User:Eml4500.f08.gravy.mmm/HW5

Principal of virtual work
Justification of eliminating rows 1,2,5,6 to obtain K2x2 in 2-bar truss:

as described in an earlier report Elimination of dofs

FD relation $$\mathbf{K}_{2X2} \mathbf{d}_{6x1}= \mathbf{F}_{6x1} $$

becomes

$$\mathbf{Kd-F}= \mathbf{0}_{6x1}$$  (Equation 1)

The Principal of virtual work gives the equation :$$ \mathbf{W}_{6x1} \cdot \overbrace{( \mathbf{Kd-F}))}^{6x1} = 0_{1x1}$$(Equation 2)

Is true for all W values Where \mathbf{W}_{6x1} is the weighting matrix.

So (Equation 1)$$\iff$$ (Equation 2)

Proof:

A)(Equation 1)$$\Rightarrow$$ (Equation 2) is trivial

B)To prove (Equation 1)$$\Rightarrow$$ (Equation 2) we must choose different values for W

Choice 1: Select 'W' such that W1=1, W2=W3=W4=W5=W6=0 so

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

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

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$ (First Equation)

Choice 2: Select 'W' such that W2=1, W1=W3=W4=W5=W6=0 so

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

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

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$ (Second Equation)

Choice 3: Select 'W' such that W3=1, W1=W2=W4=W5=W6=0 so

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

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

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$ (Third Equation)

Choice 4: Select 'W' such that W4=1, W1=W2=W3=W5=W6=0 so

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

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

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$ (Fourth Equation)

Choice 5: Select 'W' such that W5=1, W1=W2=W3=W4=W6=0 so

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

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

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$ (Fifth Equation)

Choice 5: Select 'W' such that W5=1, W1=W2=W3=W4=W6=0 so

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

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

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

Hence the equation becomes $$\mathbf{Kd}= \mathbf{F} $$ which is Equation 1 = For the Principal of virtual work when you account for the boundary conditions d1 =  d2 =  d5=  d6=0 The $$\mathbf{K}$$  matrix becomes the  $$\mathbf{\overline{K}}$$  matrix.

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

The $$\overline{ \mathbf{K}}$$ is created as described in an eariler report Elimination of dofs

$$\mathbf{W}_{6x1} \cdot (

\underbrace{\mathbf{Kd}}_{\Downarrow}-\mathbf{F}) $$

$$\begin{bmatrix} && \\&& \end{bmatrix}_{6x2} \begin{Bmatrix} \\ \\ \end{Bmatrix}_{2x1}

$$

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

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

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

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

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

Fixing the bug in the 2-bar truss code Sean
There was a bug in the code solving the 2-bar truss solved in and earier report inserst link. The code was orgnialy written for a 6-bar truss and was modified for the 2-bar truss. Unfornualy an error was created. The 2-bar truss had different values for the E's and A's for the elements,but the 6-bar truss had the same E's and A's for all of the elements.

Principal of virtual work with axial dofs
The Principal of virtual work can also be expressed with the equation $$q^{(1)}_{2x1}= \underline{T}_{2x4} d^{(i)}_{4x1}$$

This is much like how FEA course grades are calculated, $$Grade= \alpha_o \cdot HW grade + \sum^3_{i=1} Exam_i $$

Deriving $$\mathbf{K}^{(e)}_{4x4}= \mathbf{T}^{(e)T}_{4x2} \mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{T}^{(e)}_{2x4} $$

Remember from previous reports [http://en.wikiversity.org/w/index.php?title=User:Eml4500.f08.gravy.jad/Notes3.g&oldid=348085#Similarly_for_node_2_:_HW. Axial dofs equation]and Force Displacement

That:

$$\mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1}= \mathbf{p}^{(e)}_{2x1} \Longrightarrow \mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}= \mathbf{0}^{(1)}_{2x2}$$ (Equation 1)

$$ \mathbf{\hat{W}}^{(e)}_{2x1}\underbrace{(\mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1})= \mathbf{p}^{(e)}_{2x1} \Longrightarrow \mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}}_{2x1}= \mathbf{0}_{1x1} $$ (Equation 2)

We showed (Equation 1)$$\iff$$ (Equation 2) earlier in this report

Recall: $$\mathbf{q}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{d}^{(e)}_{4x1} $$

Similarly

$$ \mathbf{\hat{W}}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{W}^{(e)}_{4x1} $$