User:Egm4313.s12.team14.turano

From Wikiversity
Jump to: navigation, search

R2.1[edit]

Given the two roots and the initial conditions:

\displaystyle 
\lambda_1=-2, \lambda_2=+5

\displaystyle (Eq. 1a)

\displaystyle 
y(0)=1,y'(0)=0

\displaystyle (Eq. 1b)

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation \displaystyle r(x)

Consider no excitation:


\displaystyle 
r(x)=0

\displaystyle (Eq. 2)

Plot the solution.

Characteristic equation:

\displaystyle 
(\lambda-\lambda_1)(\lambda-\lambda_2)=0

\displaystyle (Eq. 3)


Substituting \displaystyle(Eq. 1a) into \displaystyle(Eq. 3):

\displaystyle 
(\lambda+2)(\lambda-5)=0

\displaystyle (Eq. 4)


\displaystyle 
\lambda^2-3\lambda-10=0

\displaystyle (Eq. 5)


Non-homogeneous solution:

\displaystyle 
y''-3y'-10y=0

\displaystyle (Eq. 6)


Homogeneous solution:

\displaystyle 
y_h(x)=C_1e^{-2x}+C_2e^{5x}

\displaystyle (Eq. 7)


Overall solution:

\displaystyle 
y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x)

\displaystyle (Eq. 8)


No excitation:


\displaystyle 
r(x)=0 => y_p(x)=0

\displaystyle (Eq. 9)


From intitial conditions:

\displaystyle 
y(0)=1=C_1+C_2

\displaystyle (Eq. 10)


\displaystyle 
y'(0)=0=-2C_1+5C_2

\displaystyle (Eq. 11)


Solving for \displaystyle C_1 and \displaystyle C_2 :

\displaystyle 
C_1=\frac{5}{7}

\displaystyle

and

\displaystyle 
C_2=\frac{2}{7}

\displaystyle


So, the final solution is:

\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} 


Team14R22diagram1.png

R2.2[edit]

\displaystyle 
 y'' - 10y' +25y = r(x)

\displaystyle (Eq. 1)

Initial conditions:

\displaystyle 
y(0) = 1, y'(0) = 0

\displaystyle (Eq. 2)


No excitation:

\displaystyle 
r(x) = 0

\displaystyle (Eq. 3)

Find and plot the solution for \displaystyle (Eq. 1)

Due to no excitation, \displaystyle (Eq. 1) becomes:

\displaystyle 
 y'' - 10y' +25y = 0

\displaystyle (Eq. 4)

Substituting \displaystyle \lambda into \displaystyle (Eq. 4) :

\displaystyle 
 \lambda^2-10\lambda+25=0

\displaystyle (Eq. 5)

Factoring, and solving for \displaystyle \lambda :

\displaystyle 
 (\lambda-5)^2=0

\displaystyle (Eq. 6)


\displaystyle 
 \lambda_2=\lambda_1=\lambda=5

\displaystyle

Since \displaystyle \lambda is a double root, the general solution:

\displaystyle 
y=C_1e^{5x}+C_2xe^{5x}

\displaystyle (Eq. 7)

From intitial conditions:

\displaystyle 
y(0)=1=C_1

\displaystyle (Eq. 10)


\displaystyle 
y'(0)=0=5C_1+C_2

\displaystyle (Eq. 11)


Solving for \displaystyle C_1 and \displaystyle C_2 :

\displaystyle 
C_1=1

\displaystyle

and

\displaystyle 
C_2=-5

\displaystyle


So, the final solution is:

\displaystyle y(x)=e^{5x}-5xe^{5x} 


Team14R22diagram2.png

R1.6G[edit]

G) Deformation of a Beam

\displaystyle{EIy''=f(x)}

\cdot This equation is of SECOND order and LINEAR.

\cdot To prove the applicability of Superposition, the following is done:

Homogeneous

\displaystyle{EIy_h''=0}

\displaystyle (Eq. 1)

Particular:

\displaystyle{EIy_p''=f(x)}

\displaystyle (Eq. 2)

Add (Eq. 1) and (Eq. 2):

\displaystyle{EIy_h''+EIy_p''=f(x}

It could be said that:

\displaystyle{y_{h}+y_{p}=Y}

and due to the fact that,

\displaystyle{EI(y_h''+y_p)=EIY}

It can be said that Y is a solution, confirming that superposition CAN be used.


R1.1[edit]

Team14P1R16Diagram1.png‎

File:Team14P1R11Diagram2.png
Free Body Diagram

y=y_k=y_c

Spring Force, \vec{F}_s=k(y)

Dashpot Force, \vec{F}_D=c(y')

Applied Force, \vec{F}(t)

\displaystyle 
\Sigma\vec{F}=\vec{F}(t)-\vec{F}_s-\vec{F}_D

\displaystyle (Eq. 1)


\displaystyle 
m\vec{a}=\vec{F}_i=my''

\displaystyle (Eq. 2)


From Newton's 2nd Law:

\displaystyle 
\Sigma\vec{F}=m\vec{a}

\displaystyle (Eq. 3)


Substituting  (Eq. 1) and  (Eq. 2) into  (Eq. 3):

\displaystyle 
\vec{F}(t)-\vec{F}_s-\vec{F}_D=\vec{F}_i

\displaystyle (Eq. 4)

Solving for \vec{F}(t):

\vec{F}(t)=\vec{F}_s+\vec{F}_D+\vec{F}_i

Substituting \vec{F}_s=ky, \vec{F}_D=cy', and \vec{F}_i=my'' into  (Eq. 4):

\vec{F}(t)=ky+Cy'+my''

Putting into standard form:

my''+Cy'+ky=\vec{F}(t)