User:EML4500.f08.delta 6.ramirez/hw5

Justification of eliminating rows 1,2,5,6 to obtain the K2x2 in 2 bar truss system

Force Displacement Relationship gives,

$$ Kd - F = 0 $$ (1)

Equation 2 is given as,

$$ w(Kd - F) = 0 $$ (2)

for all w.

Choice 1 : Select w so that w1 = 1, w2 = 0, ..., w6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 1 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{1j}d_j - F_1} = 0 $$

Choice 2 : Select w so that w2 = 1, w1,3,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 1 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{2j}d_j - F_2} = 0 $$

Choice 3 : Select w so that w3 = 1, w1,2,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{3j}d_j - F_3]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{3j}d_j - F_3} = 0 $$

Choice 4 : Select w so that w4 = 1, w1,2,3,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{4j}d_j - F_4]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{4j}d_j - F_4} = 0 $$

Choice 5 : Select w so that w5 = 1, w1,2,3,4,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{5j}d_j - F_5]} + 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{5j}d_j - F_5} = 0 $$

Choice 6 : Select w so that w6 = 1, w1,2,3,4,5 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 1 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{6j}d_j - F_6} = 0 $$