Materials Science · Contact, friction, hardness and wear

Wear mechanisms and changing surfaces

Explain how abrasion, adhesion, repeated contact and chemical reactions change a surface. Distinguish displaced material, transfer and loss from a specified body; calculate interval wear rates and test competing explanations for running-in.

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

Explain a causal sequence from local deformation or reaction to material detachment, rather than identify a mechanism from one surface mark alone.
Distinguish abrasion, adhesive transfer, contact fatigue and chemical contributions while recognising that several can interact.
Interpret debris, transfer layers and running-in as parts of an evolving contact.
Calculate volume from a stipulated surface profile or mass loss, with the assumptions and units stated.
Compare cumulative loss, interval loss per distance and load-normalised loss without treating a measured rate as a universal material constant.
Use surface, subsurface, debris and time-resolved evidence to discriminate explanations and limit extrapolation.

Material can leave by several routes

Abrasive particles can cut or plough a surface; adhesive junctions can transfer material; repeated contact can initiate fatigue damage. Chemical reactions can create or remove films. Several mechanisms can operate together and change during a test.

Start by naming the body and the observation. Material can move from a pin to a disc, remain as loose debris in the contact, or leave the contact altogether. Loss from the pin is not necessarily loss from the combined pin–disc–debris system. Wear is also used broadly for damaging surface change; when this lesson reports a wear volume V, it means volume lost from the specified specimen under the stated measurement definition. Permanent displacement and cracking can matter before a loose particle appears.

In abrasion, a harder protrusion or particle indents and moves across a surface. A fixed protrusion can cut a chip or plough material into side ridges. A loose particle between the two main bodies can slide, roll or indent repeatedly; its shape, confinement and motion change the damage. These are commonly called two-body and three-body situations. A groove shows displacement, but only the fraction that detaches is immediate material loss. Ridges can be flattened and deformed again, eventually cracking into debris. Indentation resistance can reduce penetration in some comparisons, while fracture resistance matters when particles or a brittle surface chip. Hardness alone does not settle all these outcomes.

Adhesive wear begins at a local junction between the contacting surfaces or their interfacial layers. Sliding may shear the junction with little detachment, or fracture may pass through one body and transfer a fragment to the other. That fragment may later detach, become a new protrusion or join a transfer layer. Interfacial shear therefore does not mean every contact creates a particle of fixed size. Composition mapping that finds material from one partner on the other is useful evidence of transfer; it does not by itself establish the entire sequence of fracture and removal.

Damage can accumulate beneath a changing film

In contact fatigue, a region experiences repeated stress changes as rolling or sliding contacts pass. Local plastic deformation, microstructural stress concentrations and existing defects can contribute to crack initiation. Cracks extend and join one another or reach the free surface; a sheet or flake then separates, leaving a pit or spall. Delamination describes separation of a near-surface layer, often after subsurface deformation and cracking. It is a possible wear route, not a claim that every wear particle has the same origin. An apparently quiet early volume-loss record can coexist with accumulating subsurface damage.

Chemical reactions can change the surface layer as well as the underlying material. An oxide or another reaction film can separate the main bodies and reduce direct junction formation if it remains supported and continuous. If sliding removes or breaks it, newly exposed material can react and the cycle can repeat. A detached hard reaction product can also act as an abrasive. Whether a film protects, is consumed, or becomes debris depends on its formation, adhesion, mechanical support and removal under those conditions. It is not enough to label a surface oxidised.

A historical National Bureau of Standards study of carbon steels illustrates that changing state. Rosenberg and Jordan (1934) reported cases where films appeared with a lower wear rate, then disappeared as the rate rose again. They also examined near-surface cracks and debris composition. Their proposed explanations included film support and deformation of the substrate. These observations support a conditional film-and-damage explanation; they do not prescribe a universal oxide benefit, a modern test procedure or a numerical rate for another steel.

The categories can interact. A transferred fragment can become a third-body abrasive; repeated ploughing can produce fatigue damage; reaction films can change adhesion. Fretting names small repeated relative motion, often at an apparently fixed joint, rather than a unique microscopic removal mechanism. Its debris and damage may involve adhesion, oxidation, abrasion and cracking together. Choose a causal account that explains the observations rather than force the whole contact into one exclusive label.

The contact after running-in is a different contact

Running-in is the early evolution of the contacting surfaces and interfacial material. High regions may flatten or be removed, the load-bearing geometry can change, and transfer or reaction layers can form. These changes can lower a later interval wear rate without requiring an increase in bulk hardness. Local work hardening is another possible contributor, but it needs evidence rather than being inferred from a falling rate alone.

Debris is a third body when it participates between the two main bodies. It may cut or indent, roll, become trapped, or compact into a layer that carries some load and separates the original surfaces. Removing it, retaining it or changing contact orientation can change later wear. Cleaning is therefore part of the test history, not a neutral act that can always be inserted between two otherwise identical measurements.

A stable-looking interval is an observation over that interval. Continued accumulation of cracks, exhaustion of a transfer layer, a change in debris circulation or penetration through a coating can produce a later transition. Running-in need not lead to a permanent steady rate. Friction and wear must also be followed separately: tangential force measures resistance to motion, whereas wear measurements follow surface change or material loss.

A measurement needs a reference and a material balance

Profilometry measures surface height relative to a reference. To infer volume, compare with an unworn or otherwise established original surface and integrate the height differences across an area. For a deliberately uniform straight track of length ℓ, a transverse missing cross-sectional area A gives V = Aℓ. A real track need not be uniform; one attractive profile cannot establish its whole volume. Multiple positions or a measured three-dimensional surface are needed to represent variation.

Below-reference groove area A− and above-reference pile-up area A+ answer different questions. If all raised material came from this specimen, density is unchanged, the whole displaced region is captured and the profile is uniform along ℓ, the net missing volume is (A− − A+)ℓ. The below-reference volume alone includes material displaced into the shoulders. Foreign transfer, reaction products, compaction, material outside the measured field or an uncertain reference can invalidate that simple subtraction. Report the definition actually used; a deep scar can matter to function even when much of its volume is displaced rather than detached.

For a homogeneous specimen, measured mass loss Δm can give V = Δm/ρ if ρ is the density of the lost material and the mass change represents that loss. For example, 8 g/cm³ is 8 mg/mm³: both numerator and denominator change by a factor of 1000. Oxidation or foreign transfer can add mass while the original body loses material, so zero net mass change does not prove zero wear. Mixed coating and substrate loss cannot in general be converted using one arbitrary density.

Cleaning, repositioning and temperature changes can alter both the measurement and the contact. A displacement sensor that follows two approaching bodies can include wear of both partners, thermal expansion and mechanical movement. Check which quantity a sensor measures before calling it specimen volume loss. State the measurement uncertainty and resolution; an unresolved change is not an exact zero. The profile geometry below is an original teaching model with an exact reference and specified assumptions, not a measured surface or a calibration procedure.

An original transverse profile uses the same six pixels per micrometre in both directions. A triangular groove 40 micrometres wide and 6 deep has area 120 square micrometres below the reference. Two native shoulders each 10 micrometres wide and 10 high have total area 100 square micrometres above it. For a uniform 2 millimetre length, below-reference volume is 0.00024 cubic millimetres, while net missing volume is 0.00004 cubic millimetres under the stated complete material-balance assumptions. The figure compares displacement and loss, not a measured wear scar.

Volume loss needs context

Record distance, load, geometry, environment and how wear was measured. Debris and transfer layers can alter the later contact. A wear rate averaged over a running-in period may differ from one measured after a more stable state develops.

Let s be accumulated sliding distance and V(s) the cumulative specimen loss measured from the start. Over an interval from s₁ to s₂, define Δs = s₂ − s₁ and ΔV = V(s₂) − V(s₁). The interval loss per distance is r = ΔV/Δs, the slope of the straight segment joining the two observations. It is an average over that interval, not an instantaneous rate at every point. With V in mm³ and s in m, r has units mm³/m. The whole-test average is V(s)/s when the initial measured loss is zero.

If the normal force magnitude N is constant, a useful additional normalisation is q = ΔV/(NΔs), with units mm³/(N m). Dividing by load allows a stated comparison; it does not prove that loss is proportional to load or that q remains constant after a mechanism changes. This dimensional q is not a friction coefficient and is not the dimensionless coefficient in a hardness-containing wear model. No universal wear law is being fitted here.

Distance is not time. If a constant sliding speed v is known and the same interval response applies, the average volume loss per time is r v; its units are mm³/s when v is in m/s. Changing speed may also change temperature, film response and mechanism, so a distance normalisation alone does not remove speed dependence.

The original teaching record used below holds N = 10 N, the counterface, geometry, speed, environment and measurement definition fixed for each comparison. Distance s (m): 0, 100, 300, 600, 1000. Specimen A loss V (mm³): 0, 0.030, 0.042, 0.060, 0.084. Specimen B loss V (mm³): 0, 0.010, 0.022, 0.040, 0.070. These chosen exact values expose the reasoning; they are not literature measurements, uncertainty-free experimental evidence or a ranking of named materials.

Use a changing rate to ask a better mechanism question

For specimen A, the first 100 m gives r = 0.030/100 = 3.0 × 10⁻⁴ mm³/m. The next 200 m gives (0.042 − 0.030)/200 = 6.0 × 10⁻⁵ mm³/m. Both later intervals also give 6.0 × 10⁻⁵ mm³/m. The early interval rate is five times the later one, while the whole-test average is 0.084/1000 = 8.4 × 10⁻⁵ mm³/m. Equal later interval averages are compatible with a more stable state over those intervals; they do not prove no smaller fluctuations occurred or identify its mechanism.

Mechanism evidence is stronger when independent observations agree. Directional grooves and matching hard particles support abrasion; compositional transfer supports movement from the counterface; cracks beneath a surface and detached flakes support a cracking route. Film composition and continuity before and after a rate transition test a film explanation. Each observation has alternatives: a groove does not uniquely identify whether the cutter was fixed or loose, and an oxide-rich particle may have reacted before or after detachment.

For a concrete example of discriminating evidence, Maich, Gronsky and Komvopoulos (2024) examined A216 carbon steel sliding against 410 stainless steel in small oscillations. The stainless partner contained much more chromium, so the researchers used chromium maps on the carbon-steel track to identify transferred counterface material. They combined such surface maps with cross-sectional microscopy and debris analysis to follow changing oxidation and damage. This is stronger evidence for transfer than the appearance of a bright patch alone; it remains evidence for that tested pair and history.

Compare interrupted companion specimens or validated in-situ observations at selected stages when possible, recognising that stopping and restarting one contact can change it. Examine both bodies, their cross-sections and collected debris. If debris removal is the proposed cause, compare a controlled removal condition with a matched retention condition, rather than change load and cleaning together. Replicates and measurement uncertainty are needed to decide whether an observed difference is repeatable.

For the chosen record, B has less total loss at 1000 m but a greater final interval rate than A. A claim about accumulated loss and a claim about current interval behaviour can therefore rank the same records differently. Neither establishes service life: a failure criterion, representative conditions, uncertainty and evidence about later mechanisms would still be needed. The next lessons examine lubrication and surface treatments, then the design of whole-system comparisons.

An original teaching record compares A and B at constant normal force 10 newtons and matched conditions. At distances 0, 100, 300, 600 and 1000 metres, A volumes are 0, 0.030, 0.042, 0.060 and 0.084 cubic millimetres; B volumes are 0, 0.010, 0.022, 0.040 and 0.070. Joined points are interval summaries, not a fitted law. The lower equal-width groups show interval rates for A of 30, 6, 6, 6 and B of 10, 6, 6, 7.5 in units of 10 to the minus fifth cubic millimetres per metre. B has lower total loss at 1000 metres but higher final interval rate. The lower categories represent unequal distance intervals with equal group width; bar height, not group width, encodes rate. These exact chosen values do not show experimental uncertainty.

Worked examples

Test a running-in explanation

Wear falls after an initially high rate. Does this prove the material became intrinsically harder?

For the chosen specimen A record, calculate the early and later interval rates: 3.0 × 10⁻⁴ and 6.0 × 10⁻⁵ mm³/m. The observed change is a factor of five.
Smoothing, changed load-bearing geometry and formation of a supported transfer or reaction layer can each reduce later loss. Subsurface work hardening is possible, but the rate record alone cannot discriminate these explanations.
Compare stage-specific surface profiles, film composition and near-surface structure. Use a suitable local hardness comparison only to test the hardening hypothesis, while accounting for the film and the affected depth.
Separate the observation, lower interval loss, from the proposed cause. A later low interval also does not rule out accumulated cracks that might cause a subsequent spall.

No. Evolving contact structure can reduce wear without that change. The data establish a rate transition in the chosen record; hardness and the responsible mechanism require additional evidence.

How much of a groove is actually missing?

A uniform 2 mm long straight track has a triangular depression 40 µm wide and 6 µm deep. Two triangular shoulders each have a 10 µm base and 10 µm height above the exact original surface. All raised material is native material displaced from this track, density is unchanged, and all displaced material is captured. Find the below-reference volume and net missing volume.

The depression area is A− = ½ × 40 × 6 = 120 µm². The two shoulders have total area A+ = 2 × ½ × 10 × 10 = 100 µm².
Convert track length to 2000 µm. Below-reference volume is 120 × 2000 = 240,000 µm³. Since 1 mm³ = 10⁹ µm³, this is 2.4 × 10⁻⁴ mm³.
Under the stated material-balance assumptions, net missing area is 120 − 100 = 20 µm², so net missing volume is 20 × 2000/10⁹ = 4.0 × 10⁻⁵ mm³.
Counting the depression alone would give six times the net missing volume in this model. If a shoulder contained transferred material or oxide, the same subtraction would no longer measure native specimen loss without additional information.

Below-reference volume: 2.4 × 10⁻⁴ mm³. Net missing volume under the stated assumptions: 4.0 × 10⁻⁵ mm³. The remaining 2.0 × 10⁻⁴ mm³ is accounted for by the raised shoulders.

A lower total can accompany a higher later rate

Using the chosen A and B records, compare total loss at 1000 m, the 600–1000 m interval rates, and their load-normalised q values. What does each comparison establish?

At 1000 m, B has lost 0.070 mm³ versus A’s 0.084 mm³. B’s total is 0.014 mm³ smaller; that is about 16.7% less than A’s total, using A as the denominator.
For A’s last 400 m, ΔV = 0.084 − 0.060 = 0.024 mm³, so r_A = 0.024/400 = 6.0 × 10⁻⁵ mm³/m. For B, r_B = (0.070 − 0.040)/400 = 7.5 × 10⁻⁵ mm³/m.
At N = 10 N, q_A = 6.0 × 10⁻⁶ and q_B = 7.5 × 10⁻⁶ mm³/(N m). B’s final interval rate is 25% higher than A’s, although its total to date is smaller.
The cumulative comparison answers how much was lost so far. The interval comparison answers how rapidly loss accumulated over the last measured distance. Neither comparison permits replacing the whole record with one universal material property.

B has lower measured cumulative loss in the chosen record; A has lower final interval loss per distance and per load-distance. The conclusion must name the quantity and interval.

Keep mass, volume and depth distinct

A homogeneous teaching specimen loses 0.160 mg of its own material over 200 m at N = 10 N. Its density is 8 mg/mm³, with no mass gain or contamination. If the measured lost volume is distributed over a specified 4 mm² track area, find V, q and the mean depth. Does this give the deepest pit?

Mass-to-volume conversion gives V = 0.160/8 = 0.0200 mm³. This uses the lost material’s density and the stipulated absence of other mass changes.
The load-distance normalisation is q = 0.0200/(10 × 200) = 1.00 × 10⁻⁵ mm³/(N m).
Mean depth over the specified area is V/A = 0.0200/4 = 0.00500 mm = 5.00 µm. It is a geometric average of that volume over that area.
An uneven track can have deeper pits and shallower regions. Maximum depth needs spatial information; the average alone cannot decide whether a local depth limit was exceeded.

V = 0.0200 mm³, q = 1.00 × 10⁻⁵ mm³/(N m), and mean depth = 5.00 µm. The maximum depth is undetermined.

Try the reasoning

Is debris always irrelevant waste?
Show the answer and reasoning
No. Debris can remain an active part of the contact.
It may cut or indent, roll, carry load after compaction, or contribute to a transfer layer. Removing it can change the mechanism rather than simply tidy the apparatus.
Does low friction guarantee low wear?
Show the answer and reasoning
No. Tangential resistance and material removal are distinct responses.
A contact can dissipate energy through junction shear or deformation without detaching a particle on each pass. Conversely, an interfacial layer may shear easily while underlying material continues to detach. Measure both quantities under the stated conditions.
A new specimen has V = 0.012 mm³ at 50 m and 0.027 mm³ at 250 m, at constant N = 5 N. Calculate the interval r and q. A learner divides 0.027 by 250 to report this interval rate: what did that division actually measure?
Show the answer and reasoning
r = 7.5 × 10⁻⁵ mm³/m; q = 1.5 × 10⁻⁵ mm³/(N m). The learner calculated a whole-test average of 1.08 × 10⁻⁴ mm³/m if the initial V was zero.
Subtract both endpoints for the requested interval: ΔV = 0.015 mm³ and Δs = 200 m. Then r = 0.015/200 and q = r/5. The initial 50 m belongs in the whole-test average but not this interval.
Use the profile example’s 120 µm² depression, 100 µm² raised area and 2 mm length. Half of the raised area is independently identified as foreign counterface material; the other half is displaced native material. With all other model assumptions retained, what is the native specimen’s missing volume?
Show the answer and reasoning
1.4 × 10⁻⁴ mm³.
Only 50 µm² of the raised area accounts for displaced native material. The native deficit is therefore (120 − 50) × 2000/10⁹ = 1.4 × 10⁻⁴ mm³. Subtracting all 100 µm² would mistake imported material for native material still present.
A worn metal surface gains a small net mass and contains oxygen-rich patches. Does this establish a protective film and zero removal? Give a discriminating next observation.
Show the answer and reasoning
No. Reaction or transfer can add mass while native material is removed, and patch composition alone does not establish protection.
Measure track geometry against a reference, inspect film coverage through the rate transition and analyse debris or the counterface. A supported continuous film accompanying repeatably reduced removal supports protection more strongly than an oxygen signal alone; reaction after detachment remains a possible alternative for debris.
A rolling track shows little measurable early volume loss but later releases flakes. Cross-sections of companion specimens show growing subsurface cracks. Is the early record inconsistent with contact fatigue?
Show the answer and reasoning
No. Crack accumulation can precede detectable detachment.
Repeated stress can initiate and extend cracks while the material is still attached. Joining cracks can subsequently release a flake. This explains a delay in loss; it does not prove that all flakes in every contact arise from fatigue.
Suppose, only as an algebraic extension of the chosen A/B record, their final interval slopes stayed unchanged after 1000 m. At what total distance would their cumulative losses become equal? Why is this not a service-life prediction?
Show the answer and reasoning
About 1933 m total, or 933 m beyond the last observation, under that stipulated continuation.
Solve 0.084 + (6.0 × 10⁻⁵)x = 0.070 + (7.5 × 10⁻⁵)x, giving x = 0.014/(1.5 × 10⁻⁵) = 933.3 m. This extrapolates unverified slopes and defines only equality of two cumulative losses, not failure. Mechanism changes, scatter and a functional limit have not been established.
For a chosen interval with r = 6.0 × 10⁻⁵ mm³/m, a constant speed is v = 0.20 m/s. Find its average loss per second. Can you assume the same r if speed is doubled?
Show the answer and reasoning
1.2 × 10⁻⁵ mm³/s for the stated interval. The same r at doubled speed requires evidence.
Multiply distance-normalised loss by distance per time: r v = 6.0 × 10⁻⁵ × 0.20. A speed change can affect heating, film development and particle motion, so doubling v does not alone establish that the same mechanism and r remain applicable.

What to carry forward

Explain how deformation, junction fracture, repeated cracking or reaction leads to surface change; several routes can operate together.
Distinguish loss from a specified body, displaced native material and transfer within the whole contact.
Debris and films help determine the next contact; running-in is an evolving state rather than proof of a changed bulk property.
State the volume reference, material-balance assumptions, units, distance interval and constant-load condition before comparing rates.
Use multiple kinds of evidence to test a mechanism and keep observed interval behaviour separate from an assumed future continuation.

References

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