User:Eml4500.f08.team.figuerrez/hw7/meeting39

Meeting 39 - Principle Axial & Transverse Dislocations Of A Beam (Shape Functions)
From Meeting 37: $$\tilde{\boldsymbol{d}}_{6x1}^{(e)}=\tilde{\boldsymbol{T}}_{6x6}^{(e)}*{\boldsymbol{d}}_{6x1}^{(e)}$$, where $${\boldsymbol{d}}_{6x1}^{(e)}$$ is known after solving the finite element system.

Now we will discuss how to compute $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$

We compute $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$ using equations (1) & and (2) from meeting (38-3).

We Compute $${\boldsymbol{U_{x}}}\tilde{\boldsymbol{(x)}},{\boldsymbol{U_{y}}}\tilde{\boldsymbol{(x)}}$$ from $${\boldsymbol{u}}\tilde{\boldsymbol{(x)}},{\boldsymbol{v}}\tilde{\boldsymbol{(x)}}$$

$$\begin{Bmatrix}U_{x}\tilde{(x)} \\ U_{x}\tilde{(x)}\end{Bmatrix}=R^{T}\begin{Bmatrix}u\tilde{(x)} \\ v\tilde{(x)}\end{Bmatrix}$$    (1)

$$\begin{Bmatrix}u\tilde{(x)} \\ v\tilde{(x)}\end{Bmatrix}=\begin{bmatrix}N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6\end{bmatrix} \begin{Bmatrix}\tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)}\end{Bmatrix} $$     (2)

We will denote $$\begin{bmatrix}N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6\end{bmatrix}$$ by N, and the matrix $$\begin{Bmatrix}\tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)}\end{Bmatrix}$$ as $$\tilde{d}^{(e)}$$ therefore:

$$\begin{Bmatrix}U_{x}\tilde{(x)} \\ U_{x}\tilde{(x)}\end{Bmatrix}=R^{T}*{\boldsymbol{N}}\tilde{(x)}*\tilde{T}^{(e)}d^{(e)}$$    (3)