Pressure vessel wall thickness ASME BPVC VIII vs AS1210

[LT Judd]

TL;DR Summary

ASME BPVC VIII uses a slightly different formula to Australian Standard 1210 for calculating the wall thickness of cylindrical shells in pressure vessels. I am trying to figure out why.

The ASME BPVC formula is t=PR/(SE-0.6P) where P = pressure , R = inside radius, S = allowed material design stress and E = joint efficiency factor. the AS1210 formula ,(equivalent nomenclature) is t=PR/(SE-P). This makes 1210 more conservative when using the inside radius . I suspect the pure theoretical formula would use R as the mid wall radius and the formula would be t=PR/SE. Does anyone have any idea why the difference between ASME and AS and what's with the 0.6 factor? Is that some empirical thing that ASME has patented? Does any one know what the EN or DIN formula is? I asked Chatbot GPT and it gave me a PC non answer , so this is your chance to prove humans are still smarter than computers.

 

[jack action]

Reverse-engineering the equation:

The equation comes from the force from the pressure inside being equal to the force acting within the wall:
$$Pd_i L = (ES)(2t)L$$
Where $d_i$ is the inside diameter and $L$ is the cylinder length. This leads to:
$$t = \frac{PR}{ES}$$
Where $R$ is the inside radius.

Rewriting both of your equations in terms of the original equation:
ASME:
$$P(d_i + 2t)L = (ES)(2t)L$$
AS1210:
$$P(d_i + 0.6 \times 2t)L = (ES)(2t)L$$
So ASME assumes the pressure acts on a surface having a width equivalent to the outside diameter and AS1210 somewhere closer to the average diameter. Both are most likely empirical. It seems irrelevant to be more conservative than ASME.

Take that ChatGPT!

 

[erobz]

TL;DR Summary: ASME BPVC VIII uses a slightly different formula to Australian Standard 1210 for calculating the wall thickness of cylindrical shells in pressure vessels. I am trying to figure out why.

If you derive the Hoop Stress using the mean wall diameter:

$$ \mathrm{E} \sigma 2t \Delta x - P( d_i + t ) \Delta x = 0 $$

You get:

$$ t = \frac{P r_i}{ \mathrm{E} \sigma - \frac{1}{2}P}$$

The ASME BPVC formula is t=PR/(SE-0.6P) where P = pressure , R = inside radius, S = allowed material design stress and E = joint efficiency factor.

So, the $0.6$ in the ASME is a further increase in wall thickness adjustment to that.

The AS standard is apparently more conservative yet; using the pressure applied to the outside diameter.

$$ \mathrm{E} \sigma 2t \Delta x - P( d_i + 2t ) \Delta x = 0 $$

$$ t = \frac{P r_i}{ \mathrm{E} \sigma - P}$$

 

[LT Judd]

Okay, I get it, ....I think. So obviously in real life the pressure only acts on the internal surface, i.e the inner diameter, not the mid-wall diameter or the whole diameter. Unless they are saying there is some kind of diffusion through the metal ?? Otherwise its just some kind of empirical safety factor thing.

 

[erobz]

Okay, I get it, ....I think. So obviously in real life the pressure only acts on the internal surface, i.e the inner diameter, not the mid-wall diameter or the whole diameter. Unless they are saying there is some kind of diffusion through the metal ?? Otherwise its just some kind of empirical safety factor thing.

In my Strength of Materials text its simply:

$$ t = \frac{Pr_i}{\sigma} $$

The "joint efficiency" is applied like a safety factor (it could have factors that are related to the actual construction method\geometry of a tank - perhaps accounting for transformations of stress and strain along a weld in a particular direction - I haven't looked it up). The increase in radius of the effective force from the pressure accounts for stress distribution across the "thin wall" which was neglected for simplicity of the analysis. In reality, stresses will be higher at the outside tank wall compared to the inside. Thats why they apply pressure to the mean wall + some bit, for the ASME, and the outside wall for the AS standard.

 

[LT Judd]

Thanks for your help. Your post got me pointed in the right direction, and I investigated the "thick wall" equations known as Lame's equations. These can be regarded as the "true" case but are quite a bit more complex that the ASME formulas , thus more liable to produce human error mistakes.
One small error in your above post , Lame's equations show the maximum stress is at the inner wall , not the outer. Other than that is what the equations show, I am not sure what the underlying reason is. The distribution of the stresses from inner wall to outer wall is a gently descending parabola , so I am guessing it may be some manifestation of St Venant's principle.
The "0.6" factor checks out using the formula derivation in the above post, using pressure applied at the inside radius plus 3/5 t.

 

[erobz]

One small error in your above post , Lame's equations show the maximum stress is at the inner wall , not the outer. Other than that is what the equations show, I am not sure what the underlying reason is.

Thanks for the correction. I should have tried to verify my belief.