Let $a$ and $b$ be the axis of the ellipse, with $0 < b \le a$.
The warping function is given
$$
\omega(x, \ y) = - \left(\dfrac{a^2-b^2}{a^2+b^2}\right) \cdot xy
$$
Note that $a=b=0$, meaning a circle, gives a constant function
Then, the torsion constant $J$ can be computed as
$$J = I_{xx}+I_{yy} - \int_{\Omega} \left(y \dfrac{\partial \omega}{\partial x} - x \dfrac{\partial \omega}{\partial y}\right) \ dx \ dy$$
$$J = \dfrac{\pi ab^3}{4} + \dfrac{\pi a^3b}{4} - \dfrac{\pi ab \left(a^2-b^2\right)^2}{4\left(a^2+b^2\right)}= \dfrac{\pi a^3 b^3}{a^2+b^2}$$
The stress field is given
$$\sigma_{xz}(x, \ y) = \dfrac{M_z}{J}\left(\dfrac{\partial \omega}{\partial x} - y\right);\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sigma_{yz}(x, \ y) = \dfrac{M_z}{J}\left(\dfrac{\partial \omega}{\partial y} + x\right)$$
$$\sigma_{xz}(x, \ y) = M_{z} \cdot \dfrac{-2y}{\pi ab^3}$$
$$\sigma_{yz}(x, \ y) = M_{z} \cdot \dfrac{2x}{\pi a^3b}$$
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