Answer
The forces that produce a deflection of .24 cm at the second point on the beam, with zero deflections at the other three points are;
Point 1=-104N
Point 2=167N
Point 3=-113N
Point 4=56N
Work Step by Step
The flexibility matrix is given as;
$\mathbf{D} = \begin{bmatrix}.0040&.0030&.0010&.0005\\.0030&.0050&.0030&.0010\\.0010&.0030&.0050&.0030\\.0005&.0010&.0030&.0040\end{bmatrix}$
We are required to 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
Deflections $\mathbf{y}$ at the four points are;
$\mathbf{y} = \begin{bmatrix}0\\0.24\\0\\0\end{bmatrix}$
Given that $\mathbf{y}=Df$
Then,
$\mathbf{f}=D^{-1}y$
$\mathbf{D^{-1} }= \begin{bmatrix}533.3333&-433.3333&233.3333&-133.3333\\-433.3333&695.8333&-470.8333&233.3333\\233.3333&-470.8333&695.8333&-433.3333\\-133.3333&233.3333&-433.3333&533.3333\end{bmatrix}$
Hence;
$\mathbf{f}
=\begin{bmatrix}533.3333&-433.3333&233.3333&-133.3333\\-433.3333&695.8333&-470.8333&233.3333\\233.3333&-470.8333&695.8333&-433.3333\\-133.3333&233.3333&-433.3333&533.3333\end{bmatrix} \begin{bmatrix}0\\0.24\\0\\0\end{bmatrix}$
$\mathbf{f}=\begin{bmatrix}-104\\167\\-113\\56\end{bmatrix}$
Hence, The forces at the four points are;
Point 1=-104N
Point 2=167N
Point 3=-113N
Point 4=56N
