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| anna_blakemore |
 Posted: Sun Mar 16, 2008 11:57 am Post subject: 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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| Harold14370 |
| Posted: Sun Mar 16, 2008 1:44 pm Post subject: |
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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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| Bunbury |
| Posted: Sun Mar 16, 2008 2:26 pm Post subject: |
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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 Sc for circumferential stress and instead of "long" we'd use Sl for longitudinal stress.
The longitudinal joint is subject to circumferential stress, so you would use 0.75 for nl. The circumferential joint is subject to longitudinal stress so you would use 0.5 for nh.
I would guess you may have mixed up the joint efficiencies
Hope this helps. |
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| anna_blakemore |
| Posted: Mon Mar 17, 2008 7:54 am Post subject: |
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| Bunbury wrote: |
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 Sc for circumferential stress and instead of "long" we'd use Sl for longitudinal stress.
The longitudinal joint is subject to circumferential stress, so you would use 0.75 for nl. The circumferential joint is subject to longitudinal stress so you would use 0.5 for nh.
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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| Bunbury |
| Posted: Mon Mar 17, 2008 8:02 am Post subject: |
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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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| Harold14370 |
| Posted: Mon Mar 17, 2008 8:49 am Post subject: |
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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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| anna_blakemore |
| Posted: Mon Mar 17, 2008 12:12 pm Post subject: |
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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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| Bunbury |
| Posted: Mon Mar 17, 2008 3:10 pm Post subject: |
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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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| matteo_1234 |
| Posted: Thu May 01, 2008 1:42 am Post subject: |
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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?.
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| Harold14370 |
| Posted: Thu May 01, 2008 3:19 am Post subject: |
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| matteo_1234 wrote: |
| 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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| Bunbury |
| Posted: Thu May 01, 2008 6:16 am Post subject: |
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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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| matteo_1234 |
| Posted: Wed May 07, 2008 1:48 am Post subject: |
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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?...
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| KALSTER |
| Posted: Wed May 07, 2008 2:37 am Post subject: |
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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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| Bunbury |
| Posted: Wed May 07, 2008 1:56 pm Post subject: |
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| matteo_1234 wrote: |
| 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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