Beam Deformation Load
Beam Deformation Load
(OP)
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"






RE: Beam Deformation Load
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
Mike McCann
MMC Engineering
RE: Beam Deformation Load
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
My guess is you are trying to manufacture some part.
You will likely need to do some test runs
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
My guess is that based on all the loads described above - you will need to probably double it!!! But as noted - testing will verify!!
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
Or testing would be probably easier, as said above.
tg
RE: Beam Deformation Load
- 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/
RE: Beam Deformation Load
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?
RE: Beam Deformation Load
BA
RE: Beam Deformation Load
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 ...
RE: Beam Deformation Load
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Beam Deformation Load
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!!
RE: Beam Deformation Load
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.
RE: Beam Deformation Load
You can't really extend theory & design procedures exactly into practice.
I agree with Spats.
Maybe the OP can tell us why he is doing this?
If it is for the manufacture of some bent bars, like I said, you're going to have to test it anyway. Some of this theory may help you size whatever mechanism you are using to bend the bar, but that's about it.
RE: Beam Deformation Load
True, but that applies to everything. The calculation isn't that hard, and given reasonably accurate material properties it will give reasonably accurate answers. I don't see people here suggesting that calculating reinforced concrete deflections is too hard to bother with, even though there can be just as much variability with that as with this case.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Beam Deformation Load
BA
RE: Beam Deformation Load
In its simplest form, the way the steel responds is seen in Paddington's stress-strain curve (his 13DEC post and attachment) which shows this for a tension specimen. The tension specimen is loaded beyond its yield strength and when unloaded from that point on the stress-strain curve. It follows the same slope, as the virgin proportional limit slope, back down to zero load (stress), hitting the stain axis at some residual strain or offset. It is a bit more complicated for the bending specimen because the whole section is not loaded with a uniform stress, and then unloaded in a uniform stress manner. And, once the outer fibers start yielding, the stress or strain are no longer linear about the neutral axis. After the unloading, there are residual compressive and tensile stresses near the top and bottom portions/fibers of the bar.
A number of the lower strength steels show a very distinct yield point and a long, almost horizontal yielding plateau (plastic region) before strain-hardening starts, and then their stress-strain curve slopes up to their tensile strength, and ASTM A36 is one of these. Most of the higher strength alloyed steels and heat treated steels do not have a yield point as such, but rather they have a yield strength which is set at a .2% strain offset, and they go directly into a fairly long strain-hardening range on their stress-strain curve before reaching their tensile strength. As long as the load step-up operation is done over a short period of time, to finally achieve the 7/8" permanent set, the bar continues to act within its original (virgin) stress-strain curve. If the bar is allowed to age at room temp. for several days or age at a slightly elevated temp. for a shorter period of time it will exhibit a higher yield stress for any added deformation, and a higher tensile strength too, but a reduced ductility. This is called strain aging.