User:Eml4500.f08.qwiki.bishop/Lecture 16

=Lecture 16 - Wednesday, October 1, 2008=

Closing the Loop: continued

 * The infinitesimal displacement of point global node 2 (point A to point D) is related to the concept of virtual displacement from statics.
 * The figure below shows the displacement of global node 1:
 * The displacement in the y direction, yD, is equivalent to the displacement, d4
 * The displacement in the x direction, xD, is equivalent to the displacement, d3
 * The vector summation of these two displacement provide the displacement vector $$\vec{AD}$$
 * To determine the axial forces, P, we must first find the axial displacements, q
 * $$q^{(1)}_{2}=l*d^{(1)}_{3}+m*d^{(1)}_{4}=cos(30)*4.352+sin(30)*6.1871=6.8625$$
 * $$q^{(2)}_{1}=l*d^{(2)}_{1}+m*d^{(2)}_{2}=cos(-45)*4.352+sin(-45)*6.1871=-1.2976$$
 * Then the axial forces
 * $$P^{(1)}_{2}=k^{(1)}*q^{(1)}_{2}=(0.75)*(6.8625)=5.145$$
 * $$P^{(2)}_{1}=k^{(2)}*q^{(2)}_{1}=(5)*(-1.2976)=-6.488$$


 * The length of line AC can be found by:
 * $$AC=\frac{\left|P^{(1)}_2 \right|}{k^{(1)}}=\frac{5.145}{.75}=6.8625$$
 * The length of line AB can be found by:
 * $$AB=\frac{\left|P^{(2)}_1 \right|}{k^{(2)}}=\frac{-6.488}{5}=-1.2976$$
 * Note 1: The negative value the length of AB is due to the assigned local axes of element 2.
 * Note 2: The lengths of AB and AC match the displacements for Elements 2 and 1 respectively.
 * So the x and y coordinates for points B and C are (if point A is the origin):
 * $$\left(x_{B},y_{B} \right)=(-.92,.92)$$
 * $$\left(x_{C},y_{C} \right)=(5.94,3.43)$$


 * So, we still have the 2 unknowns (xD, yD)
 * The figure below shows the xy axes and also the $$\tilde{x}\tilde{y}$$ axes
 * Let's analyze the vector PQ
 * The equation derived from this figure is: $$y-y_{P}=tan\theta (x-x_{P})$$
 * And the equation for the line perpendicular to this line is: $$y-y_{P}=tan(\theta +\frac{\Pi }{2}) (x-x_{P})$$
 * So, by applying some trigonometry and algebra $$(x_D, y_D)=(4.352, 6.187)$$
 * And the vector AD by definition is:
 * $$\vec{AD}=d_{3}\vec{i}+d_{4}\vec{j}$$

3-Bar Truss System

 * The force, P, has a value of 50
 * Make sure the local node numbering is selected so that it is convenient to assemble the stiffness matrix, K
 * The image below shows each Element and its local nodes as well as the global nodes