Aluminum Yielding pressure

[MAGS_R_COOL]

My question is how much pressure can one inch of aluminum stand in PSI terms. Im guessing a thousand or so but I'm probably wrong.

[phyregod]

My question is how much pressure can one inch of aluminum stand in PSI terms. Im guessing a thousand or so but I'm probably wrong.

Alot more information is needed. What type of aluminum? There are several different types that would have a major effect however; I can tell you that one inch of solid aluminum in just about any form will handle a heck of alot more than 1000 psi.

Non Heat Treatable Alloys

1100  - Commercially pure aluminum. Excellent corrosion resistance, workability and weldability.  14,000 to 24,000 psi .

3003 - Alloyed with 1,2% manganese.  Very Good workability, weldability and corrosion resistance.  Tensile strength range 17,000 to
30,000 psi .

5005 - Alloyed with .8% magnesium. Excellent workability, weldability and corrosion resistance.  Tensile strength range 18,000 to 30,000
psi .

5052 - Alloyed with 2.5% magnesium.  Very good corrosion resistance, good workability, weldability and strength. Tensile strength between
31,000 to 44,000 psi.

5083 - Alloyed with 4.45% magnesium, .65 % manganese and .15% chromium.  Excellent weldability, light weight and good corrosion
resistance.  Tensile strength between 40,000 to 59,000 psi .

5086 - Alloyed with 4.0% magnesium, .45% manganese and .15% chromium.  Very good corrosion resistance, good workability.  Tensile
strength between 40,000 to 54,000 psi.

5454 - Alloyed with 2.7% magnesium, 0.8% manganese and 0.12% chromium.  Good formability, weldability and corrosion resistance. 
Often used for pressure vessels.  Tensile strength between 36,000 to 47,000
psi.

Heat Treatable Alloys

2024 - Alloyed with 4.5% copper. Fair workability and corrosion resistance. Used for structural applications.  Tensile
strength between 30,000 to 63,000 psi.

6061 - alloyed with 1.0% magnesium and 0.6% silicon.  Good formability, weldability and corrosion resistance.  Very good
machine-ability.  Yield between 7,000 to 39,000 psi.

6063 -  Good Formability

7075 - Alloyed with zinc, magnesium, copper and chromium.  Poor formability, good machine ability.  Yield between 32,000
and 76,000 psi.

cool? cool.

[CoolHand]

You should mention that the UTS (Ultimate Yield Strength) is measured in tensile (IE pulling the piece apart).

In compression, the alum will still yield, but not in the same way. Usually compression failures are from extrusion or buckling if the section is thin enough.

When doing the stress calcs for pressurized containers, you need to look at the tensile stresses in the wall of the material (tangential stresses), not the compressive force exerted radially by the gas.

Any decent mechanics of materials text will have the equations to check the tangential stresses. Do not exceed 1/3 of the rated UTS of the material, and you should be good to go.

Please note: By giving you advice, I take no responsibility for you blowing yourself or anyone else up. IF the **** hits the fan, no responsibility lies here.

Good luck, and above all else, be safe.

[MAGS_R_COOL]

coolhand, im not going to blow myself up, i'm not even making anything, I was just wondering cuz ppl say that like those Aluminum 3k tanks will explode (given the BD doesnt work) at like 5k. I figured it wouldn't, but I just wanted a more concrete answer.

But thx phyregod and coolhand for answering w/ great posts!

[MikeRosenthal]

coolhand, im not going to blow myself up, i'm not even making anything, I was just wondering cuz ppl say that like those Aluminum 3k tanks will explode (given the BD doesnt work) at like 5k. I figured it wouldn't, but I just wanted a more concrete answer.

But thx phyregod and coolhand for answering w/ great posts!

It shouldn't explode at 5k PSI, but i wouldn't recommend trying it.

During Hydro testing the vessel is brought up to such High Pressure. That's how it's known to be safe at 3k, they test it to much higher pressure. However, during the testing the tank is filled with water, not air as well as being submersed in water. If it fails during the test it will crack. If it fails while filled with air it will act like a grenade because the compressed air will still push on the broken vessel pieces whereas the water's PE is gone once the vessel breaks.

-MR

You can get a lot of good information on the Hydro Testing process from hydro guys website www.hydrotester.com

[MikeRosenthal]

Numbers

Taken from hydro guys's scuba site:
The cylinder is going to be pressurized to 5/3 of the DOT pressure stamped on the neck of the tank. If the stamp read 2250 PSI, then the tank would be pressurized to 3750 PSI. Likewise, for a standard 3000 PSI scuba tank, it would be pressurized to 5000 PSI.

So, during the hydro testing process your 3k tank is filled to 5k. Assuming it doesn't fail the test by flexing too much from the pressure, it's good for another 3/5 years.

-MR

Source: http://www.sounddive.com/hydrotest2l.htm

[RazorMonkey]

Taken from hydro guys's scuba site:
The cylinder is going to be pressurized to 5/3 of the DOT pressure stamped on the neck of the tank. If the stamp read 2250 PSI, then the tank would be pressurized to 3750 PSI. Likewise, for a standard 3000 PSI scuba tank, it would be pressurized to 5000 PSI.

So, during the hydro testing process your 3k tank is filled to 5k. Assuming it doesn't fail the test by flexing too much from the pressure, it's good for another 3/5 years.

-MR

Source: http://www.sounddive.com/hydrotest2l.htm

Hence the 5k burst disk, as well...

[DaveSM]

If you want to get into the math here is the formula.

Circonferencial constraint = ( pressure * Inside diameter) / ( 2 * wall thickness)

Longitudnal constraint ≈ ( pressure * Inside diameter) / ( 4 * wall thickness)

You must use a security factor when you design something using this formula. You must not design something in wich the working tensile strenght would be the ultimate tensile strenght except if it's an application in which you want want the piece to blow up (eg : a burst disk).

Always choose a security factor considering the situatin in wich you are. For a paintball HPA container I wouldn't design anything whitout a safety factor lower then 5. You can calculate the safety factor y dividing the maximum tensile strenght by the working tensile strenght.

As you can see in these formula a sphrerical container is less likely to fail because you use the longitudinal constrait for them.

Aso for your information here are some metric values in MPa (these may vary and should only used as reference value) (Sory for the metric values but I'm from the province of Quebec and we run with the metric system here)

___________________Tension, Compression, Shearing (Ultimate strenght before failure)
Carbon steel 1020____448______448_________345
Carbon steel 1045____655______655_________483
Carbon steel 1095____779______979_________724
Steel 4140___________621______621
Sainless Steel 304____579______579
Forged steel__________324______324_________262
Cast Iron grade 20____138______552_________221
Cast Iron grade 30____276______862_________379
Cast Iron grade 40____414_____1172_________448
Aluminium 3003-H14__152___________________97
Aluminium 6061-T6____310__________________207
Aluminium 7075-T6____538__________________317
Concrete_______________________21
Wood (average value for dry wood)____60___50___6
Hard copper___________393__________________255
Brass_________________421__________________255
ABS___________________41

I won't get into the details of the maximum elastic constraint, the Yong modulus and all the rest but these are basics info. Contact me if you want any more details :P.

[phelix]

you can't just use the yield strength - the amount of pressure that will rip a tank apart depends on the size and wall thickness of the tank, as well as the type of aluminum. Either way, the threads that attach the tank to the regulator will yield long before the tank itself.

[nippinout]

For thin-walled cylindrical pressure vessels, you look at the hoop stresses and the radial stresses.

Hoop stresses are the stresses in the walls running perpindicular to a lengthwise cut, and is
(pressure * radius ) / (thickness)

Radial stresses are the stresses perpindicular to a cut across the vessel (Like a circle) and is
(pressure * radius ) / (2 * thickness)

As we can see here, you will have the greatest stresses in the hoop. This disregards the stresses in the neck and hemispherical portions, which has some internal stress already due to manufacturing.

You'll need your strength, working pressure, wall thickness, and radius. If you are comfortable with the Factor of Safety, you've just designed a pressure vessel.

[LorneCash]

blah blah blah that's way too much math and it won't be entirely accurate anyways since the geometry of a tank isn't an ideal equation. Plus it's way too much math and stuff this is comming from an engineer... Trust me I could do the math but FEA can tell me the "exact" answer in 10min of modeling and 3min of meshing solving. And then if you wanna change the material or a dim it's just the 3min to remesh and resolve.

For thin-walled cylindrical pressure vessels, you look at the hoop stresses and the radial stresses.

Hoop stresses are the stresses in the walls running perpindicular to a lengthwise cut, and is
(pressure * radius ) / (thickness)

Radial stresses are the stresses perpindicular to a cut across the vessel (Like a circle) and is
(pressure * radius ) / (2 * thickness)

As we can see here, you will have the greatest stresses in the hoop. This disregards the stresses in the neck and hemispherical portions, which has some internal stress already due to manufacturing.

You'll need your strength, working pressure, wall thickness, and radius. If you are comfortable with the Factor of Safety, you've just designed a pressure vessel.

[nippinout]

blah blah blah that's way too much math and it won't be entirely accurate anyways since the geometry of a tank isn't an ideal equation. Plus it's way too much math and stuff this is comming from an engineer... Trust me I could do the math but FEA can tell me the "exact" answer in 10min of modeling and 3min of meshing solving. And then if you wanna change the material or a dim it's just the 3min to remesh and resolve.

Plugging and chugging is not complex math.

In regards to idealization... The Universe is unitized!

[Pneumagger]

to the OP...

the amount of pressure a vessel can hold is a function of the wall thickness, the inner radius of the container, and the material strength.

when working with real world pressure vessels a "thin walled solution" will be inaccurate. The wall lthickness and stress variance as you travel through the wall need to be accounted for. This calls for the "Elasticity Solution" or a finite element analysis using a modeling program. Lornecash has the right idea, however it would be wise to verify your solution with an approximated solution by hand.

[AirAssault]

you can't just use the yield strength - the amount of pressure that will rip a tank apart depends on the size and wall thickness of the tank, as well as the type of aluminum. Either way, the threads that attach the tank to the regulator will yield long before the tank itself.

Internet connection $65

Tuition bill per semester where big brains crunch Aluminum Yielding pressure $15,000

Digging up a two year old thread to add your $0.02 Priceless.

[LinearGoose]

My question is how much pressure can one inch of aluminum stand in PSI terms. Im guessing a thousand or so but I'm probably wrong.

Well the Xvalve is made of aluminum and it can take about 3k of pressure and its not even an inch thick so I would imagine an inch would take more.