Sizing a radially ribbed plate 1

[sarasudhar]

Hi all,
This is my first post in Eng tips

This is a problem in one of my design.(please refer the snaps for details).Initially I had a thick solid plate which is now replaced with two thin plates with ribs in between them.The plate is welded to a shell which is fixed and in operation,the plate is subjected to pressure due to which it tends to hog.For the thick plates,I would have used Roark's formula to arrive on the thickness of the plate for expected deflection.Now the problem is that since I am using radial ribs,I am not able to fit the problem into Roark's formula(Or else can it be fitted ?).I tried to solve the problem as beam(hence finding the Z value),But I don't have the clarity of Z value for the problem(In centre there are ribs but in some parallel section,there are no ribs).My boss wants me to perform some calculation to approximately size the ribs so iterations in FEA can be reduced.To be frank,I am not even sure where and how to approach the problem.I did some searching in Internet but could only find some empirical relations.
I have only 3 months experience in this job.I can manage some things but this seems way over my head.

Please guide me in this problem and I can learn and practice.Books,papers or approaches will be very helpful.

Please forgive me If I have misstated something.Thanks for going through this post

 

[BAretired]

Welcome to Eng-Tips, sarasudhar.

Your sketch seems to indicate four ribs 90 degrees apart. I would have preferred to see eight ribs, although I'm not sure that you need them.

What is this assembly you are designing? What is the diameter of the plates? What is the diameter of the hole in the middle? How does the assembly become attached to the cylindrical shell without access to weld all parts? What was wrong with using a thick shell as first envisioned?

Total upward force on the assembly is F = P*Π*D²/4 where P is pressure. With n ribs, assuming a simple span, each rib has a reaction of F/n where it meets the shell. Each diametrically opposite rib pair carries a total load of 2*F/n and has maximum moment at the center of the circle. The central pipe, together with a portion of the top and bottom plates must carry this moment as the rib is not available at the point of maximum moment. The center pipe wall probably has to be at least as thick as each rib.




BA

 

[sarasudhar]

@BAretired Thanks for replying
I have attached the document with more detailed design that i made yesterday.Sorry for not attaching at first itself.Hope this provides more detail.
I'm still confused with the zone where there are no ribs .Is there some book for this ?.Please recommend me if there is one for this.
How to find the Z value of the section? I used PRO E to calculate IXX and IYY values for sections with and without ribs .I have found ymax and ymin for both sections.So I have the Z value of sections with and without ribs..There are two section moduli and i am not sure how to manipulate them.

If I am missing something very elementary,Please tell me so I can improve.

 

[BAretired]

1. Before worrying about the S or Z values of the rib, I suggest you start with the shear strength of the rib and the welds to the shell. I don't know the pressure or the diameter but if you were originally contemplating a six inch thick plate, the pressure must be very high.

2. I suggest placing eight ribs radially at 45^o to each other. That will give better access for welding and will equalize the span for the lower plate which must sustain the same high pressure.

3. If the total force on the plate is F, each rib will have a maximum shear of F/8. The area of the rib and the welds must be adequate to safely sustain that shear force.

4. If the assembly is to be used many times, you must decide on a reasonable safety factor under cyclic loading.

5. It would be helpful if you could provide some idea of the magnitude of the pressure to be expected and the diameter of the plates.


BA

 

[dhengr]

Sarasudhar:
I agree with BA on the 8 ribs at 45̊, as a minimum. You need/want symmetry for this structure and I believe this type of upper structure is called a press platen. How are you applying the max. force, and what is it, and how much can the platen plate pressure vary? You also need to improve your drawing presentation methods, in that you need to show things much more in their true proportions. You would be surprised at how important these correct proportions is for an experienced engineer in making his first judgements and comments. Is this a large tire press? You have to give more detail on what you are trying to do and how.

BA look at his last attachment. This thing has a 1" thick outer shell 25.6" high, and has an O.D. of 74.6". The hub is 25.6" high, 2" thick, with an O.D. of 25.6". The platen plate is 1.25" thick. Nothing like the proportions of his diagrams.

This is a two part problem: one is structural strength and welding design problem; and the other is likely a flatness range for the platen plate under load. And, this latter will likely require secondary ribs, btwn. the 8 primary radial ribs, circular in shape, at increasing radii; or a thicker platen plate. Many times these platens are done in layers of structure to try to achieve both objectives. In a fairly crude first form of structure, consider: 1.25 - 2" platen plate with a 74.6" O.D.; atop the platen plate are WF beams, side by side, and lengths trimmed to about 70" O.D., and concentric with the platen plate; atop the WF’s are two more beams, the loading beams, perpendicular to the WF’s, and maybe 18 or 24" apart. These load beams might vary in depth to adjust their deflections, as they apply a lesser load to the WF’s as you move further from the center of the platen. The trick is to adjust the WF and load beam stiffness’, or their defections, to maintain a flat platen plate. I think this is sometimes called an elastic matching problem.

 

[BAretired]

Sorry, I looked at the attachment but missed the dimensions. Blind as a bat. I will take another look tomorrow.

I'm not quite clear on the loading arrangement and I don't understand what happens inside the 550 dia. hub.



BA

 

[dhengr]

BA... no need for you to be sorry, what you said was right on the money, given his proportions and lack of meaningful info. That was one point that I was trying to make. By his proportions, the ribs would just be members loaded in shear, not much bending involved, but welding access and detailing would have been a dog. He needs to give us much more detail if he wants more help, because I have the same questions you do.

 

[sarasudhar]

@ BAretired and dhengr Thanks for replying to the post.
Iam exteremely sorry for my non informative pictures.Hope I can improve in future.
This is actually not relevant but I am a "her" not "him" and its ok that you didn't notice because my handle name is a bit foggy.
I have attached the snaps which show why 45 deg apart ribs cannot be used.And dhengr you are right it a 66" mould tire curing press.The reason for providing the hub is also shown in the snaps.
I have started with shear load calculations for the ribs and the welds.Diametrical ribs cannot be provided because of the mould mounting holes that will be clear from the snaps.
We are on a preliminary stage of design and I am allocated to the upper housing subsystem.It is the reason I am not able to show many details fully.

 

[sarasudhar]

I missed another two attachments.Sorry

 

[prex]

Assuming the force of 521 T is evenly distributed on the annular area (though this doesn't seem to be what your sketch says), the pressure on the lower plate should be 2.3 MPA or 23 bar or 330 psi.
Now the first check would be the thickness of the lower plate. The behavior of a stiffened plate is dictated by the largest circle you can inscribe between the ribs. In your case the diameter D of this circle should be close to 600 mm.
You can take the required thickness of this plate as t=0.45D[√](P/S). Assuming S=200 MPa (30 ksi), the minimum thickness is not far from 29 mm, so here you seem to be OK, possibly a little too close to the allowable limit.
If you want to stay with ASME VIII Div.1, we can take the formula for stayed flat surfaces in UG-47: t=p[√](P/SC) where C is a coefficient that ranges between 2.1 and 3.2, and p is the pitch between stays that, in a square arrangement, would give D=1.41p, so we end up with something like t=(0.4 to 0.49)D[√](P/S), quite close to the formula above.
Note however that, if stiffness is a concern, the local deflection of this plate will be much larger than the deflection of a 150 mm thick plate.
Your placement of the ribs is not bad in my opinion, as my main concern would be the bending in the center hub caused by the ribs: as these are quite tangential to the hub walls, that is OK.
For the calculation of the ribs you can take 1/8th of the load on each one and check it as a beam with a distributed load (or somewhat concentrated in the middle if you consider this more realistic), having a T section, where the flange is formed by the lower plate extended 10t on one side and midway of the parallel ribs on the other side.
The upper edge of the ribs should not go compressive, so buckling is not a concern here.
Now you are left with checking the welds. Good luck!

prex
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[prex]

Oh, and the beam representing the ribs is a cantilever if you take a single rib, or a simply supported one if you take two ribs as a beam spanning over the diameter of the outer shell (and the result in the two cases will be quite the same, of course).

prex
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[dhengr]

Sarasudhar:
Sara is not at all foggy, if that is how we should read your handle. I’m sorry for the her/him mixup, I rather like seeing women in our profession, they certainly add something special to the profession of engineering. Keep at the learning thing, that’s a life long effort, and you’ve just barely scratched the surface having finished college.

I think you want more symmetry in your stiffener layout, to manage the platen plate deflection and flatness. The major loading is the 521 tons when the top is locked down and under pressure, right? The secondary loading is the inner/upper mold half bolted to the platen plate, probably only a few tons. Why not look at the 8 stiffeners at 45̊ and just rotate them 22.5̊ w.r.t. the bolt layout. Actually, this upper platen structure is supported at the center hub by the hydraulic cylinder and at the outer edge of the platen plate by the splined locks, is that right? Thus, it is not cantilevered from the hub, but rather simply supported out at the 74" + O.D. (the splined locking means) and semi-fixed at the hub, is that right? Or, does it span the full 74" dia.? Consider making your stiffeners out of split WF beam sections (ripped down the middle of the web) or fabricated tee sections. With today’s cutting equipment, they can cut the top flanges to a radius to match the hub O.D. and the outer shell I.D. Also, make your stiffener ribs a constant depth or a straight line sloped top. Don’t do the several sloped steps unless you have a reason for this. I assume that the 512 tons is the total force on the platen plate, and not multiplied by 4. Does the pressure vary with the radius from the hub to the outer shell, or is it uniform over the platen plate? These are all things you have to think about to refine your design, and we need them to start to understand and help you. It seems that Prex has some empirical rules-of-thumb for plate stresses, but you might also want a platen plate deflection map, and you should be able to get that from your FEA.

 

[BAretired]

Let us review the situation.

The pressure inside the bladder is 30 bars which is equivalent to 30*10⁵Pa or 3000kPa or 3MPa or 435psi.

The upward force on the ribbed plate is given as 521T where I assume T represents a Metric Ton (tonne) and is a unit of mass, not force. So the equivalent force would be 521,000kgf or about 5,094kN or 1,146 kips.

The upward force is uniformly distributed over an annular ring with inside diameter of 1302mm and outside diameter of 1542mm, the area of which is about 536,000mm². The pressure on the ring is approximately 9.5MPa or about 1378psi.

Do I have this right? I ask because it seems to be quite different from the pressure found by prex. It is not clear how such a high pressure could be developed, so this needs to be reviewed.

The ribs should be spaced equally around the circle, i.e. at 45 degrees. The closely spaced holes that connect the mould and the top plate seriously weaken the plate in the vicinity of the load. Their location needs to be reviewed.



BA

 

[BAretired]

sara, are you still with us?

How did you come up with 521T? It seems that the total force upward (or downward) should be F = P*A where P is 30 bars or 3 MPa and A is (pi/4)*1845². So F = 8020kN which would equate to about an 820 Tonne-Force.

BA

 

[sarasudhar]

Sorry For Late reply.I took some days off from work.
@Dhengr
The Shell is rolled and welded member.
The Top plate is welded to the shell and the hub is welded to the top plate.
The ribs are welded to the hub,Plate and the shell.The hydraulic cylinder is mounted to the hub,and its piston rod is connected to a male spline.The mould has female splines.The splines are interlocked so,when the cylinder actuates the segments of the mould move axially resulting also in radial movement because of dovetails.
The platens are fastened to the top plate.
Also the pressure in the annular area is uniform.
Sorry if this is still confusing.I can try explaning again if this is not clear.

@ BAretired
The total upward(downward) force is calculated from the pressure inside the mould and area inside the mould.The max. mould o.D is 1710mm.This is the O.D. and the I.D is actually the Cured tire O.D.The cured tire O.D is around 1470mm.
The calculation is a bit complicated.But based on the mould geometry,the customer has given this value as input 521T for which we should design locking and squeezing mechanisms.
Also the Pressure you found out was right.(I think prex took the diameters as radii).
I have modified the rib pattern to be equally spaced around the hub.(I have attached the images).This now has welding accessibility and has been accepted by manufacturing team.Sizing the ribs is the next task ahead.
Can I take one set of rib (Diametrically Opposite) as a simply supported beam ?
How can I account for the hub then ?
I know I have been asking the same question.But I cannot find any references for this.
Thanks in Advance

 

[sarasudhar]

@dhengr
The male part in the piston rod and female part in the mould are bayonets.These provide a positive locking."Spline" was not the correct word there.

 

[prex]

No, I didn't mess up diameters and radii, I simply took the full area of the top (or bottom?) plate: OD=1845 ID=650 with 521 T (taken as 5.21 MN, don't know what kind of tons this is) gives 22.2 bar. This is to make the very first check about plate thickness, after that you would need to check more local effects taking into account the actual distribution of pressure (that is not necessarily uniform over the area with OD=1542.4 and ID=1302.4). Another consideration is that the heating plate could also help in redistributing the pressure, but I don't know what this plate is made of.
Now that you have displaced the ribs (but see also below) we can check again the thickness of the top plate also taking into account the position of the load: the diameter D of the circle that you can inscribe between the ribs and the outer diameter of the loaded area (OD=1542.4) is 427 mm. Pressure is now: 4*5210000/[π](1542.4²-1302.4²)=97 bar
So, using the formula in my previous post, the minimum required thickness is t=0.45D[√](P/S)=42 mm , so the thickness of 32 mm you quoted somewhere is not sufficient (however this depends also on the allowable stress S, taken here as 200 MPa).
Concerning the hub, that you are correctly worried about, you need to calculate it as a ring subject to 8 concentrated radial loads. Roark has the formula, that gives, for a load W: M=0.27WR/2 where R=300 mm is the mean radius of the hub. To evaluate W we can observe that the load on the hub from the ribs varies linearly along the height of the hub from zero (?not so sure of this) to a maximum and that the moment of this distribution must equate the moment in the middle of a simply supported beam representing a couple of opposed ribs.
As you see things become complex here, and I cannot go on with all the details here.
My gut feeling is that the hub will not resist those loads and also (this is my humble opinion) that the previous arrangement of the ribs was better, as there was much less bending in the hub.
What you could do now is to add another plate on top of the ribs (as you showed in your first sketch) or even just an annulus around the hub (but don't ask then me how to calculate it, it requires a bit of effort).



prex
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[BAretired]

Can I take one set of rib (Diametrically Opposite) as a simply supported beam ?
How can I account for the hub then ?
I know I have been asking the same question.But I cannot find any references for this.

I believe you can take one pair of ribs diametrically opposed, together with the hub as a simple beam spanning the diameter of the shell. You need to know where the load is acting for a proper analysis (and I guess I am still a bit unclear about this).

The ribs are deep beams which means that conventional beam theory does not strictly apply. Primarily, the ribs are acting in shear but there are bending stresses which are not linear (as they would be in beam theory). The hub feels radial compression near the bottom and radial tension near the top which is applied by eight ribs acting 45 degrees apart. This results in bending and torsional stress in the wall of the hub.

Design of the plate is another issue and the way the plate reacts to the load will determine the load distribution to the ribs and shell. It is not clear to me how pressure on an annular ring of the plate can be three times the pressure in the bladder. In fact, I do not believe that can be true, so this needs clarification.

There are four groups of five holes spaced close together, 90 degrees apart. These will provide a radial yield line through the center of each group of holes, seriously weakening the plate in bending. Can they be relocated to a more favorable location?


BA

 

[dhengr]

We have a catch-22 here, and I believe I might be seeing the error of my ways here. Prex has been arguing for the sets of two ribs at 90̊, and this does protect the bottom plate at the bolt holes better, that starts to solve BA’s issue with a yield line forming on those bolt lines. This rib arrangement probably also passes the rib moments and shears through the hub a bit more cleanly/clearly (although not so simply). My dislike was the large area of bottom plate left unstiffened by this rib arrangement, and the resulting higher plate deflections, and thus my want for more symmetry in the rib layout. I think I would be inclined to go back to the double ribs at 90̊, and then to stiffen the bottom plate, I would add 4 more single ribs at 45̊ to the double ribs. To gain more rib symmetry at the hub, and to improve welding clearance, I might splay the double ribs so they were a few inches closer together at the hub than they are out at the outer shell. Maybe even dividing both the hub and the shell into 12 equal arc lengths for the rib locations. There are still plenty of unknowns about how this thing is loaded and handled and what the loads are and where and where the reaction points are. Is there a cylinder load, downward, at the hub when this thing is under pressure, so the ribs only span dia./2, not the full dia.?

 

[BAretired]

If the total force to be carried by eight ribs is 521T (5094 kN) then each rib has a reaction of 5094/8 = 637kN. Each rib is 20mm x 650mm with an area of 13000mm², so the shear stress is 49MPa, not a very high shear stress.

If the load on each rib pair is considered concentrated at the mid diameter of the annular ring (1710 + 1470)/2 or 1590 and the span is considered to be 1870mm (c/c shell), then:
M = 637(1870 - 1590)/2 - = 89,180kN-mm or 89.2 kN-m
For each rib without bottom plate:
S = 20(650)²/6 = 1,408,333mm³
Z = 20(650)²/4 = 2,112,500mm³

So f_b = M/S = 63 MPa using beam theory and neglecting the plate as a flange in the bending of the rib.

Or alternatively, Mp of plate = 0.9*Z*Fy = 570 kN-m >> 89.2

However, I am still not convinced that the load acts at the mid-diameter of the annular ring. What holds the hub down? It must have a pressure of 30bars pushing it up. If so, the design force becomes much greater and it is more nearly centered which makes the above calculations wishful thinking.





BA