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.

Torsion boundary condition

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

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

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

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

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

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.)

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!

Integration of torque

\[\begin{align} dT =& \, x {\tau_{yz}}dA - y {\tau_{xz}}dA \\ =& \, \left[ -x {\frac{\partial ϕ}{\partial x}} - y {\frac{\partial ϕ}{\partial y}} \right] dA \end{align}\]

The total torque must be integrated (by parts): \[\begin{align} T =& \, - \iint_A \left[ x {\frac{\partial ϕ}{\partial x}} + y {\frac{\partial ϕ}{\partial y}} \right] dx dy \\ % =& \, - \iint_A \left[ {\frac{\partial }{\partial x}} (x ϕ) - ϕ \right] dx dy - \iint_A \left[ {\frac{\partial }{\partial y}} (y ϕ) - ϕ \right] dx dy \\ =& \, 2 \iint_A ϕ dx dy - \int [x ϕ]_{x_1}^{x_2} dy - \int [y ϕ]_{y_1}^{y_2} dx \\ =& \, 2 \iint_A ϕ dx dy +0 +0\\ \end{align}\]

• The last two are zero because the \(ϕ\) is constant and it is a closed path integral.
• This indicates that the solution of the torsion problem lies in finding a stress function that is constant along the lateral boundary of the bar.

Notes

• After \(ϕ\) is determined, the location of the center of twist is also determined.
• Out of plane displacement (warping) can be obtained by integrating \(\partial w / \partial x\) and \(\partial w / \partial y\)

Summary of key equations

• Compatibility and stress equillibrium \[\begin{align} {\frac{\partial^2 ϕ}{\partial x \partial x}} + {\frac{\partial^2 ϕ}{\partial y \partial y}} =& \, - 2 G θ \\ \end{align}\]

• Torque equillibrium \[\begin{align} T =& \, 2 \iint_A ϕ dx dy \\ \end{align}\]

• Boundary condition \[\begin{align} ϕ = \mathrm{constant} \left( =0 \right) \end{align}\]

• Find \(ϕ\) that satisfies all

Examples

Example of a bar with a circular cross section

The equation of a circular cross section is: \[\begin{equation}x^2 + y^2 = a^2\end{equation}\] where \(a\) is the radius of the circular boundary

Assume the following stress function: \[\begin{equation}ϕ = C \left( \frac{x^2}{a^2} + \frac{y^2}{a^2} -1\right)\end{equation}\]

Using our compatibility equation: \[\begin{align} {\frac{\partial^2 ϕ}{\partial x \partial x}} + {\frac{\partial^2 ϕ}{\partial y \partial y}} =& \, - 2 G θ \\ \frac{2C}{a^2} + \frac{2C}{a^2} =& \, - 2 G θ \\ \end{align}\]

Leads to: \[\begin{equation}C = - \frac{1}{2} a^2 G θ\end{equation}\]

In general, we would maximize \(ϕ\) (once it is known) to find the center.

Integrating the torque: \[\begin{align} T =& \, 2 \iint_A ϕ dx dy \\ =& \, 2 C \iint_A \left( \frac{x^2}{a^2} + \frac{y^2}{a^2} -1\right) dx dy \\ =& \, 2 C \iint_A \left( \frac{r^2}{a^2} -1\right) dA \\ \end{align}\]

Note the above suggests we define a polar second moment of the area (a section property called \(J\)):

\[\begin{align} J \equiv & \, \iint_A r^2 dA \\ =& \, \iint_A r^2 r \, dr \, dθ\\ =& \, \iint_A r^3 \, dr \, dθ\\ J =& \, \frac{1}{2} \pi a^4 \\ \end{align}\]

Thus: \[\begin{align} T=& \, 2 C \left( \frac{J}{a^2} -A\right) \\ \end{align}\]

The cross sectional area is: \[\begin{align} A =& \, \pi a^2 \\ a^2 A =& \, 2 J \\ \end{align}\]

Thus: \[\begin{equation}T = -\frac{2 C J}{a^2} = G J θ\end{equation}\]

We recognize \(GJ\) as the torsional rigidity and is analogous to \(EI\): \[\begin{align} M =& \, EI v'' \\ T =& \, GJ α' \\ \end{align}\]

Therefore, the shear stress is: \[\begin{align} {\tau_{xz}}=& \, + {\frac{\partial ϕ}{\partial y}} = + 2 \, C \frac{y}{a^2} = - G θ y \\ {\tau_{yz}}=& \, - {\frac{\partial ϕ}{\partial x}} = - 2 \, C \frac{x}{a^2} = + G θ x \\ \end{align}\]

Example of stress distribution in a circular bar

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

The values of the traction components are: \[\begin{align} t_x =& \, 0 \\ t_y =& \, 0 \\ t_z =& \, {\tau_{xz}}n_x + {\tau_{yz}}n_y \\ \end{align}\]

We also know that: \[\begin{align} n_x =& \, \cos β \\ n_y =& \, \sin β \\ n_z =& \, 0 \\ \end{align}\]

Plugging in the known values of shear stress: \[\begin{align} t_z =& \, -G \, θ \, y \; n_x +G \, θ \, x \; n_y\\ =& \, -G \, θ \, y \cos β +G \, θ \, x \sin β \\ =& \, -G \, θ \, \frac{y x}{r} +G \, θ \, \frac{x y}{r} \\ =& \, 0 \\ \end{align}\]

\[\begin{align} t_z =& \, 0 \\ \end{align}\]

Thus the radial shear stress vanishes since it is equal to \(t_z\) \[\begin{equation}\tau_z = t_z\end{equation}\]

This time the normal is : \[\begin{align} n_x =& \, \sin β = \frac{y}{r} \\ n_y =& \, -\cos β = -\frac{x}{r} \\ n_z =& \, 0 \\ \end{align}\]

From: \[\begin{align} t_z =& \, {\tau_{xz}}n_x + {\tau_{yz}}n_y \\ \end{align}\]

This leads to: \[\begin{equation}t_z = -G \, θ \, r\end{equation}\]

From this we can define a tangential shear stress: \[\begin{align} \tau =& \, -t_z \\ =& \, G \, θ \, r \\ \tau =& \, \frac{T r}{J} \\ \end{align}\]

It can be shown that: \[\begin{equation}w =0\end{equation}\] for the circular cross section. There is no warping for a circular cross section.