User:Eml4500.f08.bike.viale/homework report 3

=October 1=

"Closing the Loop" continued
The method of infinitesimal displacement, which is related to virtual displacement, is used to analyze the two bar truss system.

Some sample calculations for the lengths of AC and AB were performed as follows:

$$AC = \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{6.276}{5} = 1.255$$

The (x,y) coordinates $$(x_{B}, y_{B})$$ and $$(x_{C}, y_{C})$$ must then be found. This can be done using simple trigonometry and the values of the magnitude of the line and the values of theta obtained in HW 2. This method is further explained using the figure on the left, along with the intermediate calculation steps shown below.

2 unknowns $$(x_{D}, y_{D})$$

The equations for lines AB and BC are then needed.

$$ \vec{PQ} = (PQ)\vec{\tilde{i}} = (PQ)[cos\Theta \vec{i} + sin\theta \vec{j}]$$

$$= (x - x_{p})\vec{i} + (y - y_{p})\vec{j}$$

$$\Rightarrow x - x_{p} = (PQ)cos\Theta$$

and $$y - y_{p} = (PQ)sin\Theta$$

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

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

Using this equation along with $$\Theta^{1} = 30^{o}$$, it can be found that $$(x_{C}, y_{C})$$ = $$(5.92, 3.416)$$

Also, the equation for a line perpendicular to the above line, passing P is as follows:

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

This is useful in finding the value of $$(x_{B}, y_{B})$$, and using the value of $$\Theta^{2} = 135^{o}$$ it can be found that $$(x_{B}, y_{B})$$ = $$(-.877, .877)$$

To find the value of $$(x_{D}, y_{D})$$ one can use the following equation:

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

Where the value of $$(x_{A}, y_{A})$$ are zero due to choosing the axis' origin to be point A.

By definition:

$$\vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

This means that, using the values obtained in HW 2, the line AD can be expressed as:

$$\vec{AD} = 4.35\vec{i} + 6.127\vec{j}$$

Which is the displacement vector of A from FEM. This ends closing the loop.

3-Bar Truss System
The three bar truss system shown above will now be analyzed using the values below as parameters.

$$ E^{(1)} = 2$$ $$ E^{(2)} = 4$$ $$ E^{(3)} = 3$$

$$ A^{(1)} = 3$$ $$ A^{(2)} = 1$$ $$ A^{(3)} = 2$$

$$ L^{(1)} = 5$$ $$ L^{(2)} = 5$$ $$ L^{(3)} = 10$$

$$P = 30$$ $$\Theta ^{(1)} = 30^{o}$$ $$\Theta ^{(2)} = 30^{o}$$ $$\Theta ^{(3)} = 45^{o}$$

Care should be taken when numbering the local nodes in order to make the assembly of the $$K$$ matrix convenient.