# Help:: Elasticity Problem

Recently 2 day's back I studied Elasticity Chapter , and while solving a problem , i get stuck in that . will You guy's please help ?

Problem : On a smooth horizontal surface a rod is clamped at one end and other end is free. Young's modulus of rod varies linearly with length of rod , has ${ Y }_{ o }$ at the one end and $2{ Y }_{ o }$ at the other end which is free. A variable Force is applied at the free end. Then Find the work done by external agent's in Slowly extending the rod by $\Delta L$ .

Thank's !

Note by Karan Shekhawat
6 years, 4 months ago

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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

- 6 years, 4 months ago

It's interesting ! Will You Please Tell what inspire you to take it as combination of spring's ? I mean rod and spring's are different thing's ? Also Can You Please Help me in Force Method !

- 6 years, 4 months ago

First note- Rod and spring are exactly identical, infact the truth is , rods are usually more elastic than springs, thats the whole point of hooks law

assuming process is slow,, meaning that no part of rod is accelerating, all parts can be thought to be in equillibrium and hence tension must be same everywhere,

so i can simply find the equivalent spring constant and proceed

Basically the difference between mine and your method is that i used the additional fact that tension must be constant to simplify calculation,

And why must it be constant, because the force is only at the edge and not on the whole rod, so any other rod element must face equal tension from either side or it will move,

@deepanshu

$\sum { { x }_{ i } } =\quad \Delta l---a\\ \\ { k }_{ i }{ x }_{ i }=\quad constant\quad for\quad all\quad i\quad --b\\ \\ hence\quad using\quad a\quad and\quad b\\ so\quad { k }_{ 1 }{ x }_{ 1 }(\sum { \frac { 1 }{ { k }_{ i } } } )=\Delta l\\ =T(\sum { \frac { 1 }{ { k }_{ i } } } )\quad =\quad \Delta l\\ now\\ \\ Force=\quad Tension\quad at\quad free\quad end\quad \\ \\ { k }_{ eq }=\frac { T }{ \Delta l } =\quad \frac { 1 }{ \sum { \frac { 1 }{ { k }_{ i } } } } =\quad (in\quad continuous\quad limit)\quad =\quad \frac { 1 }{ \int { \frac { dx }{ { AY }_{ o }(1+\frac { x }{ L } ) } } } =\quad \frac { A{ Y }_{ 0 } }{ L\quad ln(2) }$

Additionally a very importance difference in our approaches is that i am only worrying about how far the other end has moved since i am only worrying about the applied force and the Total corresponding restoring force which is only applied at the other end,,

Whereas you are trying to find the corresponding restoring force in different parts of rod and find the work done due to them which is no doubt complicated.

- 6 years, 4 months ago

Thanks For the Help and explaining ! Your approach is really Nice ! thank's for sharing with us .

- 6 years, 4 months ago

Man This is awesome ! Have You encountered Problem's by this method earlier ? and actually It was subjective question I found in a study material, So currently I don't have the answer . But I think Your Answer is correct. !

- 6 years, 4 months ago

well, not this problem specifically, but as long as we can linearly approximate force, we can quadratically approximate energy as kx^2/2, so use it :)

- 6 years, 4 months ago

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.

- 6 years, 4 months ago

I wan't to invite , @Mvs Saketh @Ronak Agarwal @Sudeep Salgia @Karthik Kannan for discussion. will you guy's please join us.

- 6 years, 4 months ago

Just Curious to know Have you studied: " Elasticity " chapter officially ? Since I studied it just before 2-3 day's . Also when did your course is finishing in your institute officially ?

- 6 years, 4 months ago

No , I have not studied Elasticity yet , But yes It will Start probably on Saturday , Currently I only Know Hook's Law in this chapter . And we had completed Math's and chemistry ( except surface chemistry ) course in my coaching. But two chapter's of 11th class Physic's are left ,which includes gravitation (currently running) and Elasticity.

- 6 years, 4 months ago

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 :

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

- 6 years, 4 months ago

how did you integrate ronak ? I'am not getting , will you please explain a bit more ! Thank's and How did you got constant force ? I mean it is given variable in question ?

- 6 years, 4 months ago

- 6 years, 4 months ago

- 6 years, 4 months ago

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 $L$.

- 6 years, 4 months ago

Thank's a ton for replying , but can you please explain little bit more about extension in element 'dx' ?
I mean how you take linear variation of of extension ? If I want to proceed as taking 'dy' extension in rod then how should I proceed ?
And Can I don't use Stress -stress energy relation per unit volume formula and then integrate ?

- 6 years, 4 months ago

Oh! Yeah! You are correct. It may be non linear. Well, it won't be linear. Use $Y=\dfrac{FL}{A\Delta L}$ to calculate $\Delta L$. Oh but its going dirty. Let me think. What's stress-stress energy per unit volume formula? You mean $Stress\ Energy \ per\ unit\ volume=\dfrac{stress^2}{Y}$?

- 6 years, 4 months ago

I mean since Process is done slowly so:

work done = change in elastic potential energy !

${ w }_{ ext }=\int { u.dv } \\ u=energy\quad density\\ u=\cfrac { 1 }{ 2 } \times stress\times strain$

- 6 years, 4 months ago

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

- 2 years, 4 months ago

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