Surface modification and gradients

A surface can have its own state

Diffusion treatments, implantation, oxidation, mechanical working or chemical modification can create gradients in composition, defects or stress. The surface response can differ from the interior without replacing the whole material.

A polished part initially meets a surface requirement. Finishing removes some material. The outside may look better, yet the newly exposed material came from a deeper part of the treatment profile. To decide what remains, first ask what changed with depth, where depth zero was, and what the requirement actually measures.

A surface-modified region need not be a separate sheet placed on a substrate. Substrate means the underlying material that supports the region being considered. A composition gradient is a systematic change of composition with position; a defect or stress gradient describes a different changing quantity. Equal outside dimensions or equal average composition do not make those profiles identical.

Follow what changes in the existing material

In a diffusion treatment, an environment exchanges a species with the surface and atoms redistribute within the solid. For example, nitrogen can enter a suitable steel, occupy interstitial sites and participate in nitride formation. Solute interactions and fine particles can impede dislocation motion and change resistance to plastic deformation. The result depends on alloy, phases, temperature and history; a concentration alone does not supply a universal hardness. Diffusion and phase transformations develops the transport conditions, while Grains, dislocations and strengthening explains the obstacles.

Such a treatment need not produce only one smooth single-phase gradient. A nitrided tool steel can have an iron-nitride compound region above a nitrogen-rich diffusion zone containing dissolved nitrogen and fine precipitates. Composition measurements, phase-sensitive diffraction and microscopy answer complementary questions. A small particle population need not be resolved by an ordinary image or produce a separate resolved diffraction peak. A smooth-looking image is not proof of one phase.

Mechanical working near a surface can instead change dislocation structure and leave residual stress without requiring a large change in elemental composition. Impacts used in peening locally deform material; surrounding material constrains its recovery, so an unloaded part can retain a stress profile. The sign and depth pattern require evidence. A reported near-surface compressive region does not imply compression everywhere or remove the need for force and moment equilibrium.

Depositing material adds an overlayer and usually introduces a layer/substrate combination to assess. Diffusion, reaction and deposition can coexist: a deposited species can react or mix with the original surface. Use the actual structure rather than a rule that every modified region has an abrupt bonded interface or that every chemical treatment lacks one. Coatings and deposition concepts treats that separate route in more detail.

Composition, defects, stress, roughness and phase identity are different observables. A method that establishes one does not automatically establish the others. The numerical model below deliberately follows only an invented solute-composition profile. It is not a nitriding schedule or a prediction for the cited steel.

Read a profile with a defined origin and unit

Consider an original hypothetical flat body, much thicker than 240 µm. Its composition is laterally uniform in the region considered. Let z be depth measured inward from the original treated surface. One micrometre, µm, is 10⁻⁶ m. Let c(z) be the local solute mass percentage: c = 0.800 mass% means a solute mass fraction of 0.00800, not 0.800 of the material mass.

The supplied profile is a straight-line fit: c(z) = 0.800 − 0.00300 z for 0 ≤ z ≤ 200, when the numerical value of z is in µm; c = 0.200 mass% at greater depth. The slope is −0.00300 mass-percentage points per µm. These stipulated values describe one state, not a solved diffusion history. No diffusivity, treatment duration or hardness calibration is supplied.

There is no abrupt composition jump at 200 µm: the line meets the 0.200 mass% interior value. Calling 200 µm the enriched depth means only that c exceeds that reference above it. If an invented composition screen instead requires c ≥ 0.350 mass%, its limiting original depth is (0.800 − 0.350)/0.00300 = 150 µm. An enriched depth and a depth defined by a selected threshold are different quantities.

Move the origin when material is removed

Stipulate uniform removal r = 80.0 µm from this face. The remaining composition is unchanged: no further diffusion, reaction, selective loss or mixing occurs. Let x now measure depth from the finished surface. A point at x was originally at z = r + x, so c_finished(x) = c_original(r + x). Removing material changes which point is called the surface; it does not move the original high surface concentration into the remaining material.

The new surface is c_original(80) = 0.560 mass%. The profile reaches its interior value at x = 200 − 80 = 120 µm. The 0.350 mass% crossing moves to x = 150 − 80 = 70.0 µm. Thus the finished surface passes that particular local composition screen, but a requirement to meet it everywhere through the first 100 µm fails: at x = 100 µm, z = 180 µm and c = 0.260 mass%.

Before removal, the first 100 µm of the original profile had a minimum 0.500 mass% and met the same screen. Finishing changed the outcome without changing the stipulated remaining composition. In a real machining operation, heating, plastic deformation, roughness, contamination and stress redistribution may change the state as well. A coordinate shift is then only part of the assessment. In particular, a residual-stress profile cannot generally be transferred by deleting its top values and ignoring the remaining body’s equilibrium.

A finite sample answers a different question

Use a matched companion piece to assess the finished first 100 µm without pretending that a destructive assay leaves the service face untouched. Stipulate the same profile, constant cross-sectional area and uniform density. Collect all the material from x = 0 to 100 µm, with no selective material loss, and obtain its total solute mass divided by its total sample mass. Equal thicknesses then contain equal material masses.

Because the stipulated profile is linear over that interval, its mass-averaged composition equals the mean of its two end values: c_average = (0.560 + 0.260)/2 = 0.410 mass%. A sequence of equal-thickness slices gives the same result as the straight-line area divided by the interval width. This average exceeds 0.350 mass%, although the last 30.0 µm of the interval is below the local requirement.

A uniform 0.410 mass% interval would give the same assay and meet the local screen throughout. The one average cannot distinguish those two profiles. If density or area varies, weight by the actual sampled masses; an unweighted depth average need not equal a chemical mass fraction. Sampling, scale and extrapolation limits develops the corresponding location and population problem.

This chemical collection model is not a general law for surface instruments. An indentation response involves a stressed volume and the support below it; it is not normally a simple arithmetic average of local hardness. Spectroscopic signals also have method-dependent depth sensitivity. Record preparation, probe scale, calibration and the quantity inferred. Hardness measures and their test dependence develops the distinctions among hardness measurements.

Retained depth and retained solute are not the same fraction

The same profile supports a separate mass account. Stipulate density ρ = 5000 kg/m³ everywhere and a constant unit face area of 1 m². Define excess solute relative to the 0.200 mass% reference in the same amount of material. This is a reference subtraction, not the total solute in the body or the amount absorbed during treatment.

Before removal, the excess mass fraction falls linearly from (0.800 − 0.200)/100 = 0.00600 to zero over 200 µm. Its depth-average is 0.00300. The enriched region contains ρ × 200 × 10⁻⁶ m = 1.00 kg of material per square metre, so its excess solute is 1.00 × 0.00300 = 0.00300 kg/m², or 3.00 g/m².

After 80.0 µm is removed, the remaining 120 µm region contains 0.600 kg/m². Its excess mass fraction falls from (0.560 − 0.200)/100 = 0.00360 to zero, so the mean is 0.00180. Retained excess is 0.600 × 0.00180 = 0.00108 kg/m² = 1.08 g/m². That is 36.0% of the original excess, although 60.0% of the enriched depth remains. The removed outer region contained a disproportionate share of the excess.

These are conserved-composition accounts under the stated removal assumptions. Retaining 36% of this defined excess does not mean retaining 36% hardness, wear resistance or service life. Those require their own mechanisms and matched finished-state evidence.

Specify depth and function

A treatment effective over a shallow depth may resist one wear condition but be removed under another. Assess layer depth, transition region and substrate support. Adhesion is especially relevant where a distinct bonded layer or coating exists; a diffusion gradient need not contain such an interface. Report whether a measurement samples the modified layer, the substrate or both.

Keep the original material, treatment and finishing history with the result. Measure the final exposed region and enough of the transition and substrate to answer the actual question. A composition traverse can test enrichment; diffraction or microscopy can test phases and local structure; a properly interpreted mechanical test can assess the response at its relevant scale. These observations complement one another.

Compare depth definitions before comparing reported treated depths: enrichment above a reference, a selected hardness criterion, a phase boundary and a measured signal’s penetration are not interchangeable. State the surface origin, preparation, method, depth resolution and uncertainty. The exact numbers above isolate the reasoning; they are not tolerance limits for a real part.

Worked examples

Try the reasoning

1. Does a surface measurement always describe the bulk?

   Show the answer and reasoning

   No.

   Its sampling depth can be limited.

2. Why use a gradual transition?

   Show the answer and reasoning

   It can alter stress and transfer between regions.

   The benefit depends on the actual gradient and loading.

3. Instead remove 40.0 µm from the same original profile. Find the new surface composition and the minimum in the finished first 100 µm. Does that interval meet the invented 0.350 mass% screen?

   Show the answer and reasoning

   New surface: 0.680 mass%. The interval ends at original z = 140 µm, with minimum 0.380 mass%; it meets this particular screen.

   Use z = 40 + x. The remaining enriched depth is 160 µm, but the decision follows the minimum of the stated 100 µm interval, not that depth alone.

4. For the same decreasing profile and a required first-100-µm minimum of 0.350 mass%, what is the greatest uniform removal r allowed by the ideal exact composition model?

   Show the answer and reasoning

   r ≤ 50.0 µm, with equality just meeting the invented screen.

   The deepest required point is at original z = r + 100 µm. It must not exceed the 150 µm threshold crossing. This is an exact-model bound, not a machining tolerance or safe process allowance.

5. After the 80.0 µm removal, a different screen requires c ≥ 0.600 mass% at the new surface. Can the 120 µm remaining enriched depth establish compliance?

   Show the answer and reasoning

   No. The actual new surface is only 0.560 mass%, below 0.600 mass%.

   The original 0.600 mass% crossing was at about 66.7 µm, which has already been removed. A thickness defined by enrichment above 0.200 mass% does not establish a higher threshold.

6. A separate assay combines equal-area, equal-density slices: 20.0 µm has uniform c = 0.600 mass% and 80.0 µm has c = 0.300 mass%. Find the combined mass percentage. Is the unweighted mean 0.450 mass% appropriate?

   Show the answer and reasoning

   c_average = (20 × 0.600 + 80 × 0.300)/100 = 0.360 mass%; the unweighted mean is incorrect.

   The thicker slice contributes four times as much material mass. A passing average still would not establish a 0.350 mass% local minimum because the 0.300 mass% slice is below it.

7. Remove 100 µm from the original profile. The enriched depth is halved. What fraction of the original excess-solute inventory remains under the constant-density model?

   Show the answer and reasoning

   25.0%, or 0.750 g/m².

   The remaining thickness is half, and its surface excess mass fraction is also half. The triangular area is therefore one-quarter. The depth and inventory fractions are different even before asking about a property.

8. Remove 220 µm from the ideal original profile without changing the remaining chemistry. What is the newly exposed composition? Does the old 0.800 mass% surface value still describe it?

   Show the answer and reasoning

   0.200 mass%, the supplied interior reference. The original surface value no longer applies.

   The removed thickness exceeds the 200 µm enriched region. Recheck the finished state; remaining reference material is not evidence of the old surface condition.

9. An indentation extends into a graded region and its underlying substrate. May the chemical slice-average formula be used to calculate its measured hardness from a few local hardness values?

   Show the answer and reasoning

   Not without an appropriate mechanical model or validated measurement interpretation.

   Indentation produces a nonuniform stressed and possibly plastic volume; layer/substrate support and probe geometry matter. A chemical mass account is not a general contact-response mixing law.

10. A peened face is machined. A technician deletes the removed portion of the old stress graph and keeps the remaining numbers unchanged. Is that justified by the composition-shift example?

    Show the answer and reasoning

    No. Remaining stresses can redistribute when the body and boundary conditions change.

    The composition example explicitly assumes unchanged remaining chemistry. Residual stress also has to satisfy mechanical compatibility and equilibrium in the new geometry. Use finished-state stress evidence or a suitable redistribution model.

What to carry forward

• State modification depth and gradient.
• Assess the final exposed region.
• Identify the changed quantity: composition, defects, phase, stress and roughness need different evidence.
• After ideal uniform removal, finished depth x corresponds to original depth z = r + x.
• Enriched depth, threshold-defined depth, a finite sampled average and a local minimum answer different questions.
• Use actual material-mass weights for composition; do not apply that average as a generic indentation law.
• Retained excess solute is a defined inventory, not a fraction of retained hardness or service life.