User:Eml4500.f08.wiki1.schaet/hw2

=Meeting 11 Lecture Notes (19 Sep 08)=

Element 1: k(e)d(e)=f(e)

\mathbf{k} ^{(1)} = \begin{bmatrix} (\frac{9}{16})        & (\frac{3\sqrt{3}}{16})    & (-\frac{9}{16})    & (-\frac{3\sqrt{3}}{16})  \\ (\frac{3\sqrt{3}}{16}) & (\frac{3}{16})   & (-\frac{3\sqrt{3}}{16})    & (-\frac{3}{16})  \\

(-\frac{9}{16})       & (-\frac{3\sqrt{3}}{16})    & (\frac{9}{16})    & (\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})& (-\frac{3}{16})   & (\frac{3\sqrt{3}}{16})    & (\frac{3}{16}) \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{bmatrix} (-\frac{9}{16})   & (-\frac{3\sqrt{3}}{16})  \\ (-\frac{3\sqrt{3}}{16})   & (-\frac{3}{16})  \\ (\frac{9}{16})   & (\frac{3\sqrt{3}}{16})  \\ (\frac{3\sqrt{3}}{16})   & (\frac{3}{16}) \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} $$ where f1 and f2 are the reaction forces and f3 and f4 are the internal forces for element 1. Element 1 is in equilibrium so ΣFx=f1(1) + f3(1) = 0     (1) ΣFy=f2(1) + f4(1) = 0     (2)

Method 2: Statics for a 2-bar truss Use Euler Cut Method Back to FE solution. The question arises, how do you bring P back into the picture. Find the equilibrium at node 2. This is part of the homework.

P^{(1)} = [(f_1^{(1)})^2 + (f_2^{(1)})^2]^{\frac{1}{2}}$$

P^{(2)} = [(f_1^{(2)})^2 + (f_2^{(2)})^2]^{\frac{1}{2}}$$ Therefore,

P = [((f_1^{(1)})+ (f_1^{(2)}))^2 + ((f_2^{(1)}) + (f_2^{(2)}))^2]^{\frac{1}{2}} $$ Which equals our value for P. Therefore, node 2 is in equilibrium.