User:Eml4500.f08.bike.viale/homework report 5

=Principal of Virtual Work=

To justify the elimination of rows 1, 2, 5, and 6 in order to obtain K as a 2x2 matrix in a 2-bar truss, the FD relation

$$\mathbf{Kd} = \mathbf{F}$$

Can be rearranged to be:

$$ \mathbf{Kd} - \mathbf{F} = \mathbf{0}$$ (1)

Which is designated to be equation 1.

Utilizing the principle of virtual work, equation 1 becomes:

$$ \mathbf{W}\cdot (\mathbf{Kd} - \mathbf{F}) = \mathbf{0}$$ (2)

Where W is a 6x1 matrix and Kd - F is a 6x1, and 0 is a scalar number. This equation is designated as equation 2. This equation is true for all values of W and W is called the weighting matrix.

Considering this, Equation 2 being true implies Equation one is also true, and vice versa.

The proof going from Equation 1 to 2 is trivial. However, going from two to one is not trivial and to do this we must choose different values for W

1st Choice:

Choose W so that

$$W_{1} = 1, W_{2} = W_{3} = W_{4} = W_{5} = W_{6} = 0$$

Therefore

$$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

and

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 1[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$

Which is designated to be equation (a).

2nd Choice:

Choose W such that W2=1, W1=W3=W4=W5=W6=0 so that:

$$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

Therefore:

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 1[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes

$$\sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$

Which is designated to be equation (b)

3rd Choice: C hoose W such that W3=1, W1=W2=W4=W5=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 1 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$

Which is designated to be equation (c)

Choice 4:

Choose W such that W4=1, W1=W2=W3=W5=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 1 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$

This is designated to be equation (d)

Choice 5:

Choose W such that W5=1, W1=W2=W3=W4=W6=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 1 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

Thus, the result becomes  $$\sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$

This is designated to be equation (e)

Choice 6:

Choose W such that W6=1, W1=W2=W3=W4=W5=0 so that

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 1 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so that the result becomes  $$\sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$

This equation is designated to be equation (f)

Thus, when the equations (a) through (f) are combined, they become $$\mathbf{Kd}= \mathbf{F} $$ which is equation (1) from before.

The weighting coefficients must be kinetically admissible which means they cannot violate the boundary conditions so W1 = W2 =  W5=  W6=0  where the weighting coefficients represent the virtual displacement.

Now,

$$\mathbf{W} \cdot (\mathbf{Kd}-\mathbf{F})$$

Where Kd is a 6x2 matrix minus a 2x1 vector.

Thus

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K}_{2x2} \overline{d}_{2x1}- \overline{F}_{2x1} })$$ (3)

Which is designated as Equation 3.

and is true for all values of $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

and the K matrix is reduced to $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

as well as:

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$\overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

being the reduced F and d matrices.