Materials Science

4% of exam

Chemical, electrical, mechanical, and physical properties and the effects of temperature, pressure, stress, and strain, material types and compatibilities for ferrous, nonferrous, and engineered materials, corrosion mechanisms and control, and polymers, ceramics, and composites.

3 concepts

Exam areas

A. Chemical, electrical, mechanical, and physical properties

Material Properties and Corrosion Control

Read the tensile curve, apply Hooke's law and the elastic constants, account for temperature and time, and set up the electrochemical cell that drives corrosion.

A chemical plant is a museum of materials choices that went right or wrong: a flange that yielded under a thermal transient, a heat-exchanger tube that pitted through in a chloride stream, a polymer gasket that crept and leaked. The Materials Science questions on the FE Chemical reward the engineer who can read a tensile curve, pull the right elastic constant, anticipate how temperature and time degrade a property, and recognize the electrochemical cell hiding inside an ordinary bolted joint. This concept follows the FE Reference Handbook — Materials Science/Structure of Matter and Mechanics of Materials chapters, where stress, strain, the elastic constants, and the corrosion half-reactions all live. The single most expensive habit to break is reaching for a number from memory: the handbook gives you

\sigma=E\varepsilon

, the galvanic potential table, and

\delta_t=\alpha L\,\Delta T

— your job is to know which to use and to keep the units honest.

Reading the stress-strain curve

A uniaxial tensile test is the source of nearly every mechanical property you will quote. Engineering stress is the load divided by the ORIGINAL area,

\sigma=F/A_0

, and engineering strain is the elongation per original length,

\varepsilon=\Delta L/L_0

. The initial straight portion is elastic and reversible; its slope is the elastic modulus

E

(Young's modulus) and the relation along it is Hooke's law. The curve then bends at the yield point — defined by the

0.2\%

offset because most metals have no sharp knee — climbs to the ultimate tensile strength (the peak engineering stress), and falls through necking to fracture. Ductility is the permanent strain left after fracture (percent elongation); the area under the whole curve is toughness, the energy absorbed per unit volume.

\sigma = E\,\varepsilon \qquad (\text{elastic region only})

The elastic constants and how they link

An isotropic material has only two independent elastic constants; the handbook gives the relations that connect the four you will meet. Young's modulus

E

governs axial stiffness, the shear modulus

G

governs twist and shear, the bulk modulus

K

governs volumetric (hydrostatic) compression, and Poisson's ratio

\nu

is the negative of lateral-to-longitudinal strain — about

0.3

for steels, near

0.5

for rubbers. Knowing any two fixes the rest, so an exam can hand you

E

and

\nu

and expect

G

K

. The axial deformation of a bar follows directly:

\delta=PL/(AE)

G = \dfrac{E}{2(1+\nu)}, \qquad K = \dfrac{E}{3(1-2\nu)}, \qquad \delta = \dfrac{PL}{AE}

Strength, ductility, hardness, toughness

Yield strength sets the working stress; ultimate tensile strength sets the margin to fracture; ductility tells you whether a part deforms visibly before it breaks (ductile) or shatters without warning (brittle). Hardness — resistance to indentation — correlates with strength for plain-carbon steels through the handbook's rule of thumb

\text{TS(MPa)}\approx 3.5\,\text{BHN}

(or

\text{TS(psi)}\approx 500\,\text{BHN}

), a fast way to estimate strength from a nondestructive test. Fracture toughness

K_{IC}=Y\sigma\sqrt{\pi a}

ranks a material's tolerance of a crack of length

a

; high-strength alloys often trade toughness for strength, which is why a brittle material can fail far below its tensile strength once a flaw is present.

K_{IC} = Y\,\sigma\sqrt{\pi a}, \qquad \text{TS(MPa)} \approx 3.5\,\text{BHN}

Temperature, time, and cyclic loading

Properties are not constants — they drift with temperature and degrade with time and cycling. A free member changes length by

\delta_t=\alpha L(T-T_0)

; restrain that member and the same expansion becomes a thermal stress

\sigma=E\alpha\,\Delta T

that can yield a pipe or pop an anchor. At high temperature metals creep — they deform slowly under constant load, with a rate that follows an Arrhenius form

\dot{\varepsilon}\propto e^{-Q/RT}

. Under cyclic load they fail by fatigue at stresses below yield, characterized by the endurance limit and the

S

N

(Basquin) curve. And ferritic steels lose toughness below a ductile-to-brittle transition temperature — the lesson the Liberty ships taught in cold water.

\delta_t = \alpha L\,(T - T_0), \qquad \sigma_{\text{thermal}} = E\,\alpha\,\Delta T \;\;(\text{fully restrained})

Electrical and thermal properties

Beyond the mechanical, the FE samples physical properties. Electrical resistivity

\rho

relates a resistor's resistance to its geometry by

R=\rho L/A

; conductivity is its reciprocal. Dielectric (insulating) behavior is captured by the parallel-plate capacitance

C=\varepsilon A/d

, with

\varepsilon=\kappa\varepsilon_0

and

\varepsilon_0=8.85\times10^{-12}\,\text{F/m}

. Thermal transport mirrors this: thermal conductivity

\lambda

sets the steady heat flux, while thermal diffusivity (conductivity over volumetric heat capacity) sets how fast a temperature front moves. Specific heat is per unit mass; heat capacity is per mole — a basis distinction the handbook is explicit about.

\rho = \dfrac{RA}{L}, \qquad C = \dfrac{\varepsilon A}{d}, \qquad \varepsilon = \kappa\,\varepsilon_0

Corrosion: the electrochemical cell in a joint

Corrosion is an electrochemical process: it needs an anode (where metal oxidizes and is lost), a cathode (where a reduction reaction consumes the electrons), a metallic path between them, and an electrolyte. The metal dissolves at the anode as

M\rightarrow M^{n+}+ne^-

; in neutral aerated water the dominant cathode reaction is oxygen reduction,

\tfrac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-

, while in acid it is hydrogen evolution. Remove any leg — keep the surface dry, break the circuit, or coat it — and corrosion stops. The handbook's Standard Oxidation Potentials table ranks metals; in its sign convention a more POSITIVE oxidation potential is the more active (anodic) metal.

\begin{aligned} \text{Anode (oxidation):}\quad & M \rightarrow M^{n+} + ne^- \\ \text{Cathode (reduction):}\quad & \tfrac{1}{2}O_2 + H_2O + 2e^- \rightarrow 2OH^- \end{aligned}

Forms of corrosion and how to control it

Uniform corrosion thins a surface evenly and is the easy case to predict and allow for. Galvanic corrosion appears when two dissimilar metals touch in an electrolyte: the more active metal becomes the anode and corrodes faster, the noble metal is protected, and a small anode coupled to a large cathode corrodes fastest of all. The same battery can form on ONE metal — cold-worked regions are anodic to annealed, oxygen-poor crevices are anodic to oxygen-rich surfaces (crevice and pitting attack) — note that the handbook's own Corrosion paragraph (Materials Science, p.117) prints the opposite wording, 'oxygen-deficient regions ... cathodic to oxygen-rich regions', so if an exam question quotes that convention, answer in it. Control follows the cell logic: pick metals close together in the galvanic series, coat or insulate to break the circuit, add inhibitors, or deliberately install a sacrificial anode (zinc, magnesium) or impressed current so the structure becomes the protected cathode. The cell EMF combines the two half-cell potentials.

E_{cell} = E^\circ_{ox,\,\text{anode}} + E^\circ_{red,\,\text{cathode}}

Exam strategy

Sort the question first: is it mechanical (use

\sigma=E\varepsilon

\delta=PL/AE

, or an elastic-constant link), thermal (free expansion

\delta_t=\alpha L\Delta T

versus restrained stress

E\alpha\Delta T

— they differ by a factor of

E

and the answer doubles if you forget the restraint), or corrosion (build the anode/cathode cell and read the potential table)? For corrosion couples, identify the anode as the metal with the more positive oxidation potential in the handbook's table, remembering that table is written as oxidation half-cells. Keep stress in consistent units (

1\,\text{MPa}=10^6\,\text{Pa}=1\,\text{N/mm}^2

), and never confuse engineering stress (original area) with true stress (instantaneous area).

Key equations

Engineering stress and strain

\sigma = \dfrac{F}{A_0}, \qquad \varepsilon = \dfrac{\Delta L}{L_0}

Based on ORIGINAL area $A_0$ and length $L_0$ . $\sigma$ in Pa (or MPa), $\varepsilon$ dimensionless.

Hooke's law

\sigma = E\,\varepsilon

Linear elastic region only. $E$ = elastic (Young's) modulus, Pa; slope of the tensile curve's straight portion.

Axial deformation

\delta = \dfrac{PL}{AE}

Elastic elongation of a bar of length $L$ , area $A$ , modulus $E$ under axial load $P$ . Units m.

Shear and bulk modulus

G = \dfrac{E}{2(1+\nu)}, \qquad K = \dfrac{E}{3(1-2\nu)}

Isotropic elastic links. $\nu$ = Poisson's ratio ( $\approx 0.3$ for steel). Only two of $E,G,K,\nu$ are independent.

Poisson's ratio

\nu = -\dfrac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{longitudinal}}}

Negative ratio of transverse to axial strain. Dimensionless; $0 \le \nu \le 0.5$ for stable isotropic solids.

True stress and true strain

\sigma_T = \dfrac{F}{A}, \qquad \varepsilon_T = \ln(1+\varepsilon)

Use instantaneous area $A$ ; needed in plastic/necking analysis. Differs from engineering values beyond yield.

Thermal deformation and stress

\delta_t = \alpha L\,(T - T_0), \qquad \sigma_{\text{thermal}} = E\,\alpha\,\Delta T

Free expansion (first) vs fully restrained stress (second). $\alpha$ = expansion coefficient, 1/K or 1/°C.

Thermal expansion coefficient

\alpha = \dfrac{\varepsilon}{\Delta T}

Engineering strain per degree of temperature change. Units 1/K (equivalently 1/°C for a difference).

Fracture toughness

K_{IC} = Y\,\sigma\sqrt{\pi a}

Critical stress intensity; $a$ = crack length, $Y$ = geometry factor ( $1$ interior, $1.1$ surface). Units MPa·m $^{1/2}$ .

Hardness–strength estimate

\text{TS (MPa)} \approx 3.5\,\text{BHN}

Plain-carbon steel approximation (or TS(psi) ≈ 500·BHN). BHN = Brinell hardness number.

Resistivity and capacitance

\rho = \dfrac{RA}{L}, \qquad C = \dfrac{\varepsilon A}{d}

$\rho$ = resistivity (Ω·m); $C$ = parallel-plate capacitance, $\varepsilon=\kappa\varepsilon_0$ , $\varepsilon_0=8.85\times10^{-12}\,$ F/m.

Diffusion coefficient (Arrhenius)

D = D_0\,e^{-Q/(RT)}

Temperature dependence of diffusion/creep. $Q$ = activation energy, $R=8.314\,$ J/(mol·K), $T$ in K.

Galvanic cell EMF

E_{cell} = E^\circ_{ox,\,\text{anode}} + E^\circ_{red,\,\text{cathode}}

Positive $E_{cell}$ means the couple corrodes spontaneously. Use handbook oxidation potentials (reverse sign for the cathode).

Nernst equation (25 °C)

E = E^\circ - \dfrac{0.0592}{n}\log_{10} Q

Concentration correction to the standard potential; $n$ = electrons transferred, $Q$ = reaction quotient.

Worked examples

Elongation and elastic check of a tie rod

Problem. A $316$ stainless tie rod ( $E = 193\,\text{GPa}$ , yield strength $\approx 290\,\text{MPa}$ ) is $20\,\text{mm}$ in diameter and $2.5\,\text{m}$ long. It carries a steady axial tension of $60\,\text{kN}$ . Find the stress, the strain, and the elastic elongation, and confirm the rod stays elastic.

Solution. Cross-sectional area:

A = \tfrac{\pi}{4}(0.020)^2 = 3.142\times10^{-4}\,\text{m}^2

. Engineering stress:

\sigma = F/A = 60{,}000/3.142\times10^{-4} = 1.910\times10^{8}\,\text{Pa} = 191\,\text{MPa}

. Since

191\,\text{MPa} < 290\,\text{MPa}

yield, the rod is elastic and Hooke's law applies. Strain:

\varepsilon = \sigma/E = 1.910\times10^{8}/1.93\times10^{11} = 9.90\times10^{-4}

. Elongation:

\delta = \varepsilon L = (9.90\times10^{-4})(2.5) = 2.47\times10^{-3}\,\text{m} = 2.47\,\text{mm}

(equivalently

\delta = PL/AE

). Sanity check: strain

\sim 0.1\%

is typical for a metal near

\tfrac{2}{3}

of yield, and a

2.5\,\text{m}

rod stretching a couple of millimeters is reasonable. Answer:

\sigma = 191\,\text{MPa}

\delta = 2.47\,\text{mm}

\delta = \dfrac{PL}{AE} = \dfrac{(60{,}000)(2.5)}{(3.142\times10^{-4})(193\times10^{9})} = 2.47\times10^{-3}\,\text{m}

Thermal stress in a restrained pipe

Problem. A carbon-steel pipe ( $E = 200\,\text{GPa}$ , $\alpha = 12\times10^{-6}\,/^\circ\text{C}$ ) is anchored rigidly at both ends with no slack. Process startup raises its temperature by $60\,^\circ\text{C}$ . What axial stress develops, and which sign?

Solution. If free, the pipe would expand by strain

\varepsilon = \alpha\,\Delta T = (12\times10^{-6})(60) = 7.2\times10^{-4}

. Full restraint suppresses that expansion, producing a compressive stress of equal magnitude:

\sigma = E\,\alpha\,\Delta T = (200\times10^{9})(12\times10^{-6})(60) = 1.44\times10^{8}\,\text{Pa} = 144\,\text{MPa}

(compression). Sanity check:

144\,\text{MPa}

is a large fraction of a mild-steel yield (

\sim 250\,\text{MPa}

) — exactly why piping needs expansion loops or bellows. Answer:

144\,\text{MPa}

compressive. Note the trap: free elongation here would have been

\delta_t = \alpha L\,\Delta T

(depends on length); the restrained STRESS does not depend on length at all.

\sigma_{\text{thermal}} = E\,\alpha\,\Delta T = 144\,\text{MPa}\ (\text{compression})

Galvanic couple: which metal corrodes, and the driving voltage

Problem. Carbon-steel (iron) piping is joined to a copper fitting in an aerated water service. Using the handbook oxidation potentials $E^\circ_{ox}(\text{Fe}\rightarrow\text{Fe}^{2+}) = +0.440\,\text{V}$ and $E^\circ_{ox}(\text{Cu}\rightarrow\text{Cu}^{2+}) = -0.337\,\text{V}$ , identify the anode and compute the standard cell EMF.

Solution. The metal with the more POSITIVE oxidation potential is the more active and becomes the anode.

+0.440 > -0.337

, so IRON is the anode and corrodes; copper is the protected cathode. Cell EMF combines the iron oxidation with the copper reduction (reverse the sign of copper's oxidation potential to get its reduction potential,

+0.337\,\text{V}

E_{cell} = E^\circ_{ox}(\text{Fe}) + E^\circ_{red}(\text{Cu}) = 0.440 + 0.337 = 0.777\,\text{V}

. A positive

E_{cell}

confirms the couple corrodes spontaneously. Sanity check: copper is well known to accelerate steel corrosion downstream of a copper component, matching the result. Worse still if the steel (anode) area is small relative to the copper. Answer: iron is the anode;

E_{cell} = 0.777\,\text{V}

E_{cell} = E^\circ_{ox}(\text{Fe}) + E^\circ_{red}(\text{Cu}) = 0.440 + 0.337 = 0.777\,\text{V}

Common pitfalls

•Using restrained thermal STRESS ( $E\alpha\Delta T$ ) when the member is free to expand (answer is then a length change $\alpha L\Delta T$ , not a stress) — or vice versa. Decide first whether the ends are anchored.
•Confusing engineering stress (original area $A_0$ ) with true stress (instantaneous area $A$ ). They agree only at small strain; beyond necking they diverge sharply.
•Mis-reading the handbook potential table: it lists OXIDATION potentials as anode half-cells. The more POSITIVE value is the more active (anodic) metal here — the opposite of the standard reduction-potential convention.
•Forgetting the small-anode/large-cathode rule: galvanic attack concentrates on a small anode, so a steel bolt in a copper plate corrodes far faster than copper rivets in a steel plate.
•Treating $E$ , $G$ , $K$ , $\nu$ as four independent numbers. For an isotropic material only two are independent; use $G=E/[2(1+\nu)]$ and $K=E/[3(1-2\nu)]$ .
•Unit slips in stress: $1\,\text{MPa}=10^6\,\text{Pa}=1\,\text{N/mm}^2$ . Mixing GPa for modulus with MPa for stress drops a factor of $1000$ .
•Quoting yield strength when the curve asks for ultimate tensile strength (the peak) or ductility (permanent strain after fracture) — three different points on the same curve.

References

NCEES FE Reference Handbook — Materials Science/Structure of Matter
NCEES FE Reference Handbook — Mechanics of Materials — Stress-strain, Hooke's law, elastic-constant links, thermal deformation.
NCEES FE Reference Handbook — Chemistry and Biology — Standard Oxidation Potentials for Corrosion Reactions; Nernst equation.
Callister & Rethwisch, Materials Science and Engineering: An Introduction — Tensile properties, corrosion forms and control — background.

B. Material types and compatibilities

Material Types, Selection and Compatibility

Sort ferrous from nonferrous metals, read the galvanic series for compatibility, and select chemically resistant alloys for process service.

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D. Polymers, ceramics, and composites

Polymers, Ceramics and Composites

Distinguish thermoplastics from thermosets, place ceramics and glasses, and apply the composite rule of mixtures for stiffness and strength.

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