Parallel key calculation per DIN 6892
Verify parallel key joints per DIN 6892: torque, shaft diameter and key geometry yield the surface pressures on shaft, hub and key together with safety factors.
Parallel key calculator (DIN 6885 / DIN 6892 method C)
DIN 6885: b × h = 12 × 8 mm, t1 = 5 mm, t2 = 3.3 mm
Key material: C45+C per DIN 6880 with R_e ≥ 430 N/mm² (fixed). Permissible pressure per contact: 0.9 · R_e of the weakest partner (grey cast iron: R_m).
- Equivalent torque T_eq = K_A · T
- 500 Nm
- Circumferential force F_u = 2 · T_eq / d
- 25,000 N
- Effective length l_tr
- 68 mm
- Shear in the key τ (informative)
- 30.6 N/mm²
- Effective length above 1.3 · d: the additional length hardly increases the capacity because of the uneven load distribution.
Preliminary sizing per simplified method C of DIN 6892 (constant pressure, unidirectional torque only). Effective height convention: shaft t1, hub h − t1. For alternating torque or load peaks use method B of the standard.
Note: the keyway weakens the shaft considerably as a notch – fatigue verification in the shaft calculator per DIN 743.
Cross-section: shaft with keyway, key and hub
Look up application factor K_A (per DIN 3990-1)
| Drive \ driven | uniform | light shocks | moderate shocks | heavy shocks |
|---|---|---|---|---|
| uniform | 1 | 1.25 | 1.5 | 1.75 |
| light shocks | 1.1 | 1.35 | 1.6 | 1.85 |
| moderate shocks | 1.25 | 1.5 | 1.75 | 2 |
| heavy shocks | 1.5 | 1.75 | 2 | 2.25 |
Rows: driving machine, columns: driven machine. Examples: electric motor = uniform, piston pump = moderate shocks, stone crusher = heavy shocks.
Formulas and fundamentals
The calculator uses the simplified method C of DIN 6892: the pressure is assumed constant over the keyway length and height, chamfers and radii are neglected. The method applies to unidirectional torque and is intended for preliminary sizing. The key size b × h and the keyway depths t1 (shaft) and t2 (hub) follow automatically from the shaft diameter d according to the selection table of DIN 6885-1 (high form).
First the nominal torque is increased by the application factor to the equivalent torque: T_eq = K_A · T_nom. This yields the circumferential force at the shaft radius F_u = 2 · T_eq / d. The force is supported by the keyway flanks; only the effective length l_tr of the key carries load: for the round-ended form A, l_tr = l − b, for the square-ended form B the full length l carries.
The surface pressure is calculated separately for both contacts: at the shaft with the effective height t1, at the hub with h − t1 (the conservative convention common in the literature). The pressure is p = F_u / (h_eff · l_tr · i · φ) with the number of keys i and the load share factor φ (one key: φ = 1.0; two keys never carry equally because of manufacturing tolerances, hence φ = 0.75). Two keys therefore only increase the capacity by a factor of 1.5.
Following the method C rule of thumb, the permissible pressure is p_perm = 0.9 · R_e,min, where R_e,min is the smallest yield strength of the partners involved in the respective contact (shaft or hub against the key steel C45+C); for brittle materials such as grey cast iron the tensile strength R_m replaces the yield strength. The safety factor per contact is S = p_perm / p; the hub usually governs, because the effective height is smallest there and its material is often the weakest.
The required effective length follows by rearranging the pressure formula: l_tr,req = 2 · T_eq / (d · h_eff · p_perm · i · φ), evaluated for both contacts with the maximum taken as the result. After adding the end-shape allowance (form A: + b), the value is rounded up to the next standard length of the DIN 6885 series and checked against the available range of the size. Effective lengths above 1.3 · d hardly increase the capacity because of the uneven load distribution and trigger a warning.
Worked example
Given: gearbox shaft with d = 40 mm, nominal torque T = 500 Nm (static, K_A = 1.0), one parallel key form A with l = 80 mm. Shaft C45E quenched and tempered (R_e = 490 N/mm²), hub E295 (R_e = 295 N/mm²), key steel C45+C (R_e ≥ 430 N/mm²). The DIN 6885 table gives size 12 × 8 with t1 = 5.0 mm and t2 = 3.3 mm for d = 40 mm.
Calculation: T_eq = 500,000 Nmm, circumferential force F_u = 2 · 500,000 / 40 = 25,000 N, effective length l_tr = 80 − 12 = 68 mm. Pressure at the shaft: p = 25,000 / (5.0 · 68) = 73.5 N/mm². Pressure at the hub: p = 25,000 / (3.0 · 68) = 122.5 N/mm² – the hub governs.
Verification: at the hub contact E295 is the weakest partner, so p_perm = 0.9 · 295 = 265.5 N/mm². The safety factor is S = 265.5 / 122.5 = 2.17, the verification is fulfilled. Result: parallel key DIN 6885 – A 12 × 8 × 80.
Frequently asked questions
Which component is usually critical in parallel key joints?
Usually the hub, because the effective key height is smallest there and hubs are often made of the weaker material. Shaft, hub and key are verified separately.
Why does only the effective key length count?
For round-ended keys the rounded ends transmit no circumferential force. The total length is therefore reduced by the key width when calculating the pressure.
Do two keys double the load capacity?
No. Because of manufacturing tolerances two keys never carry equally; a load share factor of φ = 0.75 is applied. Two keys therefore only increase the transmittable load by a factor of 1.5.
Does the calculator also apply to alternating torque?
No. The simplified method C of DIN 6892 only applies to unidirectional torque. For continuously alternating torque direction the more accurate method B with the load direction reversal factor is required.
Does the calculator account for the weakening of the shaft by the keyway?
No, only the surface pressure of the joint is verified here. However, the keyway acts as a severe notch and considerably reduces the fatigue strength of the shaft – this is covered by the shaft calculator per DIN 743, where the keyway is included as a notch case.
Why is there no shear verification of the key?
For standard dimensions per DIN 6885 the shear stress in the key cross-section practically never governs compared with the surface pressure. The calculator reports it for information; the design is always determined by the pressure at the weaker keyway flank.