# Elasticity/Torsion of noncircular cylinders
## Torsion of Non-Circular Cylinders

* Torsion of a noncircular cylinder

### About the problem

* Solution first found by St. Venant.
* Tractions at the ends are statically equivalent to equal and opposite torques $\pm \mathbf {T} =\pm T{\widehat {\mathbf {e} }}{3}\,$.
* Lateral surfaces are traction-free.

#### Assumptions:

* An axis passes through the center of twist ($x_{3}\,$ axis).
* Each c.s. projection on to the $x_{1}-x_{2}\,$ plane rotates,but remains undistorted.
* The rotation of each c.s. ($\phi \,$) is proportional to $x_{3}\,$.
    $\phi =\alpha x_{3}\,$
where $\alpha \,$ is the twist per unit length.
* The out-of-plane distortion (warping) is the same for each c.s. and is proportional to $\alpha \,$.

#### Find:

* Torsional rigidity ($T/\alpha \,$).
* Maximum shear stress.

### Solution:
#### Displacements

    ${\begin{aligned}u_{1}&=r\cos(\phi +\theta )-r\cos \theta =x_{1}(\cos \phi -1)-x_{2}\sin \phi \\u_{2}&=r\sin(\phi +\theta )-r\sin \theta =x_{1}\sin \phi +x_{2}(\cos \phi -1)\\u_{3}&=\alpha \psi (x_{1},x_{2})\end{aligned}}$
where $\psi (x_{1},x_{2})\,$ is the warping function.
If $\phi =\alpha x_{3}<<1\,$ (small strain),
    ${\text{(10)}}\qquad {u_{1}\approx -\alpha x_{2}x_{3}~;~~u_{2}\approx \alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}$

#### Strains

    $\varepsilon _{ij}={\frac {1}{2}}\left(u_{i,j}+u_{j,i}\right)$
Therefore,
    ${\begin{aligned}\varepsilon _{11}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{22}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{33}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{kk}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0\\\varepsilon _{12}&={\frac {1}{2}}\left(-\alpha x_{3}+\alpha x_{3}\right)=0\\\varepsilon _{23}&={\frac {1}{2}}\left(\alpha \psi _{,2}+\alpha x_{1}\right){\text{(11)}}\qquad \\\varepsilon _{31}&={\frac {1}{2}}\left(\alpha \psi _{,1}-\alpha x_{2}\right){\text{(12)}}\qquad \end{aligned}}$

#### Stresses

    $\sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}$
Therefore,
    ${\begin{aligned}\sigma _{11}&=0\\\sigma _{22}&=0\\\sigma _{33}&=0\\\sigma _{kk}&=0\\\sigma _{12}&=0\\\sigma _{23}&=\mu \alpha (\psi _{,2}+x_{1}){\text{(13)}}\qquad \\\sigma _{31}&=\mu \alpha (\psi _{,1}-x_{1}){\text{(14)}}\qquad \end{aligned}}$

#### Equilibrium

    $\sigma _{ji,j}=0~~~~{\text{no body forces.}}$
Therefore,
    ${\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}=0&\Rightarrow ~~0=0\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}=0&\Rightarrow ~~0=0\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}=0&\Rightarrow ~~\mu \alpha (\psi _{,11}+\psi _{,22})=\mu \alpha \nabla ^{2}{\psi }=0{\text{(15)}}\qquad \end{aligned}}$

#### Internal Tractions

* Normal to cross sections is ${\widehat {\mathbf {n} }}={\widehat {\mathbf {e} }}{3}\,$.
* Normal traction $t^{n}=\mathbf {t} \bullet {\widehat {\mathbf {n} }}=0\,$.
* Projected shear traction is $t^{s}={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}\,$.
* Traction vector at a point in the cross section is tangent to the cross section.

#### Boundary Conditions on Lateral Surfaces

* Lateral surface traction-free.
* Unit normal to lateral surface appears as an in-plane unit normal to the boundary $\partial S$.
We parameterize the boundary curve $\partial S$ using
    $\mathbf {x} ={\tilde {\mathbf {x} }}(s)~,~~0\leq s\leq l~~;~~~{\tilde {\mathbf {x} }}(0)={\tilde {\mathbf {x} }}(l)$
The tangent vector to $s$ is
    ${\widehat {\boldsymbol {\nu }}}={\frac {d\mathbf {x} }{ds}}~{\text{and}}~~{\widehat {\mathbf {n} }}={\widehat {\boldsymbol {\nu }}}\times {\widehat {\mathbf {e} }}_{3}~~\Rightarrow ~~~{\widehat {\mathbf {n} }}={\frac {dx_{2}}{ds}}{\widehat {\mathbf {e} }}{1}-{\frac {dx_{1}}{ds}}{\widehat {\mathbf {e} }}_{2}$
The tractions $t_{1}\,$ and $t_{2}\,$ on the lateral surface are identically zero. However, to satisfy the BC $t_{3}=0\,$, we need
    $t_{3}=n_{1}\sigma _{13}+n_{2}\sigma _{23}=0~~\Rightarrow ~~~\left(\psi _{,1}-x_{2}\right)n_{1}+\left(\psi _{,2}+x_{1}\right)n_{2}=0$
or,
    ${\text{(16)}}\qquad \left(\psi _{,1}-x_{2}\right){\frac {dx_{2}}{ds}}+\left(\psi _{,2}+x_{1}\right){\frac {dx_{1}}{ds}}=0$

#### Boundary Conditions on End Surfaces

The traction distribution is statically equivalent to the torque $\mathbf {T} \,$. At $x_{3}=L\,$,
    $t_{1}=\sigma _{13}~;~~t_{2}=\sigma _{23}~;~~t_{3}=\sigma _{33}=0\,$
Therefore,
    $F_{1}=\int _{S}\sigma _{13}~dS=\mu \alpha \int _{S}(\psi _{,1}-x_{2})~dS$
From equilibrium,
    ${\begin{aligned}\nabla ^{2}{\psi }=0~~\Rightarrow ~~~\psi _{,1}-x_{2}&=(\psi _{,1}-x_{2})+x_{1}(\psi _{,11}+\psi _{,22})\\&=\psi _{,1}+x_{1}\psi _{,11}-x_{2}+x_{1}\psi _{,22}\\&=(x_{1}\psi _{,1}-x_{1}x_{2})_{,1}+(x_{1}\psi _{,2}+x_{1}x_{1})_{,2}\\&=\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}\end{aligned}}$
Hence,
    ${\text{(17)}}\qquad F_{1}=\mu \alpha \int _{S}\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}dS$

#### The Green-Riemann Theorem

If $P=f(x_{1},x_{2})\,$ and $Q=q(x_{1},x_{2})\,$ then
    $\int _{S}(Q_{,1}-P_{,2})dS=\oint _{\partial S}(Pdx_{1}+Qdx_{2})$
with the integration direction such that $S$ is to the left.
Applying the Green-Riemann theorem to equation (17), and using equation (16)
    ${\text{(18)}}\qquad F_{1}=\mu \alpha \oint _{\partial S}-x_{1}(\psi _{,2}+x_{1})dx_{1}+x_{1}(\psi _{,1}-x_{2})dx_{2}=0$
Similarly, we can show that $F_{2}=0\,$. $F_{3}=0\,$ since $t_{3}=0\,$.
The moments about the $x_{1}\,$ and $x_{2}\,$ axes are also zero.
The moment about the $x_{3}\,$ axis is
    $M_{3}=\int _{S}(x_{1}\sigma _{23}-x_{2}\sigma _{13})dS=\mu \alpha \int _{S}(x_{1}\psi _{,2}+x_{1}^{2}-x_{2}\psi _{1}+x_{2}^{2})dS=\mu \alpha {\tilde {J}}$
where $J\,$ is the torsion constant. Since $M_{3}=T\,$, we have
    $\alpha ={\frac {T}{\mu {\tilde {J}}}}$
If $\psi =0\,$, then ${\tilde {J}}=J\,$, the polar moment of inertia.

## Summary of the solution approach

* Find a warping function $\psi (x_{1},x_{2})\,$ that is harmonic. and satisfies the traction BCs.
* Compatibility is not an issue since we start with displacements.
* The problem is independent of applied torque and the material properties of the cylinder.
* So it is just a geometrical problem. Once $\psi$ is known, we can calculate
** The displacement field.
** The stress field.
** The twist per unit length.