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Materials Technology: free practice, theory and problems

A material's properties – strength, ductility, conductivity, density – are largely determined by how its atoms are arranged. Most metals are crystalline: the atoms sit in a regular, repeating pattern called a crystal lattice. Which structure a metal has (BCC, FCC or HCP) determines, among other things, how tightly the atoms are packed and how easily dislocations can move – and therefore how ductile the material is. For an engineer this is the foundation for understanding why steel behaves differently from aluminum, and why a material can become stronger or more brittle after processing.

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Structure
Mechanical properties
Heat treatment and corrosion

1. Structure

What is it about?

Concepts and formulas

Unit cell: the smallest repeating building block that makes up the whole crystal lattice.
Atoms per unit cell $n$ : a corner atom counts 1/8, a face atom 1/2, an edge atom 1/4, and a center atom counts as 1. BCC: $n=2$ , FCC: $n=4$ , HCP: $n=6$ (for the full hexagonal cell).
Atomic packing factor (APF) = volume of atoms / volume of the unit cell. Simple cubic: 0.52. BCC: 0.68. FCC and HCP: 0.74 (the densest possible packing of spheres).
Theoretical density: \rho = \frac{nM}{V_c N_A} where $M$ is the molar mass (g/mol), $V_c$ is the unit cell volume, and $N_A = 6.022\cdot10^{23}$ /mol is Avogadro's number.
Point defects: vacancy (an empty lattice site), interstitial atom (an extra atom squeezed into a gap), substitutional atom (a foreign atom replacing a host atom).
Coordination number: the number of nearest neighbors of an atom (BCC: 8, FCC and HCP: 12).

How to solve the problems

Identify the structure (BCC, FCC, HCP or simple cubic) and read off $n$ and APF from the table above.
To find the density: set up $\rho = nM/(V_cN_A)$ . Compute the unit cell volume $V_c = a^3$ for cubic cells, with $a$ in cm if you want the density in g/cm³.
To find the void fraction: use $1 - \mathrm{APF}$ , usually converted to a percentage.
Watch the units throughout – mixing nm and cm gives an answer off by many powers of ten.
Check: the density of most metals lies between 1 and 20 g/cm³ – an answer far outside that range should be double-checked.

Example

Chromium has a BCC structure with lattice parameter $a = 0.2885$ nm and molar mass $M = 52.00$ g/mol. What is the theoretical density?

BCC has $n = 2$ atoms per unit cell.
The lattice parameter in cm: $a = 0.2885\cdot10^{-7}$ cm, so $V_c = a^3 \approx 2.402\cdot10^{-23}$ cm³.
\rho = \frac{nM}{V_cN_A} = \frac{2\cdot 52.00}{2.402\cdot10^{-23}\cdot 6.022\cdot10^{23}} \approx 7.19\ \mathrm{g/cm^3}

Answer: about 7.19 g/cm³, close to the measured value for chromium.

Common mistakes

Mixing units for the lattice parameter (nm, pm, angstroms) without converting to the same unit as the rest of the formula.
Using the wrong number of atoms per unit cell, for example counting corner atoms as whole atoms.
Believing the APF depends on which element it is – it depends only on the structure (BCC, FCC, etc.), not on which atom is being packed.
Confusing a vacancy (a missing atom) with an interstitial atom (an extra atom).

The density of a crystalline substance follows directly from its structure:

\rho = nM/(V_cN_A)

– atoms per cell, molar mass and cell volume are all you need.

Concepts in this part

Crystal structures and packingMost metals are crystalline, with atoms arranged in a regular lattice. The most common structures are BCC (bod…Grain size and Hall–PetchMetals consist of many small crystals (grains) with different orientations, separated by grain boundaries. Gra…Defects and dislocationsCrystal lattices always have defects. A vacancy (empty lattice site) is a point defect, while a dislocation is…Alloys and phase diagramsAn alloy is a mixture of a metal with other elements to achieve desired properties, like carbon steel (iron an…

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2. Mechanical properties

What is it about?

When an engineer selects a material for a structure, she needs to know how much load it can carry before it yields, breaks, or fails after many load cycles. The tensile test is the most important test: a specimen is pulled to fracture while force and elongation are measured, and the result is plotted as a stress–strain curve. This curve, together with hardness and impact toughness, provides the numbers that go into every design calculation.

Concepts and formulas

Stress $\sigma = F/A_0$ (MPa), strain $\varepsilon = \Delta L/L_0$ (often in %).
In the elastic region: Hooke's law $\sigma = E\varepsilon$ , where $E$ is the modulus of elasticity (the stiffness).
Yield strength $R_e$ (or $R_{p0.2}$ for materials without a distinct yield plateau): the stress at which plastic deformation begins.
Tensile strength $R_m$ : the highest stress the curve reaches.
Elongation at break $A$ : the total plastic strain at fracture, a measure of ductility.
Hardness (Brinell HB, Vickers HV, Rockwell HRC): resistance to local indentation from an indenter; correlates roughly with $R_m$ for steel.
Impact toughness: the energy absorbed in a sudden fracture, measured with a Charpy test (J).
Safety factor $n = R_e/\sigma_{allow}$ : how much margin the structure has against yielding.
Fatigue: repeated load below $R_e$ can still cause failure after many cycles; the Wöhler (S–N) curve shows stress amplitude versus number of cycles to failure.

How to solve the problems

Find the stress and/or strain from the given measurements: $\sigma=F/A_0$ , $\varepsilon=\Delta L/L_0$ .
Are you in the elastic region (stress below $R_e$ )? Use $E=\sigma/\varepsilon$ .
To find the safety factor: compare $R_e$ (or $R_m$ ) with the actual stress, $n=R_e/\sigma$ .
Check units: cross-sectional area is often in mm², force in N or kN, stress in MPa ( $1\ \mathrm{MPa}=1\ \mathrm{N/mm^2}$ ).
Sanity check: the modulus of elasticity for steel is about 200 GPa, for aluminum about 70 GPa; yield strengths for structural steel typically lie between 235 and 500 MPa.

Example

An aluminum rod has a diameter of 12 mm and a yield strength of $R_e = 240$ MPa. It is loaded axially with 18 kN. What is the safety factor against yielding?

Area: $A = \pi d^2/4 = \pi\cdot 12^2/4 \approx 113.1$ mm².
Stress: $\sigma = F/A = 18\,000/113.1 \approx 159.2$ MPa.
Safety factor: $n = R_e/\sigma = 240/159.2 \approx 1.51$ .

Answer: the safety factor is about 1.51.

Common mistakes

Mixing N and kN, or mm² and m², in the same calculation.
Using $R_m$ when the question actually asks about safety against yielding ( $R_e$ ), or the reverse.
Using Hooke's law ( $\sigma=E\varepsilon$ ) outside the elastic region.
Assuming harder always means more ductile – it is often the opposite.

Stress is force divided by area, strain is elongation divided by original length, and in the elastic region the modulus of elasticity ties them together:

\sigma = E\varepsilon

Concepts in this part

Stress–strain curveThe stress–strain curve shows how a material responds to load. In the elastic region, stress and strain are pr…Ductility and brittlenessA ductile material, like copper or structural steel, deforms plastically a lot before fracture, while a brittl…Fatigue and the S–N curveFatigue failure occurs under varying (cyclic) load, even when the stress stays below the yield strength — crac…Work hardening (cold working)Work hardening is that plastic deformation at low temperature increases strength, because more dislocations fo…

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3. Heat treatment and corrosion

What is it about?

Steel and other alloys can end up with very different properties from the same chemical composition, depending on how they are heat treated. By controlling temperature and cooling rate, steel can be made hard and wear-resistant, or soft and formable. Materials also degrade over time – through corrosion in a humid or salty environment, or through creep under high temperature and load. Composites, where two or more materials are combined, let the engineer "tailor" properties that none of the constituents have on their own.

Concepts and formulas

Hardening (quenching): steel is heated to austenite and cooled rapidly (in water, oil or air). Fast cooling produces martensite – hard, but brittle.
Tempering: controlled reheating of hardened steel to a lower temperature, to reduce brittleness at the cost of some hardness.
Normalizing: heating above the austenite range, cooling in air, gives a fine-grained, uniform structure.
Full annealing: slow cooling (in the furnace) gives a softer, more ductile structure than normalizing.
Hardenability: how deep the steel hardens; increases with alloying elements such as Cr, Mo, Ni (not with carbon content alone, which mainly controls the maximum hardness).
Corrosion: an electrochemical process. In a galvanic pair, the least noble metal (the anode) corrodes fastest.
Creep: slow, time-dependent plastic strain under constant load, important at high temperature (typically above about 0.4 times the melting point in kelvin).
Composites: strength/stiffness along the fiber direction can be estimated with the rule of mixtures E_c = V_fE_f + V_mE_m where $V_f$ and $V_m$ are the volume fractions of fiber and matrix ( $V_f+V_m=1$ ).
Linear thermal expansion: $\Delta L = \alpha L\Delta T$ , where $\alpha$ is the coefficient of linear expansion (1/K).

How to solve the problems

Identify the process (hardening, tempering, normalizing, annealing) from how the question describes the temperature and cooling method.
For corrosion: find which metal is least noble – that is the anode, and it corrodes.
For composites: substitute the volume fractions into the rule of mixtures, remembering that $V_f + V_m = 1$ .
For thermal expansion: make sure $\Delta T$ and $L$ use consistent units, and that $\alpha$ is given per kelvin.
Sanity check: martensite is always the hardest and most brittle of the steel structures; the slower the cooling, the softer and more ductile the result.

Example

A composite material has a 60% fiber volume fraction of carbon fiber ( $E_f = 230$ GPa) in an epoxy matrix ( $E_m = 3$ GPa). What is the modulus of elasticity along the fiber direction?

$V_f = 0.60$ , so $V_m = 1 - 0.60 = 0.40$ .
Rule of mixtures: $E_c = V_fE_f + V_mE_m = 0.60\cdot 230 + 0.40\cdot 3$ .
$E_c = 138 + 1.2 = 139.2$ GPa.

Answer: about 139.2 GPa – much stiffer than the matrix alone, because the fiber carries most of the load.

Common mistakes

Assuming more carbon always means deeper hardenability – it is alloying elements like Cr and Mo that mainly control hardenability.
Confusing tempering (done after hardening, to reduce brittleness) with annealing (a separate, slower process to soften the material).
Assuming the most noble metal corrodes in a galvanic pair – it is actually the least noble one.
Applying the rule of mixtures across the fiber direction – it applies along the fiber direction; across it, the material is much softer.

Fast cooling gives hard and brittle (martensite), slow cooling gives soft and ductile – and in a galvanic pair, it is always the least noble metal that is sacrificed.

Concepts in this part

Quenching and temperingWhen steel is rapidly cooled (quenched) from austenite, martensite forms — a hard and brittle structure becaus…Normalizing and carburizingNormalizing heats the steel above the austenite range and cools it in air, giving a fine-grained and uniform s…Galvanic corrosion and stainless steelWhen two different metals are in contact in an electrolyte, the less noble one (the anode) corrodes while the …Creep and glass transitionCreep is time-dependent deformation under constant load at high temperature, important for example in turbines…Thermal expansionMost materials expand when temperature increases. The length increase is approximately proportional to the ori…

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Example problems with solutions

Here are some of the problems in materials Technology. In the app, calculation problems get new numbers every time, so you can practise until it sticks – and take a graded practice exam before the real one.

Structure: What crystal structure does iron (ferrite) have at room temperature?

Answer: BCC (body-centred cubic)

Above about 912 °C iron becomes FCC (austenite).

Mechanical properties: What is the tensile strength

R_m

Answer: The highest stress on the stress–strain curve

$R_e$ or $R_{p0.2}$ is the yield strength.

Heat treatment and corrosion: What forms when steel is quenched from austenite?

Answer: Martensite

Martensite is hard and brittle because the carbon is trapped in the lattice.

Structure: What crystal structure does aluminium have?

Answer: FCC

FCC gives many slip systems and hence good ductility.

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Matches these university courses

The content covers the syllabus found in engineering degrees, for example:

MATS1500 (OsloMet)
TMT4185 (NTNU)
TBM200 (NMBU)

Materials Technology: free practice, theory and problems

Contents

1. Structure

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

2. Mechanical properties

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

3. Heat treatment and corrosion

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

Example problems with solutions

Matches these university courses