# Elasticity/Torsion of triangular cylinder

## Example: Equilateral triangle

 Torsion of a cylinder with a triangular cross section

The equations of the three sides are

{\displaystyle {\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

${\displaystyle \phi =Cf_{1}f_{2}f_{3}\,}$

Clearly, ${\displaystyle \phi =0\,}$  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 ${\displaystyle \phi \,}$ , all we have to do is satisfy the compatibility condition to get the value of ${\displaystyle C\,}$ . If we can get a closed for solution for ${\displaystyle C\,}$ , then the stresses derived from ${\displaystyle \phi \,}$  will satisfy equilibrium.

Expanding ${\displaystyle \phi \,}$  out,

${\displaystyle \phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)}$

Plugging into the compatibility condition

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

Therefore,

${\displaystyle C=-{\frac {\mu \alpha }{6a}}}$

and the Prandtl stress function can be written as

${\displaystyle \phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})}$

The torque is given by

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

Therefore, the torsion constant is

${\displaystyle {\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}}$

The non-zero components of stress are

{\displaystyle {\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

${\displaystyle \tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}}$

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 ${\displaystyle (a,0)}$ ,

${\displaystyle \tau _{\text{max}}={\frac {3\mu \alpha a}{2}}}$

The out-of-plane displacements can be obtained by solving for the warping function ${\displaystyle \psi }$ . For the equilateral triangle, after some algebra, we get

${\displaystyle u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})}$

The displacement field is plotted below

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