10.2 Shear flow

Thin beam in shear @AriGur2007 Equilibrium element @AriGur2007

  • Define the shear flow
    • \(q = {\tau_{zs}}\cdot t = q(z,s)\)

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

The \(s\) direction equilibrium gives: \[\begin{align} \sum F_s =& 0 \\ \Delta {\sigma_{ss}}\cdot \Delta z \cdot t + \Delta q \cdot \Delta s =& \; 0\\ \frac{\Delta {\sigma_{ss}}}{\Delta s} \cdot t + \frac{\Delta q}{\Delta z} =& \; 0\\ \end{align}\]

If we apply the shear forces such that no twisting occurs (i.e., through the shear center), then \({\sigma_{ss}}=0\) and the remaining relevant equation is:

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

Recall also our stress equation from beam bending, directly from equation (??): \[\begin{align} {\sigma_{zz}}=& E \left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) \\ {{\sigma_{zz}} {}_{,z}} =& E {\left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) {}_{,z}} \\ =& \frac{E}{\det EI} \left[ -x {\left(EI_{xy} M_x + EI_{xx} M_y \right) {}_{,z}} + \right. \\ & \left. y {\left(EI_{yy} M_x + EI_{xy} M_y\right) {}_{,z}} \right] \\ =& \frac{E}{\det EI} \left[ x \left(EI_{xx} S_x - EI_{xy} S_y \right) \right. \\ & \left. + y \left(EI_{yy} S_y - EI_{xy} S_x\right) \right] \\ \end{align}\]

Lets reexamine the relationship between \(q\) and \(s\): \[\begin{align} {\frac{\partial q}{\partial s}} =& -t {\frac{\partial {\sigma_{zz}}}{\partial z}} \\ \end{align}\]

To determine the shear flow \(q\), we must integrate with respect to the area: \[\begin{equation}\begin{split} q(s) - q (s=0) =& - \frac{I_{xx} S_x - I_{xy} S_y}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s x t(s) ds \\ & - \frac{I_{yy} S_y - I_{xy} S_x}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s y t(s) ds \\ \end{split}\end{equation}\]

If we have a symmetric cross section: \[\begin{align} q(s) - q (s=0) =& - \frac{S_x}{I_{yy}} \int_0^s x t(s) ds - \frac{S_y}{I_{xx}} \int_0^s y t(s) ds \\ \end{align}\]

10.2.1 Example: past homework problem

Thin beam in shear

  • For the hat section and applied shear resultant shown (\(S_y\)), compute the shear flow and shear stress in the section as function of the given variables.
  • Plot the shear flow and shear stress in the section as a function of \(s\)
  • Assume these values of the variables: \[\begin{equation} \begin{split} l =&\, 300 ~mm\\ t_a =&\, 20 ~mm\\ t_b =&\, 30 ~mm\\ S_y =&\, 200 N \\ \end{split} \end{equation}\]

Hint: You should use \(s\) as your abscissa, \(q\) and \(\tau\) as your ordinate (on separate plots).

Thin beam in shear

Solution: Recall for an asymmetric cross section (as noted above):

\[\begin{equation}\small\begin{split} q\left(s\right)=&\frac{\left(I_{\it xy}\,\int_{s_{0}}^{s}{t\left(s \right)\,x\left(s\right)\;ds}-I_{\it yy}\,\int_{s_{0}}^{s}{t\left(s \right)\,y\left(s\right)\;ds}\right)\,S_y}{I_{\it xx}\,I_{\it yy}- I_{\it xy}^2}\\ &+\frac{\left(I_{\it xy}\,\int_{s_{0}}^{s}{t\left(s \right)\,y\left(s\right)\;ds}-I_{\it xx}\,\int_{s_{0}}^{s}{t\left(s \right)\,x\left(s\right)\;ds}\right)\,S_x}{I_{\it xx}\,I_{\it yy}- I_{\it xy}^2}+q_{0}\\ \end{split}\end{equation}\]

Thin beam in shear

We know the cross section is symmetric, therefore \(I_{xy}=0\). Also, there is no load in the \(x\) direction therefore. Therefore, we have the following: \[\begin{equation} q\left(s\right)=q_{0}-\frac{\left(\int_{s_{0}}^{s}{t\left(s\right) \,y\left(s\right)\;ds}\right)\,S_y}{I_{\it xx}}. \end{equation}\]

Thin beam in shear

We choose to integrate the different sub-sections of the beam independently. Before performing the integration, we observe the following known values for the 5 different sub-sections:

  • \[\begin{equation}\small \left.\begin{pmatrix}t\left(s_1\right)=t_{a} \\ y\left(s_1\right)=-y_c \\ q_{0}=0 \\ \end{pmatrix}\right|_{\mathrm{Sub-section 1}} \end{equation}\]
  • \[\begin{equation}\small \left.\begin{pmatrix}t\left(s_2\right)=t_{b} \\ y\left(s_2\right)=-y_c+s_2 \\ q _{0}=q_1\left(l\right) \\ \end{pmatrix}\right|_{\mathrm{Sub-section 2}} \end{equation}\]

Thin beam in shear

  • \[\begin{equation}\small \left.\begin{pmatrix}t\left(s_3\right)=t_{a} \\ y\left(s_3\right)=l-y_c \\ q _{0}={\it q_2}\left(l\right) \\ \end{pmatrix}\right|_{\mathrm{Sub-section 3}} \end{equation}\]
  • \[\begin{equation}\small \left.\begin{pmatrix}t\left(s_4\right)=t_{b} \\ y\left(s_4\right)=l-y_c-s_4 \\ q_{0}={\it q_3}\left(l\right) \\ \end{pmatrix}\right|_{\mathrm{Sub-section 4}} \end{equation}\]
  • \[\begin{equation}\small \left.\begin{pmatrix}t\left(s_5\right)=t_{a} \\ y\left(s_5\right)=-y_c \\ q_{0 }={\it q_4}\left(l\right) \\ \end{pmatrix}\right|_{\mathrm{Sub-section 5}} \end{equation}\]

Thin beam in shear

  • Note that the above equations each use a local definition of \(s_i\) and set an appropriate boundary condition \(q_0\).
  • The global and local \(s\) coordinates are shifted by the position (along the beam) of the local coordinate system.