Question

A shaft carries a thin pulley of radius \( r = 0.4\ \text{m} \) at end C. Taut and slack belt tensions are \( T_1 = 300\ \text{N} \) and \( T_2 = 100\ \text{N} \). Allowable shear stress in the shaft is 80 MPa. The shaft length segments are \( AB = BC = L = 0.5\ \text{m} \). Neglecting shock and fatigue and using maximum shear stress theory, find the minimum shaft diameter (in mm).

Hint:

For belt-driven shafts, torque is created by the difference in belt tensions, not the sum. Always convert torque to N·mm before applying shear stress formulas.

Answer 1

Correct Answer: 23.6

The pulley applies a torque on the shaft due to the difference in belt tensions: \[ T = (T_1 - T_2)\, r = (300 - 100)(0.4) = 200 \times 0.4 = 80\ \text{N·m} \] Convert torque to N·mm: \[ 80\ \text{N·m} = 80 \times 1000 = 80000\ \text{N·mm} \] For a solid circular shaft, the maximum shear stress is: \[ \tau_{\max} = \frac{16T}{\pi d^3} \] Allowable shear stress: \[ \tau_{\text{allow}} = 80\ \text{MPa} = 80\ \text{N/mm}^2 \] Set the maximum shear stress equal to allowable: \[ 80 = \frac{16(80000)}{\pi d^3} \] Rearrange: \[ d^3 = \frac{16(80000)}{80 \pi} \] \[ d^3 = \frac{1280000}{80\pi} = \frac{16000}{\pi} \] Using \( \pi \approx 3.142 \): \[ d^3 = \frac{16000}{3.142} \approx 5092.5 \] \[ d = \sqrt[3]{5092.5} \approx 23.8\ \text{mm} \] Thus, the minimum required shaft diameter is approximately \(23.8\ \text{mm}\).