User:EGM6341.s11.TEAM1.WILKS/Mtg25

=EGM6321 - Principles of Engineering Analysis 1, Fall 2010=

[[media: 2010_10_14_14_56_10.djvu | Mtg 25:]] Thu, 14 Oct 10

= Euler-Bernoulli beam (cont'd) =

[[media: 2010_10_14_14_56_10.djvu | Page 25-1]]
NOTE: Finite Element (FE) discretization (continued)

[[media: 2010_10_14_14_02_27.djvu | Eq.(2)p.24-2]] : $$ \mathbf{M} \mathbf{\ddot{d}} + \mathbf{K} \mathbf{d} = \mathbf{F} \ $$

the origin of each term is given below:

$$ \mathbf{M} \mathbf{\ddot{d}} \leftrightarrow m \frac{\partial ^2u}{\partial t^2} \ $$

$$ \mathbf{K} \mathbf{d} \leftrightarrow +EI \frac{\partial ^4u}{\partial x^4} \ $$ where the positive sign in front of $$ EI \ $$ is changed twice due to 2 integration by parts

$$ \mathbf{F} \leftrightarrow f \ $$

END NOTE

Free vibration
Back to free vibration: [[media: 2010_10_14_14_02_27.djvu | Eq.(1)p.24-3]]

Separation of variables; consider the trial solution of the form

Assume:

where $$ i^2=-1 \ $$.

is an L4-ODE-CC.

Definition:

Leads to

Method of trial solution
Method of trial solution (undetermined coefficient) [http://uf.catalog.fcla.edu/uf.jsp?Ntt=king+differential+equations&I=1&Submit=Find&N=20&S=0781249404236215&Ntk=Keyword&V=D&Nty=1#top Ref:K. p513 Sec A5.3]

Assume

where r is an undetermined coefficient; r=root of a characteristic equation.

Use Eq.(1) in Eq.(6) p.25-1:

$$ K^4r^4=1 \Rightarrow \ r^4=\frac{1}{K^4} \ $$

$$ r_{1,2} = \pm \ \frac{1}{K} \ $$

$$ r_{3,4} = \pm \ \frac{i}{K} \ $$, where $$ i= \sqrt{-1} \ $$

HW  (4.3): Find $$ X(x) \ $$ in terms $$ \sin , \cos  ,\sinh  , \cosh   \ $$ where $$ e^{i \theta\ } = \cos \theta\ + i \sin \theta\ \ $$ similar relationship for $$ \sinh \theta\ \ $$ and $$ \cosh \theta\ \ $$ END HW

= Generalization: Euler equation, special Ln-ODE-VC =

Generalization: Euler equation; special Homogeneous Ln-ODE-VC and $$ \equiv \ \ $$ Euler Ln-ODE-VC

[[media: 2010_10_14_14_56_10.djvu | Page 25-3]]
where $$ y'=y^{(1)} \ $$ and $$ y=y^{(0)} \ $$, and the ODE is homogeneous because its r.h.s. is equal to 0.

Euler equation with constant coefficient

is an Euler Ln-ODE-CC (which includes Eq.(6)p.25-1 as particular case).

Two methods for solving Eq.(2).

Method 1
Method 1:

Stage 1: Transformation of variables

$$ x=e^t \Rightarrow \ \ $$ Ln-ODE-CC

Stage 2: Trial Solution $$ e^{rt} \ $$

Method 2
Method 2: Trial solution $$ x^r \ $$

Application: Homogeneous L2-ODE-VC (Euler L2-ODE-VC)
Method 1:

Stage 1: Let

[[media: 2010_10_14_14_56_10.djvu | Page 25-4]]
current variable $$ x \rightarrow \ \ $$ new variable $$ t  \ $$

$$ y(x(t)) \Rightarrow \ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} \ $$

Using

$$ \frac{dy}{dt} =: y_t \ $$

$$ \frac{dy}{dx} =: y_x \ $$

and

$$ \frac{dx}{dt} = e^t \ $$

we obtain

Another way:

$$ y_x= \frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx} = \left ( \frac{dt}{dx} \frac{d}{dt} \right ) y \ $$

where $$ \left ( \frac{dt}{dx} \frac{d}{dt} \right ) = \frac{d}{dx} \ $$

$$ \frac{dy}{dx} =: y_x \ $$

$$ \frac{dy}{dt} =: y_t \ $$

$$ \frac{dt}{dx} = \left ( \frac{dx}{dt} \right )^{-1} \ $$

Then $$ y_x = y_t \underbrace{\left ( \frac{dx}{dt} \right )^{-1}}_{\displaystyle =e^{-t}} \ $$

Hence:

$$ y_{xx}= \frac{d^2y}{dx^2} = \frac{d}{dx} \left ( \frac{d}{dx}y \right ) = \frac{dt}{dx} \frac{d}{dt} \left ( \frac{dt}{dx}\frac{d}{dt}y \right ) \ $$

with

$$ \frac{dt}{dx} = e^{-t} \ $$

and

$$ \frac{d}{dt}y =: y_t \ $$

and

$$ \frac{d}{dt} ( e^{-t} y_t ) = - e^{-t}y_t+e^{-t}y_{tt} \ $$

we obtain: