User:Eml4500.f08.wiki1.aguilar/hw2

= Meeting 7 - 10 September 2008= Note: The director cosines are the components of ĩ (unit vector along the x-axis) with respect to basis (î,ĵ) î=cosθ(e)î + sinθ(e)ĵ Model 2-bar truss system (continued from previous lecture) Element 1:

\ \theta^{(1)} = 30^{o} $$



\ l^{(1)} = cos\theta^{(1)}=cos30^{o}=\frac\sqrt{3}{2} $$



\ m^{(1)} = sin\theta^{(1)}= sin30^{o}=\frac{1}{2} $$



\ k^{(1)} = \frac{E^1 A^1}{L^1}=\frac{(3)(1)}{(4)} = \frac{3}{4} $$



\mathbf{k} ^{(1)} = \begin{bmatrix} (k_{11}^{(1)}) & (k_{12}^{(1)})    & (k_{13}^{(1)})    & (k_{14}^{(1)})  \\ (k_{21}^{(1)}) & (k_{22}^{(1)})    & (k_{23}^{(1)})    & (k_{24}^{(1)})  \\

(k_{31}^{(1)}) & (k_{32}^{(1)})    & (k_{33}^{(1)})    & (k_{34}^{(1)})  \\ (k_{41}^{(1)}) & (k_{42}^{(1)})    & (k_{43}^{(1)})    & (k_{44}^{(1)}) \end{bmatrix} $$

where

\mathbf{k^{(e)}} = \left[k_{ij}^{(e)}\right] $$ where i = row, j = column, and e = element number. Therefore,

k_{11}^{(1)} = k^{(1)}\left (l^{(1)}\right)^2 = \left (\frac{3}{4} \right)\left(\frac\sqrt{3}{2}\right)=\frac{9}{16}$$
 * $$k_{12}^{(1)} = k^{(1)}\left (l^{(1)}m^{(1)}\right) = \frac{3\sqrt{3}}{16}$$
 * $$k_{42}^{(1)} = -k^{(1)}\left (m^{(1)}\right)^2 = -\frac{3}{16}$$

Two observations: 1. Only need to compute three values, l2, m2, and l*m. The other coefficients have the same absolute value. They just differ by a (+) or (-). 2. The matrix k(e) is symmetric. i.e. kij(e) = kji(e) or k(e)T = '''k(e)

Element 2:

\ \theta^{(2)}= \frac{-\pi}{4} $$



\ l^{(2)} = cos\theta^{(2)}=cos\frac{-\pi}{4}=\frac\sqrt{2}{2} $$



\ m^{(2)} = sin\theta^{(2)}= sin\frac{-\pi}{4}=\frac\sqrt{2}{2} $$



\ k^{(2)} = \frac{E^1 A^1}{L^1}=\frac{(5)(2)}{(2)} = 5 $$



\mathbf{k} ^{(2)} = \begin{bmatrix} (k_{11}^{(2)}) & (k_{12}^{(2)})    & (k_{13}^{(2)})    & (k_{14}^{(2)})  \\ (k_{21}^{(2)}) & (k_{22}^{(2)})    & (k_{23}^{(2)})    & (k_{24}^{(2)})  \\

(k_{31}^{(2)}) & (k_{32}^{(2)})    & (k_{33}^{(2)})    & (k_{34}^{(2)})  \\ (k_{41}^{(2)}) & (k_{42}^{(2)})    & (k_{43}^{(2)})    & (k_{44}^{(2)}) \end{bmatrix} $$

Therefore, the three values needed to compute the matrix k(2) are as follows:
 * $$ k_{11}^{(2)} = k^{(2)}\left (l^{(2)}\right)^2 = \left (5 \right)\left(\frac\sqrt{2}{2}\right)^2=2.5$$
 * $$k_{12}^{(2)} = k^{(2)}\left (l^{(2)}m^{(2)}\right) = 2.5$$
 * $$k_{42}^{(2)} = -k^{(2)}\left (m^{(2)}\right)^2 = -2.5$$

The given homework was to compute both k(1) and k(2). They are as follows:



\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} $$

\mathbf{k} ^{(2)} = \begin{bmatrix} 2.5 & 2.5    & -2.5    & -2.5  \\  2.5  & 2.5 & -2.5    & -2.5 \\

-2.5 & -2.5 & 2.5    & 2.5 \\  -2.5 & -2.5 & 2.5    & 2.5 \end{bmatrix} $$  Element FD relationship: k(e)d(e)=f(e) k is a 4x4 matrix. d is a 4x1 matrix. $$\mathbf{d} = \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix}$$ f therefore is a 4x1 matrix. $$\mathbf{f} = \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Global FD relationship: Kd=F K is a nxn matrix. d is a nx1 matrix. F is a nx1 matrix. Here, n is the number of dof's (degrees of freedom). For this problem, n=6.

$$ \begin{bmatrix} K_{11} & K_{12} & \cdots & K_{16}\\ \vdots & & & \vdots \\ K_{61} & \cdots & \cdots & K_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ \vdots \\ d_2 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ \vdots \\ F_2 \end{Bmatrix} $$