Strength calculation

The calculations are made according to the GOST 34233.2-2017 standard: "Vessels and apparatus. Norms and methods of strength calculation. Calculation of cylindric and conic shells, convex and flat bottoms and covers"

Cylindrical shell

Symbols:

$D$ - diameter of the cylinder, mm
$s$ - thickness of the cylinder's wall, mm
$p$ - overpressure, MPa
$\sigma$ - allowable stress in the material, MPa
$\psi$ - welding factor (1 if no welding)
$R_{m/20}$ - minimum yield strength at 20°C
$n_x$ - safety factor for ultimate strength

Assumptions:

$D \leq 200$
$s/D \leq 0.3$

Allowable stress:

$\sigma = R_{m/20} / n_x$

Thickness of the wall:

$s = p \times D / \left( 2 \times \sigma \times \psi - p \right)$

Pure aluminium (alloys have grater strength)

$n_x$ = 3.5
$R_{m/20}$ = 90 MPa

$\sigma = 90/3.5 = 25.71 MPa$

$s = 1 \times D / \left( 2 \times 25.71 \times 1 - 1 \right) = D / 50.42$

Example: for a cylinder ⌀ 50mm the wall's thickness must be at least 1 mm.

Lid

Symbols:

$K$ - coefficient, 0.4 for a flat lid
$K_0$ - coefficient, 1 if the lid has no holes

If the lid has holes ⌀ $d$ on a chord of the circle:

$K_0 = \sqrt{(1 - (\sum d / D)^3)/(1 - (\sum d / D))}$

Thickness of a lid:

$s = K \times K_0 \times D \sqrt{ 1 / (\psi \times \sigma)}$

Pure aluminium

$s = 0.4 \times 1 \times D \sqrt{ 1 / (1 \times 90 / 3.5)} = 0.079 \times D$

Example: for a lid ⌀ 50mm the thickness must be at least 3.95 mm.

Quartz/fuzed glass

The Design Tensile Strength is 48 MPa source, source. Some recommend using 6.89 MPa/1000 psi (i.e. 7 times less) in practice, as for a normal glass source.

The thickness of a lid:

$s = 0.4 \times 1 \times D \sqrt{ 1 / (1 \times 6.89)} = 0.152 \times D$

Example: for a glass window ⌀ 32mm (at DIFA/UNIBO) the thickness must be at least 4.87 mm.

PEEK

The thinkness of a lid:

$s = 0.4 \times 1 \times D \sqrt{ 1 / (1 \times (95/1.5))} = 0.05 \times D$

Example: for a lid ⌀ 50mm (at DIFA/UNIBO) the thickness must be at least 2.51 mm.

Screws

Lid screws

Assumptions:

Pressure is applied to a lid's surface ⌀ 50mm
Screw/bolt M5 made of steel

The force applied to the lid, N:

$F = p \times S = 1e6 \times 3.14 \times (0.05/2)^2 = 1963.5$

According to the design properties for metric hexagonal bolts:

Tensile resistance = 1700 N for the lowest material class 4.3, and safety factor 3

Number of the screws: $round(1963.5 / 1700) = 2$

The same results provides the online calculator (in Russian).

Minimal thread engagment length, according to the online calculator = 3.11 mm.

G1/4 push-in pneumatic fitting, S6510 from Camozzi

Parameters of the fitting' S6510 G1/4 thread:

$D$ - major diameter, 13.156 mm
$D2$ - pitch diameter, 11.445 mm
$H1$ - depth, 0.856 mm
$z$ - number of lines, 2

The surface of the fitting:

$A = \pi \times (13.156 / 2)^2 = 136$ mm2

The compressing forse from the gas at pressure $p$ = 1 MPa:

$F = p \times A = 1 \times 136 = 136$ N

The tension of the threads:

$\tau = F / (\pi \times D_2 \times H_1 \times z) = 136 / (\pi \times 11.445 \times 0.856 \times 2) = 2$ MPa

Example: the PEEK has tension strength 95 MPa. The actual safety factor is $95 / 2 = 42.5$.

DmitryMyadzelets/strength.md