Technical Explanation: Parallel Key and Keyway Design
A parallel key is one of the simplest and most widely used methods of transmitting torque between a shaft and a hub. The key sits in a machined slot—the keyway—cut into both the shaft and the hub, and it carries the load primarily in shear across its width and in compression (crushing) against the keyway sidewalls. The purpose of this calculator is to determine the required key length so that both failure modes are avoided, and to recommend standard key cross-sections based on shaft diameter.
Governing Formulas
Torque transmitted through the key produces a tangential force at the shaft surface:
F = 2 · T / d
where T is the transmitted torque (N·mm), d is the shaft diameter (mm), and F is the tangential force (N) acting at the key. This force is resisted by the key over its length L. Two independent strength criteria must be satisfied:
τ = F / (w · L) ≤ τ_allow
σ_c = F / ((h/2) · L) ≤ σ_c,allow
Here w is the key width, h is the key height, and h/2 is the effective bearing depth (the key is assumed embedded half in the shaft and half in the hub). τ is the shear stress across the key cross-section, and σ_c is the crushing (bearing) stress on the contact faces. Rearranging gives the required length for each criterion:
L_shear = 2 · T / (d · w · τ_allow)
L_crush = 4 · T / (d · h · σ_c,allow)
The governing key length is the larger of the two.
Derivation and Physical Basis
The tangential force F is found by equating the torque to the moment of the force about the shaft axis: T = F · (d/2). The key resists this force in two distinct ways. In shear, the failure plane is the rectangular cross-section w × L inside the shaft or hub, so the shear area is w·L. In crushing, the contact area is the key's side face, of height h/2 (the embedded portion) and length L. Crushing is typically the more severe criterion for short keys because the allowable bearing stress is not proportionally higher than the shear allowable, while the effective area is smaller. For this reason, keys are often made longer than the pure shear requirement to keep bearing stress within limits.
Input Parameters
- Transmitted Torque (T): The design torque in N·mm or N·m. For pulsating or shock loads, apply a service factor (typically 1.5–2.5) so the key is sized for the worst-case torque, not the nominal value.
- Shaft Diameter (d): The nominal shaft diameter at the keyway. It determines both the tangential force and, through standard tables, the recommended key cross-section.
- Key Width (w) and Height (h): Standard values from ISO 773 or DIN 6885 based on shaft diameter. Using a non-standard or undersized key reduces the available shear and bearing area.
- Allowable Shear Stress (τ_allow): Typically 40–70 MPa for steel keys under steady load. Shock or reversing loads require lower values.
- Allowable Crushing Stress (σ_c,allow): Typically 80–150 MPa for steel keys. This is the bearing stress the shaft and hub materials can tolerate without local yielding.
Engineering Interpretation
The calculated key length is the minimum required for the chosen cross-section and allowable stresses. In practice, engineers round up to the next standard key length and verify that the keyway does not extend beyond the hub. The result is a strength check only; it does not address fatigue, fretting, keyway stress concentration in the shaft, or the torsional capacity of the shaft itself. A complete design also checks the shaft for combined torsion and bending, the hub for hoop stress and keyway bearing, and the assembly for fit and tolerance. When load reversals or vibration are present, fatigue of the keyway fillet and fretting corrosion at the key–hub interface often govern over simple static shear.
Units and Conversions
- 1 N·m = 1,000 N·mm
- 1 MPa = 1 N/mm²
- 1 kW at N rpm: T (N·m) = 9550 · P / N
- 1 hp at N rpm: T (N·m) = 7127 · P / N
Always keep torque in N·mm, length in mm, and stress in MPa (N/mm²) for dimensional consistency.
Alternative and General Cases
This calculator addresses parallel keys under steady torque. Other cases include:
- Taper keys and Woodruff keys: Different geometry and standard tables.
- Splines: Multiple keys integral with the shaft, used for higher torque and sliding joints.
- Pulsating or reversing torque: Requires fatigue-based allowable stresses and possibly a larger key or a different coupling method.
- Interference fits and friction couplings: Transmit torque without a key and may be preferred where keyways would weaken the shaft.
Assumptions and Limitations
- The load is distributed uniformly along the key length.
- The key is a standard parallel key with a rectangular cross-section.
- The key material is ductile and homogeneous, and the shaft and hub materials are at least as strong as the key.
- Clearance and fit effects, stress concentrations, and fatigue are neglected.
- Torque is steady; shock and reversing loads require a service factor.
- Self-weight, centrifugal effects, and misalignment are not considered.
- The result is a preliminary sizing aid, not a substitute for a full design check.
Practical Engineering Use
Key and keyway calculations appear throughout mechanical power transmission: electric motor and gearbox shafts, pump and fan drives, conveyor pulleys, coupling hubs, chain sprockets, and machine tool spindles. Selecting the correct key size and length ensures the joint transmits the design torque without shearing or crushing, while staying within standard dimensions that are easy to manufacture and replace.
Frequently Asked Questions
What is the difference between shear and crushing of a key?
Shear is the tendency of the key to be cut across its width. Crushing is the compressive bearing stress on the contact faces between key, shaft and hub.
How do I choose standard key dimensions?
Use ISO 773 or DIN 6885 tables. The calculator suggests the appropriate width and height based on shaft diameter.
What allowable stresses should I use?
Typical values for steel keys: shear 40–70 MPa, crushing 80–150 MPa depending on material and load type (steady / shock).
Why is the crushing length usually longer than the shear length?
Because the effective bearing depth is only h/2 while the shear area uses the full width w. Since w and h/2 are often similar in magnitude, the smaller bearing area frequently makes crushing the governing criterion.