User:EML4500.f08.JAMAMA/Homework7

Meeting 35
Mtg. 5 $$ \Rightarrow $$ 2-bar truss
 * $$ \downarrow $$

Element 1: $$ E_1^{(1)} = 2;\ A_1^{(1)} = 0.5 $$
 * $$ E_2^{(1)} = 4;\ A_2^{(1)} = 1.5 $$

Element 2: $$ E_1^{(2)} = 3;\ A_1^{(2)} = 1 $$
 * $$ E_2^{(2)} = 7;\ A_2^{(2)} = 3 $$

HOMEWORK

NOTE, THE ELECTRIC PYTHON MATLAB CODE AND THE MODES ARE PRESENTED IN MEETING #6

Frame elem = truss(bar) elem + beam elem
 * $$ \uparrow \qquad \qquad \qquad \nwarrow $$
 * $$  {\color{Blue} \begin{matrix}

axial\ deformation & & transverse\ deformation \end{matrix}}$$

Model frame problem with 2 elements:



FBDs



In general $$,\ \underbrace{\mathbf{d_i^{(e)}}}\qquad \Rightarrow \qquad \underbrace{\mathbf{f_i^{(e)}}}$$
 * $$ {\color{blue} \begin{matrix}

generalized & & generalized \\ displacement & & force \end{matrix}}$$

e = 1,2

i = 1,...,6

$$ \begin{matrix}

d_3^{(e)}\\ d_6^{(e)} \end{matrix} \}\ contains\ dofs \Rightarrow \begin{matrix} f_3^{(e)}\\ f_6^{(e)} \end{matrix} \}\ bending\ forces $$

2-D Frame Global dofs



2 element stiffness matrix >> $$ \mathbf{k_{6 \times 6}^{(e)}}$$, e = 1,2

Global stiffness matrix >> $$ \mathbf{k_{9 \times 9}} = \mathbf{\underset{e=1}\overset{e=2}Ak_{6 \times 6}}$$



Meeting 36


$$\tilde{k}^{(e)}\tilde{d}^{(e)}=\tilde{f}^{(e)}$$

$$\tilde{d}^{(e)}=\left\{ \begin{align} & \tilde{d}_{1}^{(e)} \\ & \tilde{d}_{2}^{(e)} \\ & \tilde{d}_{3}^{(e)} \\ & \tilde{d}_{4}^{(e)} \\ & \tilde{d}_{5}^{(e)} \\ & \tilde{d}_{6}^{(e)} \\ \end{align} \right.$$



Note: The underlined displacements are distinguished because they are the bending moment displacements (rotations).

$$EI = $$Bending Stiffness

and

$$EA = $$Axial Stiffness

When you multiply the row by the column in this stiffness matrix, the dimensions for each now match each other. (dimensional analysis)

Dimensional Analysis:




 * Note that the brackets signify "dimension of", or in other words, the brackets signify that you are defining the dimensional parameter for that variable.



Meeting 37
$$\sigma =E\varepsilon \Rightarrow \left[ \sigma \right]=\left[ E \right]\left[ \varepsilon  \right]$$

$$\left[\epsilon \right]=\frac{\left[du \right]}{\left[dx \right]}=\frac{L}{L}=1$$

Note that $$\varepsilon $$ $$= 1$$ because dimensionally, the lengths cancel out (as shown in the previous equation);

Leaving no parameters (no parameters $$=$$ dimensionless $$= 1$$)

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

$$\left[A \right]=L^{2}$$, $$\left[I \right]=L^{4}$$

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

$$\left[\tilde{k}_{11} \tilde{d}_{11} \right]=\left[\tilde{k}_{11} \right]\left[\tilde{d}_{11} \right]=F$$

$$\left[\tilde{k}_{23} \tilde{d}_{23} \right]=\left[\tilde{k}_{23} \right]\left[\tilde{d}_{23} \right]=\frac{\left[6 \right]\left[E \right]\left[I \right]}{\left[L^{2} \right]}=\frac{1(\frac{F}{L^{2}})L^{4}}{L^{2}}=F$$

$$\left[\tilde{k}_{11} \tilde{d}_{11} \right]=-\left[\tilde{k}_{14} \tilde{d}_{14} \right]=-\left[\tilde{k}_{41} \tilde{d}_{41} \right]=\left[\tilde{k}_{44} \tilde{d}_{44} \right]=F$$

$$\left[\tilde{k}_{22} \tilde{d}_{22} \right]=-\left[\tilde{k}_{25} \tilde{d}_{25} \right]=-\left[\tilde{k}_{52} \tilde{d}_{52} \right]=\left[\tilde{k}_{55} \tilde{d}_{55} \right]=\frac{12EI}{L^{3}}=\frac{(\frac{F}{L^{2}})(L^{4})}{L^{3}}=\frac{F}{L}$$

$$\left[\tilde{k}_{33} \tilde{d}_{33} \right]=\left[\tilde{k}_{66} \tilde{d}_{66} \right]=\frac{4EI}{L^{3}}=\frac{(\frac{F}{L^{2}})(L^{4})}{L}=FL$$

$$\left[\tilde{k}_{36} \tilde{d}_{36} \right]=\left[\tilde{k}_{63} \tilde{d}_{63} \right]=\frac{2EI}{L^{3}}=\frac{(\frac{F}{L^{2}})(L^{4})}{L}=FL$$

Element force displacement relationship in global coordinates from element Force Displacement relationship in Local coordinates.

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

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

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

$$ \begin{bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0 \\ 0 & 0 & 0 & -m^{(e)} & l^{(e)} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{bmatrix} $$

Meeting 38
To get motivated lets consider the deformed shape of a truss element interp. of transverse displacement $$ V(s) with s = \tilde{x}$$.

Principal of Virtual Work (PVW): Eq. (1)

$$\int_{0}^{1}{w(x)}\left[-\frac{\partial{}^{2}}{\partial{x^{2}}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}} + f_{T} - m\ddot{v}\right]dx = 0$$ for all possible $$w(x) $$

For the first order term we will integrate by parts:

Let us recall the integration by parts method:

$$ r(x), s(x) $$

$$ (rs)' = r's + rs'$$

$$ r' = \frac{dr}{dx}  ;   s' = \frac{ds}{dx}$$

$$ \underbrace{\int {(rs)'}}_{(rs)}=\int{r's} + \int{rs'}$$

$$ =>\int{r's} = rs - \int{rs'}$$

Therefore if we let (α) = $$\int_{0}^{1}{\underbrace{w(x)}_{s(x)}}\frac{\partial{}^{2}}{\partial{x^{2}}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx $$

then we can write that $$\underbrace{\frac{\partial{}}{\partial{x}}\left(\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{r(x)}\right)}_{r^'(x)}$$

$$= \left[\underbrace{W\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{\beta^1}\right]_{0}^{L} - \int_{0}^{L}\underbrace{\frac{dw}{dx}}_{s'(x)}\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx}_{r(x)}$$

$$= \beta^{1} - \underbrace{\left[\frac{dw}{dx}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]_{0}^{L}}_{\beta^{2}} + \underbrace{\int_{0}^{L}\frac{d^{2}w}{dx^{2}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}dx}_{\gamma}$$ The symmetry should be noted!

Then Eq. (1) becomes:

$$- \beta^{1} + \beta^{2} - \gamma + \int_{0}^{L}{wf_{t}}dx - \int_{0}^{L}{wm\ddot{v}}dx = 0$$ for all possible $$w(x)$$

We will now focus on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.



$$V(\tilde{x}) = N_{2}(\tilde{x})\tilde{d}_{2} + N_{3}(\tilde{x})\tilde{d}_{3} + N_{5}(\tilde{x})\tilde{d}_{5} + N_{6}(\tilde{x})\tilde{d}_{6}$$

Let us recall: $$U(\tilde{x}) = N_{1}(\tilde{x})\tilde{d}_{1} + N_{4}(\tilde{x})\tilde{d}_{4}$$











From the book p.246, the following functions can be found:

$$N_{2}(\tilde{x}) = 1 - \frac{3\tilde{x}^{2}}{L^{2}} + \frac{2\tilde{x}^{3}}{L^{3}}$$ corresponding to $$ \tilde{d}_{2}$$

$$N_{3}(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^{2}}{L} + \frac{\tilde{x}^{3}}{L^{2}}$$ corresponding to $$ \tilde{d}_{3}$$

$$N_{5}(\tilde{x}) = \frac{3\tilde{x}^{2}}{L^{2}} - \frac{2\tilde{x}^{3}}{L^{3}}$$ corresponding to $$ \tilde{d}_{5}$$

$$N_{6}(\tilde{x}) = -\frac{\tilde{x}^{2}}{L} + \frac{\tilde{x}^{3}}{L^{2}}$$ corresponding to $$ \tilde{d}_{6}$$

Meeting 39


$$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$

6x1 6x6 6x1 (known after solving finite element system)

Compute $$u(\tilde{x}),v(\tilde{x})$$



$$u(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}$$

$$=U_{x}(\tilde{x})\vec{i}+U_{y}(\tilde{x})\vec{j}$$

Compute $$U(\tilde{x}), V(\tilde{x})$$ using eq.(1) and (2) from page 38-3.

*Compute $$U_{x}(\tilde{x}), V_{y}(\tilde{x})$$ from $$U(\tilde{x}), V(\tilde{x})$$

$$\begin{Bmatrix} U_{x}(\tilde{x})\\ U_{y}(\tilde{x}) \end{Bmatrix}=R^{T}\begin{Bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{Bmatrix}$$ Where R is from P 37-3 $$\begin{Bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{Bmatrix}=\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\   N_{1} & 0 & 0 & N_{4} & 0 & 0\\ 0& N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ .\\   .\\    .\\    .\\    \tilde{d}_{6}^{(e)} \end{Bmatrix}$$ $$N(\tilde{x})$$$$\tilde{d}^{(e)}$$ $$\begin{Bmatrix} U_{x}(\tilde{x})\\ U_{y}(\tilde{x}) \end{Bmatrix}=R^{T}N(\tilde{x})\tilde{T}^{(e)}d^{(e)}$$

Meeting 40
Note: This is the dimensional analysis of the follow from lecture 39

$$ \begin{Bmatrix} u \left ( \tilde{x} \right ) \\ v \left ( \tilde{x} \right ) \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \begin{Bmatrix} \tilde{d_1^e} \\ \vdots \\ \tilde{d_6^e} \end{Bmatrix} $$

From lecture 31:

$$ \left [ N_1 \right ] = \left [ N_4 \right ] = 1 $$

$$\left [ u \right ] = L \qquad$$ is shown by the following analysis

$$ \left [ N_1 \right ] \left [ \tilde{d_1} \right ] \oplus \left [ N_4 \right ] \left [ \tilde{d_4} \right ] $$

$$ \Downarrow \qquad \Downarrow \qquad \Downarrow \qquad \Downarrow $$

$$ 1 \qquad L \qquad 1 \qquad L $$

$$ \left [ N_2 \right ] \left [ \tilde{d_2} \right ] = L \qquad \big\} $$ the dimensions are equal to length (transverse displacement)

$$ \Downarrow \qquad \Downarrow$$

$$1 \qquad L$$

$$ \left [ N_3 \right ] \left [ \tilde{d_3} \right ] = L \qquad \big\}$$ the dimensions are equal to rotational displacement

$$ \Downarrow \qquad \Downarrow$$

$$ L \qquad 1$$

 Derivation of Beam Shape Functions  Refer to lecture 38 for plots of $$N_2, N_3, N_5, N_6$$

Recall Governing PDE for beams:

$$ - \frac{\partial ^2}{\partial x^2 \;} \left ( EI \frac{\partial ^2 v}{\partial x^2\;} \right ) + f_t \left ( x,t \right ) = m \left ( x \right ) \ddot{v} \qquad \Leftarrow \;$$ Transverse displacement (1)

$$ \frac{\partial ^2}{\partial x^2 \;} \left ( EI \frac{\partial ^2 u}{\partial x^2\;} \right ) + f_t \left ( x,t \right ) = m \left ( x \right ) \ddot{u} \qquad \Leftarrow \;$$ Axial displacement (2)

Now equation (1) without a distributed transverse load $$f_t \left ( x,t \right )$$ and without an inertial force $$ m \ddot{v}$$ (static cases) becomes:

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

Now consider EI to be constant:

$$ \frac{\partial ^4}{\partial x^\;}v=0$$ Then when you integrate four time you get four constants

$$ \Rightarrow \; v \left( x \right ) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 \Rightarrow \;$$ This is the equation used to obtain the equations for $$ N_2 \left ( x \right ), N_3 \left ( x \right ), N_5 \left ( x \right ), N_6 \left ( x \right )$$

The constants, and therefore the equation, for $$N_2 \left ( x \right )$$ are found by first determining the boundary conditions form the plots in lecture 38:

$$\begin{matrix} v \left ( 0 \right ) = 1 & v \left ( L \right ) = 0 \\ v^\prime \left ( 0 \right ) = 0 & v^\prime \left ( L \right ) = 0 \end{matrix}$$

$$v \left ( 0 \right ) = 1 = c_0$$

$$v^\prime \left ( 0 \right ) = 0 = c_1$$

$$v \left ( L \right ) = 0 = 1 + c_2 L^2 + c_3 L^3$$

$$ v^\prime \left ( L \right ) = 0 = 2 c_1 L + 3 c_3 L^3$$

There are two equations and two unknown. By solving for one constant in one equation then replacing it in the other, the following values are determined:

$$c_2 = - \frac{3}{L^2}$$

$$c_3 = \frac{2}{L^3}$$

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

Meeting 41
$$\tilde{k}_{23}=\frac{6EI}{L^{2}}=\int_{0}^{l}{\frac{d^{2N_{2}}}{dx^{2}}(EI)\frac{d^{2N_{3}}}{dx^{2}}dx}$$

In general,  $$\tilde{k}_{ij}=\frac{6EI}{L^{2}}=\int_{0}^{l}{\frac{d^{2N_{i}}}{dx^{2}}(EI)\frac{d^{2N_{j}}}{dx^{2}}dx}$$ i,j=2,3,5,6

Elastodynamics (trusses, frames, 2-D, 3-D, elasticity, etc.)

P. 31-1: Model pb

Discrete PVW: $$\mathbf{\bar{w}}\bullet [\bar{M}\bar{\ddot{d}}+\bar{K}\bar{d}-\bar{F}]=0$$ for all w

(boundary conditions already applied) EQ 1: $$\Rightarrow \bar{M}\bar{\ddot{d}}+\bar{K}\bar{d}=\bar{F}(t)$$ $$\bar{d}(0)=\bar{d_{0}}$$ $$\bar{\dot{d}}(0)=\bar{V_{0}}$$

Complete ordinary diff eqs (ODE's) (second order in time) + initial conditions governing the elastodynamic of discretized cont pb (MDOF)

Solving equation 1

1) Consider unforced vibration problem:

$$\bar{M}\ddot{v}+\bar{K}v=0$$

nxn nx1 nxn nx1  nx1 (unforced)

Assume $$v(t)=(sin\omega t)\phi $$

nx1 nx1

$$\phi $$ is not time dependent

$$\ddot{v}=-\omega ^{2}sin\omega t\phi $$ $$-\omega ^{2}sin\omega t\bar{M}\phi +sin\omega t\bar{K\phi} =0$$ $$\Rightarrow \bar{K}\phi =\omega ^{2}\bar{M}\phi $$

Generalized igenvalue problem: In general form

$$Ax=\lambda Bx$$ where lambda is the eigen value

Standard eval problem: $$Ax=\lambda x$$

(B=I,identity matrix) $$\begin{bmatrix} 1 & & 0\\  &  ..& \\  0&  & 1 \end{bmatrix}$$

$$\lambda =\omega ^{2} $$ Eigen Value

$$(\lambda _{i},\phi _{i})$$ Eigen pairs

i=1,....,n

mode $$i\Rightarrow V_{i}(t)=(sin\omega _{i}t)\phi _{i}$$ where i=1,...,n (animation)

2) Model Superposition method: Orthogonal prop of eigenpairs:

$$\phi _{i}^{T}\bar{M}\phi _{j}=\delta _{ij}=\begin{cases} 1 & \text{ if } i=j \\ 0 & \text{ if } i\neq j \end{cases}$$

1xn nxn  nx1 (Kronecker delta)

Mass orthog. of Eigen Vector

Eq(2) p 41-2: Eq(1) p 41-3:

$$\bar{M}\phi _{j}=\lambda _{j}\bar{K}\phi _{j}$$ $$ \phi _{i}^{T}\bar{M}\phi _{j}=\lambda _{j}\phi _{i}^{T}\bar{K}\phi _{j}$$ where $$ \phi _{i}^{T}\bar{M}\phi _{j}= \delta _{ij}$$ $$\Rightarrow \phi _{i}^{T}\bar{K}\phi _{j}=\frac{d}{\lambda _{j}}\delta _{ij}$$

Eq 1 (P 41-2): $$\bar{d}(t)=\sum_{i=1}^{n}{\xi _{i}(t)\phi _{i}}$$

nx1 1x1   nx1

$$\bar{M}\sum_{j}^{}{\ddot{\xi}_{j}\phi _{j} }+\bar{K}\sum_{j}^{}{\ddot{\xi}_{j}\phi _{j} }=F$$

$$\ddot{\bar{d}}$$      $$\bar{d}$$

$$\sum_{j}^{}{\ddot{\xi }(\phi _{i}^{T}\bar{M}\phi _{j})}+\sum_{j}^{}{\xi _{j}(\phi _{i}^{T}\bar{K}\phi _{j})}=\phi _{i}^{T}F$$

$$\delta \ddot{y}$$ $$x_{i}\delta _{ij}$$

$$\Rightarrow \ddot{\xi} _{i}+\lambda _{i}\xi _{i}=\phi _{i}^{T}F $$ where i=1,....,n

Comparing MediaWiki vs. E-learning: MediaWiki and E-learning are two very different programs. MediaWiki is a fully accessible collection of people's posts about a variety of topics. These posts can be on any topic, have many types of media within them, and the site is run by massive servers that can support millions of viewers at   a time. E-learning on the other hand is a Web Site owned and run by the University of Florida. You need a user name and password to access anything, and anything on it can only be viewed by students or teachers at the university. Because it is only supposed to be for students and faculty, the much smaller servers that E-learning runs on support fewer viewers at a time. For the type of assignments done in this class, MediaWiki is a better medium for the following reasons: You can type a web page! E-learning is a tool used by teachers to send assignments to students and post grades. Students CANNOT edit the pages, they can only turn in assignments, so to use it for this class would be pointless. That is not to say that MediaWiki is not without flaws though! One major problem is any outside viewer can edit, and CHANGE what students upload to the MediaWiki. Also, because it is   accessible by all users there is no way to control plagiarism from other people trying to do the same assignment! E-learning is much more widely used across campus by all faculty members, but this is because they are actually giving students assignments, not telling telling them to type notes. For typing notes MediaWiki is obviously the better choice.

Contributing Team Memebers
Arunas Janulevicius --Eml4500.f08.jamama.jan 14:32, 8 December 2008 (UTC)

Justin McIntire --Eml4500.f08.jamama.justin 00:09, 9 December 2008 (UTC)

Cedric Adam--Eml4500.f08.jamama.adam 08:00, 9 December 2008 (UTC)

William Mueller --Eml4500.f08.jamama.mueller 16:00, 9 December 2008 (UTC)

Chris Alford --Eml4500.f08.jamama.chris 17:38, 9 December 2008 (UTC)

Megan Alvarez --EML4500.f08.jamama.megan 21:14, 9 December 2008 (UTC)