a scuba tank is being designed for an internal pressure of 2640 psi with a factor of safety of 2.0 with respect to yielding. the yield stress of the steel is 65,000 psi in tension and 32,000 psi in shear.

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The scuba tank should be designed to withstand an internal pressure of 2640 psi with a factor of safety of 2.0, considering the yield stress of the steel, which is 65,000 psi in tension and 32,000 psi in shear.

To design a scuba tank that can safely withstand the specified internal pressure, we need to consider the factor of safety and the yield stress of the steel. The factor of safety is a measure of how much stronger the tank is compared to the expected load, and it ensures that the tank can handle unexpected variations or stress concentrations without failure.

Given a factor of safety of 2.0, we can calculate the maximum stress that the tank should experience without yielding. To do this, we divide the yield stress by the factor of safety:

Maximum stress = Yield stress / Factor of safety

For tension, the maximum stress would be 65,000 psi / 2.0 = 32,500 psi, and for shear, it would be 32,000 psi / 2.0 = 16,000 psi.

Therefore, the scuba tank should be designed to withstand a maximum internal pressure of 32,500 psi in tension and 16,000 psi in shear, ensuring that the stresses exerted on the steel do not exceed the yield limits. This design will provide a factor of safety of 2.0, meaning that the tank can handle twice the specified internal pressure before the material starts to yield.

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