User:Eml4500.f08.lulz.abcd/HW2

We are asked to prove that P1 = P2. From previous calculations we have a full FD calculation for element 1:

$$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix}\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{Bmatrix} = \begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$

The root sum squared method can be used for the proof required. Generally the root sum squared equation is:

$$P_1^{(e)} = \sqrt{\left(f_1^{(e)}\right)^2 + \left(f_2^{(e)}\right)^2}$$

We can then expand that to our condition and create a system of equations for P1 and P2.

$$P_1^{(1)} = \sqrt{\left(f_1^{(1)}\right)^2 + \left(f_2^{(1)}\right)^2}$$

$$P_2^{(1)} = \sqrt{\left(f_3^{(1)}\right)^2 + \left(f_3^{(1)}\right)^2}$$

From the FD matrix we have values for all of the local forces used in this system so P1 and P2 can be calculated directly and their equality proven.

$$P_1^{(1)} = \sqrt{(-4.4378)^2 + (-2.5622)^2} = 5.1243$$

$$P_2^{(1)} = \sqrt{(4.4378)^2 + (2.5622)^2} = 5.1243$$