Engineering Materials
Q. Which of the following parameters can be obtained by tension test of a standard specimen?
a) Proportional Limit
b) Yield Strength
c) Percentage Reduction in area
d) All of the mentioned
.
Answer: A
Explanation: Tension test determines the parameters related to stress strain curve and also reduction and elongation in area and length can be found respectively.
Q. Proportional Limit is defined as the stress at which the stress strain curves begins to deviate from the straight line.
a) True
b) False
.
Answer: A
Explanation: Proportional limit is the point of deviation in stress strain curve.
Q. Which of the following is the definition of Compliance?
a) Inverse of Rigidity
b) Inverse of Stiffness
c) Proportional to elastic Limit
d) None of the mentioned
.
Answer: B
Explanation: Compliance is a parameter which is the inverse of stiffness.
Q. Yield strength is defined as the maximum stress at which a marked increase in elongation occurs without increase in________
a) Load
b) Strength
c) Toughness
d) Hardness
.
Answer: A
Explanation: Yield strength is the strength at which strain increases. Now as the strength is constant, therefore load is constant.
Q. For the materials which do not exhibit a well-defined yield points, the yield strength is defined as the stress corresponding to a permanent set of how much percentage of gauge length?
a) 0.1
b) 0.2
c) 0.3
d) 0.4
.
Answer: B
Explanation: Generally for all the practical purpose, an offset of 0.2% of gauge length is considered.
Q. Proof strength is defined as the stress which will produce a permanent extension of how much percentage in the gauge length of the standard test specimen.
a) 0.1
b) 0.2
c) 0.3
d) 0.4
.
Answer: A
Explanation: Proof strength is used in design of fasteners. It corresponds to 0.1% permanent deformation of gauge length.
Q. All engineering materials are plastic.
a) True
b) False
.
Answer: A
Explanation: Yes all the existing engineering materials are plastic to some extent.
Q. Which of the following are true about plasticity?
a) Permanent Deformation
b) Ability to retain deformation under load or after removal of load
c) Plastic deformation is greater than elastic deformation
d) All of the mentioned
.
Answer: D
Explanation: This is the basic definition of plasticity.
Q. Which of the following is measure of stiffness?
a) Modulus of elasticity
b) Modulus of plasticity
c) Resilience
d) Toughness
.
Answer: A
Explanation: Stiffness is the ability of material to resist deformation under external load. Hence it is measured by modulus of elasticity.
Q.. Which of the following facts are true for resilience?
a) Ability of material to absorb energy when deformed elastically
b) Ability to retain deformation under the application of load or after removal of load
c) Ability of material to absorb energy when deformed plastically
d) None of the mentioned
.
Answer: A
Explanation: Toughness is ability to store energy till proportional limit during deformation and to release this energy when unloaded.
1Q. Modulus of resilience is defined as
a) Strain energy per unit volume
b) Strain energy per unit area
c) Independent of strain energy
d) None of the mentioned
.
Answer: A
Explanation: Modulus of resilience is strain energy per unit volume.
1Q. Which of the following are true for toughness
a) Ability of material to absorb energy before fracture
b) Measured by Izod & Charpy test
c) Decreases with the increase in temperature
d) All of the mentioned
.
Answer: D
Explanation: Toughness is measure of energy stored till fracture occurs. With rise of temperature, molecular bonds weakens and fracture occurs at a lesser load, hence area under stress strain curve decreases.
1Q. Malleability increases with temperature while ductility decreases with temperature.
a) True
b) False
.
Answer: A
Explanation: Malleability is the deformation under compressive load and hence with rise in temperature, the deformation increases and hence malleability increases.
1Q. Hardness is
a) Surface property
b) Resistance to abrasion
c) Depends upon resistance to plastic deformation of the material
d) All of the mentioned
.
Answer: D
Explanation: Hardness is a surface property which is the resistance of material to surface indentation. It depends upon resistance to plastic deformation.
Q. Ductile cast iron is
a) Also called nodular cast iron
b) Also called spheroidal graphite cast iron
c) Carbon is present in form of spherical nodules
d) All of the mentioned
.
Answer: D
Explanation: Ductile cast iron, also called as nodular cast iron or spheroidal cast iron consists of carbon in the form of spherical nodules in a ductile matrix.
Q. Grey Cast iron is formed when
a) Carbon content in the alloy exceeds the amount that can be dissolved
b) Carbon content in the alloy is less than the amount that can be dissolved
c) Carbon content in the alloy is equal to the amount that can be dissolved in the alloy
d) None of the mentioned
.
Q. White cast iron is formed when
a) Graphite flakes are formed
b) Most of the carbon content in the alloy forms iron carbide
c) No iron carbide is formed
d) None of the mentioned
.
Answer: B
Explanation:White cast iron is formed when most carbon in alloy forms iron carbide and there are no graphite flakes.
Q. The term high alloy steels is used for alloy steels containing more than ___ of alloying elements.
a) Q.%
b) 20%
c) 50%
d) 70%
.
Answer: A
Explanation: The term high alloy steels is used when alloying elements exceed Q.%.Low and medium alloy steels are those when this figure is less than Q.%.
Q. Which of the following are true about low carbon steels?
a) Carbon content < 0.3%
b) Also called as mild steel
c) Are soft and ductile
d) All of the mentioned
.
Answer: D
Explanation: Low carbon steels contain carbon <0.3%. Because of low carbon, it is soft and ductile.
Q. Which of the following are true for high carbon steels?
a) Carbon content=0.4%
b) Respond readily to heat treatments
c) Have much ductility as compared to low and medium carbon steels
d) Easy to weld
.
Answer: B
Explanation: High carbon steel have carbon % >0.5 and hence have less ductility. They are difficult to weld.
Q. Which of the following are true?
a) 7C4 grade steel is more ductile than Q.C4
b) Yield strength of 40C8 is greater than of 30C8
c) Hardness of 30C8 is greater than of 40C8
d) None of the mentioned
.
Answer: A
Explanation: The number before “C” indicates measure of carbon. Hence a) is true as higher the carbon, lower the ductility.
Q. In the designation “xCb” which of the following are true?
a) x indicates Q.0 times the average % of carbon
b) b indicates Q.0 times the average percentage of manganese
c) x indicates Q. times the average percentage of carbon while b indicates Q.0 times the average percentage of manganese
d) b indicates Q.0 times the average % of carbon
.
Answer: A
Explanation: X indicates Q.0 times carbon % and b indicates Q. times manganese %.
Q. Which of the steel in given options is best suited for auto mobile bodies and hoods?
a) 7C4
b) Q.C4
c) 30C8
d) 40C8
.
Answer: A
Explanation: For automobile application like manufacturing bodies, ductile material is preferred.
Q.. Which of the following steel given in options respond fastest and easily to heat treatment?
a) 7C4
b) Q.C4
c) 30C8
d) 40C8
.
Answer: D
Explanation: Alloys with higher carbon % respond willingly to heat treatment.
1Q. Which of the following given steels is popular with the name of machinery steel?
a) Low carbon steel
b) Medium carbon steel
c) High carbon steel
d) None of these
.
Answer: B
Explanation: It has ductility and can be heat treated easier than low carbon steel.
Q. Which of the following property is affected by heat treatment?
a) Hardness
b) Strength
c) Ductility
d) All of the mentioned
.
Answer: D
Explanation: Heat treatment involves changes in the micro structure and hence all the internal properties are effected.
Q. Annealing involves heating the component to a temperature
a) Slightly above the critical temperature
b) Equal to critical temperature
c) Slightly less than critical temperature
d) None of the mentioned
.
Answer: A
Explanation: In annealing, component is heated to a temperature above than critical temperature.
Q. Which of the following is true?
a) Rate of cooling in normalising is faster then in annealing
b) Annealing improves ductility
c) Normalising improves grain structure
d) All of the mentioned
.
Answer: D
Explanation: In annealing, the furnace is switched off and component cools slowly. In normalising, component is air cooled.
Q. Quenching
a) Consists of heating the component to critical temperature
b) Cooling rapidly
c) Increases hardness
d) All of the mentioned
.
Answer: D
Explanation: During quenching, component is rapidly cooled which leads to formation of martensite. Hence hardness increases.
Q. Tempering involves
a) Reheating the quenched component to a temperature greater than critical temperature
b) Increases the brittleness
c) Reheating the quenched component to a temperature equal to critical temperature
d) None of the mentioned
.
Answer: D
Explanation: Tempering involves reheating the quenched product to a temperature less than transformation range. It improves ductility and reduces brittleness.
Q. Silicon addition in spring steel increases its toughness.
a) True
b) False
.
Answer: A
Explanation: Silicon addition increases strength without lowering the ductility.
Q. Nickel addition in alloys
a) Increases toughness
b) Increases hardenability and impact resistance
c) Limit grain growth during heat treatment process
d) All of the mentioned
.
Answer: D
Explanation: Nickel addition increases toughness by limiting grain growth.
Q. Flame hardening involves
a) Heating the surface above the trAnswerformation range
b) Quenching after heating
c) Minimum case depth is 1mm
d) All of the mentioned
.
Answer: D
Explanation: Flame Hardening is a process of heating the surface with a flame above critical temperature and then quenching it.
Q. Induction hardening process involves
a) Heating surface by induction in field of invariable current
b) Case depth minimum of 2mm are produced
c) Heating surface by induction in field of alternating current
d) None of the mentioned
.
Answer: C
Explanation: Heating can only be done in presence of alternating current and not constant current.
Q.. Case carburising involves
a) Introducing carbon at surface layer
b) Heating range 880 to 980’C
c) Case depths up to 2mm are possible
d) All of the mentioned
.
Answer: D
Explanation: Case carburising involves introducing carbon at surface layer. Medium can be liquid, solid or gas and high case depths are possible.
1Q. Which of the following are not true for carbo nitriding?
a) Introducing carbon and nitrogen at surface layer
b) Component is heated in range of 650 to 920’C
c) Cyaniding is similar to carbo nitriding except that the medium is liquid
d) This process gives a lower wear resistance compared to case carburising process
.
Answer: D
Explanation: Carbo nitriding gives wear resistance greater than compared to case carburising.
1Q. Which of the following are true for nitriding?
a) Nascent oxygen is involved
b) Temperature range 490 to 590’C
c) Gaseous or liquid medium
d) All of the mentioned
.
Answer: D
Explanation: In nitriding nascent oxygen is acted on the surface of the product at a temperature of 490::590’C in a gaseous or liquid medium.
Q. Cast steel components do not exhibit the effect of directionality on mechanical properties.
a) True
b) False
.
Answer: A
Explanation: In casting process, there is no control over the fibre structure, thus non directional characteristics for cast steel components.
Q. During solidification, cast steel shrinks to quite a great extent
a) True
b) False
.
Answer: A
Explanation: Excessive contractions observed in cast steel.
Q. Relative density of aluminium is roughly _______ of steel
a) one-third
b) one-fifth
c) one-tenth
d) equal
.
Answer: A
Explanation: Aluminium alloy=Q.7 Steel=Q.Q.
Q. Which of the following statement is true
a) Cast aluminium alloys are specified by a four digit system while wrought alloys by a five digit system
b) Cast aluminium alloys are specified by a five digit system while wrought alloys by a four digit system
c) Cast aluminium alloys are specified by a six digit system while wrought alloys by a five digit system
d) Cast aluminium alloys are specified by a five digit system while wrought alloys by a six digit system
.
Answer: A
Explanation: Standard set for nomenclature.
Q. Which of the following are true for aluminium
a) Low specific gravity
b) Corrosion resistance
c) High thermal conductivity
d) All of the mentioned
.
Answer: D
Explanation: All are properties of aluminium.
Q. Is it possible to completely relieve the residual stresses in a cast steel product?
a) True
b) False
.
Answer: B
Explanation: Solidification results in great amount of residual stresses.
Q. Poor fluidity and contraction are compulsory to be taken into consideration while designing a cast steel product.
a) True
b) False
.
Answer: A
Explanation: Contraction affects to quite a extent in cast steel.
Q. In alloy 4450, 2nd digit represents?
a) average percentage of major alloying elements, halved and rounded off
b) average percentage of major alloying elements
c) average percentage of minor alloying elements, halved and doubled
d) average percentage of minor alloying elements
.
Answer: A
Explanation: Second digit is used to represent major alloying element.
Q. In alloy 4450, 4 represents?
a) Silicon
b) Aluminium
c) Manganese
d) Zinc
.
Answer: A
Explanation: In alloy nomenclature,1=Al;2=Cu;3=Mn;4=Si;5=Mg.
Q.. An aluminium alloy casting with Q.8% Cu, Q.0% Fe and 0.25% Mg
a) Alloy 2585
b) Alloy 3586
c) Alloy 2686
d) Alloy 3584
.
Answer: A
Explanation: 1st digit=2(Copper) ; 2nd digit=Q.8/2=5 ;3rd digit=8(Iron); 4th digit=5(Magnesium).
Q. Which of the following are true for copper alloys
a) Posses excellent thermal and electrical conductivity
b) Can be easily cast and machined
c) Has good corrosion resistance
d) All of the mentioned
.
Answer: D
Explanation: Properties of copper.
Q. Which of the following contain copper
a) bronze
b) gunmetal
c) monel metal
d) each of the mentioned
.
Answer: D
Explanation: All the mentioned alloys contain copper to some extent.
Q. Which of the following are true
a) Brass is costlier then copper
b) Brass has excellent corrosion resistance
c) Brass has good machinability
d) Brass has poor thermal conductivity
.
Answer: B
Explanation: Brass is a good thermal conductor,is cheaper and has poor machinability.
Q. As the amount of zinc increases
a) strength of brass increases and ductility decrease
b) strength of brass increases and ductility increase
c) strength of brass decreases and ductility increases
d) strength of brass decreases and ductility decreases
.
Answer: A
Explanation: Brass contains zinc and copper. Zinc affects strength with direct increase in strength with increase of zinc content.
Q. Which of the following are true for aluminium bronze
a) contain 5-Q.% aluminium
b) excellent corrosion and is also called imitation gold
c) difficult to cast
d) all of the mentioned
.
Answer: D
Explanation: Color of aluminium bronze is similar to gold. Other two are the properties of alloys.
Q. Which of the following statements for phosphorus bronze is true?
a) contains 0.2% phosphorus
b) phosphorus increases tensile strength
c) use for worm wheels and bearings
d) all of the mentioned
.
Answer: D
Explanation: It has good tensile strength, corrosion resistance and is used for bearings.
Q. Gunnmetal is an alloy of copper which __ % tin and __% zinc.
a) Q.,2
b) 2,Q.
c) 4,5
d) 5,4
.
Answer: A
Explanation: Zinc is used to improve fluidity during casting and hence is kept lesser than tin.
Q. Monel metal is a copper nickel alloy of __% nickel and __%copper.
a) 32,65
b) 65,32
c) 20,80
d) 80,20
.
Answer: B
Explanation: Composition.
Q. Which of the following are true for die casting alloys?
a) this process consists of forcing the molten metal into a closed metal die
b) process used for a metal with low melting point
c) surface finish is excellent
d) all of the mentioned
.
Answer: D
Explanation: Characteristics of die casting alloys.
Q. Ceramics consist of
a) ionic & covalent bonding
b) covalent bonding
c) hydrogen bonding
d) none of the mentioned
.
Answer: A
Explanation: Ceramics consists of ionic and covalent bonds only.There is no hydrogen compound so no case of H boding.
Q. All ceramics contain clay.
a) True
b) False
.
Answer: B
Explanation: No this may sound convincing but all ceramics don’t contain clay.
Q. The densities of ceramics are low.
a) True
b) False
.
Answer: A
Explanation: In ceramics, different oxides and carbides are not compactly bonded.
Q. There are small voids in ceramics structure.
a) True
b) False
.
Answer: A
Explanation: Compounds are less compactly bonded and hence voids are formed.
Q. Which of the following properties are true for ceramics.
a) light weight construction with low inertia force
b) excellent wear
c) lower friction loss
d) all of the mentioned
.
Answer: D
Explanation: Properties of ceramics.
Q. Paraffin wax is represented by
a) C18H35
b) C18H40
c) C18H36
d) None of the mentioned
.
Answer: D
Explanation: Paraffin wax in a semi solid stage is represented by C18H3Q.
Q. A thermoplastic material has a linear polymer chain while a thermosetting plastic has a crosslinked polymer chain.
a) True
b) False
.
Answer: A
Explanation: Given statement is true thats why thermoplastic can be remoulded while thermosettings can’t be.
Q. Which of the following is a thermosetting polymer?
a) PVC
b) PTFE
c) Nylon
d) Polyesters
.
Answer: D
Explanation: a), b) and c) consist of linear structures while polyester consists of cross linked and hence is a thermosetting polymer.
Q. Teflon has very high coefficient of resistance.
a) True
b) False
.
Answer: B
Explanation: Teflon is very smooth and has coefficient of friction as low as 0.0Q.
Q.. Viscoelastic materials at low temperatures show no change in its strain with change in stress.
a) True
b) False
.
Answer: B
Explanation: Viscoelastic materials follow Hooke’s law at intermediate temperatures.
Manufacturing Considerations in Machine Design
1. Stressed parts are always kept in tension.
a) True
b) False
Answer: a
Explanation: Cast iron has more compressive strength.
2. Shot blasting process improves the endurance limit of the component.
a) True
b) False
Answer: a
Explanation: Shot blasting employs throwing wheel which imparts strength to the material.
Answer: A
Explanation: Shot blasting employs throwing wheel which imparts strength to the material.
3. Which process allows controlling grain structure of the product?
a) Casting
b) Forging
c) None of the mentioned
d) Die Casting
.
Answer: B
Explanation: In casting metal is in fluid state and hence impossible to control the grain structure.
4. While designing a forging, the profile is selected such that the fibrous lines are parallel to the tensile forces and perpendicular to shear forces.
a) True
b) False
.
Answer: A
Explanation: Tensile load is applied parallel to the grain structures.
5. Cast surfaces have good finishing.
a) True
b) False
.
Answer: B
Explanation: Cavity is surrounded by sand and hence not good finish is obtained.
6. In forging, metal is in which of the following stage?
a) Elastic
b) Plastic
c) Can be in any stage
d) Rigid
.
Answer: B
Explanation: Forging is the working of metal in plastic range.
7. Forged components can be held between close limits.
a) True
b) False
.
Answer: A
Explanation: There is no shrinkage problems and hence less tolerances.
8. Among casting, forging and forging which has the slowest rate of production?
a) Casting
b) Forging
c) Machining
d) All have equal rate of production
.
Answer: C
Explanation: Machining each part is very time consuming.
9. In manual assembly, cost of screw is higher than the cost of driving a screw.
a) True
b) False
.
Answer: B
Explanation: Generally manufacturing screw is cheaper than driving it.
10. Maximum carbon content in welding is usually limited to
a) 0.22%
b) 0.8%
c) 1.5%
d) 2%
.
Answer: A
Explanation: With increase of carbon, welding becomes difficult as weld becomes susceptible to cracks.
11. Which of the following is not a classification of fit?
a) Clearance
b) Transition
c) Interference
d) Enjoining
.
Answer: D
Explanation: Simple classification of fits is down in 3 categories.
12. Which of the following always provides a positive clearance between the hole and the shaft over the entire range of tolerances?
a) Clearance
b) Transition
c) Interference
d) None of the mentioned
.
Answer: A
Explanation: Clearance positive imparts a positive clearance.
13. In this case, tolerance zone of hole is entirely below that of the shaft.
a) Clearance
b) Interference
c) Enjoining
d) Non of the mentioned
.
Answer: B
Explanation: Interference fit provides a positive clearance and hence tolerance zone of hole is below that of shaft.
14. Which of the following are true for system of tolerances to the shaft and the hole?
a) In hole-basis system, various shafts are associated with a single hole
b) In hole-basis system as well as shaft-basis system, various shafts are associated with multiple holes
c) In shaft-basis system, various shafts are associated with a single hole
d) None of the mentioned
.
Answer: A
Explanation: As per the standards.
15. The description of tolerances consists of two parts namely?
a) Fundamental deviation and magnitude of tolerance
b) Magnitude of deviation and magnitude of tolerance
c) Mean deviation and magnitude of tolerance
d) Fundamental deviation and tolerance
.
Answer: A
Explanation: Fundamental deviation and magnitude of tolerance are decrypted in the tolerance.
16. Which of the following are true for fundamental deviation?
a) Gives location of tolerance zone
b) Capital letter is used for both holes and shafts
c) Small letter is used for both holes and shaft
d) None of the mentioned
.
Answer: A
Explanation: Tolerance zone is depicted by deviation while magnitude is depicted by grade.
17. How many grades of tolerances are there according to BIS system fits and tolerances?
a) 12
b) 14
c) 16
d) 18
.
Answer: D
Explanation: As per the standard.
18. Tolerance for a shaft of 50mm diameter as the basic size, with the fundamental deviation denoted by g and tolerance of grade 7 is represented as?
a) g50,7
b) 50g7
c) 7g50
d) None of the mentioned
.
Answer: B
Explanation: First fundamental deviation is written followed by diameter and grade.
19. The temperature at which new stress free grains are formed in the metal is called the ______ temperature.
a) Recrystallization
b) Crystallization
c) Solidification
d) None of the mentioned
.
Answer: A
Explanation: Definition of recrystallization temperature.
20. Hot working or cold working, which reduces strain hardening and residual stresses.
a) Hot working
b) Cold working
c) Both have equal effect
d) Impossible to detect
.
Answer: A
Explanation: In hot working grains are re arranges as per requirement and no residual stresses.
21. Hot rolled components have better toughness and ductility.
a) True
b) False
.
Answer: A
Explanation: Hot working refines grain structure an hence improves ductility and toughness.
22. Hot rolled components have better surface finish than cold rolled components.
a) True
b) False
.
Answer: B
Explanation: Cold working gives better surface finish.
Design Against Static Load
Q. A mechanical component may fail as a result of which of the following
A. elastic deflection
B. general yielding
C. fracture
D. each of the mentioned
.
Answer: A
Explanation: Failing simply means unable to perform its function satisfactorily.
Q. Type of load affects factor of safety.
A. True
B. False
.
Answer: A
Explanation: Dynamic load has higher factor of safety as compared to static loading.
Q. For cast iron components, which of the following strength are considered to be the failure criterion?
A. Ultimate tensile strength
B. Yield Strength
C. Endurance limit
D. None of the mentioned
.
Answer: A
Explanation: Ultimate tensile strength is the highest stress a component can undergo before failingand hence is used as a criterion.
Q. For components made of ductile materials like steel, subjected to static loading which of the following strength is used as a failure of criterion?
A. Yield strength
B. Ultimate strength
C. Endurance limit
D. None of the mentioned
.
Answer: A
Explanation: In elastic material there is considerable plastic deformation at yielding point.
Q. Pitting occurs on _____ of the component.
A. Surface
B. Inner body
C. Inside or on surface
D. outside
.
Answer: A
Explanation: Pitting is a process in which small holes occur on a surface of component.
Q. Buckling is elastic instability which leads to sudden large lateral deflection.
A. True
B. False
.
Answer: A
Explanation: Definition of buckling.
Q. The critical buckling load depends upon which of the following parameters?
A. Yield strength
B. Modulus of elasticity
C. Radius of gyration
D. Each of the mentioned
.
Answer: D
Explanation: It depends on moment of inertia(which further depends on radius of gyration), elasticity and yield strength.
Q. If there are residual stresses in the material, than lower factor of safety is used.
A. True
B. False
.
Answer: B
Explanation: Residual stress increases the chance of failure.
Q. Which of the following relationship is true? (p=Poisson’s ratio)
A. E=2G (1+p)
B. E=G (2+p)
C. E= 2(G+ p)
D. No relation exist between E, G and p
.
Answer: A
Explanation: Formula.
Q.. Modulus of rigidity for carbon steels is greater than that of grey cast iron.
A. True
B. False
.
Answer: A
Explanation: Modulus of rigidity is double for carbon steels as compared to grey cast iron.
Q.. According to principal stress theory, which option represents the correct relation between yield strength in shear (YSS) and the yield strength in tension (YST)?
A. YSS=0.5YST
B. YSS=0.577YST
C. YST=0.5YSS
D. YST=0.577YSS
.
Answer: A
Explanation: Shear diagonal is at 45’ and by equation of shear stress theory, the required relation is obtained.
Q.. A beam subjected to bending moment undergoes which of the following stresses?
A. Compressive
B. Tensile
C. Both compressive & tensile
D. None of the mentioned
.
Answer: C
Explanation: The portion above the neutral axis is under compression and the portion below it is under tensile stress.
Q. The bending stress varies _______ with the distance from the neutral axis.
A. Linearly
B. Inversely
C. Squarely
D. Bending stress is independent of distance from the neutral axis
.
Answer: A
Explanation: Bending Stress= (Bending Moment x distance from neutral axis/ Moment of inertiA..
Q. Cotter joint is used when the members are subjected to which type of stresses?
A. Axial tensile
B. Axial compressive
C. Axial tensile or compressive
D. None of the mentioned
.
Answer: C
Explanation: Cotter joint is used when axial forces are applied.
Q. The principle of wedge action is used in cotter joint.
A. True
B. False
.
Answer: A
Explanation: Wedge action imparts tightening to the cotter joint.
Q. Can the cotter joint be used to connect slide spindle and fork of valve mechanism?
A. True
B. False
.
Answer: A
Explanation: As long as axial forces act, cotter joint can be employed.
Q. Which of the following is not a part of cotter joint?
A. Socket
B. Spigot
C. Cotter
D. Collar
.
Answer: D
Explanation: There is no point of mentioning collar alone in a cotter joint. It has to be a spigot collar or socket collar.
Q. Cold riveting holds the connected parts better than hot riveting.
A. True
B. False
.
Answer: B
Explanation: The compression of connected parts in hot riveting causes friction, which resist sliding of one part with respect to other. This force is greater in hot riveting.
Q. In cold riveting like hot riveting shank is subjected to majorly tensile stress.
A. True
B. False
.
Answer: B
Explanation: In cold riveting there is no reduction in length and hence no tensile stress. Shear stress dominates.
Q. Determine the width of the cotter used in cotter joint connecting two rods subjected to axial load of 50kN and permissible shear stress in cotter is 50N/(mm² ). Given thickness of cotter=Q.mm
A. 5omm
B. Q.0mm
C. 150mm
D. 25mm
.
Answer: A
Explanation: Cotter is subjected to double shear hence width=P/(2*τ*t).
Q. If joint is to fail by crushing of socket collar then estimate the diameter of socket collar. Given Permissible compressive stress= Q.Q.67 N/mm².; Spigot dia=65mm; thickness 0f collar=15mm
A. 131mm
B. 139mm
C. 141mm
D. 149mm
.
Answer: A
Explanation: Compressive stress= P/ [(socket dia-spigot diA.*thickness].
Q. Cotter joint can be used to connect two rods for torque transmission purpose.
A. True
B. False
.
Answer: B
Explanation: Cotter Joint is never used to connect two rods for torque transmission purpose.
Q.. Thickness of plate is required more in welding than in riveting.
A. True
B. False
.
Answer: B
Explanation: In riveting, cross section is weakened due to the holes and to compensate this, thicker plates are required in riveting.
Q. Knuckle Joint can’t be used to connect two intersecting rods.
A. Yes
B. No, it can’t be used
C. It can be used with some modificatios
D. It is expensive and hence isn’t used
.
Answer: B
Explanation: Knuckle Joint is used to connect two rods whose axes coincide or intersect and lie in a same plane.
Q. A knuckle joint is unsuitable for two rotating shafts, which transmit torque
A. True
B. False
.
Answer: A
Explanation: Knuckle joint can’t be used for torque transmission.
Q. A maximum of how many roads may be connected using a knuckle joint?
A. 2
B. 3
C. 4
D. 5
.
Answer: B
Explanation: In rare explanation, two rods with forks and one rod with eye is connected.
Q. A knuckle joint is also called socket pin joint.
A. True
B. False
.
Answer: B
Explanation: A knuckle joint is also called a Forked Pin Joint.
Q. Which of the following are important parts of knuckle joint?
A. Eye
B. Pin
C. Fork
D. Each of the mentioned
.
Answer: D
Explanation: All the mentioned parts are important components of knuckle joint.
Q. Calculate the diameter of pin from shear consideration with maximum shear stress allowed is 40NN/mm² and an axial tensile force of 50kN is acting on the rod.
A. 39mm
B. 44mm
C. 49mm
D. 52mm
.
Answer: A
Explanation: As the pin is subjected to double shear diameter (D. = √(2P/π x τ) = 3Q.80mm.
Q. If knuckle joint is to fail by crushing failure of pin in fork, then determine the diameter of knuckle pin when 50kN axial tensile force act on rods. Given: Max allowable compressive stress=25N/mm², thickness of each eye of fork=25mm.
A. 40mm
B. 50mm
C. 60mm
D. 70mm
.
Answer: A
Explanation: d=P/2aσ = 40mm.
Q. If any cross section is subjected to direct tensile stress and bending stress, then find the dimension of cross section. Given length & breadth are t and 2t respectively. F=25kN acts on the top fibre of the cross section, M=F x t . Also maximum allowable tensile stress =Q.0N/mm².
A. 2Q.5mm
B. 30.2mm
C. 2Q.55mm
D. None of the mentioned
.
Answer: A
Explanation: σ= [P/A] + [My/I], where y=t & I=t(2t)ᴲ/Q..
Q. A knuckle joint can be used in valve mechanism of a reciprocating engine.
A. Yes
B. No
C. Yes but there are stress probles
D. No as it is very dangerous to use
.
Answer: A
Explanation: Knuckle joint can be used till the rods coincide or intersect in a plane.
Q.. Distortion energy theory is slightly liberal as compared to maximum shear stress theory.
A. True
B. False
.
Answer: A
Explanation: The hexagon of maximum shear theory falls completely inside the ellipse of distortion energy theorem.
Q. The normal stress is perpendicular to the area under considerations, while the shear stress acts over the area.
A. True
B. False
.
Answer: A
Explanation: This is the convention used.
Q. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm²Find the maximum amount of shear stress to which the body is subjected.
A. 2Q.4mm
B. 25mm
C. 2Q.3mm
D. 2Q.2mm
.
Answer: A
Explanation: τ(max)=√( [σ(x)-σ(y) ]²/2² + τ²).
Q. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is Q.N/mm². Find the inclination of the plane in which shear stress is maximal.
A. 45’
B. 30’
C. 60’
D. 15’
.
Answer: A
Explanation: tan (2Ǿ)=2τ/[σ(x) – σ(y)].
Q. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm². Find the maximum normal stress.
A. 90
B. 9Q.4
C. 9Q.2
D. 96
.
Answer: B
Explanation: σ=[σ(x) +σ(y)]/2 + √( [σ(x)-σ(y) ]²/2² + τ²).
Q. If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm². Find the minimum normal stress.
A. 4Q.4
B. 4Q.6
C. 4Q.2
D. 50.6
.
Answer: B
Explanation: σ=[σ(x) +σ(y)]/2 – √( [σ(x)-σ(y) ]²/2² + τ²).
Q. If compressive yield stress and tensile yield stress are equivalent, then region of safety from maximum principal stress theory is of which shape?
A. Rectangle
B. Square
C. Circle
D. Ellipse
.
Answer: B
Explanation: The equation of four lines is given by σ1=± S(yt), σ2=±S(yC. Now given S(yt)=S(yC., hence the region of safety is of square shape.
Q. Maximum Principal Stress Theory is not good for brittle materials.
A. True
B. False
.
Answer: B
Explanation: Experimental investigations have shown that maximum principle stress theory gives good results for brittle materials.
Q. The region of safety in maximum shear stress theory contains which of the given shape
A. Hexagon
B. Rectangle
C. Square
D. None of the mentioned
.
Answer: A
Explanation: In maximum shear stress theory we have the following equations: σ1= ±S(yt)
σ2= ±S (yt), σ1 – σ2 =±S (yt) assuming S(yt)=S(yC..
Q. How many classes of levers are there?
A. 2
B. 3
C. 4
D. 5
.
Answer: B
Explanation: The three classes are Fulcrum in the middle, Resistance in the middle and Effort in the middle.
Q. A bottle opener belongs to which class of the levers.
A. Effort in the middle
B. Fulcrum in the middle
C. Resistance in the middle
D. None of the mentioned
.
Answer: C
Explanation: In a bottle opener fulcrum can be considered at the end and resistance in middle as the actual work is done on the bottle cap which lies in middle of opener.
Q. Fulcrum can be located at one end of the lever.
A. True
B. False
.
Answer: A
Explanation: Fulcrum can be located in middle or at one end of the lever.
Design Against Fluctuating Load
1. Stress concentration is defined as the localization of high stresses due to irregularities present in the component and no changes of the cross section.
A. True
B. False
.
Answer: B
Explanation: The given statement is true apart from the fact that there is no change in the cross section.
2. Stress Concentration Factor is the ratio of nominal stress obtained by elementary equations for minimum cross-section and highest value of actual stress near discontinuity.
A. True
B. False
.
Answer: B
Explanation: The stress concentration factor is just the reciprocal of that cited in the question.
3. If a flat plate with a circular hole is subjected to tensile force, then its theoretical stress concentration factor is?
A. 2
B. 3
C. 4
D. 1
.
Answer: B
Explanation: For any ellipse, K=1+2 x (semi major axis/semi minor axis).
4. For an elliptical hole on a flat plate, if width of the hole in direction of the load decrease, Stress Concentration Factor will______
A. Increase
B. Decrease
C. Remains constant
D. Can’t be determined. Varies from material to material
.
Answer: A
Explanation: K=1+2 x (semi major axis/semi minor axis. Hence K is inversely proportional to the semi minor axis.
5. In which of the following case stress concentration factor is ignored?
A. Ductile material under static load
B. Ductile material under fluctuating load
C. Brittle material under static load
D. Brittle material under fluctuating load
.
Answer: A
Explanation: In ductile materials under static load, there is plastic deformation near yielding point and hence redistribution of stresses take place. The plastic deformation is restricted to a smaller area and hence no perceptible damage take place.
6. Is it logical to use fluid analogy to understand the phenomenon of stress concentration?
A. True
B. False
.
Answer: A
Explanation: There is a similarity between velocity distribution in fluid flow in a channel and the stress distribution in an axially loaded plate. The equations for flow potential and in fluid mechanics and stress potential in solid mechanics are same.
7. Use of multiple notches in a V shaped flat plate will
A. Reduce the stress concentration
B. Increase the stress concentration
C. No effect
D. Cannot be determined
.
Answer: A
Explanation: The sharp bending of a force flow line is reduced due to multiple notches.
8. Which of the following reduces the stress concentration?
A. Use of multiple notches
B. Drilling additional holes
C. Removal of undesired material
D. Each of the mentioned
.
Answer: D
Explanation: All the mentioned options reduce the sharp bending of a force flow line.
9. A flat plate 30mm wide and “t”mm wide is subjected to a tensile force of 5kN. The plate has a circular hole of diameter 15mm with the centre coinciding with the diagonal intersection point of the rectangle. If stress concentration factor=2.16, find the thickness of the plate if maximum allowable tensile stress is 80N/mm².
A. 8mm
B. 9mm
C. 10mm
D. 12mm
.
Answer: B
Explanation: σ=P/ (w-D. x t or σ=5000/(30-15)xt ; σ(max)=K x σ or σ(max)=2.16 x σ or σ(max)=720/t; Also σ(max)=80.
10. The stress represented by sin (t) + 1 belongs to which category?
A. Fluctuating Stresses
B. Alternating stresses
C. Repeated Stresses
D. Reversed Stresses
.
Answer: C
Explanation: The minimum stress value is zero and hence is belongs to Repeated Stress category.
11. The stress represented by sin (t) + 2 belongs to which category?
A. Fluctuating Stresses
B. None of the mentioned
C. Repeated Stresses
D. Reversed Stresses
.
Answer:a
Explanation: The mean as well as the amplitude value is non zero hence it belongs to Fluctuation Stress Category.
12. The stress represented by sin (t) + 4 belongs to which category?
A. Alternating Stresses
B. None of the mentioned
C. Repeated Stresses
D. Reversed Stresses
.
Answer: A
Explanation: The mean as well as the amplitude value is non zero hence it belongs to Alternation Stresses category which is the other name of Fluctuation Stresses.
13. The stress represented by cos (t) belongs to which category?
A. Fluctuating Stresses
B. Alternating Stresses
C. Repeated Stresses
D. Reversed Stresses
.
Answer: D
Explanation: Half cycle is in tensile stress and other half in compressive stress, hence it belongs to Reversed Stresses Category.
14. If the mean stress value for a sinusoidal stress function is zero, then this type of stress falls in which category?
A. Fluctuating Stresses
B. Alternating Stresses
C. Repeated Stresses
D. Reversed Stresses
.
Answer: D
Explanation: If mean is to be zero, then there must be compressive as well as tensile stresses and hence belongs to reversed stresses category.
15. The phenomenon of decreased resistance of the materials to fluctuating stresses is the main characteristic of _____ failure.
A. Fracture
B. Fatigue
C. Yielding
D. None of the mentioned
.
Answer: B
Explanation: Fatigue failure of the material is the failure at low stress levels under fluctuating syresses.
16. Fatigue failure is time dependent failure.
A. True
B. False
.
Answer: A
Explanation: Fatigue failure is defined as time delayed fracture under cyclic loading.
17. There is sufficient plastic deformation prior to fatigue failure, which gives a warning well in advance.
A. True
B. False
.
Answer: B
Explanation: Fatigue cracks are not visible until they reach the surface and by that time the failure has already taken place. Material nevers enters in the plastic range
Machine Design MCQ
1. A power screw is only used to convert rotary motion into linear motion and not for transmitting power.
A. True
B. False
.
Answer: B
Explanation: Power screw converts the motion from rotary to linear and is used for power transmission.
2. Depending upon the holding arrangement, power screws operate in how many different arrangements.
A. 2
B. 3
C. 4
D. 5
.
Answer: A
Explanation: There are two types of arrangements. In one screw rotates while nut remains stationary and vice versa.
3. A power screw has no problem of wear as there is very less amount of friction associated.
A. True
B. False
.
Answer: B
Explanation: Wear is a serious problem in power screws as there is high friction in threads.
4. V threads are highly recommended for fastening as well as power transmission purpose.
A. Yes
B. Never
C. In some cases
D. Can’t be stated
.
Answer: B
Explanation: In fastening, high frictional force is required and hence V threads are used whereas in power transmission, reduction in frictional forces is required.
5. Trapezoidal threads are better than square threads as there is radial pressure or side thrust on the nut.
A. True
B. False
.
Answer: B
Explanation: Trapezoidal and not square threads suffer from the problem of bursting.
6. Trapezoidal threads screws have less load carrying capacity as compared to square thread screws.
A. True
B. False
.
Answer: B
Explanation: Square threads have less thickness at the core diameter and hence lower load carrying capacity.
7. Which of the following are true for buttress threads?
A. Combination of square and trapezoidal threads
B. Transmit motion in one direction only
C. They are used in vices
D. All of the mentioned
.
Answer: D
Explanation: As force is applied only in one direction in a vice so buttress threads are used.
8. Tr 40 x 14(P 7), here 14 indicates
A. Pitch
B. Lead
C. Diameter
D. None of the mentioned
.
Answer: B
Explanation: Tr=Trapezoidal threads,14=Lead(mm),7=Pitch(mm).
9. Nominal diameter of the screw thread is the same as core diameter.
A. True
B. False
.
Answer: B
Explanation: Nominal diameter is the largest diameter while core diameter is the smallest diameter of the screw thread.
10. If nominal diameter of screw thread=50mm and pitch=10mm then the mean diameter of the screw thread will be?
A. 40mm
B. 45mm
C. 60mm
D. 55mm
.
Answer: B
Explanation: Diameter(mean)=Diameter(nominal) – 0.5P .
11. If the load itself begin to the screw and descend down, unless a restraining torque is applied then the condition is termed as
A. Halting
B. Overhaulting
C. Front driving
D. None of the mentioned
.
Answer: B
Explanation: Overhaulting is the condition when load itsel begin to turn the screw.
12. Self-locking takes place when
A. Coefficient of friction is equal to or greater than the tangent of the helix angle
B. Coefficient of friction is lesser than or equal to the tangent of the helix angle
C. Coefficient of friction is equal to or greater than the tangent of the helix angle
D. None of the mentioned
.
Answer: C
Explanation: Self locking takes place if load does not descend on its own and that is possible only in c condition.
13. Efficiency of screw depends upon lead of the screw.
A. True
B. False
.
Answer: A
Explanation: Efficiency= WL/[Pπd].
14. Efficiency of the screw depends upon helix angle but does not depend on friction angle.
A. True
B. False
.
Answer: B
Explanation: Efficiency=tan(ἀ)/tan(Ǿ+ἀ) where ἀ=Helix angle and Ǿ=Friction angle.
15. Efficiency of the screw _______ with increase of coefficient of friction.
A. decreases
B. increases
C. has no effect
D. cannot be determined
.
Answer: A
Explanation: Efficiency is inversely proportional to tan of the sum of helix and efficiency angle.
16. Maximum efficiency of a square threaded is given by
A. 1-sinǾ/1+sinǾ
B. 1+sinǾ/1-sinǾ
C. 1-2sinǾ/1+2sinǾ
D. 1+2sinǾ/1-2sinǾ
.
Answer: A
Explanation: Efficiency=Sin(2ἀ+Ǿ)-Sin Ǿ/Sin(2 ἀ+Ǿ+Sin Ǿ),For max efficiency, sin(2ἀ+ Ǿ)=1.
17. If friction angle is 30’ then the maximum efficiency of the screw is
A. 33%
B. 66%
C. 50%
D. Noe of the mentioned
.
Answer: A
Explanation: Maximum efficiency=1-sinǾ/1+sinǾ.
18. Maximum possible efficiency of a self-locking screw is
A. 50%
B. 75%
C. 66%
D. 33%
.
Answer: A
Explanation: For self locking screw Ǿ(friction angle)>ἀ (helix angle), hence efficiency < tanǾ/tan(Ǿ+ἀ) or efficiency < tanǾ(1-tan²Ǿ)/2tanǾ.
19. In trapezoidal threads, f (coefficient of friction) can be taken as
A. f sec θ
B. f cos θ
C. f sin θ
D. f cosec θ
.
Answer: A
Explanation: The normal force acting on the thread is W sec θ therefore the effect of the thread angle is to increase the frictional force by a term sec θ.
20. Clutch and coupling can be considered to be same.
A. True
B. False
.
Answer: B
Explanation: Clutch is a temporary join while coupling is a permanent joint.
21. Cold rolled components have higher strengths and hardness than hot rolled components.
A. True
B. False
.
Answer: A
Explanation: During grain re-structuring, original strength of material is lost.
22. Which of the following are true?
A. Cold working reduces toughness and ductility
B. Cold worked components have poor resistance to shocks and vibrations
C. Tooling for cold working is cheaper as compared to hot working
D. All of the mentioned
.
Answer: D
Explanation: Properties of cold working.
MCQ’s On Threaded Joints in Machine Design
Q. Threaded joints are non-separable joints.
A. True
B. False
Answer: B
Explanation: Threaded joints are separable joints of machine parts that are held together by means of a threaded fastening such as nut and bolt.
Q. Which type of joints is better when the product is subjected to large vibrations: welded or threaded?
A. Welded
B. Threaded
C. Both have same results
D. Depends on the magnitude of the vibrational force
Answer: A
Explanation: Threaded joints loosen when subjected to vibrations.
Q. If a fastener is threaded into a tapped hole, then the fastener is likely to be called as
A. Screw
B. Bolt
C. Washer
D. Screw or bolt
Answer: A
Explanation: Bolt is threaded into a nut while screw is threaded into a tapped hole.
Q. In nut and bolt fastener, nut is generally held stationary while bolt is rotated.
A. True
B. False
Answer: B
Explanation: Bolt is always held stationary while nut is rotated.
Q. Which of the following is not a function of washer?
A. Distributes load over a large area of clamped parts
B. Provides bearing surface over large clearance bolts
C. Prevents marring of the bolt head and nut surface
D. Helps in locking of the fastener
.
Answer: D
Explanation: Washer doesn’t help in locking of fasteners.
Q. Shank is the portion of bolt between the head and the thread.
A. True
B. False
Answer: A
Explanation: Shank is the portion between head and the thread of the bolt.
Q. If there is no place to accommodate the nut, then one would choose the
A. Through Bolts
B. Tap Bolts
C. Studs
D. None of the mentioned
.
Answer: B
Explanation: Tap bolts are directly threaded into the clamped parts and does not require any nut.
Q. Which of the following are good for parts that are seldom dismantled.
A. Through Bolts
B. Tap Bolts
C. Studs
D. Cannot be stated
Answer: B
Explanation: Tap bolts, if removed frequently will leave the threads in the part worn out and hence are used where dismantling is rare.
Q. Which of the following requires more space for the rotation of spanner?
A. Square Head
B. Hexagonal Head
C. Both require equal space of rotation
D. Cannot be stated
Answer: A
Explanation: Angle of rotation for hexagonal head is one sixth of a revolution to enable the next pair of flats to be engaged while it is one fourth of a revolution in case of square head.
Q. Fillister, button, flat and hexagonal heads are all tightened externally.
A. True
B. False
.
Answer: B
Explanation: All of these are internal drive screws having a slot for the screw driver of different shape on the head.
Q. Set screws are subjected to tensile forces only.
A. Yes
B. No, they are subjected to compressive forces only
C. Both compressive and tensile
D. Can’t be determined
Answer: B
Explanation: Set screws are subjected to compressive forces only.
Q. If the lateral force acting on the set screw is large and tends to displace one part with respect to other than which point is recommended?
A. Dog Point
B. Cone Point
C. Window Point
D. Cut Point
.
Answer: A
Explanation: Dog point is used when the lateral force is large provided the part has sufficient thickness to accommodate a cylindrical hole for dog point.
Q. Find the torque required to raise the load of 15kN and mean diameter of triple threaded screw being 46mm. Also given pitch=8mm and coefficient of friction is 0.15.
A. 11831.06N-mm
B. 11813.06N-mm
C. 12811.06N-mm
D. None of the listed
.
Answer: A
Explanation: tanἀ=l/πd or ἀ=9.429’ as l=3p. tan Ǿ=0.15 or Ǿ=8.531’. M=W x d x tan (Ǿ+ἀ)/2.
Q. What is the collar friction torque if outer and inner diameters are 100mm and 65mm respectively. Coefficient of friction is 0.15 and the load acting is of 15kN. Consider uniform wear theory.
A. 89886.6N-mm
B. 38796.5N-mm
C. 92812.5N-mm
D. 87645.5N-mm
Answer: C
Explanation:M=0.15 x W(D+D./4.
Q. For a double threaded screw, what will be the tangent of helix angle if nominal diameter and pitch are 100mm and 12mm respectively?
A. 0.045
B. 0.081
C. 0.094
D. 0.023
.
Answer: B
Explanation: tan ἀ= l/πd where l=2p, d=D-0.5p.
Q. If for a trapezoidal thread, angle of thread is 15’ then what will be the replacement of the coefficient of friction which is 0.15.
A. 0.1553
B. 0.1335
C. 0.1667
D. 0.1776
Answer: A
Explanation: f=0.15 x sec 15’.
Q. What will be efficiency of the screw in case of raising the load when coefficient of friction is 0.1553 and tangent of helix angle is 0.0813?
A. 34%
B. 45%
C. 54%
D. 43%
.
Answer: A
Explanation: Efficiency= tanἀ(1-0.1553 x tan ἀ)/(0.1553+tanἀ).
Q.A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the screw torque in terms of clamping force W if nominal diameter=22mm and pitch=5mm.
A. 3.567W N-mm
B. 2.286W N-mm
C. 3.564W N-mm
D. None of the mentioned
.
Answer: B
Explanation: M₁=Wd x tan(Ǿ+ἀ)/2 or M₁=W x (22-0.5×2.5)x tan(4.66’+8.531’)/2 as tan(ἀ)=l/πd and tanǾ=0.15.
Q.A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the collar torque in terms of clamping force W assuming uniform wear theory if nominal diameter=22mm and pitch=5mm.
A. 4.5W
B. 5.4W
C. 4.25W
D. 3.37W
.
Answer: C
Explanation: M₂=0.17 x W x (55+45)/4 or M₂=4.25W N-mm.
Q.A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the clamping force W if nominal diameter=22mm and pitch=5mm.
A. 4539.45N
B. 2868.73N
C. 3657.56N
D. 2134.34N
.
Answer: B
Explanation: Net torque=M₁+M₂ or 125 x 150=2.286W + 4.25W or W=2868.73N.
Q.A machine vice whose length of the handle is 150mm and the coefficient of friction for thread and collar are 0.15 and 0.17 respectively has a force applied at handle of 125N. Also the outer and inner diameters of collar are 55mm and 45mm respectively. Find the overall efficiency if nominal diameter=22mm and pitch=5mm.
A. 18.12%
B. 12.18%
C. 21.23%
D. 23.21%
.
Answer: B
Explanation: Efficiency= W l/2π(M₁+M₂) where
M₁=2.286W [M₁=Wd x tan (Ǿ+ἀ)/2 or M₁=W x (22-0.5×2.5) x tan (4.66’+8.531’)/2 as tan(ἀ)=l/πd and tan Ǿ=0.15].
M₂=4.25W [M₂=0.17 x W x (55+45)/4 or M₂=4.25W N-mm],
l=5mm,
W=2868.73N [Net torque=M₁+M₂ or 125 x 150=2.286W + 4.25W or W=2868.73N].
Q. Find the bending stress to which a screw of nominal diameter 22mm is subjected when the clamp exerts a force of 5kN acts on it. The screw is double threaded and pitch of screw is 5mm.Given: Coefficient of friction is 0.15.It is assumed operator exerts a force of 250N at the handle of length 275mm.
A. 123.45N/mm²
B. 132.54N/mm²
C. 142.54N/mm²
D. 124.45N/mm²
.
Answer: C
Explanation: Bending stress=32M/πdᵌ where M=250 x 275N-mm,d=22-2 x 0.5 x 5mm.
Q. Find the torsional shear stress to which a screw of nominal diameter 22mm is subjected when the clamp exerts a force of 5kN on it. The screw is double threaded and pitch of screw is 5mm.Given: Coefficient of friction is 0.15.
A. 25.45N/mm²
B. 21.29N/mm²
C. 31.21N/mm²
D. 35.65N/mm²
.
Answer: B
Explanation:=16M/πd₂ᵌ where M=Wd₁ tan (Ǿ+ἀ)/2 where d₁=22-0.5 x 5,W=5000N,
tanἀ=2 x 5/[πx19.5] and d₂=19.5-0.5 x 5.
Q. If bending stress in a screw is of magnitude 95N/mm² and torsional shear stress of magnitude 15N/mm², then what is the principal shear stress?
A. 48.7N/mm²
B. 54.3 N/mm²
C. 51.4 N/mm²
D. 49.8 N/mm²
.
Answer: D
Explanation:τ(max)=√(95/2)²+15².
Q. The nominal diameter of the screw is 22mm and the allowable bearing pressure for the nut is 15N//mm². Find the length of the nut if pitch of the screw is
A. 15mm
B. 20mm
C. 25mm
D. 30mm
.
Answer: A
Explanation:z=4W/π S(d₁²-d₂²) where S=15N/mm²,d₁=22mm and d₂=17mm. Hencez=2.2 or z=Length of nut=zp or l=3 x 5= 15mm.
Q.A but joint can be defined as a joint between two components lying approximately in a different plane.
A. True
B. False
.
Answer: B
Explanation: Butt joint is defined a joint between two plates lying in the same plane.
Q.A fillet weld is an approximately triangular cross section joining two surfaces lying parallels to each other.
A. True
B. False
.
Answer: B
Explanation: Two surfaces lie perpendicular to each other.
Q. How many types of jacks are there which can be used in a screw jack?
A. 2
B. 3
C. 4
D. 5
.
Answer: A
Explanation: Hydraulic jack and mechanical jack.
Q. Which of the following does not form the important part of the screw jack?
A. Frame
B. Nut
C. Cup
D. Coupling
.
Answer: D
Explanation: There is no specific requirement of coupling in the screw jack.
Q. The transverse shear stress at the root of the threads in the nut can be given by?(symbols have their usual meaning, z=number of threads in nut)
A. 4W/πdz²
B. W/πdtz
C. 4W/πtd²
D. None of the mentioned
.
Answer: B
Explanation: At root of threads, the area parallel to direction of force is considered which is equal to circumference x thickness x no. of threads.
Q. What type of friction in cup design is recommended for the set screw?
A. Sliding
B. Rolling
C. Static
D. None of the mentioned
.
Answer: B
Explanation: Sliding friction, the total force require to raise the load is quite large as compared to that required in rolling friction. Hence length of handle in some cases becomes extremely large and thus it is impractical to use cups with sliding friction.
Q.A differential screw is defined as a mechanical device consisting of two screws connected in parallel.
A. True
B. False
.
Answer: B
Explanation: The two screws are connected in series.
Q.In a differential screw, two screws are connected in series such that the resultant motion is the summation of the motion of the two screws.
A. True
B. False
.
Answer: B
Explanation: The resultant motion is the difference of the individual motion of two screws.
Q.A compound screw consist of two screws in parallel such that the resultant motion is the summation of the individual motion of screws.
A. True
B. False
.
Answer: B
Explanation:The two screws are connected in series with the resultant motion being the sum of individual motion.
Q. What is the output from differential screws when pitch of the two screws is 12mm and 8mm? Also the nut is rotated by applying a force of 120N at a radius of 300mm and the two screws remain stationary. The torque of raising and lowering for the two screws is 5k N-mm and 2.5k N-mm where k is the effective axial weight on the screw.
A. 13200 N-mm
B. 15200 N-mm
C. 19200 N-mm
D. 17200 N-mm
.
Answer: C
Explanation:120 x 300=5k+2.5k or k=4800N. Output=k x (12-8) or output=19200N-mm.
Q. What is the efficiency of differential screws when pitch of the two screws is 12mm and 8mm? The nut is rotated by applying a force of 120N at a radius of 300mm and the two screws remain stationary. The torque of raising and lowering for the two screws is 5k N-mm and 2.5k N-mm where k is the effective axial weight on the screw.
A. 6.48%
B. 8.48%
C. 23.1%
D. 42.8%
.
Answer: B
Explanation:120 x 300=5k+2.5k or k=4800N. Output=k x (12-8) or output=19200N-mm. Efficiency=output/2πx120x300.
Q. Efficiency of conventional power screw is greater than the efficiency of recirculating ball screw.
A. True
B. False
.
Answer: B
Explanation: In recirculating ball screw, sliding friction is replaced by rolling friction and efficiency increases from 40% to 90%.
Q. In which case is the wear more?
A. Conventional power screw
B. Recirculating power screw
C. Equal in both the cases
D. Cannot be determined
.
Answer: A
Explanation: In recirculating ball screw, there is a lubricant film between the contacting surfaces and hence lesser wear.
Q. Resilience is defined as the property of material to absorb energy when deformed _________ and to release this energy when unloaded.
A. Elastically
B. Plastically
C. Up to fracture point
D. None of the listed
.
Answer: A
Explanation:It is the definition of resilience.
Q.The shock absorbing capacity of bolt can be increased by making the shank diameter ________ the root diameter.
A. Lesser than
B. Equal to
C. Greater than
D. No effect
.
Answer: B
Explanation:If shank diameter is equal to the thread diameter than, the shank is subjected to higher stress and thus absorbs a great portion of strain energy and relieves the thread portion of high stress.
Q.The shock absorbing capacity of bolts can be increased by
A. Increasing the shank diameter
B. Decreasing the length of shank portion of bolt
C. Increase the shank diameter and decreasing the length of shank
D. None of the listed
.
Answer: D
Explanation:Shock absorbing capacity can be increased by decreasing the shank diameter or increasing the length of shank.
Q.If the length of the shank is doubled, then strain energy absorbed by shank
A. Doubles
B. Remains same
C. Increases 4 time
D. Become half
.
Answer: A
Explanation: Strain energy absorbed by shank is linearly proportional to its length.
Q. of the following help in accomplishing the locking?
A. Creating supplementary friction
B. Using special locking devices like split pins
C. By plastic deformation
D. All of the listed
.
Answer: D
Explanation: All the methods can be employed to lock the screw.
Q. Crests and roots of the threads may lead to leakage in fluid tight threads if flattened or rounded to a circular arc.
A. True
B. False
.
Answer: B
Explanation: As there is engagement in the threads, chances of leakage are reduced.
Q.The static load carrying capacity of fine threads is greater than that of coarse threads.
A. True
B. False
.
Answer: B
Explanation: Coarse threads have a higher load carrying capacity.
Q. Coarse threads have helix angle _____ fine threads.
A. Equal to
B. Greater than
C. Lesser than
D. There is no relation
.
Answer: C
Explanation: Finer threads have lower helix angle.
Q. Which threads have greater resistance to unscrewing?
A. Coarse
B. Fine
C. They have equal resistance
D. Cannot be determined
.
Answer: B
Explanation: Fine threads have greater resistance to unscrewing as a result of lower helix angle.
Q. Which type of threads is recommended for fluctuating loads?
A. Fine threads
B. Coarse Threads
C. Either of fine or coarse
D. None of the listed
.
Answer: A
Explanation: Fine threads have lesser helix angle and hence greater resistance for unscrewing and thus are recommended for fluctuating loads.
Q. Which of the following statements are true about eccentrically loaded circular base?
A. All bolts are symmetrically spaced with equal angle on either side of vertical
B. Bolts are not preloaded
C. Bearing is rigid
D. Dowel pins are used to relieve stresses
.
Answer: A
Explanation: Bolts are symmetrically placed only in a special case.
Q. Certain types of radial bearings can also take thrust load.
A. True
B. False
.
Answer: A
Explanation: Sometimes radial bearings can take thrust load and vice versa.
Q. Balls and the races of the deep groove ball may roll freely without any sliding.
A. True
B. False
.
Answer: A
Explanation: There is a point contact between the balls and the races.
Q.A flange of radius 200mm is fastened to the machine screw by means of four cap screws of pitch circle radius 150mm. The external force P is 20kN which is loaded at 160mm from the machine screw. There are two dowel pins to take shear load. Determine the maximum effective tensile force acting on the bolt. The four bolts are equally spaced radially.
A. 4777.6N
B. 5777.6N
C. 6777.6N
D. 7777.7N
.
Answer: A
Explanation: Maximum force F=2Pl[a+bCos(90/2)] / 4(2a²+b²).
Q.A flange of radius 200mm is fastened to the machine screw by means of four cap screws of pitch circle radius 150mm. The external force P is 20kN which is loaded at 160mm from the machine screw. There are two dowel pins to take shear load. Determine the core diameter of cap screw if permissible tensile stress in cap screw is 40N/mm².
A. None of the listed
B. 11.21mm
C. 12.33mm
D. 14.12mm
.
Answer: C
Explanation: Maximum tensile force=4777.6N or πd² x σ/4=4777.6N.
Welded and Riveted Joints
Q.All welding processes require pressure along with heat.
A. Yes
B. No, fusion doesn’t require
C. Can’t be stated
D. None of the listed
.
Answer: B
Explanation: Welding processes involving only heat and no pressure are called the fusion welding processes.
Q.The capacity of welded structures to damp vibrations is quite good.
A. True
B. False
.
Answer: B
Explanation: Capacity of welded joints to damp vibrations is poor.
Q.Themit consists of a finely divided mixture of iron oxide and copper.
A. True
B. False
.
Answer: B
Explanation: Thermite is a mixture of iron oxide and copper.
Q.Rails in the field are generally welded by using
A. Thermit welding
B. Gas welding
C. Electric arc welding
D. Forge welding
.
Answer: A
Explanation: Wherever it is uneconomical to carry welding equipments, thermit welding is used.
Q.In gas welding which of the following is generally used?
A. Oxygen-Hydrogen
B. Oxygen-Acetylene
C. Oxygen-Hydrogen or Oxygen-Acetylene
D. None of the mentioned
.
Answer: C
Explanation: Intense heat is released I a controlled way and at a moderate temperature.
Q.Among gas and electric arc welding, which has the higher rate of heating?
A. Gas welding
B. Electric arc welding
C. Gas welding and electric arc welding have equal rate of heating
D. Cannot be determined
.
Answer: B
Explanation: Gas welding has a gas temperature of about 3200’C while arc temperature is about 40000’C.
Q.Forge welding involves heating of parts to elastic stage and joint is prepared by impact force.
A. Yes
B. No, it is done up to plastic stage
C. Heating is done up to boiling point
D. None of the listed
.
Answer: B
Explanation: Heating is done up to plastic stage.
Q.Filler material is used in electric resistance welding.
A. Yes
B. No filler material used
C. Depends on the type of welding
D. None of the listed
.
Answer: B
Explanation: No filler material is used. Only the heat released from resistance of metallic parts to current is used for melting the adjoining parts.
Q.Which type of welding is generally used in automobile sector?
A. Electric arc welding
B. Electric resistance welding
C. Gas welding
D. Forge welding
.
Answer: B
Explanation: Electric resistance welding can be easily automated and hence is used in automobile sector.
Q.Hard peening is
A. Hammering the weld across the length while the joint is hot
B. Hammering the weld along the length while the joint is hot
C. Hammering the weld along the length while the joint is cold
D. Hammering the weld across the length while the joint is cold
.
Answer: B
Explanation: Hammering is doe to relieve stresses and inducing compressive stresses to improve the fatigue strength of the joint.
Q.Strength of parallel fillet weld is greater than strength of transverse fillet weld.
A. True
B. False
.
Answer: B
Explanation: Strength of transverse fillet weld is 1.17 times the strength of parallel fillet weld.
Q.If force act in a direction parallel to the direction of weld, then fillet weld is called as?
A. Transverse
B. Longitudinal
C. Parallel
D. Longitudinal or Parallel
.
Answer: D
Explanation: Parallel fillet weld consist of force acting in a direction of weld.
Q.Convex weld is generally preferred over normal weld.
A. True
B. False
.
Answer: B
Explanation: There is more stress concentration in convex weld and hence is less preferred.
Q.Parallel fillet weld and transverse fillet weld both have the plane in which maximum shear stress occurs at 45’ to the leg dimension.
A. True
B. False
.
Answer: B
Explanation: Transverse fillet weld has that plane inclined at 67.5 .
Q.The length of each of the two equal sides of a parallel fillet weld is called
A. Leg
B. Throat
C. Arm
D. None of the listed
.
Answer: A
Explanation: Leg is the length of each of the two equal sides of a parallel fillet weld.
Q.Relation between throat and leg for a parallel fillet weld is
A. t =h Cos (45’)
B. h =t Cos (45’)
C. h= t
D. None of the listed
.
Answer: A
Explanation: Throat is the minimum cross section of the weld and that plane lies at 45’ for a parallel fillet weld.
Q.If length of weld is l and leg h, then area of throat can be given by
A. 0.707 hl
B. 1.414hl
C. hl
D. None of the listed
.
Answer: A
Explanation: Area=t x l where t=h Cos (45’).
Q.Which of the following isn’t a main part of rivet?
A. Head
B. Shank
C. Point
D. Thread
.
Answer: D
Explanation: There aren’t any threads in rivets.
Q.A rivet is specified as a 20mm rivet. What does it mean?
A. None of the mentioned
B. Shank dia 20mm
C. Head dia 20mm
D. Both head and shank dia 20mm
.
Answer: B
Explanation: Terminology is this way.
Q.In hand riveting die is a part of hammer.
A. True
B. False
.
Answer: B
Explanation: Die is a part of hammer in machine riveting.
Q.In hot riveting the shank portion is subjected to compressive stress.
A. True
B. False
.
Answer: B
Explanation: It is subjected to tensile stress as the head rests against the connected members to prevent reduction in length.
Q.The amount by which the two rods to be joined are drawn together is called as?
A. Draw
B. Portray
C. Lead
D. Pitch
.
Answer: A
Explanation: Draw measures the amount of distance advancing after which spigot rests on socket.
Q.Two steel rods connected by cotter joint are subjected to 50 kN load each. What is the minimum diameter required of the rods? (Given: Yielding Stress= 400N/mm² ; Factor of Safety=6)
A. 31mm
B. 35mm
C. 36mm
D. 40mm
.
Answer: A
Explanation: d=√(4P/πσ) where σ= Yielding Stress/F.S ; d=30.90mm or 31 mm.
Q.Among punching and drilling, which is cheaper?
A. Punching
B. Drilling
C. Equally expensive
D. Cannot be determined
.
Answer: A
Explanation: Drilling has more accuracy and is more expensive.
Q.Among punching and drilling, punching is safer.
A. True
B. False
.
Answer: B
Explanation: Punching injures the metal in the vicinity of the hole.
Q.Riveting is not recommended for aluminium alloys.
A. True
B. False
.
Answer: B
Explanation: Aluminium alloys have poor weldability and hence riveting is preferred.
Q.In joining steel plate and asbestos, welding is preferred over riveting.
A. True
B. False
.
Answer: B
Explanation: Riveting is preferred in joining heterogeneous materials.
Q.Quality checking of riveted joint is much expensive than that of welded joint.
A. True
B. False
.
Answer: B
Explanation: In welding, inspections like radiographic inspections are quite costly.
Q.Which of the following rivet head consist of frustum of cone attached to the shank?
A. Pan head rivet
B. Countersunk head rivet
C. Flat head rivet
D. Cone head rivet
.
Answer: D
Explanation: Cone head rivet consist of frustum of cone attached to shank.
Q.Among flat head and snap head rivet, which has the higher head height?
A. Snap
B. Flat
C. Equal
D. There is no such relation
.
Answer: A
Explanation: Flat head has lesser height of protruding head and thus it does not weaken the plates being assembled.
Q.Which of the following are used in light sheet metal work?
A. Tinmen’s rivets
B. Snap head rivets
C. Button head rivets
D. Each of the mentioned
.
Answer: A
Explanation: Tinmen’s rivets are flat head rivets of small sizes.
Q.Two plates each of thickness t are to be riveted together. If length of shank portion necessary to form the closing head is a, then length of rivet shank is given by?
A. 2t+a
B. 2(t+A.
C. t+2a
D. Can’t be determined
.
Answer: A
Explanation: l=t₁+t₂+a.
Q.A strap is used in a lap joint which is riveted to each of the two plates.
A. True
B. False
.
Answer: B
Explanation: That joint is called butt joint.
Q.What is the other name for pitch?
A. Transverse pitch
B. Back pitch
C. Row pitch
D. None of the listed
.
Answer: D
Explanation: Transverse, back and row pitch are all same. It is the distance b/w two consecutive rows of pitch.
Q.Are the rivets subjected any bending moment in case of lap joint?
A. Yes
B. No
.
Answer: A
Explanation: The rivets are subjected to bending moment which causes distortion.
Q.Lozenge joint is a kind of lap joint.
A. True
B. False
.
Answer: B
Explanation: Lozenge joint is another name for diamond joint and it is a kind of butt joint.
Q.Can we use rivets of the same materials as the parts to be joined?
A. True
B. False
.
Answer: A
Explanation: Heterogeneous material may lead to formation of galvanic pairs which can cause corrosion.
Q.Failure in rivet occurs by which mode?
A. Shear
B. Compression
C. Tensile
D. Each of the mentioned
.
Answer: D
Explanation: Rivet may fail by shearing, plates between two rivets can undergo tensile failure and plates might fail by crushing.
Q.Fullering is done with a pointed tool.
A. True
B. False
.
Answer: B
Explanation: Fullering is done to close the joint with a tool of thickness same as that of the plate.
Q.Calculate the diameter of the rivets by shear considerations if permissible shear stress in rivets is 60N/mm² and P=15kN.
A. 6mm
B. 7mm
C. 9mm
D. 8mm
.
Answer: C
Explanation: P=4x[τ x πd²/4].
Q.Calculate the diameter of the rivets by crushing consideration if permissible compression stress in rivets is 120N/mm², thickness of plate 3mm and P=15kN.
A. 10.4mm
B. 11.5mm
C. 9.2mm
D. 8.6mm
.
Answer: A
Explanation: P=4x[d t σ].
Q.Calculate width of the band if permissible tensile stress is 80N/mm². Force P=15kN and diameter of the rivet can be taken as 9mm. Given thickness=3mm
A. 80.5mm
B. 79.5mm
C. 76.66mm
D. 54.6mm
.
Answer: A
Explanation: [w-2d]tσ=P.
Q.If diameter of rivets is 9mm, then margin can be taken as?
A. 13.5mm
B. 12.5mm
C. 11.5mm
D. 9mm
.
Answer: A
Explanation: m=1.5d.
Shafts, Keys and Couplings in Machine Design
Q.The shaft is always stepped with ________ diameter at the middle portion and __________ diameter at the shaft ends.
a) Minimum, maximum
b) Maximum, minimum
c) Minimum, minimum
d) Zero, infinity
.
Answer: B
Explanation: Maximum diameter is in the middle portion while it is minimum at the ends.
Q.______ is used for a shaft that supports rotating elements like wheels, drums or rope sleaves.
a) Spindle
b) Axle
c) Shaf
d) None of the listed
.
Answer: A
Explanation: That is axle and not a spindle.
Q.Axle is frequently used in torque transmission.
a) True
b) False
.
Answer: B
Explanation: Axle is a shaft supporting rotating elements.
Q.Is it necessary for an axle to be ______ with respect to rotating element?
a) Stationary
b) Moving
c) Moving or stationary
d) None of the listed
.
Answer: C
Explanation: The axle may be stationary or rotate with the element.
Q.Counter shaft is a secondary shaft.
a) True
b) False
.
Answer: A
Explanation: It is a secondary shaft used to counter the direction of main shaft.
Q.Hot rolling produces a stronger shaft then cold rolling.
a) True
b) False
.
Answer: B
Explanation: Cold rolling produces stronger shaft as grain structure isn’t deformed in cold working.
Q.Shafts are subjected to ______ forces.
a) Compressive
b) Tensile
c) Shear
d) None of the listed
.
Answer: B
Explanation: Shafts are subjected to tensile forces.
Q.Which of the following act on shafts?
a) Torsional moment
b) Bending Moment
c) Both torsional and bending
d) None of the mentioned
.
Answer: C
Explanation: Shaft is subjected to torsional moment as well as bending moment.
Q.When the shaft is subjected to pure bending moment, the bending stress is given by?
a) None of the listed
b) 32M/πdᵌ
c) 16M/πdᵌ
d) 8M/πdᵌ
.
Answer: B
Explanation: Stress =My/I where y=d/2 and I=πd⁴/64.
Q.When the shaft is subjected to pure torsional moment, the torsional stress is given by?
a) None of the listed
b) 32M/πdᵌ
c) 16M/πdᵌ
d) 8M/πdᵌ
.
Answer: C
Explanation: Stress=Mr/J where r=d/2 and J=πd⁴/64.
Q.Calculate the shaft diameter on rigidity basis if torsional moment is 196000N-mm, length of shaft is 1000mm. Permissible angle of twist per meter is 0.5’ and take G=79300N/mm².
a) None of the listed
b) 41.2mm
c) 35.8mm
d) 38.8mm
.
Answer: B
Explanation: d⁴=584Ml/Gθ.
Q.If yielding strength=400N/mm², the find the permissible shear stress according to ASME standards.
a) 72 N/mm²
b) 76 N/mm²
c) 268 N/mm²
d) 422 N/mm²
.
Answer: A
Explanation: 0.18×400.
Q.The stiffness of solid shaft is more than the stiffness of hollow shaft with same weight.
a) True
b) False
.
Answer: B
Explanation: Hollow shaft is more stiff.
Q.The strength of hollow shaft is more than the strength of solid shaft of same weight.
a) True
b) False
.
Answer: A
Explanation: Outer fibers are more effective in resisting the applied moments. In hollow shafts material is removed and spread on a larger radius.
Q.Solid shaft is costlier than hollow shaft of same weight.
a) True
b) False
.
Answer: B
Explanation: Hollow shaft cost is more as material is to be selectively emplaced.
Q.Solid shafts are used in epicyclic gearboxes.
a) True
b) False
.
Answer: B
Explanation: In epicyclic gears, one shaft rotates inside other and hence hollow shafts are used.
Q.Flexible shafts have _______ rigidity in torsion making them flexible.
a) Low
b) High
c) Very high
d) Infinitely small
.
Answer: B
Explanation: Flexible shafts have high rigidity in torsion making then capable to transmit torque.
Q.Flexible shafts have ______ rigidity in bending moment.
a) High
b) Low
c) Very high
d) Extremely low
.
Answer: B
Explanation: Flexible shafts have low rigidity in bending moments making them flexible.
Q.While designing shaft on the basis of torsional rigidity, angle of twist is given by?
a) Ml/Gd⁴
b) 584Ml/Gd⁴
c) 292 Ml/Gd⁴
d) None of the mentioned
.
Answer: B
Explanation: θ=(180/π)xMl/GJ where J=πd⁴/32.
Q.According to the ASME code, maximum allowable shear stress is taken as X% of yield strength or Y% of ultimate strength.
a) X=30 Y=18
b) X=30 Y=30
c) X=18 Y=18
d) X=18 Y=30
.
Answer: A
Explanation: ASME Standard. The lesser value is taken among the two.
Q.Does ASME Standard take into consideration shock and fatigue factors?
a) Yes
b) No
.
Answer: A
Explanation: Moment is multiplied by a number to consider these factors while designing the shaft.
Q.A force 2P is acting on the double transverse fillet weld. Leg of weld is h and length l. Determine the shear stress in a plane inclined at θ with horizontal.
a) PSinθ(Sinθ+Cosθ)/hl
b) P(Sinθ+Cosθ)/hl
c) Pcosθ(Sinθ+Cosθ)/hl
d) None of the listed
.
Answer: A
Explanation: F=PSinθ and width=h/(Sinθ+Cosθ).
Q.Maximum shear stress in transverse fillet weld of leg h and length l is
a) P/hl
b) 1.21P/hl
c) P/1.21hl
d) None of the listed
.
Answer: B
Explanation: τ= PSinθ(Sinθ+Cosθ)/hl, by maximising it θ=67.5’ and hence find corresponding τ.
Q.A sunk key fits in the keyway of the _____ only.
a) Hub
b) Sleeve
c) Both hub and sleeve
d) Neither hub nor sleeve
.
Answer: A
Explanation: Sunk key fits halfway in the hub and halfway in the shaft.
Q.Hollow saddle key is superior to flat saddle key as far as power transmitting capability is concerned.
a) True
b) False
.
Answer: B
Explanation: The resistance to slip in case of flat key is more.
Q.Saddle key is more suitable than sunk key for heavy duty applications.
a) True
b) False
.
Answer: B
Explanation: In sunk key, relative motion is also prevented by shear resistance of sunk key and hence sunk key is recommended.
Q.The main advantage of sunk key is that it is a _____ drive.
a) Positive
b) Negative
c) Neutral
d) None of the listed
.
Answer: A
Explanation: Sunk key is a positive drive and no slip occurs.
Q.Woodruff key permits _____ movement b/w shaft and the hub.
a) Axial
b) Radial
c) Eccentric
d) None of the listed
.
Answer: B
Explanation: Woodruff key is a sunk key and doesn’t permit axial moment.
Q.Determine the length of kennedy key required to transmit 1200N-m and allowable shear in the key is 40N/mm². The diameter of shaft and width of key can be taken as 40mm and 10mm respectively.
a) 49mm
b) 36mm
c) 46mm
d) 53mm
.
Answer: D
Explanation: l=M/[dbτ√2].
Q.Splines are keys.
a) True
b) False
.
Answer: A
Explanation: Splines are keys made with shafts.
Q.Involute splines have stub teeth with a pressure angle of ___
a) 30
b) 45
c) 60
d) Can’t be determined
.
Answer: B
Explanation: Pressure angle is 30’ and not 60’.
Q.A coupling is a mechanical device that temporarily joins two rotating shafts to each other.
a) True
b) False
.
Answer: B
Explanation: A coupling permanently joins two rotating shafts.
Q.In distortion energy theorem, if a unit cube is subjected to biaxial stress, then S(yt) is given by which of the following?
a) √ (σ₁²- σ₁σ₂ +σ₂²)
b) σ₁²- σ₁σ₂ +σ₂²
c) √ (σ₁²+ σ₁σ₂ +σ₂²)
d) σ₁²+ σ₁σ₂ +σ₂²
.
Answer: A
Explanation: S(yt)=√[(σ₁ -σ₂)²+(σ₂ -σ₃)²+(σ₃ -σ₁)²]/2, for σ₃=0; expression a is obtained.
Q.Oldham coupling i used to connect two shafts having intersecting axes.
a) True
b) False
.
Answer: B
Explanation: Oldham coupling is used to connect two parallel shafts separated by a small distance.
Q.A machine started malfunctioning due to some issues with the coupling. The coupling emplaced in the machine was Oldham. The only coupling available in the workshop is Hooke’s coupling. So Oldham coupling can be replaced by Hooke’s coupling.
a) True
b) False
.
Answer: B
Explanation: It is not possible to replace Oldham by Hooke’s coupling as Hooke’s is used to connect intersecting axes and not parallel axes shafts.
Q.A muff coupling is connecting two shafts. The torque involved is 650N-m. The shaft diameter is 45mm with length and breadth of the key being 14mm and 80mm respectively. Find the shear stress induced in the key.
a) 30.2N/mm²
b) 25.8N/mm²
c) 34.4N/mm²
d) None of the listed
.
Answer: B
Explanation: τ =2M/dbl.
Q.The region of safety for biaxial stresses is of which shape in the case of maximum distortion energy theorem.
a) Ellipse
b) Circle
c) Rectangle
d) Square
.
Q.Distortion energy theorem is not recommended for ductile materials.
a) True
b) False
.
Answer: B
Explanation: Experimental investigations have shown that distortion energy theorem gives good results for failure of ductile component.
Q.Among maximum shear stress theory and distortion energy theory, which gives the higher value shear yield strength?
a) Maximum shear stress theory
b) Distortion energy theory
c) Both give equal values
d) Vary from material to material
.
Answer: B
Explanation: Maximum shear stress theory gives S(sy)=0.5S(yt) while Distortion energy theorem gives S(sy)=0.577(Syt).
Q.A muff coupling is connecting two shafts. The torque involved is 650N-m. The shaft diameter is 45mm with length and height of the key being 14mm and 80mm respectively. Find the compressive stress induced in the key.
a) 70.1 N/mm²
b) 51.6N/mm²
c) 45.5N/mm²
d) None of the listed
.
Answer: B
Explanation: σ=4M/dhl.
Q.Clamp coupling employ bolts.
a) True
b) False
.
Answer: A
Explanation: The two halves of sleeve are clamped together by bolts.
Q.Entire torque is transmitted by friction in the clamp coupling.
a) True
b) False
.
Answer: B
Explanation: Some torque is transferred by friction and some by key.
Q.Does shaft has to be shifted i axial direction to remove clamp coupling.
a) Yes
b) No
.
Answer: B
Explanation: Clamp coupling unlike muff coupling is easily removed without shifting the shaft in axial direction.
Q.If shaft diameter is 40mm, calculate the diameter of sleeves in clamp coupling.
a) 100mm
b) 80mm
c) 60mm
d) 40mm
.
Answer: A
Explanation: Dia=2.5d.
Q.If shaft diameter is 40mm, calculate the length of sleeves in clamp coupling.
a) 80mm
b) 140mm
c) 100mm
d) 120mm
.
Answer: B
Explanation: L=3.5d.
Q.If 8 bolts are emplaced in a clamp coupling with shaft diameter 80mm d, calculate the tensile force on each bolt if coefficient of friction is 0.3 and torque transmitted is 4000N-m.
a) 51234.4N
b) 45968.3N
c) 41666.7N
d) None of the listed
.
Answer: C
Explanation: P=2M/fdn.
Q.If 8 bolts are emplaced in a clamp coupling with shaft diameter 80mm d, calculate the diameter of each bolt if coefficient of friction is 0.3 and torque transmitted is 4000N-m. Permissible tensile stress is 80N/mm².
a) 27mm
b) 25mm
c) 23mm
d) 21mm
.
Answer: A
Explanation: Ax80=2M/fdn.
Q.Power is transmitted by only in rigid flange coupling.
a) True
b) False
.
Answer: B
Explanation: Power is transmitted by key as well as bolts.
Q.If shaft diameter is 60mm, how many bolts are recommended for rigid flange coupling?
a) 2
b) 3
c) 4
d) 5
.
Answer: C
Explanation: If shaft diameter is between 40mm and 100mm, 4 bolts are used.
Q.Determine the diameter of the bolts used in rigid flange coupling if transmitted torque is 270N-m, pitch circle diameter=125mm and four bolts are emplace in the coupling. Permissible shear stress in the bolts is 70N/mm².
a) 3.8mm
b) 3.6mm
c) 4.4mm
d) 4mm
.
Answer: C
Explanation: d²=8M/πDNτ.
Q.The hub is treated as a solid shaft while calculating torsional shear stress in the hub.
a) True
b) False
.
Answer: B
Explanation: Hub is treated as a hollow shaft.
Q.Find the shear stress in a flange at the junction of hub in rigid flanged coupling if torsional moment is 2980N-m and diameter of hub being 125mm. Also the thickness of flange is 25mm.
a) 6.77N/mm²
b) 10.24N/mm²
c) 4.84N/mm²
d) 4.22N/mm²
.
Answer: C
Explanation: τ=2M/πd²t.
Q.If shaft diameter is 30mm and number of pins emplaced are 6, then the diameter of the pin will be?
a) 6.4mm
b) 5.6mm
c) 6.1mm
d) 5.9mm
.
Answer: C
Explanation: d₁=.5d/√N.
Q.Calculate the force acting on the each pin in flexible coupling if torque transmitted is 397N-m and PCD=120mm with number of pins 6.
a) 1400.3N
b) 1102.8N
c) 1320.3N
d) None of the listed
.
Answer: B
Explanation: P=2M/DN.
Springs
Q. Which of the following function can the spring perform?
A. Store energy
B. Absorb shock
C. Measure force
D. All of the mentioned
.
Answer: D
Explanation: Spring can easily perform all the listed functions.
Q. The helix angle is very small about 2⁰. The spring is open coiled spring.
A. Yes
B. It is closed coiled spring
C. That small angle isn’t possible
D. None of the listed
.
Answer: B
Explanation: When the helix angle is small, the plane containing each coil is almost at right angles and hence it is called closed coiled spring.
Q. The helical spring ad wire of helical torsion spring, both are subjected to torsional shear stresses.
A. True
B. False
.
Answer: B
Explanation: The wire of helical torsion sprig is subjected to bending stresses.
Q. The longest leaf in a leaf spring is called centre leaf.
A. It is called middle leaf
B. It is called master leaf
C. Yes
D. None of the listed
.
Answer: B
Explanation: It is called master leaf.
Q. Multi leaf springs are not recommended for automobile and rail road suspensions.
A. True
B. False
.
Answer: B
Explanation: They are highly used in automobile and rail road suspensions.
Q. The spring index is the ratio of wire diameter to mean coil diameter.
A. True
B. False
.
Answer: A
Explanation: It is the ratio of mean coil diameter to wire diameter.
Q. If spring index=Q.5, what can be concluded about stresses in the wire?
A. They are high
B. They are negligible
C. They are moderate
D. Cannot be determined
.
Answer: A
Explanation: If indexis <3 then stresses are high due to curvature effect.
Q. A spring with index=15 is prone to buckling.
A. True
B. False
.
Answer: A
Explanation: Due to large variation, such a spring is prone to buckling.
Q. If the spring is compressed completely and the adjacent coils touch each other,the the length of spring is called as?
A. Solid length
B. Compressed length
C. Free length
D. None of the mentioned
.
Answer: A
Explanation: Terminology.
Q. If number of coils are 8 and wire diameter of spring 3mm, then solid length is given by?
A. None of the listed
B. 27mm
C. 24mm
D. 21mm
.
Answer: C
Explanation: Solid length=8×Q.
Q. Compressed length is smaller than the solid length.
A. True
B. False
.
Answer: B
Explanation: Compressed is length of spring under maximum compressive force. There is some gap between the coils under maximum load.
Q. Pitch of coil is defined as axial distance in compressed state of the coil.
A. Yes
B. It is measured in uncompressed state
C. It is same in uncompressed or compressed state
D. None of the listed
.
Answer: B
Explanation: Pitch is measured in uncompressed state.
Q. If uncompressed length of spring is 40mm and number of coils 10mm, then pitch of coil is?
A. 4
B. 40/9
C. 40/11
D. None of the mentioned
.
Answer: B
Explanation: Pitch=Uncompressed length/N-Q.
Q. Active and inactive, both types of coils support the load although both don’t participate in spring action.
A. Active coils don’t support the load
B. Inactive coils don’t support the load
C. Both active and inactive don’t support the load
D. Both active and inactive support the load
.
Answer: B
Explanation: Inactive coils don’t support the load.
Q. If a spring has plain ends then number of inactive coils is?
A. 1
B. 2
C. 3
D. 0
.
Answer: D
Explanation: There are no inactive coils in plain ends.
Q. Spring having square ends has 1 inactive coil.
A. True
B. False
.
Answer: B
Explanation: There are 2 inactive coils.
Q. Martin’s factor compensates for curvature effect in springs.
A. True
B. False
.
Answer: B
Explanation: Wahl’s factor accommodates curvature effect while designing spring.
Q. The angle of twist for the equivalent bar to a spring is given by? (Symbols have their usual meaning)
A. 8PD²N/Gd⁴
B. 16PD²N/Gd⁴
C. 16PDN/Gdᵌ
D. 8PDN/Gdᵌ
.
Answer: B
Explanation: θ=Ml/GJ where M=PD/2, l=πDN and J=πd⁴/3Q.
Q. The axial deflection of spring for the small angle of θ is given by?
A. 328PDᵌN/Gd⁴
B. 8PDᵌN/Gd⁴
C. 16PDᵌN/Gd⁴
D. 8PD²N/Gdᵌ
.
Answer: B
Explanation: Deflection=θxD/Q.
Q. A spring of stiffness constant k is cut in two equal parts. The stiffness constant of new spring will be k/Q.
A. True
B. False
.
Answer: B
Explanation: k=Gd⁴/8DᵌN, hence k is inversely proportional to number of coils. Thus result will be 2k.
Q. For two spring connected in series, the force acting on each spring is same and equal to half of the external force.
A. True
B. False
.
Answer: B
Explanation: Te force on each spring is equal to the external force.
Q. For two springs connected in series, the net deflection is equal to the sum of deflection in two springs.
A. True
B. False
.
Answer: A
Explanation: The net deflection is sum of the deflection of sprigs connected in series.
Q. For two springs connected in parallel, net force is equal to the sum of force in each spring.
A. True
B. False
.
Answer: A
Explanation: Net force applied is distributed in the two springs.
Q. Patenting is defined as the cooling below the freezing point of water.
A. True
B. False
.
Answer: B
Explanation: Patenting is heating steel above critical range followed by rapid cooling.
Q. Find the Wahl’s factor if spring index is Q.
A. Q.2020
B. Q.2424
C. Q.2525
D. Q.5252
.
Answer: C
Explanation: K=[4C-1/4C-4]+0.615/C.
Q. Find the shear stress in the spring wire used to design a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
A. 45Q.2N/mm²
B. 46Q.6N/mm²
C. 5Q.2N/mm²
D. None of the listed
.
Answer: B
Explanation: τ=K x 8PC/πd² where K=[4C-1/4C-4]+0.615/C.
Q. Find the mean coil diameter of a helical compression sprig if a load of 1200N is applied on the spring. Spring index is 6, and wire diameter 7mm.
A. 7/6mm
B. 42mm
C. 1200×6/7 mm
D. None of the listed
.
Answer: B
Explanation: D=Cd.
Q. Find total number coils in a spring having square and ground ends. Deflection in the spring is 6mm when load of 1100N is applied. Modulus of rigidity is 81370N/mm². Wire diameter and pitch circle diameter are 10mm and 50mm respectively.
A. 7
B. 6
C. 5
D. 4
.
Answer: A
Explanation: Deflection=8PDᵌN/Gd⁴ or N=Q.4 or Q. Total coils=5+2(square grounded ends).
Q. A railway wagon moving with a speed of Q.5m/s is brought to rest by bumper consisting of two springs. Mass of wagon is 100kg. The springs are compressed by 125mm. Calculate the maximum force acting on each spring.
A. 1200N
B. 1500N
C. 1800N
D. 2000N
.
Answer: C
Explanation: mv²/2=Pxdeflection/Q.
Q. In concentric springs, vibrations called surge are amplified.
A. True
B. False
.
Answer: B
Explanation: Surge are eliminated.
Q. Can concentric springs be used to obtain a force which is not proportional to its deflection?
A. True
B. False
.
Answer: A
Explanation: Two sprigs having different free length can be meshed to obtain such a result.
Q. The load shared by each spring is inversely proportional to the cross section of wire.
A. Yes
B. No, it is directly proportional
C. It is proportional to its square
D. It is proportional to its square root
.
Answer: B
Explanation: It is directly proportional to the cross section of wire.
Q. A concentric spring consists of 2 sprigs of diameter 10mm and 4mm. The net force acting on the composite spring is 5000N. Find the force acting on each of the two springs.
A. 123Q.2N and 376Q.8N
B. 78Q.4N and 42Q.6N
C. 68Q.7N and 43Q.3N
D. 64Q.3N and 435Q.7N
.
Answer: C
Explanation: P₁/P₂=d₁²/d₂² and P₁+P₂=5000.
Q. If the spring have same solid length and number of coils in the two springs are 8 and 10, then find the diameter of the spring with 8 coils. It is given diameter of spring with 10 coils is 12mm.
A. Q.6mm
B. 9mm
C. 12mm
D. 15mm
.
Answer: B
Explanation: N₁d₁=N₂d₂.
Q. Two spring having stiffness constants of 22N/mm and 25N/mm are connected in parallel. They are to be replaced by a single spring to have same effect. The stiffness of that spring will be?
A. None of the mentioned.
B. 3N/mm
C. 47N/mm
D. Q.7N/mm
.
Answer: C
Explanation: k=22+2Q.
Q. What will happen if stresses induced due to surge in the spring exceeds the endurance limit stress of the spring.
A. Fatigue Failure
B. Fracture
C. None of the listed
D. Nipping
.
Answer: A
Explanation: If endurance limit is passed, fatigue failure will follow.
Q. Surge is caused by resonance effect in the spring.
A. True
B. False
.
Q. Surge is a desirable effect in the sprigs.
A. True
B. False
.
Answer: B
Explanation: Surge means vibratory motion which isn’t welcomed anywhere in the machinery.
Q. For a helical torsion sprig, the stress concentration factor at inner fibre is? Give spring index=Q.
A. Q.005
B. Q.175
C. Q.223
D. Q.545
.
Answer: B
Explanation: K=4C²-C-1/4C(C-1).
Q. For a helical torsion sprig, the stress concentration factor at outer fibre is? Give spring index=Q.
A. 0.78
B. 0.87
C. Q.87
D. 0.69
.
Answer: B
Explanation: K=4C²+C-1/4C(C+1).
Q. Spiral spring is quite rigid.
A. Yes
B. No it is flexible
C. It is of moderate rigidity
D. Rigidity can’t be determined
.
Answer: B
Explanation: Spiral spring is made of very thin wire which imparts great flexibility.
Q. The strip of spiral spring is never subjected to pure bending moment.
A. True
B. False
.
Answer: B
Explanation: The strip of spiral spring is subjected to pure bending moment.
Q. Calculate the bending stress induced in the strip of the helical spring. The spring is subjected to a moment of 1250N-mm with breadth and thickens of the strip being 11mm and Q.5mmm respectively.
A. 50Q.8N/mm²
B. 6Q.2N/mm²
C. 60Q.1N/mm²
D. 56Q.3N/mm²
.
Answer: C
Explanation: σ=12M/bt².
Q. Angle of rotation o arbour with respect to drum is given by?
A. None of the listed
B. 12ML/Ebtᵌ
C. 8 ML/Ebtᵌ
D. 16ML/Ebtᵌ
.
Answer: B
Explanation: θ=ML/EI where I=btᵌ/Q.
Q. Bending stress in graduated length leaves are more than that in full length leaves.
A. Yes
B. No
C. In some cases
D. Can’t be stated
.
Answer: B
Explanation: Bending stress in full length leaves is around 50% higher than graduated length leaves.
Q. Nip is the initial gap between extra full length leaf and the graduated length leaf before the assembly.
A. True
B. False
.
Answer: A
Explanation: Nipping is done to balance the bending stress in the full length leaf and graduated length leaf.
Q. Nipping is defined as leaving the gap between full length leaf and graduated length leaf.
A. True
B. False
.
Answer: B
Explanation: Nipping is pre stressing achieved by a difference in radii of curvature.
Q. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is Q.2m. Width and thickness of the leaves are 100mm and 12mm respectively. If modulus of elasticity is 207000N/mm², calculate the initial nip.
A. 2Q.8mm
B. 2Q.9mm
C. 2Q.5mm
D. 2Q.1mm
.
Answer: B
Explanation: C=2PLᵌ/Enbtᵌ where L=Q.2/2, n=3+14,P=70/Q.
Q. A leaf spring consists of 3 extra full length leaves and 14 graduated length leaves. The maximum force that can act on the spring is 70kN and the distance between eyes of the spring is Q.2m. Width and thickness of the leaves are 100mm and 12mm respectively. Calculate the initial pre load required to close the nip.
A. 433Q.2N
B. 467Q.1N
C. 498Q.4N
D. Can’t be determined
.
Answer: B
Explanation: P=2x3x14x35000/17(3×3+2×14).
Q. Belleville spring can be used in clutch application.
A. True
B. False
.
Answer: A
Explanation: When h/t=Q.1, the Belleville spring can be used in clutches.
Q. Belleville spring can only produce linear load deflection characteristics.
A. Only linear
B. Linear as well as non linear
C. Non-linear
D. None of the mentioned
.
Answer: B
Explanation: Beliville spring can provide any linear or non-linear load deflection characteristics.
Q. When two Belleville sprigs are arranged in series, half deflection is obtained for same force.
A. One fourth deflection
B. Double deflection
C. Four time deflection
D. None of the listed
.
Answer: B
Explanation: Double deflection is obtained.
Q. When two Belleville springs are in parallel, half force is obtained for a given deflection.
A. Half force
B. Double force
C. Same force
D. Can’t be determined
.
Answer: B
Explanation: Double force is obtained.
Q. Propagation of fatigue failure is always due to compressive stresses.
A. Due to bending
B. Due to tensile
C. Due to fatigue
D. None of the listed
.
Answer: B
Explanation: Propagation is always due to tensile stresses.
Q. The strain energy stored in a spiral spring is given by?
A. 12M²L/Ebtᵌ
B. 6M²L/Ebtᵌ
C. 8M²L/Ebtᵌ
D. None of the listed
.
Answer: B
Explanation: U=Mθ/2 where θ=12ML/Ebtᵌ.
MCQ on Friction Clutches, Brakes, Belt Drives and Chain Drives
Q. Clutch and coupling perform the same action.
A. Both being permanent joints
B. No they are different type of joints
C. Both being temporary joints
D. None of the listed
.
Answer: B
Explanation: Clutch is a temporary joint while coupling is a permanent joint.
Q. Eddy current clutch is a type of friction clutch.
A. Yes
B. No, it is an electromagnetic type clutch
C. It is a mechanical clutch
D. None of the listed
.
Answer: B
Explanation: Eddy current clutch is a type of electromagnetic clutch.
Q. In positive contact clutches, power transmission is achieved by means of friction.
A. Yes
B. It is achieved by shear contact
C. Major part is achieved by friction
D. None of the listed
.
Answer: B
Explanation: Power transmission is achieved by interlocking of jaws or teeth.
Q. The jaw clutches show great amount of slip.
A. Yes
B. Zero slip
C. Used in non-synchronous applications
D. None of the listed
.
Answer: B
Explanation: Jaw clutches do not slip and are used in synchronous applications.
Q. For a new friction lining, uniform wear theory is used.
A. True
B. False
.
Answer: B
Explanation: Uniform pressure theory is used.
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform wear theory, calculate the operating force in the clutch.
A. 15546N
B. 12344N
C. 23562N
D. 24543N
.
Answer: C
Explanation: P=πpd(D-D./Q.
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform wear theory, calculate the torque transmitting capacity of the clutch.
A. 4Q.23N-m
B. 35Q.43N-m
C. 33Q.53N-m
D. 39Q.34N-m
.
Answer: B
Explanation: P=πpd(D-D./2 and M=μP(D+D./Q.
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform wear theory, calculate the power transmitting capacity of the clutch at 80rad/s.
A. 2Q.8kW
B. 3Q.4kW
C. 2Q.2kW
D. 3Q.5kW
.
Answer: A
Explanation: P=πpd(D-D./2 and M=μP(D+D./Q.Power=Mxω.
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform pressure theory, calculate the operating force in the clutch.
A. 15546N
B. 12344N
C. 23562N
D. 35343N
.
Answer: D
Explanation: P=πp(D²-d²)/Q.
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform pressure theory, calculate the torque transmitting capacity of the clutch.
A. 4Q.23N-m
B. 54Q.78N-m
C. 56Q.54N-m
D. 67Q.86N-m
.
Answer: B
Explanation: P= πp(D²-d²)/4 and M=μP(Dᵌ-dᵌ)/3(D²-d²).
Q. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the friction disk is Q.0mm and 200mm respectively. The coefficient of friction is 0.2 and permissible intensity of pressure is Q.5N/mm². Assuming uniform wear theory, calculate the power transmitting capacity of the clutch at 80rad/s.
A. 27kW
B. 32kW
C. 39kW
D. 44kW
.
Answer: D
Explanation: P= πp(D²-d²)/4 and M=μP(Dᵌ-dᵌ)/3(D²-d²).Power=Mxω.
Q. If number of contacting surfaces are 5, then number of disks required in multi disk clutch are?
A. 4
B. 5
C. 6
D. Can’t be determined
.
Answer: C
Explanation: Disks=Contacting surfaces+Q.
Q. Multi disk clutches are dry clutches.
A. Plasma clutches
B. Wet clutches
C. Yes
D. Depends on the lubrication used
.
Answer: B
Explanation: They are wet clutches as a lot of heat is dissipated due to more contacting surfaces.
Q. In scooters, generally single plate clutches are used.
A. True
B. False
.
Answer: B
Explanation: Multi disk plate clutches which are compact are used.
Q. The coefficient of friction is high in multi disk plate clutch.
A. Yes
B. Coefficient of friction is less
C. Coeffficient of friction is high
D. None of the listed
.
Answer: B
Explanation: In multi disk clutch, due to oil cooling, coefficient of friction decrease.
Q. A cone clutch consists of inner conical surface and outer cylindrical surface.
A. Both cylindrical
B. Both conical
C. Outer conical and inner cylindrical
D. True
.
Answer: B
Explanation: It consists of both inner and outer conical surfaces.
Q. Power is transmitted only by key and friction in the cone clutch.
A. Only by spline
B. By key, spline and friction
C. By friction only
D. By key only
.
Answer: B
Explanation: Power is transmitted via key, friction and spline.
Q. The torque transmitting capacity of cone clutch increases as its semi vertical angle increase.
A. True
B. Decreases
C. Remains constant
D. None of the listed
.
Answer: B
Explanation: Torque transmitted is inversely proportional to the sin of the semi vertical angle.
Q. When semi vertical angle is greater than angle of static friction, clutch results in self-engagement.
A. True
B. False
.
Answer: B
Explanation: Self engagement takes place when semi vertical angle is reduced so low to increase torque capacity that it become smaller than static friction angle.
Q. A cone clutch with a small semi cone angle requires a relatively large force to engage and small force to disengage.
A. True
B. False
.
Answer: B
Explanation: It requires large force to disengage.
Q. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 3Q.mm.Assuming uniform wear theory; find the maximum torque which is transmitted.
A. 50Q.4N-m
B. 54Q.3N-m
C. 46Q.72N-m
D. 45Q.5N-m
.
Answer: C
Explanation: M=PowerxQ.ᵌx60/2πN.
Q. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 3Q.mm.Assuming uniform wear theory; find the inner diameter.
A. 275mm
B. 300mm
C. 290mm
D. 280mm
.
Answer: B
Explanation: M=πpμd(D²-d²)/8Sinἀ.
Q. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 3Q.mm.Assuming uniform wear theory; find the face width of friction lining.
A. None of the listed
B. 4Q.1mm
C. 5Q.2mm
D. 5Q.8mm
.
Answer: B
Explanation: b=D-d/2Sinἀ.
Q. A cone clutch transmits 24kW at 490rpm. The coefficient of friction is 0.2 and allowable intensity of pressure is 0.35N/mm². The semi cone angle is 12⁰. The outer diameter is fixed as 3Q.mm.Assuming uniform wear theory; calculate force required to engage clutch.
A. 354Q.9N
B. 223Q.5N
C. 324Q.5N
D. 435Q.5N
.
Answer: C
Explanation: P=4MSinἀ/μ(D+D..
Q. Centrifugal clutches are not recommended for IC engine applications.
A. True
B. False
.
Answer: B
Explanation: IC engine is not able to start under load ad hence centrifugal clutches are used.
Q. In a centrifugal clutch, when the centrifugal force is slightly more than the spring force, shoe begins to move in a radially inward direction.
A. Radially outwards
B. Radially upwards
C. Radially downwards
D. Radially inwards
.
Answer: A
Explanation: It moves in radially outward direction.
Q. Chain saws, lawnmowers, golf carts etc. never use centrifugal clutches.
A. True
B. No, it is highly used
C. Depends on the load required
D. None of the listed
.
Answer: B
Explanation: The electric motor has time to accelerate to reach the operating speed before it takes the load and hence centrifugal clutches are used.
Q. The centrifugal clutches are used in light duty vehicles and not in heavy duty vehicles.
A. True
B. False
.
Answer: B
Explanation: Centrifugal clutches are used in heavy uty applications as they provide a time delay that is sufficient to permit the prime mover to gain momentum before taking over the load.
Q. Find number of contacting surfaces for a multi disk clutch plate transmitting torque of Q.N-m and inner and outer diameters of friction lining are 70mm and Q.0mm respectively. The operating force is of magnitude 305N and coefficient of friction is 0.Q.
A. 5
B. 2
C. 4
D. 3
.
Answer: C
Explanation: z=4M/μP(D+D..
Q. The energy absorb by brake is always kinetic.
A. No, potential
B. Kinetic or potential
C. Potential
D. Strain Energy
.
Answer: B
Explanation: It can be either kinetic or potential.
Q. Pneumatic brakes are same as electrical brakes.
A. Yes both are concerned with electricity
B. No, one deals with pressure and other with electricity
C. Yes both deals with pressure
D. None of the listed
.
Answer: B
Explanation: Pneumatic brakes are operated by fluid pressure.
Q. Disc brakes are radial brakes.
A. True
B. False
.
Answer: B
Explanation: Disc brakes are axial brakes.
Q. Internal shoe brakes are radial while external shoe brakes are axial brakes.
A. True
B. False
.
Answer: B
Explanation: Both internal and external shoe brakes are radial brakes.
Q. A solid cast iron disk of mass Q.00kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is Q.6sec.Calculate energy absorbed by the brake if square of radius of gyration is 0.Q.
A. None of the mentioned
B. 13Q.3kJ
C. 16Q.3kJ
D. 13Q.2kJ
.
Answer: B
Explanation: E=mk²ω²/Q.
Q. A solid cast iron disk of mass Q.00kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is Q.6sec.Square of radius of gyration is 0.Q. Calculate the average angular velocity during braking period.
A. Q.45rad/s
B. Q.32rad/s
C. 1Q.32rad/s
D. None of the mentioned
.
Answer: C
Explanation: ω(avg)=[ω()initial+ω(final)]/Q.
Q. A solid cast iron disk of mass Q.00kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is Q.6sec.Square of radius of gyration is 0.Q. Calculate the angle through which disk rotated during braking period.
A. 2Q.5rad
B. 2Q.6rad
C. 2Q.3rad
D. 3Q.4rad
.
Answer: C
Explanation: ω(avg)=[ω()initial+ω(final)]/2 and θ=ωt.
Q. A solid cast iron disk of mass Q.00kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is Q.6sec.Square of radius of gyration is 0.Q. Calculate the torque capacity of the brake.
A. 8Q.4N-m
B. 458Q.6N-m
C. 6Q.4N-m
D. Noe of the listed
.
Answer: B
Explanation: M=E/θ.
Q. An automobile brake is only used to reduce the speed or bring the vehicle to hault.
A. Yes
B. No, it also be used to hold the car
C. Brake acts only onmoving vehicles
D. None of the mentioned
.
Answer: B
Explanation: It can also be used to hold the car stationary.
Q. Block brakes are never used in railway trains.
A. True
B. False
.
Answer: B
Explanation: Blocks brakes are commonly employed in railway trains.
Q. If torque capacity of a block brake is 240N-m and radius of drum is 2Q.mm with coefficient of friction 0.Q.
A. None of the listed
B. 285Q.14N
C. 344Q.42N
D. 233Q.62N
.
Answer: B
Explanation: N=M/μR.
Q. If initially drum of radius 200mm is rotating at Q.0rpm, then calculate the heat generated if frictional force acting is 1140N.
A. 3345W
B. 4456W
C. 5969W
D. 11938W
.
Answer: C
Explanation: Heat=Frictional forcexaverage velocity.
Q. In a block brake with short shoe, it is assumed that friction force and normal reaction are concentrated at the midpoint of the shoe.
A. True
B. False
.
Answer: A
Explanation: The angle of contact is less than 45⁰ and hence the assumption is valid.
Q. The pressure intensity at an element on long shoe brake varies directly with the sin of the pressure angle of the element of friction lining.
A. True
B. False
.
Answer: B
Explanation: p=Cxcos(Ǿ).
Q. The equivalent coefficient of friction in case of block brake with long shoe is denoted by μ[4Sinθ/2θ+sin4θ].
A. True
B. False
.
Answer: B
Explanation: μ[4Sinθ/2θ+sin2θ].
Q. In pivoted block brakes, moment of frictional force about pivot is non zero.
A. True
B. False
.
Answer: B
Explanation: Pivot point is selected such that moment of frictional force about pivot point is zero.
Q. Shoe is constrained to move towards the drum to compensate for the moment acting.
A. Yes
B. It compensates for wear
C. It compensates for both wear and moment
D. None of the listed
.
Answer: B
Explanation: It compensates for the wear which is constant for all points.
Q. A pivoted double block brake has a drum radius of 280mm with two shoes subtending an angle of Q.0⁰. Calculate the distance of pivot from axis of drum.
A. 32Q.5mm
B. 3Q.3mm
C. 28Q.5mm
D. None of the mentioned
.
Answer: B
Explanation: h=4Rsinθ/2θ+sin2θwhere θ=Q.0/Q.
Q. A pivoted double block brake has a drum radius of 280mm with two shoes subtending an angle of Q.0⁰. Maximum pressure intensity is 0.5N/mm². If the width of friction lining is 90mm, find the torque capacity f each shoe. Assume coefficient of friction as 0.Q.
A. 789N-m
B. Q.81N-m
C. 945N-m
D. None of the listed
.
Answer: B
Explanation: M=2μR²wpsinθ.
Q. A pivoted double block brake has a drum radius of 280mm with two shoes subtending an angle of Q.0⁰. Maximum pressure intensity is 0.5N/mm². If the width of friction lining is 90mm, find the reaction at pivot in horizontal direction. Assume coefficient of friction as 0.Q.
A. None of the listed
B. 21123N
C. 17200N
D. 16789N
.
Answer: C
Explanation: R(x)=Rwp(2θ+sin2θ)/Q.
Q. A pivoted double block brake has a drum radius of 280mm with two shoes subtending an angle of Q.0⁰. Maximum pressure intensity is 0.5N/mm². If the width of friction lining is 90mm, find the reaction at pivot in horizontal direction. Assume coefficient of friction as 0.Q.
A. None of the listed
B. 2113N
C. 1720N
D. 3440N
.
Answer: D
Explanation: R(y)=μRwp(2θ+sin2θ)/Q.
Q. The intensity of normal pressure between the friction lining and the brake drum at any point is proportional to square of the vertical distance from the pivot.
A. It is independent
B. Proportional to vertical distance linearly
C. Inversely proportion to vertical distance
D. None of the listed
.
Answer: B
Explanation: It is proportional to the vertical distance linearly.
Q. The coefficient of friction in internally expanding brakes is constant.
A. True
B. No, it varies with increase in speed linearly
C. It decreases with increase in speed linearly
D. None of the listed
.
Answer: A
Explanation: Coefficient of friction is constant in these types of brakes.
Q. The centrifugal force acting on the shoe can never be omitted.
A. True
B. Always omitted
C. Depends on the magnitude
D. There is no centrifugal force acting
.
Answer: B
Explanation: Centrifugal force on the internal expanding brakes is generally neglected as it is very small.
Q. These kinds of brakes require little maintenance.
A. Yes
B. No
C. Very much maintenance is required
D. Depends on the working environment
.
Answer: A
Explanation: Very little maintenance is required as there is less wear.
Q. In internally expanding brake, a large actuating force produces a small braking torque and hence this brake isn’t used generally.
A. True
B. No, small actuating braking torque is produces
C. Zero actuating force is required
D. None of the listed
.
Answer: B
Explanation: Small actuating force produce large braking torque.
Q. Heat dissipating capacity of internally expanding brakes is a problem.
A. True
B. No
C. There is no heat generated
D. Heat is convected away on its own
.
Answer: A
Explanation: It has relatively poor dissipation of heat.
Q. Internally expanding brakes never suffer the problem of self-locking.
A. True
B. Wear might lead to self-locking
C. Brakes are never self-locked
D. None of the listed
.
Answer: B
Explanation: Due to improper design, wear may be caused which might lead to self-locking of brakes.
Q. Moment of normal force and frictional forces about the pivot axis are 640000N-mm and 250000N-mm respectively. If force acts at a distance of 190mm from the pivoted point, calculate the actuating force.
A. Q.7Q.6N
B. 205Q.6N
C. 322Q.5N
D. 445Q.5N
.
Answer: B
Explanation: P=M(n)-M(f)/C.
Q. Preloading is recommended for bearings.
A. True
B. False
.
Answer: A
Explanation: Preloading removes the internal clearances.
Q. Needles bearings have not much load capacity as they are quite small.
A. True
B. False
.
Answer: B
Explanation: Needle bearings have quite large loading capacity as compared to their size.
Q. In both simple band brake and differential band brake, at least one end of the band passes through fulcrum.
A. Both the ends
B. Neither end
C. One end
D. None of the listed
.
Answer: B
Explanation: In differential, neither end of the brake passes through fulcrum.
Q. Is it possible to prevent back rotation of drum?
A. Yes by using back stop brake
B. No
C. Depends on the angular velocity
D. Only in some specific cases
.
Answer: A
Explanation: Back stop brake can be used to prevent rotation of drum.
Q. A self-locking differential brake and back stop brakes is same thing.
A. True
B. False
.
Answer: A
Explanation: Back stop brake is other name of self locking differential brake.
Q. Band brake has complex construction with large number of parts.
A. Yes
B. It has simple construction
C. It has simple constructions but large number of parts
D. It has complicated construction but small number of parts
.
Answer: B
Explanation: Band brake has simple construction with small number of parts.
Q. Band brakes require _____ maintenance.
A. Little
B. Zero
C. Much
D. None of the listed
.
Answer: A
Explanation: There is relatively less maintenance required as there are small number of parts and hence chances of brake going out of order is less.
Q. The wear of band brake is even.
A. True
B. False
.
Answer: B
Explanation: There is uneven wear from one side to other side of the friction lining.
Q. Calculate the braking torque if tension in the two sides is 3500N and 1980N with radius of drum being 160mm.
A. None of the mentioned
B. 24Q.20N-m
C. 22Q.4N-m
D. 27Q.6N-m
.
Answer: B
Explanation: Braking torque=(3500-1980)x160/Q.00.
Q. Disk brakes have low torque transmitting capacity in high volumes.
A. True
B. False
.
Answer: B
Explanation: They have high torque transmitting capacity in low volumes.
Q. In drum brakes, as the temperature increases coefficient of friction ______
A. increases
B. decreases
C. remains same
D. Can’t be determined
.
Answer: D
Explanation: Coeffecient of friction varies inversely with the temperature.
Q. Loss of torque transmitting capacity at high temperatures is called
A. Fading
B. Rolling
C. Drifting
D. Planking
.
Answer: A
Explanation: Terminology.
Q. Disk brakes largely suffer from the problem of self-locking.
A. Yes
B. No
C. Self-locking but it depends on radius of the wheel rotating
D. Material dependent
.
Answer: B
Explanation: There is no problem of self-locking in disk brakes.
Q. Disk brakes can be used for opposite rotation of disk also.
A. No they are effective for one direction of motion only
B. They can be used
C. Poor efficiency in opposite direction
D. None of the listed
.
Answer: A
Explanation: Disk brakes are equally effective for both direction of motion.
Q. The disk brake torque is linearly proportional to the actuating force.
A. Yes
B. No, it is proportional to its square
C. Proportional to its cube
D. Independent of force
.
Answer: A
Explanation: M=μPr.
Q. Belt, chain and rope are called rigid drives.
A. True
B. Belt is a flexible drive only
C. All three are flexible drives
D. None of the listed
.
Answer: B
Explanation: They all are flexible drive examples.
Q. In flexible drives, rotary motion of driving shaft is directly converted to rotary motion of driven shafts
A. Yes
B. No, translator motion is involved
C. No, Cylindrical motion is involved
D. No, Parabolic motion is involved
.
Answer: B
Explanation: It is first converted into translator and then into rotary.
Q. Flexible drive can absorb shock loads and damp vibrations.
A. True
B. No
C. Depends on the load applied
D. Doesn’t damp vibrations
.
Answer: A
Explanation: Intermediate link is long and flexible.
Q. Velocity ratio in both flexible and rigid drive is constant.
A. True
B. False
.
Answer: B
Explanation: In flexible drive, velocity ratio is not constant.
Q. Is there any worry for overloading conditions in flexible belt drives?
A. Yes
B. No
C. Only after a particular threshold limit
D. Depends on external factrors
.
Answer: B
Explanation: In case of overloading, belt slips over the pulley and hence protect it from the overload.
Q. V belts have v shaped cross section.
A. No, rectangular
B. No, trapezium
C. No, circular
D. No, spherical
.
Answer: B
Explanation: They have trapezoidal cross section.
Q. The force of friction between belt and V grooved pulley is high.
A. Yes, supported by wedge action
B. No
C. There is no wedge action involved
D. None of the listed
.
Answer: A
Explanation: Due to wedge action, force of friction is high.
Q. V belts result in smooth and quite operation even at high speeds.
A. Yes
B. No they are very noisy
C. They are not endless and hence not smooth motion
D. None of the listed
.
Answer: A
Explanation: V belts are endless.
Q. The efficiency of flat belt drive is 35%. If all the parameters are same and flat belt is replaced by V belt, than the efficiency of V belt will be?
A. <35% B. >35%
C. =35%
D. Cant be determined
.
Answer: A
Explanation: The efficiency of flat belt drive is more that V belt drive.
Q. There is problem of bending stress in the V belt drive.
A. True
B. False
.
Answer: A
Explanation: The ratio of cross sectional height to pulley diameter is high.
Q. The layer of a belt is generally called as
A. Ply
B. Layer
C. Segment
D. Sediment
.
Answer: A
Explanation: Terminology.
Q. Velocity ratio for chain drive is lesser than that for belts.
A. Yes
B. No
C. In some cases
D. Can’t be determined
.
Answer: B
Explanation: Velocity ratio for chain drives is about 15:1 while for belts it is around 7:Q.
Q. Fabric rubber belts are not widely used as they can’t be operated at high speeds.
A. They can’t be used at velocities 50m/s
B. They can be used at high velocities
C. Limiting velocity is 20m/s
D. None of the listed
.
Answer: B
Explanation: These belts can be operated at 300m/s.
Q. Power transmitting capacity of V belts is more than that of flat belt.
A. Yes
B. No
C. Only for V angle > 15
D. None of the listed
.
Answer: A
Explanation: Coefficient of friction is Q.92times of flat belts in V belts for identical materials.
Q. The optimum velocity of the belt for maximum transmission is given by √P/6m.
A. Yes
B. No, by replacing 6 by 3
C. Replacing 6 by 9
D. None of the listed
.
Answer: B
Explanation: It is given by √P/3m.
Q. Creep is the slight absolute motion of the belt as it passes over the pulley.
A. Yes
B. No, it is a relative motion
C. None of the listed
D. It is absolute motion
.
Answer: B
Explanation: It is a relative and not absolute motion.
Q. In horizontal belt, the loose side is generally kept on the bottom.
A. True
B. False
.
Answer: B
Explanation: Loose side is kept on the top so that arc of contact increases and hence efficiency of the drive increases.
Q. Vertical drives are preferred over horizontal.
A. True
B. False
.
Answer: B
Explanation: In vertical drive, gravitational force on the belt causes slip which reduces the efficiency.
Q. The law of belting states that the centreline of the belt when it approaches a pulley must not lie in the midplane of the pulley.
A. True
B. False
.
Answer: A
Explanation: The centreline must lie in the midplane.
Q. It is possible to use the belting reverse direction without violating the law of belting.
A. True
B. False
.
Answer: B
Explanation: Law of belting is violated if belt is used in reverse direction.
Q. A shorter centre distance is always preferred in belt drives.
A. Yes due to stability
B. No due to stability
C. Yes due to instability
D. No due to instability
.
Answer: A
Explanation: It is more stable and compact.
Q. If velocity ratio is less than 3, then centre distance is given by D+.5d.
A. True
B. False
.
Answer: B
Explanation: It is given by D+Q.5d.
Q. If velocity ratio is more than 3, then centre distance is given by 2D.
A. By 3D
B. By D
C. By 4D
D. By 5D
.
Answer: B
Explanation: It is given by D.
Q. Is it possible to reduce the centre distance as much as we want?
A. Yes power transmission increases
B. No
C. Can’t be cited
D. None of the listed
.
Answer: B
Explanation: It depends on physical dimension and the minimum angle of wrap required to transfer the required power.
Q. The diameter of the shorter pulley in leather belt drive is 265mm. It is rotating at 1440 rpm. Calculate the velocity of the belt.
A. 25m/s
B. 20m/s
C. 30m/s
D. None of the mentioned
.
Answer: B
Explanation: v=πdn/60×Q.00.
Q. Calculate the angle of wrap if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
A. 17Q.8⁰
B. 16Q.8⁰
C. 15Q.3⁰
D. None of the mentioned
.
Answer: A
Explanation: ὰ=180 – 2sin¯¹(D-d/2C..
Q. Calculate the arc of contact if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
A. Q.04
B. Q.03
C. Q.01
D. Q.02
.
Answer: D
Explanation: ὰ=180 – 2sin¯¹(D-d/2C.. Factor=1+ (Q.04-1)(180-17Q.8)/(180-170).
Q. Calculate the belt length if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
A. Q.5m
B. Q.66m
C. Q.94m
D. Q.26m
.
Answer: C
Explanation: L=2C + π(D+D./2 + (D-D.²/4C.
Q. Crowns are never mounted on the pulley.
A. True
B. False
.
Answer: B
Explanation: Crowns are used to avoid slip in case of misalignment or non-parallelism.
Q. In a cast iron pulley minor axis is generally kept in the plane of rotation.
A. True
B. False
.
Answer: B
Explanation: Keeping minor axis in plane of rotation increases the cross section.
Q. The number of V belts required for a given application are given by (ignoring correction factor for arc of contact and belt length) Transmitted power/kW rating of single belt x Industrial Service Factor.
A. True
B. False
.
Answer: B
Explanation: It is given by Transmitted power x Industrial Service Factor /kW rating of single belt.
Q. The pitch diameter of bigger pulley D in terms of small diameter d is given by
A. dx[speed of smaller pulley/speed of bigger pulley].
B. dx[speed of bigger pulley/speed of smaller pulley].
C. d
D. None of the mentioned
.
Answer: A
Explanation: Product of diameter and speed of pulley is constant.
Q. If maximum tension in the belt is 900N and allowable belt load is 500N. Calculate the number of belts required to transmit power.
A. 2
B. 3
C. 4
D. 5
.
Answer: A
Explanation: No of belts=900/500.
Q. The belt tension is maximum when velocity of belt is 0.
A. It is max at v=infinity
B. True
C. It is velocity independent
D. It has a constant value
.
Answer: A
Explanation: P₁-mv²/P₂-mv²=e^(fa/sinθ/2). Hence belt tension is maximum when v=0.
Q. If belt tension in the two sides is 730N and 140N and belt is moving with a velocity of Q.m/s, calculate the power transmitted.
A. Q.5kW
B. Q.9kW
C. Q.2kW
D. None of the mentioned
.
Answer: B
Explanation: Power=(P₁-P₂)xv.
Q. If tensions in the belt are P₁ and P₂, then find P₁-mv²/P₂-mv². Contact angle for smaller pulley is 156⁰, Groove angle is 36⁰ and coefficient of friction is 0.Q.
A. Q.21
B. Q.83
C. Q.66
D. Q.36
.
Answer: B
Explanation: P₁-mv²/P₂-mv²=e^(fὰ/sinθ/2).
Q. Is it possible to remove the slack from the belt?
A. Yes
B. No
C. It can be done but at the cost of strength
D. None of the listed
.
Answer: A
Explanation: A short length of belt is cut to remove slack provided it is not an endless belt.
Q. A ribbed V belt is a negative drive.
A. Yes
B. No, it’s a positive drive
C. Zero drive
D. None of the listed
.
Answer: B
Explanation: It is a positive drive.
Q. The cross section of V belt consist of rubber backing to help engagement of teeth with sprocket.
A. Yes
B. No, it is used for protection
C. There is no rubber packing
D. None of the listed
.
Answer: B
Explanation: It is used for protecting the load carrying tension members.
Q. Ribbed V belt is not generally recommended as the output speed keeps fluctuating.
A. True
B. No, it is generally recommended
C. No variation in output speeds
D. None of the listed
.
Answer: B
Explanation: There is no variation in output speed in case of ribbed V belts.
Q. The ribbed V belts don’t require any type of lubrication like V belts.
A. True
B. Intense lubrication is required
C. No belts require any lubrication
D. Ribbed V belts can be operated with or without lubrication
.
Answer: A
Explanation: Ribbed V belts are free from lubrication problem.
Q. Ribbed V belts are very stable and can tolerate some misalignment without any malfunctioning.
A. True
B. False
.
Answer: B
Explanation: Ribbed V belts are very sensitive.
Q. Toothed belt is preferred over ribbed V belt.
A. Yes
B. No
C. They both imply to same thing
D. None of the listed
.
Answer: C
Explanation: Toothed belt and ribbed V belt is same thing.
Q. The idler pulley increases the power transmitting capacity of the belt drive.
A. True
B. False
.
Answer: A
Explanation: Idler pulley increases the arc of contact which further increases the power transmitting capacity.
Q. Adjusting screw can be used for adjusting belt tension?
A. Yes
B. No
C. Distance can never be changed
D. None of the listed
.
Answer: A
Explanation: Adjusting screw varies the distance between electric motor and bedplate.
Q. A chain drive consist of an endless chain wrapped around two sprockets.
A. True
B. False
.
Answer: A
Explanation: This is the basic definition of chain drive.
Q. The chain drive is same as gear drive.
A. Yes
B. No, it is intermediate between belt and gear drive
C. It is superior to gear drive
D. It is inferior to gear drive
.
Answer: B
Explanation: Chain drive is intermediate between belt drive and gear drive.
Q. Gear drives are preferred over chain drives for long centre distances.
A. Yes
B. No, they require additional idle gears.
C. No idler gears are required
D. None of the listed
.
Answer: B
Explanation: Gear drive will require additional idler gears.
Q. Belt drives have compact construction as compared to chain drives.
A. Yes
B. No, chain drives have compact construction
C. Both have same construction
D. Can’t be cited
.
Answer: B
Explanation: Chain drives have overall small dimensions and hence resulting in compact constructions.
Q. The maximum achievable efficiency of the chain drives is about 80%.
A. It is about 50%
B. It is about 96-98%
C. It is about 75%
D. It is always <40%
.
Answer: B
Explanation: Maximum achievable efficiency is around 96-98%.
Q. Chain drives suffer from wear problems.
A. Yes, chain might leave the sprocket
B. No, there is no wear problem
C. Wear is insignificant in chain drives
D. None of the listed
.
Answer: A
Explanation: Wear increases pitch of the chain which leads to chain leaving the sprocket.
Q. For non-parallel shafts, chain drives are highly recommended.
A. True
B. Gear drive is preferred
C. Depends on the transmission ratio
D. Can’t be stated
.
Answer: B
Explanation: Bevel or worm gears can be used in this case. Chain drives isn’t suitable for this application.
Q. The velocity of the chain is always constant in case of chain drive.
A. Yes
B. No, it can never be constant
C. It is constant above a particular value
D. Can’t be stated
.
Answer: B
Explanation: Velocity isn’t constant in chain drive which results in non-uniform speed of the driven shaft.
Q. Chain drives require little or no maintenance and creates no noise.
A. True
B. False
.
Answer: B
Explanation: Chain drives require lubrication at proper periods and also generates noise.
Q. In roller chains, there is sliding friction between roller and sprocket teeth.
A. True
B. False
.
Answer: B
Explanation: Rollers are freely fitted on the bushes and during engagement turn with teeth. This results in rolling friction.
Q. In the designation ‘04B’, pitch of the chain is?
A. Q.7mm
B. Q.35mm
C. 4mm
D. None of the mentioned
.
Answer: B
Explanation: Pitch=[04/16]x2Q.4mm.
Q. The designation ‘04B’ is as per ANSI series.
A. Yes
B. No, it is as par British standards
C. One digit is missing
D. None of the listed
.
Answer: B
Explanation: The letter B indicates British Standard Series.
Q. Wear ultimately results in ride in of the chain on sprocket teeth.
A. True
B. Ride out
C. Ride up
D. Ride down
.
Answer: B
Explanation: Wear causes ride out of chain.
Q. Galling is a stick slip phenomenon.
A. True
B. No, it is only a stick phenomenon
C. No, it is only a slip phenomenon
D. Can’t be stated
.
Answer: A
Explanation: During high tension, welds are formed at the high spot of contacting area which are subsequently broken due to relative motion.
Q. A gear meshes with another gear and a sprocket meshes with another sprocket.
A. True
B. False
.
Answer: B
Explanation: A sprocket meshes with with an intermediate link which further meshes with another sprocket.
Q. The number of links of the chain should be always ______
A. Odd
B. Even
C. Multiple of 3
D. None of the listed
.
Answer: B
Explanation: The odd number of teeth are meshed with even number of links.
Q. In chain drives, vertical drives are generally avoided.
A. True
B. False
.
Answer: B
Explanation: Sag causes the chain to leave the profile of the teeth at lower side.
Q. Sum of number of links and number of teeth of the sprocket is _____
A. Odd
B. Even
C. Odd or Even
D. None of the listed
.
Answer: A
Explanation: Number of links are even while number of teeth on the sprocket are odd.
Q. The driving sprocket has 19 teeth and rotates at 1Q.0rpm. The pitch of the chain is 18mm. Calculate the velocity of the chain.
A. Q.6m/s
B. Q.3m/s
C. Q.8m/s
D. Q.1m/s
.
Answer: B
Explanation: v=zpn/60×Q.00.
Q. Calculate the pitch circle diameter of the driving pulley is pitch is given as 18mm and sprocket has 19 teeth.
A. 1Q.33mm
B. Q.Q.36mm
C. Q.Q.66mm
D. Q.Q.33mm
.
Answer: B
Explanation: D=p/sin(180/z).
Q. Driving and driven pulleys are rotating at Q.00rpm and 500rpm. If number of teeth on driving sprocket are 20, find number of teeth in the driven sprocket.
A. 41
B. 40
C. Q.
D. 11
.
Answer: B
Explanation: n₁z₁=n₂z₂.
Q. If centre distance for a chain drive is 750mm with number of teeth on the driving and driven sprockets being 40 and 20 respectively, then calculate the number of links required. Given: Pitch is taken as 19mm.
A. 112
B. 1Q.
C. 111
D. Q.9
.
Answer: B
Explanation: 2(a/p) + (z₁+z₂)/2 +( z₂- z₁/2π)²xp/a.
Q. In silent chain sprocket teeth have a rectangular profile.
A. Yes rectangular
B. No trapezoidal
C. No circular
D. No hyperbolic
.
Answer: B
Explanation: They have a trapezoidal profile.
Q. Roller chains have less reliability as compared to roller chains.
A. True
B. False
.
Answer: B
Explanation: Silent chains have laminated structure and hence are more reliable.
Rolling Contact Bearings, Spur Gears and Helical Gears in Machine Design
Q. Which of the following are functions of bearings?
A. Ensure free rotation of shaft with minimum friction
B. Holding shaft in a correct position
C. Transmit the force of the shaft to the frame
D. All of the listed
.
Answer: D
Explanation: Bearings are used for all the above listed purposes.
Q. A radial bearing supports the load that acts along the axis of the shaft.
A. True
B. False
.
Answer: B
Explanation: Radial bearing supports the load acting perpendicular to the axis of the shaft.
Q. A_______ bearing supports the load acting along the axis of the shaft.
A. Thrust
B. Radial
C. Longitudinal
D. Transversal
.
Answer: A
Explanation: Thrust bearing supports load acting along axis of shaft.
Q. Sliding contact bearings, also called plain bearings have no problem of wear.
A. True
B. False
.
Answer: B
Explanation: Surface of shaft slide over surface of the bush resulting in friction and wear.
Q. In steam and gas turbines, rolling contact bearings are used.
A. True
B. False
.
Answer: B
Explanation: Sliding contact bearings are generally used.
Q. Which of the following are true about plasticity?
A. Permanent Deformation
B. Ability to retain deformation under load or after removal of load
C. Plastic deformation is greater than elastic deformation
D. All of the mentioned
.
Answer: D
Explanation: This is the basic definition of plasticity.
Q. Which of the following is measure of stiffness?
A. Modulus of elasticity
B. Modulus of plasticity
C. Resilience
D. Toughness
.
Answer: A
Explanation: Stiffness is the ability of material to resist deformation under external load. Hence it is measured by modulus of elasticity.
Q. Which of the following facts are true for resilience?
A. Ability of material to absorb energy when deformed elastically
B. Ability to retain deformation under the application of load or after removal of load
C. Ability of material to absorb energy when deformed plastically
D. None of the mentioned
.
Answer: A
Explanation: Toughness is ability to store energy till proportional limit during deformation and to release this energy when unloaded.
Q. Modulus of resilience is defined as
A. Strain energy per unit volume
B. Strain energy per unit area
C. Independent of strain energy
D. None of the mentioned
.
Answer: A
Explanation: Modulus of resilience is strain energy per unit volume.
Q. In gear boxes and small size motors, rolling contact bearings are used.
A. True
B. False
.
Answer: A
Explanation: In small size applications, rolling contact bearings are preferred.
Q. Deep groove ball bearings creates a lot of noise.
A. Yes
B. They create very less noise
C. Depends on the application
D. No reference frame for comparison is mentioned
.
Answer: B
Explanation: They create very less noise due to point contact.
Q. There is problem of alignment in deep groove ball bearings.
A. Yes
B. No, it is self-aligning
C. It aligns itself only in some particular cases
D. Can’t be determined
.
Answer: A
Explanation: It is not self-aligning.
Q. Deep groove ball bearing has immense rigidity.
A. True
B. No it has point contact and hence low rigidity
C. It has surface contact
D. It has line contact
.
Answer: B
Explanation: Due to point contact, rigidity is not so good.
Q. Cylindrical load bearing has lower load capacity as compared to deep groove ball bearing.
A. True
B. False
.
Answer: B
Explanation: Cylindrical load bearing has a line contact and hence higher load capacity.
Q. Angular contact bearing can take thrust as well as radial loads.
A. True
B. False
.
Answer: A
Explanation: The line of reaction at the contact surfaces makes an angle with axis of bearing and thus has two components, hence allowing it to take both type of loads.
Q. In angular contact bearings, ____ bearings are required to take thrust load in both directions.
A. 1
B. 4
C. 2
D. 3
.
Answer: C
Explanation: Angular contact bearing has this disadvantage of needing two bearings.
Q. The angular play bearing must be mounted without axial play.
A. Yes
B. Little tolerance is adjusted
C. Little tolerance is necessary
D. Can’t be stated
.
Answer: A
Explanation: There is no tolerance of misalignment.
Q. Taper rolling supports
A. Axial loads
B. Thrust loads
C. Both axial and thrust loads
D. None of the mentioned
.
Answer: C
Explanation: The line of reaction makes an angle with the axis of bearing and hence both type of loads can be carried.
Q. Which of the following isn’t the property of taper roller?
A. High rigidity
B. Easy dismantling
C. Take low radial and heavy loads
D. All are the properties of tapper roller
.
Answer: C
Explanation: Due to line contact it can take high radial and thrust loads.
Q. Which of the following cannot take radial load?
A. Cylindrical Roller bearing
B. Taper roller bearing
C. Thrust ball bearing
D. None of the listed
.
Answer: C
Explanation: There is no inclination in the line of reaction and hence only thrust loads can be carried.
Q. Which of the following cannot tolerate misalignment?
A. Angular contact bearing
B. Cylinder roller bearing
C. Thrust ball bearings
D. All of the listed
.
Answer: D
Explanation: All of these require précised alignment.
Q. Cylinder roller creates lesser noise than deep groove ball bearing.
A. True
B. False
.
Answer: B
Explanation: Due to line contact, cylinder roller creates far more noise
Q. Static load is defined as the load acting on the bearing when shaft is _____
A. Stationary
B. Rotating at rpm<10
C. Rotating at rpm<5
D. None of the listed
.
Answer: A
Explanation: As name suggests, static load means load during stationary position of the shaft.
Q. A total permanent deformation of 0.0001 of the ball diameter is taken for considering static load capacity of the shaft.
A. True
B. False
.
Answer: A
Explanation: Permissible magnitude of deformation is set as per the experimental data.
Q. Stribeck equation gives dynamic load capacity of the bearing.
A. True
B. False
.
Answer: B
Explanation: It gives static load capacity of the bearing.
Q. Which of the following is expression for stribeck equation?(Number of balls=z)?
A. C=kd²z/5
B. C=kd²z/15
C. C=kd²z/10
D. None of the listed
.
Answer: A
Explanation: C=(1/5)zP and P=kd².
Q. The life of an individual ball bearing is the time period for which it works without any signs of failures.
A. True
B. False
.
Answer: B
Explanation: It is measured by number of revolutions and not by time.
Q. For majority of bearings, actual life is considerably greater than rated life.
A. True
B. False
.
Answer: A
Explanation: Stastically it has been proved that life which 50% of a group of bearings will exceed is approximately five times rating of rated life.
Q. The dynamic load carrying capacity of a bearing is defined as the radial load in radial bearings that can be carried for a minimum life of 1000 revolutions.
A. True
B. False
.
Answer: B
Explanation: It is for 1 million revolutions.
Q. In the expression of dynamic load capacity P=XVF(r) + YF(A., V stands for ?
A. Race rotation factor
B. Radial factor
C. Thrust factor
D. None of the listed
.
Answer: A
Explanation: It represent race rotation factor which depends upon whether inner race is rotating or outer race.
Q. Calculate the bearing life if expected life for 90% bearings is 9000h and shaft is rotating at 1500rpm.
A. 850 million rev
B. 810 million rev
C. 810 h
D. 850h
.
Answer: B
Explanation: L=60x1500x9000/10⁶ million rev.
Q. The bearing is subjected to a radial load of 4000N. Expected life for 90% bearings is 9000h and shaft is rotating at 1500rpm. Calculate the dynamic load capacity.
A. 4Q.21kN
B. 3Q.29kN
C. 2Q.33kN
D. 3Q.22kN
.
Answer: B
Explanation: C=4000x(810)⅓.
Q. A rolling contact bearing is specified as X30Q. Calculate the bearing diameter.
A. 35mm
B. 28mm
C. 21mm
D. 7mm
.
Answer: A
Explanation: Diameter=5 times the last two digits.
Q. A rolling contact bearing is specified as X30Q. Determine the series.
A. Extra light series
B. Light series
C. Medium series
D. Heavy series
.
Answer: C
Explanation: 1- Extra light series,2-Light series,3-Medium series,4-heavy series.
Q. Needles bearing consist of rollers of small diameter and very small length.
A. True
B. False
.
Answer: B
Explanation: Also called as quill bearings, they consist of small diameter and comparatively longer b=rollers.
Q. Extreme pressure causes ______ wear in the bearing parts.
A. Abrasive
B. Corrosive
C. Pitting
D. All of the mentioned
.
Answer: B
Explanation: Extreme pressure elements in EP additives are added in lubricating oils.
Q. Scoring is a ________ phenomenon.
A. Stick-slip
B. Fracture
C. Fatigue
D. In-out
.
Answer: A
Explanation: Alternative welding and shearing takes place.
Q. Thick film lubrication describes a phenomenon where two surfaces are _______ separated.
A. Completely
B. Partially
C. Not
D. None of the mentioned
.
Answer: A
Explanation: There is no contact between the two surfaces.
Q. Hydrodynamic bearing is a self-acting bearing.
A. True
B. False
.
Answer: A
Explanation: The pressure in the bearing is created within the system due to rotation of the shaft.
Q. A journal bearing is a ______ contact bearing working on the hydrodynamic lubrication and which supports load in____ direction.
A. Sliding, Axial
B. Rolling, Radial
C. Sliding, Radial
D. Rolling, Axial
.
Answer: C
Explanation: This is how journal bearing works. It derives its name from the portion of the shaft inside the bearing.
Q. Partial bearing is preferred over journal bearing.
A. True
B. No
C. More friction losses
D. Can’t be determined
.
Answer: A
Explanation: Friction losses are lesser and construction is simple.
Q. Temperature rise in partial bearing is ____ than full bearing.
A. Lesser
B. Greater
C. Equal
D. Undeterminable
.
Answer: A
Explanation: Due to lesser friction, temperature rise is less.
Q. A clearance bearing is design accurately to keep the radius of journal and bearing equal.
A. Journal radius is kept larger
B. Journal radius is kept smaller
C. True
D. Can’t be determined
.
Answer: B
Explanation: Journal radius is kept smaller.
Q. Fitted bearing must be partial bearing.
A. True
B. No
C. No lubricating space is required
D. Can’t be stated
.
Answer: A
Explanation: To provide space for lubricating oil.
Q. Footstep bearing is an axial load bearing.
A. True
B. Thrust load
C. Shear load
D. None of the listed
.
Answer: B
Explanation: It is a thrust bearing in which shaft end is in contact with bearing surface.
Q. Hydrostatic and hydrodynamic lubrication are the same thing.
A. True
B. False
.
Answer: B
Explanation: In hydrodynamic, motion is provided by the shaft while in hydrostatic the motion is provided by an external source.
Q. If we exclude the cost factor, which bearing is preferred?
A. Hydrostatic
B. Hydrodynamic
C. Both are equally preferred
D. Cannot be determined
.
Answer: A
Explanation: High load capacity, no starting friction and no rubbing action.
Q. If fluid film pressure is high and surface rigidity is low than mode of lubrication is called as elastohydrodynamic lubrication.
A. True
B. False
.
Answer: A
Explanation: In such cases, elastic deflections of the parts causes development of the film.
Q. Viscosity is defined as the external resistance offered by a fluid to change its shape or relative motion of its parts.
A. Yes
B. It is internal resisting force
C. It is not offered but exerted on the fluids
D. None of the listed
.
Answer: B
Explanation: It is an internal resisting force.
Q. Stream line flow happens when intermediate layers move with velocities proportional to the square of distance from the stationary plate.
A. True
B. False
.
Answer: B
Explanation: It is proportional to distance and not square of it.
Q. Newton law of viscosity states that shear stress is proportional to rate of shear at any point in the fluid.
A. True
B. False
.
Answer: A
Explanation: P/A proportional to U/h.
Q. Calculate the kinematic viscosity if Saybolt viscosity is 400cSt.
A. 400SUS
B. 40.25SUS
C. 8Q.32SUS
D. 8Q.55SUS
.
Answer: D
Explanation: z=0.22t-[180/t] where t=400.
Q. Viscosity of lubricating oil decrease with increasing temperature.
A. Yes
B. It increases linearly
C. It increases hyperbolically
D. it remains constant
.
Answer: A
Explanation: Intermolecular forces decrease on the increase of temperature.
Q. Which of the following lubricant has least rate of change of viscosity w.r.t temperature.
A. VI=20
B. VI=30
C. VI=40
D. VI=50
.
Answer: D
Explanation: Greater the VI, lesser is the rate of change w.r.t temperature.
Q. Which of the following are not true for petroff’s equation?
A. Shaft is considered concentric with the bearing
B. Bearing is subjected to light load
C. Is used to find coefficient of friction
D. Frictional torque is given by fpr²l
.
Answer: D
Explanation: M=fWr=f(2prl)r, W=projected area of bearing x pressure.
Q. In hydrodynamic lubrication, film thickness remains unaffected by change in speeds.
A. True
B. Increase with increase in speed
C. Decrease with increase in speed
D. Disappear as the speed tends to infinty
.
Answer: B
Explanation: As speed increases more and more lubricant is forces and pressure builds up thus separating the two surfaces. There is transition from thin film thick film.
Q. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm, film thickness=0.15mm, viscosity of lubricant=160SUS and specific gravity=0.8Q.
Calculate supply pressure
A. Q.2Pa
B. Q.01Pa
C. Q.01Mpa
D. Q.2Mpa
.
Answer: C
Explanation: P=2Wln(225/155)/[π(225²-155²)] N/mm².
Q. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm, film thickness=0.15mm, viscosity of lubricant=160SUS and specific gravity=0.8Q.
Calculate flow requirement
A. 0.89l/s
B. 3Q.94l/min
C. 2Q.8l/min
D. None of the mentioned
.
Answer: B
Explanation: Q=πPhᵌ/6µln(225/155) whereµ=z/10⁹ and z=0.86x[0.22×160-180/160]. µ=2Q.3 x 10¯⁹N-s/mm².
Q. For a hydrostatic thrust bearing,
Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm,film thickness=0.15mm,viscosity of lubricant=160SUS and specific gravity=0.8Q.
Calculate power loss in pumping.
A. Q.68kW
B. Q.35kW
C. Q.6kW
D. Q.2kW
.
Answer: C
Explanation: kW=Q(P-0)x10¯⁶.
Q. For a hydrostatic thrust bearing, Thrust load=450kN, shaft speed=730rpm, shaft diameter=450mm, recess diameter=310mm,film thickness=0.15mm,viscosity of lubricant=160SUS and specific gravity=0.8Q.
Calculate frictional power loss.
A. None of the listed
B. Q.3kW
C. Q.56kW
D. Q.2kW
.
Answer: C
Explanation: kW=µn²(225⁴-155⁴)/hx5Q.05×10⁶.
Q. The fundamental equation for viscous flow is given by ∆pbhᵌ/6µl where symbols have their usual meanings.
A. 6 is to be replaced by 3
B. 6 is to be replaced by 12
C. 6 is to be replaced by 9
D. 6 is to be replaced by 15
.
Answer: B
Explanation: It is givenby ∆pbhᵌ/12µl.
Q. A single transverse weld is preferred over double transverse fillet weld.
A. True
B. False
.
Answer: B
Explanation: A single transverse weld is not preferred because the edge of the plate which is not welded can warp out of shape.
Q. Transverse fillet weld can be designed using the same equations as of parallel fillet weld.
A. True
B. False
.
Answer: B
Explanation: Vice versa is true as strength of transverse fillet weld is greater than that of parallel fillet weld.
Q. The common normal to the curves of the two teeth must not pass through the pitch point.
A. True
B. It must pass
C. It may or may not pass
D. None of the listed
.
Answer: B
Explanation: The common normal must pass through the point where two mating gears meet.
Q. Which of the following can be used for power transmission in intersecting shafts.
A. Spur Gear
B. Helical Gear
C. Bevel Gear
D. None of the listed
.
Answer: C
Explanation: Bevel gears are used for power transmission in case of intersecting shafts.
Q. Is it possible to transmit power between shafts lying in different planes using gears?
A. Yes
B. No
.
Answer: A
Explanation: Worm or crossed helical gears can be used in this case for power transmission.
Q. The two gears are said to have conjugate motion if
A. They have constant angular velocity ratio
B. Variable angular velocity ratio
C. Infinitely small angular velocity ratio
D. None of the mentioned
.
Answer: A
Explanation: Two gear are said to have conjugate motion and tooth profiles are said to have conjugate curves if they have constant angular velocity ratio.
Q. Which of the following is not true about gears?
A. Positive drive
B. Constant velocity ratio
C. Transmit large power
D. Bulky construction
.
Answer: D
Explanation: They have compact construction.
Q. Gear drive don’t require precise alignment of shafts.
A. True
B. False
.
Answer: B
Explanation: A minute level of misalignment isn’t tolerated in gears.
Q. Spur gears can be used only when the two shafts are parallel.
A. True
B. False
.
Answer: A
Explanation: The teeth are cut parallel to the axis of shaft.
Q. The teeth of the helical gears are cut parallel to the shaft axis.
A. True
B. False
.
Answer: B
Explanation: They are cut at an angle with the shaft axis.
Q. Herringbone gear can be used in
A. Intersecting shafts only
B. Parallel shafts only
C. Both intersection and parallel shafts
D. None of the mentioned
.
Answer: B
Explanation: It consist of two helical gears with the opposite hand of the helix.
Q. Bevel gears impose ____ loads on the shafts.
A. Radial and thrust
B. Radial
C. Thrust
D. Neither radial nor thrust
.
Answer: A
Explanation: Bevel gears have the shape of a truncated cone and tooth is cut straight or spiral.
Q. Which of the following are true for worm gears?
A. Worm is in the shape of threaded screw
B. Threads on the worm have small lead
C. Worm imposes high thrust loads
D. Characterised by low speed reduction ratio
.
Answer: D
Explanation: They are characterised by high speed reduction ratio.
Q. Greater the velocity ratio, smaller the gearbox.
A. True
B. Greater the gearbox
C. Size of gearbox remains unaffected
D. None of the listed
.
Answer: B
Explanation: Greater velocity leads to increase in size of gear wheel which results in size of gearbox.
Q. Required velocity ratio is 60:1, which of the following are recommended?
A. Worm
B. Spur
C. Bevel
D. None of the mentioned
.
Answer: A
Explanation: For high speed reduction ratio, worm gears are recommended.
Q. For a constant velocity ratio, the common normal to the tooth profile at point of contact must pass through a continuously variable point.
A. True
B. It pass through a fixed point
C. Constant velocity ratio isn’t required, hence variable point is preferred
D. None of the listed
.
Answer: B
Explanation: It must pass through a fixed point called pitch to maintain a constant velocity ratio.
Q. In cycloidal gears contact area is
A. Comparatively smaller
B. Comparatively larger
C. Can’t be determined
D. None of the listed
.
Answer: B
Explanation: Convex flank on one tooth meets with concave on the other thus increasing the contact area.
Q. Involute gears have greater contact area as compared to cycloidal gears.
A. True
B. False
.
Answer: B
Explanation: There is mating of two convex surfaces and hence lesser contact area.
Q. Cycloidal teeth consist of
A. Hypocycloid curve
B. Epicycloid gear
C. Both hypocycloid curve and epicycloid curve
D. None of the mentioned
.
Answer: C
Explanation: It consist of both and thus are hard to manufacture.
Q. Pressure angle remains constant in case of involute profile.
A. True
B. False
.
Answer: A
Explanation: The common normal always passes through the pitch point and thus maintain the constant inclination.
Q. Pressure angle is _____ in case of cycloidal teeth.
A. Constant
B. Variable
C. zero
D. None of the listed
.
Answer: B
Explanation: Cycloidal teeth consist of two profiles.
Q. Velocity ratio is the ratio angular velocity of driving gear to that of driven gear.
A. True
B. False
.
Answer: A
Explanation: Velocity ratio is simply the angular velocities ratio.
Q. Velocity ratio and transmission ratio are the same thing.
A. True
B. False
.
Answer: B
Explanation: Transmission ratio is measured between first and last gear.
Q. Contact ratio is always
A. =1
B. >1
C. <1
D. Can’t be determined
.
Answer: B
Explanation: Some overlapping is essential for continuous transfer of power.
Q. Product of diametric pitch and circular pitch is?
A. π
B. 1/π
C. None of the listed
D. 2
.
Answer: A
Explanation: CP=πd/z and circular pitch=z/d.
Q. Diameteral pitch is 5, then calculate module of the gear.
A. 0.2
B. 0.4
C. 5
D. 10
.
Answer: A
Explanation: Module is the inverse of diameteral pitch.
Q. If centre distance between the two gears on same shaft is unequal to the centre distance on the other two gears on the second shaft, then this gear train is called reverted gear train.
A. True
B. False
.
Answer: B
Explanation: The centre distance is equal in both the shafts.
Q. If one gear is fixed while the other gear has motion of two types i.e. rotary about its own axis and rotation about axis of fixed gear, than the gear train is _____
A. Epicyclic gear train
B. Reverted gear train
C. Kepler gear train
D. None of the mentioned
.
Answer: A
Explanation: Definition of epicyclic gear train.
Q. Which of the following are true about epicyclic gear train?
A. Fixed gear is called sun gear
B. Rotating gear are called earth gear
C. Crank is called sun carrier
D. None of the listed
.
Answer: A
Explanation: Rotating gears are called planet gears.
Q. Epicyclic gears are not generally recommended due to bulky construction.
A. True
B. False
.
Answer: B
Explanation: Epicyclc gears have compact construction.
Q. A compound gear train consists of at least 3 shafts connected to each other.
A. True
B. False
.
Answer: B
Explanation: A compound gear train is characterised by one shaft carrying two gears atleast.
Q. Interference is caused by?
A. Overlapping of tooth profiles
B. Large size of dedendum
C. Meshing of involute and no-involute profiles
D. All of the mentioned
.
Answer: D
Explanation: In some cases, portion of tooth below base circle is not involute due to large dedendum. This leads to overlapping of involute portion of teeth with the non-involute portion of the teeth.
Q. Interference can be removed by under cutting.
A. True
B. False
.
Answer: A
Explanation: Undercutting removes the interfering portion of the tooth.
Q. Interference is maximum when the largest pinion is in mesh with the smallest gear.
A. True
B. False
.
Answer: B
Explanation: It is maximum when smallest pinion is in mesh with the largest gear.
Q. Undercutting is consider healthy for the tooth as it eliminates undercutting.
A. True
B. False
.
Answer: B
Explanation: Although it eliminates undercutting but it also weakens the strength of the tooth and removes a small involute portion near to the base circle.
Q. Backlash is defined as
A. Difference in the width of tooth space and engaging tooth thickness
B. Amount by which engaging tooth thickness exceeds the tooth space
C. Other name for interference
D. None of the listed
.
Answer: A
Explanation: Backlash is the amount by which tooth space exceeds the engaging tooth thickness.
Q. The amount of backlash depends on
A. Diameteral pitch
B. Module
C. Centre distance
D. All of the mentioned
.
Answer: D
Explanation: Module and diameteral pitch are the same thing, and centre distance is also a factor while deciding the magnitude of the backlash.
Q. The two teeth have thickness t₁ and t₂. If backlash of t₁ is greater than that of t₂, than
A. t₁ > t₂
B. t₁ < t₂
C. t₁ = t₂
D. None of the mentioned
.
Answer: B
Explanation: Backlash=Tooth space-tooth thickness, greater the backlash lower is the tooth thickness.
Q. There are ____ standard systems for the shape of gear teeth.
A. 1
B. 2
C. 3
D. 4
.
Answer: C
Explanation: Q.5⁰ full depth involute system, 20⁰ full depth involute system and 20⁰ stub involute system.
Q. When the number of teeth reaches infinity, circle radius approaches infinity the gear becomes an infinite loop.
A. True
B. False
.
Answer: B
Explanation: It becomes a rack with straight sided teeth.
Q. Which of the following statements are not true?
A. Increasing pressure angle improves the tooth strength
B. Contact duration is decreased with increase in pressure angle
C. 20⁰ pressure angle has quieter operation then Q.5⁰
D. All of the statements are true
.
Answer: C
Explanation: Lower the pressure angle, quieter is the operation. Lower the pressure angle, lower is the breadth of the tooth at root.
Q. 20⁰ stub involute system have comparatively smaller interference.
A. True
B. False
.
Answer: A
Explanation: They have shorter addendum and shorter dedendum.
Q. Which of the following have stronger teeth?
A. Stub teeth
B. Full depth teeth
C. Both have equal strength
D. Can’t be determined
.
Answer: A
Explanation: Smaller moment arm of bending force leads to stronger stub teeth.
Q. As the module increases, index of size of gear decreases.
A. True
B. False
.
Answer: B
Explanation: Module is the measure of size of index of the gear tooth.
Q. Crowning is an abrasive process that debars the gear strength.
A. True
B. False
.
Answer: B
Explanation: Crowningis used to strengthen the tooth.
Q. Inaccuracies in tooth profile lead to concentration of pressure on the middle of tooth.
A. True
B. False
.
Answer: B
Explanation: Inaccuracies lead to shift of pressure at the end of tooth which can be improved by crowning.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the centre distance.
A. 280mm
B. 269mm
C. 350mm
D. 305mm
.
Answer: C
Explanation: C = m(z(p)+z(g))/Q.
C = 5(25 + 115)/2 => 700/2 => 350mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate pitch circle diameter of the pinion.
A. 95mm
B. 105mm
C. 115mm
D. 125mm
.
Answer: D
Explanation: D = 5×25 = 125mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the pitch circle diameter of the gear.
A. Cannot be determined
B. 31mm
C. 475mm
D. 575mm
.
Answer: D
Explanation: D = 5×115 = 575mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the addendum.
A. None of the listed
B. Q.75mm
C. Q.25mm
D. 5mm
.
Answer: D
Explanation: H=m=5mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the dedendum.
A. Q.75mm
B. 5mm
C. Q.25mm
D. Q.68mm
.
Answer: C
Explanation: H = Q.25xm = Q.25x5mm = Q.25mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the tooth thickness.
A. Q.23mm
B. Q.44mm
C. Q.854mm
D. Q.16mm
.
Answer: C
Explanation: T = Q.5708xm = Q.5708x5mm = Q.854mm.
Q. A pair of spur gears consist of 25 teeth pinion meshing with a 115 teeth gear. The module is 5mm. Calculate the bottom clearance.
A. None of the listed
B. Q.75mm
C. Q.5mm
D. Q.25mm
.
Answer: D
Explanation: C = 0.25m = 0.25x5mm = Q.25mm.
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse module.
A. Q.3mm
B. Q.1mm
C. Q.9mm
D. Q.7mm
.
Answer: A
Explanation: m=4/Cos(22⁰).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the transverse pressure angle in degrees.
A. 1Q.9
B. 20.4
C. 1Q.6
D. 1Q.4
.
Answer: B
Explanation: tanᾰ=tan(19⁰)/Cos(22⁰).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the axial pitch.
A. None of the listed
B. 3Q.2mm
C. 3Q.4mm
D. 2Q.6mm
.
Answer: C
Explanation: p=πx(transverse module)/tan(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of pinion.
A. 6Q.7mm
B. 5Q.6mm
C. 5Q.6mm
D. 6Q.8mm
.
Answer: A
Explanation: d=zxm/Cos(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the pitch circle diameter of the gear.
A. 17Q.6mm
B. 14Q.6mm
C. 180.3mm
D. 20Q.4mm
.
Answer: A
Explanation: d=zxm/Cos(22).
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the centre distance.
A. 12Q.4mm
B. 13Q.6mm
C. 11Q.65mm
D. 14Q.4mm
.
Answer: C
Explanation: C=Sum of diameter of pinion and gear/Q.
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate addendum circle diameter of the pinion.
A. 5Q.2mm
B. 7Q.7mm
C. 6Q.4mm
D. None of the listed
.
Answer: B
Explanation: D(A.=m[z/Cos(22) + 2].
Q. A pair of parallel helical gears consist of 15 teeth pinion meshing with a 40 teeth gear. The helix angle is 22⁰ and normal pressure angle 19⁰. The normal module is taken as 4mm. Calculate the dedendum circle diameter of the pinion.
A. 6Q.5mm
B. 5Q.7mm
C. 5Q.2mm
D. None of the listed
.
Answer: B
Explanation: D(f)=m[z/Cos(22) – Q.5].
Q. The direction of tangential component for a driving gear is same to the direction of rotation.
A. True
B. False
.
Answer: B
Explanation: The direction is opposite and not same.
Q. If tangential component of force on tooth is 200N and helix angle is 25⁰, calculate the axial component of the force.
A. 200N
B. 30Q.5N
C. 9Q.26N
D. 2Q.6N
.
Answer: C
Explanation: P(A.=200xtan(25).
Q. Among spur gear and helical gear, which has smooth engagement and thus lesser noise?
A. Helical Gears
B. Spur Gears
C. Both have equal noises
D. Can’t be determined
.
Answer: A
Explanation: There is a gradual pick up of load in helical gears and hence smooth operation.
Q. There is same type of tooth meshing in helical and spur gear.
A. True
B. False
.
Answer: B
Explanation: In spur gears contact occurs along entire face width which leads to impact condition while in helical contact begins from a single point and then there is gradual increase in load.
Q. Among the normal module and transverse module, which one has greater value?
A. Normal Module
B. Transverse Module
C. Both have equal module
D. Insufficient information
.
Answer: B
Explanation: Normal Module=Transverse modulexCos(helix angle).
Q. The net axial force acting on bearing is zero in case of double helical gears while none zero in case of herringbone gears.
A. True
B. False
.
Answer: B
Explanation: It is zero in both the cases.
Q. Helix angle of herringbone and double helical gears is relatively higher.
A. True
B. False
.
Answer: A
Explanation: There is no thrust force and hence higher angles are permitted.
Q. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.Q. If permissible bending stress is 500N/mm², then calculate the beam strength.
A. 15000N
B. 12000N
C. 8000N
D. 10000N
.
Answer: B
Explanation: S=mbσY.
Q. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.Q. Calculate the ratio factor Q.
A. Q.2
B. Q.4
C. Q.7
D. Q.4
.
Answer: C
Explanation: Q=2×100/100+20.
Q. A herringbone speed reducer consist of 20 teeth pinion driving a 100 teeth gear. The normal module of gear is 2mm. The face width of each half is 30mm and Lewis factor is 0.Q. Calculate the material constant K if surface hardness is 400BHN.
A. Q.25 N/mm²
B. Q.05 N/mm²
C. Q.25N/mm²
D. Q.56N/mm
.
Answer: C
Explanation: K=0.16x[BHN/100]².
Q. Helical gears mounted on parallel shafts are called crossed helical gears.
A. True
B. False
.
Answer: B
Explanation: Crossed helical gears are the helical gears mounted on non-parallel shafts.
Q. Crossed helical gears have very low load carrying capacity.
A. True
B. False
.
Answer: B
Explanation: There is point contact and hence very less area and thus wear is comparatively rapid.
Q. Calculate the shaft angle for same hand of helix if helix angle of two gears are 20⁰ and 17⁰.
A. 17⁰
B. 20⁰
C. 37⁰
D. 3⁰
.
Answer: C
Explanation: For same hand of helix, shaft angle=sum of helix angles of two gears.
Q. Lewis form factor is based on real number of teeth.
A. True
B. False
.
Answer: B
Explanation: It is based on virtual umber of teeth only.
Q. Beam strength indicates the maximum value of radial force that a tooth can transmit without fatigue failure.
A. True
B. False
.
Answer: B
Explanation: It indicates maximum force for bending failure and not fatigue failure.
Q. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the ratio factor.
A. 0.74
B. 0.88
C. Q.57
D. Q.44
.
Answer: C
Explanation: Ratio factor Q=2×90/90+2Q.
Q. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the material constant k If surface hardness is 260BHN.
A. 0.64N/mm²
B. 0.88N/mm²
C. Q.08N/mm²
D. Q.66N/mm²
.
Answer: C
Explanation: K=0.16x[BHN/100]².
Q. A pair of helical gears consist of 25 teeth pinion gear meshing with a 90 teeth gear. Calculate the wear strength If surface hardness is 260BHN. Also face width=35mm, module=4mm and helix angle=25⁰.
A. 44Q.5N
B. 112Q.6N
C. 797Q.9N
D. 10Q.2N
.
Answer: C
Explanation: S=bQdK/cos²Ɯ where Ɯ=25⁰, d=zm/cosƜ, Q=2×90/90+25, K=0.16x[BHN/100]².
MCQ on Bevel Gears, Worm Gears and FlyWheel in Machine Design
1. There are ____ common types of bevel gears.
a) 1
b) 2
c) 3
d) 4
.
Answer: b
Explanation: Spiral and straight.
2. Straight bevel gears are easy to design and manufacture and give reasonably good service with quieter operation.
a) True
b) No, noise problem
c) No, vibration problem
d) None of the listed
.
Answer: b
Explanation: They produce a lot of noise while working.
3. Which of the following creates smoother motion?
a) Straight bevel gears
b) Spiral bevel gears
c) Equal for straight and spiral
d) None of the mentioned
.
Answer: b
Explanation: Spiral bevel gear has smooth engagement which results in quieter operation.
4. When the pitch angle is greater than 90⁰, it is called external bevel gear.
a) True
b) Internal
c) Helical
d) Herringbone
.
Answer: b
Explanation: For external bevel gear, pitch angle is less than 90⁰.
5. For crown bevel gear, pitch angle is
a) <90 b) =90 c) >90
d) None of the listed
.
Answer: b
Explanation: Crown gears are characterised by pitch angle of 90⁰.
6. If pitch angle is >90, the bevel gear is?
a) Internal
b) External
c) Crown
d) Static
.
Answer: a
Explanation: Nomenclature.
7. Miter gears are the bevel gears mounted on shaft which are intersecting at angle greater than 90.
a) True
b) False
.
Answer: b
Explanation: The shafts which are intersecting at 90.
8. Crown gears having a pitch angle of 90⁰ are mounted on shafts intersecting at an angle
a) =90
b) <90 c) >90
d) None of the listed
.
Answer: c
Explanation: Application of crown gear according to its structure.
9. Which of the following are characteristics of skew bevel gears?
a) Straight teeth
b) Mounted on parallel shafts
c) Mounted on intersecting shafts
d) All of the mentioned
.
Answer: a
Explanation: They are used in non-parallel and non-intersecting shafts.
10. Hypoid gears are based on surfaces that are paraboloids of revolution.
a) True
b) False
.
Answer: b
Explanation: They are hyperboloids of revolution.
11. When two hyperboloids are rotated, the resulting motion is
a) Sliding
b) Turning
c) Combination of turning and sliding
d) Rotary
.
Answer: c
Explanation: There is turning as well as sliding motion.
12. In hypoid bevel gears, shafts may continue past each other.
a) True
b) False
.
Answer: a
Explanation: Offset of the shaft is quite considerable in case of hypoid gears.
13. Hypoid gears have the following characteristics.
a) Curved teeth
b) Mounted on non -parallel non intersecting shafts
c) Efficiency of 97%
d) All of the mentioned
.
Answer: d
Explanation: It has curved teeth and are mounted on non-parallel non intersecting shafts with efficiency of 96-98%.
14. Zerol gears are straight bevel gears with zero spiral angle.
a) True
b) False
.
Answer: b
Explanation: They happen to fall in the category of spiral bevel gears.
15. Zerol gears give lesser contact ratio.
a) True
b) No, larger contact ratio
c) Zero contact
d) None of the listed
.
Answer: b
Explanation: Zerol gears have more gradual contact and slighter larger contact ratio.
16. If pitch angle and addendum angles are 5⁰ and 12⁰ respectively, then face angle is equal to?
a) 17⁰
b) 7⁰
c) 5⁰
d) 12⁰
.
Answer: a
Explanation: Face angle=pitch angle+ addendum angle.
17. If pitch angle is 8⁰ and dedendum angle is 4⁰, then find root angle.
a) 12⁰
b) 4⁰
c) 8⁰
d) None of the listed
.
Answer: b
Explanation: Root angle=pitch angle-dedendum angle.
18. If back cone distance is 12mm and module at large end of the tooth is 4mm, then formative number of teeth will be?
a) 3
b) 6
c) 4
d) 12
.
Answer: b
Explanation: Formative number=2r/m.
19. If back cone distance is 12mm and module at large end of the tooth is 4mm, and virtual number of teeth is 12 then find the diameter of the tooth.
a) 5
b) 4
c) 3
d) 2
.
Answer: d
Explanation: 12/z = 2×12/4.
20. Calculate the cone distance of in a pair of bevel gears if pitch circle diameter of pinion and gear are 20mm and 24mm respectively.
a) 44mm
b) 22mm
c) 15.6mm
d) 20.2mm
.
Answer: c
Explanation: A=√(20/2)²+(24/2)².
21. Calculate the pitch angle if pitch circle diameter of the pinion and gear are 150mm and 210mm.
a) 28.14⁰
b) 35.54⁰
c) 36.22⁰
d) 63.22⁰
.
Answer: b
Explanation: tanϒ=D(p)/D(g)=150/210.
22. Calculate the radius of pinion at midpoint along the face width if PCD of pinion is 150mm and of gear is 210mm. Also face width of the tooth is 35mm.
a) 56.35mm
b) 64.83mm
c) 66.57mm
d) 58.69mm
.
Answer: b
Explanation: r=(Dp/2)-(bsinϒ/2).
23. Calculate the tangential component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm.
a) 1668N
b) 2946N
c) 3000N
d) 3326N
.
Answer: b
Explanation: P=M/r. r=(Dp/2)-(bsinϒ/2) where ϒ is ptch angle and is calculated by tanϒ=D(p)/D(g)=150/210.
24. Calculate the radial component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm. Also pressure angle is 20⁰.
a) 996.6N
b) 332.6N
c) 489.2N
d) 739.2N
.
Answer: d
Explanation: P radial=P tangential x[tan20 Cos ϒ].
25. Calculate the axial component of gear tooth force if power transmitted is 6kW and diameters of pinion and gear are 150mm and 210 mm with face width of tooth being 35mm. Power is transmitted at 3000rpm. Also pressure angle is 20⁰.
a) 660.05N
b) 528.06N
c) 448.21N
d) 886.6N
.
Answer: b
Explanation: P radial=P tangential x[tan20 Sin ϒ].
26. The pinion of a face gear is a
a) Spur gear
b) Helical gear
c) Either spur or helical
d) None of the mentioned
.
Answer: c
Explanation: Face gear consist of a spur or helical gear mating with a conjugate gear of disk form.
27. Lewis equation is used to obtain ____ strength of bevel gears.
a) Beam
b) Abrasive
c) Wear
d) Corrosive
.
Answer: a
Explanation: According to Lewis equation, Beam strength=product of module, face width of elemental section, permissible bending stress and Lewis form factor.
28. Beam strength of the bevel gear is independent of the cone distance.
a) True
b) False
.
Answer: b
Explanation: In other form S=mbσY[1- b/A] where A= cone distance.
29. Beam strength indicates the maximum value of the ___________ force at the large end of the tooth that the tooth can transmit without bending.
a) Tangential
b) Radial
c) Axial
d) None of the listed
.
Answer: a
Explanation: Beam strength is analysed by using the pitch radius at the larger end of the tooth.
30. Wear strength of the bevel gear can be calculated by using ________ equation.
a) Buckingham
b) Lewis
c) Newtonian
d) Rayleigh
.
Answer: a
Explanation: Bevel gear is considered to be equivalent to a formative spur gear in a plane which is perpendicular to the large end and hence Buckingham equation is applied.
31. If back cone distance is 10mm, then pitch circle diameter of the formative pinion is given by?
a) 10mm
b) 20mm
c) 5mm
d) 10√2 mm
.
Answer: b
Explanation: PCD=2xr.
32. The wear strength indicates the maximum value of radial force at the large end of the tooth that the tooth can transmit without pitting failure.
a) True
b) False
.
Answer: b
Explanation: It indicates the maximum value of tangential force.
33. If velocity is 5m/s, then velocity factor for a cut teeth is
a) 0.55
b) 0.66
c) 1.55
d) 1.66
.
Answer: a
Explanation: C=6/6+v.
34. If velocity is 5m/s, then velocity factor for generated teeth is
a) 0.71
b) 1.1
c) 0.9
d) 1.71
.
Answer: a
Explanation: C=5.6/5.6+√v.
35. Calculate the ratio factor Q if a pair of bevel gear consist of 25 teeth pinion meshing with a 40 teeth gear.
a) 1.964
b) 1.438
c) 1.554
d) 0.998
.
Answer: b
Explanation: tanϒ=25/40 or ϒ=32⁰. Q=2×40/ [40+25xtan(32)].
36. If surface hardness for a par of bevel gears is 400BHN, then material constant is
a) 3N/mm²
b) 2.56N/mm²
c) 0.98N/mm²
d) 1.44N/mm²
.
Answer: b
Explanation: K=0.16x[BHN/100]².
37. Calculate the wear strength for a pair of bevel gears having face width=20mm, module=5mm, No of teeth on pinion and gear 25 and 40 respectively and PCD of pinion=75mm.
a) 3668.5N
b) 4884.5N
c) 5126.6N
d) 4117.3N
.
Answer: b
Explanation: S=0.75xbxQxDxK/Cos ϒ. Tan ϒ=25/40 or ϒ=32⁰. Q=2×40/ [40+25xtan(32)] K=0.16x[BHN/100]².
38. If service factor is 1.4 & tangential ad dynamic load is 1180N and 1500N respectively, then calculate the effective load.
a) 3662N
b) 2889N
c) 3152N
d) 2236N
.
Answer: c
Explanation: Effective load=1.4×1180 + 1500.
39. Worm gear drives are used to transmit power between two non-intersecting shafts which are generally at right angles to each other.
a) True
b) False
.
Answer: a
Explanation: This is how the worm gear design fits into use.
40. The worm and worm wheel both are threaded screw.
a) True
b) Worm wheel is a toothed gear
c) Worm is a toothed gear
d) None of the listed
.
Answer: b
Explanation: Worm wheel is a toothed gear.
41. Which of the following is not true about worm gears?
a) Compact
b) Smooth and silent operation
c) Low speed reduction
d) All the mentioned are true
.
Answer: c
Explanation: Speed reduction can be high up to 100:1.
42. Is it possible to use worm gears in cranes for lifting purpose?
a) True
b) No self-locking and hence not possible
c) Possible up to a threshold load
d) None of the listed
.
Answer: a
Explanation: Worm gears support self-locking operation and hence are advantageous to use in lifting operations.
43. The power transmitting capacity of worm gears is high although efficiency is low.
a) True
b) False
.
Answer: b
Explanation: Both power transmitting capacity and efficiency of worm gears are low.
44. Can worm gears be used in steering mechanism?
a) True
b) False
.
Answer: a
Explanation: In steering mechanism, efficiency is of little importance but major requirement is of large mechanical advantage.
45. The worm helix angle is the _____ of the worm lead angle.
a) Complement
b) Half
c) Double
d) Supplement
.
Answer: a
Explanation: Worm helix angle+worm lead angle=90⁰.
46. If worm helix angle is 30⁰, then worm should have at least ___ threads.
a) 5
b) 6
c) 7
d) 8
.
Answer: a
Explanation: The permissible helix angle is 6⁰ and hence there should be at least five threads i.e. 30/6.
47. A pair of worm gear is written as 2/40/12/6. Calculate the centre distance.
a) 40mm
b) 156mm
c) 200mm
d) 80mm
.
Answer: b
Explanation: C=m(q+z)/2 where m=6mm, q=12 and z=40.
48. A pair of worm gear is written as 2/40/12/6. Calculate the speed reduction.
a) 2
b) 20
c) 15
d) 6
.
Answer: b
Explanation: i=40/2.
49. A pair of worm gear is written as 2/40/12/6. Calculate the pitch circle diameter of worm wheel.
a) 72mm
b) 240mm
c) 260mm
d) 320mm
.
Answer: b
Explanation: d=mxz where m=6mm and z=40.
50. A pair of worm gear is written as 2/40/12/6. Calculate the throat diameter of the worm wheel.
a) 220.5mm
b) 246.4mm
c) 190.44mm
d) 251.7mm
.
Answer: d
Explanation: d(t)=m[z+4cosϒ-2] where ϒ=9.46⁰ is the lead angle. tanϒ=2/12, z=40 and m=6mm.
51. A pair of worm gear is written as 2/40/12/6. Calculate the root diameter of the worm wheel.
a) 186.22mm
b) 250.4mm
c) 225.6mm
d) 250.44mm
.
Answer: c
Explanation: d=m[z-2-0.4cosϒ] where ϒ=9.46⁰ is the lead angle. tanϒ=2/12, z=40 and m=6mm.
52. If tangential force on worm is 1500N, then axial force on worm wheel will be?
a) 1500N
b) 3000N
c) 1500√2 N
d) 750N
.
Answer: a
Explanation: P₂(axial)=P₁(tangential).
53. Is use of a flywheel recommended when a large motor is required only for a small instant of time?
a) True
b) False
.
Answer: a
Explanation: Fly wheel allows the large motor to be replaced by a smaller motor as the peak power is required only for a small instance of time and not through the entire operation.
54. Flywheels are used in punching and shear operation.
a) True
b) False
Answer: a
Explanation: Flywheels store the kinetic energy imparted during idle positions and delivers this energy during actual shearing or punching.
55. Which of the following are functions of flywheel?
a) Store and release energy during work cycle
b) Reduce power capacity of the electric motor
c) Reduce amplitude of speed fluctuations
d) All of the listed
.
Answer: d
Explanation: Flywheel provides uniform motion by storing energy during idle positions and using this at actual operational time and thus reducing the power capacity of the electric motor.
56. When comes down to stress reduction, which one is preferred?
a) Solid flywheel
b) Split flywheel
c) Both have equal stresses
d) Cannot be determined
.
Answer: b
Explanation: The arms are free in split flywheel to contract and hence are better for stress reduction.
57. Flywheel and governor can be interchanged.
a) True
b) False
.
Answer: b
Explanation: Flywheel influences cyclic speed fluctuations while governor control mean speed.
58. If load on the engine is constant, the mean speed will be constant and ___ will not operate.
a) Flywheel
b) Governor
c) Both flywheel and governor
d) None of the mentioned
.
Answer: b
Explanation: Governor is used to control the mean speed and if mean speed is held constant then the governor will not operate.
59. The operation of flywheel is continuous.
a) True
b) False
.
Answer: a
Explanation: Flywheel is used to control the fluctuations and thus is continuously working.
60. Which of the following doesn’t waste energy?
a) Flywheel
b) Governor
c) Both flywheel and governor
d) Neither flywheel nor governor
.
Answer: a
Explanation: Flywheel has kinetic energy which is 100% convertible while governor suffer from friction losses.
61. Which of the following is not true for cast iron flywheels?
a) Excellent damping
b) Cheap
c) Given complex shape
d) Sudden failure
.
Answer: d
Explanation: Due to low tensile strength failure is sudden.
62. When the driving torque is more than load torque, flywheel is ______
a) Accelerated
b) Decelerated
c) Constant velocity
d) Can’t be determined
.
Answer: a
Explanation: I[dω/dt]=Driving torque-load torque.
63. Feather key can be used to prevent axial motion between two elements.
a) True
b) False
.
Answer: b
Explanation: All types of key but feather key can be used to prevent axial motion between two machine elements.
64. Keyed joints never lead to stress concentration on shafts.
a) True
b) False
.
Answer: b
Explanation: Keyway results in stress concentration in the shaft and makes the part weak.
Cylinders and Pressure Vessels and Design of IC Engine Components
1. A cylinder is considered thin when the ratio of inner diameter to wall thickness is more than 5.
A. True
B. False
.
Answer: B
Explanation: A cylinder is considered thin when the ratio of inner diameter to wall thickness is more than 15.
2. Tangential stress in a cylinder is given by [symbols have their usual meanings].
A. PD/2t
B. 2PD/t
C. PD/4t
D. 4PD/t
.
Answer: A
Explanation: Considering equilibrium in half portion of cylinder of unit length, DP=2σt.
3. Longitudinal stress in a cylinder is given by [symbols have their usual meanings].
A. PD/2t
B. 2PD/t
C. PD/4t
D. 4PD/t
.
Answer: C
Explanation: Considering equilibrium PxπD²/4=σxπDt.
4. A seamless cylinder of storage capacity of 0.03mᵌis subjected to an internal pressure of 21MPa. The ultimate strength of material of cylinder is 350N/mm².Determine the length of the cylinder if it is twice the diameter of the cylinder.
A. 540mm
B. 270mm
C. 400mm
D. 350mm
.
Answer: A
Explanation: 0.03=πd²L/4 and L=2d.
5. A seamless cylinder of storage capacity of 0.03mᵌis subjected to an internal pressure of 21MPa. The ultimate strength of material of cylinder is 350N/mm².Determine the thickness of the cylinder if it is twice the diameter of the cylinder.
A. 12mm
B. 4mm
C. 8mm
D. 16mm
.
Answer: C
Explanation: t=PD/2σ.
6. Cylinder having inner diameter to wall thickness ratio less than 15 are
A. Thin cylinders
B. Thick Cylinders
C. Moderate cylinders
D. None of the listed
.
Answer: B
Explanation: Smaller dia to thickness ratio implies more thickness and hence these are classified under thick cylinder.
7. Lame’s equation used to find the thickness of the cylinder is based on maximum strain failure.
A. True
B. False
.
Answer: B
Explanation: It is based on maximum principal stress theory.
8. Lame’s equation is generally used for ductile materials.
A. True
B. False
.
Answer: B
Explanation: Lame’s equation is used to determine thickness of the brittle as it used principal stress theory.
9. The piston rod of a hydraulic cylinder exerts an operating force of 10kN. The allowable stress in the cylinder is 45N/mm². Calculate the thickness of the cylinder using Lame’s equation. Diameter of the cylinder is 40mm and pressure in cylinder is 10MPa.
A. 2.05mm
B. 4.2mm
C. 5.07mm
D. None of the listed
.
Answer: C
Explanation: t=D/2[√[σ+ P /σ-P] -1 ].
10. The piston rod of a hydraulic cylinder exerts an operating force of 10kN. The allowable stress in the cylinder is 70N/mm². Calculate the thickness of the cylinder using Clavarinoe’s equation. Diameter of the cylinder is 240mm.μ=0.3 and pressure in cylinder is 15MPa.
A. 35mm
B. 30mm
C. 27mm
D. None of the listed
.
Answer: C
Explanation: t=D/2[√[σ+(1-2μ) P /σ-(1+μ)P] -1 ].
11. Autofrettage is beneficial for the high pressure cylinder.
A. True
B. False
.
Answer: A
Explanation: It increases pressure capacity of the cylinder and reduces compressive stresses.
12. Autogreggate is a process of ___ stressing the cylinder.
A. Pre
B. Post
C. Over
D. None of the listed
.
Answer: A
Explanation: It is a pre stressing phenomenon to improve pressure capacity.
13. Can we pre-stress the cylinder by subjecting cylindrical portion near inner diameter in plastic range and outer diameter is still in the elastic range.
A. True
B. False
.
Answer: A
Explanation: On releasing the pressure, outer portion contracts exerting pressure on the inner portion which has undergone permanent deformation. This induces residual compressive stresses at the inner surface.
14. A compound cylinder consists of
A. 2 cylinders
B. Cylinder and a jacket
C. 2 jackets
D. At least two cylinders
.
Answer: B
Explanation: Inner diameter of jacket increase and outer diameter of cylinder decreases when the jacket is heated.
15. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the shrinkage pressure.
A. 5.88N/mm²
B. 2.28N/mm²
C. 4.56N/mm²
D. 3.66N/mm²
.
Answer: B
Explanation: σ=P[D₃²+D₂²]/[D₃²-D₂²] where D₂=40mm and D₃=60mm.
16. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the interference [E=210kN/mm²].
A. 2.8mm
B. 4.6mm
C. 5.4mm
D. 4.8mm
.
Answer: A
Explanation: Δ=PD₂[2D₂²(D₃²-D₁²)] / Ex[(D₃²-D₂²)(D₂²-D₁²)].
17. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the radial stresses due to shrink shift in jacket.
A. +2.56[(45/r) ² – 1]
B. 1.824[(45/r) ² – 1]
C. -1.824[(45/r) ² – 1]
D. None of the listed
.
Answer: C
Explanation: σ(r)=-PD₂²[D₃²/4r² – 1]/ [D₃²-D₂²].
18. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the tangential stresses due to shrink shift in jacket.
A. +2.56[(45/r) ² – 1]
B. 1.824[(45/r) ² – 1]
C. -1.824[(45/r) ² – 1]
D. None of the listed
.
Answer: B
Explanation: σ(t)=+PD₂²[D₃²/4r² – 1]/ [D₃²-D₂²].
19. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the radial stress due to shrink shift in inner tube.
A. +3.04[1-(15/r) ²]
B. -3.04[1-(15/r) ²]
C. -3.04[1-(10/r) ²]
D. +3.04[1-(10/r) ²]
.
Answer: B
Explanation: σ(r)= σ(r)=-PD₂²[1-D₁²/4r² ]/ [D₂²-D₁²].
20. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². Calculate the tangential stress due to shrink shift in inner tube.
A. +3.04[1+ (15/r) ²]
B. -3.04[1+ (15/r) ²]
C. -3.04[1-(10/r) ²]
D. +3.04[1-(10/r) ²]
.
Answer: B
Explanation: σ(r)= σ(r)=-PD₂²[1+D₁²/4r² ]/ [D₂²-D₁²].
21. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². In service the cylinder is further subjected to an internal pressure of 25MPa. Calculate the radial stress in compound cylinder.
A. -3.75[(45/r) ² – 1]
B. +3.75[(45/r) ² – 1]
C. -3.75[(45/r) ² -+1]
D. +3.75[(45/r) ² +1]
.
Answer: A
Explanation: σ(r)=-PD₁²[D₃²/4r² – 1]/ [D₃²-D₁²]. Here P=30.
22. A high pressure cylinder consists of a steel tube with inner and outer diameters 30 and 60mm respectively. It is jacketed by an outer steel tube, having an outer diameter of 90mm. Maximum principal stress induced is 80N/mm². In service the cylinder is further subjected to an internal pressure of 25MPa. Calculate the tangential stress in compound cylinder.
A. -6.75[(45/r) ² + 1]
B. +3.75[(45/r) ² +1]
C. -3.75[(45/r) ² +1]
D. -3.75[(45/r) ² -1]
.
Answer: B
Explanation: σ(t)=+PD₁²[1+ D₃²/4r² ]/ [D₃²-D₁²]. Here P=30.
23. A gasket is a device to allow easy diffusion of fluids across mating surfaces of a mechanical assembly.
A. True
B. False
.
Answer: B
Explanation: A gasket acts as a barrier between two mating surfaces to prevent fluid flow.
24. Asbestos gaskets like other non-metallic gaskets can be used only up to a temperature of 70⁰C.
A. True
B. False
.
Answer: B
Explanation: Asbestos can be used upto a temperature of 250⁰C while all other non-metallic gaskets can be used only upto a temperature of 70⁰C.
25. Compression Ratio in diesel engine is lesser than that in Spark Ignition engine.
A. True
B. No
C. They are equal
D. Doesn’t matter
.
Answer: B
Explanation: In diesel engine, there is no spark to ignite, charge is compressed to greater extent to support self-ignition.
26. Petrol engine are more economical than diesel engine.
A. True
B. False
.
Answer: B
Explanation: Diesel engine have better thermal efficiency and hence are more economical.
27. There are ____ types of liners.
A. 1
B. 2
C. 3
D. 4
.
Answer: B
Explanation: Dry liners and wet liners.
28. If no re boring allowance is to be given, then thickness of cylinder wall is max pressure is 3.5N/mm², bore diameter=200mm and permissible tensile stress is 40N/mm².
A. 7mm
B. 8mm
C. 9mm
D. 10mm
.
Answer: C
Explanation: t=pD/2σ=8.75mm.
29. If diameter of cylinder of bore is 120mm, then thickness of the cylinder will be
A. Information not sufficient
B. 7mm
C. 12mm
D. 6mm
.
Answer: B
Explanation: t=0.045D + 1.6.
30. If diameter of cylinder bore is 120mm, then thickness of dry liner will be
A. 2.2mm
B. 3.6mm
C. 4.8mm
D. 6mm
.
Answer: B
Explanation: t= 0.03D to 0.035D.
31. If diameter of cylinder bore is 120mm, then thickness of water jacket wall will be
A. 4.26mm
B. 5.25mm
C. 2.56mm
D. All of the listed
.
Answer: D
Explanation: p=t/3 to 3t/4, where t= thickness of cylinder wall=7mm.
32. If diameter of cylinder bore is 120mm, then thickness of water cylinder flange will be
A. 8.8mm
B. 10.2mm
C. 7.8mm
D. 12mm
.
Answer: A
Explanation: q=1.2t to 1.4t.
33. If nominal diameter of bolt used is 20mm, then find the radial distance between outer diameter of flange and pitch circle diameter of studs.
A. 24mm
B. 28mm
C. 32mm
D. 36mm
.
Answer: B
Explanation: r=d+6 to 1.5d i.e. 26mm to 30mm.
34. The circumferential hoop stress and longitudinal stress are both shear stress.
A. True
B. False
.
Answer: B
Explanation: They both are tensile stress.
35. The circumferential hoop stress and longitudinal stress act in same direction and hence are straight away added.
A. True
B. False
.
Answer: B
Explanation: They both act perpendicularly to each other and thus net stress in these directions is reduced.
Miscellaneous Machine Elements
1. An unfired pressure vessel is used to carry stem, gases or fluids at pressure
A. Less than atmospheric pressure
B. Greater than atmospheric pressure
C. Equal to atmospheric pressure
D. None of the mentioned
.
Answer: B:
Explanation: Terminology.
2. Class 1 pressure vessels are used to contain
A. Lethal substances
B. Light duties applications
C. None of the listed
D. LPG
.
Answer: A:
Explanation: Class 1 pressure vessels are used exclusively for lethal substances. LPG doesn’t fall in the category of lethal substances.
3. An oil seal is a mechanical device made of elastic material which is used to prevent leakage of fluid between two machine components.
A. True
B. False
.
Answer: B:
Explanation: It is made of elastomer material.
4. Does oil seal consist of any spring in its structure?
A. Yes
B. No
.
Answer: A:
Explanation: It consist of garter spring which exerts a radial pressure on the rotating shaft and prevents leakage.
5. Which of the following isn’t true about oil seals?
A. Cheap
B. Can be used over a wide range of lubricating oils
C. Can’t tolerate misalignment
D. All of the mentioned
.
Answer: C:
Explanation: Oil seals can tolerate the misalignment to some extent.
6. If magnitude of contact pressure between sealing lip and rotating shaft is very high then
A. There is excessive friction
B. High temperature
C. Rapid wear
D. All of the listed
.
Answer: D:
Explanation: With high pressure, friction increases leading to high temperatures and increasing wear.
7. Wire ropes have
A. High strength to weight ratio
B. Solent operation even at high velocities
C. Greater reliability
D. All of the mentioned
.
Answer: D:
Explanation: All of the listed are true for wire ropes.
8. Wire rope specified as 8×10. Calculate the number of strands.
A. 8
B. 10
C. 16
D. 20
.
Answer: A:
Explanation: First number indicates number of strands.
9. Wire rope is specified as 8×10. Calculate the number of steel wires in each strand.
A. 10
B. 20
C. 5
D. 8
.
Answer: A:
Explanation: Second number represents the number of steel wires in each strand.
10. If the wires in the strand are twisted in the same direction as the strands, then rope is called a _____ lay rope.
A. Lang
B. Regular Lay
C. Ordinary
D. None of the listed
.
Answer: A:
Explanation: The lay of the rope tells the manner in which the wires are helically laid into strands and strands into rope. Same direction is indicated by Lang’s lay.
11. Which among regular lay and Lang lay are popular?
A. Regular Lay
B. Lang Lay
C. Both are equally used
D. Can’t be stated
.
Answer: A:
Explanation: In regular lay ropes, there is a balance resulting from opposite direction of twisting the strands to that of the wires.
12. In Lang lay ropes, both inner and outer wires are bent on same arc of radius.
A. True
B. False
.
Answer: B:
Explanation: Due to same direction of rotation, outer wires are bent on an arc of larger radius.
13. The wire rope is subjected to
A. Tensile Stress
B. Bending Stress
C. Both bending and tensile stress
D. None of the listed
.
Answer: C:
Explanation: There is direct tensile stress due to load being raised as well as bending stress.
14. In design of wire ropes, bending stress is converted into an equivalent bending load which is given by [d= diameter of individual wire and D= diameter of sheave].
A. AEd/D
B. 2AEd/D
C. AEd/2D
D. None of the listed
.
Answer: A:
Explanation: P=σA and σ=Ed/D.
15. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500. Number of wire ropes can be taken 2. Calculate the weight of material raised by each load.
A. None of the listed
B. 2450N
C. 4900N
D. 9800N
.
Answer: B:
Explanation: Weight of material raised by each rope= 4.9/2.
16. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Calculate the weight of wire if weight of 50m wire is 18kg.
A. 19.36N
B. 90.25N
C. 77.70N
D. 56.66N
.
Answer: C:
Explanation: W=mass per unit length x length x g.
17. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Number of rope wires is assumed to be 2. Calculate the weight of wire if weight of 50m wire is 18kg. Calculate the force due to acceleration.
A. 1000N
B. 309.144N
C. 504.225N
D. 102.225N
.
Answer: B:
Explanation: Mass of material raised by each wire rope= 2450/9.81 = 249.7kg. Mass of each wire rope= 77.7/9.81 =7.92kg. Force due to acceleration= [249.7+7.92] x 1.2.
18. An elevator is assembled to raise equipment to a height of 22m. It is estimated that maximum weight of the material to be raised is 4.9kN. It is observed acceleration in such cases is 1.2m/s².10mm diameter wire ropes with fibre core are used. The tensile designation of the wire is 1500.Number of rope wires is assumed to be 2. Number of rope wires is assumed to be 2. Calculate the total load on wire rope neglecting the bending load if weight of 50m wire is 18kg.
A. 1054.55N
B. 3504.55N
C. 2836.84N
D. 5678.6N
.
Answer: C:
Explanation: F=4900/2 + 77.7 + 249.7×1.2 + 7.92×1.2.
19. Small sheaves are preferred over large sheaves on what parameters?
A. Increasingly centrifugal force in large sheaves
B. Higher cost of small sheaves
C. Preferring centrifugal force reduction even for more money
D. None of the listed
.
Answer: A:
Explanation: Larger sheaves have higher centrifugal forces and higher cost because more material is used in their construction.
20. Rope operating over steel sheaves wear slowly than those used in conjunction with cast iron sheaves.
A. True
B. False
.
Answer: B:
Explanation: Rope operating over steel sheaves wear faster. If wear on cast iron sheave is 100, than wear on steel sheave is 110.
21. Which type of rope drum is preferred?
A. Drums with helical grooves
B. Plain cylindrical drums
C. Both are equally preferred
D. None of the listed
.
Answer: A:
Explanation: They have more bearing surface of the drum and prevent friction between adjacent turns of the rope.
22. If nominal diameter of the rope is 20mm, then pitch of the groove is around?
A. 2mm
B. 22mm
C. 40mm
D. 10mm
.
Answer: B:
Explanation: p= d + 2 mm.
23. Piston rod is an example of column.
A. True
B. False
.
Answer: A:
Explanation: An slender machine component having its length considerable in proportion to its width qualifies for a column criterion.
24. Bucking of column means
A. Lateral deflection
B. Axial deflection
C. Torsional deflection
D. None of the listed
.
Answer: D:
Explanation: Buckling is characterised by lateral deflection but it is different from lateral deflection as there is sudden lateral deflection in buckling unlike lateral deflection where there is gradual deflection.
25. Slenderness ratio is [l= length of column and k= least radius of gyration of cross section about its axis].
A. l/k
B. k/l
C. l/2k
D. k/2l
.
Answer: A:
Explanation: It is a ratio of length to least radius of gyration.
26. Columns with what slenderness ratio are not designed with respect to buckling but are designed for compressive stresses.
A. >1
B. <1 C. >30
D. <30
.
Answer: D:
Explanation: When slenderness ratio is <30, there is no effect of buckling.
27. If slenderness ratio=45, which mode of failure will dominate?
A. Buckling
B. Compressive Stresses
C. Both buckling and compressive stress
D. Can’t be stated
.
Answer: A:
Explanation: If slenderness ratio>30, column shall be more prone to buckling.
28. Short column and long column are classified on the basis of
A. Slenderness ratio
B. Diameter
C. Length
D. None of the listed
.
Answer: A:
Explanation: Slenderness ratio takes into consideration length and radius of gyration and thus is preferred.
29. Cast iron column with a slenderness ratio of 75 are
A. Short Columns
B. Long Columns
C. Very short columns
D. None of the listed
.
Answer: A:
Explanation: Cast iron column with a slenderness ratio <80are short columns.
30. Steel columns with a slenderness ratio of 95 are
A. Short Columns
B. Long columns
C. Very long columns
D. None of the listed
.
Answer: A:
Explanation: Steel columns with slenderness ratio <100 are classified in short columns.
31. Which of the following are true for End fixity coefficient
A. Dimensionless number
B. Used in Euler’s equation
C. Provides condition of restraint at two ends
D. All of the listed
.
Answer: D:
Explanation: It is a dimensionless number used in Euler’s equation to take into account the restraints at the two ends.
32. Value of end fixity coefficient for both hands fixed is
A. 1
B. 4
C. 2
D. 0.25
.
Answer: B:
Explanation: Lower the mobility higher is the end fixity component.
