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From the book "Design of Liquid Propellant Rocket Engines". Under stress analysis related pages, the author used an equation for finding the thickness of the wall needed for sphere under stress.

$$\text{Thickness of sphere wall} = \dfrac{\text{Yield pressure}\times\text{diameter}}{4\times\text{yield strength at 300°F}}$$

I would like to know the equation for tubes.

Raze
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3 Answers3

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Assuming a very long cylinder (to neglect effects at the top and bottom of the cylinder) with a small wall thickness $t$ (to assume that the stress is constant across the thickness of the wall) and with mean diameter $d_{\text{m}}=d_{\text{inner}}+t$ you can easily derive the so-called Barlow's formula.

$$\sigma_{\text{tangetial}}=\frac{d_m}{2t}(p_{\text{inner}}-p_{\text{outer}})$$ $$\sigma_{\text{axial}}=\frac{\sigma_{\text{tangetial}}}{2}$$

MrYouMath
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Had to post this as an answer: enter image description here

Which deals with thick-walled cylinders : Source : http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html Or here, from Lamé's equations : http://www.mydatabook.org/solid-mechanics/stress-for-thick-walled-cylinders-and-spheres-using-lames-equations/#cylinder_axial_stress

Solar Mike
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As mentioned by @MrYouMath, you can use Barlow's Formula (see 1, [2] with a calculator for US units). In short the equation is $$P = \frac{2 S_y t}{d}$$

where:

  • P: pressure
  • $S_Y$: allowed yield stress
  • $d$ : diameter
  • $t$ : thickness.

if you are solving for the thickness:

$$t = \frac{d}{2 S_y P} = \frac{r}{ S_y P} $$

keep in mind the following:

  • In (the unlikely) case you end up with $d<20 t$, then the assumptions do not hold and you shouldn't use the results. You should use either FE, or analysis that can't be covered in this paragraph.
  • Regarding the selection diameter, you are better off selecting the outer diameter. Especially if $d>20 t$, there won't be any noteworthy difference. (consider the selection of sheet metal thickness for the pipe).
  • It is highly recommended, to use a safety factor N(>1). The factor should be proportional to the hazard caused by the failure of the pipe. (e.g. if you carrying a toxic fluid, or if the pressure in the pipe is so high that essentially its a bomb). In that case the equation turns into:

$$t = \frac{d\cdot N}{2 S_y P} = \frac{r\cdot N}{ S_y P} $$

NMech
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