Question

(M) Compute the stifness matrix \({D^{ - {\bf{1}}}}\) for \(D\) in Exercise 39. List the forces needed to produce a deflection of .04 in. at point, with zero deflection at other points.

Short answer

\(\left( {0,\, - 5,\;10} \right)\;{\rm{pounds}}\)

Step-by-step solution

Find the inverse of matrix D

Let \(D = \left( {\begin{aligned}{*{20}{c}}{.005}&{.002}&{.001}\\{.002}&{.004}&{.002}\\{.001}&{.002}&{.005}\end{aligned}} \right)\).

Use the code inMATLAB to find the inverse of D.

\( > > \,\,D = \left( {\begin{aligned}{*{20}{c}}{.005}&{.002}&{.001;\,\begin{aligned}{*{20}{c}}{.002}&{.004}&{.002;\,\,\begin{aligned}{*{20}{c}}{.001}&{.002}&{.005}\end{aligned}}\end{aligned}}\end{aligned}} \right)\)

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

So, the inverse matrix is \({D^{ - 1}} = 125\left( {\begin{aligned}{*{20}{c}}2&{ - 1}&0\\{ - 1}&3&{ - 1}\\0&{ - 1}&2\end{aligned}} \right)\).

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

The deflection matrix is \(\left( {\begin{aligned}{*{20}{c}}0\\0\\{.04}\end{aligned}} \right)\).

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

\(\begin{aligned}{c}f = 125\left( {\begin{aligned}{*{20}{c}}2&{ - 1}&0\\{ - 1}&3&{ - 1}\\0&{ - 1}&2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0\\0\\{.04}\end{aligned}} \right)\\ = 125\left( {\begin{aligned}{*{20}{c}}0\\{ - .04}\\{.08}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\{10}\end{aligned}} \right)\end{aligned}\)

So, the force required to produce the deflection at point 3 is \(\left( {0,\, - 5,\;10} \right)\;{\rm{pounds}}\).