I have an application where a vertical tube (2 in OD) is subjected to combined presure and axial loading. At one point, the tube has an opening for a branch (also of 2 in. OD) to form a tee. This tee is of a welded construction and formed from tubes. How do I analyse the effect of the opening?
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Combined axial loading and pressure induced stresses 1
- Thread starter scsekhar
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The simplest way would be to evaluate in accordance with one of the piping codes. Pressure qualification via area replacement rules, and axial loading in accordance with flexibility rules. Note that the rules will not directly consider axial loading, so you need to do a little more than Code rules. Suggest you consider the stress to be the stress in the cylinder due to axial load (F/A) times the greater of the full value of the in plane or out plane stress intensification factor (this will be conservative).
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I would suggest to check the following:
stress1 = (F/A) + pressure + [i * (Mbending / Z )]
stress2 = (F/A) + pressure - [i * (Mbending / Z )]
with:
1/ F = axial force , > 0 , if tensile and negative if compressive.
2/ i = stress intensification factor , between 1 to about 5. (But there are tables in B31.1 or B31.3)
3/ Mbending = total bendig moment,
Allowable is : S from code if sigma1 and/or sigma 2 are > 0.
Otherways , there is buckling ( which is never checked by ANSI 31.3, it is the design responsability to check it , by adding extra guides,.. ), and one has to use the ASME VIII div 1 , procedure, i.e, calculate A and find out the allowable compressive stress B in psi.
stress1 = (F/A) + pressure + [i * (Mbending / Z )]
stress2 = (F/A) + pressure - [i * (Mbending / Z )]
with:
1/ F = axial force , > 0 , if tensile and negative if compressive.
2/ i = stress intensification factor , between 1 to about 5. (But there are tables in B31.1 or B31.3)
3/ Mbending = total bendig moment,
Allowable is : S from code if sigma1 and/or sigma 2 are > 0.
Otherways , there is buckling ( which is never checked by ANSI 31.3, it is the design responsability to check it , by adding extra guides,.. ), and one has to use the ASME VIII div 1 , procedure, i.e, calculate A and find out the allowable compressive stress B in psi.
Hi , Guys,
My apologies , forgot to tell you in my above answer
to use this:
( pressure * Mean radius)/ ( 4* thickness )
[INSTEAD of ( pressure) alone !!!]
this is the longitudinal component due to pressure= 0.5* the circumferential one.
----------------
For: mean radius , Area= A ,and modulus of resistance
Z , introduce Corroded thickness and deduct also mill tolerance of pipe/tube which is -12.5 % or -10 % of the
pipe/ tube Nominal thickness.
For (i) = keep the nominal thickness, this is more conservative..
--------------------
Note:
Another more elegant (but time consuming) approach if system looks overstressed is to use stress classification as per ASME VIII div.2.
e.g. segregate stresses into primary , secondary , peak , etc..Each type or combination therof has a different allowable...
Caesar or FE pipe could do this , if so instructed.
My apologies , forgot to tell you in my above answer
to use this:
( pressure * Mean radius)/ ( 4* thickness )
[INSTEAD of ( pressure) alone !!!]
this is the longitudinal component due to pressure= 0.5* the circumferential one.
----------------
For: mean radius , Area= A ,and modulus of resistance
Z , introduce Corroded thickness and deduct also mill tolerance of pipe/tube which is -12.5 % or -10 % of the
pipe/ tube Nominal thickness.
For (i) = keep the nominal thickness, this is more conservative..
--------------------
Note:
Another more elegant (but time consuming) approach if system looks overstressed is to use stress classification as per ASME VIII div.2.
e.g. segregate stresses into primary , secondary , peak , etc..Each type or combination therof has a different allowable...
Caesar or FE pipe could do this , if so instructed.
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