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A large spherical tank contains gas at a pressure of 400 psi. The tank is constructed of high-strength steel having a yield stress in tension of 80 ksi. For volume requirements, the inner diameter of the tank is 45-ft. Determine the required thickness (to the nearest ¼ inch) of the wall of the tank if a factor of safety of 3.0 with respect to yielding is required.

Respuesta :

Answer:

t=2.025 inches

Explanation:

Given that

P = 400 Psi

Yield stress ,σ = 80 ksi

Diameter ,d= 45 ft

We know that

1 ft = 12 inches

d= 540 inches

Factor of safety ,K= 3

The required thickness given as

[tex]\dfrac {Pd}{4t}=\dfrac{\sigma}{K}[/tex]

t=thickness

[tex]\dfrac {PdK}{4\sigma}=t[/tex]

[tex]\dfrac {400\times 540\times 3}{4\times 80\times 1000}=t[/tex]

t=2.025 inches

Therefore thickness will be 2.025 inches.

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