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Topic review (newest first)

Mrwhy
2013-04-24 03:50:10

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".

BeamReacher
2013-04-24 01:10:49

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

Mrwhy
2013-04-22 04:36:14

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)

bob bundy
2013-04-22 03:51:14

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

Mrwhy
2013-04-22 03:29:18

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!.

bobbym
2013-04-21 23:28:51

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.

bob bundy
2013-04-21 23:19:43

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

Mrwhy
2013-04-20 22:22:05

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.

bobbym
2013-04-20 19:09:49

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.

Mrwhy
2013-04-20 19:07:31

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.

bob bundy
2013-04-20 08:04:59

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

bobbym
2013-04-20 07:09:18

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.

BeamReacher
2013-04-20 06:36:44

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

bob bundy
2013-04-20 05:40:11

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

bobbym
2013-04-20 03:17:08

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.