Elasticity/Torsion of triangular cylinder

Example: Equilateral triangle

File:Torsion of triangle cylinder.png Torsion of a cylinder with a triangular cross section

The equations of the three sides are

$\begin{array}{cc}\text{side}\partial S^{(1)}:& f_1(x_1,x_2)=x_1-\sqrt{3}{x}_2+2a=0\\ \text{side}\partial S^{(2)}:& f_2(x_1,x_2)=x_1+\sqrt{3}{x}_2+2a=0\\ \text{side}\partial S^{(3)}:& f_3(x_1,x_2)=x_1-a=0\end{array}$

Let the Prandtl stress function be

$\varphi =C{f}_1{f}_2{f}_3$

Clearly, $\varphi =0$ 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 $\varphi$, 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 $\varphi$ will satisfy equilibrium.

Expanding $\varphi$ out,

$\varphi =C(x_1-\sqrt{3}{x}_2+2a)(x_1+\sqrt{3}{x}_2+2a)(x_1-a)$

Plugging into the compatibility condition

${\mathrm{\nabla }}^2\varphi =12Ca=-2\mu \alpha$

Therefore,

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

and the Prandtl stress function can be written as

$\varphi =-\frac{\mu \alpha }{6a}(x_1^3+3a{x}_1^2+3a{x}_2^2-3x_1{x}_2^2-4a^3)$

The torque is given by

$T=2\underset{S}{\int }\varphi dA=2\underset{-2a}{\overset{a}{\int }}\underset{-(x_1+2a)/\sqrt{3}}{\overset{(x_1+2a)/\sqrt{3}}{\int }}\varphi d{x}_{2}d{x}_1=\frac{27}{5\sqrt{3}}\mu \alpha a^4$

Therefore, the torsion constant is

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

The non-zero components of stress are

$\begin{array}{cc}{\sigma }_{13}={\varphi }_{,2}& =\frac{\mu \alpha }{a}(x_1-a)x_2\\ {\sigma }_{23}=-{\varphi }_{,1}& =\frac{\mu \alpha }{2a}(x_1^2+2a{x}_1-x_2^2)\end{array}$

The projected shear stress

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

is plotted below

File:Torsion triangle cs stress.png 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

File:Torsion triangle cs disp.png Displacements $u_3$ in a cylinder with a triangular cross section.

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