Maximum Torque Calculation for Steel Shaft

This solution determines the maximum torque ($T_e$) a solid steel shaft can withstand while remaining within its elastic limit.

Shaft Specifications

• Diameter ($d$): 8 cm = 0.08 m
• Radius ($r$): $d/2$ = 4 cm = 0.04 m
• Elastic Limit in Shear ($\tau_0$): 145 MPa = $145 \times 10^6$ Pa

Torsion Formula for Maximum Torque

The relationship between torque ($T$), maximum shear stress ($\tau$), polar moment of inertia ($J$), and radius ($r$) in a circular shaft is:

$ \frac{T}{J} = \frac{\tau}{r} $

The maximum elastic torque ($T_e$) is reached when the shear stress at the surface equals the elastic limit ($\tau = \tau_0$):

$ T_e = \frac{\tau_0 J}{r} $

Calculating Polar Moment of Inertia ($J$)

For a solid cylindrical shaft, the polar moment of inertia is:

$ J = \frac{\pi d^4}{32} = \frac{\pi r^4}{2} $

Determining Maximum Torque ($T_e$)

Substitute $J$ into the torque equation:

$ T_e = \frac{\tau_0}{r} \times \left( \frac{\pi r^4}{2} \right) = \frac{\tau_0 \pi r^3}{2} $

Insert the given values:

$ T_e = \frac{(145 \times 10^6 \, \text{Pa}) \times \pi \times (0.04 \, \text{m})^3}{2} $

$ T_e = \frac{145 \times 10^6 \times \pi \times 0.000064}{2} \, \text{N-m} $

$ T_e = 145 \times \pi \times 32 \, \text{N-m} $

$ T_e = 4640 \pi \, \text{N-m} $

Calculating the numerical value:

$ T_e \approx 4640 \times 3.14159 \approx 14577.4 \, \text{N-m} $

Final Result

The calculated maximum elastic torque is approximately 14,577.4 N-m. This is closest to the value 14,580 N-m.