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(M) Let \(D\) as in Exercise 41, determine the forces that produce a deflection of .24 cm at the second point on the beam, with zero deflections at the other three points. How is the answer related to the enteries in \({D^{ - {\bf{1}}}}\)? (Hint: First answer the question when the deflection is 1 cm at the second point.)

Short Answer

Expert verified

\(\left( { - 104,\,167,\, - 113,\,56} \right)\;\;{\rm{newtons}}\)

The forces are 0.24 times the second column of \({D^{ - 1}}\).

Step by step solution

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01

Find the inverse of matrix D

So, the inverse of the flexibility matrix is

\({D^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{533.333}&{ - 433.333}&{233.333}&{ - 133.333}\\{ - 433.333}&{695.833}&{ - 470.833}&{233.333}\\{233.333}&{ - 470.833}&{695.833}&{ - 433.333}\\{ - 133.333}&{233.333}&{ - 433.333}&{533.333}\end{aligned}} \right)\).

02

Solve the equation \(f = {D^{ - 1}}y\)

The deflection matrix is \(\left( {\begin{aligned}{*{20}{c}}0\\{.24}\\0\\0\end{aligned}} \right)\).

Use the equation \(f = {D^{ - 1}}y\).

\(\begin{aligned}{c}f = \left( {\begin{aligned}{*{20}{c}}{533.333}&{ - 433.333}&{233.333}&{ - 133.333}\\{ - 433.333}&{695.833}&{ - 470.833}&{233.333}\\{233.333}&{ - 470.833}&{695.833}&{ - 433.333}\\{ - 133.333}&{233.333}&{ - 433.333}&{533.333}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0\\{.24}\\0\\0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 104}\\{167}\\{ - 113}\\{56}\end{aligned}} \right)\end{aligned}\)

So, the forces required to produce deflections of \(\left( {0,\,\,0.24,\,0,\,0} \right)\;{\rm{cm}}\) at points 1, 2, 3, and 4, respectively, are \(\left( { - 104,\,167,\, - 113,\,56} \right)\;\;{\rm{newtons}}\).

03

Interpret the applied force

Forces required to produce zero deflection at all points except point 2 are 0.24 times of the second column of the inverse flexibility matrix.

As \(y \mapsto {D^{ - 1}}y\) represents the linear transformation, the forces that produce the deflection of 0.24 cm at the second point are 0.24 times the forces required to produce a deflection of 1 cm at the second point.

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Most popular questions from this chapter

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

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