Question

The power transmitted by a shaft is 300 kW at 450 rpm. The maximum torque is 25% more than mean torque. The permissible shear stress for the shaft material is 42 N/mm². The twist in the shaft should not exceed 1.5° per meter length. Determine the diameter required, if a) the solid shaft b) the shaft is hollow with internal diameter is 0.5 times external diameter.​

Answer 1

The mean torque is given by:

T = (300 kW) / (2π x 450 rpm) = 8.2 Nm

The maximum torque is 25% more than the mean torque, so the maximum torque is:

Tmax = 1.25 x 8.2 Nm = 10.25 Nm

The maximum shear stress in the shaft is given by:

τmax = (Tmax) / (π x d^3 / 32)

where d is the diameter of the shaft.

Solving for d, we get:

d = (32 x Tmax) / (π x τmax)

d = (32 x 10.25 Nm) / (π x 42 N/mm²)

d = 28.3 mm

The twist in the shaft should not exceed 1.5° per meter length. The twist in the shaft is given by:

θ = (32 x Tmax x L) / (π x d^4)

where L is the length of the shaft.

For the given conditions, the maximum length of the shaft is:

Lmax = (π x d^4 x 1.5°) / (32 x Tmax)

Lmax = (π x (28.3 mm)^4 x 1.5°) / (32 x 10.25 Nm)

Lmax = 0.9 m