Designing a solid shaft for torque transmission

First find torque from P = 2πNT/60, then use τ = 16T/(πd³).

Worked answer.

Step 1: Torque. Power P = 2πNT/60, so
T = 60P / (2πN) = (60 × 20,000) / (2π × 300) = 1,200,000 / 1884.96 ≈ 636.6 N·m.
In N·mm: T ≈ 636,620 N·mm.

Step 2: Torsion equation. For a solid circular shaft, the maximum shear stress at the surface is
τ = 16T / (πd³).
Rearranging: d³ = 16T / (πτ).

Step 3: Diameter.
d³ = (16 × 636,620) / (π × 40) = 10,185,920 / 125.66 ≈ 81,057 mm³.
d = (81,057)^(1/3) ≈ 43.3 mm.

Step 4: Standard size. Round up to the next preferred size, 45 mm.

Check. With d = 45 mm, τ = 16 × 636,620 / (π × 45³) = 10,185,920 / 286,278 ≈ 35.6 MPa, which is below the 40 MPa allowable. A 42 mm shaft would give about 43.8 MPa, which is too high, so 45 mm is the right choice.

Practical notes

• Real shafts also carry bending moments from gears or pulleys, so the design would use combined loading, for example the equivalent torque T_e = √(M² + T²) (maximum shear stress theory).
• Keyways reduce strength. A common textbook practice is to reduce the allowable stress by about 25% for shafts with keyways.
• Angle of twist (rigidity) may control the design for long shafts.

Common mistake. Using P = T × N without converting rpm to rad/s. That gives T = 20,000/300 ≈ 66.7, which is wrong by a factor of 60/(2π) ≈ 9.55.

Likely follow-up. "If the speed doubles to 600 rpm at the same power, how does the diameter change?" Torque halves, and d ∝ T^(1/3), so d falls by a factor of 2^(1/3) ≈ 1.26, to about 34.4 mm. This is why high-speed shafts can be thinner.