Example: Equilateral triangle [ edit | edit source ]
Torsion of a cylinder with a triangular cross section

The equations of the three sides are

${\begin{aligned}{\text{side}}~\partial S^{(1)}~:~~&f_{1}(x_{1},x_{2})=x_{1}-{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(2)}~:~~&f_{2}(x_{1},x_{2})=x_{1}+{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(3)}~:~~&f_{3}(x_{1},x_{2})=x_{1}-a=0\end{aligned}}$
Let the Prandtl stress function be

$\phi =Cf_{1}f_{2}f_{3}\,$
Clearly, $\phi =0\,$ at the boundary of the cross-section (which is what we need for solid cross sections).

Since, the traction-free boundary conditions are satisfied by $\phi \,$ , all we have to do is satisfy the compatibility condition to get the value of $C\,$ . If we can get a closed for solution for $C\,$ , then the stresses derived from $\phi \,$ will satisfy equilibrium.

Expanding $\phi \,$ out,

$\phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)$
Plugging into the compatibility condition

$\nabla ^{2}{\phi }=12Ca=-2\mu \alpha$
Therefore,

$C=-{\frac {\mu \alpha }{6a}}$
and the Prandtl stress function can be written as

$\phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})$
The torque is given by

$T=2\int _{S}\phi dA=2\int _{-2a}^{a}\int _{-(x_{1}+2a)/{\sqrt {3}}}^{(x_{1}+2a)/{\sqrt {3}}}\phi dx_{2}dx_{1}={\frac {27}{5{\sqrt {3}}}}\mu \alpha a^{4}$
Therefore, the torsion constant is

${\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}$
The non-zero components of stress are

${\begin{aligned}\sigma _{13}=\phi _{,2}&={\frac {\mu \alpha }{a}}(x_{1}-a)x_{2}\\\sigma _{23}=-\phi _{,1}&={\frac {\mu \alpha }{2a}}(x_{1}^{2}+2ax_{1}-x_{2}^{2})\end{aligned}}$
The projected shear stress

$\tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}$
is plotted below

Stresses in a cylinder with a triangular cross section under torsion

The maximum value occurs at the middle of the sides. For example,
at $(a,0)$ ,

$\tau _{\text{max}}={\frac {3\mu \alpha a}{2}}$
The out-of-plane displacements can be obtained by solving for the
warping function $\psi$ . For the equilateral triangle, after some
algebra, we get

$u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})$
The displacement field is plotted below

Displacements

$u_{3}\,$ in a cylinder with a triangular cross section.