Forces on Bridges Exercises 
Using MATLAB to Solve a System of EquationsInvoke MATLAB and type the following: (Click here to download these commands as text) 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 nonzero 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.
Student ExercisesExercises 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.
Answer: 18.8 tons at joint 6 is the maximum force allowed.
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.
Answer: c = cos 40 and s = sin 40 in the matrix for #1. 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.
Answer: c = cos 50 and s = sin 50 in the matrix for #1.
Answer: 20 tons since fm and fi must support the most downward force.
Answer: Enter the following matrix in the Matlab program with c = cos 45 and s = sin
45

