User:Eml4500.f08.qwiki.bishop/Lecture 22

=Meeting 22, Wednesday October 15, 2008=

Formal Justification of Assembly Process of Element Stiffness Matrix into Global Stiffness Matrix
For a formal justification, we will consider the two-bar truss discussed in meeting 9. We must also recall that the Force-Displacement relationship for each element is $$\mathbf{k}^{(e)}_{(4x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$. This relationship allows us to compute the nodal displacements and forces, unlike statics which only considers element displacements and forces. Therefore, if we follow the Finite Element Method we can isolate and analyze each node and element in the system according to the Euler Cut Principle.

In terms of the DOFs and their naming for this justification, we refer to the Free Body Diagrams of Element 1 and 2 from Meeting 4. Notice that global node 2 DOFs correspond to node 2 of element 1 and node 1 of element 2.


 * Global Node 1:
 * $$d_1=d_1^{(1)}$$
 * $$d_2=d_2^{(1)}$$


 * Global Node 2:
 * $$d_3=d_3^{(1)}=d_1^{(2)}$$
 * $$d_4=d_4^{(1)}=d_2^{(2)}$$


 * Global Node 3:
 * $$d_5=d_3^{(2)}$$
 * $$d_6=d_4^{(2)}$$

Since global node 2 is the intersecting node of element 1 and 2, we can examine the equilibrium of global node 2



The next step of the justification process is to invoke the element FD relationship $$\mathbf{k}_{(4x4)}^{(e)}\mathbf{d}_{(4x1)}^{(e)} = \mathbf{f}_{(4x1)}^{(e)}$$ for each element. We will replace each internal element force in equations 1 and 2 shown above with its respective internal displacements, where $$f_i^{(e)} = \sum_{j=1}^{4}{k_{ij}^{(e)}d_{j}^{(e)}}$$
 * $$\sum{F}_x = -f_3^{(1)} - f_1^{(2)} = 0$$ Eqn. 1
 * $$\sum{F}_y = \mathbf{P} - f_1^{(4)} - f_2^{(2)} = 0$$   Eqn. 2