User:Eml4500.f08.delta 6.castrillon/HW 5

=Justification=

We can justify the elimination of rows 1,2,5,6 to obtain $$\bar{\mathbf{K}}$$ in a 2-bar truss.


 * We begin with the FD relationship, writing it as $$K$$$$d$$ = $$F$$. Based on the model 2-bar system, we can generalize this method for any kind of application.


 * $$K_{6x6}$$$$d_{6x1}$$ = $$F_{6x1}$$

Therefore,
 * $$K$$$$d$$-$$F$$=$$0_{6x1}$$


 * Principle of Virtual Work:


 * $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

Where w is the weighting matrix, and the equation is equalized to zero as a sacalar (or 1x1 matrix). In other words, the weighting matix is dotted with the FD relationship matrix to obtain a scalar. Also, according to the wise Prof. Vu-Quoc, "$$w$$ was not selected out of frivolity," as it is called the weighting coefficient or matrix.We say that the FD relationship is equivalent to the PVW.

=Proof of FD relationship and PVW equality=


 * 1) FD relationship to PVW is trivial because if FD relationship is true, then $$w$$(0)=0, this being obvious
 * 2) We now want to show how to go from PVW to FD relationship.


 * We have the PVW: $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$


 * Choice 1: Select $$w$$ such that $$w_{1}$$=1, $$w_{2}$$=...=$$w_{6}$$=0


 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 1•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+...+0•$$\left[{\sum_{j=1}^{6}K_{6j}d_{j}-F_6} \right]$$=0
 * $$\sum_{j=1}^{6}{K_{1j}d_j}=F_1$$ (1$$^{st}$$equation)


 * Choice 2: Select $$w$$ such that $$w_{1}$$=0, $$w_{2}$$=1, $$w_{3}$$=...=$$w_{6}$$=0


 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 0•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+1•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{3j}d_{j}-F_3} \right]$$+...=0
 * $$\sum_{j=1}^{6}{K_{2j}d_j}=F_2$$ (2$$^{nd}$$equation)


 * Choices 3 through 6 repeat this pattern.

Ultimately, we will have achieved the FD relationship.

=PVW: Furthermore=

According to the boundary conditions (BCs) of the 2-bar truss, $$d_1=d_2=d_5=d_6=0$$.

The weighting cofficient must be kinematically admissible. It cannot violate the BCs: $$w_1=w_2=w_5=w_6=0$$. Therefore, weighting coefficients go hand-in-hand with virtual displacements, as explained by calculus of variations.

For the 2-bar truss at hand, it is evident that $$w$$•($$K$$$$d$$-$$F$$) can be reduced to a rectangular matrix multiplied by a 2x1 matrix with a 6x1 matrix as a result.


 * $$w$$•($$K$$$$d$$-$$F$$) =$$\begin{bmatrix}

w_3\\ w_2 \end{bmatrix}$$•($$\bar{K}_{2x2}$$$$\bar{d}_{2x1}$$-$$\bar{F}_{2x1}$$)=0

Where,
 * $$\bar{K}$$=$$\begin{bmatrix}

K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix}$$
 * $$\bar{d}$$=$$\begin{Bmatrix}

d_3\\ d_4 \end{Bmatrix}$$
 * $$\bar{F}$$=$$\begin{Bmatrix}

F_3\\ F_4 \end{Bmatrix}$$