User:Eml4500.f08.jamama.adam/homework3

Meeting 15
The previous simplification that was applied to the Stiffness Matrix was done by application of the boundary conditions as follows:

$$\left[\begin{array}{cccccc} K_{11}&K_{12}&K_{13}&K_{14}&K_{15}&K_{16}\\ K_{21}&K_{22}&K_{23}&K_{24}&K_{25}&K_{26}\\ K_{31}&K_{32}&K_{33}&K_{34}&K_{35}&K_{36}\\ K_{41}&K_{42}&K_{43}&K_{44}&K_{45}&K_{46}\\ K_{51}&K_{52}&K_{53}&K_{54}&K_{55}&K_{56}\\ K_{61}&K_{62}&K_{63}&K_{64}&K_{65}&K_{66}\end{array}\right] \left[\begin{array}{c}0\\0\\d_3\\d_4\\0\\0\end{array}\right]= \left[\begin{array}{cc} K_{13}&K_{14}\\K_{23}&K_{24}\\K_{33}&K_{34}\\K_{43}&K_{44}\\K_{53}&K_{54}\\K_{63}&K_{64} \end{array}\right] \left[\begin{array}{c}d_3\\d_4\end{array}\right]= \left[\begin{array}{c}F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{array}\right]$$

F3 and F4 being the known applied loads, the computation for rows 3 and 4 is no longer needed for it would only return F3 and F4. The only needed computations would be for rows 1, 2, 5, 6 to get F1, F2, F5, and F6

To close the loop between Statics and Finite Element Methods, Virtual Displacement can be used as follows:

Two bar Truss syst:



By statics, the reactions forces are known leading to the member forces P1(1), P2(1), component axial displacement degrees of freedom:

$$\begin{array}{l} q_1^{(1)}=\dfrac{P_1^{(1)}}{k^{(1)}}=\dfrac{P_2^{(1)}}{k^{(1)}}=AC\\ q_1^{(1)}=0,\, fixed\, node\, (1)\\ q_1^{(2)}=-\dfrac{P_2^{(2)}}{k^{(2)}}\\ q_2^{(2)}=0,\, fixed\, node\, (3)\\ \end{array}$$

Such a situation makes us wonder about how it would possible to back out the displacement degrees of freedom of node (2) from the results above?





Uy = R sin(α) ≈ Rα (if α small)

Ux = R(1 - cos(α)) ≈ 0 (first Order)