Question

(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

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

Step-by-step solution

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

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

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