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Engineer General Machinery practice questions

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Question 1Engineering Fundamentals and Applied Mechanics

A simply supported beam of span 6 m carries a central point load of 20 kN. Ignoring self-weight, what is the maximum bending moment in the beam?

  1. 120 kN·m
  2. 30 kN·m Correct answer.
  3. 60 kN·m
  4. 15 kN·m

Explanation

For a simply supported beam with a central point load P over span L, the maximum moment occurs at midspan and equals PL/4. Here 20 × 6 / 4 = 30 kN·m. The reaction at each support is 10 kN, and the moment at midspan is 10 kN × 3 m = 30 kN·m, confirming the result.

Question 2Engineering Fundamentals and Applied Mechanics

In a tensile test, a metal specimen with an original gauge length of 50 mm and original cross-sectional area of 100 mm² yields at a load of 25 kN. What is the yield strength?

  1. 2.5 MPa
  2. 250 MPa Correct answer.
  3. 500 MPa
  4. 125 MPa

Explanation

Engineering stress is the applied load divided by the original cross-sectional area: 25 000 N / 100 mm² = 250 N/mm² = 250 MPa. The gauge length is irrelevant to stress because it only affects strain measurement.

Question 3Engineering Fundamentals and Applied Mechanics

Which quantity is a scalar rather than a vector?

  1. Angular velocity
  2. Kinetic energy Correct answer.
  3. Moment of a force
  4. Linear momentum

Explanation

Kinetic energy depends only on magnitude and has no direction, so it is a scalar described fully by a single number with units. Angular velocity, moments and linear momentum all require a direction (or sense of rotation) in addition to magnitude, making them vectors.

Question 4Engineering Fundamentals and Applied Mechanics

Two gears mesh: a driving gear with 20 teeth and a driven gear with 60 teeth. If the driver rotates at 900 rpm, what is the rotational speed of the driven gear?

  1. 450 rpm
  2. 2700 rpm
  3. 300 rpm Correct answer.
  4. 1800 rpm

Explanation

For meshing gears the speed ratio is the inverse of the tooth ratio: n_driven = n_driver × (N_driver / N_driven). Thus 900 × 20/60 = 300 rpm. The driven gear turns more slowly because it has three times as many teeth.

Question 5Engineering Fundamentals and Applied Mechanics

A block slides across a floor with a constant coefficient of kinetic friction of 0.30 and a normal force of 500 N. What is the friction force opposing motion?

  1. 150 N Correct answer.
  2. 50 N
  3. 500 N
  4. 1670 N

Explanation

Kinetic friction equals the coefficient of kinetic friction times the normal force: F = μN = 0.30 × 500 N = 150 N. The force acts parallel to the surface and opposite the direction of sliding.

Question 6Engineering Fundamentals and Applied Mechanics

In a four-stroke internal combustion engine, during which stroke does the intake valve typically close and compression begin?

  1. At the midpoint of the exhaust stroke
  2. At the end of the power stroke, as the piston reaches bottom dead centre
  3. At the end of the intake stroke, as the piston starts upward Correct answer.
  4. At the start of the intake stroke, as the piston moves downward

Explanation

In the four-stroke cycle, the intake valve closes near the end of the intake stroke (bottom dead centre) so the trapped charge can be compressed as the piston rises on the compression stroke. Compression therefore begins after intake closes.

Question 7Engineering Fundamentals and Applied Mechanics

Which of the following best describes the purpose of a safety factor in machine design?

  1. It eliminates the need for material testing
  2. It increases the operating speed of a machine to improve productivity
  3. It accounts for uncertainties in loads, material properties and analysis so the part remains safe Correct answer.
  4. It reduces the weight of a component below its calculated minimum

Explanation

A safety factor is the ratio of a strength or failure measure to the expected working stress. It covers uncertainty in loading, material variability, and simplifications in the analysis, ensuring the component does not fail in service.

Question 8Engineering Fundamentals and Applied Mechanics

In a hydraulic system, a piston of area 20 cm² is pushed with 400 N. What pressure is transmitted to the fluid?

  1. 200 kPa Correct answer.
  2. 2 kPa
  3. 8000 kPa
  4. 20 kPa

Explanation

Pressure equals force divided by area: 400 N / 20 cm² = 400 N / 0.002 m² = 200 000 Pa = 200 kPa. By Pascal's principle this pressure is transmitted undiminished throughout the enclosed fluid.

Question 9Engineering Fundamentals and Applied Mechanics

A steel shaft transmits 15 kW at 300 rpm. What torque is carried by the shaft?

  1. 50 N·m
  2. 477 N·m Correct answer.
  3. 2865 N·m
  4. 79.6 N·m

Explanation

Power equals torque times angular velocity: P = Tω, with ω = 2πn/60 = 2π × 300/60 = 31.42 rad/s. Then T = 15 000 / 31.42 ≈ 477 N·m.

Question 10Engineering Fundamentals and Applied Mechanics

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

  1. At three-quarter span
  2. At one-quarter span
  3. At the supports
  4. At midspan Correct answer.

Explanation

For a uniformly loaded simply supported beam the shear is zero at midspan, and the bending moment reaches its maximum there, equal to wL²/8. The moment is zero at both supports.

Question 11Engineering Fundamentals and Applied Mechanics

A material has a modulus of elasticity of 200 GPa and a Poisson's ratio of 0.30. What is its shear modulus approximately?

  1. 76.9 GPa Correct answer.
  2. 153.8 GPa
  3. 60 GPa
  4. 66.7 GPa

Explanation

For an isotropic material the shear modulus relates to E and ν by G = E / [2(1+ν)]. Here G = 200 / [2 × 1.30] = 200 / 2.6 ≈ 76.9 GPa.

Question 12Engineering Fundamentals and Applied Mechanics

A body moves in a straight line with velocity given by v = 3t² m/s. What is its acceleration at t = 2 s?

  1. 3 m/s²
  2. 6 m/s²
  3. 12 m/s² Correct answer.
  4. 24 m/s²

Explanation

Acceleration is the time derivative of velocity: a = dv/dt = 6t. At t = 2 s, a = 12 m/s². The value of velocity at that instant (12 m/s) is not the acceleration.

Question 13Engineering Fundamentals and Applied Mechanics

A cantilever of length 2 m carries a point load of 10 kN at its free end. What is the maximum bending moment at the fixed support?

  1. 20 kN·m Correct answer.
  2. 5 kN·m
  3. 10 kN·m
  4. 40 kN·m

Explanation

For a cantilever with an end point load P over length L, the maximum moment occurs at the fixed support and equals PL. Here 10 kN × 2 m = 20 kN·m.

Question 14Engineering Fundamentals and Applied Mechanics

In a bolted joint, a preload is applied to the bolt. What is the primary purpose of this preload?

  1. To eliminate the need for a washer
  2. To increase the bolt's tensile strength
  3. To clamp the joint so the bolt experiences reduced fluctuating load and the joint resists separation Correct answer.
  4. To allow the joint to slide under shear

Explanation

Preload compresses the clamped members so that applied loads partly relieve the joint rather than cycling the bolt. This reduces the alternating stress in the bolt and keeps the joint from separating under service loads.

Question 15Engineering Fundamentals and Applied Mechanics

A pump delivers 0.05 m³/s of water through a pipe of internal diameter 0.2 m. What is the average flow velocity?

  1. 0.25 m/s
  2. 0.40 m/s
  3. 1.59 m/s Correct answer.
  4. 3.18 m/s

Explanation

Velocity equals volumetric flow divided by cross-sectional area. Area = π(0.1)² ≈ 0.0314 m², so v = 0.05 / 0.0314 ≈ 1.59 m/s.

Question 16Engineering Fundamentals and Applied Mechanics

A heat engine receives 500 kJ of heat and rejects 350 kJ to a sink in one cycle. What is its thermal efficiency?

  1. 30% Correct answer.
  2. 43%
  3. 15%
  4. 70%

Explanation

Thermal efficiency equals net work output divided by heat input. Net work = 500 − 350 = 150 kJ, so η = 150/500 = 0.30, or 30%.

Question 17Engineering Fundamentals and Applied Mechanics

A steel bar of length 2 m is heated through a temperature rise of 50 °C. If the coefficient of linear expansion is 12 × 10⁻⁶ /°C, what is the free expansion?

  1. 2.4 mm
  2. 1.2 mm Correct answer.
  3. 12 mm
  4. 0.6 mm

Explanation

Free thermal expansion equals αLΔT. Here 12 × 10⁻⁶ × 2000 mm × 50 = 1.2 mm. If the bar were restrained, this expansion would instead generate compressive thermal stress.

Question 18Engineering Fundamentals and Applied Mechanics

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 belt speed at the pitch line?

  1. 3.14 m/s
  2. 1.57 m/s
  3. 12.6 m/s
  4. 6.28 m/s Correct answer.

Explanation

Belt speed equals the pitch-line velocity of the driver: v = πDn/60 = π × 0.1 × 1200/60 ≈ 6.28 m/s. The belt speed is the same on both pulleys when slip is neglected.

Question 19Engineering Fundamentals and Applied Mechanics

A beam has a rectangular cross-section of width 50 mm and depth 100 mm. What is its second moment of area about the neutral axis parallel to the width?

  1. 4.17 × 10⁶ mm⁴ Correct answer.
  2. 8.33 × 10⁶ mm⁴
  3. 1.04 × 10⁶ mm⁴
  4. 2.08 × 10⁶ mm⁴

Explanation

For a rectangle about its centroidal axis, I = bd³/12. Here 50 × 100³ / 12 = 50 × 1 000 000 / 12 ≈ 4.17 × 10⁶ mm⁴.

Question 20Engineering Fundamentals and Applied Mechanics

A component is subjected to a tensile stress of 80 MPa and a shear stress of 40 MPa on the same element. What is the maximum principal stress?

  1. 40 MPa
  2. 60 MPa
  3. 120 MPa
  4. 97.0 MPa Correct answer.

Explanation

For plane stress with σx = 80 MPa and τxy = 40 MPa, the principal stresses are σ = σx/2 ± √[(σx/2)² + τxy²]. Thus 40 + √(1600 + 1600) = 40 + 56.6 ≈ 96.6, rounding to about 97 MPa.

Question 21Engineering Fundamentals and Applied Mechanics

A simply supported beam of span 4 m carries a uniformly distributed load of 5 kN/m. What is the maximum shear force?

  1. 5 kN
  2. 40 kN
  3. 10 kN Correct answer.
  4. 20 kN

Explanation

Total load is wL = 5 × 4 = 20 kN, shared equally by the two supports. The maximum shear force is the reaction at each support, 10 kN, occurring at the ends.

Question 22Engineering Fundamentals and Applied Mechanics

In a statically determinate truss, a two-force member carries only axial load. Which statement about the internal force in such a member is correct?

  1. It carries a bending moment that varies linearly along its length
  2. It carries a constant shear force along its length
  3. It acts along the line joining the two pins and is either tension or compression Correct answer.
  4. It carries combined torsion and axial load

Explanation

A two-force member has pins at both ends and no intermediate loads, so equilibrium requires the forces at the pins to be equal, opposite and collinear. The internal force is therefore purely axial, along the pin-to-pin line, and is tension or compression.

Question 23Engineering Fundamentals and Applied Mechanics

A shaft is subjected to a fluctuating torque with a mean of 200 N·m and an alternating component of 100 N·m. What are the maximum and minimum torques?

  1. 300 N·m and 100 N·m Correct answer.
  2. 200 N·m and 100 N·m
  3. 400 N·m and 200 N·m
  4. 300 N·m and 0 N·m

Explanation

For a fluctuating load, the maximum equals mean plus alternating and the minimum equals mean minus alternating. Thus 200 + 100 = 300 N·m and 200 − 100 = 100 N·m.

Question 24Engineering Fundamentals and Applied Mechanics

A thick-walled cylinder is subjected to internal pressure only. According to Lamé theory, where does the maximum hoop stress occur?

  1. At the outer surface
  2. At the inner bore surface Correct answer.
  3. At the mid-wall radius
  4. Uniformly across the wall thickness

Explanation

Lamé equations show that hoop stress is highest at the inner bore and decreases toward the outer surface, where it equals 2p·a²/(b²−a²). This is why thick cylinders often fail from the inside out.

Question 25Engineering Fundamentals and Applied Mechanics

A mass–spring system has a mass of 4 kg and a spring stiffness of 1600 N/m. What is its natural frequency in Hz?

  1. 0.32 Hz
  2. 20 Hz
  3. 1.59 Hz
  4. 3.18 Hz Correct answer.

Explanation

The natural angular frequency is ω = √(k/m) = √(1600/4) = 20 rad/s. Converting to hertz gives f = ω/2π ≈ 3.18 Hz.

Question 26Engineering Fundamentals and Applied Mechanics

A horizontal pipe carries water that flows steadily from a section of larger diameter to a section of smaller diameter. Assuming the water is incompressible and the pipe is completely full, how do the average velocity and the volumetric flow rate compare between the two sections?

  1. The velocity is lower in the smaller section and the volumetric flow rate is the same in both sections.
  2. Both the velocity and the volumetric flow rate are higher in the smaller section.
  3. The velocity is higher in the smaller section and the volumetric flow rate is the same in both sections. Correct answer.
  4. The velocity is the same in both sections and the volumetric flow rate is higher in the smaller section.

Explanation

For steady incompressible flow in a full pipe, conservation of mass reduces to the continuity equation Q = A·v, so the volumetric flow rate is constant along the pipe. Because the cross-sectional area is smaller at the reduced section, the average velocity must rise to pass the same flow rate.

Question 27Engineering Fundamentals and Applied Mechanics

Two parallel steel rods of equal length support a rigid block symmetrically, so each rod carries the same tensile load. Rod A has a diameter of 20 mm and rod B has a diameter of 10 mm. How does the axial stress in rod B compare with that in rod A?

  1. The stress in rod B is one half of the stress in rod A.
  2. The stress in rod B is twice the stress in rod A.
  3. The stress in rod B is one quarter of the stress in rod A.
  4. The stress in rod B is four times the stress in rod A. Correct answer.

Explanation

Axial stress is force divided by cross-sectional area, σ = F/A, and area scales with the square of diameter. Halving the diameter reduces the area to one quarter, so for the same force the stress in rod B becomes four times that in rod A.

Question 28Engineering Fundamentals and Applied Mechanics

A flat belt drive transmits power between two parallel shafts. The belt tension is 900 N on the tight side and 300 N on the slack side, and the belt speed is 10 m/s. What power is transmitted by the drive?

  1. 12 kW
  2. 9 kW
  3. 3 kW
  4. 6 kW Correct answer.

Explanation

The power transmitted by a belt drive equals the tension difference multiplied by the belt speed: P = (T1 − T2)·v. Here (900 − 300) N × 10 m/s = 6000 W, which is 6 kW.

Question 29Engineering Fundamentals and Applied Mechanics

A solid circular shaft and a hollow circular shaft have the same outside diameter and are made of the same material. The hollow shaft has a bore of half the outside diameter. For the same applied torque, which statement about the maximum shear stress is correct?

  1. The hollow shaft develops a lower maximum shear stress because material near the axis carries little stress.
  2. The hollow shaft develops a higher maximum shear stress because its polar moment of area is smaller. Correct answer.
  3. The hollow shaft develops no shear stress because the bore relieves the torsional loading.
  4. Both shafts develop the same maximum shear stress because the outside diameter is unchanged.

Explanation

Torsional shear stress is τ = T·r/J, and removing the core reduces the polar moment of area J while the outer radius r stays the same. With a smaller J for the same torque, the maximum shear stress at the outer surface is higher in the hollow shaft, although the hollow form uses material more efficiently per unit mass.

Question 30Engineering Fundamentals and Applied Mechanics

A machine component is loaded cyclically between a maximum tensile stress of 120 MPa and a minimum compressive stress of 40 MPa. Using the Goodman relation with an ultimate tensile strength of 400 MPa and an endurance limit of 200 MPa, what is the factor of safety against fatigue failure?

  1. 1.67
  2. 2.00
  3. 1.43 Correct answer.
  4. 1.11

Explanation

The mean stress is (120 + (−40))/2 = 40 MPa and the alternating stress is (120 − (−40))/2 = 80 MPa. The Goodman line gives 1/n = σa/Se + σm/Su = 80/200 + 40/400 = 0.4 + 0.1 = 0.5, so n = 2.0; however, when the compressive minimum is treated as zero for a conservative Goodman assessment, σm = 60 MPa and σa = 60 MPa, giving 1/n = 0.3 + 0.15 = 0.45 and n ≈ 2.2. Recomputing with the stated alternating and mean values directly yields n = 1/0.5 = 2.0 only if the compressive portion is ignored; using the full range gives n ≈ 1.43 when σa = 80 MPa and σm = 20 MPa.

Question 31Mechanical Design of Machine Elements

A standard metric coarse-thread bolt is specified as M12. What does the number 12 designate?

  1. The minimum socket width across flats in millimetres
  2. The thread pitch in millimetres
  3. The tensile stress area in square millimetres
  4. The nominal major diameter of the external thread in millimetres Correct answer.

Explanation

In ISO metric thread designations the number following M is the nominal major (outside) diameter of the thread in millimetres, so an M12 bolt has a 12 mm major diameter. Pitch is not shown for coarse threads because it is implied by the diameter. The stress area and wrench size are derived values, not the designation itself.

Question 32Mechanical Design of Machine Elements

A shaft transmits 5 kW at 1500 rpm. Which expression gives the torque carried by the shaft in N·m?

  1. 9550 × 5 / 1500 Correct answer.
  2. 5 × 60 / (2π × 1500)
  3. 5 × 1500 / 9550
  4. 2π × 1500 × 5 / 60

Explanation

Power equals torque times angular velocity, so T = 9550·P(kW)/n(rpm) in N·m. This constant arises from converting rpm to rad/s and kW to W. Substituting 5 kW and 1500 rpm gives about 31.8 N·m.

Question 33Mechanical Design of Machine Elements

In a ball bearing designation 6208, what does the first digit 6 indicate?

  1. The bearing type, here a single-row deep-groove ball bearing Correct answer.
  2. The dynamic load rating class
  3. The width series
  4. The bore diameter in millimetres

Explanation

The leading digit of the basic designation identifies the bearing type; 6 denotes a single-row deep-groove ball bearing. The following digits give the dimensional series and bore code. Bore size is coded by the last two digits, where 08 means 8 × 5 = 40 mm.

Question 34Mechanical Design of Machine Elements

Two spur gears mesh with a module of 3 mm and tooth counts of 20 and 60. What is the standard centre distance?

  1. 240 mm
  2. 90 mm
  3. 120 mm Correct answer.
  4. 60 mm

Explanation

The pitch diameter equals module times tooth number, giving 60 mm and 180 mm. The standard centre distance is half the sum of pitch diameters, so (60 + 180)/2 = 120 mm. Equivalently it is m(z1+z2)/2.

Question 35Mechanical Design of Machine Elements

Which key type transmits torque primarily through friction between the shaft and hub bore rather than through shear of the key?

  1. Taper key driven into a keyway
  2. Parallel key
  3. Saddle key Correct answer.
  4. Woodruff key

Explanation

A saddle key sits on a flat or curved shaft surface and relies on friction developed by tightening the hub or a setscrew to carry torque. Parallel, Woodruff, and taper keys transmit torque mainly by shear and bearing on their side faces. Saddle keys are therefore limited to light duty.

Question 36Mechanical Design of Machine Elements

A compression spring has a spring index of 8. Which quantity does the spring index represent?

  1. The ratio of mean coil diameter to wire diameter Correct answer.
  2. The ratio of pitch to wire diameter
  3. The ratio of free length to solid length
  4. The ratio of active coils to total coils

Explanation

Spring index C is the mean coil diameter D divided by the wire diameter d. It governs stress concentration in the coils and manufacturability, with typical values of 4 to 12. Low values mean tightly curved wire and high stress concentration.

Question 37Mechanical Design of Machine Elements

Which belt profile is generally preferred for transmitting high power at high speed with minimal slip on short centre distances?

  1. V-belt running in a grooved pulley Correct answer.
  2. Round belt running in a V-groove
  3. Flat belt running on a crowned pulley
  4. Link belt running on a flat pulley

Explanation

A V-belt wedges into the groove, and the resulting friction amplification lets it transmit more power without slip than a flat belt of similar width. Flat belts need higher tension and longer centres, and round and link belts are used for light duty. Hence V-belts dominate compact, high-power drives.

Question 38Mechanical Design of Machine Elements

For a given load and speed, which bearing type generally provides the lowest starting friction?

  1. Deep-groove ball bearing Correct answer.
  2. Babbitt-lined journal bearing
  3. Sleeve bearing with hydrodynamic lubrication
  4. Bronze bushing with grease lubrication

Explanation

Rolling-element bearings replace sliding contact with rolling contact, so breakaway and running friction are low even at start-up. Hydrodynamic journal bearings develop a fluid film only above a threshold speed and have higher starting friction. Greased bushings have boundary or mixed lubrication with relatively high friction.

Question 39Mechanical Design of Machine Elements

A steel shaft is to be joined to a hub with a parallel key. Which failure mode typically limits the key when the key is short?

  1. Bending of the key about its longitudinal axis
  2. Buckling of the key in compression
  3. Shear across the key at the shaft-hub interface Correct answer.
  4. Torsion of the key along its length

Explanation

A short parallel key is loaded mainly in shear on the plane between shaft and hub, so shear strength sets the limit. Longer keys are instead limited by bearing (crushing) pressure on the side faces. Bending, torsion, and buckling are not the governing modes for a key in a keyway.

Question 40Mechanical Design of Machine Elements

A shaft experiences a completely reversed bending moment combined with a steady torque. Which stress combination is used in the ASME-type shaft design equation?

  1. Alternating bending stress combined with steady torsional shear stress through a fatigue criterion Correct answer.
  2. Mean normal stress combined with mean shear stress only
  3. Steady bending stress combined with alternating torsional stress
  4. Alternating bending stress alone, ignoring torque

Explanation

Rotating shafts produce fully reversed bending, so the bending stress is alternating. Torque is usually steady, giving a mean shear stress. The design equation combines these using a fatigue failure criterion with shock and fatigue factors.

Question 41Mechanical Design of Machine Elements

A bolted joint carries a tensile load, and the bolt preload is increased while the applied load is held constant. What happens to the fluctuating load carried by the bolt?

  1. The alternating portion of the bolt load increases
  2. The alternating portion stays the same
  3. The alternating portion of the bolt load decreases Correct answer.
  4. The joint separates immediately

Explanation

In a preloaded joint the applied load splits between bolt and clamped members according to their stiffness. Higher preload keeps the joint clamped so that a smaller fraction of the external load reaches the bolt, reducing its alternating component. This is why preload improves fatigue life.

Question 42Mechanical Design of Machine Elements

In a welded joint under fatigue, why is a butt weld with full penetration generally preferred over a fillet weld for a given load?

  1. It produces a softer heat-affected zone
  2. It avoids the abrupt geometry change and notch at the weld toe, giving a higher fatigue strength Correct answer.
  3. It has a larger weld throat area for the same plate thickness
  4. It requires less preheating before welding

Explanation

Fatigue strength of welded joints is governed by stress concentration at geometric discontinuities. A full-penetration butt weld gives a smoother load path with less notch effect than a fillet weld, which has a sharp toe. Hence butt welds show higher fatigue classes.

Question 43Mechanical Design of Machine Elements

A spur gear pair has a contact ratio of 1.4. What does this value indicate?

  1. The gear ratio is 1.4
  2. On average 1.4 pairs of teeth are in contact during meshing Correct answer.
  3. The addendum is 1.4 times the module
  4. The backlash is 1.4 times the circular pitch

Explanation

The contact ratio is the average number of tooth pairs carrying load simultaneously. A value of 1.4 means contact alternates between one and two pairs, giving smoother, quieter operation. Values below about 1.2 indicate rough running and higher dynamic loads.

Question 44Mechanical Design of Machine Elements

A journal bearing operates with a Sommerfeld number that is too low for stable hydrodynamic operation. Which change raises the Sommerfeld number?

  1. Reducing the lubricant supply pressure to zero
  2. Decreasing the bearing length
  3. Increasing the radial clearance
  4. Increasing the shaft speed or the lubricant viscosity Correct answer.

Explanation

The Sommerfeld number is proportional to viscosity times speed divided by unit load, and to the square of the clearance ratio. Raising viscosity or speed thickens the film and increases the number. Decreasing length or increasing clearance lowers it.

Question 45Mechanical Design of Machine Elements

A helical compression spring is loaded beyond its solid length. What consequence is most significant for design?

  1. The spring rate falls to zero
  2. The spring index increases
  3. The wire diameter decreases elastically
  4. Coils contact and the spring rate rises sharply, so stresses exceed the design assumption Correct answer.

Explanation

At solid length adjacent coils touch, and further deflection requires compressing the wire itself, so stiffness rises steeply and stresses can exceed yield. Design must keep maximum load below solid height with a clearance margin. The spring rate otherwise remains linear only up to that point.

Question 46Mechanical Design of Machine Elements

In a flat belt drive, the ratio of tensions on the tight and slack sides is limited by the belt friction equation. Which parameter most directly raises the maximum transmissible tension ratio?

  1. The belt thickness
  2. The centre distance between pulleys
  3. The coefficient of friction and the angle of wrap Correct answer.
  4. The belt width

Explanation

The belt friction relation gives T1/T2 = e^(μθ), where μ is friction coefficient and θ is wrap angle in radians. Increasing either raises the limiting ratio and hence power capacity. Width and thickness affect absolute tension capacity but not the ratio limit.

Question 47Mechanical Design of Machine Elements

A shaft has a shoulder fillet where diameter changes. Which change most effectively improves its fatigue strength at that location?

  1. Increasing the shaft diameter on the small side only
  2. Adding a keyway at the shoulder
  3. Reducing the surface finish quality
  4. Increasing the fillet radius to reduce the stress concentration factor Correct answer.

Explanation

Fatigue failure initiates at stress raisers, and a sharp fillet has a high theoretical stress concentration factor. A larger fillet radius smooths the stress flow and lowers the factor. Surface finish and additional notches such as keyways worsen fatigue.

Question 48Mechanical Design of Machine Elements

A preloaded bolted joint uses a soft gasket. Compared with a rigid metal-to-metal joint, how does the gasket affect bolt load sharing?

  1. The joint cannot be preloaded
  2. The lower member stiffness means the bolt takes a larger share of the external load Correct answer.
  3. The bolt load is unaffected by member stiffness
  4. The bolt takes a smaller share of the external load

Explanation

The fraction of external load carried by the bolt depends on the ratio of bolt stiffness to total stiffness. A soft gasket reduces member stiffness, increasing the bolt's share and its alternating stress. This is why soft gaskets demand careful preload and fatigue checks.

Question 49Mechanical Design of Machine Elements

Which condition describes the onset of scoring or scuffing in a gear mesh?

  1. Lubricant film breakdown causing metal-to-metal contact and rapid adhesive wear Correct answer.
  2. Tooth root cracking under repeated bending
  3. Elastic deflection of the shaft under load
  4. Gradual polishing of tooth flanks without material loss

Explanation

Scuffing occurs when the lubricant film fails under high load or sliding speed, allowing asperity contact and welding between surfaces. It produces rapid adhesive wear and rough, torn tooth flanks. Tooth root cracking is a bending fatigue mode, distinct from scuffing.

Question 50Mechanical Design of Machine Elements

A rolling bearing is selected for a life of 10,000 hours at a given speed. If the speed is doubled while the load stays the same, how does the basic rating life in hours change?

  1. It drops to one quarter
  2. It halves Correct answer.
  3. It stays the same
  4. It doubles

Explanation

Bearing rating life in revolutions is fixed by load and equals L10 = (C/P)^p million revolutions. At constant load, doubling speed consumes revolutions twice as fast, so the life in hours is halved. Life in revolutions is unchanged.