User:Eml4500.f08.FEABBQ.koby/HW7

Notes for Nov. 21
Continuation of Dimensional Analysis

$$[ \varepsilon ] = \frac{du}{dx} = \frac{L}{L}=1$$

$$[ \sigma ] = [E] = \frac{F}{L^2}$$

$$\left[ \frac{EA}{L}\right] = \left[ \tilde{k_{11}}\right]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$\left[ \tilde{k_{11}} \tilde{d_1} \right] = [\tilde{k_{11}}][\tilde{d_1}]=\frac{[6][E][I]}{[L^2]} = \frac{1(F/L^2)L^4}{L^2} = F$$

Element F-d relation in global coordinates from element F=d relations in local coordinates, i.e. obtain

$$\mathbf{k}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)}=\mathbf{f}_{6x1}^{(e)}$$

$$\mathbf{k}_{6x6}^{(e)}=\tilde{\mathbf{T}}_{6x6}^{(e)T}\tilde{\mathbf{k}}_{6x6}^{(e)}\tilde{\mathbf{T}}_{6x6}^{(e)}$$

Notes for Dec. 5
$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \mathbf{R}^T\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}$$  (1)

$$\begin{Bmatrix} u_x(\tilde{x})\\ u_y(\tilde{x}) \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d_1}^{(e)}\\ \vdots\\ \tilde{d_1}^{(e)} \end{Bmatrix}$$  (2)

Perform dimensional analysis of equation (2):

$$[u]=L$$
 * First, a look at the normal linear degrees of freedom:

$$[N_1]=[N_4] = 1$$

$$[N_1][\tilde{d_1}],[N_4][\tilde{d_4}]$$

$$1\cdot L,1\cdot L = L$$

The dimensional analysis for DOF's 2 and 5


 * Now a look at the dimensional analysis of a rotational DOF:

$$[v]=L$$

$$[\tilde{d_3}]=[1]$$

$$[N_3][\tilde{d_3}],[N_6][\tilde{d_6}]$$

$$L\cdot 1,L\cdot 1 = L$$

Derivation of Shape Functions

Recall: Governing PDE for beams. Look at governing equations without internal forces of transverse forces or distributed truss loads. This is the static case:

$$\frac{\delta ^2}{\delta x^2} \left\{(EI) \frac{\delta v^2}{\delta x^2}\right\}=0$$

If EI is constant along the beam: $$\frac{\delta ^2}{\delta x^2}v=0$$

After integrating this equation four times, you arrive at: $$\Rightarrow v(x)=C_0+C_1x^1+C_2x^2+C_3x^3$$

Now the boundary conditions can be applied to find $$C_0, C_1, C_2, C_3$$ for each shape function.


 * $$\mathbf{N_2}$$

$$v(0)=1=C_0$$

$$v(L)=0=1+C_1L+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(0)=0=C_1$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={-2C_2}/{3L}$$

plugging $$C_3$$ back into the equation for v(L): $$0=1+C_2L^2+{-2C_2}/{3L}L^3$$

$$C_2=-3/L, C_3=2/L^3$$

$$N_2(x)=1-\left( \frac{3}{L^2}\right)x+\left( \frac{2}{L^3}\right)x^3$$


 * $$\mathbf{N_3}$$

change... $$v(0)=0=C_0$$

$$v'(0)=1=C_1$$

$$v(L)=0=L+C_2L^2+C_3L^3$$

solving for $$C_3=-1/L^2-C_2/L$$

$$v'(L)=0=1+2C_2L+3C_3L^2$$

solving for $$C_3=-1/3L^2-2C_2/3L$$

Now solve for $$C_2, C_3$$ by first setting $$C_3$$ equal to itself.

$$C_3=-1/3L^2-2C_2/3L=-1/L^2-C_2/L$$

$$C_2=-2/L$$

$$C_3=1/L^2$$

$$N_2(x)=x-\left( \frac{2}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$


 * $$\mathbf{N_5}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=1=0+C_2L^2+C_3L^3$$

$$v'(x)=C_1+2C_2x+C_3x^2$$

$$v'(L)=0=2C_2L+3C_3L^2$$

Now solve for $$C_2, C_3$$:

$$C_3={1-C_2L^2}/{L^3}$$ from v(L)

$$C_3={-2C_2}/{3L}$$ from v'(L)

Setting $$C_3$$ equal to itself: $${(1-C_2L^2)}/{L^3}={-2C_2}/{3L}$$

$$C_2={3}/{L^2}, C_3={-2}/{L^3}-2L/3$$

$$N_5(x)=\left( \frac{3}{L^2}\right)x^2-\left( \frac{2}{L^3}\right)x^3$$


 * $$\mathbf{N_6}$$

$$v(0)=0=C_0$$

$$v'(0)=0=C_1$$

$$v(L)=0=C_2L^2+C_3L^3$$

Solve for $$C_3$$... $$C_3=-C_2/L$$

$$v'(L)=1-2C_2L+3C_3L^2$$

Solving again for $$C_3$$... $$C_3 = 1/3L^2-{2C_2}/{3L}$$

Setting $$C_3$$ equal to itself yields $$C_2$$: $$C_2=-1/L$$

$$C_3=1/L^2$$

$$N_6(x)=-\left( \frac{1}{L}\right)x^2+\left( \frac{1}{L^2}\right)x^3$$