… for truss structures#
The general concept of the force method is shown in Apply force method and covered in chapter 2.1 while the force method for truss structures is treated in in chapter 2.2.8 - 2.2.10 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (in Dutch) (Hartsuijker and Welleman, 2007).
We’ll cover the application to truss structures with the following example, which includes a williot diagram to calculate displacements.
Example
Fig. 27 Example structure. Although this is not a true truss structure, deformation only are caused by extension, not by bending.#
Determine the degree of statical determinacy.
Example
For our example, we might be interested in the internal force distribution, so we need to evaluate the degree of internal statical determinacy.
Fig. 28 There are 17 unknown forces.#
Fig. 29 There are 16 equilibrium equations#
So this structure is 1st order internally statically indeterminant.
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 here, of which a few are shown below:
The last option is chosen.
Solve for the displacement in terms of the unknown indeterminate forces as you would normally do for a statically determinate structure.
Example
We’ve chosen the following statically determinate structure with displacement constraint \(w_{\rm{B}}\left( B_{\rm{v}} \right) = 0\):
Fig. 30 The statically determinate structure with displacement constraint#
Since \(\rm{AE}\) is infinitely stiff, all deformations will be the result of bars extending / compressing. To calculate this, first the normal forces can be evaluated as a function of \(B_{\rm{v}}\) using i.e. a moment equilibrium around \(\rm{A}\) for the member \(\text{ADE}\):
\(N_{\rm{CD}}\left( B_{\rm{v}} \right) = 210 - 2.5 B_{\rm{v}}\)
\(N_{\rm{BE}} \left( B_{\rm{v}} \right) = - B_{\rm{v}}\)
This leads to the following extension of the elements, using \(\Delta L = \cfrac{N \ L}{EA}\):
\(\Delta L_{\rm{CD}}\left( B_{\rm{v}} \right) = \cfrac{1400}{EA} - \cfrac{50 B_{\rm{v}}}{3 EA}\)
\(\Delta L_{\rm{BE}}\left( B_{\rm{v}} \right) = -\cfrac{5 B_{\rm{v}}}{EA}\)
This leads to the following displacement, using a Williot diagram:
Fig. 31 The displacement of \(\rm{D}\) is \(\cfrac{5}{4} \Delta L_{\rm{CD}} \)#
\(w_{\rm{D}}\left( B_{\rm{v}} \right) = \cfrac{1750}{EA} - \cfrac{125 B_{\rm{v}}}{6 EA} \left( \downarrow \right) \)
\(w_{\rm{E}}\left( B_{\rm{v}} \right) = \cfrac{3500}{EA} - \cfrac{125 B_{\rm{v}}}{3 EA} \left( \downarrow \right) \)
\(w_{\rm{B}}\left( B_{\rm{v}} \right) = \cfrac{3500}{EA} - \cfrac{140 B_{\rm{v}}}{3 EA} \left( \downarrow \right) \)
Use your displacement constraints to solve for the statically indeterminate forces
Example
\[\begin{split} \begin{align*} w_{\rm{B}}\left( B_{\rm{v}} \right) &= 0 \\ \cfrac{3500}{EA} - \cfrac{140 B_{\rm{v}}}{3 EA} &= 0 \\ B_{\rm{v}} &= 75 \ \rm{kN} \end{align*} \end{split}\]This leads to the following other results:
\(N_{\rm{CD}} = 22.5 \ \rm{kN}\)
\(N_{\rm{BE}} -75 \ \rm{kN} \)
\(w_{\rm{D}} = \cfrac{375}{2EA} \left( \downarrow \right) \)
\(w_{\rm{E}} = \cfrac{375}{EA} \left( \downarrow \right) \)
Exercises#
Exercises 2.31 - 2.41, in chapter 2.3 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (in Dutch) (Hartsuijker and Welleman, 2007). Answers are available on this website.