plastic deformation of beams

[corus]

Is there a formula for the plastic deformation of a beam given a bending moment, say for a cantilevered beam, when that moment will induce stresses above the yield point and assuming perfectly elastic plastic material?
Presumably for full plasticity the answer will be infinite while for perfectly elastic stresses the formula would be given using standard formulae given in Roark, etc. I need the expression for stresses inbetween these two.
Any references would be useful.

corus

 

[GregLocock]

The method is the same as that used for working out the moment of plasticity in steel reinforced concrete beams. That is, you assume that plane sections remain plane, strain is proportional to distance from the neutral axis (which may move) and that the total axial force on the section is zero (this is how you find the neutral axis). The bending moment is then easy to work out.

Is that enough to go on? It really is not very difficult, but I do need a whiteboard!



Cheers

Greg Locock

 

[rowe]

I'm not clear on what you call a "perfectly elastic plastic material".

In my experience, the yield stress is considered the maximum that a steel beam would develop (strain hardening to ultimate failure is ignored).

However, for a steel beam that can safely develop a plastic hinge (remains stable - no collapse), the bending moment can be increased beyond the yield strength of the extreme fibers. In design/analysis of the plastic hinge, the stress is still considered to be the yield stress. The difference is that at the plastic moment, the entire x-section has reached the yield stress instead of just the extreme fibers.

If you are trying to predict the strain or deformation after yield, then the elastic equations are no good, because, as greg stated, they are based upon geometry that does not change during elastic behavior. I don't recall any references/software that can predict the amount of plastic deformation after yield.

Good luck.

 

[corus]

Thanks greg, but it's not the moment I require. The moment is the value that is given but this gives a stress above yield when, as rowe says, elastic deformation doesn't apply. The deformation or displacement at the end of the beam at which the force or moment is applied will lie between inifnity for a plastic hinge and the elastic calculated value of displacement.
For rowe, a perfectly elastic-plastic material in my book is one in which the part of the stress strain curve above yield is flat, and is an approximation to the true stress-strain curve.
Perhaps a white board may be necessary.

corus

 

[GregLocock]

You have to think bass ackwards on this one. You have a given moment you need to support, what mixture of linear elastic behaviour (sigma=E*strain) and plastic behaviour (sigma=sigmayield) will support that moment, while satisfying the no axial force requirement (which controls where the neutral axis will lie).

It is called the partially plastic moment, except in the limiting cases. Obviously a fully plastic moment will not have a limiting deflection in simple thoery, then you have to use a work equation.

So the strain diagram looks like, with o at the origin
x
x
x
o
x
x
x
x
x

and the stress diagram will look like

x
x
x
o
x
x
x
x
x

note the neutral axis will no longer be central

M=integral y*stress*width*dy

where y is measured from the neutral axis

and

F=integral stress*width*dy=0

I'd work it out for a cantilever first.

Then the angular deformation of that slice can be calculated in the usual fashion.





Cheers

Greg Locock

 

[prex]

corus, can't really understand what you mean when speaking about infinity.
The behaviour of your cantilever will be qualitatively as follows:
- up to the point where the extreme fiber of beam section is below yield, the behavior is of course elastic
- from this point the end section, and progressively the rest of the beam, will start to yield: the relationship between load and deformation is no more linear, but, just as a starting point, you could assume it is grossly linear up to the point where the end section becomes fully plastic
- the plastic moment resisted by a fully plastic section greatly depends on its shape: it is 50% more than the elastic moment for a solid rectangular section, but only some 10% more for common I shapes
- from the point of full plasticity, the resistance of your beam remains constant, so that, if you continue to increase the load, it will instantly fail.

prex

http://www.xcalcs.com

Online tools for structural design

 

[corus]

Thanks to all.
From greg's and prex's comments can it be assumed that once the distance y to the plastic region is determined the deformation can be determined by assuming that only the depth y of elasticity is used in calculating the inertia, so that for fully elastic behaviour the normal equations are used whereas for fully plastic behaviour the inertia is zero (y=0) so that the deformation is infinite, and other values of y would give a deformation inbetween the two? For instance for a rectangular beam of depth 2h, the value of y is calculated of the elastic region and the inertia I=b.y^3/12 is used?

corus

 

[rowe]

It seems that you have an actual problem to analyze beyond just a theoretical curiosity.

If that's the case, I would suggest that for a cantilever, the bending moment causing initial yield of the extreme fibers would be the max moment allowed (i.e. the formation or development of a plastic hinge is not allowed because the cantilever would collapse). Subjecting the cantilever beyond the initial yield would be lowering the factor of safety below usual thresholds.

Otherwise - For an elastic/perfectly plastic modulus curve the slightest increase of stress beyond yield indicates infinite strain - no way to interpolate.

The analysis of the max moment allowed at a plastic hinge is not based upon strain at all, but merely the geometry of the section and the statics of tension and compression.

Between the initial yield at the extreme fiber and the fully platic section, you could estimate the strain based upon the linear stress/strain of the unyielded areas of the x-section, however, that estimate would be less accurate the closer you get to the fully plastic section.

 

[GregLocock]

"From greg's and prex's comments can it be assumed that once the distance y to the plastic region is determined the deformation can be determined by assuming that only the depth y of elasticity is used in calculating the inertia, so that for fully elastic behaviour the normal equations are used whereas for fully plastic behaviour the inertia is zero (y=0) so that the deformation is infinite, and other values of y would give a deformation inbetween the two? For instance for a rectangular beam of depth 2h, the value of y is calculated of the elastic region and the inertia I=b.y^3/12 is used? "

No. You have to split the contribution up into the elastic and plastic areas. You cannot use I=1/12bd^3, by itself, and it is not applicable to the plastic region anyway.



Cheers

Greg Locock