User:Eml4500.f08.delta 6.lopez/HW6

=FEM via PVW (cont.)=

Apply same interpolation for W(X). For example:

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

=Elemental Stiffness Matrix for Element i=

Now the element stiffness matrix can be written as:

$$ \beta = \int_{x_{i}}^{x_{i+1}}{[N_{i}'W_{i}+N_{i+1}'W_{i+1}](EA)}[N_i'd_i+N_{i+1}'d_{i+1}]dx$$ where $$N_{i}'W_{i}+N_{i+1}'W_{i+1}=W'(x)$$ and $$N_i'd_i+N_{i+1}'d_{i+1}=U'(x)$$

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

Likewise, the relation is the same for Ni+1':

$$ N_{i+1}':=\frac{dN_{i+1}(x)}{dx} $$


 * NOTE

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

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

Similarly:

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

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

Now, let us recall the element degrees of freedom:



$$ \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 now becomes: $$ \beta =\int_{x_{i}}^{x_{i+1}}{\left(\textbf{BW}^{(i)}\right)\left(EA \right)\left(\textbf{Bd}^{(i)}\right)}dx$$

$$ =\textbf{W}^{(i)}* \left(\textbf{k}^{(i)} \textbf{d}^{(i)}\right) $$

$$ \beta =\int_{x_i}^{X_{i+1}}(EA)(\mathbf{BW^{(i)}})*(\mathbf{Bd^{(i)}})dx-(\mathbf{BW^{(i)}})^T(\mathbf{Bd^{(i)}}) $$

where

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

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

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

=Transformation of variable coordinate system from $$x$$ to $$ \tilde{x} $$=

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

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

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

=Homework=

Show that, if EA is constant,

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

Knowing that

$$ \beta = \int_{0}^{L}{\frac{dw}{dx}EA\frac{du}{dx}dx}\; \; \; \; \; \; \; \; \; \; Equation (1) $$

$$ \frac{dW}{dx}=[N_i'W_i + N_{i+1}'W_{i+1}] $$

$$ \frac{du}{dx}=[N_i'd_i + N_{i+1}'d_{i+1}] $$

Equation 1 reduces to

$$ \beta = \int_{x_i}^{x_{i+1}}{\mathbf{B}W^{(i)}EA\mathbf{B}d^{(i)}dx}\; \; \; \; \; \; \; \; \; \; Equation (2) $$

where

$$ W^{(i)} = \begin{bmatrix} W_i\\ W_{i+1} \end{bmatrix},

d^{(i)} = \begin{bmatrix} d_i\\ d_{i+1} \end{bmatrix},

\mathbf{B} = \begin{bmatrix} N_i' &N_{i+1}' \end{bmatrix} $$

Manipulating Equation (2) with transpose definitions, we determine:

$$ \mathbf{B} = W^{(i)}*\left(\int_{x_i}^{x_{i+1}}{\mathbf{B^T}EA\mathbf{B}dx} \right)d^{(i)}\; \; \; \; \; \; \; \; \; \; Equation (3) $$

Equation (3) shows that

$$ \mathbf{k^{(i)}}= \int_{x_i}^{x_{i+1}}{\mathbf{B^T}EA\mathbf{B}dx} $$

Since EA remains constant, this reduces to

$$ \mathbf{k^{(i)}}= EA\int_{x_i}^{x_{i+1}}{\mathbf{B^T}\mathbf{B}dx} $$

Now $$\mathbf{B(x)}$$ must be analyzed.

$$ \mathbf{B(x)}= \begin{bmatrix} N_i'(x) & N_{i+1}'(x) \end{bmatrix} = \begin{bmatrix} \frac{1}{x_i-x_{i+1}} & \frac{1}{x_{i+1}-x_i} \end{bmatrix}= \begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix} $$

$$ \mathbf{B^T}=\begin{bmatrix} \frac{1}{-L^i}\\ \frac{1}{L^i} \end{bmatrix} $$

Hence,

$$ \mathbf{k^{(i)}}= EA\int_{x_i}^{x_{i+1}}{\begin{bmatrix} \frac{1}{-L^i}\\ \frac{1}{L^i} \end{bmatrix}\begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix}dx} $$

Multiplying matrices gives

$$ \mathbf{k^{(i)}} = EA\begin{bmatrix} \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} & \int_{x_i}^{x_{i+1}}{\frac{-1}{L^2}dx}\\ \int_{x_i}^{x_{i+1}}{\frac{-1}{L^2}dx} & \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} \end{bmatrix} $$


 * NOTE

One integer will be used for example. Others differ by a positive or negative sign.

$$ \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} = \frac{x_{i+1}-x_i}{L^2} = \frac{L}{L^2} = \frac{1}{L} $$

Thus,

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

Extracting $$\frac{1}{L}$$ gives us

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

=Homework 2=



where

$$ A(\tilde{x})=N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2} $$

and

$$ E(\tilde{x})=N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2} $$

We will find an expression for $$\mathbf{k^{(i)}}$$ using

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

By plugging the above equations in for $$E(\tilde{x})$$ and $$A(\tilde{x})$$, we get

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\mathbf{B^T(\tilde{x})}(N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2})(N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2})\mathbf{B(\tilde{x})}d\tilde{x} $$