Category: Manufacturing Engineering Objective Questions

Manufacturing Processes Multiple Choice Questions on “Electron Beam Machining – 4”.

1. In electron beam machining, the plane of _____ is on the surface of the workpiece.
a) focusing
b) finishing
c) heating
d) drilling
Answer: a
Clarification: The plane of focusing would be on the surface of the workpiece or just below the surface of the workpiece. This controls the kerf shape or the shape of the hole.

2. _____ can manoeuvre the electron beam.
a) Nozzles
b) Magnetic lens
c) Electromagnetic coils
d) Deflector coils
Answer: d
Clarification: In electron beam gun assembly, the final deflection coil can manoeuvre the electron beam which helps in generating holes of non-circular cross-section as required.

3. Electron beam machining process can machine holes of diameters in the range of _____
a) 10 μm to 80 μm
b) 50 μm to 100 μm
c) 100 μm to 2 mm
d) 2 mm to 5 mm
Answer: c
Clarification: Electron beam machining can provide holes of diameter in the range of 100 μm to 2 mm with a depth up to 15 mm, i.e., with a length/diameter ratio of around 10.

4. Which of the following is true about electron beam machining (EBM)?
a) By EBM process, tapered holes can be generated
b) Electro-magnetic coils are used to change the direction of the electron beam
c) Electron beam gun works under high pressure
d) Increasing the current density increases the spot size
Answer: a
Clarification: Electron beam machining can generate a hole which can be tapered along the depth or barrel shaped. By focusing the beam below the surface a reverse taper can also be obtained.

5. Which of the following holds true for electron beam machining?
a) This process does not generate burr
b) Holes having length/diameter ratio as high as 50 can be machined by this process
c) In electron beam gun, magnetic lens is used to diverge the beam
d) Electron beam is accelerated by electromagnetic coils
Answer: a
Clarification: Generally burr formation does not occur in electron beam machining. There would be an edge rounding at the entry point along with a presence of recast layer. Also, magnetic lens is used to focus the beam at a single point. Annular anodes are used to accelerate the beam.

6. Which of the following materials is not machined by the EBM process?
a) Titanium
b) Wood
c) Plastic
d) Leather
Answer: b
Clarification: Materials which can be machined by EBM are:
• Commercial grade steel
• Stainless steel
• Ti and Ni super alloys
• Aluminium plastics
• Ceramics
• Leathers.

7. For EBM process, heat affected zone is about _____
a) 10 μm to 80 μm
b) 20 μm to 30 μm
c) 100 μm to 1 mm
d) 2 mm to 5 mm
Answer: b
Clarification: In electron beam machining, the heat-affected zone is rather narrow due to the shorter pulse duration. Typically, the heat affected zone is around 20 μm to 30 μm.

8. Which of the following materials are easy to a machine by EBM process?
a) Aluminium
b) Steel
c) Plastic
d) Wood
Answer: a
Clarification: EBM can machine steel titanium alloys, aluminium, plastics, nickel alloys, etc., but some of the materials like Al and Ti alloys are more readily machined compared to steel.

9. Number of holes drilled per second depends on the holes diameter.
a) True
b) False
Answer: a
Clarification: Number of holes drilled per second depends on:
• hole diameter
• power density
• depth of the hole
• material type.

10. While machining, there are chances of thermal damage associated with EBM.
a) True
b) False
Answer: a
Clarification: As the mechanism of material removal is thermal in nature as for example in electro-discharge machining, there would be thermal damages associated with EBM. On the contrary, heat affected zone is narrow.

Advanced Manufacturing Processes Questions and Answers on “Laser Welding – 5”.

1. In keyhole mode of laser welding, hole is stabilized by the _____
a) weld puddle
b) shielding gases
c) pressure of the vapour
d) laser beam
Answer: c
Clarification: In “keyhole” welding in which there is sufficient energy/unit length to cause evaporation and hence a hole in the melt pool. This hole is stabilized by the pressure from the vapour being generated.

2. In keyhole mode of laser welding, the keyhole behaves like _____
a) optical black body
b) an energy reflector
c) an energy amplifier
d) a shielding
Answer: a
Clarification: The “keyhole” behaves like an optical black body in that the radiation enters the hole and is subject to multiple reflections before being able to escape. Nearly all the beam energy is absorbed once the keyhole is formed.

3. Which of the following has the highest joining efficiency?
a) Acetylene flame welding
b) TIG welding
c) Electron beam welding
d) Laser welding
Answer: c
Clarification: The joining efficiencies for different welding processes is listed below;
Oxy-acetylene flame welding — 0.2-0.5
Manual arc welding – 2-3
TIG welding – 0.8-2
Electron beam welding – 20-30
Laser beam welding – 15-25.

4. Flow structures can directly affect the ____
a) laser wave formation
b) frozen bead
c) weld puddle
d) porosity
Answer: b
Clarification: The two principle areas of interest in laser beam welding are;
– Flow structures which directly affects the wave formation in the weld pool and frozen bead geometry
– Absorption mechanism
• Fresnel absorption (absorption during reflection from surface)
• Inverse Bremmstrahlung, leading to plasma re-radiation.

5. During laser welding, any hump on the surface can cause _____
a) increase in weld porosity
b) flow instability
c) higher energy absorption
d) decrease in the weld thickness
Answer: c
Clarification: Any hump on the surface will cause localized higher absorption and an explosion due to instantaneous evaporation. The keyhole walls are fluctuating with flow velocities up to 0.4 m/s.

6. The hot plasma vapour emerging from the keyhole may ionize _____
a) shielding gas
b) shroud gas
c) lasing material
d) workpiece material
Answer: b
Clarification: The keyhole contains considerable metal vapour, which is partially absorbing and hence capable of forming a plasma on further heating. This hot plasma vapour emerging from the keyhole may ionize the shroud gas. Ionized gas has free electrons and is thus capable of absorbing or even blocking the beam.

7. How many of the following process parameters can affect the welding process?
• Power pulses
• Wavelength
• Gap
• Joint geometries
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: Following are the parameters that can affect the welding process:
• Beam Properties:
– Power, pulsed or continuous
– Spot size and mode
– Polarisation – Wavelength
• Transport Properties:
– Speed
– Focal position
– Joint geometries
– Gap tolerance.

8. Which of the following does not affect the operating characteristics in laser welding?
a) shroud gas composition
b) material surface conditions
c) component orientation
d) shroud design
Answer: c
Clarification: Following are the parameters that can affect the welding process:
• Shroud/shielding Gas Properties:
– Composition
– Shroud design
– Pressure/velocity
• Material Properties:
– Composition
– Surface condition.

9. The maximum welding speed varies directly with the power
a) True
b) False
Answer: a
Clarification: The maximum welding speed for a given thickness rises with an increase in power. The fall off shown at the higher power levels of 2kW could be attributed to the poorer mode structure given by most lasers when working at their peak power.

10. Pulse repletion factor is considered while determining the welding speed.
a) True
b) False
Answer: a
Clarification: The use of pulsed power allows two more variables:
– pulse repetition frequency (PRF),
– % overlap to be considered.
The welding speed is decided by the following relation:
– spot size x PRF x (1% overlap).

Manufacturing Processes Multiple Choice Questions on “Hardening – 1”.

1. How many types of hardening techniques are there?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Different techniques to improve the hardness of the steels are as follows:
• conventional hardening
• martempering
• austempering.

2. In a conventional hardening of hypereutectoid steels, the component is heated above _____
a) 228⁰C
b) 363⁰C
c) 585⁰C
d) 738⁰C
Answer: d
Clarification: The first step in hardening involves heating the steel to above 910⁰C for hypoeutectoid steels and above 738⁰C for hypereutectoid steels by approximately 50⁰C.

3. After hardening, the component is cooled ______
a) at the critical cooling rate
b) at the rate just more than the critical cooling rate
c) in still air
d) at rate at which it was heated
Answer: b
Clarification: After austenization, the next step in hardening involves cooling of hot steel components at a rate just exceeding the critical cooling rate of the steel to room temperature or below room temperature.

4. In conventional hardening _______ transforms to _______
a) austenite, martensite
b) cementite, martensite
c) perlite, α-iron
d) austenite, ferrite
Answer: a
Clarification: In this conventional hardening process, the austenite transforms to martensite. The final step in the hardening process is the tempering of the martensite to achieve the desired hardness. This martensite structure improves the hardness of the component.

5. Martempering uses _______ quenching.
a) water
b) oil
c) brine
d) interrupted
Answer: d
Clarification: Martempering process overcomes the limitation of the conventional hardening process. This process follows interrupted quenching operation. In other words, the cooling is stopped at a point above the martensite transformation region to allow sufficient time for the centre to cool to the temperature as the surface. Further cooling is continued through the martensite region, followed by the usual tempering.

6. Austempering is used to overcome the limitations of ______
a) martempering
b) conventional hardening
c) quenching
d) annealing
Answer: b
Clarification: This process is also used to overcome the limitation of the conventional hardening process. Here the quench is interrupted at a higher temperature than for martempering to allow the metal at the centre of the part to reach the same temperature as the surface.

7. During austempering, steel is transformed to ______
a) bainite
b) bauxite
c) α-ferrite
d) α-martensite
Answer: a
Clarification: In austempering, both the centre and surface are allowed to transform to bainite and are then cooled to room temperature. Austempering causes less distortion and cracking than that in the case of martempering.

8. Austempering improves ______
a) corrosion resistance
b) toughness
c) hardness
d) tensile strength
Answer: b
Clarification: Austempering improves the impact toughness and the ductility of the metal than that in the case of martempering and conventional hardening. It also avoids the tempering operation.

9. Alloy steels can be hardened by simple air cooling.
a) True
b) False
Answer: a
Clarification: Alloy steels have less critical cooling rate and hence some of the alloy steels can be hardened by simple air cooling. Proper quenching medium should be used such that the component gets cooled at a rate just exceeding the critical cooling rate of that steel.

10. High carbon steels are hardened by oil quenching.
a) True
b) False
Answer: a
Clarification: High carbon steels have slightly more critical cooling rate and has to be hardened by oil quenching. Medium carbon steels have still higher critical cooling rates and hence water or brine quenching is necessary.

Manufacturing Engineering Multiple Choice Questions on “Heat Treatment”.

1. Which of the following is the hardest constituent of steel?
a) Ledeburite
b) Austenite
c) Bainite
d) Martensite
Answer: d
Clarification: Martensite is the hardest constituent of steel. The primary reasons accounting for this could be, the internal strains within BCC iron due to the excess carbon presence and due to the plastic deformation of parent FCC iron (austenite) surrounding the martensitic plate. Rate of cooling and the amount of carbon percentage in steel are directly proportional to the amount of hardness achieved in martensitic transformation.

2. Iron possesses BCC crystal structure up to (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
Answer: b
Clarification: Pure iron possess either BCC or FCC crystal structure as its temperature is increased from room temperature to its melting point. At room temperature to 910^oC, it is having BCC, between 910^oC and 1410^oC it is having face centered cubic, and from 1410^oC to its melting point (1539^oC) it returns to its BCC crystal structure.

3. Iron possesses BCC crystal structure above (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
Answer: d
Clarification: From 1410^oC to its melting point (1539^oC) iron is having BCC crystal structure.

4. Iron possesses FCC crystal structure above (in degree centigrade)?
a) 1539
b) 768
c) 910
d) 1410
Answer: c
Clarification: Between 910^oC and 1410^oC iron is having face centered cubic crystal structure.

5. Which of the following form of iron is magnetic in nature?
a) α
b) δ
c) γ
d) λ
Answer: a
Clarification: The alpha form of iron is magnetic and stable at all temperatures below 910^oC.

6. For steel, which one of the following properties can be enhanced upon annealing?
a) Hardness
b) Toughness
c) Ductility
d) Resilience
Answer: c
Clarification: A furnace cooling technique, annealing will enhance the ductility of steel, due to the formation of coarse pearlite.

7. In Annealing, cooling is done in which of the following medium?
a) Air
b) Water
c) Oil
d) Furnace
Answer: d
Clarification: In annealing, after solutionising, material is used to furnace cool, means furnace is switched off and the steel sample inside is let cool down.

8. In normalizing, cooling is done in which of the following medium?
a) Air
b) Water
c) Oil
d) Furnace
Answer: a
Clarification: In normalizing, steel is heat treated above its critical temperature, solutionised, and then allowed to cool for a long time by keeping it in air. In steel, it forms fine pearlite, which imparts strength to steel.

9. Mild steel can be converted into high carbons steel by which of the following heat treatment process?
a) Annealing
b) Normalizing
c) Case hardening
d) Nitriding
Answer: c
Clarification: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

10. Upon annealing, eutectoid steel converts to which of the following?
a) Perlite
b) Cementite
c) Austenite
d) Martensite
Answer: a
Clarification: Eutectoid steels upon annealing produces pearlite (coarse pearlite). Pearlite is an alternate lamellae of ferrite and cementite.

Manufacturing Engineering Multiple Choice Questions on “Tool Life: Wear and Failure”.

1. Crater wear occurs mainly on the
a) nose part, front relief face and side relief face of the cutting tool
b) face of the cutting tool at a short distance from the cutting edge only
c) cutting edge only
d) front face only
Answer: b
Clarification: Crater wear occurs on the rake face of the tool, while flank wear occurs on the relief (flank) face of the tool.

2. Flank wear depends upon the
a) hardness of the work and tool material at the operating temperature
b) amount and distribution of hard constituents in the work material
c) degree of strain hardening in the chip
d) none of the mentioned
Answer: b
Clarification: Flank wear occurs as a result of friction between the progressively increasing contact area on the tool flank.

3. Crater wear is predominant in
a) carbon steels
b) tungsten carbide tools
c) high speed steel tools
d) ceramic tools
Answer: b
Clarification: Crater wear is usually found while machining brittle materials and tungsten carbide tools favour this phenomenon.

4. Flank wear is due to the abrasive action of hard mis-constituents.
a) True
b) False
Answer: a
Clarification: Flank wear is due to the abrasive action of hard mis-constituents including debris from built up edge as the work material rubs the work surface.

5. Crater wear is mainly due to the phenomenon is known as
a) adhesion of metals
b) oxidation of metals
c) diffusion of metals
d) none of the mentioned
Answer: c
Clarification: Flank wear is due to the abrasive action and crater wear is due to diffusion of metals.

6. Crater wear leads to
a) increase in cutting temperature
b) weakening of tool
c) friction and cutting forces
d) all of the mentioned
Answer: d
Clarification: None

7. Crater wear is usually found while machining ductile materials.
a) True
b) False
Answer: b
Clarification: Crater wear is usually found while machining brittle materials.

8. The tool may fail due to
a) cracking at the cutting edge due to thermal stresses
b) chipping of the cutting edge
c) plastic deformation of the cutting edge
d) all of the mentioned
Answer: d
Clarification: None

9. Flank wear occurs mainly on the
a) nose part, front relief face and side relief face of the cutting tool
b) face of the cutting tool at a short distance from the cutting edge only
c) cutting edge only
d) front face only
Answer: a
Clarification: Crater wear occurs on the rake face of the tool, while flank wear occurs on the relief (flank) face of the tool.

10. Tool life is measured by the
a) number of pieces machined between tool sharpenings
b) time the tool is in contact with the job
c) volume of material removed between tool sharpenings
d) all of the mentioned
Answer: d
Clarification: None

11. The tool life is said to be over if
a) poor surface finish is obtained
b) there is sudden increase in cutting forces and power consumption
c) overheating and fuming due to heat of friction starts
d) all of the mentioned
Answer: d
Clarification: None

12. Tool life is generally better when
a) grain size of the metal is large
b) grain size of the metal is small
c) hard constituents are present in the micro structure of the tool material
d) none of the mentioned
Answer: a
Clarification: None

13. The relation between the tool life(T) in minutes and cutting speed (V) in m/min is
a) VⁿT = C
b) VTⁿ = C
c) Vⁿ/T = C
d) V/Tⁿ = C
Answer: b
Clarification: None

14. Using the Taylor Equation for tool life and letting n = 0.5 and C = 120, calculate the percentage increase in tool life when the cutting speed is reduced by 50%.
a) 100%
b) 200%
c) 300%
d) 400%
Answer: c
Clarification: Since n = 0.5, the Taylor equation can be rewritten as VT^0.5 = 120.

Let’s denote V₁ as the initial speed and V₂ the reduced speed; thus, V₂ = 0.5 V₁. Because C is the constant 120, we have the relationship
0.5V₁ sqrt T₂ = V₁sqrt T₁

Simplifying this equation, T₂/T₁ = 1/0.25 = 4. This
indicates that the change in tool life is
(T₂ – T₁/ T₁) = (T₂/T₁) – 1 = 4 – 1 = 3,

or that tool life is increased by 300%. Thus, a reduction in cutting speed has resulted in a major increase in tool life. Note also that, for this problem, the magnitude of C is not relevant.