User:Eml4500.f08.FEABBQ.koby/HW3

Sept 29

Class Notes: September 29, 2008
Using the Global Force-Displacement Relation

Use the global force-displacement relationship matrix from lecture 10:

$$ \left[ \begin{array}{cc} k_{13}^{(1)} & k_{14}^{(1)} \\ k_{23}^{(1)} & k_{24}^{(1)} \\ (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) \\  (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)})  \\  k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)}  \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2  \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right]$$ (6x2) x (2x1)  =  (6x1)

Only do computations for rows 1,2,5,6 to get F1,F2,F5,F6. ....

Closing the loop between the Finite Element Method and statics: Virtual Displacement

For the two bar truss system:



by statics, reactions are known and therefor the member forces $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute the axial displacement of the degrees of freedom (extension of bars):

$$q_2^{(1)} = \frac{P_1^{(1)}}{k^{(1)}} = \frac{P_2^{(1)}}{k^{(1)}}$$

$$ q_1^{(1)} = 0$$, fixed global node 1

$$q_1^{(2)} = \frac{-P_2^{(2)}}{k^{(2)}} = -AB$$

$$q_2^{(2)} = 0 $$, fixed global node 3

Question: How to get the displacement DOF's from the above results for node 2?