huangyhg

pressure in a cylinder
yes. the stress is proportional to the diameter/thickness ratio. to keep the stress the same, the wall thickness must be increased by the same percentage as the diameter.
can i get some oppinions:
we are a aerospace sheet metal house that is using microcadum 2d software to develope all of our flat pattern and lofting data. we want to go to solid modeling for the sheet metal, i use catia v-5 for mechanical part modeling i.e. forgings, castings. how well does catia aerospace sheet metal package work? we are looking at solidworks also but are not sure, sw does look much more intuitive to use and learn than catia.
any input is valuable
thank you
what you are refering to is the hoop stress. it is = pr/t.
p= pressure
r= radius
t= thickness
nigel waterhouse & associates
aeronautical consulting engineers
transport canada and f.a.a approval & certification of fixed and rotor wing aircraft alterations: structures, systems, powerplants and electrical. faa pma, tc pda.
pebs
cylinders have a complex stress condition of hoop stress as given by nigel above and axial or longitudinal stress pr/(8t). i suggest you consider using complex stress failure criteria to analyse the cylinder. von mises is widely used for ductile materials. the attachment of the "dome" ends of the cylinder are also important and can give rise to local bending stresses at the interface to the cylinder. i suggest you look at den hartog's book "advanced strength of materials" it has quite a good section on this topic of complex membrane stresses.
ed
pebs: if you are interested in deformation rather than strength, then the answer is still "yes," but how much to increase wall thickness by depends on which deformation you are interested in, e.g. the change in diameter half way along the cylinder, change in length, etc. it also depends to a degree on the form of the cylinder ends (flat vs. hemispherical vs. other).
however, if strength is your concern the above pretty much covers it.
nb: for a relatively modest outlay a copy of roark's formulas for stress and strain (also has deflections!) can answer many fundamental questions like this. for your purposes you'd probably be best off finding a second hand copy of ed. 4.
-r.
pressure vessels must be designed to protect against several different modes of failure including plastic collapse, fast fracture, and fatigue failure. the hoop stress in a thin walled cylindrical pressure vessel containing gas at a pressure p is given by
hoop stress = pr/t
where r is the radius of the pressure vessel and t is the wall thickness. as you can see from this equation, if you hold the pressure and wall thickness constant as the radius of the vessel increases, the the hoop stress increases as well. if you increase the thickness of the walls in proportion to the amount you increase the radius so that the ratio r/t doesn't change, then the hoop stress will remain unchanged. so your instincts are correct. if you increase the cylinder diameter, then you will also have to increase the wall thickness to maintain the same level of hoop stress.
but there are other failure modes that may need to be considered. for failure by general yielding we have
hoop stress = yield stress
for failure by fast fracture we have
fracture stress = kic/[(3.14*crack length)^0.5]
if we plot stress versus crack length, the criteria for general yielding and for fast fracture meet at a critical crack length, call it ac. for crack lengths greater than ac, the structure will fail by fast fracture at a stress less than the yield stress without warning and with potentially catastrophic consequences. if the largest flaw size in the vessel is less than ac, the vessel will be safe, assuming an appropriate safety factor has been built in.
if the pressure vessel is subjected to cyclic loading, cracks can grow by fatigue, and a vessel that was inspected and passed as safe may become unsafe due to crack growth. we can protect against this failure mechanism by designing the vessel to leak before it breaks. we do this by specifying that the critical flaw size ac must be greater than the wall thickness. then gas will leak out through the crack before the crack is big enough to run. to be on the safe side take
ac = 2t
and since
kic = stress*[(3.14*ac)^0.5]
the maximum stress is given by
max stress = kic/[(3.14*2t)^0.5]
in order to attain this extra safety, either the pressure must be lowered, or the wall thickness substantially increased.
maui

__________________
借用达朗贝尔的名言：前进吧，你会得到信心!
[url="http://www.dimcax.com"]几何尺寸与公差标准[/url]