Diameter of steel ds = 50mm
Total torque applied = 9KNm
Torque sheared by steel rod Ts = 9 × 60/100 = 5.4KN-m
Torque sheared by brass rod Tb = 9 × 40/100 = 3.6KN-m
GS = 85GPa, Gb = 45GPa
Let; Outer diameter of brass = dob = ?, Inner diameter of brass = dib = 50mm, Maximum shear stress induced in steel = ιs Maximum shear stress induced in brass = ιb, Since shaft are parallel i.e.; θs = θb & T = Ts + Tb,Now applied θs = θb
TS.LS/GS.JS = Tb.Lb/Gb.Jb; LS = Lb
We get; TS/GS.JS = Tb/Gb. Jb
Or; Jb/JS = Tb.GS/TS.Gb
Putting all the values
{(π/32)(dob4 – 504)}/ {(π/32).504} = (3.6 × 85)/(5.4 × 45)
dob = 61.29mm
Again T/J = ι/R = TS/JS = ιS/RS
ιS = TS.RS/JS = {(5.4 × 103)(25 × 10-3)}/{(π/32)(50 × 10–3)4}
ιS = 2.2 × 108 N/m2
ιS = 220 GPa