User:Eml4500.f08.jamama.jan/Meeting7

Model of the bar turss system (continued)

Elem. 1
θ(1) = 30°

l(1) = cos(1)(θ) = cos(1)(30°) = $$ \frac{\sqrt{3}}{2} $$

m(1) = sin(1)(θ) = sin(1)(30°) = $$ \frac{1}{2} $$

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

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}$$ = [k$$_{ij}^{(e)}]$$4 &times; 4, where $$_{j = 1,2,3,4\  \Rightarrow \ column\ index}^{i = 1,2,3,4\ \Rightarrow \ row\ index}$$  and $$e $$ = number of element

$$k_{11}^{(1)} = k^{(1)} [l^{(1)}]^2 = \frac {3}{4} * [\frac{\sqrt{3}}{2}]^2 = \frac {9}{16}$$

$$k_{12}^{(1)} = k^{(1)} [l^{(1)}m^{(1)}] = \frac {3}{4} * \frac{\sqrt{3}}{2}]*\frac{1}{2} = \frac {3\sqrt{3}}{16}$$

$$k_{42}^{(1)} = k^{(1)} (-[m^{(1)}]^2) = \frac {3}{4} * (-\frac{1}{4}) = -\frac {3}{16}$$

Note: this matrix is symetircal with respect to the diagnol $$k_{xx}^{(1)}, where\; x= 1,2,3... \to k_{13}^{(1)}=k_{31}^{(1)}. $$

In genral, $$ k_{ij}^{(e)}= k_{ji}^{(e)}\ or\ [{\mathbf{k}^{(e)}}]^T = \mathbf{k}^{(e)}$$

Elem. 2


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

θ(2) $$= -45^\circ \Rightarrow _{m^{(2)} =\ sin(-45^\circ)\ =-\frac{\sqrt{2}}{2}}^{l^{(2)} =\ cos(-45^\circ)\ =\frac{\sqrt{2}}{2}} $$

$$\mathbf{k}^{(e)}$$ = [k$$_{ij}^{(e)}]$$4 &times; 4$$,\qquad \begin{matrix} k_{11}^{(2)} = k^{(2)} [l^{(2)}]^2 = 5 * [\frac{\sqrt{2}}{2}]^2 = \frac {5}{2}\\ k_{12}^{(2)} = k^{(2)} l^{(2)}m^{(2)} = 5 * \frac{\sqrt{2}}{2}* (-\frac{\sqrt{2}}{2}) = -\frac {5}{2}\\ k_{42}^{(2)} = k^{(2)} (-[m^{(2)}]^2) = 5 * (-[\frac{\sqrt{2}}{2}]^2) = -\frac {5}{2} \end{matrix} $$

Observe: Absolute values for all the coeficients are the same [k$$_{ij}^{(e)}]$$4 &times; 4, where $$ e =\ 2,and\ _{j = 1,2,3,4}^{i = 1,2,3,4}$$. All that differs are the positive/negative signs in front of the magnitudes of the coeficients.

Element F-D relation: $$ \mathbf{k}_{4 \times 4}^{e}*\mathbf{d}_{4\times1}^{e} = \mathbf{f}_{4\times1}^{e} $$

$$ \mathbf{d}^{e}\ ={ \begin{Bmatrix} d_1^{(e)}\\ \vdots\\ d_4^{(e)} \end{Bmatrix}}_{4 \times 1};\quad

\mathbf{f}^{e}\ = {\begin{Bmatrix} f_1^{(e)}\\ \vdots\\ f_4^{(e)} \end{Bmatrix}}_{4 \times 1} $$

Global F-D Relation("free-free" structure)


$$ \mathbf{K}_{n \times n}*\mathbf{d}_{n\times1} = \mathbf{F}_{n\times1} \Longrightarrow $$ there are $$\mathbf{6} $$ degrees of freedom (6 forces are acting upon the sturcture) $$\quad \Rightarrow \mathbf{n=6} $$