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?

the factor of safety is 3.87

Explanation:

Outside diameter, d₀ = 3.5 in

Thickness, t = 0.065 in

Poission ratio, v = 0.334

Yield strength, = 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' = √[(σ₁ - σ₁)² + (σ₁ - σ₁)² + (σ₁ - σ₁)²]

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

=11875.1986 psi

Factor of safety Л =    / σ'

                              = 46000 / 11875.1986

                              = 3.8736

Л  = 3.87

1.75 or if it’s a fraction 1 3/4

Step-by-step explanation: