User:Eml4500.f08.gravy.bje/Notes2

Lecture 9 Notes
$$\ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

$$\ \ = \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3428 & 0 & 0 \\  0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & (0.5625 + 2.5) & (0.3248 + 2.5) & -2.5 & -2.5 \\ -0.3248 & -0.1875 & (0.3248 + 2.5) & (0.1875 + 2.5) & -2.5 & -2.5 \\  0 & 0 & -2.5 & -2.5 & 2.5 & 2.5 \\ 0 & 0 & -2.5 & -2.5 & 2.5 & 2.5  \end{bmatrix}_{6X6}\,$$

$$\ \begin{bmatrix} -0.5 \\ 0 \\ 1 \\ 0 \end{bmatrix}$$ $$\ \begin{bmatrix} 0.333 \\ -1.333 \\ 0 \\ 1 \end{bmatrix}$$ For λ = 13

$$\ \begin{bmatrix} 0.286 \\ 1 \\ 1 \\ 0 \end{bmatrix}$$ $$\ \begin{bmatrix} 0 \\ -1 \\ 0 \\ 1 \end{bmatrix}$$ For λ = -12

$$\ \begin{bmatrix} 5.6 \\ 5.6 \\ 7.2 \\ 1 \end{bmatrix}$$ For λ = 24

$$\ \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix}$$ $$\ \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ For λ = -4

$$\ \begin{bmatrix} 1 \\ 1 \\ -2 \\ 1 \end{bmatrix}$$ For λ = 1

x(t) = c1 $$\ \begin{bmatrix} 7 \\ -2 \\ 3 \end{bmatrix}$$ e4t + c2 $$\ \begin{bmatrix} -3 \\ 1 \\ -1 \end{bmatrix}$$ e3t + c3 $$\ \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$$ e2t

EX:

$$\ K_{11} = K_{11}^{(1)} \,$$

$$\ K_{33} = K_{33}^{(1)} + K_{11}^{(2)} \,$$

$$\ K_{43} = K_{44} = K_{34}^{(1)} + K_{12}^{(2)} = K_{43}^{(1)} + K_{21}^{(2)} \,$$

$$\ K_{11}^{(1)} = \frac{9}{16} \,$$

$$\ K_{12}^{(1)} = \frac{3\sqrt{3}}{16}$$

$$ K_{33} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$

$$ K_{43} = \frac{3\sqrt{3}}{16} + \frac{-5}{2} = 2.1752 $$

$$ K_{44} = K_{44}^{(1)} + K_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$

4.) Elimination of known dof's


 * Reduce global FD relations (p. 5-3)


 * Elimination of column 1, 2, 5, 6 because d's are zero

Lecture 11 Notes
$$ Elem 1: \underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)} $$

$$\ d^{(1)} = \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} 0.5625 & 0.32476 & ... & ... \\ & & & \\ & & & \\ & & & \end{bmatrix}_{4X4} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -0.5625 & -0.32476 \\ ... & ... \\ & \\ & \end{bmatrix}_{4X2} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{bmatrix}_{4X1}\,$$


 * f1(1) and f2(1) = Reactions


 * f3(1) and f4(1) = Internal Forces


 * Observation: Element 1 is in equilibrium


 * ΣFx = f1(1) + f3(1)


 * ΣFy = f2(1) and f4(1)




 * ΣMat any point = 0


 * $$P_{1}^{(1)} = [(f_{1}^{(1)})^2 + (f_{2}^{(1)})^2]^ \frac{1}{2}$$

$$ Elem 2: \underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)} $$


 * The same method to solve for element one can also be used to solve for element 2