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GCE A/L Engineering Technology study materials
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GCE A/L Engineering Technology practice questions

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Question 1Engineering Graphics and Design Communication

In engineering drawing practice, which projection symbol indicates that the object is placed in the first angle relative to the planes of projection?

  1. A truncated cone with the smaller circle drawn on the side of the larger circle Correct answer.
  2. A square divided by two diagonal lines forming four triangles
  3. Two concentric circles with the inner circle drawn as a hidden line
  4. A truncated cone with the larger circle drawn on the side of the smaller circle

Explanation

The first-angle symbol shows a truncated cone whose smaller end faces the viewer, with the front view (small circle) appearing beside the side view (large circle). This reflects the object being placed between the observer and the plane of projection.

Question 2Engineering Graphics and Design Communication

A drawing is produced at a scale of 1:20. A feature measures 35 mm on the drawing. What is its actual length?

  1. 0.7 mm
  2. 1.75 mm
  3. 700 mm Correct answer.
  4. 175 mm

Explanation

A scale of 1:20 means one unit on the drawing represents twenty units on the object, so the drawing length is multiplied by 20. Here 35 mm × 20 = 700 mm, which is 0.7 m.

Question 3Engineering Graphics and Design Communication

Which line type is used in a technical drawing to represent the centre line of a symmetrical component?

  1. A continuous thick line
  2. A short-dashed thin line
  3. A continuous thin line
  4. A long-dashed thin line with short dashes between Correct answer.

Explanation

Centre lines are drawn as a chain of long dashes separated by short dashes, in a thin line weight. They mark axes of symmetry and centres of circles, and must extend slightly beyond the outline.

Question 4Engineering Graphics and Design Communication

A rectangular block 80 mm long, 50 mm wide and 30 mm high is drawn in orthographic projection. Which view shows the 80 mm × 50 mm face true size?

  1. The end view
  2. The front view
  3. The auxiliary view
  4. The plan view Correct answer.

Explanation

The plan view is projected onto the horizontal plane and shows the length and width of the block true size. The front view would show 80 mm × 30 mm and the end view 50 mm × 30 mm.

Question 5Engineering Graphics and Design Communication

In a drawing, the title block is normally placed in which position on the sheet?

  1. Centred along the top edge of the frame
  2. In the top left-hand corner of the frame
  3. In the bottom right-hand corner of the frame Correct answer.
  4. Along the left-hand binding margin

Explanation

Standard drawing sheet layouts place the title block in the bottom right-hand corner, where it is read when the sheet is folded with the title panel outermost. It carries the drawing number, title, scale, and approval details.

Question 6Engineering Graphics and Design Communication

A dimension of 45 mm is to be placed on a drawing produced at a scale of 2:1. Which value should be written on the dimension line?

  1. 90
  2. 22.5
  3. 2:1
  4. 45 Correct answer.

Explanation

Dimensions always state the true size of the object regardless of the scale used to draw it. At 2:1 the feature is drawn 90 mm long, but the dimension text remains 45.

Question 7Engineering Graphics and Design Communication

Which type of pictorial drawing keeps all three principal axes equally inclined to the plane of projection so that all three can be measured to the same scale?

  1. Oblique projection
  2. Isometric projection Correct answer.
  3. Perspective projection
  4. Orthographic projection

Explanation

In isometric projection the three axes are equally inclined to the picture plane, so foreshortening is equal and a single scale serves all three. In isometric drawing the same full scale is applied along each axis as a simplified convention.

Question 8Engineering Graphics and Design Communication

A line 120 mm long is to be divided into five equal parts using the geometric construction with a subsidiary line. What is the first practical step after drawing the given line?

  1. Draw a line from one end at any convenient acute angle and step off five equal divisions on it Correct answer.
  2. Construct a 60° angle at each end of the line
  3. Bisect the line with a perpendicular at its midpoint
  4. Draw a semicircle on the line as diameter

Explanation

The standard construction draws a subsidiary line at a convenient angle from one end and marks off five equal steps with dividers. Lines drawn parallel to the last division then cut the given line into five equal parts.

Question 9Engineering Graphics and Design Communication

A regular hexagon is to be constructed on a circle of 60 mm diameter using the compass method. What distance is stepped around the circumference to locate the vertices?

  1. 30 mm Correct answer.
  2. 60 mm
  3. 51.9 mm
  4. 42.4 mm

Explanation

For a regular hexagon the side length equals the radius of the circumscribing circle. The radius is half of 60 mm, so the compass is set to 30 mm and stepped six times around the circle.

Question 10Engineering Graphics and Design Communication

Which statement correctly describes the difference between a first-angle and a third-angle layout of the same object?

  1. The plan appears below the front view in first angle and above it in third angle Correct answer.
  2. The two layouts are identical because projection angle affects only the symbol
  3. The front view is omitted in third angle because the plan replaces it
  4. The plan appears above the front view in first angle and below it in third angle

Explanation

In first-angle projection the object lies between observer and plane, so the plan is projected onto the plane below and appears under the front view. In third angle the plane lies between observer and object, so the plan appears above the front view.

Question 11Engineering Graphics and Design Communication

A sectional view is required for a component with a web. According to standard practice, how is the web treated when the cutting plane passes along its length?

  1. It is removed entirely from the section
  2. It is left unhatched to avoid giving a false impression of thickness Correct answer.
  3. It is hatched in the same direction as the adjacent body
  4. It is drawn with a double chain line

Explanation

Thin features such as webs and ribs are not hatched when cut lengthwise, because hatching would suggest a solid mass rather than a thin stiffening feature. Their outline is still drawn in continuous thick line.

Question 12Engineering Graphics and Design Communication

A drawing carries the tolerance 40 ±0.2. Which pair of limits correctly describes the permissible sizes?

  1. Upper 39.8, lower 40.2
  2. Upper 40.0, lower 39.8
  3. Upper 40.2, lower 39.8 Correct answer.
  4. Upper 40.2, lower 40.0

Explanation

A bilateral tolerance of ±0.2 adds 0.2 to and subtracts 0.2 from the nominal size. The limits are therefore 40.2 maximum and 39.8 minimum, giving a tolerance zone of 0.4 mm.

Question 13Engineering Graphics and Design Communication

In the ISO system of limits and fits, what does the capital letter H designate in a hole basis assembly?

  1. A hole with a clearance fit guaranteed for all shaft letters
  2. A tolerance grade of high precision
  3. A hole whose fundamental deviation is zero at the lower limit Correct answer.
  4. A shaft whose upper deviation is zero

Explanation

The capital letter denotes the hole and the number denotes the tolerance grade. Letter H fixes the lower deviation of the hole at zero, so the minimum hole size equals the basic size.

Question 14Engineering Graphics and Design Communication

A drawing shows a surface finish symbol consisting of a tick with a horizontal bar and the value Ra 3.2. What does the horizontal bar indicate?

  1. The surface requires a protective coating after machining
  2. The surface is to be left as cast and not machined
  3. The surface must be machined by a specified production process Correct answer.
  4. The surface is to be produced by any method at the discretion of the manufacturer

Explanation

Adding a horizontal bar to the basic tick means material removal by machining is required, with the process stated elsewhere or implied. The bare tick alone leaves the production method to the manufacturer's discretion.

Question 15Engineering Graphics and Design Communication

An oblique pictorial drawing is made with the receding axis at 45° and the front face true shape. Which statement about the receding dimensions is correct?

  1. They are halved to compensate for foreshortening
  2. They are drawn at two-thirds scale
  3. They are drawn at full scale along the 45° axis Correct answer.
  4. They are projected to a vanishing point

Explanation

In cavalier oblique drawing the receding edges are drawn at full scale, which is the simplest convention. Cabinet oblique differs by halving the receding dimensions to reduce apparent distortion.

Question 16Engineering Graphics and Design Communication

A drawing of a stepped shaft shows a diameter of 25 mm with a total tolerance of 0.05 mm. Which tolerance expression is written correctly to BS/ISO convention?

  1. 25 +0.05/-0.05
  2. 25 +0.025/-0.025 Correct answer.
  3. 25 ±0.05
  4. 25 H7

Explanation

A total tolerance of 0.05 mm split symmetrically gives deviations of plus and minus 0.025 mm about the nominal size. Writing ±0.05 would give a total zone of 0.1 mm, twice the requirement.

Question 17Engineering Graphics and Design Communication

A cutting plane passes fully through a symmetrical component along its axis, and the resulting view is placed in the normal front-view position. Which section is this?

  1. A half section
  2. A removed section
  3. A full section Correct answer.
  4. An offset section

Explanation

A full section results when the cutting plane passes completely through the object, and the whole view is shown in section. The cutting plane line and arrows indicate the direction of viewing.

Question 18Engineering Graphics and Design Communication

A drawing must show the true shape of an inclined face on a component. Which construction is required?

  1. An auxiliary view projected perpendicular to the inclined face Correct answer.
  2. An enlarged detail view of the inclined face
  3. A sectional view taken through the inclined face
  4. A second plan view rotated through 90°

Explanation

An auxiliary view is projected onto a plane parallel to the inclined face, so that face appears true shape and true size. Rotating or sectioning does not remove the foreshortening caused by inclination.

Question 19Engineering Graphics and Design Communication

In a drawing, a circular hole of 20 mm diameter is shown in the plan as an ellipse with a major axis of 26 mm. What does this indicate about the hole?

  1. The hole is actually elliptical in cross-section
  2. The hole is countersunk on the far side
  3. The hole is drawn at a reduced scale in that view only
  4. The hole lies in a face inclined to the plane of the view Correct answer.

Explanation

A circle appears as an ellipse when its plane is inclined to the plane of projection, with the major axis equal to the true diameter only in a true-shape view. The apparent 26 mm major axis shows the view is not perpendicular to the hole axis.

Question 20Engineering Graphics and Design Communication

A component is to be dimensioned so that a chain of three dimensions locates three holes from a common datum. Which dimensioning approach avoids the accumulation of tolerances?

  1. Dimensioning each hole with its own bilateral tolerance only
  2. Baseline dimensioning from a single datum Correct answer.
  3. Combined dimensioning using both methods equally
  4. Chain dimensioning from hole to hole

Explanation

Baseline dimensioning locates every feature from one common datum, so errors and tolerances do not accumulate along the chain. Chain dimensioning adds each successive tolerance, enlarging the possible error at the far end.

Question 21Engineering Graphics and Design Communication

A hole and shaft are specified as 30 H7/g6. Which statement best describes the resulting fit?

  1. An interference fit in which the shaft is always larger than the hole
  2. A line fit in which the limits coincide exactly
  3. A transition fit that may be either clearance or interference
  4. A clearance fit in which the hole is always larger than the shaft Correct answer.

Explanation

H7 gives a hole whose lower deviation is zero, while g6 gives a shaft whose upper deviation is negative. The shaft is therefore always smaller than the hole, producing clearance under all permitted sizes.

Question 22Engineering Graphics and Design Communication

A drawing is to be produced at A3 size from an original A1 drawing using standard ISO A-series reduction. Which scale should be stated in the title block?

  1. 1:2
  2. 4:1
  3. 1:8
  4. 1:4 Correct answer.

Explanation

Each step down the A series halves the area, so A1 to A3 is two steps and the linear scale is 1:2 twice, giving 1:4. The stated scale is the ratio of drawing length to true length.

Question 23Engineering Graphics and Design Communication

A tolerance is specified as 50 +0.03/-0.01. Which statement about the tolerance zone is correct?

  1. The zone is 0.02 mm wide and lies wholly below the basic size
  2. The zone is 0.04 mm wide and is centred on the basic size
  3. The zone is 0.04 mm wide and lies wholly above the basic size Correct answer.
  4. The zone is 0.03 mm wide and extends from basic size upward

Explanation

The zone width is the sum of the magnitudes of the deviations, 0.03 + 0.01 = 0.04 mm. Because both deviations are positive, the whole zone lies above the basic size of 50 mm.

Question 24Engineering Graphics and Design Communication

A drawing note specifies that a feature is controlled by a geometric tolerance with a cylindrical tolerance zone. Which characteristic requires the diameter symbol in the feature control frame?

  1. Parallelism of two plane faces
  2. Position of a hole axis Correct answer.
  3. Angularity of a surface
  4. Flatness of a face

Explanation

A cylindrical tolerance zone is appropriate for an axis, so the position tolerance for a hole is preceded by the diameter symbol. Tolerances applied to plane surfaces use two parallel planes instead.

Question 25Engineering Materials and Their Properties

In a tensile test on a low-carbon steel specimen, which point on the engineering stress–strain curve marks the onset of localised necking and the start of the fall in load?

  1. The proportional limit
  2. The upper yield point
  3. The 0.2% proof stress point
  4. The ultimate tensile strength point Correct answer.

Explanation

The ultimate tensile strength is the maximum engineering stress the specimen sustains; beyond it, deformation becomes localised in a neck, so the recorded load falls even though the true stress in the neck keeps rising. Necking therefore begins at the UTS point, not at yielding or at the elastic limit.

Question 26Engineering Materials and Their Properties

A Brinell hardness test is carried out with a 10 mm ball and a 3000 kgf load on a thick steel block. Which combination correctly describes the indenter and the nature of the resulting impression?

  1. A steel ball producing a rhombic impression
  2. A diamond pyramid producing a square impression
  3. A diamond cone producing a shallow scratch
  4. A hardened steel or tungsten-carbide ball producing a circular impression Correct answer.

Explanation

The Brinell method presses a hardened ball (steel or tungsten carbide) into the surface and measures the diameter of the resulting circular impression with a microscope; hardness is calculated from load and impression area. Cones and pyramids belong to other hardness scales, and a ball never leaves a rhombic impression.

Question 27Engineering Materials and Their Properties

Which statement correctly describes the crystal structure of austenite (gamma iron)?

  1. Hexagonal close-packed with atoms at the corners of a hexagonal prism
  2. Face-centred cubic with atoms at corners and at each face centre Correct answer.
  3. Body-centred cubic with an atom at each corner and one at the body centre
  4. Simple cubic with atoms only at the eight corners

Explanation

Austenite has a face-centred cubic lattice, which is why it can dissolve far more carbon (up to about 2.1%) than ferrite. The higher packing efficiency and larger interstitial sites of the FCC cell explain this greater solubility.

Question 28Engineering Materials and Their Properties

A material is described as having high toughness. Which property combination does this term represent?

  1. High hardness together with high brittleness
  2. High stiffness together with very low density
  3. High strength together with good ductility, so much energy is absorbed before fracture Correct answer.
  4. High electrical conductivity together with low thermal expansion

Explanation

Toughness is the energy a material absorbs up to fracture, which requires both reasonable strength and the ability to deform plastically. A hard but brittle material absorbs little energy and therefore has low toughness despite its strength.

Question 29Engineering Materials and Their Properties

In a binary phase diagram, what is the significance of the eutectic point?

  1. It is the composition and temperature at which a liquid solidifies simultaneously into two different solid phases. Correct answer.
  2. It is the composition and temperature at which two solid phases transform into a single solid phase.
  3. It is the maximum temperature at which a single solid solution can exist.
  4. It is the temperature at which a pure metal melts at constant pressure.

Explanation

At the eutectic point a liquid of fixed composition solidifies at a single, constant temperature into two distinct solid phases that form a lamellar mixture. This is the lowest melting temperature for any composition in that system.

Question 30Engineering Materials and Their Properties

Which class of polymer can be repeatedly softened by heating and remoulded, because its chains are held together by secondary forces rather than a covalent network?

  1. Elastomer with permanent crosslinks
  2. Thermoplastic polymer Correct answer.
  3. Fully crosslinked phenolic resin
  4. Thermosetting polymer

Explanation

Thermoplastics consist of long chains held by weak van der Waals forces and chain entanglement, so heating allows them to flow and they can be reshaped many times. Crosslinked networks cannot be remelted without degrading the covalent bonds.

Question 31Engineering Materials and Their Properties

Which of the following best explains why aluminium is preferred over steel for aircraft skin panels?

  1. Aluminium alloys offer a high strength-to-weight ratio, giving stiff, light structures. Correct answer.
  2. Aluminium is harder than steel and resists abrasion better.
  3. Aluminium has a higher melting point and higher modulus than steel.
  4. Aluminium has a higher density, which improves stability in flight.

Explanation

The key advantage of aluminium alloys in airframes is their favourable strength-to-weight ratio, achieved with densities about one third that of steel. Lower mass reduces the structural loads and fuel burn for a given strength requirement.

Question 32Engineering Materials and Their Properties

A steel shaft must be surface-hardened while retaining a tough, shock-resistant core. Which treatment is most appropriate?

  1. Full annealing to relieve stresses and soften the steel
  2. Case carburising followed by quenching and tempering Correct answer.
  3. Electroplating the shaft with a thin layer of zinc
  4. Through-hardening by quenching the whole section from the austenitising temperature

Explanation

Carburising enriches the surface with carbon so that, after quenching, only the outer case transforms to hard martensite while the low-carbon core remains tough. This combination resists wear at the surface and absorbs shock in the core.

Question 33Engineering Materials and Their Properties

A copper–nickel alloy is required to have a yield strength of about 200 MPa while remaining ductile. The engineer notes that increasing nickel content raises strength but reduces ductility. Which approach best reflects sound material selection practice?

  1. Select the highest nickel content available, since strength is the only criterion.
  2. Choose the lowest nickel content, ignoring the strength requirement.
  3. Choose the composition that meets the strength requirement with the least loss of ductility. Correct answer.
  4. Select the composition with the highest ductility regardless of strength.

Explanation

Material selection involves balancing competing properties against the design requirement, so the alloy should just meet the strength target while preserving as much ductility as possible. Maximising a single property usually sacrifices another needed characteristic.

Question 34Engineering Materials and Their Properties

A specimen of steel with a gauge length of 50 mm extends to 62 mm before fracture. What is the percentage elongation?

  1. 20%
  2. 12%
  3. 31%
  4. 24% Correct answer.

Explanation

Percentage elongation equals the increase in length divided by the original gauge length, multiplied by 100. The increase is 12 mm on a 50 mm gauge, giving 12/50 × 100 = 24%.

Question 35Engineering Materials and Their Properties

A 20 mm diameter steel rod carries a tensile load of 31.4 kN. Taking pi as 3.14, what is the engineering tensile stress in the rod?

  1. 200 MPa
  2. 50 MPa
  3. 100 MPa Correct answer.
  4. 400 MPa

Explanation

Stress equals load divided by cross-sectional area. The area is 3.14 × (10 mm)^2 = 314 mm², so 31 400 N / 314 mm² = 100 N/mm², which is 100 MPa.

Question 36Engineering Materials and Their Properties

Which statement correctly distinguishes a dislocation from a grain boundary in a metal?

  1. Both are point defects caused by missing atoms in the lattice.
  2. A dislocation is a planar defect, whereas a grain boundary is a line defect.
  3. A grain boundary is a vacancy cluster, whereas a dislocation is an interstitial atom.
  4. A dislocation is a line defect that allows slip at stresses far below the theoretical shear strength, whereas a grain boundary separates crystals of different orientation. Correct answer.

Explanation

Dislocations are one-dimensional line defects whose motion produces plastic deformation at stresses well below the theoretical shear strength. Grain boundaries are two-dimensional interfaces between neighbouring crystals with different orientations.

Question 37Engineering Materials and Their Properties

A plain-carbon steel contains 0.8% carbon and is slowly cooled from the austenite region. Which microstructure is expected at room temperature?

  1. Pure ferrite with no carbide phase
  2. Free cementite network surrounding martensite
  3. Coarse pearlite, consisting of alternating ferrite and cementite lamellae Correct answer.
  4. Single-phase austenite retained at room temperature

Explanation

A steel of eutectoid composition (about 0.8% carbon) transforms on slow cooling into pearlite, a lamellar mixture of ferrite and cementite. No pro-eutectoid ferrite or cementite forms because the composition matches the eutectoid point.

Question 38Engineering Materials and Their Properties

Which mechanism best explains why fine-grained metals are generally stronger than coarse-grained metals at room temperature?

  1. Finer grains contain more vacancies, which pin dislocations.
  2. Grain boundaries act as barriers to dislocation motion, and a finer grain size provides more barriers per unit volume. Correct answer.
  3. Finer grains have a lower elastic modulus, which increases the yield strength.
  4. Grain boundaries lower the melting point, which raises the yield stress.

Explanation

Slip must transfer across grain boundaries, and boundaries of differing orientation obstruct dislocation motion. Refining the grain size increases the boundary area per unit volume, so a higher stress is needed to propagate slip, as described by the Hall–Petch relationship.

Question 39Engineering Materials and Their Properties

Which set of properties is characteristic of ceramic materials such as alumina?

  1. High density, high ductility and low compressive strength
  2. High ductility, low hardness and high thermal conductivity
  3. High hardness, high compressive strength and low electrical conductivity Correct answer.
  4. Low melting point, high toughness and high electrical conductivity

Explanation

Ionic and covalent bonding in ceramics produces high hardness, good compressive strength and resistance to heat, with electrons localised so electrical conductivity is low. Their brittleness and low tensile strength follow from the absence of slip systems.

Question 40Engineering Materials and Their Properties

A component made from a thermoplastic is found to creep excessively under a sustained load at 60 °C. Which change is most likely to improve its long-term performance?

  1. Adding a plasticiser to increase chain mobility
  2. Selecting a polymer with a higher glass transition temperature and greater crystallinity Correct answer.
  3. Increasing the applied stress to stiffen the chains
  4. Reducing the molecular weight of the polymer

Explanation

Creep resistance improves when the service temperature is well below the glass transition temperature and when crystalline regions restrain chain sliding. A higher Tg and greater crystallinity therefore reduce time-dependent deformation.

Question 41Engineering Materials and Their Properties

Which statement correctly describes the difference between hardness and hardenability of steel?

  1. Hardness is the resistance to indentation, whereas hardenability is the depth to which a steel can be hardened by quenching. Correct answer.
  2. Hardness and hardenability are interchangeable terms for the same property.
  3. Hardness depends only on carbon content, whereas hardenability depends only on the quenching medium.
  4. Hardness is the depth of the hardened case, whereas hardenability is the resistance to scratching.

Explanation

Hardness measures resistance to local indentation and is governed mainly by the carbon content and microstructure. Hardenability describes how readily a steel forms martensite through the section during quenching and depends largely on alloying elements.

Question 42Engineering Materials and Their Properties

A galvanised steel sheet is exposed to a marine atmosphere. Which mechanism protects the steel at a scratched area where the coating is removed?

  1. Zinc is anodic to steel, so it corrodes preferentially and protects the exposed steel cathodically. Correct answer.
  2. Steel is anodic to zinc, so the steel corrodes preferentially at the scratch.
  3. The zinc coating blocks oxygen but provides no protection once broken.
  4. The zinc oxide layer is electrically insulating and stops all current flow.

Explanation

Zinc has a more negative electrode potential than steel, so in the presence of an electrolyte it becomes the anode and dissolves while the exposed steel acts as the cathode. This sacrificial protection continues even where the coating is damaged.

Question 43Engineering Materials and Their Properties

Which non-destructive test is best suited to detecting internal planar defects such as lack-of-fusion in a thick steel weld?

  1. Visual inspection
  2. Ultrasonic testing Correct answer.
  3. Magnetic particle inspection
  4. Dye penetrant inspection

Explanation

Ultrasonic testing transmits high-frequency waves through the material and detects echoes from internal reflectors, making it effective for subsurface planar defects in thick sections. Surface methods cannot reach internal flaws.

Question 44Engineering Materials and Their Properties

A steel component in service at 400 °C is found to have failed progressively under a constant stress well below its room-temperature yield strength. Which mechanism is responsible?

  1. Hydrogen embrittlement at room temperature
  2. Corrosion fatigue caused by a marine environment
  3. Fatigue failure caused by cyclic loading
  4. Creep deformation and rupture at elevated temperature Correct answer.

Explanation

Creep is time-dependent plastic deformation under sustained stress at elevated temperature, and it can lead to rupture even when the stress is below the yield strength. The constant load and high temperature are the defining conditions.

Question 45Engineering Materials and Their Properties

In a Charpy V-notch impact test, a steel specimen absorbs 15 J at 20 °C and 90 J at 60 °C. Which conclusion is best supported?

  1. The steel shows a ductile-to-brittle transition, becoming more brittle at lower temperature. Correct answer.
  2. The absorbed energy increases because the specimen's yield strength rises with temperature.
  3. The steel is notch-insensitive at all temperatures.
  4. The test measures hardness rather than toughness.

Explanation

The large fall in absorbed energy as temperature decreases is the classic ductile-to-brittle transition of ferritic steels, indicating a risk of brittle fracture in cold service. Impact energy, not hardness, is the measured quantity.

Question 46Engineering Materials and Their Properties

An alloy must combine high specific stiffness with good fatigue resistance for a bicycle frame tube. Which material choice best meets these requirements?

  1. Soda-lime glass
  2. Gray cast iron
  3. Aluminium alloy or carbon-fibre composite Correct answer.
  4. Unreinforced polypropylene

Explanation

Aluminium alloys and carbon-fibre composites offer high stiffness-to-density ratios and good fatigue performance, which is why they dominate lightweight frame construction. The other materials are either too dense, too weak or too brittle.

Question 47Engineering Materials and Their Properties

A steel is quenched from the austenitising temperature fast enough to avoid the pearlite and bainite noses of the TTT diagram. Which transformation product forms, and why is it hard?

  1. Spheroidite, because cementite particles coarsen during the quench.
  2. Coarse pearlite, because slow diffusion allows thick cementite plates to form.
  3. Ferrite with dispersed graphite, because carbon diffuses out during rapid cooling.
  4. Martensite, because carbon is trapped in a supersaturated body-centred tetragonal lattice that resists dislocation motion. Correct answer.

Explanation

Rapid quenching suppresses diffusion-controlled transformations, so austenite transforms displacively to martensite with carbon trapped in a strained tetragonal lattice. The distorted lattice and high dislocation density impede slip, giving high hardness.

Question 48Engineering Materials and Their Properties

A component is designed with a safety factor of 2 on yield strength and operates under a fully reversed bending stress. Cracks are observed to initiate at a sharp fillet after many cycles. Which explanation is most consistent with the failure?

  1. The static safety factor guarantees infinite life, so the cracks must be a material defect.
  2. Stress concentration at the fillet raises the local cyclic stress, accelerating fatigue crack initiation despite the nominal safety factor. Correct answer.
  3. Fatigue cannot occur below the yield strength, so the observation is impossible.
  4. The failure is caused by creep, because the load is fully reversed.

Explanation

Fatigue is driven by local cyclic stress, and geometric discontinuities concentrate stress well above the nominal value. A safety factor on yield strength does not protect against fatigue, so a sharp fillet can initiate cracks after many cycles.

Question 49Engineering Mechanics and Applied Mechanics

A force of 80 N acts at 30° above the horizontal. What is the horizontal component of this force?

  1. 69.3 N Correct answer.
  2. 40.0 N
  3. 46.2 N
  4. 80.0 N

Explanation

The horizontal component is F cos θ = 80 × cos 30° = 80 × 0.866 = 69.3 N. Resolving a force into perpendicular components uses cosine for the component adjacent to the given angle.

Question 50Engineering Mechanics and Applied Mechanics

A body moves in a straight line with constant acceleration. Which quantity remains unchanged throughout the motion?

  1. Displacement
  2. Speed
  3. Velocity
  4. Acceleration Correct answer.

Explanation

Constant acceleration means the rate of change of velocity is fixed, so acceleration itself does not vary with time. Velocity and speed change continuously, and displacement accumulates as the body moves.