User:Eml4500.f08.team.allen.oca/hw2-2

P. 10-1 Previously, the global matrices were found to be

$$\mathbf{K}_{6x6} \; \begin{Bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0\\ \end{Bmatrix}_{6x6} = \begin{bmatrix} K_{13} &K_{14} \\ K_{23} &K_{24} \\ K_{33} &K_{34} \\ K_{43} &K_{44} \\ K_{53} &K_{54} \\ K_{63} &K_{64} \end{bmatrix}_{6x6}

\begin{Bmatrix} d_{3}\\ d_{4}\\ \end{Bmatrix}_{2x1} =\mathbf{F}_{6x1} $$ More rows in the K matrix can be removed by using the principle of virtual work. Also, incorporating the fixed boundary conditions leds to the deletion of the corresponding columns in the global stiffness matrix K. By principle of virtual work (PVW) the corresponding rows are also deleted, here rows 1, 2, 5, and 6. The corresponding rows in F are deleted as well. P. 10-2 The resulting Force Displacement relationship becomes $$ \begin{bmatrix} K_{3x3} &K_{3x4} \\ K_{4x3} &K_{4x4} \end{bmatrix}_{2x2} \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}_{2x1} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}_{2x1} $$ which can alternately be written $$ \mathbf{\bar{K}}_{2x2} \mathbf{\bar{d}}_{2x1} = \mathbf{\bar{F}}_{2x1} $$ $$ \mathbf{\bar{F}}= \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ P \end{Bmatrix} $$ where F3 is known to be 0 and P is 7. Recall that for $$\mathbf{\bar{K}}_{2x2}$$ above, the determinate is $$det\; \mathbf{\bar{K}}=k_{33}k_{44}-k_{34}k_{43}$$ and also that the inverse of $$\mathbf{\bar{K}}_{2x2}$$ is $$ \mathbf{k^{-1}}=\frac{1}{det \; \mathbf{k}} \begin{bmatrix} k_{44} &-k_{34} \\ -k_{43} & k_{33} \end{bmatrix} $$ P. 10-3

Verifying the above two properties $$ \mathbf{\bar{K}}\; \mathbf{\bar{K}}^{-1} =\mathbf{\bar{K}}^{-1}\; \mathbf{\bar{K}} = \mathbf{I}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \frac{1}{det \boldsymbol{\bar{k}}} \begin{bmatrix} det \boldsymbol{\bar{k}} &0 \\ 0 & det \boldsymbol{\bar{k}} \end{bmatrix} $$

For finite element method, $$ k_{43} = k_{34}$$? Note: $$ \mathbf{\bar{k}}^{T}=\begin{bmatrix} k_{33} &k_{43} \\ k_{34} & k_{44} \end{bmatrix} $$ $$ \mathbf{\bar{k}}^{-1}\neq \frac{1}{det\mathbf{\bar{k}}}\mathbf{\bar{k}}^{T} $$ Row echelon method

Back to 2D bar truss system: $$ \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \mathbf{\bar{k}}^{-1} \begin{Bmatrix} 0\\ \mathbf{P} \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \mathbf{\bar{k}}^{-1} \begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix} $$ which is the displacement of node 2.