User:Eml4500.f08.Ateam.paquette/Homework2

 Global Force Displacement Relation 

The Global Force Displacement Relation is known as K d = F. Matrix K is the element stiffness matrix in global coordinates. In our example, the global stiffness matrix is a 6 x 6 grid, the global displacement matrix is 6 x 1, and the global force matrix is 6 x 1, and is shown below:


 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} = \begin{bmatrix}

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{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}$$

Since the stiffness matrix is 6 x 6, there are six degrees of freedom in the problem. The Global Force Displacement Relation can be shown in a more compact notation:
 * $$\begin{bmatrix} K_{ij} \end{bmatrix} \begin{Bmatrix} d_{j} \end{Bmatrix} = \begin{Bmatrix} F_{i} \end{Bmatrix}$$

or    $$\sum_{j=1}^6 $$ Kij dj  = Fi , i=1,2,3,4,5,6

K or $$\begin{bmatrix} K_{ij} \end{bmatrix}$$n x n is the global stiffness matrix. d or $$\begin{Bmatrix} d_{j} \end{Bmatrix}$$n x 1 is the global displacement matrix. F or $$\begin{Bmatrix} F_{i} \end{Bmatrix}$$n x 1 is the global force matrix.

These global matrices are different than the element force displacement relation k(e) d(e) = f(e). In this relationship, recall the k(e) is a 4 x 4 matrix, called the element stiffness matrix where (e) is the element number. d(e) and f(e) are the element displacement and force matrices, both of which are 4 x 1. These element matrices are smaller than the global ones because each one pertains to an element rather than the entire picture. To go from an element matrix to a global matrix, the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom need to be identified.



At the global level, the degrees of freedom are $$d_1,\ d_2,\ d_3,\ d_4,\ d_5,\ d_6$$. At the element level, there is a set of degrees of for each element.

In Element 1, the degrees of freedom are $$d_1^{(1)},\ d_2^{(1)},\ d_3^{(1)},\ d_4^{(1)}$$. Element 2 is $$d_1^{(2)},\ d_2^{(2)},\ d_3^{(2)},\ d_4^{(2)}$$.

The schematic above shows that some of the global degrees of freedom are the same as the element. At node 1:
 * d1 = d1(1)
 * d2 = d2(1)

Node 2 on the global schematic has the same displacement degrees of freedom as on both the first and second element; therefore:
 * d3 = d3(1) = d1(2)


 * d4 = d4(1) = d2(2)

Node 3 is only:
 * d5 = d3(2)
 * d6 = d4(2)

The conceptual step of assembly has a small matrix for each element stiffness matrix and two zero matrices all within the global force displacement relation. The basic matrix is shown below:

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$$