User:EML4500.F08.RAMROD.B/Homework 3

 Referring back to the picture of the truss system, it can be seen that the displacement of point $$\displaystyle D$$ is simply the displacement of the global node $$\displaystyle 2$$ at point $$\displaystyle A$$. In other words,

$$\displaystyle Y_{D} = d_{4} $$ and

$$\displaystyle X_{D} = d_{3} $$

We can now proceed to calculate the length of lines $$\displaystyle AC $$ and $$\displaystyle AB $$ be computed which as we already know are defined to be the axial forces. In other words it is the amount the bars extend or elongate due to the applied force. As previously stated the formula to find line $$\displaystyle AC $$ is as follows,

$$\displaystyle AC = \frac{\left|P_{2}^1 \right|}{k^1}$$

where

$$\displaystyle \left|P_{2}^1 \right| = 5.1243 $$ $$\displaystyle k^1 = \frac{3}{4}$$

Therefore,

$$\displaystyle AC = 6.8324$$

Remember that $$\displaystyle \left|P_{2}^1 \right|$$ is the internal force $$\displaystyle 2$$ at element $$\displaystyle 1 $$. Remember also that in order to obtain the value of $$\displaystyle \left|P_{2}^1 \right|$$, it is necessary to take the Root Sum Squared, RSS for short, of the two forces making acting at that node. The formula for the RSS of line AC extending from element 1 is,

$$\displaystyle \sqrt{(f_3)^2 + (f_4)^2}$$

The same can be done for $$\displaystyle AB $$ with the axial force from element two.

$$\displaystyle AB = \frac{\left|P_{1}^2 \right|}{k^2}$$ $$\displaystyle \left|P_{1}^2 \right| = 6.276 $$

$$\displaystyle k^2 = 5$$

Therefore

$$\displaystyle AB = 1.2552$$

From this point it is now possible to find the coordinates of points $$\displaystyle B $$ and $$\displaystyle C $$. Simple trigonometry will be used to carry out these calculations. First we will find the coordinates for point C, $$\displaystyle (x_C,y_C) $$. If you recall from previous lessons

$$\displaystyle \theta^1 = 30^{\circ} $$

and we already know the length of $$\displaystyle AC $$ so therefore the coordinates of $$\displaystyle C $$ can be expressed as

$$\displaystyle x_C = AC \cos (30^{\circ}) = 6.8324 *\frac{\sqrt{3}}{2} = 5.917$$

$$\displaystyle y_C = AC \sin (30^{\circ}) = 6.8324 *\frac{1}{2} = 3.4162$$

The coordinates of point $$\displaystyle B $$ can be found in the same way. the angle used is the angle the line $$\displaystyle AB $$ makes with the negative x-axis.

$$\displaystyle x_B = AB \cos (45^{\circ}) = 1.2552 *\frac{\sqrt{2}}{2} = -.88756 $$

$$\displaystyle y_B = AB \sin (30^{\circ}) = 1.2552*\frac{\sqrt{2}}{2} = .886756$$

We next need to find the coordinates of the point $$\displaystyle D $$. There are two unknowns at point $$\displaystyle D $$, the x and the y coordinate $$\displaystyle (x_D,y_D) $$. Since there are two unknowns, we need to find two equations in which these two variables are being utilized. The equation for the lines $$\displaystyle AB $$ and $$\displaystyle AC $$ are the two equations we will utliize. In order to do so we will work with the following figure.



The line $$\displaystyle PQ $$ can be represented in the following way;

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

The cosine and the sine of the angle can then be expressed as;

$$\displaystyle \vec{PQ} = (x-x_p) \vec{i} + (y-y_p) \vec{j} $$

Dividing the previous two equations will yield the following;

$$\displaystyle \frac{(y-y_p)}{(x-x_p)} = \tan(\theta) $$

Multiplying through by $$\displaystyle (x-x_p)$$ and the equation of the line $$\displaystyle \vec{PQ} $$ is given by the following;

$$\displaystyle y-y_p= \tan(\theta)(x-x_p) $$

This equation for line $$\displaystyle \vec{PQ} $$ is equivalent to the equation for line $$\displaystyle \vec{AC} $$. The equation for a line that is perpendicular to line $$\displaystyle \vec{AC} $$ and passing throught the point $$\displaystyle A $$ is given by the following;

<p style="text-align:center;">$$\displaystyle y-y_p= \tan(\theta +\frac{\pi}{2})(x-x_p) $$

This equation is equivalent to the equation for line $$\displaystyle \vec{AC} $$. When the coordinates for lines $$\displaystyle \vec{BD} $$ and $$\displaystyle \vec{CD} $$ are put into the above equation then it is possible to solve for the coordinates of point $$\displaystyle D $$.

<p style="text-align:center;">$$\displaystyle y_D-y_B= \tan(\theta +\frac{\pi}{2})(x_D-x_B) $$ <br \> <br \> $$\displaystyle y_D-y_C= \tan(\theta +\frac{\pi}{2})(x_D-x_C)$$ <br \> <br \> $$\displaystyle y_D-.886756= \tan(\frac{\pi}{4})(x_D+.886756) $$ <br \> <br \> $$\displaystyle y_D-3.4162= \tan (\frac{2\pi}{3}) - (x_D-5.917)$$

Solving for $$\displaystyle y_D $$ for each equation will allow you to then solve for $$\displaystyle x_D$$. Plug this value back into one of the equations will allow you to get the value of $$\displaystyle y_D$$. When done, the coordinate of point $$\displaystyle D $$ becomes;

<p style="text-align:center;">$$\displaystyle D = (4.352, 6.1271) $$

Using the Finite Element Method (FEM), we have already found the displacement of point $$\displaystyle D $$. In other words;

<p style="text-align:center;">$$\displaystyle \vec{AD} = (x_D-x_A) \vec{i} + (y_D-y_A) \vec{j} $$

In this equation the variables $$\displaystyle x_A $$ and $$\displaystyle y_A $$ are zero and the only variables that remain are the coordinates of point $$\displaystyle D $$. By definition;

<p style="text-align:center;"> $$\displaystyle x_D = d_3 $$ and $$\displaystyle y_D = d_4 $$

And therefore,

<p style="text-align:center;">$$\displaystyle \vec{AD} = d_3 \vec{i} + d_4 \vec{j} $$

where the coefficients are the global displacements of point $$\displaystyle A $$ to point $$\displaystyle D $$.

=Three Truss Bar System=

We will now look at the process of analyzing a three bar truss system as shown below.

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It is important to note the helpfulness in numbering the local nodes of each element. Doing this in a certain manner make the assembly of the global stiffness matrix more convenient. The following numbering is the best and most convenient system.

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