# Antiplane shearing: singularities of $\vert\nabla\Phi\vert$

Suppose we wish to find the displacement field $(0, 0, \Phi(x,y))$ in an antiplane shear problem with a fracture along $0 < x < k, y = 0$. Then, according to my notes, $\vert \nabla\Phi\vert$ has "an inverse square root singularity at the crack tips, so that the displacement is finite."

Can anyone please explain this sentence? In particular, why does the inverse square root singularity arise and why would this then imply a finite displacement?

• What are the assumptions that go into the material? Is it isotropic for example, where this reduces to the Neumann boundary condition problem for the Laplace equation? Maybe you could provide the governing equations for the deformation if the assumptions are different? – DinosaurEgg Sep 17 '18 at 19:58

Very roughly speaking, the gradient of the displacement field (also called the strain) has the form $$\nabla \phi = C x^{-1/2} + \dots$$ At $x = 0$ (the crack tip), the strain is infinite. If we integrate the above to find $\phi$ (again very roughly) $$\phi = \int C x^{-1/2} dx + \int \dots \sim A x^{1/2} + \dots$$ So $\phi$ is finite at the crack tip.