Torsion

Torsion of Circular Cylinders

Torsion of a cylinder with a circular cross section

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

$\text{(8)} \qquad {\phi = \frac{TL}{JG}}$

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

Solution:

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

• 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

$\mathbf{x} = \tilde{\mathbf{x}}(s) ~,~~ 0 \le s \le l~~;~~~ \tilde{\mathbf{x}}(0) = \tilde{\mathbf{x}}(l)$

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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