User:Eml4500.f08.FEABBQ.Jayma/notes2

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 (reaction) matrix is unknown.

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



Element 2 (e=2)

Similarly, for $$\underline{d}^{(2)}$$, the stiffness matrix is already known and the following equation can be used to solve for the reactions of element 2.

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

$$\underline{d}^{(2)}=\left[ \begin{array}{c} 4.352\\6.1271\\0\\0 \end{array} \right]$$

Class Notes: September 19, 2008
Revisiting element 1 and the known $$\underline{k}^{(1)}$$ and $$\underline{d}^{(1)}$$ matrices, we can solve for the reactions.

$$\left[ \begin{array}{cccc} 0.5625 & 0.32476 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 0\\ 0\\ 4.352\\ 6.1271 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(1)}_{1}\\ \underline{f}^{(1)}_{2}\\ \underline{f}^{(1)}_{3}\\ \underline{f}^{(1)}_{4} \end{array} \right]$$

Matrix multiplication results in the following values for the reactions of element 1...

$$ \left[ \begin{array}{c} -4.4378\\-2.5622\\4.4378\\2.5622 \end{array} \right] $$ where $$\underline{f}^{(1)}_{1}$$ and $$\underline{f}^{(1)}_{2}$$ are the reactions at the constrained node and $$\underline{f}^{(1)}_{3}$$ and $$\underline{f}^{(1)}_{4}$$ are the internal forces.

Similarly, for the stiffness and displacement matrices of element 2

$$\left[ \begin{array}{cccc} 2.5 & -2.5 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 4.352\\ 6.1271\\ 0\\ 0 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(2)}_{1}\\ \underline{f}^{(2)}_{2}\\ \underline{f}^{(2)}_{3}\\ \underline{f}^{(2)}_{4} \end{array} \right] = \left[ \begin{array}{c} -4.4377\\4.4377\\4.4377\\-4.4377 \end{array} \right]$$

Through the sum of the forces we can see that $$\Sigma F_{x} = \underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{3}=-4.4378+4.4378=0$$ and $$\Sigma F_{y} = \underline{f}^{(2)}_{1}+\underline{f}^{(1)}_{4}=-2.5622+2.5622=0$$. We also note that $$ \Sigma M_{any point}=0$$. These three conditions satisfy the criteria for self-equilibrium.

HW for September 19, 2008
For element 1, if we call the reactions at local node 1, $$ -P^{(1)}_{1} $$, that is

$$ -P^{(1)}_{1}=\underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{2}=-4.4378x-2.5622y$$

and if we also call the internal forces at node 2, $$ P^{(1)}_{2} $$ and $$ P^{(1)}_{2}=\underline{f}^{(3)}_{1}+\underline{f}^{(4)}_{2}=4.4378x+2.5622y$$

then $$ P^{(1)}_{2}=-(-P^{(1)}_{1})=P^{(1)}_{1} $$, further demonstrating equilibrium of element 1.

Similarly, for element 2, if we call local node 2, $$P^{(2)}_{2}$$, that is

$$ P^{(2)}_{2}=\underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=4.4377x-4.4377y$$

then $$ P^{(2)}_{1}= -P^{(2)}_{2} = \underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=-4.4377x+4.4377y$$

5.2) The second method for solving the two-bar truss problem involves using Statics. Here we use the Cut Principal by Euler.


 * Image of 2 different Euler Cut methods go here.

To bring the Finite Element Solution into the picture, we verify the equilibrium of global node 2.

HW for September 19, 2008

 * Image of global node 2 with three P forces goes here.