Keys and couplings

Why it matters

Every gear, pulley, sprocket and coupling hub must be locked to its shaft, and every motor must be joined to the shaft it drives. Keys and couplings do that job, and they are deliberately among the weakest links so that an overload shears a cheap key rather than a costly shaft. In servo and stepper drives, the coupling also decides positioning accuracy through its backlash and torsional stiffness.

Key ideas

Keys transmit torque between a shaft and a hub through a key seated partly in a shaft keyway and partly in a hub keyway.

• Sunk keys (rectangular, square, parallel) sit half in the shaft and half in the hub; most common. Older proportions: square key b = h = d/4; rectangular (flat) key b = d/4, h = 2b/3. Modern practice takes b and h from the standard key table for the shaft diameter.
• Taper and gib-head keys (taper 1:100) give a tight drive; the gib head aids removal.
• Feather keys are fixed to one member and allow axial sliding of the hub (gear shifting).
• Woodruff key - a semicircular disc in a deep seat; self-aligning, used on tapered shafts and for light torque.
• Saddle keys (no keyway in the shaft; flat or hollow) transmit torque by friction only - light duty.
• Tangent keys - a pair at 90° for heavy reversing torque. Round keys - pins in drilled holes.
• Splines - many integral keys; used where torque is high or the hub must slide.

Forces on a sunk key. The torque creates a tangential force F = 2T/d at the shaft surface. The key can fail by:

• shear across its width along the shaft/hub interface: τ = F/(b·l),
• crushing (bearing) on the side faces, with half the height (h/2) in contact with the hub: σc = F/((h/2)·l) = 2F/(h·l). For a square key (b = h), shear and crushing strengths are equal when σc = 2τ. The key length is found from both and the larger value used; it is usually limited to about the hub length (and at least about 1.5·d for full shaft strength).

Key as strong as the shaft. Equating shaft torque π·d³·τ/16 with key shear torque l·b·τ·d/2 for the same material gives l = π·d²/(8·b) = 1.571·d when b = d/4.

Couplings join two shafts end to end.

• Rigid couplings (muff or sleeve, split muff, flange) need accurate alignment and transmit bending; flange couplings transmit torque by bolts in shear (fitted bolts) or by friction.
• Flexible couplings accommodate misalignment and absorb shock: bushed-pin flange (rubber bushes), jaw/spider, tyre, gear, Oldham (parallel offset), universal (Hooke's) joint (angular).
• Servo couplings in mechatronic axes - bellows, disc, beam and zero-backlash jaw couplings - must be torsionally stiff and backlash-free while allowing small misalignment, because the encoder is often on the motor and any wind-up shows up as position error.

Flange coupling bolts. With n bolts on pitch circle diameter D, each bolt shank carries shear F = 2T/(n·D). Also check the hub, the key, the flange web in shear at the hub, and crushing of the bolts against the flange.

Formulas

T = 60·P / (2π·N)

• T: torque (N·m); P: power (W); N: speed (rpm).

F = 2T / d

• F: tangential force on the key (N); d: shaft diameter (m).

τ = F / (b·l) · σc = 2F / (h·l)

• τ: shear stress in the key (Pa); σc: crushing stress (Pa); b: key width (m); h: key height (m); l: key length (m).

l = π·d² / (8·b) (key as strong as the shaft in shear, same material); = 1.571·d when b = d/4

T = n·(π/4)·db²·τb·(D/2)

• n: number of bolts; db: bolt shank diameter (m); τb: permissible bolt shear stress (Pa); D: bolt pitch circle diameter (m).

T = (π/16)·τ·(D_h⁴ − d⁴)/D_h (hollow hub or muff in torsion)

• D_h: outer diameter of hub/muff (m).

Worked examples

Example 1 (standard). A 40 mm shaft transmits 400 N·m through a key 12 mm wide and 8 mm high. Permissible stresses: τ = 50 MPa, σc = 100 MPa. Find the key length.

1. Tangential force: F = 2T/d = 2 × 400 × 10³ N·mm / 40 mm = 20,000 N.
2. Shear: l = F/(b·τ) = 20,000 / (12 × 50) = 33.3 mm.
3. Crushing: l = 2F/(h·σc) = 2 × 20,000 / (8 × 100) = 50.0 mm.
4. Crushing governs: l = 50 mm (check that the hub is at least this long).

Example 2 (GATE level). A flange coupling transmits 15 kW at 100 rpm through 6 fitted bolts on a pitch circle of 150 mm. Permissible shear stress in the bolts is 40 MPa. Find the bolt diameter.

1. Torque: T = 60P/(2πN) = 60 × 15,000 / (2π × 100) = 1432.4 N·m.
2. Torque capacity of bolts: T = n·(π/4)·db²·τb·(D/2).
3. db² = T / (n·(π/4)·τb·(D/2)) = 1432.4 × 10³ / (6 × 0.7854 × 40 × 75) = 101.3 mm².
4. db = 10.07 mm → use M12 fitted bolts (shank in shear), then check bolt crushing against the flange thickness.

Common mistakes

• Writing τ = T/(b·h·l): torque must first be converted to force at the shaft radius, F = 2T/d.
• Using the full key height for crushing; only h/2 bears on the hub (or shaft).
• Using the pitch circle diameter instead of the radius (D/2) in the bolt-torque equation.
• Forgetting the 60/(2π) conversion when torque comes from power and rpm.
• Choosing a rigid coupling where alignment cannot be maintained, or a compliant elastomer coupling in a servo axis where wind-up causes position error.
• Making the key stronger than the shaft - the key should fail first.

For GATE ME

Expect numericals on key length or width from shear and crushing, torque capacity of a flange coupling from bolt shear, and torque from power and speed. Concept questions ask which key or coupling suits a duty (feather key for sliding hubs, Oldham for parallel offset, universal joint for angular misalignment). Practise converting torque to the force on a key or bolt quickly.

Quick check

1. Torque 300 N·m on a 50 mm shaft. Force on the key?
2. Which key allows the hub to slide along the shaft?
3. Which coupling suits two parallel shafts with a small lateral offset?
4. Crushing stress for F = 12 kN, h = 8 mm, l = 40 mm? Answers: 1. 12 kN. 2. Feather key (or splines). 3. Oldham coupling. 4. σc = 2 × 12,000 / (8 × 40) = 75 MPa.