Design of Shafts

Why it matters

Shafts are critical components in mechanical systems, transmitting power and rotational motion between different parts. Proper design ensures reliability, efficiency, and safety in applications ranging from automotive to industrial machinery.

Key ideas

• Types of Shafts: Shafts can be classified as transmission shafts (e.g., line shafts, counter shafts) and machine shafts (e.g., crankshafts, spindles).
• Materials: Common materials include steel, alloy steel, and sometimes aluminum for lightweight applications.
• Stresses in Shafts: Shafts experience bending, torsional, and axial stresses. The design must account for combined stresses.
• Design Considerations: Factors include material selection, loading conditions, deflection limits, and critical speed.
• Standards and Codes: Design often follows standards such as IS codes for dimensions and tolerances.

Model and limits

The following equations describe nominal stresses in a circular elastic shaft. The combined expression is von Mises equivalent stress when the local stress state consists of one normal stress and one shear stress. A rotating shaft under stationary transverse loading can experience alternating bending stress, requiring a fatigue check.

Formulas

• Torsional stress: τ = T·r / J
  • τ: Shear stress (Pa)
  • T: Torque (N·m)
  • r: Radius of the shaft (m)
  • J: Polar moment of inertia (m⁴)
• Bending stress: σ = M·y / I
  • σ: Bending stress (Pa)
  • M: Bending moment (N·m)
  • y: Distance from the neutral axis (m)
  • I: Moment of inertia (m⁴)
• Combined stress: σ_combined = √(σ² + 3τ²)

Worked example

A smooth solid circular shaft has diameter d = 0.05 m, torque T = 500 N m and bending moment M = 300 N m. Neglect axial force and stress concentrations.

J = pi d^4/32 = 6.135923e-7 m^4. I = pi d^4/64 = 3.067962e-7 m^4.

The outer-surface torsional shear stress is T(d/2)/J = 20.372 MPa. The largest bending normal-stress magnitude is M(d/2)/I = 24.446 MPa.

At the extreme bending fibre, von Mises equivalent stress is sqrt(24.446^2 + 3(20.372)^2) ≈ 42.93 MPa. The absolute maximum shear stress at that point under combined loading is sqrt((24.446/2)^2 + 20.372^2) ≈ 23.76 MPa. This differs from the torsional shear component alone.

A final diameter selection also needs allowable strength, fatigue history, deflection and twist limits, stress concentrations and critical-speed checks.

Common mistakes

• Neglecting the effects of stress concentrations at keyways or shoulders.
• Incorrectly calculating the polar moment of inertia for non-circular shafts.
• Overlooking the combined effect of bending and torsional stresses.

For GATE ME

Questions often involve calculating stresses in shafts under combined loading conditions. Practice problems on determining critical speeds and deflection limits.

Quick check

1. What is the primary function of a shaft in mechanical systems?
2. Name two types of stresses that shafts commonly experience.
3. What is the formula for calculating torsional stress in a shaft?

Answers: 1. Transmit power and rotational motion. 2. Bending and torsional stresses. 3. τ = T·r / J.