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Rufina [12.5K]
3 years ago
8

A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65

mm 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?
Engineering
1 answer:
Andru [333]3 years ago
7 0

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[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}]

2). factor of safety, m = \frac{strength}{stress_{max}}

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[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}]

  Axial Stress_{max} = 3.5\times 26.78 =93.74 MPa

Now, Using formula (2)

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

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