May I ask you how to derive the following formula. It is based on the Kolosov-Mushkelisvili formulae in mechanics, but I could not get what is written in this book i.e. Savin, G. N. (1968). Stress Distribution around Holes.
The Kolosov-Muskhelisvili formulae is,
$\sigma_{r}+\sigma_{\theta} = 2[\Phi(\zeta)+\overline{\Phi(\zeta)}]$
$\sigma_\theta-\sigma_{r}+2i\tau_{\rho\theta}=\dfrac{2\zeta}{\rho^2\overline{\omega'(\zeta)}}[\overline{\omega(\zeta)}\Phi'(\zeta)+\omega'(\zeta)\Psi(\zeta)]$
where $\Phi(\zeta)$ and $\Psi(\zeta)$ are defined as:
$\Phi(\zeta)=\dfrac{\phi'(\zeta)}{\omega'(\zeta)}$
$\Psi(\zeta)=\dfrac{\psi'(\zeta)}{\omega'(\zeta)}$
in which the prime symbol is for derivative and the overline is the complex conjugate.
For a circle, the two functions can be defined as:
$\phi(\zeta)=\dfrac{pR}{4}\left(\dfrac{1}{\zeta}+2\zeta\right)$
$\psi(\zeta)=-\dfrac{pR}{2}\left(\dfrac{1}{\zeta}+\zeta-\zeta^3 \right)$
and $\zeta^k = (\rho e^{i\theta})^k = \rho^k(\cos k\theta + i\sin k\theta$)
And,
$\omega(\zeta)=\dfrac{R}{\zeta}$
$R=1$ is the radius of the circle.
Using all of the definition mentioned above, the $\sigma_\theta$ can be obtained from the Kolosov-Muskhelisvili formulae by inserting the corresponding value from all of the above definitions,
$$\sigma_\theta=\dfrac{p}{2}\left[(1+\rho^2)-(1+3\rho^4)\cos 2\theta\right]$$
$$\sigma_r=\dfrac{p}{2}\left[(1-\rho^2)+(1-4\rho^4+3\rho^4)\cos 2\theta\right]$$
$$\sigma_{r\theta}=\dfrac{p}{2}(1+2\rho^2-3\rho^4)\sin 2\theta$$
My question is, how do we arrive that definition of $\sigma_\theta$? Yes, I am only interested in $\sigma_\theta$. I am not able to get it from the Kolosov-Muskhelisvili formulae. I have done it using substitution and elimination, still cannot it exactly as what is written in the book.
Can anyone help me to prove it?
Suppose that $a$ is an integer, $a>0$. Let $$I(a)=\int_\gamma t^{-a}\frac{t+\zeta}{t-\zeta}\frac{dt}{t}=2\pi i(\mbox{sum of the residues}),$$ where residues are taken at the poles in the unit disk. The function has two poles, one at $0$ another at $\zeta$. The pole at $\zeta$ is simple and the residue at this pole is $2\zeta^{-a}$. Sum of all residues, including the one at $\infty$ is always $0$.
When $|\zeta|<1$, both poles are in the unit disk, and the sum of their residues is $0$ because the residue at $\infty$ is zero. If $|\zeta|>1$, there is only one pole in the unit disk and the residue at $0$ is negative of the residue at $\zeta$, so it is $-2\zeta^{-a}$.
The case $a\leq 0$ is easier because there is only one pole at $\zeta$.
Then put $a=k$ or $a=1-k$, or whatever you want.