here is the question:

Much as in exercise 8.1, the end of a .75-in-diameter rod is fixed to an immovable plate. the length of the rod is 18.0-in. A 50-lb weight is attached to a 16.0-in lever at the opposite end of the rod so the weight is hanging perpendicular to the longitudinal axis of the rod.

Reference problem 8.1 is this:

The end of a .75-in-diameter rod is fixed to an immovable plate as shown. The length of the rod is 18.0 in. A 50-lb weight is attached to a 16.0-in lever at the opposite end of the rod so that the weight is hanging perpendicular to the longitudinal axis of the rod, For steel, E=29,000,000 lb/in² and μ=.303. The poplar moment of inertia for a solid circular shaft may be found by J=πD⁴/32, where D is the shaft diameter.

A. Determine the minimum diameter ro so that the maximum torsional stress does not exceed 8,000 lb/in². (I have solved this problem)

To solve this I did the following:

Tmax= 8,000 lb/in²

T=FL = 50 lb * 16.0 in = 800 in-lb

Tmax solid = TR/J where J=πD⁴ / 32 and R=D/2 =

T(D/2) / πD⁴ / 32 so Tmax= 16T/π³ I took Tmax to both sides and D³ to get my final formula of

D³=16T/ π (Tmax solid) =

D³= 16(800 in-lb) / π (8,000 lb-in²) = .5092958179 in then took the invert by ∛(.5092958179 in) =

.7985890849 in thus D= .7985890849

To check I used 16T/ πD³ = 16 (800 in-lb) / π(.7985890849 in³) = 8,000.000 lb-in²

B. Determine the minimum diameter rod so that the maximum torsional stress does not exceed 8,000 lb/in². (this is the one I am stuck on)

I think I need G which I used E/2(1+μ) = 29,000,000 lb/in² / 2(1 +.303) = 11,128,165.77 lb/in²

please note π is meant to represent pi