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

Question

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.

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Ben Gia 3 years 2021-08-13T18:26:39+00:00 1 Answers 2 views 0

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  1. thugiang
    0
    2021-08-13T18:27:43+00:00
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    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:

    \sigma_{\phi} = \frac{pr}{2t_{\phi}}

    where p = 100 psi is the internal pressure, t is the wall thickness \sigma_{\phi} = 55000 psi is the maximum tensile stress[tex]55000 = \frac{100*6}{2t_{\phi}}

    t_{\phi} = \frac{100*6}{2*55000} = 0.0054 in

    The hoop stress can be calculated as the following

    \sigma_{\theta} = \frac{pr}{t_{\theta}}

    55000 = \frac{100*6}{55000} = 0.011 in

    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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