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 Yo{ Y }_{ o } at the one end and 2Yo2{ 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 ΔL\Delta L .

Thank's !

Note by Karan Shekhawat
4 years, 7 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 = 0LdxY0A(1+xL)=LAY0ln(2)sok=AYLln(2)andenergykx22=AY2Lln(2)Δl2\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

Mvs Saketh - 4 years, 7 months ago

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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 !

Deepanshu Gupta - 4 years, 7 months ago

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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

xi=Δlakixi=constantforallibhenceusingaandbsok1x1(1ki)=Δl=T(1ki)=ΔlnowForce=Tensionatfreeendkeq=TΔl=11ki=(incontinuouslimit)=1dxAYo(1+xL)=AY0Lln(2)\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.

Mvs Saketh - 4 years, 7 months ago

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

Deepanshu Gupta - 4 years, 7 months ago

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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. !

Karan Shekhawat - 4 years, 7 months ago

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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 :)

Mvs Saketh - 4 years, 7 months ago

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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+xL+x
:Consider an small element of width dydy at the distance 'y' from.

Let at t=t+dt

:Length of rod becomes L+x+dxL+x+dx :extension in element is dzdz . 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 : wext=0ΔLF.dx=(AYo(1+yL+x)dzdy)dx\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.

Deepanshu Gupta - 4 years, 7 months ago

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

Deepanshu Gupta - 4 years, 7 months ago

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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 ?

Karan Shekhawat - 4 years, 7 months ago

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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.

Deepanshu Gupta - 4 years, 7 months ago

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Here's how's I did it.

Let the elongation be performed by a force FF.

Let at a distance xx from the clamped end we take a element dxdx of rod.

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

Let elongation of that element be dy dy

Hence we have :

FA=dydxY0(1+xL) \dfrac { F }{ A } =\dfrac { dy }{ dx } { Y }_{ 0 }(1+\dfrac { x }{ L } )

Integrating it out we have :

F=Y0AΔlLln(2) F=\dfrac { { Y }_{ 0 }A\Delta l }{ Lln(2) }

Now we find the energy contained in the rod. Again we take at a distance xx a dxdx element of rod is taken :

dE=12×Stress×Strain×dV dE = \dfrac{1}{2} \times Stress \times Strain \times dV

Also strain=StressY0,dV=Adx,Stress=FAstrain =\dfrac{Stress}{{Y}_{0}} , dV=Adx , Stress=\dfrac{F}{A}

dE=F22AY0dx1+x/L dE = \dfrac { { F }^{ 2 } }{ 2A{ Y }_{ 0 } } \dfrac { dx }{ 1+x/L }

E=F2Lln(2)2AY0 E = \dfrac { { F }^{ 2 }Lln(2) }{ 2A{ Y }_{ 0 } }

Put the value of FF to get :

E=Y0AΔl22Lln(2) E=\dfrac { { Y }_{ 0 }A\Delta { l }^{ 2 } }{ 2Lln(2) }

Ronak Agarwal - 4 years, 7 months ago

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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 ?

Karan Shekhawat - 4 years, 7 months ago

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will you Please tell your Procedure . Thank's !

Deepanshu Gupta - 4 years, 7 months ago

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Well, since it varies linearly, young's modulus at distance xx from free end can be written as Y(x)=Y0+xLY0Y(x)=Y_0+\dfrac{x}{L}Y_0

Now take a small element dxdx at a distance xx. Since the rod is extended by ΔL\Delta L, this element will be extended by ΔLLdx\dfrac{\Delta L}{L}dx. Calculate work done for this element and then integrate with lower limit 00 and upper limit LL.

Pranjal Jain - 4 years, 7 months ago

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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 ?

Karan Shekhawat - 4 years, 7 months ago

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Oh! Yeah! You are correct. It may be non linear. Well, it won't be linear. Use Y=FLAΔLY=\dfrac{FL}{A\Delta L} to calculate ΔL\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=stress2YStress\ Energy \ per\ unit\ volume=\dfrac{stress^2}{Y}?

Pranjal Jain - 4 years, 7 months ago

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@Pranjal Jain I mean since Process is done slowly so:

work done = change in elastic potential energy !

wext=u.dvu=energydensityu=12×stress×strain{ w }_{ ext }=\int { u.dv } \\ u=energy\quad density\\ u=\cfrac { 1 }{ 2 } \times stress\times strain

Karan Shekhawat - 4 years, 7 months ago

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d(ΔL)=FY(x)Adxd(\Delta L) = \frac{F}{Y(x)A}dx
d(ΔL)d(\Delta L) the elongation in the infinitesimal element dxdx due to the applied force FF at the area AA of the infinitesimal element with Young Modulus Y(x)Y(x)
ΔL=LFA1Yo(1+xL)dx\Delta L = \int{L} \frac{F}{A}\frac{1}{Y_o(1+\frac{x}{L})}dx

Ahmed Aljayashi - 7 months ago

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