Question

A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 3.50 in. OD, a 0.065 in. wall thickness, and ν = 0.340. The purchase order specifies a minimum yield strength of 46 kpsi. Using the distortion-energy theory, determine the factor of safety if the pressure-release valve is set at 500 psi?

Answer 1

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Step-by-step explanation:

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Answer 2

the factor of safety is 3.87

Step-by-step explanation:

Outside diameter, d₀ = 3.5 in

Thickness, t = 0.065 in

Poission ratio, v = 0.334

Yield strength,
`S_y` = 46 kpsi = 46000 psi

Pressure release load, d₁ = d₀ - 2t = 3.5 - (2 x 0.065) = 3.37 in

Tangential stress, σ₁ = P(d₁ + t) / 2t

= 500 (3.37 + 0.065) / 2 x 0.065

= 13211.5385 psi

σ₁ = P(d₁) / 4t = 500(3.37) / 4 x 0.065 = 6480.769 psi

Radial stress, σ₀ = -P ==> -500 psi

Principal stresses, σ' =
`(1)/(√(2))` √[(σ₁ - σ₁)² + (σ₁ - σ₁)² + (σ₁ - σ₁)²]

σ' =
`(1)/(√(2))` √[(13211.5385 - 6480.769)² + (6480.769 - (-500))² + (-500 - 13211.5385)²]

=11875.1986 psi

Factor of safety Л =
`S_y` / σ'

= 46000 / 11875.1986

= 3.8736

Л = 3.87