10.3

Thin beam in shear

Integrate in the local coordinate system:

• \[\begin{equation}\small {\it q_1}\left(s_{1}\right)=\frac{s_{1}\,t_{a}\,S_y\,y_c}{ I_{\it xx}} \end{equation}\]
• \[\begin{equation}\small {\it q_2}\left(s_{2}\right)=\frac{t_{b}\,S_y\,\left(s_{2}\,y_c- \frac{s_{2}^2}{2}\right)}{I_{\it xx}}+\frac{t_{a}\,l\,S_y\,y_c}{ I_{\it xx}} \end{equation}\]

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Thin beam in shear

• \[\begin{equation}\small\begin{split} {\it q_3}\left(s_{3}\right)=&\frac{t_{a}\,S_y\,\left(l\,y_c+s_{3}\, y_c-s_{3}\,l\right)}{I_{\it xx}} \\ & +\frac{t_{b}\,S_y\,\left(l\,y_c-\frac{l^2}{2}\right)}{I_{\it xx}} \end{split}\end{equation}\]
• \[\begin{equation}\small\begin{split} {\it q_4}\left(s_{4}\right)=&\frac{t_{b}\,S_y\,\left(l\,y_c+s_{4}\,y_c-\frac{l^2}{2}-s_{4}\,l+\frac{s_{4}^2}{2}\right)}{I_{\it xx}} \\ & +\frac{t_{a}\,S_y\,\left(2\,l\,y_c-l^2 \right)}{I_{\it xx}} \end{split}\end{equation}\]

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Thin beam in shear

• \[\begin{equation}\small\begin{split} {\it q_5}\left(s_{5}\right)=&\frac{t_{a}\,S_y\,\left(2\,l\,y_c+s_{5} \,y_c-l^2\right)}{I_{\it xx}} \\ & +\frac{t_{b}\,S_y\,\left(2\,l\,y_c-\frac{l^2}{2}\right)}{I_{\it xx}} \end{split}\end{equation}\]

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• Now we must calculate the centroid and the moment of inertia.
• Due to symmetry we know that the centroid is on the \(x=0\) axis. Calculate the \(y\) value of the centroid:

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Thin beam in shear

\[\begin{equation} A=2\,t_{b}\,l+3\,t_{a}\,l \end{equation}\] \[\begin{equation} \displaystyle\sum_{i=1}^{n}{y\left(i\right)\,A\left(i\right)}=2\,\left(\frac{t_{ b}\,l\,l}{2}\right)+t_{a}\,l\,l \end{equation}\] \[\begin{equation} y_c=\frac{\displaystyle\sum_{i=1}^{n}{y\left(i\right)\,A\left(i\right)}}{A} \end{equation}\] \[\begin{equation} y_c=\frac{l\,\left(t_b+t_a\right)}{2\,t_b+3\,t_a} \end{equation}\]

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Thin beam in shear

Calculate the moment of inertia using the parallel axis theorem: \[\begin{equation}\begin{split} I_{\it xx}= & 2\,\left(\frac{l\,t_a^3}{12}+t_a\,l\,y_c^2\right) +2\, \left(\frac{t_b\,l^3}{12}+t_b\,l\,\left(\frac{l}{2}-y_c\right)^2 \right) \\ & +1\,\left(\frac{l\,t_a^3}{12}+t_a\,l\,\left(l-y_c\right)^2 \right) \end{split}\end{equation}\]

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The final item is the shear stress. For sub-section \(i\), the shear stress is: \[\begin{equation} \tau_i = \frac{q_i}{t_i}. \end{equation}\]

The shear flow is plotted against the global \(s\) coordinate system in figure 10.1. The shear stress is plotted in figure 10.2.

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• Let’s check to see if we have a reasonable solution.
• We can do this by examining shear flows at points where the shear flow may be known without significant computation. At the edges, the shear flow is zero:

\[\begin{align} {\it q_1}\left(0\right)=&\, 0, \\ {\it q_5}\left(l\right)=&\, 0. \\ \end{align}\]

• The beam is symmetric about its center and loaded symmetrically, therefore we know that there can be no shear flow at the point of symmetry. \[\begin{equation} {\it q_3}\left(\frac{l}{2}\right)=0 \end{equation}\]