User:Eml4500.f08.team.foskey.ckf/hw4b

 HW Redo: A better figure for Node 2 was added. Further explanation of the math and equations were provided.

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Eml4500.f08.team.foskey.ckf 02:01, 3 November 2008 (UTC)

Justifying the Global Stiffness Matrix
The assembly of the elemental stiffness matrices, k(e) into a global stiffness matrix, K, can be justified by reviewing the two bar truss problem.

If we look back at the Two bar truss problem, and recall the element force-displacement relation $$ k^{(e)}d^{(e)} = f^{(e)}$$, we can use the Euler Cut Principle to give the equilibrium of the node relationship.

Then, if the free body diagrams of elements 1 and 2 are considered, and the global degrees of freedom for global node 2 are compared to the degrees of freedom for elements 1 and 2, the equilibrium of the node can be confirmed.

Looking at the equilibrium of node 2 for the two bar truss pictured above, the addition of k values is because the forces in the x and y direction sum up to equal zero. Using a statics approach and calculating the forces in each direction gives:

$$ \Sigma F_x = 0= -f^{(1)}_3 - f^{(2)}_1=0$$     (1) $$ \Sigma F_y = 0= P-f^{(1)}_4 - f^{(2)}_2=0$$     (2)

When these two equations are rewritten they become:

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

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

The next step is to involve the force displacement relationship ( $${k^{(e)}} {d^{(e)}} = {f^{(e)}} $$ ). Replacing f with k*d will get the middle of the 6 by 6 global stiffness matrix.

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 for the sum of the forces in the x-direction becomes:

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

The sum of the forces in the y-direction can be found in a similar manner.

Thus, statics and equilibrium evaluation of global node 2 has verified the middle of the global stiffness matrix.