EXAMPLE Matlab Help

A Statically Indetenninate Problem

Determine the forces in the three equally spaced supports that hold up a light fixture. The. supports are 5 ft apart. The fixture weighs 400 lb, and its mass center is 4 ft from the right end. (a) Solve the problem by hand. (b) Obtain the solution using the MATLAB left-division method and the pseudoinverse method
solution
(a) Figure 6.4-1 shows the fixture and the free-body diagram, where T1• T2,and T3are the ‘on forces in the supports. For the fixture to be in equilibrium. the vertical forces must I, and the total moments about an arbitrary fixed point-say. the right endpointbe zero. These conditions give the two equations:

TJ+T2+T3-400=O
4,00(4) – IOTJ – 5T2 = 0

A light fixture and its free-body diagram.

A light fixture and its
free-body diagram.

Capture

Substitute this expression into (6.4-3) and solve for T\ to find that T\ = T3 – 80. Thus the solution in terms of T3 is
T\ = T3 – 80

T2 = 320 – 2T\ = 320 – 2(T3 – 80) = 480 – 2T3

We cannot determine numerical values for any of the -forces, Such a problem, in which the equations of statics do not give enough equations to find all of the unknowns, is called statically indeterminate.

(b) A MATLAB session to check the ranks and to solve this problem using left division looks like

»A = [1, 1, 1; 10, 5, 0) ;
»b= [400;1600);
»rank(A)
ans =
·2
»rank ([A b))
ans =
2
»A\b
ans
160.0000
o
240.0000

The answer corresponds to T\ = 160, T2 = 0, and T3 = 240 lb. This example illustrates  how the MATLAB left-division operator produces a solution with one or more variables set to zero, for underdeterrnined sets having more unknowns than equations.

To use the pseudoinverse operator, type the command pinv (A) *b. The result is T\ = 93.3333, T2 = 133.3333, and T3 = 173.3333 lb. This answer is the minimum norm solution for real values of the variables. The minimum norm solution consists of the real values of T\, T2′ and T3 that minimize

Capture

The smallest value N can have is zero. This result occurs when T3 = 173 ± 97i, which corresponds to T\ = 93 ± 97i and T2 = 827 ± 194i. This result is a valid solution of the original equations, but is not the minimum norm solution where T\ T2′ and T3 are restricted to real values (we know that the forces cannot be complex).

We can find the real value of T3 that minimizes N by.plotting N versus T3 or by using calculus. The answer is T3 = 173.3333, which gives T\ = 93.3333 and T2 = 133.3333. These values are the minimum norm solution given by the pseudoinverse method.

The smallest value N can have is zero. This result occurs when T3 = 173 ± 97i, which corresponds to T\ = 93 ± 97i and T2 = 827 ± 194i. This result is a valid solution of the original equations, but is not the minimum norm solution where T\, T2′ and T3 are restricted to real values (we know that the forces cannot be complex).
We can find the real value of T3 that minimizes N by.plotting N versus T3 or by using calculus. The answer is T3 = 173.3333, which gives T\ = 93.3333 and T2 = 133.3333. These values are the minimum norm solution given by the pseudoinverse method.

Posted on July 30, 2015 in Linear Algebric Equations

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