User:Eml4500.f08.qwiki.bishop/Lecture 9

=Lecture 9 - Monday, September 15, 2008= $$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\\ \end{bmatrix}\begin{pmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{pmatrix}=\begin{pmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6\\ \end{pmatrix}$$
 * Global Force/Displacement Relationship
 * (Global stiffness matrix, K) X (global displacement, d) = (global force, F)
 * $$\mathbf{K_{(6x6)}d_{(6x1)} = F_{(6x1)}}$$
 * The global stiffness matrix, K, is found by combining the element stiffness matrices $$k^{(1)}$$ & $$k^{(2)}$$ as seen below.
 * NOTE: There is an overlap for displacements $$d_1$$ & $$d_1$$, which results from Element 1 and Element 2 sharing a node; and their k values are then summed together.
 * The values for the global stiffness matrix are as follows:
 * $$K_{11} = k_{11}^{(1)} = \tfrac{9}{16} = 0.5625$$
 * $$K_{12} = k_{12}^{(1)} = \tfrac{3\sqrt{3}}{16} = 0.3248$$
 * $$K_{13} = k_{13}^{(1)} = -\tfrac{9}{16} = -0.5625$$
 * $$K_{14} = k_{14}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{15} = 0$$
 * $$K_{16} = 0$$
 * $$K_{21} = k_{21}^{(1)} = \tfrac{3\sqrt{3}}{16} = 0.3248$$
 * $$K_{22} = k_{22}^{(1)} = \tfrac{3}{16} = 0.1875$$
 * $$K_{23} = k_{23}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{24} = k_{24}^{(1)} = -\tfrac{3}{16} = 0.1875$$
 * $$K_{25} = 0$$
 * $$K_{26} = 0$$
 * $$K_{31} = k_{31}^{(1)} = -\tfrac{9}{16} = -0.5625$$
 * $$K_{32} = k_{32}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{33} = k_{33}^{(1)}+k_{11}^{(2)} = \tfrac{9}{16}+\tfrac{5}{2} = 3.0625$$
 * $$K_{34} = k_{34}^{(1)}+k_{12}^{(2)} = \tfrac{3\sqrt{3}}{16}-\tfrac{5}{2} = -2.1752$$
 * $$K_{35} = k_{13}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{36} = k_{14}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{41} = k_{41}^{(1)} = -\tfrac{3\sqrt{3}}{16} = -0.3248$$
 * $$K_{42} = k_{42}^{(1)} = -\tfrac{3}{16} = 0.1875$$
 * $$K_{43} = k_{43}^{(1)}+k_{21}^{(2)} = \tfrac{3\sqrt{3}}{16}-\tfrac{5}{2} = -2.1752$$
 * $$K_{44} = k_{44}^{(1)}+k_{22}^{(2)} = \tfrac{3}{16}+\tfrac{5}{2} = 2.6875$$
 * $$K_{45} = k_{23}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{46} = k_{24}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{51} = 0$$
 * $$K_{52} = 0$$
 * $$K_{53} = k_{31}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{54} = k_{32}^{(2)} = \tfrac{5}{2} = -2.5$$
 * $$K_{55} = k_{33}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{56} = k_{34}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{61} = 0$$
 * $$K_{62} = 0$$
 * $$K_{63} = k_{41}^{(2)} = \tfrac{5}{2} = 2.5$$
 * $$K_{64} = k_{42}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{65} = k_{43}^{(2)} = -\tfrac{5}{2} = -2.5$$
 * $$K_{66} = k_{44}^{(2)} = \tfrac{5}{2} = 2.5$$
 * Step 4: Eliminate known degrees of freedom
 * $$d_1 = d_2 = d_5 = d_6 = 0$$
 * This reduces the global stiffness matrix from a (6x6) matrix to a (2x2) matrix