Nepal Engineering Council · Mechanical Engineering · Chapter 6
Mechanical Design
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182 questions in 6 syllabus topics.
6.1 Design classification
31 questions · AMeE0601
1. In the usual engineering design process, which step immediately follows the recognition of a need and the definition of the problem?
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Think about what must happen before any analysis can be done.
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Answer: D. Synthesis, i.e. generating alternative concepts
After the need and problem are defined, designers generate (synthesise) alternative solution concepts, then analyse, evaluate and finally document the chosen design.
2. A design requirement is best described as a statement of
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It is about 'what', not 'how'.
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Answer: A. what the product must do and the limits it must satisfy
Requirements state the needed functions, performance and constraints, not how they will be achieved; the how is decided later in design.
3. Which of the following is NOT normally a design consideration for a machine component?
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Look for the option unrelated to the component itself.
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Answer: A. Share price of the manufacturing company
Strength, rigidity, wear, corrosion and manufacturability are technical design considerations; a company's share price is not.
4. In design, a requirement that must be satisfied or the design is rejected is called a
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Mandatory versus optional.
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Answer: A. constraint (demand)
Demands or constraints are mandatory; wishes are desirable but can be traded off.
5. The factor of safety for a ductile material under static load is generally defined as
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Which stress marks failure in a ductile material?
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Answer: D. yield strength divided by working (allowable) stress
For ductile materials the failure criterion is yielding, so FOS = σy / σworking; for brittle materials ultimate strength is used.
6. The factor of safety should be chosen higher when
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Risk and uncertainty go together with margin.
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Answer: C. loads are uncertain and failure would be dangerous
Uncertainty in loads or material properties and severe consequences of failure justify larger factors of safety.
7. The main benefit of interchangeability achieved through standardization is that
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Think about spare parts.
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Answer: D. parts made at different places can be assembled or replaced without fitting
Standard dimensions and tolerances let any part of a type replace another, easing assembly, repair and mass production.
8. Which of the following best describes a 'code' in engineering practice?
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Safety rules for a whole class of equipment.
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Answer: B. A set of rules for design, manufacture and inspection to ensure safety
Codes (e.g. boiler or pressure vessel codes) lay down minimum requirements for safe design, fabrication and inspection; standards give specifications for items, tests and dimensions.
9. Using standard components such as bolts and bearings in a design mainly
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Think about availability and price.
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Answer: B. reduces cost and makes replacement easy
Standard parts are mass-produced, cheap and readily available, which simplifies design and maintenance.
10. ISO stands for
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The 'S' is for Standardization.
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Answer: A. International Organization for Standardization
ISO is the worldwide federation of national standards bodies and publishes international standards.
11. ISO 9001 specifies requirements for a
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Think of the most famous ISO 'quality' standard.
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Answer: B. quality management system
ISO 9001 deals with quality management; ISO 14001 with the environment and ISO 45001 with occupational health and safety.
12. ISO 14001 is a standard for
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The number 14 (000) series is associated with the environment.
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Answer: D. environmental management systems
ISO 14001 sets requirements for an environmental management system.
13. The ISO system of limits and fits for holes and shafts is given in
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It is a three-digit standard number.
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Answer: A. ISO 286
ISO 286 defines the system of tolerance grades (IT) and fundamental deviations used for limits and fits.
14. ISO 2768 is mainly used to specify
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Think of tolerances not written on every dimension.
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Answer: A. general tolerances for dimensions without individual tolerance indication
ISO 2768 gives general tolerances for linear and angular dimensions that carry no individual tolerance note on the drawing.
15. A metric thread designated M10 indicates a
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'M' stands for metric.
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Answer: B. nominal major diameter of 10 mm
In the ISO metric thread designation, M is followed by the nominal (major) diameter in mm; pitch is given separately for fine threads.
16. Which organisation is the national standards body of Nepal?
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The word 'Standards' is in its name.
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Answer: C. Nepal Bureau of Standards and Metrology
NBSM prepares and promotes Nepal Standards and metrology services.
17. AGMA standards are chiefly concerned with
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The first letter G stands for the component.
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Answer: A. gears and gear drives
The American Gear Manufacturers Association publishes gear rating and design standards.
18. Which pair correctly matches a standards organisation with its country?
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Match each acronym to the language of its name.
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Answer: C. DIN and Germany
DIN is the German standards body; ANSI is American, BS British, JIS Japanese.
19. In a hole-basis system of fits, the hole has
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The H hole has its lower limit at the basic size.
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Answer: D. a lower deviation of zero for all fits
In the hole-basis system the hole is H (lower deviation zero) and the shaft is varied to get the required fit.
20. Which of the following is an interference fit?
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Compare the sizes of shaft and hole.
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Answer: D. A shaft always larger than the mating hole
In an interference fit the maximum hole size is smaller than the minimum shaft size, so assembly needs force.
21. A long slender shaft must be designed so that its deflection under load is limited. The design is based on
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Deflection is the governing quantity.
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Answer: C. rigidity (stiffness) rather than strength
When deformation limits govern, the part is designed for stiffness (rigidity) and checked for strength afterwards.
22. Design for manufacture and assembly (DFMA) aims to
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Fewer parts, simpler operations.
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Answer: A. simplify the design so it is cheaper to make and assemble
DFMA reduces part count and simplifies operations to reduce cost and errors.
23. Reliability of a system is defined as the probability that it will
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It is a probability, tied to time.
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Answer: C. perform its intended function for a stated time under stated conditions
Reliability is a probability of satisfactory performance for a specified period under specified conditions.
24. A steel has a yield strength of 250 MPa. If a factor of safety of 2.5 is used, the allowable stress is
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Divide, do not multiply.
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Answer: C. 100 MPa
σallow = σy / FOS = 250 / 2.5 = 100 MPa.
25. A tie rod carries an axial load of 20 kN. If the allowable stress is 100 MPa, the minimum cross-sectional area is
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Area = force / stress (use N and mm).
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Answer: B. 200 mm²
A = P / σ = 20 000 N / 100 N/mm² = 200 mm².
26. A component of ultimate strength 400 MPa works at a stress of 80 MPa. The factor of safety based on ultimate strength is
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Strength over working stress.
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Answer: C. 5
FOS = 400 / 80 = 5.
27. Two components of reliability 0.95 and 0.90 work in series. The system reliability is
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Multiply for series.
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Answer: B. 0.855
Series reliability is the product: 0.95 × 0.90 = 0.855.
28. A hole is 50.000 +0.025 mm and the shaft is 50.000 −0.025/−0.050 mm (i.e. limits 49.975 and 49.950 mm). The maximum clearance is
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Largest hole minus smallest shaft.
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Answer: C. 0.075 mm
Max clearance = largest hole − smallest shaft = 50.025 − 49.950 = 0.075 mm.
29. Three parts, each 20 ± 0.05 mm, are stacked in a line. The worst-case tolerance on the total length is
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Worst case means adding all tolerances.
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Answer: B. ± 0.15 mm
Worst-case tolerances add: 3 × 0.05 = 0.15 mm, so total length 60 ± 0.15 mm.
30. In the R5 series of preferred numbers starting at 10 (ratio ≈ 1.6), the terms are 10, 16, 25, 40, ... The next term after 40 is
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Multiply by about 1.6.
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Answer: B. 63
Next term = 40 × 10^(1/5) = 40 × 1.585 ≈ 63.
31. The step ratio of the R10 series of preferred numbers is approximately
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Ten steps per decade.
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Answer: D. 1.26
R10 has ten steps per decade: 10^(1/10) ≈ 1.26 (R5: 1.58, R20: 1.12, R40: 1.06).
6.2 Mechanical components
30 questions · AMeE0602
32. The fundamental law of gearing states that for a constant velocity ratio the common normal at the point of contact must
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The fixed point is called the pitch point.
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Answer: D. pass through the pitch point on the line of centres
A constant angular velocity ratio requires the common normal at every contact point to cut the line of centres at a fixed point, the pitch point.
33. The standard pressure angle commonly used for involute spur gears is
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It is the modern standard value.
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Answer: D. 20°
20° full-depth involute teeth are the usual standard (14.5° is the older one).
34. Which gear type produces axial thrust on its bearings while running?
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Look for inclined teeth.
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Answer: B. Helical gear
Inclined teeth of helical gears give an axial force component that bearings must carry; spur gears have none.
35. A worm gear drive is most suitable when
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One stage, big ratio, crossed shafts.
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Answer: A. a large speed reduction is needed in one stage between shafts at 90°
Worm drives give high single-stage ratios on perpendicular, non-intersecting shafts, with lower efficiency than spur or helical gears.
36. Interference in involute spur gears occurs when
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Few teeth on the pinion is the cause.
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Answer: A. the tip of the gear tooth touches the non-involute part of the pinion flank
Interference arises with too few pinion teeth, so the gear tooth tip cuts into the pinion's non-involute flank below the base circle.
37. A spur gear has a pitch circle diameter of 120 mm and 40 teeth. Its module is
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Module = pitch diameter / number of teeth.
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Answer: B. 3 mm
m = d / z = 120 / 40 = 3 mm.
38. Two meshing spur gears of module 4 mm have 20 and 45 teeth. The centre distance is
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Sum of pitch radii.
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Answer: C. 130 mm
a = m(z1 + z2)/2 = 4 × 65 / 2 = 130 mm.
39. In a belt drive the tight side tension is 1200 N, the slack side tension 400 N and the belt speed 10 m/s. The power transmitted is
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Use the difference of tensions.
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Answer: C. 8 kW
P = (T1 − T2) v = 800 × 10 = 8000 W = 8 kW.
40. For a flat belt on a pulley with coefficient of friction 0.3 and angle of contact 180°, the ratio T1/T2 (e^μθ) is approximately
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Convert 180° to radians first.
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Answer: D. 2.57
θ = π rad, μθ = 0.942, e^0.942 ≈ 2.57.
41. Compared with a flat belt, a V-belt can transmit more power for the same tension because
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Think of the groove shape.
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Answer: A. of the wedging action in the groove
The wedge action increases the normal force on the sides of the groove and hence the friction grip.
42. A timing (toothed) belt is used when
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Think of an engine's camshaft drive.
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Answer: B. positive drive with no slip is needed
Teeth engage grooves in the pulley so the velocity ratio is fixed, as in an engine cam drive.
43. A flexible coupling is used mainly to
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Think of imperfect alignment.
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Answer: D. accommodate small misalignment between connected shafts
Flexible couplings tolerate small misalignment and reduce shock and vibration transmission.
44. An Oldham coupling is used to connect
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Parallel but not collinear.
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Answer: B. parallel shafts with a small lateral offset
An Oldham coupling has a sliding central disc and joins parallel shafts whose axes are slightly offset.
45. A universal (Hooke's) joint is used to connect shafts that
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The axes are not parallel.
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Answer: C. intersect at an angle
A Hooke's joint transmits rotation between shafts with intersecting axes at an angle.
46. Which bearing is best suited to carry combined radial and axial loads?
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The cone-shaped element helps.
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Answer: C. Tapered roller bearing
Tapered rollers carry both radial and thrust loads; cylindrical roller and needle bearings are mainly radial.
47. The basic rating life L10 of a rolling bearing is the life that
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10 is the percentage failing.
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Answer: A. 90% of a group of identical bearings complete or exceed
L10 life is based on 10% failure probability, i.e. 90% reliability.
48. A deep groove ball bearing has the designation 6205. The bore diameter is (bore = last two digits × 5 mm)
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Multiply the last two digits by 5.
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Answer: A. 25 mm
Last two digits 05 × 5 = 25 mm bore.
49. A ball bearing has a dynamic load rating C = 30 kN and works under an equivalent load P = 10 kN. The L10 life is
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Ball bearings use exponent 3.
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Answer: A. 27 million revolutions
L10 = (C/P)^3 = 3^3 = 27 million revolutions.
50. A hydrodynamic journal bearing carries load mainly by
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The film pressure is self-generated.
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Answer: A. a pressurised oil wedge created by shaft rotation
Rotation drags oil into the converging gap and generates pressure that separates the surfaces.
51. An M12 bolt (tensile stress area 84.3 mm²) carries an axial load of 10 kN. The tensile stress is approximately
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Use the stress area in mm².
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Answer: A. 119 MPa
σ = 10 000 / 84.3 ≈ 118.6 ≈ 119 MPa.
52. A helical spring has wire diameter 4 mm, mean coil diameter 32 mm and 10 active coils (G = 80 000 N/mm²). Its stiffness k = Gd⁴/(8D³N) is
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Use the formula directly in N and mm.
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Answer: D. 7.8 N/mm
k = 80000 × 256 / (8 × 32768 × 10) = 7.81 N/mm.
53. A viscous damper produces a force that is
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Think F = c·(something).
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Answer: D. proportional to the velocity
In a viscous damper F = c v, where c is the damping coefficient in N·s/m.
54. A power screw is self-locking when
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Compare tan λ with μ.
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Answer: B. the friction angle is greater than or equal to the lead angle
Self-locking needs μ ≥ tan λ; such screws have efficiency below 50%.
55. A double-start screw has a pitch of 5 mm. Its lead is
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Lead = starts × pitch.
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Answer: B. 10 mm
Lead = number of starts × pitch = 2 × 5 = 10 mm.
56. A block brake has a normal force of 2 kN on a drum of radius 0.25 m, with coefficient of friction 0.3. The braking torque is
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Friction force times radius.
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Answer: C. 150 N·m
T = μ N r = 0.3 × 2000 × 0.25 = 150 N·m.
57. A single-plate clutch with both sides effective has μ = 0.3, axial force 2 kN and mean radius 0.1 m. The torque capacity is
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Two friction surfaces double the single-surface value.
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Answer: D. 120 N·m
T = μ W n Rm = 0.3 × 2000 × 2 × 0.1 = 120 N·m.
58. In friction clutch design, the uniform wear assumption gives
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The more conservative theory is used.
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Answer: C. a lower torque capacity than uniform pressure
Uniform wear (mean radius (R1+R2)/2) is more conservative than uniform pressure, so it is used for design.
59. A solid shaft of diameter 40 mm has an allowable shear stress of 40 MPa. The maximum torque it can transmit is
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Use T = π/16 · τ · d³.
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Answer: B. 503 N·m
T = (π/16) τ d³ = 0.1963 × 40 × 64000 = 502 655 N·mm ≈ 503 N·m.
60. A shaft transmits 20 kW at 600 rpm. The torque is
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Convert rpm to rad/s.
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Answer: C. 318 N·m
T = 60P/(2πN) = 60 × 20000/(2π × 600) = 318 N·m.
61. An axle differs from a shaft in that an axle
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Think of a railway wagon axle.
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Answer: B. carries bending loads but does not transmit torque
A shaft transmits power (torque); an axle supports rotating or stationary parts and carries mainly bending.
6.3 Loads and tools
32 questions · AMeE0603
62. A force of 200 N acts at a perpendicular distance of 0.3 m from a point. Its moment about that point is
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Moment is force times perpendicular distance.
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Answer: D. 60 N·m
M = F × d = 200 × 0.3 = 60 N·m.
63. Two equal and opposite parallel forces of 50 N each act 0.4 m apart. The moment of this couple is
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The resultant force is zero but the moment is not.
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Answer: C. 20 N·m
Couple moment = F × d = 50 × 0.4 = 20 N·m.
64. Two forces of 30 N and 40 N act at a point at right angles to each other. The magnitude of the resultant is
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Use Pythagoras.
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Answer: A. 50 N
R = √(30² + 40²) = 50 N.
65. A simply supported beam of span 5 m carries a point load of 10 kN at 2 m from support A. The reaction at B is
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Take moments about A.
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Answer: A. 4 kN
Taking moments about A: RB × 5 = 10 × 2, so RB = 4 kN (RA = 6 kN).
66. The moment of a couple
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Think of a free vector.
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Answer: A. is the same about every point in its plane
The moments of the two forces about any point always sum to F × d, so a couple's moment is independent of the reference point.
67. For a rigid body in plane equilibrium under coplanar forces, the necessary conditions are
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Three scalar equations in a plane.
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Answer: D. ΣFx = 0, ΣFy = 0 and ΣM = 0
A coplanar rigid body is in equilibrium when the force components in two directions and the moment about any point are all zero.
68. In a single-point cutting tool, increasing the positive rake angle generally
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Think of a sharper wedge.
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Answer: B. reduces the cutting force but weakens the cutting edge
A larger positive rake eases chip flow and lowers force and power, but the wedge angle becomes smaller and the edge weaker.
69. The clearance (relief) angle on a cutting tool is provided to
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Which face is next to the finished surface?
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Answer: D. prevent the flank from rubbing against the machined surface
Clearance keeps the tool flank from rubbing on the finished surface.
70. The composition of 18-4-1 high speed steel is
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Tungsten is the main alloying element.
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Answer: A. 18% tungsten, 4% chromium, 1% vanadium
18-4-1 HSS means 18% W, 4% Cr and 1% V (plus carbon and often some cobalt or molybdenum in variants).
71. Cemented carbide tool inserts consist mainly of tungsten carbide particles bonded with
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The binder is a ferromagnetic metal.
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Answer: A. cobalt
Cobalt is the usual binder in WC-Co cemented carbides.
72. Which type of chip is normally produced when machining brittle materials such as cast iron at low speed?
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Brittle material cracks rather than flows.
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Answer: D. Discontinuous chip
Brittle materials fracture ahead of the tool, giving segmented (discontinuous) chips.
73. A 50 mm diameter bar is turned at 400 rpm. The cutting speed is approximately
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Use πDN with D in mm and divide by 1000.
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Answer: B. 62.8 m/min
V = πDN/1000 = π × 50 × 400 / 1000 = 62.8 m/min.
74. In the Taylor tool life relation VT^0.25 = C, if cutting speed is increased by 20%, the tool life becomes about
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Exponent is 1/n = 4.
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Answer: C. 48% of the original
T2/T1 = (V1/V2)^(1/n) = (1/1.2)^4 = 0.482.
75. A cylindrical job is turned for a length of 150 mm at a feed of 0.25 mm/rev and 300 rpm. The machining time is
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Feed per minute = f × N.
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Answer: B. 2 min
t = L / (f N) = 150 / (0.25 × 300) = 2 min.
76. In orthogonal cutting the chip thickness ratio is 0.4 and the rake angle is 10°. The shear angle (tanφ = r cosα/(1 − r sinα)) is approximately
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Use the Merchant geometric relation.
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Answer: C. 23°
tanφ = 0.4 × 0.985/(1 − 0.4 × 0.174) = 0.423, so φ ≈ 23°.
77. In a blanking operation, the clearance is provided by making the
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The blank is the useful part.
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Answer: D. punch smaller than the die opening, with the die size equal to the blank
In blanking the slug is the product, so the die opening is the blank size and the punch is smaller by the clearance.
78. In a piercing operation, the die opening is made
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The hole is the useful part.
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Answer: B. larger than the punch, with the punch equal to the required hole size
In piercing the hole is the product, so the punch has the hole size and the die is larger by the clearance.
79. A circular blank of 40 mm diameter is cut from 2 mm sheet with shear strength 300 MPa. The blanking force is approximately
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Shear area = perimeter × thickness.
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Answer: D. 75 kN
F = π D t τ = π × 40 × 2 × 300 = 75 400 N ≈ 75 kN.
80. Providing shear on the face of a punch or die mainly
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Cut a little at a time.
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Answer: C. reduces the maximum press force required
Angled cutting faces make the cut progressive, lowering the peak force though the stroke needed is longer.
81. A die in which several operations are performed in successive stations as the strip advances is called a
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The strip is moved forward step by step.
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Answer: B. progressive die
A progressive die performs different operations at successive stations at each stroke while the strip is fed forward.
82. The usual point angle of a standard twist drill for general-purpose drilling of mild steel is
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It is close to, but less than, 120°.
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Answer: C. 118°
A standard drill has a 118° point angle; harder materials use 130–140°.
83. A hole 30 mm deep is drilled at a feed of 0.2 mm/rev and 500 rpm (ignore approach and overrun). The drilling time is
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Convert minutes to seconds at the end.
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Answer: C. 18 s
t = L / (f N) = 30 / (0.2 × 500) = 0.3 min = 18 s.
84. The flutes of a twist drill serve to
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Think about what leaves the hole.
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Answer: B. remove chips and let coolant reach the cutting edges
Helical flutes carry chips out of the hole and allow cutting fluid in.
85. In closed-die forging, the excess metal that escapes between the die faces is called
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It is trimmed off afterwards.
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Answer: D. flash
Flash forms in the gap at the parting line and fills the die cavity by building up back-pressure; it is trimmed later.
86. Draft angles are provided on forging dies to
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It helps ejecting the part.
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Answer: A. allow the forged part to be removed from the die cavity
Taper on vertical walls lets the part be ejected easily.
87. The projected area of a hot forging is 2500 mm² and the average flow stress at forging temperature is 100 MPa. The approximate forging load, ignoring friction and flash, is
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Force = stress × area.
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Answer: C. 250 kN
F = σ A = 100 × 2500 = 250 000 N = 250 kN.
88. Oil of viscosity 0.1 Pa·s fills a 1 mm gap between a fixed plate and a plate moving at 2 m/s. The shear stress in the oil is
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Newton's law of viscosity with a linear profile.
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Answer: B. 200 Pa
τ = μ U / h = 0.1 × 2 / 0.001 = 200 Pa.
89. The viscosity index of a lubricating oil indicates
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Related to temperature sensitivity.
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Answer: D. how little its viscosity changes with temperature
A high viscosity index means viscosity varies less over a temperature range.
90. Boundary lubrication occurs when
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It happens at start-up and at high load.
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Answer: A. the surfaces are separated only by a very thin adsorbed film
At low speed or high load the film is molecular and surface contact partially occurs.
91. Which of the following is the main function of a cutting fluid in machining?
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What is the problem with high temperature?
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Answer: B. To cool the tool and work and reduce friction
Cutting fluids carry away heat, reduce friction and flush chips.
92. A soluble-oil cutting fluid is mainly used for
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Water gives the cooling.
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Answer: C. cooling at high cutting speeds, as it is an oil-in-water emulsion
Water-based emulsions have good cooling ability, suitable for high speed light cuts, with enough lubrication.
93. Grease is preferred to oil for a rolling bearing when
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Think of the lubricant staying in place.
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Answer: A. it is hard to supply oil and leakage must be avoided
Grease stays in place, needs little attention and seals out dirt; oil is better for high speeds and heat removal.
6.4 Static analysis of systems
30 questions · AMeE0604
94. The maximum principal stress (Rankine) theory is best suited for predicting failure of
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Fracture rather than yielding.
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Answer: D. brittle materials
Brittle materials fail by fracture when the maximum principal stress reaches the ultimate strength.
95. For ductile materials under static loading, the theory that agrees best with experiments is the
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Named after von Mises.
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Answer: B. distortion energy (von Mises) theory
The von Mises distortion energy criterion fits yield data for ductile metals most closely.
96. According to the maximum shear stress (Tresca) theory, yielding in a simple tension test occurs when the maximum shear stress equals
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In uniaxial tension τmax = σ/2.
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Answer: B. half of the yield strength in tension (Sy/2)
In tension τmax = σ/2, so yielding occurs at τmax = Sy/2.
97. At a point in a plane stress state σ1 = 100 MPa and σ2 = 50 MPa (no shear). The von Mises equivalent stress is
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Use the biaxial von Mises formula.
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Answer: D. 86.6 MPa
σ' = √(σ1² − σ1σ2 + σ2²) = √(10000 − 5000 + 2500) = 86.6 MPa.
98. A plate has a stress concentration factor Kt = 2.5 at a hole and a nominal stress of 80 MPa. The maximum stress at the hole is
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Multiply nominal stress by Kt.
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Answer: C. 200 MPa
σmax = Kt σnom = 2.5 × 80 = 200 MPa.
99. At a point σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. The maximum principal stress is
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Mohr circle: centre 50, radius 50.
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Answer: C. 100 MPa
σ1 = 50 + √(30² + 40²) = 50 + 50 = 100 MPa.
100. A shaft carries a bending moment of 300 N·m and a torque of 400 N·m. The equivalent torque Te = √(M² + T²) is
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A 3-4-5 triangle.
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Answer: B. 500 N·m
Te = √(300² + 400²) = 500 N·m.
101. For a ductile component loaded statically, a stress concentration is usually
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Think of ductility.
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Answer: A. neglected in design because local yielding relieves the peak
Ductile material yields locally and redistributes the stress, so Kt is generally ignored for static loads (not for brittle or fatigue loads).
102. Design of a machine part using a yield-based approach for a ductile material takes the allowable stress as
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Division by FOS is always used.
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Answer: A. yield strength divided by the factor of safety
Allowable = Sy/FOS for ductile materials; Su/FOS for brittle ones.
103. A solid shaft transmits a torque of 500 N·m with allowable shear stress 40 MPa. The minimum diameter required is about
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Use T = π τ d³/16.
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Answer: B. 40 mm
d = (16T/πτ)^(1/3) = (16 × 500 000/(π × 40))^(1/3) = 39.9 mm ≈ 40 mm.
104. A spur gear of pitch diameter 200 mm transmits a torque of 250 N·m with a 20° pressure angle. The radial force on the shaft is
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Radial = tangential × tan(pressure angle).
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Answer: C. 910 N
Ft = 2T/d = 2 × 250/0.2 = 2500 N; Fr = Ft tan20° = 910 N.
105. A 12 mm wide key on a 40 mm shaft transmits 400 N·m. If allowable shear stress of the key is 60 MPa, the minimum key length is about
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Force at shaft surface = T / radius.
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Answer: B. 27.8 mm
F = T/(d/2) = 400/0.02 = 20 kN; L = F/(w τ) = 20000/(12 × 60) = 27.8 mm.
106. A sunk key can fail by
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Two checks: shear and bearing.
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Answer: C. shearing across its width plane or crushing on its side face
The two key checks are shear across the key-shaft interface and compressive crushing of the side face.
107. A shaft carrying a pulley subjected to a belt pull between its supports experiences
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More than one load acts.
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Answer: D. combined bending and torsion
The belt pull bends the shaft while the transmitted power twists it.
108. For the same weight and material, a hollow shaft compared with a solid shaft has
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Material at the centre does little.
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Answer: A. greater torsional strength and rigidity
Material far from the axis is more effective in torsion, so a hollow shaft is stronger per unit weight.
109. A rivet of 16 mm diameter in single shear carries a load of 20 kN. The shear stress is about
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Area of one rivet cross-section.
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Answer: C. 99.5 MPa
τ = P/(πd²/4) = 20 000/201.06 = 99.5 MPa.
110. A parallel fillet weld of leg 6 mm and length 100 mm has an allowable shear stress of 80 MPa on the throat. The load capacity is about
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The throat is 0.707 times the leg.
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Answer: B. 33.9 kN
Throat = 0.707 × 6 = 4.24 mm; P = 4.24 × 100 × 80 = 33 900 N.
111. In a riveted joint, pitch 50 mm, rivet hole diameter 20 mm, plate thickness 10 mm. The tearing efficiency of the plate is
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Compare the strength per pitch with the solid plate.
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Answer: C. 60%
η = (p − d)/p = 30/50 = 0.6 = 60%.
112. Which of the following is NOT a failure mode of a riveted lap joint?
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Look at standard load-induced failures.
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Answer: A. Bending of the rivet head by thermal expansion
Rivet shear, crushing and plate tearing (or edge shearing) are the standard failure modes; the other is not.
113. The main purpose of tightening a bolted joint to a high preload is to
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Clamping force.
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Answer: D. keep the joint faces clamped so external load does not separate them
Preload keeps the members in compression; the bolt sees only a fraction of external load so fatigue life improves.
114. The effective throat of a fillet weld with leg size s (equal legs, 45° face) is
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Think of an isosceles right triangle.
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Answer: A. 0.707 s
Throat is s sin45° = 0.707 s, the shortest distance from the root to the hypotenuse.
115. A bracket riveted to a column carries a load offset from the rivet group centroid. Each rivet then experiences
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Eccentricity adds a moment.
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Answer: C. primary direct shear plus secondary shear due to the moment
Eccentric load = direct load at the centroid + moment, producing primary and secondary shear on the rivets.
116. A simply supported beam of span 2 m with a rectangular section 50 mm wide × 100 mm deep carries a central point load of 5 kN. The maximum bending stress is
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M = PL/4 for central load.
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Answer: A. 30 MPa
M = PL/4 = 2.5 kN·m; Z = bh²/6 = 83 333 mm³; σ = 2.5×10⁶/83 333 = 30 MPa.
117. A pin-ended steel column (E = 200 GPa), solid circular of 30 mm diameter and 1 m long. The Euler buckling load is about
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Use π²EI/L².
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Answer: B. 78 kN
I = πd⁴/64 = 39 761 mm⁴; Pcr = π²EI/L² = π² × 200 000 × 39 761/10⁶ = 78.5 kN.
118. A thin cylindrical vessel of 500 mm diameter, wall thickness 5 mm, carries an internal pressure of 2 MPa. The hoop stress is
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The hoop stress is twice the longitudinal stress.
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Answer: B. 100 MPa
σh = pd/2t = 2 × 500/(2 × 5) = 100 MPa.
119. The effective length of a column fixed at both ends is
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The more fixed the shorter.
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Answer: A. 0.5 L
For fixed-fixed ends the effective length is L/2; pinned-pinned is L; fixed-free is 2L.
120. In pure bending of a beam, the neutral axis passes through
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Zero bending stress there.
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Answer: D. the centroid of the cross-section
For linear elastic bending of a symmetric/homogeneous section, the neutral axis passes through the centroid.
121. Among sections of equal area, which is most efficient in bending?
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Material far from the axis.
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Answer: A. I-section, as it has larger I and Z
I-sections place material far from the neutral axis, giving the greatest section modulus per unit area.
122. A rectangular beam section 40 mm × 60 mm carries a transverse shear force of 12 kN. The maximum shear stress (at the neutral axis) is
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Maximum is 1.5 times average for a rectangle.
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Answer: D. 7.5 MPa
τavg = 12000/2400 = 5 MPa; τmax = 1.5 τavg = 7.5 MPa.
123. A double-cover butt joint is stronger than a lap joint of the same plates and rivets mainly because the rivets
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Count the shear planes.
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Answer: D. are in double shear and the load is axial (no eccentricity)
A double-strap butt joint has no eccentric bending and each rivet resists the load across two shear planes.
6.5 Mechanical vibrations
30 questions · AMeE0605
124. The number of degrees of freedom of a vibrating system is
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Independent coordinates.
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Answer: C. the minimum number of independent coordinates needed to describe its configuration
DoF equals the number of independent coordinates required to fix the position of all parts of the system at any time.
125. A rigid body moving in a plane (free of constraints) has how many degrees of freedom?
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2 translations + 1 rotation.
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Answer: B. 3
Two translations and one rotation give 3 DoF; a body in space has 6.
126. A simple pendulum swinging in a plane has
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One angle is enough.
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Answer: C. one degree of freedom
The angle from the vertical alone fixes the position, so DoF = 1.
127. Two masses connected by springs and constrained to move along one straight line have
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One coordinate per mass.
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Answer: A. two degrees of freedom
Each mass needs one displacement coordinate, hence 2 DoF.
128. Generalized coordinates are
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They need not be Cartesian.
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Answer: C. independent coordinates that completely describe the configuration of a system
They can be lengths, angles or other quantities; their number equals the DoF for a holonomic system.
129. The undamped free vibration of a single-DoF system is
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No energy is lost.
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Answer: A. simple harmonic with constant amplitude
m ẍ + k x = 0 gives sinusoidal motion at ωn with constant amplitude set by the initial conditions.
130. The natural frequency of a spring–mass system depends on
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Free of amplitude.
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Answer: D. stiffness and mass only
ωn = √(k/m), independent of the initial conditions.
131. A mass of 5 kg is attached to a spring of stiffness 20 000 N/m. The natural frequency is
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Square root of k/m.
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Answer: B. 63.2 rad/s
ωn = √(k/m) = √(20000/5) = 63.2 rad/s (fn = 10.07 Hz).
132. A system has a static deflection of 10 mm under its own weight on a spring. Its natural frequency is about (g = 9.81 m/s²)
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Use static deflection formula.
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Answer: D. 31.3 rad/s
ωn = √(g/δst) = √(9.81/0.01) = 31.3 rad/s.
133. If the mass of a spring–mass system is increased four times with the same spring, the natural frequency
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Square root relation.
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Answer: A. becomes one-half
ωn ∝ 1/√m, so 4× mass halves the frequency.
134. A simple pendulum of length 1 m (g = 9.81 m/s²) has a time period of about
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Use T = 2π√(L/g).
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Answer: D. 2.0 s
T = 2π√(L/g) = 2π × 0.319 = 2.006 s.
135. A disc of polar mass moment of inertia 0.5 kg·m² is on a shaft of torsional stiffness 1000 N·m/rad. The torsional natural frequency is
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Torsional analogue of √(k/m).
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Answer: A. 44.7 rad/s
ωn = √(kt/J) = √(1000/0.5) = 44.7 rad/s.
136. A single-DoF system has m = 5 kg, k = 20 000 N/m and c = 200 N·s/m. The damping ratio is
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ζ = c/cc.
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Answer: D. 0.316
cc = 2√(km) = 2√(100 000) = 632 N·s/m; ζ = 200/632 = 0.316.
137. The critical damping coefficient of a system is
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ζ = 1.
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Answer: D. the value of c that just prevents oscillation (ζ = 1)
cc = 2√(km) = 2mωn; at c = cc the system returns to rest in the shortest time without oscillating.
138. A system with damping ratio ζ = 1.5 is
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ζ > 1.
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Answer: A. overdamped, with non-oscillatory return to equilibrium
ζ > 1 gives two real negative roots: no oscillation, slower return than critical damping.
139. In an underdamped free vibration, the successive peak amplitudes decrease
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Constant ratio, not constant difference.
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Answer: B. in a geometric progression
Peaks fall by a constant ratio e^(2πζ/√(1−ζ²)).
140. A system has ωn = 63.25 rad/s and ζ = 0.2. The damped natural frequency is
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Slightly less than ωn.
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Answer: C. 62.0 rad/s
ωd = ωn√(1 − ζ²) = 63.25 × 0.9798 = 62.0 rad/s.
141. In free damped vibration, the amplitude of successive cycles reduces by half each cycle (ratio 2). The logarithmic decrement is
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Use a natural log.
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Answer: B. 0.693
δ = ln(x1/x2) = ln 2 = 0.693.
142. For a logarithmic decrement of 0.693 the damping ratio ζ = δ/√(4π² + δ²) is about
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Use the exact formula.
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Answer: B. 0.11
ζ = 0.693/√(39.48 + 0.48) = 0.693/6.32 = 0.110.
143. The damping that produces a force proportional to velocity and opposite to it is called
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Fluid-like damping.
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Answer: C. viscous damping
Viscous damping F = c ẋ; Coulomb damping is constant dry friction.
144. In free vibration with Coulomb (dry friction) damping, the amplitude decays
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Constant friction force.
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Answer: B. linearly with time
Friction removes a constant energy amount per cycle, so the envelope is a straight line.
145. A system with initial displacement 10 mm and initial velocity 0.6325 m/s starts undamped free vibration at ωn = 63.25 rad/s. The amplitude is
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v0/ωn = 10 mm.
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Answer: D. 14.1 mm
X = √(x0² + (v0/ωn)²) = √(10² + 10²) = 14.1 mm.
146. In steady-state forced harmonic vibration, the system vibrates at
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Driven, not free.
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Answer: C. the frequency of the excitation force
After transients decay, the response follows the forcing frequency.
147. In the undamped forced vibration of a single-DoF system, resonance occurs when
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Frequency ratio equal to one.
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Answer: A. the forcing frequency equals the natural frequency
At ω = ωn (r = 1), the undamped amplitude theoretically becomes infinite.
148. A damped system has ζ = 0.05. At resonance, the magnification factor 1/(2ζ) is
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Use 1/(2ζ).
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Answer: C. 10
M = 1/(2 × 0.05) = 10.
149. For an undamped system the magnification factor at a frequency ratio r = 0.5 is
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1/(1 − r²).
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Answer: D. 1.33
M = 1/(1 − r²) = 1/(1 − 0.25) = 1.33.
150. A damped system has ωn = 100 rad/s and ζ = 0.1. The forcing frequency at which the displacement amplitude is a maximum (constant force amplitude) is ωn√(1 − 2ζ²), about
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Distinguish from damped natural frequency.
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Answer: A. 99.0 rad/s
ωn√(1 − 2ζ²) = 100 × √0.98 = 99.0 rad/s, slightly below ωd = 99.5 rad/s.
151. The phase angle between displacement and the exciting force in a forced damped system at r = 1 is
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A quarter cycle.
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Answer: A. 90°
At resonance the response lags the force by 90° irrespective of ζ.
152. The transmissibility of a vibration isolator is less than 1 only when the frequency ratio is
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The crossover value is √2.
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Answer: B. greater than √2
TR < 1 requires r > √2, where isolation begins; at r = √2, TR = 1 regardless of damping.
153. For an undamped isolator at frequency ratio r = 3, the transmissibility is 1/|1 − r²| =
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Evaluate 1 − r².
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Answer: B. 0.125
TR = 1/|1 − 9| = 0.125.
6.6 Problem solving and decision-making
29 questions · AMeE0606
154. In the Wallas model of creative problem solving, the correct order of stages is
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It starts by gathering information.
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Answer: D. preparation, incubation, inspiration (illumination), verification
Preparation (study the problem) → incubation (unconscious work) → inspiration/illumination (insight) → verification (test the idea).
155. The preparation stage of creative problem solving involves
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The first stage.
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Answer: A. collecting information and studying the problem consciously
Preparation is the conscious effort to define and understand the problem and gather relevant data.
156. During the incubation stage, the problem is
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A rest from active effort.
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Answer: C. set aside while the mind works on it unconsciously
Incubation is a pause in conscious effort that lets the mind process the problem in the background.
157. A sudden flash of understanding of a possible solution corresponds to the
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The 'Aha!' moment.
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Answer: C. inspiration (illumination) stage
Inspiration is the 'Aha!' moment, after incubation.
158. Checking a proposed idea by analysis, experiment or prototype corresponds to the
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Testing the idea.
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Answer: A. verification stage
In verification, the idea is tested and refined to see if it works.
159. Which is a basic rule of a brainstorming session?
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Postpone judgment.
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Answer: D. Criticism of ideas is withheld until generation has finished
Deferring judgment encourages free flow of ideas; evaluation comes after.
160. In brainstorming, the emphasis is on
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Think numbers.
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Answer: C. quantity of ideas, including unusual ones
More ideas increase the chance of good ones; wild ideas are welcome and can be built on by others.
161. Brainstorming was popularised by
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An advertising executive.
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Answer: B. Alex Osborn
Alex F. Osborn introduced brainstorming in the 1940s–1950s.
162. A good problem statement should
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Does it say 'what' or 'how'?
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Answer: C. describe the need and constraints without prescribing the solution
Stating the need and criteria without implying a solution keeps the design space open.
163. Which of the following is the best problem statement for a design task?
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Look for the one with no named device.
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Answer: A. Provide a means to lift a 500 kg load through 2 m safely
The first statement defines the need and constraints and leaves the solution open; the others name specific solutions.
164. The first step of the general problem-solving process is to
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You cannot solve what is not defined.
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Answer: C. define the problem clearly
Without a clear definition, later steps may solve the wrong problem.
165. An 'invention' differs from an 'innovation' in that an invention is
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One creates, the other commercialises.
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Answer: B. the creation of a new idea, device or process, while innovation puts it to successful use
Invention is the new idea; innovation is its successful introduction into practice or market.
166. Divergent thinking in design is used to
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It opens up options.
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Answer: D. generate many alternative ideas
Divergent thinking widens the set of options; convergent thinking narrows it by evaluation.
167. A weighted decision matrix is used to
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A multi-criteria tool.
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Answer: A. compare alternatives against weighted criteria and rank them
Each alternative is scored on criteria multiplied by weights, and total scores rank the options.
168. In a Pugh decision chart each alternative is compared with
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One option is the baseline.
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Answer: B. a reference (datum) concept using +, − or S
A datum design is chosen, and others are rated better (+), worse (−) or same (S) for each criterion.
169. In a decision tree, a decision node is conventionally drawn as a
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Squares and circles.
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Answer: A. square
Squares represent decisions and circles represent chance (uncertain) events.
170. A circle at a node in a decision tree represents a
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Probabilities attach to it.
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Answer: A. chance (uncertain) event
Chance nodes branch into possible events with probabilities.
171. In a decision tree, expected values are calculated by working
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Rollback.
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Answer: A. from the right (outcomes) towards the left (first decision)
Rollback: evaluate end values, take probability-weighted sums at chance nodes and best values at decision nodes.
172. The probabilities of all branches leaving a chance node must
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Probabilities are normalised.
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Answer: B. sum to 1
The branches are exhaustive and mutually exclusive, so their probabilities total 1.
173. A fishbone (Ishikawa) diagram is mainly used to
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Cause-and-effect.
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Answer: D. identify and sort possible causes of a problem
It organises potential causes of an effect into categories such as man, machine, material, method.
174. A morphological chart helps the designer to
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Functions versus means.
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Answer: D. combine alternative means for each function to generate concepts
It lists sub-functions in rows and possible means in columns; combining one per row yields concepts.
175. A weighted decision matrix has criteria weights 0.4, 0.3, 0.2 and 0.1. Design B scores 7, 9, 6 and 8 respectively. B's weighted total is
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Multiply each score by its weight, then add.
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Answer: B. 7.5
0.4×7 + 0.3×9 + 0.2×6 + 0.1×8 = 2.8 + 2.7 + 1.2 + 0.8 = 7.5.
176. Criteria weights in a decision matrix are 0.35, 0.25, 0.15 and w4. The weights must sum to 1, so w4 is
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Normalise to one.
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Answer: A. 0.25
w4 = 1 − (0.35 + 0.25 + 0.15) = 0.25.
177. A project has a 60% chance of earning Rs 100 000 and a 40% chance of losing Rs 40 000. The expected monetary value is
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Probability-weighted outcomes.
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Answer: D. Rs 44 000
EMV = 0.6 × 100 000 − 0.4 × 40 000 = 60 000 − 16 000 = Rs 44 000.
178. An option costs Rs 30 000 and then gives Rs 200 000 with probability 0.3, otherwise Rs 50 000. The net expected value (in thousands of rupees) is
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Subtract the cost from the expected return.
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Answer: C. 65
Expected return = 0.3×200 + 0.7×50 = 95; net = 95 − 30 = 65.
179. A decision tree compares option A (50% chance of 80 and 50% chance of 20, in thousands) with option B (a sure 45). The better choice is
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Compute A's EMV, then compare.
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Answer: D. A, with expected value 50
EMV(A) = 0.5×80 + 0.5×20 = 50 > 45, so choose A on expected value.
180. A Pugh chart rates a concept as 4 pluses, 1 minus and 2 'same' against the datum. Its net score is
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Same counts as zero.
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Answer: B. +3
Net score = pluses − minuses = 4 − 1 = +3 (S counts zero).
181. A machine costs Rs 60 000 and saves Rs 15 000 per year. The simple payback period is
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Divide the investment by the annual saving.
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Answer: B. 4 years
Payback = investment / annual saving = 60 000/15 000 = 4 years.
182. Fixed cost of a product is Rs 50 000, variable cost Rs 20 per unit and selling price Rs 45 per unit. The break-even quantity is
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Contribution per unit = price − variable cost.
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Answer: C. 2000 units
Q = F/(p − v) = 50 000/25 = 2000 units.