Crushing large pipe with diagram

[y061q]

I am having a bit of a problem. I have a 30" OD pipe (1/2" wall) on a 17 degree angle and have a 100ton weight on it. The pipe is fully supported, so I am assuming that the deflection is minimal.

I can calculate the bending stress and shear but is there a way to calculate the rupture stress (not sure if that is a proper term). So under the assumption that there is minimal bending would the 100t weight (assumed to be a point load for simplicity) crush the pipe. How would you prove or dis prove this.

Thanks for the help.

http://img269.imageshack.us/i/20110328155000.jpg/

 

[IDS]

Why would the deflection and bending be negligible? Even if the pipe is fully encased (which isn't clear from your sketch) you will need to look at the possibility of buckling.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[msquared48]

I would very stgrongly consider bridging over the pipe with that level of load to protect it from splaying out and failing.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[y061q]

If I may try to simplify it (or explain it better). How would I calculate the max point load before failure of the 1/2" wall pipe that is supported at the bottom only.

http://img717.imageshack.us/i/20110329074023.jpg/

 

[ishvaaag]

To ascertain the rupture of reinforced or precast-prestressed concrete you need to closely observe the nature of the stresses in the wall of the pipe, upon interaction with the passive and/or active reinforcement. On these premises you can deal with your question with the proper understanding of reinforced concrete and prestressed concrete technology.

If the solicitation is mainly of the flexural type the limit is usually called modulus of rupture of the concrete. Notional (reduced) modulus of rupture are prescribed as limits not to be surpassed for factored loads to get a safe design.

If the longitudinal tensile stresses are predominant in some thickness instead of the modulus of rupture of the concrete you need to look for the limit tensile stresses for the concrete. Again, reduced values are prescribed to be compared with solicitations from factored forces to get proper designs.

Upon tridimensional states of stress with the 3 axes are being significantly stressed it is usual to define some condition that is understood to signify the limit strength of the material; for concrete criteria like by Coulomb-Mohr or Willam are sometimes used.

Since the highest tensile principal stress is readily obtained by FEM one quite likely would get a quick appraisal of the situation by comparing it against the limit tensile stress for the concrete material.

It must also be understood that sections having significant part of the same well over the limit tensile stress of the concrete at factored levels are entirely common in reinforced concrete design, where the concrete cracks and transverse reinforcement restrain the crack in frequency and width; not so much in pipe design where normally is wanted the prestress keep the concrete without cracks.

In short, the generic answer to your question implies some knowledge of RC, PC and Pipe technologies. The values above named for the limits are in the general handbooks for the same.

Also to remark that if one wants some particular concrete item without cracks, one must never allow the actual stresses in the concrete to surpass their (at least probabilistic) limit values or comparison stresses. A safe design would require the imposition of factors to the load and notional material strength reductions. But if you allow the stresses go above the limits as described in the first part of this paragraph, you will be exceeding with the actual forces the actual strength and you will have essentially cracked portions of concrete (hopefully) restrained in place by the rebar if present. This leads to catastrophic results when pressure is involved at the interface, such outwards in pipes or inwards as in marine concrete.

 

[a2mfk]

What is the material supporting it? Other than concrete, I doubt any soil will not greatly compress/settle and you would have all kinds of problems. Agree with Mike...

 

[y061q]

The steel pipe would supported by a layer of 6"-8" shot rock over bedrock. This area is inter tidal so smaller bedding material can't be used

 

[ishvaaag]

By misreading I was understanding it was concrete pipe.

For steel you use a von Mises comparison stress that is usually readily given by the FEM program used in the analysis, against the yield value of the steel.

1. Analyze in FEM the pipe for the support and load you are surmising it will have
2. Go to the von Mises stress graphics in the FEM package looking for the highest value.
3. See if bigger than Fy.

All this supposing slow application of the loads. Otherwise you would have a dynamical interaction simulation problem also solvable in FEM but somewhat more difficult to set.

 

[prex]

You can treat it as a ring beam loaded by a top load and supported at bottom. You should define a contributing length of pipe, presumably the loaded length plus 10 times the thickness on each side.
This situation is treated by Roark and is also found in the first site below, under Beams -> Curved -> Circular rings -> Top load.
It is of course a rough approximation, but there isn't a closed form method for the actual condition, where a long pipe is locally crushed, except that WRC107 and WRC297 might provide something more realistic.

prex

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[csd72]

A few other items that need to be considered.

1. What other loads are applied to the pipe i.e. water loads, wave loads, internal pressure, soil loads, temperature, thrust, and shrinkage loads.
2. What implications are there for fatigue.
3. also if it is a sloping pipe then you may induce some axial loads that will increase the stress in the axial loads.
4. Where is the weld?

100 tons is a heck of a lot of load for that pipe and I would be very surprised if it could take it.

 

[ishvaaag]

In my last post, do not forget that appropriate safety factors for the coastal use must be included, for a proper check, i.e., you compare the von Mises stresses from factored loads against Fy.

 

[a2mfk]

I stand corrected, not a geotech but that is more substantial than what I was thinking. However, even if the pipe is OK I still may want a geotech to "bless" this...

 

[JStephen]

Some of the AWWA standards include methods for design of buried pipes due to external loading. I'm not sure if they're applicable to your case, but that would be preferred to starting from scratch on the problem.

 

[BAretired]

Let w be the line load on the top and bottom of the pipe. Then wL = P = 100 tons = 200,000# where is the length of pipe under the load.

Simple span moment = PD/4 = 200,000*30/4 = 1,500,000"#

Plastic moment of pipe wall, Mp = phi*Z*Fy = 0.9*0.5^2/4*50,000 = 2812"#/" (assuming Fy = 50 ksi).

If a plastic moment develops at top, bottom and both sides, 2Mp = 1,500,000 or Mp = 750,000"# so L required = 750,000/2812 = 267" or 22.2' to just cause yielding.

If a load factor of 2 is required, L should be increased to 44'. Needless to say, this is an extremely approximate solution.

BA

 

[y061q]

BA,

I am just running through your response and I thought:

Z = (d^3 - di^3)/6

= (30"^3 - 29"^3)/6 = 435.2 "^3

 

[BAretired]

y061q,

I am talking about the plastic modulus of the 1/2" plate per inch of length, so the dimension of the section is 1" x 1/2".

You are talking about the plastic modulus of a ring section which is not relevent in the present context. The load will try to squash the pipe causing four yield lines to occur. They will run parallel to the pipe axis, one top, one bottom and one each side of the pipe.

BA

 

[y061q]

That's great. Thanks everyone for all of the help.