10.7 Closed section vs open section

10.7.1 Closed section vs open section

Equal values of mean width, equal thickness

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

Assume: \[\begin{equation}t=1.0\end{equation}\] \[\begin{equation}d=10.0\end{equation}\]

\[\begin{equation}J_{\mathrm{open}} = 10.4\end{equation}\] \[\begin{equation}J_{\mathrm{ef}}^{\mathrm{cir}}= 78.5\end{equation}\]

The ratio is: \[\begin{align} \frac{J_{\mathrm{ef}}^{\mathrm{cir}}}{J_{\mathrm{open}}} =\frac{3}{4} \left(\frac{d}{t}\right)^2=75.0 \end{align}\]

We see how crucial the shear transfer is to resisting torsion.

Shear in a closed and open section, from Rockwood and Green, 4th Ed

Recall that shear flow requires a shear stress field that can be assumed constant through the thickness. Clearly, shear flow theory cannot be used for torsion of a thin walled section.

10.7.2 Recall these equations

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

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

10.7.3 Multicell thin walled torsion

@AriGur2007 @AriGur2007

  • \(q_1\) and \(q_2\) are skin shear flows
  • \(q_{12}\) is a web shear flow

\[\begin{align} \sum F_z =& \, 0 \\ q_1 dz + q_2 dz + q_{12} dz =& \, 0 \\ q_1 + q_2 + q_{12} =& \, 0 \\ \end{align}\]

At the junction: \[\begin{equation}\displaystyle \sum q_i = 0\end{equation}\]

@AriGur2007

  • In a multi-cell beam, each cell carries its own shear flow \(q_i\).
  • The shear flow in the walls is calculated from the cellular shear flows \[\begin{equation} q_{\mathrm{web}} = q_{i+1} - q_i\end{equation}\]
  • Torque of a single cell: \(T_i = 2 q_i A_i\)
  • For a multiple cell beam: \(T = \sum T_i = 2 \displaystyle \sum_{i=1}^n q_i A_i\)
  • Extending the assumption of undistorted shape of the cross-section, all the cells rotate together:
  • \(\theta_1 = \theta_2 = ... = \theta_n = \theta\)
  • \(\frac{1}{2 A_i} \oint_{\mathrm{cell_i}} \frac{q}{G \cdot t} ds = \frac{1}{2 A_1} \oint_{\mathrm{cell_1}} \frac{q}{G \cdot t} ds\)
  • The equation for \(T\) and the \(n-1\) equations \(\theta_i = \theta_1\) provide \(n\) equations for the \(n\) unknowns \(q_i\)

10.7.4 Example:

  • \(T = 2 \times 10^4\) \[in \cdot lb\]
  • \(L = 100\) in
  • \(G_A = 5 \times 10^6\) psi ; \(G_B = G_C = 12\times 10^6\) psi
  • \(t_A = 0.1\) in ; \(t_B = t_C = 0.05\) in
  • Required: \(\tau_A = ?\); \(\tau_B = ?\) ; \(\tau_C = ?\) ; \(GJ = ?\)

The solution is:

  • The torque:
    • \(T = 2 \times 10^4 = 2 \cdot (q_2 \cdot A_2 + q_1 \cdot A_1)\)
  • The areas:
    • \(A_1 = {\frac{1}{2}}\times 10 \times 5 = 25\) in^2
    • \(A_2 = {\frac{1}{2}}\pi 5^2 = 39.27\) in^2
  • Therefore:
    • \(T = 2 \times 10^4 = 2 \cdot (39.27 \cdot q_2 + 25 \cdot q_1)\)
    • This provides one of two equations needed for the shear flows.
  • Shear flow in the web section:
    • \(q_2+q_3-q_1=0\)
    • \(q_3=q_1-q_2\)
  • Recall we need two equations and two unknowns to solve for the two cell flows:
    • All cells twist at the same rate: \[\begin{equation} \theta_2 = \theta_1 \end{equation}\]
  • \(\theta_i=\frac{1}{2 A_i} \oint_{\mathrm{cell_i}} \frac{q}{G \cdot t} ds\)
  • Therefore:
    • \(\theta_1 = \frac{1}{(2A_1)} \left[\frac{q_1 \cdot 2 \cdot 5 \sqrt{2}}{(G_A \cdot t_A)} + \frac{(q_1-q_2) \cdot 10}{(G_C \cdot t_C)} \right]\)
    • \(\theta_2 = \frac{1}{(2A_2)} \left[\frac{q_2 \cdot 5 \pi}{(G_B \cdot t_B)} + \frac{-(q_1-q_2) \cdot 10}{(G_C \cdot t_C)} \right]\)
    • \(\theta_1 = 8.990 \times 10^{-7} \cdot q_1 - 3.333 \times 10^{-7} \cdot q_2\)
    • \(\theta_2 = 5.455 \times 10^{-7} \cdot q_2 - 2.122 \times 10^{-7} \cdot q_1\)
  • Using equation (10.2):
    • \(\theta_2 = \theta_1\)
    • \(\Longrightarrow(5.455+3.333) \cdot q_2 = (8.990+2.122) \cdot q_1\)
    • This provides the second necessary equation.
    • \(q_2 = 1.2645 \cdot q_1\)
  • Thus, the torque becomes:
    • From above: \(T = 2 \times 10^4 = 2 \cdot (39.27 \cdot q_2 + 25 \cdot q_1)\)
    • \(2 \times 10^4 = 2 \cdot [39.27 \cdot (1.2645 \cdot q_1) + 25 \cdot q_1]\)
    • \(2 \times 10^4 = 2 \cdot [74.655 \cdot q_1]\)
    • \(q_1 = 133.95\) lb/in
    • \(q_2 = 169.38\) lb/in
  • The stress:
    • \(\tau = \frac{q}{t} \Longrightarrow\)
      • \(\tau_A = \frac{q_1}{t_A} = \frac{133.95}{0.1} = 1339.5\) psi
      • \(\tau_B = \frac{q_2}{t_B} = \frac{169.38}{0.05} = 3387.6\) psi
      • \(\tau_C = \frac{q_3}{t_C} = \frac{(q_1-q_2)}{t_C} = \frac{-35.43}{0.05} = -708.6\) psi (opposite to the direction assumed)
  • Using one of the rate of twist equations to establish the rate of twist:
    • \(\theta_2 = 5.455 \times 10^{-7} \cdot q_2 - 2.122 \times 10^{-7} \cdot q_1 = 639.73 \times 10^{-7}\) rad/in
  • Thus, the torsional rigidity:
    • \(GJ = \frac{T}{\theta} = 2 \cdot \frac{10^4}{6.3973} \times 10^{-5} = 312.6 \times 10^6\) \[lb \cdot in^2/rad\]

Note: When one cell is open is it likely negigible!

Hence, set \(q=0\).

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