(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.)

So, the inverse of the flexibility matrix is

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

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

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.