9.3 Integration of torque

@Sun2006 Figure 3.4: Shear stress on a cross section

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

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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.

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9.3.1 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 \(\partial w / \partial y\)

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9.3.2 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

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9.3.3 Examples

9.3.3.1 Example of a bar with a circular cross section

Circular Shaft

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

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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.

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

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

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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 \(EI\): \[\begin{align} M =& \, EI v'' \\ T =& \, GJ α' \\ \end{align}\]

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

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9.3.3.2 Example of stress distribution in a circular bar

Shear stress on a cylindrical cut (section) with radius \(r\)

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

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

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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 \(t_z\) \[\begin{equation}\tau_z = t_z\end{equation}\]

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Shear stress on a radial cut (section) \(r\)

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

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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: \[\begin{align} \tau =& \, -t_z \\ =& \, G \, θ \, r \\ \tau =& \, \frac{T r}{J} \\ \end{align}\]

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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.

Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture 0.