User:Eml4500.f08.team.allen.oca/hw3

Solving the two bar truss by Infinitesimal Displacement
In order to complete the analysis of the two bar truss problem, the final configuration of the truss after the load has been applied must be found. To accomplish this, analysis must be done on the individual elements.

The length of segment AC in the figure can be found by treating the element as a spring, where the elongation length is equal to the force applied divided by the spring constant. In this manner, AC can be found by: $$ AC= \frac{P^{(1)}_2}{k^{(1)}}$$ $$ AC=\frac{5.1243}{\frac{3}{4}}=6.8324$$

The length of segment AB, as well as the coordinates of points B and C are solved as homework problems.HW from Oct 1

With the locations of points B and C known, the location of point D can be found by using the equations for the lines AB and BC.

To find the equation for a line containing a generic point P, we need to know the slope of the line, and the coordinates of P. If the angle of the line is known, the slope of the line is given as the "rise over run", or the y/x. Since the angle is $$\theta$$, the y portion of the line is $$sin\theta$$ and the x portion is $$cos\theta$$. y/x gives us $$ \frac{sin\theta}{cos\theta} = tan\theta$$. Using the point-slope formula, we arrive at $$ y-y_p = tan\theta(x-x_p)$$ where x_p and y_p are the x and y coordinates of point P. Since the lines BD and CD are perpendicular to AB and AC respectively, the slopes of BD and CD are the negative inverses of the lines they are perpendicular to. Thus the equations for lines BD and CD become: BD: $$ y-y_B = (tan\theta^{(2)})^{-1}(x-x_B)$$ CD:$$ y-y_C = (tan\theta^{(1)})^{-1}(x-x_C)$$

The location of point D is found in the homework problem. HW from Oct 1

The three bar truss
In the three bar truss diagram, the truss has the following parameters: $$     E^{(1)} = 2 $$ $$      A^{(1)} = 3 $$ $$      L^{(1)} = 5 $$ $$      E^{(2)} = 4 $$ $$      A^{(2)} = 1 $$ $$      L^{(2)} = L^{(1)} = 5 $$ $$      E^{(3)} = 3 $$ $$      A^{(3)} = 2 $$ $$      L^{(3)} = 10 $$ $$      P = 30 $$  $$\theta^{(1)} = 30 $$ $$   \theta^{(2)} = -30 $$ $$   \theta^{(3)} = 45 $$

In order to make solving the three bar truss system the most convenient, the local nodes are labeled as follows: