User:Eml4500.f08.team.figuerrez/hw4

Justification of assembly of element stiffness matrix ( k (e)) into global stiffness matrix ( K )

Recall element FD relation: $$\underset{4 \times 4}{k^{(e)}}\underset{4 \times 1}{d^{(e)}} = \underset{4 \times 1}{f^{(e)}} $$     {e = 1, 2}

For the two bar truss shown above:

$$\sum{F_{x}}=0=-f_{3}^{(1)}-f_{1}^{(2)}=0\;\;$$

$$\sum{F_{y}}=0=P-f_{4}^{(1)}-f_{2}^{(2)}=0\;\;$$

Next, we use the element FD relation: $${k^{(e)}} {d^{(e)}} = {f^{(e)}} $$

When these two equations are rewritten they become:

$$f^{(1)}_{3} + f^{(2)}_{1}= 0$$

$$f^{(1)}_{4} + f^{(2)}_{2}= 0$$

Therefore, the element forces can then be written in terms of K and d.

$$f_{3}^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$f_{1}^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Then, the equation becomes:

$$ \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$