Introduction to Elasticity/Torsion of noncircular cylinders

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[edit] Torsion of Non-Circular Cylinders

Torsion of a noncircular cylinder

[edit] 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.

[edit] 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\,.

[edit] Find:

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

[edit] Solution:

[edit] 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) }

[edit] 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}

[edit] 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}

[edit] 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}

[edit] 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.

[edit] 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

[edit] 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

[edit] 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.

[edit] 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 ψ is known, we can calculate
    • The displacement field.
    • The stress field.
    • The twist per unit length.
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