9.1 Torsion of uniform bars (shafts)
Straight bar with constant cross section subjected to equal and opposite end torques.
- We will set the origin of the coordinate system at the center of twist.
- In plane displacements \(u\) and \(v\) are zero along the \(z\)-axis.
- define \(α\) is the twist angle (the total angle of rotation from the \(z\) origin)
- Cross section can warp (\(w\) displacement), however the projection onto the \(x\)-\(y\) plane rotates as a rigid body.
- The \(w\) displacement is assumed to be independent of \(z\):
Define our twist per unit length \[\begin{equation}θ =\frac{α}{z}\end{equation}\]
The displacement components at point \(P\) (as \(P\) goes to \(P'\)) are: \[\begin{align} u(x,y,z) =& \, - r α \sin β = -α y = -θ z y\\ v(x,y,z) =& \, + r α \cos β = +α x = +θ z x\\ \end{align}\]
The w displacement is assumed to a function independent of \(z\): \[\begin{equation}w(x,y) = θ \psi (x,y)\end{equation}\]
\(\psi\) is the warping function.
Using these assumptions, several of the strains are shown to be zero (by their definition, i.e., take appropriate derivatives of displacement).
\[\begin{align} u(x,y,z) =& \, - r α \sin β = -α y = -θ z y\\ v(x,y,z) =& \, - r α \cos β = α x = θ z x\\ w(x,y) =& \, + θ \psi (x,y) \\ \end{align}\]
\[\begin{align} {\varepsilon_{xx}}&= {\frac{\partial u}{\partial x}} = 0 \\ {\varepsilon_{yy}}&= {\frac{\partial v}{\partial y}} = 0 \\ {\varepsilon_{zz}}&= {\frac{\partial w}{\partial z}} = 0 \\ {\gamma_{xy}}&= {\frac{\partial u}{\partial y}} + {\frac{\partial v}{\partial x}} = 0 \\ \end{align}\]
Assume the body forces are zero. Our equilibrium equations become: \[\begin{equation}{\frac{\partial {\tau_{xz}}}{\partial x}} + {\frac{\partial {\tau_{yz}}}{\partial y}} =0\end{equation}\]
Prandtl introduced a stress function \(ϕ(x,y)\) such that: \[\begin{align} {\tau_{xz}}=& \, + {\frac{\partial ϕ}{\partial y}} \\ {\tau_{yz}}=& \, - {\frac{\partial ϕ}{\partial x}} \\ \end{align}\]
At this point, we don’t know what \(ϕ\) is, however we know that is satisfies equilibrium.
We know by the definition of strain that: \[\begin{align} {\gamma_{xz}}=& \, {\frac{\partial w}{\partial x}} +{\frac{\partial u}{\partial z}} \\ {\gamma_{yz}}=& \, {\frac{\partial w}{\partial y}} +{\frac{\partial v}{\partial z}} \\ \end{align}\]
Using our deformation assumption: \[\begin{align} {\gamma_{xz}}=& \, {\frac{\partial w}{\partial x}} - θ y \\ {\gamma_{yz}}=& \, {\frac{\partial w}{\partial y}} + θ x \\ \end{align}\]
Taking partials of the above equation leads to: \[\begin{align} {\frac{\partial {\gamma_{xz}}}{\partial y}} =& \, {\frac{\partial^2 w}{\partial x \partial y}} - θ \\ {\frac{\partial {\gamma_{yz}}}{\partial x}} =& \, {\frac{\partial^2 w}{\partial y \partial x}} + θ \\ \end{align}\]
Carried from the last slide: \[\begin{align} {\frac{\partial {\gamma_{xz}}}{\partial y}} =& \, {\frac{\partial^2 w}{\partial x \partial y}} - θ \\ {\frac{\partial {\gamma_{yz}}}{\partial x}} =& \, {\frac{\partial^2 w}{\partial y \partial x}} + θ \\ \end{align}\]
Subtracting the 1st equation from the second: \[\begin{align} {\frac{\partial {\gamma_{yz}}}{\partial x}} - {\frac{\partial {\gamma_{xz}}}{\partial y}} =& \, 2 θ \\ \end{align}\] This equation defines the compatibility equation for torsion of the shaft.
Using our constitutive relationship: \[\begin{align} {\gamma_{yz}}=& \, \frac{1}{G} {\tau_{yz}}\\ {\gamma_{xz}}=& \, \frac{1}{G} {\tau_{xz}}\\ \end{align}\]
Leads to: \[\begin{align} {\frac{\partial {\tau_{yz}}}{\partial x}} - {\frac{\partial {\tau_{xz}}}{\partial y}} =& \, 2 G θ \\ \end{align}\]
In the above equation, we can replace the stresses with our stress function:
\[\begin{align} {\frac{\partial^2 ϕ}{\partial x \partial x}} + {\frac{\partial^2 ϕ}{\partial y \partial y}} =& \, - 2 G θ \\ \end{align}\]
The solution of a torsional problem now reduces to finding an appropriate function \(ϕ\) that satisfies the boundary conditions.