Burst pressure in a filament/wire wound cylinder
Posted: Sun May 17, 2015 12:39 am
Attempting to determine an estimated burst pressure for a barrel that is made from an inner cylinder sleeve reinforced with thin fibre filament or wire winded and wrapped uniformly over the cylinder sleeve. This method of construction is used for composite oxygen/gas tanks and other high-pressure vessels. I'm attempting to apply the concept into making a barrel.
Right now, I'm assuming I can utilize the filament's tensile breaking strength and dimensions to find an overall tensile strength of the material in pounds per square inch (psi), than applying this number into Barlow's formula for thin-walled cylinder vessels. Ignoring the hoop strength of the inner sleeve for now.
So given a fibre filament with a round cross-section, and a diameter of 0.018" and a breaking strength of 50 pounds:
filament area = 0.018^2 x pi <-- assuming rectangular/square cross-section due to space usage of wound filaments.
=~ 0.000324 sq.in
Number of strands that fits in 1 sq.in of cross-section = 1/0.000324 =~ 3086 strands
Hence, axial tensile strength in psi = 3086 strands x 50 pounds =~ 154,320 psi
Apply to Barlow's formula, 0.08" wall-thickness and 1" outer-diameter cylinder:
(154,320 psi x 0.08" x 2) / 1 = 24,691.2 psi (bursting pressure)
This value seems abnormally high, I can't help feel I'm missing something in my considerations. Can someone shed some light on this?
Right now, I'm assuming I can utilize the filament's tensile breaking strength and dimensions to find an overall tensile strength of the material in pounds per square inch (psi), than applying this number into Barlow's formula for thin-walled cylinder vessels. Ignoring the hoop strength of the inner sleeve for now.
So given a fibre filament with a round cross-section, and a diameter of 0.018" and a breaking strength of 50 pounds:
filament area = 0.018^2 x pi <-- assuming rectangular/square cross-section due to space usage of wound filaments.
=~ 0.000324 sq.in
Number of strands that fits in 1 sq.in of cross-section = 1/0.000324 =~ 3086 strands
Hence, axial tensile strength in psi = 3086 strands x 50 pounds =~ 154,320 psi
Apply to Barlow's formula, 0.08" wall-thickness and 1" outer-diameter cylinder:
(154,320 psi x 0.08" x 2) / 1 = 24,691.2 psi (bursting pressure)
This value seems abnormally high, I can't help feel I'm missing something in my considerations. Can someone shed some light on this?