User:Eml4500.f08.ramrod.c/11-20-08

Justification of Eliminated rows 1, 2, 5, and 6 to obtain K 2-bar truss:
Force Displacement Relation: K6x6d6x1 = F6x1

Moving F to the other side of the equation gives us Equation (1):

 Equation (1)

$$\mathbf{Kd}-\mathbf{F}=\mathbf{0}_{6x1} $$

Using the Principle of Virtual Work, we can arrive at Equation (2):

 Equation (2)

$$\mathbf{w}_{6x1}\bullet(\mathbf{Kd}-\mathbf{F})=\mathbf{0}_{1x1} $$
 * This is true for all w6x1
 * w is known as the "weighting mat."
 * Note that the dot product is used in this equation.

Equation (1)↔ Equation (2)

Using Principle of Virtual Work
Taking account for the boundary conditions of the 2 bar truss:
 * d1=d2=d5=d6=0

Weighting coefficient must be "kinematically admissible," ie cannot violate the boundary conditions
 * w1=w2=w5=w6=0

Weighting coefficients = virtual displacement


 * (calculus of variations)

From before:

$$\begin{Bmatrix} w_3\\ w_4\end{Bmatrix}\bullet(\mathbf{Kd}-\mathbf{F})=0$$

$$\mathbf{w}\bullet(\mathbf{Kd}-\mathbf{F})$$

$$\begin{Bmatrix} w_3\\ w_4 \end{Bmatrix}\bullet[\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix} - \begin{Bmatrix} F_1\\ F_2\\ F_3\\ F_4\\ F_5\\ F_6 \end{Bmatrix}] = 0$$
 * This is true for all $$\begin{Bmatrix}

w_3\\ w_4\end{Bmatrix}$$

$$\mathbf{\overline{K}} = \begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$

$$\mathbf{\overline{d}} = \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}$$

$$\mathbf{\overline{F}} = \begin{Bmatrix} F_3\\ F_4 \end{Bmatrix} $$

Comparing Equations (3) and (4)
Equation (3) : $$(kd-F) = 0$$

Equation (4) : $$w(kd-F) = 0$$

Equation (3) → Equation (4) is trivial

Equation (4) → Equation (3)
 * Since Equation (4) is valid for all w, select w=1


 * $$1(kd-F) = 0$$ Thus proving Equation (3) is true