User:Caan L/sandbox

Solution to the Williams Free Edge Problem

The Free Edge Problem Williams considered can be modeled as a tendon-to-bone attachment. The asymptotic stress field in the cylindrical coordinate system of this model is represented by the following equations:

$\sigma _{rr}=r^{\lambda -1}(\Phi '')(\theta ;\lambda )+(\lambda +1)\Phi (\theta ;\lambda )$

$\sigma _{\theta \theta }=r^{\lambda -1}(\lambda (\lambda +1)\Phi (\theta ;\lambda ))$

$\sigma _{r\theta }=-r^{\lambda -1}\Phi '(\theta ;\lambda )$

Where $\Phi (\theta ;\lambda )$ is a function whose form, given by Williams, contains four unknown constants that must be found along with a set of eigenvalues using the boundary conditions.

The unknown constants can be used to find the displacements by using the equations below:

$\sigma _{ij}={\frac {E}{1+v}}\epsilon _{ij}+{\frac {E}{(1+v)(1-2v)}}\delta _{ij}\epsilon _{kk}$

where $\epsilon _{ij}$ represents the 3 x 3 matrix of components of the strain tensor $\epsilon$ in a particular coordinate system, $u_{i}$ represents the three components of the displacement vector u in that coordinate system, and $u_{i,j}=dy/dx$ and

$\epsilon _{ij}={\frac {1}{2}}(u_{i,j}+u_{j,i})$

where $\delta _{ij}$ is the Kronecker's delta function that equals 1 when $i=j$ and 0 otherwise, and $\epsilon _{kk}=\epsilon _{11}+\epsilon _{22}+\epsilon _{33}$.

Displacements can be written in terms of the constraints:

$u_{r}=r^{\lambda }\left({\frac {1+v}{E}}\right)(-(\lambda +1)\Phi (\theta ;\lambda )+(1-n)\Psi '(\theta ;\lambda ))$

$u_{0}=r^{\lambda }\left({\frac {1+v}{E}}\right)(-\Phi '(\theta ;\lambda )+(1-n)(\lambda -1)\Psi (\theta ;\lambda ))$

where $n-v/(1-v)$, and $\Psi (\theta ;\lambda )$.

The four boundary conditions used to determine the unknown constants result in a system of equations with a nontrivial solution when the equation below is satisfied:

$sin^{2}(\lambda \theta _{T})=\left({\frac {1}{3-4n}}\right)(4(1-n)^{2}-(\lambda \theta _{T})^{2}sin^{2}\theta _{T})$

The stress field near the free edge of the body becomes singular at a critical insertion angle $\theta _{T}$ where $\lambda$ reaches its lowest positive value.