Lesson Wednesday Oktober 1st

During today’s lesson it’s demonstrated how you to use the force method for statically indeterminate problems which involve bending.

Attribution

This demonstration is adapted from https://oit.tudelft.nl/CTB2210/TOZ/opgave.html

Demonstration

Given a structure as shown below:

Fig. 93 Structure

First, the degree of statically indeterminacy must be determined.

Fig. 94 Free body diagrams of hinged parts

There are 7 unknown support reactions and 6 unknown internal forces. This gives an external degree of statical indeterminacy of 1. Since this structure is not a closed structure, the internal degree of statical indeterminacy is also equal to 1.

Now we can apply the force method. For this we need to transform the structure into a statically determinate structure. There are several ways to do this, of which some are incorrect when the transformed structure becomes a (partial) mechanism.

Some correct ways to transform the structure are:

Replace the fixed support at \(\rm{A}\) with a hinged support

Split the two-force member \(\rm{BD}\)

Add hinge between \(\rm{A}\) and \(\rm{D}\)

Some incorrect ways to transform the structure are:

Split two two-force member \(\rm{CD}\)

This creates a partial mechanism!

Add hinge between \(\rm{D}\) and \(\rm{E}\)

This creates a partial mechanism!

Add hinge between \(\rm{E}\) and \(\rm{G}\)

This creates a partial mechanism!

Replaced hinged support at \(\rm{C}\) with a vertical roller support

This creates a partial mechanism!

The following statically determinate structure is chosen, including a statically indeterminate force \(B_{\rm{v}}\) and a displacement constraint at point \(\rm{B}\), \(w_{\rm{B}} = 0\):

Now, the force distribution and displacement can be solved for.

The normal force in member \(\rm{BD}\) is found with:

\[\begin{split} \begin{align} \sum F_{\rm{v}}^{\rm{BD}} &= 0 \\ B_{\rm{v}} + N_{\rm{BD}}&= 0 \\ N_{\rm{BD}} &= -B_{\rm{v}} \end{align} \end{split}\]

The normal force in member \(\rm{CG}\) is found with:

\[\begin{split} \begin{align} \sum \left. T \right|_{\rm{E}}^{\rm{EG}} &= 0 \\ 10.5 \cdot 8 \cdot 4 - N_{\rm{CG}} \cdot 8 &= 0 \\ N_{\rm{CG}} &= 42 \ \rm{kN} \end{align} \end{split}\]

The shear force and bending moment just left of point \(\rm{D}\) is found with:

\[\begin{split} \begin{align} \sum F_{\rm{v}}^{\rm{DG}} &= 0 \\ V_{\rm{D}}^{\rm{AD}}-B_{\rm{v}} - 84 + 42 &= 0 \\ V_{\rm{D}}^{\rm{AD}} &= B_{\rm{v}} + 42 \end{align} \end{split}\]

\[\begin{split} \begin{align} \sum \left. T \right|_{\rm{D}}^{\rm{DG}} &= 0 \\ M_{\rm{D}} + 84 \cdot 8 - 42 \cdot 12 &= 0 \\ M_{\rm{D}} &= -168 \ \rm{kNm} \end{align} \end{split}\]

The displacement at point \(\rm{D}\) is found with a forget-me-not:

\[\begin{split} \begin{align} w_{\rm{D}} &= \cfrac{168 \cdot 4^2}{2 \cdot 64000} + \cfrac{\left( B_{\rm{v}} + 42 \right) \cdot 4^3}{3 \cdot 64000} \\ w_{\rm{D}} &= \cfrac{1}{3000} \cdot B_{\rm{v}} + 0.035 \\ w_{\rm{D}} & \approx 0.000333 \cdot B_{\rm{v}} + 0.035 \end{align} \end{split}\]

The displacement of \(\rm{B}\) can be found with the extension of a member due to axial forces:

\[\begin{split} \begin{align} w_{\rm{B}} &= - w_{\rm{D}} - \Delta L_{\rm{BD}} \\ w_{\rm{B}} &= - w_{\rm{D}} - \cfrac{-B_{\rm{v}} \cdot 8}{4000} \\ w_{\rm{B}} &= \cfrac{7}{3000} \cdot B_{\rm{v}} + 0.035 \\ w_{\rm{B}} & \approx 0.00233\cdot B_{\rm{v}} + 0.035 \\ \end{align} \end{split}\]

Now, the displacement constraint at point \(\rm{B}\) can be used to solve for the statically indeterminate force \(B_{\rm{v}}\):

\[\begin{split} \begin{align} w_{\rm{B}} &= 0 \\ \cfrac{7}{3000} \cdot B_{\rm{v}} + 0.035 &= 0 \\ B_{\rm{v}} &= -15 \ \rm{kN} \end{align} \end{split}\]

Finally, with the found value for \(B_{\rm{v}}\), the internal forces and displacements can be calculated. For example, the displacement at \(\rm{E}\). The rotation at \(\rm{D}\) is found with a forget-me-not:

\[\begin{split} \begin{align} \varphi_{\rm{D}} &= \cfrac{168 \cdot 4}{64000} + \cfrac{\left( -15 + 42 \right) \cdot 4^2}{2 \cdot 64000} \\ \varphi_{\rm{D}} &= 0.03 \ \rm{rad} \end{align} \end{split}\]

The shear force at \(\rm{E}\) is found with:

\[\begin{split} \begin{align} \sum F_{\rm{v}}^{\rm{EG}} &= 0 \\ V_{\rm{E}}- 84 + 42 &= 0 \\ V_{\rm{E}} &= 42 \ \rm{kN} \end{align} \end{split}\]

The displacement at \(\rm{E}\) is found with a forget-me-not:

\[\begin{split} \begin{align} w_{\rm{E}} &= w_{\rm{D}} + \varphi_{\rm{D}} \cdot 4 + \cfrac{ 42 \cdot 4^3}{3 \cdot 64000} \\ w_{\rm{E}} &= 0.0995 \ \rm{m} \\ \end{align} \end{split}\]