User:Eml4500.f08.wiki1.aguilar/hw6

FEM via PVW (cont'd)
We have just showed how we got to the relation:

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

Page 31.3 of our notes shows the interpolation for u(x). We can apply this same interpolation for w(x), i.e.,

$$ w(x)=N_i(x)\; W_i\; +\; N_{i+1}(x)\; W_{i+1}$$

Element Stiffness Matrix for Element i
In order to make the leap from the discrete version of the Principle of Virtual Work to the continuous version, the stiffness matrix must involve an integral.

The form of this integral is in the form below:

$$\int_{x_i}^{x_{i+1}}[N'_i\; W_i\; +\; N'_{i+1}\; W_{i+1}](EA)[N'_i\;  d_i\; +\; N_{i+1}\; W_{i+1}]dx        \;\;\;\;\;\;\;\;\;   (eqn.\; \beta) $$

with    $$N'_i:= \frac {dN'_i(x)}{dx}$$      and     $$N'_{i+1}:= \frac {dN'_{i+1}(x)}{dx} $$.

Note:

$$u(x)= \underbrace{ \begin{bmatrix}N_i(x)&N_{i+1}(x)\end{bmatrix}}_{\mathbf{N}(x)_{1\times 2}}  \begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}_{2\times 1}$$

and

$$\frac{du(x)}{dx} = \begin{bmatrix}N'_i(x)&N'_{i+1}(x)\end{bmatrix}_{2\times 1}  \begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}_{2\times 1} $$

Similarly:

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

and

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

Recall the element degrees of freedom:



$$\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix} = \begin{Bmatrix}d_1^{(i)}\\d_2^{(i)}\end{Bmatrix} = \mathbf{d}^{(i)}$$

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

We're looking to take $$ eqn. \beta$$ and move some terms around so that it corresponds to the relation: $$\mathbf{w}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)})$$.

So how do we get: $$\int_{x_i}^{x_{i+1}} (\mathbf{B}\mathbf{w}^{(i)})_{1\times1}(EA)_{1\times1}(\mathbf{B}\mathbf{d}^{(i)})_{1\times1} dx = \mathbf{w}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)}) $$ ?

Well,

$$\int_{x_i}^{x_{i+1}} (EA)(\mathbf{B}\mathbf{w}^{(i)})(\mathbf{B}\mathbf{d}^{(i)})dx - (\mathbf{B}\mathbf{w}^{(i)})^T(\mathbf{B}\mathbf{d}^{(i)}) $$

As a 1x1 matrix, $$(\mathbf{Bw}^{(i)})$$ is equal to its transpose, $$(\mathbf{Bw}^{(i)}))^T$$.

$$(\mathbf{Bw}^{(i)})^T = (\mathbf{w}^{(i)})^T \cdot (\mathbf{B})^T = (\mathbf{w}^{(i)}) \cdot (\mathbf{B})^T$$

So we now have: $$\int (\mathbf{w}^{(i)}\mathbf{B}^T)(EA)(\mathbf{B}\mathbf{d}^{(i)}) dx $$

We can also move parts out of the integral if they are not dependent on 'x', ($$\mathbf{w}^{(i)},\mathbf{d}^{(i)}$$):

Therefore we get: $$ \mathbf{w}^{(i)} \cdot ( \int \mathbf{B}^T (EA) \mathbf{B} dx) \;\mathbf{d}^{(i)}$$

We can now see that:

$$\mathbf{k}^{(i)}_{2\times2}=\int_{x_i}^{x_{i+1}} \mathbf{B}_{2\times1}^T (EA)_{1\times1} \mathbf{B}_{1\times2} dx$$

where $$ \mathbf{B}^T$$, $$(EA)$$, and $$\mathbf{B}$$ are all functions in terms of 'x'.

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

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

Transfer of Coordinates
We now do a basic transfer of $$x$$ to $$\tilde x$$.

$$\tilde x := x - x_i$$ and $$d\tilde x = dx$$

We get a new equation based on our old for the element stiffness matrix as: $$\mathbf{k}^{(i)}=\int_{\tilde x=0}^{\tilde x=L^{(i)}} \mathbf{B}(\tilde x) (EA)(\tilde x) \mathbf{B}(\tilde x) d\tilde x$$

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