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**BeamReacher****Member**- Registered: 2013-04-17
- Posts: 6

Hi,

A friend on the fencing team posed this problem to me, 30 years ago; I've considered it off and on since then. I have an APPROXIMATE answer, but I've never been able to solve it rigorously.

Say you had a perfectly straight steel bar a mile long. You push in on one side by one inch, so it bows in the middle. Would the resulting gap at the center of the bar be high enough to ...

- slide a penny under it?

- roll a baseball under it?

- drive a truck under it?

I answered "all of the above," and my friend said I was correct. Then came the sticking point -- PROVE IT!

Since then, I've tried a variety of approaches on my own; made great circles in which the bar was an arc and the ground was a chord; assume it's a parabola and try to solve for H at the midpoint (but that leads to a complex expression using hyperbolic sin to get the length of the parabola). The best I can do is to break the bar into two straight line segments, and solve for the height of the isosceles triangle - hypotenuse 2640 feet, one side 2639.96 feet:

2640^2 - 2639.96^2 = H^2

211.2 ~= H^2

H ~= 14.5

That's a little higher than the actual answer, since the peak of the triangle will be higher than of an arc of the same length. But not much.

How would you get a more precise answer?

Thanks!

Beam

50 year old Marathon runner whose heart goes out to the runners, families, and friends at the Boston Marathon.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi BeamReacher

Welcome to the forum.

I feel as though this ought to be an inverted catenary. See

[url]http://en.wikipedia.org/wiki/Catenary[url]May take a while to factor in the fixed length of the beam. Too late now. I'll have a go tomorrow.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Hi Bob

I do not think it is a catenary. I tried to set up the equations and it turns out that's not it.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Hi anonimnystefy;

You should know this one, it is a variation of a problem from Forman's book and the solution is there.

A friend on the fencing team posed this problem to me, 30 years ago;

Back in the old days it was a foot not an inch.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**BeamReacher****Member**- Registered: 2013-04-17
- Posts: 6

Thanks for the responses! Do you have a link to Forman's book?

Beam

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

You can google for it. The title is

Numerical methods That Usually Work by Forman S. Acton.

Mind you it is still tough because this is a problem in numerical accuracy disguised as a geometry problem. If your friend had the correct answer 30 years ago I salute him. With those old 6-8 digit calculators this one is a bear. Of course today we have much better tools.

Let me know if you find the book.

Also your estimate is not as close as you think.

**In mathematics, you don't understand things. You just get used to them.**

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**BeamReacher****Member**- Registered: 2013-04-17
- Posts: 6

Aha. I had always assumed it was one of those problems which seems complex on the outside but has a simple answer if you could just see it. Looks like it's just the opposite - seems simple but it ain't.

I last took numerical analysis in 1984; didn't expect to need it for this problem! But even in those days, we had access to some pretty decent computers; didn't have to rely on the Radio Shack calculator.

Thanks,

Jim

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Yes, I had one of them too. If I remember they had a single precision of about 6-7 and a double precision of about 15-16. But the trig functions were single precision. The amount of round off error on this problem could chew that machine right up.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Oh, yeah, it's the one from the exact beginning of the book. I haven't had enough time to go through the whole book yet, though.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Yep, that problem has a foot instead of an inch.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Okay. But, for some reason, I didn't think it would be a circle arc.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

I did not think about it. It really should be stated in the problem.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

I think there is only way a steel bar can be when bent like that, but I will leave that up to physicists.

Have you seen my answer in the other thread?

*Last edited by anonimnystefy (2013-04-19 05:46:21)*

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

The amazing thing is 44+ ft for a 1ft bar!

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

1 foot bar?

Have you seen my answer in the other thread?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Yes, the Forman problem is a 1 ft piece but it is easily modified for this problem.

What thread?

**In mathematics, you don't understand things. You just get used to them.**

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**BeamReacher****Member**- Registered: 2013-04-17
- Posts: 6

anonimnystefy wrote:

I think there is only way a steel bar can be when bent like that, but I will leave that up to physicians.

What do medical doctors know about steel bars?

An actual steel bar would most likely just distort or compress if you were able to apply sufficient pressure to one end to push it inward by a foot; or bend in the middle so it was on the ground anyway. Unless you did it in zero G.

Anonim, where do I look for your answer?

Beam

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Hi;

He meant a physicist. Although his English is excellent, English is not his primary language.

He also meant the solution to my rhombus problem in another thread.

Do you require the answer to this question? That I have on hand. If you need the working that will take some time to latex up, so please tell me now.

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi

bobbym wrote:

Numerical methods That Usually Work by Forman S. Acton.

This is available from Amazon. The problem is shown in the 'Look Inside' advert. The solution is not.

Forman has the curvature as an arc of a circle.

I think I can generate an answer if you ask me nicely.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**BeamReacher****Member**- Registered: 2013-04-17
- Posts: 6

Mr. Moderator, if it please you, I would be very grateful if you could show me how to solve this problem.

(Otherwise, I guess I could buy/borrow the book.)

Merci beaucoup,

Beam

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

bobbym wrote:

Do you require the answer to this question? That I have on hand. If you need the working that will take some time to latex up, so please tell me now.

I guess my solution is not necessary, okay.

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,536

hi bobbym,

Sorry, I missed the bit where you had the answer. You're saying 44 ft ?

At the moment I'm getting 63 ft, so I'm thinking there's a flaw in my working. Difficult to get theta.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

This question seems to be about bending of a bar.

But such a bar will bend under its own weight if horizontal or buckle if vertical

There is no limit to the ways it could bend depending upon

Which DIRECTION the one inch is - sideways or lengthwise.

HOW the bar ends are held - clamped angle or free to pivot

How easily the steel of the bar compresses under axial force.

The bending moment at any position x of the bar is the sum of the axial forces times the y deflection at that point. In this we should add any forces due to the weight of the bar.

A useful formula is the deflection of the bar at point x is

Bending moment M at point x is E I d2ydx2 , where E is Young's modulus and I is the moment of area of the beam's cross-section at position x.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,753

Hi Bob;

≈ 44.5 ft is for a 1 foot bar, this is for 1 inch so the method is the same but the answer is different.

Also there is the difference that one problem has 1 ft added to the bar while the other has the bar pushed in by an inch.

**In mathematics, you don't understand things. You just get used to them.**

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**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

If the bar were, by some immense stroke of luck, to remain straight the axial pressure to reduce its length by one inch would be a small fraction of a pound per square inch.

In practice it would buckle under gravity if vertical or sag under its own weight if horizintal, without requiring any axial pressure at all.

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