User:EML4500.f08.Ateam.Miller/homework3

Closing the Loop between FEM and Statics - continued
We continue to back out the displacement DOF's of global node 2. From the previous figure, we can calculate AB and AC.

$$ AB=\frac{|P_1^{(2)}|}{K^{(2)}}=\frac{6.276}{5}=1.2552 $$

$$ AC=\frac{|P_2^{(1)}|}{K^{(1)}}=\frac{5.1243}{0.75}=6.8324 $$

Using these results and the following figure, the location, or (x,y) coordinates of each point can be calculated with trigonometry.



$$ \overrightarrow{PQ}=(\underline{P}Q)\hat{\tilde{i}}=(PQ)[cos\theta\hat{i}+sin\theta\hat{j}] =(x-x_{\underline{p}})\hat{i}+(y-y_{\underline{p}})\hat{j} $$

$$\Rightarrow x-x_{\underline{p}}=(\underline{P}Q)cos\theta $$

$$ y-y_{\underline{p}}=(\underline{P}Q)sin\theta $$

$$\Rightarrow y-y_{\underline{p}}$$

$$\Rightarrow \frac{y-y_{\underline{p}}}{x-x_{\underline{p}}}=tan\theta$$

$$y-y_{\underline{p}}=(tan\theta)(x-x_{\underline{p}})$$

Therefore, the equation for the line perpendicular to the above line passing through P is:

$$y-y_{\underline{p}}=(tan(\theta+\frac{\pi }{2}))(x-x_{\underline{p}})$$

Now, using θ(1}=30° and θ(2}=135° (xB,yB)=(-0.88756,0.88756) and (xC,yC)=(5.91703,3.4162). Finally, it is important to find (xD,yD) to verify that it is equal to displacements calculated with the FEM method. We can calculate (xD,yD) with the equation below.

$$\overrightarrow{AD}=(x_{D}-x_{A})\hat{i}+(y_{D}-y_{A})\hat{j}$$

By choosing the origin to be at point A, xA=yA=0 and the above equation reduces to

$$\overrightarrow{AD}=(x_D)\hat{i}+(y_D)\hat{j}$$

and (xD,yD)=(4.135,6.127).Finally, by the definition of AD,

$$\overrightarrow{AD}=d_3\hat{i}+d_4\hat{j}=4.135\hat{i}+6.127\hat{j}$$

from the FEM method and the loop is closed.

3-Bar Truss System
We now extend the FEM method to a three-bar truss system as shown below.



The properties of the truss system are listed below.

E(1) = 2 ;  A(1) = 3  ;  L(1) = 5

E(2) = 4 ;  A(2) = 1  ;  L(2) = 5

E(3) = 3 ;  A(3) = 2  ;  L(3) = 10

P =30

In order simplify the assembly of the K matrix, a convenient local numbering system must be assigned. A convenient local node numbering system is shown in the figure below.