9.2 Torsion boundary condition

@Sun2006 Figure 3.3: Cross section tangents and normal

For a lateral surface of the bar: \[\begin{equation}\{t\} = [\sigma] \{n\}\end{equation}\]

This leads to: \[\begin{equation}\left\{ \begin{array}{c} t_x \\ t_y \\ t_z \end{array} \right\} = \left[ \begin{array}{ccc} 0 & 0 & {\tau_{xz}}\\ 0 & 0 & {\tau_{yz}}\\ {\tau_{xz}}& {\tau_{yz}}& 0\\ \end{array} \right] \left\{ \begin{array}{c} n_x \\ n_y \\ 0 \end{array} \right\}\end{equation}\]

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From the prior slide: \[\begin{equation}\left\{ \begin{array}{c} t_x \\ t_y \\ t_z \end{array} \right\} = \left[ \begin{array}{ccc} 0 & 0 & {\tau_{xz}}\\ 0 & 0 & {\tau_{yz}}\\ {\tau_{xz}}& {\tau_{yz}}& 0\\ \end{array} \right] \left\{ \begin{array}{c} n_x \\ n_y \\ 0 \end{array} \right\}\end{equation}\]

Hence the values of the traction components are: \[\begin{align} t_x =& \, 0 \\ t_y =& \, 0 \\ t_z =& \, {\tau_{xz}}n_x + {\tau_{yz}}n_y = {\frac{\partial ϕ}{\partial y}} n_x - {\frac{\partial ϕ}{\partial x}} n_y \\ \end{align}\]

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Define a tangent-normal coordinate system \(s,n\): \[\begin{align} n_x =& \, 1 \cdot \sin η = + {\frac{\partial y}{\partial s}} \\ n_y =& \, 1 \cdot \cos η = - {\frac{\partial x}{\partial s}} \\ \end{align}\]

A change of variables (into tangent and normal directions)

and the chain rule

lead to: \[\begin{align} t_z =& \, {\frac{\partial ϕ}{\partial y}} {\frac{\partial y}{\partial s}} + {\frac{\partial ϕ}{\partial x}} {\frac{\partial x}{\partial s}} \\ t_z =& \, \frac{d ϕ}{d s} \\ \end{align}\]

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On any lateral surface, we also know that: \[\begin{align} t_z =& \, 0 \end{align}\]

due to the traction free boundary.

Therefore: \[\begin{equation}ϕ = \text{constant} \end{equation}\]

Note: is it typically best to force \(ϕ = 0\) on a boundary. (This will simplify the mathematics.)

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A boundary condition is identified for a stress function \(ϕ\) which will satisfy equilibrium if it is found, however we are interested in the stresses on the cross section. This must be related to the torque!

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