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Simple Beam Questions 4

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Spoonful

Mechanical
Oct 18, 2008
175
Hi All,

For a simple been both end fixed,under uniform distributed load, as per Roark's table 8.1 case 2d. Max moment is at edge of Wl^2/12, and hence max stress at edge. If we apply a load which will result a stress at edge exceed the allowable stress, or even exceed material yield stress, the beam will starts to plastic bending.

The question is as long as the load is applied, and the beam will start to bend, no longer straight,The beam geometry changed. will the formula M max = wl^2 / 12 remain true? Or the max moment will be shifted to elsewhere along the beam.

My point is how to find out the true load that will break the beam, or cause into plastic bending.

Thanks in advance for any comment.

Regards

Spoonful
 
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Hi Greg,

I am not trying to get someone else to solve my problem, just want to get some light on how to solve this, I am just confused that once the beam is no longer straight what to do.
 
If the beam were to bend such that it is fully plastic then the resulting 3 hinged mechanism would see a maximum moment of WL^2/12/4 (I think, by inspection in your terminology) at the quarter span points, so you need to decide exactly how plastic you are.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
Hi Greg,

If we see it as a initial 3 hinged mechanism, and the moment is reduced, and them later becomes infinite number of hinged system, as your bending moment is infinite small? hence its bending stress?

 
Yes, ultimately a fully plastic member, fully plastic for its entire length and depth, can support virtually no moment and ultimately supports the load by axial tension. It is a cable. They hold lots of bridges up.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
Agree, but then how do you workout the max load can a beam support?
 
Agree, but then how do you workout the max load can a beam support?

You refer to the limits on moment redistribution in the applicable design code, or if working from first principles you calculate the curvature at failure, which puts a limit on rotation.

Doug Jenkins
Interactive Design Services
 
"how do you workout the max load can a beam support?" the best way is to test it, 'cause all calcs have assumptions in them.

take your uniformly loaded beam.

1) as the beam deflects, is the loading still uniform (or does it peak towards the ends) ? still vertically down (or normal to the deflected beam) ?

2) as the beam starts to go non-linear, the deflection (and curvature) become significant. the solution for a curved beam is radically different to "plane sections remain plane" and "small displacements" assumed in the linear solution (wL2/12).

3) another analysis thing to consider is that the beam is now no longer staight, so there is stress at the neutral axis and the beam is reacting some of the applied load with axial load (like membrane reaction in a plate).

4) i think this leads to your observation that with an infinite number of plastic hinges the moment goes to zero ... 'cause now all of the load is reacted by axial loads.

5) full NL FEA might get you close

Quando Omni Flunkus Moritati
 
Using plastic theory, the beam forms three hinges, one at each end and one at midspan. If the beam is capable of developing plastic moment at all locations, the beam fails when the plastic moment, Mp at each hinge is wL2/16 which is half the simple span moment. The maximum load the beam could sustain is 16Mp/L2.

BA
 
Perhaps I should have added that a uniformly loaded beam, fixed at both ends, while in the elastic range, will have a negative bending moment of wL2/12 at each end and wL2/24 at midspan.

When the negative moment reaches Mp (the plastic moment of the beam), any increase in load has no effect on the negative moment but increases the positive moment. When the positive moment reaches Mp, the beam becomes a mechanism and fails. This occurs when Mp = wL2/16.

BA
 
BA has given a good description of how redistribution works in the theory of plastic design. One thing for the OP...the title of this thread "Simple Beam Questions" is misleading, because the question and the discussion are not about simple beams, but rather about beams with their ends fixed. The question may be simple, but the beams are not.
 
Thanks All for the kind responses, especially BA.

What I am try to find out is: I have situation where a round bar is seated across the span of a large hole(both ends can be assumed as fixed), the diameter of the hole would be the the span of the beam. And fluid will be flowed across the hole, or on other other world, there will be a load(assume uniformly distributed across the span length of the beam) applied onto the beam. It is still OK, if the beam deflects under the load into plastic region(but preferably not), as long as the load will be not break the beam. The question is what is the max load can be applied?

As BA suggest use Max as wL^2/12 (where L is entire span) for elastic bending, and wL^2/16 (where L is half of the span, assume 3 hinges) for plastic bending. Am I understand this right?

 
No, that is wrong. L is the span in both cases which, in your case is the diameter of the hole. The simple span moment is wL2/8 but that occurs with a hinge at one end and a roller at the other.

The fixed end moment is wL2/12 and under elastic conditions, the simultaneous midspan moment is wL2/24. Notice that the sum of positive and negative moments are wL2/8, the simple span moment.

If the load is gradually increased until the beam yields, the end moments will be equivalent to Mp, the plastic moment of the beam. The beam cannot resist any more negative moment than Mp, so any increase in load will result in a redistribution to the middle of the span. When the third hinge forms at midspan at the ultimate load wu, each end moment and the midspan moment are equal to Mp and the beam fails because it has become a three hinged mechanism. The sum of the negative and positive moments still must be wuL2/8, same as the simple span moment. So each of the three moments is numerically equal to Mp = wuL2/16 where wu is the ultimate uniform load on the beam when it fails.

BA
 
I have just got myself confused again,

For example Span L = 50mm and beam (round Bard) dia = 10mm, Load = 250 N/mm, Sy = 170 Mpa, -----> I = 7.85E3 , Z= 166.625



Max Elastic bending moment = Wl^2/12 = 52084 N mm
and stress casued by this moment is My/I = 33.17 Mpa, which this beam is not yielded.

BUT when check for plastic bending
Mp=Z*Sy = 28326.25
Max load Wu = 16*Mp/L^2 = 181.3 n/mm, Where it is less than applied load.

Could someone see where I have went wrong?

Regards

 
Given this is a pipe, I think you should stay in the elastic range. When taken into the plastic range, the section tends to close, the outer fibers, tension and compression, move toward the center reducing flow, but more importantly, reducing the properties, section modulus, plastic modulus etc.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
 

For example Span L = 50mm and beam (round Bard) dia = 10mm, Load = 250 N/mm, Sy = 170 Mpa, -----> I = 7.85E3 , Z= 166.625
Fy (not Sy) = 170 MPa is low...it is likely about 248Mpa or more.

S = πR3/4 = 98.2 mm3
I = πR4/4 = 490.8 mm4


Max Elastic bending moment = Wl^2/12 = 52084 N mm
and stress casued by this moment is My/I = 33.17 Mpa, which this beam is not yielded.

fb = M/S = 52084/98.2 = 530MPa which means beam has yielded

BUT when check for plastic bending
Mp=Z*Sy = 28326.25 (Mp = Z.Fy = 166.7*248 = 41,300N-mm)
Max load Wu = 16*Mp/L^2 = 181.3n/mm (264N-mm), Where it is less (a little more) than applied load.

Simple beam theory does not apply to very deep beams, so it is not accurate for a span to depth ratio of 5 which is what you have here. Usually the span to depth ratio is about 20 or more, so the theory is not strictly applicable here.


BA
 
paddington,
As I understand it, the beam consists of a round bar fully fixed at each end. It is not a pipe, but it spans across a 50mm pipe.

BA
 
Once again, I stand corrected BA, I have to slow up on my reading. I read "fluid flowing across wrongly".

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
 
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