Question

A steel shaft as shown below is subjected to equal and opposite torque at the ends. Find the maximum permissible value of d for the maximum shearing stress in AB not to exceed that in CD. Calculate the total angle of twist, if the torque applied is 500N-m. G = 80,000 N/mm2.

Answer 1

We know that:

T/J = ι/r = ι/D/2 = 2τ/D

For shaft AB

τAB = 16.T.dO/π(dO4 – di4) = 16.T.4/π(44 – d4)

τAB = 64.T/π(256 – d4) ...(i)

For shaft CD

τCD = 16.T/π(3.53) = 16.T/π(42.875)τCD = 16.T/π.42.875 ...(ii)

Equating equation (i) and (ii); we get

d = 3.03cm

Now all shafts are in series i.e.; θ = θ1 + θ2 + θ3

θ = [TAB.LAB/JAB.G + TBC.LBC/JBC.G + TCD.LCD/JCD.G]

Since given that; TAB = TBC = TCD = 500 N.m

G = 80,000 N/mm2

LAB = 12cm

LBC = 15cm

LCD = 24cm

Putting all the values we get

θ = (500/80000)[12 / {(π/32)(44 – 3.54)} + 15/(π/32)44 + 24/(π/32)(3.5)4]

θ = 1.05