Elasticity Problem: Uniform Rod Stretched Under Its Own Weight

[Dr.Brain]

A Uniform rod is fixed from top and becomes stretched under its own weight.Then which of the following will be true on its elongation:

a) Diameter at the top is smaller than at the top
b) Diameter at the bottom is larger than that of top
c) Diameter is uniform throughout
d) Dimater becomes smaller than previous one uniformly

I know the answer but i want your views and the way you think...I tried to check it out with the forumla for Young's Modulus and Poisson's Ratio...But I think my description for the right answer wasnt right...so contribute please...

 

[HallsofIvy]

What? You know the answer but you won't tell us? Is this a test?

As the rod stretches vertically, it radius has to become smaller (conservation of mass, basically). Since it is stretching under its own weight, the bottom, subjected to relatively little weight, will only stretch a small amount compared with the top. The top has smaller diameter than the bottom.

 

[turin]

As the rod stretches vertically, it radius has to become smaller (conservation of mass, basically).

How is this conservation of mass? It sounds more like conservation of mass-density, but that doesn't sound like a law that I've heard. There's got to be another explanation.

 

[cookiemonster]

If it's stretching and the radius doesn't change, then there's more volume. If the density is constant, then that implies more mass. Stretching doesn't create mass, so the radius must decrease in order to suitably decrease the volume and maintain the mass.

cookiemonster

 

[turin]

If the density is constant, ...

OK, but that's what I don't get. Why assume this? (Other than the obvious, "because we observe the affect.")

 

[cookiemonster]

Because of the word "uniform" in the description of the rod. A uniform rod.

cookiemonster

 

[turin]

Because of the word "uniform" in the description of the rod.

I think that is a weak justification. The question can easily be interpretted as the rod starting out uniform. When it stretches, it either becomes non-uniform, or non-rod. Though, I guess that this does seem to be the best we can do with this problem. I really hate these first year physics questions.

 

[cookiemonster]

It's not how I would personally phrase the problem, that's for sure.

cookiemonster

 

[Dr.Brain]

I really hate these first year physics questions. : turin

Even I do...but none of you have been able to interpret the question or even try to get near to the answer...

"DIAMETER AT TOP IS SMALLER THAN THAT OF BOTTOM"

Can anyone explain me ..why?...

U have made fun of "Conservation Of Mass" .. You are applying This law within body itself? Wow! ..You people really innovative

The explanation to this problem has something to do with Young's Modulus or Poisson's Ratio as far as I am concerned. and remember it is extending under it own weight"...I am trying to use that fact...

 

[Doc Al]

Poisson's Ratio is positive

I know the answer but i want your views and the way you think...I tried to check it out with the forumla for Young's Modulus and Poisson's Ratio...But I think my description for the right answer wasnt right...so contribute please...

The tension in the rod is greater near the top (since it must support the weight of the section below), so the amount of axial stress (and thus strain) is greater towards the top. The lateral strain is related to the axial strain by Poisson's Ratio. Since most materials have a positive Poisson's Ratio, they shrink (get thinner) under axial stress. So it makes sense that the rod will be thinner at the top.

Now if you are asking why most materials have a positive Poisson's ratio: I haven't a clue.

 

[turin]

The lateral strain is related to the axial strain by Poisson's Ratio. Since most materials have a positive Poisson's Ratio, they shrink (get thinner) under axial stress.

Now we're getting somewhere! Excellent suggestion Doc Al! I will now proceed to investigate the meaning and canonical (first year?) applications of this "Poisson's Ratio."

 

[cookiemonster]

What's wrong with HallsofIvy's explanation again?

cookiemonster

 

[turin]

What's wrong with HallsofIvy's explanation again?

I have yet to look into this Poisson's Ratio, but I got the impression that it is not just a statement of conservation of mass.

As the rod stretches vertically, it radius has to become smaller (conservation of mass, basically).

...

 

[Dr.Brain]

The tension in the rod is greater near the top (since it must support the weight of the section below), so the amount of axial stress (and thus strain) is greater towards the top. The lateral strain is related to the axial strain by Poisson's Ratio. Since most materials have a positive Poisson's Ratio, they shrink (get thinner) under axial stress. So it makes sense that the rod will be thinner at the top.

Now if you are asking why most materials have a positive Poisson's ratio: I haven't a clue.

Poisson's Ratio varies between [-1] to [+0.5]

any say?

 

[Dr.Brain]

Doc Al

That was a good reply. But I am not even 10% satisified with that
cuz'

Poisson's Ratio= negative of fractional change in diameter/fractional change in longitudinal length

 

[Doc Al]

Poisson's Ratio varies between [-1] to [+0.5]

This is true. So?

 

[Doc Al]

That was a good reply. But I am not even 10% satisified with that
cuz'

Poisson's Ratio= negative of fractional change in diameter/fractional change in longitudinal length

This is also true. What is your point?

 

[turin]

OK, here's what I found:

1. Poisson's Ratio is not discussed in my old physics book (I don't think this is used any more, so it might not be relevant): Serway's "Physics: For Scientists and Engineers" 4th ed. Since this is the case, my investigation consisted of a terse internet search, resulting in information extracted and compiled from unqualified/unofficial sources. The justification for this is that I found 100% consistency out of the 10 pages I viewed that gave information of the subject.

2. Poisson's Ratio is as DocAl has described: it can certainly be used to relate the axial strain to the radial strain.

3. For a material with isotropic elastic properties, Poisson's Ratio must be positive. Since the rod is uniform, I think it is reasonable to apply this qualification to directional uniformity of the elastic properties. In other words, I think it is reasonable to assume that Poisson's Ratio is positive according to the wording of the problem.

4. Therefore, I conclude that the rod will be narrower at the top.

 

[cookiemonster]

I hate to say it, but that's not convincing at all, either...

We're still skirting around the "why" of the issue.

cookiemonster

 

[turin]

We're still skirting around the "why" of the issue.

Can you pose the "why" question, please? I will try to address it, because I think I am comfortable with it now, and your question would help me test that comfort zone.

 

[cookiemonster]

Building off your last post, the next question should be "Why must Poisson's Ratio be positive for materials with isotropic elastic properties?"

cookiemonster

 

[turin]

Why must Poisson's Ratio be positive for materials with isotropic elastic properties?

That's what I thought you meant.

If the elastic properties of a material are isotropic, then it would rather change shape than volume under an axial stress. This is seen most easily by a central force atomic model in which the configuration energy is minimized by decreasing the lateral separations of the atoms in the lattic to accommodate the increase in energy due to the increase in axial separation. I suppose this does assume something about the form of the central force. I'll have to think about that some more.

 

[Dr.Brain]

Ok ppl hear this:

consider an uniform rod as i have put it, Noiw when we stretch it , that is I apply a deforminmg force, No the rod will stretch vertically (if hanged vertically) as natural , and as it stretches vertically ..there has to be a change in its diameter.

Now Change in Diameter= Final diameter- Initial diameter

as final diameter will be smalleer than initial one...there change in diameter ios ALWAYS NEGATIVE ...

How do u explain that?

 

[Doc Al]

as final diameter will be smalleer than initial one...there change in diameter ios ALWAYS NEGATIVE ...

How do u explain that?

What do you think we've been talking about throughout this thread?

If your issue is: Why is Poisson's Ratio positive for typical isotropic materials? Turin is working on explaining that.

If your issue is: Why do you get a negative change in diameter when Poisson's ratio is positive? Realize that Poisson's ratio has a built in negative, so reductions in diameter mean a positive Poisson ratio.

 

[cookiemonster]

Yeah, I'm still not really satisfied with turin's reason, but I'm too lazy to go search for it on my own. So I just wait until he decides to come back.

cookiemonster

 

[turin]

Yeah, I'm still not really satisfied with turin's reason, but I'm too lazy to go search for it on my own. So I just wait until he decides to come back.

Sorry, that's all I have the motivation for. The explanation has satiated me, but I do admit there seems to be something missing. Sorry, I just don't feel the urge to investigate further; I have more interesting things on my plate, and by the time they're done, I'll have forgotten about this.