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The atmospheric pressure at the surface of the earth is about 101 $kPa$.

For Carbon fiber reinforced plastic (70/30 fibre/matrix, unidirectional, along grain):

  • Young's Modulus is: $181\ GPa$
  • Density - $1.6\ g/m^3$

With these parameters in mind:

How do we calculate the minimum thickness/weight of a vacuum sealed (i.e. inside the container is vacuum) spherical container/vessel that would withstand the standard atmospheric pressure without much change in shape with its inner volume of $k\ m^3$?

And is there a general formula to calculate this, just based on the external surface area (independent of shape of the container)?

My background isn't in Physical Engineering sciences thus an apology beforehand.

Fred
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TheoryQuest1
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  • Typically you look at a formula for *stess in a thick walled sphere/cylinder under internal/external pressure*. Cylinders under internal pressure are the most common case, so this is a less common one. Poisson's ratio is also important, and really you want a 2-dimensional treatment because of the anisotropy. – Pete W Dec 25 '21 at 17:04
  • @PeteW Thanks a lot. But, again with due apologies by background is CS so I don't understand much of the background. I was looking for a black box formula (in spherical or cylindrical cases. I do understand the anisotropy part, but I am not sure how to take that into account for a 3D chamber (whatever shape it might be). Can we assume an ideal case scenario, and have a rough approximation/formula? would be of great help. – TheoryQuest1 Dec 25 '21 at 17:14
  • I don't know it offhand for sphere, maybe someone else does! If not, the italicized part above would be the search term to start with. With symmetry it should be reducable to 2D. The "easy" formulas don't account for the grain of the material, which is pretty important for CFRP. – Pete W Dec 25 '21 at 17:17
  • I understand. For the sake of comparison (in a cylindrical container) do we have an idea for steel (the kind which would make lightest vessel)? That i guess would take care of the anisotropy and grain problem? – TheoryQuest1 Dec 25 '21 at 17:27
  • By the way, FRP/CFRP tanks can lay the fiber in cris-crossing directions. [Example youtube](https://youtu.be/1A3vaJaNDLY?t=122) ... note there are 2 patterns of layers shown in the video, with 3 different fiber directions between them! There is still more anisotropy besides "in-plant", because of the third direction normal to the surface! – Pete W Dec 25 '21 at 22:21

2 Answers2

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Typically the maximum stress for a thin walled spherical pressure vessel for a given diameter, thickness and pressure is:

$$\sigma = \frac{P\cdot R}{2\cdot t}$$

where:

  • $\sigma$ is the observed pressure (in Pa)
  • $P$ is the pressure (in Pa)
  • $R$ is the radius of the vessel (in m)
  • $t$ : is the thickness of the vessel (in m)

using this it is possible to find the minimum mass of a pressure vessel is (Wikipedia)

$$M={3 \over 2}PV{\rho \over \sigma }$$, where:

  • M is mass, (kg)
  • P is the pressure difference from ambient (the gauge pressure), (Pa) (100000Pa = 1 Atm)
  • V is volume,
  • $\rho$ is the density of the pressure vessel material, (expressed in $kg/m^3$)
  • $\sigma_{all}$ is the maximum allowable working stress that material can tolerate. ($Pa$)

However, an interesting parameter in the question which is without changing much its shape. That can be interpreted as having a very small deformation. Usually a common value which is used to denote small deformation within the elastic range is 0.2% or (0.002). So the value that you could use for the allowable stress is $\sigma_{all}= E\cdot 0.002 = 181 GPa \cdot 0.002= 362 MPa$

Compression buckling

An very important additional check is that of buckling (because the shell will be under compressive loads).

The buckling pressure $q_C$ of a elastic thin spherical shell was obtained by Zoelly (1915) and Van der Neut (1932):

$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$

where

  • E is the elastic modulus
  • $\nu$ is the Poisson Ratio
  • $h$ is the shell thickness
  • $R$ is the shell radius. Assuming a volume V, the radius of the sphere will be $R= \sqrt[3]{\frac{3V}{4\pi}}$

Therefore, given that the critical pressure needs to be 1 atm ($q_C= 1Atm = 100000Pa$), the minimum thickness can be estimated:

$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$

$$ h = \sqrt[3]{\frac{3V}{4\pi}} \sqrt{\frac{q_c}{2 E}\sqrt{3(1-\nu^2)}} $$

After having calculated the minimum thickness, it is possible to obtain the total required volume of material $V_m$ and weight $M$ by:

$$V_m = 4\pi R^2 h $$ $$M = V_m\cdot \rho = 4\pi R^2 h \cdot \rho $$

NMech
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0

In a sphere container with the volume K

$$r= (\frac{3K}{4\pi} )^{1/3}$$

Assuming the strength, which you have not provided, $\sigma_c = 2.02×10^5 psi = 1.39 GPa$

The pressure transferred by the atmosphere is equivalent to the pressure over the surface of the circle as compression with a factor of safety. And in a thin-walled sphere, thickness<< r:

$$C_{stress/m}= 101kPa *\pi r^2/2\pi r=101kPa* r/2= 101kPa*(\frac{3K}{8\pi} )^{1/3}$$

The thikness is $thicknes =\frac{1.39GPa}{C_{stress}}$

kamran
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  • Thanks a lot! Wikipedia lists the (Specific Strength) in units $10^6 m^2 s^{-2}$ of this material as: 113. (https://en.wikipedia.org/wiki/Specific_modulus) Do we need specific stiffness or the strength? – TheoryQuest1 Dec 25 '21 at 18:02
  • I have no experience and guile to correct you! just a humble doubt from a noob:) : We as the Question states do not want any deformation of the chamber thus shouldn't we take stiffness or specific stiffness into account rather than strength (the capacity to deform and return to original shape by definition if i am correct) ? – TheoryQuest1 Dec 25 '21 at 18:10
  • a thin container even of steal is handled like a ballon if its a sphere. otherwise details are needed as to geometry. – kamran Dec 25 '21 at 18:30