Materials Science · Contact, friction, hardness and wear

Wear comparisons require an interacting system

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

Define a wear comparison by the two bodies, their preparation, motion, environment, history and measured endpoint.
Interpret a controlled counterface comparison and distinguish a ranking interaction from an isolated cause.
Use load–distance normalisation as a conditional descriptor without assuming that it removes changed test conditions.
Assess loss from both partners against a stated functional criterion and distinguish mean from local damage.
Design fresh-pair repeats and matched comparisons that separate candidate effects from order, counterface and preparation.
Explain what an accelerated screening test establishes and what additional evidence a service prediction needs.

The counterface belongs in the result

Compare test pair, roughness, load, motion, speed, environment, duration and measurement method. A material ranking can reverse when one changes. Wear is therefore an interaction outcome rather than one universal scalar property.

Suppose two candidate pins slide against a replaceable track. Calling one pin more wear resistant leaves an important question unanswered: against which track, during which motion, and judged by whose damage? The pin can protect itself while abrading the track. A transferred layer may change what touches on later passes. The result belongs to the interacting pair and its evolving interface.

Changing the counterface changes more than a material name. It can alter surface chemistry, resistance to indentation, compliance, heat removal and the ability to retain a transfer layer. Each can affect the next contact. If only the counterface is deliberately substituted, the experiment tests the effect of that substitution on the whole pair; it does not automatically isolate one of those physical causes.

A concrete research example is Tian and colleagues’ study of carbon-fibre-filled polyethersulfone. At matched applied load and sliding speed, they compared steel and alumina counterfaces and reported different wear responses and counterface temperatures. The authors explicitly noted that counterface hardness and heat dissipation also changed and were not separately analysed in detail. That comparison supports a dependence on the tested partner; it does not uniquely prove that chemistry alone caused the difference.

Name the response before choosing a winner

A comparison starts with a decision. Screening pin materials for low pin-volume loss is different from keeping a track within a dimensional tolerance, limiting frictional power, or delaying a defined event such as coating breakthrough. State the body, quantity, observation interval and criterion. A smaller number is meaningful only after its denominator and endpoint are known.

For a fresh contact, total native material loss V over a specified sliding distance answers how much of that body was removed. At constant normal force magnitude N, the dimensional descriptor q = V/(Ns) divides by accumulated absolute sliding distance s and load. With V in mm³, N in newtons and s in metres, q has units mm³/(N m). The preceding wear lesson develops interval versions; here the issue is whether the records being divided describe comparable systems.

Calculating q does not establish the law V = qNs with one constant q at every load, speed or surface state. That extra use assumes the same coefficient remains applicable. Changes in contact pressure, lubrication, debris or film failure can invalidate it. This q is neither a friction coefficient nor the dimensionless coefficient used in a hardness-containing wear model.

A measurement definition also belongs to the comparison. A below-reference groove volume, a net native-material deficit and a specimen’s net mass change need not coincide. Use a common justified definition, account for transfer or reaction where relevant, and retain the reference surface and measurement uncertainty. Repeating an accurate calculation cannot remove a bias shared by every measurement.

Read a complete counterface comparison

The following is an original invented dataset, not measurements of named products. A and B are two fictional bulk pin materials. Each fresh flat pin has a 2.00 mm × 2.00 mm face. It slides on a fresh flat steel or ceramic coupon. Both counterfaces use the same documented finish target and checks of directional texture; their actual properties remain part of the substitution. Load is 10.0 N throughout.

Every run uses the same sinusoidal reciprocating motion: an 8.00 mm end-to-end stroke at 6.25 complete out-and-back cycles per second, for 62,500 cycles. One cycle travels 16.0 mm, so total absolute sliding distance is 1000 m. Duration is 10,000 s and mean absolute speed is 0.100 m/s; speed is zero at reversals, not constant throughout a cycle. The chamber is held at 23 ± 1 °C and 45 ± 5% relative humidity, without liquid lubrication. Matching chamber temperature does not establish equal local contact temperature.

Preparation, alignment, motion history, debris-retention arrangement, observation times and volume-definition method are held fixed. Each combination has three independently prepared fresh pin–coupon pairs. All four combinations occur once in each of three test blocks, with order randomised within a block. Recorded volumes are taken to represent native loss after accounting for retained foreign material and surface displacement. These conditions are teaching stipulations, not a standard procedure or evidence that three repeats are always sufficient.

The table gives all three pin results and the arithmetic means for both partners. The plotted individual pin values show the observed repeat spread; their range is not a confidence interval or the measurement uncertainty. Actual experimental reporting would retain each coupon result, instrument evidence and uncertainty as well. The small exact dataset teaches how to read a comparison, rather than certifying a population ranking.

Invented fresh-pair results after 1000 m at 10.0 N. All volumes are native loss in mm³; means average three pairs.
Pin	Counterface	Three pin volumes	Mean pin loss	Mean counterface loss
A	Steel	0.110, 0.120, 0.130	0.120	0.900
B	Steel	0.180, 0.200, 0.220	0.200	0.050
A	Ceramic	0.280, 0.300, 0.320	0.300	0.020
B	Ceramic	0.090, 0.100, 0.110	0.100	0.025

A reversed ranking is an interaction, not an explanation

Against steel, A’s mean pin loss is 0.120 mm³ versus B’s 0.200 mm³: A is 40% lower, using B as the reference. Against ceramic, A loses 0.300 mm³ versus B’s 0.100 mm³: A is three times B. The direction of the A–B difference changes with counterface. This is a ranking interaction: the effect of choosing A rather than B depends on the partner.

A single candidate ranking averaged over the two counterfaces would hide that decision. Equal weighting gives A a mean of 0.210 mm³ and B 0.150 mm³, but no actual test used an average counterface. Such a summary only answers a deliberately defined mixture of applications. For an application with steel, the steel comparison is the relevant starting evidence; the ceramic result is not a substitute.

To investigate a cause, connect the volumes to independent observations. Suppose the invented A–steel contacts show grooves in the steel and pin-derived particles within the track, while A–ceramic contacts show a discontinuous transferred layer and exposed pin regions. These observations motivate cutting/third-body and film-retention hypotheses. A groove alone cannot identify the particle source, and an end-of-test layer does not establish when it formed or that it caused low loss.

Compare stage-specific composition and profiles, both bodies and debris, or a controlled change that discriminates the proposed mechanism. For example, a matched debris-retention versus removal comparison tests the role of recirculating particles more directly than changing debris removal and load together. A mechanism claim should explain the evidence and the alternatives it excludes; the crossed ranking by itself only establishes the conditioned response in these records.

Original invented dataset with a normal force of 10 newtons and 1000 metres of total absolute sliding per run. There are three independent fresh pin–counterface pairs per combination. Against steel, pin A loses 0.110, 0.120 and 0.130 cubic millimetres, with a mean of 0.120 cubic millimetres; pin B loses 0.180, 0.200 and 0.220 cubic millimetres, with a mean of 0.200 cubic millimetres. Against ceramic, pin A loses 0.280, 0.300 and 0.320 cubic millimetres, with a mean of 0.300 cubic millimetres; pin B loses 0.090, 0.100 and 0.110 cubic millimetres, with a mean of 0.100 cubic millimetres. Open dots show individual pin records and short bars show their means. Solid blue guides connect pin A means; dashed violet guides connect pin B means. The counterfaces are categories, and the connecting guides do not interpolate a material. Pin A has lower loss against steel, while pin B has lower loss against ceramic. The raw repeat spread is not a confidence interval. The crossed ranking does not identify a unique physical mechanism.

The best pin may damage the other component

Return to the steel results. A loses less pin volume but its steel partner loses 0.900 mm³, compared with 0.050 mm³ beside B. The track’s response therefore changes the choice if track damage matters. Adding the two volumes is arithmetically possible, but total pair volume is not automatically the functional criterion: a small loss at a critical seal or electrical contact can matter more than a larger loss elsewhere.

For a defined geometric comparison, divide a native recession volume by its stated reference area: mean recession h̄ = V/A_ref. This average is valid when the volume represents recession over that area with other deformation and retained deposits accounted for. It does not supply the deepest pit. The pin area here is 4.00 mm². The swept coupon rectangle is 2.00 mm wide and 10.0 mm long: the 8.00 mm motion plus the pin’s 2.00 mm length. Its reference area is 20.0 mm².

Now stipulate a decision for this teaching assembly after the stated 1000 m exposure: mean pin recession must not exceed 60.0 µm and mean track recession must not exceed 10.0 µm. These invented endpoint limits are not material standards. A gives 30.0 µm pin recession and 45.0 µm track recession; B gives 50.0 and 2.50 µm. A meets the pin criterion but fails the track criterion. B meets both mean criteria for the supplied mean records.

That conclusion does not establish that every future B pair will pass. It also does not answer a maximum-depth, fatigue, friction, debris-emission or service-life criterion. The reference areas must belong to the actual damage regions, and nonuniform damage needs spatial evidence if the decision concerns a local defect. Naming the functional endpoint makes the inference narrower and more useful.

Original invented records for pin A against steel and pin B against steel. Mean recession is native recession volume divided by reference area, with other deformation and retained deposits assumed accounted for. Each pin has a reference area of 4.00 square millimetres. Pin A loses a mean volume of 0.120 cubic millimetres, giving 30.0 micrometres of mean recession. Pin B loses 0.200 cubic millimetres, giving 50.0 micrometres. The pin mean-recession limit is 60.0 micrometres. Each steel track has a reference area of 20.0 square millimetres. The steel track paired with pin A loses 0.900 cubic millimetres, giving 45.0 micrometres of mean recession. The steel track paired with pin B loses 0.050 cubic millimetres, giving 2.50 micrometres. The track mean-recession limit is 10.0 micrometres. Both panels use the same scale from 0 to 60 micrometres. Dashed orange lines show the different body-specific limits. The A–steel pair meets the pin criterion but exceeds the track criterion. The B–steel pair meets both mean criteria for these supplied mean records. These comparisons establish neither maximum depth, the chance that a new pair passes, nor future lifetime.

Separate the candidate from test order and prior history

If every A run uses a new coupon and every B run reuses A’s track, candidate and surface history are confounded: they change together, so their effects cannot be separated from those records. B may encounter a smoother surface, a transferred film or trapped debris produced by A. That sequence can be a legitimate investigation of a serviced track, but it is not a matched fresh-pair comparison.

Use independently prepared pairs to compare fresh systems. Randomising run order prevents a fixed order from automatically following candidate identity. A block groups tests under a shared nuisance condition, such as one preparation batch or one machine session; placing every candidate–counterface combination within each block lets the comparison be made without assigning one candidate to only one batch. Record the block and conditions. Randomisation and blocking do not repair a biased measurement or guarantee identical local temperatures.

Repeatedly scanning one scar improves knowledge of that scar; it does not create additional independently prepared wear contacts. Likewise, consecutive intervals from the same run share its developing surface history. Fresh-pair repeats reveal repeat variation, while repeated measurements help assess measurement variation. Both can matter, but they answer different questions. A range from three pairs does not by itself quantify confidence in a population mean.

Keep the entire running-in and later record. If the decision concerns an exposure from new, do not discard its early loss merely because a later interval is quieter. If the question instead concerns an established running state, specify how that state was reached and verify it for each system. Interrupted companion contacts or validated in-situ measurements can distinguish stages, while repeated stopping and cleaning of one contact can alter the very history being compared.

Match the engineering decision

An accelerated test should reproduce relevant mechanisms, not merely remove material quickly. Examine worn surfaces and debris to check transferability. State whether a comparison screens candidates or supports a specific service-life prediction.

A faster test can compress sliding distance into less clock time while changing heating, chemical exposure and lubricant supply. Even equal total frictional work does not imply equal temperature history or surface chemistry. A change from film-supported contact to repeated coating fracture would make the accelerated record a different damage process, rather than a simple faster version of the service process.

Simple tribometers can separate candidates under controlled conditions; component tests add the actual geometry and motion, while system tests include further coupled functions. Jones and Jansen’s NASA report distinguishes these testing levels and explains why short accelerated lubricant rankings cannot generally be converted directly to component life. This is a transfer argument, not a requirement to copy a spacecraft test.

Build the inference in stages: identify what the screening endpoint measured, check the relevant surface state and mechanism over its observation window, compare the component’s contact and environment, and validate any acceleration or continuation model against suitable evidence. A matched mechanism is necessary for the proposed transfer but is not sufficient evidence that a rate remains constant indefinitely. The lifetime-evidence lesson develops the separate population and extrapolation questions.

Worked examples

Calculate a descriptor without losing the counterface

Use the invented 10.0 N, 1000 m comparison. Calculate the pin q for A and B against each counterface. What survives normalisation, and what causal conclusion does it permit?

Every denominator is Ns = 10.0 × 1000 = 10,000 N m. Against steel, q_A = 0.120/10,000 = 1.20 × 10⁻⁵ and q_B = 2.00 × 10⁻⁵ mm³/(N m). Against ceramic, they are 3.00 × 10⁻⁵ and 1.00 × 10⁻⁵.
Because all four load–distance denominators match, dividing does not change the ranking. Against steel, (0.200 − 0.120)/0.200 = 0.400, or 40.0% less pin loss for A. Against ceramic, q_A/q_B = 3.00.
The A-minus-B mean difference is −0.080 mm³ with steel and +0.200 mm³ with ceramic. Changing the partner changes that difference by +0.280 mm³ in this dataset. It is not a constant offset associated with A alone.
This establishes a conditioned ranking interaction in the chosen records. It does not separate chemistry, compliance, heat removal or transfer-layer stability, all of which may change when a counterface is substituted. The three raw values per combination also do not replace an uncertainty or population analysis.

Normalisation preserves the crossed ranking: A has lower pin q against steel, B against ceramic. The partner matters; the specific physical cause needs discriminating evidence.

Choose by both stated dimensional criteria

For the steel contacts, use the supplied mean native volumes, reference areas and 60.0 µm pin/10.0 µm track mean-recession limits. Which pair meets this particular decision? Would pin loss alone select it?

For A, h̄_pin = 0.120/4.00 = 0.0300 mm = 30.0 µm. Its track gives 0.900/20.0 = 0.0450 mm = 45.0 µm.
For B, h̄_pin = 0.200/4.00 = 0.0500 mm = 50.0 µm. Its track gives 0.050/20.0 = 0.00250 mm = 2.50 µm.
A’s pin is within 60.0 µm but its track exceeds 10.0 µm. Both B means are within their respective limits. Pin-volume ranking alone would choose A and miss the track failure.
The arithmetic evaluates the supplied mean records against invented mean criteria after 1000 m. It does not bound a deepest pit, calculate a future failure time, or establish the probability that another pair meets both criteria.

B–steel meets both stated mean-recession criteria in the supplied records; A–steel does not. The choice changes when the whole pair’s functional endpoint is included.

Two valid tests can ask different questions

Coating A wins in a dry pin test but loses in a lubricated bearing. Is that necessarily contradictory?

Make the contrast concrete with separate invented records. A dry screen slides each coating against an alumina pin at 5.00 N, 0.100 m/s and 23 °C for 100 m, using matched geometry and preparation within that screen. Measured coating-volume losses are 0.030 mm³ for A and 0.090 mm³ for B. A wins that defined removal endpoint.
A lubricated bearing comparison instead holds its own geometry, oil supply, 200 N load, 100 rpm motion and 60 °C oil inlet temperature fixed for A versus B. Its endpoint is the first measured torque rise of 2.00 mN m above that bearing’s run-in baseline. In these invented individual records, A reaches the event at 80 h and B at 200 h. B lasts longer to this bearing endpoint.
The two within-test comparisons can both be valid while ranking different interactions and responses. Counterface, contact geometry, lubrication, motion, temperature and endpoint changed between tests. A coating that resists dry removal may not maintain the same attachment or compatible film under the bearing’s repeated loading and oil chemistry.
Inspect film state, substrate deformation, coating cracks/detachment and debris near each endpoint. A torque rise by itself does not identify the cause. These observations can test a proposed explanation; the records alone neither prove one mechanism nor supply population lifetimes. Do not divide the two event times into a universal coating improvement factor.

No. The tests can rank different interactions.

Equal distance and work do not mean equal exposure

In a separate bounded model, take constant N = 10.0 N and sliding-friction ratio μ = 0.300. Compare 1000 m traversed at mean absolute speeds 0.100 and 0.500 m/s. Calculate duration, mean frictional power and total frictional work. Does this justify a fivefold service-life acceleration?

Duration is t = s/v̄: 1000/0.100 = 10,000 s, or 2.78 h; the faster case takes 2000 s, or 0.556 h. This uses total absolute travel, so back-and-forth motion does not cancel.
The friction-force magnitude is μN = 3.00 N. With that stipulated constant force, mean mechanical power dissipated is P̄ = μN v̄: 0.300 W and 1.50 W. Power is five times larger in the faster record.
Total frictional work is μNs = 3000 J in either case. Equivalently P̄t gives the same result. Equal work delivered at different rates need not produce equal temperature histories; predicting temperature requires heat storage and transfer information.
The faster run also supplies only one fifth of the clock time for processes such as reaction, replenishment or ageing. Real μ, film state and contact dynamics may change too. Same distance and assumed work do not validate a fivefold rate or life mapping.

The faster record has one fifth the duration and five times the mean power, despite equal 3000 J work in this stipulated model. A service-life acceleration factor remains unvalidated.

Try the reasoning

What makes a wear test relevant to service?
Show the answer and reasoning
A justified relationship of geometry, conditions and mechanisms.
Name the component endpoint and test whether the relevant pair, contact, motion, surface history and mechanism support the transfer. Matching one nominal number or causing rapid damage is insufficient; mechanism matching alone does not validate an unlimited time extrapolation.
Why retain running-in records?
Show the answer and reasoning
The contact state evolves.
For exposure from new, early loss belongs in the endpoint. For a later-state comparison, the route to that state must be matched and documented. Discarding the early record can hide different film formation, damage or time spent reaching the reported state.
Two unmatched records give A: V = 0.060 mm³ at 5 N over 300 m; B: V = 0.180 mm³ at 20 N over 300 m. Calculate q for each. Does B’s smaller q prove it would lose less at 5 N?
Show the answer and reasoning
q_A = 4.00 × 10⁻⁵ and q_B = 3.00 × 10⁻⁵ mm³/(N m). No; B’s lower descriptor was measured at a different load.
The denominators are 1500 and 6000 N m. Predicting B at 5 N would require the extra assumption that its q remains applicable there; the records do not test that. A fair next comparison includes both candidates at the same specified load with matched preparation and fresh counterfaces.
A new reciprocating comparison uses a 3.00 mm end-to-end stroke for 20,000 complete out-and-back cycles at 4.00 Hz. Find distance, duration and mean absolute speed. A learner reports zero distance because the pin returns to its starting position: what quantity was confused?
Show the answer and reasoning
Distance = 120 m; duration = 5000 s; mean absolute speed = 0.0240 m/s. The learner confused net displacement with accumulated travel.
One cycle travels 2 × 3.00 mm = 0.00600 m. Thus s = 0.00600 × 20,000 and t = 20,000/4.00. Wear exposure follows the path length even though the net displacement over a complete cycle is zero.
Design a fresh-pair comparison of two pins and two counterfaces using three independent test blocks. Every combination is to occur once per block. How many contacts are tested? If each scar is scanned five times, are there five times as many independent wear contacts?
Show the answer and reasoning
Twelve fresh contacts; sixty scans still describe twelve contacts.
There are 2 × 2 × 3 = 12 combinations across blocks. Randomise the four run identities within each block and record preparation and conditions. Five measurements of each existing scar can assess measurement variation; they do not recreate the surfaces or add independent wear histories.
A lab tests A first on a new coupon and B afterwards on the same track. B loses less. Give a fair fresh-pair follow-up, and a different question for which the original sequence could be useful.
Show the answer and reasoning
Test A and B on independently prepared fresh coupons with matched conditions and randomised order. The original sequence can instead study how B performs on a track previously run with A.
Candidate identity and prior track state changed together. Fresh-pair tests separate that history from candidate choice. If serviced-track performance is the intended question, include controlled prior histories and state the sequence rather than labelling the result a universal material ranking.
A coupon loses 0.0400 mm³ over a stated 20.0 mm² recession area. Its criterion is a maximum local depth of 8.00 µm. Does the mean depth establish a pass? Give two simple depth distributions consistent with the same volume.
Show the answer and reasoning
Mean recession is 2.00 µm, but maximum depth is undetermined. Uniform 2.00 µm loss meets the depth limit; 10.0 µm loss over 4.00 mm² and zero over the remaining 16.0 mm² does not.
Both volumes are 0.0400 mm³: 0.00200 × 20.0 and 0.0100 × 4.00. A spatial profile or surface map is required to distinguish these cases. A mean criterion and a maximum-depth criterion are different decisions.
A matched material pair loses 0.120 mm³ at 10 N over 1000 m, then a fresh pair tested at 20 N loses 0.600 mm³ over the same distance. What would constant q have predicted? Compare the measured q values and identify what can be concluded.
Show the answer and reasoning
Constant q predicts 0.240 mm³ at 20 N. The descriptors are 1.20 × 10⁻⁵ and 3.00 × 10⁻⁵ mm³/(N m), a factor of 2.50 apart.
The simple proportional-load assumption does not describe these supplied records. The change could reflect altered contact state, a mechanism transition or other uncontrolled effects; it does not by itself identify the cause. Examine and repeat matched tests before assigning a mechanism or a transferable law.
An accelerated test shows repeated coating detachment, while representative component observations show gradual loss of a continuous boundary film without coating detachment. May test hours simply be multiplied by the speed ratio to predict the component endpoint?
Show the answer and reasoning
No. The observed damage processes differ, so that speed ratio is not a validated time mapping.
The accelerated test may still screen resistance to its coating-fracture condition. To support component transfer, establish a representative contact state and endpoint and validate the proposed mapping over appropriate observations. Equal sliding distance, hardness or total frictional work does not repair the mechanism mismatch.

What to carry forward

A comparison belongs to both contacting bodies, their preparation, conditions and evolving history.
A counterface can reverse the candidate ranking; a changed ranking does not uniquely identify its cause.
Load–distance normalisation describes a record; a constant-coefficient prediction requires additional evidence.
Choose an endpoint that represents the assembly’s function and inspect damage to both partners.
Fresh-pair repeats, run order and test blocks distinguish candidate effects from preparation and history.
Screening, component evidence and validated service prediction support different claims.

References

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