Take a hollow aluminium cylinder with outer radius $r$ and length $h$, capped with two circular endcaps. How thick does the aluminium have to be, that is what is the inner diameter of the cylinder, to withstand 1 atm on the outside and 0 mbar on the inside, without crumpling?
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$\begingroup$ So you are just looking for an equation? There are also various strengths and types of aluminum. $\endgroup$– hazzey ♦Commented Sep 29, 2015 at 1:39
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$\begingroup$ @hazzey Any necessary info not given in the question could be either specified or parameterized by the answerer. $\endgroup$– user1717828Commented Sep 29, 2015 at 1:55
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1 Answer
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I found this formula here:
$$p_{crit}=\frac{2\,E\,t}{D}\left(\frac1{(n^2-1)\left(1+\left(\frac{2\,n\,L}{\pi\,D}\right)^2\right)^2}+\frac{t^2}{3(1-\nu^2)D^2}\left(n^2-1+\frac{2\,n^2-1-\nu}{\left(\frac{2\,n\,L}{\pi\,D}\right)^2-1}\right)\right)$$
So you should set $p_{crit}$ to about 3 atm to be safe, then find $E$ and $\nu$ for your aluminum, plug those in along with your diameter, and length. Then plug in a few integer values of $n>1$ to find a thickness $t$ that is strong enough for all values of $n$.
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$\begingroup$ $p_{crit}$ is the buckling pressure, $t$ is the wall thickness, $D$ is the (inner?) diameter, $L$ is length, $n$ is a positive integer. Do you know what $\nu$ and $E$ are? $\endgroup$ Commented Oct 8, 2015 at 21:01
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1$\begingroup$ $\nu$ and $E$ are probably the Poisson Ratio and Tensile (Young's) Modulus of the material, respectively. The "shape" of the equation terms and factors suggest that, especially the factor $(1-\nu^2)$. It is also a "thin-walled" formula, from the link. That suggests $D$ is assumed to be approximately both diameters, i.e. the wall is assumed to have relatively small thickness. In other words, $t \ll L$ and $t \ll D$. $\endgroup$ Commented Oct 9, 2015 at 0:26