Beam displacement 1

[marmilew]

Hello

Is there moderately easy way for hand-calcuations of the beam dispacement after plasticity occurs? My model is beam clamped on left and simply supported on right side. Force P in the middle of beam and plastic moment is Mpl (plastic resistance). I use elastic perfectly plastic material model. I used unit-load method to determi displacement in the middle of beam but there is a big diffrence with FEM results.
My calculations and model are in attachment or under link:

https://www.dropbox.com/s/5vzo994gjtrzwli/Scan job (3).pdf?dl=0

I will appreaciate any help

 

[KootK]

Upon reread, I believe that the OP is simply trying to verify FEM output for a model that reflects elastic / perfectly plastic material behavior and may indeed include distributed plasticity. In light of that, I'll refine my position to this:

1) In the solution that I have proposed, I chose to consider binary "point" section plasticity only and ignore the effects of distributed plasticity. Within the bounds of the assumptions that I've made, I believe my proposed solution to be correct.

2) If the OP's FEM model includes distributed plasticity effects, then my proposed solution will serve as merely a lower bound estimate on expected deflection output.

3) Within the realm of what I would consider to be "practical hand calculations", I contend that my solution is both reasonable and valuable as a benchmark.

4) My proposed solution ought to yield the same answer as the unit load method given that both are correctly applied. To my knowledge, the unit load method would also not account for distributed plasticity. My method might serve as a check that the OP's unit load calculation was done properly.



The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[BAretired]

The problem states that the material is elastic perfectly plastic. That means it has a particular stress/strain diagram which may be found using Google.

If a plastic hinge has formed at Point A, the sections immediately adjacent to point A will be nearly plastic with only a small portion of the web behaving elastically; thus the effective elastic section is severely reduced resulting in increased curvature. Sections further away will have a reduction in effective section until the point is reached where the entire section is elastic.

Because of the increased curvature near A, the moment at B will tend to increase beyond the theoretical value of 5PL/32 (83% Mp) which was determined assuming elastic conditions throughout. But even if M[sub]B[/sub] remains at 83% Mp, any section in the immediate vicinity of B will be only partially within the elastic range.

Deflections can be calculated for this ideal material, but it will not be by superposition as described above. Offhand, I would not like to suggest how it could be done.

BA

 

[KootK]

@BA: Per my last post, we agree regarding distributed plasticity and it's plethora of complicating influences. I still very much contend that superposition applies to the solution that I proposed, within the parameters that I set out for that solution. We'll just have to agree to disagree on that.

@OP: if this is an FEM verification study -- as I suspect -- rerun your FEM with a beam that is long and slender. This will result in more of the beam remaining truly elastic and better correlation to hand calculations that do not account for distributed plasticity.



The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[IDS]

I am reminded of my favourite quote from that well known engineer, Albert Einstein:
"Everything should be simplified as far as possible, but no further".

An approach that should give a reasonable estimate of FEA output deflections:

- Find a bi-linear-plastic moment-curvature diagram for the beam; i.e. linear1 up to first yield, linear2 up to full plasticity, then fully plastic.
- Find the moment distribution along the beam from a linear combination of the simply supported moments and the plastic moment applied at the fixed end.
- Divide the beam into short sections and find the effective EI at the mid-point of each section.
- Find the deflection as suggested by KookT from combination of simply supported deflections, and deflections due to plastic moment at the fixed end.

I think I might give that a go and see how close I get.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

Well, I hope that the OP resurfaces to confirm or deny some this theory chatter for us. Even at a fully plastic section, you can expect to be back to fully elastic within two beam depths or so. I feel that having a reduced effective section over such a short length should not impact deflections all that much. Certainly, I would expect it to have a small impact relative to the formation of a full rotational hinge next door.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[IDS]

KootK - a quick check suggests you are right. So long as the mid-span moment is less than the plastic moment your method seems to give good agreement with a non-linear FEA analysis.

I was using beam elements in the FEA, so it could be that it was using a simplified moment-curvature response as well, but I don't have time for anything more elaborate right now.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

Well, I've been validating FEA for decades. It's about time that FEA returned the favour. Thanks for stepping up Doug. I think that we should each insist on at least A-'s for this.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[BAretired]

Don't strain your deltoids patting yourself on the back, KootK.

I should correct an erroneous statement I made earlier. The formation of a plastic hinge at A will not affect the moment at B unless the load is increased beyond P (the load required to produce Mp at A).

Doug's method of dividing the beam into short sections is good for a computer solution; the OP asked for a moderately easy hand calculation; I would just say that Doug's method could be used in a hand calculation (similar to the Newmark Method of calculating deflections in an elastic beam). It would be prudent to take smaller sections in the immediate vicinity of points A and B.

BA

 

[KootK]

Don't strain your deltoids patting yourself on the back, KootK.

Oh no, I'm going to party like it's 1999. Any minute now someone will surface with a contradictory FEM result and I'll be back to eating humble pie.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[Denial]

Unless I've missed it, the OP hasn't given us the beam's cross-section type.[ ] If it is an "I" section, then the moment-curvature relationship will be very close to being a sloping elastic leg followed by a horizontal plastic leg.[ ] At the other extreme, if it is (say) a "+" section, then the M-C relationship will have a very protracted middle portion where the section's marginal bending stiffness gradually decreases as the applied moment increases.
»[ ] In the former case KootK's simplifications are fully valid.
»[ ] In the latter, the complications raised by BA are highly influential.
»[ ] IDS proposes a compromise approach, but in a really extreme case (a "+" cross-section with a fat "-" and a thin "|"?) I suspect his approach might fail his own Einstein test.

 

[dhengr]

I’d tackle that type of problem using The Conjugate Beam or the Moment Area Methods and Newmark’s Method. As I recall, I’ve done this problem in the distant past. The last two are pretty much what Doug suggests in his 13DEC14, 22:40 post; with the short lengths of beam elements and EI, a numerical integration method. I’d do these problems in a stepwise fashion, up to the FEMoment starting to yield, I could use the std. formulas for moments and deflections found in any steel manual, textbook, or ref. book. As the FEMoment yielded more and more, I would calc. a fixed end moment, and would adjust the moment at the center point load accordingly, as P changed. I would use superposition at this FEMoment and new load step point to draw a moment diagram. Then I could use those moments, that moment diagram on the beam to find the deflection at that point in the process, using Newmark’s Method. It is true that we don’t know quite what to do with the St. Venant affects in the immediate vicinity of the fixed end moment or at the large moment under the load. We also don’t have any convenient way of accounting for strain hardening effects as the actual conditions climb the stress/strain curve. I would also assume that with the proper modeling, element types and mesh size, the FEA could be refined to better represent what goes on in the vicinity of the FEMoment and the centerline, point load moment. These are areas and stress/strain/rotation conditions which we just don’t have good closed form, long hand, methods for, so I would tend to want to (try to) understand what a proper FEA model was telling me.

 

[KootK]

The method that I proposed at the top works for the case where distributed plasticity is ignored. It can also be expected to provide a reasonable lower bound estimate of deflection when distributed plasticity is to be included and proportions are reasonable.

In keeping with the goal of keeping things simple, I think that it would be expeditious to have an upper bound estimate of the deflection as well. If the two solutions suitable bookend the FEM result, then all is well.

I propose that that one could make a simple estimate of deflection assuming that the plastic hinge has a finite length of twice the member depth. This should be manageable as a hand calculation and suffice as an upper bound deflection estimate.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[IDS]

I have now had a closer look at the effect of different beam cross sections and different methods of analysis (see attachment).

I looked at a 10 m long beam, fixed at the left hand end and simply supported at the right, with a point load at mid-span, and plotted deflections under the load from just before first yield of the top and bottom surfaces at the fixed support up to near full plastic moment at both the support and mid-span.

The top graph is for an I-beam approx. 300x150 mm with 10 mm flange and 5.5 mm web. The green and blue lines are from a frame analysis with either a linear-perfectly plastic stress-strain curve for the steel (green line), or a moment-curvature curve calculated for the same steel properties (blue line). These two analyses gave near-identical results. The purple line is from a spreadsheet based linear-elastic analysis up to the support moment reaching the full plastic moment, then a simply supported analysis with the plastic reaction moment applied at the left support. This has given pretty good results, up until the load at which yielding at mid span becomes significant.

The bottom graph is for a rectangular beam with approximately the same moment at first yield as the I beam (approx. 300 deep x 25 mm). In this case the red line is from a frame analysis program with calculated moment-curvature diagram. The green line is from a spreadsheet analysis with 10 segments, each with a stiffness based on the same moment-curvature diagram, and the moment at the start of the segment. This has given reasonably good agreement up until the mid-span moment is close to the full plastic moment. The two blue lines are using the applied end moment technique, using the full plastic moment for the light blue line, and the first yield moment for the light blue line. The former gives a reasonably good estimate of the deflection up to about 85% of the maximum load, and the latter gives a conservative upper bound estimate up to about 95% of the maximum load, which is really as good as can be expected for a simple analysis.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Denial]

This thread has tweaked my interest in the moment-curvature graphs for various beam cross-sections when, under increasing curvature, more and more of the material in the cross-section is strained beyond its elastic limit.

Consider a RECTANGULAR cross-section, width b and depth d.[ ] The moment-curvature relationship is linear until the curvature reaches
C[sub]y[/sub] = 2σ[sub]y[/sub]/(dE)
at which point the bending moment is
M[sub]y[/sub] = (bd[sup]2[/sup]/6) * σ[sub]y[/sub]

At larger curvatures, it can be shown that the moment M when the curvature is C is given by
M = (bd[sup]2[/sup]/4)*σ[sub]y[/sub] - b/(3E[sup]2[/sup]C[sup]2[/sup])*σ[sub]y[/sub][sup]3[/sup]

This can be made simpler and more meaningful by using C[sub]y[/sub] and M[sub]y[/sub] to achieve non-dimensional measures for the curvature and the moment.[ ] Define
C[sub]n[/sub] = C/C[sub]y[/sub] as the non-dimensional curvature
and
M[sub]n[/sub] = M/M[sub]y[/sub] as the non-dimensional bending moment.
The above moment-curvature relationship then becomes
M[sub]n[/sub] = 3/2 - 1/(2C[sub]n[/sub][sup]2[/sup])

Now consider a DIAMOND-SHAPED cross-section, width b and depth d.[ ] With this shape the width of the cross-section decreases linearly with distance from the neutral axis.[ ] The algebra is slightly more voluminous than for the rectangular case, but leads to the following results:
C[sub]y[/sub] = 2σ[sub]y[/sub]/(dE) [ ][ ][ ] (same as before)
M[sub]y[/sub] = (bd[sup]2[/sup]/24) * σ[sub]y[/sub]
M[sub]n[/sub] = 2 - (2C[sub]n[/sub]-1)/C[sub]n[/sub][sup]3[/sup]

Now consider an HOURGLASS-SHAPED cross-section, width b and depth d.[ ] (This shape comprises a downwards-pointing isosceles triangle sitting atop an upwards pointing, congruent, isosceles triangle.)[ ] With this shape the width of the cross-section increases linearly with distance from the neutral axis.[ ] The algebra now leads to the following results:
C[sub]y[/sub] = 2σ[sub]y[/sub]/(dE) [ ][ ][ ] (same as before)
M[sub]y[/sub] = (3bd[sup]2[/sup]/24) * σ[sub]y[/sub]
M[sub]n[/sub] = 4/3 - 1/(3C[sub]n[/sub][sup]3[/sup])

The shapes of the three resulting moment-curvature graphs are shown in the attachment.