Question

A solid steel shaft is supported between 2 rigid supports; the diameter of AB is .625 in. and the diameter of BC is .75 in.

Answer 1

Answer:

The maximum allowable torque that can be applied at B without exceeding an allowable shear stress in the shaft of 10.0 ksi is 20.92 klb.in

Explanation:

The maximum torque will be equal to the torque on both ends A, and C, due to rods AB, and BC, respectively.

First we find polar moments of inertia of both rods:

(J)ab = π/2(0.625 in)^4

(J)ab = 0.2396 in^4

(J)bc = π/2(0.75 in)^4

(J)bc = 0.497 in^4

Now,

Tmax = (T)ab + (T)bc

Tmax = τ(J)ab/(r)ab + τ(J)bc/(r)cb

Tmax = (10000 psi)(0.2396 in^4)/(0.3125 in)  +  (10000 psi)(0.497 in^4)/(0.375 in)

Tmax = 7667.2 lb.in + 13253.33 lb.in

Tmax = 20.92 klb.in

Answer 2

I have attached the question with diagram for better understanding

Answer:

T = 976 lb.in

Explanation:

Please refer to the attached file below for detailed explanation.