The solid spindle ab is made of a steel with an allowable shearing stress of 12 ksi, and sleeve cd is made of a brass with an allowable shearing stress of 7 ksi. determine (a) the largest torque t that can be applied at a if the allowable shearing stress is not to be exceeded in sleeve cd, (b) the corresponding required value of the diameter d, of spindle ab.

solution:

solid spindle ab:

c= 1/2ds = ½(1.5) = 0.75in.

j= π/2c1= π/2(0.75)2 = 0.49701 in4

jmax = Tc/j

tab=jtal/c = (0.49701)(12)/0.75 = 7.952

sleeve cd = c2 = 1/2d2 = ½(3.0) = 1.5in

j = π/2(c24 – c14) = π/2(1.54-1.254) = 4.1172 in4

tab = jtl/c2 = (4.1172)(7)/1.5 = 19.213 kip.in

2? if not the correct answer you might need to provide more information

step-by-step explanation:

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