… for truss structures

The general concept of the force method is shown in Apply force method and discussed in section 2.1. Specifically, the force method for truss structures is covered in sections 2.2.8 - 2.2.10 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (Hartsuijker and Welleman, 2007).

We discuss the application to truss structures with the following example. This example includes a Williot diagram to calculate the displacements.

Example

Fig. 27 Example structure. Although this is not a truss structure, deformation is only caused by extension, not by bending.

1. Determine the degree of static determinacy.

   Example

   For our example, we are interested in the internal force distribution, so we need to evaluate the degree of internal static indeterminacy.

   

   Fig. 28 There are 17 unknown forces.

   

   Fig. 29 There are 16 equilibrium equations

   This structure is therefore first-order internally statically indeterminate.

2. Transform the structure into a statically determinate system by removing supports, splitting the structure at a two-force mmembers, or adding hinges: add unknown statically indeterminate forces and deformation conditions for each support you have removed and hinge you have added. Be careful not to transform the structure into a (partially) mechanism!

   Example

   There are many options here, some of which are shown below:

   Split the structure at a hinge

   

   Add a hinge

   

   Release the horizontal movement direction of a support

   

   Remove the vertical support

   

   The last option is chosen.

3. Solve 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 deformation condition \(w_{\rm{B}}\left( B_{\rm{v}} \right) = 0\):

   

   Fig. 30 The statically determinate structure with deformation condition

   Because \(\rm{AE}\) is infinitely stiff, all deformations will be the result of bars stretching/compressing. To calculate this, the normal forces can first be evaluated as a function of \(B_{\rm{v}}\) using, for example, a moment equilibrium around \(\rm{A}\) for the element \(\text{ADE}\):

   • \(N_{\rm{CD}}\left( B_{\rm{v}} \right) = 210 - 2.5 B_{\rm{v}}\)

   • \(N_{\rm{BE}} \left( B_{\rm{v}} \right) = - B_{\rm{v}}\)

   This leads to the following elongation 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 \(\cfrac{5}{4} \Delta L_{\rm{CD}} \)

   • \(w_{\rm{D}}\left( B_{\rm{v}} \right) = \cfrac{1750}{EA} - \cfrac{125 B_{\rm{v}}}{6 EA} \left( \downarrow \right) \)

   • \(w_{\rm{E}}\left( B_{\rm{v}} \right) = \cfrac{3500}{EA} - \cfrac{125 B_{\rm{v}}}{3 EA} \left( \downarrow \right) \)

   • \(w_{\rm{B}}\left( B_{\rm{v}} \right) = \cfrac{3500}{EA} - \cfrac{140 B_{\rm{v}}}{3 EA} \left( \downarrow \right) \)

4. Use your deformation conditions 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:

   • \(N_{\rm{CD}} = 22.5 \ \rm{kN}\)

   • \(N_{\rm{BE}} -75 \ \rm{kN} \)

   • \(w_{\rm{D}} = \cfrac{375}{2EA} \left( \downarrow \right) \)

   • \(w_{\rm{E}} = \cfrac{375}{EA} \left( \downarrow \right) \)

Instructions from lecture

This topic is presented in a lecture available to 42:50 here in Dutch for TU Delft students.

Exercises

• Exercises 2.31 - 2.41, in section 2.3 of the book Mechanica, Statisch onbepaalde constructies en bezwijkanalyse (Hartsuijker and Welleman, 2007) Answers are available here.