Rating a pipe stand

[DrewJ88]

I work out at a mine site, and we have a number of home made stands which are used on our heavy machinery. Most of the stands are constructed using a pipe as the column and then they have a base plate and a top plate. I've analyzed the main column using the Euler, JBJ, and Secant methods for column buckling.

However, I'm struggling analyze the top plate. Most of the stands have a diameter of 6 inches or larger and a wall thickness of 0.5 inches or larger. Then the top plate is simple a 0.75 inch thick plate welded to the top. Some have a caved in top plate where the plate has begun to yield and deform. What is the best method to analyze that top plate to give the stand an overall load rating?

Any suggestions would be helpful.

 

[Ussuri]

You have a vertical pipe with a flat plate welded over the end? You place a load on the plate from machinery so it transfers through the plate to the stand? Correct

A simplistic approach would be to consider it as a flat circular plate. Depending on whether edge weld is sufficient to transfer any moment you could look at the edge conditions as fixed or simply supported. Roarks and Timoschenko would give you some guidance.

You state that the plate has yielded. In which case the applied load must have transfered to the pipe wall (assuming the object supported is larger than the OD of the pipe). Does this mean that they are still used as a support? In which case the capacity of the end plate is irrelevant. The axial stiffness of the pipe wall will be way greater than the bending stiffness of the end plate.

 

[DrewJ88]

My concern is not for the column itself, but rather having the load punch through the top plate. Perhaps yielded wasn't the proper term, but rather plastically deformed. If the load had a footprint large enough to spread across the top plate, there wouldn't be an issue. However, the stands are getting point loaded which causes the plate to cave in. How do I determine the maximum point load in center of that plate?

 

[swearingen]

As Ussuri said, Roark or Timoschenko is the way to go. Roark's has this exact case listed.


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[MiketheEngineer]

What kind of loads are we talking about??

 

[DrewJ88]

Anywhere from 20 to 50 Tons depending on specific stand design. Also, I don't have a Roark's or Timoshenko reference book handy.

 

[swearingen]

You'll need to find somebody with a Roark's. Looking in mine, they have a case that would probably suit you nicely (Table 24, Case No. 9), but the formulae and tables are prohibitively large for me to post here.

If you give me all of the pertinent data, I could bang out the calc for you pretty quickly (years ago I wrote a spreadsheet with all of the L's, F's, C's, & G's - for those of you that know what I'm talking about, yes, I was bored).

I have the (outer?) diameter of the pipe at 6", the pipe thickness at 0.5" and the plate thickness at 0.75". Exactly how is the load applied? You said it is a point load, but there is no such thing for 50 tons. I need some sort of footprint to put on the plate. Also, what is the rate of loading? Is it slammed down or let down slowly?


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[DrewJ88]

Ok, so my Roark's (7th Addition) just arrived and I see my case in Table 11.2 Case 19. However, the formulas are in terms of y and M based geometric variables and the load W. I want to find out the maximum that W can be, so how do I relate the M or y back to the maximum allowable stress of the material.

Usually for a beam I would use stress=(M*c)/Ix and then knowing the equation for M I could solve for a maximum load. How do I relate the Roark's formula for M to determine the maximum load W? Any suggestions?

 

[DrewJ88]

Anybody have any suggestions or guidance?

 

[swearingen]

The moments given are in force-length per unit length. That means you can take a unit width of your plate, calculate the I of that unit width, and get a stress based on the given moment using Mc/I.

I don't have access to a 7th edition Roark's, so to be sure we're on the same page, Table 24, Case 9 in my 6th edition is entitled:

Table 24 - "Formulas for flat circular plates of constant thickness"
Case 9 - "Uniform annular line load"

With Case 9a being simply supported and Case 9b being fixed at the edges.


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[DrewJ88]

Ok, so if I had a circular plate of diameter = 8" and a thickness of 0.5" then the I would be 0.083 in^4 (I=(b*h^3)/12) with b=8" and h=0.5". Then I can plug in the Roark's equation for Mmax and solve for W. Correct?

The case I'm looking at is:
Uniform load over a small eccentric circular area of radius r0; edge fixed

 

[swearingen]

No, that is not correct. Again, the moments are given in force-length per unit width. To put some actual units to it, if you're in the US, your moment would come out in in-lb/in. Since you're working with a single inch width, your b in the I equation would be 1, giving an I of 1/96 in^4.

Also, the case you're looking at is less conservative than the "Uniform load over a very small central circular area of radius ro, edge fixed" case. If the load is in the center, the moments will be a little higher. Unless, of course, you can guarantee that the load will never be in the center...


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[DrewJ88]

Thanks for all your help on this swearingen, I really appreciate it. I've taken you last bit of advice and built myself an Excel sheet to calculate the maximum allowed load. I've attached it. Could you take a quick look at it for me? Any mistakes or false assumptions?

Thanks again in advance.

 

[swearingen]

You show Wr & Wt in lb*in and qr & qt in lb/in. W is just a load, in this case measured in lb. q is a load per unit area, in this case measured in psi. I understand why you are breaking it out to say that Wr is the load that would provide maximum Mr, but don't lose sight of what they mean. In this case, you don't need qr or qt at all - you're just looking for a load rating. Your answer is the min of the absolute value of the -9451 and -31502, measured in lbs, or 9451 lb. Note that the logic in F34 is backwards; it is taking the absolute value of the min.

So - the corrections:

1. Eliminate the qr and qt calc lines (B28:G28 and B31:G31)
2. Change the units in G27 and G30 to lb.
3. Formula in F34 should read =MIN(ABS(Wr),ABS(Wt))
4. A small, inconsequential error - the formula in F23 should have 0.675 instead of 0.65




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[swearingen]

Also note that the case you're using in your sheet is for fixed edge. We know that your case is better than free edge, but not as good as fixed, so the answer it gives will be somewhat unconservative. However, your safety factor of 2 should eclipse that.

Also, as you would expect from the equation for I, your rating will vary greatly by the plate thickness. Capacity more than doubles for a 3/4" plate and a 1" plate gives 4 times the capacity.


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[DrewJ88]

Awesome! Thanks again. I've just got one item that I'm confused on. That would be the units. When I look at the formula for Mt and simply look at the units, I consider moments to be in lb*in which then breaks down the W to have to be in lb*in as well for the formula to work. So if W is indeed in lbs, then that means Mt is also in lbs.

You said before that moments are given in force-length per unit width. That would of course be (lb*in)/in which is lbs. I guess I'm just hung up on that concept, and I haven't yet found anywhere in the Roark's book that explains this concept.

I guess I'm still missing something. Perhaps you can clear that up or direct me a chapter or passage within Roark's that discusses this.

Anyway, I've just got to say one more time...thanks for all your help and guidance.

 

[swearingen]

In the 6th edition, at the beginning of Table 24 (pg 398), all of the variables used are defined:

"W = total applied load (force)
q = load per unit area
Mr = unit radial bending moment
Mt = unit tangential bending moment

Bending moments can be found from the moments Mr and Mt
by the expression sigma = 6M/t^2"

So, W is indeed in lb and Mr and Mt are in lb as well (lb-in/in as you said - "unit bending moment"). The 6M/t^2 comes from sigma = Mc/I with c = t/2 and I = bh^3/12 with b = 1 and h = t.

The units do work out - you just have to get used to their convention.






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[DrewJ88]

I'm with you, it's just unusual for me and I will just have to get use to it.

Thanks.