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Description: Consider a pressure vessel that is made of two half spherical domes with 1 ft in diameter, and the cylinder portion is 3 ft in length and is subjected to 100 psi pressure. The maximum tensile strength of the material is 55 ksi The factor of safety for this design is 1.5. Problem: Determine the minimum thickness for this design.

Answer :

Answer:

0.016 in

Explanation:

1 ft = 12 in. So the radius r = 6 in

3 ft = 36 in

55 ksi = 1000 psi

The longitudinal stress of the vessel can be calculated as the following:

[tex]\sigma_{\phi} = \frac{pr}{2t_{\phi}}[/tex]

where p = 100 psi is the internal pressure, t is the wall thickness [tex]\sigma_{\phi} = 55000 psi is the maximum tensile stress

[tex]55000 = \frac{100*6}{2t_{\phi}}[/tex]

[tex]t_{\phi} = \frac{100*6}{2*55000} = 0.0054 in[/tex]

The hoop stress can be calculated as the following

[tex]\sigma_{\theta} = \frac{pr}{t_{\theta}}[/tex]

[tex]55000 = \frac{100*6}{55000} = 0.011 in[/tex]

As 0.011 > 0.0054 we will pick t  = 0.011 as design to withstand the maximum stress. Taking into account of factor of safety design, the appropriate thickness is 0.011*1.5 = 0.016 in

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