Materials and Material Properties: Lesson Test

Answer: C

Correct answer: Hardness. Hardness is usually measured by pressing a hard indenter, such as a steel ball or diamond, into the surface and measuring the dent, as in the Rockwell and Brinell tests; a harder material leaves a smaller dent. Hard materials resist wear and scratching, which is why cutting tools and files are hardened. Toughness is the common confusion: a very hard material such as glass can still be brittle and shatter easily. Hard does not mean tough.

• A: Ductility describes plastic stretching before rupture, not surface scratch resistance.
• B: Density compares mass with volume and does not directly describe scratching.
• C (correct): Hardness measures scratch or indentation resistance under a defined test.
• D: Toughness concerns energy absorbed before fracture, not the ease of scratching.

Answer: B

Correct answer: Ductility. A ductile material can be stretched a great deal under tension and deformed permanently without breaking. Wire drawing pulls metal through progressively smaller dies, so the metal must lengthen and thin out under tensile stress. Copper, gold and aluminum are highly ductile, which is why they make good wire. Watch for the close cousin, malleability: it refers to shaping under compression, such as hammering or rolling, while ductility refers to stretching.

• A: Heat transfer behavior does not explain the large tensile shape change.
• B (correct): Wire drawing uses useful tensile plastic deformation, the hallmark of ductility.
• C: Brittleness would tend to cause limited deformation before rupture instead.
• D: Corrosion behavior addresses chemical environments, not the drawing process.

Answer: C

Correct answer: Malleability. Malleability is the ability to be hammered, pressed or rolled into thin shapes without cracking. Rolling squeezes metal between heavy rollers, a compressive process, so malleability is the key property. Gold is the classic example and can be beaten into leaf thinner than paper; aluminum foil is another. A memory hook: malleable comes from the Latin word for hammer, while ductile relates to drawing out, as in drawing wire.

• A: Transparency concerns passage of light rather than metal shaping.
• B: Magnetic response is not the property that allows sheet forming.
• C (correct): Sheet rolling involves useful permanent shaping, often under compressive loading.
• D: Insulation describes resistance to energy flow, not permanent sheet formation.

Answer: C

Correct answer: Toughness. Toughness is the ability to absorb energy, by deforming, before cracking through. Impact tests such as the Charpy test swing a pendulum into a notched sample and measure the energy it absorbs as it breaks. Hammers, axles and vehicle parts need toughness to survive shock. Do not confuse it with hardness: a hard material such as glass can resist scratching yet shatter from a sudden blow. Tough materials usually combine good strength with some ductility.

• A: Appearance does not directly establish impact fracture behavior.
• B: Greater mass per volume alone does not guarantee shock tolerance.
• C (correct): Toughness concerns the ability to absorb energy before fracture under relevant conditions.
• D: Change in size with temperature is different from resisting sudden blows.

Answer: B

Correct answer: Density. Density tells how much mass is packed into a given volume, written as density = mass ÷ volume, often in g/cm³ or kg/m³. It explains why a small lead weight feels heavier than a larger block of wood. Each pure material has a characteristic density, so it can help identify materials. The tempting distractor is stress, which is also a ratio but divides force by area. Specific gravity is density compared with that of water.

• A: Engineering strain is relative length change and does not use mass.
• B (correct): Density is mass per unit volume, regardless of shape if volume is determined correctly.
• C: Stress is force divided by area, not mass divided by volume.
• D: Yield strength is a stress limit associated with plastic deformation.

Answer: A

Correct answer: Copper. A good wire material needs low electrical resistance and enough ductility to be drawn into long, thin strands. Copper has the second-highest conductivity of any metal, after silver, yet costs far less than silver, and it bends without cracking. That combination makes it the standard for household and vehicle wiring. Glass, ceramic and dry wood are insulators, which is why glass and ceramic are used to support power lines rather than to carry current.

• A (correct): Copper combines useful electrical conductivity with ductility for wire manufacture.
• B: Dry wood is not a conventional high-conductivity wire material.
• C: A conventional ceramic tile is brittle and typically not used as conductive wire.
• D: Ordinary dry glass is generally an electrical insulator rather than a wire conductor.

Answer: C

Correct answer: The names refer to tree groups, not actual hardness. Hardwoods come from broadleaf, mostly deciduous trees such as oak and maple, while softwoods come from conifers such as pine and fir. The labels describe the type of tree, not how hard the wood is. Balsa, used in model airplanes, is botanically a hardwood yet is one of the softest woods, while some softwoods such as yew are quite hard. So never judge hardness or strength from the name alone.

• A: Density varies across species and can overlap between the two groups.
• B: Properly graded softwood lumber is commonly used in structural construction.
• C (correct): The terms classify trees botanically, so some hardwoods are softer than some softwoods.
• D: All wood can be affected by grain orientation and splitting conditions.

Answer: D

Correct answer: It breaks with little or no bending or stretching first. A brittle material has almost no ability to deform plastically, so once stress exceeds its strength, a crack runs through it suddenly. The broken pieces usually fit back together cleanly because the shape barely changed. Glass, gray cast iron, ceramics and chalk behave this way. A ductile metal, by contrast, bends, stretches and necks down before tearing, giving visible warning. Brittleness is a fracture behavior and says nothing about conductivity or buoyancy.

• A: Buoyancy depends on density relative to water, not on brittleness.
• B: Large stretching and thinning before fracture describe ductile behavior, the opposite of brittle.
• C: Electrical conductivity is a separate physical property unrelated to the fracture behavior described.
• D (correct): Brittle materials fracture with very little permanent deformation beforehand.

Answer: B

Correct answer: Elastic deformation. When a load is small enough, the atoms in the metal are pulled slightly apart but not rearranged, so they spring back to their original positions once the load is removed. That recoverable stretch is elastic deformation, and springs are designed to stay within this range. If the load exceeds the elastic limit, some deformation becomes permanent, called plastic deformation, and the spring stays bent. Compare a rubber band with a bent paper clip.

• A: Corrosion is a chemical or electrochemical process, not spring recovery.
• B (correct): Elastic strain is recovered when a load is removed within the applicable range.
• C: Plastic deformation would leave a permanent set after the load was removed.
• D: Creep involves time-dependent strain during sustained loading, not instant recovery.

Answer: D

Correct answer: Creep. Creep is slow, continuing deformation of a material held under a constant load, even when that load is below what would cause immediate yielding. Plastics can creep noticeably at room temperature, while most metals creep significantly only at high temperatures, as in turbine blades and boiler tubes. The clue in the question is time: the load did not change, yet the bracket kept sagging for months. Fatigue, by contrast, requires repeated loading cycles rather than a steady load.

• A: A surface scratch test would not explain gradual sagging under constant load.
• B: Color change is unrelated to this specified mechanical deflection.
• C: A density ratio alone cannot explain gradual deformation under constant stress.
• D (correct): Time-dependent deformation under sustained load is called creep.

Answer: B

Correct answer: Grain orientation. Wood is made of long fibers that run mostly along the length of the trunk, so its properties depend on direction, a behavior called anisotropy. Wood is much stronger and stiffer along the grain than across it, splits easily parallel to the grain, and shrinks and swells far more across the grain than along it. That is why boards cup or warp as moisture changes. Glass, by contrast, is isotropic and behaves the same in every direction.

• A: Wood does not behave as a simple melting metal and grain is structural.
• B (correct): Wood fibers and growth structure cause direction-dependent properties called anisotropy.
• C: Isotropy would imply similar properties in different directions rather than the observed differences.
• D: Common wood grain behavior is not governed by magnetic domains.

Answer: D

Correct answer: Polymers. Polymers are materials built from long chain molecules, and many are commonly called plastics. Thermoplastics, such as polyethylene and nylon, soften when heated and harden when cooled, so they can be remelted and recycled. Thermosets, such as epoxy and phenolic resins, undergo a chemical cure that links the chains into a permanent network, so they char rather than melt. Both groups belong to the polymer family, while ferrous and copper alloys are metals.

• A: Ferrous alloys are iron-based metals, not polymer network categories.
• B: Copper-base alloys are metallic, not thermoplastic or thermoset systems.
• C: These categories describe polymer processing behavior rather than mineral identity.
• D (correct): Thermoplastics and thermosets are broad categories of polymer-based materials.

Answer: A

Correct answer: It cannot be remelted and reshaped like a thermoplastic. During curing, a thermoset's polymer chains form chemical crosslinks that tie them into one rigid three-dimensional network. Heat cannot undo those bonds, so reheating makes the material char or decompose instead of melting. Thermoplastics lack such crosslinks, so their chains slide past one another when warmed. A helpful comparison: a thermoplastic is like candle wax, which remelts, while a thermoset is like a baked cake, which cannot be unbaked.

• A (correct): Crosslinks formed during curing lock the structure, so heat degrades it instead of remelting it.
• B: Thermosets are polymers, not metals.
• C: Gentle reheating does not normally remelt a cured crosslinked thermoset.
• D: Strength varies widely, and thermoset classification implies no universal ranking above steel.

Answer: A

Correct answer: Its lower density makes the cover lighter than steel. Aluminum weighs about one third as much as steel for the same volume, roughly 2.7 g/cm³ versus 7.8 g/cm³. A lighter cover is easier to lift, open and close, and puts less load on hinges. Aluminum is not stronger than steel in general, so designers may use a thicker section, yet the part usually still ends up lighter. Many alloys also resist corrosion well, though they can corrode in harsh or galvanic conditions.

• A (correct): A lower-density metal gives a lighter part, which helps a cover that must be moved.
• B: Aluminum is ordinarily an electrical conductor rather than an insulator.
• C: Strength varies by grade; aluminum cannot be ranked universally above steel.
• D: Aluminum can corrode in unsuitable environments or galvanic pairings.

Answer: B

Correct answer: Iron. Steel is an alloy made mostly of iron with a small amount of carbon, typically under about 2 percent, plus other elements in some grades. Carbon makes steel much harder and stronger than pure iron. Stainless steel adds at least about 10.5 percent chromium for corrosion resistance, but iron is still the main ingredient. Zinc is the tempting distractor because galvanized steel is coated with it, yet that coating is only a thin surface layer.

• A: Zinc may protect steel galvanically but is not the principal base metal.
• B (correct): Steel is an iron-based alloy with carbon and sometimes further alloying elements.
• C: Silicon can be an alloying element but does not define ordinary steel as base.
• D: Tin may be used in coatings or certain alloys but is not steel’s primary constituent.

Answer: C

Correct answer: Documented corrosion data for that grade in that setting. Corrosion depends on both the material and its surroundings. Salt spray, industrial pollution, humidity and contact with other metals can make a grade that lasts decades inland fail quickly near the coast. Even within one family, grades differ: 316 stainless resists chlorides better than 304. Tested data for the specific grade and environment is the reliable basis for choosing. Appearance misleads because a fresh, shiny surface says nothing about long-term exposure.

• A: Color and finish do not specify chemical compatibility or corrosion rate.
• B: Surface shine does not prove resistance to specific chemical or moisture exposure.
• C (correct): Corrosion behavior depends on composition, grade and exposure conditions, so tested data is the reliable guide.
• D: Dimensions and weight do not establish a metal’s environmental durability.

Answer: A

Correct answer: High hardness with relatively limited tolerance of tensile impact. Technical ceramics such as alumina, silicon carbide and zirconia are held together by strong ionic and covalent bonds, which make them very hard, wear resistant and able to withstand high temperatures. Those same rigid bonds leave almost no way to deform plastically, so a small flaw can grow into a crack under tension or a sharp blow. That is why ceramics excel as cutting inserts and furnace linings but can shatter if dropped.

• A (correct): Many ceramic materials are hard yet show brittle behavior under impact or tension.
• B: Ceramics are not defined by a universal low melting temperature.
• C: Technical ceramics do not generally permit large tensile plastic wire drawing.
• D: Most familiar insulating ceramics do not outperform copper as electrical conductors.

Answer: D

Correct answer: A material’s elastic stiffness in tension or compression. Young’s modulus is the slope of the straight, elastic part of a stress-strain curve: stress divided by strain. A high value means the material barely stretches under load. Steel’s modulus is about three times that of aluminum, so a steel bar of the same size deflects about one third as much under the same load. Stiffness is not strength: modulus tells how much a part flexes, while strength tells how much load it takes before yielding or breaking.

• A: Fracture energy and toughness are distinct from the elastic modulus.
• B: Mass per volume defines density, not the elastic modulus.
• C: Corrosion rate comes from chemical behavior in an environment, not from elastic stiffness.
• D (correct): Young’s modulus is the ratio of stress to strain in the elastic range, a measure of stiffness.

Answer: A

Correct answer: Fatigue. Fatigue failure happens when a part is loaded and unloaded many times at stresses below what would break it in a single pull. Each cycle can grow a tiny crack, often starting at a scratch, hole or sharp corner, until the remaining metal suddenly gives way. Bending a paper clip back and forth until it snaps is a quick example. The clue is repeated cycles; a steady load that causes slow deformation over time points to creep instead.

• A (correct): Repeated cyclic stresses can initiate and grow cracks through fatigue.
• B: Color change does not account for the specified repeated mechanical cycles.
• C: Relative density does not determine cyclic crack-growth behavior.
• D: Light transmission is unrelated to repeated mechanical bending failure.

Answer: D

Correct answer: 3 g/cm³. Density is mass divided by volume, so 480 g ÷ 160 cm³ = 3 g/cm³. A quick check is that

The units also guide you: grams over cubic centimeters means mass goes on top. Flipping the ratio gives about 0.33, which would mean the sample is lighter than water, while multiplying or adding the numbers produces meaningless units. A density of 3 g/cm³ is a little more than that of aluminum.

• A: Multiplying the values is not how density is calculated.
• B: This reverses the density ratio and gives volume divided by mass.
• C: Adding mass and volume does not yield a meaningful density.
• D (correct): The calculation is 480 divided by 160, giving 3 grams per cubic centimeter.

Answer: C

Correct answer: 0.004, or 0.4%. Engineering strain is the change in length divided by the original length: ΔL/L₀ = 0.4 mm ÷ 100 mm = 0.004. Because both lengths carry the same unit, strain is dimensionless, and multiplying by 100 expresses it as a percent. The most common slip is reporting the elongation itself, 0.4, as the strain and forgetting to divide. Strains in ordinary tensile tests are usually small fractions of a percent up to a few percent, so a value like 40% should look suspicious.

• A: This fails to divide the 0.4 mm elongation by initial length.
• B: This overstates the relative extension by a factor of one thousand.
• C (correct): Length change is 0.4 mm; dividing by original 100 mm gives 0.004.
• D: Final length itself is not strain; strain is fractional length change.

Answer: A

Correct answer: 3 MPa. Average normal stress is force divided by the area it acts on, σ = F/A. Here 1,200 N ÷ 400 mm² = 3 N/mm². A handy fact is that 1 N/mm² equals exactly 1 MPa, because 1 MPa is 1,000,000 N spread over 1,000,000 mm², which is one square meter. The common error is flipping the ratio to area over force. Ask whether more force should raise the answer; with force on top, it does.

• A (correct): Stress is 1200 N divided by 400 mm², which equals 3 N/mm² or 3 MPa.
• B: Adding force and area gives neither the proper value nor compatible units.
• C: This multiplies force by area rather than dividing force by area.
• D: This reverses the ratio of force to area in the simple model.

Answer: B

Correct answer: 200 GPa. Young's modulus E is the slope of the elastic part of the stress-strain curve: E = stress ÷ strain. Since strain has no units, E carries the units of stress: 80 MPa ÷ 0.0004 = 200,000 MPa, and dividing by 1,000 converts that to 200 GPa. This value is typical of steel, which makes a good sanity check. A frequent mistake is multiplying stress by strain; a stiff material needs a large stress to produce a tiny strain, so E must be far larger than the stress.

• A: Dividing strain by stress inverts the required modulus relation.
• B (correct): 80 MPa divided by 0.0004 is 200,000 MPa, or 200 GPa.
• C: This mistakes the given stress magnitude for the much larger modulus.
• D: Multiplying stress by strain does not produce the elastic modulus.

Answer: B

Correct answer: 270 g. Density is mass per unit volume, so rearranging ρ = m/V gives m = ρ × V. With 2.7 g/cm³ and 100 cm³, the cubic centimeters cancel and leave grams:

A density of 2.7 g/cm³ matches aluminum, a useful reference value. Dividing volume by density is the typical mix-up, and tracking units catches it, because that division leaves cm⁶/g instead of grams.

• A: This result incorrectly adds an extra factor of ten to the product.
• B (correct): Multiplying density by volume gives 2.7 times 100, equal to 270 g.
• C: Dividing volume by density is not the mass relation.
• D: Adding the numerical measures is not physically meaningful here.

Answer: C

Correct answer: 0.6 mm. Free thermal expansion follows ΔL = α × L₀ × ΔT. Here 20 × 10⁻⁶ per °C × 1,000 mm × 30 °C = 0.6 mm. The per-degree unit cancels against the temperature change, leaving the same length unit as L₀. Most errors come from mishandling the power of ten, so check the thermal strain first: 20 × 10⁻⁶ × 30 = 0.0006, or 0.06%, and 0.06% of 1,000 mm is 0.6 mm. Small as it is, this movement is why long bridges and pipelines need expansion joints.

• A: This is a factor of ten smaller than the stated product.
• B: This is a factor of ten larger than the stated product.
• C (correct): Multiply coefficient, original length and temperature change: 20e-6 × 1000 × 30.
• D: This confuses a small thermal strain with a huge length change.

Answer: A

Correct answer: It ranks relative scratch resistance, not equal numerical increments of strength. Friedrich Mohs ranked ten reference minerals, from talc at 1 to diamond at 10, by which ones can scratch which. The numbers show order only, not equal steps: diamond is far harder than corundum at 9 even though they are one step apart. To place an unknown, you test whether it scratches or is scratched by a reference mineral. The tempting mistake is treating the numbers as ratios, as if 8 meant twice the strength of 4; the scale says nothing about tensile strength or toughness.

• A (correct): The Mohs scale is ordinal and compares scratching among reference minerals.
• B: Energy absorption prior to fracture is associated with toughness, not Mohs ranking.
• C: That calculation defines density, unrelated to a mineral scratch scale.
• D: Scratch ranking does not determine tensile strength or a two-to-one ratio.

Answer: D

Correct answer: Hardness and toughness. Hardness resists indentation and abrasion, which keeps a cutting edge from wearing down. Toughness is the ability to absorb energy before fracturing, which keeps teeth from chipping when they strike a knot or a hard spot. The two often trade off: making steel harder usually makes it more brittle. That is why many saw blades use hard carbide tips brazed onto a tougher steel body. Choosing maximum hardness alone is the common misconception, since a very hard but brittle edge chips easily.

• A: Neither surface color nor electrical resistance directly establishes edge durability.
• B: Mass per volume and light transmission do not directly ensure a durable edge.
• C: Sheet-forming ability and large expansion do not solve both wear and chipping concerns.
• D (correct): Wear resistance can need suitable hardness while impact and edge integrity need toughness.

Answer: A

Correct answer: Geometry and boundary support can make a compression member unstable. A slender column under compression can bow sideways at a load far below the stress needed to crush the material. The critical buckling load rises with the material's stiffness and the cross-section's moment of inertia, and falls with the square of the effective length, so doubling the length cuts it to about one quarter. End conditions matter too: fixed ends resist buckling better than pinned ends. The trap is assuming strength alone governs; for long members, stiffness and geometry usually control first.

• A (correct): Buckling depends strongly on slenderness and support conditions as well as material stiffness.
• B: Solid columns can buckle; liquids are not required for this failure.
• C: Color has no bearing on compressive strength or on whether a column buckles.
• D: A properly sized and supported rod can carry compressive load; its length only affects how much before buckling.

Answer: A

Correct answer: Direction-dependent wood swelling and shrinkage. Wood is hygroscopic, meaning it absorbs and releases moisture until it balances with the surrounding air. As it does, it swells and shrinks unevenly: movement along the grain is very small, while tangential movement is roughly one and a half to two times radial movement. Those unequal changes across a board cause cupping, bowing and twisting. The misconception is that wood moves the same in every direction; it is anisotropic, which is why lumber is dried before use and furniture panels are allowed room to move.

• A (correct): Wood exchanges moisture and changes dimensions differently relative to the grain.
• B: Species, grain, geometry and conditions change moisture-related movement.
• C: Drying does not turn wood into a metal or remove anisotropy.
• D: Wood is hygroscopic and exchanges moisture with its environment.

Answer: B

Correct answer: Alternating veneer grain directions distribute directional movement. Each plywood veneer moves much more across its grain than along it. By gluing layers with grain directions at right angles, every layer's large cross-grain movement is restrained by its neighbor's nearly stable lengthwise grain. The panel still exchanges moisture, but its overall swelling and shrinkage in width and length are much smaller than in a solid wide board. A common misconception is that the glue seals out moisture; the stability comes mainly from cross-lamination, not waterproofing.

• A: Adhesive layers do not make plywood impervious; the panel still gains and loses moisture with the surrounding air.
• B (correct): Cross-oriented bonded layers change how the panel responds to grain-direction changes.
• C: Panel ratings and environments still determine proper use and safety.
• D: Plywood is made of wood veneers, and their fiber directions still govern how the panel moves.

Answer: D

Correct answer: Galvanic corrosion. Galvanic corrosion needs three things: two different metals, an electrical connection between them, and an electrolyte such as saltwater. Together they form a small battery: the more active metal becomes the anode and corrodes faster, while the more noble metal becomes the cathode and is protected. Removing any one condition, for example by insulating the joint, stops the cell. Boats carry sacrificial zinc anodes for this reason. A tempting error is thinking contact alone is harmless; the electrolyte completes the circuit.

• A: Elastic recovery concerns unloading of a stressed body, not the electrochemical interaction of connected metals.
• B: Wood grain fracture cannot explain the stated metal corrosion mechanism.
• C: Remolding concerns polymers rather than galvanic attack on metallic parts.
• D (correct): An electrolyte and conductive path can form a galvanic couple between dissimilar metals.

Answer: C

Correct answer: Conductivity concerns heat flow; heat resistance concerns surviving the operating temperature. Thermal conductivity measures how readily heat passes through a material, in watts per meter-kelvin. Heat resistance describes whether a material keeps its strength, shape and chemistry at high temperature. The two can differ sharply: copper conducts heat extremely well yet softens at temperatures many ceramics tolerate easily, while refractory firebrick conducts heat poorly and still withstands intense heat. The common mistake is assuming a good conductor must also be heat resistant. Engineers often choose one material to move heat and another to survive it.

• A: Fast heat transfer does not guarantee mechanical or chemical stability at high temperature; a good conductor can still soften or oxidize.
• B: Thermal durability and passage of electrical current are distinct properties.
• C (correct): A heat sink can conduct heat quickly while another component is selected for temperature tolerance.
• D: The terms describe different behaviors, and thermal conductivity has units of W/(m·K), so it is not unitless.

Answer: D

Correct answer: Properties and strength in the transverse loading direction. Unidirectional fiber composites are strongest and stiffest along the fibers, because the fibers carry the load. Across the fibers, the load must pass mostly through the resin and the fiber-resin bond, so transverse strength and stiffness are often many times lower. This direction dependence is called anisotropy. A designer should check data for the actual load direction or add plies at other angles. The tempting mistake is assuming the impressive lengthwise strength applies in every direction, as it roughly would for a metal.

• A: That assumption ignores the composite’s potential anisotropy and layup.
• B: Appearance does not specify load resistance across the fiber orientation.
• C: No universal ranking follows without material data and load definitions.
• D (correct): Fiber composites are often direction-dependent and may be weaker across reinforcement.

Answer: C

Correct answer: 10,000 J. The heat needed to change a material's temperature is

× ΔT, where c, the specific heat, is the energy to warm 1 kg by 1 °C. Here 2 kg × 500 J per kg per °C × 10 °C = 10,000 J; the kilograms and degrees cancel, leaving joules. A common slip is dropping one factor, such as the temperature change, which gives 1,000 J. A specific heat near 500 J per kg per °C is typical of steel, while water needs about 4,200, which is why water warms so slowly.

• A: This omits the specified ten-degree temperature rise from the product.
• B: This is not obtained by multiplying all three required factors.
• C (correct): Q = m c ΔT = 2 × 500 × 10, giving 10,000 joules.
• D: This value understates the given energy requirement by two orders of magnitude.

Answer: A

Correct answer: Fatigue resistance under cyclic loads. Fatigue is damage from many repeated load cycles. Tiny cracks can start at scratches, holes or sharp corners where stress concentrates, then grow a little with each cycle even when the peak stress stays below the yield strength. Eventually the remaining section fractures suddenly, often with little warning. That is why a single static test does not prove a vibrating shaft is safe; designers use fatigue data such as S-N curves, which plot stress against cycles to failure. The misconception is that no yielding means no damage.

• A (correct): Fatigue damage can accumulate during repeated loads even below a single-test yield criterion.
• B: One-time elastic behavior does not guarantee survival under unlimited cycles.
• C: Surface flaws and stress concentrations can influence fatigue performance.
• D: Coating appearance is not sufficient to assess repeated-load cracking.

Answer: B

Correct answer: At least some permanent plastic deformation likely occurred. When a metal is loaded within its elastic range, atomic bonds stretch and spring back, so the part returns to its original shape once the load is removed. Beyond the yield point, planes of atoms slip past one another through dislocation movement, and that change does not reverse. A bend that remains after unloading, called permanent set, is therefore the signature of plastic deformation. The tempting error is concluding the response was entirely elastic; elastic strain would have recovered. Some springback still occurs, which is why parts are slightly over-bent.

• A: Pure elastic deformation would recover rather than leave the specified set.
• B (correct): Remaining permanent set is evidence the response exceeded purely recoverable elasticity.
• C: Plastic bending does not require the metal to reach its melting temperature.
• D: Electrical properties are not established by a visible permanent bend.

Answer: B

Correct answer: Half as large. Average normal stress is σ = F/A, so for a fixed force, stress is inversely proportional to area. Doubling the area spreads the same load over twice as much material, cutting the stress to half. This is why heavier loads call for thicker cables and larger bolts. The tempting misconception is that equal forces mean equal stress; force and stress are different quantities, and stress always accounts for how much area carries the load. The change is a simple factor of two, not four, because the area itself doubled.

• A: Equal forces do not create equal stress when areas differ.
• B (correct): Under the simple axial model, sigma = F/A, so doubling area halves stress.
• C: Increasing area does not multiply stress in the stated inverse relation.
• D: Stress decreases with larger area for a fixed axial force.

Answer: D

Correct answer: Magnetic attraction is a clue, but not proof of a particular metal grade. Iron, nickel, cobalt and many of their alloys, including most carbon steels and some stainless steels, are attracted to magnets. Meanwhile, common austenitic stainless steels are usually only weakly magnetic or nonmagnetic, though cold working can increase their response. A magnet test therefore narrows the possibilities but cannot identify a specific grade. The tempting conclusion is that attraction proves pure iron or a certain alloy; confirming a grade requires markings, records or laboratory analysis. Shiny plating can also hide the base metal.

• A: Steels, nickel, cobalt and many alloys are also attracted to magnets, so attraction does not prove pure iron.
• B: Magnetic response says nothing about resistance to moisture or chemical attack.
• C: A magnetically attracted part can be ferrous; the label is not excluded.
• D (correct): Magnetic behavior depends on composition and condition, and appearance can mislead.

Answer: C

Correct answer: Some metal, wood and plastic dusts can be combustible or explosible in suitable conditions. Combustible dust consists of fine particles with a huge surface area relative to their mass, so they can burn extremely fast when dispersed in air. A dust explosion needs fuel, oxygen, an ignition source, dispersion and confinement. Materials that are hard to ignite as solid pieces, including aluminum, wood and many plastics, can burn violently as a dust cloud. The misconception is that a material safe in bulk is safe as powder. Good housekeeping, approved dust collection and control of ignition sources reduce the hazard.

• A: Unapproved vacuum equipment can introduce ignition or dust-handling hazards.
• B: Some metals can form combustible dust even when larger pieces do not ignite easily.
• C (correct): OSHA warns that finely divided combustibles can become hazardous under appropriate conditions.
• D: Color does not define inhalation or combustible dust hazards.

Answer: D

Correct answer: Check both material modulus and beam geometry/support conditions. Beam stiffness depends on both material and shape. For a simple beam, deflection grows with the load and with the span raised to a power, the cube for a single point load, and shrinks as the modulus E times the area moment of inertia I increases. Making a beam deeper raises I sharply, which is why I-beams place material far from the center. The misconception is that a hard or dense material alone solves stiffness; Mohs hardness and density do not measure resistance to bending.

• A: Scratch hardness is not a direct predictor of whole-beam stiffness.
• B: Length and cross-section can strongly alter beam deflection.
• C: Density does not substitute for stiffness and shape requirements.
• D (correct): Deflection depends on modulus together with dimensions, span and support/loading arrangement.

Answer: B

Correct answer: About 149 MPa. Stress is force divided by cross-sectional area. With a radius of 8 mm,

mm².

Always halve the diameter before squaring.

• A: This comes from using the 16 mm diameter as the radius, which makes the area four times too large.
• B (correct): The radius is 8 mm, so the area is about 201 mm² and 30,000 ÷ 201 is about 149 MPa.
• C: This divides by 16 × 16 = 256 mm², the area of a square, and forgets the pi/4 factor for a circle.
• D: This divides by the circumference (about 50 mm), which is a length, not the cross-sectional area.

Answer: C

Correct answer: Resistance to corrosion in salt water. Chemical properties describe how a material reacts with its environment, such as rusting or attack by salt water, acids or solvents. Melting point and conductivity are physical properties, and hardness is a mechanical property.

• A: Melting point can be found without changing what the metal is made of, so it is a physical property.
• B: Hardness describes the response to an indenting force, so it is a mechanical property.
• C (correct): Corrosion is a reaction between the metal and its surroundings, which makes it a chemical property.
• D: How fast heat flows through a material is observed without changing it, so it is a physical property.

Answer: C

Correct answer: 0.9; it floats. Find density first: 45 g ÷ 50 cm³ = 0.9 g/cm³. Divide by water's 1.0 g/cm³ to get SG = 0.9, a pure number. Because it is less than 1, the block floats.

• A: The number is right, but specific gravity is a ratio and carries no units such as grams.
• B: This divides volume by mass instead of mass by volume, flipping the ratio.
• C (correct): Density is 45 ÷ 50 = 0.9 g/cm³, so SG = 0.9 with no units, and a value below 1 means it floats.
• D: The SG is right, but a value below 1 means the block is less dense than water, so it floats.

Answer: B

Correct answer: 10.2 mm. Convert the length to millimeters, then use

• A: This is ten times too small, a decimal slip in the multiplication.
• B (correct): With 12 m = 12,000 mm, the growth is 0.000017 × 12,000 × 50 = 10.2 mm.
• C: This uses 12 for the length (meters) but labels the answer in millimeters; 0.0102 m is really 10.2 mm.
• D: This is ten times too large, a decimal slip when handling the $10^{-6}$ factor.

Answer: B

Correct answer: Anneal it by heating and cooling slowly. Bending metal cold makes it harder, stronger and less ductile; this is work hardening. Annealing, heating the metal and letting it cool slowly, softens it again so it can be formed without cracking.

• A: Quenching hardens steel and makes it more brittle, which would make cracking worse.
• B (correct): Cold bending work-hardened the metal; annealing restores softness and ductility so forming can continue.
• C: More cold work hardens the metal further and drives the cracks to failure.
• D: A coating is a surface layer; it does not change the hardness or ductility of the steel underneath.

Answer: D

Correct answer: 0.9 mm. Step 1: stress = 24,000 N ÷ 200 mm² = 120 MPa.

Step 2: strain = stress ÷

MPa = 0.0006.

Step 3: stretch = 0.0006 × 1,500 mm = 0.9 mm. Keep E in MPa and the length in millimeters.

• A: This divides 120 MPa by 200 without converting GPa to MPa, giving an impossible strain of 0.6.
• B: The strain of 0.0006 is right, but it was multiplied by 1.5 (meters) and then labeled as millimeters.
• C: This uses 20,000 MPa for the modulus, a slip of one zero in the GPa-to-MPa conversion.
• D (correct): Stress is 120 MPa, strain is 120 ÷ 200,000 = 0.0006, and 0.0006 × 1,500 mm = 0.9 mm.

Answer: A

Correct answer: Chlorides break down the protective chromium oxide film in spots. Stainless steel stays clean because chromium forms a self-repairing oxide film. Chlorides from salt water or pool chemicals can break that film in small spots, causing pitting, so the grade must be matched to the environment.

• A (correct): Stainless resists rust through a thin oxide film, but chlorides from salt or pool chemicals can attack it locally and cause pitting.
• B: Stainless steel is corrosion resistant, not rust-proof; real stainless grades can still pit in chlorides.
• C: Fatigue produces cracks that grow from stress raisers under cycling loads, not scattered corrosion pits.
• D: Chromium is added for corrosion resistance; surface hardness does not explain pits that form only in the pool area.

Answer: C

Correct answer: 3.93 kg. Identical shape means identical volume. Volume = 1,350 g ÷ 2.7 g/cm³ = 500 cm³.

Steel mass = 7.85 g/cm³ × 500 cm³ = 3,925 g, or about 3.93 kg.

• A: This flips the density ratio, multiplying by 2.7 ÷ 7.85, which wrongly makes the steel part lighter.
• B: This is only the extra mass the steel version adds (5.15 g/cm³ × 500 cm³), not its total mass.
• C (correct): The volume is 1,350 ÷ 2.7 = 500 cm³, and 500 × 7.85 = 3,925 g, about 3.93 kg.
• D: This multiplies the aluminum mass by steel density, skipping the step of finding the volume first.

Answer: A

Correct answer: Nitrile. Each elastomer resists different things: nitrile handles oil and fuel, silicone handles heat, and EPDM handles weather and ozone. For an oil and gasoline seal, nitrile is the right choice.

• A (correct): Nitrile rubber is the elastomer known for handling oil and fuel.
• B: EPDM is chosen for weather and ozone resistance, but it swells in oil and fuel.
• C: Silicone is valued for heat resistance, not for resistance to fuel and oil.
• D: Natural rubber is stretchy and tough but is attacked by oils and fuels.

Answer: D

Correct answer: Shear. Shear makes one layer of material move sideways past the next. A pin or bolt joining two parts that are pulled in opposite directions across it is loaded in shear, like a pair of scissors.

• A: Tension would stretch the pin along its length, but here the force acts across the pin.
• B: Compression squeezes a part along its length, as in a column, which is not how this pin is loaded.
• C: Torsion is a twisting load like that on a drive shaft; the pin is not being twisted.
• D (correct): The two parts push opposite ways across the pin, trying to slice it, which is shear.

Answer: D

Correct answer: Brass. Ferrous metals have iron as their main ingredient, which includes cast iron and every kind of steel. Brass is a copper alloy, so it is nonferrous.

• A: Cast iron is mostly iron with 2 to 4 percent carbon, so it is ferrous.
• B: Stainless steel is iron-based with added chromium, so it is ferrous even when it is not magnetic.
• C: Tool steel is an iron-carbon alloy, so it belongs to the ferrous family.
• D (correct): Brass is copper plus zinc and contains little or no iron, so it is nonferrous.

Answer: B

Correct answer: Tempering. Hardening by quenching leaves steel hard but brittle. Tempering reheats it to a lower temperature, trading a little hardness for the toughness a struck tool needs.

• A: Annealing heats and cools slowly to make the steel soft for machining, not to fine-tune a hardened tool.
• B (correct): Tempering reheats hardened steel to a lower temperature, giving up some hardness to gain toughness.
• C: Galvanizing is a zinc coating for corrosion protection and does not change the steel inside.
• D: Work hardening comes from bending or hammering cold metal, not from reheating it.

Answer: A

Correct answer: 324 kJ. Use

with the temperature change, not the final temperature:

• A (correct): The rise is 180 − 20 = 160 °C, and 4.5 × 450 × 160 = 324,000 J = 324 kJ.
• B: This uses the final temperature of 180 °C as the temperature change instead of the 160 °C rise.
• C: This enters the mass as 4,500 g while c is per kilogram, making the answer 1,000 times too large.
• D: This adds the two temperatures (200 °C) instead of subtracting them to find the change.

Answer: D

Correct answer: Reject it or use it where it carries little load. A knot is the base of a branch; the grain bends around it and makes a weak spot. On the tension edge of a loaded member it is especially dangerous, so the piece should be rejected or placed where it carries little load.

• A: Knots bend the grain around them and create a real weak spot, so they are not just an appearance issue.
• B: Paint may seal the surface, but it does not restore the strength lost where the grain bends around the knot.
• C: Drilling leaves a hole in the tension edge, which removes even more material and weakens the joist further.
• D (correct): A large knot on a tension edge is a serious weak point, so the joist should be rejected or moved to a low-load spot.