Plastic Deformations Question 2

[Once20036]

I`m currently looking at a situation where there is an A-36 steel rod embedded in concrete and cantilevering out. My experience to date has been with steel in the elastic range and I`m trying to wrap my head around what happens to this rod as it becomes partially inelastic & then fully plastic.

Phase 1 - As load is applied, compression forces develop on one side of the rod and tensile stresses develop on the opposite side. Stresses vary linearly across the section, and any deflections will straighten out once the load is removed.
As additional load is applied, tensile stress = compressive stress = fy.

Phase 2 - Past this point, the extreme fibers cannot take additional stress. The steel will be stressed to fy for some distance in from the extreme fibers, and stress will vary linearly between these "plastic" zones.

Question 1) To calculate the deflection at the tip of the cantilever, I believe that a virtual work method is required. Correct?
Question 2) Once the load is removed, I believe that there will be residual stresses left in the steel, in only the zones that became plastic. How can these stressed be quantified?
Question 3) One the load is removed, the member will not return to it's original undeformed shape, nor will it maintain the deflected shape it took while the load was applied. Is it possible to determine the new unloaded shape?

Phase 3 - If enough load is applied to the undeflected shape, it will become fully plastic. Each location of the cross section will be at fy in either tension or compression. I understand that at this point, you'd have a collapse in a gravity beam, but what if it was a hydraulic ram pushing a cantilever up 3"? You'd never have more than 3" of movement.
Question 4) Once this load is removed, as above, I believe that there will be residual stresses in the cross section and a deflected shape. Can these be determined theoretically?

Thanks in advance.

 

[BAretired]

Question 1) To calculate the deflection at the tip of the cantilever, I believe that a virtual work method is required. Correct?

Not necessarily. If you know the shape of the stress/strain diagram for the material, you can estimate the rotation and deflection at every section using moment-area theorems.

Question 2) Once the load is removed, I believe that there will be residual stresses left in the steel, in only the zones that became plastic. How can these stressed be quantified?

The plastic portion will try to shrink or stretch to its original length, but will be restrained by the elastic portion. After all load has been released, the elastic portion will have residual stresses as well. Perhaps they can be quantified by numerical integration.

Question 3) One the load is removed, the member will not return to it's original undeformed shape, nor will it maintain the deflected shape it took while the load was applied. Is it possible to determine the new unloaded shape?

I think so but testing would be needed to confirm the theory.

Question 4) Once this load is removed, as above, I believe that there will be residual stresses in the cross section and a deflected shape. Can these be determined theoretically?

I think you could get pretty close.


BA

 

[Once20036]

Thanks for the quick response BA,
Re Question 1: The material is A-36, so stress/strain diagrams are available.
I had to go back to a structural analysis text to brush up on the moment-area theorum. Looking through the equations, they are all based on M/EI, which does not include anything regarding the stresses/strains and whether or not they are in the elastic or plastic range. Virtual work is the only process listed (in this specific text) that specifically indicates it is applicable for inelastic displacements. Is there a modification to the moment-area theorum to include plastic effects?

Re Question 2: Suppose it were an easier problem, with a tensile force applied to the same steel rod, enough to make the entire cross section plastic. Suppose that elongation were limited to one inch. Once the load was released, I believe that the rod would not stay elongated one inch, nor do i believe would it return to it's original length. Based on this, I think there's more going on than the elastic "core" preventing the plastic "shell" from returning to it's original position. I think there's some permanent strain in the shell, but how can that be quantified? Once it's quantifiable, elastic stressed in the core could be balanced with the shell to determine deflected shape. It sounds like some heavy math, but manageable.

 

[Once20036]

I forgot one item in the original post:
Question 5) At some point the tensile face of the rod should split/crack from the tension. Is that crack expected to occur when the theoretical stress in the tensile fibers (assuming they weren't limited to fy) reach fult?

 

[dik]

If it splits or cracks, the material is too brittle and you may want to reconsider plastic design.

Dik

 

[BAretired]

Re Question 1) In the elastic range, curvature is defined as M/EI. In the plastic range, that definition will not suffice, but if you assume that plane sections remain plane, it should be possible to calculate the curvature at any point in the beam based on geometry.

If the load is a concentrated load at the free end, the moment diagram is a straight line varying from 0 to P*L. The curvature diagram will be simply M/EI for the portion of beam which is fully elastic. Closer to the support, it will deviate from a straight line because curvature is increasing at a geometric rate as a result of plastic deformation.

The area under the curvature diagram is the rotation of the tip. The area under the curvature diagram times the distance from its center of gravity to the tip is the deflection at the tip (assuming small angle theory).

Re Question 2) If you start with a bar with an area of one square inch and stress it to Fy in pure tension, the bar has followed Hooke's Law until you increase the stress to something greater. Let's say you stress it to 1.5Fy, hold it for a while, then release it. Use the stress/strain curve for the material and locate the point corresponding to 1.5Fy. Note the strain at that point, then take a straight edge and draw a line through the point parallel to the elastic portion of the curve, that is at slope E. Where that line intersects the zero stress line is the strain you will have after the load has been released.

The plastic material will try to do the same in a beam, but it will be restrained by the adjoing elastic material.

BA

 

[BAretired]

Question 5) If the material is stressed to F_ult, it may be expected to split or crack. In fact, if there are many repetitions (cyclic loading), it will split or crack at much lower stress levels than F_ult. That is called fatigue.

BA

 

[TERIO]

For the math to answer most of these questions try looking at section 4.4 of Plasticity Theory by Lubliner:

http://www.ce.berkeley.edu/~coby/plas/pdf/

Also, regarding phase 3, you are correct that there is no collapse if the stress is displacement controlled (also referred to as secondary stress).

 

[BAretired]

Returning to the question of calculating curvature, it is M/EI in the elastic region of the cantilever.

For simplicity, let's say that the cantilevered rod is rectangular in cross section with width b and depth d and it is loaded with a concentrated load P at the tip.

Within the elastic range, the moment is M = Fb*S and the curvature is M/EI or Fb*S/EI. For a rectangular rod, that would be 2*Fb/dE.

Closer to the support, the outer fibers are strained beyond yield strain but a central portion of the cross section is still elastic. Assume that portion is b x d' where d' is less than d. The total moment resistance of the section is Fy*Z + Fy*S' where Z = b(d^2 - d'^2)/4 and S = bd'^2/6.

If plane sections remain plane, then curvature in the partially plastic zone can be calculated from the elastic central part of the section. It is M'/EI' = Fy*S'/EI' or 2*Fy/d'E. As d' approaches zero, curvature increases without limit.

Deflections can be calculated from the curvature diagram provided d'>0.

BA

 

[Once20036]

dik - if this were a new design I would absolutely pick a different material or use a larger rod to eliminate all of these issues. However, I`m just trying to understand the physics behind plastic steel sections/failures

Terio - thanks for the link. Looks like I`ll have some light reading for this weekend.

BA - So the permanent strain left in the rod after plastic bending is equal to the final strain less the elastic portion of the strain. This makes a lot of sense intuitively. Same with Fult causing splits or cracks. Regarding curvature in the plastic range, I had come to the same conclusion myself, re:calculating Ix based on the portion of the cross section that remains elastic. It took a little time for me to realize that strain is linear across the cross section, but starts to increase as more material goes plastic and Ix decreases.

This afternoon's project will be figuring out a way to calculate I for a rod with some variable amount of steel trunicated in the plastic zone & automating that process so excel can replicate the calculations at a couple hundred cross sections along the length of the rod.

 

[dhengr]

I think you’re on the right track, but I don’t see the need to do calcs. at many sections along the length of the canti. beam. We know the plastic hinge will form at the max. moment, at the fixed end. The question is over what length of beam does this plastic action take place, what is the length extent of the hinge and how is it located near the fixed end. And, in our simplified world, there is no easy way to account for shear deformations in this hinge action or development. As you move beyond the hinge toward the tip of the canti. you can go back to treating the beam in our standard elastic fashion, that won’t change. You just get a very concentrated rotation, over a short length of beam, right at the hinge, near the fixed end.

If you look at the stress-strain curve for A36 you have a very distinct upper yield point, then a slightly lower yield point as you continue to strain the member a small amount, then a long horiz. plateau or plastic range, where strain increases without any appreciable increase in stress; then a long strain hardening range where stress does increase again with increasing strain, and finally Fult.. Many other structural steels don’t have a yield point, they have a yield strength defined by a .2% strain offset and no plateau, but rather they go right into a strain hardening slope or range, and then to Fult.

I’m not really sure what you are trying to accomplish, but it seems to me with the long plastic plateau of A36, you have failure shortly after plasticity starts at the hinge. As BA suggested if you remove the load out at the tip, the stress-strain curve will back-off, back down at a slope E, and you will see a residual strain which represents the plastic material working against the material which is still elastic. You can reload the member and the stress-strain curve will climb up that second slope E, back to the point you were at on the plateau or strain hardening curve, and you can do this a number of times, stepwise, climbing the strain hardening curve to Fult.

I have to think about this a bit more, but I think that at any step, you can plot the strain and stress over the depth of the member, and calc. the existing moment at the hinge, adjusting the depth of the stress blocks, lever arms and I’s, and then calc. the hinge rotation. But, this is a fairly localized moment and rotation, not something which is happening over an appreciable length of beam. I would feel comfortable starting with BA’s formulation from his last post. But, given the variability in exact materials specs., our assumption of hinge area length, neglect of shear deformations, etc., for our hand calcs., I wonder if I wouldn’t just pick some canti. beam tip deflection limit and call that the failure point. With a structure like this you have a failure mechanism once the first hinge starts to form, you aren’t redistributing loads and moments to a second hinge formation point. This would be a wonderful problem for FEA modeling, with a fine mesh over the hinge length, and watch the stress and strains change with increased load. I’ll bet this horse has already been beat to death.

 

[Once20036]

Thinking about this a little bit more, it's not stress that remains linear over the cross section, it's strain.

So for A36 steel there is:
1) an elastic range (where stress = E*strain), followed by
2) a relatively level area (where stress = constant & strain increases with deflection), followed by
3) an area of strain hardening (where stress and strain both increase), followed by
4) an area of necking (stress decreases, strain increases), followed by,
5) rupture/complete failure.

To figure out the moment that causes rupture/failure, I need to
1) assume strain is linear across the width of the section (plane sections remain plane)
2) figure out the stress in each little slice of the cross section the corresponds to the strain in that cross section
3) distance from centroid * area of little slice * stress = moment contribution for that little slice.
4) sum of these little slices gives total moment on the section.

From there, per Dhengr, I can look at strains in a couple key spots in the plastic range to get plastic deflections, plus traditional elastic equations to get elastic deflections. Done.

The big missing piece of the puzzle right now is item 2, above. I have charts schematically showing the stress/strain relationship for steel, but does anyone have more exact info available? equations? charts for a number of strain values & their corresponding stress values? I`m sure this information is out there somewhere based on years of testing.

 

[TERIO]

Regarding your item 2 above it would be reasonable just to assume elastic-perfectly plastic (i.e., not strain hardening), at least as a starting point.

Back at the beginning it sounded like you have a real structure to qualify, and that this is not solely a theoretical discussion. If so, a few questions:

1. How is the structure loaded (you mentioned imposed displacements)?
2. What criteria do you intend to use to qualify it?

 

[gobsmacked]

Once the load is removed, there are no residual stresses arising from that applied load, although there may be some from a previous history of differential heating/cooling. But you do have residual strain (plastic deformation). When you bend a paper clip, there is no residual stress once you have stopped the applied moment. What you have done is use up some of the plateau of ductility.

 

[BAretired]

gobsmacked,

Residual stresses in steel can result from heating and cooling during manufacture and also as a result of cold bending. Check out this link:

http://www.nordicsteel2009.se/pdf/120.pdf

BA