Mohr's Circle, Principal Stress, Thin Cylinders

At a point in a loaded body, the normal and shear stresses depend on the plane you look at. The planes with zero shear carry the principal stresses, the largest and smallest normal stresses at that point.

Stress transformation (2D)

With stresses σx, σy and τxy on the x and y faces:

• Principal stresses: σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]
• Principal planes: tan 2θp = 2τxy/(σx − σy)
• Maximum in-plane shear stress: τmax = √[((σx − σy)/2)² + τxy²] = (σ₁ − σ₂)/2, on planes at 45° to the principal planes.
• σx + σy = σ₁ + σ₂ (the sum is invariant).

Absolute maximum shear stress in 3D is (σmax − σmin)/2 using all three principal stresses (the third being 0 for plane stress). If σ₁ and σ₂ have the same sign, it is σ₁/2.

Mohr's circle

Plot each face as a point (σ, τ). The circle has:

• Centre at ((σx + σy)/2, 0)
• Radius R = √[((σx − σy)/2)² + τxy²]

The principal stresses are where the circle cuts the σ-axis: centre ± R. An angle of θ on the element is 2θ on the circle. Pure shear τ gives a circle centred at the origin, so σ₁ = τ and σ₂ = −τ.

Thin cylinders and spheres

For a thin vessel (t < d/20) with internal pressure p, diameter d and wall thickness t:

The hoop stress is twice the longitudinal stress, so cylinders split along their length. Volumetric strain of a cylinder: ε_v = (pd/(4tE))(5 − 4ν).

Worked example

At a point σx = 120 MPa, σy = 40 MPa and τxy = 30 MPa. Find the principal stresses and the maximum shear stress.

• Centre = 80 MPa, R = √(40² + 30²) = 50 MPa.
• σ₁ = 130 MPa, σ₂ = 30 MPa, in-plane τmax = 50 MPa.
• Both principal stresses are tensile, so the absolute maximum shear = 130/2 = 65 MPa.

Practice