User talk:Eml5526.s11.team5.Dahiya

Hello everybody, My name is Lokesh Dahiya

Solution
Free Body Diagram:



 Figure 2 

 Theory: 

In elastodynamic case, there are 2 major contributors that affect body's motion:


 * Static Forces
 * Body's Inertial effects (resistance to motion caused by Static forces)

In accord to conservation laws as well as Newton's 3rd law, these 2 opposing contributors must be equal to each other.


 * {| style="width:100%" border="0"

$$F_{static} = F_{dynamic} \Rightarrow \vartriangle F = ma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.1)
 * }

where:


 * {| style="width:100%" border="0"

$$ F = force \quad (N) $$ $$ 
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ m = mass \quad (kg) $$ $$ 
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ a = acceleration \quad(\frac{m}{s^2}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

The assumed differential element shape is of trapezoid with force acting on both of its bases. Dynamic equation that resulted is shown below:


 * {| style="width:100%" border="0"

$$ -p(x) + f(x + \frac {\vartriangle x}{2})\vartriangle x + p(x+\vartriangle x) = ( \hat m(x) * \vartriangle x) a(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.2)
 * }

where:


 * {| style="width:100%" border="0"

$$ \hat m = mass \ per \ unit \ length \ (\frac{kg}{m}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }


 * {| style="width:100%" border="0"

$$ \frac {p(x+\vartriangle x) - p(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \frac {( \hat m(x) *\vartriangle x)}{\vartriangle x} a(t) = \hat m(x) a(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }


 * {| style="width:100%" border="0"

$$ \Rightarrow a(t) = \frac{\vartriangle v(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 1.3)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \vartriangle v(t) = \frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.4)
 * }

Therefore:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac {p(x+\vartriangle x) - p(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \hat m(x) \frac{1}{\vartriangle t}\frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.5)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \lim_{\vartriangle x \to 0} \frac {p(x+\vartriangle x) - p(x)}{\vartriangle x} + f(x + \frac {\vartriangle x}{2}) = \lim_{\vartriangle t \to 0} \hat m(x) \frac{1}{\vartriangle t}\frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \lim_{\vartriangle x \to 0} \frac {p(x+\vartriangle x) - p(x)}{\vartriangle x} = \frac {\partial p}{\partial x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \lim_{\vartriangle t \to 0} \frac{1}{\vartriangle t} = \frac{1}{\partial t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \lim_{\vartriangle t \to 0} \hat m \frac{u(t + \vartriangle t) - u(t)}{\vartriangle t} = \hat m(x) \frac {\partial u}{\partial t} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Therefore, substituting back into equation we get:

$$ \frac {\partial p}{\partial x} + f(x) = \hat m(x) \frac{1}{\partial t} \frac {\partial u}{\partial t} \Rightarrow \frac {\partial p}{\partial x} + f(x) = \hat m(x) \frac {\partial^2 u}{\partial t^2} $$ $$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 1.6)

Force = Stress * Area,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \begin{matrix} p(x) = \sigma(x) * A(x) \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.7)
 * }

Stress = Strain * Modulus of Elasticity,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \begin{matrix} \sigma(x) = \epsilon(x) * E(x) \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.8)
 * }

where

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \epsilon(x) = \frac {\partial u}{\partial x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.9)
 * }

From Equations 1.7-1.9:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ p(x) = E(x) * A(x) * \frac {\partial u}{\partial x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.10)
 * }

Mass / Unit Length = Density * Area

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \hat m(x) = \rho(x) * A(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 1.11)
 * }

Substituting the above relations we get

$$ \frac {\partial}{\partial x} \left[E(x)A(x)\frac{\partial u}{\partial x}\right] + f(x) = \rho(x) A(x) \frac {\partial^2 u} {\partial t^2} $$ $$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 1.12)

b) If we consider a case in which the bar has a rectangular cross section



 Figure 3 

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \begin{matrix} A(x) = h(x) * b \end{matrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

where

<span id="(1)">
 * {| style="width:100%" border="0"

$$ h(x) = height \ of \ bar \ at \ x \ (m), $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ b = width \ of \ the \ bar \ (constant)\ (m) \ [i.e. \ Independent \ of \ x,t] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Replacing $$ A(x) $$ with $$ h(x)*b $$ in our equation we get:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac {\partial}{\partial x} \left[E(x)h(x)*b*\frac{\partial u}{\partial x}\right] + f(x) = \rho(x) h(x)*b* \frac {\partial^2 u} {\partial t^2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

$$ b $$ is independent and can be brought outside the partial derivative of the first term

The equation is divided by $$ b $$ to give

$$ \frac {\partial}{\partial x} \left[E(x)h(x)\frac{\partial u}{\partial x}\right] + \frac{f(x)}{b} = \rho(x) h(x) \frac {\partial^2 u} {\partial t^2} $$ $$ $$
 * <p style="text-align:right"> $$ \displaystyle
 * <p style="text-align:right;"> $$ \displaystyle (Eq. 1.13)

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MaintenanceBot (discuss • contribs) 03:46, 11 November 2013 (UTC)