User:Eml4500.f08.delta 6.guzman/Lecture Notes 09/12

Relationships between Global Diagrams and Element Diagrams
It is necessary to understand the relationships between the global free diagrams and the element free diagrams in order to solve the truss system of matrices going from the element matrices to the global matrices.

Global Free Diagram Relationship
The global free diagram relationship is given by the equation: $$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$ where the d and F  elements are composed by 6 x 6 matrices as following:

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$ where i goes from 1 through 6 since there are six equations
 * Global Stiffness Matrix:

$$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$
 * Global Displacement Matrix:

$$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$
 * Global Force Matrix:

$$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$

Element Free Diagram Relationship
The element free diagrams relationship is given by the equation: $$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ where e indicates the element number. The d and f elements are composed by 4 x 1 matrices as following:

$$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ $$d_{4x1}^{(e)}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$ $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ $$f_{4x1}^{(e)}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ $$k_{4x4}^{(e)}=\mathbf \begin{bmatrix} K_{11}^{(e)} & K_{12}^{(e)} & K_{13}^{(e)} & K_{14}^{(e)}\\K_{21}^{(e)} & K_{22}^{(e)} & K_{23}^{(e)} & K_{24}^{(e)}\\K_{31}^{(e)} & K_{32}^{(e)} & K_{33}^{(e)} & K_{34}^{(e)}\\K_{41}^{(e)} & K_{42}^{(e)} & K_{43}^{(e)} & K_{44}^{(e)}\end{bmatrix}$$
 * Element Displacement Matrix:
 * Element Force Matrix:
 * Element Stiffness Matrix:

Since the matrix k is symmetric in general the upper triangular part of the matrix should transpose to the lower triangular part, i.e. $$k_{ij}^{(e)} = k_{ji}^{(e)}$$

Assembly Process
From the free body general relationships it is possible to generate new associations in order to solve the stiffness, displacement and forces matrices:


 * Global level: $$\mathbf\begin{Bmatrix} d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & d_{6}\end{Bmatrix}$$


 * Element level:
 * Element 1: $$\mathbf\begin{Bmatrix} d_{1}^{(1)} & d_{2}^{(1)} & d_{3}^{(1)} & d_{4}^{(1)} & d_{5}^{(1)} & d_{6}^{(1)}\end{Bmatrix}$$
 * Element 2: $$\mathbf\begin{Bmatrix} d_{1}^{(2)} & d_{2}^{(2)} & d_{3}^{(2)} & d_{4}^{(2)} & d_{5}^{(2)} & d_{6}^{(2)}\end{Bmatrix}$$


 * Nodes:
 * Node 1: $$d_{2} = d_{1}^{(1)}$$
 * Node 2: $$d_{3} = d_{2}^{(1)}$$
 * $$d_{3} = d_{3}^{(1)} = d_{1}^{(2)}$$
 * $$d_{4} = d_{4}^{(1)} = d_{2}^{(2)}$$

Note: These points are from the same node therefore must move in the same direction =>> (same global number = same direction)
 * Node 3: $$d_{5} = d_{3}^{(2)}$$
 * $$d_{6} = d_{4}^{(2)}$$