# Introduction to Elasticity/Torsion of triangular cylinder

## Example: Equilateral Triangle

 Torsion of a cylinder with a triangular cross section

The equations of the three sides are

\begin{align} \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{align}

Let the Prandtl stress function be

$\phi = C f_1 f_2 f_3 \,$

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

$\nabla^2{\phi} = 12 C a = -2\mu\alpha$

Therefore,

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

and the Prandtl stress function can be written as

$\phi = -\frac{\mu\alpha}{6a} (x_1^3+3ax_1^2+3ax_2^2-3x_1x_2^2-4a^3)$

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

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

The non-zero components of stress are

\begin{align} \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{align}

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.