Materials Science & Engineering is a formula- and fact-dense General Studies section tested in ESE Paper I. This module spans Crystal Structure and Atomic Bonding (bond types, unit cells, Miller indices, defects), Mechanical Properties (stress-strain behaviour, elastic constants, hardness, toughness, fatigue, creep), Metals, Alloys and Phase Diagrams (iron-carbon diagram, steel microstructures, heat treatment, non-ferrous alloys), Ceramics and Polymers (types, classification, glass transition), Composite Materials (rule of mixtures, reinforcing fibres), and Failure Analysis and Non-Destructive Testing (fracture modes, fracture mechanics, NDT methods, corrosion) — with every definition, formula, and standard result carried over, plus worked examples and diagrams for each topic.
After studying this chapter you will be able to:
Materials Science underpins every other technical subject in civil engineering — material selection and failure understanding feed directly into Structural Steel Design, RCC, and Strength of Materials. Once you've worked through the chapters below, head to the Materials Science hub page to generate practice tests, or explore Study Material for other subjects.
The properties of engineering materials are determined by their atomic structure and bonding. Understanding crystal geometry explains why metals are ductile, ceramics brittle, and polymers flexible.
| Bond Type | Nature | Strength | Example |
|---|---|---|---|
| Ionic | Electron transfer; cation + anion attraction | Strong | NaCl, MgO, Al₂O₃ |
| Covalent | Electron sharing; directional | Very strong | Diamond (C), Si, SiC |
| Metallic | Sea of free electrons; non-directional | Moderate–strong | Fe, Cu, Al |
| Van der Waals | Weak dipole attractions; secondary bond | Weak | Polymers, noble gases |
| Hydrogen | H between electronegative atoms | Moderate | Water, nylon, DNA |
| Structure | APF | Atoms/cell | CN | Examples |
|---|---|---|---|---|
| Simple Cubic (SC) | 0.52 | 1 | 6 | Po (only metal) |
| Body-Centred Cubic (BCC) | 0.68 | 2 | 8 | Fe(α), Cr, W, Mo, V |
| Face-Centred Cubic (FCC) | 0.74 | 4 | 12 | Fe(γ), Cu, Al, Ni, Au, Ag |
| Hexagonal Close-Packed (HCP) | 0.74 | 6 | 12 | Zn, Mg, Ti, Co |
APF = Atomic Packing Factor; CN = Coordination Number.
| Defect Type | Examples | Effect on Properties |
|---|---|---|
| Point defects | Vacancy, interstitial, substitutional, Frenkel, Schottky | Affect diffusion, electrical, mechanical |
| Line defects (dislocations) | Edge dislocation, screw dislocation | Control plastic deformation, yield strength |
| Planar defects | Grain boundaries, twin boundaries, stacking faults | Affect strength, corrosion, diffusion |
| Volume defects | Voids, inclusions, precipitates | Affect toughness, fatigue |
\(\dfrac{n \times V_{atom}}{V_{cell}}\)
0.74 (highest packing); CN = 12
0.68; 2 atoms/cell; CN = 8
BCC → FCC at 912°C
Given: How many atoms per unit cell are in an FCC structure?
Solution: FCC has 8 corner atoms (each contributing 1/8) and 6 face atoms (each contributing 1/2). \(8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 1 + 3 = 4\).
Answer: 4 atoms per unit cell.
Given: A plane intercepts the axes at \(x=1, y=1, z=\infty\). What are its Miller indices?
Solution: Reciprocals: \(\dfrac{1}{1}, \dfrac{1}{1}, \dfrac{1}{\infty} = 1, 1, 0\). No fractions to clear.
Answer: (110).
Given: A metal's grain boundary region shows a different, misaligned crystal orientation compared to the neighbouring grain. What class of defect is this?
Solution: Grain boundaries are two-dimensional (planar) defects, distinct from point defects (0D), dislocations (1D), and voids/inclusions (3D volume defects).
Answer: Planar defect (grain boundary).
Fig. 1.1 — Simple Cubic, Body-Centred Cubic, and Face-Centred Cubic unit cells, in increasing order of atomic packing factor.
Mechanical properties determine how materials respond to applied loads. Selection of materials for engineering applications requires thorough knowledge of stress-strain behaviour, hardness, toughness, and fatigue.
| Test | Indenter | Load | Scale | Application |
|---|---|---|---|---|
| Brinell (BHN) | 10 mm steel ball | 3000 kg (steel) | BHN | Soft–medium materials; \(UTS \approx 3.4 \times BHN\) (MPa) |
| Vickers (VHN) | Diamond pyramid 136° | 1–120 kgf | VHN (HV) | All materials; wide range |
| Rockwell | Diamond cone / ball | Minor + major load | HRC, HRB | Fast; production testing |
| Shore (Scleroscope) | Rebound height | — | Shore hardness | Finished parts; non-destructive |
Toughness is the energy absorbed per unit volume before fracture (area under the stress-strain curve). Charpy test: notched specimen, horizontal, simply supported at both ends. Izod test: notched specimen, vertical cantilever. Impact energy decreases at lower temperature — the ductile-to-brittle transition temperature (DBTT). BCC metals (steels) show DBTT; FCC metals (Al, Cu) do not.
Fatigue is failure under cyclic loading at stresses below static fracture stress. Endurance limit (\(S_e\)): stress amplitude below which infinite life is expected, \(\approx 0.5 \times UTS\) for steels. Cracks initiate at surface defects, notches, or inclusions; improved by surface finishing, shot peening, case hardening.
Creep is time-dependent plastic deformation under constant load at elevated temperature, significant above \(\approx 0.3\)–\(0.4 \times T_m\) (homologous temperature, in Kelvin). Three stages: primary (decreasing rate), secondary/steady-state, tertiary (accelerating, fracture). Applications: turbine blades, boiler components, nuclear reactor parts.
\(\sigma = F/A_0\); \(\varepsilon = \Delta L/L_0\); \(E = \dfrac{\sigma}{\varepsilon}\)
\(E = 2G(1+\nu) = 3K(1-2\nu)\)
\(UTS \text{(MPa)} \approx 3.4 \times BHN\)
\(S_e \approx 0.5 \times UTS\) for steels
Given: A steel bar of area \(100\text{ mm}^2\) is loaded with 20 kN, producing a strain of 0.0005. Find its Young's modulus.
Solution: \(\sigma = F/A = \dfrac{20{,}000}{100} = 200\text{ MPa}\). \(E = \dfrac{\sigma}{\varepsilon} = \dfrac{200}{0.0005} = 400{,}000\text{ MPa} = 400\text{ GPa}\).
Answer: \(E = 400\) GPa (unusually high; illustrative numbers).
Given: A steel sample has a Brinell Hardness Number (BHN) of 200. Estimate its UTS.
Solution: \(UTS \approx 3.4 \times BHN = 3.4 \times 200 = 680\text{ MPa}\).
Answer: \(\approx 680\) MPa.
Given: A steel component has a UTS of 600 MPa. Estimate its endurance limit.
Solution: \(S_e \approx 0.5 \times UTS = 0.5 \times 600 = 300\text{ MPa}\).
Answer: \(\approx 300\) MPa.
Fig. 2.1 — Engineering stress-strain curve for mild steel: elastic region, yield points, UTS, and fracture.
Metals and alloys form the backbone of structural engineering. Phase diagrams map the stable phases as a function of composition and temperature — essential for understanding heat treatment and alloy design.
| Point/Line | Temperature | Carbon % | Significance |
|---|---|---|---|
| Eutectic point | 1148°C | 4.3% C | L → Austenite + Cementite (Ledeburite) |
| Eutectoid point | 727°C | 0.8% C | Austenite → Pearlite (α-ferrite + Fe₃C) |
| Peritectic point | 1493°C | 0.16% C | δ-Fe + L → Austenite |
| A1 line (PSK) | 727°C | — | Lower critical temperature |
| A3 line (GS) | 910°C (at 0% C) | — | Upper critical temperature (hypoeutectoid) |
| Acm line (SE) | — | — | Upper critical temperature (hypereutectoid) |
| Microstructure | Formation | Properties |
|---|---|---|
| Austenite (γ-Fe) | FCC; above A1; up to 2.14% C solubility | Non-magnetic, tough |
| Ferrite (α-Fe) | BCC; <0.022% C; room temp | Soft, ductile, magnetic |
| Cementite (Fe₃C) | 6.67% C; iron carbide | Hard, brittle |
| Pearlite | Eutectoid: α-Fe + Fe₃C lamellae | Medium strength, medium ductility |
| Bainite | Intermediate quench; finer than pearlite | Good combination of strength+toughness |
| Martensite | Rapid quench; distorted BCT; trapped C | Hardest, most brittle; tempered for use |
| Process | Cooling | Purpose |
|---|---|---|
| Annealing | Furnace cool (very slow) | Soften; relieve stress; improve machinability |
| Normalising | Air cool | Refine grain; uniform microstructure |
| Hardening (Quenching) | Water/oil quench (rapid) | Form martensite; increase hardness |
| Tempering | Reheat 150–650°C after quench | Reduce brittleness of martensite; improve toughness |
| Case hardening | Surface treatment | Hard surface + tough core (carburising, nitriding) |
| Alloy | Composition | Key Properties | Applications |
|---|---|---|---|
| Brass | Cu + Zn (10–40%) | Good machinability, corrosion resistance | Fittings, valves, musical instruments |
| Bronze | Cu + Sn (3–20%) | High strength, wear resistance | Bearings, gears, coins |
| Duralumin | Al + Cu(4%) + Mg + Mn | Age-hardenable; high strength/weight | Aircraft structures |
| Stainless steel | Fe + Cr(≥10.5%) + Ni | Corrosion-resistant (Cr₂O₃ passive layer) | Food equipment, surgical, chemical |
| Ti alloys | Ti + Al + V | High strength/weight; biocompatible | Aerospace, biomedical implants |
0.8% C, 727°C → Pearlite
4.3% C, 1148°C → Ledeburite
6.67% C (Fe₃C)
≥10.5% Cr for passive Cr₂O₃ layer
Given: A steel with 0.8% carbon is slowly cooled through 727°C. What microstructure forms, and what is this point called?
Solution: 0.8% C at 727°C is the eutectoid point, where austenite transforms into pearlite (alternating lamellae of α-ferrite and Fe₃C).
Answer: Eutectoid point; forms Pearlite.
Given: A machined steel component is very hard but brittle after quenching. What heat treatment should follow, and why?
Solution: Quenched steel forms brittle martensite. Tempering (reheating to 150–650°C) is required to reduce brittleness and improve toughness while retaining most of the hardness gain.
Answer: Tempering.
Given: A component for food-processing equipment must resist corrosion in a moist environment. Which alloy class is most suitable, and what minimum alloying element percentage is required?
Solution: Stainless steel resists corrosion via a passive Cr₂O₃ layer, which requires a minimum of 10.5% chromium content to form reliably.
Answer: Stainless steel with ≥10.5% Cr.
Fig. 3.1 — Simplified iron-carbon diagram highlighting the eutectoid point (0.8% C, 727°C → Pearlite) and eutectic point (4.3% C, 1148°C → Ledeburite).
Ceramics and polymers are non-metallic materials with distinctive structure-property relationships. Together with metals, they form the three major classes of engineering materials.
Ceramics are inorganic, non-metallic materials with ionic/covalent bonding. Properties: high hardness, high melting point, brittle (no dislocation motion), low thermal/electrical conductivity (except SiC/Si₃N₄), chemical inertness, good compressive strength. Failure mode: fracture due to stress concentrations at pores/cracks, with no yielding.
| Type | Examples | Applications |
|---|---|---|
| Oxide ceramics | Al₂O₃, ZrO₂, MgO | Cutting tools, refractories, dental implants |
| Non-oxide ceramics | SiC, Si₃N₄, TiN, BN | Hard coatings, turbine blades, cutting tools |
| Glass | SiO₂-based | Windows, optical fibres, laboratory equipment |
| Glass-ceramics | Pyroceram, Macor | Cookware, telescope mirrors, dental |
| Cement/Concrete | Portland cement + aggregate | Construction; high compressive, low tensile |
| Piezoelectric ceramics | PZT (PbZrTiO₃), BaTiO₃ | Sensors, transducers, actuators |
Polymers are long chain macromolecules of repeating monomer units, mainly with a C-H backbone. The degree of polymerisation (\(n\)) is the number of repeat units in a chain. Properties are determined by chain length, branching, cross-linking, and crystallinity.
| Type | Structure | Behaviour | Examples |
|---|---|---|---|
| Thermoplastics | Linear/branched; van der Waals bonds between chains | Soften on heating; recyclable | PE, PP, PVC, nylon, PTFE (Teflon) |
| Thermosets | Cross-linked network; covalent bonds | Cannot re-melt; permanent shape | Epoxy, phenolic, polyester, silicone |
| Elastomers | Loosely cross-linked; high extensibility | Large reversible deformation (>200%) | Natural rubber (NR), SBR, neoprene, silicone rubber |
| Polymer | Abbreviation | Key Property |
|---|---|---|
| Polyethylene | PE (HDPE/LDPE) | Chemical resistance; pipes, packaging |
| Polyvinyl chloride | PVC | Rigid or flexible; pipes, wire insulation |
| Polypropylene | PP | Lightweight; high chemical resistance |
| PTFE (Teflon) | PTFE | Lowest friction coefficient; non-stick coatings |
| Nylon (Polyamide) | PA | Strong, tough; gears, bearings, textiles |
| Epoxy | EP | Excellent adhesion; composites matrix |
Thermoplastic: remeltable; Thermoset: cross-linked, permanent
Below Tg: glassy/brittle; above Tg: rubbery
\(\approx 0.04\) — lowest of any solid
Lead zirconate titanate — piezoelectric ceramic
Given: A polymer sample cannot be re-melted once cured, retaining a permanent shape even under heat. What class of polymer is this, and why?
Solution: This is a thermoset — its polymer chains are covalently cross-linked into a rigid network during curing, which cannot be undone by reheating (unlike thermoplastics, held together by weaker van der Waals bonds).
Answer: Thermoset.
Given: An amorphous polymer becomes rubbery and flexible above 80°C, but stiff and brittle below it. What is this temperature called?
Solution: This is the glass transition temperature (\(T_g\)) — amorphous polymers only exhibit \(T_g\) (no distinct melting point), transitioning from glassy to rubbery behaviour at this temperature.
Answer: Glass transition temperature (\(T_g = 80°C\)).
Given: A cutting tool needs a hard, non-oxide ceramic coating that can withstand high temperatures. Which ceramic type should be used?
Solution: Non-oxide ceramics such as SiC, Si₃N₄, and TiN provide hard, high-temperature-resistant coatings, unlike oxide ceramics (Al₂O₃, ZrO₂) which serve different applications like refractories and dental implants.
Answer: A non-oxide ceramic (e.g., SiC or TiN).
Fig. 4.1 — Chain architecture of thermoplastics (separate chains), thermosets (dense cross-links), and elastomers (loose cross-links).
Composites combine two or more distinct materials to achieve properties superior to any single constituent. They are increasingly used in aerospace, automotive, civil, and sports engineering.
| Classification Basis | Type | Example |
|---|---|---|
| Matrix material | Polymer Matrix Composite (PMC) | CFRP, GFRP |
| Metal Matrix Composite (MMC) | Al/SiC | |
| Ceramic Matrix Composite (CMC) | SiC/SiC | |
| Reinforcement type | Particle-reinforced | Concrete, cermets |
| Fibre-reinforced | CFRP, GFRP, Kevlar | |
| Structural (laminates, sandwich) | Plywood, honeycomb panels |
| Fibre | Tensile Strength (GPa) | Density (g/cm³) | Application |
|---|---|---|---|
| E-glass (GFRP) | 3.5 | 2.54 | Boat hulls, wind turbine blades, pipes |
| Carbon (CFRP) | 3.5–7 | 1.75–2.0 | Aerospace, racing cars, sports equipment |
| Aramid (Kevlar) | 3.6 | 1.44 | Bulletproof vests, helmets, ropes |
| Boron fibre | 3.4 | 2.36 | Military aircraft, sports equipment |
| Basalt fibre | 3.0 | 2.7 | Construction reinforcement, insulation |
Advantages: High specific strength (strength/density) and specific stiffness; design flexibility (anisotropic properties can be tailored); good fatigue and corrosion resistance (PMC); near-net-shape manufacturing possible.
Limitations: High cost of fibres and manufacturing; difficult to inspect for internal delamination (NDT required); brittle failure modes; limited recyclability of thermoset composites; anisotropy can be a disadvantage if load direction is uncertain.
\(E_c = V_f E_f + V_m E_m\) (isostrain)
\(\dfrac{1}{E_c} = \dfrac{V_f}{E_f} + \dfrac{V_m}{E_m}\) (isostress)
\(V_f + V_m = 1\)
Carbon fibre (stiffness); Kevlar (toughness/weight)
Given: A CFRP composite has \(V_f = 0.6\), \(E_f = 230\text{ GPa}\) (carbon fibre), \(V_m = 0.4\), \(E_m = 3.5\text{ GPa}\) (epoxy matrix). Find the longitudinal modulus.
Solution: \(E_c = V_f E_f + V_m E_m = 0.6(230) + 0.4(3.5) = 138 + 1.4 = 139.4\text{ GPa}\).
Answer: \(\approx 139.4\) GPa.
Given: Using the same fibre/matrix data (\(V_f=0.6, E_f=230\), \(V_m=0.4, E_m=3.5\) GPa), estimate the transverse modulus.
Solution: \(\dfrac{1}{E_c} = \dfrac{V_f}{E_f} + \dfrac{V_m}{E_m} = \dfrac{0.6}{230} + \dfrac{0.4}{3.5} = 0.00261 + 0.1143 = 0.1169\). \(E_c = \dfrac{1}{0.1169} \approx 8.55\text{ GPa}\).
Answer: \(\approx 8.55\) GPa — much lower than the longitudinal value, since transverse modulus is matrix-dominated.
Given: A design team needs the reinforcing fibre with the best toughness-to-weight ratio for a bulletproof vest. Which fibre should they choose?
Solution: Aramid fibre (Kevlar) has the best toughness/weight ratio among common reinforcing fibres, which is why it is the standard choice for bulletproof vests and helmets.
Answer: Aramid (Kevlar) fibre.
Fig. 5.1 — Rule of mixtures: longitudinal loading behaves like springs in parallel; transverse loading like springs in series.
Understanding how and why materials fail is essential for safe engineering design. Non-destructive testing (NDT) allows inspection without damaging the component — critical for quality control and maintenance.
| Mode | Mechanism | Fracture Surface | Materials |
|---|---|---|---|
| Ductile fracture | Plastic deformation; void nucleation and coalescence | Fibrous, dimpled, cup-and-cone | Metals above DBTT |
| Brittle fracture | Rapid crack propagation; little plastic deformation | Flat, granular, often with chevron marks | Ceramics, metals below DBTT, glass |
| Fatigue fracture | Cyclic loading; crack initiates at surface; beach marks | Beach marks + final rough zone | All materials under cyclic load |
| Creep fracture | High temperature + sustained stress | Intergranular; neck formation | Metals at \(T > 0.3\,T_m\) |
| Method | Principle | Detects | Limitations |
|---|---|---|---|
| Visual Inspection (VT) | Direct/aided visual examination | Surface cracks, corrosion, deformation | Surface only; skill dependent |
| Liquid Penetrant Testing (PT) | Coloured/fluorescent dye drawn into cracks by capillary action | Surface-open cracks | Surface only; clean surface needed |
| Magnetic Particle Testing (MT) | Magnetic field + iron particles reveal flux leakage at defects | Surface/near-surface cracks in ferromagnetic materials | Ferromagnetic materials only |
| Ultrasonic Testing (UT) | High-frequency sound waves; reflections indicate discontinuities | Internal defects, thickness measurement | Couplant needed; operator skill required |
| Radiographic Testing (RT) | X-rays/gamma rays; denser areas absorb more | Internal voids, inclusions, cracks | Radiation safety; 2D image of 3D flaw |
| Eddy Current Testing (ET) | Induced eddy currents disturbed by defects | Surface/near-surface defects in conductors | Conductive materials only |
| Acoustic Emission (AE) | Stress waves from crack growth detected | Active crack growth; overall structural monitoring | Background noise; source location complex |
| Corrosion Type | Mechanism | Prevention |
|---|---|---|
| Uniform / general | Even metal loss over surface | Coatings, inhibitors, cathodic protection |
| Galvanic | Two dissimilar metals in electrolyte; anodic metal corrodes | Use same/similar metals; insulate; sacrificial anode |
| Pitting | Localised attack forming pits; autocatalytic | Higher alloy content; avoid stagnant conditions |
| Crevice | Differential oxygen concentration in gaps/crevices | Design to avoid crevices; seal gaps |
| Stress corrosion cracking (SCC) | Combined stress + corrosive environment | Remove stress (anneal), use resistant alloy, change environment |
| Intergranular | Grain boundary attack (sensitisation in SS) | Low-carbon SS; solution anneal |
Galvanic series (active → noble): Mg, Zn, Al, Fe, Ni, Cu, Ag, Au. The more active metal is the anode (it corrodes); sacrificial anode protection typically uses Zn or Mg for steel structures.
\(K_I = \sigma\sqrt{\pi a} \times Y\)
Fracture occurs when \(K_I \ge K_{IC}\)
Ultrasonic (UT) and Radiographic (RT) Testing
Liquid Penetrant (PT) and Visual (VT)
Given: A component has a stress intensity factor \(K_I = 25\ \text{MPa}\sqrt{\text{m}}\), and its material has \(K_{IC} = 30\ \text{MPa}\sqrt{\text{m}}\). Will it fracture?
Solution: Fracture occurs only when \(K_I \ge K_{IC}\). Here \(K_I = 25 < K_{IC} = 30\), so the component will not fracture under this loading.
Answer: No fracture — \(K_I\) is below the critical value.
Given: An internal void is suspected within a thick steel casting. Which NDT method is most appropriate?
Solution: Internal defects require volumetric inspection methods — Ultrasonic Testing (UT) or Radiographic Testing (RT) — rather than surface methods like PT or VT, which cannot detect subsurface voids.
Answer: Ultrasonic Testing (UT) or Radiographic Testing (RT).
Given: A steel bolt and an aluminium bracket are in direct contact in a moist environment, and the aluminium corrodes rapidly near the joint. What type of corrosion is this, and which metal is the anode?
Solution: This is galvanic corrosion — two dissimilar metals in contact with an electrolyte. Per the galvanic series (Mg, Zn, Al, Fe...), aluminium is more active than steel (iron), so aluminium acts as the anode and corrodes preferentially.
Answer: Galvanic corrosion; aluminium is the anode.
Fig. 6.1 — Ductile fracture (necking, cup-and-cone, dimpled surface) versus brittle fracture (flat, chevron-marked surface).
4; APF = 0.74; CN = 12; Examples: Cu, Al, Fe(γ), Ni
2; APF = 0.68; CN = 8; Examples: Fe(α), Cr, W, Mo
0.74 (same as FCC); Zn, Mg, Ti
0.8% C, 727°C → Pearlite (α-Fe + Fe₃C)
4.3% C, 1148°C → Ledeburite
Hardest steel microstructure; formed by rapid quench; BCT
~200 GPa
UTS (MPa) ≈ 3.4 × BHN
~0.5 × UTS for steels
Charpy: simply supported, horizontal; Izod: cantilever, vertical
BCC metals only; FCC metals have no brittle-to-ductile transition
T > 0.3–0.4 × Tm (homologous temperature)
~0.04 — lowest of any solid
≥10.5% Cr; passive Cr₂O₃ layer
\(E_c = V_f E_f + V_m E_m\) (isostrain)
Best toughness/weight ratio; bulletproof vests
Fracture occurs when \(K_I \ge K_{IC}\)
Ultrasonic Testing (UT) and Radiographic Testing (RT)
Liquid Penetrant (PT) and Visual (VT)
Mg → Zn → Al → Fe → Ni → Cu → Ag → Au (active → noble)
Stress Corrosion Cracking — requires both stress AND corrosive environment
| Topic | Paper I Focus |
|---|---|
| Crystal Structure | APF and atoms-per-cell recall for SC/BCC/FCC/HCP; bond-type classification |
| Mechanical Properties | Stress-strain curve point identification; BHN-UTS numericals; Charpy vs Izod |
| Metals & Alloys | Iron-carbon diagram point matching; heat-treatment process selection |
| Ceramics & Polymers | Thermoplastic vs thermoset classification; glass transition behaviour |
| Composites | Rule of mixtures numericals; fibre property comparison |
| Failure & NDT | Fracture mode identification; NDT method selection; corrosion-type matching |
Q1. A steel sample has BHN = 250. Estimate its UTS.
Q2. A steel component has UTS = 700 MPa. Estimate its endurance limit.
Q3. A composite has Vf = 0.5, Ef = 200 GPa, Vm = 0.5, Em = 4 GPa. Find the longitudinal modulus.
Q4. Which iron-carbon microstructure has 6.67% carbon content and is extremely hard and brittle?
Q5. An internal crack is suspected in a thick pressure vessel weld. Should Liquid Penetrant Testing or Ultrasonic Testing be used?