10.5 Shear of single cell thin-walled cross sections
Closed cell under shear line load.
Recall: \[\begin{equation*} {\frac{\partial q}{\partial s}} = -t {\frac{\partial {\sigma_{zz}}}{\partial z}} \end{equation*}\]
\[\begin{equation}\begin{split} q =& - \frac{I_{xx} S_x - I_{xy} S_y}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s x t(s) ds \\ & - \frac{I_{yy} S_y - I_{xy} S_x}{I_{xx}I_{yy}-I_{xy}^2} \int_0^s y t(s) ds \\ & + q (s=0) \\ \end{split}\end{equation}\]
- These bending equations were derived using an open thin walled segment, however, the only restriction is that the walls are thin.
- The equations are valid for open or closed sections.
- In an open section, it is easy to identify a point where the shear is zero, therefore the s coordinate system is defined from that point. This is not as easy in a closed section.
Closed cell under shear line load.
Where do we start our \(s\) coordinate system, what is the value of the shear flow at that point?
We have a symmetric cross section therefore: \[\begin{equation*} \begin{pmatrix}I_{\it xy}=0 \\ S_x=0 \\ \end{pmatrix} \end{equation*}\]
We also know some information in the following sections:
Section 1: \[\begin{equation*} \begin{pmatrix}t\left(s_1\right)=t \\ y\left(s_1\right)=s_1-\frac{h}{2} \\ q_{s_1=0}=q_0 \\ \end{pmatrix} \end{equation*}\]
Section 2: \[\begin{equation*} \begin{pmatrix}t\left(s_2\right)=t \\ y\left(s_2\right)=\frac{h}{2} \\ q_{s_2=0}=q_1\left(h\right) \\ \end{pmatrix} \end{equation*}\]
Section 3: \[\begin{equation*} \begin{pmatrix}t\left(s_3\right)=t \\ y\left(s_3\right)=\frac{h}{2}-s_3 \\ q_{s_3=0}=q_2\left(b\right) \\ \end{pmatrix} \end{equation*}\]
Section 4: \[\begin{equation*} \begin{pmatrix}t\left(s_4\right)=t \\ y\left(s_4\right)=-\frac{h}{2} \\ q_{s_4=0}=q_3\left(h\right) \\ \end{pmatrix} \end{equation*}\]
We integrate to get the shear flows:
Section 1: \[\begin{equation*} q_1\left(s_1\right)=- \frac{S_y \, t}{{I_{xx}}} \left[ \frac{\left(s_1^2-h\,s_1\right)}{2} \right] + q_0 \end{equation*}\]
Section 2: \[\begin{equation*} q_2\left(s_2\right)= - \frac{S_y \, t}{{I_{xx}}} \left[ \frac{h\,s_2}{2} \right] + 0 + q_0 \end{equation*}\]
Section 3: \[\begin{equation*} q_3\left(s_3\right)= \frac{S_y \, t}{{I_{xx}}} \left[ \frac{\left(s_3^2-h\,s_3\right)}{2}-\frac{b\,h}{2} \right] +q_0 \end{equation*}\]
Section 4: \[\begin{equation*} q_4\left(s_4\right)= \frac{S_y \, t}{{I_{xx}}} \left[ \frac{h\,s_4}{2}-\frac{b\,h}{2} \right] +q_0 \end{equation*}\]
Recall the integration to obtain forces: \[\begin{equation*} F=\int_{s_0}^{s_1}{q\left(s\right)\;ds} \end{equation*}\]
Integrating each arm over its limits of integration:
Section 1: \[\begin{equation*} F_1=\frac{S_y \, t}{{I_{xx}}} \left[ \frac{h^3}{12} \right]+h\,q_0 \end{equation*}\]
Section 2: \[\begin{equation*} F_2= \frac{S_y \, t}{{I_{xx}}} \left[ -\frac{b^2\,h}{4} \right]+ b\,q_0 \end{equation*}\]
Section 3: \[\begin{equation*} F_3= \frac{S_y \, t}{{I_{xx}}} \left[ -\frac{h^3}{12}-\frac{b\,h^2}{2} \right]+h\,q_0 \end{equation*}\]
Section 4: \[\begin{equation*} F_4= \frac{S_y \, t}{{I_{xx}}} \left[ -\frac{b^2\,h}{4} \right] + b\,q_0 \end{equation*}\]
Examine equillibrium: \[\begin{align} \displaystyle\sum F_x =&\; 0 \\ =&\; F_2 - F_4 = 0 \\ \end{align}\]
\(F_2\) and \(F_4\) are identical, therefore, this above equation does not add to our understanding.
\[\begin{align} \displaystyle\sum F_y =&\; S_y \\ =&\; F_1 - F_3 \\ \end{align}\]
When \({I_{xx}}\) is calculated, the above equation leads to \(S_y=S_y\) (\(q_0\) is elliminated from the equation by the subtraction). Again, this does not add to our understanding.
To find \(q_0\), moment equillibrium is used.
Sum the moments about a convenient location. For example, the sum of the moments about the shear center (box center) is zero: \[\begin{equation*} \displaystyle\sum M_{\mathrm{s.c.}} = \frac{h\,\left(F_4+F_2\right)}{2}+\frac{b\,\left(F_3+F_1\right)}{2}=0 \end{equation*}\]
This allows calculation of the value of shear flow at the point chosen as origin of the \(s_1\) coordinate system. \[\begin{equation*} q_0=\frac{S_y \, t}{{I_{xx}}} \left[ \frac{b\,h}{4} \right] \end{equation*}\]
In general, this process can be expressed as: \[\begin{equation*} q(s) = q_0 + \hat{q}(s) \end{equation*}\] where \(\hat{q}(s)\) is evaluated as an open tube.
To find the shear center through it, since \(\theta=0\) \[\begin{equation*} \theta = \oint \frac{q(s)}{2GAt} ds = 0 \end{equation*}\] when \(t\) is a constant.
\[\begin{equation*} \oint q(s) ds = \oint q_0 ds + \oint \hat{q}(s) ds \end{equation*}\] But \(q_0\) is a constant offset over the integration domain and can be pulled out of the corresponding equation. Thus:
\[\begin{equation*} q_0 = \frac{\oint \hat{q}(s) ds}{\oint ds} \end{equation*}\]
Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture 0
Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture -1
Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture -2