Linear Algebra for Team-Based Inquiry Learning

2022 Edition

Steven Clontz Drew Lewis
University of South Alabama University of South Alabama

September 14, 2022

Section A.1: Civil Engineering: Trusses and Struts

Definition A.1.1

In engineering, a truss is a structure designed from several beams of material called struts, assembled to behave as a single object.

Figure 1. A simple truss
Figure 2. A simple truss

Activity A.1.2

Consider the representation of a simple truss pictured below. All of the seven struts are of equal length, affixed to two anchor points applying a normal force to nodes \(C\) and \(E\text{,}\) and with a \(10000 N\) load applied to the node given by \(D\text{.}\)

Figure 3. A simple truss

Which of the following must hold for the truss to be stable?

  1. All of the struts will experience compression.

  2. All of the struts will experience tension.

  3. Some of the struts will be compressed, but others will be tensioned.

Observation A.1.3

Since the forces must balance at each node for the truss to be stable, some of the struts will be compressed, while others will be tensioned.

Figure 4. Completed truss

By finding vector equations that must hold at each node, we may determine many of the forces at play.

Remark A.1.4

For example, at the bottom left node there are 3 forces acting.

Figure 5. Truss with forces

Let \(\vec F_{CA}\) be the force on \(C\) given by the compression/tension of the strut \(CA\text{,}\) let \(\vec F_{CD}\) be defined similarly, and let \(\vec N_C\) be the normal force of the anchor point on \(C\text{.}\)

For the truss to be stable, we must have:

\begin{equation*} \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 \end{equation*}

Activity A.1.5

Using the conventions of the previous remark, and where \(\vec L\) represents the load vector on node \(D\text{,}\) find four more vector equations that must be satisfied for each of the other four nodes of the truss.

Figure 6. A simple truss
\begin{equation*} A: \unknown \end{equation*}
\begin{equation*} B: \unknown \end{equation*}
\begin{equation*} C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 \end{equation*}
\begin{equation*} D:\unknown \end{equation*}
\begin{equation*} E:\unknown \end{equation*}

Remark A.1.6

The five vector equations may be written as follows.

\begin{equation*} A: \vec F_{AC}+\vec F_{AD}+\vec F_{AB}=\vec 0 \end{equation*}
\begin{equation*} B: \vec F_{BA}+\vec F_{BD}+\vec F_{BE}=\vec 0 \end{equation*}
\begin{equation*} C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 \end{equation*}
\begin{equation*} D: \vec F_{DC}+\vec F_{DA}+\vec F_{DB} +\vec F_{DE}+\vec L=\vec 0 \end{equation*}
\begin{equation*} E: \vec F_{EB}+\vec F_{ED}+\vec N_E=\vec 0 \end{equation*}

Observation A.1.7

Each vector has a vertical and horizontal component, so it may be treated as a vector in \(\IR^2\text{.}\) Note that \(\vec F_{CA}\) must have the same magnitude (but opposite direction) as \(\vec F_{AC}\text{.}\)

\begin{equation*} \vec{F}_{CA} = x\begin{bmatrix} \cos(60^\circ) \\ \sin(60^\circ) \end{bmatrix} = x\begin{bmatrix} 1/2 \\ \sqrt{3}/2\end{bmatrix} \end{equation*}
\begin{equation*} \vec{F}_{AC} = x\begin{bmatrix} \cos(-120^\circ) \\ \sin(-120^\circ) \end{bmatrix} = x\begin{bmatrix} -1/2 \\ -\sqrt{3}/2\end{bmatrix} \end{equation*}

Activity A.1.8

To write a linear system that models the truss under consideration with constant load \(10000\) newtons, how many scalar variables will be required?

  • \(7\text{:}\) \(5\) from the nodes, \(2\) from the anchors

  • \(9\text{:}\) \(7\) from the struts, \(2\) from the anchors

  • \(11\text{:}\) \(7\) from the struts, \(4\) from the anchors

  • \(12\text{:}\) \(7\) from the struts, \(4\) from the anchors, \(1\) from the load

  • \(13\text{:}\) \(5\) from the nodes, \(7\) from the struts, \(1\) from the load

Figure 7. A simple truss

Observation A.1.9

Since the angles for each strut are known, one variable may be used to represent each.

Figure 8. Variables for the truss

For example:

\begin{equation*} \vec F_{AB}=-\vec F_{BA}=x_1\begin{bmatrix}\cos(0)\\\sin(0)\end{bmatrix} =x_1\begin{bmatrix}1\\0\end{bmatrix} \end{equation*}
\begin{equation*} \vec F_{BE}=-\vec F_{EB}=x_5\begin{bmatrix}\cos(-60^\circ)\\\sin(-60^\circ)\end{bmatrix} =x_5\begin{bmatrix}1/2\\-\sqrt{3}/2\end{bmatrix} \end{equation*}

Observation A.1.10

Since the angle of the normal forces for each anchor point are unknown, two variables may be used to represent each.

Figure 9. Truss with normal forces
\begin{equation*} \vec N_C=\begin{bmatrix}y_1\\y_2\end{bmatrix} \hspace{3em} \vec N_D=\begin{bmatrix}z_1\\z_2\end{bmatrix} \end{equation*}

The load vector is constant.

\begin{equation*} \vec L = \begin{bmatrix}0\\-10000\end{bmatrix} \end{equation*}

Remark A.1.11

Each of the five vector equations found previously represent two linear equations: one for the horizontal component and one for the vertical.

Figure 10. Variables for the truss
\begin{equation*} C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 \end{equation*}
\begin{equation*} \Leftrightarrow x_2\begin{bmatrix}\cos(60^\circ)\\\sin(60^\circ)\end{bmatrix}+ x_6\begin{bmatrix}\cos(0^\circ)\\\sin(0^\circ)\end{bmatrix}+ \begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \end{equation*}
\(\sqrt{3}/2\approx 0.866\)
\begin{equation*} \Leftrightarrow x_2\begin{bmatrix}0.5\\0.866\end{bmatrix}+ x_6\begin{bmatrix}1\\0\end{bmatrix}+ y_1\begin{bmatrix}1\\0\end{bmatrix}+ y_2\begin{bmatrix}0\\1\end{bmatrix}= \begin{bmatrix}0\\0\end{bmatrix} \end{equation*}

Activity A.1.12

Expand the vector equation given below using sine and cosine of appropriate angles, then compute each component (approximating \(\sqrt{3}/2\approx 0.866\)).

Figure 11. Variables for the truss
\begin{equation*} D:\vec F_{DA}+\vec F_{DB}+\vec F_{DC}+\vec F_{DE}=-\vec L \end{equation*}
\begin{equation*} \Leftrightarrow x_3\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ x_4\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ x_6\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ x_7\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}= \begin{bmatrix}\unknown\\\unknown\end{bmatrix} \end{equation*}
\begin{equation*} \Leftrightarrow x_3\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ x_4\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ x_6\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ x_7\begin{bmatrix}\unknown\\\unknown\end{bmatrix}= \begin{bmatrix}\unknown\\\unknown\end{bmatrix} \end{equation*}

Observation A.1.13

The full augmented matrix given by the ten equations in this linear system is given below, where the elevent columns correspond to \(x_1,\dots,x_7,y_1,y_2,z_1,z_2\text{,}\) and the ten rows correspond to the horizontal and vertical components of the forces acting at \(A,\dots,E\text{.}\)

\begin{equation*} \left[\begin{array}{ccccccccccc|c} 1&-0.5&0.5&0&0&0&0&0&0&0&0&0\\ 0&-0.866&-0.866&0&0&0&0&0&0&0&0&0\\ -1&0&0&-0.5&0.5&0&0&0&0&0&0&0\\ 0&0&0&-0.866&-0.866&0&0&0&0&0&0&0\\ 0&0.5&0&0&0&1&0&1&0&0&0&0\\ 0&0.866&0&0&0&0&0&0&1&0&0&0\\ 0&0&-0.5&0.5&0&-1&1&0&0&0&0&0\\ 0&0&0.866&0.866&0&0&0&0&0&0&0&10000\\ 0&0&0&0&-0.5&0&-1&0&0&1&0&0\\ 0&0&0&0&0.866&0&0&0&0&0&1&0\\ \end{array}\right] \end{equation*}

Observation A.1.14

This matrix row-reduces to the following.

\begin{equation*} \sim \left[\begin{array}{ccccccccccc|c} 1&0&0&0&0&0&0&0&0&0&0&-5773.7\\ 0&1&0&0&0&0&0&0&0&0&0&-5773.7\\ 0&0&1&0&0&0&0&0&0&0&0&5773.7\\ 0&0&0&1&0&0&0&0&0&0&0&5773.7\\ 0&0&0&0&1&0&0&0&0&0&0&-5773.7\\ 0&0&0&0&0&1&0&0&0&-1&0&2886.8\\ 0&0&0&0&0&0&1&0&0&-1&0&2886.8\\ 0&0&0&0&0&0&0&1&0&1&0&0\\ 0&0&0&0&0&0&0&0&1&0&0&5000\\ 0&0&0&0&0&0&0&0&0&0&1&5000\\ \end{array}\right] \end{equation*}

Observation A.1.15

Thus we know the truss must satisfy the following conditions.

\begin{align*} x_1=x_2=x_5&=-5882.4\\ x_3=x_4&=5882.4\\ x_6=x_7&=2886.8+z_1\\ y_1&=-z_1\\ y_2=z_2&=5000 \end{align*}

In particular, the negative \(x_1,x_2,x_5\) represent tension (forces pointing into the nodes), and the postive \(x_3,x_4\) represent compression (forces pointing out of the nodes). The vertical normal forces \(y_2+z_2\) counteract the \(10000\) load.

Figure 12. Completed truss