= Solution
Use the <torsion stress function for a simply connected bar>. The only nonzero <strain> components are $e_{xz}=(w_x-\Omega y)/2$ and $e_{yz}=(w_y+\Omega x)/2$. Thus
$$
\tau_{xz}=\mu(w_x-\Omega y),\qquad\tau_{yz}=\mu(w_y+\Omega x).
$$
The axial <equilibrium> equation is $\partial_x\tau_{xz}+\partial_y\tau_{yz}=0$. Since the section is <simply connected>, the planar <divergence-free> <stress> field has a <stream function>:
$$
\boxed{\tau_{xz}=\phi_y,\quad\tau_{yz}=-\phi_x,\quad w_x=\Omega y+\phi_y/\mu,\quad w_y=-\Omega x-\phi_x/\mu.}
$$
Equality of the mixed <torsional warping>[warping] derivatives gives $\boxed{\Delta\phi=-2\mu\Omega}$. On the lateral boundary the axial <traction> is $\phi_y n_x-\phi_xn_y$, the tangential derivative of $\phi$. <Traction>-free walls therefore make $\phi$ constant on the connected boundary, and its arbitrary additive constant can be chosen so $\boxed{\phi=0\text{ on }\partial D}$.
The axial <torque> is
$$
M=\int_D(x\tau_{yz}-y\tau_{xz})\,dA=-\int_D(x\phi_x+y\phi_y)\,dA=\boxed{2\int_D\phi\,dA}.
$$
The <divergence theorem> removes the boundary term because $\phi=0$.
Tresca's yield criterion requires the maximum <shear stress>, half the largest principal-<stress> difference, not to exceed the material's shear <yield stress> $\tau_Y$. The <stress> <eigenvalues> here are $0,\pm\sqrt{\tau_{xz}^2+\tau_{yz}^2}$, so the maximum shear is $|\nabla\phi|$ and
$$
\boxed{|\nabla\phi|\leq\tau_Y.}
$$
If yield is instead quoted as a uniaxial tensile <stress>, that parameter is $2\tau_Y$; the present convention is the source's shear parameter.
For the <ellipse>, the proposed quadratic vanishes on its boundary. Its Laplacian fixes
$$
\boxed{C=\frac{\mu\Omega a^2b^2}{a^2+b^2},\qquad\phi=C(1-x^2/a^2-y^2/b^2).}
$$
The corresponding <torsional warping>[warping] is $w=\Omega(b^2-a^2)xy/(a^2+b^2)+\text{constant}$. Elliptic polar coordinates $x=ar\cos\vartheta$, $y=br\sin\vartheta$ give $\int_D\phi=C\pi ab/2$, hence
$$
\boxed{M=\frac{\pi\mu\Omega a^3b^3}{a^2+b^2},\qquad R=M/\Omega=\frac{\pi\mu a^3b^3}{a^2+b^2}.}
$$
To locate first yield, maximize $|\nabla\phi|=2|C|\sqrt{x^2/a^4+y^2/b^4}$. It increases radially to the boundary. For $a>b$, the maximum there occurs at $(0,\pm b)$ and equals $2|C|/b$. Therefore
$$
\boxed{|\Omega|_Y=\frac{(a^2+b^2)\tau_Y}{2a^2b\mu}.}
$$
The positive-twist value is the requested expression. The ends of the minor axis yield first, not the ends of the longer axis. This is <elliptical-bar torsion and first yield>.
Back to article page
