… for frame structures

The general concept of the force method is covered in chapter 2.1 while the force method for framestructures is treated in in chapter 2.2.5 - 2.2.7 and the more specific ‘hoekveranderingsvergelijkingen’ in chapter 3.1 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (in Dutch) (Hartsuijker and Welleman, 2007).

Again, the method of ‘hoekveranderingsvergelijkingen’ has the advantage that it’s very easy to calculate the required rotations using forget-me-nots. However, not for all structures forget-me-nots might be available.

Whenever the book mentions the ‘momentenvlakstelling’ in example 2.2.6 and 2.2.7, you can also find the displacements using forget-me-nots. The method with moveable nodes (‘hoekveranderingsvergelijkingen met verplaatsbare knopen’) which was taught is the past is not treated anymore.

We’ll cover the application to bending structures with the following example.

Example

Fig. 36 Example structure

1. Determine the degree of statical determinacy.

   Example

   For our example, we are interested in the distribution of internal forces, so we must evaluate the degree of internal static indeterminacy.

   

   Fig. 37 There are 21 unknown forces.

   

   Fig. 38 There are 19 equilibrium equations

   This structure is therefore 2nd degree internally statically indeterminate.

2. Transform the structure in a statical determinant system by releasing releasing a support, splitting the structure at a two-force member or adding hinges: add unknown statically indeterminate forces and displacement constraints for each of the support you released and hinges you added. Be aware that you don’t transform the structure in a (partial) mechanism!

   Example

   There are many options, some of which are possible options:

   Release horizontal support at \(\rm{B}\) and add hinge at \(\rm{B}\)

   

   Release horizontal supports at \(\rm{B}\) and \(\rm{C}\)

   

   Release horizontal and vertical supports at \(\rm{A}\)

   

   Release horizontal support at \(\rm{A}\) and add hinge at \(\rm{B}\)

   

   The third option is chosen.

3. Solve for the displacement in terms of the unknown indeterminate forces as you would normally do for a statically determinate structure.

   Example

   We have chosen the following statically determinate structure with displacement conditions \(w_{\rm{A,v}}\left( A_{\rm{v}}, A_{\rm{h}} \right) = 0 \) and \(w_{\rm{A,h}}\left( A_{\rm{v}}, A_{\rm{h}} \right) = 0 \):

   

   Fig. 39 The statically determinate structure with displacement conditions

   The force distribution can be found with equilibrium:

   • \(M_{\rm{C}} = 90 \ \rm{kNm}\) (◠/ᑐ)

   • \(M_{\rm{B}} = 6A_{\rm{v}}\) (◡/ᑐ)

   Using the forget-me-nots method, the rotations can now be evaluated:

   • \(\varphi_{\rm{B}} = 0.0012 A_{\rm{v}} - 0.018\)

   • \(w_{\rm{A}} = 0.0216 A_{\rm{v}} - 0.108\)

   For the horizontal displacement applies: \(w_{\rm{A,h}} = \cfrac{6A_{\rm{h}}}{EA} \)

4. Use your displacement constraints to solve for the statically indeterminate forces

   Example

   \[\begin{split} \begin{align*} w_{\rm{A,v}}\left( A_{\rm{v}}, A_{\rm{h}} \right) &= 0 \\ 0.0216 A_{\rm{v}} - 0.108 &= 0 \\ A_{\rm{v}} &= 5 \ \rm{kN} \\ \\ w_{\rm{A,h}}\left( A_{\rm{v}}, A_{\rm{h}} \right) &= 0 \\ \cfrac{6A_{\rm{h}}}{EA} &= 0 \\ A_{\rm{h}} &= 0 \ \rm{kN} \end{align*} \end{split}\]

Exercises

• Exercises 2.15 - 2.22, 2.24, 2.26 - 2.29, 2.42 - 2.48 in chapter 2.3 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (in Dutch) (Hartsuijker and Welleman, 2007).

• Exercises 3.11 - 3.15, 3.22, 3.23, 3.25 - 3.33/1, 3.35, 3.36, 3.45, 3.47-1, 3.47-2, 3.47-4, 3.50, 3.51 in chapter 3.4 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (in Dutch) (Hartsuijker and Welleman, 2007).

Answers are available on this website for chapter 2 and here for chapter 3.