Question

A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65mm wall thickness, and Poisson's ratio 0.334. The purchase order specifies a minimum yield strength pf 320 MPa. What is the factor of safety if the pressure-release valve is set at 3.5 MPa?

Answer 1

Given:

Outer Diameter, OD = 90 mm

thickness, t = 0.1656 mm

Poisson's ratio,
`\mu` = 0.334

Strength = 320 MPa

Pressure, P =3.5 MPa

Formula Used:

1).
`Axial Stress_(max)` =
`P[(r_(o)^(2) + r_(i)^(2))/(r_(o)^(2) - r_(i)^(2))]`

2). factor of safety, m =
`(strength)/(stress_(max))`

Step-by-step explanation:

Now, for Inner Diameter of cylinder, ID = OD - 2t

ID = 90 - 2(1.65) = 86.7 mm

Outer radius,
`r_(o)` = 45mm

Inner radius,
`r_(i)` = 43.35 mm

Now by using the given formula (1)

`Axial Stress_(max)` =
`3.5[(45^(2) + 43.35^(2))/(45^(2) - 43.35^(2))]`

`Axial Stress_(max)` =
`3.5* 26.78` =93.74 MPa

Now, Using formula (2)

factor of safety, m = \frac{320}{93.74} = 3.414