User:Eml4500.f08.echo.sell.cm/HW6

Wednesday, November 12th
I was working on this date during class one day to kill time. I'll let someone else take over from here though. I'll go ahead and start working on this section and let someone else work on the homework portions. Bradley LaCroix 21:20, 20 November 2008 (UTC)

FEM via PVW (continued)

HW: Show that $$u^{ }_{}(x_{i+1})=d_{i+1}$$

$$u(x_{i+1})=N_i^{}(x_{i+1})d_i+N_{i+1}(x_i)d_{i+1}$$

In this case, $$N_i^{}(x_{i+1})=0$$ and $$N_{i+1}^{}(x_{i+1})=1$$

Substituting these values into the equation, we get $$u(x_{i+1}^{})=(0)\cdot d_i+(1)\cdot d_{i+1}$$

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

END HW

From page 31-3: Interpolate for u(x) Apply same interpolation for u(x), i.e.,

$$w(x)=N_i^{ }(x)w_i+N_{i+1}(x)w_{i+1}$$

Element Stiffness Matrix for element i

$$\beta=\int_{x_i}^{x_{i+1}} \left [ N_i^'w_i + N_{i+1}^'w_{i+1} \right ] (EA) \left [ N_i^'d_i+N_{i+1}^'d_{i+1} \right ]\, dx$$

Where $$w'(x)=N_i^'w_i + N_{i+1}^'w_{i+1}$$

and $$u'(x)=N_i^'d_i+N_{i+1}^'d_{i+1}$$

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

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

Note $$u(x)=\begin{bmatrix} N_i^'(x) & N_{i+1}(x) \end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1}$$

$$\frac{du(x)}{dx}=\begin{bmatrix} N_i^'(x) & N_{i+1}^'(x) \end{bmatrix}_{1x2} \begin{Bmatrix} d_i \\ d_{i+1}\end{Bmatrix}_{2x1}$$

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

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

Recall the elemental degrees of freedom:



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

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

$$\beta=\int_{x_i}^{x_{i+1}}(\mathbf{Bw}^{(i)})_{1x1}(EA)_{1x1}(\mathbf{Bd}^{(i)})_{1x1}\,dx=\mathbf{w}^{(i)}\cdot (\mathbf{k}^{(i)} \mathbf{d}^{(i)})$$

$$\beta=\int_{x_i}^{x_{i+1}}(EA)(\mathbf{Bw}^{(i)}) \cdot (\mathbf{Bd}^{(i)})\, dx$$

$$(\mathbf{Bw}^{(i)})^T (\mathbf{Bd}^{(i)})$$

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

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

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

HW $$B(x)=\begin{bmatrix} \mathbf{HW} & \frac{1}{L^{(i)}} \end{bmatrix}$$

L(i)=xi+1-xi (length of element i)

HW6: Consider EA=const $$\mathbf{k}^{(i)}=\frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Transfer of variable coordinates from x to $$\tilde x$$

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

$$d \tilde x = dx$$

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

HW6: Find expression for $$\mathbf{k}^{(i)}$$ using the above forumula

HW6



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

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

Find $$\mathbf{k}^{(i)}$$