The 60-mm-diameter shaft is made of 6061-t6 aluminum having an allowable shear stress of τallow = 80 mpa. determine the maximum allowable torque t . also, find the corresponding angle of twist of disk a relative to disk c .

The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:

Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ

As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.

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