Elasticity/Torsion of noncircular cylinders

Torsion of Non-Circular Cylinders

File:Torsion of noncircular cylinder.png 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 $\pm 𝐓=\pm T\widehat{𝐞}3$.
• 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. ($\varphi$) is proportional to $x_3$.

$\varphi =\alpha x_3$

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

$\begin{array}{cc}{u}_1& =r\cos (\varphi +\theta )-r\cos \theta =x_1(\cos \varphi -1)-x_2\sin \varphi \\ u_2& =r\sin (\varphi +\theta )-r\sin \theta =x_1\sin \varphi +x_2(\cos \varphi -1)\\ u_3& =\alpha \psi (x_1,x_2)\end{array}$

where $\psi (x_1,x_2)$ is the warping function.

If $\varphi =\alpha x_3<<1$ (small strain),

$\text{(10)}{u}_1\approx -\alpha x_2{x}_3;u_2\approx \alpha x_1{x}_3;u_3=\alpha \psi (x_1,x_2)$

Strains

${\epsilon }_{ij}=\frac{1}{2}(u_{i,j}+u_{j,i})$

Therefore,

$\begin{array}{cc}{\epsilon }_{11}& =\frac{1}{2}(0+0)=0\\ {\epsilon }_{22}& =\frac{1}{2}(0+0)=0\\ {\epsilon }_{33}& =\frac{1}{2}(0+0)=0\\ {\epsilon }_{kk}& ={\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33}=0\\ {\epsilon }_{12}& =\frac{1}{2}(-\alpha x_3+\alpha x_3)=0\\ {\epsilon }_{23}& =\frac{1}{2}(\alpha {\psi }_{,2}+\alpha x_1)\text{(11)}\\ {\epsilon }_{31}& =\frac{1}{2}(\alpha {\psi }_{,1}-\alpha x_2)\text{(12)}\end{array}$

Stresses

${\sigma }_{ij}=2\mu {\epsilon }_{ij}+\lambda {\epsilon }_{kk}{\delta }_{ij}$

Therefore,

$\begin{array}{cc}{\sigma }_{11}& =0\\ {\sigma }_{22}& =0\\ {\sigma }_{33}& =0\\ {\sigma }_{kk}& =0\\ {\sigma }_{12}& =0\\ {\sigma }_{23}& =\mu \alpha ({\psi }_{,2}+x_1)\text{(13)}\\ {\sigma }_{31}& =\mu \alpha ({\psi }_{,1}-x_1)\text{(14)}\end{array}$

Equilibrium

${\sigma }_{ji,j}=0\text{no body forces.}$

Therefore,

$\begin{array}{cc}{\sigma }_{11,1}+{\sigma }_{21,2}+{\sigma }_{31,3}=0& \Rightarrow 0=0\\ {\sigma }_{12,1}+{\sigma }_{22,2}+{\sigma }_{32,3}=0& \Rightarrow 0=0\\ {\sigma }_{13,1}+{\sigma }_{23,2}+{\sigma }_{33,3}=0& \Rightarrow \mu \alpha ({\psi }_{,11}+{\psi }_{,22})=\mu \alpha {\mathrm{\nabla }}^2\psi =0\text{(15)}\end{array}$

Internal Tractions

• Normal to cross sections is $\widehat{𝐧}=\widehat{𝐞}3$.
• Normal traction $t^n=𝐭\bullet \widehat{𝐧}=0$.
• 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

$𝐱=\widetilde{𝐱}(s),0\le s\le l;\widetilde{𝐱}(0)=\widetilde{𝐱}(l)$

The tangent vector to $s$ is

$\hat{\mathit{\nu }}=\frac{d𝐱}{ds}\text{and}\widehat{𝐧}=\hat{\mathit{\nu }}\times {\widehat{𝐞}}_3\Rightarrow \widehat{𝐧}=\frac{d{x}_2}{ds}\widehat{𝐞}1-\frac{d{x}_1}{ds}{\widehat{𝐞}}_2$

The tractions $t_1$ and $t_2$ on the lateral surface are identically zero. However, to satisfy the BC $t_3=0$, we need

$t_3=n_1{\sigma }_{13}+n_2{\sigma }_{23}=0\Rightarrow ({\psi }_{,1}-x_2)n_1+({\psi }_{,2}+x_1)n_2=0$

or,

$\text{(16)}({\psi }_{,1}-x_2)\frac{d{x}_2}{ds}+({\psi }_{,2}+x_1)\frac{d{x}_1}{ds}=0$

Boundary Conditions on End Surfaces

The traction distribution is statically equivalent to the torque $𝐓$. At $x_3=L$,

$t_1={\sigma }_{13};t_2={\sigma }_{23};t_3={\sigma }_{33}=0$

Therefore,

$F_1=\underset{S}{\int }{\sigma }_{13}dS=\mu \alpha \underset{S}{\int }({\psi }_{,1}-x_2)dS$

From equilibrium,

$\begin{array}{cc}{\mathrm{\nabla }}^2\psi =0\Rightarrow {\psi }_{,1}-x_2& =({\psi }_{,1}-x_2)+x_1({\psi }_{,11}+{\psi }_{,22})\\ & ={\psi }_{,1}+x_1{\psi }_{,11}-x_2+x_1{\psi }_{,22}\\ & =(x_1{\psi }_{,1}-x_1{x}_2{)}_{,1}+(x_1{\psi }_{,2}+x_1{x}_1{)}_{,2}\\ & ={[x_1({\psi }_{,1}-x_2)]}_{,1}+{[x_1({\psi }_{,2}+x_1)]}_{,2}\end{array}$

Hence,

$\text{(17)}{F}_1=\mu \alpha \underset{S}{\int }{[x_1({\psi }_{,1}-x_2)]}_{,1}+{[x_1({\psi }_{,2}+x_1)]}_{,2}dS$

The Green-Riemann Theorem

If $P=f(x_1,x_2)$ and $Q=q(x_1,x_2)$ then

$\underset{S}{\int }(Q_{,1}-P_{,2})dS={\oint }_{\partial S}(Pd{x}_1+Qd{x}_2)$

with the integration direction such that $S$ is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)

$\text{(18)}{F}_1=\mu \alpha {\oint }_{\partial S}-x_1({\psi }_{,2}+x_1)d{x}_1+x_1({\psi }_{,1}-x_2)d{x}_2=0$

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

$M_3=\underset{S}{\int }(x_1{\sigma }_{23}-x_2{\sigma }_{13})dS=\mu \alpha \underset{S}{\int }(x_1{\psi }_{,2}+x_1^2-x_2{\psi }_1+x_2^2)dS=\mu \alpha \tilde{J}$

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.