Math 2.0

Mechanics of deformable solids Definition

Stress intensity factor

Builds on Modes of fractureCauchy stress tensorLinear elasticity

Definition

Let a linear elastic isotropic body contain a crack with traction-free faces. Let PP be a point of the crack front where the front is smooth. Use the frame at PP of the modes of fracture: x1x_1 ahead of the crack, x2x_2 normal to the crack plane, x3x_3 along the front. Let (r,θ)(r, \theta) be polar coordinates in the x1x2x_1 x_2-plane, with θ=0\theta = 0 ahead of the crack. Let σij\sigma_{ij} be the components of the Cauchy stress. The stress intensity factors at PP are KI=lim⁡r→02πr σ22(r,0),KII=lim⁡r→02πr σ12(r,0),KIII=lim⁡r→02πr σ23(r,0),K_I = \lim_{r \to 0} \sqrt{2 \pi r} \, \sigma_{22}(r, 0), \quad K_{II} = \lim_{r \to 0} \sqrt{2 \pi r} \, \sigma_{12}(r, 0), \quad K_{III} = \lim_{r \to 0} \sqrt{2 \pi r} \, \sigma_{23}(r, 0), whenever these limits exist.

Remarks

Example: a straight crack of length 2a2a in an infinite plate under remote tension σ\sigma perpendicular to the crack has, ahead of each tip, σ22(r,0)=σ(a+r)/r(2a+r)\sigma_{22}(r, 0) = \sigma (a + r) / \sqrt{r (2a + r)}. Then 2πr σ22(r,0)→σ2π a/2a=σπa\sqrt{2 \pi r} \, \sigma_{22}(r,0) \to \sigma \sqrt{2 \pi} \, a / \sqrt{2a} = \sigma \sqrt{\pi a}, so KI=σπaK_I = \sigma \sqrt{\pi a}, while KII=KIII=0K_{II} = K_{III} = 0. Non-example: for a circular hole in a plate the stress at the edge is finite, so 2πr σ22→0\sqrt{2\pi r}\,\sigma_{22} \to 0; a hole has no stress intensity, only a stress concentration factor. The units of KK are stress times the square root of length, for example MPam\mathrm{MPa}\sqrt{\mathrm{m}}.

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  1. Cauchy stress tensor concept
  2. Linear elasticity concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Stress_intensity_factor
About this page
TagHX9B38KD2X
URIm2://HX9B38KD2X
Kinddefinition
WritingW0 stub
MSC74R10, 74G70, 74B05
DefinesStress intensity factor
UsesModes of fracture, Cauchy stress tensor, Linear elasticity
Changed3 November 2026

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