User:Eml4500.f08.ateam.mcnally/Week 4

Step 4 Cont. The Theory behind the Elimination of known degrees of freedom (dofs)
Meeting 10 on Wednesday September the 17th of 2008 was a continuation of elimination of the known dofs to reduce the Force Displacement relationship. By applying the fixed boundary conditions it becomes apparent that d1=d2=d5=d6=0. With elements d1=d2=d5=d6 of the global displacement matrix d =0 computation time can reduced by realizing that columns 1,2,5, and 6 can be removed from the global displacement matrix K. See the example below for further explanation.

\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}

\begin{bmatrix} d_{3}\\ d_{4}\\ \end{bmatrix} =

\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}

$$

From the principle of virtual work the corresponding rows 1, 2, 5, and 6 of the global stiffness and global force matrix K and F may also be deleted. This allows the force displacement relation to reduced further. The resulting force displacement relation is shown below.

\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \\ \end{bmatrix}

\begin{bmatrix} d_{3}\\ d_{4}\\ \end{bmatrix} =

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

F3 =0 and F4=P by inspection of Global Force pictures shown below.

Global displacements d₃ and d₄ can then solved for by inverting K and multiplying by F.

\begin{bmatrix} d_{3}=4.352\\ d_{4}=6.127\\ \end{bmatrix} =K^{-1}

\begin{bmatrix} F_{3}\\ F_{4}\\ \end{bmatrix} $$

Displacements d3 and d4 correspond to X and Y displacement of Node 2 in the global picture. See the following example for a review on inverting matrices.



K^{-1}=1/det(K) \begin{bmatrix} K_{44} & K_{-34} \\ K_{-43} & K_{33} \\ \end{bmatrix} $$ With

det(K)=(K_{33}*K_{44})-(K_{34}*K_{43}) $$

Please see the MATLAB solution at the end of this test for the full solution.