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M 221 — Introduction to Linear Algebra — Spring 2018

TR 9:25-10:40, Cheever Hall 215

1 Linear Algebra for Structural Engineering

Linear Algebra is a major tool of structural engineering. This is because the analysis of a structure in equilibrium involves writing down a large collection of equations in many unknowns. Often, these equations are linear (or, can be linearized). This is where linear algebra and computing techniques immediately apply. Consider, for example, a two dimensional truss: a structure of beams joined together by smooth pins, and supports that are fastened so that they cannot move. Since the pins are smooth, no torque is transmit- ted. We assume that beams are perfectly rigid, and only longitudinal forces (along each beam) appear (compression/extension). External forces act at the joints (arrows). The truss is stable if and only if the vertical and horizontal components of the forces at each joint sum to zero.

1.1 Single joint

We start with a single joint, as shown below:

A

B

Figure 1: Two beams, A and B, joined at a common joint by a smooth pin, subject to an external force.

The sum of all forces acting on the joint must equal zero for the joint to be at equilibrium. For example, a compressed beam A will push onto the joint, while an extended beam B will pull on it. Denote by A and B the two compressive forces acting along the bars at the joint, in the direction of the joint, and let F denote the external force vector. Negative forces indicate extension: beams pulling away from the joint, instead. All three vectors are in R2. We assume 45◦ and 90◦ angles, resp. The three relevant force vectors at the joint are

A = a

[ 1√ 2 1√ 2

] B = b

[ −1 0

] F = f

[ 1√ 2

− 1√ 2

]

At equilibrium, these three force vectors need sum to zero: A + B + F = 0, thus a and b must satisfy:

1√ 2 a− b + 1√

2 f = 0

1√ 2 a− 1√

2 f = 0

Typically, the forces a and b along each beam are unknowns, while the external load f is given. We can write and solve this as a linear algebra problem:[

1√ 2 −1

1√ 2

0

] [ a b

] =

[ − f√

2 f√ 2

] Thus for stability: a = −f and b = −

√ 2f.

1

A

B

R1

R2

Figure 2: Two bars, A and B, are joined at a common joint by a smooth pin, and subject to an external force. The bars rest on supports R1 and R2 that absorb forces in horizontal and vertical direction, but exert no torque.

1.2 A simple complete system

Now consider a more complete system that includes anchors (supports), in figure 1. Reaction forces will appear in each support (the support pushes back against a compressed beam, or pulls on an extended beam): both supports will exert vertical and horizontal forces to counter the force of the beam; again, at each support forces need to cancel out. This adds 4 more unknowns,

R1 =

[ r1x r1y

] R2 =

[ r2x r2y

] and 4 more equations. The forces at the end of a beam push in opposite directions (otherwise the beam would move): As much as beam A pushes against the upper joint, it exerts the exact opposite force onto its support R1. At the supports, the relevant force vectors are thus −A and −B, and therefore R1 +(−A) = 0 and R2 + (−B) = 0:

− 1√ 2 a + r1x = 0 b + r2x = 0

− 1√ 2 a + r1y = 0 r2y = 0

In total, stability requires that there exist a solution to:

1√ 2 −1 0 0 0 0

1√ 2

0 0 0 0 0

− 1√ 2

0 1 0 0 0

− 1√ 2

0 0 1 0 0

0 1 0 0 1 0 0 0 0 0 0 1





a b r1x r1y r2x r2y

 = 

− f√ 2

f√ 2

0 0 0 0

 (1)

The properties of the matrix are very informative. If the matrix is invertible, then we know that a solution is guaranteed to exists. (Later, we will see how a non-invertible matrix is indicative of other things…). To determine whether a matrix is invertible, we do not need to actually invert it or solve the system: We can just compute its pivots, e.g., by an LU-decomposition and checking the diagonal of U. If we do solve the system, then we find out the compression/extension forces that each beam is subject to, and the reaction forces appearing at each support. These numbers can then be compared against material properties of beams and supports, so the calculation are essential in the determination if a structure will fail. But even for such a simple setup, the system involves 6 equations and 6 unknowns; not something we want to solve by hand.

2

1.3 A complete truss

We now consider a more complete truss, illustrated in figure 3. The system includes a total of 8 beams, 4 joints, and 2 supports. Each beam contributes an unknown force (its compression/extension), each support contributes two unknown forces (horizontal and vertical component of the reactive force), for a total of 12 unknowns. Further, each support and joint also provides two linear equations (stability requires forces to cancel in both horizontal and vertical direction), for a total of 12 equations. Here, the system matrix is square and thus has a chance of being invertible, and a stability solution might exist.

A

B C

D E

FG H

R1

R2

Figure 3: A complete truss. The system consists of 8 beams (8 unknowns) and 2 supports (4 unknowns), for a total of 12 unknowns; 4 joints and 2 supports for a total of 12 equations. Vertical load w is applied at the joint between beams B, C, and H.

Again, we collect all equations required by stability at each joint and support, and put them together into one big linear system. There are two equations for each of the four joints, ordered from left to right and top down, then two equations for R1 and R2, each. The resulting linear system is the following:

1√ 2 −1 0 0 0 0 0 0 0 0 0 0

1√ 2

0 0 0 0 0 1 0 0 0 0 0

0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 1 − 1√

2 0 0 0 0 0 0

0 0 0 0 0 − 1√ 2

0 −1 0 0 0 0 − 1√

2 0 0 −1 0 0 0 0 1 0 0 0

− 1√ 2

0 0 0 0 0 0 0 0 1 0 0

0 0 1 0 0 1√ 2

0 0 0 0 1 0

0 0 0 0 0 1√ 2

0 0 0 0 0 1





a b c d e f g h r1x r1y r2x r2y



=



0 0 0 w 0 0 0 0 0 0 0 0



(2)

Again, the properties of the system matrix are essential. An invertible matrix will ensure that no matter what external forces are loaded onto the joints (encoded in the right hand side vector), a beam force distribution exists that achieves overall stability of the system (no net forces and nothing will move). Clearly, nobody wants to solve this 12× 12 system by hand.

3

Student Name: Student GID:

1. Load the matrix and right-hand-side vector of the system (1) corresponding to figure 1 into MATLAB (assuming f = 1):

s = 1/sqrt(2);

A = [

s -1 0 0 0 0

s 0 0 0 0 0

-s 0 1 0 0 0

-s 0 0 1 0 0

0 1 0 0 1 0

0 0 0 0 0 1

];

b = [ -s s 0 0 0 0 ]’;

2. Solve for the unknown beam and support forces, e.g., using Gauss-Jordan elimination on the aug- mented matrix rref( [ A b ] ). If A is invertible, then we get a [ I c ] as output, and imme- diately find x = c as the solution vector.

3. In the figure below, label each beam with the force thus calculated for stability.

4. Which beam(s) are compressed, which are extended? (mark them)

5. Without redoing any MATLAB calculation: in the second figure, label the forces if the external load was f = −2, instead, i.e., the joint is pulled out, twice as hard.

6. Now, load the matrix and RHS vector for the system in (2) corresponding to figure 3, with w = 1.

7. Solve for the beam and support forces, and label them in the figure below.

8. Mark the beams subjected to the largest extension, and largest compression, respectively.

3)

A

B

R1

R2

5)

A

B

R1

R2

7)

A

B C

D E

FG H

R1

R2

1

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