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cylindrical shell stresses

  • Thread starter Thread starter luigi.paiano
  • Start date Start date
luigi.paiano said:
but look at that you are talking to one who makes the square bouquet every day and if known I never ask simple questions but at the most I look for opinions!!
However solved the mystery, we can consider longitudinal welding.
the welding is solicited (in the worst case, i.e. when located at distance r from the black axis) by bending + axial effort due to p in its cross section and by a perpendicular effort due to the circumferential one.all this if there is no torque moment,otherwise you must add the tangential effort in the longitudinal section of the welding
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Oh, don't warm up, I was just joking! If you do it in the square, it's normal....there are those who then do friselle sciences or worse than calendars and who maybe becomes minister. .
Hi.
 
on the longitudinal welding look at this exercise. It solves it taking into consideration the welding to complete penetration.The thing that does not understand is why it did not consider the axial stress due to the internal pressure
 

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luigi.paiano said:
on the longitudinal welding look at this exercise. It solves it taking into consideration the welding to complete penetration.The thing that does not understand is why it did not consider the axial stress due to the internal pressure
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because perhaps it is forgotten, or perhaps because the so-called axial sigma due to the pressure is distorted into a tangential component variable from the intradox to the extradox becoming

maximum intradox = [(Pres*Dint)/(4*Spes)] tau mass extradosima = [(Pres*Dint)/(4*Spes)]-(pres/2)

that has not analyzed
 

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luigi.paiano said:
on the longitudinal welding look at this exercise. It solves it taking into consideration the welding to complete penetration.The thing that does not understand is why it did not consider the axial stress due to the internal pressure
Click to expand...
neutral axis of the moment stinging on the welding??!!!? ? ? : eek:: eek:: eek:: eek:
Where did you find him?
for the axial p is the "philosophical" speech of yesterday..
ahhhh... I had not seen the axis of the moment ehehhehe
 
This is material of the university of bs.
However it is a fact that the flexional sigma is worth zero in points 3 and 9 (watch reference) which becomes a neutral axis
 
for the flexional sigma okkkkk but for the axial one due to p?????? ? There's something that doesn't fit. If the welding had been a corner cord, I wouldn't have had to consider it because it's a stress of traction/compression in the cross plane of the cord but this is a complete penetration
 
to complete penetration is part of the base material, that is to be calculated as it was a piece of tube only that could have different resistive characteristics.
If it were angle cord, I think it is subject to the axial force of pressure.
 
then for corner strings the norm says:Eventual tensions of traction or compression present in the cross section of the cord should not be taken into account for the purpose of the sizing of the cord itself.So for the axial sigma due to p what is done?
If in fact you see the drawings of the legislation I think it is understood that the axial stress should not be considered if you were talking about angle strings. but here no!Also if you look at the design of the prof you see the parallel sigma? He designed it.I don't want him to have confused the law and he didn't think it by accident.
 
on the uni en 10011 cf. 5.1.2.2 is indicated not to use parallel sigma in order to assess the verification of the cord "as a durable part of the section". This is referred to the corner strings.

5.1.1 is the calculation of the ideal sigma with parallel sigma. this is referred to the joints of head, to t and to complete penetration.

According to one in 1993-1-8 (Eurocode3) cf.4.5.3.2 is not considered parallel in the directional motion.

Although not reported in legislation it is said that it is scrupulous to verify the static resistance to yielding or breaking including the parallel sigma term as reported by old norms uni en 10011 no longer in force.
 
then we continue the exercise. assume that the thickness is about 34 mm.
c.ecco I tried to do it, but the alternating stresses equivalent (considering the average sigma also) seem insignificant, that is they are all below the limit of fatigue.
What do you say?
 
I ignore the gassner curves, what are they? the auditions? I don't think so.

then with what loads? mf and p?
Can't you make miner?
 
:d..cmq I have simply considered the axial stress given by the sum of that due at the moment of bending and that due to the pressure and circumferential stress.poi I have applied von mises, aspect your news:d.
 
luigi.paiano said:
:d..cmq I have simply considered the axial stress given by the sum of that due at the moment of bending and that due to the pressure and circumferential stress.poi I have applied von mises, aspect your news:d.
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But you did von mises with what? the two sigma are added or subtracted ( worse and better). have you considered mt that is not there but have we put it x do the prevented case?

on the part of gassner I really don't know where to put my hand, I still have to study:
 
as they sum up or subtract. They are two tensions on two different planes. the torque moment I have not considered it.
Circum­ferential sig­ma­ture = pd/2t
Axial sigma = pd/4t + m/j d/2
 
the gassner curve is a particular curve that is built when a load story is known in which the alternating sigma varies!
 

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