3: Forces in Statically Determinate Trusses
3.6: Matrix Methods of Determinate Truss Analysis
3.6.1: Introduction
Figure 3.6-1a: Three Bar Truss
This section introduces a more mechanical method of solvinf statically determinat etrusses, and thus one that is more amenable to computer-based solutions. Many of the concepts introduced here will be used in more advanced matrix methods later in your studies.
The method is based on the method of joints, and simply involves writing all possible joint equilibrium equations and expressing the resulting system in matrix form.
As an example, consider the simple three-bar truss shown in Fig. 3.6-1a.
Figure 3.6-1b: Joint FBDs
The free body diagrams of all three joints are shown in Fig. 3.6-1b. We will write the two equilibrium equations at each joint as follows.
Joint a:
\[\begin{split} \sum F_x &= 0.8 T_{ab} + T_{ac} + H_a = 0\\ \sum F_y &= 0.6 T_{ab} + V_a = 0 \end{split}\]Joint b:
\[\begin{split} \sum F_x &= -0.8 T_{ab} + 60 = 0\\ &~~~-0.8 T_{ab} = -60\\ \sum F_y &= -0.6 T_{ab} - T_{bc} = 0\\ \end{split}\]Joint c:
\[\begin{split} \sum F_x &= -T_{ac} = 0\\ \sum F_y &= T_{bc} + V_c = 0 \end{split}\]When doing this manually, we take pains to write the equations in the correct order, with as few unknowns in each as possible, and solve as we go. But here, we simply assemble all six equilibrium equations into one matrix expression:
\[\left[ \begin{array}{ccccc} 0.8 & 1 & 0 & 1 & 0 & 0\\ 0.6 & 0 & 0 & 0 & 1 & 0\\ -0.8 & 0 & 0 & 0 & 0 & 0\\ -0.6 & 0 & -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right] \left\{\begin{array}{c}T_{ab}\\T_{ac}\\T_{bc}\\H_a\\V_a\\V_c\end{array}\right\} = \left\{\begin{array}{c}0\\0\\-60\\0\\0\\0\end{array}\right\}\]or, in symbolic form:
\[\left[C\right] \left\{Q\right\} = \left\{P\right\}\]Solving, we have:
\[\left\{Q\right\} = \left[C\right]^{-1} \left\{P\right\} = \left\{\begin{array}{r}75\\0\\-45\\-60\\-45\\45\end{array}\right\} = \left\{\begin{array}{c}T_{ab}\\T_{ac}\\T_{bc}\\H_a\\V_a\\V_c\end{array}\right\}\]Fortunately, the creation of the matrices is easy to mechanize, as shown in the following sections.
3.6.2: Development of General Method
Figure 3.6-2: Truss Portion
Fig. 3.6-2 part a) shows two joints in a truss, i and j, and the member k that connects between them. Also shown are other members that frame into each joint, as well as the components of the loads, P, that are applied to each joint.
To develop the equilibrium equations, we shall refer to part b) of that figure which shows some dimensions, the forces corresponding to member k, and the indices for the equilibrium equations to be developed: 2i-1 and 2i at joint i, and 2j-1 and 2j at joint j.
We first define some of the geometric quantities:
\[\begin{split} \Delta_x &= x_j - x_i\\ \Delta_y &= y_j - y_i\\ L_k &= \sqrt{\Delta_x^2 + \Delta_y^2}\\ l_{ij} &= \cos \theta_x = \Delta_x / L_k\\ m_{ij} &= \cos \theta_y = \Delta_y / L_k \end{split}\]The equilibrium equation that corresponds to $\sum F_x=0$ at joint i is equation number $2i-1$. The equation itself, showing only the contribution of member k and the applied load, is:
\[\cdots + \cos\theta_x Q_k + \cdots + P_{ix} = 0\]or, rewriting to have only unknowns on the left and knowns on the right:
\[\begin{split} &\cdots + l_{ij} Q_k + \cdots = -P_{ix}~~~~~~~~~~~~~~~~&(\sum F_x @ i) \end{split}\]Similarly, the other three equilibrium equations can be written directly:
\[\begin{split} &\cdots + m_{ij} Q_k + \cdots = -P_{iy}~~~~~~~~~~~~~~~~&(\sum F_y @ i)\\ &\cdots - l_{ij} Q_k + \cdots = -P_{jx}~~~~~~~~~~~~~~~~&(\sum F_x @ j)\\ &\cdots - m_{ij} Q_k + \cdots = -P_{jy}~~~~~~~~~~~~~~~~&(\sum F_y @ j) \end{split}\]We note that these four equations are the only ones in which the force in member k participates. As there is one column in the [C] matrix for each unknown, this gives us a way to easily and mechanically construct the [C] matrix. For each member, k, we calculate the two directional cosine, and place them in 4 locations in the column corresponding to member k, all other values in that column being zero. This is shown in Fig. 3.6-3.
Figure 3.6-3: Matrix Equations
In this figure, we show the four rows corresponding to the four equilibrium equations developed above. Column k is the only column that contains any data relevant to unknown k (i.e., member k). In that column we show the four non-zero values derived from the geometry of member k; all other entries in that column are zero.
To handle the unknowns corresponding to the reaction forces, we observe that each of those impinges on only one joint and so will be included in no more than two equilibrium equations. We simply determine the directional cosines of each reaction force vector, and enter those in the column corresponding to that unknown, in the rows appropriate for the joint on which it acts (rows 2i-1 and 2i for joint i). This is easily seen in the example in the following section.
3.6.3: Example M-1
Figure 3.6-4: Example M-1
We are to solve the truss of Fig. 3.6-4 by the matrix method of joints. The figure shows the joints and unknowns numbered sequentially starting from 1. Joints are numbered thus: 1 and unknowns thus: 1.
Figure 3.6-5: Example M-1 - Table of Member Data
Fig. 3.6-5 shows the tabular data used to compute the directional cosines of all the unknowns. Unknowns 1 through 7 are member forces, and so the geometry calculations developed in the previous section are used.
Unknowns 8, 9 and 10 are reaction forces. Their directional cosines are computed directly from the figure. Each of these, of course, touch only one joint.
The values for unknown 3 are highlighted so we can more easily observe how the values are used to construction the [C] matrix.
Figure 3.6-6: Example M-1 - Matrix Equations [C]{Q}={P}
The resulting matrix equations are shown in Fig. 3.6-6.
Note column 3 of the [C] matrix. The directional cosines from row three of the table were placed there and are shown highlighted. The pair of directional cosines are placed directly in rows 1 and 2; those are the rows corresponding to the i joint of member 3 (joint 1). The negatives of the directional cosines are placed in rows 7 and 8; these are the rows corresponding to the equilibrium equations for the j joint of member 3 (joint 4).
Fig. 3.6-7, below, shows input prepared for Octave (i.e., Matlab) to perform the solution.
Figure 3.6-7: Example M-1 - Input to Octave
Fig. 3.6-7, below, shows the results displayed by Octave. In particular, the member and reaction forces are shown.
Figure 3.6-8: Example M-1 - Results from Octave
Figure 3.6-9: Example M-1 - Summary of Forces
The summary of the forces are shown in Fig. 3.6-9.
Note that in interpreting the results of Fig. 3.6-8, all development was based on tension forces in members being positive. Therefore, negative values for these means the forces are compressive (members 4 and 7). Negative values for the reactions simply means they act in a direction opposite to that given by the directional cosines (reaction 9).