Forces on Bridges Exercises

#### Using MATLAB to Solve a System of Equations

```
n = 5;				% n is the number of variables
A = zeros(n,n);			% A is first filled with zeros
B = zeros(n,1);			% B is filled with zeros
angle = pi/4;			% The angle is stored in radians
s = sin(angle);			% The coefficients are calculated
c = cos(angle);
A(1,1) = -c;			% Enter in the non-zero elements of A and B
A(1,5) = c;
A(2,1) = -s;
A(2,3) = -1;
A(2,5) = -s;
A(3,2) = -1;
A(3,4) = 1;
A(4,3) = 1;
A(5,4) = -1;
A(5,5) = -c;
B(4,1) = 10;			% The solution to A*X = B could be:

X= inv(A)*B			% This uses the inverse
X= A\B				% This uses Gaussian Elimination

```

These solutions will appear on the screen as text.

```
-7.0711
5
10
5
-7.0711

```

This code segment may be saved as a script file to be run again at a later time. The coefficient matrix could be entered into a data file with a text editor and then loaded into the program. The documentation with MATLAB explains those techniques.

#### Exercises 1 - 8:

1. On the bridge above, a 10 ton vehicle is located at joints 6, 7, and 8. Find the forces on each member if joint 1 is fixed and all angles are 45 degrees.

Answer: Enter the following matrices into Matlab. c = cos 45 and s = sin 45.

2. Suppose that the maximum force supported by each member is 20 tons. If a single vehicle crosses the bridge without collapse, what is the maximum tonnage it may be?

Answer: 18.8 tons at joint 6 is the maximum force allowed.

3. Suppose the maximum force supported by each member is 20 tons. If vehicles are located at joints 6, 7 and 8, what are their maximum tonnages? Can the bridge hold more by spreading the forces over the bridge or by concentrating most of the force in one place?

Answer: 9.4 tons at joints 6, 7 and 8 is supported by the bridge. Distributing the weight allows 28.2 tons total versus 18.8 tons at a single joint.

4. Suppose that the bridge is changed by using 40 degree angles at the bottom joints, rather than 45 degree angles. Place a 10 ton vehicle at joints 6, 7, and 8. What are the forces on the members now?

Answer: c = cos 40 and s = sin 40 in the matrix for #1.
Also, use:

5. Using the changed bridge, suppose that the maximum force supported by each member is 20 tons. If a single vehicle crosses the bridge without collapse, what is the maximum tonnage it may be?

Answer: 17.1 tons at joint 6 is the maximum force allowed.

6. Using 50 degree angles at the bottom joints, place a 10 ton vehicle at joints 6, 7, and 8. What are the forces on the members?

Answer: c = cos 50 and s = sin 50 in the matrix for #1.
Also, use:

7. Using this bridge, suppose that the maximum force supported by each member is 20 tons. If a single vehicle crosses the bridge without collapse, what is the maximum tonnage it may be?

Answer: 20 tons since fm and fi must support the most downward force.

8. On the bridge above, if forces of 10 tons are at joint 3, 15 tons at joint 5 and 10 tons at joint 9, find the forces on the members of the bridge.

Answer: Enter the following matrix in the Matlab program with c = cos 45 and s = sin 45

Continue on to the Background Information or back to the Bridge Model Main Page.

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