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A thin-walled cylinder has been designed to resist internal pressure with a factor of safety of FS = 2 gad failure due to shear.

A thin-walled cylinder has been designed to resist internal pressure with a factor of safety of FS = 2 gad

A thin-walled cylinder has been designed to resist internal pressure with a factor of safety of FS = 2 gad failure due to shear. The cylinder is made of A36 steel with a shear yield strength Ty = 44 ksi, and tensile yield strength oy = 36 ksi. The cylinder has an inner radius, 7 = 10 in and thickness t = 0.1 in. (a) Determine the maximum pressure p that can be applied within the desired factor of safety. (15 pts) HINT: you will have to determine the absolute maximum shear stress Tabs max and in-plane shear stress Timpl in terms of p to compare against Ty. (b) If the pressure in part (a) is p = 0.44 ksi, and suppose that in addition to the pressure p, a torque T = 1,000 kip. in is applied at the ends of the cylinder as shown below, evaluate the maximum principal stress for this scenario and indicate if the walls of the cylinder will yield by comparing against dy. (20 pts) HINT: you will have to determine the polar moment of inertia, J first and then calculate the shear stress due to torsion. Subsequently, you will need to consider all normal and shear stresses (including on. O, and t) to determine the principal stresses (ap1, p2, Tmax etc.). For part (a) For part (b) J T= 1,000 kip. in J T= 1,000 kip. in

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Part a Determine the maximum pressure p that can be applied within the desired factor of safety FS 25 against the failure due to shear Given that the absolute maximum shear stress tauabsmax and inplan... blur-text-image

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