What is stress? Stress is internal force per unit area. Imagine cutting through a body along a plane with normal nˆ. The traction vector t = σ · nˆ gives the force per area on that cut face.
In 3D: 6 independent components: 3 normal stresses (σxx, σyy, σzz) and 3 shear stresses (τxy, τxz, τyz). Arrows show force components on each face of the element.
Normal stresses (MPa):
σxx0—
σyy0—
σzz0—
Shear stresses (MPa):
τxy0—
τxz0—
τyz0—
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Normal Stresses
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σii = Fn/A | ΔV/V ≈ tr(T)/3K
σxx: force in x on x-face (tension/compression) Hydrostatic: p = −⅓(σxx+σyy+σzz) Mean stress: σm = ⅓ tr(T) = −p
Normal stresses act perpendicular to the face. Positive = tension (pulling), negative = compression (pushing). Each face of the cube has its own normal stress.
Volume change: Normal stresses change the volume. The trace σxx+σyy+σzz determines the mean stress. Positive mean stress = expansion, negative = compression.
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Shear Stresses
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τij = Ft/A | τij = τji
Complementary shear: τxy = τyx (moment equilibrium) No volume change: shear stresses cause only distortion Principal at 45°: pure shear = tension + compression at 45°
Shear stresses act tangent to the face. τxy means: force in x-direction on a face with y-normal. By moment equilibrium, τxy = τyx. This makes the stress tensor symmetric.
Key insight: Pure shear τxy is equivalent to tension + compression at 45° (rotate 45° and all shear vanishes).
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Principal Stresses & Mohr's Circle
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Eigenvalue problem: det(T − λI) = 0
σn(α) = ½(σxx+σyy) + ½(σxx−σyy)cos2α + τxysin2α
τ(α) = −½(σxx−σyy)sin2α + τxycos2α
3D: three principal stresses σI ≥ σII ≥ σIII Tresca: τmax = ½(σI − σIII) 3 nested Mohr circles for the three principal planes
3D Mohr's circles: In 3D, there are three Mohr circles — one for each pair of principal stresses. The outermost circle (σI–σIII) gives the absolute maximum shear. All stress states on any cut plane lie between the three circles.
Tresca criterion: τmax = (σI−σIII)/2. Yielding when τmax ≥ σY/2.
Cut angle α0°
Eigenvalues & Eigenvectors:
det(T − λI) = 0 gives σI ≥ σII ≥ σIII. These are the stresses on faces where ALL shear vanishes. The eigenvectors define the principal directions.
Why decompose? The hydrostatic part changes volume but NOT shape. The deviatoric part changes shape and drives yielding. Materials rarely fail under pure hydrostatic pressure.
von Mises criterion: σvM ≥ σY means yielding. This depends ONLY on the deviatoric stress s.
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Coordinate Transformation
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T' = Q T QT
Invariants preserved: I1 = tr(T), I2, I3 = det(T) Eigenvalues unchanged: σI, σII, σIII same in any basis Cylindrical coords: σrr, σθθ, σzz, τrθ, τrz, τθz
Why transform? For symmetric problems (pipes, shafts), cylindrical coordinates make the stress tensor diagonal or nearly so. The components have physical meaning: hoop stress σθθ tells you if the pipe splits along its length.
Thin-walled pressure vessel:σθθ = pR/t (hoop) σzz = pR/(2t) (axial) σrr ≈ 0
Hoop stress is 2× axial — this is why pipes burst along their length!
Thick-walled cylinder (Lamé):σrr = A − B/r² σθθ = A + B/r²
At the inner wall, σθθ is maximum (tensile), σrr = −p (compressive).
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