Simple Beam Questions
Simple Beam Questions
(OP)
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
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






RE: Simple Beam Questions
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
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.
RE: Simple Beam Questions
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
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?
RE: Simple Beam Questions
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
RE: Simple Beam Questions
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
http://newtonexcelbach.wordpress.com/
RE: Simple Beam Questions
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
RE: Simple Beam Questions
BA
RE: Simple Beam Questions
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
RE: Simple Beam Questions
RE: Simple Beam Questions
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?
RE: Simple Beam Questions
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
RE: Simple Beam Questions
Thank you for the explanation, I found this in Roarks book (see attached), where my case a=0, then Max load=16Mp/L^2. Where Mp is Z* Sy.
RE: Simple Beam Questions
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
RE: Simple Beam Questions
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Simple Beam Questions
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
RE: Simple Beam Questions
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
RE: Simple Beam Questions
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Simple Beam Questions
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.
RE: Simple Beam Questions
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
RE: Simple Beam Questions
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.
RE: Simple Beam 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
RE: Simple Beam Questions
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
RE: Simple Beam Questions
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: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
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
RE: Simple Beam Questions
RE: Simple Beam Questions
BA
RE: Simple Beam Questions
RE: Simple Beam Questions
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
RE: Simple Beam Questions
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?
RE: Simple Beam Questions
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
RE: Simple Beam Questions
A = area of wire
w = spacing of wires
2 for both directions (two wires)
Quando Omni Flunkus Moritati
RE: Simple Beam Questions
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Simple Beam Questions
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
RE: Simple Beam Questions
BA
RE: Simple Beam Questions
RE: Simple Beam Questions
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
RE: Simple Beam Questions
RE: Simple Beam Questions
The midspan beam will interact and be supported by other shorter beams and cannot be considered in isolation. Depending on the magnitude of sag, cable action will play a role in carrying load. Cable tension will stretch the entire length of cable, not just the portion covering the 3mm hole. That strain will contribute to the sag and tend to make cable action more dominant than beam action.
It is not a simple problem and I cannot suggest an easy way of solving it.
BA
RE: Simple Beam Questions
I'm not sure I could run this on an FE program. The wires are continuous and linked to the perpendicular wires only perpendicular to the plane of the grid.
Do a thought experiment on paper. Draw your circle and grid in isometric, assume all joints are supported to start with, then think what happens when you put a load on the members connected at the central point and remove its support. Next imagine the removal of the joints at the far end of the original members with their connected members. Follow in stages, the removal of the supports until you reach the edges of the circle. Do it again with loads on all members. You will get only a general idea from this, what doesn't show here is the effect of redistribution to achieve mutually consistent joint deflections.
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Simple Beam Questions
People who don’t have a good idea how to approach their problem, would be so much better off, if they did a good job of laying out and describing their problem, just the facts, but all the facts; and then letting some of the smart people here, who are willing to help and have the experience and judgement to make good sound determinations about the tack to take on solving the problem, what assumptions to make and how to approach their problem. Then, the OP’er. should read what they have to say, a couple times, for the deeper meaning, the whole meaning and thought process. Why is it so difficult to get engineers to show a sketch, with dimensions, loads, etc. and to do a good job of describing their basic problem without a bunch of their preconceived notions which just hiding the real facts of the matter?
Now consider, a woven steel screen with .16mm wires and .26mm spacing btwn. wires and wires woven at 90̊ is fairly fine mesh. The wire is probably a pretty hard, brittle and notch sensitive, with pretty high yield and tensile strengths, cold drawn, cold worked in the weaving,, etc.; probably won’t weld worth a darn either. The top edges of the 3mm holes better be pretty finely finished and shaped so that as the screen it bent over these edges into its dished shape, it is not kinked and fractured.
BA.... You would have made one hell of a good teacher (Prof.) if you hadn’t become such a good Structural Engineer and mentor instead. The patience and fortitude of a saint, the ability to boil things down to their basics, posts 17MAY13 @ 14:43 and 21MAY13 @ 2:13, nice presentation on, 20MAY13 @ 12:21, even checking the algebra, etc.
GregL.... You must be clairvoyant if you saw the wire mesh net idea coming, by the third or forth post.
Spoonful.... I think BA’s post 21MAY13 @ 23:11 suggests the proper way to analyze this problem, but you seem to be quite a difficult person to convince. And, his post 22MAY13 @ 11:02 adds some significant detail to the thought process. He’s spent (wasted?) a fair amount of time trying to help you and offer sound advice, but in almost ever cycle you have only half read what he said and meant, or misinterpreted it in some way. Paddington is also offering some good advice and a thought process/method that we often use on these kinds of problems. What is your engineering educational and experience background, please tell us. If you won’t do that, I think you should take some engineering courses, so you gain some better fundamental understanding of what the guys have been talking about. They can’t solve the problem for you. They can explain and suggest, but you have to grasp what they’re telling you from the engineering standpoint. They can’t make you an engineer in one difficult post, and Roark’s book won’t make you an engineer either, if you don’t vaguely understand where that stuff came from. You need to have a good fundamental understanding of the engineering concepts involved if you want to do what you are trying to do.
RE: Simple Beam Questions
I do not claim any expertise with FEA, but FE programs can analyze 3 dimensional objects, so I'm not sure why they would have difficulty analyzing this. It is like analyzing the strings of a tennis racket except that the strings are not fastened to the frame of the racket. Instead, they run through the frame and fasten to a square frame further out.
BA
RE: Simple Beam Questions
First appreciate that you have spend all the time reading through all the previous posts. It is the fact that I do not have extensive experience/knowledge in structural, that is why I am seeking advise here. It is Ok that you not helping, and you may save your precious time to spend on something else.
I appreciate all your kind helpful posts, especially BA, and I do due to my limited knowledge, having trouble understand some posts. I don't see anything inappropriate here by asking further questions for things that I don't clearly understand, if there are people willing to help.
Paddingtongreen,
Good point, the wire would have already been yielded during woven process. To just have a conservative result, could we assume there are 14 wire in the same direction? rather than 7 by 7 wire at 90 degrees? and ignore the effect of its joints?
RE: Simple Beam Questions
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
BA
RE: Simple Beam Questions
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Simple Beam Questions
Typical Wire Spanning 3mm
Diameter D = 0.16mm
Area A = 0.020mm2
Yield strength Fy = 170MPa
Yield tension Ty = 3.4N
Unit Load w = 0.21N/mm
Factored Unit Load wf = 0.42 (using S.F. of 2)
Factored Moment Simple Span Mf = 0.42*32/8 = 0.472 N-mm
Sag required = 0.472/3.4 = 0.139mm
Neglecting bending moments and considering only cable action, the longest wire can resist its tributary factored load wf if it can strain enough to sag 0.139mm at midspan. The remainder of the wires will be less critical.
Ultimate failure will occur when the wire reaches its ultimate stress which will occur when it reaches its ultimate strain (a property of the wire). Sag and failure load will vary accordingly.
Assumptions made above
1. For a cable T = wL2/8s where s = sag
2. Cables are anchored adequately at the ends to develop not less than yield strength and preferably ultimate strength.
BA