User:Eml4500.f08.team.Bengtson/teamHW6

FEM via PVW (Continued)
Prove: $$ u(x_{i+1})=d_{i+1} $$ Citing the previous days notes $$\displaystyle u(x_{i+1})=N_i(x_{i+1})d_i+N_{i+1}(x_{i+1})d_{i+1}=(0)d_i+(1)d_{i+1}=d_{i+1}$$ $$\displaystyle u(x_{i+1})=d_{i+1}$$ The two expressions are equal

Apply same interpolation for w(x), ie

$$ w(x)=N_{i}(x)w_{i}+ N_{i+1}(x)w_{i+1} $$

Element Stiffness matrix for element:

$$ (B)=\int_{x_{i}}^{x_{i+1}}[N_{i}(x)w_{i}+ N_{i+1}(x)w_{i+1}][N_{i}(x)d_{i}+ N_{i+1}(x)d_{i+1}]dx $$ This integral represents the elemental stiffness matrix for the element.

$$ N_{i}=\frac{\partial N_{i}(x) }{\partial x} $$

Likewise for $$ N_{i+1}$$

Note: $$ u(x)=[N_{i}(x)N_{i+1}(x)]\begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} $$

$$N_{i}= [N_{i}(x)N_{i+1}(x)]$$ and is a 1x2 matrix

$$ \frac{\partial N_{i}(x) }{\partial x}=[N'_{i}(x)N'_{i+1}(x)]\begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} $$

$$ B(x)= [N'_{i}(x)N'_{i+1}(x)]$$ and is a 1x2 matrix.

Similarly $$w(x)= N(x)\begin{Bmatrix} w_i\\ w_{i+1} \end{Bmatrix} $$

$$ \frac{\partial \mathbf{w}(x)}{\partial x}= \mathbf{B}(x) \begin{Bmatrix} w_i\\ w_{i+1} \end{Bmatrix} $$

Recall the element defintions These are the axial displacements of the element.

$$ \begin{Bmatrix} w_i\\ w_{i+1} \end{Bmatrix}=\begin{Bmatrix} w_1^i\\ w_2^i \end{Bmatrix}=\mathbf{w}^{(i)} $$

$$ (\boldsymbol{B})=\int_{x_i}^{x_{i+1}}{(\mathbf{B}\mathbf{W}^{(i)})(EA)\mathbf{B}\mathbf{d}^{(i)}}dx $$

$$ =w^{(i)}\bullet (\mathbf{k}^{(i)}\mathbf{d}^{(i)}) $$

From this we want to identify the stiffness matrix.

$$ (\boldsymbol{B})=\int_{x_i}^{x_{i+1}}{(\mathbf{B}\mathbf{W}^{(i)})(EA)(\mathbf{B}\mathbf{W}^{(i)}) \bullet \mathbf{B}\mathbf{d}^{(i)}}dx $$

$$ (\mathbf{B}\mathbf{W}^{(i)}) \bullet \mathbf{B}\mathbf{d}^{(i)}= (\mathbf{B}\mathbf{W}^{(i)})^T \mathbf{B}\mathbf{d}^{(i)}$$

As proven before, the transpose of the first matrix multiplied by the second is equivalent to the dot product of both together. $$(\mathbf{B} \mathbf{W}^{(i)})^T=$$ $$\mathbf{W}^{(i)T} $$ $$\mathbf{B}^T= \mathbf{W}^{(i)}\bullet \mathbf{B}^T$$

$$ B= w^{i}\bullet (\int B^T(x)(EA)\mathbf{B}dx)\mathbf{d}^{(i)} $$

$$ (\boldsymbol{K})^i=\int_{x_i}^{x_{i+1}}{(\mathbf{B}(x))(EA)\mathbf{B}(x)}dx $$

HW $$B(x)= \begin{bmatrix}Hw & \frac{1}{L^{i}}\end{bmatrix} $$

$$ L^{(i)}=x_{i+1}-x_i$$ length of element i

Transference of Var(coordinates) from x to $$\tilde{x}$$

$$ \tilde{x}=x_i-x_o $$

$$ d\tilde{x}=dx $$