Math 2.0

Mechanics of deformable solids Fracture and damage Theorem

Williams crack-tip expansion

Builds on Stress intensity factorModes of fractureLinear elasticityAiry stress function

Statement

Let Ω⊂ℝ2\Omega \subset \R^2 contain the slit {(x1,0):−δ<x1≤0}\set{(x_1, 0) : -\delta < x_1 \le 0} and let (r,θ)(r, \theta), −π<θ<π-\pi < \theta < \pi, be polar coordinates centred at its tip. Let the stress σ\sigma solve the equations of plane linear elasticity for an isotropic material in Bδ(0)B_\delta(0) minus the slit, without body forces. Assume that the crack faces θ=±π\theta = \pm \pi are traction-free and that the elastic energy near the tip is finite. Then there are real numbers ana_n, bnb_n such that, for r<δr < \delta, σij(r,θ)=∑n=1∞rn/2−1(anfij(n)(θ)+bngij(n)(θ)).\sigma_{ij}(r, \theta) = \sum_{n=1}^{\infty} r^{n/2 - 1} \left( a_n f_{ij}^{(n)}(\theta) + b_n g_{ij}^{(n)}(\theta) \right). Here f(n)f^{(n)} and g(n)g^{(n)} are fixed trigonometric functions, symmetric (mode I) and antisymmetric (mode II) with respect to the crack line. The leading term is the inverse square root singularity with the stress intensity factors; in mode I it reads σ11=KI2πrcos⁡θ2(1−sin⁡θ2sin⁡3θ2),σ22=KI2πrcos⁡θ2(1+sin⁡θ2sin⁡3θ2),σ12=KI2πrcos⁡θ2sin⁡θ2cos⁡3θ2.\sigma_{11} = \frac{K_I}{\sqrt{2 \pi r}} \cos\frac{\theta}{2} \Big(1 - \sin\frac{\theta}{2} \sin\frac{3\theta}{2}\Big), \quad \sigma_{22} = \frac{K_I}{\sqrt{2 \pi r}} \cos\frac{\theta}{2} \Big(1 + \sin\frac{\theta}{2} \sin\frac{3\theta}{2}\Big), \quad \sigma_{12} = \frac{K_I}{\sqrt{2 \pi r}} \cos\frac{\theta}{2} \sin\frac{\theta}{2} \cos\frac{3\theta}{2}. The term with n=2n = 2 is a uniform stress σ11=T\sigma_{11} = T parallel to the crack, the T-stress.

Remarks

The proof writes the stress with an Airy stress function ϕ\phi, which is biharmonic, and looks for separated solutions ϕ=rλ+1F(θ)\phi = r^{\lambda + 1} F(\theta). The traction-free conditions on θ=±π\theta = \pm \pi give a homogeneous linear system for the coefficients of FF, which has nontrivial solutions only when sin⁡(2πλ)=0\sin(2\pi\lambda) = 0, that is λ=n/2\lambda = n/2. Finite energy near the tip excludes λ≤0\lambda \le 0, which leaves λ=1/2,1,3/2,…\lambda = 1/2, 1, 3/2, \dots. Without the finite energy hypothesis, solutions with stronger singularities such as r−3/2r^{-3/2} would be admitted. At θ=0\theta = 0 the mode I formulas give σ22=KI/2πr\sigma_{22} = K_I / \sqrt{2 \pi r}, consistent with the definition of KIK_I.

Going deeper

For a student

The expansion justifies the one-parameter description of linear elastic fracture mechanics: close to the tip, the first term dominates, so KIK_I, KIIK_{II} and KIIIK_{III} characterize the crack-tip loading. The second term, the T-stress, measures the constraint at the tip and affects the size and shape of the plastic zone and the stability of the crack path. The same method of separated solutions gives the singularities at the vertex of a wedge or a notch, where the exponents are no longer half-integers. The modes of fracture correspond to the symmetric and antisymmetric parts of the expansion, and mode III arises from the analogous antiplane problem.

Reading path

To understand this page from the ground up, read these 5 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Airy stress function concept
  2. Linear elasticity concept
  3. Modes of fracture concept
  4. Stress intensity factor concept
  5. Williams crack-tip expansion statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Fracture_mechanics
About this page
TagSEG9TKP2KQ
URIm2://SEG9TKP2KQ
Kindstatement, theorem
WritingW0 stub
MSC74R10, 74G70, 74B05
StatusThis copy of the site has no build data. Run m2 status SEG9TKP2KQ.
AboutStress intensity factor
UsesStress intensity factor, Modes of fracture, Linear elasticity, Airy stress function
Changed3 November 2026

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