User:Eml4500.f08.team.dwyer.jd/HW3 (9-29)

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

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

Only need to compute rows 1,2,5,6 to get F1, F2, F3, & F4.

Closing the loop between FEM and statics:

Virtual Displacement

 Two-Bar truss system: |center



By statics, reactions known and therefore the member forces; $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute axial displacement degree of freedoms (amount of 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)