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In their 2006 paper Christiansen et al. describe the "stuffed Whipple shield" wall of the ISS's Destiny Module. The outer skin, the Whipple shield's "bumper" layer, is a thin layer of aluminum (alloy 6061-T7). Inside that is the "stuffing" (appropriate for a Thanksgiving question! ;-) layers of Kevlar fabric, Nextel fabric, and multi-layer insulation (MLI). The innermost layer is another aluminum layer (alloy 2219-T87 for this layer, slightly denser than the 6061-T7) that serves as both the pressure vessel wall and for the inner layer of the Whipple shield. Its thickness is given as 4.8 mm.

Question: for that inner layer, which of the two functions, either withstanding the internal atmospheric pressure or providing sufficient areal mass density for the Whipple shield, set that thickness?

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Although it's an older question I just wanted to say that above answer uses the Young's modulus incorrectly. It is not supposed to be in the strength calculation. The sigma refers to a maximum strength (same units though force/area).

One typical value you could use is the yield strength of the material but you could also use another value the designer is comfortable with. For high strength alloys it should be in the range of 200 MPa.

Then the thickness for your parameters would be 0.11 mm for tangential stress and obviously half for the axial stress. For pressure vessels strength is rarely the limiting factor.

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The thickness is set by MMOD protection requirements.

Simplified calculation for the wall thickness required by the pressure vessel:

$ \sigma_\theta = \dfrac{Pr}{t} $

$\sigma$ = Young's modulus: 70 GPa
p = pressure: 105 Pa
r = radius: 2200 mm
t = thickness

$ t = \dfrac{Pr}{\sigma_\theta} $

t = 0.003 mm which seems too low to me. Apply a safety factor of 10 (far higher than used in practice) and you still get a paper-thin skin. Haven't found a better calculation though. Falcon 9 first stage tanks are something like 6 mm thick for 6 bar pressure plus the weight of the upper stage+payload.

Other elements of the ISS have a pressure shell of 1.4 mm (FGB) with thicker Whipple shields in front. So 4.8 mm is far thicker than necessary just to contain the pressure.

The Whipple shield config is designed to protect against debris with these specs:

  • 1.3 mm diameter aluminium sphere
  • impact speed 7 km/s

This requirement dictates a thickness of 4.8 mm.

Rule of thumb: at 7km/s, aluminum sphere can penetrate completely through an aluminum plate with thickness 4 times the sphere’s diameter

A multi-layer spaced shield provides more effective protection from hypervelocity impact than single layer (total shield thickness < projectile diameter)

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  • $\begingroup$ The pressure vessel is calculated using Barlow's formula. But you have to use a design factor of about 0.4. Some excess for corrosion and tolerances are necessary. The force in axial direction of the cylinder is unattended. But the result is still less than 1 mm. Bending and warping stress is not considered. $\endgroup$
    – Uwe
    Commented Nov 28, 2018 at 13:58
  • $\begingroup$ Excuse me, it protects against 7km/s MMODs?! Not sure I understood the quoted part, how big can the MMOD be before this rule doesn't apply? Is it that the shield is thicker than the sphere is in diameter? $\endgroup$
    – Magic Octopus Urn
    Commented Nov 28, 2018 at 19:59
  • $\begingroup$ I'll see if I can find a source that has more detailed calculations. $\endgroup$
    – Hobbes
    Commented Nov 28, 2018 at 20:05
  • $\begingroup$ @Hobbes Yes, I did the same calculation and wound up with a similar indicated minimum thickness (used a slightly different value for Young's modulus of 2219-T87 I found on the web), but I couldn't find anything about what safety factor NASA applies, and I couldn't rule out anything, even if I considered it absurdly high. $\endgroup$
    – Tom Spilker
    Commented Dec 3, 2018 at 5:30
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    $\begingroup$ @MagicOctopusUrn I think the quoted part isn't saying that thickness protects against a 7 km/s impactor, but that a 7 km/s aluminum sphere completely penetrates through a plate of that thickness. You can have interior damage even if the impactor doesn't penetrate the plate (wall). At those speeds, the collision launches a shock wave through the wall. When that shock wave hits the wall's inner surface it spalls off pieces of wall material—at high speeds, just smaller and slower than the initial impactor— that then cause shrapnel damage... $\endgroup$
    – Tom Spilker
    Commented Dec 3, 2018 at 5:38

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