Crushing large pipe with diagram
Crushing large pipe with diagram
(OP)
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/
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/






RE: Crushing large pipe with diagram
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Crushing large pipe with diagram
Mike McCann
MMC Engineering
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RE: Crushing large pipe with diagram
http://img717.imageshack.us/i/20110329074023.jpg/
RE: Crushing large pipe with diagram
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.
RE: Crushing large pipe with diagram
RE: Crushing large pipe with diagram
RE: Crushing large pipe with diagram
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.
RE: Crushing large pipe with diagram
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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RE: Crushing large pipe with diagram
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.
RE: Crushing large pipe with diagram
RE: Crushing large pipe with diagram
RE: Crushing large pipe with diagram
RE: Crushing large pipe with diagram
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
RE: Crushing large pipe with diagram
I am just running through your response and I thought:
Z = (d^3 - di^3)/6
= (30"^3 - 29"^3)/6 = 435.2 "^3
RE: Crushing large pipe with diagram
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
RE: Crushing large pipe with diagram