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Engineering & Technology · Mechanisms, Machine Elements & Mechanical Design

Shafts, bearings, joints and load paths

A gear is attached to a shaft between two bearings. The shaft transmits torque, but the gear contact also pushes sideways. Those forces continue through bearing contacts into the housing. A shaft free-body diagram is therefore an interface map as well as a calculation.

Use Statics, Stress & Deformation; Dynamics, Vibration & Mechanical Response; and Materials Selection, Durability & Degradation. Retain force and moment balances, small elastic models, torque-speed power and units. The slider case supplies elementary trigonometry and a square-root geometry relation. All numbers, records, coefficients and efficiencies are constructed teaching inputs, not hardware selections, permissible loads, maintenance intervals or operating instructions.

Back to the chapter

After reading, you can

Follow force and torque through a shaft, bearings and housing, and avoid assuming equal support sharing.

01

Torque transfer is only one shaft function

A shaft can transmit torque while also carrying bending from gears, pulleys or an attached load. An axial force along the shaft introduces another path. For a tangential contact force Ft acting at pitch radius r, the associated torque magnitude is Ft r. Other gear-force components depend on geometry and cannot be inferred from torque alone.

A hub-to-shaft connection must transfer the required interactions while locating the component as intended. A key, clamp, interference connection or other joint has its own stiffness, contact and failure questions. A connection that can transmit torque is not automatically an adequate axial locator or alignment reference.

02

Constrain the needed motion without hiding thermal effects

A bearing supports selected relative motions and loads. A radial action is perpendicular to the shaft axis; an axial action is along it. Plain and rolling-element bearings provide different contact arrangements, and specific types differ in which directions they support. The desired rotation is permitted while other motions are constrained to the needed degree.

One common function scheme assigns axial location to one region while another region accommodates shaft growth. This is a concept, not a universal mounting recipe. Constraining every bearing axially can create unintended forces when shaft and housing expand differently. Fits, alignment, lubrication, seals and the intended temperature range belong to the actual arrangement.

03

Find the reactions before selecting a bearing

Support reactions depend on load positions and constraint assumptions. Two bearings do not necessarily share a transverse load equally. Draw one load plane clearly; another perpendicular plane may require its own balance. If more supports or elastic constraints are added, equilibrium alone may no longer determine every reaction.

A reaction result is one input to further evaluation, including speed, life, contact, stiffness and housing deformation. A shaft may meet a simple average-stress comparison yet deflect enough to disturb gear alignment. The selected bearing model also needs to match how the actual races, shoulders and housing carry the reactions rather than treating a catalogue label as the complete load path.

The shaft transfers transverse loads into unequal support reactionsA two-support shaft has two perpendicular loads; their positions rather than equal sharing determine the reactions in the selected plane.One transverse plane; rigid reaction model200 N down100 N downA: 175 N upB: 125 N up050150200Axial position / mm; force arrows 0.5 units/NSupport span 200 mm; shaft weight neglected
Support A at 0mm, B at 200mm. Downward loads 200N at 50mm and 100N at 150mm. Moment balance gives By×200=200×50+100×150=25000N·mm, so By 125N and Ay 175N. Geometry 2.2 units/mm and force arrows.5units/N. Torques, axial forces, elastic deflection and detailed bearing selection are not shown.

Solve the original transverse support case

Supports A and B are 200 mm apart. A 200 N downward load is 50 mm from A, and a 100 N load is 150 mm from A. Neglect shaft weight. Taking moments about A gives By×200=200×50+100×150=25000 N·mm. Thus By=125 N upward, and Ay=300−125=175 N upward.

The force sum is 175+125=300 N; the opposite moment check about B gives 175×200=200×150+100×50=35000 N·mm. The reactions are unequal because the load distribution is unequal. Separately, a 200 N tangential gear component at 20 mm pitch radius corresponds to 4 N·m torque. That torque relation does not supply the missing axial or other transverse components.

Try the reasoning

  1. Why do two bearings not each carry 150 N in this case?

    Show the answer and reasoning

    Moment equilibrium depends on where the loads act, not just their total.

  2. Does a torque value determine every gear contact force?

    Show the answer and reasoning

    No. Pitch radius and contact geometry are needed, and other force components may remain.

  3. Why allow for differential thermal growth?

    Show the answer and reasoning

    Unintended constraints can turn relative expansion into additional bearing or housing loads.

What to carry forward

Trace every relevant interaction through shaft, joints, bearings and housing before evaluating individual parts.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. Massachusetts Institute of Technology2 sources
    1. FUNdaMENTALS of Design, Topic 6 Power Transmission Elements II (2008)Alexander H. Slocum · MITTeaching resource · PDFpergatory.mit.eduRead source
    2. FUNdaMENTALS of Design, Topic 10 Bearings — mounting and load pathsAlexander H. Slocum · MITTeaching resource · PDFpergatory.mit.eduRead source
  2. 5.4 2D Rigid Body EquilibriumDaniel Baker and William Haynes · Engineering StaticsTeaching resourceengineeringstatics.orgRead source