User:Eml4500.f08.FEABBQ.Jayma/notes1

Class Notes: September 17, 2008
Certain degrees of freedom in our example problem are already known; the boundary conditions set these displacements to equal zero, while the displacements of global node 2 are to be determined. Setting $$ d_1 = d_2 = d_5 = d_6 =0 $$ is the same as deleting columns 1, 2, 5 and 6 in the stiffness matrix. Shown below are columns 3 and 4 and the simplified displacement vector..

$$\underline{K}\left[ \begin{array}{l} d_1=0 \\ d_2=0 \\ d_3=? \\ d_4=? \\ d_5=0 \\ d_6=0 \end{array} \right]= \left[ \begin{array}{ll} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right] \left[ \begin{array}{ll} d_3 \\ d_4 \end{array} \right] = \underline{F} $$

Since the stiffness matrix reduces down to a 6-by-2 size and the displacement reduces to a 2-by-1 matrix, the resulting force matrix will have dimensions 6-by-1.

Further reduction of this system of matrices can be performed by the Principal of Virtual Work, which allows us to delete the same rows from the stiffness matrix as we did the columns (1, 2, 5 and 6) as shown below.

$$\left[ \begin{array}{ll} K_{33} & K_{34} \\ K_{43} & K_{44} \end{array} \right] \left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{c} F_3=0 \\ F_4=P=7 \end{array} \right] $$

Since the goal now is to determine the displacements $$d_3$$ and $$d_4$$, we can rearrange the equation above to solve for these displacements.

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \underline{\overline{K}}^{-1} \left[ \begin{array}{c} 0 \\ P \end{array} \right] $$

We now have an inverse matrix. As a review...

$$\underline{\overline{K}}^{(-1)} = \dfrac{1}{det(\underline{\overline{K}})} \left[ \begin{array}{cc} K_{44} & -K_{34} \\ -K_{43} & K_{33} \end{array} \right]; det(\underline{\overline{K}}) = K_{33}K_{44}-K_{34}K_{43}$$

With the values $$K_{34} = K_{43} = -2.175$$, $$K_{33} = 3.0625$$ and $$ K_{44} = 2.6875 $$, we have

$$det(\underline{\overline{K}}) = 3.0625(2.6875)-(-2.175)(-2.175) = 3.4998$$

$$\underline{\overline{K}}^{(-1)} = \left[ \begin{array}{cc} 2.6875/3.4998 & 2.175/3.4998 \\ 2.175/3.4998 & 3.0625/3.4998 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] $$

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] \left[ \begin{array}{c} 0 \\ 7 \end{array} \right] = \left[ \begin{array}{cc} 4.35022 \\ 6.12535 \end{array} \right]$$

Since finding the inverse stiffness matrix was the prominent calculation, it is worthwhile to check the solution, especially given such a simple check as shown below.

$$\underline{\overline{K}}(\underline{\overline{K}}^{-1})=\underline{\overline{K}}^{-1}(\underline{\overline{K}})=\underline{I}=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

The $$\underline{I}$$ above is called the Identity Matrix and results when a matrix is multiplied by it's own inverse.


 * A special note about the transpose of a matrix...

$$\underline{\overline{K}}^{T} = \left[ \begin{array}{cc} K_{33} & K_{43} \\ K_{34} & K_{44} \end{array} \right] $$

The transpose of the matrix $$\underline{\overline{K}}$$ is not the same as the matrix appearing in the $$\underline{\overline{K}}^{-1}$$ equation.

In order to compare to Statics, we compute the reaction forces.

5) Compute Reactions
We use two methods:

5.1) Use the element Force-Displacement relationships; $$\underline{k}^{(e)}\underline{d}^{(e)}=\underline{f}^{(e)}$$, where $$e=1,2$$.

Element 1 (e=1)

$$\underline{k}^{(1)}\underline{d}^{(1)}=\underline{f}^{(1)}$$. The contents of the stiffness matrix were previously calculated, and the displacement matrix is complete from our calculations above for the global displacements, while the force matrix is only partially known.

$$\underline{d}^{(1)}=\left[ \begin{array}{c} 0\\0\\4.352\\6.1271 \end{array} \right]$$


 * Image goes here

Element 2 (e=2)

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