FW = √(F₁² + F₂² − 2F₁F₂ cos β)
Shaft load is the vector resultant of both span tensions, not their effective-force difference.
Shaft load is the vector resultant of both span tensions, not their effective-force difference.
Select a target and calculate.
Shaft load is the vector resultant of both span tensions, not their effective-force difference.
F₁ = 1,000 N, F₂ = 400 N and β = 165° give FW ≈ 1,390.2 N.
Planar static force resultant at the pulley center; belt weight, axial forces, shock factors and pulley overhang to the bearing are excluded.
This calculator combines tight-side tension, slack-side tension and span angle into the resultant force on a pulley. That shaft load governs bearing and shaft bending and is fundamentally different from the effective tension difference that transmits torque.
Both belt spans pull along their own directions. The law of cosines gives FW = √(F₁²+F₂²−2F₁F₂ cos β), where β is the angle between span-force vectors directed away from the pulley center.
Two ropes leaving the same ring in nearly opposite directions load the ring by almost the sum of their magnitudes. If they point nearly the same way, the resultant approaches their difference. β captures this directional effect.
FW = √(F₁² + F₂² − 2F₁F₂ cos β)
FW = √(F₁²+F₂²−2F₁F₂ cos β)β = arccos[(F₁²+F₂²−FW²)/(2F₁F₂)]at β = 180°: FW = F₁+F₂at β = 0°: FW = |F₁−F₂|| Symbol / input | Meaning |
|---|---|
| Resultant shaft load FW | Force resultant acting on pulley hub, shaft and bearings. |
| Tight-side tension F₁ | Force in the more highly loaded belt span. |
| Slack-side tension F₂ | Force in the less highly loaded belt span. |
| Included span angle β | Angle between the two span-force vectors directed away from the pulley center; for an open two-pulley drive it equals wrap angle at the pulley considered. |
F₁ and F₂ are operating span tensions. β is the included angle of outward force vectors; for a simple open two-pulley drive it equals the wrap angle at the pulley. Confirm the convention with a free-body diagram.
Select FW and enter both tensions and β. Use the result as an external radial load in subsequent shaft and bearing analysis, adding actual pulley overhang, load direction, other pulleys and service factors.
For F₁ = 1,000 N, F₂ = 400 N and β = 165°, FW = 1,390.2 N. At β = 180° it would be exactly 1,400 N; at β = 0°, 600 N.
A 1.39 kN resultant acts at the pulley center. An overhung pulley also creates a bending moment at the bearing. The load direction follows from vector geometry and is not supplied by this magnitude-only result.
Forces use newtons and angles are evaluated in radians internally; degrees can be selected. For a later bending moment, multiply FW by pulley-to-bearing offset in metres.
FW is an external force at the pulley center. With a pulley between two bearings, reactions split according to distances. With an overhung pulley outside the bearings, reactions and shaft bending rise. Bearing-life input therefore comes only after resolving all external loads into each bearing reaction.
Shaft load supports bearing-life, shaft-deflection and gearbox overhung-load checks, and shows the effect of changed pretension. Gates explicitly recommends comparing calculated overhung load with the gearbox manufacturer's limit.
This is a planar static resultant. It excludes misalignment axial force, weight, shock/service factors and spatial superposition of other elements. Actual bearing reactions also require bearing locations and static equilibrium.
Common mistake: Do not use F₁−F₂ as bearing load. Do not assume 180° without checking geometry, and do not confuse wrap angle with the smaller angle between tangent lines. Treat forces for multiple belts consistently.
Only at β = 180°. In general, calculate the vector resultant with the law of cosines.
The angle between both span-force vectors directed away from the pulley center. In a simple open drive it equals pulley wrap angle.
Fᵤ is the torque-producing difference, while both pretensioned spans pull on the shaft and largely add geometrically.
Only if one bearing truly carries the complete resultant. Usually the force must be combined with other loads and split into two bearing reactions.
Depending on the drive, larger sheaves, appropriate rather than excessive tension, suitable wrap and reduced overhang can help; changes must follow the manufacturer selection.