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Materials Science · Chapter 10 of 10 · Intermediate

Material Classes and Selection

Metals are not the whole story. Ceramics, polymers, and composites each own a corner of the property space, and choosing well means matching the loading to the right material index.

01

Readiness check

This capstone draws on bonding, properties, and the specific properties of Chapter 1. Tick only what you can do closed-notes.

  • Recall the four material classes and their bonding.
  • Define stiffness, strength, and density.
  • Compute specific properties (per unit mass).
  • Work with powers and roots.
  • Read a log-log property chart.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview specific properties in Chapter 1.
3 or more weak itemsRevisit bonding (Chapter 2) and properties (Chapter 5) first.
02

The core idea

Each material class occupies a region of the property space; the best choice maximises a performance index built from the properties the loading actually rewards.

Ec = EfVf + EmVmtie: maximise E/ρ · beam: maximise E1/2/ρselection = property need + index

Metals are tough and conductive; ceramics are stiff, hard, and heat-resistant but brittle; polymers are light, cheap, and flexible; composites blend a stiff reinforcement with a light matrix to reach property combinations no single class can. Selecting a material is not picking the strongest or stiffest, but maximising the index that the geometry and loading make decisive, while respecting cost, corrosion, and manufacturability.

The skill works when: you derive the right index for the load case and read the property chart for the class that maximises it.
The skill breaks down when: a single property is chased, or anisotropy and corrosion are ignored in service.
The concept. An Ashby chart plots stiffness against density, with each class as a region. A guideline of constant performance index sweeps across it; the best material sits furthest up that line, not simply highest in E.
03

The skills, taught in order

This chapter surveys the non-metal classes and turns selection into a method. Five skills cover ceramics, polymers, composites, corrosion, and the Ashby index.

10.1 The four classes in property space

Each class clusters in a region of the stiffness-density chart, a direct consequence of its bonding. Knowing roughly where a class sits lets you rule options in or out at a glance before any calculation.

ClassStiffness, densityBehaviourExamples
Metalshigh E, high ρductile, tough, conductivesteel, Al, Ti
Ceramicsvery high E, medium ρhard, brittle, heat-resistantalumina, SiC, glass
Polymerslow E, low ρflexible, cheap, low-temperaturePE, nylon, epoxy
Compositestunable E, low-medium ρanisotropic, high specific propertiesCFRP, GFRP

10.2 Ceramics and glasses

Their ionic and covalent bonds make ceramics stiff, hard, chemically stable, and heat-resistant, but brittle: strong in compression, weak in tension because flaws cannot be blunted by plastic flow. They serve as cutting tools, refractories, and wear surfaces, where their weakness in tension is designed around.

10.3 Polymers

Polymers are long covalent chains held to one another by weak secondary bonds, hence light, flexible, and low-melting. Thermoplastics soften and reflow on heating; thermosets are cross-linked and do not. They are viscoelastic (time- and temperature-dependent) and are chosen for low cost, low density, and corrosion resistance.

10.4 Composites and the rule of mixtures

A composite combines stiff, strong fibres with a light, tough matrix. Loaded along the fibres, stiffness follows the rule of mixtures Ec = EfVf + EmVm (the upper bound); loaded across them it is far lower (the lower bound), so composites are strongly anisotropic. This tailorability gives the highest specific stiffness and strength available.

10.5 Corrosion and selection by index

Materials degrade in service, metals by electrochemical corrosion (galvanic couples, where the less noble metal sacrifices itself), polymers by UV and solvents. Selection then becomes a method: write the performance index the loading rewards, and pick the class that maximises it.

Load caseMinimise mass forMaximise index
Tie (tension)stiffnessE/ρ
Tie (tension)strengthσ/ρ
Beam (bending)stiffnessE1/2/ρ
Beam (bending)strengthσ2/3/ρ

Engineering connection: this chapter closes the loop opened in Chapter 1, turning the structure-property reasoning of the whole course into material choices for real structures and products.

04

Worked example 1: stiffness of a fibre composite

A carbon-fibre composite has 60% fibre by volume (Ef = 230 GPa) in an epoxy matrix (Em = 3 GPa). Find the longitudinal and transverse elastic moduli, and comment on the anisotropy.

Figure 1. Loaded along the fibres, the stiff carbon dominates (rule of mixtures, upper bound). Loaded across, the soft matrix governs (lower bound), so the composite is about twenty times stiffer one way than the other.
  1. ProblemFind the longitudinal and transverse moduli of the composite in Figure 1.
  2. Given / findEf = 230 GPa, Em = 3 GPa, Vf = 0.60, Vm = 0.40. Find Ec,long and Ec,trans.
  3. AssumptionsAligned continuous fibres, perfect bonding, isostrain along and isostress across.
  4. ModelRule of mixtures for the longitudinal (upper-bound) and the inverse rule for the transverse (lower-bound) modulus.
  5. EquationsEc,long = EfVf + EmVm Ec,trans = EfEm/(EfVm + EmVf)
  6. SolveEc,long = 230(0.6) + 3(0.4) = 138 + 1.2 = 139 GPa. Ec,trans = (230 × 3)/(230 × 0.4 + 3 × 0.6) = 690/93.8 = 7.4 GPa.
  7. CheckAlong the fibres the composite nearly matches steel (200 GPa) at a quarter the density; across, it is barely stiffer than the epoxy. The 19-fold anisotropy is exactly why fibres are laid up in multiple directions.
  8. ConclusionComposites are designed, not just chosen: orienting fibres puts stiffness where the load is. The longitudinal value gives the high specific stiffness that made carbon composites dominate aerospace and racing.
Result. Ec,long = 139 GPa, Ec,trans = 7.4 GPa (about 19 times anisotropy).
05

Worked example 2: selecting a light, stiff beam

A beam must be as light as possible for a required bending stiffness, which rewards the index E1/2/ρ. Rank steel (E = 210 GPa, ρ = 7.85), aluminium (70, 2.70), wood along the grain (10, 0.5), and a CFRP laminate (100, 1.6).

Figure 2. For a light stiff beam, wood and CFRP win and aluminium beats steel, even though steel has the highest modulus. The right index, not raw stiffness, decides.
  1. ProblemRank the four materials in Figure 2 for a light, stiffness-limited beam.
  2. Given / findSteel (210, 7.85), aluminium (70, 2.70), wood (10, 0.5), CFRP (100, 1.6), all E in GPa and ρ in g/cm³. Find and rank E1/2/ρ.
  3. AssumptionsBending stiffness governs (the beam index applies); properties along the loading direction.
  4. ModelCompute the beam material index M = E1/2/ρ for each and rank, largest is lightest.
  5. EquationsM = E1/2/ρ
  6. SolveWood: √10/0.5 = 6.3. CFRP: √100/1.6 = 6.3. Aluminium: √70/2.70 = 3.1. Steel: √210/7.85 = 1.9. Ranking: wood ≈ CFRP > aluminium > steel.
  7. CheckSteel has by far the highest E yet ranks last, because the index rewards low density strongly (E appears only as a square root). This is why aircraft, bicycles, and floor joists avoid solid steel for stiffness-limited parts.
  8. ConclusionThe performance index, derived from how the part is loaded, can completely reorder a ranking based on a single property. Selecting well means choosing the index first, then the material, the method that closes this course.
Result. Beam index: wood 6.3 ≈ CFRP 6.3 > aluminium 3.1 > steel 1.9.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Pick the highest single propertySteel chosen for a light stiff part"Which index does the loading reward?"Use the performance index (often a specific property), not E or σ alone.
Wrong index for the load caseTie index used for a beam"Is it tension or bending?"Tie uses E/ρ; beam uses E1/2/ρ. The geometry sets the exponent.
Ignoring composite anisotropyTransverse load on aligned fibres"Along or across the fibres?"Stiffness drops sharply across fibres; lay them up for the load directions.
Forgetting corrosionGalvanic couple corrodes in service"Are dissimilar metals in contact, in a wet environment?"Account for corrosion and degradation, not just mechanical properties.
07

Practice ladder

Level 1 · Direct skill

A glass-fibre composite has Vf = 0.40 (Ef = 72 GPa) in a polyester matrix (Em = 3.4 GPa). Find the longitudinal modulus.

Show answer

Ec,long = 72(0.4) + 3.4(0.6) = 28.8 + 2.0 = 30.8 GPa. The stiff fibres dominate even at 40% volume, lifting the modulus far above the matrix.

Level 2 · Mixed concept

For a light, strong tie (index σ/ρ), would the ranking of materials match the beam ranking of Worked Example 2? Why or why not?

Show answer

Not necessarily: a tie rewards σ/ρ, a different index from the beam's E1/2/ρ, and uses strength rather than stiffness. A material strong but not especially stiff (or vice versa) can move up or down. The load case and the property both change the order.

Level 3 · Independent problem

Why are ceramics used for cutting tools and engine liners despite being brittle? Frame the answer in terms of which property the application rewards.

Show answer

Those applications load the ceramic mainly in compression and demand hardness, wear resistance, and high-temperature stability, exactly where ceramics excel. The brittleness (weakness in tension) is designed around, so the rewarded properties win.

Level 4 · Transfer to real engineering

Pick a real product and reverse-engineer its material choice: identify the load case, write the performance index, and explain why its material class maximises that index.

What good work looks like

The loading identified, the correct index derived, the chosen class shown to maximise it, and secondary factors (cost, corrosion, manufacturability) acknowledged.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that the performance index I derived matches the load case."
"Give me five products; I will name the class and the index that drove each choice."
"Which material should I use?" Deriving the index and reading the chart is the skill.
"Compute the composite modulus." Applying the rule of mixtures yourself is the point.

Portfolio task

Write a one-page selection note for a real part: state the load case, derive the index, rank two or three candidate classes, and justify the winner including a corrosion or cost caveat.

Must include: the index derived from the loading, a quantitative ranking, and a non-mechanical factor considered.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Name the four classes and one defining property of each.

Metals (tough, conductive), ceramics (hard, brittle, heat-resistant), polymers (light, flexible), composites (high specific properties, anisotropic).

2. Write the rule of mixtures for longitudinal stiffness.

Ec = EfVf + EmVm.

3. Give the indices for a light stiff tie and a light stiff beam.

Tie: E/ρ; beam: E1/2/ρ.

4. Why are ceramics strong in compression but weak in tension?

Brittle bonds cannot blunt flaws by plastic flow, so tensile cracks run; compression closes them.

5. What is galvanic corrosion?

When two dissimilar metals contact in an electrolyte, the less noble one corrodes preferentially.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive both worked examples from a blank page.
+3 daysSelect a material for one new load case using its index.
+7 daysConnect selection back to Mechanics of Materials.
+30 daysReview the whole course from the Materials Science hub.
10

Textbook mapping

ItemMapping
Primary sourceCallister and Rethwisch, Materials Science and Engineering: An Introduction, Chapters 12 to 17 (ceramics, polymers, composites, corrosion)
Cross-referenceAshby, Materials Selection in Mechanical Design · Askeland, Ch. 14 to 19 · Shackelford, Ch. 12 to 15
Core topics10.1 Property space · 10.2 Ceramics · 10.3 Polymers · 10.4 Composites · 10.5 Corrosion and selection
Engineering connectionTurns the whole course into real material choices for structures and products.
Read nextYou have completed the Materials Science course. Return to the course hub or continue to Manufacturing Processes.