Engineering Questions with Answers - Multiple Choice Questions

Manufacturing Engineering MCQs – Tool Wear Failure

1 - Question

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

View Answer

Answer: b
Explanation: Crater wear occurs on the rake face of the tool, while flank wear occurs on the relief (flank) face of the tool.




2 - Question

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

View Answer

Answer: b
Explanation: Flank wear occurs as a result of friction between the progressively increasing contact area on the tool flank.




3 - Question

Crater wear is predominant in
a) carbon steels
b) tungsten carbide tools
c) high speed steel tools
d) ceramic tools

View Answer

Answer: b
Explanation: Crater wear is usually found while machining brittle materials and tungsten carbide tools favour this phenomenon.




4 - Question

Flank wear is due to the abrasive action of hard mis-constituents.
a) True
b) False

View Answer

Answer: a
Explanation: 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 - Question

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

View Answer

Answer: c
Explanation: Flank wear is due to the abrasive action and crater wear is due to diffusion of metals.




6 - Question

Crater wear leads to
a) increase in cutting temperature
b) weakening of tool
c) friction and cutting forces
d) all of the mentioned

View Answer

Answer: d
Explanation: None




7 - Question

Crater wear is usually found while machining ductile materials.
a) True
b) False

View Answer

Answer: b
Explanation: Crater wear is usually found while machining brittle materials.




8 - Question

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

View Answer

Answer: d
Explanation: None




9 - Question

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

View Answer

Answer: a
Explanation: Crater wear occurs on the rake face of the tool, while flank wear occurs on the relief (flank) face of the tool.




10 - Question

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

View Answer

Answer: d
Explanation: None




11 - Question

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

View Answer

Answer: d
Explanation: None




12 - Question

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

View Answer

Answer: a
Explanation: None




13 - Question

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

View Answer

Answer: b
Explanation: None




14 - Question

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%

View Answer

Answer: c
Explanation: Since n = 0.5, the Taylor equation can be rewritten as VT0.5 = 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.

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