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Chapter 2: Q41E (page 93)

(M) Let \(D = \left( {\begin{aligned}{*{20}{c}}{.{\bf{0040}}}&{.{\bf{0030}}}&{.{\bf{0010}}}&{.{\bf{0005}}}\\{.{\bf{0030}}}&{.{\bf{0050}}}&{.{\bf{0030}}}&{.{\bf{0010}}}\\{.{\bf{0010}}}&{.{\bf{0030}}}&{.{\bf{0050}}}&{.{\bf{0030}}}\\{.{\bf{0005}}}&{.{\bf{0010}}}&{.{\bf{0030}}}&{.{\bf{0040}}}\end{aligned}} \right)\) be a flexibility matrix for an elastic beam with four points at which force is applied. Units are centimeters per newton of force. Meaurements at the four points show deflections of .08, .12, and .12 cm. Determine the forces at the four points.

Short Answer

Expert verified

\(\left( {12,\,1.5,\,21.5,\,12} \right)\;\;{\rm{newtons}}\)

Step by step solution

01

Find the inverse of matrix D

Let \(D = \left( {\begin{aligned}{*{20}{c}}{.0040}&{.0030}&{.0010}&{.0005}\\{.0030}&{.0050}&{.0030}&{.0010}\\{.0010}&{.0030}&{.0050}&{.0030}\\{.0005}&{.0010}&{.0030}&{.0040}\end{aligned}} \right)\).

Use the code in MATLAB to find the inverse of D.

\( > > \,\,D = \left( \begin{aligned}{l}\begin{aligned}{*{20}{c}}{.0040}&{.0030}&{.0010}&{.0005;\,\,\begin{aligned}{*{20}{c}}{.0030}&{.0050}&{.0030}&{.0010;\,\,}\end{aligned}}\end{aligned}\\\begin{aligned}{*{20}{c}}{.0010}&{.0030}&{.0050}&{.0030;\,\,\begin{aligned}{*{20}{c}}{.0005}&{.0010}&{.0030}&{.0040}\end{aligned}}\end{aligned}\end{aligned} \right)\)

\( > > {\rm{U}} = {\rm{Inv}}\left( D \right)\)

So, the inverse 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 \(\left( {\begin{aligned}{*{20}{c}}{.08}\\{.12}\\{.16}\\{.12}\end{aligned}} \right)\).

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

\(\begin{aligned}{c}f = \left( {\begin{aligned}{*{20}{c}}{533.333}&{ - 433.333}&{233.333}&{ - 133.333}\\{ - 433.333}&{695.833}&{ - 470.833}&{233.333}\\{233.333}&{ - 470.833}&{695.833}&{ - 433.333}\\{ - 133.333}&{233.333}&{ - 433.333}&{533.333}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{.08}\\{.12}\\{.16}\\{.12}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{12}\\{1.5}\\{21.5}\\{12}\end{aligned}} \right)\end{aligned}\)

So, the forces required to produce deflection at points 1, 2, 3 and 4 are \(\left( {12,\,1.5,\,21.5,\,12} \right)\;\;{\rm{newtons}}\).

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Most popular questions from this chapter

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.
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