User:Eml4500.f08.Ateam.Slevinski/Homework6

HW 6 - Meeting 33 - 11/12/08
Continuing the finite element method via the principle of virtual work:

$$u(x_{i+1}) = d_{i+1}$$

Using the same method for the interpolation of $$u(x)$$ we can apply this to $$ W(x)$$:

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

Element Stiffness Matrix for Element i:

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

In the above equation, $$[N'_{i}W_{i}  N'_{i+1}W_{i+1}]$$ represents $$W'(x)$$, and $$u'(x) = [N'_{i}d_i   N_{i+1}d_{i+1}]$$

Also, $$N'_i$$ is the derivative of $$N_i(x)$$ which is represented as: $$N'_i = \frac{dN_i(x)}{dx}$$ The same applies for $$N'_{i+1}$$

Note: From the previous equation, $$u(x) = \begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix}_{1x2} \begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$ where $$\begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix}_{1x2}$$ is the 1x2 matrix $$\underline{N}(x)$$

Now, the derivative of this is: $$\frac{du(x)}{dx} = \begin{bmatrix} N'_i(x) & N'_{i+1}(x)\end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1}$$ and in this equation, $$\begin{bmatrix} N'_i(x) & N'_{i+1}(x)\end{bmatrix}_{1x2}$$ is the 1x2 matrix $$\underline{B}(x)$$

Now, with the matrices $$\underline{N}(x)$$ and $$\underline{B}(x)$$ we can develop the following relationships:

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

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

Now we need to recall the element degrees of freedom:



From the figure above, we have: $$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^{(i)} \\ d_{2}^{(i)} \end{Bmatrix} = \underline{d}^{(i)}$$ and also the relationship for the weighting coefficient $$\begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^{(i)} \\ W_{2}^{(i)} \end{Bmatrix} = \underline{W}^{(i)}$$

We can now rewrite equation (1) using the previous relationships. Equation (1) becomes:

$$\beta = \int_{x_i}^{x_{i+1}}{(\underline{B}\underline{W}^{(i)})(EA)(\underline{Bd}^{(i)})dx}$$ (2)

$$ = \underline{W}^{(i)} \cdot (\underline{k}^{(i)} \underline{d}^{(i)})$$

Now, rearranging equation (2) with the dot product we have:

$$\beta = \int_{x_i}^{x_{i+1}}{(EA)(\underline{BW}^{(i)})\cdot(\underline{Bd}^{(i)})dx}$$ (3)

From equation (3) we can simplify even further using the transpose of weighting matrix:

$$(\underline{BW}^{(i)})^T (\underline{Bd}^{(i)}) = $$

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

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

Once again, we can rewrite equation (3) to form:

$$\beta = \underline{W}^{(i)} \cdot (\int {\underline{B}^T(EA)\underline{B}}dx)\underline{d}^{(i)}$$ (4)

In equation (4), the integral represents the 2x2 element stiffness matrix:

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

Transform of variable from $$x$$ to $$ \tilde{x}$$:

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

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

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