10.6 Torque and shear flow in a closed cell

---

The content of this section was delivered over several lectures:

• Part 1
• Part 2

---

10.6.1 Torque of a closed cell

@AriGur2007

• Thin wall, \(t \ll r_{\text min}\)
• \(s\) = tangential coordinate (measured along the mid-thickness line)
• \(t = t(s)\); (measured perpendicular to \(s\))

---

Assumptions:

1. \(T\) causes only \(\tau_{zs}\) at the wall (pure torsion)
2. \(\tau_{zs}\) is uniform through the wall thickness (average)

---

We have deduced an equilibrium equation for shear flow:

Equilibrium element

The axial stress is: \[\begin{align} \sum F_z =& 0 \\ \Delta {\sigma_{zz}}\cdot \Delta s \cdot t + \Delta q \cdot \Delta z =& \; 0\\ \frac{\Delta {\sigma_{zz}}}{\Delta z} \cdot t + \frac{\Delta q}{\Delta s} =& \; 0\\ \end{align}\]

---

Therefore the shear flow and axial stress are related through: \[\begin{equation}{\frac{\partial q}{\partial s}} = -t {\frac{\partial {\sigma_{zz}}}{\partial z}}\end{equation}\]

Importantly, this implies that if there are no resulting bending moments and no resulting shear loads (ie pure torsion), we know that the value of \(q\) is constant.

\[\begin{equation}\sum F_z = 0 \Longrightarrow \tau_{sz2} \cdot t_2 \cdot \Delta z = \tau_{sz1} \cdot t_1 \cdot \Delta z\end{equation}\]

\[\begin{equation}\tau_{sz2} \cdot t_2 = \tau_{sz1} \cdot t_1 =\tau_{zs}(s) \cdot t\end{equation}\] Define \(q\) as the shear flow. \(\tau_{zs}(s) \cdot t =q\)

---

The shear flow (\(q\)) is constant along the wall of a cross-section of a monocoque beam under pure torsion

@AriGur2007

\[\begin{equation}\displaystyle \sum F_z = 0 \rightarrow \tau_{sz2} \cdot t_2 \cdot \Delta z = \tau_{sz1} \cdot t_1 \cdot \Delta z\end{equation}\] \[\begin{equation}\tau_{sz2} \cdot t_2 = \tau_{sz1} \cdot t_1 = \tau_{sz} \cdot t = q = \text{shear flow}\end{equation}\]

The shear flow (\(q\)) is constant along the wall of a cross-section of a monocoque beam under pure torsion

---

Differential torque contribution due to \(q \cdot ds\) @AriGur2007

• \(q =\) shear flow \[force/length\]
• \(q \cdot ds\) is the tangential force (vector)
• \(r\) is the position vector of \(q \cdot ds\) (measured from a point in the cross-section)

---

• Note: \(r\) and \(q \cdot ds\) are vectors that are not (necessarily) perpendicular
• The resultant (shear) force on the cross-section is zero: \[\begin{align} \oint q \cdot ds =& \;0 \\ \end{align}\]
• The resultant torque is: \[\begin{align} \oint r \times (q \cdot ds) =& \; T \end{align}\]

---

• For constant shear flow:
  • \(q \oint r \times ds = T\)
  • \(d A = {\frac{1}{2}}ds \cdot d = {\frac{1}{2}}ds \cdot r \cdot sin(\beta)\)
  • \(d A = {\frac{1}{2}}\left(r \times ds\right)\)
  • \(r \times ds = 2 \cdot dA\)

---

• Thus, the total torque is:
  • \(\vec{T} = T \hat{k} = q \oint r \times ds =\)
  • \(= q \cdot 2 A = 2 q A \hat{k}\)
  • \(A\) = cross-section enclosed area (average)
• Thus, the shear flow is:
  • \(q = \frac{T}{2 A}\)
  • Shear flow is a constant!!!

---

10.6.2 Torque twist rate relationship

@AriGur2007

• We’ve shown that: \[\begin{equation}q = \frac{T}{2A}\end{equation}\]

---

• Recall: \[\begin{equation}\tau_{sz} \cdot t = q\end{equation}\]

• Thus: \[\begin{equation}\tau_{sz} = \frac{T}{(2A \cdot t)}\end{equation}\] \[\begin{equation}\gamma_{sz} = \frac{\tau_{sz}}{G} = \frac{T}{(2G \cdot A \cdot t)} = \frac{q}{(G \cdot t)}\end{equation}\]

• Shear strain depends on shear flow, the modulus, and \(t\)!

• Not constants but a function of \(s\).

---

• Due to the torque \(T\), the cross-section rotates about \(z\)-axis: \(\alpha(z)\)

@AriGur2007

---

• Recall the definition of strain, now related in our tangent-normal-z coordinate system: \[\begin{equation}\gamma_{sz} = \frac{\partial u_s}{\partial z} + \frac{\partial w}{\partial s}\end{equation}\]

• \(u_s\) is the displacement along the arc-length direction. Warping is small, thus: \[\begin{equation}\frac{\partial w}{\partial s} \approx 0\end{equation}\]

• The two arc-lengths are the same, thus:

\[\begin{equation}\gamma_{sz} \cdot \Delta z = \Delta \alpha \cdot r\end{equation}\]

\[\begin{equation}\frac{\Delta \alpha}{\Delta z} = \frac{\gamma_{sz}}{r}\end{equation}\]

\[\begin{equation}\theta \equiv \lim_{\Delta \rightarrow 0} \frac{\Delta \alpha}{\Delta z} = \frac{\gamma_{sz}}{r}\end{equation}\]

---

• Recall by prior geometric argument: \[\begin{equation}2A = \oint r \times ds\end{equation}\]

• Thus: \[\begin{equation}2A = \oint \frac{\gamma_{sz}}{\theta} ds\end{equation}\]

• and:

\[\begin{equation} \theta = \oint \frac{\gamma_{sz} ds}{(2A)} \end{equation}\]

---

There are other ways to write equivalent mathematical statements:

• Recall also that: \(\gamma_{sz} = \frac{q}{(G \cdot t)}\)

• Substitute into (10.1) and pull out the constants:

• Leading to: \[\begin{equation}\theta = ({\frac{1}{2}}\frac{q}{A}) \oint \left(G \cdot t\right)^{-1} ds\end{equation}\]

• Recall also the constant shear flow is: \[\begin{equation}q = {\frac{1}{2}}\frac{T}{A}\end{equation}\]

• Where \(A\) = enclosed cell area (average)

---

• Thus:

\[\begin{equation}\theta = \frac{T}{(4 A^2)} \oint \left(G \cdot t\right)^{-1} ds\end{equation}\]

• We can use the above to compute an effective torsional rigidity \(GJ\) because we know: \[\begin{equation}GJ \theta=T\end{equation}\]

---

10.6.3 Example: torsional rigidity of a circle and a square

A hollow square and circle

---

10.6.3.1 Same thickness, same geometric width.

\[\begin{equation}b=d\end{equation}\]

\[\begin{equation}A^{\mathrm{sq}}\left(d\right)=d^{2}\end{equation}\] \[\begin{equation}A^{\mathrm{cir}}\left(d\right)=\frac{d^{2} \,\pi}{4}\end{equation}\] \[\begin{equation}\theta=\frac{T \oint{\frac{1}{t\left(s\right) \,G\left(s\right)}\;ds}}{4 \,{A_{\mathrm{Enc}}}^{2}}\end{equation}\]

\[\begin{equation}GJ_{\mathrm{ef}}^{\mathrm{sq}}=\frac{T}{\theta}=d^{3} \,t \,G\end{equation}\] \[\begin{equation}GJ_{\mathrm{ef}}^{\mathrm{cir}}=\frac{T}{\theta}=\frac{d^{3} \,\pi \,t \,G}{4}\end{equation}\]

\[\begin{equation}\frac{GJ_{\mathrm{ef}}^{\mathrm{sq}}}{GJ_{\mathrm{ef}}^{\mathrm{cir}}}=\frac{4}{\pi} = 1.27\end{equation}\]

---

10.6.3.2 Same material area (cross section area), same thickness

---

10.6.4 Same material area (cross section area), same thickness

\[\begin{equation}b \neq d\end{equation}\]

\[\begin{equation}A_{\mathrm{sec}}^{\mathrm{sq}}\left(b\right)=4 \,b \,t\end{equation}\] \[\begin{equation}A_{\mathrm{sec}}^{\mathrm{cir}}\left(d\right)=d \,\pi \,t\end{equation}\]

Equate the cross sectional areas \[\begin{equation}4 \,b \,t=d \,\pi \,t\end{equation}\]

Solve for the dimension of the square \[\begin{equation}b=\frac{d \,\pi}{4}\end{equation}\]

---

Now what are the enclosed areas? \[\begin{equation}A^{\mathrm{sq}}=\frac{d^{2} \,\pi^{2}}{16}\end{equation}\] \[\begin{equation}A^{\mathrm{cir}}=\frac{d^{2} \,\pi}{4}\end{equation}\]

\[\begin{equation}GJ_{\mathrm{ef}}^{\mathrm{sq}}=\frac{T}{\theta}=\frac{d^{3} \,\pi^{4} \,t \,G}{256}\end{equation}\] \[\begin{equation}GJ_{\mathrm{ef}}^{\mathrm{cir}}=\frac{T}{\theta}=\frac{d^{3} \,\pi \,t \,G}{4}\end{equation}\]

\[\begin{equation}\frac{GJ_{\mathrm{ef}}^{\mathrm{sq}}}{GJ_{\mathrm{ef}}^{\mathrm{cir}}}=\left(\frac{\pi^{3}}{64}\right)\end{equation}\] \[\begin{equation}\frac{GJ_{\mathrm{ef}}^{\mathrm{sq}}}{GJ_{\mathrm{ef}}^{\mathrm{cir}}}=0.484\end{equation}\]

---

10.6.5 Example:

• \(T = 2 \times 10^4 {\; \mathrm{in\cdot lb} \;}\)
• \(L = 100\) in
• \(G_A = 5 \times 10^6\) psi; \(G_B = 12 \times 10^6\) psi
• \(t_A = 0.1\) in ; \(t_B = 0.05\) in
• Required: \(\tau_i = ?\) ; \(\theta = ?\) ; \(GJ = ?\)

---

Solution:

• The stress:
  • \(\tau = \frac{q}{t}\) ; \(q = \frac{T}{2A}\)
• The area:
  • \(A = {\frac{1}{2}}\pi 5^2 + {\frac{1}{2}}\times 10 \times 5 = 64.27\) in^2
• The shear flow:
  • \(q = \frac{T}{2A} = \frac{2 \times 10^4}{(2 \times 64.27)} = 155.594\) \[lb/in\]
• Therefore:
  • \(\tau_A = \frac{q}{t_A} = 1555.94\) psi
  • \(\tau_B = \frac{q}{t_B} = 3111.88\) psi

---

• The rate of twist:
  • \(\theta = ( {\frac{1}{2}}\frac{q}{A}) \oint (G \cdot t)^{-1} ds\)
  • \(\theta = ( {\frac{1}{2}}\frac{q}{A}) [\frac{2 \times 5 \sqrt{2}}{(G_A \cdot t_A)} + \frac{5 \pi}{(G_B \cdot t_B)}]\)
  • \(\theta = 6.593 \times 10^{-5}\) \[rad/in\]

For the overall length \(L=100\) in:

• the total twist angle \(\alpha = \theta \cdot L = 6.593 \times 10^{-3}\) rad

• \(GJ = \frac{T}{\theta} = \frac{2 \times 10^4}{6.593 \times 10^{-5}} = 303.352 \times 10^6\) \[lb \cdot in^2/rad\]

---

10.6.6 Torsion of thin open sections

Open thin beam

\[\begin{equation}{\tau_{yz}}=0\end{equation}\]

\[\begin{equation}{\frac{\partial \phi}{\partial x}} = -{\tau_{yz}}= 0\end{equation}\]

Recall compatibility equation for the Prandtl stress function: \[\begin{align} {\frac{\partial^2 \phi}{\partial x \partial x}} + {\frac{\partial^2 \phi}{\partial y \partial y}} =& - 2 G \theta \\ {\frac{\partial^2 \phi}{\partial y \partial y}} =& - 2 G \theta \\ \end{align}\]

---

Therefore: \[\begin{equation}\phi = -G \theta y^2 + C_1 y + C_2\end{equation}\]

Boundary conditions at \(y=\pm\frac{t}{2}\): \[\begin{equation}\phi = 0\end{equation}\]

Therefore: \[\begin{align} C_1 =& \, 0 \\ C_2 =& \, G \theta \frac{t^2}{4} \\ \end{align}\]

Therefore: \[\begin{equation}\phi = -G \theta \left(y^2 - \frac{t^2}{4} \right)\end{equation}\]

---

The shear stress is therefore: \[\begin{align} {\tau_{xz}}= \; {\frac{\partial \phi}{\partial y}} \\ {\tau_{xz}}= \; -2 G \theta y \\ \end{align}\]

The shear stress is parallel to the surface of the thin member and varies linearly across it’s cross section. \[\begin{equation}T=2 \iint \phi dA\end{equation}\] \[\begin{equation}T=-2 G \theta \int_{-b/2}^{b/2} \int_{-t/2}^{t/2} \left(y^2 - \frac{t^2}{4} \right) dx \, dy\end{equation}\] \[\begin{equation}T=\frac{b t^3}{3} G \theta\end{equation}\]

---

Recall the torsion differential equation: \[\begin{equation}T= G J\theta\end{equation}\]

\[\begin{equation}J= \frac{b t^3}{3}\end{equation}\] What is the big deal?