Simple Beam Questions 4

[Spoonful]

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

 

[Spoonful]

Hi BA,

Thank you very much for your help, I found that in my I calculation forgot to divide the diameter into radius, then I got the same result as yours.

FYI, ASTM A276 316L Round bar Yield strength is 170Mpa

Also, I made an assumption by doing above calculation to real case, which is woven wire mesh over perforated plate. I treat it as multiple length round bar over span of the hole(by approximation and conservatively treated as a square), even the wire can never be straight because of the woven and the wires are 90 degrees across each other. I tend to think wire un-straightness and cross of wire should not play an importance rule here.

Could anyone see any problem with above assumptions?

Thanks in advance.

 

[BAretired]

Yes, I do. Bars or wires lying over holes in a plate do not have fixed ends. I would assume hinged ends unless each bar is welded to the plate each side of each hole.

Maybe it would be better if you described the actual problem instead of attempting to determine an equivalent problem.

What is the size and spacing of the holes in the plate? What is the size and spacing of wire in the mesh? How did you determine the uniform load on the bar?

BA

 

[Spoonful]

Hi BA,

Please see below, appreciate for you kind advise.


The problem is to determine if high differential pressure on different side of wire mesh over perforated plate configuration. I tend to think the wire edge at hole as fixed, because the wire mesh will be tightly fit onto the perforated plate, and at the edge, the wire could have a moment due to load.

Hole size 3mm
mesh wire dia 0.16
mesh opening 0.26mm
Load P=1Mpa

Treat hole as a square, then A= 9mm^2
Load = 9N
There will be 3/(0.16+0.26)=7.14, say 7 wire each way across the span, so in total 14 wires over the area,
so the load / length of wire = 0.214N/mm

Elastic:
M max = wl^2/12 = 0.16 N mm
Stress caused = 400 Mpa <--- yielded

Plastic:
Mp= Z* yeild = 0.00682*170 = 0.1116 N mm
Max load = 16Mp/l^2 = 0.206 N/mm <---- less than applied load, this wire mesh will break.

 

[BAretired]

I see almost no similarity between the problem you have just described and the original post. I don't have the answer to the new problem but let me ask you a few questions.

1. How is the pressure applied? By a liquid?
2. Will the liquid leak through the mesh?
3. Even if it leaks through the mesh, is the rate of leaking slow enough that the mesh supports a height of liquid equivalent to 1 MPa over the 3mm dia. openings in the plate?
4. In the space between openings, there is no pressure on the wire, so how can the wires be considered fixed at the edge of the holes?
5. Is the mesh fastened to the edges of the plate? Is it fastened by welding or other means?
6. Is it justifiable to conclude that the mesh will fail by bending?
7. Is the mesh capable of acting like a tensile membrane over the 3mm holes?
8. What are the stress vs. strain characteristics of the material?

My guess is that you have not found the correct mode of failure. I don't believe bending is it.


BA

 

[Spoonful]

Hi BA,
It is a screen. Load is applied by differential pressure on each size of the screen, please ignore the fact that flow can pass through the hole and equalize the pressure on both side, In fact assume the hole completely blocked (worse case scenario when screen is fully clogged), hence full load by differential pressure must be supported by the wire mesh.

Wire mesh is welded on top of the perforated plate.

At the edge of hole, mesh wire extend beyond the hole area, but also subject to pressure from the liquid to push it against the un-perforated area of the plate. Hence the wire at the edge can develop a bending moment.

Another failure mode could be direct shear, but for given load and wire thickness, I won't think it will fail by shear first.

Thanks

 

[GregLocock]

When the screen is clogged you more or less have a diaphragm, with plastic bending at the circumference.

So another approach would be to simply support the smeared properties of the gauze across the disk (needs some thought, by you, not me) and check that. In reality gauzes support amazing loads because they deform plastically and become catenarys, which is roughly where we were several zillion posts back.








Cheers

Greg Locock


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

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[BAretired]

I disagree. The screen may be fully clogged over the perforations, but not necessarily between perforations. So it is not clear why the mesh between perforations is subject to any pressure at all. Downward pressure is balanced by upward pressure and does not push the mesh against the un-perforated plate.

I believe the mesh will behave as a membrane carrying its load primarily by tensile forces in orthogonal directions. Its failure load cannot be estimated without knowing the distance between openings and the stress/strain characteristics of the material. It is not a simple problem.

BA

 

[Spoonful]

I agree the wire will be acting as round bars, crossed each other to support a load. Three of the possible failure mode, Bending, Shear and tensile. Bending we have just discussed, and shear failure is less likely. And how to work out its tensile stress at along the round bar caused by load? as the bending angle constantly changing when load increases

 

[BAretired]

I think you can forget about bending and shear. The failure mode will be two way tension. At this time, there is insufficient information to solve the problem so perhaps we should let it go at that.

BA

 

[Spoonful]

BA, Any light on how to solve the tension problem? From elastic bending we can work our the max deflection at mid span, hence the angle then from force equilibrium to workout the tension. but how about once it bend beyond elastic?

http://www.eng-tips.com/viewthread.cfm?qid=345051

 

[BAretired]

spoonful, you are missing the point. Forget about bending and consider the catenary action of each wire in the mesh.

When tension is applied to a wire, it strains. When it strains, it sags over the 3mm diameter openings in the plate. If the wire happens to be in the middle of the opening, it spans 3mm, otherwise it spans a lesser distance.

If there are seven wires crossing an opening in each direction, it is possible to compute the sag at each node inside the circular opening under elastic conditions.

When the first wire reaches yield stress, it continues to strain without increase in tension. This increases its sag which allows it to resist larger applied loads. Meanwhile, neighboring wires can reach and exceed yield strain without failing.

If load is increased further, all wires are strained far beyond yield. As a result of their increased sag, they carry considerably more load, but at some point one wire reaches its ultimate stress and it fails in pure tension. This starts a chain reaction and the mesh eventually fails.

Failure may be defined in a variety of ways. If you define failure as the point where the first wire reaches yield, then the problem is soluble, but in actual fact, the failure load is much higher. This is one of the inherent advantages of a cable system. As the cable stretches, the sag increases and allows it to carry more load.



BA

 

[Spoonful]

BA,
Understood what you said and agrees, but the question remains how to find the failure load(until wire breaks). I think the key is to find sag angle to determine the tensile reaction force from force equilibrium, But how do we find the wire sag angle? Before the wire yield we have a formula. how about after it yields into plastic?

 

[paddingtongreen]

This request has morphed into an anything-but-simple question.

I haven't seen the word "strainer" mentioned but it now looks like one. I thought the fluid was running parallel to the perforated plate, but now it appears to flow through the strainer. The wires would only be fixed ended if they were welded to the plate, precisely at the edge of the hole.

You now have a grid of continuous wires, if an up/down wire wants to bend, a left/right wire will probably have to bend with it. These will drive the parallel wires on both sides of them to deflect almost as much, nd these will drive....

No one wire acts alone! they work together, community action rather than libertarian.

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

 

[rb1957]

you could consider the mesh to be a diaphragm ... thk = 2*A/w ...
A = area of wire
w = spacing of wires
2 for both directions (two wires)

Quando Omni Flunkus Moritati

 

[paddingtongreen]

rb1957, That only works if each wire is welded at the edge of the hole, if not, the edge of the hole becomes a fulcrum and the mesh bends away from the plate and then back to the weld points.

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

 

[rb1957]

and this is a "simple beam" question ?? (i know, you've already noted such to the OP).

agreed there would be lift up around the hole (the mesh inside the hole would slope (away from the pressure) and this slope would be carried over to the mesh outside the hole (lifting up from the holey plate, into the pressure) ... clearly not a simple question.

Quando Omni Flunkus Moritati

 

[BAretired]

It is not a simple beam and it is not a simple question.



BA

 

[Spoonful]

In fact it is a strainer screen, and at the beginning I only thought need to model the wire as a simple both end fixed beam over a span. but apparently it wasn't quite correct. And seems most of you agree it is not a simple problem, but any suggestion on where I should start my research/study on, or any better way of modeling this case?

 

[BAretired]

There are 14 "beams" over each opening, 7 each way. If you assume one beam is in the middle of the circle, there are 3 each side with a known span, namely the chord length of the circle at the beam location.

If the holes are set out in a square array, the center to center distance of the holes is known. Call it L. That is also the length of each beam (not the span but the overall length to be analyzed). Ensure that you apply the correct boundary conditions to all beams.

Apply a concentrated load P at every node. Since the beams are spaced at 0.16 + 0.26 = 0.42mm and design pressure is 1MPa, P = 1(0.42)² = 0.176 N.

Using a simple frame program, solve for the axial loads, shears and moments for each node point of all beams. If you want to take into account deflection, modify node co-ordinates after each run to include deflection of all beams. Repeat until subsequent runs yield the same result at all points of all beams.

BA

 

[Spoonful]

BA, please see attached sketch page one, if I have understood you correctly. If correct, then it can be treated as page two? then we are back to the beginning of both end fixed beam under uniform load? But only to solve for axial loads? If all true, only need to solve for the mid span beam, as it will fail first.