The following exercise shows you how to set up and solve a (fairly) large linear equation system from mechanics. You will also learn how to use Matlab's data structure sparse to define sparse matrices. The goals of this tutorial are to set up and solve a large system of linear equations.

NOTE: This tutorial is available in Swedish: MATLAB_TUTO3_Sv.pdf (discard the mention about the Dugga 2).

Framework

A truss is a construction where rods / beams are assembled to give a light but stable construction. Normally it is a three-dimensional assembly, but for the sake of simplicity we study only flat trusses. In order to easily obtain stability, the rods are assembled so that the truss consists of a number of triangles. The points where the rods are put together are called nodes. The simplest truss consists of a single triangle.

We will now see how to determine the forces in the truss rods when it is loaded by an external force F = 8 kN (same unit in the following).

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All rod forces are considered as tensile forces. A compressive force is seen as a negative tensile force. When the truss is exposed, all bar forces will act from the node in the direction of the long bar, towards the opposite node. We also choose the driving forces that are targeted.

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In each node, the force sum must be equal to the zero vector, 0.

In node 1, this means that $LaTeX: x_1 \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} + x_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + H_1 \begin{bmatrix} -1 \\ 0 \end{bmatrix} + V_1 \begin{bmatrix} 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ .

This vector equation corresponds to two coordinate equations:

$LaTeX: \begin{cases} \frac{1}{\sqrt{2}}x_1 &+ x_2 &-H_1 &+0V_1 &=0 \\ \frac{1}{\sqrt{2}}x_1 &+ 0x_2 &+0H_1 &-V_1 &=0 \end{cases}$

In node 2 we get the equations:

$LaTeX: \begin{cases} -\frac{1}{\sqrt{2}}x_1 &+ \frac{1}{\sqrt{2}}x_3 & &=0 \\ -\frac{1}{\sqrt{2}}x_1 &-\frac{1}{\sqrt{2}}x_3 &-8 &=0 \end{cases}$

In node 3 we get the equations:

$LaTeX: \begin{cases} -x_2 &- \frac{1}{\sqrt{2}}x_3 &+H_2 &+0V_2 &=0 \\ 0x_2 &+\frac{1}{\sqrt{2}}x_3 &+0H_2 &-V_2 &=0 \end{cases}$

The variable support in node 3 means that H2 = 0.

We now have a total of six equations and six unknowns, LaTeX: x_1, x_2, x_3. H_1, V_1, V_2 :

$LaTeX: \begin{cases} \frac{1}{\sqrt{2}} &x_1 &+ &x_2 &+0 &x_3 &- &H_1 &+0 &V_1 &+0 &V_2 &=0 \\ \frac{1}{\sqrt{2}} &x_1 &+ 0 &x_2 &+0 &x_3 &+0 &H_1 &- &V_1 &+0 &V_2 &=0 \\ -\frac{1}{\sqrt{2}} &x_1 &+ 0 &x_2 &+\frac{1}{\sqrt{2}} &x_3 &+0 &H_1 &+0 &V_1 &+0 &V_2 &=0 \\ -\frac{1}{\sqrt{2}} &x_1 &+ 0 &x_2 &-\frac{1}{\sqrt{2}} &x_3 &+0 &H_1 &+0 &V_1 &+0 &V_2 &=0 \\ 0 &x_1 &- &x_2 &-\frac{1}{\sqrt{2}} &x_3 &+0 &H_1 &+0 &V_1 &+0 &V_2 &=0 \\ 0 &x_1 &+0 &x_2 &+\frac{1}{\sqrt{2}} &x_3 &+0 &H_1 &+0 &V_1 &- &V_2 &=0 \end{cases}$

The coefficient matrix of this system is:

$LaTeX: \begin{bmatrix} \frac{1}{\sqrt{2}} &1 &0 &-1 &0 & 0 \\ \frac{1}{\sqrt{2}} &0 &0 &0 &-1 & 0 \\ -\frac{1}{\sqrt{2}} &0 &\frac{1}{\sqrt{2}} &0 &0 & 0 \\ -\frac{1}{\sqrt{2}} &0 &-\frac{1}{\sqrt{2}} &0 &0 & 0 \\ 0 &-1 &-\frac{1}{\sqrt{2}} &0 &0 & 0 \\ 0 &0 &\frac{1}{\sqrt{2}} &0 &0 &-1 \end{bmatrix}$

and the total matrix becomes:

$LaTeX: \begin{bmatrix} \frac{1}{\sqrt{2}} &1 &0 &-1 &0 & 0 &| &0 \\ \frac{1}{\sqrt{2}} &0 &0 &0 &-1 & 0 &| &0 \\ -\frac{1}{\sqrt{2}} &0 &\frac{1}{\sqrt{2}} &0 &0 & 0 &| &0 \\ -\frac{1}{\sqrt{2}} &0 &-\frac{1}{\sqrt{2}} &0 &0 & 0 &| &8 \\ 0 &-1 &-\frac{1}{\sqrt{2}} &0 &0 & 0 &| &0 \\ 0 &0 &\frac{1}{\sqrt{2}} &0 &0 &-1 &| &0 \end{bmatrix}$

Using rref, we find the solution: LaTeX: x_1=-5.66, x_2=4, x_3=-5,66, H_1=0, V_1=-4, V_2=-4 . We can also use x=inv(A)*b or x=A\b; the latter is more effective.

In problems of this type, the coefficient matrix will always have many zeros. It will then be a bit boring to enter all of these. One way to reduce the amount of work is to start by setting all the elements in the matrix be zeros, A=zeros(6,6), and then change the non-zero elements: A(1,1)=1/sqrt(2); A(1,2)=1; A(1,4)=-1; and so on.

This method will also be cumbersome if the matrix is large. Instead a sparse matrix can be used. With the command A=sparse(R,K,E,6,6) the arguments R, K, E are row matrices that contain row indices, column indices and elements for all zero-separated positions in A, a 6 × 6 matrix with zeros is created in all other positions. In the example above, A has 13 zero-separated elements. The row matrices R, K, E must therefore contain 13 elements that satisfy that E(i)=A(R(i),K(i)). To simplify, we start with u=1/sqrt(2); so that we do not have to enter the entire fraction over and over again. Then we assign the following:

Screenshot 2022-02-02 at 08.38.44.png

However, you can not see the matrix A if you define it with A=sparse(R,K,E,6,6)and omit the semicolon. You only see a list of positions and zero-separated elements. With the command full(A) you get to see the matrix in the usual way and with spy(A) you get a graphical picture of where the zero-separated elements are. This can sometimes be enough to check that you have written correctly. With training, being systematic and care, you can enter the matrices R, K, E without writing down the entire system of equations. Each row in A corresponds to a node, and constrain either the horizontal or the vertical forces in the node. Each column corresponds to a specific support force.

Practice exercise:

Determine the support and support forces in the planar trusses below. The horizontal and vertical bars are all the same length.

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