User:Eml4500.f08.RAMROD.A/MyContributionHW4

A Closer Look at the Force Displacement Relationship
Equation One:   $$\displaystyle F_3^{(1)}+F_1^{(2)}=0 $$

Equation Two:   $$ F_4^{(1)}+F_2^{(2)}=P $$

From the Force Displacement relationship: k(1)*d(1) = F(1), We can break the forces above into their k*d form:

$$ f_3^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} $$

$$ f_1^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} $$

Equation One Becomes:

$$ \begin{bmatrix} k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)} \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)}+k_{13}^{(2)}d_3^{(2)}+k_{14}^{(2)}d_4^{(2)} \end{bmatrix} = 0 $$

And when we convert to Global dof's:

$$ \begin{bmatrix} k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4 \end{bmatrix} +  \begin{bmatrix} k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6 \end{bmatrix} = 0 $$

We can take the k coefficients of d3 and d4 and reduce the Matrices

This Proves our Global Stiffness K, from our earlier learning's of the Finite Element Method:

$$ K = \begin{bmatrix} k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} \\ k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} \end{bmatrix} $$

Another way to arrive at Global K
K= Ae * ke, where K is an n x n Matrix.

n is the total # of global dof's before eliminating the boundary conditions.

ned is the # of element dof's

Then:      Kn x n = A(e)undefined * k(e)ned x ned

A is the Assembly Operator

Virtual Work
The General Equation $$ P * \upsilon * \omega  $$

Define q:

q2x1 = T(e) * d(e)

Define k

k(e) = T(e) T * k(e) * T(e)

Deriving Finite Element Method PDE's
Equation (3) is Implied: Equation (3): $$ k * d - F = 0 $$

Equation (4) is derived and is not tirvial:

Equation (4): $$ w(k * d -F) = 0 $$, for all w