User:Radinadd/FEA Homework 2

Problem 1
$$ k = \frac {EA}{L} \ \begin{bmatrix} C^2 & CS & -C^2 & -CS \\ CS & S^2 & -CS & -S^2 \\ -C^2 & -CS & C^2 & CS \\ -CS & -S^2 & CS & S^2 \end{bmatrix} $$

$$ k_1 = \frac {(1x10^6 psi)(5 in^2)}{(100 in)(\frac{1}{2})} \ \begin{bmatrix} \frac{1}{4} & \frac {\sqrt {3}}{4} & -\frac{1}{4} & -\frac{\sqrt{3}}{4} \\ \frac {\sqrt {3}}{4} & \frac{3}{4} & -\frac {\sqrt {3}}{4} & -\frac{3}{4} \\ -\frac{1}{4} & -\frac {\sqrt {3}}{4} & \frac{1}{4} & \frac {\sqrt {3}}{4} \\ -\frac {\sqrt {3}}{4} & -\frac{3}{4} & \frac {\sqrt {3}}{4} & \frac{3}{4} \end{bmatrix} $$

$$ k_2 = \frac {(1x10^6 psi)(5 in^2)}{100 in} \ \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

$$ k_3 = \frac {(1x10^6 psi)(5 in^2)}{(100 in)(\frac{\sqrt{3}}{2})} \ \begin{bmatrix} \frac{3}{4} & -\frac {\sqrt {3}}{4} & \frac {\sqrt {3}}{4} & \frac {\sqrt {3}}{4} \\ -\frac {\sqrt {3}}{4} & \frac{1}{4} & \frac {\sqrt {3}}{4} & -\frac{1}{4} \\ -\frac{3}{4} & \frac {\sqrt {3}}{4} & \frac{3}{4} & -\frac {\sqrt {3}}{4} \\ \frac {\sqrt {3}}{4} & -\frac{1}{4} & -\frac {\sqrt {3}}{4} & \frac{1}{4} \end{bmatrix} $$

$$ K = (50,000) \ \begin{bmatrix} 1.650 & -.375 & - & - & - & - & - & - \\ -.375 & 0.217 & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & - \\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ \end{bmatrix} $$

$$ (50,000) \ \begin{bmatrix} 1.650 & -.375 & - & - & - & - & - & - \\ -.375 & 0.217 & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & - \\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ - & - & - & - & - & - & - & -\\ \end{bmatrix} $$ x $$  \begin{bmatrix} d_{x1} \\ d_{y1} \\ d_{x2} = 0 \\ d_{y2} = 0 \\ d_{x3} = 0\\ d_{y3} = 0 \\ d_{x4} = 0 \\ d_{y5} = 0 \end{bmatrix} $$ = $$  \begin{bmatrix} 1000 \\ 1000 \\ F_{x2} \\ F_{y2} \\ F_{x3} \\ F_{y3} \\ F_{x4} \\ F_{y5} \end{bmatrix} $$

$$ \begin{bmatrix} 1000 \\ 1000 \end{bmatrix} $$ = $$ (50,000) \ \begin{bmatrix} 1.650 & -.375 \\ -.375 & 0.217 \end{bmatrix} $$ x $$  \begin{bmatrix} d_{x1} \\ d_{y1} \end{bmatrix} $$

$$ d_{1x} = 0.186 in ; d_{1y} = 0.0544 in $$

$$ \sigma^{(1)}= \frac {(1x10^6 psi)(5 in^2)}{(100 in)(\frac{1}{2})} [-\frac {\sqrt {3}}{4}  -\frac{3}{4}   \frac {\sqrt {3}}{4}   \frac{3}{4}] $$ $$ \begin{bmatrix} 0.186 \\ 0.0544 \\ 0 \\ 0 \end{bmatrix} $$

$$ \sigma^{(2)}= \frac {(1x10^6 psi)(5 in^2)}{(100 in)} [-1 0  1  0] $$ $$ \begin{bmatrix} 0.186 \\ 0.0544 \\ 0 \\ 0 \end{bmatrix} $$

$$ \sigma^{(3)}= \frac {(1x10^6 psi)(5 in^2)}{(100 in)(\frac{ \sqrt{3} }{2})} [\frac {\sqrt {3}}{4} -\frac{1}{4}  -\frac {\sqrt {3}}{4}  \frac{1}{4}] $$ $$ \begin{bmatrix} 0.186 \\ 0.0544 \\ 0 \\ 0 \end{bmatrix} $$

$$ \sigma^{(1)}= -3034 psi ; \sigma^{(2)}= -9300 psi  ;  \sigma^{(3)}= 2899 psi$$

Problem 2
$$ k_1 = \frac {(210x10^9 Pa)(5.0 x 10^{-4} m^2)}{(5m)(\frac{1}{2})} \ \begin{bmatrix} \frac{1}{4} & \frac {\sqrt {3}}{4} & -\frac{1}{4} & -\frac{\sqrt{3}}{4} \\ \frac {\sqrt {3}}{4} & \frac{3}{4} & -\frac {\sqrt {3}}{4} & -\frac{3}{4} \\ -\frac{1}{4} & -\frac {\sqrt {3}}{4} & \frac{1}{4} & \frac {\sqrt {3}}{4} \\ -\frac {\sqrt {3}}{4} & -\frac{3}{4} & \frac {\sqrt {3}}{4} & \frac{3}{4} \end{bmatrix} $$

$$ k_2 = \frac {(210x10^9 Pa)(5.0 x 10^{-4} m^2)}{(5m)(\frac{1}{2})} \ \begin{bmatrix} \frac{1}{4} & \frac {\sqrt {3}}{4} & -\frac{1}{4} & -\frac{\sqrt{3}}{4} \\ \frac {\sqrt {3}}{4} & \frac{3}{4} & -\frac {\sqrt {3}}{4} & -\frac{3}{4} \\ -\frac{1}{4} & -\frac {\sqrt {3}}{4} & \frac{1}{4} & \frac {\sqrt {3}}{4} \\ -\frac {\sqrt {3}}{4} & -\frac{3}{4} & \frac {\sqrt {3}}{4} & \frac{3}{4} \end{bmatrix} $$

$$ k_3 = 4000 \frac{N}{m} \ \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 \end{bmatrix}  $$

$$ \begin{bmatrix} 100 x 10^3 \\ 0 \end{bmatrix} $$ = $$ (4,000) \ \begin{bmatrix} 656.3 & 1137 \\ 1137 & 1969 \end{bmatrix} $$ x $$  \begin{bmatrix} d_{x1} \\ d_{y1} \end{bmatrix} $$

$$ d_{1x} = 9.626 m ; d_{1y} = 16.67 m $$

$$ \sigma^{(1)}= \frac {(210x10^9 Pa)(5.0 x 10^{-4} m^2)}{(5m)(\frac{1}{2})} [-\frac {\sqrt {3}}{4} -\frac{3}{4}  \frac {\sqrt {3}}{4}  \frac{3}{4}] $$ $$ \begin{bmatrix} 9.626 \\ 16.67 \\ 0 \\ 0 \end{bmatrix} $$

$$ \sigma^{(2)}= \frac {(210x10^9 Pa)(5.0 x 10^{-4} m^2)}{(5m)(\frac{1}{2})} [-\frac {\sqrt {3}}{4} -\frac{3}{4}  \frac {\sqrt {3}}{4}  \frac{3}{4}] $$ $$ \begin{bmatrix} 9.626 \\ 16.67 \\ 0 \\ 0 \end{bmatrix} $$

$$ \sigma^{(1)}= -175 MPa ; \sigma^{(2)}= -175 MPa $$