User:Egm4313.s12.team2.williams.ccw

I am a brand new Wikiversity user that has created this account for my Intermediate Engineering Analysis class. This account will be used to share information between myself and members of my team, team number 2. I am looking forward to learning more advanced mathematical techniques. The reports contained in my Wikiversity page will hopefully benefit other engineering students.Egm4313.s12.team2.williams.ccw

About Me
My name is Corey Williams and I am a mechanical engineering student at the University of Florida.University of Florida home pageEgm4313.s12.team2.williams.ccw My hobbies include, being with my friends, playing guitar and singing, learning new things, anything dealing with soccer, and do-it-yourself projects. I am about to start my senior year in college. I earned my associates degree from Daytona State College in Daytona Beach, Florida, in 2009.--Egm4313.s12.team2.williams.ccw 23:20, 15 January 2012 (UTC) The link for the main page for team 2 can be found here:Team 2 --Egm4313.s12.team2.williams.ccw 02:41, 22 January 2012 (UTC)

Side Note

 * The wiki format is something that I am not used to using, especially not in a college classroom. Is this the new future of education? Is the vast, faceless internet destined to replace the face to face, personal transfer of knowledge that we have all been instinctually drawn to since the beginning of our species? Will education become a case of more quantity and less quality? It is up to educators and students alike to ensure that we use the internet responsibly in our lives, and in the context of our education.--Egm4313.s12.team2.williams.ccw 03:17, 22 January 2012 (UTC)

= R 1.1 =

Problem Statement

 * Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$f(t)$$.

Free Body Diagram
Fig. 1.
 * In Fig. 1. above, the spring (in green) is represented by K, the dashpot (in blue) is represented by C, the mass (in red) is represented by M, and the applied force (in black) is

represented by $$\bar{r}(t)$$.

Kinematics
$$\bar{r}=y\bar{E}_y$$ $$\bar{v}={y}'\bar{E}_y$$ $$\bar{a}=y''\bar{E}_y$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 1. are derived, first, from the position function of the mass with respect to time ($$r(t)$$). The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$r(t)$$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

Kinetics- Free Body Diagram
Fig. 2.

Using Newton's Second Law
$$F=m\bar{a}$$ Summing Forces $$\bar{F}_k+\bar{F}_c+m\bar{g}=r(t)$$ $$ky\bar{E}_y+cy'\bar{E}_y+my''\bar{E}_y=r(t)$$ --Corey Williams 20:42, 27 January 2012 (UTC) =R2.2=

Problem Statement
Given initial conditions $$ y(0) = 1 \ $$, $$ y'(0) = 0 \ $$ and excitation $$ r(x) = 0 \ $$ , find the solution to $$ y'' - 10y' + 25y = r(x) \ $$

Solution

 * $$ y'' - 10y' + 25y = r(x) \ $$
 * $$ \lambda^2 - 10\lambda + 25 = 0 \ $$
 * $$ (\lambda - 5)(\lambda - 5) = 0 \ $$
 * $$ \lambda = 5 \ $$

Double real root
 * $$ \lambda = -\frac 1 2 a $$
 * $$ y = (C_1 + C_2 x)e^{- \frac {ax} 2} \ $$
 * $$ y = C_1 e^{5x} + C_2 xe^{5x} \ $$

Apply initial conditions
 * $$ 1 = C_1 e^{5(0)} + C_2 (0)e^{5(0)} \ $$
 * $$ 1 = C_1 e \ $$
 * $$ C_1 = 1 \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 (1*e^{5x}) + x*5xe^{5x} \ $$
 * $$ y' = 5C_1 e^{5x} + C_2 e^{5x} + 5x^2 e^{5x} \ $$
 * $$ y' = 5(1) e^{5(0)x} + C_2 e^{5(0)} + 5(0)^2 e^{5(0)} \ $$
 * $$ 0 = 5+ C_2 + 0 \ $$
 * $$ C_2 = -5 \ $$
 * $$ y = (1) e^{5x} + (-5) xe^{5x} \ $$

$$ y = e^{5x}-5xe^{5x} \ $$ --Corey Williams 17:55, 6 February 2012 (UTC)

Matlab Code
x = 0:0.01:1; >> ft = exp(5*x) - 5*x.*exp(5*x); >> plot(x,ft)

Effort Distribution
--Corey Williams 15:58, 31 January 2012 (UTC)

Archived Link to Reports
|Archived Report 1 | Report 2