User:Eml4500.f08.ateam.mcnally/Week 5

Meeting 16 on Wednesday October 1st 2008 was a continuation of the “closing the loop” infinitesimal displacement lecture given during meeting 15 on September 29 2008. The image below is the infinitesimal displacement of the two bar truss system outlined in previous lectures. . The infinitesimal displacements are very closely related to virtual displacement from the principal of virtual work.

It is important to note in the image the coordinate system (in red) is taken to be at point A which correlates to the Global node 2 of the two bar truss system. The black vector ending at point D is the virtual displacement vector.

The length of the displacements AC and AB of the deformed shape are then solved for using the following equations.

$$

AC = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$

$$ AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552

$$

The next step is to solve for the (x,y) coordinates of point B and Point C using the following equations. $$ X_{B} = ABcos(135) = 1.2552cos(135) = -.88756 $$

$$ Y_{B} = ABsin(135) = 1.2552sin(135) = .88756 $$

$$ X_{C} = ACcos(30) = 6.8324cos(30) = 5.91703 $$

$$ Y_{C}=ACsin(30) = 6.8324sin(30) = 3.4162 $$

The final two unknowns are XD and YD which can be solved for with the following image and equations.

The two unknowns are (XD, YD) so we need an equation for the line AB and the line BC

Note:Θ=a

$$

PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$

$$

(PQ)i^{~} = PQ[ cos(a)i + sin(a)j ] $$

Therefore

$$ X-X_{P} = PQcos(a) $$

$$ Y-Y_{P} = PQsin(a) $$

$$ \frac {Y-Y_{P}}{X-X_{P}} = tan(a) $$

$$ Y-Y{P} = (tan(a))(X-X_{P}) $$

and the equation for a line perpendicular to the above line passing through P

$$ Y-Y_{P} = tan(a + pi/2)(X-X{P}) $$

And by definition and from FEM in which we solved for d3 and d4

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

Therefore $$ X_{D} = 4.35 $$ and $$ Y_{D} = 6.125 $$

The lecture concluded with the start of the Finite Element Method ‘’’FEM’’’ for a three bar truss. The following data corresponds the Global Picture of the three bar truss.

$$ P=30 $$

$$ \begin{matrix} 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 \end{matrix}

$$

The corresponding element picture is a follows.

--Eml4500.f08.ateam.mcnally 17:33, 7 October 2008 (UTC)