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Question 1Engineering Fundamentals and Applied Science

A steel shaft transmits 15 kW at 300 rpm. Which expression gives the torque carried by the shaft in newton-metres?

  1. 15000 / (300 / 60)
  2. 15000 × 300 / 60
  3. 15000 / (2π × 300 / 60) Correct answer.
  4. 15000 × 2π × 300

Explanation

Rotational power equals torque multiplied by angular velocity in rad/s, so torque equals power divided by angular velocity. The angular velocity is 2π times revolutions per second, giving 2π × 300/60 rad/s. Dividing 15000 W by that value yields torque in N·m.

Question 2Engineering Fundamentals and Applied Science

In a simple gear train, gear A with 20 teeth drives gear B with 60 teeth. If A rotates at 900 rpm, what is the rotational speed of B?

  1. 1800 rpm
  2. 300 rpm Correct answer.
  3. 900 rpm
  4. 2700 rpm

Explanation

For meshing gears the speed ratio is inversely proportional to the tooth ratio. Gear B has three times as many teeth as A, so it turns at one third of A's speed. Thus 900 divided by 3 gives 300 rpm.

Question 3Engineering Fundamentals and Applied Science

Which quantity is a scalar in classical mechanics?

  1. Velocity
  2. Acceleration
  3. Energy Correct answer.
  4. Force

Explanation

A scalar has magnitude only, whereas a vector has both magnitude and direction. Energy is described fully by a magnitude and unit, such as joules. Velocity, force and acceleration all require a stated direction to be complete.

Question 4Engineering Fundamentals and Applied Science

A beam is simply supported at both ends and carries a single concentrated load at midspan. Where does the maximum bending moment occur?

  1. At midspan, directly under the load Correct answer.
  2. At the quarter points of the span
  3. At the left support
  4. At the right support

Explanation

For a simply supported beam with a central point load, the bending moment diagram is triangular, peaking where the shear changes sign. That point is directly beneath the load at midspan. The moment is zero at both simple supports.

Question 5Engineering Fundamentals and Applied Science

Which material property describes the stress at which a ductile metal begins to deform permanently?

  1. Modulus of elasticity
  2. Ultimate tensile strength
  3. Toughness
  4. Yield strength Correct answer.

Explanation

Yield strength marks the end of elastic behaviour, where further stress produces permanent plastic strain. Ultimate tensile strength is the maximum stress reached before necking. The elastic modulus describes stiffness within the elastic region, not the onset of yielding.

Question 6Engineering Fundamentals and Applied Science

In a series DC circuit, two resistors of 4 Ω and 6 Ω are connected across a 20 V supply. What current flows?

  1. 0.5 A
  2. 10 A
  3. 20 A
  4. 2 A Correct answer.

Explanation

In series the resistances add, giving a total of 10 Ω. Ohm's law then gives current as voltage divided by resistance, so 20 V divided by 10 Ω equals 2 A. The same current passes through both resistors.

Question 7Engineering Fundamentals and Applied Science

Which thermodynamic process occurs with no heat transfer across the system boundary?

  1. Isothermal
  2. Isothermal compression
  3. Adiabatic Correct answer.
  4. Isobaric

Explanation

An adiabatic process is defined by zero heat exchange with the surroundings, so any work done changes internal energy. Isothermal, isobaric and isochoric processes are defined by constant temperature, pressure or volume instead. A well-insulated rapid compression is a common approximation of an adiabatic process.

Question 8Engineering Fundamentals and Applied Science

What is the SI unit of pressure?

  1. Joule
  2. Watt
  3. Pascal Correct answer.
  4. Newton

Explanation

The pascal is defined as one newton per square metre, which is the SI unit of pressure and stress. The newton is the unit of force, the joule of energy and the watt of power. Pressure is force distributed over an area.

Question 9Engineering Fundamentals and Applied Science

A simply supported beam of span 6 m carries a uniformly distributed load of 10 kN/m over its full length. What is the magnitude of the vertical reaction at each support?

  1. 60 kN
  2. 15 kN
  3. 10 kN
  4. 30 kN Correct answer.

Explanation

The total load is the intensity multiplied by the span, giving 60 kN. By symmetry each support carries half of that total. Half of 60 kN is 30 kN per support.

Question 10Engineering Fundamentals and Applied Science

A block of mass 5 kg is pulled along a horizontal surface by a horizontal force of 30 N against a constant friction force of 10 N. What is its acceleration?

  1. 8 m/s²
  2. 4 m/s² Correct answer.
  3. 6 m/s²
  4. 2 m/s²

Explanation

The net horizontal force is the applied force minus friction, giving 20 N. Newton's second law states acceleration equals net force divided by mass. Dividing 20 N by 5 kg yields 4 m/s².

Question 11Engineering Fundamentals and Applied Science

In a parallel circuit, two resistors of 6 Ω and 3 Ω are connected across a 12 V supply. What is the total current drawn from the supply?

  1. 6 A Correct answer.
  2. 4 A
  3. 18 A
  4. 2 A

Explanation

Each branch sees the full 12 V, so the 6 Ω branch draws 2 A and the 3 Ω branch draws 4 A. Parallel branch currents add, giving 6 A total. Equivalently the equivalent resistance is 2 Ω and 12 V divided by 2 Ω equals 6 A.

Question 12Engineering Fundamentals and Applied Science

A hydraulic cylinder has a piston area of 0.005 m² and must exert a force of 10 kN. What fluid pressure is required?

  1. 0.5 MPa
  2. 2 MPa Correct answer.
  3. 200 kPa
  4. 50 kPa

Explanation

Pressure equals force divided by area, so 10000 N divided by 0.005 m² equals 2 000 000 Pa. That is 2 MPa. This follows directly from the definition of pressure as force per unit area.

Question 13Engineering Fundamentals and Applied Science

A heat engine absorbs 500 kJ of heat from a hot reservoir and rejects 300 kJ to a cold reservoir per cycle. What is its thermal efficiency?

  1. 167%
  2. 40% Correct answer.
  3. 60%
  4. 80%

Explanation

Useful work equals heat absorbed minus heat rejected, giving 200 kJ. Thermal efficiency is useful work divided by heat absorbed, so 200 divided by 500 equals 0.40, or 40%. Efficiency can never exceed 100% for a real cycle.

Question 14Engineering Fundamentals and Applied Science

A belt drive has a driver pulley of 100 mm diameter running at 1200 rpm and a driven pulley of 250 mm diameter. What is the driven pulley speed, assuming no slip?

  1. 250 rpm
  2. 600 rpm
  3. 3000 rpm
  4. 480 rpm Correct answer.

Explanation

For a belt drive the surface speed is the same on both pulleys, so speed is inversely proportional to diameter. The diameter ratio is 250/100, which is 2.5. Dividing 1200 rpm by 2.5 gives 480 rpm.

Question 15Engineering Fundamentals and Applied Science

A cantilever beam of length 2 m carries a point load of 5 kN at its free end. What is the magnitude of the bending moment at the fixed support?

  1. 10 kN·m Correct answer.
  2. 2.5 kN·m
  3. 5 kN·m
  4. 20 kN·m

Explanation

For a cantilever with an end point load, the maximum moment occurs at the fixed support and equals load multiplied by length. Multiplying 5 kN by 2 m gives 10 kN·m. The moment decreases linearly to zero at the free end.

Question 16Engineering Fundamentals and Applied Science

Which statement correctly describes the difference between a stress and a strain?

  1. Stress and strain are both measured in pascals
  2. Stress is a dimensionless ratio, whereas strain is force per unit area
  3. Stress is force per unit area, whereas strain is a dimensionless ratio of deformation to original length Correct answer.
  4. Stress is measured in newtons, whereas strain is measured in metres

Explanation

Stress expresses internal force intensity in newtons per square metre, while strain expresses geometric deformation as a ratio. Because strain is a ratio of two lengths, it has no units. Their relationship in the elastic region is given by the elastic modulus.

Question 17Engineering Fundamentals and Applied Science

A resistor of 220 Ω is connected across a 12 V supply. What power is dissipated?

  1. 2.64 W
  2. 26.4 W
  3. 0.055 W
  4. 0.65 W Correct answer.

Explanation

Power can be found from voltage squared divided by resistance. Squaring 12 V gives 144, and dividing by 220 Ω gives about 0.655 W. The current is about 54.5 mA, which confirms the result.

Question 18Engineering Fundamentals and Applied Science

A steel rod has an original length of 2 m and extends by 1.2 mm under load. What is the longitudinal strain?

  1. 0.0012
  2. 0.0006 Correct answer.
  3. 0.6
  4. 0.00024

Explanation

Strain is the change in length divided by the original length, with both in the same units. Converting 1.2 mm to 0.0012 m, dividing by 2 m gives 0.0006. Strain is dimensionless and typically small for metals.

Question 19Engineering Fundamentals and Applied Science

In a four-stroke internal combustion engine, how many complete revolutions of the crankshaft occur per thermodynamic cycle?

  1. One
  2. Two Correct answer.
  3. Four
  4. Half

Explanation

A four-stroke cycle spans intake, compression, power and exhaust strokes, each occupying half a crankshaft revolution. Together they require two full revolutions. A two-stroke engine instead completes a cycle in one revolution.

Question 20Engineering Fundamentals and Applied Science

A simply supported beam of span 4 m carries a point load of 8 kN at 1 m from the left support. What is the vertical reaction at the right support?

  1. 8 kN
  2. 6 kN
  3. 2 kN Correct answer.
  4. 4 kN

Explanation

Taking moments about the left support, the load of 8 kN acts at 1 m, giving 8 kN·m clockwise. The right reaction acts at 4 m, so it equals 8 divided by 4, which is 2 kN. The left reaction then carries the remaining 6 kN.

Question 21Engineering Fundamentals and Applied Science

A copper conductor of length 2 m and cross-sectional area 1 × 10⁻⁶ m² has a resistivity of 1.7 × 10⁻⁸ Ω·m. What is its resistance?

  1. 0.034 Ω Correct answer.
  2. 0.0034 Ω
  3. 3.4 Ω
  4. 0.34 Ω

Explanation

Resistance equals resistivity multiplied by length divided by area. Multiplying 1.7 × 10⁻⁸ by 2 gives 3.4 × 10⁻⁸, and dividing by 1 × 10⁻⁶ gives 0.034 Ω. The result is consistent with a short, thick conductor having low resistance.

Question 22Engineering Fundamentals and Applied Science

A hollow circular shaft has an outer diameter of 100 mm and an inner diameter of 60 mm. Which expression gives its polar second moment of area in mm⁴?

  1. (π/4)(100² − 60²)
  2. (π/32)(100⁴ − 60⁴) Correct answer.
  3. (π/32)(100³ − 60³)
  4. (π/64)(100⁴ − 60⁴)

Explanation

The polar second moment of area for a solid circle is πd⁴/32, and for a hollow section the bore contribution is subtracted. Because it depends on the fourth power of diameter, the expression uses 100⁴ and 60⁴. The π/64 factor applies to the second moment about a diameter, not the polar value.

Question 23Engineering Fundamentals and Applied Science

A mass–spring system has stiffness 200 N/m and mass 2 kg. What is its natural angular frequency?

  1. 100 rad/s
  2. 20 rad/s
  3. 10 rad/s Correct answer.
  4. 1 rad/s

Explanation

The natural angular frequency is the square root of stiffness divided by mass. Dividing 200 N/m by 2 kg gives 100 s⁻², and its square root is 10 rad/s. This is angular frequency, not the value in hertz.

Question 24Engineering Fundamentals and Applied Science

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its maximum theoretical efficiency?

  1. 50% Correct answer.
  2. 100%
  3. 33%
  4. 67%

Explanation

Carnot efficiency depends only on absolute reservoir temperatures and equals one minus the cold temperature divided by the hot temperature. Dividing 300 by 600 gives 0.5, so efficiency is 1 − 0.5, or 50%. No engine operating between these reservoirs can exceed this value.

Question 25Engineering Fundamentals and Applied Science

A beam of rectangular cross-section has width b and depth d. If the depth is doubled while the width is unchanged, how does the second moment of area about the neutral axis change?

  1. It remains unchanged
  2. It increases by a factor of two
  3. It increases by a factor of eight Correct answer.
  4. It increases by a factor of four

Explanation

For a rectangle the second moment of area about the neutral axis is bd³/12, so it varies with the cube of depth. Doubling d multiplies the value by 2³, which is eight. Width enters linearly, so only the depth change matters here.

Question 26Engineering Fundamentals and Applied Science

In a first-angle orthographic projection drawing of a machined bracket, where is the view from above placed relative to the front view?

  1. Below the front view Correct answer.
  2. Above the front view
  3. To the right of the front view
  4. To the left of the front view

Explanation

First-angle projection places the object between the observer and the projection plane, so each view is projected onto the far side. The top view therefore lands below the front view, and the left-side view lands on the right. This is the reverse of third-angle projection, where the top view sits above the front view.

Question 27Engineering Fundamentals and Applied Science

A component is specified as a nominal 40.00 mm shaft with a tolerance of ±0.05 mm. Which statement correctly describes the permissible limits of size?

  1. The shaft may measure anywhere from 39.95 mm to 40.05 mm, but only if the surface roughness is also within 0.05 µm.
  2. The shaft may measure anywhere from 39.95 mm to 40.05 mm and still conform. Correct answer.
  3. The shaft must measure exactly 40.00 mm with a permissible deviation of 0.05 mm in one direction only.
  4. The shaft may measure anywhere from 39.90 mm to 40.10 mm, since the tolerance applies to each side of the diameter.

Explanation

A bilateral tolerance of ±0.05 mm means the permitted deviation from the nominal size is 0.05 mm in either direction. Adding and subtracting this from the 40.00 mm nominal gives limits of 39.95 mm and 40.05 mm. Any size within these limits, inclusive, satisfies the drawing specification.

Question 28Engineering Fundamentals and Applied Science

A simply supported beam of span L carries a uniformly distributed load w over its full length. A point load of the same total magnitude (wL) is then applied at midspan instead. How does the maximum bending moment change?

  1. It decreases from wL²/8 to wL²/16, so it halves.
  2. It stays the same at wL²/8 because the total load is unchanged.
  3. It increases from wL²/8 to wL/4, so it doubles. Correct answer.
  4. It increases from wL²/8 to 3wL²/16, a rise of 50 %.

Explanation

For a uniformly distributed load the maximum moment is wL²/8 at midspan. For a central point load P the maximum moment is PL/4; substituting P = wL gives wL²/4. The moment therefore doubles when the same total load is concentrated at midspan, because the load is applied further from the supports on average.

Question 29Engineering Fundamentals and Applied Science

A mild steel specimen is loaded in tension and the stress is held within the elastic region. Which combination of changes occurs as the applied stress is increased?

  1. Longitudinal strain increases while lateral strain increases with the same sign and magnitude.
  2. Longitudinal strain increases while lateral strain decreases in magnitude as stress rises.
  3. Both longitudinal and lateral strain increase in magnitude, with the lateral strain opposite in sign. Correct answer.
  4. Longitudinal strain increases while lateral strain remains zero until yield is reached.

Explanation

Within the elastic range, axial tension produces a positive longitudinal strain and a negative lateral strain, linked by Poisson's ratio. Both magnitudes grow in proportion to the applied stress while the material remains linearly elastic. The lateral contraction is a fixed fraction of the axial extension, not an independent or vanishing quantity.

Question 30Engineering Fundamentals and Applied Science

A closed-loop control system has a forward path transfer function G(s) = K/s and unity negative feedback. A step input is applied. Which statement about the steady-state error is correct?

  1. The steady-state error to a step input is K/(1+K) and rises as K is increased.
  2. The steady-state error to a step input is infinite because the plant contains an integrator.
  3. The steady-state error to a step input is zero for any K greater than zero, because the open-loop system is type 1. Correct answer.
  4. The steady-state error to a step input is 1/(1+K) and falls as K is increased.

Explanation

The open-loop transfer function K/s has one pole at the origin, making it a type 1 system. A type 1 system tracks a step input with zero steady-state error, provided the closed loop is stable for the chosen K. The integrator accumulates until the error is driven to zero, so the value of K does not affect the final error for a step.

Question 31Engineering Design and Analysis

In a design review for a bracket assembly, an engineer specifies a safety factor of 2.0 on yield strength. Which statement best describes what this safety factor provides?

  1. The ratio of ultimate strength to applied working stress only
  2. A margin covering uncertainties in loading, material properties, and analysis assumptions Correct answer.
  3. A direct measure of the component's stiffness under service loads
  4. A guarantee that the component will never fail under any conceivable load

Explanation

A design safety factor expresses how much the allowable stress exceeds the expected working stress, absorbing uncertainty in loads, material data, and modeling. It is a margin against unknowns, not a proof of absolute safety, and it is distinct from stiffness or a single strength ratio.

Question 32Engineering Design and Analysis

A simply supported beam of span L carries a uniformly distributed load w per unit length over its full length. What is the maximum bending moment?

  1. wL²/8 Correct answer.
  2. wL²/4
  3. wL²/12
  4. wL²/2

Explanation

For a simply supported beam with a full uniform load, the maximum moment occurs at midspan and equals wL²/8. The end moments are zero because the supports are simple pins or rollers.

Question 33Engineering Design and Analysis

An engineer needs to select a material for a lightweight beam where bending stiffness is critical and mass must be minimized. Which material index should be maximized?

  1. E^(1/2)/ρ Correct answer.
  2. σ_y/ρ
  3. E/ρ
  4. ρ/E

Explanation

For a stiffness-limited beam of free cross-section, the mass scales inversely with E^(1/2)/ρ, so maximizing this index minimizes weight. The exponent arises because section geometry can grow as stiffness demands increase.

Question 34Engineering Design and Analysis

A shaft transmits torque T with a solid circular cross-section of diameter d. Which expression gives the maximum shear stress?

  1. 4T/(πd²)
  2. 32T/(πd⁴)
  3. 8T/(πd²)
  4. 16T/(πd³) Correct answer.

Explanation

Torsional shear stress is τ = Tc/J, where c = d/2 and J = πd⁴/32 for a solid shaft. Substituting gives τ = 16T/(πd³).

Question 35Engineering Design and Analysis

A component experiences a fully reversed bending stress with alternating amplitude 120 MPa and mean zero. Using the Goodman relation with ultimate strength 600 MPa and endurance limit 240 MPa, what equivalent fully reversed stress should be compared to the endurance limit?

  1. 480 MPa
  2. 120 MPa Correct answer.
  3. 240 MPa
  4. 360 MPa

Explanation

With zero mean stress, the Goodman line reduces to the alternating stress itself, so the equivalent fully reversed stress equals 120 MPa. The mean-stress correction only changes the result when a nonzero mean is present.

Question 36Engineering Design and Analysis

In a tolerance stack-up analysis, what does the worst-case method assume about individual component tolerances?

  1. Tolerances are negligible compared with nominal dimensions
  2. Tolerances follow a normal distribution and combine by root-sum-square
  3. Only the largest single tolerance affects the assembly
  4. All tolerances occur simultaneously at their extreme limits in the unfavorable direction Correct answer.

Explanation

Worst-case (arithmetic) stack-up sums all tolerances at their extremes, giving the largest possible variation. It is conservative and does not rely on statistical assumptions about distributions.

Question 37Engineering Design and Analysis

A pressure vessel is being designed as a thin-walled cylinder with internal pressure p, radius r, and wall thickness t. Which stress is the larger principal membrane stress?

  1. Longitudinal stress, pr/(2t)
  2. Torsional shear stress pr/t
  3. Hoop stress, pr/t Correct answer.
  4. Radial stress equal to p at the inner surface

Explanation

In a closed thin-walled cylinder the hoop stress is pr/t and the longitudinal stress is pr/(2t), so hoop stress is twice the longitudinal value. Radial stress is small compared with these and varies through the wall.

Question 38Engineering Design and Analysis

Which of the following best distinguishes a design constraint from a design criterion?

  1. A constraint is optional; a criterion is mandatory
  2. A constraint applies only to cost; a criterion applies only to strength
  3. A constraint is a limit that any acceptable solution must satisfy; a criterion is a measure used to compare solutions Correct answer.
  4. A constraint is quantitative; a criterion must be qualitative

Explanation

Constraints are hard limits that eliminate infeasible designs, such as maximum envelope or minimum strength. Criteria (objectives) are the measures by which feasible designs are ranked, such as mass or cost.

Question 39Engineering Design and Analysis

A bolted joint carries a shear load through friction between clamped plates. Which change most directly increases the friction capacity of the joint?

  1. Increasing the bolt diameter without changing preload
  2. Reducing the plate surface roughness
  3. Adding a lubricant between the plates
  4. Increasing the clamp preload on the bolts Correct answer.

Explanation

Friction capacity equals the friction coefficient times the normal clamp force, so increasing preload raises the normal force and thus the friction capacity. Surface condition and coefficient also matter, but preload is the direct lever.

Question 40Engineering Design and Analysis

A cantilever beam of length L has a point load P at its free end. What is the maximum deflection at the free end (E = modulus, I = second moment of area)?

  1. PL³/(8EI)
  2. PL³/(3EI) Correct answer.
  3. PL³/(48EI)
  4. PL²/(2EI)

Explanation

The standard result for an end-loaded cantilever is δ = PL³/(3EI) at the free end. The cube on length reflects that deflection grows rapidly with span.

Question 41Engineering Design and Analysis

In a kinematic analysis of a four-bar linkage, Grashof's condition states that a crank can rotate fully if which relationship holds?

  1. All four links are of equal length
  2. The sum of the shortest and longest links is less than or equal to the sum of the other two links Correct answer.
  3. The sum of the two shortest links equals the sum of the two longest links
  4. The shortest link is adjacent to the fixed link only

Explanation

Grashof's criterion is s + l ≤ p + q, where s and l are the shortest and longest links. When satisfied with the shortest link adjacent to the frame, a crank-rocker results with full rotation.

Question 42Engineering Design and Analysis

A control system's plant has a transfer function with a right-half-plane pole. What does this imply about the open-loop system?

  1. It is unstable and will diverge without feedback correction Correct answer.
  2. It is stable but has excessive overshoot
  3. It has zero steady-state error
  4. It is marginally stable and oscillates indefinitely

Explanation

A right-half-plane pole corresponds to an exponentially growing response, meaning the open-loop system is unstable. Feedback may stabilize it, but the plant itself diverges.

Question 43Engineering Design and Analysis

During a design for manufacture review, an engineer adds draft angles to a part intended for injection molding. What is the primary purpose of draft?

  1. To improve the surface finish of the mold cavity
  2. To allow easy ejection of the part from the mold Correct answer.
  3. To increase the part's stiffness
  4. To reduce the required wall thickness

Explanation

Draft angles taper vertical walls so the molded part releases cleanly from the tool without dragging or scuffing. They are a manufacturability feature tied to ejection, not a structural or finishing requirement.

Question 44Engineering Design and Analysis

A strain gauge with gauge factor 2.0 is bonded to a steel member and experiences a strain of 500 microstrain. What is the fractional change in resistance?

  1. 0.001 Correct answer.
  2. 0.01
  3. 0.002
  4. 0.0005

Explanation

The fractional resistance change is ΔR/R = GF × ε = 2.0 × 500×10⁻⁶ = 0.001. This small change is why bridge circuits and amplification are used.

Question 45Engineering Design and Analysis

Which failure theory is most appropriate for predicting yielding of a ductile isotropic material under a general three-dimensional stress state?

  1. von Mises yield criterion Correct answer.
  2. Maximum principal stress criterion
  3. Maximum normal strain criterion
  4. Mohr-Coulomb criterion

Explanation

The von Mises (distortion energy) criterion predicts yielding of ductile metals well under multiaxial stress. It compares an equivalent stress to the uniaxial yield strength.

Question 46Engineering Design and Analysis

A heat sink is being designed to dissipate power from an electronic component. Which combination of properties is most desirable for the fin material?

  1. High thermal conductivity and high surface area Correct answer.
  2. High thermal conductivity and low surface area
  3. Low thermal conductivity and high heat capacity
  4. Low thermal conductivity and high emissivity only

Explanation

Fin performance improves with high thermal conductivity to spread heat and large surface area to convect it to the surroundings. Both properties work together to lower thermal resistance.

Question 47Engineering Design and Analysis

In a fatigue context, what does the endurance limit represent for a steel alloy?

  1. The maximum stress a material can sustain once before fracture
  2. The stress amplitude below which the material has effectively infinite fatigue life Correct answer.
  3. The stress amplitude at which fatigue life equals 10³ cycles
  4. The stress at which the first microcrack always initiates

Explanation

For steels, the endurance limit is the stress amplitude below which the S-N curve becomes horizontal, indicating very long life. It is not a crack-initiation threshold or a static strength.

Question 48Engineering Design and Analysis

A tolerance analysis shows a clearance fit is required between a shaft and a bore. Which condition must hold for the assembly to always have clearance?

  1. The minimum bore size exceeds the maximum shaft size Correct answer.
  2. The nominal bore equals the nominal shaft
  3. The bore tolerance equals the shaft tolerance
  4. The maximum bore size exceeds the minimum shaft size

Explanation

Guaranteed clearance requires the smallest possible hole to be larger than the largest possible shaft, i.e., minimum bore minus maximum shaft is positive. Equal nominal sizes or equal tolerances do not ensure clearance.

Question 49Engineering Design and Analysis

An engineer is comparing two candidate designs and wants to account for the time value of money over a multi-year service life. Which method is most appropriate?

  1. Lowest initial purchase price
  2. Simple payback based on first cost only
  3. Highest immediate return on sales
  4. Net present value of lifecycle costs Correct answer.

Explanation

Net present value discounts future costs and benefits to a common basis, correctly reflecting the time value of money over the service life. First-cost or payback comparisons ignore later costs and discounting.

Question 50Engineering Design and Analysis

A stepped shaft has a fillet at a change in diameter. What is the purpose of the stress concentration factor Kt in design?

  1. To account for the weight reduction from the smaller diameter
  2. To specify the surface finish required at the shoulder
  3. To convert torque into an equivalent bending moment
  4. To relate the local peak stress at the fillet to the nominal stress in the smaller section Correct answer.

Explanation

Kt multiplies the nominal stress to estimate the local peak stress caused by geometric discontinuity such as a fillet. Designers use it, often with notch sensitivity, to check fatigue and yield at the feature.