User:Eml4500.f08.ateam.rivero/w7

Meeting 38: November 24,2008
Motivation: deformed shape of truss, interpertation of transverse displacements $$ v(\tilde x) $$ element.

Principle of Virtual Work: Beams $$ \int\limits_{0}^{L} w(\tilde x) \left \{ \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] + f_{t} - m \ddot v \right \} dx = 0 \qquad \qquad (1) \qquad$$ for all possible $$ w(\tilde x) $$

Integration by Parts of 1st Term: $$ \alpha = \int\limits_{0}^{L} w(\tilde x) \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] dx = 0 \qquad \qquad  \qquad$$ $$ \frac{\partial}{\partial x} \left \{\frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] \right \} $$ $$ = \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} - \int\limits_{0}^{L} \frac{\partial w}{\partial x} \frac{\partial}{\partial x} \left \{(EI) \frac{\partial^2 v}{\partial x^2}\right \} dx  $$ Note: Where $$ \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} $$ will be noted as $$ \beta_{1} $$ $$ \beta_{1} - \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} + \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx$$ Note: Where $$ \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} $$ will be noted as $$ \beta_{2} $$ and $$ \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx $$ will be noted as $$ \gamma $$ Note: Also notice the symmetry in the expression above Therefore Equation (1) becomes: $$ -\beta_{1} + \beta_{2} - \gamma + \int\limits_{0}^{L} wf_{t}dx-\int\limits_{0}^{L}wm \ddot v dx $$, for all possible $$ w(x) $$

Now let us focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.



$$ v(\tilde x) = N_{2}(\tilde x)\tilde d_{2} + N_{3}(\tilde x)\tilde d_{3} + N_{5}(\tilde x)\tilde d_{5} + N_{6}(\tilde x)\tilde d_{6} $$

Recall: $$ u(\tilde x) = N_{1}(\tilde x)\tilde d_{1} + N_{4}(\tilde x)\tilde d_{4} $$

Shape Functions: $$ N_{2}(\tilde x) = 1 - \frac{3\tilde x^3}{L^2} + \frac{2\tilde x^3}{L^3} => \tilde d_{2} $$ $$ N_{3}(\tilde x) = \tilde x - \frac{2\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{3} $$ $$ N_{5}(\tilde x) = \frac{3\tilde x^2}{L^2} - \frac{2\tilde x^3}{L^3} => \tilde d_{5} $$ $$ N_{6}(\tilde x) = \frac{-\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{6} $$

Meeting 41: December 8, 2008
$$ \tilde k_{23} = \frac {6EI}{L^2} = \int\limits_{0}^{L} \frac{d^2N_{2}}{dx^2}(EI) \frac{d^2N_{3}}{dx^2}dx$$

In general, $$ \tilde k_{ij} = \int\limits_{0}^{L} \frac{d^2N_{i}}{dx^2}(EI) \frac{d^2N_{j}}{dx^2}dx $$, where $$ i,j = 2,3,5,6 $$

Structural Dynamics
Elastodynamics (truss, frames, 2D & 3D elasticity)

Discrete Principle of Virtual Work:

Note: Bold Face implies that boundry conditins are already applied; Therefore the expression below is the reduced version.

$$ \mathbf w \cdot \left [\mathbf M \mathbf \ddot d + \mathbf K \mathbf d - \mathbf F \right ] = 0 $$, for all  $$  \mathbf w $$ Equation (1): $$ \mathbf M \mathbf \ddot d + \mathbf K \mathbf d = \mathbf F_{t} $$ $$ \mathbf d(0) = \mathbf d_{0} $$ $$ \mathbf \dot d_{0} = \mathbf v_{0} $$ Note: Complete ordinary differential equations and initial conditions governing the elastodynamics of the discretized problem.

Solving Equation (1):

1. Consider unforced vibrations problem

$$ \mathbf M \underline \ddot v + \mathbf K \underline v = \underline 0 $$

Assume:

$$ \underline v(t)_{nx1} = (sin(wt)) \phi_{nx1(not line dependent)} $$

Then:

$$ \underline \ddot v = w^2sin(wt) \underline \phi $$

Then:

$$ -w^2sin(wt) \mathbf M \phi + sin(wt) \mathbf K \phi = \underline 0 $$

Then giving us the generalized eigen value problem below:

$$ \mathbf K \phi = w^2 \mathbf M \phi $$

Which we can compare to the general form below:

$$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ \lambda $$ is the eigen value

Where in your standard eigen value problem $$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ ( \underline B = \underline I ) $$, where $$ \underline I $$ is your identity matrix shown below:

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

Where:

$$ \lambda = w^2 $$  eigen value

and,

$$ (\lambda_{i}, \underline \phi_{i}) $$  eigen pairs, where i = 1,. . ., n

where,

Mode $$ i => animation => \underline V_{i}(t) - (sin(w_{i}t))d_{i} $$

2. Model Superposition Method

Orthogonal problem of eigen pairs:

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 1 $$ if $$ i=j $$

or,

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 0 $$ if $$ i $$ does not equal $$ j $$

Mass orthogonal of eigenvalue.

$$ \mathbf M \phi_{j} = \lambda_{j}^{-1} \mathbf K \phi_{j} $$

$$ \phi_{i}^T \mathbf M \phi_{j} = \lambda_{j}^{-1} \phi_{j}^{-1} \mathbf K \phi_{j} $$

Giving us:

=> $$ \phi_{i}^T \mathbf K \phi_{j} = \lambda_{j} \delta_{ij} $$

$$ \mathbf d(t) = \sum_{i=1}^{n} \xi_{i}(t) \phi_{i} $$

$$ \mathbf M (\sum_{j} \xi_{j} \underline \phi_{j}) + \mathbf K(\sum_{j} \xi_{j} \underline \phi_{j}) = \underline F $$

Where,

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \ddot d $$

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \dot d $$

$$ \sum_{j} \xi_{j}(\phi_{i}^{T} \mathbf M \phi_{j}) + \sum_{j} \xi_{j} \phi_{i}^{T} \mathbf K \phi_{j} = \phi_{i}^{T}F $$

Finally Giving:

$$ => \ddot \xi_{i} + \lambda_{i} \xi_{i} = \phi_{i}^T \underline F $$, where $$ i=1,....,n $$