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"

 

[ToadJones]

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.

 

[IDS]

You can't really extend theory & design procedures exactly into practice.

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/

 

[BAretired]

I agree with Doug. The calculation should be no less accurate than the stress-strain curve for the A36 bar.

BA

 

[dhengr]

This is just a cold forming operation caused by inelastic deformation, with the bar retaining some wished for deformed shape. Any shop with a press brake, bulldozer or some such forming equip. can do this for you and tell you what force was required to achieve the 7/8" permanent deformation. They will just keep applying more force, in steps, until they to achieve the desired result. Maybe they could record the forces and permanent deformations, thus allowing you to draw a M/ or P/Delta curve for that bar. Obviously, to start to approach this problem form the analytical standpoint, you must know the mechanical properties of the steel involved, and the shape of the/its stress-strain curve. It’s a little more complicated than just applying Delta = PL^3/48EI, because you are talking about a permanent deformation of 7/8," a little more than half the depth of the bar. It is a fairly basic Structural Engineering problem/concept, unless the pinned latch and the hinge prevent one of the support points from moving/sliding along the length of the beam, as if on a roller. But, it’s not a simple or exact calculation process. The deflection calc., Delta = PL^3/48EI is pretty straight forward until the extreme fibers start to yield; then the rate of change of deflection or localized rotation per unit of load applied starts to increasing, it’s no longer linear, since the beam is starting to form a plastic hinge. That’s the NL part in NL-FEA, and I believe this is done in a stepwise fashion also. It seems to me that over the years I’ve seen and done some calcs. and analysis methods which relate the formation of a plastic hinge to the moment and angular rotation of the yielding section of the member. I think it was an internal work vs. external work or energy method we used. The analytical difficulty is in this moment, angular rotation, approximation and then a conversion to a deflection, but I suspect this is the iterative approach BA was suggesting. At best, it’s still just a good approximation. It’s the locked in (residual) strain in the upper and lower quarter +/- of the bar’s depth that causes/retains the permanent deformation or deflection.

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.