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  1. #1 Science question 
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    I am hoping that I have come to the right place for a push in the right direction. I have got the following question and I think I know the equation but when I put the values in it doesnt seem to give the correct answer, the equation I have are

    long= pd/4tnh
    hor= pd/2tnl

    A thin cylinder compressed air vessel has an internal diameter of 1,5m and is made of plate 10mm thick. The joint efficency of the longitudinal joint is 75% and the circumferential joint 50%. The plate has a tensile strength of 400MPa. What is the maximum internal pressure to which the vessel should be used if there is to be a factor of safety of 5?

    Any help welcome
    Thanks
    Anna


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  3. #2  
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    I think I can help you with this but first please explain the equations you are using and what the symbols represent, the answer you got and what the answer is supposed to be.


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  4. #3  
    Forum Isotope Bunbury's Avatar
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    These are the equations for longitudinal and circumferential stress in a cylinder with internal pressure. They are a bit confusing as written because more usually instead of "hor" we would use something like S<sub>c</sub> for circumferential stress and instead of "long" we'd use S<sub>l</sub> for longitudinal stress.

    The longitudinal joint is subject to circumferential stress, so you would use 0.75 for n<sub>l</sub>. The circumferential joint is subject to longitudinal stress so you would use 0.5 for n<sub>h</sub>.

    I would guess you may have mixed up the joint efficiencies

    Hope this helps.
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  5. #4  
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    Quote Originally Posted by Bunbury
    These are the equations for longitudinal and circumferential stress in a cylinder with internal pressure. They are a bit confusing as written because more usually instead of "hor" we would use something like S<sub>c</sub> for circumferential stress and instead of "long" we'd use S<sub>l</sub> for longitudinal stress.

    The longitudinal joint is subject to circumferential stress, so you would use 0.75 for n<sub>l</sub>. The circumferential joint is subject to longitudinal stress so you would use 0.5 for n<sub>h</sub>.

    I would guess you may have mixed up the joint efficiencies

    Hope this helps.
    Thats correct, I have not yet got used to codes on forums yet.

    I think the problem that I am having is that I am not calculating the pressure correctly.
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  6. #5  
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    The maximum allowable stress is the tensile strength/5 (in this particular case). You would calculate this value and use it as the left hand side of the equations you wrote, then rearrange to get the p corresponding to that stress.
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    The stress you have labeled "horizontal" is also called the "hoop stress" because an example is the stress applied to the hoops on a wooden barrel.

    Hoop stress = PD/2t
    Longitudinal stress = PD/4t

    Hoop stress is the stress that would tend to split the cylinder along the longitudinal seam. The longitudinal stress would tend to blow out the heads that are welded onto the ends of the tank with a circumferential joint.

    As you can see, the hoop stress is twice as big as the longitudinal stress so the tank will probably fail by splitting out the longitudinal seam. Even though the longitudinal seam is stronger than the circumferential seam, it is not twice as strong.

    Does this help get you started?
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  8. #7  
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    Ok I think I have an answer, hopefully it is somewere close!!!!!!

    I have the hoop stress = 0,5MN/m^2 (533N/m^2)
    And the longitudinal = 1,6MN/m^2.

    Harold, thanks for the explanation of hoop stress, it is always interesting to know were the names come from.
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  9. #8  
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    I'm afraid you still have the joint efficiencies backwards. Remember that hoop stress applies to the longitudinal joint and longitudinal stress applies to the circumfential seams.

    The allowable stress is 400/5 = 80 Mpa

    Rearranging the equations,

    for the longitudinal joint the allowable pressure p = 80x2x10x0.75/1500 = 0.8 MPa

    and for the circumferential joint p = 80x4x10x0.5/1500 = 1.067 MPa.

    This means the maximum pressure that the cylinder can take is limited by the longitudinal joint, and is 0.8 MPa.
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    NOW one question: If your "pressure vessel".. doesn't have ANY TOP and BOTTOM CAPS,, (basically your pressure vessel is a thin cylinder without caps under internal pressure), have we got longitudinal stress?.
    Thanks.
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    Quote Originally Posted by matteo_1234
    NOW one question: If your "pressure vessel".. doesn't have ANY TOP and BOTTOM CAPS,, (basically your pressure vessel is a thin cylinder without caps under internal pressure), have we got longitudinal stress?.
    How did you manage to build up pressure without any end caps?
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  12. #11  
    Forum Isotope Bunbury's Avatar
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    He is probably envisaging a short section of a long pipe, and forgetting that if the pipe is holding pressure, however long it is, it is still sealed at each end by some means.
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  13. #12  
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    This is just a "theoric" case,in other words: if you got an anular ring under internal pressure, the only stress is circunferencial?....what happen if the pressure changes from the inside to the outside (as an external pressure)... the ring see the same stress?...
    Thanks.
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  14. #13  
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    I would guess that a resultant external pressure would tend to force the seems together. The only way I can think of that the seems would then come into play is if the external pressure is of such great magnitude that the cylinder would collapse and fold inward along the seem, but given the fact that it is a cylinder (the pressure would be constant all over the surface) the required pressure for that to happen would be MUCH greater. IMHO :?
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  15. #14  
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    Quote Originally Posted by matteo_1234
    This is just a "theoric" case,in other words: if you got an anular ring under internal pressure, the only stress is circunferencial?....what happen if the pressure changes from the inside to the outside (as an external pressure)... the ring see the same stress?...
    What would it be like if there were no hypothetical questions?

    Yes the stress value is the same if the pressure is the same, but the stress will be compressive instead of tensile. As Kalster points out, the joint efficiency of the welded seam is not an issue when the shell is under compressive stress.

    For large diameter thin walled vessels failure under external pressure is more likely to occur by buckling due to out of out-of-roundness or flat spots, and design codes set limits on these manufacturing variables.
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  16. #15  
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    Are the ends of the cylinder flat plate, or concave and or convex end caps?
    Length of Pressure Vessel, unknown

    1.5 M. I.D. is 4'- 11 7/16" (59 7/16" I.D.) with a Circumference of 188.50 inches, = 15.70833 ft.

    10 mm. is not thin plate ! but = .3937" thick and is between standard plate thicknesses of 3/8" and 7/16" and also between 25/64" and 13/32" thick.

    What grade of steel? (Pressure Vessel Quality Plate)
    ASTM A-285 grade C
    ASTM A-515 grade 70
    ASTM A-515 grade 55

    Plates High Carbon-Hot Rolled .40/.50 Killed Steel

    What type of weld rod? (Beam Welded, Submerged arc )
    Bare wire and Flux, equivalent of;
    MAW rod; LH-7018, LH-8018, LH-12518
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  17. #16  
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    In pressure vessel engineering thin-walled is usually taken to mean any vessel in which the wall thickness is less than or equal to one tenth of the diameter.
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  18. #17  
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    Quote Originally Posted by Bunbury
    In pressure vessel engineering thin-walled is usually taken to mean any vessel in which the wall thickness is less than or equal to one tenth of the diameter.
    Thanks for that input! I've been out of the trades, a number of years now. A lot of things don't come to mind as fast as they once did. I was a welder for over 25 yrs. and a AWS welding inspector for another 10. On nuclear job sites, during the construction phase, and Welding Shops.

    I know you stated it's a Hypothectical, but in my minds eye, I like to see and consider all parameters, visualy before drafting it out, and see its construction mentaly.

    She stated thickness, I.D., and safety factor. No end caps or length.
    A ten foot long vessel with flat ends would not have the same volume and psi, as a ten ft. long vessel with domed end caps.
    I would also want to know the plate, A.I.S.I. #, to find carbon and manganese content %, to determine tensile and yield strengths per psi's

    Where an average could be; 100,000 psi yield strength, and 115,000 > 135,000 psi tensile strength



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  19. #18  
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    I was a welder for over 25 yrs. and a AWS welding inspector for another 10. On nuclear job sites, during the construction phase, and Welding Shops.
    Good skills to have.
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