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Design of shafts, keys, bearings and springs

Mechanical Engineering (B.E./B.Tech) · Machine design · free study notes for exams, placements and interviews

In short

Shafts carry torque and bending between prime movers and driven machines; keys lock hubs to shafts; bearings support the shaft and locate it; springs store energy and apply controlled force. These four are the "machine elements" core of machine design — every gearbox, pump, conveyor, motor mounting and vehicle suspension is a combination of them. Design means sizing each element for strength, rigidity and life against standard allowable stresses, catalogue ratings and Indian standards (IS 1570 steels, IS 2048 keys, ISO 281/IS rolling-bearing ratings, IS 4454 and IS 7906 springs). This block is heavily examined in GATE and in university Design of Machine Elements papers, and it is the usual first technical round at Kirloskar, Thermax, TVS, Tata Motors, Mahindra, L&T, Godrej, BHEL, SKF, Schaeffler and Rane.

Explained

Shafts: types, loads and the design approach

Rigidity, critical speed and standard sizing

Keys, keyways and splines

Bearings: types, rating and selection

Springs: types, design and practice

Key points

Formulas and examples

T = 60P/(2πN) = 9550 P(kW)/N(rpm) → T in N·m; P in W for the first form
Torsion: τ = 16T/(πd³) (solid); τ = 16T·do/(π(do⁴ − di⁴)) (hollow). Bending: σ = 32M/(πd³)
Te = √((Kb M)² + (Kt T)²) = (π/16) τ d³; Me = ½[Kb M + Te] = (π/32) σb d³ (moments in N·mm, stresses in N/mm²)
With axial load F: replace Kb M by (Kb M + α F d/8); α = column factor = 1 in tension, and in compression 1/(1 − 0.0044 L/k) for L/k < 115 (Johnson's formula above 115)
Twist: θ = T L/(G J), J = πd⁴/32 (mm⁴), G ≈ 80,000 N/mm² for steel, θ in radians
Whirling speed: Nc ≈ 946/√δ rpm, δ = static deflection in mm
Key: F = 2T/d; shear l = F/(w τ); crushing l = 2F/(t σc); keyway factor e = 1 − 0.2(w/d) − 1.1(t/2d)
Bearing: L10 = (C/P)a × 10⁶ rev (a = 3 ball, 10/3 roller); Lh = L10/(60 N); C = P (60 N Lh/10⁶)1/a
P = (X V Fr + Y Fa) Ks; variable duty Pm = [Σ(Pia ni)/Σni]1/a; journal p = W/(L d); Petroff f = 2π²(ηN/p)(r/c)
Spring: τ = K·8FD/(πd³) = K·8FC/(πd²); K = (4C − 1)/(4C − 4) + 0.615/C
δ = 8FD³n/(Gd⁴) = 8FC³n/(Gd); k = F/δ = Gd⁴/(8D³n); U = ½Fδ = ½kδ²
Lf = nt·d + δmax + clearance; pitch = Lf/(nt − 1); fn = ½√(k/m) Hz (both ends on flat seats)
Torsion spring: σb = K·32M/(πd³), θ = 64 M D n/(E d⁴) rad
Leaf spring (half-span L, load P per half, nf full + ng graduated leaves, n = nf + ng):
σg = 12PL/(bt²(3nf + 2ng)); σf = 1.5 σg; δ = 12PL³/(E b t³(3nf + 2ng)); nipped: σf = σg = 6PL/(n b t²)

Example 1 — Shaft under combined bending and torsion. A shaft transmits 20 kW at 300 rpm; bending moment 800 N·m; rotating shaft with minor shock, Kb = 1.5, Kt = 1.0; τ = 40 N/mm² (keyway present), σb = 80 N/mm².

Example 2 — Hollow shaft for the same duty, di/do = 0.6, same Te and τ.

SolidHollow (k = 0.6)
Outside diameter55.7 mm58.4 mm
Cross-section area2438 mm²1712 mm²
Relative mass1.000.70

Example 3 — Key for the 60 mm shaft, τ = 40 N/mm², σc = 80 N/mm², T = 636,700 N·mm.

Example 4 — Deep groove ball bearing selection. Fr = 3000 N, Fa = 1500 N, 1200 rpm, required life 10,000 h, X = 0.56 and Y = 1.5 from the catalogue table, service factor Ks = 1.2, rotating inner ring (V = 1).

Example 5 — Helical compression spring. Max load 500 N, deflection 25 mm, C = 5, allowable τ = 420 N/mm², G = 81,370 N/mm², squared and ground ends.

Deep groove ballTaper rollerJournal (plain)
Load typeRadial + light thrustHeavy radial + thrustRadial, very heavy
Starting frictionLowLowHigh (boundary)
Speed capabilityHighModerateHigh with film
Shock/dirt tolerancePoorModerateGood
Space neededModerateLargeSmall radially
Typical useMotors, pumpsWheel hubs, gearboxesEngine crankshafts, turbines

Where it is used

Interview questions

Common mistakes

Related topics

Machine design: failure theories and fatigue · Bearings, lubrication and tribology · Gear design · Clutches and brakes · Bolted, riveted and welded joints · Strength of materials: torsion and bending of beams

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