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\includegraphics[width=10cm]{simiode-wordheader-Color}
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\begin{center}
{\Huge \bf Spring Design to Meet Specs at Minimum Cost}

\vspace*{.9 true cm}

Brian Winkel, Director@simiode.org

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SIMIODE, Cornwall NY USA
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\frametitle{Overview of Class}
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\centerline{\Large \bf Spring Cost}
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We put together a  model for determining  spring constant in a commercial spring to meet certain specifications at low costs.

\vspace*{.4 true cm}

\begin{enumerate}
\item Model single spring used in some application.
\vspace*{.1 true cm}\item Ascertain spring constant which gives overdamped solution.
\vspace*{.1 true cm}\item Determine spring constant and  minimum cost to meet specifications:
\begin{itemize}
\item [*] From same initial condition spring reaches given distance from static equilibrium in a given time.
\item [*] For each set of specifications (a) given distance from static equilibrium and (b)  given time render minimum cost.
\end{itemize}
\vspace*{.1 true cm}\item Discussions
\end{enumerate}

Source

\vspace*{.2 true cm}
{\bf \href{https://www.simiode.org/resources/7553}{3-031-SpringCost }} SIMIODE Modeling Scenario

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Now let us think of a vertically hanging spring with mass $m$  at the bottom of the  spring, whose displacement from its static equilibrium is measured by $y(t)$ in meters at time $t$ in seconds.

\begin{figure}
\centering
\includegraphics[width=3.5cm]{SpringMass.png}
\end{figure}


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Newton's Second Law of Motion says that the mass times the acceleration of that mass, $m\cdot y''(t)$,  is equal to the sum of all the external forces acting on the mass.



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Consider spring mass system below.


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\begin{figure}
\centering
\includegraphics[width=3.5cm]{SpringMass.png}
\end{figure}

\vspace*{-.9 true cm}
 Place the mass at static equilibrium so the force due to gravity is balanced by  force needed to pull the spring up from its  stretched state. The spring is  at rest.  
 \vspace*{.2 true cm}
 
 If we now further extend the spring and release it what are  the forces  acting on  mass?  We display them in a Free Body Diagram.

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{\bf Options for a Free Body Diagram}. All constants, $m$, $c$, and $k$ are positive and  $y'(t) >0$ and $y(t)>0$ stand  for downward velocity and position or expansion of the spring from static equilibrium. $y'(t) <0$ and $y(t)<0$ stand  for upward velocity and position or compression. Choose  correct Free Body Diagram!



\begin{figure}
\centering
\includegraphics[width=7cm]{SpringMassFBD.png}
\end{figure}


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From correct Free Body Diagram build a differential equation of the motion of the mass using Newton's Second Law of Motion.
\vspace*{.1 true cm}
\begin{figure}
\centering
\includegraphics[width=3.5cm]{SpringRight.jpg}
\end{figure}

\begin{equation*}
m\cdot y''(t) = \rule{3 true cm}{.02 cm}\, ?
\end{equation*}



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From correct Free Body Diagram build a differential equation of the motion of the mass using Newton's Second Law of Motion.
\vspace*{.1 true cm}
\begin{figure}
\centering
\includegraphics[width=3.5cm]{SpringRight.jpg}
\end{figure}

\begin{equation*}
m\cdot y''(t) = \rule{3 true cm}{.02 cm}\, ?
\end{equation*}



\begin{equation}\label{eq:springmassDE}
m\cdot y''(t) =  - c y'(t) - k y(t)
\end{equation}

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Now using Mathematica we see the  general solution of (\ref{eq:springmassDE}) with initial conditions, say $y(0) = 0.4$.
 and $y'(0) = 0$
 \vspace*{.2 true cm}
 
\begin{eqnarray}\label{eq:gensol}\nonumber
y(t) &=& \frac{0.2 c \exp \left(\frac{t \sqrt{c^2-4 k
   m}}{m}-\frac{t \left(\sqrt{c^2-4 k
   m}+c\right)}{2 m}\right)}{\sqrt{c^2-4. k
   m}}\\ \nonumber
   && +0.2 \exp \left(\frac{t \sqrt{c^2-4 k
   m}}{m}-\frac{t \left(\sqrt{c^2-4 k
   m}+c\right)}{2 m}\right)
   \\ \nonumber
   && -\frac{0.2 c
   e^{-\frac{t \left(\sqrt{c^2-4 k
   m}+c\right)}{2 m}}}{\sqrt{c^2-4. k m}}
   +0.2
   e^{-\frac{t \left(\sqrt{c^2-4 k
   m}+c\right)}{2 m}}\, .
   \end{eqnarray}
  \vspace*{.2 true cm}
    
   What common expression do we see in all terms?
   
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 What common expression do we see in all terms?
  

  $$\sqrt{c^2-4 k m} $$
   
   \large If $c^2-4 k m>0$  all exponential forms in our solution, e.g., 
   $$ e^{\left(\frac{t \sqrt{c^2-4 k
   m}}{m}-\frac{t \left(\sqrt{c^2-4 k
   m}+c\right)}{2 m}\right)}$$
   
   \noindent  exhibit exponentially decay and  solution  goes to zero.
 \vspace*{.4 true cm}
   
\noindent This means the spring displacement will return to 0, BUT with no oscillation for there would be no sine or cosine terms as in the case when $c^2-4 k m <0$.

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\vspace*{-.5 true cm}
We consider a spring with $m = 1.2$ kg, $c = 46.4$ N/(m/s).

\vspace*{.3 true cm}  $k$ N/m is  to be determined in order to meet certain specifications. Thus we have a model of our specific spring mass system as a differential equation with initial conditions (\ref{eq:springmassSpec}):
\vspace*{.1 true cm} 
\begin{equation}\label{eq:springmassSpec}
1.2 y''(t) +46.4 y'(t) + k y(t) = 0\,  , \,  \, y(0) = 0.4\, ,\,  y'(0) = 0\, .
\end{equation}


\noindent We have been tasked to determine spring values $k$ so that the spring meets certain specifications.

\vspace*{.2 true cm} Further, we have been advised that to produce a spring with spring constant $k$ there is a cost of  Cost$(k) = 13.70 + 0.01 k^2$.

\vspace*{.7 true cm} \hfill So what are these specification?
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We examine some cases for values of $k = 50, 100, 200, 400$. 
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All cases where  spring is overdamped, i.e.  $c^2-4 k m > 0$.

\begin{figure}
\hspace*{-.7 true cm}
\includegraphics[width=12cm]{FourKImages.png}
\end{figure}


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\begin{frame}
{\bf Specifications}

\vspace*{-.4 true cm}
We  seek a value of $k$ such that for the spring design  in (\ref{eq:springmassSpec})
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\setcounter{equation}{1}
\begin{equation}
1.2 y''(t) +46.4 y'(t) + k y(t) = 0\,  ,  y(0) = 0.4\,  , y'(0) = 0\,  ,
\end{equation}

\begin{enumerate}
\item spring is overdamped,
\item spring returns to within $a$ meters of  static equilibrium of 0,
\item spring does so within $b$ seconds from time $t = 0$.
\end{enumerate}

\vspace*{.4true cm}
 Further, we wish to know the cost of such a spring to meet these specifications for all combinations of $a$ and $b$ which are reasonable.

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For $a = 0.2$ m and $b = 0.5$ s draw a sketch of displacement of the spring  over time and explain in the sketch   the role of $a$ and $b$. 
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For this sketch there is no need to ascertain $k$, just  a sketch.

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\centerline{\Large Assignment}

\begin{enumerate}
\item Explain what it means for a spring  to return to within $a$ m of its static equilibrium of 0
in $b$ seconds from  $t = 0$.
\item Using  cost function,  Cost$(k) = 13.70 + 0.01 k^2$ find the cheapest spring with spring constant $k$   which returns to within $a = 0.1$ m of its static equilibrium of 0
in $b = 0.5$ seconds from time $t = 0$.

\item Provide others a tool through which they can determine, generally, the cost function for a spring  which is overdamped,
returns to within $a$ m of its static equilibrium of 0, and does so within $b$ seconds from time $t = 0$.

\item Provide  a representation of level curve pricing in which for a fixed cost, say \$600, all possible spring configurations (i.e.  returns to within $a$ m of its static equilibrium of 0  and does so within $b$ seconds from time $t = 0$) are displayed. 
\end{enumerate}

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\noindent {\bf CONGRATULATIONS!!!}
\vspace*{.3 true cm}

\noindent We went from
 
\begin{itemize}
\item rebuilding spring-mass model with differential equation, 
\item to solving   equation with spring constant $k$ as parameter, 
\item to determining  parameter values for overdamped solution,  
\item to using our model to determine spring constant which meets  specifications on distance from static equilibrium in set time, 
\item to determining cost of resulting spring, 
\item to custom designing spring for given specifications, 
\item to examining all spring configurations for fixed costs.
\end{itemize}
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