User:Eml4500.f08.RAMROD.A/MyContributionHW7

Dimensional Analysis
Dimensions of common parameters

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

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

$$ [A]=L^2 $$

$$ [I]=L^4 $$

$$ [\frac{EA}{L}]=[K_{11}^{N}]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L} $$

$$ [\tilde{K}_{11}\tilde{d}_1]=[\tilde{K}_{11}][\tilde{d}_1]=F $$

$$[\tilde{K}_{23}\tilde{d}_3]=[\tilde{K}_{23}][\tilde{d}_3]=F $$

FD relationship and the Transpose
Converting the element FD relationship in global coordinates from element FD in local coordianates

The relationship is presented as follows:

$$ K^{(e)}=\tilde{T}^{(e)T}\tilde{k}^{(e)}\tilde{T}^{(e)} $$

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

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

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

where

$$ \tilde{d} = \begin{Bmatrix}\tilde{d}_1\\\tilde{d}_2\\\tilde{\underline{d}}_3\\\tilde{d}_4\\\tilde{d}_5 \\\tilde{\underline{d}}_6\end{Bmatrix}$$

And

$$ \tilde{T} = \begin{bmatrix}\underline{R}_{2x2} & & 0 & 0 & 0 & 0\\ &  & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & \underline{R}_{2x2} & & 0\\0 & 0 & 0 &  &  & 0\\0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Friday Dec 5
Equation One: $$ \begin{Bmatrix}U_x(\tilde{x})\\\mu_y(y)\end{Bmatrix} = R^T \begin{Bmatrix}U(\tilde{x})\\V(\tilde{x}) \end{Bmatrix}$$

Equation Two: $$ \begin{Bmatrix}U(\tilde{x}\\V(\tilde{x})\end{Bmatrix} = \begin{bmatrix}N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6\end{bmatrix} \begin{Bmatrix}\tilde{d}_1\\\tilde{d}_2\\\tilde{d}_3\\\tilde{d}_4\\\tilde{d}_5 \\\tilde{d}_6\end{Bmatrix}^{(e)} $$

Equation Three: $$ \begin{Bmatrix}U_x(\tilde{x})\\U_y(\tilde{x})\end{Bmatrix} = R_{2x2}^TN_{2x6}(\tilde{x})\tilde{T}^{(e)}d^{(e)} $$

Once again, we check the dimensional analysis:

$$ [U]=[V]=L $$

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

$$ [N_1][d_1]+[N_4][d_4]=L+L $$

d = displacement or rotation

displacement:

$$ [d_1]=[d_2]=[d_4]=[d_5]=L $$

rotation:

$$ [d_3]=[d_6]=\frac{L}{[\theta] } =\frac{L}{rad}=L $$

Equations of V and V' of Ni
Looking at N5:

$$V(0)=0,V'(L)=-1$$

$$V'(0)=0,V(L)=0 $$

Looking at N6:

$$V(0)=0,V(L)=0 $$

$$V'(0)=0,V'(L)=1 $$

Derived Coefficient in $$ \tilde{k}$$:

$$ \tilde{k}_{22}=\frac{12EI}{L^3}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx} $$

$$ \tilde{k}_{23}=\frac{6EI}{L^2}=\int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx} $$

Elastodynamics
Studying frames, beams and trusses in 3-D

discrete PVW:

$$w\cdot{(m\ddot{d}+kd-F)}=0$$, for all w

$$d(0)$$ is denoted d0

and $$\dot{d}(0)$$ is denoted V0

Method:
1) Complete ordinary diff. equations (ODE's)

(2nd order in time) and initial conditions are used

Next: Governing the elastodynamics of discrete and continuous problems

For Unforced Vibration:

general equation:

$$ M\ddot{V}+kV=0$$

Assume:

$$ v(t)=\sin(wt)\phi$$

$$-w^2\sin(wt)M\phi+\sin(t)k\phi=0$$

$$ -w^2k\phi =w^2M\phi $$ general Eigenvalue

General Form Eigenvalue:

$$ A_x=\lambda B_x$$

$$I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$

And:

$$ \lambda=w^2$$

2) Model Superposition

$$ M\phi=\lambda^{-1}k\phi $$

finally,

$$ M(\varepsilon,\ddot{\xi},\phi)+k(\varepsilon,\dot{\xi},\phi)=F $$