Lesson Friday September 20th

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

Demonstration¶

Given a structure as shown below:

We can find the normal force distribution using the force method. Therefore, we need to know the degree of statically determinacy. The free-body-diagram of the full structure is as follows:

There are 4 unknown support reactions. With only 3 equilibrium equations for this self-contained structure, it can be concluded that this structure is a first-order statically determinant structure.

To apply the force method we need to replace a part of the structure by a statically indeterminate force, which turns the structure into a statically determinate structure. To solve the statically indeterminate structure we need to include a compatibility condition.

Define statically determinate structure with compatibility condition¶

In this structure, we choose to replace the right support with a rolling hinged support. This leads to a horizontal statically indeterminate force $B_\text{v}$ with the compatibility conditions $u_{h,\text{B}} = 0$:

Solve displacements statically determinate structure in terms of statically indeterminate force¶

The force distribution and displacements in this statically determinate structure is now solved for in terms of $B_\text{h}$.

Support reactions¶

First, the support reactions are solved for

Leading to:

Section forces¶

The section forces are solved for, starting with the forces in $\text{BE}$ and $\text{BD}$:

Now, let’s continue with a section through beams $\text{AD}$, $\text{CD}$ and $\text{CE}$:

Thirdly, let’s continue with the joint $\text{D}$:

And finally joint $\text{C}$:

Shortening/lengthening of elements¶

Now, for each element the shortening / lengthening can be calculated:

Displacement structure due to $20 \text{ kN}$¶

For now we’ll ignore the shortening/lengethening due to $B_\text{h}$:

This gives the following Williot-diagram with a fixed $\text{AC}$:

Leading to the following displacements if $\text{AC}$ doesn’t rotate:

$\text{B}$ shouldn’t move vertically, so this structure has to be rotated back with $\theta \approx \cfrac{{103}}{{12000}} \approx 8.5476 \cdot {10^{ - 3}}{\text{ rad}}$ ⟳, leading to:

Resulting in total displacements of:

Displacement structure due to $B_\text{h}$¶

For the displacement due to $B_\text{h}$ in $\text{kN}$, the following Williot-diagram with a fixed $\text{AD}$ can be drawn:

Leading to the following displacements if $\text{AD}$ doesn’t rotate:

Again, $\text{B}$ Shouldn’t move vertically, so this structure has to be rotated back with $\theta = \cfrac{2.4{B_{\rm{h}}}}{{12000}} = 0.0002{B_{\rm{h}}}{\text{ rad}}$ ⟲, leading to:

Resulting in total displacements of:

Solve statically indeterminate structure with compatibility conditions¶

Now, we can fill in the compatibility conditions:

Section forces statically indeterminate structure¶

The section forces can be calculated by filling in the resulting $B_\text{h}$ in our previous expressions:

Displacements statically indeterminate structure¶

Now, the displacements can be found as well by filling in the resulting $B_\text{h}$ in our previous expressions: