Help:: Elasticity Problem

Treat the rod as composed of several several springs in series each having a spring constant of

k = YA/dl (where dl is not elongation but the infetismal spring length)

now , since all in series so equivalent spring constant is the reciprocal of the sum of reciprocals

or

1/ k = $\int _{ 0 }^{ L }{ \frac { dx }{ { Y }_{ 0 }A(1+\frac { x }{ L } ) } } \quad =\quad \frac { L }{ { { AY } }_{ 0 } } ln(2)\quad \\ \\ \\ so\quad k\quad =\quad \frac { AY }{ L\quad ln\quad (2) } \\ \\ and\quad energy\\ \\ \frac { k{ x }^{ 2 } }{ 2 } =\quad \frac { AY }{ 2L\quad ln(2) } { \Delta l }^{ 2 }$

is my solution correct , @KARAN SHEKHAWAT

This really interesting situation , and quite difficult one !
Okay Let's I give a try : Let at t=t

: length of rod become's $L+x$
:Consider an small element of width $dy$ at the distance 'y' from.

Let at t=t+dt

:Length of rod becomes $L+x+dx$ :extension in element is $dz$ . Using Hook's law on small elemen't :

Here Come's quite confusing Point : "while extending young's modulus remain's follow linear relation ship ? " Okay I assume Yes , to made this question meaningfull , which is rasonable when extension by external agent is small

So : $\displaystyle{{ w }_{ ext }=\int _{ 0 }^{ \Delta L }{ F.dx } =\int { (A{ Y }_{ o }(1+\cfrac { y }{ L+x } )\cfrac { dz }{ dy } )dx } }$ From Now I'am unable to proceed . I think It need's triple Integration.

Here's how's I did it.

Let the elongation be performed by a force $F$.

Let at a distance $x$ from the clamped end we take a element $dx$ of rod.

Now stress remains uniform through out and it is : $\dfrac{F}{A}$

Let elongation of that element be $dy$

Hence we have :

$\dfrac { F }{ A } =\dfrac { dy }{ dx } { Y }_{ 0 }(1+\dfrac { x }{ L } )$

Integrating it out we have :

$F=\dfrac { { Y }_{ 0 }A\Delta l }{ Lln(2) }$

Now we find the energy contained in the rod. Again we take at a distance $x$ a $dx$ element of rod is taken :

$dE = \dfrac{1}{2} \times Stress \times Strain \times dV$

Also $strain =\dfrac{Stress}{{Y}_{0}} , dV=Adx , Stress=\dfrac{F}{A}$

$dE = \dfrac { { F }^{ 2 } }{ 2A{ Y }_{ 0 } } \dfrac { dx }{ 1+x/L }$

$E = \dfrac { { F }^{ 2 }Lln(2) }{ 2A{ Y }_{ 0 } }$

Put the value of $F$ to get :

$E=\dfrac { { Y }_{ 0 }A\Delta { l }^{ 2 } }{ 2Lln(2) }$

@Ronak Agarwal @Deepanshu Gupta @Mvs Saketh @Pratik Shastri @Azhaghu Roopesh M @Pranjal Jain @Sudeep Salgia @Anish Puthuraya @Krishna Sharma @Vraj Mehta @megh choksi and any other please help

Well, since it varies linearly, young's modulus at distance $x$ from free end can be written as $Y(x)=Y_0+\dfrac{x}{L}Y_0$

Now take a small element $dx$ at a distance $x$. Since the rod is extended by $\Delta L$, this element will be extended by $\dfrac{\Delta L}{L}dx$. Calculate work done for this element and then integrate with lower limit $0$ and upper limit $L$.

$d(\Delta L) = \frac{F}{Y(x)A}dx$
$d(\Delta L)$ the elongation in the infinitesimal element $dx$ due to the applied force $F$ at the area $A$ of the infinitesimal element with Young Modulus $Y(x)$
$\Delta L = \int{L} \frac{F}{A}\frac{1}{Y_o(1+\frac{x}{L})}dx$