Materials Science · Contact, friction, hardness and wear

Lubrication, coatings and surface treatments

Distinguish fluid pressure support from boundary-film protection, and follow how speed, viscosity, load and supply change a contact. Calculate simple pressure and shear balances, then test why a hard coating still needs support, strain tolerance and a durable interface.

Useful background: Elasticity, plasticity and tensile strength; Processing, structure and useful properties. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish full fluid-film, mixed and boundary lubrication by load support and the role of the surface contacts.
Explain how motion, geometry, external pressure and supply can establish or lose a separating film.
Use a supplied pressure profile and a bounded viscous-shear model without confusing normal support with tangential drag.
Explain why coating hardness, film cohesion, interface adhesion and substrate response answer different questions.
Distinguish an added coating from a surface region modified by a treatment, and connect depth and environment to durability.
Transfer a stated compatibility or failure-event model to an unfamiliar case while identifying the evidence still needed.

Separating surfaces changes the mechanism

A lubricant can carry load through a fluid film or modify boundary interactions when direct contact remains. Viscosity, speed, load, roughness and temperature influence the regime. A coating can alter hardness, chemistry or adhesion but must remain attached and supported.

Start with a loaded slider supplied with oil. The oil can transmit normal force through pressure while its layers resist relative tangential motion. These are different actions. A visible pool beside the slider does not establish that a separating film exists inside the loaded region, and a lubricant need not remove all load-bearing surface junctions to be useful.

In full fluid-film lubrication, fluid pressure supports the applied normal load and the two main solid surfaces remain separated in the model. In mixed lubrication, fluid pressure and load-bearing surface contacts both contribute. In boundary lubrication, the response of thin interfacial films and surface junctions dominates; bulk fluid pressure provides little of the support. Thin adsorbed or reacted films can still separate the immediate chemical surfaces at a junction, even though there is no thick continuous fluid gap.

Write the normal-force balance for a quasi-static contact as N = N_f + N_s. Here N is the applied normal force magnitude, N_f is the resultant supported by fluid pressure above the surrounding pressure, and N_s is the contribution transmitted through load-bearing surface junctions and their thin films. This is a balance, not a formula for predicting the two shares. The original cross-sections below use chosen shares at N = 100 N to illustrate the regimes; asperity count in a cartoon is not a measurement of load.

Three original cross-sections use chosen force shares at an applied normal force of 100 newtons. In the boundary illustration the fluid contribution is 5 newtons and the surface-junction contribution 95. In the mixed illustration these shares are 40 and 60 newtons. In the full-fluid-film illustration the fluid supports 100 newtons and the main solid surfaces do not contact, so the surface share is zero. Blue marks fluid and orange marks thin films at loaded surface junctions. The gap shapes and number of asperities are qualitative cartoons, not a measured roughness, thickness or load distribution. The full-film state still permits viscous drag; boundary junctions need not be chemically bare solids.

Pressure requires a mechanism and a supply

A moving surface drags nearby viscous liquid with it. In a converging gap, the volume that motion would carry through the wider inlet differs from that through the narrower region. Pressure-driven flow adjusts to satisfy the flow balance. The resulting pressure distribution can push the surfaces apart and balance the normal load. This is the central idea of a self-acting hydrodynamic bearing: an appropriate gap, motion and lubricant supply are all needed. Two ideal parallel faces sliding at a fixed gap with no imposed pressure difference do not acquire a normal load capacity just because the liquid has viscosity.

There are other ways to support a fluid film. In hydrostatic lubrication an external pressure supply supports the load, so sliding speed need not be nonzero. Approaching surfaces can also develop transient squeeze-film pressure because viscous liquid must escape from the closing gap. In a concentrated rolling contact, elastic deformation of the solids and pressure-dependent fluid properties can strongly affect the film; this is elastohydrodynamic lubrication. These names identify mechanisms, not interchangeable thickness formulas.

For a self-acting contact of otherwise comparable geometry and supply, reducing speed or liquid viscosity, or increasing load, can reduce separation and increase the importance of surface contacts. A liquid lubricant often becomes less viscous as it warms. Temperature can therefore change both drag and the ability to maintain a film, while also changing surface reactions and the solids. These tendencies do not determine a unique regime without geometry, roughness, actual temperature and supply information.

Roughness supplies another relevant length scale. A mean clearance between reference surfaces does not establish the smallest local separation: opposed high regions may still meet while fluid fills most of the apparent contact area. A thicker film relative to the relevant height variations generally reduces that opportunity, but the actual distribution, deformation and film state matter. There is no universal absolute thickness that divides all contacts into these regimes.

Startup, stopping and reversal deserve attention. At zero instantaneous sliding speed, a steady motion-generated wedge cannot be assumed to keep its running-state support. An external supply or transient squeeze effect may still act. Starvation means that too little lubricant reaches the inlet or loaded region; a filled reservoir alone does not rule it out. Misalignment, contamination and changing roughness can also alter separation. A larger viscosity is not always an improvement: it can increase drag and make delivery more difficult under some conditions.

Calculate pressure support and shear separately

If gauge pressure p varies over a projected contact area, the fluid normal resultant is the sum of pressure times each small area. For a rectangular region of width b and length L with pressure uniform across b, this is N_f = b × area under the p-versus-distance curve. A chosen triangular profile that is zero at both ends and reaches p_max has mean pressure p_max/2, so N_f = ½ p_max bL. The triangle here is supplied teaching information, not a solution of a bearing-flow equation.

For a separate shear model, take a steady Newtonian liquid between parallel plates separated by a uniform gap h. One plate is stationary and the other moves at relative speed U. Assume constant dynamic viscosity η, no slip at either wall, negligible inertia and edge effects, no pressure gradient along sliding, and a temperature maintained uniform by the apparatus. The layers then have a linear velocity change from 0 to U, giving velocity gradient U/h. Newtonian viscosity relates shear stress to this gradient: τ = ηU/h.

For plate area A, the tangential drag magnitude is F = τA = ηUA/h. The mechanical power supplied against that drag is P = FU = ηAU²/h. Dynamic viscosity η has units Pa s; (Pa s)(m/s)/m gives Pa for τ. This simple model predicts shear resistance in an already maintained gap. It does not predict the gap, the normal load capacity or the force in a rough mixed contact.

Increasing U or η at fixed h increases this drag. In an actual self-acting bearing the gap and temperature can change too. A friction-versus-operating-condition curve can therefore fall as surface contacts diminish, then rise when viscous losses matter. This familiar Stribeck behaviour is conditional; a friction minimum is not a universal film-thickness boundary or a wear-life guarantee. The worked pressure and shear cases below are distinct models and must not be silently combined.

A thin film or a modified surface can protect differently

When thick-film separation is incomplete, lubricant molecules or additives can adsorb on the surface or react to form a boundary film. Adsorption is attachment at a surface; reaction produces a chemically changed layer. A useful film must survive or be replenished where contacts pass, while permitting comparatively easy shear. Its chemistry, cohesion, attachment and compatibility with the solids matter alongside the bulk oil viscosity. Boundary lubrication does not simply mean dry bare-metal contact, and the presence of an additive does not establish that the required film has formed.

A deposited coating deliberately adds a material layer. It can alter resistance to indentation, junction chemistry, shear or chemical attack. A hard wear-resistant layer and an easily sheared solid-lubricant layer may serve different purposes. Protection also needs continuity, thickness appropriate to the damage scale, and attachment to the underlying substrate. A hard coating over a deforming support does not turn the support into a rigid body.

A surface treatment can instead modify the original near-surface region. For example, nitriding introduces nitrogen, which diffuses inward and can enter solution or form nitride particles and a compound layer, depending on the material and process. The resulting depth-dependent structure differs from simply gluing on a uniform hard sheet. An H13 tool-steel study by Zagonel and colleagues used composition profiles, microscopy and indentation to investigate such a modified region. It is evidence for that treated material, not a universal nitriding recipe or a numerical hardness-to-life conversion.

A graded treated region still needs adequate depth and a suitable response beneath it. Removal of the modified material can expose a different state; a poorly supported or brittle outer region can fail. Surface preparation, contamination and processing history can also alter a deposited interface. The relevant comparison is among the complete surface–support–lubricant combinations under the intended conditions, rather than coating names alone.

The coating must follow or resist the support’s deformation

A contact can elastically bend a compliant substrate or leave a permanent depression if it yields. A bonded layer must accommodate that deformation. Tensile strain can open cracks within a brittle coating; separation can instead occur at the interface. Failure within the coating is cohesive failure; failure of attachment at the coating–substrate interface is adhesive failure. Both can matter even when an indentation test reports high hardness.

A small compatibility model makes one part of this clear. Consider a thin coating bonded to a straight substrate strip. Over a gauge length L, the substrate is stretched uniformly by δ. Suppose the intact interface prevents slip and separation and transfers the same longitudinal extension to the coating. The demanded coating strain is then ε_req = δ/L. Compare it with an independently supplied tensile crack-onset strain ε_c for that film under the stated conditions. This model says nothing about indentation hardness, the force needed to deform the substrate, or the interface’s own failure load.

If ε_req exceeds ε_c, an intact coating cannot remain below the stipulated crack threshold while following the support. A more supportive substrate can reduce the deformation under a given contact, but changing support can also change contact stresses, so the benefit must be checked. Improving adhesion alone does not remove a film’s tensile-strain limit. If debonding occurs first, the no-slip intact-bond model no longer predicts the film strain. Residual and thermal-mismatch strains would also need accounting in an actual coated part.

Thickness is not a one-way solution. It changes compliance, stresses and the energy available to drive a crack or separation. Chai and Lawn’s model ceramic/polymer coating study reports different crack modes as coating thickness changes under spherical contact. The appropriate lesson is that thickness and elastic mismatch can change the failure route, rather than that thicker, harder or more strongly bonded always means a more durable system.

An original uniform-extension model shows a thin film bonded to a straight substrate. Original gauge length is 5 millimetres; an intact no-slip bond transfers the substrate extension to the film, giving required strain equal to extension divided by gauge length. A plot with horizontal axis 0–20 micrometres extension and vertical axis 0–0.4 percent strain shows the demand. A supplied crack-onset strain of 0.20 percent is reached at 10 micrometres. Support A’s 5 micrometres extension requires 0.10 percent, while B’s 15 requires 0.30 percent. Beyond the threshold, a dashed line denotes the strain demand if the film remained intact, not an actual post-crack strain. The layer drawing has schematic thicknesses and is not a contact stress calculation. Passing the strain check does not establish interface or overall component survival.

Protection is a system property

A coating may crack if the substrate deforms excessively, while a lubricant may fail to maintain separation under changed conditions. Evaluate replenishment, contamination and wear products. Do not infer durable protection from an initial low-friction value alone.

Use evidence that distinguishes the proposed failure route. Film-thickness or suitable contact-detection evidence helps test separation. Pressure or load-share evidence tests support. Surface chemistry can test boundary-film formation; a viscosity measurement alone cannot. Cross-sections and failure-surface examination can distinguish a cracked coating, a deformed substrate and interfacial separation.

In a ramped scratch test, a moving indenter is pressed with increasing normal force until a defined event occurs. The first film crack and the first coating detachment can occur at different loads. A recorded critical load belongs to a particular event, tip, loading rate, speed, coating thickness, substrate, environment and observation method. It is not automatically an intrinsic adhesion stress. Friction or acoustic-emission changes can help locate events but do not alone identify whether fracture was within the coating or at its interface.

Finally, follow durability rather than only the first pass. Replenishment, starvation, particles, chemical change, repeated deformation and loss of a layer can change the mechanism. The next lesson develops matched whole-system comparisons; here the main question is which support or protection mechanism operates now, what could change it, and what observation would reveal that change.

Worked examples

A pressure profile is not a peak-pressure-times-area load

A chosen gauge-pressure profile is triangular along a 10 mm by 5 mm rectangular region and uniform across its width. It rises from zero to 4 MPa then returns to zero. Find N_f. In a second, independently observed mixed-contact state, the profile has the same shape and footprint but a 1.6 MPa peak, while N = 100 N. What N_s is required for quasi-static balance?

The projected area is bL = 5 × 10 = 50 mm². The first profile has mean pressure 4/2 = 2 MPa, so N_f = 2 N/mm² × 50 mm² = 100 N. Using the 4 MPa peak over the entire region would incorrectly give 200 N.
For the supplied second profile, the mean is 1.6/2 = 0.8 MPa, so N_f = 0.8 × 50 = 40 N.
Because the second state is independently established as a quasi-static mixed contact with no other normal support, N_s = N − N_f = 100 − 40 = 60 N.
A lower generated pressure at one trial gap alone would not prove that the actual contact has this 60 N share. A freely moving body could close the gap or accelerate; its new equilibrium and any surface contact must be established. The profile change here is supplied, not predicted from temperature alone.

First fluid support: 100 N. In the specified second mixed-contact state, fluid supports 40 N and surface junctions transmit 60 N.

Complete separation still costs power

In the separate maintained-gap shear apparatus, η = 0.050 Pa s, U = 0.20 m/s, h = 20 µm and A = 200 mm². Find τ, F and P. Then double U while the apparatus holds h and η fixed.

Convert h = 20 × 10⁻⁶ m and A = 200 × 10⁻⁶ m² = 2.00 × 10⁻⁴ m². The velocity gradient is U/h = 0.20/(20 × 10⁻⁶) = 10,000 s⁻¹.
The shear stress is τ = ηU/h = 0.050 × 10,000 = 500 Pa. Drag is F = τA = 500 × 2.00 × 10⁻⁴ = 0.100 N.
Power is P = FU = 0.100 × 0.20 = 0.0200 W. At U = 0.40 m/s, τ becomes 1000 Pa, F = 0.200 N and P = 0.0800 W: twice the drag and four times the power.
These changes follow from fixed gap, constant viscosity and the stated laminar no-slip model. They do not establish what a self-acting bearing’s film thickness or total friction would become at the new speed.

At the initial speed: 500 Pa, 0.100 N and 0.0200 W. At doubled speed with the stated controls: 1000 Pa, 0.200 N and 0.0800 W.

A hard film can fail by following a deforming substrate

A hard coating fails on a soft substrate under load. Why can greater coating hardness be insufficient? Use a uniform-extension analogue: an intact bonded film spans L = 5.00 mm and has supplied ε_c = 0.20%. Support A extends by 5 µm; support B by 15 µm under their respective stipulated loading states. Compare the strain demands.

The substrate deforms beneath contact, and a bonded film must accommodate the deformation. The simple strip analogue isolates longitudinal compatibility; it does not calculate the strain field under an indenter.
Use L = 5000 µm. A demands ε_req = 5/5000 = 0.001 = 0.10%, below the supplied 0.20% crack-onset threshold. B demands 15/5000 = 0.003 = 0.30%, above it.
The threshold extension is δ_c = ε_c L = 0.002 × 5000 = 10 µm. An intact film following B cannot remain below that tensile threshold. Being below it for A only passes this one check; interface separation, defects or other loading modes could still matter.
Cracking or debonding can govern the system. Hardness does not supply ε_c, adhesion or support stiffness. If the interface separates first, the intact-bond compatibility assumption must be revised.

Hardness cannot replace substrate support and interface durability. In this chosen analogue, A demands 0.10% strain and B 0.30%; B exceeds the supplied film crack threshold even though the coating hardness has not changed.

Keep a scratch event with its load

A teaching scratch test starts at 2 N and ramps normal load at 4 N/s while the tip moves at 0.50 mm/s. Microscopy identifies first film cracking at 1.50 mm and first interfacial detachment at 3.00 mm. Find the corresponding loads. Does the first crack load directly measure adhesion?

Time from the start is distance divided by constant speed. First cracking occurs at 1.50/0.50 = 3.00 s; detachment at 3.00/0.50 = 6.00 s.
The load schedule is N(x) = N_start + (load ramp/speed)x = 2 + (4/0.50)x with x in mm. Thus cracking occurs at 2 + 4 × 3 = 14 N and detachment at 2 + 4 × 6 = 26 N.
The 14 N event is a cohesive crack in the film, not the stated interface event. The 26 N result belongs to the observed detachment in this system and method; it is not an interfacial stress obtained by dividing by a convenient nominal area.
To compare another coating or treatment, define the same observed failure event and control the relevant tip, load schedule, traverse speed, thickness, substrate and environment. A signal spike without microscopy would not establish which event occurred.

First film cracking: 14 N. First interfacial detachment: 26 N. The first event is not an adhesion measurement; the second is a conditional test-event load, not a universal interface property.

Try the reasoning

Does fluid lubrication always eliminate all asperity contact?
Show the answer and reasoning
No. Mixed or boundary-dominated lubrication can retain load-bearing surface junctions.
Oil may occupy the rest of the gap while contacts through thin films still transmit part of the normal force. Distinguish presence of lubricant from complete fluid-pressure separation.
Why test temperature changes?
Show the answer and reasoning
Temperature can change liquid viscosity, film formation, surface reactions and material response.
A warm lubricant may give less viscous drag at a maintained gap yet provide less self-acting separation under the actual load. Test the operating state and supply rather than ranking protection from viscosity or starting friction alone.
A different supplied triangular gauge-pressure profile acts over 12 mm by 4 mm and peaks at 3 MPa. In a confirmed quasi-static mixed state N = 120 N and there is no other support. Find N_f and N_s.
Show the answer and reasoning
N_f = 72 N and N_s = 48 N.
Mean pressure is 1.5 MPa and projected area is 48 mm², giving 1.5 × 48 = 72 N. The surface share is the remaining 120 − 72 = 48 N. No friction coefficient or microscopic pressure follows from this balance.
In the maintained-gap shear example, temperature changes η from 0.050 to 0.020 Pa s while U = 0.20 m/s, h = 20 µm and A = 200 mm² remain controlled. Find F and P. Does the lower power prove better protection in an otherwise unspecified bearing?
Show the answer and reasoning
F = 0.0400 N and P = 0.00800 W. No protection ranking follows for the unspecified bearing.
The same gradient is 10,000 s⁻¹, so τ = 200 Pa, F = 200 × 0.0002 and P = F × 0.20. These apparatus controls do not establish a bearing’s film thickness, load share, starvation or chemistry at its new temperature.
A hydrodynamic slider stops under load. One learner says its normal fluid force must instantly be zero; another assumes the running film remains indefinitely. What mechanisms are missing from both claims?
Show the answer and reasoning
Transient squeeze pressure, possible external pressurisation, drainage and evolving surface contact must be considered.
The steady motion-generated wedge does not persist merely by retaining the old running speed in a formula. But a closing gap can generate transient pressure, and an externally supplied hydrostatic film need not require sliding. The actual time-dependent state is not given by either extreme claim.
A new intact-film analogue has L = 8.00 mm and supplied tensile crack-onset strain 0.15%. Its substrate extension is 16 µm. Find the demanded strain and threshold extension. Would stronger adhesion alone make the demand fall below the threshold?
Show the answer and reasoning
Demand is 0.20%; threshold extension is 12 µm. Stronger adhesion alone does not reduce the imposed strain demand.
16/8000 = 0.002 = 0.20%; δ_c = 0.0015 × 8000 = 12 µm. Maintaining the bond transfers the support’s extension rather than eliminating it. Film strain tolerance, support deformation and other failure modes require separate evidence.
In the scratch example, an acoustic signal changes at the 14 N first-crack position. A report calls 14 N the “adhesion strength” and compares it with another test’s first-detachment load of 26 N. Identify two errors.
Show the answer and reasoning
It assigns an interface event from a nonunique signal and compares different failure endpoints as though they were the same property.
The example’s microscopy identifies 14 N as cracking within the film. Signal changes can have multiple causes; even the 26 N interface event is a load under a specified tip, substrate and schedule, not an intrinsic stress. Compare matching observed events under relevant matched conditions.
A treated region is harder near the surface, but a component is later ground deeply enough to remove that region. Separately, a boundary additive is present in an oil sample, but no protective film is found in the loaded track. What inference fails in each case?
Show the answer and reasoning
The first assigns the removed surface state to the finished surface; the second equates an available chemical ingredient with a film that actually formed and survived.
Use the final treatment-depth/profile and grinding history to identify the exposed state. For the lubricant, examine the actual loaded surface and film formation/removal conditions. Neither a treatment name nor an ingredient label substitutes for the operating structure.

What to carry forward

Fluid pressure support and viscous shear drag are different; a lubricant’s presence alone does not establish separation.
Identify the mechanism that generates pressure and the supply that maintains it, including startup and changed conditions.
Boundary films, deposited coatings and modified surface regions protect through different structures that must form and survive.
Coating hardness does not replace strain tolerance, substrate support, cohesion or a durable interface.
Keep a supplied model, observed failure event and measurement conditions with each number before comparing protection.

References

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