User:Eml4500.f08.ateam.mcnally/Week 8

Continueing the FEM via PVW
From a previous lecture on meeting 31 about the linear interpolation of U(x), W(x) can be shown by applying the same principle.

i.e.

$$ W(x)=N_{i}(x)W_{i}+N_{i+1}(x)W_{i+1} $$

Now the element stiffness matrix for element I can be written as:

$$ \beta = \int_{x_{i}}^{x_{i+1}}{[N_{i}'W_{i}+N_{i+1}'W_{i+1}](EA)} $$

with

$$[N_{i}'W_{i}+N_{i+1}'W_{i+1}]=W'(x)$$

$$[N_i'd_i+N_{i+1}'d_{i+1}]=U'(x)$$

$$ N_i':=\frac{dN_i(x)}{dx} $$

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

NOTE

$$ u(x)=\begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} =\mathbf{N(x)_{(1x1)}} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

$$ \frac{du(x)}{dx}=\begin{bmatrix} N_i'(x) & N_{i+1}'(x) \end{bmatrix} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} =\mathbf{B(x)} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

Similarly W(x) can be shown as:

$$ W(x)=\mathbf{N(x)}\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

$$ \frac{dW(x)}{dx}=\mathbf{B(x)}\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

Now, let us recall the element degrees of freedom as discussed in a previous lecture



therefore

$$ \begin{Bmatrix} d_i\\ d_{i+1}

\end{Bmatrix} = \begin{Bmatrix} d_1^{(i)}\\ d_2^{(i)} \end{Bmatrix} = \mathbf{d^{(i)}} $$

$$ \begin{Bmatrix} W_i\\ W_{i+1}

\end{Bmatrix} = \begin{Bmatrix} W_1^{(i)}\\ W_2^{(i)} \end{Bmatrix} = \mathbf{W^{(i)}} $$

The element stiffness matrix can now be written as

$$ \beta =\int_{x_{i}}^{x_{i+1}}{{BW}^{(i)}_{(1x1)}(EA)_{(1x1)}{Bd}^{(i)}_{(1x1)}}dx = {W}^{(i)}\cdot({k}^{(i)}{d}^{(i)}) $$

$$ \beta =\int_{x_i}^{X_{i+1}}(EA)({BW^{(i)}})_{(1x1)}\cdot({Bd^{(i)}}){(1x1)}dx

$$

$$ ({BW^{(i)}})^T({Bd^{(i)}}) $$

where

$$ {BW^{(i)T}} = {W^{(i)T}B^T} = {W^{(i)} \cdot B^T} $$

$$ \beta= \mathbf{W^{(i)}}\cdot\int \mathbf{B^T}EA\mathbf{B}dx\mathbf{d^{(i)}} $$

$$ \mathbf{k^{(i)}}_{(2x2)}=\int_{x_i}^{x^{i+1}}{\mathbf{B(x)^T}}_{(2x1)}(EA)_{1x1}\mathbf{B(x)}_{(1x2)}dx $$

If we consider $$ EA $$ to constant the matrix $$ {k^{(i)}} $$ can be written as

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

The next step is the transformation of variable coordinate system from:

$${x}$$ to $$ \tilde{x} $$

Given

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

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

Now $$ {k^{(i)}} $$ can be written as

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\mathbf{B^T(\tilde{x})}}EA(\tilde{x})\mathbf{B(\tilde{x}})d(\tilde{x}) $$

The following problem outlined by the image below can be solved using the transformed coordinate system.

Given:



With the following functions for the modulus elasticity and cross sectional area $$ {k^{(i)}} $$ matrix can be written as

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\mathbf{B^T(\tilde{x})}}{[N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2]}(\tilde{x})\mathbf{B(\tilde{x}})d(\tilde{x}) $$