Lesson 4.3: Design of Shafts, Keys & Couplings

Shafts, keys, and couplings are fundamental machine elements used to transmit power between rotating components. GATE and PSU exams often test strength, torsion, shear, and design calculations.

🔹 1. Shafts

• Definition: A shaft is a rotating machine element that transmits power from one part to another.

• Types:

  1. Solid Shaft

  2. Hollow Shaft → lighter, less material, high strength-to-weight ratio

• Design Considerations:

  • Torque (T): Shaft transmits torque → torsional stress

τ=16Tπd3(solid shaft)\tau = \frac{16 T}{\pi d^3} \quad \text{(solid shaft)}τ=πd316T​(solid shaft)

• Bending Moment: Combine bending and torsion using equivalent twisting moment for fluctuating loads

• Materials: Mild steel, carbon steel, alloy steel

🔹 2. Keys

• Definition: A key is a machine element used to connect shaft and hub to transmit torque.

• Types of Keys:

  1. Saddle Key

  2. Rectangular Parallel Key

  3. Gib Key

  4. Taper Key

• Design Considerations:

  • Shear stress: $\tau =T/(0.5dL)$

  • Crushing stress on hub and shaft

• Example: Shaft diameter 40 mm, torque 500 N·m, material allowable shear stress 60 MPa → design key dimensions

🔹 3. Couplings

• Definition: Coupling connects two shafts to transmit power while accommodating misalignment and vibrations.

• Types:

  1. Rigid Coupling: Sleeve, flange → shafts perfectly aligned

  2. Flexible Coupling: Helical, Oldham → accommodates minor misalignment

  3. Gear Coupling: High torque transmission

• Design Considerations:

  • Torque transmission capacity

  • Shear and bending stress on coupling elements

  • Misalignment accommodation

• Example: Shaft transmitting 20 kW at 300 rpm, material allowable shear stress 60 MPa → select appropriate rigid flange coupling

🔹 4. Solved Examples (PYQ Style)

Example 1 (GATE ME 2018):
Design a solid shaft to transmit 15 kW at 200 rpm. Material allowable shear stress 60 MPa.

• Torque: $T=P/\omega =15000/(2\pi 200/60)\approx 716N\cdot m$

• Shaft diameter: d=(16T/πτ)1/3≈28mmd = (16 T / π τ)^{1/3} ≈ 28 mmd=(16T/πτ)1/3≈28mm

Example 2 (PSU Exam):
Design rectangular key for shaft diameter 40 mm, torque 500 N·m, allowable shear stress 60 MPa. Find key width and length.

Example 3:
Select a rigid coupling for two shafts transmitting 20 kW at 300 rpm. Check torsional stress.

🔹 5. Practice Exercises

1. Calculate shaft diameter for solid and hollow shafts under given torque.

2. Design rectangular and gib key for given torque and shaft diameter.

3. Select suitable rigid and flexible coupling for a given application.

4. Compute equivalent torque for shafts subjected to combined bending and torsion.

5. Explain advantages of hollow shaft over solid shaft.

🔹 6. Summary

• Shafts: Transmit power, torsion and bending considerations, solid vs hollow

• Keys: Connect shaft and hub, shear and crushing stress critical

• Couplings: Connect shafts, transmit torque, accommodate misalignment

• Applications: Motors, gearboxes, turbines, machine tools