Beam Deformation Load

[Stl63]

If you have an A36 bar 1" wide and 1-1/2" deep. 29" long between a hinge and a pinned latch. What force is required to permanently deform the bar, in the strong axis, by 7/8" at the center? The load is applied by a point at center span.

"Saving the World One Beam at a Time"

 

[JAE]

Isn't this basic beam theory? Delta = PL^3 / 48EI

 

[woodman88]

Delta = PL^3 / 48EI is the elastic deflection of a steel beam with a point load in the center.
For permanent deformation of the steel beam, the beam must go into the plastic zone at the center. Once there the steel will elongate at the tension face relaxing the elastic sections at the ends so they won't go plastic. Now how much deflection do you apply to get enough plastic deformed steel at the center to resist the elastic ends and the elastic steel in the center of the beam at the plastic section to have a certain amount of bending in the unloaded steel beam?
Unfortunately, I have neither the time or the data to work out the problem. I am sure that somebody somewhere has done it.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.

 

[paddingtongreen]

Why didn't you do a web search?

Delta plus 7/8". It should spring back by delta when the load is removed.

I found a picture for you, page 2:

http://www.me.mtu.edu/~mavable/Book/Chap3.pdf

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[msquared48]

The key word in the question was "permanent". Woodman has it.

Mike McCann
MMC Engineering

 

[JAE]

Yes - I missed the "permanent" word as well. That does make for a more difficult solution.

 

[paddingtongreen]

Damn, I missed out specifying Delta. I meant to write:

Delta plus 7/8". Where Delta = PL^3 / 48EI. It should spring back by Delta when the load is removed leaving 7/8".

That didn't answer the question though, OP wanted the force.


Woodman isn't quite right, it does not have to go plastic all the way to the center. A thin layer will still be in the elastic range.

This cannot be calculated exactly without having the mill certs for the bar, yield stresses vary considerably from heat to heat.




Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[ToadJones]

You will not be able to calculate this accurately.
My guess is you are trying to manufacture some part.
You will likely need to do some test runs

 

[woodman88]

Sorry paddingtongreen, my part about "...and the elastic steel in the center of the beam at the plastic section..." was to imply that the center span section of the beam would have the lower (depth) part going plastic with the center and maybe upper (depth) part still in the elastic.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.

 

[rb1957]

i guess if you want to do this analytically you'll need NL FEA.

it's easy to calc the load to yield the beam's outer fiber; if you need precision, you'll need the yield of the bar (not of the material). Part of the trouble is i don't think this starts to create permanent deformation of the bar (ie the bar will return to straight after unloading at a (slightly?) higher load 'cause the material around the yielded material can strain to accomodate the plastic deformation).

if you really need to know, then testing is an easy solution.

 

[MiketheEngineer]

Most steels have a safety factor of about 1.5 to 2

My guess is that based on all the loads described above - you will need to probably double it!!! But as noted - testing will verify!!

 

[spats]

Everybody needs to think in terms of going back to the stress-strain curve for steel. The modulus of elasticity is the slope of the stress-strain curve. The modulus of elasticity is not constant for all stress ranges, and there are different stress at different points along the length of the bar,

Young's modulus is based on an idealized stress-strain diagram, and is not valid for stresses above yield, where you get into the calculation of tangent modulii. Think in terms of a bar with a varying modulus along most of it's length. Stresses would be in the elatics range only near the supports.

When all is said and done, I don't know how you would ever calculate such a problem. If you load steel to stresses beyond the proportional limit, then unload it, the response is essentially linearly elastic, parallel with the original slope line. When the load reaches zero, the permanent set can be measured along the strain axis. However, this tells you what's happening only at the point of maximum stress. Maybe somebody who remembers how integrate a curve could figure this out, but it's beyond me.

 

[trainguy]

It has to be done with NL FEA, and make sure to include material and geometric nonlinearity considering the length to depth ratio you are working with.

Or testing would be probably easier, as said above.

tg

 

[IDS]

Non-linear FEA is probably the easisest way for someone who has a suitable package available and knows how to drive it, but it wouldn't be too hard to do this from first principles with a spreadsheet:

- Generate a moment-curvature diagram based on the best available information for the steel stress-strain curve.

- Use this to generate a force displacement diagram, based on standard beam bending theory.

- On this diagram plot a line starting at 7/8 in deflection, parallel to the initial straight line part of the curve.

- Find the intersection point, that is the force for 7/8 in plastic deflection.

This doesn't take geometric non-linearity into consideration, but for the proportions giving, and assuming the bar isn't restrained longitudinally, I don't think that will be very significant.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[yakpol]

The force shall be large to yield the section, assuming the strain hardening curve is not ver steep:
Fy = PL/(bh^2/6) or P = Fybh^2/6L
For A36 realistic Fy=40-44 ksi
P = 44(1)1.5(1.5)/6/29 = 0.57 kips

Is it reasonable?

 

[BAretired]

It is really not a very difficult problem. The bending moment is known in terms of the applied load, i.e. PL/4. It is simply a matter of calculating how much load is required to produce plasticity in the beam which results in a total permanent deflection of 7/8". It could be easily done iteratively.

BA

 

[rb1957]

i thought it'd become difficult when you start yielding a portion of the beam, which'll change it's stiffness, etc ...

and calc'ing deflection under load is also easy, calc'ing the permanent deflection (without load) seems to me to be quite a bit more work.

how do you calc the perment kink/deflection in a beam ? your beam has some "small" permanent strain, partially reacted by elastic strains ...

 

[IDS]

rb1957 - Calculate the total deflection and subtract the elastic deflection to get the plastic deflection. To get the load for a specific plastic deflection you can do it graphically as I suggested, or iteratively as BA suggested.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[MiketheEngineer]

Every thing every body says makes sense - but you won't know till you test it.

I have seen many beams, plates, etc sold as A36 but were really A50 when tested. And many engineers will say that the change in Fb shouldn't make a difference - but it does!!

Trying to predict these things with any hint of accuracy is difficult!!

 

[paddingtongreen]

"Trying to predict these things with any hint of accuracy is difficult!!"

Have you ever tried to explain to a manager why we have difficulty designing something to fail without having the actual mill certs and exact dimensions?

Michael.
Timing has a lot to do with the outcome of a rain dance.