User:Eml4500.f08.delta 6.krueger/Notes Monday September 29th

Use global force displacement relation. $$ \begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$ This matrix is a 6x2 multiplied by a 2x1 matrix and equals a 6x1 matrix. The forces $$F_{3}$$ and $$F_{4}$$are the known applied loads. Therefore, there is no need to compute them. The computations that need to be done are for rows 1,2,5 and 6. This will give The reactions $$F_{1}$$, $$F_{2}$$, $$F_{5}$$ and $$F_{6}$$. Closing the loop between The finite element method (FEM) and statics: Virtual Displacement Sloving the two bar truss system uses a process shown in the figure below. Where side 1 is the finite element method and side two is the statics approach. The statics approach is used to confirm the finite element method and is known as closing the loop.  The process of determinig the displacements using statics was discussed above in "Determination of Element FD (wrt global Coordinates). In the statics process the reactions are known. Therefore, the member forces $$P^{(1)}_{1}$$ and $$P^{(2)}_{2}$$ are known. To compute the axial displacement dof's ($$q$$) or the amount of extension of the bars the member forces and the element stiffiness's need to be used. The following formulas show the relationship between the member forces, element stiffness's and the axial displacements. $$ P=Kq$$ so, $$ q=P/K$$ $$q^{(1)}_{2}= \frac{P^{(1)}_{1}}{K^{(1)}}= \frac{P^{(1)}_{2}}{K^{(1)}}= AC$$ $$q^{(2)}_{1}= \frac{-P^{(2)}_{2}}{K^{(2)}}= \frac{-P^{(2)}_{2}}{K^{(2)}}= -AB$$ The axial displacements dof's at node 1 and 3 are equal to zero.  $$q^{(1)}_{2}=q^{(2)}_{2}= 0$$  Question: How does one back out from the displacement dof's of node 2?  Where: A = node 2, AC is the extension of 1A and AB is the extension of 3A. First an extensions must be drawn perpendicular to AC and AB. Then a line can be drawn between the point D, where the two extensions meet, and node 1. A line is also drawn between node 3 and point D. These two lines are the dotted line and denote the deformed shape of the two-bar truss system. The vector R is the displacement of A or node 2. It is found by taking the square root of the sum of the squared magnitudes of AC and AB. $$R= (AC^{2} + AB^{2})^{\frac{1}{2}}$$ Review of the small angle approximation.

Where A is the angle between the two bars. $$U_{y}= Rsin(A)$$ $$U_{x}= R(1-cos(A))$$ If the angle is small then $$sin(A)$$ is approximately equal to A and $$cos(A)$$ is approximately equal to 1. The above equations become: $$U_{y}= R*A$$ $$U_{x}= 0$$