Locking Assembly / Keyless Locking Device Calculator
Locking assemblies transmit torque and axial force between shaft and hub purely by friction, without a keyway. From the operating torque, the application factor and the catalogue values of the chosen locking assembly (nominal torque, allowable axial force), the tool derives the required size, the combined utilization under simultaneous axial force, the required interface pressure as an informational value, and the torque transmittable by this locking assembly, live with every input.
Calculation
Model: catalogue sizing per Roloff/Matek (friction-fit connection). Combined utilization from torque and axial force via conservative linear interaction. Sizing tool for mechanical engineering, does not replace manufacturer approval.
Results
Calculating …
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Formulas and fundamentals
Friction-fit principle and application factor
A locking assembly (keyless locking device) clamps shaft and hub together purely by friction using one or two opposing conical rings: free of backlash, adjustable to any position on the shaft, and without the notch effect of a keyway. Because shocks and non-uniform running additionally load the friction connection, the operating torque is scaled up by an application factor K_A to a required nominal torque:
Typical values: K_A = 1.0 for uniform running, 1.25 for light shocks, 1.5 for moderate shocks and 2.0 for heavy shocks (depending on the application, drive type and duty). The chosen locking assembly must have a catalogue nominal torque M_KN ≥ M_KN,req.
Combined utilization from torque and axial force
If an axial force F_ax also acts in addition to the torque (e.g. from helical gearing or an axial process force), it is added conservatively and linearly to the torque utilization:
A_F is only computed when both the axial force F_ax and the locking assembly's allowable axial force F_axN are given; without these, the tool only evaluates the torque utilization A_M. A combined load always reduces the transmission capacity available for the torque alone.
Required interface pressure (informational value)
As additional information, the tool computes the interface pressure between the locking assembly and the shaft that is mathematically required for the torque:
with the friction coefficient µ (default 0.12, typical range 0.08 to 0.15), the shaft diameter d_W and the locking assembly's friction face length l_F, all in mm; the factor 1000 converts M_t from Nm to Nmm. If p_req exceeds roughly 150 N/mm² (rule of thumb for steel hubs), the hub strength or allowable bearing pressure should be checked separately, see the interference fit calculator.
Transmittable torque and traffic-light rating
Conversely, the catalogue nominal torque and the application factor give the operating torque that this locking assembly can transmit permanently:
The tool rates the total utilization A: up to 0.8 pass (green), up to 1.0 marginal (amber), above that fail (red) - in that case the locking assembly is too small and the next larger size should be chosen.
Model limitations
This calculator is a sizing tool based on the locking assembly manufacturer's catalogue values (nominal torque, allowable axial force), not a derivation of the interface pressure from material and geometry data as with an interference fit. Runout tolerance, allowable shaft misalignment and the number of locking assemblies required per connection are governed by the catalogue and are not part of this calculation. For hub strength at a high required interface pressure, see the interference fit calculator.
Worked example
Given: a locking assembly is to transmit an operating torque M_t = 500 Nm under moderate shocks (K_A = 1.5) onto a shaft with d_W = 40 mm, with an additional axial force F_ax = 2000 N. Friction face length l_F = 50 mm, friction coefficient µ = 0.12 (default). Catalogue values of the chosen locking assembly: nominal torque M_KN = 900 Nm, allowable axial force F_axN = 40,000 N.
Required nominal torque M_KN,req = 500 · 1.5 = 750 Nm. Torque utilization A_M = 750/900 = 0.833. Axial force utilization A_F = 2000 · 1.5/40,000 = 0.075.
Total utilization A = 0.833 + 0.075 = 0.908 - above 0.8 but below 1.0: amber rating (marginal but still acceptable). Required interface pressure p_req = 2 · 500 · 1.5 · 1000/(0.12 · pi · 40² · 50) ≈ 49.7 N/mm², well below the 150 N/mm² threshold - hub strength is not critical.
Without the axial force, this locking assembly would allow an operating torque up to M_t,allow = M_KN/K_A = 900/1.5 = 600 Nm; the additional axial force reduces the remaining margin to 9.2 percent. For more frequent or larger load peaks, the next larger locking assembly from the catalogue is recommended.
Frequently asked questions
Locking assembly or keyed shaft connection - what is the difference?
A parallel key connection transmits torque positively through the keyway and is tied to a fixed axial position; it always has some backlash and the keyway weakens shaft and hub through the notch effect. A locking assembly transmits purely by friction through radial clamping force, is free of backlash, adjustable to any position on the shaft, and needs no keyway. In exchange, the transmittable load is limited by the friction force and is more sensitive to manufacturing tolerances on shaft and hub.
How does a simultaneous axial force affect the sizing?
Torque and axial force both load the same friction connection together. The tool therefore adds the two utilization ratios A_M (torque) and A_F (axial force) linearly to the total utilization A - a conservative but common approximation. A locking assembly that just barely transmits the torque alone can already be overloaded once an axial force is added; in that case only a larger locking assembly or several locking assemblies in series (to be checked against the catalogue) will help.
Why is an application factor K_A used?
The measured or calculated operating torque is usually a mean or nominal value; in practice, shocks, starting events or non-uniform running cause briefly higher load peaks. The application factor K_A adds a flat margin for these peaks onto the nominal torque so that the locking assembly holds safely under real operating conditions. The more shock-prone the operation, the larger K_A should be.
What if the hub is too thin-walled for the required interface pressure?
The required interface pressure p_req reported here is an informational value; it says nothing about whether the hub can withstand that pressure. For thin-walled hubs or p_req above roughly 150 N/mm², the hub strength should be verified separately, e.g. with the interference fit calculator (equivalent stress and allowable interface pressure from the hub diameter ratio and material yield strength).
When does it make sense to use several locking assemblies on one shaft?
If a single catalogue locking assembly is not sufficient or the available hub length is limited, manufacturers place several locking assemblies axially in series; the transmittable nominal torques then add up approximately. The exact number, minimum spacing and any derating factors depend on the design and should be taken from the manufacturer's catalogue.
How is the friction coefficient µ for the interface pressure determined?
The friction coefficient µ between the locking assembly's conical surfaces and the shaft depends on surface condition, lubrication and material pairing; for dry to lightly greased steel pairings it is usually in the range 0.08 to 0.15, and the tool uses 0.12 as the default. Manufacturer data for the specific locking assembly takes precedence; assuming too high a friction coefficient understates the required interface pressure.