When Does a Beam Start to Behave Like a String Under Tension?

AI Thread Summary
A beam cannot behave like a string due to its ability to support both tension and compression, while a string only supports tension. The transition point where a beam starts to act like a string occurs when sufficient tension is applied to eliminate all compression within the beam. This requires calculating the force difference between the top and bottom of the beam based on its dimensions and loading conditions. The stress in a beam under axial tension and bending is complex and does not equate to simple string tension. Understanding this behavior involves examining combined axial and bending stresses in structural engineering literature.
Excom
Messages
58
Reaction score
0
Hallo

I have been thinking about when beam starts to behave like string.

My questions are: How much tension do you need to put on a beam before it starts to behave like a string? And when does the behaviour become purely string like?

Thanks
 
Physics news on Phys.org
A beam can never behave like a string.

A string can only support tension; it cannot support compression. This is fundamental.

It is also fundamental to beam action that there is both tension and compression.

So anything capable of acting as a beam can support compression and cannot be a string.
 
Excom said:
Hallo
I have been thinking about when beam starts to behave like string.

My questions are: How much tension do you need to put on a beam before it starts to behave like a string? And when does the behaviour become purely string like?

The beam acts as a beam when the upper half is under compression, and the bottom half is under tension.

If you take a long enough section of beam, and you can apply enough tension, then there will be a tension level where no part of the beam is under compression any more.

My hunch is that you would need to compute the force difference between top and bottom. Given a particular length of the beam, a particular mass per length unit, and a particular height of the beam you can compute how much force difference there must be between top and bottom of the beam.
Then you have the amount of tension that must be applied at the ends so that there is no longer any compression in the beam.
 
Cleonis you are quite correct that you can eliminate the compressions by 'prestressing'

The beam loading then becomes one of combined axial tension plus bending moment.

For a simply supported beam, breadth b and depth d, loaded at the most stressful section (the middle) by a point load Q and under an axial tension P the stress at any point A in the beam is given by

{S_A} = \frac{P}{{bd}} \pm \frac{{6M}}{{b{d^2}}} = \frac{1}{{bd}}\left\{ {\frac{P}{1} \pm \frac{{6Qx}}{{2d}}} \right\}

We can certainly calculate the point at which the prestress just eliminates the compression, but since the moment varies with distance, x along the beam, from zero at the support to a max at the load point, this can only occur at one section at a time.

Equally the equation is nothing like the simple tension in a string. So I will leave it up to otheres to decide if this is 'string action' or not.
 
Thanks for the answers.

Do any of you know some literature about the subject?
 
Look up combined axial and bending stresses

Here for instance

http://shjwc.sau.edu.cn/jpk/cllx/50/4/Chapter%208-Combined%20Bending%20Stresses.doc

Note download the target and rename to docx.

read paragraph 8.2
 
Last edited by a moderator:
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Hello! I am generating electrons from a 3D gaussian source. The electrons all have the same energy, but the direction is isotropic. The electron source is in between 2 plates that act as a capacitor, and one of them acts as a time of flight (tof) detector. I know the voltage on the plates very well, and I want to extract the center of the gaussian distribution (in one direction only), by measuring the tof of many electrons. So the uncertainty on the position is given by the tof uncertainty...
Back
Top