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Shafts, Keys, Couplings and Bearings

Shaft design for strength and rigidity, keys and couplings, rolling contact bearing life and equivalent load, and sliding bearings with Petroff's equation and Sommerfeld number.

Shafts carry power; keys and couplings connect them to hubs and to other shafts; bearings support them. GATE asks for shaft sizes, key stresses and especially bearing life.

Shafts

  • Strength: with moment M and torque T, use the equivalent torque √(M² + T²) (Tresca) or √(M² + 0.75T²) (von Mises). See static design.
  • Rigidity: limit the angle of twist θ = TL/(GJ), often to about 0.25° per metre.
  • ASME code adds shock factors K_m and K_t to M and T.

Keys

A rectangular sunk key of width w, height h and length l on a shaft of diameter d carries torque T through a tangential force F = 2T/d.

  • Shear of the key: τ = F/(w l).
  • Crushing of the key: σ_c = F/(l h/2).
  • A square key (w = h) is equally strong in shear and crushing when σ_c = 2τ.

Couplings

  • Rigid (sleeve, flange) couplings need accurately aligned shafts.
  • Flexible couplings (bush-pin, Oldham, universal joint) allow some misalignment. A universal (Hooke's) joint connects intersecting shafts; its output speed varies within each turn.

Rolling contact bearings

  • Rating life (90% reliability): L₁₀ = (C/P)^a million revolutions, with a = 3 for ball bearings and 10/3 for roller bearings. C is the dynamic load rating.
  • Equivalent dynamic load: P = X V F_r + Y F_a (radial and axial loads with factors X, Y; V = rotation factor).
  • Life in hours = L₁₀ × 10⁶/(60 N).
  • Doubling the load on a ball bearing cuts its life by 2³ = 8.

Sliding (journal) bearings

  • Petroff's equation (lightly loaded): friction coefficient f = 2π² (μN/p)(r/c), where μ is viscosity, N speed (rev/s), p bearing pressure, c radial clearance.
  • Sommerfeld number S = (r/c)² (μN/p): the main design number of hydrodynamic bearings.

Worked example

A ball bearing has C = 30 kN and carries P = 3 kN at 1000 rpm. Find its life in hours.

  • L₁₀ = (30/3)³ = 1000 million revolutions.
  • Hours = 1000 × 10⁶/(60 × 1000) = 16,667 h.

Practice

GATE-styleNAT · numerical answer

A roller bearing has C = 40 kN and P = 4 kN. What is its L₁₀ life in million revolutions? (Round to the nearest integer.)

Show answer and explanationAnswer: 2154

L₁₀ = 10^(10/3) ≈ 2154 million revolutions.

GATE-styleNAT · numerical answer

A shaft of diameter 50 mm transmits 500 N·m through a key 15 mm wide and 60 mm long. What is the shear stress in the key in MPa? (Give 2 decimal places.)

Show answer and explanationAnswer: 22.22

F = 2T/d = 2 × 500,000/50 = 20,000 N; τ = 20,000/(15 × 60) = 22.22 MPa.

GATE-styleMCQ · one correct answer

If the load on a ball bearing is halved, its rated life becomes:

Show answer and explanationAnswer: C

L ∝ (1/P)³ for ball bearings, so halving P gives 2³ = 8 times the life.

GATE-styleMCQ · one correct answer

Which coupling connects two shafts whose axes intersect at an angle?

Show answer and explanationAnswer: C

A Hooke's joint joins intersecting shafts; an Oldham coupling joins parallel shafts with a small offset.