The content of this section was delivered over several lectures:

• Part 1
• Part 2
• Part 3
• Part 4

Thin beam in shear

• Define a new coordinate system \(s\).
• Assume a thin wall:
  • \(t = t(s)\)
• Assume the shear stress does not vary through the thickness
  • \({\tau_{zs}}= {\tau_{zs}}(z,s)\)

Recall that the shear force and moment resultants are related through:

\[\begin{align} \frac{d M_y}{d z} =& -S_x \\ \frac{d M_x}{d z} =& +S_y \\ \end{align}\]

Note for thin walled sections, warping of the cross section cannot be avoided. “Plane sections remain plane...” is no longer valid, we need to develop a new formulation!

Shear flow

• 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 @ref(eq:beamstressequation): \[\begin{align} {\sigma_{zz}}=& E \left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) \\ {{\sigma_{zz}} {}_{,z}} =& E {\left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) {}_{,z}} \\ =& \frac{E}{\det EI} \left[ -x {\left(EI_{xy} M_x + EI_{xx} M_y \right) {}_{,z}} + \right. \\ & \left. y {\left(EI_{yy} M_x + EI_{xy} M_y\right) {}_{,z}} \right] \\ =& \frac{E}{\det EI} \left[ x \left(EI_{xx} S_x - EI_{xy} S_y \right) \right. \\ & \left. + y \left(EI_{yy} S_y - EI_{xy} S_x\right) \right] \\ \end{align}\]

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

Example: past homework problem

• 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).

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

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

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: \[\begin{equation} q\left(s\right)=q_{0}-\frac{\left(\int_{s_{0}}^{s}{t\left(s\right) \,y\left(s\right)\;ds}\right)\,S_y}{I_{\it xx}}. \end{equation}\]

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:

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

• 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.

Integrate in the local coordinate system:

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

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

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

• 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:

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

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

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 @ref(fig:shearflow). The shear stress is plotted in figure @ref(fig:shearstress).

• 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. \[\begin{equation} {\it q_3}\left(\frac{l}{2}\right)=0 \end{equation}\]

Shear center

\[\begin{equation*} q\left(s\right)=q(s=0)-\frac{\left(\int_{s_{0}}^{s}{t\left(s\right) \,x\left(s\right)\;ds}\right)\,S_x}{{I_{yy}}} \end{equation*}\]

For different sections of our cross section, these have constant values

• \[\begin{equation*}\small \begin{pmatrix}t\left(s_1\right)=t \\ x\left(s_1\right)=-\frac{b}{2} \\ q_1(s_1=0)=0 \\ \end{pmatrix} \end{equation*}\]
• \[\begin{equation*}\small \begin{pmatrix}t\left(s_2\right)=t \\ x\left(s_2\right)=s_2-\frac{b}{2} \\ q_2(s_2=0)=q_1\left(s_1=h\right) \\ \end{pmatrix} \end{equation*}\]
• \[\begin{equation*}\small \begin{pmatrix}t\left(s_3\right)=t \\ x\left(s_3\right)=\frac{b}{2} \\ q_3(s_3=0)=q_2\left(s_2=b\right) \\ \end{pmatrix} \end{equation*}\]

We can perform the integrations to obtain the shear flows: \[\begin{equation*} q_1\left(s_1\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,s_1\,t}{2} \right] \end{equation*}\]

\[\begin{equation*} q_2\left(s_2\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,h\,t}{2}-\frac{ \left(s_2^2-b\,s_2\right)\,t}{2} \right] \end{equation*}\]

\[\begin{equation*} q_3\left(s_3\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,h\,t}{2}-\frac{b \,s_3\,t}{2} \right] \end{equation*}\]

We can perform the integrations again to obtain the resultant forces: \[\begin{equation*} F=\int_{s_0}^{s_1}{q\left(s \right)\;ds} \end{equation*}\]

For sub-section 1: \[\begin{equation*} F_1=\int_{s_1=0}^{s_1=h}{q_1\left(s_1 \right)\;ds_1} \end{equation*}\]

\[\begin{equation*} F_1= \frac{S_x}{{I_{yy}}} \left. \frac{b\,\left(\frac{s_1^2}{2} \right)\,t}{2} \right|_0^h \end{equation*}\]

\[\begin{equation*} F_1=\frac{S_x}{{I_{yy}}} \left[\frac{b\,h^2\,t}{4}\right] \end{equation*}\]

For sub-section 2: \[\begin{equation*} F_2=\int_{s_2=0}^{s_2=b}{q_2\left(s_2 \right)\;ds_2} \end{equation*}\]

\[\begin{equation*} F_2= \left. \frac{S_x}{{I_{yy}}} \frac{\left(-2\,s_2^3+3\,b\,s_2^2-6\,b\,h\,s_2\right)\,t}{12} \right|_0^b \end{equation*}\]

\[\begin{equation*} F_2=\frac{S_x}{{I_{yy}}} \frac{\left(6\,b^2\,h+b^3\right)\,t}{12} \end{equation*}\]

For sub-section 3: \[\begin{equation*} F_3=\int_{s_3=0}^{s_3=h}{q_3\left(s_3 \right)\;ds_3} \end{equation*}\]

\[\begin{equation*} F_3= \frac{S_x}{{I_{yy}}} \left. \frac{\left(b\,s_3^2-2\,b\,h\, s_3\right)\,t}{4} \right|_0^h \end{equation*}\]

\[\begin{equation*} F_3=\frac{S_x}{{I_{yy}}} \frac{b\,h^2\,t}{4} \end{equation*}\]

The moment of inertia about the \(y\) axis is (by inspection using parallel axis theorem): \[\begin{equation*} {I_{yy}}=\frac{t\,b^3}{12}+2\,h\,t\,\left(\frac{b}{2}\right)^2 + 2 \frac{h t^3}{12} \end{equation*}\]

\[\begin{equation*} {I_{yy}}=\frac{t\,b^3}{12}+2\,h\,t\,\left(\frac{b}{2}\right)^2 + 0 \end{equation*}\]

Now: \[\begin{equation*} F_1=S_x \frac{3\,h^2}{6\,b\,h+b^2} \end{equation*}\]

\[\begin{equation*} F_2=S_x \end{equation*}\]

\[\begin{equation*} F_3=S_x \frac{3\,h^2}{6\,b\,h+b^2} \end{equation*}\]

The resultant forces must be the static equivalent of the shear forces applied through the shear center.

• Shear Center
  • The point at which shear loads must be applied in order to no have twisting of the beam to occur
    • Location depends on geometry, not loads
    • Always located on a line of symmetry

Moments about the shear center are zero, shear center assumed to be below the channel.

\[\begin{equation*} \frac{b\,F_1}{2}-F_2\,e_y+\frac{F_3\,b}{2}=0 \end{equation*}\]

\[\begin{equation*} e_y = \frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about the shear center are zero, shear center assumed to be above the channel \[\begin{equation*} F_2\,e_y+\frac{F_3 \,b}{2}+\frac{F_1\,b}{2}=0 \end{equation*}\]

\[\begin{equation*} e_y=-\frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about corner A is equivalent to resultant shear moment, shear center is assumed below the channel \[\begin{equation*} F_1\,b=e_y\,S_x \end{equation*}\]

\[\begin{equation*} e_y=\frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about corner A is equivalent to resultant shear moment, shear center is assumed above the channel \[\begin{equation*} F_1\,b=-e_y\,S_x \end{equation*}\]

\[\begin{equation*} e_y=-\frac{3\,h^2}{6\,h+b} \end{equation*}\]

Steps to take to find the shear center

1. Assume a shear load is applied at the shear center
   • If symmetry, assume the load lies on the line of symmetry
   • Assume a convenient location, it is often easiest to assume the load is outside the geometry.
2. Compute the shears flows and resultant shear forces
3. Sum the moments about a convenient point.
   • If the point is the shear center, the moments need to sum to zero.
   • If the point is off the shear center, the moments must sum to the same moment as the resultant force would impart.

Remember: the moment from the shear flows must be the same as that of the resultant shear vector that is assumed. The computed and assumed resultants are statically equivalent! If they are not, you have made an error.

Shear of single cell thin-walled cross sections

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.

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:

\[\begin{equation*} q_0 = \frac{\oint \hat{q}(s) ds}{\oint ds} \end{equation*}\]