Question 1

Allotropic Forms of Iron — Temperature Ranges Which allotropic form of iron is stable between 910°C and 1403°C under equilibrium conditions?

Accepted Answer

Introduction / Context: Pure iron exhibits several allotropic forms that appear over different temperature ranges. These phases—designated α, γ, and δ under modern nomenclature—have distinct crystal structures and magnetic properties, dictating transformation behavior in steels. Given Data / Assumptions: Equilibrium temperature ranges for pure iron (no alloying additions). Interest in the phase stable from 910°C up to 1403°C. Concept / Approach: Below ~910°C, α-iron (ferrite, BCC) is stable; between ~910°C and ~1403°C, γ-iron (austenite, FCC) is stable; from ~1403°C up to the melting point (~1538°C), δ-iron (BCC) reappears. The obsolete term β-iron referred to nonmagnetic α-iron just above the Curie point (~770°C) and is not a distinct crystal structure in modern usage. Step-by-Step Solution: Identify the temperature interval: 910–1403°C.Match to phase: γ-iron (austenite, FCC) is stable in this range.Therefore, the correct selection is γ-iron. Verification / Alternative check: Fe–C phase diagrams and metallurgy texts universally show the austenitic γ field occupying this interval in pure iron. Why Other Options Are Wrong: α-iron: stable below 910°C (magnetic below ~770°C). β-iron: outdated term (nonmagnetic α), not a separate phase. δ-iron: stable only above ~1403°C up to melting. Common Pitfalls: Confusing magnetic transition (Curie point) with structural transitions; the latter define α/γ/δ fields. Final Answer: γ-iron

Question 2

Low-Carbon Steel Alloying — Effect on Yield Point and Atmospheric Corrosion In low carbon steels, which element, when added in small amounts, raises the yield point and improves resistance to atmospheric corrosion (weathering)?

Accepted Answer

Introduction / Context: Minor alloying additions in low-carbon steels can significantly modify mechanical properties and environmental resistance. Phosphorus, while often controlled due to embrittlement concerns at higher levels, provides beneficial effects in small amounts for structural sheet and weathering steels. Given Data / Assumptions: Carbon content is low (typical mild steel range). Small phosphorus additions, typically below levels that cause cold-shortness. Atmospheric (not marine or chloride) corrosion considered. Concept / Approach: Phosphorus in small quantities strengthens ferrite by solid-solution effects, raising yield point and improving bake hardenability. It also enhances resistance to atmospheric corrosion by promoting the formation of a more adherent rust layer, a principle leveraged in weathering steels (with additional Cu, Cr, Ni). Conversely, sulphur usually harms ductility (hot shortness) unless balanced by Mn; Mn primarily improves hardenability and deoxidation; Si acts as a deoxidizer/strengthener but is not chiefly responsible for the atmospheric corrosion improvement in mild steels. Step-by-Step Solution: Target effects: higher yield point + improved atmospheric corrosion resistance.Match element: phosphorus provides both benefits in small additions.Therefore, select phosphorus. Verification / Alternative check: Automotive sheet steels (BH/IF grades) and certain weathering steels use controlled P to raise yield and enhance corrosion resistance in air exposure. Why Other Options Are Wrong: Sulphur: causes hot shortness unless neutralized; not used to raise yield/corrosion resistance. Manganese: improves hardenability and sulphur control; not the principal atmospheric corrosion improver. Silicon: strengthens and deoxidizes but is not the primary contributor to weathering behavior in mild steels. Common Pitfalls: Assuming any strengthener will also improve atmospheric corrosion; specific chemical effects and passivation behavior matter. Final Answer: phosphorus

Question 3

Impact Loading — Property Definition The ability of a material to resist fracture when subjected to high impact or shock loads is called:

Accepted Answer

Introduction / Context: Designers must distinguish among mechanical properties to ensure safe, reliable components. Impact-prone applications such as tools, automotive components, and structural members in seismic zones require materials that absorb energy without fracturing. This question clarifies the property name associated with resistance to sudden loads. Given Data / Assumptions: Impact or shock loading scenarios are considered. Standard property definitions from materials science apply. No specific material system is required; the concept is general. Concept / Approach: Toughness is the capacity of a material to absorb energy up to fracture, commonly assessed by Charpy or Izod impact tests and represented by the area under the stress–strain curve. It requires both strength and ductility. Strength alone (high ultimate stress) does not guarantee toughness if elongation is low, and stiffness (modulus) is independent of failure mechanisms. Brittleness is the opposite of toughness—fracture occurs with minimal plastic deformation. Step-by-Step Solution: Identify key requirement: survive high-rate loading without fracturing.Map to property: toughness quantifies energy absorption before fracture.Therefore, the correct term is toughness.Corollary: tough materials typically show higher impact energies and greater ductility at service temperature. Verification / Alternative check: Impact testing standards (e.g., Charpy V-notch) directly report absorbed energy, an indicator of toughness; ductile-to-brittle transition in BCC steels highlights how temperature alters toughness. Why Other Options Are Wrong: Strength: resistance to permanent deformation or failure under static loads, not necessarily impact. Stiffness: elastic response (modulus), unrelated to fracture toughness. Brittleness: propensity to fracture with little plastic deformation, the inverse concept. Common Pitfalls: Using “strong” and “tough” interchangeably; a glass can be strong yet brittle, failing catastrophically under impact. Final Answer: toughness

Question 4

Cast Irons — Hyper-Eutectic Definition at the Eutectic Composition of 4.3% C Iron–carbon alloys in the cast-iron range are termed hyper-eutectic when their carbon content is __________ 4.3%.

Accepted Answer

Introduction / Context: Cast irons solidify through a eutectic reaction near 4.3% carbon (with silicon affecting the equivalent carbon). Classifying compositions as hypo- or hyper-eutectic helps predict primary phases formed during solidification and the resulting microstructure and properties. Given Data / Assumptions: Eutectic composition for Fe–C (no modifiers) is about 4.3% C. Terminology refers to composition relative to the eutectic point. Silicon can shift “carbon equivalent,” but the definition here addresses nominal carbon. Concept / Approach: Hyper-eutectic means the composition lies to the right of the eutectic point on the Fe–C diagram, i.e., carbon content higher than 4.3%. Such melts tend to precipitate primary graphite (in grey irons) prior to eutectic solidification, influencing shrinkage, fluidity, and final properties. Hypo-eutectic (below 4.3%) favors primary austenite formation before the eutectic reaction. Step-by-Step Solution: Identify reference: eutectic at 4.3% C.Hyper-eutectic → composition greater than eutectic.Therefore, fill blank with “more than.”Implication: expect primary graphite/austenite differences in as-cast structures. Verification / Alternative check: Foundry practice uses “carbon equivalent” (C + Si/3 + P/3) to estimate solidification path; hyper-eutectic behavior corresponds to higher CE values leading to graphite flotation risk. Why Other Options Are Wrong: Equal to: exactly eutectic, not hyper-eutectic. Less than: defines hypo-eutectic, the opposite region. Common Pitfalls: Confusing eutectic (liquid→two solids) with eutectoid (solid→two solids at ~0.8% C in steels). The 4.3% figure pertains to the cast-iron eutectic, not steel. Final Answer: more than

Question 5

Copper Alloys — Naming by Composition An alloy consisting of copper, tin, and zinc is commonly called:

Accepted Answer

Introduction / Context: Copper alloys are often categorized by their primary alloying additions. The correct naming—brass, bronze, gun metal, or specialized brasses—matters for selecting materials with the right combination of strength, wear resistance, and corrosion behavior in marine and industrial environments. Given Data / Assumptions: We consider traditional naming conventions. Alloy contains copper (base), tin, and zinc. Concept / Approach: Gun metal is a classic copper alloy containing copper with tin and zinc (typical compositions around Cu 85–88%, Sn 5–10%, Zn 2–5% with small Pb sometimes). It is widely used for bearings, valves, and marine fittings due to a balance of strength and corrosion resistance. Brass is primarily Cu–Zn; bronze traditionally refers to Cu–Sn without significant zinc; Muntz metal is a specific 60/40 brass (Cu–Zn) used for sheathing and hardware. Step-by-Step Solution: Identify elements present: Cu + Sn + Zn.Match to conventional name: Gun metal corresponds to Cu–Sn–Zn alloys.Therefore, select “gun metal.”Confirm: Other names do not include all three elements in traditional definitions. Verification / Alternative check: Materials handbooks list gun metal compositions and applications distinct from brasses and binary bronzes. Why Other Options Are Wrong: Brass: Cu–Zn, generally without tin as a main addition. Bronze: Cu–Sn, typically with minimal or no zinc. Muntz metal: a 60/40 brass (Cu–Zn), again lacking tin as a principal alloying element. Common Pitfalls: Using “bronze” generically for any copper alloy; precise composition determines the correct traditional name and properties. Final Answer: gun metal

Question 6

Blast Furnace Burden — Typical Proportions by Weight The charge (burden) fed to a blast furnace typically consists of calcined iron ore, coke, and limestone in approximately which ratio by weight?

Accepted Answer

Introduction / Context: The blast furnace burden must supply iron units, fuel, and flux in balanced proportions to achieve efficient smelting. Typical textbook ratios provide a baseline for understanding material balances, although modern plants optimize these values based on ore quality, sinter/basicity, and pulverized coal injection (PCI). Given Data / Assumptions: Classical burden comprising calcined ore (or sinter/pellets), coke as fuel and permeable support, and limestone as flux. Approximate, illustrative ratio used for teaching purposes. Concept / Approach: Ore provides Fe2O3/Fe3O4; coke supplies carbon for reduction and heat; limestone (CaCO3) decomposes to CaO for slag formation, capturing silica and other impurities. A common illustrative mix is roughly 8:4:1 (ore:coke:limestone) by weight, reflecting the dominant mass share of iron units, the substantial but lesser fuel requirement, and a smaller flux portion sufficient to achieve target basicity of the slag. Step-by-Step Solution: Identify roles: ore → iron, coke → fuel/reductant/structure, limestone → flux.Recall the didactic ratio: 8 (ore) : 4 (coke) : 1 (limestone).Select the option matching this ratio exactly.Hence, choose option with “8, 4, 1.” Verification / Alternative check: Metallurgy texts often present this 8:4:1 ratio to introduce burden composition before detailing real-plant adjustments (sinter/pellet chemistry, PCI rates, flux additions). Why Other Options Are Wrong: Swapped or exaggerated proportions do not reflect iron unit dominance or realistic flux levels. Equal parts ignore practical mass balances for reduction and slag formation. Common Pitfalls: Assuming fixed ratios across all plants; in practice, burden composition is optimized continuously based on feed chemistry and operational goals. Final Answer: calcined ore (8 parts), coke (4 parts), limestone (1 part)

Question 7

Metallurgy fundamentals: A plain carbon steel is termed as hypo-eutectoid steel when its carbon content is __________ relative to the eutectoid composition (approximately 0.8% C).

Accepted Answer

Introduction / Context: This question tests core steel metallurgy terminology related to the iron–carbon phase diagram. Correctly identifying hypo-eutectoid steels is essential for predicting phase transformations (ferrite plus pearlite formation) and for choosing heat-treatment schedules in manufacturing and design. Given Data / Assumptions: Eutectoid composition in plain carbon steel is taken as about 0.8% carbon. Only the carbon percentage varies; other alloying effects are ignored for the definition. Room-temperature microstructures are considered after slow cooling. Concept / Approach: By definition, steels are classified relative to the eutectoid point: hypo-eutectoid (less carbon than eutectoid), eutectoid (at the eutectoid carbon), and hyper-eutectoid (greater than eutectoid). Hypo-eutectoid steels transform on cooling to a mixture of proeutectoid ferrite plus pearlite, because the ferrite field extends to the left of the eutectoid point. Step-by-Step Solution: Identify eutectoid carbon content reference: approximately 0.8% C.Define classes: hypo-eutectoid < 0.8% C; eutectoid ≈ 0.8% C; hyper-eutectoid > 0.8% C.Apply definition: a steel is hypo-eutectoid if its carbon content is below 0.8%.Therefore, the correct completion is “below 0.8%”. Verification / Alternative check: Indicator microstructure: presence of proeutectoid ferrite along grain boundaries with pearlite colonies confirms hypo-eutectoid character. Why Other Options Are Wrong: Above 0.8% indicates hyper-eutectoid steel (proeutectoid cementite + pearlite). Exactly 0.8% describes eutectoid steel (fully pearlitic after slow cooling). Between 1.0% and 1.5% is plainly hyper-eutectoid. Indeterminate is incorrect because composition alone is sufficient for the classification. Common Pitfalls: Confusing 0.77% vs 0.8% values; for exam purposes, 0.8% is the standard rounded reference for the eutectoid composition. Final Answer: below 0.8%

Question 8

Steel integrity and gases: The presence of hydrogen in steel primarily causes which detrimental effect during service or post-weld conditions?

Accepted Answer

Introduction / Context: Hydrogen is one of the most insidious impurities in steels. Even in very small amounts, atomic hydrogen can diffuse into stressed regions and cause delayed cracking. Recognizing hydrogen embrittlement is vital for welding, heat treatment, plating, and high-strength steel applications. Given Data / Assumptions: Base material: carbon or low-alloy steels. Hydrogen can be introduced during welding, pickling, electroplating, or service exposure. High-strength steels are particularly susceptible. Concept / Approach: Hydrogen embrittlement results from hydrogen atoms accumulating at trap sites (dislocations, inclusions, interfaces) and decohesion ahead of a crack tip. This lowers ductility and fracture toughness, leading to sudden failure under static or cyclic loads, often after a delay. Mitigation includes low-hydrogen welding consumables, preheat/interpass temperature control, and post-weld baking to diffuse hydrogen out. Step-by-Step Solution: Identify the key effect of hydrogen: it weakens metallic bonds locally.Correlate with observed failures: delayed cracking, brittle fracture patterns.Select the term that captures this phenomenon: embrittlement.Conclude that option “embrittlement” is correct. Verification / Alternative check: Practical evidence: weldments show underbead or toe cracking relieved by hydrogen bake-out or by adopting low-hydrogen procedures. Why Other Options Are Wrong: Reduced neutron absorption is irrelevant to mechanical integrity here. Hydrogen does not improve weldability; it increases cracking risk. Hydrogen does not enhance corrosion resistance or fatigue limit in steels. Common Pitfalls: Confusing hydrogen embrittlement with temper embrittlement (a different phenomenon related to alloy segregation). Final Answer: embrittlement

Question 9

Material property definitions: Evaluate the statement: “Stiffness is the ability of a material to resist deformation under applied stress.”

Accepted Answer

Introduction / Context: Accurate use of mechanical property terminology prevents design errors. Stiffness, strength, hardness, and toughness are often confused, yet they represent distinct behaviors under load. Here, stiffness refers to deformation response within the elastic range. Given Data / Assumptions: Linear elastic behavior is assumed near the origin of the stress–strain curve. Small-strain regime is relevant for defining stiffness. No time-dependent effects (like creep) are considered in the definition. Concept / Approach: Stiffness quantifies resistance to elastic deformation. At the material level, it is captured by modulus of elasticity (Young’s modulus) in tension/compression and shear modulus in shear. A higher modulus means less elastic strain for a given stress. This differs from strength (stress at failure or yield), toughness (energy absorbed before fracture), and hardness (resistance to indentation or scratching). Step-by-Step Solution: Define stiffness: minimal deformation for a given applied stress.Relate to modulus: E = stress / strain in the linear elastic region.Compare to strength: high strength does not guarantee high stiffness if modulus is low.Conclude that the given statement is a correct description of stiffness. Verification / Alternative check: Example: Steel (high E) is stiffer than aluminum (lower E) even if a particular aluminum alloy may exhibit high strength. Why Other Options Are Wrong: Limiting to elastomers or to the yield point confuses stiffness with nonlinearity and plasticity. Equating stiffness with strength is a classic misconception; stiffness is about deformation, not failure. Common Pitfalls: Using stiffness and strength interchangeably; ignoring anisotropy where directional moduli vary. Final Answer: True

Question 10

Aluminium alloy families — Y-alloy composition: Identify the composition range that correctly represents the classic Y-alloy (an Al–Cu–Ni–Mg high-temperature casting alloy).

Accepted Answer

Introduction / Context: Y-alloy is a well-known aluminum casting alloy formulated for better strength retention at elevated temperatures (e.g., in some engine components). Recognizing its distinctive Ni-bearing composition differentiates it from duralumin and other Al–Cu–Mg wrought alloys used at lower temperatures. Given Data / Assumptions: We seek a composition that includes copper and nickel with minor elements improving castability and strength at temperature. Ranges are approximate and typical of handbook values. Balance of the alloy is aluminum. Concept / Approach: Classic Y-alloy contains copper for precipitation strengthening and nickel to improve hot strength and creep resistance. Small additions of manganese, silicon, magnesium, and iron aid castability, strength, and control of intermetallics. Duralumin-like compositions lack nickel and are primarily wrought; tin-bearing options relate to bearing alloys, not Y-alloy. Step-by-Step Solution: Identify the signature element: nickel at about 1.8 to 2.3% is characteristic of Y-alloy.Confirm supporting elements: copper ~3.5 to 4.5%, manganese ~1.2 to 1.7%, with small silicon, magnesium, and iron additions.Match the listed range to these features.Select option (b) as the correct Y-alloy composition. Verification / Alternative check: Handbooks categorize Y-alloy among Al–Cu–Ni–Mg casting alloys with good strength at moderately elevated temperatures. Why Other Options Are Wrong: Option (a) corresponds to duralumin-type wrought alloy (no nickel). Option (c) is magnesium-rich and does not represent Y-alloy. Option (d) is a tin–copper bearing alloy family, not Y-alloy. Option (e) resembles a low-alloy Al–Zn–Mg composition, not Y-alloy. Common Pitfalls: Confusing wrought duralumin with cast Y-alloy; overlooking the critical role of nickel in Y-alloy for high-temperature performance. Final Answer: 3.5 to 4.5% copper, 1.2 to 1.7% manganese, 1.8 to 2.3% nickel, 0.6% each of silicon, magnesium and iron, balance aluminium

Question 11

Heat treatment objectives — normalising of steel: Normalising is performed primarily to achieve which of the following outcomes?

Accepted Answer

Introduction / Context: Normalising is a widely used heat treatment for steels that improves microstructural uniformity, removes internal stresses, and refines grains. Designers and manufacturing engineers rely on normalising to obtain predictable mechanical properties before machining or further heat treatment. Given Data / Assumptions: Process: austenitise above the critical range followed by air cooling. Applicable to carbon and low-alloy steels. No surface enrichment (like carburising) is intended during normalising. Concept / Approach: Normalising dissolves non-uniform microconstituents into austenite and, upon air cooling, forms a fine pearlitic or bainitic structure with reduced segregation. It relieves stresses from cold or hot working and refines grains compared with slower furnace cooling (annealing). The combined benefits justify the selection of option covering all listed outcomes. Step-by-Step Solution: Heat the steel to a temperature above A3 or Acm (depending on composition) to obtain homogeneous austenite.Hold sufficiently to dissolve carbides and equalize temperature.Air cool to produce a fine, uniform microstructure with reduced internal stresses.Results include grain refinement, removal of cold-work strains, and mitigation of hot-working defects. Verification / Alternative check: Mechanical tests show improved toughness and machinability after normalising compared with as-forged or as-rolled conditions. Why Other Options Are Wrong: Choosing any single outcome underestimates the multifaceted benefits of normalising. Increasing carbon content by diffusion is not a feature of normalising; that would be case hardening like carburising. Common Pitfalls: Equating normalising with annealing; the former generally produces finer grains and higher strength relative to full annealing. Final Answer: all of the above

Question 12

High-temperature behaviour — duralumin vs Y-alloy: Assess the statement: “Duralumin has better strength than Y-alloy at high temperature.”

Accepted Answer

Introduction / Context: Duralumin (Al–Cu–Mg wrought alloy) is optimized for room-temperature to moderately elevated temperatures, whereas Y-alloy (Al–Cu–Ni–Mg casting alloy) was designed specifically to retain strength at higher temperatures. Understanding this distinction informs material selection for engine parts and hot-service components. Given Data / Assumptions: Duralumin typically contains copper and magnesium with minor manganese. Y-alloy contains copper and nickel with minor additions for hot strength and stability. Comparison is at elevated service temperatures typical of engine environments. Concept / Approach: Nickel in Y-alloy stabilizes strengthening phases and improves creep resistance and hot strength, allowing Y-alloy to outperform duralumin as temperature increases. Duralumin suffers greater overaging and softening at high temperature. Therefore the statement that duralumin has better high-temperature strength than Y-alloy is incorrect. Step-by-Step Solution: Identify alloy families: duralumin (wrought), Y-alloy (casting).Relate alloying to performance: Ni in Y-alloy enhances high-temperature strength retention.Compare softening behavior: duralumin loses strength more rapidly with temperature.Conclude the statement is false. Verification / Alternative check: Handbook property charts show better tensile and creep strength for Y-alloy at elevated temperatures compared with duralumin. Why Other Options Are Wrong: Any qualified “True” options contradict the documented superiority of Y-alloy at high temperature. Removing copper makes no sense in this context; copper contributes to precipitation hardening in both systems. Common Pitfalls: Assuming a stronger room-temperature alloy remains stronger at high temperature without considering overaging and phase stability. Final Answer: False

Question 13

Aluminium alloy identification — duralumin composition: Select the composition range that corresponds to classic duralumin (a precipitation-hardenable Al–Cu–Mg wrought alloy).

Accepted Answer

Introduction / Context: Duralumin is a foundational wrought aluminum alloy system that achieves strength through precipitation hardening. Correctly identifying its composition helps distinguish it from casting alloys like Y-alloy and from other series such as 7xxx Al–Zn–Mg. Given Data / Assumptions: Duralumin emphasizes copper and magnesium with a small amount of manganese. It is a wrought, not a casting, alloy in its classic form. Balance is aluminum with trace impurities. Concept / Approach: The classic duralumin composition is approximately 4% copper, about 0.5% magnesium, and about 0.5% manganese. This balance enables solution treatment and aging for strength. The listed range in option (a) matches this profile, while the Ni-bearing composition in option (b) is characteristic of Y-alloy used for cast parts at higher temperatures. Step-by-Step Solution: Look for Cu ~4%, Mg ~0.5%, Mn ~0.5% in a wrought alloy.Match to option (a) exactly.Eliminate options featuring nickel (Y-alloy) or zinc (7xxx series).Select option (a) as duralumin. Verification / Alternative check: Historical standards and textbooks list duralumin within these composition ranges and describe its age-hardening behavior. Why Other Options Are Wrong: Option (b) is Y-alloy; option (c) is magnesium-rich and not classic duralumin; option (d) represents bearing alloys; option (e) corresponds to Al–Zn–Mg–Cu 7xxx family, not duralumin. Common Pitfalls: Mixing cast and wrought alloy families; assuming the presence of nickel in duralumin, which is incorrect. Final Answer: 3.5 to 4.5% copper, 0.4 to 0.7% magnesium, 0.4 to 0.7% manganese and balance aluminium

Question 14

Cupola practice — charge composition control: Evaluate the statement: “In a cupola furnace, the proportion of steel scrap in the charge controls the grade of cast iron produced.”

Accepted Answer

Introduction / Context: The cupola is a shaft furnace used to melt and refine cast iron. The final composition and hence the grade of iron depend strongly on the charge makeup, including pig iron, steel scrap, foundry returns, coke, and flux. Understanding how steel scrap proportion influences chemistry is essential for controlling carbon and silicon levels. Given Data / Assumptions: Cupola charge may include pig iron, steel scrap, and returns, plus coke and flux. Steel scrap is relatively low in carbon and silicon compared to pig iron. Target grades vary by required carbon, silicon, and mechanical properties. Concept / Approach: Increasing steel scrap in the charge dilutes carbon and silicon, pushing the melt toward lower carbon and silicon contents unless compensated by carburisers and ferrosilicon additions. This directly affects microstructure (e.g., pearlite/ferrite balance) and mechanical properties, thereby controlling the grade produced. Operators tune additions and the scrap ratio to hit specification. Step-by-Step Solution: Recognize that steel scrap introduces low C and low Si to the bath.More scrap → lower base C and Si levels.Adjustments via carburiser/ferrosilicon set final composition and grade.Thus, the steel scrap proportion is a key lever in grade control; the statement is correct. Verification / Alternative check: Foundry practice guides explicitly recommend controlling scrap/pig ratio to meet grade chemistry and hardness targets. Why Other Options Are Wrong: Limiting to specific iron types is unnecessary; scrap proportion matters broadly. Flux influences slag chemistry and sulphur but does not alone fix carbon and silicon to set grade. Common Pitfalls: Assuming coke rate or blast alone determines grade; while important, charge chemistry via scrap ratio is fundamental. Final Answer: Correct

Question 15

Stainless steels — martensitic family identification: A steel containing about 12 to 14% chromium and 0.12 to 0.35% carbon is classified as which stainless steel type?

Accepted Answer

Introduction / Context: Stainless steels are grouped by structure and composition: martensitic, ferritic, austenitic, duplex, and precipitation-hardening. Chromium around 12 to 14% with moderate carbon is characteristic of martensitic grades used for cutlery, turbine blades, and wear-resistant parts. Given Data / Assumptions: Chromium content: ~12 to 14%. Carbon content: ~0.12 to 0.35% (hardenable range). Heat treatability by quenching and tempering is intended. Concept / Approach: Martensitic stainless steels are hardenable due to sufficient carbon and display a body-centred tetragonal martensitic structure after quench. Ferritic grades have similar chromium but very low carbon and are not hardenable by quench. Austenitic grades require higher nickel to stabilise the face-centred cubic structure and usually contain 16 to 20% chromium plus 8 to 10% nickel. Duplex and precipitation-hardening grades have different composition balances. Step-by-Step Solution: Match composition window: 12–14% Cr with 0.12–0.35% C.Recognize hardenability and martensitic transformation potential.Select the family that fits: martensitic stainless steel.Confirm that other families require different alloy balances. Verification / Alternative check: Examples include types similar to 410/420 families that fit this Cr–C window and are quench-and-temper hardenable. Why Other Options Are Wrong: Austenitic needs high Ni and higher Cr; ferritic has very low carbon; duplex blends ferrite and austenite with notable Ni and Mo; P.H. grades rely on age-hardening precipitates and different compositions. Common Pitfalls: Assuming all stainless steels are non-magnetic (martensitic and ferritic can be magnetic) or equating stainless solely with high nickel content. Final Answer: martensitic stainless steel

Question 16

Cast iron classification — white iron characteristics: Which statements correctly describe white cast iron?

Accepted Answer

Introduction / Context: White cast iron is a cast iron in which carbon exists predominantly as cementite (iron carbide) rather than as free graphite. This microstructure confers specific mechanical and machining characteristics that distinguish it from grey and ductile irons. Given Data / Assumptions: Rapid cooling or appropriate composition suppresses graphite formation. Microstructure: ledeburite and cementite networks with little to no free graphite. Mechanical behavior considered under static loading. Concept / Approach: Cementite is extremely hard and brittle. Consequently, white iron exhibits very high hardness and compressive strength but poor tensile strength and virtually no ductility. Its lack of graphite makes it difficult to machine. These features make white iron suitable for abrasion-resistant applications such as liners and wear plates, but unsuitable where tensile loads or machining are dominant considerations. Step-by-Step Solution: Identify carbon state: present as carbide rather than graphite → white fracture surface.Infer mechanical properties: hard and brittle → low tensile strength, high compressive strength.Compare to grey iron: grey iron has graphite flakes that aid machinability; white iron does not.Therefore, all listed statements (a), (b), and (c) correctly describe white cast iron. Verification / Alternative check: Metallography reveals cementite networks; hardness tests show high values; machining is extremely difficult without special tooling. Why Other Options Are Wrong: Option (e) describes grey cast iron where graphite flakes improve machinability and damping; it does not apply to white iron. Common Pitfalls: Confusing white iron with malleable iron (which is produced by heat-treating white iron to convert carbides to temper carbon nodules) or assuming all cast irons machine easily. Final Answer: all of these

Question 17

Polymer classification in materials engineering Polyvinyl chloride, commonly abbreviated as PVC and widely used in pipes, wire insulation, and flooring, belongs to which class of polymeric materials with respect to its thermal softening and recyclab…

Accepted Answer

Introduction / Context: Knowing whether a polymer is a thermoplastic or thermoset is fundamental for processing, recycling, and service-temperature decisions. Polyvinyl chloride (PVC) is one of the most common engineering plastics used in civil and electrical applications, so its classification is frequently tested in materials science and manufacturing exams. Given Data / Assumptions: PVC is a vinyl polymer based on the monomer vinyl chloride. Question asks for material class by softening behavior and reprocessability. No numerical data are required; this is a conceptual identification. Concept / Approach: Thermoplastics soften on heating and harden on cooling repeatedly because their molecular chains are not permanently cross-linked. Thermosets, by contrast, form irreversible cross-linked networks during curing and will not soften to a melt on reheating. PVC’s structure consists of linear (or lightly branched) chains that can be softened with heat and shaped by extrusion, injection molding, or calendaring, and then cooled to retain form. Step-by-Step Solution: Identify PVC’s processing: routinely extruded into pipes and profiles → reheatable and remoldable.Relate behavior to definition: reheatable + remoldable → thermoplastic.Cross-check: thermosets (e.g., epoxy, Bakelite) do not melt on reheating; elastomers exhibit rubber-like elasticity.Therefore, PVC is a thermoplastic. Verification / Alternative check: Industrial recycling streams treat PVC as a thermoplastic; plastic identification codes and processing guides list PVC under melt-processable materials, confirming the classification. Why Other Options Are Wrong: Thermosetting plastics irreversibly cure; elastomers prioritize large elastic strain; “thermoelastic composite” and “semi-metallic polymer” are not standard classes for PVC. Common Pitfalls: Confusing “rigid PVC” mechanical stiffness with thermoset behavior; stiffness does not imply cross-linking. Final Answer: thermoplastic

Question 18

Bearing materials — service domain of tin-base white metals (Babbitts) Tin-base white-metal (Babbitt) bearings are primarily selected for which operating regime considering load, pressure, and wear characteristics?

Accepted Answer

Introduction / Context: White metals (Babbitts) are classic bearing linings used on shafts and journals. They offer excellent conformability, embeddability, and seizure resistance. Understanding their best-use window—especially relative to load and speed—is crucial for reliable bearing selection in machines and turbines. Given Data / Assumptions: Tin-base Babbitt is a soft, low-friction alloy with hard intermetallics dispersed in a soft matrix. Hydrodynamic lubrication is assumed during normal operation. The question targets typical design guidance rather than extreme specialty variants. Concept / Approach: Tin-base Babbitts excel in low to moderate unit pressures and high surface speeds, where an oil film can be maintained. Their softness protects the shaft, accommodates misalignment, and embeds debris. However, they are not intended for sustained very high loads or shock; harder bronzes or tri-metal bearings are preferred in those regimes. Step-by-Step Solution: Identify characteristic: soft matrix + good conformability → tolerant of slight misalignment and contaminants.Map to operating envelope: best at light load/pressure with adequate lubrication.Exclude high-pressure, high-shock duty, where fatigue strength must be higher.Thus select “light load and pressure.” Verification / Alternative check: Design handbooks specify allowable PV (pressure * velocity) values; tin-base Babbitts show relatively low allowable pressure compared with bronzes but handle good speeds, aligning with the chosen option. Why Other Options Are Wrong: “High pressure and load” overstates capability; “elevated temperatures” risk softening; “large surface wear” suggests boundary/dry regime, the opposite of intended hydrodynamic use; “intermittent dry rubbing” is inappropriate for Babbitts. Common Pitfalls: Picking a bearing solely for low friction without checking allowable unit pressure and temperature limits. Final Answer: light load and pressure

Question 19

Alloying effects in steels — role of chromium When chromium (Cr) is added to carbon steel in conventional alloying amounts, what is the general effect on the tensile strength of the steel?

Accepted Answer

Introduction / Context: Chromium is a widely used alloying element in steels for structural, tool, and stainless applications. Understanding its broad influence on mechanical properties, especially tensile strength, helps with material selection and performance prediction after heat treatment. Given Data / Assumptions: Cr is added in moderate amounts typical of low-alloy and stainless steels. Carbon content and heat treatment are within practical ranges. Question asks for the general trend, not a specific grade curve. Concept / Approach: Chromium increases hardenability, refines pearlite, and promotes formation and stability of carbides (e.g., Cr carbides) in appropriate conditions. These effects tend to raise strength levels after standard heat treatments. In stainless steels, high Cr also aids corrosion resistance via passive films; mechanical effects still include higher strength relative to plain-carbon steels at similar processing histories. Step-by-Step Solution: Note metallurgical action: improved hardenability → deeper martensite/tempered martensite after quench and temper.Carbide strengthening and grain refinement add to tensile strength.Therefore, chromium broadly increases tensile strength for comparable conditions.Select the option reflecting this general increase. Verification / Alternative check: Property charts for Cr-Mo or Cr steels show higher ultimate tensile strength versus plain-carbon analogs after similar heat treatments. Why Other Options Are Wrong: “Does not effect” is incorrect; “decreases” contradicts data; the remaining distractors are exaggerated or non-general behaviors. Common Pitfalls: Confusing effect of excessive carbide formation on toughness with tensile strength trends; strength can rise while toughness may require careful tempering. Final Answer: increases

Question 20

Historic steelmaking routes — naming by process Steel produced by the traditional cementation process (carburising wrought iron bars in carbonaceous material and later forging) is commonly known as which type of steel?

Accepted Answer

Introduction / Context: Before modern basic oxygen and electric arc processes, steel was produced by several historic methods. The cementation process diffused carbon into wrought iron bars, producing a distinctive surface appearance and composition gradient. Recognizing the correct name helps link process history with microstructural characteristics. Given Data / Assumptions: Cementation packs wrought iron with charcoal and heats for long durations. Carbon diffuses inward, raising surface carbon significantly. The product exhibits surface blisters from gas evolution and structural changes. Concept / Approach: The output of cementation is called blister steel because of the blistered surface on the carburised bars. In some routes, blister steel was subsequently melted in crucibles to homogenize composition, creating “crucible steel” (a different, later step). Thus, cementation → blister steel; melting blister in a crucible → crucible steel. Step-by-Step Solution: Identify process: solid-state carburising of wrought iron bars.Recognize product appearance: characteristic blisters on the bar surface.Map process to name: cementation product = blister steel.Therefore choose “blister”. Verification / Alternative check: Metallurgical histories consistently label cementation output as blister steel, later refined by crucible melting for uniformity. Why Other Options Are Wrong: “Crucible” is a subsequent refining method; Bessemer, open-hearth, and EAF are unrelated modern processes. Common Pitfalls: Assuming “crucible” because it is often mentioned alongside cementation; remember crucible is the secondary melting step. Final Answer: blister

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