Power transmission between two rotating components requires a secure connection between the shaft of the driving component, and the hub of the mating part. One way to ensure torque transmission without slipping is by using a keyed connection. For most automation applications (motors, gearboxes, pulleys, etc.), a parallel key, also known as a straight key, with a square cross-section is used.

Keyed connections, however, do have a drawback—they require a reduction in shaft diameter of the torque-transmitting component. And of course, the smaller the diameter, the less torque can be transmitted before failure occurs. To demonstrate, below is an example of the maximum torque that can be transmitted by a keyless shaft, compared with the torque that can be transmitted by the same shaft when a key (per DIN 6885) is added.

Torque transmitted by a plain shaft

τ = yield stress; typically 50% of tensile strength; for this example, τ = 390 x 10⁶ (N/m²)

J = polar moment of inertia (m⁴)

r = shaft radius (m)

For a solid steel shaft of 18mm diameter:

J = (3.1415 * (.009)⁴) / 2

J = 1.03 x 10^-8 m⁴

T = (390 x 10⁶ * 1.03 x 10^-8 ) / .009

T = 446.6 Nm

Torque transmitted by a keyed shaft

τ = yield stress; 390 x 10⁶ for this example (N/m²)

d = shaft diameter (m)

l = keyway effective length (m)

h = keyway depth (m)

For a solid steel shaft of 18mm diameter, with keyway 6 wide x 32 long by 3.5 deep:

T = (390 x 10⁶ * 0.018 * 0.032 * 0.0035) / 2

T = 393.1 Nm

In this case, the torque that can be transmitted when a key is added to the shaft is approximately 12 percent lower than the torque that can transmitted by the plain shaft (393 Nm vs 447 Nm). However, these calculations do not include any reduction factors to account for shock loads, reversing effects, or fatigue life—all of which can be a greater concern for keyed shafts than for plain shafts.  Also, the above equations assume that the shaft will fail before the key. But there are two failure modes for parallel keys that must be considered: shear failure and compressive failure.

Shear Failure

When the connection rotates, the shaft and hub (mating part) exert opposing forces on the key, attempting to shear the key. The maximum torque that can be transmitted before exceeding the yield strength of the key is calculated as:

F = shear force acting on key (N)

r = shaft radius (m)

τ = yield stress; 390 x 10⁶ for this example (N/m²)

w = key width (m)

l = keyway effective length (m)

F = 390 x 10⁶ * 0.006 * .032

F = 74,880 N

T = 74,880 * .009

T = 673.9 Nm

Crushing Failure

The second possible mode of failure, also caused by forces exerted on the key by the shaft and hub, is based on compressive forces, which can induce permanent deformation and crush the key. In this case, the maximum torque that can be transmitted is based on the height, rather than the width, of the key.

F = compressive force acting on key (N)

r = shaft radius (m)

τ = yield stress; 390 x 10⁶ for this example (N/m²)

l = effective keyway length (m)

h = key height (m)

F = (390 x 10⁶ * 0.032 * 0.006) / 2

F = 37,440 N

T = 37,440 * .009

T = 337.0 Nm

Based on the scenarios above, the limiting factor is actually the crushing force on the key, and the maximum torque that can be transmitted is reduced by approximately 25 percent (337 Nm based on crushing failure with a key versus 447 Nm for a plain shaft). Again, it’s important to note that the calculations for a keyed shaft do not take into account any service, application, or safety factors. So the recommended maximum torque may be lower than the examples above, depending on other application parameters.

Feature image credit: 4mechtech.com