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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi bobbym and BeamReacher,

Firstly, I agree with MrWhy. There are too many variables to correctly determine the nature of the curve. But let's assume it is part of a circle.

In my diagram, I've converted the distances to feet.

The following equations apply:

From these last two

I used trial and improvement to find a = 0.03370775.... radians.

This gives the radius as

From the top two equations

Using the value of R and the quadratic formula

d = 44.498509.....

(My earlier answer of 63 was because I had used 2641 rather than 2640.5 for the arc length.)

Bob

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,309

Hi;

For the sake of solving the problem we are forced to assume that the steel bar is a hypothetical one whose curve we know.

That is close to the answer for the problem in the book. But the OP's problem is different. You will have to recompute for it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

An arc of a circle is into what shape a rod gets bent if the bending moment (and cross section) are the SAME all along its length - which they can never be if the forces are only at the ends!.

*Last edited by Mrwhy (2013-04-21 06:05:46)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

If the compression is one inch then the revised figures are

angle = 0.0097313... radians

R = 271293.83203...

d = 12.84541736

LATER EDIT: I need to modify this calc. It is slightly wrong. Only a few microns out I hope. Back soon.

EVEN LATER EDIT.

Forman's problem is not the same as BeamReacher's. He adds a foot. The Op here has the beam pushed in by an inch. So the arc length is 1 mile; the straight line distance is 1 mile less one inch. The diagram below shows the re-worked version. As I suspected the difference wouldn't come to much and it doesn't. The angle is unchanged.

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

Bob, here is something you might have fun finding out for yourself.

Your bent bar is still a mile long: it has been bent, not compressed.

(Well to bend it we compressed it one side and stretched it the other)

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

Thanks much! An answer I can understand. Also gratified to see that my approximation of 14.5' was not THAT far off (I did say it would be "a little higher" than the actual value).

I think I previously laid out the problem in a similar way, but got stuck on the "trial and improvement" step.

Glad I can close out this issue which has been left unfinished since college. Now, to find a ghost writer for my novel ... :-)

Beam

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

An interesting thing about this question:-

If you pull the ends of a string bent into a circle it will stretch and go straight.

But if you PUSH the ends of a piece of string you have NO IDEA what shape you will get.

It is amazing how the sort of maths we are taught at school says "everything is linear and reversible" whereas we all know that is nonsense (try mending a broken window by "reversing the forces".

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