Torsion

About the problem:

 * Circular Cylinder.
 * Centroidal axis thru the center of each c.s.
 * Length $$L$$, Outer radius $$c$$.
 * Applied torque $$T$$.
 * Angle of twist $$\phi$$.

Assumptions:

 * Each c.s. remains plane and undistorted.
 * Each c.s. rotates through the same angle.
 * No warping or change in shape.
 * Amount of displacement of each c.s. is proportional to  distance from end.

Find:

 * Shear strains in the cylinder ($$\gamma$$).
 * Shear stress in the cylinder ($$\tau$$).
 * Relation between torque ($$T$$) and angle of twist ($$\phi$$).
 * Relation between torque ($$T$$) and shear stress ($$\tau$$).

Solution:
If $$\gamma$$ is small, then
 * $$\text{(1)} \qquad

L\gamma = r\phi \Rightarrow {\gamma = \frac{r\phi}{L}} $$ Therefore,
 * $$\text{(2)} \qquad

\gamma_{\text{max}} = \frac{c\phi}{L} \Rightarrow \gamma = \frac{r} \gamma_{\text{max}} $$ If the deformation is elastic,
 * $$\text{(3)} \qquad

\tau = G\gamma \Rightarrow {\tau = \frac{r\phi G}{L}} $$ Therefore,
 * $$\text{(4)} \qquad

\tau_{\text{max}} = \frac{r\phi G}{L} \Rightarrow \tau = \frac{r} \tau_{\text{max}} $$ The torque on each c.s. is given by
 * $$\text{(5)} \qquad

T = \int_A \tau r dA = \frac{\phi G}{L}\int_A r^2 dA = \frac{G\phi J}{L} $$ where $$J$$ is the polar moment of inertia of the c.s.
 * $$\text{(6)} \qquad

J = \begin{cases} \frac{1}{2} \pi c^4 & \text{solid circular c.s.} \\ \frac{1}{2} \pi (c_2^{~4}-c_1^{~4}) & \text{ annular circular c.s.} \end{cases} $$ Therefore,
 * $$\text{(7)} \qquad

{\tau = \frac{Tr}{J}} \Rightarrow \tau_{\text{max}} = \frac{Tc}{J} $$ and
 * $$\text{(8)} \qquad

{\phi = \frac{TL}{JG}} $$

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 $$x_3$$. :$$\text{(9)} \qquad    \phi = \alpha x_3  $$  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.

Displacements

 * $$\begin{align}

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{align}$$ where $$\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{align}

\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{align}$$

Stresses


\sigma_{ij} = 2\mu\varepsilon_{ij} + \lambda\varepsilon_{kk}\delta_{ij} $$ Therefore,
 * $$\begin{align}

\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{align}$$

Equilibrium


\sigma_{ji,j} = 0 ~ \text{no body forces.} $$ Therefore,
 * $$\begin{align}

\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{align}$$

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
We parameterize the boundary curve $$\partial S$$ using
 * Lateral surface traction-free.
 * Unit normal to lateral surface appears as an in-plane unit  normal to the boundary $$\partial S$$.

\mathbf{x} = \tilde{\mathbf{x}}(s) ~, 0 \le s \le 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{align}

\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_1x_2)_{,1} + (x_1\psi_{,2} + x_1x_1)_{,2} \\ & = \left[x_1(\psi_{,1} - x_2)\right]_{,1} + \left[x_1(\psi_{,2} + x_1)\right]_{,2} \end{align}$$ Hence,
 * $$\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} (P dx_1 + Q dx_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.

The Torsion Problem Summarized

 * 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.

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