Parallel key calculation per DIN 6892

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: 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.

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, live with every input.

Calculation

Shaft diameter d in mmNominal torque T in NmApplication factor K_AKey length l in mm

Key form

Number of keys

Shaft materialR_e or R_m in N/mm²Hub materialR_e or R_m in N/mm²

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).

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.

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.

Results

Notes

Effective length above 1.3 · d: the additional length hardly increases the capacity because of the uneven load distribution.

DesignationParallel key DIN 6885 - A 12 × 8 × 80

Shaft keyway (effective height t1)

p = 73.5 / perm. 387 N/mm²S = 5.26

Hub keyway (effective height h − t1)

p = 122.5 / perm. 265.5 N/mm²S = 2.17

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

Required length43.4 mmSuggested standard length45 mm

Cross-section: shaft with keyway, key and hub

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Parallel key cross-sections DIN 6885-1 (d = 6 to 500 mm)

Width b, height h and the keyway depths t1 (shaft) and t2 (hub) of the tall form per DIN 6885-1, assigned by shaft diameter d. The row of the entered diameter is highlighted.

Shaft diameter d [mm]	b [mm]	h [mm]	t1 [mm]	t2 [mm]	Length range l [mm]
6 < d ≤ 8	2	2	1.2	1	6 … 20
8 < d ≤ 10	3	3	1.8	1.4	6 … 36
10 < d ≤ 12	4	4	2.5	1.8	8 … 45
12 < d ≤ 17	5	5	3	2.3	10 … 56
17 < d ≤ 22	6	6	3.5	2.8	14 … 70
22 < d ≤ 30	8	7	4	3.3	18 … 90
30 < d ≤ 38	10	8	5	3.3	22 … 110
38 < d ≤ 44	12	8	5	3.3	28 … 140
44 < d ≤ 50	14	9	5.5	3.8	36 … 160
50 < d ≤ 58	16	10	6	4.3	45 … 180
58 < d ≤ 65	18	11	7	4.4	50 … 200
65 < d ≤ 75	20	12	7.5	4.9	56 … 220
75 < d ≤ 85	22	14	9	5.4	63 … 250
85 < d ≤ 95	25	14	9	5.4	70 … 280
95 < d ≤ 110	28	16	10	6.4	80 … 320
110 < d ≤ 130	32	18	11	7.4	90 … 360
130 < d ≤ 150	36	20	12	8.4	100 … 400
150 < d ≤ 170	40	22	13	9.4	… 400
170 < d ≤ 200	45	25	15	10.4	… 400
200 < d ≤ 230	50	28	17	11.4	… 400
230 < d ≤ 260	56	32	20	12.4	… 400
260 < d ≤ 290	63	32	20	12.4	… 400
290 < d ≤ 330	70	36	22	14.4	… 400
330 < d ≤ 380	80	40	25	15.4	… 400
380 < d ≤ 440	90	45	28	17.4	… 400
440 < d ≤ 500	100	50	31	19.5	… 400

The length range shows the usual available lengths for this size; from b = 40 mm the lower limit is not stored. Standard length series: 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 28, 32, 36, 40, 45, 50, 56, 63, 70, 80, 90, 100, 110, 125, 140, 160, 180, 200, 220, 250, 280, 320, 360, 400 mm.

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Formulas and fundamentals

Method C per DIN 6892

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).

Torque, circumferential force and effective length

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.

Surface pressure and safety factor

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)

Here i is the number of keys and φ the load share factor (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

Here 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.

Required length

The required effective length follows by rearranging the pressure formula:

l_tr,req = 2 · T_eq / (d · h_eff · p_perm · i · φ)

It is 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.

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