ + σθ+90° = σ₁ + σ₂ for every θ. Use this to check your arithmetic.

Mohr's circle. Plot each face as a point (σ, τ). With the convention used here, plot face x at (σx, −τxy) and face y at (σy, +τxy) (shear plotted positive downwards is an equivalent alternative; be consistent). The line joining the two points is a diameter. Centre C = (σx + σy)/2 on the σ-axis; radius R = √[((σx − σy)/2)² + τxy²]. A rotation of the plane by θ on the element is a rotation of 2θ on the circle, in the same sense. The ends of the horizontal diameter are σ₁ and σ₂; the top and bottom of the circle give τmax = R.

Special cases:

• Uniaxial σ: circle through 0 and σ; τmax = σ/2 at 45°.
• Pure shear τ: circle centred at the origin; σ₁ = τ, σ₂ = −τ at 45° (why brittle shafts fail on 45° helices).
• Equal biaxial σx = σy, τxy = 0: the circle shrinks to a point; every plane is principal.

Formulas

σθ = (σx + σy)/2 + (σx − σy)/2·cos 2θ + τxy·sin 2θ τθ = −(σx − σy)/2·sin 2θ + τxy·cos 2θ Stresses (Pa) on a plane whose normal is at θ to x (counter-clockwise positive).

σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²] Principal stresses (Pa), plane stress.

tan 2θp = 2·τxy / (σx − σy) Principal plane angle θp (rad or degrees).

τmax,in-plane = √[((σx − σy)/2)² + τxy²] = (σ₁ − σ₂)/2 Maximum in-plane shear (Pa), on planes at θp ± 45°.

τabs,max = max(|σ₁ − σ₂|, |σ₁|, |σ₂|) / 2 Absolute maximum shear in plane stress (σ₃ = 0).

σx + σy = σ₁ + σ₂ First stress invariant (Pa).

Worked examples

Example 1 (standard). At a point, σx = 80 MPa, σy = −40 MPa and τxy = 30 MPa. Find the normal and shear stress on the plane whose normal is at θ = 30° to the x-axis, and the normal stress on the perpendicular plane.

1. Average = (80 − 40)/2 = 20 MPa; half-difference = (80 + 40)/2 = 60 MPa; 2θ = 60°.
2. σθ = 20 + 60·cos 60° + 30·sin 60° = 20 + 30 + 25.98 = 75.98 MPa.
3. τθ = −60·sin 60° + 30·cos 60° = −51.96 + 15 = −36.96 MPa.
4. Perpendicular plane (θ = 120°, 2θ = 240°): σ = 20 + 60 × (−0.5) + 30 × (−0.866) = −35.98 MPa.
5. Check: 75.98 + (−35.98) = 40 = σx + σy.

Answer: σθ ≈ 76.0 MPa, τθ ≈ −37.0 MPa, σ on the perpendicular plane ≈ −36.0 MPa.

Example 2 (GATE level). A thin skin panel has σx = 60 MPa, σy = 20 MPa and τxy = −30 MPa (plane stress). Find the principal stresses and their directions, the maximum in-plane shear and the absolute maximum shear.

1. Centre C = (60 + 20)/2 = 40 MPa.
2. Radius R = √(20² + 30²) = √1300 = 36.06 MPa.
3. σ₁ = 40 + 36.06 = 76.06 MPa; σ₂ = 40 − 36.06 = 3.94 MPa.
4. tan 2θp = 2τxy/(σx − σy) = −60/40 = −1.5, so 2θp = −56.31° and θp = −28.15°.
5. Check by substitution: σ(−28.15°) = 40 + 20·cos(−56.31°) − 30·sin(−56.31°) = 40 + 11.09 + 24.96 = 76.06 MPa, so σ₁ acts on the plane at 28.15° clockwise from x; σ₂ is 90° away.
6. τmax,in-plane = R = 36.06 MPa, at 45° to the principal planes.
7. σ₃ = 0 and σ₁, σ₂ are both tensile, so τabs,max = σ₁/2 = 38.03 MPa, acting out of plane.

Answer: σ₁ ≈ 76.1 MPa at θ = −28.2°, σ₂ ≈ 3.9 MPa; τmax in-plane ≈ 36.1 MPa; absolute τmax ≈ 38.0 MPa.

Common mistakes

• Using θ on Mohr's circle instead of 2θ, or rotating in the opposite sense.
• Mixing up the angle of the plane with the angle of its normal.
• Taking the in-plane τmax as the absolute maximum when both principal stresses have the same sign.
• Picking the wrong root of tan 2θp without substituting back to see which plane carries σ₁.
• Forgetting that the planes of maximum shear also carry the average normal stress.

For GATE AE

Expect numericals on principal stresses, the principal angle, maximum in-plane and absolute shear, stress on a given inclined plane, and special cases such as pure shear or equal biaxial stress where Mohr's circle becomes trivial. Practise reading answers straight from the centre and radius, and always check with the invariant σx + σy.

Quick check

1. In pure shear τ, what are the principal stresses?
2. What is the radius of Mohr's circle for σx = 50 MPa, σy = 10 MPa, τxy = 15 MPa?
3. If σ₁ = 100 MPa and σ₂ = 40 MPa in plane stress, what is the absolute maximum shear stress?
4. A 30° rotation of the element corresponds to what angle on Mohr's circle?

Answers: 1. +τ and −τ. 2. 25 MPa. 3. 50 MPa. 4. 60°.