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3 years ago

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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?

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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