Production engg. question set with answers
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This document provides an overview of the contents of a book on metal cutting, metal forming, and metrology. It lists the sections and chapters of the book, along with the page numbers for each chapter. The sections include theory of metal cutting, metal forming, and metrology. The chapters within these sections cover topics such as basics of metal cutting, tool life/wear, cold working, rolling, forging, extrusion, sheet metal operations, limits and fits, and measurement of lines and surfaces. The document also provides sample questions from past IES, GATE, and other exams related to these topics.
![( )
GATE – 2008 (PI)
Brittle materials are machined with tools having
zero or negative rake angle because it
(a) results in lower cutting force
IES 2007
Cast iron with impurities of carbide requires a
particular rake angle for efficient cutting with single
point tools, what is the value of this rake angle, give
reasons for your answer
answer.
[ 2 marks]
(b)
( ) improves surface finish
(c) provides adequate strength to cutting tool
S 99
IAS – 1994
Consider the following characteristics
1. The cutting edge is normal to the cutting velocity.
2. Th tti f
The cutting forces occur in two directions only.
i t di ti
l
3. The cutting edge is wider than the depth of cut.
The characteristics applicable to orthogonal cutting
would include
(a) 1 and 2
(b) 1 and 3
(c) 2 and 3
(d) 1 2 and 3
1, 2 and 3
(d) results in more accurate dimensions
IES 2012
IES ‐
During orthogonal cutting, an increase in cutting speed
causes
(a) An increase in longitudinal cutting force
(b) An increase in radial cutting force
(c) An increase in tangential cutting force
( )
(d) Cutting forces to remain unaffected
g
IES‐1995
The
the face and th fl k of th
Th angle b t
l between th f
d the flank f the
single point cutting tool is known as
a) Rake angle
b) Clearance angle
g
c) Lip angle
d) Point angle
angle.
For-2014 (IES, GATE & PSUs)
IES‐2006
Which of the following is a single point cutting
tool?
(a) Hacksaw blade
(b) Milling cutter
(c) Grinding wheel
(d) P ti t l
Parting tool
IES‐2006
iron,
Assertion (A): For drilling cast iron the tool is
provided with a point angle smaller than that
required for a ductile material
material.
Reason (R): Smaller point angle results in lower
rake angle.
k
l
(a) Both A and R are individually true and R is the
correct explanation of A
(b) Both A and R are individually true but R is not the
y
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Page 2 of 78
IES 2012
IES ‐
()
g
g
p
g
Statement (I): Negative rake angles are preferred on rigid set‐
ups for interrupted cutting and difficult‐to machine
materials.
Statement (II):Negative rake angle directs the chips on to the
machined surface
(a) Both Statement (I) and S
( ) B h S
d Statement (II) are i di id ll
individually
true and Statement (II) is the correct explanation of
Statement (I)
(b) Both Statement (I) and Statement (II) are individually
( )
p
true but Statement (II) is not the correct explanation of
Statement (I)
(c) Statement (I) is true but Statement (II) is false
(d) Statement (I) is false but Statement (II) is true
IES‐2002
Consider the following statements:
The strength of a single point cutting tool depends
upon
1. Rake angle
2. Clearance angle
3. Lip angle
Which of these statements are correct?
(a)
( ) 1 and 3
d
(b) 2 and 3
d
(c) 1 and 2
(d) 1, 2 and 3](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-3-320.jpg)
![GATE – 1995 ‐Conventional
While turning a C steel rod of 160 mm di
C‐15 t l d f 6
diameter at
Whil t
i
t
t
315 rpm, 2.5 mm depth of cut and feed of 0.16
mm/rev b a t l of geometry 00, 100, 80, 90,150, 750,
/
by tool f
t
0(mm), the following observations were made.
Tangential component of the cutting force = 500 N
Axial component of the cutting force = 200 N
p
g
Chip thickness = 0.48 mm
Draw schematically the Merchant’s circle diagram
Merchant s
for the cutting force in the present case.
IAS‐2003 Main Examination
During turning process with 7 ‐ ‐ 6 – 6 – 8 – 30 – 1
D i t
i
ith
(mm) ASA tool the undeformed chip thickness of
2.0 mm and width of cut of 2.5 mm were used. Th
d idth f t f
d The
side rake angle of the tool was a chosen that the
machining operation could b approximated t b
hi i
ti
ld be
i t d to be
orthogonal cutting. The tangential cutting force and
thrust f
th
t force were 1177 N and 560 N respectively.
d 6
ti l
Calculate:
[30 marks]
(i) h d
( ) The side rake angle
k
l
(ii) Co‐efficient of friction at the rake face
(iii) The dynamic shear strength of the work material
GATE 2013
GATE‐2013
A steel bar 200 mm in diameter is turned at a feed of
0.25 mm/rev with a depth of cut of 4 mm. The
rotational speed of the workpiece is 160 rpm. The
material removal rate in mm3/s is
(a) 160 (b) 167 6 (c) 1600
167.6
(d) 1675 5
1675.5
GATE‐2007
In orthogonal turning of medium carbon steel. The
I th
l t
i f di
b t l Th
specific machining energy is 2.0 J/mm3. The cutting
velocity, feed and depth of cut are 120 m/min, 0.2
l it f d d d th f t / i
mm/rev and 2 mm respectively. The main cutting
force in N is
f
i N i
(a) 40
(b) 80
(c) 400
(d) 800
Example
When the rake angle is zero during orthogonal cutting,
show that
τs
pc
=
(1 − μ r ) r
1+ r2
Where τs is the shear strengrh of the material
p c = specific power of cutting
p
r = chip thickness ratio
μ = coefficient of friction in tool chip interface
For-2014 (IES, GATE & PSUs)
For PSU & IES
In strain gauge dynamometers the use of how
many active gauge makes the dynamometers more
effective
(a) Four
(b) Three
(c) T o
Two
(d) One
Ans. (a)
Page 8 of 78
IES 2004
IES ‐
A medium carbon steel workpiece is turned on a
lathe at 50 m/min. cutting speed 0.8 mm/rev feed
and 1.5 mm depth of cut. What is the rate of metal
removal?
(a) 1000 mm3/min
(b) 60,000 mm3/min
(c) 20,000 mm3/min
( )
(d) Can not be calculated with the given data
g
GATE 2013 (PI) C
D
Q
i
GATE‐2013 (PI) Common Data Question
A disc of 200 mm outer and 80 mm inner diameter is
faced of 0.1 mm/rev with a depth of cut of 1 mm. The
facing operation is undertaken at a constant cutting
speed of 90 m/min in a CNC lathe. The main
(tangential) cutting force is 200 N.
Neglecting the contribution of the feed force
towards cutting power the specific cutting energy
power,
in J/mm3 is
(a)
( ) 0.2
(b) 2
(c)
( ) 200
(d) 2000
GATE‐2006 Common Data Questions(1)
In an orthogonal machining operation:
I th
l
hi i
ti
Uncut thickness = 0.5 mm
Cutting speed = 20 m/min Rake angle = 15°
Width of cut 5 mm
Width of cut = 5 mm
Chip thickness 0.7 mm
Chip thickness = 0.7 mm
Thrust force = 200 N
Cutting force = 1200 N
Assume Merchant's theory.
A
M h t' th
The coefficient of friction at the tool‐chip interface is
(a) 0.23
( )
(b) 0.46
(b)
(c) 0.85
(d) 0.95](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-9-320.jpg)
![GATE 1992
The effect of rake angle on the mean f
friction angle in
h
ff
f k
l
h
l
machining can be explained by
(A) sliding (Coulomb) model of friction
(B) sticking and then sliding model of friction
(C) sticking friction
( )
(D) Sliding and then sticking model of friction
g
g
IES‐2004
Assertion (A): The ratio of uncut chip thickness to
actual chip thickness is always less than one and is
termed as cutting ratio in orthogonal cutting
g
g
g
Reason (R): The frictional force is very high due to the
occurrence of sticking friction rather than sliding
g
g
friction
( )
(a) Both A and R are individually true and R is the correct
y
explanation of A
( )
(b) Both A and R are individually true but R is not the
y
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
IAS – 2009 Main
Tool Wear, Tool Life & Machinability
Tool Wear, Tool Life & Machinability
Explain ‘sudden‐death mechanism’ of tool failure.
[ 4 – marks]
[
k ]
Show crater wear and flank wear on a single point
Sh t
d fl k
i l i t
cutting tool. State the factors responsible for wear
on a turning tool.
t
i t l
[ 2 –marks]
For-2014 (IES, GATE & PSUs)
The ff t f k
the
friction
Th effect of rake angle on th mean f i ti angle i
l
l in
machining can be explained by
(a)
( ) Sliding (
(coulomb) model of friction
)
(b) sticking and then siding model of friction
g
g
(c) Sticking friction
(d) sliding and then sticking model of friction
GATE 2008 (PI)
GATE‐2008 (PI)
During machining, the wear land (h) has been plotted
against machining time (T) as given i the f ll i
in h following
i
hi i
i
i
figure.
For a critical wear land of 1.8 mm, the cutting tool life (in
minute) is
(a) 52.00
(b) 51.67
(c) 51.50
(d) 50.00
By S K Mondal
B S K M d l
IES 2009 Conventional
GATE‐1993
IES 2010
IES 2010
Flank wear occurs on the
(a) Relief face of the tool
(b) Rake face
(c) Nose of the tool
(d) Cutting edge
Page 13 of 78
S 2007
IES – 200
Flank wear occurs mainly on which of the
following?
(a) Nose part and top face
(b) Cutting edge only
(c) Nose part, front relief face, and side relief face of the
cutting tool
(d) Face of the cutting tool at a short distance from
g g
the cutting edge](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-14-320.jpg)
![S
IAS – 2002
Consider the following actions:
1. Mechanical abrasion 2.
Diffusion
3. Pl ti d f
Plastic deformation
ti
4.
Oxidation
O id ti
Which of the above are the causes of tool wear?
(a) 2 and 3
(b) 1 and 2
(c) 1, 2 and 4 (d) 1 and 3
IES 2012
IES ‐
In Taylor s tool life equation VTn = C, the constants n
In Taylor’s tool life equation VT C, the constants n
and C depend upon
1. Work piece material
1 Work piece material
2. Tool material
3. Coolant
( )
(a) 1, 2, and 3
3
(b) 1 and 2 only
(c) 2 and 3 only
(d) 1 and 3 only
S 999
IAS – 1999
S
IAS – 2003
The type of wear that occurs due to the cutting
action of the particles in the cutting fluid is
referred to as
(a) Attritions wear
(b) Diff i wear
Diffusion
(c) Erosive wear
(d) Corrosive wear
Consider the following statements:
Chipping of a cutting tool is due to
1. T l material b i t b ittl
Tool
t i l being too brittle
2. Hot hardness of the tool material.
3. High positive rake angle of the tool.
Which of these statements are correct?
(a) 1, 2 and 3 (b) 1 and 3
(c)
( ) 2 and 3
d
(d) 1 and 2
d
S 20 0 C
i
l
IES 2010 Conventional
IFS 2009
With the help of Taylor’s tool life equation,
determine the shape of the curve between velocity
Draw tool life curves for cast alloy, High speed steel and
ceramic tools.
[2 – Marks]
Ans.
of cutting and life of the tool. Assume an HSS tool
and steel as work material
material.
[
[10‐Marks]
]
1. High speed steel
IES‐1996
Chip equivalent is increased by
(a) An increases in side‐cutting edge angle of tool
(b) An increase in nose radius and side cutting
edge angle of tool
(c) Increasing the plant area of cut
(d) Increasing the depth of cut.
For-2014 (IES, GATE & PSUs)
S 992
IES – 1992
Tool life is generally specified by
(a) Number of pieces machined
(b) V l
Volume of metal removed
f t l
d
(c) Actual cutting time
(d) Any of the above
Page 15 of 78
2. cast alloy and 3. ceramic tools.
G
200
GATE‐2004
operation,
In a machining operation doubling the
1
cutting speed reduces the tool life to 8 th of
the i i l l
th original value. Th exponent n i T l '
The
t in Taylor's
n = C, is
tool life equation VT
(a )
1
8
(b)
1
4
(c )
1
3
(d )
1
2](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-16-320.jpg)
![S 99
IAS – 1995
Assertion (A): An increase in depth of cut shortens
the tool life.
Reason(R): Increases in depth of cut gives rise to
relatively small increase in tool temperature.
(a) Both
( ) B th A and R are i di id ll t
d
individually true and R i th
d
is the
correct explanation of A
(b) Both A and R are individually true but R is not the
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
S 2009 C
i
l
IES 2009 Conventional
Determine the optimum cutting speed for an
operation on a Lathe machine using the following
information:
Tool change time: 3 min
Tool
T l regrinds ti
i d time: 3 min
i
Machine running cost Rs.0.50 per min
Depreciation of tool regrinds Rs. 5.0
The constants in the tool life equation are 60 and
0.2
GATE‐2009 Linked Answer Questions (1)
S 2006 conventional
i
l
IES – 2006
operation.
An HSS tool is used for turning operation The
tool life is 1 hr. when turning is carried at 30
m/min. Th t l lif will b reduced t 2.0 min if
/ i The tool life ill be d d to
i
the cutting speed is doubled. Find the suitable
speed in RPM for turning 300 mm diameter so
that tool life is 30 min.
3
S 200 C
i
l
ESE‐2001 Conventional
In a certain machining operation with a cutting
speed of 50 m/min, tool life of 45 minutes was
observed. Wh th cutting speed was i
b
d When the tti
d
increased
d
to 100 m/min, the tool life decreased to 10 min.
Estimate the cutting speed for maximum
p
productivity if tool change time is 2 minutes.
y
g
GATE‐2009 Linked Answer Questions (2)
In a machining experiment, tool life was found to vary
with the cutting speed in the following manner:
Cutting speed (m/min)
Tool life (minutes)
60
81
90
36
The exponent (n) and constant (k) of the Taylor's
p
( )
( )
y
tool life equation are
(a) n 0.5 and k 540
(a) n = 0.5 and k = 540
(b) n 1 and k 4860
(b) n= 1 and k=4860
(c) n = ‐1 and k = 0.74
(d) n‐0.5 and k=1.15
In a machining experiment, tool life was found to vary
with the cutting speed in the following manner:
Cutting speed (m/min)
Tool life (minutes)
60
81
90
36
What is the percentage increase in tool life when
p
g
the cutting speed is halved?
(a) 50%
(b) 200%
(c) 300%
(d) 400%
For-2014 (IES, GATE & PSUs)
Page 19 of 78
S 999 S 20 0 C
i
l
ESE‐1999; IAS ‐2010 Conventional
The following equation for tool life was obtained for HSS
tool. A 60 min tool life was obtained using the following
cutting condition VT0.13f0.6d0.3= C. v = 40 m/min, f = 0.25
mm, d = 2.0 mm. Calculate the effect on tool life if
speed, feed and depth of cut are together increased by
25% and also if they are increased individually by 25%;
where f = feed, d = depth of cut, v = speed.
IAS – 2011 Main
Determine the optimum speed for achieving
maximum production rate in a machining
operation. The data is as follows :
Machining time/job = 6 min
min.
Tool life = 90 min.
Taylor s
Ta lor's equation constants C = 100, n = 0
00
0.5
Job handling time = 4 min./job
Tool changing time = 9 min.
l h
i
i
i
[10‐Marks]
G
999
GATE‐1999
What is approximate percentage change is
the life, t, of a tool with zero rake angle used
in
i orthogonal cutting when it clearance
th
l
tti
h
its l
o to 7o?
angle, α, is changed from 10
(Hint: Flank wear rate is proportional to cot α
(a) 30 % increase (b) 30% decrease
30%,
(c) 70% increase (d) 70% decrease](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-20-320.jpg)
![S 2007
IAS – 200
Assertion (A): The optimum cutting speed for the
minimum cost of machining may not maximize the
profit.
Reason (R): The profit also depends on rate of
production.
production
(a) Both A and R are individually true and R is the
correct explanation of A
t
l
ti
f
(b) Both A and R are individually true but R is not the
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
IES 2011 C
ti
l
IES 2011 Conventional
g
Discuss the effects of the following elements on the
machinability of steels:
(i) Aluminium and silicon
(ii) Sulphur and Selenium
(iii) L d and Ti
Lead d Tin
(iv) Carbon and Manganese
(v) Molybdenum and Vanadium
[5 Marks]
ISRO‐2007
Machinablity depends on
(a)
( ) Microstructure, physical and mechanical
h
l
d
h
l
properties and composition of workpiece material.
(b) Cutting forces
( ) yp
(c) Type of chip
p
(d) Tool life
For-2014 (IES, GATE & PSUs)
S 99
IAS – 1997
In turning, the ratio of the optimum cutting speed
for minimum cost and optimum cutting speed for
maximum rate of production is always
(a) Equal to 1
(b) I th
In the range of 0.6 to 1
f 6 t
(c) In the range of 0.1 to 0.6
(d) Greater than 1
S 992
IES – 1992
Ease of machining is primarily judged by
(a) Life of cutting tool between sharpening
(b) Ri idit f
Rigidity of work ‐piece
k i
(c) Microstructure of tool material
(d) Shape and dimensions of work
S
IES – 2003
( )
y
p
Assertion (A): The machinability of steels improves
by adding sulphur to obtain so called 'Free
Machining Steels‘.
Reason (R): Sulphur in steel forms manganese
sulphide inclusion which helps to produce thin
ribbon like continuous chip.
(a) Both A and R are individually true and R is the
correct explanation of A
(b) Both A and R are individually true but R is not the
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Page 21 of 78
IES 2012
IES ‐
The usual method of defining machinability of a
material is by an index based on
(a) Hardness of work material
(b) Production rate of machined parts
(c) Surface finish of machined surfaces
( )
(d) Tool life
S 2007, 2009
IES – 200 2009
Consider the following:
1. Tool life
2. C tti f
Cutting forces
3. Surface finish
Which of the above is/are the machinability
criterion/criteria?
(a) 1, 2 and 3
(b) 1 and 3 only
(c) 2 and 3 only
(d) 2 only
S
IES – 2009
The elements which, added to steel, help in chip
formation during machining are
(a) Sulphur lead and phosphorous
Sulphur,
(b) Sulphur, lead and cobalt
(c) Aluminium, lead and copper
( )
(d) Aluminium, titanium and copper
pp](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-22-320.jpg)
![IAS 1996
IAS ‐
IES 1999
IES ‐
Given that
S = feed in mm/rev. and
R = nose radius i mm,
di in
the maximum height of surface roughness Hmax
produced by a single‐point turning tool is given by
( )
(a) S2/2R
(b) S2/4R
(c) S2/4R
(d) S2/8R
In turning operation, the feed could be doubled to
increase the metal removal rate. To keep the same
level of surface finish, the nose radius of the tool
should be
(a) Halved
(b) Kept unchanged
(c) doubled
(d) Made four times
GATE 2007 (PI)
GATE – 2007 (PI)
GATE 2005
GATE ‐
30o
A tool with Side Cutting Edge angle of
and
o is used for fine
End Cutting Edge angle of 10
turning with a feed of 1 mm/rev Neglecting nose
mm/rev.
radius of the tool, the maximum (peak to valley)
height f
h i h of surface roughness produced will b
f
h
d d ill be
( )
(a) 0.16 mm
( )
(b) 0.26 mm
(c) 0.32 mm
(d) 0.48 mm
IES 2006
IES ‐
In the selection of optimal cutting conditions, the
requirement of surface finish would put a limit on
which of the following?
(a) The maximum feed
(b) Th maximum d th of cut
The
i
depth f t
(c) The maximum speed
(d) The maximum number of passes
Two tools P and Q have signatures 5 5 6 6 8 30
5°‐5°‐6°‐6°‐8°‐30°‐
0 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively.
They are used to turn components under the same
machining conditions. If hp and hQ denote the peak‐
to valley
to‐valley heights of surfaces produced by the tools P
and Q, the ratio hp/hQ will be
tan 8o + cot15o
tan 8o + cot 30o
tan15o + cot7o
(c )
tan 30o + cot7o
(a)
A cutting tool has a radius of 1.8 mm. The feed rate
for a theoretical surface roughness of Ra = 5 m is
μ
(a) 0 36 mm/rev
0.36 mm/rev
(b) 0.187 mm/rev
(c) 0.036 mm/rev
( )
(d) 0.0187 mm/rev
7
IES 1993 ISRO 2008
IES – 1993, ISRO‐2008
For achieving a specific surface finish in single point
turning the most important factor to be controlled
is
(a) Depth of cut
(b) Cutting speed
(c) Feed
( ) F d
(d) T l rake angle
Tool k
l
tan15o + cot 8o
tan 30o + cot 8o
tan7o + cot15o
(d )
tan7o + cot 30o
(b)
GATE 2010 (PI)
GATE ‐2010 (PI)
During turning of a low carbon steel bar with TiN coated
carbide insert, one need to improve surface finish
without sacrificing material removal rate. To achieve
h
f
l
l
h
improved surface finish, one should
(a) decrease nose radius of the cutting tool and increase
depth of cut
(b) Increase nose radius of the cutting tool
(c) Increase feed and decrease nose radius of the cutting
tool
For-2014 (IES, GATE & PSUs)
GATE 1997
GATE ‐
(d) Increase depth of cut and increase feed
Page 23 of 78
IAS 2009 Main
IAS ‐2009 Main
What are extreme‐pressure lubricants?
[ 3 – marks]
g
pressures and rubbing action are
g
Where high p
encountered, hydrodynamic lubrication cannot be
maintained; so Extreme Pressure (EP) additives must be
added to the l b i
dd d
h lubricant. EP l b i i i provided b a
lubrication is
id d by
number of chemical components such as boron,
phosphorus, sulfur, chlorine,
phosphorus sulfur chlorine or combination of these
these.
The compounds are activated by the higher temperature
resulting from extreme pressure As the temperature
pressure.
rises, EP molecules become reactive and release
derivatives such as iron chloride or iron sulfide and
forms a solid protective coating.](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-24-320.jpg)
![GATE 2013
GATE‐2013
S 2008
IES‐2008
Which one of the following is correct?
Malleability is the property by which a metal or
alloy can be plastically deformed by applying
(a) Tensile stress
(b) Bending stress
(c) Shear stress
(d) Compressive stress
ISRO‐2006
In
the t t
I a rolling process, th state of stress of th
lli
f t
f the
material undergoing deformation is
g
g
Which of the following processes would produce
(a) pure compression
strongest components?
(b) pure shear
(a) Hot rolling
(c) compression and shear
(b)
( ) Extrusion
(d) tension and shear
(c) Cold rolling
(d) Forging
ISRO‐2009
Ring rolling is used
(a) To decrease the thickness and increase
diameter
(b) To increase the thickness of a ring
(c) For producing a seamless tube
(d) For producing large cylinder
IFS – 2010
Calculate the neutral plane to roll 250 mm wide
annealed copper strip from 2 5 mm to 2 0 mm
2.5
2.0
thickness with 350 mm diameter steel rolls. Take µ =
0.05
0 05 and σ’o =180 MPa
σ
MPa.
[10‐marks]
For-2014 (IES, GATE & PSUs)
S
IES – 2006
Which one of the following is a continuous bending
process in which opposing rolls are used to produce
long sections of formed shapes from coil or strip
stock?
(a) Stretch forming
(b) Roll forming
(c) Roll bending
(d) Spinning
( )
GATE – 2009 (PI)
Anisotropy in rolled components is caused by
(a) changes in dimensions
(b) scale formation
(c) closure of defects
(d) grain orientation
i
i
i
G
2008
GATE‐2008
In a single pass rolling operation, a 20 mm thick
plate with plate width of 100 mm, is reduced to 18
mm. The roller radius is 250 mm and rotational
speed is 10 rpm. The average flow stress for the plate
material is 300 MPa. The power required for the
rolling operation in kW is closest to
(a) 15 2
15.2
(b) 18.2
(c) 30.4
(d) 45.6
45
Page 27 of 78
G
200
GATE‐2007
The thickness of a metallic sheet is reduced from an
initial value of 16 mm to a final value of 10 mm in
one single pass rolling with a pair of cylindrical
rollers each of diameter of 400 mm. The bite angle
in degree will be
(a) 5.936
(b) 7.936
6
(c) 8.936
(d) 9.936](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-28-320.jpg)
![IES‐2013
G
20 0 ( )
GATE ‐2010 (PI)
( )
GATE‐1989(PI)
Hot die t l
forging,
H t di steel, used f l
d for large solid di i d
lid dies in drop f i
At the last hammer stroke the excess material from
should necessarily have
y
the finishing cavity of a forging die is pushed
(a) high strength and high copper content
into……………..
(b) high hardness and low hardenability
(c) high toughness and low thermal conductivity
(d) high hardness and high thermal conductivity
IFS‐2011
Statement (I): The dies used in the forging process are
p
made in pair.
Statement (II): The material is pressed between two
surfaces and the compression force applied, gives it a
shape.
(a) Both Statement (I) and Statement (II) are individually
true and Statement (II) is the correct explanation of
Statement (I)
(b) Both Statement (I) and Statement (II) are individually
true b t St t
t
but Statement (II) i not th correct explanation of
t
is t the
t
l
ti
f
Statement (I)
(c)
( ) Statement ( ) is true b Statement ( ) is f l
(I)
but
(II) false
(d) Statement (I) is false but Statement (II) is true
IAS‐2011 Main
IES ‐ 2007
What advantages does press forging have over drop
Compare Smith forging, drop forging, press
Sometimes the parting plane between two forging
forging ? Why are pure metals more easily cold worked
forging and upset forging. Mention three points
dies is not a horizontal plane, give the main reason
than ll
th alloys ?
for each.
for this design aspect, why is parting plane
[5 marks]
[5‐marks]
[10 – Marks]
[ M k ]
provided,
provided in closed die forging?
[
[ 2 marks]
]
G
200
GATE‐2007
open die
In open‐die forging, a disc of diameter 200 mm and
height 60 mm is compressed without any barreling
effect. The final diameter of the disc is 400 mm. The
true strain is
(a) 1 986
1.986
(b) 1 686
1.686
(c) 1.386
(d) 0.602
For-2014 (IES, GATE & PSUs)
GATE‐1992, ISRO‐2012
G
20 2
GATE‐2012 Same Q GATE ‐2012 (PI)
The true strain for a low carbon steel bar which is
doubled in length by forging is
(a) 0.307
(b) 0.5
(c) 0.693
(d) 1.0
A solid cylinder of diameter 100 mm and height 50 mm
A lid li d f di
t
d h i ht
Page 31 of 78
is forged between two frictionless flat dies to a height of
g
g
25 mm. The percentage change in diameter is
(a) 0
(b) 2.07
(c) 20.7
(d) 41.4](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-32-320.jpg)
![IES 2007 C
ti
l
IES – 2007 Conventional
S 2005 Conventional
i
l
IES – 200 C
A strip of l d with i iti l di
t i f lead ith initial dimensions 24 mm x 24
i
mm x 150 mm is forged between two flat dies to a
5
g
final size of 6 mm x 96 mm x 150 mm. If the
coefficient of friction is 0.25, determine the
maximum f i f
i
forging force. Th average yield stress of
The
i ld t
f
lead in tension is 7 N/mm2
A cylinder of height 60 mm and diameter 100 mm is
forged at room temperature between two flat dies. Find
the die load at the end of compression to a height 30
mm, using slab method of analysis. The yield strength of
the work material is given as 120 N/mm2 and
the
coefficient of friction is 0.05. Assume that volume is
constant after deformation. There is no sticking. Also
find mean die pressure.
[20‐Marks]
IES 2006 C
ti
l
IES – 2006 ‐ Conventional
A certain disc of lead of radius 150 mm and thickness 50
mm is reduced to a thickness of 25 mm by open die
forging. If the co‐efficient of friction between the job and
co efficient
die is 0.25, determine the maximum forging force. The
average shear yield stress of lead can be taken as 4
N/mm2.
[10 – Marks]
[10]
GATE‐1987
P ti P bl
1
Practice Problem‐1
Ch‐15: Forging
Q. No
Option
A
6
B
2
3
A
A
7
8
C
C
A
9
C
5
occurs near the……(Centre/ends)
Option
1
4
In forging operation the sticking friction condition
Q. No
B
A strip of metal with initial dimensions 24 mm x 24 mm
x 150 mm is forged between two flat dies to a final size of
6 mm x 96 mm x 150 mm. If the coefficient of friction is
0.05, determine the maximum forging force. Take the
average yield strength in tension is 7 N/mm2
[Ans. 178.24 kN]
P ti P bl
2
Practice Problem‐2
P ti P bl
3
Practice Problem‐3
P ti P bl
4
Practice Problem‐4
A circular disc of 200 mm in diameter and 100 mm in
A cylindrical specimen 150 mm in diameter and 100 mm
A circular disc of 200 mm in diameter and 70 mm in
height is compressed between two flat dies to a height of
in height is upsetted by open die forging to a height of 50
height is forged to 40 mm in height. Coefficient of
50 mm. Coefficient of friction is 0.1 and average yield
mm. Coefficient of friction is 0.2 and flow curve
friction is 0.05. The flow curve equation of the material
strength in compression is 230 MPa. Determine the
equation is
σ f = 1030ε
0.17
MPa . Calculate the maximum
is given by
σ f = 200(0.01 + ε ) 0.41 MP
MPa
. Determine maximum
[Ans. 405 MPa]
For-2014 (IES, GATE & PSUs)
forging force.
forging load, mean die pressure and maximum pressure.
[Ans. 46.26 MN]
maximum die pressure
pressure.
[ Ans. 9.771 MN, 178 MPa, 221 MPa]
[Hint. First calculate true strain ε and put the value in
the equation
σ f = 1030ε
0.17
=σy
]
Page 36 of 78
[Hint. First calculate true strain ε and put the value in
the equation
σ f = 200(0.01 + ε ) 0.41 = σ y
]](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-37-320.jpg)
![Practice Problem ‐5 {GATE‐2010 (PI)}
Practice Problem 5 {GATE 2010 (PI)}
During open die forging process using two flat and parallel dies,
a solid circular steel disc of initial radius (R IN ) 200 mm and initial
lid i l t l di
f i iti l di
d i iti l
height (H IN ) 50 mm attains a height (H FN ) of 30 mm and radius of R FN .
Practice Problem ‐5 {GATE‐2010 (PI)}
Extrusion & Drawing
iii.In the region 0 ≤ r ≤ R SS ,sticking condition prevails
The value of R SS (in mm), where sticking condition changes to sliding
Along the die-disc interfaces.
⎛
⎞
i. the coefficient of friction (μ ) is: μ = 0.35 ⎜ 1 + e
⎟
⎜
⎟
⎝
⎠
ii. in h
ii i the region R ss ≤ r ≤ RFN ,sliding friction prevails, and
i
lidi f i i
il
d
R
− IN
RFN
2μ
Contd…….
friction, is
(a) 241.76
(b) 254.55
(c) 265.45
(d) 278.20
( RFN − r )
p = 3K H FN
Ke
and τ = μ p,
d
where p and τ are the normal and shear stresses, respectively;
K is the shear yield strength of steel and r is the radial distance
of any point
(contd ........)
IAS‐2010 Main
How
are
By S K Mondal
B S K M d l
metal
tooth‐paste
IES 2009 Conventional
tubes
made
( )
GATE‐1994(PI)
A moving mandrel is used in
commercially ? Draw the tools configuration with
(a) Wire drawing
[30‐Marks]
IES – 2011 Conventional
12.5
A 12 5 mm diameter rod is to be reduced to 10 mm
diameter by drawing in a single pass at a speed of 100
/
g
g
m/min. Assuming a semi die angle of 5o and coefficient
of friction between the die and steel rod as 0.15,
calculate:
(i) The power required in drawing
( )
(ii) Maximum possible reduction in diameter of the rod
p
(iii) If the rod is subjected to a back pressure of 50
/
N/mm2 , what would be the draw stress and maximum
possible reduction ?
4
/
Take stress of the work material as 400 N/mm2 .
[15 Marks]
For-2014 (IES, GATE & PSUs)
GATE – 2011 (PI) Common Data‐S1
GATE 2011 (PI) Common Data S1
In a multi‐pass drawing operation, a round bar of 10 mm
diameter and 100 mm length is reduced in cross‐section
by drawing it successively through a series of seven dies
of decreasing exit diameter. During each of these
drawing operations, the reduction in cross‐sectional area
is 35%. The yield strength of the material is 200 MPa.
Ignore strain hardening.
The total true strain applied and the final length (in
), p
y,
mm), respectively, are
(a) 2.45 and 8 17
(b) 2.45 and 345
(c) 3 02 and 2043
3.02
(d) 3 02 and 3330
3.02
Page 37 of 78
(b) Tube drawing
(c) Metal Cutting
the help of a neat sketch.
(d) Forging
GATE – 2011 (PI) Common Data‐S2
GATE 2011 (PI) Common Data S2
In a multi‐pass drawing operation, a round bar of 10 mm
diameter and 100 mm length is reduced in cross‐section
by drawing it successively through a series of seven dies
of decreasing exit diameter. During each of these
drawing operations, the reduction in cross‐sectional area
is 35%. The yield strength of the material is 200 MPa.
Ignore strain hardening.
Neglecting friction and redundant work, the force (in
) q
g
g
,
kN) required for drawing the bar through the first die, is
(a) 15.71
(b) 10.21
(c) 6.77
(c) 6 77
(d) 4.39
(d) 4 39](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-38-320.jpg)
![IAS‐2011 Main
For punching a 10 mm circular hole, and cutting a
rectangular bl k of 50 x 200 mm f
t
l blank f
from a sheet of 1
h t f
mm thickness (mild steel, shear stress = 240
N/mm2) C l l t i each case :
N/
), Calculate, in
h
(i) Size of punch
(ii) Size of die
( )
(iii) Force required.
q
[
[10‐Marks]
]
S 999
IES – 1999
S O 2008 20
ISRO‐2008, 2011
A hole is to be punched in a 15 mm thick plate
having ultimate shear strength of 3N‐mm‐2. If the
allowable crushing stress in the punch is 6 N‐mm‐2,
N mm
the diameter of the smallest hole which can be
punched is equal to
(a) 15 mm
(b) 30 mm
(c) 60
( ) 6 mm
(d) 120 mm
With a punch for which the maximum crushing
stress is 4 times the maximum shearing stress of
the l t the biggest h l th t can b punched i
th plate, th bi
t hole that
be
h d in
the plate would be of diameter equal to
1
× Thickness of p
plate
4
1
(b) × Thickness of plate
2
(c) Plate thickness
( )
(a)
(d) 2 × Plate thickness
IES 2013
IES‐2013
A hole of diameter d is to be punched in a plate of
thickness t. For the plate material, the maximum
crushing stress is 4 times the maximum allowable
shearing stress. For punching the biggest hole, the ratio
of diameter of hole to plate thickness should be equal to:
1
1
2
( )
(a) 4
( )
(b)
(c) 1
E
l
Example
Example
A hole, 100 mm diameter, is to be punched in steel plate
5.6 mm thi k Th ultimate shear stress i 550 N/
6
thick. The lti t h
t
is
N/mm2 .
With normal clearance on the tools, cutting is complete
,
g
p
at 40 per cent penetration of the punch. Give suitable
shear angle for the punch to bring the work within the
(d) 2
capacity of a 30T press.
it f
T
G
20 0 S
i k d
GATE‐2010 Statement Linked 1
Statement for Linked Answer Questions:
In a shear cutting operation, a sheet of 5 mm thickness
is cut along a length of 200 mm. The cutting blade is 400
mm long and zero‐shear (S = 0) is provided on the edge.
The ultimate shear strength of the sheet is 100 MPa and
g
penetration to thickness ratio is 0.2. Neglect friction.
G
20 0 S
i k d2
GATE‐2010 Statement Linked 2
Q
Statement for Linked Answer Questions:
In a shear cutting operation, a sheet of 5mm thickness
is cut along a length of 200 mm. The cutting blade is 400
mm long and zero shear (S = 0) is provided on the edge
zero‐shear
edge.
The ultimate shear strength of the sheet is 100 MPa and
penetration to thickness ratio is 0.2. Neglect friction.
400
400
S
Assuming force vs displacement curve to be rectangular,
the work done (in J) is
(a) 100 (b) 200 (c)
250 (d) 300
For-2014 (IES, GATE & PSUs)
S
A shear of 20 mm (S = 20 mm) is now provided on the
blade. Assuming force vs displacement curve to be
trapezoidal, the maximum force (in kN) exerted is
(a) 5
(b) 10 Page 44 of 78
(c)
20
(d) 40
A washer with a 12.7 mm internal hole and an outside
diameter of 25.4 mm is to be made from 1.5 mm thick
strip. The ultimate shearing strength of the material of
p
g
g
the washer is 280 N/mm2.
( )
(a) Find the total cutting force if both punches act at
g
p
the same time and no shear is applied to either punch
or the die.
(b) What will be the cutting force if the punches are
staggered, so that only one punch acts at a time.
(c) Taking 60% penetration and shear on punch of 1
mm, what will be the cutting force if both punches act
g
p
together.
GATE 2011
MPa.
The shear strength of a sheet metal is 300 MPa The
blanking force required to produce a blank of 100
mm diameter from a 1 5 mm thick sheet is close to
1.5
(a) 45 kN
(b) 70 kN
(c) 141 kN
4
(d) 3500 kN](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-45-320.jpg)
![S
IAS – 2003
S 99
IES – 1997
spring back
The 'spring back' effect in press working is
(a) Elastic recovery of the sheet metal after removal of
the load
(b) Regaining the original shape of the sheet metal
(c) Release of stored energy in the sheet metal
( )
(d) Partial recovery of the sheet metal
y
A cup of 10 cm height and 5 cm diameter is to be
made from a sheet metal of 2 mm thickness. The
number of deductions necessary will be
(a) One
(b) T
Two
(c) Three
(d) Four
G
2008
GATE‐2008
IFS ‐ 2009
What is deep drawing process for sheet metal
forming? Explain the function of a blank holder.
What is drawing ratio and how is the drawing ratio
increased ?
[
[10 – marks]
]
G
2003
GATE‐2003
A shell of 100 mm diameter and 100 mm height with
the corner radius of 0.4 mm is to be produced by
cup drawing. The required blank diameter is
(a) 118 mm
(b) 161 mm
(c)
( ) 224 mm
(d) 312 mm
In the deep drawing of cups, blanks show a tendency to
wrinkle up around the periphery (flange).
The most likely cause and remedy of the phenomenon are,
respectively,
(A) Buckling due to circumferential compression; Increase
blank holder pressure
(B) High blank holder pressure and high friction; Reduce
blank holder pressure and apply lubricant
(C) High temperature causing increase in circumferential
length: Apply coolant to blank
(D) Buckling due to circumferential compression; decrease
blank holder pressure
ISRO‐2011
The initial blank diameter required to form
a cylindrical cup of outside di
li d i l
f
id diameter 'd‘ and
d
total height 'h' having a corner radius 'r' is
g
g
obtained using the formula
(a ) Do = d 2 + 4dh − 0 5r
0.5
(b) Do = d + 2h + 2r
(c) Do = d 2 + 2h 2 + 2r
For-2014 (IES, GATE & PSUs)
(d ) Do = d 2 +Page− 0.5r78
4dh 48 of
E
l
Example
A symmetrical cup of 80 mm diameter and 250 mm
height is to be fabricated on a deep drawing die.
How many drawing operations will be necessary if
no intervening annealing is done.
Also find the drawing force
G
999
GATE‐1999
Identify the stress ‐ state in the FLANCE portion of a
PARTIALLYDRAWN CYLINDRICAL CUP when deep ‐
drawing without a blank holder
(a) Tensile in all three directions
(b) N stress i th fl
No t
in the flange at all, b
t ll because th
there i no
is
blank‐holder
(c) Tensile stress in one direction and compressive in
the one other direction
(d) Compressive in two directions and tensile in the
third direction
G
2006
GATE‐2006
Match the items in columns I and II.
Column I
Column II
P.
P Wrinkling
1.
1
Yield point elongation
Q. Orange peel
2.
Anisotropy
R. Stretcher strains 3.
R S
h
i
Large grain size
L
i i
S. Earing
4.
Insufficient blank holding
force
5.
Fine grain size
6.
Excessive blank holding force
(a) P – 6, Q – 3, R – 1, S – 2 (b) P – 4, Q – 5, R – 6, S – 1
3
5
(c) P – 2, Q – 5, R – 3, S – 4 (d) P – 4, Q – 3, R – 1, S – 2](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-49-320.jpg)
![IES 2011
IFS‐2011
Compare metal spinning with press work.
[2‐marks]
[
k ]
IES 2010
IES 2010
( )
g
gy
g
Assertion (A) : In the high energy rate forming
method, the explosive forming has proved to be an
g
g
g
excellent method of utilizing energy at high rate and
utilizes both the high explosives and low explosives.
Reason (R): The gas pressure and rate of detonation
( )
g p
can be controlled for both types of explosives.
(a) Both A and R are individually true and R is the correct
explanation of A
(b) Both A and R are individually true but R is NOT the
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
S 2005
IES – 200
Magnetic forming is an example of:
(a) Cold forming
(b) Hot forming
(c) High
( ) Hi h energy rate f
t forming
i
(d) R ll f
Roll forming
i
For-2014 (IES, GATE & PSUs)
High energy rate forming process used for forming
components from thin metal sheets or deform thin
tubes is:
(a) Petro‐forming
(b) Magnetic pulse forming
(c) Explosive forming
p
g
(d) electro‐hydraulic forming
S 2007
IES – 200
Which one of the following metal forming
processes is not a high energy rate forming process?
(a) Electro mechanical forming
Electro‐mechanical
(b) Roll‐forming
(c) Explosive forming
( )
(d) Electro‐hydraulic forming
y
g
IFS‐2011
Write four advantages of high velocity forming process.
[2‐marks]
[
k ]
Page 50 of 78
20 0
JWM 2010
( )
g
p
g
,
Assertion (A) : In magnetic pulse‐forming method,
magnetic field produced by eddy currents is used to
p
create force between coil and workpiece.
Reason (R) : It is necessary for the workpiece
material to have magnetic properties.
(a) Both A and R are individually true and R is the
correct explanation of A
(b) Both A and R are individually true but R is NOT the
correct explanation of A
t
l
ti
f
(c) A is true but R is false
(d) A is false but R is true
S
IES – 2009
Which one of the following is a high energy rate
forming process?
(a) Roll forming
(b) Electro‐hydraulic forming
(c) Rotary forging
( )
(d) Forward extrusion
G
2000
GATE‐2000
A 1.5 mm thick sheet is subject to unequal biaxial
stretching and the true strains in the directions of
stretching are 0.05 and 0.09. The final thickness of
the sheet in mm is
(a) 1 414
1.414
(b) 1 304
1.304
(c) 1.362
(d) 289](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-51-320.jpg)
![Ch‐18: Sheet Metal Forming
Q. No
Option
Q. No
( )
GATE ‐2011 (PI)
Option
1
C
10
C
2
3
B
A
11
12
C
C
4
A
13
C
5
D
14
D
6
7
A
A
15
16
A
B
8
9
A
A
17
D
Powder Metallurgy
gy
Which of the following powder production
methods produces spongy and porous particles?
th d
d
d
ti l ?
(a) Atomization
(b) Reduction of metal oxides
( )
(c) Electrolytic deposition
y
p
(d) Pulverization
By S K Mondal
IES 2012
IES ‐
In electrolysis
(a) For making copper powder, copper plate is made
cathode in electrolyte tank
(b) For making aluminum powder, aluminum plate is
made anode
d
d
(c) High amperage produces powdery deposit of cathode
metal on anode
( )
(d) Atomization process is more suitable for low melting
p
g
point metals
IES – 2007 Conventional
Metal powders are compacted by many methods, but
In powder metallurgy, sintering of a component
sintering is required to achieve which property? What
(a) Improves strength and reduces hardness
is hot iso‐static pressing?
i h ti
t ti
i ?
(c) Improves hardness and reduces toughness
(d) R d
Reduces porosity and i
i
d increases b i l
brittleness
( )
GATE – 2009 (PI)
Which of the following process is used to
manufacture products with controlled porosity?
(a) Casting
(b)
( ) welding
(c) formation
(d) Powder metallurgy
For-2014 (IES, GATE & PSUs)
(b) Reduces brittleness and improves strength
[
[ 2 Marks]
]
IES – 2011 Conventional
What is isostatic pressing of metal powders ?
What are its advantage ?
h
d
[ 2 Marks]
( )
GATE ‐2010 (PI)
Page 52 of 78
( )
GATE – 2011 (PI)
The binding material used in cemented carbide
cutting t l i
tti tools is
(a) graphite
(b) tungsten
( )
(c) nickel
(d) cobalt](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-53-320.jpg)
![IES 2005
IES ‐
GATE 2000
GATE ‐
p
y
g
The tolerance specified by the designer for the
diameter of a shaft is 20.00 ± 0.025 mm. The shafts
produced by three different machines A, B and C
have mean diameters of 19∙99 mm, 20∙00 mm and
20.01 mm respectively, with same standard
deviation. Wh t will b th percentage rejection f
d i ti
What ill be the
t
j ti
for
the shafts produced by machines A, B and C?
(a) Same f th machines A B d C since th standard
( ) S
for the
hi
A, Band
i
the t d d
deviation is same for the three machines
(b) L t f machine A
Least for
hi
(c) Least for machine B
(d) Least for machine C
GATE 2007
GATE ‐
0 .0 5 0
A slot is to be milled centrally on a block with a
dimension of 40 ± 0.05 mm. A milling cutter of 20
mm width is located with reference to the side of
the block within ± 0.02 mm. The maximum offset in
mm between the centre lines of the slot and the
block is
(a) ± 0 070
0.070
(b) 0 070
0.070
(c) ± 0.020
(d) 0.045
IES 2013
IES‐2013
IES 2011
GATE 2005
GATE ‐
Which of the following is a joint formed by
Interference fit joints are provided for:
(a) Assembling bush bearing in housing
(b) Mounting heavy duty gears on shafts
( )
(c) Mounting pulley on shafts
gp
y
(d) Assembly of flywheels on shafts
A hole is specified as 4 0 0 . 0 0 0
mm. The mating
shaft has a clearance fit with minimum clearance of
0.01 mm. The tolerance on the shaft is 0.04 mm. The
maximum clearance in mm between the hole and
the shaft is
(a) 0.04
(b) 0.05
(c) 0.10
(d) 0.11
interference fits?
(a) Joint of cycle axle and its bearing
(b) Joint between I.C. Engine piston and cylinder
In order to have interference fit, it is essential that
the lower limit of the shaft should be
(a) Greater than the upper limit of the hole
(b) Lesser than the upper limit of the hole
(c) Greater than the lower limit of the hole
( )
(d) Lesser than the lower limit of the hole
(c) Joint between a pulley and shaft transmitting power
(d) J i of l h spindle and i b i
Joint f lathe i dl
d its bearing
GATE 2011
A hole is of dimension φ 9
+0 015
0.015
+0
GATE ‐2012 Same Q in GATE‐2012 (PI)
mm. The
In an interchangeable assembly, shafts of size
+0.010
corresponding shaft is of dimension
p
g
The resulting assembly has
(a) loose running fit
(b) close running fit
(c) transition fit
( )t
iti
(d) interference fit
φ 9 +0.001
For-2014 (IES, GATE & PSUs)
mm.
mm mate with holes of size
25+0.03
+0 02
0.02
25
+0.04
−0.01
mm.
The maximum interference (in microns) in the assembly
is
(a)
( ) 40
(b) 30
(c)
( ) 20
(d) 10
IAS‐2011 Main
An interference assembly, of nominal diameter 20
mm, is of a unilateral holes and a shafts. The
manufacturing t l
f t i
tolerances f th h l are t i
for the holes
twice
that for the shaft. Permitted interference values are
0.03 to 0.09 mm. Determine the sizes, with limits,
for the two mating parts.
[10‐Marks]
[10 Marks]
Page 57 of 78](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-58-320.jpg)
![IES 2012
IES ‐
Clearance in a fit is the difference between
(a) Maximum hole size and minimum shaft size
(b) Mi i
Minimum h l size and maximum shaft size
hole i
d
i
h ft i
(c) Maximum hole size and maximum shaft size
(d) Minimum hole size and minimum shaft size
ISRO‐2008
Basic shaft and basic hole are those whose upper
deviations and lower deviations respectively are
(a) +ve, ‐ve
(b) ‐ve, +ve
(c)
( ) Zero, Zero
(d)
( ) None of the above
IES 2008
IES ‐
S 2006 C
i
l
IES‐2006 Conventional
Find the limit sizes, tolerances and allowances for a
100 mm diameter shaft and hole pair designated by
F8h10. Also specify the type of fit that the above pair
belongs to.
Given: 100 mm diameter lies in the diameter step
range of 80‐120 mm. The fundamental deviation for
shaft designation ‘f’ is ‐5.5 D0.41
f
55
The values of standard tolerances for grades of IT 8
and IT 10 are 25i and 6 i respectively.
d
i d 64i
ti l
Also, indicate the limits and tolerance on a diagram.
[15‐Marks]
GATE 2009
GATE ‐
pp
What are the upper and lower limits of the shaft
represented by 60 f8?
Use the following data:
Diameter 60 lies in the diameter step of 50‐80 mm.
Fundamental tolerance unit,
i, in μ m= 0.45 D1/3 + 0.001D, where D is the
representative size in mm;
Tolerance value f lT8 = 25i.
T l
l for
i
Fundamental deviation for 'f shaft = ‐5.5D0.41
(a)
( ) Lower l
limit = 59.924 mm, Upper Limit = 59.970 mm
(b) Lower limit = 59.954 mm, Upper Limit = 60.000 mm
(c)
( ) Lower limit = 59.970 mm, Upper Limit = 60.016 mm
(d) Lower limit = 60.000 mm, GATELimit = 60.046 mm
For-2014 (IES, Upper & PSUs)
Consider the following statements:
A nomenclature φ 50 H8/p8 denotes that
1. H l di
Hole diameter i 50 mm.
t is
2. It is a shaft base system.
3. 8 indicates fundamental deviation.
Which of the statements given above is/are incorrect?
(a) 1, 2 and 3
(b) 1 and 2 only
d
l
(c) 1 and 3 only
(d) 3 only
GATE 2008 (PI)
GATE – 2008 (PI)
Following data are given for calculating limits of
dimensions and tolerances for a hole: Tolerance unit i (in
µm) = 0.45 ³√D + 0.001D. The unit of D is mm. Diameter
step is 18‐30 mm. If the fundamental deviation for H
hole is zero and IT8 = 25 i the maximum and minimum
i,
limits of dimension for a 25 mm H8 hole (in mm) are
(a) 24.984, 24.967
(b) 25.017, 24.984
(c) 25.033, 25.000
(d) 25.000, 24.967
Page 59 of 78
IES 2005
IES ‐
Assertion (A): Hole basis system is generally
preferred to shaft basis system in tolerance design
for getting the required fits.
Reason (R): Hole has to be given a larger tolerance
band than the mating shaft
shaft.
(a) Both A and R are individually true and R is the
correct explanation of A
t
l
ti
f
(b) Both A and R are individually true but R is not the
correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
IES 2002
IES ‐
In the tolerance specification 25 D 6, the letter D
represents
(a) Grade of tolerance
(b) Upper deviation
(c) Lower deviation
( ) yp
(d) Type of fit
GATE 2000
GATE ‐
A fit is specified as 25H8/e8. The tolerance value for
a nominal diameter of 25 mm in IT8 is 33 microns
and fundamental deviation for the shaft is ‐ 40
microns. The maximum clearance of the fit in
microns is
(a) ‐7
(b) 7
(c) 73
(d) 106](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-60-320.jpg)
![GATE 2003
GATE ‐
The dimensional limits on a shaft of 25h7 are
(a) 25.000, 25.021 mm
(b) 25.000, 24.979 mm
(c) 25.000, 25.007 mm
(d) 25.000, 24.993 mm
GATE 1996 IES 2012
GATE – 1996, IES‐2012
GATE‐2010 (PI)
A small bore is designated as 25H7. The lower
(minimum) and upper (maximum) limits of the bore
are 25.000 mm and 25.021 mm, respectively. When the
bore is designated as 25H8, then the upper (maximum)
limit is 25 033 mm When the bore is designated as
25.033 mm.
hole shaft
H7
The fit on a hole‐shaft system is specified as H7‐
s6.The type of fit is
(a) Clearance fit
(b) Running fit (sliding fit)
(c) Push fit (transition fit)
( )
(d) Force fit (interference fit)
(
)
25H6, then the upper (maximum) limit of the bore (in
mm) is
(a) 25.001 (b) 25.005 (c) 25.009 (d) 25.013
IES 2000
IES ‐
Which one of the following tolerances set on inner
diameter and outer diameter respectively of headed
jig bush for press fit is correct?
(a) G7 h 6
(b) F7 n6
(c)
( ) H 7h 6
h
(d) F j6
F7j6
ISRO‐2008
IAS‐2010 main
Interchangeability can be achieved by
What is the difference between hole basis system and
(a) Standardization
shaft basis system ? Why is hole basis system the more
extensive i use ?
t i in
(b) Better process planning
What are the differences between interchangeability
(c) Simplification
and selective assembly ?
(d) B
Better product planning
d
l
i
GATE 2003
GATE ‐ 2003
[12‐Marks]
GATE – 2008 (PI)
GATE 2008 (PI)
GATE 1997
GATE ‐
A three‐component welded cylindrical assembly is shown
below. The mean length of the three components and their
respective tolerances (both in mm) are given in the table
below.
Three blocks B1 , B2 and B3 are
to be inserted in a channel of
width S maintaining a
minimum gap of width T =
i i
f idth
0.125 mm, as shown in Figure.
For P = 18 75 ± 0 08;
18.
0.08;
Q = 25.00 ± 0.12;
R = 28 125 ± 0 1 and
28.125 0.1
S = 72.35 + X, (where all
dimensions are in mm) the
mm),
tolerance X is
(a) + 0.38
(a) + 0 38
For-2014 (IES, GATE & PSUs)
(b) 0.38
(b) ‐ 0 38
(c) + 0.05
(c) + 0 05
Page 60 of 78
(d) 0.05
(d) ‐0 05
Assuming a normal distribution for the individual
g
component dimensions, the natural tolerance limits for the
length (Y) of the assembly (mm) is
(a) 65
( ) 6 ±2.16
6
(b) 6 ±1.56 ( ) 6 ±0.96 (d) 6 ±0.36
65
6 (c) 65
6
65
6](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-61-320.jpg)
![IAS‐2011 Main
GATE ‐2012 (PI)
ISRO‐2011
A sine bar has a length of 250 mm. Each roller has
Draw a self explanatory sketch of Sigma
A sine bar is specified by
mechanical comparator. Explain how
(a) Its total length
(i) shock load is avoided,
(b) The size of the rollers
(ii) oscillations of the pointer are damped.
( )
(c) The
( ) Th centre di t
t distance b t
between th t rollers
the two ll
[10 –
[10 marks]
( )
GATE – 2011 (PI)
The best wire size (in mm) for measuring effective
diameter of a metric th d (i l d d angle i 6 o)
di
t
f
t i thread (included
l is 60
of 20 mm diameter and 2.5 mm pitch using two
wire method i
i
th d is
(a) 1.443
(b) 0.723
( )
(c) 2.886
(d) 2.086
IES 1992
IES ‐
Which grade symbol represents surface rough of
broaching?
(a) N12 (b) N8
(c) N4 (d) N1
a di
diameter of 20 mm. D i
t
f
During t
taper angle
l
measurement of a component, the height from the
p
,
g
surface plate to the centre of a roller is 100 mm.
The calculated taper angle (in degrees) is
( )
(d) The distance between rollers and upper surface
pp
(a) 21.1
GATE 2013
GATE‐2013
method. The diameter of the best size wire in mm is
(a) 0.866
(b) 1.000
(c) 1.154
(d) 2.000
IFS‐2011
y
g
What are they ? Define the terms 'roughness
height', 'waviness width' and 'lay' in connection
ISRO‐2011
CLA value and RMS values are used for measurement
of
(a) Metal hardness
(b) Sharpness of tool edge
with surface irregularities.
[10‐marks]
[
k ]
(c) Surface dimensions
(d) Surface roughness
For-2014 (IES, GATE & PSUs)
Page 63 of 78
(d) 68.9
To measure the effective diameter of an external
metric thread (included angle is 60o) of 3 5 mm
3.5
pitch, a cylindrical standard of 30.5 mm diameter
a d t o
and two wires of 2 mm diameter each are used.
es o
d a ete eac a e
The micrometer readings over the standard and
over the wires are 16.532 mm and 15.398 mm,
respectively. The effective diameter (in mm) of the
thread is
(a) 33.366
(b) 30.397
(c) 29.366
(d) 26.397
What is
Wh t i meant b i t h
t by interchangeable manufacture?
bl
f t ?
Laser light has unique advantages for inspection
inspection.
(c) 23.6
( )
GATE – 2011 (PI)
A metric th d of pitch 2 mm and th d angle 6
t i thread f it h
d thread
l 60
inspected for its pitch diameter using 3‐wire
p
p
g 3
(b) 22.8](https://image.slidesharecdn.com/production2014questionsetwithanswer-140223052853-phpapp01/85/Production-engg-question-set-with-answers-64-320.jpg)
Production engg. question set with answers
- 1. Metal Cutting, Metal Forming & Metrology Questions & Answers-2014 (All Questions are in Sequence) IES-1992-2013 (22 Yrs.), GATE-1992-2013 (22 Yrs.), GATE (PI)-2000-2012 (13 Yrs.), IAS-19942011 (18 Yrs.), some PSUs questions and conventional questions are added. Section‐I: Theory of Metal Cutting Chapter-1: Basics of Metal Cutting Chapter-2: Force & Power in Metal Cutting Chapter-3: Tool life, Tool Wear, Economics and Machinability Page-1 Page-7 Page-13 Section‐II: Metal Forming Chapter-4: Cold Working, Recrystalization and Hot Working Chapter-5: Rolling Chapter-6: Forging Chapter-7: Extrusion & Drawing Chapter-8: Sheet Metal Operation Chapter-9: Powder Metallurgy Page-24 Page-27 Page-30 Page-37 Page-43 Page-52 Section‐III: Metrology Chapter-10: Limit, Tolerance & Fits Chapter-11: Measurement of Lines & Surfaces Chapter-12: Miscellaneous of Metrology Section‐IV: Cutting Tool Materials For‐2014 (IES, GATE & PSUs) Page-56 Page-61 Page-65 Page‐67
- 2. IES 2013 IES‐2013 Carbide tool is used to machine a 30 mm diameter Theory of Metal Cutting steel shaft at a spindle speed of 1000 revolutions per minute. The cutting speed of the above turning operation is: (a) ( ) 1000 rpm By S K Mondal B S K M d l S 200 IES‐2001 single‐point For cutting of brass with single point cutting tool on a lathe, tool should have (a) Negative rake angle ( ) N ti k l (b) Positive rake angle (c) Zero rake angle (d) Zero side relief angle (b) 1570 m/min (c) 94.2 m/min (d) 47.1 m/min IES‐1995 GATE‐1995; 2008 Cutting power consumption in turning can be C tti ti i t i b Single point thread cutting tool should ideally have: a) Zero rake ) b) Positive rake c) Negative rake d) Normal rake significantly reduced by g y y (a) Increasing rake angle of the tool (b) Increasing the cutting angles of the tool (c) Widening the nose radius of the tool (d) Increasing the clearance angle (d) I i h l l S 2005 IES – 200 Assertion (A): Carbide tips are generally given negative rake angle. Reason (R): Carbide tips are made from very hard materials. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES, GATE & PSUs) S IES – 2002 Assertion (A): Negative rake is usually provided on carbide tipped tools. Reason (R): Carbide tools are weaker in compression. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 1 of 78 IES‐1993 Assertion (A): For a negative rake tool, the specific cutting pressure is smaller than for a positive rake tool under otherwise identical conditions. Reason (R): The shear strain undergone by the chip in the case of negative rake tool is larger larger. (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES 2011 Which one of the following statement is NOT correct with reference to the purposes and effects of rake angle of a cutting tool? (a) To guide the chip flow direction (b) To reduce the friction between the tool flanks and the machined surface (c) To add keenness or sharpness to the cutting edges. (d) To provide better thermal efficiency.
- 3. ( ) GATE – 2008 (PI) Brittle materials are machined with tools having zero or negative rake angle because it (a) results in lower cutting force IES 2007 Cast iron with impurities of carbide requires a particular rake angle for efficient cutting with single point tools, what is the value of this rake angle, give reasons for your answer answer. [ 2 marks] (b) ( ) improves surface finish (c) provides adequate strength to cutting tool S 99 IAS – 1994 Consider the following characteristics 1. The cutting edge is normal to the cutting velocity. 2. Th tti f The cutting forces occur in two directions only. i t di ti l 3. The cutting edge is wider than the depth of cut. The characteristics applicable to orthogonal cutting would include (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1 2 and 3 1, 2 and 3 (d) results in more accurate dimensions IES 2012 IES ‐ During orthogonal cutting, an increase in cutting speed causes (a) An increase in longitudinal cutting force (b) An increase in radial cutting force (c) An increase in tangential cutting force ( ) (d) Cutting forces to remain unaffected g IES‐1995 The the face and th fl k of th Th angle b t l between th f d the flank f the single point cutting tool is known as a) Rake angle b) Clearance angle g c) Lip angle d) Point angle angle. For-2014 (IES, GATE & PSUs) IES‐2006 Which of the following is a single point cutting tool? (a) Hacksaw blade (b) Milling cutter (c) Grinding wheel (d) P ti t l Parting tool IES‐2006 iron, Assertion (A): For drilling cast iron the tool is provided with a point angle smaller than that required for a ductile material material. Reason (R): Smaller point angle results in lower rake angle. k l (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 2 of 78 IES 2012 IES ‐ () g g p g Statement (I): Negative rake angles are preferred on rigid set‐ ups for interrupted cutting and difficult‐to machine materials. Statement (II):Negative rake angle directs the chips on to the machined surface (a) Both Statement (I) and S ( ) B h S d Statement (II) are i di id ll individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually ( ) p true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true IES‐2002 Consider the following statements: The strength of a single point cutting tool depends upon 1. Rake angle 2. Clearance angle 3. Lip angle Which of these statements are correct? (a) ( ) 1 and 3 d (b) 2 and 3 d (c) 1 and 2 (d) 1, 2 and 3
- 4. IES 2012 IES ‐ IES‐2009 Tool life increase with increase in (a) Cutting speed (b) Nose radius (b) N di (c) Feed (d) Depth of cut Consider the following statements with respect to the effects of a large nose radius on the tool: 1. It d t i deteriorates surface fi i h t f finish. 2. It increases the possibility of chatter. 3. It improves tool life. Which of the above statements is/are correct? (a) 2 only (b) 3 only (c) ( ) 2 and 3 only d l (d) 1, 2 and 3 d IES 1994 IES‐1994 IES 2009 IES‐2009 Tool geometry of a single point cutting tool is specified by the following elements: 1. Back rake angle 2. Side rake angle 3. End cutting edge angle 4. Side cutting edge angle 5. Side relief angle 6. End relief angle 7. Nose radius The correct sequence of these tool elements used for correctly specifying the tool geometry is (a) ( ) 1,2,3,6,5,4,7 (b) ( ) 1,2,6,5,3,4,7 (c) 1,2,5,6,3,4,7 (d) 1, 2, 6, 3, 5, 4,7 GATE‐2008 ISRO‐2011 A cutting tool having tool signature as 10, 9, 6, 6, 8, 8, 2 will have side rake angle (a) 10o (b) 9o (c) 8o single‐ The following tool signature is specified for a single point cutting tool in American system: 10, 12, 8 6 15, 20, 3 8, 6, What does the angle 12 represent? (a) Side cutting‐edge angle (b) Side rake angle (c) Back rake angle (d) Sid clearance angle Side l l (d) 2o For-2014 (IES, GATE & PSUs) In a single point turning tool, the side rake angle and orthogonal rake angle are equal. Φ is the principal cutting edge angle and its range is 0o ≤ φ ≤ 90o . The chip flows in the orthogonal plane. The value of Φ is closest to (a) 00 (b) 450 0 (c) 60 (d) 900 Page 3 of 78 IES 1995 IES‐1995 Consider the following statements about nose radius 1. 1 It improves tool life 2. It reduces the cutting force 3. It improves the surface finish. Select the correct answer using the codes given below: g g (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1 2 and 3 1, S 993 IES‐1993 System, 8‐6‐5‐5‐ In ASA System if the tool nomenclature is 8 6 5 5 10‐15‐2‐mm, then the side rake angle will be (a) ° ( ) 5° (b) 6° (c) ( ) 8° (d) 10° ° G GATE – 20 0 ( ) 2010 (PI) The tool geometry of a single point right handed turning tool is provided in the orthogonal rake system (ORS). The sum of the principal (major) cutting edge angle and the auxiliary (minor) cutting edge angle of the above tool is 90o. The inclination angles of the principal and the auxiliary cutting edges are both 0o. The principal and auxiliary orthogonal clearance angles are 10o and 8o, respectively. The rake angle (in degree) measured on the orthogonal plane is (a) 0 (b) 2 (c) 8 (d) 10
- 5. GATE‐2001 During D i orthogonal cutting of mild steel with h l i f ild l ih a 10° rake angle tool, the chip thickness ratio was obtained as 0.4. The shear angle (in degrees) evaluated from this data is g ) (a) 6.53 (b) 20.22 (c) 22.94 ( ) (d) 50.00 ( ) IES ‐ 2009 Minimum shear strain in Mi i h i i g g g orthogonal turning with a cutting tool of zero rake angle is (a) 0 0 0.0 ( ) 5 (b) 0.5 (c) 1.0 (d) 2.0 IES‐2004 Consider the following statements with respect to the relief angle of cutting tool: 1. This affects the direction of chip flow 1 This affects the direction of chip flow 2. This reduces excessive friction between the tool and work piece d k i 3. This affects tool life 4. This allows better access of coolant to the tool work piece interface p Which of the statements given above are correct? (a) 1 and 2 (b) 2 and 3 (c) 2 and 4 For-2014 3 and 4 (d) (IES, GATE & PSUs) GATE 2011 12 A single – point cutting tool with 12° rake angle is used to machine a steel work – piece. The depth of cut, i.e. cut i e uncut thickness is 0 81 mm The chip 0.81 mm. thickness under orthogonal machining condition is 1.8 mm. 1 8 mm The shear angle is approximately (a) 22° (b) 26° (c) 56° 5 (d) 76° IES ‐ 2004 In a machining operation chip thickness ratio is 0 3 and the rake 0.3 angle of the tool is 10°. What is the value of th shear strain? l f the h t i ? ( ) 3 (a) 0.31 ( ) (b) 0.13 3 (c) 3.00 (d) 3.34 IES‐2006 Consider the following statements: 1. A large rake angle means lower strength of the cutting edge. cutting edge 2. Cutting torque decreases with rake angle. Which of the statements given above is/are correct? ( ) (a) Only 1 y ( ) (b) Only 2 y (c) Both 1 and 2 (d) Neither 1 nor 2 Page 4 of 78 IES‐1994 The following parameters determine the model of continuous chip formation: 1. T True f d feed 2. Cutting velocity g y 3. Chip thickness 4. R k angle of the cutting tool. Rake l f h i l The parameters which govern the value of shear p g angle would include (a) ( ) 1,2 and 3 (b) 1,3 and 4 d d (c) 1,2 and 4 (d) 2,3 and 4 GATE 2012 GATE ‐2012 p g g g Details pertaining to an orthogonal metal cutting process are given below. Chip thickness ratio 0.4 04 Undeformed thickness 0.6 mm Rake R k angle l +10° ° Cutting speed 2.5 m/s Mean thickness of primary shear zone 25 microns The shear strain rate in s–1 during the process is (a) 0.1781×105 (b) 0.7754×105 5 (c) ( ) 1.0104×10 (d) 4.397×105 IES‐2004 Match. List I with List II and select the correct answer using the codes given below the Lists: List I List II A. Plan approach angle 1. Tool face B. B Rake angle 2. 2 Tool flank C. Clearance angle 3. Tool face and flank D. Wedge angle D W d l 4. Cutting edge C i d 5. Tool nose A B C D A B C D (a) 1 4 2 5 5 (b) 4 1 3 3 2 (c) 4 1 2 3 (d) 1 4 3 5
- 6. IES‐2003 The angle of inclination of the rake face with respect to the tool base measured in a plane perpendicular to the base and parallel to the width of the tool is called (a) Back rake angle (b) Side rake angle (c) Side cutting edge angle ( ) (d) End cutting edge angle g g g IES‐2001 tool, If α is the rake angle of the cutting tool φ is the shear angle and V is the cutting velocity, then the velocity of chip sliding along th shear plane i l it f hi lidi l the h l is given by (a) (c) V cos α cos(φ − α ) V cos α sin(φ − α ) (b) (d) V sin φ cos (φ − α ) IES‐2004, ISRO‐2009 15 The rake angle of a cutting tool is 15°, shear angle 45° and cutting velocity 35 m/min. What is the l it Wh t i th velocity of chip along th t l f hi l the tool face? (a) 28.5 m/min (b) 27.3 m/min (c) 25 3 m/min 25.3 (d) 23 5 m/min 23.5 IES‐2003 An orthogonal cutting operation is being carried out under the following conditions: cutting speed = 2 m/s, d th of cut = 0.5 mm, tti d / depth f t chip thickness = 0.6 mm. Then the chip velocity is (a) 2 0 m/s (b) 2 4 m/s 2.0 2.4 (c) 1.0 m/s (d) 1.66 m/s IES‐2008 Consider the following statements: In an orthogonal cutting the cutting ratio is found to be 0 75. The cutting speed is 60 m/min and depth of cut 2 4 0∙75. The cutting speed is 60 m/min and depth of cut 2∙4 mm. Which of the following are correct? 1. Chip velocity will be 45 m/min 1 Chip velocity will be 45 m/min. 2. Chip velocity will be 80 m/min. 3. Chip thickness will be 1∙8 mm. 3 Chip thickness will be 1 8 mm 4. Chip thickness will be 3∙2 mm. Select the correct answer using the code given below: l h h d b l (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 IAS‐2003 In orthogonal cutting, shear angle is the angle between (a) Shear plane and the cutting velocity (b) Shear plane and the rake plane (c) Shear plane and the vertical direction (d) Sh l Shear plane and the direction of elongation of crystals in d h di i f l i f l i the chip V sin α sin(φ − α ) IAS‐2002 IAS‐2000 IAS‐1998 The cutting velocity in m/sec, for turning a work piece of diameter 100 mm at the spindle speed of 480 RPM is (a) 1.26 (b) 2.51 (c) 48 (d) 151 For-2014 (IES, GATE & PSUs) Page 5 of 78
- 7. Plain milling of mild steel plate produces (a) egu a s aped d sco t uous c ps (a) Irregular shaped discontinuous chips (b) Regular shaped discontinuous chip (c) Continuous chips without built up edge (c) Continuous chips ithout built up edge (d) Joined chips GATE‐2002 A built‐up‐edge is formed while machining A b ilt d i f d hil hi i (a) Ductile materials at high speed (b) Ductile materials at low speed p (c) Brittle materials at high speed (d) Brittle materials at low speed For-2014 (IES, GATE & PSUs) G GATE – 2009 ( ) Common Data S‐2 2009 (PI) An orthogonal turning operation is carried out at 20 An orthogonal turning operation is carried out at 20 m/min cutting speed, using a cutting tool of rake angle m/min cutting speed, using a cutting tool of rake angle 15o. The chip thickness is 0.4 mm and the uncut chip 15o. The chip thickness is 0.4 mm and the uncut chip thickness i 0.2 mm. hi k is thickness i 0.2 mm. hi k is The chip velocity (in m/min) is ( ) (a) 26.8 GATE‐1995 G GATE – 2009 ( ) Common Data S‐1 2009 (PI) The shear plane angle (in degrees) is IAS‐1995 In an orthogonal cutting, the depth of cut is halved and the feed rate is double. If the chip thickness ratio is unaffected with the changed cutting conditions, the g g , actual chip thickness will be ( ) (a) Doubled ( ) (b) halved (c) Quadrupled (d) Unchanged. ( ) (a) 8 ( ) 7 (b) 27.8 ( ) (c) 28.8 ( ) 9 (d) 29.8 ( ) (b) 10 IES 2007 IES 2007 During machining, excess metal is removed in the form of chip as in the case of turning on a lathe. Which of the following are correct? Continuous ribbon like chip is formed when turning C ti ibb lik hi i f d h t i 1. At a higher cutting speed 2. A l At a lower cutting speed i d 3. A brittle material 4. A d A ductile material il i l Select the correct answer using the code given below: (a) ( ) 1 and 3 d (b) 1 and 4 d (c) 2 and 3 (d) 2 and 4 IES‐1997 Assertion (A): For high speed turning of cast iron pistons, carbide tool bits are provided with chip breakers. Reason (R): High speed turning may produce long, ribbon type continuous chips which must be broken into small lengths which otherwise would be difficult to handle and may prove hazardous. (a) Both A and R are individually true and R is the correct explanation of A l i f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 6 of 78 ( ) (c) 12 ( ) 4 (d) 14 IAS‐1997 Consider the following machining conditions: BUE will form in (a) Ductile material. (b) High cutting speed. (c) Small rake angle. (d) Small uncut chip thickness. Ch‐1: Mechanics of Basic Machining Operation Q. No Option p Q. No Option p 1 C 11 D 2 B 12 D 3 D 13 B 4 C 14 C 5 B 15 D 6 D 16 B 7 B 17 B 8 A 18 D 9 B 19 D 10 B 20 B
- 8. ESE ‐2000 (Conventional) Force & Power in M t l C tti F P i Metal Cutting By S K Mondal GATE ‐2008 (PI) Linked S‐1 GATE 2008 (PI) Linked S 1 The following data from th orthogonal cutting t t the th Th f ll i d t f l tti test is available. Rake angle = 100, chip thickness ratio = 0.35, uncut chip thi k t hi thickness = 0.51 mm, width of cut = idth f t 3 mm, yield shear stress of work material = 285 N/mm2, mean f i ti N/ friction co‐efficient on t l f ffi i t tool face = 0.65, Determine (i) () Cutting force ( c) (F (ii) Radial force (iii) Normal force (N) on tool and (iv) Shear force (Fs ) ). GATE ‐2008 (PI) Linked S‐2 GATE 2008 (PI) Linked S 2 g g p g In an orthogonal cutting experiment, an HSS tool having g g p g In an orthogonal cutting experiment, an HSS tool having the following tool signature in the orthogonal reference the following tool signature in the orthogonal reference system ( (ORS) h b ) has been used: 0‐10‐7‐7‐10‐75‐1. Given d system ( (ORS) h b ) has been used: 0‐10‐7‐7‐10‐75‐1. Given d width of cut = 3.6 mm; shear strength of workpiece width of cut = 3.6 mm; shear strength of workpiece material = 460 N/mm2; depth of cut = 0.25 mm; material = 460 N/mm2; depth of cut = 0.25 mm; coefficient of friction at tool‐chip interface = 0.7. coefficient of friction at tool‐chip interface = 0.7. Shear plane angle (i d Sh l l (in degree) f minimum cutting f ) for i i tti force Minimum power requirement (i kW) at a cutting speed Mi i i t (in t tti d is of 150 m/min is (a) 20.5 (b) 24.5 (c) 28.5 (d) 32.5 GATE 2007 (PI) C D t 2 GATE – 2007 (PI) Common Data‐2 g g , g In an orthogonal machining test, the following observations were made Cutting force 1200 N Thrust force 500 N Tool k T l rake angle l zero Cutting speed 1 m/s Depth of cut 0.8 mm Chip thickness 1.5 mm Chip speed along the tool rake face will be (a) 8 ( ) 0.83 m/s / (b) 0.53 m/s / (c) 1.2 m/s For-2014(d) 1.88 m/s & PSUs) (IES, GATE (a) 3.15 (b) 3.25 (c) 3.35 ESE‐2005 Conventional Mild steel i b i l is being machined at a cutting hi d i speed of 200 m/min with a tool rake angle of 10. The width of cut and uncut thickness are 2 mm and 0.2 mm respectively. If the average p y g value of co‐efficient of friction between the tool and the chip is 0 5 and the shear stress of 0.5 the work material is 400 N/mm2, Determine (i) shear angle and (ii) Cutting and thrust component of the force. GATE 2007 (PI) C D t 1 GATE – 2007 (PI) Common Data‐1 g g , g In an orthogonal machining test, the following observations were made Cutting force 1200 N Thrust force 500 N Tool k T l rake angle l zero Cutting speed 1 m/s Depth of cut 0.8 mm Chip thickness 1.5 mm Friction angle during machining will be (a) 6 ( ) 22.6o (b) 32.8o 8 (c) ( ) 57.1o (d) 6 o 67.4 (d) 3.45 GATE‐2007 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, tool the cutting velocity is 90 m/min The feed m/min. is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0 48 mm If the 0.48 mm. orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle is 90 degree is (a) ( ) 20.56 6 (b) 26.56 6 6 (c) 30.56 (d) 36.56 Page 7 of 78 S 2003 Conventional i l ESE‐2003‐ C During turning a carbon steel rod of 160 mm diameter by a carbide tool of geometry; 0, 0, 10, 8 15, 75, 0 ( 8, (mm) at speed of bid l f ) d f 400 rpm, feed of 0.32 mm/rev and 4.0 mm depth of cut, the following observation were made made. Tangential component of the cutting force, Pz = 1200 N Axial component of the cutting force Px = 800 N force, Chip thickness (after cut),α 2 = 0.8 mm. For the above machining condition determine the values of (i) Friction force, F and normal force, N acting at the chip tool interface. interface (ii) Yield shears strength of the work material under this machining condition. (iii) Cutting power consumption in kW.
- 9. GATE – 1995 ‐Conventional While turning a C steel rod of 160 mm di C‐15 t l d f 6 diameter at Whil t i t t 315 rpm, 2.5 mm depth of cut and feed of 0.16 mm/rev b a t l of geometry 00, 100, 80, 90,150, 750, / by tool f t 0(mm), the following observations were made. Tangential component of the cutting force = 500 N Axial component of the cutting force = 200 N p g Chip thickness = 0.48 mm Draw schematically the Merchant’s circle diagram Merchant s for the cutting force in the present case. IAS‐2003 Main Examination During turning process with 7 ‐ ‐ 6 – 6 – 8 – 30 – 1 D i t i ith (mm) ASA tool the undeformed chip thickness of 2.0 mm and width of cut of 2.5 mm were used. Th d idth f t f d The side rake angle of the tool was a chosen that the machining operation could b approximated t b hi i ti ld be i t d to be orthogonal cutting. The tangential cutting force and thrust f th t force were 1177 N and 560 N respectively. d 6 ti l Calculate: [30 marks] (i) h d ( ) The side rake angle k l (ii) Co‐efficient of friction at the rake face (iii) The dynamic shear strength of the work material GATE 2013 GATE‐2013 A steel bar 200 mm in diameter is turned at a feed of 0.25 mm/rev with a depth of cut of 4 mm. The rotational speed of the workpiece is 160 rpm. The material removal rate in mm3/s is (a) 160 (b) 167 6 (c) 1600 167.6 (d) 1675 5 1675.5 GATE‐2007 In orthogonal turning of medium carbon steel. The I th l t i f di b t l Th specific machining energy is 2.0 J/mm3. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 l it f d d d th f t / i mm/rev and 2 mm respectively. The main cutting force in N is f i N i (a) 40 (b) 80 (c) 400 (d) 800 Example When the rake angle is zero during orthogonal cutting, show that τs pc = (1 − μ r ) r 1+ r2 Where τs is the shear strengrh of the material p c = specific power of cutting p r = chip thickness ratio μ = coefficient of friction in tool chip interface For-2014 (IES, GATE & PSUs) For PSU & IES In strain gauge dynamometers the use of how many active gauge makes the dynamometers more effective (a) Four (b) Three (c) T o Two (d) One Ans. (a) Page 8 of 78 IES 2004 IES ‐ A medium carbon steel workpiece is turned on a lathe at 50 m/min. cutting speed 0.8 mm/rev feed and 1.5 mm depth of cut. What is the rate of metal removal? (a) 1000 mm3/min (b) 60,000 mm3/min (c) 20,000 mm3/min ( ) (d) Can not be calculated with the given data g GATE 2013 (PI) C D Q i GATE‐2013 (PI) Common Data Question A disc of 200 mm outer and 80 mm inner diameter is faced of 0.1 mm/rev with a depth of cut of 1 mm. The facing operation is undertaken at a constant cutting speed of 90 m/min in a CNC lathe. The main (tangential) cutting force is 200 N. Neglecting the contribution of the feed force towards cutting power the specific cutting energy power, in J/mm3 is (a) ( ) 0.2 (b) 2 (c) ( ) 200 (d) 2000 GATE‐2006 Common Data Questions(1) In an orthogonal machining operation: I th l hi i ti Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut 5 mm Width of cut = 5 mm Chip thickness 0.7 mm Chip thickness = 0.7 mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. A M h t' th The coefficient of friction at the tool‐chip interface is (a) 0.23 ( ) (b) 0.46 (b) (c) 0.85 (d) 0.95
- 10. GATE‐2006 Common Data Questions(2) In an orthogonal machining operation: I th l hi i ti Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut 5 mm Width of cut = 5 mm Chip thickness 0.7 mm Chip thickness = 0.7 mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. A M h t' th The percentage of total energy dissipated due to friction at the tool‐chip interface is f h l h f (a) 30% (b) 42% (c) 58% (d) 70% GATE‐2003 Common Data Questions(2) is turned on a l th with orthogonal lathe ith A cylinder i t li d d th l machining principle. Spindle rotates at 200 rpm. The axial f d rate i 0.25 mm per revolution. D th of cut i i l feed t is l ti Depth f t is 0.4 mm. The rake angle is 10°. In the analysis it is found that the h th t th shear angle i 27.75° l is ° In the above problem, the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is (a) 0.18 (b) 0.36 (c) 0.71 (d) 0.98 GATE ‐2010 (PI) Linked S‐1 In orthogonal turning of an engineering alloy, it has been observed that the friction force acting at the chip‐ tool interface is 402.5 N and the friction force is also perpendicular to the cutting velocity vector. The feed velocity is negligibly small with respect to the cutting velocity. The ratio of friction force to normal force associated with the chip‐tool interface is 1. The uncut chip tool chip thickness is 0.2 mm and the chip thickness is 0.4 mm. The cutting velocity is 2 m/s. The shear force (in N) acting along the primary shear plane is (a) 180.0 (b) 240.0 (c) 360.5 (d) 402.5 For-2014 (IES, GATE & PSUs) GATE‐2006 Common Data Questions(3) In an orthogonal machining operation: I th l hi i ti Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut 5 mm Width of cut = 5 mm Chip thickness 0.7 mm Chip thickness = 0.7 mm Thrust force = 200 N Cutting force = 1200 N Assume Merchant's theory. A M h t' th The values of shear angle and shear strain, respectively, are l (a) 30.3° and 1.98 (b) 30.3° and 4.23 (c) 40.2° and 2.97 (d) 40.2° and 1.65 GATE‐2008 Common Data Question (1) Orthogonal t O th l turning i performed on a cylindrical work i is f d li d i l k piece with shear strength of 250 MPa. The following conditions are used: cutting velocity i 180 m/min. f d diti d tti l it is 8 / i feed is 0.20 mm/rev. depth of cut is 3 mm. chip thickness ratio = 0.5. Th orthogonal rake angle i 7o. A l ti The th l k l is Apply Merchant's theory for analysis. The shear plane angle (in degree) and the shear ( ) force respectively are (a) 52: 320 N (b) 52: 400N (c) 28: 400N (d) 28:320N GATE ‐2010 (PI) Linked S‐2 g g g g y, In orthogonal turning of an engineering alloy, it has been observed that the friction force acting at the chip‐ tool interface is 402.5 N and the friction force is also perpendicular to the cutting velocity vector. The feed velocity is negligibly small with respect to the cutting velocity. Th ratio of f i ti l it The ti f friction f force t normal f to l force associated with the chip‐tool interface is 1. The uncut chip thickness is 0 2 mm and the chip thickness is 0 4 0.2 0.4 mm. The cutting velocity is 2 m/s. Assume that the energy expended during machining is completely converted to heat. The rate of heat generation (in W) at the primary shear plane is (a) 180.5 (b) 200.5 (c) 302.5 (d) 402.5 Page 9 of 78 GATE‐2003 Common Data Questions(1) is turned on a l th with orthogonal lathe ith A cylinder i t li d d th l machining principle. Spindle rotates at 200 rpm. The axial f d rate i 0.25 mm per revolution. D th of cut i i l feed t is l ti Depth f t is 0.4 mm. The rake angle is 10°. In the analysis it is found that the h th t th shear angle i 27.75° l is ° The thickness of the produced chip is (a) 0.511 mm (b) 0.528 mm (c) 0.818 mm (d) 0.846 mm GATE‐2008 Common Data Question (2) Orthogonal t O th l turning i performed on a cylindrical work i is f d li d i l k piece with shear strength of 250 MPa. The following conditions are used: cutting velocity i 180 m/min. f d diti d tti l it is 8 / i feed is 0.20 mm/rev. depth of cut is 3 mm. chip thickness ratio = 0.5. Th orthogonal rake angle i 7o. A l ti The th l k l is Apply Merchant's theory for analysis. The cutting and Thrust forces, r Rees e ctively, a (a) 568N; 387N (b) 565N; 381N (c) 440N; 342N (d) 480N; 356N Linked Answer Questions GATE‐2013 S‐1 In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4 mm and cutting velocity of 90 m/min, it is observed that the main (tangential)cutting force is perpendicular to friction force acting at the chip tool interface chip‐tool interface. The main (tangential) cutting force is 1500 N. The orthogonal rake angle of the cutting tool in degree is (a) zero (b) 3.58 (c) 5 (d) 7.16
- 11. Linked Answer Questions GATE‐2013 S‐2 In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4 mm and cutting velocity of 90 m/min, it is observed that the main (tangential)cutting force is perpendicular to friction force acting at the chip tool interface chip‐tool interface. The main (tangential) cutting force is 1500 N. The normal force acting at the chip‐tool interface in N is (a) 1000 (b) 1500 (c) 20oo (d) 2500 IES 2012 IES ‐ During orthogonal cutting, an increase in cutting speed causes (a) An increase in longitudinal cutting force (b) An increase in radial cutting force (c) An increase in tangential cutting force ( ) (d) Cutting forces to remain unaffected g GATE‐1997 In a typical metal cutting operation, using a I i l l i i i cutting tool of positive rake angle = 10°, it was observed that the shear angle was 20°. The friction angle is g (a) 45° (b) 30° (c) 60° ( ) (d) 40° ( ) For-2014 (IES, GATE & PSUs) GATE – 2011 (PI) Linked S1 GATE 2011 (PI) Linked S1 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle, the following data is obtained: Uncut chip thickness = 0.25 mm Chip thickness = 0 75 mm 0.75 Width of cut = 2.5 mm Normal f N l force = 950 N Thrust force = 475 N The shear angle and shear force, respectively, are (a) 71 565o, 150 21 N 71.565 150.21 (b) 18 435o , 751 04 N 18.435 751.04 (c) 9.218o, 861.64 N (d) 23.157o , 686.66 N IES 2010 IES 2010 The relationship between the shear angle Φ, the friction angle β and cutting rake angle α is given as S 999 IAS – 1999 In an orthogonal cutting process, rake angle of the tool is 20° and friction angle is 25.5°. Using Merchant s Merchant's shear angle relationship, the value of shear angle will be (a) 39 5° 39.5 (b) 42 25° 42.25 (c) 47.75° (d) 50.5° Page 10 of 78 GATE – 2011 (PI) Linked S2 GATE 2011 (PI) Linked S2 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle, the following data is obtained: Uncut chip thickness = 0.25 mm Chip thickness = 0 75 mm 0.75 Width of cut = 2.5 mm Normal f N l force = 950 N Thrust force = 475 N The ultimate shear stress (in N/mm2) of the work material is (a) 235 (b) 139 (c) 564 (d) 380 IES‐2005 Which is h Whi h one of the f ll i f h following i the correct expression for the Merchant's machinability constant? (a) 2φ + γ − α (b) 2φ − γ + α (c) 2φ − γ − α (d) φ + γ − α (Where φ = shear angle,γ = friction angle andα = rake angle) IES‐2003 In I orthogonal cutting test, the cutting f h l i h i force = 900 N, the thrust force = 600 N and chip shear angle is 30o. Then the chip shear force is (a) 1079 4 N 1079.4 (b) 969 6 N 969.6 (c) 479.4 N (d) 69.6 N
- 12. IES‐2000 IES‐1996 In an orthogonal cutting test, the cutting force and thrust force were observed to be 1000N and 500 N respectively. If the rake angle of tool is zero, the coefficient of friction in chip‐tool interface will be 1 (a) 2 ( b) 2 ( c) 1 ( d) 2 2 IES‐1997 Consider the f ll i forces acting on a C id h following f i finish turning tool: 1. Feed force 2. 2 Thrust force 3. Cutting force. g The correct sequence of the decreasing order of the magnitudes of these forces is (a) 1, 2, 3 (b) 2, 3, 1 (c) 3, 1, 2 (d) 3, 2, 1 IES‐2002 In I a machining process, the percentage of hi i h f heat carried away by the chips is typically (a) 5% (b) 25% (c) 50% (d) 75% For-2014 (IES, GATE & PSUs) Which of the following forces are measured directly by strain gauges or force dynamometers during metal g cutting ? 1. Force exerted by the tool on the chip acting normally to the tool face. 2. Horizontal cutting force exerted by the tool on the work piece. 3. Frictional resistance of the tool against the chip flow acting along the tool face. 4. V i l f Vertical force which h l hi h helps i h ldi in holding the tool i h l in position. (a) ( ) 1 and 3 d (b) 2 and 4 d (c) 1 and 4 (d) 2 and 3 IES‐1999 The di l force i single‐point tool d i in i l Th radial f i l during turning operation varies between (a) 0.2 to 0.4 times the main cutting force (b) 0 4 to 0 6 times the main cutting force 0.4 0.6 (c) 0.6 to 0.8 times the main cutting force g (d) 0.5 to 0.6 times the main cutting force IES‐1998 In I metal cutting operation, the approximate l i i h i ratio of heat distributed among chip, tool and work, in that order is (a) 80: 10: 10 (b) 33: 33: 33 (c) 20: 60: 10 (d) 10: 10: 80 Page 11 of 78 GATE‐2007 In th I orthogonal t l turning of l i f low carbon steel pipe with b t l i ith principal cutting edge angle of 90°, the main cutting force i 1000 N and th f d f f is d the feed force i 8 N Th shear is 800 N. The h angle is 25° and orthogonal rake angle is zero. Employing M h t’ th E l i Merchant’s theory, th ratio of f i ti the ti f friction force to normal force acting on the cutting tool is (a) ( ) 1.56 (b) ( ) 1.25 (c) 0.80 (d) 0.64 IES‐1995 The Th primary tool f i l force used i calculating d in l l i the total power consumption in machining is the (a) Radial force (b) Tangential force (c) Axial force (d) Frictional force. S IAS – 2003 As the cutting speed increases (a) More heat is transmitted to the work piece and less heat is transmitted to the tool (b) More heat is carried away by the chip and less heat is transmitted to the tool t itt d t th t l (c) More heat is transmitted to both the chip and the tool ( ) (d) More heat is transmitted to both the work piece and p the tool
- 13. S 99 IAS – 1995 IES‐2001 Power consumption i metal cutting i in is P i l i mainly due to (a) Tangential component of the force (b) Longitudinal component of the force (c) Normal component of the force p (d) Friction at the metal‐tool interface IES‐1993 IES 2011 'Dynamometer' i a d i device used f A 'D ' is d for the h measurement of (a) Chip thickness ratio (b) Forces during metal cutting (c) Wear of the cutting tool g (d) Deflection of the cutting tool S IAS – 2003 The heat generated in metal conveniently be determined by (a) Installing thermocouple on the job (b) Installing thermocouple on the tool (c) Calorimetric set‐up ( ) (d) Using radiation pyrometer g py Thrust force will increase with the increase in (a) Side cutting edge angle (b) Tool nose radius (b) T l di (c) Rake angle (d) End cutting edge angle. The instrument or device used to measure the cutting forces in machining is : (a) Tachometer ( ) T h t (b) Comparator (c) Dynamometer (d) Lactometer IES‐1998 cutting For-2014 (IES, GATE & PSUs) can The factor of a resistive pick‐up of Th gauge f f i i i k f cutting force dynamometer is defined as the ratio of (a) Applied strain to the resistance of the wire (b) The proportional change in resistance to the applied strain (c) The resistance to the applied strain (d) Change in resistance to the applied strain Page 12 of 78 IES 2010 IES 2010 Consider the following statements: In an orthogonal, single‐point metal cutting, as th side‐cutting edge angle i i the id tti d l is increased, d 1. The tangential force increases. g 2. The longitudinal force drops. 3. Th radial f The di l force i increases. Which of these statements are correct? (a) 1 and 3 only (b) 1 and 2 only (c) ( ) 2 and 3 only (d) ( ) 1, 2 and 3 S 200 IAS‐2001 ( ) p Assertion (A): Piezoelectric transducers and preferred over strain gauge transducers in the dynamometers for measurement of three‐dimensional cutting forces. Reason (R): In electric transducers there is a significant leakage of signal from one axis to the other, such cross error is negligible in the case of piezoelectric transducers. (a) Both A and R are individually true and R is the correct explanation of A ( ) (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true IES‐2000 Assertion (A) I metal cutting, the normal (A): In A i l i h l laws of sliding friction are not applicable. Reason (R): Very high temperature is produced at the tool‐chip interface tool chip interface. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) is ( ) A i true b R i f l but is false (d) A is false but R is true
- 14. GATE 1992 The effect of rake angle on the mean f friction angle in h ff f k l h l machining can be explained by (A) sliding (Coulomb) model of friction (B) sticking and then sliding model of friction (C) sticking friction ( ) (D) Sliding and then sticking model of friction g g IES‐2004 Assertion (A): The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed as cutting ratio in orthogonal cutting g g g Reason (R): The frictional force is very high due to the occurrence of sticking friction rather than sliding g g friction ( ) (a) Both A and R are individually true and R is the correct y explanation of A ( ) (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS – 2009 Main Tool Wear, Tool Life & Machinability Tool Wear, Tool Life & Machinability Explain ‘sudden‐death mechanism’ of tool failure. [ 4 – marks] [ k ] Show crater wear and flank wear on a single point Sh t d fl k i l i t cutting tool. State the factors responsible for wear on a turning tool. t i t l [ 2 –marks] For-2014 (IES, GATE & PSUs) The ff t f k the friction Th effect of rake angle on th mean f i ti angle i l l in machining can be explained by (a) ( ) Sliding ( (coulomb) model of friction ) (b) sticking and then siding model of friction g g (c) Sticking friction (d) sliding and then sticking model of friction GATE 2008 (PI) GATE‐2008 (PI) During machining, the wear land (h) has been plotted against machining time (T) as given i the f ll i in h following i hi i i i figure. For a critical wear land of 1.8 mm, the cutting tool life (in minute) is (a) 52.00 (b) 51.67 (c) 51.50 (d) 50.00 By S K Mondal B S K M d l IES 2009 Conventional GATE‐1993 IES 2010 IES 2010 Flank wear occurs on the (a) Relief face of the tool (b) Rake face (c) Nose of the tool (d) Cutting edge Page 13 of 78 S 2007 IES – 200 Flank wear occurs mainly on which of the following? (a) Nose part and top face (b) Cutting edge only (c) Nose part, front relief face, and side relief face of the cutting tool (d) Face of the cutting tool at a short distance from g g the cutting edge
- 15. S 2004 IES – 200 Consider the following statements: During the third stage of tool‐wear, rapid deterioration of tool edge takes place because 1. Flank wear is only marginal 2. Flank wear is large 3 3. Temperature of the tool increases gradually p g y 4. Temperature of the tool increases drastically Which of the statements given above are correct? (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) 2 and 3 S IES – 2000 Crater wear starts at some distance from the tool tip because (a) Cutting fluid cannot penetrate that region (b) Stress on rake face is maximum at that region (c) Tool strength is minimum at that region ( ) (d) Tool temperature is maximum at that region p g S 99 IES – 1995 Crater wear is predominant in (a) Carbon steel tools (b) T Tungsten carbide tools t bid t l (c) High speed steel tools (d) Ceramic tools For-2014 (IES, GATE & PSUs) S IES – 2002 Crater wear on tools always starts at some distance from the tool tip because at that point (a) Cutting fluid does not penetrate (b) Normal stress on rake face is maximum (c) Temperature is maximum ( ) (d) Tool strength is minimum g S 996 IES – 1996 Notch wear at the outside edge of the depth of cut is due to (a) Abrasive action of the work hardened chip material (b) Oxidation (c) Slip‐stick action of the chip ( ) (d) Chipping. pp g S 99 IES – 1994 Assertion (A): Tool wear is expressed in terms of flank wear rather than crater wear. Reason (R): Measurement of flank wear is simple and more accurate. (a) Both A and R are individually true and R is the ( ) B th A d R i di id ll t d R i th correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 14 of 78 S 2007 IAS – 200 Why does crater wear start at some distance from the tool tip? (a) Tool strength is minimum at that region (b) Cutting fluid cannot penetrate that region (c) Tool temperature is maximum in that region ( ) (d) Stress on rake face is maximum at that region g S 99 IES – 1995 Match List I with List II and select the correct answer using the codes given below the lists: List I (Wear type) List II (Associated mechanism) A. Abrasive wears 1. Galvanic action B. Adhesive wears 2. Ploughing action C. Electrolytic wear y 3 3. Molecular transfer D. Diffusion wears 4. Plastic deformation 5. 5 Metallic bond Code: A B C D A B C D (a) 2 5 1 3 (b) 5 2 1 3 (c) 2 1 3 4 (d) 5 2 3 4 S IES – 2008 What are the reasons for reduction of tool life in a machining operation? 1. 1 Temperature rise of cutting edge 2. Chipping of tool edge due to mechanical impact 3. Gradual wears at tool point 4 4. Increase in feed of cut at constant cutting force g Select the correct answer using the code given below: (a) 1, 2 and 3 (b) 2, 3 and 4 (c) d ( ) 1, 3 and 4 (d) 1, 2 and 4 d
- 16. S IAS – 2002 Consider the following actions: 1. Mechanical abrasion 2. Diffusion 3. Pl ti d f Plastic deformation ti 4. Oxidation O id ti Which of the above are the causes of tool wear? (a) 2 and 3 (b) 1 and 2 (c) 1, 2 and 4 (d) 1 and 3 IES 2012 IES ‐ In Taylor s tool life equation VTn = C, the constants n In Taylor’s tool life equation VT C, the constants n and C depend upon 1. Work piece material 1 Work piece material 2. Tool material 3. Coolant ( ) (a) 1, 2, and 3 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only S 999 IAS – 1999 S IAS – 2003 The type of wear that occurs due to the cutting action of the particles in the cutting fluid is referred to as (a) Attritions wear (b) Diff i wear Diffusion (c) Erosive wear (d) Corrosive wear Consider the following statements: Chipping of a cutting tool is due to 1. T l material b i t b ittl Tool t i l being too brittle 2. Hot hardness of the tool material. 3. High positive rake angle of the tool. Which of these statements are correct? (a) 1, 2 and 3 (b) 1 and 3 (c) ( ) 2 and 3 d (d) 1 and 2 d S 20 0 C i l IES 2010 Conventional IFS 2009 With the help of Taylor’s tool life equation, determine the shape of the curve between velocity Draw tool life curves for cast alloy, High speed steel and ceramic tools. [2 – Marks] Ans. of cutting and life of the tool. Assume an HSS tool and steel as work material material. [ [10‐Marks] ] 1. High speed steel IES‐1996 Chip equivalent is increased by (a) An increases in side‐cutting edge angle of tool (b) An increase in nose radius and side cutting edge angle of tool (c) Increasing the plant area of cut (d) Increasing the depth of cut. For-2014 (IES, GATE & PSUs) S 992 IES – 1992 Tool life is generally specified by (a) Number of pieces machined (b) V l Volume of metal removed f t l d (c) Actual cutting time (d) Any of the above Page 15 of 78 2. cast alloy and 3. ceramic tools. G 200 GATE‐2004 operation, In a machining operation doubling the 1 cutting speed reduces the tool life to 8 th of the i i l l th original value. Th exponent n i T l ' The t in Taylor's n = C, is tool life equation VT (a ) 1 8 (b) 1 4 (c ) 1 3 (d ) 1 2
- 17. S IES – 2000 S 999 IES – 1999 In a tool life test, doubling the cutting speed reduces the tool life to 1/8th of the original. The Taylor s Taylor's tool life index is 1 ( a ) 2 1 ( b ) 3 1 ( c ) 4 1 ( d ) 8 In a single point turning operation of steel with a In a single‐point turning operation of steel with a cemented carbide tool, Taylor's tool life exponent is 0.25. If the cutting speed is halved, the tool life will increase by (a) Two times (b) Four times (c) Eight times (d) Sixteen times S IES – 2006 S IES – 2008 In Taylor s tool life equation is VT In Taylor's tool life equation is VTn = constant. What is the value of n for ceramic tools? (a) ( ) 0.15 to 0.25 t (b) 0.4 to 0.55 t (c) 0.6 to 0.75 (d) 0.8 to 0.9 S 999 IES – 1999 Which of the following values of index n is associated with carbide tools when Taylor's tool life equation, V.Tn = constant is applied? (a) 0∙1 to 0∙15 (b) 0∙2 to 0∙4 (c) ( ) 0.45 t 0∙6 to 6 (d) 0∙65 t 0∙9 6 to The approximately variation of the tool life exponent 'n' of cemented carbide tools is (a) 0 03 to 0 08 0.03 0.08 (b) 0 08 to 0 20 0.08 0.20 (c) 0.20 to 0.48 (d) 0.48 to 0.70 S 998 IAS – 1998 ( g ) Match List ‐ I (Cutting tool material) with List ‐ II (Typical value of tool life exponent 'n' in the Taylor's equation V.Tn = C) and select the correct answer using the codes given below the lists: th d i b l th li t List – I List – II A. A HSS 1. 0.18 8 B. Cast alloy 2. 0.12 C. Ceramic C C i 3. 0.25 D. Sintered carbide 4. 0.5 Codes: A B d C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 (c) ( ) 2 1 4 3 (d) ( ) 1 2 4 3 IES 2013 IES‐2013 ( ) GATE ‐2009 (PI) ISRO‐2011 0.25) A carbide tool(having n = 0 25) with a mild steel A 50 mm d diameter steel rod was turned at 284 rpm and l d d d In an orthogonal machining operation, the tool life work‐piece was found to give life of 1 hour 21 tool failure occurred in 10 minutes The speed was minutes. obtained is 10 min at a cutting speed of 100 m/min, minutes while cutting at 60 m/min. The value of C changed to 232 rpm and the tool failed in 60 minutes. while at 75 m/min cutting speed, the tool life is 30 in Taylor’s tool life equation would be equal to: Assuming straight line relationship between cutting min. min The value of index (n) in the Taylor’s tool life Taylor s (a) ( ) 200 speed and tool l f the value of Taylorian Exponent is d d l life, h l f l equation (b) 180 (a) 0 21 0.21 (a) 0.262 (b) 0.323 (c) 0.423 (d) 0.521 For-2014 (IES, GATE & PSUs) (c) 150 (d) 100 Page 16 of 78 (b) 0 13 0.13 (c) 0 11 0.11 (d) 0 23 0.23
- 18. IES 2010 The above figure shows a typical relationship between tool life and cutting speed for different g p materials. Match the graphs for HSS, Carbide and Ceramic tool materials and select the correct i l d l h answer using the code given below the lists: Code: HSS Carbide Ceramic (a) 1 2 3 (b) 3 2 1 (c) 1 3 2 (d) 3 1 2 Example p The following data was obtained from the tool‐life cutting test: Cutting Speed, m/min:49.74 49 23 48 6 4 6 42 8 d 49 4 49.23 48.67 45.76 42.58 Tool life, min 2.94 3.90 4.77 9.87 28.27 Determine the constants of the Taylor tool life equation VTn = C IES 2010 IES 2010 Tool life is affected mainly with (a) Feed (b) Depth of cut (c) Coolant (d) Cutting speed For-2014 (IES, GATE & PSUs) GATE 2013 GATE‐2013 G 20 0 GATE‐2010 A, Taylor s For tool A Taylor’s tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, B n = 0.3 and K = 6 Th cutting speed (i d 60. The tti d (in m/min) above which tool A will have a higher tool life than tool B is (a) 26 7 (b) 42 5 (c) 80 7 (d) 142 9 26.7 42.5 80.7 142.9 Two cutting tools are being compared for a machining operation. The tool life equations are: Carbide tool: VT 1.6 = 3000 HSS tool: VT 0.6 = 200 Where V i the cutting speed i m/min and T i the Wh is h i d in / i d is h tool life in min. The carbide tool will provide higher p g tool life if the cutting speed in m/min exceeds (a) 15.0 G 2003 GATE‐2003 A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut How many cut. components can be produced with one cutting tool at 60 rpm? (a) 29 (b) 31 (c) 37 (d) 42 S 99 IES – 1997 Consider the following elements: 1. Nose radius 2. Cutting speed 3. D th f t Depth of cut 4. Feed F d The correct sequence of these elements in DECREASING order of their influence on tool life is ( ) (a) 2, 4, 3, 1 4 3 ( ) 4 3 (b) 4, 2, 3, 1 (c) 2,4, 1, 3 (d) 4, 2, I, 3 Page 17 of 78 (b) 39.4 (c) 49.3 (d) 60.0 S 99 200 IES – 1994, 2007 For increasing the material removal rate in turning, without any constraints, what is the right sequence to adjust the cutting parameters? 1. Speed 2. Feed 3. Depth of cut Select the correct answer using the code given below: (a) 1‐ 2‐ 3 (b) 2‐ 3‐ 1 (c) 3‐ 2‐ 1 3 2 (d) 1‐ 3‐ 2 1 3 ISRO‐2012 What is the correct sequence of the following parameters i t in order of th i maximum t d f their i to minimum influence on tool life? 1. Feed rate d 2. Depth of cut 3. Cutting speed Select the correct answer using the codes given below (a) 1 2 3 1, 2, (b) 3 2 1 (c) 2 3 1 (d) 3 1 2 3, 2, 2, 3, 3, 1,
- 19. S 992 IES – 1992 Tool life is generally better when (a) Grain size of the metal is large (b) G i i f th t l i Grain size of the metal is small ll (c) Hard constituents are present in the microstructure of the tool material ( ) (d) None of the above S IAS – 2003 The tool life curves for two tools A and B are shown in the figure and they follow the tool life equation VTn = C. Consider the following statements: g 1. 2. 3. 4. Value of n for both the tools is same. Value of C for both the tools is same. Value of C for tool A will be greater than that for the tool B. Value of C for tool B will be greater than that for the tool A. a ue o C o too be g eate t a t at o t e too . S IAS – 2002 Using the Taylor equation VTn = c, calculate the percentage increase in tool life when the cutting speed is reduced by 50% (n 0 5 and c 400) speed is reduced by 50% (n = 0∙5 and c = 400) (a) 300% (b) 400% (c) ( ) 100% % (d) 50% % Which of these statements is/are correct? (a) 1 and 3 (b) 1 and 4 (c) 2 only (d) 4 only S IAS – 2002 Optimum cutting speed for minimum cost (Vc min ) i and optimum cutting speed for maximum production rate (Vr max ) have which one of the following relationships? (a) Vc min = Vr max (b) Vc min > Vr max (c) Vc min < Vr max (d) V2c min = Vr max S 99 IAS – 1997 Taylor s In the Taylor's tool life equation, VTn = C, the value of n = 0.5. The tool has a life of 180 minutes at a cutting speed of 18 m/min. If the tool life is reduced to 45 minutes, then the cutting speed will be (a) 9 m/min (b) 18 m/min (c) 36 m/min (d) 72 m/min For-2014 (IES, GATE & PSUs) IES 2010 IES 2010 With increasing cutting velocity, the total time for machining a component (a) Decreases ( )D ( ) (b) Increases (c) Remains unaffected (d) Fi d First decreases and then i d h increases S 996 IAS – 1996 The tool life increases with the (a) Increase in side cutting edge angle (b) D Decrease in side rake angle i id k l (c) Decrease in nose radius (d) Decrease in back rake angle Page 18 of 78 S IAS – 2000 Consider the following statements: The tool life is increased by 1. B ilt d f Built ‐up edge formation ti 2. Increasing cutting velocity 3. Increasing back rake angle up to certain value Which of these statements are correct? (a) 1 and 3 (b) 1 and 2 (c) d ( ) 2 and 3 (d) 1, 2 and 3 d S 99 IAS – 1995 In a single point turning operation with a cemented carbide and steel combination having a Taylor exponent of 0.25, if the cutting speed is halved, then the tool life will become (a) Half (b) Two times (c) Eight times ( ) (d) Sixteen times.
- 20. S 99 IAS – 1995 Assertion (A): An increase in depth of cut shortens the tool life. Reason(R): Increases in depth of cut gives rise to relatively small increase in tool temperature. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 2009 C i l IES 2009 Conventional Determine the optimum cutting speed for an operation on a Lathe machine using the following information: Tool change time: 3 min Tool T l regrinds ti i d time: 3 min i Machine running cost Rs.0.50 per min Depreciation of tool regrinds Rs. 5.0 The constants in the tool life equation are 60 and 0.2 GATE‐2009 Linked Answer Questions (1) S 2006 conventional i l IES – 2006 operation. An HSS tool is used for turning operation The tool life is 1 hr. when turning is carried at 30 m/min. Th t l lif will b reduced t 2.0 min if / i The tool life ill be d d to i the cutting speed is doubled. Find the suitable speed in RPM for turning 300 mm diameter so that tool life is 30 min. 3 S 200 C i l ESE‐2001 Conventional In a certain machining operation with a cutting speed of 50 m/min, tool life of 45 minutes was observed. Wh th cutting speed was i b d When the tti d increased d to 100 m/min, the tool life decreased to 10 min. Estimate the cutting speed for maximum p productivity if tool change time is 2 minutes. y g GATE‐2009 Linked Answer Questions (2) In a machining experiment, tool life was found to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 The exponent (n) and constant (k) of the Taylor's p ( ) ( ) y tool life equation are (a) n 0.5 and k 540 (a) n = 0.5 and k = 540 (b) n 1 and k 4860 (b) n= 1 and k=4860 (c) n = ‐1 and k = 0.74 (d) n‐0.5 and k=1.15 In a machining experiment, tool life was found to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 What is the percentage increase in tool life when p g the cutting speed is halved? (a) 50% (b) 200% (c) 300% (d) 400% For-2014 (IES, GATE & PSUs) Page 19 of 78 S 999 S 20 0 C i l ESE‐1999; IAS ‐2010 Conventional The following equation for tool life was obtained for HSS tool. A 60 min tool life was obtained using the following cutting condition VT0.13f0.6d0.3= C. v = 40 m/min, f = 0.25 mm, d = 2.0 mm. Calculate the effect on tool life if speed, feed and depth of cut are together increased by 25% and also if they are increased individually by 25%; where f = feed, d = depth of cut, v = speed. IAS – 2011 Main Determine the optimum speed for achieving maximum production rate in a machining operation. The data is as follows : Machining time/job = 6 min min. Tool life = 90 min. Taylor s Ta lor's equation constants C = 100, n = 0 00 0.5 Job handling time = 4 min./job Tool changing time = 9 min. l h i i i [10‐Marks] G 999 GATE‐1999 What is approximate percentage change is the life, t, of a tool with zero rake angle used in i orthogonal cutting when it clearance th l tti h its l o to 7o? angle, α, is changed from 10 (Hint: Flank wear rate is proportional to cot α (a) 30 % increase (b) 30% decrease 30%, (c) 70% increase (d) 70% decrease
- 21. G 200 GATE‐2005 S 2007 Contd… C d IAS – 200 g g A diagram related to machining economics with various cost components is given above. Match List I (Cost Element) with List II (Appropriate Curve) and select the correct answer using the code given below the Lists: List I List II (Cost Element) (Appropriate Curve) A. Machining cost 1. Curve‐l 2. Curve‐2 B. Tool cost C. Tool grinding cost 3. Curve‐3 D. Non productive cost 4. D Non‐productive cost 4 Curve 4 Curve‐4 5. Curve‐5 S 998 IES – 1998 The variable cost and production rate of a machining process against cutting speed are shown in the given figure. For efficient machining, the range of best cutting speed would be between (a) 1 and 3 (b) 1 and 5 (c) 2 and 4 ( ) (d) 3 and 5 S IES – 2000 The magnitude of the cutting speed for maximum profit rate must be (a) In between the speeds for minimum cost and maximum production rate (b) Hi h th th speed f maximum production rate Higher than the d for i d ti t (c) Below the speed for minimum cost (d) Equal to the speed for minimum cost For-2014 (IES, GATE & PSUs) S 999 IES – 1999 Consider the following approaches normally applied for the economic analysis of machining: 1. 1 Maximum production rate 2. Maximum profit criterion 3. Minimum cost criterion The correct sequence in ascending order of optimum q g p cutting speed obtained by these approaches is (a) 1, 2, 3 (b) 1, 3, 2 (c) 3, 2, 1 (d) 3, 1, 2 S 2004 IES – 200 g Consider the following statements: 1. As the cutting speed increases, the cost of production initially reduces, then after an optimum cutting speed it increases 2. As the cutting speed increases the cost of production also i l increases and after a critical value i reduces d f i i l l it d 3. Higher feed rate for the same cutting speed reduces cost of production 4. Higher feed rate for the same cutting speed increases the cost of production Which of the statements given above is/are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) 3 only Page 20 of 78 Contd Contd………. From previous slide Code:A (a) 3 (c) 3 B 2 1 C 4 4 D 5 2 (b) (d) A 4 4 B 1 2 C 3 3 D 2 5 IES 2011 The optimum cutting speed is one which should have: 1. Hi h metal removal rate High t l l t 2. High cutting tool life 3. Balance the metal removal rate and cutting tool life (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) ( ) 3 only S IES – 2002 In economics of machining, which one of the following costs remains constant? (a) Machining cost per piece (b) Tool changing cost per piece (c) Tool handling cost per piece ( ) (d) Tool cost per piece p p
- 22. S 2007 IAS – 200 Assertion (A): The optimum cutting speed for the minimum cost of machining may not maximize the profit. Reason (R): The profit also depends on rate of production. production (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES 2011 C ti l IES 2011 Conventional g Discuss the effects of the following elements on the machinability of steels: (i) Aluminium and silicon (ii) Sulphur and Selenium (iii) L d and Ti Lead d Tin (iv) Carbon and Manganese (v) Molybdenum and Vanadium [5 Marks] ISRO‐2007 Machinablity depends on (a) ( ) Microstructure, physical and mechanical h l d h l properties and composition of workpiece material. (b) Cutting forces ( ) yp (c) Type of chip p (d) Tool life For-2014 (IES, GATE & PSUs) S 99 IAS – 1997 In turning, the ratio of the optimum cutting speed for minimum cost and optimum cutting speed for maximum rate of production is always (a) Equal to 1 (b) I th In the range of 0.6 to 1 f 6 t (c) In the range of 0.1 to 0.6 (d) Greater than 1 S 992 IES – 1992 Ease of machining is primarily judged by (a) Life of cutting tool between sharpening (b) Ri idit f Rigidity of work ‐piece k i (c) Microstructure of tool material (d) Shape and dimensions of work S IES – 2003 ( ) y p Assertion (A): The machinability of steels improves by adding sulphur to obtain so called 'Free Machining Steels‘. Reason (R): Sulphur in steel forms manganese sulphide inclusion which helps to produce thin ribbon like continuous chip. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 21 of 78 IES 2012 IES ‐ The usual method of defining machinability of a material is by an index based on (a) Hardness of work material (b) Production rate of machined parts (c) Surface finish of machined surfaces ( ) (d) Tool life S 2007, 2009 IES – 200 2009 Consider the following: 1. Tool life 2. C tti f Cutting forces 3. Surface finish Which of the above is/are the machinability criterion/criteria? (a) 1, 2 and 3 (b) 1 and 3 only (c) 2 and 3 only (d) 2 only S IES – 2009 The elements which, added to steel, help in chip formation during machining are (a) Sulphur lead and phosphorous Sulphur, (b) Sulphur, lead and cobalt (c) Aluminium, lead and copper ( ) (d) Aluminium, titanium and copper pp
- 23. S 998 IES – 1998 Consider the following criteria in evaluating machinability: 1. 1 Surface finish 2 2. Type of chips 3. Tool life 4. Power consumption In modern high speed CNC machining with coated carbide tools, the correct sequence of these criteria in DECREASING order of their importance is ( ) (a) 1, 2, 4, 3 4 3 ( ) (b) 2, 1, 4, 3 4 3 (c) 1, 2, 3, 4 (d) 2, 1, 3, 4 S 99 IES – 1995 In low carbon steels, presence of small quantities sulphur improves (a) Weldability (b) Formability (c) Machinability (d) Hardenability IES‐1995 Consider the following work materials: C id th f ll i k t i l 1. Titanium 2. Mild steel 3. Stainless steel 4. Grey cast iron. The correct sequence of these materials in terms of increasing order of difficulty in machining is (a) 4 2 3 1 4,2,3,1 (b) 4 2 1 3 4,2, 1,3 (c) 2,4,3,1 (d) 2, 4, 1, 3. For-2014 (IES, GATE & PSUs) S 996 IES – 1996 Which of the following machinability? 1. 1 Smaller shear angle 2. Higher cutting forces 3. Longer tool life 4 4. Better surface finish. (a) 1 and 3 (b) 2 and 4 (c) 1 and 2 (d) 3 and 4 S 996 IES – 1996 indicate better S 992 IES – 1992 Machining of titanium is difficult due to (a) High thermal conductivity of titanium (b) Ch i l Chemical reaction between tool and work ti b t t l d k (c) Low tool‐chip contact area (d) None of the above G 2009 GATE‐2009 Friction at the tool chip interface can be Friction at the tool‐chip interface can be reduced by (a) decreasing the rake angle ( ) (b) increasing the depth of cut (c) Decreasing the cutting speed (d) increasing the cutting speed Page 22 of 78 Small amounts of which one of the following elements/pairs of elements is added to steel to increase its machinability? (a) Nickel (b) Sulphur and phosphorus (c) Silicon ( ) Sili (d) M Manganese and copper d S 996 IAS – 1996 Assertion (A): The machinability of a material can be measured as an absolute quantity. Reason (R): Machinability index indicates the case with which a material can be machined (a) Both A and R are individually true and R is the ( ) B th A d R i di id ll t d R i th correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES 2002 IES ‐ h The value of surface roughness 'h' obtained during the turning operating at a feed 'f' with a round nose tool having radius 'r' is given as r
- 24. IAS 1996 IAS ‐ IES 1999 IES ‐ Given that S = feed in mm/rev. and R = nose radius i mm, di in the maximum height of surface roughness Hmax produced by a single‐point turning tool is given by ( ) (a) S2/2R (b) S2/4R (c) S2/4R (d) S2/8R In turning operation, the feed could be doubled to increase the metal removal rate. To keep the same level of surface finish, the nose radius of the tool should be (a) Halved (b) Kept unchanged (c) doubled (d) Made four times GATE 2007 (PI) GATE – 2007 (PI) GATE 2005 GATE ‐ 30o A tool with Side Cutting Edge angle of and o is used for fine End Cutting Edge angle of 10 turning with a feed of 1 mm/rev Neglecting nose mm/rev. radius of the tool, the maximum (peak to valley) height f h i h of surface roughness produced will b f h d d ill be ( ) (a) 0.16 mm ( ) (b) 0.26 mm (c) 0.32 mm (d) 0.48 mm IES 2006 IES ‐ In the selection of optimal cutting conditions, the requirement of surface finish would put a limit on which of the following? (a) The maximum feed (b) Th maximum d th of cut The i depth f t (c) The maximum speed (d) The maximum number of passes Two tools P and Q have signatures 5 5 6 6 8 30 5°‐5°‐6°‐6°‐8°‐30°‐ 0 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively. They are used to turn components under the same machining conditions. If hp and hQ denote the peak‐ to valley to‐valley heights of surfaces produced by the tools P and Q, the ratio hp/hQ will be tan 8o + cot15o tan 8o + cot 30o tan15o + cot7o (c ) tan 30o + cot7o (a) A cutting tool has a radius of 1.8 mm. The feed rate for a theoretical surface roughness of Ra = 5 m is μ (a) 0 36 mm/rev 0.36 mm/rev (b) 0.187 mm/rev (c) 0.036 mm/rev ( ) (d) 0.0187 mm/rev 7 IES 1993 ISRO 2008 IES – 1993, ISRO‐2008 For achieving a specific surface finish in single point turning the most important factor to be controlled is (a) Depth of cut (b) Cutting speed (c) Feed ( ) F d (d) T l rake angle Tool k l tan15o + cot 8o tan 30o + cot 8o tan7o + cot15o (d ) tan7o + cot 30o (b) GATE 2010 (PI) GATE ‐2010 (PI) During turning of a low carbon steel bar with TiN coated carbide insert, one need to improve surface finish without sacrificing material removal rate. To achieve h f l l h improved surface finish, one should (a) decrease nose radius of the cutting tool and increase depth of cut (b) Increase nose radius of the cutting tool (c) Increase feed and decrease nose radius of the cutting tool For-2014 (IES, GATE & PSUs) GATE 1997 GATE ‐ (d) Increase depth of cut and increase feed Page 23 of 78 IAS 2009 Main IAS ‐2009 Main What are extreme‐pressure lubricants? [ 3 – marks] g pressures and rubbing action are g Where high p encountered, hydrodynamic lubrication cannot be maintained; so Extreme Pressure (EP) additives must be added to the l b i dd d h lubricant. EP l b i i i provided b a lubrication is id d by number of chemical components such as boron, phosphorus, sulfur, chlorine, phosphorus sulfur chlorine or combination of these these. The compounds are activated by the higher temperature resulting from extreme pressure As the temperature pressure. rises, EP molecules become reactive and release derivatives such as iron chloride or iron sulfide and forms a solid protective coating.
- 25. IES 2001 IES ‐ IES 2012 IES ‐ Ch‐3: Cutting Tools, Tool Life and Cutting Fluid Option p Q Q. No Option p Q Q. No Option p 1 B 12 C 23 A 2 A 13 A 24 C 3 A 14 A 25 C 4 D 15 B 26 B 5 D 16 B 27 B D 17 B 28 A 7 B 18 A 29 B 8 A 19 B 30 A 9 A 20 A 31 C 10 D 21 B 32 B 11 The most important function of the cutting fluid is to (a) Provide lubrication (b) Cool the tool and work piece (b) C l th t l d k i (c) Wash away the chips (d) Improve surface finish Q Q. No 6 Dry and compressed air is used as cutting fluid for machining (a) Steel (b) Aluminium (c) Cast iron (d) Brass C 22 B 33 C Ch‐4: Economics of Machining Operation Q. No Option Q. No C 6 B 2 3 B A 7 8 A C 4 C 9 A 5 A GATE‐1995 Option 1 A test specimen is stressed slightly beyond the Metal Forming yield point and then unloaded. Its yield strength (a) Decreases (b) ( ) Increases (c) Remains same By S K Mondal B S K M d l (d) Become equal to UTS IES‐2013 Statement (I): At higher strain rate and lower temperature structural steel tends to become brittle. Statement (II): At higher strain rate and lower temperature the yield strength of structural steel tends to increase increase. (a) Both Statement (I) and Statement (II) are individually true and S d Statement (II) i the correct explanation of is h l i f Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) ( ) (c) Statement (I) is true but Statement (II) is false () ( ) (d) Statement (I) is false but Statement (II) is true For-2014 (IES, GATE & PSUs) IES 2011 Lead, Assertion (A): Lead Zinc and Tin are always hot worked. Reason (R) : If th are worked i cold state R they k d in ld t t they cannot retain their mechanical properties. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT p the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 24 of 78 G 2003 GATE‐2003 Cold working of steel is defined as working (a) At its recrystallisation temperature (b) Ab Above it recrystallisation t its t lli ti temperature t (c) Below its recrystallisation temperature (d) At two thirds of the melting temperature of the metal
- 26. ISRO 2010 ISRO‐2010 G 2002 S O 20 2 GATE‐2002, ISRO‐2012 Hot rolling of mild steel is carried out (a) At recrystallisation temperature (b) B t Between 100°C t 150°C °C to °C (c) Below recrystallisation temperature (d) Above recrystallisation temperature Materials after cold working are subjected to following process to relieve stresses (a) Hot working S IES – 2006 Which one of the following is the process to refine the grains of metal after it has been distorted by hammering or cold working? (a) Annealing (b) Softening (c) Re‐crystallizing (d) N ( ) R t lli i Normalizing li i (b) Tempering (c) Normalizing (d) Annealing S 2004 IES – 200 Consider the following statements: In comparison to hot working, in cold working, 1. Hi h f Higher forces are required i d 2. No heating is required 3. Less ductility is required 4. Better surface finish is obtained Which of the statements given above are correct? (a) ( ) 1, 2 and 3 (b) 1, 2 and 4 d d (c) 1 and 3 (d) 2, 3 and 4 S IES – 2008 Cold forging results in improved quality due to which of the following? 1. 1 Better mechanical properties of the process process. 2. Unbroken grain flow. 3. Smoother finishes. 4 4. High pressure. g p Select the correct answer using the code given below: (a) 1 2 and 3 (b) 1 2 and 4 1, 1, (c) 2, 3 and 4 (d) 1, 3 and 4 For-2014 (IES, GATE & PSUs) S IES – 2009 Consider the following characteristics: 1. Porosity in the metal is largely eliminated. 2. St Strength i d th is decreased. d 3. Close tolerances cannot be maintained. Which of the above characteristics of hot working is/are correct? (a) 1 only (b) 3 only (c) 2 and 3 (d) 1 and 3 S 2004 IES – 200 Assertion (A): Cold working of metals results in increase of strength and hardness Reason (R): Cold working reduces the total number of dislocations per unit volume of the material (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 25 of 78 S IES – 2008 Consider the following statements: 1. Metal forming decreases harmful effects of impurities and improves mechanical strength strength. 2. Metal working process is a plastic deformation process. 3. Very intricate shapes can be produced by forging process as compared to casting process. Which of the statements given above are correct? g (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only S IES – 2003 Cold working produces the following effects: 1. Stresses are set up in the metal 2. G i structure gets di t t d Grain t t t distorted 3. Strength and hardness of the metal are decreased 4. Surface finish is reduced Which of these statements are correct? (a) 1and 2 (b) 1, 2 and 3 (c) ( ) 3 and 4 d (d) 1 and 4 d
- 27. S IES – 2000 Assertion (A): To obtain large deformations by cold working intermediate annealing is not required. Reason (R): Cold working is performed below the recrystallisation temperature of the work material. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 996 IES – 1996 Consider the following statements: When a metal or alloy is cold worked 1. It i It is worked below room temperature. k d b l t t 2. It is worked below recrystallisation temperature. 3. Its hardness and strength increase. 4. Its hardness increases but strength does not increase. Of these correct statements are (a) 1 and 4 (b) 1 and 3 (c) 2 and 3 (d) 2 and 4 ISRO‐2009 In the metal forming process, the stresses encountered are t d (a) Greater than yield strength but less than ultimate strength l h (b) Less than yield strength of the material (c) Greater than the ultimate strength of the material (d) Less than the elastic limit S IES – 2006 Assertion (A): In case of hot working of metals, the temperature at which the process is finally stopped should not be above the recrystallisation temperature. y p Reason (R): If the process is stopped above the recrystallisation temperature, grain growth will take y p , g g place again and spoil the attained structure. ( ) (a) Both A and R are individually true and R is the correct y explanation of A ( ) (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true S 996 IAS – 1996 S 2004 IAS – 200 For mild steel, the hot forging temperature range is (a) 4000C to 6000C (b) 7000C t 9000C to 0C to 12000C (c) 1000 (d) 13000Cto 15000C Assertion (A): Hot working does not produce strain hardening. Reason (R): Hot working is done above the re‐ Reason (R): Hot working is done above the re crystallization temperature. (a) Both A and R are individually true and R is the ( ) B th A d R i di id ll t d R i th correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES, GATE & PSUs) Page 26 of 78 S 99 IES – 1997 In metals subjected to cold working, strain hardening effect is due to (a) Slip mechanism (b) Twining mechanism (c) Dislocation mechanism ( ) (d) Fracture mechanism S 992 IES – 1992 Specify the sequence correctly (a) Grain growth, recrystallisation, stress relief (b) St Stress relief, grain growth, recrystallisation li f i th t lli ti (c) Stress relief, recrystallisation, grain growth (d) Grain growth, stress relief, recrystallisation S 2002 IAS‐2002 Assertion (A): There is good grain refinement in hot working. Reason (R): In hot working physical properties are generally improved. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
- 28. GATE 2013 GATE‐2013 S 2008 IES‐2008 Which one of the following is correct? Malleability is the property by which a metal or alloy can be plastically deformed by applying (a) Tensile stress (b) Bending stress (c) Shear stress (d) Compressive stress ISRO‐2006 In the t t I a rolling process, th state of stress of th lli f t f the material undergoing deformation is g g Which of the following processes would produce (a) pure compression strongest components? (b) pure shear (a) Hot rolling (c) compression and shear (b) ( ) Extrusion (d) tension and shear (c) Cold rolling (d) Forging ISRO‐2009 Ring rolling is used (a) To decrease the thickness and increase diameter (b) To increase the thickness of a ring (c) For producing a seamless tube (d) For producing large cylinder IFS – 2010 Calculate the neutral plane to roll 250 mm wide annealed copper strip from 2 5 mm to 2 0 mm 2.5 2.0 thickness with 350 mm diameter steel rolls. Take µ = 0.05 0 05 and σ’o =180 MPa σ MPa. [10‐marks] For-2014 (IES, GATE & PSUs) S IES – 2006 Which one of the following is a continuous bending process in which opposing rolls are used to produce long sections of formed shapes from coil or strip stock? (a) Stretch forming (b) Roll forming (c) Roll bending (d) Spinning ( ) GATE – 2009 (PI) Anisotropy in rolled components is caused by (a) changes in dimensions (b) scale formation (c) closure of defects (d) grain orientation i i i G 2008 GATE‐2008 In a single pass rolling operation, a 20 mm thick plate with plate width of 100 mm, is reduced to 18 mm. The roller radius is 250 mm and rotational speed is 10 rpm. The average flow stress for the plate material is 300 MPa. The power required for the rolling operation in kW is closest to (a) 15 2 15.2 (b) 18.2 (c) 30.4 (d) 45.6 45 Page 27 of 78 G 200 GATE‐2007 The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degree will be (a) 5.936 (b) 7.936 6 (c) 8.936 (d) 9.936
- 29. G 200 GATE‐2004 In a rolling process, sheet of 25 mm thickness is rolled to 20 mm thickness. Roll is of diameter 600 mm and it rotates at 100 rpm. The roll strip contact length will be (a) 5 mm (b) 39 mm (c) 78 mm (d) 120 mm G 998 GATE‐1998 cross section A strip with a cross‐section 150 mm x 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rolls. The angle subtended by the deformation zone at the roll centre is (in radian) (a) 0 01 (b) 0 02 0.01 0.02 (c) 0.03 (d) 0.06 GATE – 2012 Same Q in GATE – 2012 (PI) In a single pass rolling process using 410 mm diameter steel rollers a strip of width 140 mm and rollers, thickness 8 mm undergoes 10% reduction of thickness. The angle of bite in radians is (a) 0.006 (c) 0 062 0.062 G 2006 GATE‐2006 A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any change in its width. The friction coefficient at the work‐roll work roll interface is 0.1. The minimum possible thickness of the sheet that can be produced in a single pass is (a) 1.0 mm (b) 1.5 mm (c) ( ) 2.5 mm (d) 3.7 mm S IES – 2003 ( ) g g Assertion (A): While rolling metal sheet in rolling mill, the edges are sometimes not straight and flat but are wavy. Reason (R): Non‐uniform mechanical properties of the flat material rolled out result in waviness of the edges. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES, GATE & PSUs) (b) 0.031 (d) 0 600 0.600 ( ) GATE‐1992(PI) GATE – 2011 (PI) The thickness of a plate is reduced from 30 mm to 10 mm by successive cold rolling passes using identical rolls of diameter 600 mm Assume that mm. there is no change in width. If the coefficient of friction between the rolls and the work piece is 0 1 0.1, the minimum number of passes required is (a) ( )3 (b) 4 (c) 6 (d) 7 S IES – 2002 In rolling a strip between two the neutral point in the arc depend on (a) Amount of reduction (b) (c) Coefficient f friction (d) ( ) C ffi i t of f i ti rolls, the position of of contact does not Diameter of the rolls Material f the M t i l of th rolls ll Page 28 of 78 If the elongation f factor d during rolling of an ingot f h l ll f is 1 22 The minimum number of passes needed to 1.22. produce a section 250 mm x 250 mm from an ingot of 750 mm x 750 mm are (a) 8 (b) 9 (c) ( ) 10 (d) 17 S 2001 IES – 200 g p Which of the following assumptions are correct for cold rolling? 1. The material is plastic. p 2. The arc of contact is circular with a radius greater than the radius of the roll. 3. Coefficient of friction is constant over the arc of g contact and acts in one direction throughout the arc of contact. g g Select the correct answer using the codes given below: Codes: (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3
- 30. S 2001 IES – 200 A strip is to be rolled from a thickness of 30 mm to 15 mm using a two‐high mill having rolls of diameter 300 mm. The coefficient of friction for unaided bite should nearly be (a) 0 35 0.35 (b) 0 5 0.5 (c) 0.25 (d) 0.07 GATE ‐2008(PI) In thickness of a strip i reduced f I a rolling process, thi k lli f t i is d d from 4 ( ) GATE‐1990 (PI) mm to 3 mm using 300 mm diameter rolls rotating at 100 g3 g While rolling a strip the peripheral velocity of the rpm. The velocity of the strip in (m/s) at the neutral roll is ….A…..than the entry velocity of the strip point is and is ……B …..the exit velocity of the strip. (a) ( ) 1.57 (b) 3.14 (c) ( ) 47.10 (d) 94.20 (a) less th / ( )l than/greater l t less (b) Greater than/less than S 2000, GATE‐2010(PI) 20 0( ) IES – 2000 G In the rolling process, roll separating force can be decreased by (a) Reducing the roll diameter (b) Increasing the roll diameter (c) Providing back‐up rolls ( ) (d) Increasing the friction between the rolls and the g metal S 993 G 989( ) IES – 1993, GATE‐1989(PI) The blank diameter used in thread rolling will be (a) Equal to minor diameter of the thread (b) E Equal t pitch di l to it h diameter of th th d t f the thread (c) A little large than the minor diameter of the thread (d) A little larger than the pitch diameter of the thread For-2014 (IES, GATE & PSUs) S 999 IES – 1999 Assertion (A): In a two high rolling mill there is a limit to the possible reduction in thickness in one pass. Reason (R): The reduction possible in the second pass is less than that in the first pass. pass is less than that in the first pass (a) Both A and R are individually true and R is the correct explanation of A t l ti f A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 992 G 992( ) IES – 1992, GATE‐1992(PI) Thread rolling is restricted to (a) Ferrous materials (b) D til t i l Ductile materials (c) Hard materials (d) None of the above Page 29 of 78 S 993 IES – 1993 In order to get uniform thickness of the plate by rolling process, one provides (a) Camber on the rolls (b) Offset on the rolls (c) Hardening of the rolls ( ) (d) Antifriction bearings g S 2004 IAS – 200 Assertion (A): Rolling requires high friction which increases forces and power consumption. Reason (R): To prevent damage to the surface of the rolled products, lubricants should be used. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
- 31. S 2001 IAS – 200 Consider the following characteristics of rolling process: 1. 1 Shows work hardening effect 2. Surface finish is not good 3. Heavy reduction in areas can be obtained Which of these characteristics are associated with hot rolling? (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 S 2007 IAS – 200 g Match List I with List II and select the correct answer using the code given below the Lists: List I List II (Type of Rolling Mill) ( ) (Characteristic) ( ) A. Two high non‐reversing mills 1. Middle roll rotates by friction B. Th B Three hi h mills high ill 2. B small working roll, power By ll ki ll for rolling is reduced C. C Four high mills 3. 3 Rolls of equal size are rotated only in one direction D. Cluster mills 4. Diameter of working roll is g very small Code:A B C D A B C D (a) ( ) 3 4 2 1 (b) 2 1 3 4 (c) 2 4 3 1 (d) 3 1 2 4 S IAS – 2000 Rolling very thin strips of mild steel requires (a) Large diameter rolls (b) S ll di Small diameter rolls t ll (c) High speed rolling (d) Rolling without a lubricant S IAS – 2003 In one setting of rolls in a 3 high rolling mill, one 3‐high gets (a) One reduction in thickness (b) Two reductions in thickness (c) Three reductions in thickness ( ) (d) Two or three reductions in thickness depending p g upon the setting S 998 IAS – 1998 Match List ‐ I (products) with List ‐ II (processes) and select the correct answer using the codes given below the lists: List – I List ‐II A. M.S. A M S angles and channels l d h l 1. Welding W ldi B. Carburetors 2. Forging C. Roof trusses 3. Casting D. Gear wheels 4. Rolling Codes:A B C D A B C D (a) ( ) 1 2 3 4 (b) 4 3 2 1 (c) 1 2 4 3 (d) 4 3 1 2 S 2007 IAS – 200 Consider the following statements: Roll forces in rolling can be reduced by 1. R d i f i ti Reducing friction 2. Using large diameter rolls to increase the contact area. 3 3. Taking smaller reductions per pass to reduce the g p p contact area. Which of the statements given above are correct? (a) 1 and 2 only (b) 2 and 3 only (c) ( ) 1 and 3 only d l (d) 1, 2 and 3 d Rolling Ch‐14 GATE 2011 The maximum possible draft in cold rolling of sheet increases with the (a) increase i coefficient of f i ti ( )i in ffi i t f friction (b) decrease in coefficient of friction (c) decrease in roll radius (d) increase in roll velocity Q. No Option 1 2 C B 3 D 4 5 D A A 7 B 8 9 D C 10 11 For-2014 (IES, GATE & PSUs) 6 C B 12 C Forging Page 30 of 78 By S K Mondal
- 32. IES‐2013 G 20 0 ( ) GATE ‐2010 (PI) ( ) GATE‐1989(PI) Hot die t l forging, H t di steel, used f l d for large solid di i d lid dies in drop f i At the last hammer stroke the excess material from should necessarily have y the finishing cavity of a forging die is pushed (a) high strength and high copper content into…………….. (b) high hardness and low hardenability (c) high toughness and low thermal conductivity (d) high hardness and high thermal conductivity IFS‐2011 Statement (I): The dies used in the forging process are p made in pair. Statement (II): The material is pressed between two surfaces and the compression force applied, gives it a shape. (a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true b t St t t but Statement (II) i not th correct explanation of t is t the t l ti f Statement (I) (c) ( ) Statement ( ) is true b Statement ( ) is f l (I) but (II) false (d) Statement (I) is false but Statement (II) is true IAS‐2011 Main IES ‐ 2007 What advantages does press forging have over drop Compare Smith forging, drop forging, press Sometimes the parting plane between two forging forging ? Why are pure metals more easily cold worked forging and upset forging. Mention three points dies is not a horizontal plane, give the main reason than ll th alloys ? for each. for this design aspect, why is parting plane [5 marks] [5‐marks] [10 – Marks] [ M k ] provided, provided in closed die forging? [ [ 2 marks] ] G 200 GATE‐2007 open die In open‐die forging, a disc of diameter 200 mm and height 60 mm is compressed without any barreling effect. The final diameter of the disc is 400 mm. The true strain is (a) 1 986 1.986 (b) 1 686 1.686 (c) 1.386 (d) 0.602 For-2014 (IES, GATE & PSUs) GATE‐1992, ISRO‐2012 G 20 2 GATE‐2012 Same Q GATE ‐2012 (PI) The true strain for a low carbon steel bar which is doubled in length by forging is (a) 0.307 (b) 0.5 (c) 0.693 (d) 1.0 A solid cylinder of diameter 100 mm and height 50 mm A lid li d f di t d h i ht Page 31 of 78 is forged between two frictionless flat dies to a height of g g 25 mm. The percentage change in diameter is (a) 0 (b) 2.07 (c) 20.7 (d) 41.4
- 33. G 99 GATE‐1994 Match 4 correct pairs between List I and List II for the questions List I gives a number of processes and List II gives a number of products List I List II (a) Investment casting ( ) I t t ti 1. Turbine t T bi rotors (b) Die casting 2. Turbine blades (c) Centrifugal casting 3. Connecting rods (d) Drop forging 4. Galvanized iron pipe (e) Extrusion 5. Cast iron pipes (f) Sh ll moulding Shell ldi 6. 6 Carburettor body C b tt b d S 2005 IES – 200 Consider the following statements: 1. Forging reduces the grain size of the metal, which results in a decrease in strength and toughness toughness. 2. Forged components can be provided with thin sections, without reducing th strength. ti ith t d i the t th Which of the statements given above is/are correct? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 IES 2013 IES‐2013 In the forging process: 1. The metal structure is refined 2. Original unidirectional fibers are distorted. 3. Poor reliability, as flaws are always there due to intense working 4. Part are shaped by plastic deformation of material 4 Part are shaped by plastic deformation of material (a) 1, 2 and 3 (b) 1, 3 and 4 (c) 1, 2 and 4For-2014 (IES, (d) 2, 3 and 4 GATE & PSUs) G 998 GATE‐1998 (A) (B) (C) (D) List I List II Aluminium brake shoe (1) Deep drawing Plastic t bottle Pl ti water b ttl (2) Blow ( ) Bl moulding ldi Stainless steel cups (3) Sand casting Soft drink can (aluminium) (4) Centrifugal casting (5) Impact extrusion (6) U t f i Upset forging S 996 IES – 1996 Which one of the following is an advantage of forging? (a) Good surface finish (b) Low tooling cost (c) Close tolerance ( ) (d) Improved physical property. p p y p p y IES 2012 IES ‐ () Statement (I): It is difficult to maintain close tolerance in normal forging operation. ( ) g g p p Statement (II): Forging is workable for simple shapes and has limitation for parts having undercuts. ( ) (a) Both Statement (I) and Statement (II) are () ( ) individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Page 32 of 78 S IES – 2006 Assertion (A): Forging dies are provided with taper or draft angles on vertical surfaces. Reason (R): It facilitates complete filling of die cavity and favourable grain flow. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES 2012 IES ‐ Which of the following statements is correct for forging? (a) Forgeability is property of forging tool, by which forging can be done easily easily. (b) Forgeability decreases with temperature upto lower critical t iti l temperature. t (c) Certain mechanical properties of the material are influenced by forging. ( ) (d) Pure metals have good malleability, therefore, poor g y p forging properties. S 993 G 99 ( ) IES – 1993, GATE‐1994(PI) Which one of the following manufacturing processes requires the provision of ‘gutters’? (a) Closed die forging (b) Centrifugal casting (c) Investment casting ( ) (d) Impact extrusion p
- 34. S 99 IES – 1997 ( ) p g g p Assertion (A): In drop forging besides the provision for flash, provision is also to be made in the forging die for additional space called gutter. Reason (R): The gutter helps to restrict the outward flow of metal thereby helping to fill thin ribs and bases in the upper die. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S IES – 2002 Consider the following steps involved in hammer forging a connecting rod from bar stock: 1. 1 Blocking 2 2. Trimming 3. Finishing 4. Fullering 5. Edging Which of the following is the correct sequence of operations? (a) 1, 4, 3, 2 and 5 (b) 4, 5, 1, 3 and 2 (c) 5, 4, 3, 2 and 1 (d) 5, 1, 4, 2 and 3 d S 2005 IES – 200 The process of removing the burrs or flash from a forged component in drop forging is called: (a) Swaging (b) Perforating (c) Trimming (d) Fettling For-2014 (IES, GATE & PSUs) S 2004 IES – 200 ( y ) Match List I (Different systems) with List II (Associated terminology) and select the correct answer using the codes given below the Lists: List I List II A. Riveted Joints J 1. Nipping pp g B. Welded joints 2. Angular movement C. C Leaf springs 3. 3 Fullering D. Knuckle joints 4. Fusion A B C D A B C D (a) 3 2 1 4 (b) 1 2 3 4 (c) ( ) 1 4 3 2 (d) 3 4 1 2 S 999 IES – 1999 Consider the following operations involved in forging a hexagonal bolt from a round bar stock, whose diameter is equal to the bolt diameter: 1. Flattening 2. Upsetting 3. S Swaging i 4. Cambering C b i The correct sequence of these operations is (a) 1, 2, 3, 4 (b) 2, 3, 4, 1 (c) 2, 1, 3, 4 (d) 3, 2, 1, 4 IES 2011 Which of the following processes belong to forging operation ? 1. F ll i Fullering 2. Swaging 3. Welding (a) 1 and 2 only (b) 2 and 3 only (c) ( ) 1 and 3 only d l (b) 1, 2 and 3 only Page 33 of 78 S IES – 2003 A forging method for reducing the diameter of a bar and in the process making it longer is termed as (a) Fullering (b) Punching (c) Upsetting (d) Extruding S IES – 2003 Consider the following steps in forging a connecting rod from the bar stock: 1. 1 Blocking 2 2. Trimming 3. Finishing 4. Edging Select the correct sequence of these operations using the codes given below: Codes: (a) 1‐2‐3‐4 1234 (b) 2‐3‐4‐1 2341 (c) 3‐4‐1‐2 (d) 4‐1‐3‐2 S IES – 2008 The balls of the ball bearings are manufactured from steel rods. The operations involved are: 1. 1 Ground 2. Hot forged on hammers 3. Heat treated 4 4. Polished What is the correct sequence of the above operations from start? (a) 3‐2‐4‐1 (b) 3‐2‐1‐4 (c) ( ) 2‐3‐1‐4 (d) 2‐3‐4‐1
- 35. S 2001 IES – 200 In the forging operation, fullering is done to (a) Draw out the material (b) B d th t i l Bend the material (c) Upset the material (d) Extruding the material S IES – 2008 Match List‐I with List‐II and select the correct answer using the code given b l below th li t the lists: i List‐I (Forging Technique) List‐II (Process) A. Smith Forging 1. Material is only upset to get the desired shape B. Drop Forging 2. Carried out manually open dies C. Press Forging 3. Done in closed impression dies by hammers in blows D. Machine Forging 4. Done in closed impression dies by continuous squeezing force Code: A B C D (a) 2 3 4 1 (b) 4 3 2 1 (c) 2 1 4 3 (d) 4 1 2 3 IES‐2013 S i Statement (I) I hi h velocity f (I): In high l i forming process, hi h high energy can be transferred to metal with relatively small weight. i ht Statement (II): The kinetic energy is the function of mass and velocity. (a) ( ) Both Statement (I) and Statement (II) are individually () ( ) y true and Statement (II) is the correct explanation of () Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement For-2014 (IES, GATE & (II) is true (I) is false but Statement PSUs) IES 2011 Consider the following statements : 1. Any metal will require some time to undergo complete plastic d f l ti deformation particularly if d f ti ti l l deforming metal h i t l has to fill cavities and corners of small radii. 2. For larger work piece of metals that can retain toughness at forging temperature it is preferable to use forge press rather than forge hammer. (a) 1 and 2 are correct and 2 is the reason for 1 (b) 1 and 2 are correct and 1 is the reason for 2 (c) 1 and 2 are correct but unrelated (d) 1 only correct S 998 IES – 1998 Which one of the following processes is most commonly used for the forging of bolt heads of hexagonal shape? (a) Closed die drop forging (b) O Open di upset f i die t forging (c) Close die press forging (d) Open die progressive forging IES‐2013 Statement (I): In power forging energy is provided by compressed air or oil pressure or gravity. Statement (II): The capacity of the hammer is given by the total weight which the falling pans weigh weight, weigh. (a) Both Statement (I) and Statement (II) are individually true and St t t d Statement (II) i th correct explanation of t is the t l ti f Statement (I) (b) Both Statement ( ) and Statement ( ) are individually h (I) d (II) d d ll true but Statement (II) is not the correct explanation of Statement ( ) (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Page 34 of 78 S 2005 IES – 200 ( yp g g) ( p ) Match List I (Type of Forging) with List II (Operation) and select the correct answer using the code given below the Lists: List I List II A. Drop Forging 1. Metal is gripped in the dies and pressure i applied on the h is li d h heated end d d B. Press Forging 2. Squeezing action C. Upset F i C U Forging 3. M l i placed b Metal is l d between rollers and ll d pushed D. D Roll Forging 4 Repeated hammer blo s 4. blows A B C D A B C D (a) ( ) 4 1 2 3 (b) 3 2 1 4 (c) 4 2 1 3 (d) 3 1 2 4 S 99 S O 20 0 IES – 1994, ISRO‐2010 In drop forging, forging is done by dropping (a) The work piece at high velocity (b) Th h The hammer at hi h velocity. t high l it (c) The die with hammer at high velocity (d) a weight on hammer to produce the requisite p impact. S IES – 2009 Match List‐I with List‐II and select the correct answer using the code given b l below the Lists: h d h List‐I List‐II (Article) (Processing Method) A. Disposable coffee cups 1. Rotomoulding B. Large water tanks g 2. Expandable bead moulding p g C. Plastic sheets 3. Thermoforming D. Cushion pads 4. Blow moulding 5. C l d i Calendaring Code: (a) A B C D (b) A B C D 3 5 1 2 4 5 1 2 ( ) (c) A B C D ( ) (d) A B C D 4 3 3 1 3 1 5 2
- 36. S IAS – 2003 ( g g p ) ( Match List I (Forging Operation) with List II (View of the Forging Operation) and select the correct answer using the codes given below the lists: List‐I List‐II (Forging Operation) (View of the Forging Operation) (A) Edging 1. 2. (B) Fullering (C) Drawing 3. 4. g g (D) Swaging Codes:A B C D A B C D ( ) (a) 4 3 2 1 ( ) (b) 2 1 4 3 (c) 4 1 2 3 (d) 2 3 4 1 Click to see file Page 4 – 5 ‐6 S 2001 IAS – 200 Match List I (Forging operations) with List II (Descriptions) and select the correct answer using the codes given b l below d l h h d the Lists: List I List II A. Flattening 1. Thickness is reduced continuously at different sections along length B. D B Drawing i 2. Metal is displaced M l i di l d away f from centre, reducing thickness in middle and increasing length C. Fullering 3. Rod is pulled through a die D. Wire drawing 4. Pressure a workpiece between two flat dies Codes:A B C D A B C D ( ) (a) 3 2 1 4 ( ) (b) 4 1 2 3 (c) 3 1 2 4 (d) 4 2 1 3 S 998 IAS – 1998 The forging defect due to hindrance to smooth flow of metal in the component called 'Lap' occurs because (a) The corner radius provided is too large (b) Th corner radius provided i t small The di id d is too ll (c) Draft is not provided (d) The shrinkage allowance is inadequate S IAS – 2002 Consider the following statements related to forging: 1. 1 Flash is excess material added to stock which flows around parting line. 2. Fl h h l i filli f thi ib d b Flash helps in filling of thin ribs and bosses in upper i die. 3. Amount of flash depends upon forging force. Which of the above statements are correct? (a) 1, 2 and 3 (b) 1 and 2 (c) 1 and 3 (d) 2 and 3 S IAS – 2000 Drop forging is used to produce (a) Small components (b) L Large components t (c) Identical Components in large numbers (d) Medium‐size components IES 2011 Assertion (A) : Hot tears occur during forging because of inclusions in the blank material Reason (R) : Bonding between the inclusions and the parent material is through physical g and chemical bonding. (a) Both A and R are individually true and R is the p correct explanation of A (b) Both A and R are individually true but R is NOT p the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES‐2013 Consider the following statements pertaining to the open‐die forging of a cylindrical specimen between two flat dies: 1. Lubricated specimens show more surface movement than un‐lubricated ones. h l bi d 2. Lubricated specimens show less surface movement than un‐lubricated ones. 3 3. Lubricated specimens show more barrelling than un‐ p g lubricated ones. 4. Lubricated specimens shows less barrelling than un‐ un lubricated ones. Which of these statements are correct? (a) 1 and 3 For-2014 4 (c) GATE3& PSUs) 2 and 4 (b) 1 and (IES, 2 and (d) ( ) GATE ‐2008 (PI) Match the following Group ‐1 P . Wrinkling Q. Centre burst R. Barrelling g S. Cold shut Group‐2 1. Upsetting 2. Deep drawing 3 3. Extrusion 4. Closed die forging (a) P – 2, Q – 3, R – 4, S‐1 (c) P – 2, Q – 3, R – 1, S‐4 (c) P 2 Q 3 R 1 S 4 (b) P – 3, Q – 4, R – 1, S‐2 (d) P – 2, Q – 4, R – 3, S‐1 (d) P 2 Q 4 R 3 S 1 Page 35 of 78 IES 2012 IES ‐ Assumptions adopted in the analysis of open die forging are 1. 1 Forging force attains maximum value at the middle of the operation. 2. C ffi i t of f i ti i constant b t Coefficient f friction is t t between work piece k i and die 2. Stress in the vertical (Y‐direction) is zero. ( ) (a) 1 and 2 ( ) (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3
- 37. IES 2007 C ti l IES – 2007 Conventional S 2005 Conventional i l IES – 200 C A strip of l d with i iti l di t i f lead ith initial dimensions 24 mm x 24 i mm x 150 mm is forged between two flat dies to a 5 g final size of 6 mm x 96 mm x 150 mm. If the coefficient of friction is 0.25, determine the maximum f i f i forging force. Th average yield stress of The i ld t f lead in tension is 7 N/mm2 A cylinder of height 60 mm and diameter 100 mm is forged at room temperature between two flat dies. Find the die load at the end of compression to a height 30 mm, using slab method of analysis. The yield strength of the work material is given as 120 N/mm2 and the coefficient of friction is 0.05. Assume that volume is constant after deformation. There is no sticking. Also find mean die pressure. [20‐Marks] IES 2006 C ti l IES – 2006 ‐ Conventional A certain disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. If the co‐efficient of friction between the job and co efficient die is 0.25, determine the maximum forging force. The average shear yield stress of lead can be taken as 4 N/mm2. [10 – Marks] [10] GATE‐1987 P ti P bl 1 Practice Problem‐1 Ch‐15: Forging Q. No Option A 6 B 2 3 A A 7 8 C C A 9 C 5 occurs near the……(Centre/ends) Option 1 4 In forging operation the sticking friction condition Q. No B A strip of metal with initial dimensions 24 mm x 24 mm x 150 mm is forged between two flat dies to a final size of 6 mm x 96 mm x 150 mm. If the coefficient of friction is 0.05, determine the maximum forging force. Take the average yield strength in tension is 7 N/mm2 [Ans. 178.24 kN] P ti P bl 2 Practice Problem‐2 P ti P bl 3 Practice Problem‐3 P ti P bl 4 Practice Problem‐4 A circular disc of 200 mm in diameter and 100 mm in A cylindrical specimen 150 mm in diameter and 100 mm A circular disc of 200 mm in diameter and 70 mm in height is compressed between two flat dies to a height of in height is upsetted by open die forging to a height of 50 height is forged to 40 mm in height. Coefficient of 50 mm. Coefficient of friction is 0.1 and average yield mm. Coefficient of friction is 0.2 and flow curve friction is 0.05. The flow curve equation of the material strength in compression is 230 MPa. Determine the equation is σ f = 1030ε 0.17 MPa . Calculate the maximum is given by σ f = 200(0.01 + ε ) 0.41 MP MPa . Determine maximum [Ans. 405 MPa] For-2014 (IES, GATE & PSUs) forging force. forging load, mean die pressure and maximum pressure. [Ans. 46.26 MN] maximum die pressure pressure. [ Ans. 9.771 MN, 178 MPa, 221 MPa] [Hint. First calculate true strain ε and put the value in the equation σ f = 1030ε 0.17 =σy ] Page 36 of 78 [Hint. First calculate true strain ε and put the value in the equation σ f = 200(0.01 + ε ) 0.41 = σ y ]
- 38. Practice Problem ‐5 {GATE‐2010 (PI)} Practice Problem 5 {GATE 2010 (PI)} During open die forging process using two flat and parallel dies, a solid circular steel disc of initial radius (R IN ) 200 mm and initial lid i l t l di f i iti l di d i iti l height (H IN ) 50 mm attains a height (H FN ) of 30 mm and radius of R FN . Practice Problem ‐5 {GATE‐2010 (PI)} Extrusion & Drawing iii.In the region 0 ≤ r ≤ R SS ,sticking condition prevails The value of R SS (in mm), where sticking condition changes to sliding Along the die-disc interfaces. ⎛ ⎞ i. the coefficient of friction (μ ) is: μ = 0.35 ⎜ 1 + e ⎟ ⎜ ⎟ ⎝ ⎠ ii. in h ii i the region R ss ≤ r ≤ RFN ,sliding friction prevails, and i lidi f i i il d R − IN RFN 2μ Contd……. friction, is (a) 241.76 (b) 254.55 (c) 265.45 (d) 278.20 ( RFN − r ) p = 3K H FN Ke and τ = μ p, d where p and τ are the normal and shear stresses, respectively; K is the shear yield strength of steel and r is the radial distance of any point (contd ........) IAS‐2010 Main How are By S K Mondal B S K M d l metal tooth‐paste IES 2009 Conventional tubes made ( ) GATE‐1994(PI) A moving mandrel is used in commercially ? Draw the tools configuration with (a) Wire drawing [30‐Marks] IES – 2011 Conventional 12.5 A 12 5 mm diameter rod is to be reduced to 10 mm diameter by drawing in a single pass at a speed of 100 / g g m/min. Assuming a semi die angle of 5o and coefficient of friction between the die and steel rod as 0.15, calculate: (i) The power required in drawing ( ) (ii) Maximum possible reduction in diameter of the rod p (iii) If the rod is subjected to a back pressure of 50 / N/mm2 , what would be the draw stress and maximum possible reduction ? 4 / Take stress of the work material as 400 N/mm2 . [15 Marks] For-2014 (IES, GATE & PSUs) GATE – 2011 (PI) Common Data‐S1 GATE 2011 (PI) Common Data S1 In a multi‐pass drawing operation, a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section by drawing it successively through a series of seven dies of decreasing exit diameter. During each of these drawing operations, the reduction in cross‐sectional area is 35%. The yield strength of the material is 200 MPa. Ignore strain hardening. The total true strain applied and the final length (in ), p y, mm), respectively, are (a) 2.45 and 8 17 (b) 2.45 and 345 (c) 3 02 and 2043 3.02 (d) 3 02 and 3330 3.02 Page 37 of 78 (b) Tube drawing (c) Metal Cutting the help of a neat sketch. (d) Forging GATE – 2011 (PI) Common Data‐S2 GATE 2011 (PI) Common Data S2 In a multi‐pass drawing operation, a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section by drawing it successively through a series of seven dies of decreasing exit diameter. During each of these drawing operations, the reduction in cross‐sectional area is 35%. The yield strength of the material is 200 MPa. Ignore strain hardening. Neglecting friction and redundant work, the force (in ) q g g , kN) required for drawing the bar through the first die, is (a) 15.71 (b) 10.21 (c) 6.77 (c) 6 77 (d) 4.39 (d) 4 39
- 39. E l Example Calculate the drawing load required to obtain 30% reduction in area on a 12 mm diameter copper wire. The following data is given Calculate the C l l t th power of th electric motor if th f the l t i t the drawing speed is 2.3 m/s. Take efficiency of motor is 98%. 8% G 200 G 200 ( ) GATE‐2001, GATE ‐2007 (PI) For rigid perfectly plastic work material, negligible perfectly‐plastic interface friction and no redundant work, the theoretically maximum possible reduction in the wire drawing operation is (a) 0 36 0.36 (b) 0 63 0.63 (c) 1.00 (d) 2.72 20 0 JWM 2010 GATE ‐2008 (PI) Linked S‐1 GATE 2008 (PI) Linked S 1 GATE ‐2008 (PI) Linked S‐2 GATE 2008 (PI) Linked S 2 A 10 mm diameter annealed steel wire is drawn through a die at a speed of 0.5 m/s to reduce the diameter by a die at a speed of 0.5 m/s to reduce the diameter by 20%. The yield stress of the material is 800 MPa. 20%. The yield stress of the material is 800 MPa. Neglecting f i ti N l ti friction and strain h d i d t i hardening, th stress the t The Th power required f th d i d for the drawing process (i kW) i i (in is required for drawing (in MPa) is q g( ) (a) 8 97 8.97 G 2003 GATE‐2003 (c) 1287.5 GATE – 2009 (PI) extrusion process, Using direct e trusion process a round billet of 100 mm length and 50 mm diameter is extruded. Considering an ideal deformation process (no friction and no redundant work), extrusion ratio 4, and average fl k) i i d flow stress of f material 300 MPa, the pressure (in MPa) on the ram will 3 , p ( ) (a) 416 (b) 624 (b) 14 0 14.0 (c) 17 95 17.95 (d) 28 0 28.0 (d) 2575.0 be For-2014 (IES, GATE & PSUs) In a wire drawing operation, diameter of a steel wire is reduced from 10 mm to 8 mm. The mean flow stress of the material is 400 MPa. The ideal force required for drawing (ignoring friction and redundant work) is (a) 4.48 kN (b) 8.97 kN (c) ( ) 20.11 kN (d) 31.41 kN A 10 mm diameter annealed steel wire is drawn through (a) 178.5 (b) 357.0 A brass billet is to be extruded from its initial diameter of 100 mm to a final diameter of 50 mm. The working temperature of 700°C and the 700 C extrusion constant is 250 MPa. The force required for extrusion is (a) 5.44 MN (b) 2.72 MN (c) ( ) 1.36 MN 6 (d) 0.36 MN 6 G 2006 GATE‐2006 ( ) p p Assertion (A) : Extrusion speed depends on work material. Reason (R) : High extrusion speed causes cracks in the material. (a) Both A and R are individually true and R is the correct explanation of A (b) B h A and R are i di id ll true b R i not the Both d individually but is h correct explanation of A (c) A is true but R is false ( ) (d) A is false but R is true (c) 700 Page 38 of 78 (d) 832 G 996 GATE‐1996 A wire of 0.1 mm diameter is drawn from a rod of 15 mm diameter. Dies giving reductions of 20%, 40% and 80% are available. For minimum error in the final size, the number of stages and reduction at each stage respectively would be (a) 3 stages and 80% reduction for all three stages (b) 4 stages and 8 % reduction f fi t th t d 80% d ti for first three stages t followed by a finishing stage of 20% reduction (c) 5 stages and reduction of 80%, 80%.40%, 40%, 20% in a sequence (d) none of the above
- 40. G 99 GATE‐1994 The process of hot extrusion is used to produce (a) Curtain rods made of aluminium (b) St l pipes/or d Steel i / domestic water supply ti t l (c) Stainless steel tubes used in furniture (d) Large she pipes used in city water mains IES 2012 IES ‐ Which of the following are correct for an indirect hot extrusion process? 1. 1 Billet remains stationary 2. There is no friction force between billet and container walls. ll 3. The force required on the punch is more in comparison to direct extrusion. 4 4. Extrusion parts have to be provided a support. p p pp (a) 1, 2, 3 and 4 (b) 1, 2 and 3 only (c) 1 2 and 4 only 1, (d) 2 3 and 4 only 2, S IES – 2009 Which one of the following statements is correct? (a) In extrusion process, thicker walls can be obtained by increasing the forming pressure (b) Extrusion is an ideal process for obtaining rods from metal h i poor d it t l having density (c) As compared to roll forming, extruding speed is high (d) Impact extrusion is quite similar to Hooker's process g g including the flow of metal being in the same direction For-2014 (IES, GATE & PSUs) S 2007 IES – 200 g Which one of the following is the correct statement? ( ) (a) Extrusion is used for the manufacture of seamless tubes. ( ) (b) Extrusion is used for reducing the diameter of round g bars and tubes by rotating dies which open and close rapidly on the work? (c) Extrusion is used to improve fatigue resistance of the metal by setting up compressive stresses on its surface (d) Extrusion comprises pressing the metal inside a chamber to force it out by high pressure through an orifice which is shaped to provide the desired from of the finished part. S 993 IES – 1993 Assertion (A): Direct extrusion requires larger force than indirect extrusion. Reason (R): In indirect extrusion of cold steel zinc steel, phosphate coating is used. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 999 IES – 1999 Which one of the following is the correct temperature range for hot extrusion of aluminium? (a) 300 340°C (b) 350 400°C 300‐340 C 350‐400 C (c) 430‐480°C (d) 550‐650°C Page 39 of 78 S 2007 IES – 200 Assertion (A): Greater force on the plunger is required in case of direct extrusion than indirect one. Reason (R): In case of direct extrusion, the direction of the force applied on the plunger and the direction of the movement of the extruded metal are the same. (a) Both A and R are individually true and R is the correct explanation of A p (b) Both A and R are individually true but R is not the correct explanation of A p (c) A is true but R is false (d) A is false but R is true S 99 IES – 1994 Metal extrusion process is generally used for producing (a) Uniform solid sections (b) Uniform hollow sections (c) Uniform solid and hollow sections ( ) (d) Varying solid and hollow sections. y g S IES – 2000 g Consider the following statements: In forward extrusion process 1. The ram and the extruded product travel in the same p direction. 2. The ram and the extruded product travel in the opposite p pp direction. 3. The speed of travel of the extruded product is same as that of the ram. f h 4. The speed of travel of the extruded product is greater than that of the ram ram. Which of these Statements are correct? (a) ( ) 1 and 3 d (b) 2 and 3 d (c) 1 and 4 (d) 2 and 4
- 41. S IES – 2009 What is the major problem in hot extrusion? (a) Design of punch (b) Design of die (c) Wear d tear of di ( ) W and t f die (d) W of punch Wear f h S IES – 2003 The extrusion process (s) used for the production of toothpaste tube is/are 1. 1 Tube extrusion 2. Forward extrusion 3. Impact extrusion Select the correct answer using the codes given below: g g Codes: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 3 only ( ) GATE‐1990(PI) Semi brittle materials can be extruded by (a) Impact extrusion (b) Closed cavity extrusion (c) Hydrostatic extrusion (d) B k Backward extrusion d i For-2014 (IES, GATE & PSUs) IES 2012 IES ‐ Extrusion process can effectively reduce the cost of product through (a) Material saving (b) process time saving (c) Saving in tooling cost ( ) (d) saving in administrative cost g S 2001 IES – 200 g Which of the following statements are the salient features of hydrostatic extrusion? 1. It is suitable for soft and ductile material. 2. It is suitable for high‐strength super‐alloys. p 3.The billet is inserted into the extrusion chamber and pressure is applied by a ram to extrude the billet through the die. 4. The billet is inserted into the extrusion chamber where it is surrounded b a suitable li id Th bill i extruded d d by i bl liquid. The billet is d d through the die by applying pressure to the liquid. Select the correct ans er using the codes gi en belo answer given below: Codes: (a) ( ) 1 and 3 d (b) 1 and 4 d (c) 2 and 3 (d) 2 and 4 IES 2010 IES 2010 ( ) g g Assertion (A): Pickling and washing of rolled rods is carried out before wire drawing. Reason (R): They lubricate the surface to reduce friction while drawing wires. (a) Both A and R are individually true and R is the correct explanation of A (b) B h A and R are i di id ll true b R i NOT the Both d individually but is h correct explanation of A (c) A is true but R is false ( ) (d) A is false but R is true Page 40 of 78 S 2008, GATE‐1989(PI) 989( ) IES – 2008 G Which Whi h one of th f ll i methods i used f th f the following th d is d for the manufacture of collapsible tooth‐paste tubes? p p (a) Impact extrusion (b) Direct extrusion (c) Deep drawing (d) Piercing S IES – 2006 What does hydrostatic pressure in extrusion process improve? (a) Ductility (b) Compressive strength (c) Brittleness (d) Tensile strength S IES – 2009 Which one of the following stress is involved in the wire drawing process? (a) Compressive (b) Tensile (c) Shear (d) Hydrostatic stress
- 42. S 993 IES – 1993 Tandem drawing of wires and tubes is necessary because (a) It is not possible to reduce at one stage (b) Annealing is needed between stages (c) Accuracy in dimensions is not possible otherwise ( ) (d) Surface finish improves after every drawing stage p y g g S 996 IES – 1996 Match List I with List II and select the correct answer List I (Metal/forming process) List II (Associated force) A. A Wire drawing B. Extrusion C. Blanking D. Bending g Codes:A B C (a) 4 2 1 (c) 2 3 1 1. 1 2. 3. 4 4. D 3 4 (b) (d) Shear force Tensile force Compressive force Spring back force p g A B C D 2 1 3 4 4 3 2 1 S 993 S O 20 0 IES – 1993, ISRO‐2010 Match List I with List II and select the correct answer using the codes given below the lists: ( p p y) ( ) List I (Mechanical property) List II (Related to) A. Malleability 1. Wire drawing B. B Hardness 2. 2 Impact loads C. Resilience 3. Cold rolling D. D Isotropy 4. 4 Indentation 5. Direction Codes:A C d A B C D A B C D (a) 4 2 1 3 (b) 3 4 2 5 (c) 5 4 2 3 (d) 3 2 1 5 For-2014 (IES, GATE & PSUs) S IES – 2000 S 999 IES – 1999 (C p ) Match List I (Components of a table fan) with List II (Manufacturing processes) and select the correct answer using the codes given below the Lists: List I List II A. Base with stand 1. Stamping and p g pressing g B. Blade 2. Wire drawing C. Armature coil wire 3. Turning D. D Armature shaft 4. 4 Casting Codes:A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 2 3 4 1 (d) 4 1 2 3 Match List‐I with List‐II and select the correct answer using the codes given below the Lists: List‐I List‐II A. Drawing 1. Soap solution B. B Rolling 2. 2 Camber C. Wire drawing 3. Pilots D. D Sheet metal operations using 4 4. Crater progressive dies 5. Ironing Code:A C d A B C D A B C D (a) 2 5 1 4 (b) 4 1 5 3 (c) 5 2 3 4 (d) 5 2 1 3 S 996 IES – 1996 S 99 IES – 1994 In wire drawing process, the bright shining surface on the wire is obtained if one (a) does not use a lubricant (b) uses solid powdery lubricant. (c) uses thick paste lubricant ( ) (d) uses thin film lubricant Match List I with List II and select the correct answer using the codes given below the Lists: List I (Metal farming process) List II (A similar process) A. B. B C. D. Blanking Coining C i i Extrusion Cup drawing Codes:A (a) ( ) 2 (c) 3 S 2007 IES – 200 Which metal forming process manufacture of long steel wire? (a) Deep drawing (b) Forging (c) Drawing (d) Extrusion Page 41 of 78 B 3 2 C 4 1 1. 2. 3. 4. 5. D 1 5 (b) (d) Wire drawing Piercing Pi i Embossing Rolling Bending A B C 2 3 1 2 3 1 D 4 5 S 2005 IES – 200 is used for Which of the following types of stresses is/are involved in the wire‐drawing operation? (a) Tensile only (b) Compressive only (c) A combination of tensile and compressive stresses ( ) (d) A combination of tensile, compressive and shear p stresses
- 43. GATE‐1987 For wire drawing operation, the work material should essentially be (a) Ductile Which one of the following lubricants is most suitable for drawing mild steel wires? (a) Sodium stearate (b) Water (c) Lime‐water (d) Kerosene (b) Tough (c) Hard ( ) S IES – 2000 S 993 IES – 1993 A moving mandrel is used in (a) Wire drawing (b) T b d Tube drawing i (c) Metal cutting (d) Forging (d) Malleable ( ) S IES – 2002 Match List I with List II and select the correct answer: List I (Parts) List II (Manufacturing processes) A. Seamless tubes 1. Roll forming B. Accurate and smooth tubes 2. B A d h b Shot Sh peening i C. Surfaces having higher 3. Forging hardness and fatigue strength4. Cold forming Codes: A B C A B C (a) 1 4 2 (b) 2 3 1 (c) ( ) 1 3 2 (d) 2 4 1 S 2001 IAS – 200 Match List I (Products) with List II (Suitable processes) and select the correct answer using the codes given below the Lists: List I List II A. Connecting rods A C ti d 1. Welding W ldi B. Pressure vessels 2. Extrusion C. Machine tool beds 3. Forging D. Collapsible tubes 4. Casting Codes:A B C D A B C D (a) ( ) 3 1 4 2 (b) 4 1 3 2 (c) 3 2 For-20141(IES, (d) 4 & PSUs) 3 4 2 1 GATE S 2004 IAS – 200 Assertion (A): Indirect extrusion operation can be performed either by moving ram or by moving the container. Reason (R): Advantage in indirect extrusion is less quantity of scrap compared to direct extrusion extrusion. (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 99 IAS – 1997 Extrusion force DOES NOT depend upon the (a) Extrusion ratio (b) T Type of extrusion process f t i (c) Material of the die (d) Working temperature Page 42 of 78 S 99 IAS – 1995 The following operations are performed while preparing the billets for extrusion process: 1. 1 Alkaline cleaning 2. Phosphate coating 3. Pickling 4 4. Lubricating with reactive soap. g p The correct sequence of these operations is (a) 3 1 4 2 3, 1, 4, (b) 1 3 2 4 1, 3, 2, (c) 1, 3. 4, 2 (d) 3, 1, 2, 4 S IAS – 2000 Assertion (A): Brittle materials such as grey cast iron cannot be extruded by hydrostatic extrusion. Reason(R): In hydrostatic extrusion billet is extrusion, uniformly compressed from all sides by the liquid. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
- 44. S IAS – 2002 ( ) g p , Assertion (A): In wire‐drawing process, the rod cross‐section is reduced gradually by drawing it several times in successively reduced diameter dies. Reason (R): Since each drawing reduces ductility of the wire, so after final drawing the wire is normalized. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true ( ) GATE‐1991(PI) IES 2011 IES 2011 Match List –I with List –II and select the correct answer using the code given below the lists : List –I List –II A. Connecting rods 1. Welding B. Pressure vessels 2. Extrusion C. Machine tool beds C Machine tool beds 3. Forming 3 Forming D. Collapsible tubes p 4. Casting g Codes C d A (a) ( ) 2 (c) 2 B 1 4 C 4 1 D 3 3 (b) (d) A 3 3 B 1 4 Ch‐16: Drawing Option Q. No Option 1 D 8 B 2 3 C D 9 10 B A D 11 B 5 6 B C 12 13 B A 7 rolling, drawing and D 2 2 B E l Example Determine the die and punch sizes for blanking a circular disc of 20‐mm diameter from a sheet whose thickness is 1.5 mm. Sheet Metal Operation p Shear strength of sheet material = 294 MPa Also determine the die and punch sizes for punching a circular hole of 20‐mm diameter from a sheet whose thickness is 1 5 mm 1.5 mm. By S K Mondal For-2014 (IES, GATE & PSUs) g Which of the following methods can be used for manufacturing 2 metre long seamless metallic tubes? 1. Drawing 2. Extrusion 3. 3 Rolling 4. 4 Spinning Select the correct answer using the codes given below Codes: ( ) (a) 1 and 3 ( ) (b) 2 and 3 (c) 1, 3 and 4 (d) 2, 3 and 4 Ch‐17: Extrusion Q. No 4 Seamless long steel tubes are manufactured by C 4 1 IAS 1994 IAS 1994 Page 43 of 78 Q. No Option 1 A 2 C 3 4 C B 5 6 C D E l Example Estimate the blanking force t cut a bl k 25 mm wide to t blank E ti t th bl ki f id and 30 mm long from a 1.5 mm thick metal strip, if the 3 g 5 p, ultimate shear strength of the material is 450 N/mm2. Also determine the work done if the percentage penetration i 25 percent of material thi k t ti is t f t i l thickness.
- 45. IAS‐2011 Main For punching a 10 mm circular hole, and cutting a rectangular bl k of 50 x 200 mm f t l blank f from a sheet of 1 h t f mm thickness (mild steel, shear stress = 240 N/mm2) C l l t i each case : N/ ), Calculate, in h (i) Size of punch (ii) Size of die ( ) (iii) Force required. q [ [10‐Marks] ] S 999 IES – 1999 S O 2008 20 ISRO‐2008, 2011 A hole is to be punched in a 15 mm thick plate having ultimate shear strength of 3N‐mm‐2. If the allowable crushing stress in the punch is 6 N‐mm‐2, N mm the diameter of the smallest hole which can be punched is equal to (a) 15 mm (b) 30 mm (c) 60 ( ) 6 mm (d) 120 mm With a punch for which the maximum crushing stress is 4 times the maximum shearing stress of the l t the biggest h l th t can b punched i th plate, th bi t hole that be h d in the plate would be of diameter equal to 1 × Thickness of p plate 4 1 (b) × Thickness of plate 2 (c) Plate thickness ( ) (a) (d) 2 × Plate thickness IES 2013 IES‐2013 A hole of diameter d is to be punched in a plate of thickness t. For the plate material, the maximum crushing stress is 4 times the maximum allowable shearing stress. For punching the biggest hole, the ratio of diameter of hole to plate thickness should be equal to: 1 1 2 ( ) (a) 4 ( ) (b) (c) 1 E l Example Example A hole, 100 mm diameter, is to be punched in steel plate 5.6 mm thi k Th ultimate shear stress i 550 N/ 6 thick. The lti t h t is N/mm2 . With normal clearance on the tools, cutting is complete , g p at 40 per cent penetration of the punch. Give suitable shear angle for the punch to bring the work within the (d) 2 capacity of a 30T press. it f T G 20 0 S i k d GATE‐2010 Statement Linked 1 Statement for Linked Answer Questions: In a shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long and zero‐shear (S = 0) is provided on the edge. The ultimate shear strength of the sheet is 100 MPa and g penetration to thickness ratio is 0.2. Neglect friction. G 20 0 S i k d2 GATE‐2010 Statement Linked 2 Q Statement for Linked Answer Questions: In a shear cutting operation, a sheet of 5mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long and zero shear (S = 0) is provided on the edge zero‐shear edge. The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0.2. Neglect friction. 400 400 S Assuming force vs displacement curve to be rectangular, the work done (in J) is (a) 100 (b) 200 (c) 250 (d) 300 For-2014 (IES, GATE & PSUs) S A shear of 20 mm (S = 20 mm) is now provided on the blade. Assuming force vs displacement curve to be trapezoidal, the maximum force (in kN) exerted is (a) 5 (b) 10 Page 44 of 78 (c) 20 (d) 40 A washer with a 12.7 mm internal hole and an outside diameter of 25.4 mm is to be made from 1.5 mm thick strip. The ultimate shearing strength of the material of p g g the washer is 280 N/mm2. ( ) (a) Find the total cutting force if both punches act at g p the same time and no shear is applied to either punch or the die. (b) What will be the cutting force if the punches are staggered, so that only one punch acts at a time. (c) Taking 60% penetration and shear on punch of 1 mm, what will be the cutting force if both punches act g p together. GATE 2011 MPa. The shear strength of a sheet metal is 300 MPa The blanking force required to produce a blank of 100 mm diameter from a 1 5 mm thick sheet is close to 1.5 (a) 45 kN (b) 70 kN (c) 141 kN 4 (d) 3500 kN
- 46. ( ) GATE – 2009 (PI) GATE 2013 (PI) GATE‐2013 (PI) Circular blanks of 10 mm diameter are punched S O 2009 ISRO‐2009 The force required to punch a 25 mm hole in a A disk of 200 mm diameter is blanked from a strip from an aluminium sheet of 2 mm thickness. The of an aluminum alloy of thickness 3.2 mm. The shear strength of aluminium is 80 Mpa. The material shear strength to fracture is 150 MPa. The minimum punching f i i hi force required i kN i i d in is stress of the plate is 500 N/mm2 will be nearly blanking force (in kN) is (a) 2 57 2.57 (a) 78 kN (b) 393 kN (c) 98 kN (d) 158 kN ( ) 9 (a) 291 ( )3 (b) 301 ( ) 3 (c) 311 ( )3 (d) 321 mild steel plate 10 mm thi k when ultimate shear ild t l l t thick, h lti t h (b) 3.29 (c) 5.03 (d) 6.33 G 200 GATE‐2007 The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is ‘t’ and diameter of the blanked part is t ‘d’. For the same work material, if the diameter of the blanked part is increased to 1.5 d and thickness is reduced to 0.4 t, the new blanking force in kN is (a) 3 0 (b) 4 5 3.0 4.5 (c) 5.0 (d) 8.0 GATE ‐ 2012 Calculate the punch size in mm, for a circular blanking operation for which details are given below. Size of the blank 25 mm Thickness of the sheet 2 mm Radial clearance bet een punch and die 0 06 mm between 0.06 Die allowance 0.05 mm (a) ( ) 24.83 (b) 24.89 (c) 25.01 (d) 25.17 For-2014 (IES, GATE & PSUs) G 200 GATE‐2004 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N / mm2 and penetration is 40%. Shear provided on the punch is 2 mm. The blanking force during the operation will be (a) 22.6 kN (b) 37.7 kN (c) 61.6 ( ) 6 6 kN (d) 94.3 kN ( ) GATE‐2008(PI) A blank of 50 mm diameter is to be sheared from a sheet of 2.5 mm thickness. The required radial clearance between the die and the punch is 6% of sheet thickness. The punch and die diameters (in mm) for this blanking operation, respectively, are (a) 50 00 and 50 30 50.00 50.30 (b) 50 00 and 50 15 50.00 50.15 (c) 49.70 and 50.00 (d) 49.85 and 50.00 Page 45 of 78 G 2003 GATE‐2003 A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the die clearance is 3%. The required punch diameter is (a) 19.88 mm (b) 19.94 mm (c) ( ) 20.06 mm (d) 20.12 mm 6 G 2002 GATE‐2002 In a blanking operation, the clearance is provided on (a) The die (b) Both the die and the punch equally (c) The punch ( ) (d) Brittle the punch nor the die p
- 47. G 200 GATE‐2001 The cutting force in punching and blanking operations mainly depends on (a) The modulus of elasticity of metal (b) The shear strength of metal (c) The bulk modulus of metal ( ) (d) The yield strength of metal y g S IES – 2002 Consider the following statements related to piercing and blanking: 1. 1 Shear on the punch reduces the maximum cutting force 2. Sh Shear i increases th capacity of th press needed the it f the d d 3. Shear increases the life of the punch 4. The total energy needed to make the cut remains p unaltered due to provision of shear Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 S 2004 IES – 200 Which one of the following statements is correct? If the size of a flywheel in a punching machine is increased (a) Then the fluctuation of speed and fluctuation of energy will b th d ill both decrease (b) Then the fluctuation of speed will decrease and the fluctuation of energy will increase ( ) (c) Then the fluctuation of speed will increase and the p fluctuation of energy will decrease (d) Then the fluctuation of speed and fluctuation of energy both will increase For-2014 (IES, GATE & PSUs) G 996 GATE‐1996 A 50 mm diameter disc is to be punched out from a carbon steel sheet 1.0 mm thick. The diameter of the punch should be (a) 49.925 mm (b) 50.00 mm (c) ( ) 50.075 mm (d) none of th above f the b S 99 IAS – 1995 In blanking operation the clearance provided is (a) 50% on punch and 50% on die (b) O di On die (c) On punch (d) On die or punch depending upon designer’s choice S 99 IES – 1997 For 50% penetration of work material, a punch with single shear equal to thickness will (a) Reduce the punch load to half the value (b) Increase the punch load by half the value (c) Maintain the same punch load ( ) (d) Reduce the punch load to quarter load p q Page 46 of 78 S 99 IES – 1994 In sheet metal blanking, shear is provided on punches and dies so that (a) Press load is reduced (b) Good cut edge is obtained. (c) Warping of sheet is minimized ( ) (d) Cut blanks are straight. g S IES – 2006 In which one of the following is a flywheel generally employed? (a) Lathe (b) Electric motor (c) Punching machine (d) Gearbox S IAS – 2000 A blank of 30 mm diameter is to be produced out of 10 mm thick sheet on a simple die. If 6% clearance is recommended, then the nominal diameters of die and punch are respectively (a) 30 6 mm and 29 4 mm 30.6 29.4 (b) 30.6 mm and 30 mm (c) 30 mm and 29.4 mm ( ) 3 (d) 30 mm and 28.8 mm
- 48. GATE 2007 (PI) GATE – 2007 (PI) Circular blanks of 35 mm diameter are punched from a steel sheet of 2 mm thickness. If the clearance per side between the punch and die is to be kept as 40 microns, the sizes of punch and die h ld di should respectively b i l be (a) 35+0.00 and 35+0.040 (b) 35‐0.040 and 35‐0.080 (c) 35‐0.080 and 35+0.00 (d) 35+0.040 and 35‐0.080 S 2007 IAS – 200 For punching operation the clearance is provided on which one of the following? (a) The punch (b) The die (c) 50% on the punch and 50% on the die ( ) 3 (d) 1/3rd on the punch and 2/3rd on the die p 3 S IAS – 2003 Match List I (Press‐part) with List II (Function) and select the correct answer using the codes given b l below the li i h d i h lists: List‐I List‐II ( (Press‐part) p ) ( (Function) ) (A) Punch plate 1. Assisting withdrawal of the punch (B) Stripper 2. Advancing the work‐piece through correct distance di t (C) Stopper 3. Ejection of the work‐piece from die cavity ( ) (D) Knockout 4 4. Holding the small punch in the proper g p p p position Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 4 1 2 3 (d) 2 3 4 1 For-2014 (IES, GATE & PSUs) S 99 IAS – 1994 In a blanking operation to produce steel washer, the maximum punch load used in 2 x 105 N. The plate thickness is 4 mm and percentage penetration is 25. The work done during this shearing operation is (a) 200J (b) 400J (c) 600 J (d) 800 J S 99 IAS – 1995 Assertion (A): A flywheel is attached to a punching press so as to reduce its speed fluctuations. Reason(R): The flywheel stores energy when its speed increase. (a) Both A and R are individually true and R is the correct explanation of A (b) B th A and R are i di id ll t Both d individually true b t R i not th but is t the correct explanation of A (c) A is true but R is false (d) A is false but R is true S IES – 2000 Best position of crank for blanking operation in a mechanical press is (a) Top dead centre (b) 20 degrees below top dead centre (c) 20 degrees before bottom dead centre ( ) (d) Bottom dead centre Page 47 of 78 S IAS – 2002 In deciding the clearance between punch and die in press work in shearing, the following rule is helpful: (a) Punch size controls hole size die size controls blank size (b) P Punch size controls b th h l size and bl k size h i t l both hole i d blank i (c) Die size controls both hole size and blank size (d) Die size controls hole size, punch size controls blank size S IES – 2002 Which one is not a method of reducing cutting forces to prevent the overloading of press? (a) Providing shear on die (b) Providing shear on punch (c) Increasing die clearance ( ) (d) Stepping punches pp g p S 999 IES – 1999 Assertion (A): In sheet metal blanking operation, clearance must be given to the die. Reason (R): The blank should be of required dimensions. (a) Both ( ) B th A and R are i di id ll t d individually true and R i th d is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
- 49. S IAS – 2003 S 99 IES – 1997 spring back The 'spring back' effect in press working is (a) Elastic recovery of the sheet metal after removal of the load (b) Regaining the original shape of the sheet metal (c) Release of stored energy in the sheet metal ( ) (d) Partial recovery of the sheet metal y A cup of 10 cm height and 5 cm diameter is to be made from a sheet metal of 2 mm thickness. The number of deductions necessary will be (a) One (b) T Two (c) Three (d) Four G 2008 GATE‐2008 IFS ‐ 2009 What is deep drawing process for sheet metal forming? Explain the function of a blank holder. What is drawing ratio and how is the drawing ratio increased ? [ [10 – marks] ] G 2003 GATE‐2003 A shell of 100 mm diameter and 100 mm height with the corner radius of 0.4 mm is to be produced by cup drawing. The required blank diameter is (a) 118 mm (b) 161 mm (c) ( ) 224 mm (d) 312 mm In the deep drawing of cups, blanks show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively, (A) Buckling due to circumferential compression; Increase blank holder pressure (B) High blank holder pressure and high friction; Reduce blank holder pressure and apply lubricant (C) High temperature causing increase in circumferential length: Apply coolant to blank (D) Buckling due to circumferential compression; decrease blank holder pressure ISRO‐2011 The initial blank diameter required to form a cylindrical cup of outside di li d i l f id diameter 'd‘ and d total height 'h' having a corner radius 'r' is g g obtained using the formula (a ) Do = d 2 + 4dh − 0 5r 0.5 (b) Do = d + 2h + 2r (c) Do = d 2 + 2h 2 + 2r For-2014 (IES, GATE & PSUs) (d ) Do = d 2 +Page− 0.5r78 4dh 48 of E l Example A symmetrical cup of 80 mm diameter and 250 mm height is to be fabricated on a deep drawing die. How many drawing operations will be necessary if no intervening annealing is done. Also find the drawing force G 999 GATE‐1999 Identify the stress ‐ state in the FLANCE portion of a PARTIALLYDRAWN CYLINDRICAL CUP when deep ‐ drawing without a blank holder (a) Tensile in all three directions (b) N stress i th fl No t in the flange at all, b t ll because th there i no is blank‐holder (c) Tensile stress in one direction and compressive in the one other direction (d) Compressive in two directions and tensile in the third direction G 2006 GATE‐2006 Match the items in columns I and II. Column I Column II P. P Wrinkling 1. 1 Yield point elongation Q. Orange peel 2. Anisotropy R. Stretcher strains 3. R S h i Large grain size L i i S. Earing 4. Insufficient blank holding force 5. Fine grain size 6. Excessive blank holding force (a) P – 6, Q – 3, R – 1, S – 2 (b) P – 4, Q – 5, R – 6, S – 1 3 5 (c) P – 2, Q – 5, R – 3, S – 4 (d) P – 4, Q – 3, R – 1, S – 2
- 50. S IES – 2008 S 999 IES – 1999 A cylindrical vessel with flat bottom can be deep drawn by (a) Shallow drawing (b) Single action deep drawing (c) Double action deep drawing ( ) (d) Triple action deep drawing p p g S 2007 IAS – 200 In drawing operation, proper lubrication essential for which of the following reasons? 1. 1 To improve die life 2. To reduce drawing forces 3. To reduce temperature 4 4. To improve surface finish p Select the correct answer using the code given below: (a) 1 and 2 only (b) 1 3 and 4 only 1, (c) 3 and 4 only (d) 1, 2, 3 and 4 S 99 IAS – 1997 is S 998 IES – 1998 Assertion (A): The first draw in deep drawing operation can have up to 60% reduction, the second draw up to 4 40% reduction and, the third draw of about 30% only. , 3 y Reason (R): Due to strain hardening, the subsequent draws in a deep drawing operation have reduced p g p percentages. ( ) (a) Both A and R are individually true and R is the correct y explanation of A ( ) (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES, GATE & PSUs) Consider the following statements: Earring in a drawn cup can be due to non‐uniform 1. 1 Speed of the press 2. Clearance between tools 3. Material properties 4 4. Blank holding g Which of these statements are correct? (a) 1 2 and 3 (b) 2 3 and 4 1, 2, (c) 1, 3 and 4 (d) 1, 2 and 4 Which one of the following factor promotes the tendency for wrinking in the process of drawing? (a) Increase in the ratio of thickness to blank diameter of work material (b) D Decrease i th ratio thi k in the ti thickness t bl k di to blank diameter of t f work material (c) Decrease in the holding force on the blank ( ) (d) Use of solid lubricants G 992 GATE‐1992 The thickness of the blank needed to produce, by power spinning a missile cone of thickness 1.5 mm and half cone angle 30 , is and half cone angle 30°, is (a) 3.0 mm (b) 2.5 mm (c) ( ) 2.0 mm (d) 1.5 mm Page 49 of 78 S 99 IES – 1994 For obtaining a cup of diameter 25 mm and height 15 mm by drawing, the size of the round blank should be approximately (a) 42 mm (b) 44 mm (c) 6 ( ) 46 mm (d) 48 mm 8 S 99 IAS – 1994 Consider the following factors 1. Clearance between the punch and the die is too small. small 2. The finish at the corners of the punch is poor. 3. The finish at the corners of the die is poor. 4 4. The punch and die alignment is not proper. p g p p The factors responsible for the vertical lines parallel to the axis noticed on the outside of a drawn cylindrical cup would include. (a) 2 3 and 4 (b) 1 and 2 2, (c) 2 and 4 (d) 1, 3 and 4 S 99 IES – 1994 The mode of deformation of the metal during spinning is (a) Bending (b) Stretching (c) Rolling and stretching ( ) (d) Bending and stretching. g g
- 51. IES 2011 IFS‐2011 Compare metal spinning with press work. [2‐marks] [ k ] IES 2010 IES 2010 ( ) g gy g Assertion (A) : In the high energy rate forming method, the explosive forming has proved to be an g g g excellent method of utilizing energy at high rate and utilizes both the high explosives and low explosives. Reason (R): The gas pressure and rate of detonation ( ) g p can be controlled for both types of explosives. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true S 2005 IES – 200 Magnetic forming is an example of: (a) Cold forming (b) Hot forming (c) High ( ) Hi h energy rate f t forming i (d) R ll f Roll forming i For-2014 (IES, GATE & PSUs) High energy rate forming process used for forming components from thin metal sheets or deform thin tubes is: (a) Petro‐forming (b) Magnetic pulse forming (c) Explosive forming p g (d) electro‐hydraulic forming S 2007 IES – 200 Which one of the following metal forming processes is not a high energy rate forming process? (a) Electro mechanical forming Electro‐mechanical (b) Roll‐forming (c) Explosive forming ( ) (d) Electro‐hydraulic forming y g IFS‐2011 Write four advantages of high velocity forming process. [2‐marks] [ k ] Page 50 of 78 20 0 JWM 2010 ( ) g p g , Assertion (A) : In magnetic pulse‐forming method, magnetic field produced by eddy currents is used to p create force between coil and workpiece. Reason (R) : It is necessary for the workpiece material to have magnetic properties. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A t l ti f (c) A is true but R is false (d) A is false but R is true S IES – 2009 Which one of the following is a high energy rate forming process? (a) Roll forming (b) Electro‐hydraulic forming (c) Rotary forging ( ) (d) Forward extrusion G 2000 GATE‐2000 A 1.5 mm thick sheet is subject to unequal biaxial stretching and the true strains in the directions of stretching are 0.05 and 0.09. The final thickness of the sheet in mm is (a) 1 414 1.414 (b) 1 304 1.304 (c) 1.362 (d) 289
- 52. IES 1998 IES‐1998 g q g, The bending force required for V‐bending, U‐ bending and Edge bending will be in the ratio of (a) 1 : 2 : 0 5 0.5 (b) 2: 1 : 0 5 0.5 (c) 1: 2 : 1 (d) 1: 1 : 1 E l Example g Calculate the bending force for a 45o bend in aluminium blank. Blank thickness, 1.6 mm, bend length = 1200 mm, p g , , p g Die opening = 8t, UTS = 455 MPa, Die opening factor = 1.33 G 200 GATE‐2005 A 2 mm thick metal sheet is to be bent at an angle of one radian with a bend radius of 100 mm. If the stretch factor is 0.5, the bend allowance is (a) 99 mm (b) 100 mm (c) ( ) 101 mm (d) 102 mm 2mm 1 radian G 200 GATE‐2007 g Match the correct combination for following metal working processes. Associated state of stress Processes P. Blanking 1. Tension Q. Q Stretch Forming 2 2. Compression R. Coining 3. Shear S. S Deep Drawing 4. 4 Tension and Compression 5. Tension and Shear Codes:P C d P Q R S P Q R S (a) 2 1 3 4 (b) 3 4 1 5 (c) 5 4 3 1 (d) 3 1 2 4 GATE ‐2012 Same Q in GATE‐2012 (PI) Match the following metal forming processes with their associated stresses in the workpiece. Metal forming process l f i 1. Coining 2. Wire Drawing 3. Blanking 3 Blanking 4. Deep Drawing D D i (a) 1‐S, 2‐P, 3‐Q, 4‐R (c) 1‐P, 2‐Q, 3‐S, 4‐R Type of stress f P. Tensile Q. Shear R. Tensile and R Tensile and compressive S. Compressive S C i (b) 1‐S, 2‐P, 3‐R, 4‐Q (d) 1‐P, 2‐R, 3‐Q, 4‐S S 999 IAS – 1999 S 99 IAS – 1997 ( ) ( p ) Match List I (Process) with List II (Production of parts) and select the correct answer using the codes given below the lists: List‐I List‐II A. Rolling 1. Discrete parts B. Forging 2. Rod and Wire C. Extrusion 3. Wide variety of shapes with thin walls ll D. Drawing 4. Flat plates and sheets 5. Solid and h ll parts l d d hollow Codes:A B C D A B C D (a) ( ) 2 5 3 4 (b) ( ) 1 2 5 4 (c) 4 1 For-2014 2 3 (d) 4 & PSUs) 5 1 2 (IES, GATE List I List II Match List‐I (metal forming process) with List‐II (Associated feature) and select the correct answer using the codes given below the Lists: List‐l List‐ II A. Blanking A Bl ki 1. Shear angle Sh l B. Flow forming 2. Coiled stock C. Roll forming 3. Mandrel D. Embossing 4. Closed matching dies Codes:A B C D A B C D (a) ( ) 1 3 4 2 (b) 3 1 4 2 (c) 1 3 2 Page 51 of 78 4 (d) 3 1 2 4 G 200 GATE‐2004 Match the following Product Process P. Moulded luggage P M ld d l 1. Injection I j ti moulding ldi Q. Packaging containers for liquid 2. Hot rolling R. Long structural shapes 3. Impact extrusion S. Collapsible tubes 4. Transfer moulding 5. Blow moulding 6. 6 Coining C i i (a) P‐1 Q‐4 R‐6 S‐3 (b) P‐4 Q‐5 R‐2 S‐3 (c) P‐1 Q‐5 R‐3 S‐2 (d) P‐5 Q‐1 R‐2 S‐2 IES 2010 IES 2010 g Consider the following statements: The material properties which principally determine how well a metal may be drawn are 1. Ratio of yield stress to ultimate stress. 2.Rate of increase of yield stress relative to p g progressive amounts of cold work. 3. Rate of work hardening. Which f the b Whi h of th above statements i / t t t is/are correct? t? (a) 1 and 2 only (b) 2 and 3 only (c) 1 only (d) 1, 2 and 3
- 53. Ch‐18: Sheet Metal Forming Q. No Option Q. No ( ) GATE ‐2011 (PI) Option 1 C 10 C 2 3 B A 11 12 C C 4 A 13 C 5 D 14 D 6 7 A A 15 16 A B 8 9 A A 17 D Powder Metallurgy gy Which of the following powder production methods produces spongy and porous particles? th d d d ti l ? (a) Atomization (b) Reduction of metal oxides ( ) (c) Electrolytic deposition y p (d) Pulverization By S K Mondal IES 2012 IES ‐ In electrolysis (a) For making copper powder, copper plate is made cathode in electrolyte tank (b) For making aluminum powder, aluminum plate is made anode d d (c) High amperage produces powdery deposit of cathode metal on anode ( ) (d) Atomization process is more suitable for low melting p g point metals IES – 2007 Conventional Metal powders are compacted by many methods, but In powder metallurgy, sintering of a component sintering is required to achieve which property? What (a) Improves strength and reduces hardness is hot iso‐static pressing? i h ti t ti i ? (c) Improves hardness and reduces toughness (d) R d Reduces porosity and i i d increases b i l brittleness ( ) GATE – 2009 (PI) Which of the following process is used to manufacture products with controlled porosity? (a) Casting (b) ( ) welding (c) formation (d) Powder metallurgy For-2014 (IES, GATE & PSUs) (b) Reduces brittleness and improves strength [ [ 2 Marks] ] IES – 2011 Conventional What is isostatic pressing of metal powders ? What are its advantage ? h d [ 2 Marks] ( ) GATE ‐2010 (PI) Page 52 of 78 ( ) GATE – 2011 (PI) The binding material used in cemented carbide cutting t l i tti tools is (a) graphite (b) tungsten ( ) (c) nickel (d) cobalt
- 54. IES 2010 IES 2010 Consider the following parts: 1. Grinding wheel 2. Brake lining 3. Self lubricating 3 Self‐lubricating bearings Which of these parts are made by powder metallurgy technique? ll h i ? ( ) , (a) 1, 2 and 3 ( ) (b) 2 only y (c) 2 and 3 only (d) 1 and 2 only S 2001 IES – 200 Match List I (Components) with List II List‐I List‐II (Manufacturing Processes) and select the correct answer using the codes given below the lists: List I List II A. Car body (metal) A C b d ( t l) 1. Machining M hi i B. Clutch lining 2. Casting C. Gears 3. Sheet metal pressing D. Engine block 4. Powder metallurgy Codes:A B C D A B C D (a) ( ) 3 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2 S 99 IES – 1997 Which of the following components can be manufactured by powder metallurgy methods? 1. 1 Carbide tool tips 2. 2 Bearings 3. Filters 4. Brake linings Select the correct answer using the codes given below: ( ) (a) 1, 3 and 4 (b) 2 and 3 ( ) (c) 1, 2 and 4 (d) 1, 2, 3 and 4 For-2014 (IES, GATE & PSUs) IES 2010 IES 2010 Metallic powders can be produced by (a) Atomization (b) Pulverization (c) Electro‐deposition process Electro deposition (d) All of the above GATE 2011 The operation in which oil is permeated into the pores of a powder metallurgy product is known as (a) i i ( ) mixing (b) sintering (c) impregnation (d) Infiltration S 999 IES – 1999 The correct sequence of the given processes in manufacturing by powder metallurgy is (a) Blending compacting sintering and sizing Blending, compacting, (b) Blending, compacting, sizing and sintering (c) Compacting, sizing, blending and sintering ( ) (d) Compacting, blending, sizing and sintering p g g g g Page 53 of 78 S IES – 2002 The rate of production of a powder metallurgy part depends on (a) Flow rate of powder (b) Green strength of compact (c) Apparent density of compact ( ) (d) Compressibility of powder p y p S 998 IES – 1998 In powder metallurgy, the operation carried out to improve the bearing property of a bush is called (a) infiltration (b) impregnation (c) plating (d) heat treatment S 99 IES – 1997 ( p ) ( Match List‐I (Gear component) with List‐II (Preferred method of manufacturing) and select the correct answer using the codes given below the Lists: List‐I List‐II A. Gear for clocks 1. Hobbing B. Bakelite gears 2. Stamping C. Aluminium gears 3. Powder compacting D. Automobile transmission gears 4. Sand casting 5. Extrusion Code:A B C D A B C D (a) 2 3 5 1 (b) 5 3 4 2 (c) 5 1 2 3 (d) 2 4 5 3
- 55. S 2001 IES – 200 Carbide tipped Carbide‐tipped cutting tools are manufactured by powder‐ metal technology process and have a composition of (a) Zirconium‐Tungsten (35% ‐65%) (b) T Tungsten carbide‐Cobalt ( % ‐ 10%) t bid C b lt (90% %) (c) Aluminium oxide‐ Silica (70% ‐ 30%) (d) Nickel‐Chromium‐ Tungsten (30% ‐ 15% ‐ 55%) S 2007 IES – 200 What are the advantages of powder metallurgy? 1. Extreme purity product 2. L l b Low labour cost t 3. Low equipment cost. Select the correct answer using the code given below (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only IES 2012 IES ‐ () yp gy Statement (I): Parts made by powder metallurgy do not have as good physical properties as parts casted. ( ) p p gy Statement (II): Particle shape in powder metallurgy influences the flow characteristic of the powder. ( ) (a) Both Statement (I) and Statement (II) are () ( ) individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true For-2014 (IES, GATE & PSUs) S 999 IES – 1999 Assertion (A): In atomization process of manufacture of metal powder, the molten metal is forced through a small orifice and broken up by a stream of compressed p y p air. Reason (R): The metallic powder obtained by ( ) p y atomization process is quite resistant to oxidation. ( ) (a) Both A and R are individually true and R is the correct y explanation of A ( ) (b) Both A and R are individually true but R is not the y correct explanation of A (c) A is true but R is false (d) A is false but R is true S IES – 2006 Which of the following are the limitations of powder metallurgy? 1. 1 High tooling and equipment costs costs. 2. Wastage of material. 3. It cannot be automated. 4 4. Expensive metallic powders. p p Select the correct answer using the codes given below: (a) Only 1 and 2 (b) Only 3 and 4 (c) Only 1 and 4 (d) Only 1, 2 and 4 S IES – 2009 Which of the following cutting tool bits are made by powder metallurgy process? (a) Carbon steel tool bits (b) Stellite tool bits (c) Ceramic tool bits (d) HSS tool bits Page 54 of 78 S 99 IES – 1997 Match List‐I with List‐II and select the correct answer using the codes given below the Lists: List‐I List‐II (Bearing materials) (Properties) A. Babbits 1. Porous B. Bronze 2. Good Embeddability 3 3. Suitable for high loads and low g C. C.I. speeds D. Sintered powdered metal 4. Runs well with C.I. journals l Code:A B C D A B C D (a) ( ) 2 3 4 1 (b) ( ) 3 2 1 4 (c) 2 3 1 4 (d) 3 2 4 1 S 2004 IES – 200 Consider the following factors: 1. Size and shape that can be produced economically 2. P Porosity of th parts produced it f the t d d 3. Available press capacity 4. High density Which of the above are limitations of powder metallurgy? (a) 1 3 and 4 (b) 2 and 3 1, (c) 1, 2 and 3 (d) 1 and 2 S IAS – 2003 Which of the following are produced by powder metallurgy process? 1. 1 Cemented carbide dies 2. Porous bearings 3. Small magnets 4 4. Parts with intricate shapes p Select the correct answer using the codes given below: Codes: (a) 1, 2 and 3 (b) 1, 2 and 4 (c) 2, 3 and 4 (d) 1, 3 and 4
- 56. S IAS – 2003 S IAS – 2000 In parts produced by powder metallurgy process, pre‐sintering is done to (a) Increase the toughness of the component (b) Increase the density of the component (c) Facilitate bonding of non‐metallic particles ( ) (d) Facilitate machining of the part g p S 998 IAS – 1998 Throwaway tungsten manufactured by (a) Forging (c) Powder metallurgy carbide (b) (d) Consider the following processes: 1. Mechanical pulverization 2. At i ti Atomization 3. Chemical reduction 4. Sintering Which of these processes are used for powder preparation in powder metallurgy? (a) 2 3 and 4 (b) 1 2 and 3 2, 1, (c) 1, 3 and 4 (d) 1, 2 and 4 S 996 IAS – 1996 tip tools are Brazing Extrusion Which one of the following processes is performed in powder metallurgy to promote self‐lubricating properties in sintered parts? (a) Infiltration (b) Impregnation (c) Plating ( ) Pl ti (d) G hiti ti Graphitization S 99 IAS – 1997 ( ) C Assertion (A): Close dimensional tolerances are NOT possible with isostatic pressing of metal powder in powder metallurgy technique. Reason (R): In the process of isostatic pressing, the pressure is equal in all directions which permits uniform density of the metal powder. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true GATE 2008 (PI) GATE ‐2008 (PI) Match the following Group – 1 P. Mulling P Mulling Q. Impregnation R. Flash trimming l h S. Curing g Group ‐2 1. Powder metallurgy 1 Powder metallurgy 2. Injection moulding 3. Processing of FRP composites f 4. Sand casting g ( ) (a) P – 4, Q – 3, R – 2, S – 1 4, Q 3, , (c) P – 2, Q – 1, R – 4, S – 3 S 2007 IAS – 200 ( ) g Assertion (A): Mechanical disintegration of a molten metal stream into fine particles by means of a jet of compressed air is known as atomization. Reason (R): In atomization process inert‐gas or water cannot be used as a substitute for compressed air. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES, GATE & PSUs) S 2004 IAS – 200 The following are the constituent steps in the process of powder metallurgy: 1. 1 Powder conditioning 2. Sintering 3. Production of metallic powder 4 4. Pressing or compacting into the desired shape g p g p Indentify the correct order in which they have to be performed and select the correct answer using the codes given below: (a) 1 2 3 4 1‐2‐3‐4 (b) 3 1 4 2 3‐1‐4‐2 (c) 2‐4‐1‐3 (d) Page 55 of 78 4‐3‐2‐1 ( ) (b) P – 2, Q – 4, R – 3, S ‐ 1 , Q 4, 3, (d) P – 4, Q – 1, R – 2, S ‐ 3 S IAS – 2003 Assertion (A): Atomization method for production of metal powders consists of mechanical disintegration of molten stream into fine particles. p Reason (R): Atomization method is an excellent means of making powders from high temperature metals. gp g p (a) Both A and R are individually true and R is the correct explanation of A p (b) Both A and R are individually true but R is not the correct explanation of A p (c) A is true but R is false (d) A is false but R is true
- 57. S 2007 IAS – 200 Consider the following basic steps involved in the production of porous bearings: 1. 1 Sintering 2. Mixing 3. Repressing 4 4. Impregnation p g 5. Cold‐die‐compaction Which one of the following is the correct sequence of the above steps? Ch‐12: Powder Metallurgy gy Q. No 1 Option D Q. No 5 Option C 2 3 B C 6 7 B D 4 A 8 Limit, Tolerance & Fits Limit Tolerance & Fits C By S K Mondal For PSU Tolerances are specified (a) ( ) To obtain desired fits b d df (b) because it is not possible to manufacture a size exactly ( ) (c) to obtain higher accuracy g y (d) to have proper allowances ISRO‐2010 Expressing a dimension as 25.3±0.05 mm is the case of (a) Unilateral tolerance (b) Bilateral tolerance (c) Limiting dimensions GATE 2010 ISRO 2012 GATE – 2010, ISRO‐2012 −0.009 A shaft has a dimension,φ35−0.025 The Th respective values of f d ti l f fundamental d i ti and t l deviation d tolerance are (a) − 0.025, ± 0.008 (c) − 0.009, ± 0.008 (b) − 0.025,0.016 (d) − 0.009,0.016 (d) All of the above f h b GATE 1992 GATE ‐ Two shafts A and B have their diameters specified as 100 ± 0.1 mm and 0.1 ± 0.0001 mm respectively. Which of the following statements is/are true? (a) Tolerance in the dimension is greater in shaft A (b) The relative error in the dimension is greater in shaft A (c) Tolerance in the dimension is greater in shaft B (d) The relative error in the dimension is same for shaft A and shaft B For-2014 (IES, GATE & PSUs) GATE 2004 GATE ‐ 2004 S O 20 0 ISRO‐2010 In an interchangeable assembly, shafts of size +0.020 25.000 mm mate with holes of size 25.000−0.000 mm. The maximum possible clearance in the assembly will be (a) ( ) 10 microns i (b) 20 microns (c) 30 microns (d) 60 microns Dimension of the hole is 50+0.02 mm and shaft is 50+0.02 mm. −0.00 +0.00 mm +0.040 −0.0100 Page 56 of 78 The minimum clearance is (a) 0.02 mm (c) -0.02 mm 0 02 (b) 0.00 mm (d) 0.01 mm 0 01
- 58. IES 2005 IES ‐ GATE 2000 GATE ‐ p y g The tolerance specified by the designer for the diameter of a shaft is 20.00 ± 0.025 mm. The shafts produced by three different machines A, B and C have mean diameters of 19∙99 mm, 20∙00 mm and 20.01 mm respectively, with same standard deviation. Wh t will b th percentage rejection f d i ti What ill be the t j ti for the shafts produced by machines A, B and C? (a) Same f th machines A B d C since th standard ( ) S for the hi A, Band i the t d d deviation is same for the three machines (b) L t f machine A Least for hi (c) Least for machine B (d) Least for machine C GATE 2007 GATE ‐ 0 .0 5 0 A slot is to be milled centrally on a block with a dimension of 40 ± 0.05 mm. A milling cutter of 20 mm width is located with reference to the side of the block within ± 0.02 mm. The maximum offset in mm between the centre lines of the slot and the block is (a) ± 0 070 0.070 (b) 0 070 0.070 (c) ± 0.020 (d) 0.045 IES 2013 IES‐2013 IES 2011 GATE 2005 GATE ‐ Which of the following is a joint formed by Interference fit joints are provided for: (a) Assembling bush bearing in housing (b) Mounting heavy duty gears on shafts ( ) (c) Mounting pulley on shafts gp y (d) Assembly of flywheels on shafts A hole is specified as 4 0 0 . 0 0 0 mm. The mating shaft has a clearance fit with minimum clearance of 0.01 mm. The tolerance on the shaft is 0.04 mm. The maximum clearance in mm between the hole and the shaft is (a) 0.04 (b) 0.05 (c) 0.10 (d) 0.11 interference fits? (a) Joint of cycle axle and its bearing (b) Joint between I.C. Engine piston and cylinder In order to have interference fit, it is essential that the lower limit of the shaft should be (a) Greater than the upper limit of the hole (b) Lesser than the upper limit of the hole (c) Greater than the lower limit of the hole ( ) (d) Lesser than the lower limit of the hole (c) Joint between a pulley and shaft transmitting power (d) J i of l h spindle and i b i Joint f lathe i dl d its bearing GATE 2011 A hole is of dimension φ 9 +0 015 0.015 +0 GATE ‐2012 Same Q in GATE‐2012 (PI) mm. The In an interchangeable assembly, shafts of size +0.010 corresponding shaft is of dimension p g The resulting assembly has (a) loose running fit (b) close running fit (c) transition fit ( )t iti (d) interference fit φ 9 +0.001 For-2014 (IES, GATE & PSUs) mm. mm mate with holes of size 25+0.03 +0 02 0.02 25 +0.04 −0.01 mm. The maximum interference (in microns) in the assembly is (a) ( ) 40 (b) 30 (c) ( ) 20 (d) 10 IAS‐2011 Main An interference assembly, of nominal diameter 20 mm, is of a unilateral holes and a shafts. The manufacturing t l f t i tolerances f th h l are t i for the holes twice that for the shaft. Permitted interference values are 0.03 to 0.09 mm. Determine the sizes, with limits, for the two mating parts. [10‐Marks] [10 Marks] Page 57 of 78
- 59. IES 2007 IES ‐ ISRO‐2011 A shaft and hole pair is designated as 50H7d8. This assembly constitutes (a) Interference fit IES 2006 IES ‐ Which of the following is an interference fit? (a) Push fit (b) R Running fit i (c) Sliding fit (d) Shrink fit (b) Transition fit (c) Clearance fit (d) None of the above IES 2009 IES ‐ Consider the following joints: 1. Railway carriage wheel and axle 2. IC engine cylinder and li i li d d liner Which of the above joints is/are the result(s) of interference fit? ( ) (a) 1 only y (b) 2 only (c) Neither 1 nor 2 (d) Both 1 and 2 IES 2008 IES ‐ Consider the following statements: 1. The amount of interference needed to create a tight joint varies with diameter of the shaft shaft. 2. An interference fit creates no stress state in the shaft. h ft 3. The stress state in the hub is similar to a thick‐ walled cylinder with internal pressure. Which of the statements given above are correct? g (a) 1, 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only IES 2003 IES ‐ GATE 2001 GATE ‐ 2001 ( ) ( g Match List‐I (Phenomenon) with List‐II (Significant Parameters/Phenomenon) and select the correct answer using the codes given below the Lists: List‐I List I List‐II List II (Phenomenon) (Significant Parameters/Phenomenon) / ) A. Interference fit 1. Viscosity index B. Cyclic loading 2. Interference C. Gear meshing 3. Notch sensitivity D. Lubricating of bearings 4. Induced compressive stress t Codes:A B C D A B C D (a) 3 4 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 4 3 1 2 Allowance in limits and fits refers to (a) Maximum clearance between shaft and hole (b) Mi i Minimum clearance b t l between shaft and h l h ft d hole (c) Difference between maximum and minimum size of hole ( ) (d) Difference between maximum and minimum size of shaft For-2014 (IES, GATE & PSUs) Page 58 of 78 IES 2004 IES ‐ Consider the following fits: 1. I.C. engine cylinder and piston 2. B ll b Ball bearing outer race and h i t d housing i 3. Ball bearing inner race and shaft Which of the above fits are based on the interference y system? (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1, 2 and 3 GATE 1998 GATE ‐ In the specification of dimensions and fits, (a) Allowance is equal to bilateral tolerance (b) All Allowance i equal t unilateral t l is l to il t l tolerance (c) Allowance is independent of tolerance (d) Allowance is equal to the difference between p y maximum and minimum dimension specified by the tolerance.
- 60. IES 2012 IES ‐ Clearance in a fit is the difference between (a) Maximum hole size and minimum shaft size (b) Mi i Minimum h l size and maximum shaft size hole i d i h ft i (c) Maximum hole size and maximum shaft size (d) Minimum hole size and minimum shaft size ISRO‐2008 Basic shaft and basic hole are those whose upper deviations and lower deviations respectively are (a) +ve, ‐ve (b) ‐ve, +ve (c) ( ) Zero, Zero (d) ( ) None of the above IES 2008 IES ‐ S 2006 C i l IES‐2006 Conventional Find the limit sizes, tolerances and allowances for a 100 mm diameter shaft and hole pair designated by F8h10. Also specify the type of fit that the above pair belongs to. Given: 100 mm diameter lies in the diameter step range of 80‐120 mm. The fundamental deviation for shaft designation ‘f’ is ‐5.5 D0.41 f 55 The values of standard tolerances for grades of IT 8 and IT 10 are 25i and 6 i respectively. d i d 64i ti l Also, indicate the limits and tolerance on a diagram. [15‐Marks] GATE 2009 GATE ‐ pp What are the upper and lower limits of the shaft represented by 60 f8? Use the following data: Diameter 60 lies in the diameter step of 50‐80 mm. Fundamental tolerance unit, i, in μ m= 0.45 D1/3 + 0.001D, where D is the representative size in mm; Tolerance value f lT8 = 25i. T l l for i Fundamental deviation for 'f shaft = ‐5.5D0.41 (a) ( ) Lower l limit = 59.924 mm, Upper Limit = 59.970 mm (b) Lower limit = 59.954 mm, Upper Limit = 60.000 mm (c) ( ) Lower limit = 59.970 mm, Upper Limit = 60.016 mm (d) Lower limit = 60.000 mm, GATELimit = 60.046 mm For-2014 (IES, Upper & PSUs) Consider the following statements: A nomenclature φ 50 H8/p8 denotes that 1. H l di Hole diameter i 50 mm. t is 2. It is a shaft base system. 3. 8 indicates fundamental deviation. Which of the statements given above is/are incorrect? (a) 1, 2 and 3 (b) 1 and 2 only d l (c) 1 and 3 only (d) 3 only GATE 2008 (PI) GATE – 2008 (PI) Following data are given for calculating limits of dimensions and tolerances for a hole: Tolerance unit i (in µm) = 0.45 ³√D + 0.001D. The unit of D is mm. Diameter step is 18‐30 mm. If the fundamental deviation for H hole is zero and IT8 = 25 i the maximum and minimum i, limits of dimension for a 25 mm H8 hole (in mm) are (a) 24.984, 24.967 (b) 25.017, 24.984 (c) 25.033, 25.000 (d) 25.000, 24.967 Page 59 of 78 IES 2005 IES ‐ Assertion (A): Hole basis system is generally preferred to shaft basis system in tolerance design for getting the required fits. Reason (R): Hole has to be given a larger tolerance band than the mating shaft shaft. (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES 2002 IES ‐ In the tolerance specification 25 D 6, the letter D represents (a) Grade of tolerance (b) Upper deviation (c) Lower deviation ( ) yp (d) Type of fit GATE 2000 GATE ‐ A fit is specified as 25H8/e8. The tolerance value for a nominal diameter of 25 mm in IT8 is 33 microns and fundamental deviation for the shaft is ‐ 40 microns. The maximum clearance of the fit in microns is (a) ‐7 (b) 7 (c) 73 (d) 106
- 61. GATE 2003 GATE ‐ The dimensional limits on a shaft of 25h7 are (a) 25.000, 25.021 mm (b) 25.000, 24.979 mm (c) 25.000, 25.007 mm (d) 25.000, 24.993 mm GATE 1996 IES 2012 GATE – 1996, IES‐2012 GATE‐2010 (PI) A small bore is designated as 25H7. The lower (minimum) and upper (maximum) limits of the bore are 25.000 mm and 25.021 mm, respectively. When the bore is designated as 25H8, then the upper (maximum) limit is 25 033 mm When the bore is designated as 25.033 mm. hole shaft H7 The fit on a hole‐shaft system is specified as H7‐ s6.The type of fit is (a) Clearance fit (b) Running fit (sliding fit) (c) Push fit (transition fit) ( ) (d) Force fit (interference fit) ( ) 25H6, then the upper (maximum) limit of the bore (in mm) is (a) 25.001 (b) 25.005 (c) 25.009 (d) 25.013 IES 2000 IES ‐ Which one of the following tolerances set on inner diameter and outer diameter respectively of headed jig bush for press fit is correct? (a) G7 h 6 (b) F7 n6 (c) ( ) H 7h 6 h (d) F j6 F7j6 ISRO‐2008 IAS‐2010 main Interchangeability can be achieved by What is the difference between hole basis system and (a) Standardization shaft basis system ? Why is hole basis system the more extensive i use ? t i in (b) Better process planning What are the differences between interchangeability (c) Simplification and selective assembly ? (d) B Better product planning d l i GATE 2003 GATE ‐ 2003 [12‐Marks] GATE – 2008 (PI) GATE 2008 (PI) GATE 1997 GATE ‐ A three‐component welded cylindrical assembly is shown below. The mean length of the three components and their respective tolerances (both in mm) are given in the table below. Three blocks B1 , B2 and B3 are to be inserted in a channel of width S maintaining a minimum gap of width T = i i f idth 0.125 mm, as shown in Figure. For P = 18 75 ± 0 08; 18. 0.08; Q = 25.00 ± 0.12; R = 28 125 ± 0 1 and 28.125 0.1 S = 72.35 + X, (where all dimensions are in mm) the mm), tolerance X is (a) + 0.38 (a) + 0 38 For-2014 (IES, GATE & PSUs) (b) 0.38 (b) ‐ 0 38 (c) + 0.05 (c) + 0 05 Page 60 of 78 (d) 0.05 (d) ‐0 05 Assuming a normal distribution for the individual g component dimensions, the natural tolerance limits for the length (Y) of the assembly (mm) is (a) 65 ( ) 6 ±2.16 6 (b) 6 ±1.56 ( ) 6 ±0.96 (d) 6 ±0.36 65 6 (c) 65 6 65 6
- 62. GATE 2007 (PI) GATE – 2007 (PI) Tolerance on the dimension x in the two component assembly shown below is GATE ‐2007(PI) GATE 2007(PI) g The geometric tolerance that does NOT need a datum Diameter of a hole after plating needs to be controlled for its specification is (a) Concentricity (b) Runout (c) Perpendicularity ( (All dimensions in mm) ) GATE 2007 (PI) GATE – 2007 (PI) (d) Flatness (a) ( ) ±0.025 ( ) (a) ±0.030 3 (a) ±0.040 (a) ( ) ±0.045 +0 050 between 30+0.050 mm. If th plating thickness varies b t the l ti thi k i +0.010 between 10 - 15 microns, diameter of the hole before plating should be (a) 30+0.070 mm +0.030 (c) +0.080 ( ) 30+0 080 mm 0.030 GATE 2013 GATE‐2013 Cylindrical pins of 25+0.020 mm diameter are +0.010 electroplated in a shop. Thickness of the l t l t di h Thi k f th plating is 30 ±2.0 micron Neglecting gage micron. tolerances, the size of the GO gage in mm to inspect the plated components is (a) 25.042 (b) 25.052 (c) 25.074 (d) 25.084 GATE 1995 GATE ‐ GO NO GO Checking the diameter of a hole using GO‐NO‐GO gauges is an, example of inspection by …..(variables/attributes) The above statement is (a) Variables ( ) V i bl (b) Attributes (c) Cant say (d) Insufficient data For-2014 (IES, GATE & PSUs) ISRO‐2008 Plug gauges are used to (a) ( ) Measure the d h diameter of the workpieces f h k (b) Measure the diameter of the holes in the workpieces ( ) (c) Check the diameter of the holes in the workpieces (d) Check the length of holes in the workpieces (b) 30+0.065 mm +0.020 0.070 (d) 30+0 070 mm +0.040 GATE 2004 GATE ‐ GO and NO GO plug gages are to be designed for a NO‐GO 0.05 hole 200.01 mm. Gage tolerances can be taken as 10% of the hole tolerance Following ISO system of gage tolerance. design, sizes of GO and NO‐GO gage will be respectively (a) 20.010 mm and 20.050 mm (b) 20.014 mm and 20.046 mm ( ) (c) 20.006 mm and 20.054 mm 54 (d) 20.014 mm and 20.054 mm GATE 2006 VS 2012 GATE – 2006, VS‐2012 A ring gauge is used to measure (a) Outside diameter but not roundness (b) R Roundness b t not outside di d but t t id diameter t (c) Both outside diameter and roundness (d) Only external threads Page 61 of 78 Measurement of Lines & Surfaces By S K Mondal
- 63. ISRO‐2010 , ISRO‐2009, 2011 ISRO‐2008 The vernier reading should not be taken at its face value b f i f l before an actual check h l h k has been taken for (a) Zero error (b) It calibration Its lib ti ( ) (c) Flatness of measuring jaws gj (d) Temperature equalization The least count of a metric vernier caliper having In a simple micrometer with screw pitch 0.5 mm 25 divisions on vernier scale, matching with 24 and divisions of main scale (1 main scale divisions = 0.5 corresponding to 5 divisions on barrel and 12 mm) is divisions on thimble is GATE 2008 GATE – 2008 GATE 2008 td f S1 GATE – 2008 contd… from S‐1 S1 S‐1 A displacement sensor (a dial indicator) measures the lateral displacement of a mandrel mounted on the taper hole inside a drill spindle. The mandrel axis is an extension of the drill spindle taper hole axis and the protruding portion of the mandrel surface is perfectly cylindrical. Measurements are taken with the sensor placed at two positions P and Q as shown in the figure. The readings are recorded as Rx = maximum deflection minus minimum deflection, corresponding to sensor position at X, over one rotation. ISRO‐2008 Standards to be used for reference purposes in laboratories and workshops are termed as (a) Primary standards (b) ( ) Secondary standards ( ) (a) 0.005 mm 5 ( ) (b) 0.01 mm ( ) (a) 2.620 mm (c) 0.02 mm (d) 0.005mm (c) 2.120 mm the reading ( ) 5 (b) 2.512 mm (d) 5.012 mm If Rp= RQ>0, which one of the following would be consistent with the observation? (A) The drill spindle rotational axis is coincident with the drill spindle taper hole axis (B) The drill spindle rotational axis intersects the drill spindle taper hole axis at point P (C) The drill spindle rotational axis is parallel to the drill spindle taper hole axis (D) The drill spindle rotational axis intersects the drill spindle taper hole axis at point Q ISRO‐2010 A master gauge is (a) A new gauge (b) An international reference standard (c) ( ) A standard gauge f checking accuracy of t d d for h ki f gauges used on shop floors (d) A gauge used by experienced technicians GATE 2007 (PI) GATE – 2007 (PI) Which one of the following instruments is a comparator ? (a) Tool Maker’s Microscope ( ) T l M k ’ Mi ( ) (b) GO/NO GO gauge g g (c) Optical Interferometer (d) Di l G Dial Gauge (c) Tertiary standards PSU A feeler gauge is used to check the (a) Pitch of the screw (b) Surface roughness (c) Thickness of clearance (d) Flatness of a surface (d) Fl f f (d) Working standards For-2014 (IES, GATE & PSUs) divisions on thimble 50, Page 62 of 78
- 64. IAS‐2011 Main GATE ‐2012 (PI) ISRO‐2011 A sine bar has a length of 250 mm. Each roller has Draw a self explanatory sketch of Sigma A sine bar is specified by mechanical comparator. Explain how (a) Its total length (i) shock load is avoided, (b) The size of the rollers (ii) oscillations of the pointer are damped. ( ) (c) The ( ) Th centre di t t distance b t between th t rollers the two ll [10 – [10 marks] ( ) GATE – 2011 (PI) The best wire size (in mm) for measuring effective diameter of a metric th d (i l d d angle i 6 o) di t f t i thread (included l is 60 of 20 mm diameter and 2.5 mm pitch using two wire method i i th d is (a) 1.443 (b) 0.723 ( ) (c) 2.886 (d) 2.086 IES 1992 IES ‐ Which grade symbol represents surface rough of broaching? (a) N12 (b) N8 (c) N4 (d) N1 a di diameter of 20 mm. D i t f During t taper angle l measurement of a component, the height from the p , g surface plate to the centre of a roller is 100 mm. The calculated taper angle (in degrees) is ( ) (d) The distance between rollers and upper surface pp (a) 21.1 GATE 2013 GATE‐2013 method. The diameter of the best size wire in mm is (a) 0.866 (b) 1.000 (c) 1.154 (d) 2.000 IFS‐2011 y g What are they ? Define the terms 'roughness height', 'waviness width' and 'lay' in connection ISRO‐2011 CLA value and RMS values are used for measurement of (a) Metal hardness (b) Sharpness of tool edge with surface irregularities. [10‐marks] [ k ] (c) Surface dimensions (d) Surface roughness For-2014 (IES, GATE & PSUs) Page 63 of 78 (d) 68.9 To measure the effective diameter of an external metric thread (included angle is 60o) of 3 5 mm 3.5 pitch, a cylindrical standard of 30.5 mm diameter a d t o and two wires of 2 mm diameter each are used. es o d a ete eac a e The micrometer readings over the standard and over the wires are 16.532 mm and 15.398 mm, respectively. The effective diameter (in mm) of the thread is (a) 33.366 (b) 30.397 (c) 29.366 (d) 26.397 What is Wh t i meant b i t h t by interchangeable manufacture? bl f t ? Laser light has unique advantages for inspection inspection. (c) 23.6 ( ) GATE – 2011 (PI) A metric th d of pitch 2 mm and th d angle 6 t i thread f it h d thread l 60 inspected for its pitch diameter using 3‐wire p p g 3 (b) 22.8
- 65. IES 2006 IES ‐ IES 2007 IES ‐ E system The M and E‐system in metrology are related to measurement of: (a) Screw threads (b) Flatness (c) Angularity (d) Surface finish What is the dominant direction of the tool marks or scratches in a surface texture having a directional quality, called? (a) Primary texture (b) Secondary texture (c) Lay ( ) L (d) Fl Flaw Match List I with List II and select the correct answer using the code given below the lists: List I List II (Symbols for direction of lay) (Surface texture) A 4 4 B 2 1 C 1 2 D 3 3 (b) (d) A 3 3 What term is used to designate the direction of the predominant surface pattern produced by machining operation? (a) Roughness (b) Lay (c) Waviness ( ) W i (d) C t off Cut ff IES 2008 IES ‐ 2008 IES 2010 (a) (c) IES 2008 IES ‐ B 2 1 C 1 2 ISRO‐2010 Surface roughness represented b d by (a) Triangles (b) Circles (c) Squares (d) Rectangles on a drawing is D 4 4 GATE 1997 GATE ‐ 1997 List I List II (A) Surface profilometer 1. Calibration (B) Light Section Microscope 2 Form tester 2. (C) Microkater 3. Film thickness measurement (D) Interferometer 4. Centre line average 5. 5 Comparator 6. Surface lay measurement Codes:A C d A B C D A B C D (a) 4 1 2 3 (b) 4 3 5 1 (c) 4 2 1 3 (d) 3 1 2 4 For-2014 (IES, GATE & PSUs) ISRO‐2007 Gratings are used in connection with (a) Flatness measurement (b) Roundness measurement (c) Surface texture measurement (d) Li Linear di l displacement measurements Page 64 of 78 GATE 2003 GATE ‐ 1.000 Two slip gauges of 10 mm width measuring 1 000 mm and 1.002 mm are kept side by side in contact with each other lengthwise. An optical flat is kept resting on the slip gauges as shown in the figure. Monochromatic light of wavelength 0.0058928 mm is used in the inspection. The total number of straight fringes that can be observed on both slip gauges is (a) 2 (c) 8 (b) 6 (d) 13
- 66. GATE 1998 GATE ‐ 1998 GATE – 2011 (PI) Observation of a slip gauge on a fl t flatness Ob ti f li interferometer produced fringe counts numbering 10 and 14 f t d for two readings. Th second reading i di The d di is taken by rotating the set‐up by 180o. Assume that both faces of th slip gauge are fl t and th b th f f the li flat d the wavelength of the radiation is 0.5086 µm. The parallelism error (i µm) b t ll li (in ) between th t the two f faces of f the slip gauge is (a) ( ) 0.2543 (b) 1.172 (c) 0.5086 (d) 0.1272 Miscellaneous of Metrology Auto collimator is used to check (a) Roughness (b) Fl t Flatness (c) Angle (d) Automobile balance. By S K Mondal ( ) GATE – 2009 (PI) An autocollimator is used to (a) measure small angular displacements on flat surface (b) ( ) compare known and unknown dimensions (c) measure the flatness error S O 20 0 ISRO‐2010 IES 1998 IES ‐ Optical square is (a) Engineer's square having stock and blade set at 90o (b) A constant d i ti t t deviation prism h i i having th angle of the l f deviation between the incident ray and reflected ray, equal t 90o l to (c) A constant deviation prism having the angle of deviation between the incident ray and reflected ray, equal to 45o (d) Used to produce interference fringes Match List‐I with List‐II and select the correct answer using the codes given b l below the li d i h lists: List‐I List‐II ( (Measuring Device) g ) ( (Parameter Measured) ) A. Diffraction grating 1. Small angular deviations on long flat surfaces B. Optical flat B 2. 2 On‐line measurement of moving parts C. Auto collimators 3. Measurement of gear pitch D. L D Laser scan micrometer4. i t Surface t t S f texture using i t f i interferometer t 5. Measurement of very small displacements Code: A B C D A B C D (a) 5 4 2 1 (b) 3 5 1 2 (c) 3 5 4 1 (d) 5 4 1 2 GATE 2004 GATE ‐ GATE 1995 GATE ‐ (d) measure roundness error between centers GATE 1992 GATE ‐ p y q y Match the instruments with the physical quantities they measure: Instrument Measurement (A) Pilot‐tube (1) R.P.M. of a shaft g (2) Displacement p (B) McLeod Gauge (C) Planimeter (3) Flow velocity (4) Vacuum 4 (D) LVDT (5) Surface finish ( ) (6) Area Codes:A B C D A B C D ( ) (a) 4 1 2 3 ( ) (b) 3 4 6 2 (c) 4 2 1 3 (d) 3 1 2 4 For-2014 (IES, GATE & PSUs) Match the following Feature to be inspected Instrument P Pitch and Angle errors of screw thread 1 Auto Collimator 1. Q Flatness error of a surface plate 2. Optical Interferometer R Ali Alignment error of a machine slide way 3. Di idi H d f hi lid Dividing Head and Dial Gauge S P fil of a cam Profile f 4. S i i L l Spirit Level 5. Sine bar 6. Tool maker's Microscope (a) P‐6 Q‐2 R‐4 S‐6 (b) P‐5 Q‐2 R‐1 S‐6 5 (c) P‐6 Q‐4 R‐1 S‐3 (d) P‐1 Q‐4 R‐4 S‐2 Page 65 of 78 List I (Measuring instruments) (A) T l Talysurf f 1. (B) Telescopic gauge 2. (C) Transfer callipers 3. (D) Autocollimator 4. Codes:A B C D (a) ( ) 4 1 2 3 (b) (c) 4 2 1 3 (d) List II (Application) T‐slots T l t Flatness Internal diameter Roughness A B C D 4 3 1 2 3 1 2 4
- 67. GATE ‐2008 (PI) GATE 2008 (PI) GATE 2010 GATE ‐ A taper hole is inspected using a CMM, with a probe of 2 mm diameter. At a height, Z = 10 mm from the bottom, 5 points are touched and a diameter of circle (not compensated for probe size) is obtained as 20 mm. Similarly, a 40 mm diameter is obtained at a height Z = 40 mm. the smaller diameter (in mm) of hole at Z = 0 is (a) 13.334 (b) 15.334 (c) 15.442 (d) 15.542 An experimental setup is planned to determine the taper of workpiece as shown in the figure. If the two precision rollers have radii 8 mm and 5 mm and the total thickness of slip gauges i inserted b d between the rollers i 15.54 mm, the taper h ll is h angle θ is (a) degree ( )6d (b) 10 degree (c) 11 degree (d) 12 degree g Option Q. No 1 C 10 D 2 C 11 A 3 A 12 B 4 C 13 Option B 5 C 14 D 6 B 15 B 7 C 16 C 8 B 17 B 9 B For-2014 (IES, GATE & PSUs) Which f h following inevitable i the Whi h of the f ll i errors are i i bl in h measuring system and it would be vain full g y exercise to avoid them (a) Systematic errors (b) R d Random errors (c) Calibration errors (d) Environmental errors Ch‐13: Metrology gy Q. No ISRO‐2007 Page 66 of 78
- 68. Introduction Cutting Tool Materials By S K Mondal Success in metal cutting depends on selection of the proper cutting tool (material and geometry) for a given work material. A wide range of cutting tool materials is available with g g a variety of properties, performance capabilities, and cost. These include: High carbon Steels and low/medium alloy steels steels, High‐speed steels, Cast cobalt alloys, b l ll Cemented carbides, Cast carbides, Cast carbides Coated carbides, Coated high speed steels, dh h d l Ceramics, Cermets, Whisker reinforced ceramics, Whisker reinforced ceramics Sialons, Sintered polycrystalline cubic boron nitride (CBN), d l ll b b d ( ) Sintered polycrystalline diamond, and single‐crystal natural diamond. Contd… Carbon Steels FIGURE: Improvements in cutting tool materials have reduced machining time. Limited Li i d tool lif Th f l life. Therefore, not suited to mass i d production Can b f C be formed i d into complex shapes f l h for small ll production runs Low L cost Suited to hand tools, and wood working Carbon content about 0.9 to 1.35% with a hardness ABOUT 62 C Rockwell Maximum cutting speeds about 26 ft/min. dry The hot hardness value is low. This is the major factor in tool life. S 99 IAS – 1997 Assertion (A): Cutting tools made of high carbon steel have shorter tool life. Reason(R): During machining the tip of the cutting machining, tool is heated to 600/700°C which cause the teal tip to lose its hardness hardness. (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Ans. (a) For-2014 (IES, GATE & PSUs) Fig. Productivity raised by cutting tool materials High speed steel These steels are used f cutting metals at a much Th t l d for tti t l t h higher cutting speed than ordinary carbon tool steels. The high speed steels have the valuable property of retaining their hardness even when heated to red heat. Most of the high speed steels contain tungsten as the y g chief alloying element, but other elements like cobalt, chromium, vanadium, etc. may be present in some p p proportion. Page 67 of 78 Contd…
- 69. With time the effectiveness and efficiency of HSS (too s) a d t e app cat o a ge e e g adua y (tools) and their application range were gradually enhanced by improving its properties and surface condition through ‐ Refinement of microstructure Addition of large amount of cobalt and Vanadium to g increase hot hardness and wear resistance respectively Manufacture by powder metallurgical process Surface coating with heat and wear resistive g materials like TiC , TiN , etc by Chemical Vapour Deposition (CVD) or Physical Vapour Deposition (PVD) 18‐4‐1 High speed steel This t l Thi steel contains 18 per cent t t i 8 t tungsten, 4 per cent t t chromium and 1 per cent vanadium. It is considered to be one of the best of all purpose tool steels. It is widely used for drills, lathe, planer and shaper g g tools, milling cutters, reamers, broaches, threading dies, punches, etc. IES‐1993 The blade of a power saw is made of Th bl d f i d f (a) Boron steel (b) High speed steel (c) Stainless steel (d) Malleable cast iron Ans. (b) A (b) For-2014 (IES, GATE & PSUs) IES 2013 IES‐2013 Vanadium in high speed steels: ( ) (a) Has a tendency to promote decarburization y p (b) Form very hard carbides and thereby increases the wear resistance of the tool (c) Helps in achieving high hot hardness (d) H a tendency to promote retention of A Has d i f Austenite i Ans. Ans (b) IAS‐1997 Which of the following processes can be used for production thin, hard, heat resistant coating at TiN, on HSS? 1. Physical vapour deposition. . S te g u de educ g at osp e e. 2. Sintering under reducing atmosphere. 3. Chemical vapour deposition with post treatment 4. Plasma spraying. Select the correct answer using the codes given below: Codes: (a) 1 and 3 (b) 2 and 3 (c) 2 and 4 (d) 1 and 4 Ans. (a) IES‐2003 The correct sequence of elements of 18‐4‐1 HSS Th t f l t f 8 HSS tool is (a) ( ) W, Cr, V (b) Mo, Cr, V (c) Cr, Ni, C (d) Cu Zn Sn Cu, Zn, Sn Ans. (a) Molybdenum high speed steel This t l Thi steel contains 6 per cent t t i t tungsten, 6 per cent t t molybdenum, 4 per cent chromium and 2 per cent vanadium. di It has excellent toughness and cutting ability. The molybdenum high speed steels are better and p yp cheaper than other types of steels. It is particularly used for drilling and tapping operations. Page 68 of 78 IES 2007 Cutting tool material 18‐4‐1 HSS has which one of C tti t l t i l 8 HSS h hi h f the following compositions? (a) ( ) 18% W, 4% Cr, 1% V (b) ( ) 18% Cr, 4% W, 1% V (c) 18% W, 4% Ni, 1% V (d) 18% Cr, 4% Ni, 1% V Ans. (a) Ans (a) Super high speed steel This t l is l Thi steel i also called cobalt hi h speed steel ll d b lt high d t l because cobalt is added from 2 to 15 per cent, in order to increase th cutting efficiency especially at hi h t i the tti ffi i i ll t high temperatures. This steel contains 20 per cent tungsten, 4 per cent chromium, 2 per cent vanadium and 12 per cent cobalt.
- 70. IES‐1995 The compositions of some of the alloy steels are as Th iti f f th ll t l under: 1. 18 W 4 Cr 1 V 8 W C V 2. 12 Mo 1 W 4 Cr 1 V 3. 6 Mo 6 W 4 Cr 1 V 4. 18 W 8 Cr 1 V The compositions of commonly used high speed steels would include (a) 1 and 2 (b) 2 and 3 (c) 1 and 4 (d) 1 and 3 Ans. (d) IAS‐2001 Assertion (A): For high‐speed turning of magnesium Assertion (A): For high speed turning of magnesium alloys, the coolant or cutting fluid preferred is water‐ miscible mineral fatty oil. Reason (R): As a rule, water‐based oils are recommended for high‐speed operations in which high temperatures are generated due to high frictional heat. Water being a good generated due to high frictional heat Water being a good coolant, the heat dissipation is efficient. ( ) (a) Both A and R are individually true and R is the correct y explanation of A (b) Both A and R are individually true but R is not the correct explanation of A l i f A (c) A is true but R is false (d) A is false but R is true Ans. (a) Ans (a) IES‐2000 Percentage of various alloying elements present P t f i ll i l t t in different steel materials are given below: 1. 18% W; 4% Cr; 1% V; 5% Co; 0.7% C 2. 8% Mo; 4% Cr; 2% V; 6% W; 0.7% C 3. 27% Cr; 3% Ni; 5% Mo; 0.25% C 4. 18% Cr; 8% Ni; 0.15% C 4 18% Cr; 8% Ni; 0 15% C Which of these relate to that of high speed steel? (a) ( ) 1 and 3 d (b) 1 and 2 d (c) 2 and 3 (d) 2 and 4 Ans. (b) IAS 1994 Assertion (A): The characteristic feature of High speed Steel is its red hardness. Reason (R): Chromium and cobalt in High Speed promote martensite formation when the tool is cold worked. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A i b R i f l (d) A is false but R is true Ans. (b) IES‐1992 The main alloying elements in high speed Steel in Th i ll i l t i hi h d St l i order of increasing proportion are (a) ( ) Vanadium, chromium, tungsten (b) Tungsten, titanium, vanadium g (c) Chromium, titanium, vanadium (d) Tungsten chromium titanium Tungsten, chromium, titanium Ans. (a) Cast cobalt alloys/Stellite Cast cobalt alloys are cobalt‐rich, chromium‐tungsten‐ carbon C b l ll b l i h h i b cast alloys having properties and applications in the intermediate range between high‐speed steel and cemented g g p carbides. Although comparable in room‐temperature hardness to high‐ speed steel tools, cast cobalt alloy tools retain their h d d l l b l ll l i h i hardness to a much higher temperature. Consequently, they can be used at higher cutting speeds (25% higher) than HSS tools. g g p ( 5 g ) Cutting speed of up to 80‐100 fpm can be used on mild steels. Cast cobalt alloys are hard as cast and cannot be softened or y heat treated. Cast cobalt alloys contain a primary phase of Co‐rich solid solution strengthened b Cr and W and dispersion hardened b by by complex hard, refractory carbides of W and Cr. Contd… Other elements added include V, B, Ni, and Ta. Tools of cast cobalt alloys are generally cast to shape and finished to size b grinding. by f h d d They are available only in simple shapes, such as single‐ point t l and saw bl d i t tools d blades, b because of li it ti f limitations i th in the casting process and expense involved in the final shaping (grinding). The high cost of fabrication is due primarily to the high hardness of the material in the as‐cast condition. Materials machinable with this tool material include plain‐ p carbon steels, alloy steels, nonferrous alloys, and cast iron. Cast cobalt alloys are currently being phased out for cutting‐tool applications because of increasing costs, shortages of strategic raw materials (Co, W, and Cr), and the development of other superior tool materials at lower other, cost. For-2014 (IES, GATE & PSUs) IES 2011 non‐ferrous Stellite is a non ferrous cast alloy composed of: (a) Cobalt, chromium and tungsten (b) Tungsten, vanadium and chromium ( ) (c) Molybdenum, tungsten and chromium y g (d)Tungsten, molybdenum, chromium and vanadium Ans. Ans (a) Page 69 of 78 Cemented Carbide Carbides, which are nonferrous alloys, are also called, C bid hi h f ll l ll d sintered (or cemented) carbides because they are manufactured by powder metallurgy techniques techniques. Most carbide tools in use today are either straight g ( ) tungsten carbide (WC) or multicarbides of W‐Ti or W‐ Ti‐Ta, depending on the work material to be machined. Cobalt is the binder. These tool materials are much harder, are chemically more stable, have better hot hardness, high stiffness, and lower friction, and operate at hi h cutting speeds than d HSS fi i d higher i d h do HSS. They are more brittle and more expensive and use strategic metals (W T C ) more extensively. t l (W, Ta, Co) t i l Contd…
- 71. Cemented carbide tool materials based on TiC have bee de e oped, p a y o dust y been developed, primarily for auto industry applications using predominantly Ni and Mo as a binder. These are used for higher‐speed (> 1000 ft/min) finish machining of steels and some malleable cast irons. Cemented carbide tools are available in insert form in many different shapes; squares, triangles, diamonds, and rounds. d d Compressive strength is high compared to tensile strength, th f t th therefore th bit are often b the bits ft brazed t steel d to t l shanks, or used as inserts in holders. These i Th inserts may often h t ft have negative rake angles. ti k l Speeds up to 300 fpm are common on mild steels Hot hardness properties are very good Coolants d lubricants can b used t i C l t and l b i t be d to increase t l tool life, but are not required. Special alloys are needed to cut steel Contd… S 99 IAS – 1994 IES‐1995 The straight grades of cemented carbide cutting tool materials contain (a) Tungsten carbide only (b) Tungsten carbide and titanium carbide (c) Tungsten carbide and cobalt (d) T Tungsten carbide and cobalt carbide t bid d b lt bid Ans. (c) Table below shows detail grouping of cemented carbide tools ISO Application group Material Contd… Process P01 Steel, Steel Steel castings Precision and finish machining, high speed machining P10 Steel, Steel castings P20 Steel, steel castings, malleable cast iron Turning, threading, and milling high speed, small chips Turning, milling, medium speed with small chip section P30 Steel, steel castings, malleable cast iron Turning, milling, medium speed with small chip section P40 Steel and steel casting with sand inclusions Turning, planning, Turning planning low cutting speed large chip speed, section P50 Steel and steel castings Operations requiring high toughness turning, of medium or low tensile planning, shaping at low cutting speeds strength For-2014 (IES, GATE & PSUs) Assertion (A): Cemented carbide tool tips are produced by powder metallurgy. Reason (R): Carbides cannot be melted and cast cast. (a) Both A and R are individually true and R is the correct explanation of A t l ti f (b) Both A and R are individually true but R is not the correct explanation of A ( ) (c) A is true but R is false (d) A is false but R is true Ans. (a) K01 K10 K20 K30 K40 M10 M20 M30 M40 Hard grey C.l., chilled casting, Turning, precision turning and boring, milling, Al. alloys with high silicon scraping Grey C l h d C.l. hardness > 220 HB HB. Turning, milling, boring, reaming, broaching, G T i illi b i i b hi Malleable C.l., Al. alloys scraping containing Si Grey C l hardness up to 220 C.l. Turning, milling, broaching Turning milling broaching, requiring high HB toughness Soft grey C.l. Low tensile Turning, reaming under favourable conditions strength steel Soft non-ferrous metals Turning milling etc. Steel, steel castings, Turning, milling, medium cutting speed and medium manganese steel, grey C.l. chip section Steel casting, austentic steel Turning milling medium cutting speed and medium casting steel, Turning, milling, manganese steel, chip section spherodized C.l., Malleable C.l. Steel, austenitic steel, Turning, milling, planning, medium cutting speed, spherodized C.l. heat medium or large chip section resisting alloys f Free cutting steel, low tensile Turning, profile turning, specially in automatic strength steel, brass and light machines. alloy Page 70 of 78 The standards developed by ISO for grouping of carbide tools and their application ranges are given in Table below. ISO Code Colour Code Application P For machining long chip forming common materials like plain carbon and low alloy steels M For machining long or short chip forming ferrous materials like Stainless steel K For machining short chipping, ferrous and non- ferrous material and non – metals like Cast I d t l lik C t Iron, Brass etc. IES‐1999 Match List‐I (ISO classification of carbide tools) with List‐ M h Li I (ISO l ifi i f bid l ) i h Li II (Applications) and select the correct answer using the codes given below the Lists: g List‐I List‐II A. P‐10 1. Non‐ferrous, roughing cut g g B. P‐50 2. Non‐ferrous, finishing cut C. K‐10 3. Ferrous material, roughing cut D. K‐50 4. Ferrous material, finishing cut Code: A B C D A B C D (a) ( ) 4 3 1 2 (b) ( ) 3 4 2 1 (c) 4 3 2 1 (d) 3 4 1 2 Ans. (c) A ( )
- 72. Ceramics Ceramics are essentially alumina ( Al2O3 ) b d hi h based high C i i ll l i refractory materials introduced specifically for high speed machining of difficult to machine materials and cast iron. These can withstand very high temperatures are temperatures, chemically more stable, and have higher wear resistance than the other cutting tool materials materials. In view of their ability to withstand high temperatures, they can be used for machining at very high speeds of the order of 10 m/s. They can be operated at from two to three times the cutting speeds of tungsten carbide. Contd… Through last few years remarkable improvements in strength and toughness and hence overall performance of ceramic tools could have been possible by several means which include; Sinterability, microstructure, strength and toughness of Al2O3 ceramics were improved to some extent by adding TiO2 and MgO, Transformation toughening b adding appropriate T f i h i by ddi i amount of partially or fully stabilised zirconia in Al2O3 Al O powder, d Isostatic and hot isostatic pressing (HIP) – these are very effective but expensive route. It is possible to get mirror finish on cast iron using ceramic turning. i i The main problems of ceramic tools are their low strength, poor thermal characteristics, and the tendency to chipping. They are not suitable for intermittent cutting or for low g p cutting speeds. Very high hot hardness properties Often used as inserts in special holders holders. Comparison of important properties of ceramic and tungsten carbide tools Introducing nitride ceramic (Si3N4) with proper sintering technique – this material is very tough but prone to built‐up‐ edge formation in machining steels Developing SIALON – deriving beneficial effects of Al2O3 and S d Si3N4 Adding carbide like TiC (5 ~ 15%) in Al2O3 powder – to impart t i t toughness and th h d thermal conductivity l d ti it Reinforcing oxide or nitride ceramics by SiC whiskers, which enhanced strength toughness and life of the tool and thus strength, productivity spectacularly. Toughening Al2O3 ceramic by adding suitable metal like silver which also impart thermal conductivity and self lubricating property; this novel and inexpensive tool is still gp p y p in experimental stage. Contd… IES 2013 IES‐2013 Sialon ceramic is used as: ( ) (a) Cutting tool material g (b) Creep resistant (c) Furnace linens Alumina toughned by (i) Zirconia (ii) SiC whiskers (iii) Metal (Silver etc) (iii) Metal (Sil er etc) For-2014 (IES, GATE & PSUs) Cutting fluid, if applied should in flooding with fluid, copious quantity of fluid to thoroughly wet the entire machining zone, since ceramics have very poor thermal shock resistance Else it can be machined resistance. Else, with no coolant. Ceramic tools are used f machining work pieces, C i l d for hi i k i which have high hardness, such as hard castings, case hardened and h d h d d d hardened steel. d l Typical products can be machined are brake discs, brake drums, cylinder liners and flywheels. Contd… High Performance ceramics (HPC) Silicon Nitride (i) Plain (ii) SIALON (iii) Whisker toughened Contd… (d) High strength Ans. (a) Page 71 of 78 IES 2010 IES 2010 Constituents of ceramics are oxides of different materials, which are (a) Cold i d to ( ) C ld mixed t make ceramic pallets k i ll t ( ) (b) Ground, sintered and palleted to make ready , p y ceramics (c) Ground washed with acid heated and cooled Ground, acid, (d) Ground, sintered, palleted and after calcining cooled in oxygen Ans. Ans (b)
- 73. IAS‐1996 IES‐1997 IES‐1996 Match List I with List II and select the correct answer using the codes given below the lists: List I (Cutting tools) ( g ) List II (Major constituent) ( j ) A. Stellite l. Tungsten B. H.S.S. 2. Cobalt C. Ceramic 3. Alumina D. DCON 4. Columbium 5. Titanium Ti i Codes: A B C D A B C D (a) 5 1 3 4 (b) 2 1 4 3 (c) 2 1 3 4 (d) 2 5 3 4 Ans. (c) Ans (c) Assertion (A): Ceramic tools are used only for light, A i (A) C i l d l f li h smooth and continuous cuts at high speeds. Reason (R): Ceramics have a high wear resistance and high temperature resistance. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A ( ) (c) A is true but R is false (d) A is false but R is true Ans. (b) ( ) A machinist desires to turn a round steel stock of outside diameter 100 mm at 1000 rpm. The material has tensile strength of 75 kg/mm2. The depth of cut chosen is 3 mm at a feed rate of 0.3 mm/rev. Which one of the following tool h h f h f ll l materials will be suitable for machining the component under the specified cutting d h f d conditions? (a) Sintered carbides (b) Ceramic ( ) (c) HSS ( ) (d) Diamond Ans. (b) IES 2007 IAS‐2000 IAS‐2003 Consider the following cutting tool materials used for C id h f ll i i l i l d f metal‐cutting operation at high speed: 1. Tungsten carbide 2. 2 Cemented titanium carbide 3. High‐speed steel 4. C Ceramic i The correct sequence in increasing order of the range of cutting speeds for optimum use of these materials is (a) 3,1,4,2 (b) 1,3,2,4 (c) 3 1 2 4 3,1,2,4 (d) 1 3 4 2 1,3,4,2 Ans. (c) Ans (c) Which one of the following is not a ceramic? (a) Alumina (b) Porcelain (c) Whisker (d) Pyrosil Ans. (d) (d) Coated Carbide Tools Coated tools are b becoming the norm in the metalworking d l h h l k industry because coating , can consistently improve, tool life lif 200 or 300% or more. % In cutting tools, material requirements at the surface of the tool need to b abrasion resistant, h d and chemically l d be b i i hard, d h i ll inert to prevent the tool and the work material from interacting chemically with each other during cutting cutting. A thin, chemically stable, hard refractory coating of TiC, TiN, TiN or Al2O3 accomplishes this objective objective. The bulk of the tool is a tough, shock‐resistant carbide that can withstand hi h t ith t d high‐temperature plastic d f t l ti deformation and ti d resist breakage. For-2014 (IES, GATE & PSUs) Contd… The coatings must be fine grained, & free of binders porosity. and porosity Naturally, the coatings must be metallurgically bonded to the substrate. h b Interface coatings are graded to match the properties of the coating and the substrate. The coatings must be thick enough to prolong tool life g g p g but thin enough to prevent brittleness. Coatings should have a low coefficient of friction so that the chips do not adhere to the rake face. Multiple coatings are used with each layer imparting used, its own characteristic to the tool. Contd… Page 72 of 78 At room temperature, which one of the following is the correct sequence of increasing hardness of the tool materials? (a) Cast alloy‐HSS‐Ceramic‐Carbide y (b) HH‐Cast alloy‐Ceramic‐Carbide (c) HSS‐Cast alloy‐Carbide‐Ceramic (d) Cast alloy‐HSS‐Carbide‐Ceramic Ans. (c) ( ) The most successful combinations are TiN/TiC/TiCN/TiN and TiN/TiC/ Al2O3 . Chemical vapour deposition (CVD) is the technique p p ( ) q used to coat carbides. Contd…
- 74. IAS‐1999 The coating materials for coated carbide tools, includes (a) TiC, TiN and NaCN (b) TiC and TiN (c) TiN and NaCN (d) TiC and NaCN Ans. (b) A (b) TiN‐Coated High‐Speed Steel Coated high‐speed steel ( (HSS) does not routinely ) provide as dramatic improvements in cutting speeds as do coated carbides, with increases of 10 to 20% being typical. In addition to hobs, gear‐shaper cutters, and drills, g HSS tooling coated by TiN now includes reamers, taps, chasers, spade‐drill blades, broaches, bandsaw and circular saw blades, insert tooling, form tools, end mills, and an assortment of other milling cutters. Contd… Physical vapour deposition (PVD) has proved to be the HSS, best process for coating HSS primarily because it is a relatively low temperature process that does not exceed the tempering point of HSS HSS. Therefore, no subsequent heat treatment of the cutting tool i required. i l is i d The advantage of TiN‐coated HSS tooling is reduced tool wear. Less tool wear results in less stock removal during tool g regrinding, thus allowing individual tools to be g reground more times. IES‐2000 Cermets are (a) Metals for high temperature use with ceramic like g p properties (b) Ceramics with metallic strength and luster (c) Coated tool materials (d) M t l Metal‐ceramic composites i it Ans. (d) For-2014 (IES, GATE & PSUs) Cermets These sintered hard inserts are made by combining ‘cer’ from ceramics lik TiC TiN or TiCN and ‘ i like TiC, d ‘met’ f ’ from metal (bi d ) l (binder) like Ni, Ni‐Co, Fe etc. Harder, H d more chemically stable and h h i ll t bl d hence more wear resistant i t t More brittle and less thermal shock resistant Wt% of bi d metal varies f f binder t l i from 10 t 20%. to % Cutting edge sharpness is retained unlike in coated carbide inserts Can machine steels at higher cutting velocity than that used for tungsten carbide even coated carbides in case of light cuts carbide, cuts. Modern cermets with rounded cutting edges are suitable for finishing and semi‐finishing of steels at higher speeds, stainless semi finishing steels but are not suitable for jerky interrupted machining and machining of aluminium and similar materials. S IES – 2003 The correct sequence of cutting tools in the ascending order of their wear resistance is (a) HSS Cast non ferrous alloy (Stellite) Carbide HSS‐Cast non‐ferrous (Stellite)‐Carbide‐ Nitride (b) C t non‐ferrous alloy (St llit ) HSS C bid Cast f ll (Stellite)‐HSS‐Carbide‐ Nitride (c) HSS‐Cast non‐ferrous alloy (Stellite)‐Nitride‐ Carbide (d) Cast non‐ferrous alloy (Stellite)‐Carbide‐Nitride‐ Ans. (a) HSS Page 73 of 78 IES 2010 IES 2010 The cutting tool material required to sustain high temperature is (a) High carbon steel alloys (b) Composite of lead and steel (c) Cermet (d) Alloy of steel, zinc and tungsten Ans. (c) Diamonds Diamond is the hardest of all the cutting tool materials. Diamond h the f ll i properties: Di d has h following i extreme hardness, low h l thermal expansion, l high heat conductivity, and a very low co‐efficient of friction. This is used when good surface finish and dimensional accuracy are d i d desired. The work‐materials on which diamonds are successfully employed are the non ferrous one such as copper brass zinc aluminium non‐ferrous one, copper, brass, zinc, and magnesium alloys. On ferrous materials diamonds are not suitable because of the materials, diffusion of carbon atoms from diamond to the work‐piece Contd… material.
- 75. ( ) GATE – 2009 (PI) Diamond cutting tools are not recommended for machining of ferrous metals due to (a) high tool hardness (b) ( ) high thermal conductivity of work material (c) poor tool toughness (d) chemical affinity of tool material with iron Diamond tools have the applications in single point turning and g , g , ,g g , g boring tools, milling cutters, reamers, grinding wheels, honing tools, lapping powder and for grinding wheel dressing. Due to their brittle nature, the diamond tools have poor resistance to shock and so, should be loaded lightly. Polycrystalline diamond (PCD) tools consist of a thin layer (0.5 to 1.5 mm) of'fine grain‐ size diamond particles sintered together and metallurgically bonded to a cemented carbide substrate. substrate The main advantages of sintered polycrystalline tools over natural single‐crystal tools are better quality greater toughness single crystal quality, toughness, and improved wear resistance, resulting from the random orientation of the diamond grains and the lack of large cleavage g g g planes. Diamond tools offer dramatic performance improvements over carbides. Tool life is often greatly improvements over carbides Tool life is often greatly improved, as is control over part size, finish, and surface integrity. surface integrity Positive rake tooling is recommended for the vast majority of diamond tooling applications. majority of diamond tooling applications If BUE is a problem, increasing cutting speed and the use of more positive rake angles may eliminate it. f ii k l li i i Oxidation of diamond starts at about 450oC and thereafter it can even crack. For this reason the diamond tool is kept flooded by the coolant during cutting, and light feeds are used. Contd… IES‐1995 Assertion (A): Non‐ferrous materials are best A i (A) N f i l b machined with diamond tools. Reason (R): Diamond tools are suitable for high speed machining. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A ( ) (c) A is true but R is false (d) A is false but R is true Ans. (b) ( ) IES‐1992 Which of the following given the correct order of increasing hot hardness of cutting tool material? (a) Diamond, Carbide, HSS (b) Carbide, Diamond, HSS (c) HSS, carbide, Diamond (d) HSS Di HSS, Diamond, Carbide d C bid Ans. (d) For-2014 (IES, GATE & PSUs) IES‐2001 Assertion (A): Diamond tools can be used at high A i (A) Di d l b d hi h speeds. Reason (R): Diamond tools have very low coefficient of friction. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A ( ) (c) A is true but R is false (d) A is false but R is true Ans. (b) ( ) S 999 IAS – 1999 Assertion (A): During cutting, the diamond tool is kept flooded with coolant. Reason (R): The oxidation of diamond starts at about 4500C (a) Both A and R are individually true and R is the ( ) B th A d R i di id ll t d R i th correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Ans. (a) Page 74 of 78 S 999 IES – 1999 Consider the following statements: For precision machining of non‐ferrous alloys, diamond is preferred because it has 1. Low coefficient of thermal expansion 2. High wear resistance 3 3. High compression strength g p g 4. Low fracture toughness Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 Ans. (a) Cubic boron nitride/Borazon Next to diamond, cubic boron nitride is the hardest material presently available. It is made by bonding a 0.5 – 1 mm layer of p y y polycrystalline cubic boron nitride to cobalt based carbide substrate at very high temperature and pressure. It remains inert and retains high hardness and fracture oug ess a e e a ed ac g toughness at elevated machining speeds. It shows excellent performance in grinding any material of high hardness and strength strength. Contd…
- 76. The operative speed range for cBN when machining 400 grey cast iron is 300 ~400 m/min Speed ranges for other materials are as follows: Hard H d cast i t iron ( 400 BHN) : 8 – 300 m/min (> 80 / i Superalloys (> 35 RC) : 80 – 140 m/min Hardened steels (> 45 RC) : 100 – 300 m/min It is best to use cBN tools with a honed or chamfered t s c too s t o ed o c a e ed edge preparation, especially for interrupted cuts. Like , y ceramics, cBN tools are also available only in the form of indexable inserts. The only limitation of it is its high cost cost. IES‐1994 CBN is less reactive with such materials as hardened steels, hard‐chill cast iron, and nickel‐ and cobalt‐ based superalloys. CBN can be used efficiently and economically to y y machine these difficult‐to‐machine materials at higher g speeds (fivefold) and with a higher removal rate (fivefold) than cemented carbide, and with superior accuracy, finish, and surface integrity. Consider the following tool materials: 1. Carbide 2. Cermet 3. Ceramic 4. Borazon. Correct sequence of these tool materials in increasing order of their ability to retain their hot hardness is (a) ( ) 1,2,3,4 (b) 1,2,4,3 (c) 2, 1, 3, 4 (d) 2, 1, 4, 3 Ans. (a) Contd… IES‐2002 IES‐1996 Which one of the following is the hardest cutting tool material next only to diamond? (a) Cemented carbides (b) Ceramics (c) Silicon (d) C bi b Cubic boron nitride it id Ans. (d) Cubic boron nitride (a) Has a very high hardness which is comparable to y g p that of diamond. (b) Has a hardness which is slightly more than that of HSS (c) Is used for making cylinder blocks of aircraft engines (d) I d f ki ti l l Is used for making optical glasses. Ans. (a) IAS‐1998 Which of the following tool materials have cobalt as a constituent element? 1. Cemented carbide 2. CBN 3. Stellite 4. UCON Select the correct answer using the codes given below: Codes: C d (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 (d) 2 and 3 Ans. (b) ( ) For-2014 (IES, GATE & PSUs) Coronite Coronite is made basically by combining HSS for strength and toughness and tungsten carbides f h h d bid for heat and wear resistance. d i Microfine TiCN particles are uniformly dispersed into the matrix. Unlike a solid carbide, the coronite b d lk ld bd h based tool is made of three l d f h layers; the central HSS or spring steel core th t l HSS i t l a layer of coronite of thickness around 15% of the tool diameter a thin (2 to 5 μm) PVD coating of TiCN The coronite tools made b hot e trusion follo ed b PVD by extrusion followed by PVD‐ coating of TiN or TiCN outperformed HSS tools in respect of cutting forces tool life and surface finish. forces, Page 75 of 78 IES‐1994 Cubic boron nitride is used (a) As lining material in induction furnace g (b) For making optical quality glass. (c) For heat treatment (d) For none of the above. Ans. (d) (d) IES‐1993 Match List I with List IT and select the correct answer using the M t h Li t I ith Li t IT d l t th t i th codes given below the lists: List ‐ I (Cutting tool Material) List ‐ I I(Major characteristic constituent) h t i ti tit t) A. High speed steel 1. Carbon y B. Stellite 2. Molybdenum C. Diamond 3. Nitride D. Coated carbide tool 4. Columbium 5. 5 Cobalt Codes: A B C D A B C D (a) 2 1 3 5 (b) 2 5 1 3 (c) 5 2 4 3 (d) 5 4 2 3 Ans. (b)
- 77. IES‐2003 IES‐2000 Which one of the following is not a synthetic abrasive material? (a) Silicon Carbide (b) Aluminium Oxide (c) Titanium Nitride (d) Cubic Boron Nitride Ans. (b) Consider the following tool materials: 1. HSS 2. Cemented carbide 3. Ceramics 4. Diamond The correct sequence of these materials in decreasing order of their cutting speed is (a) ( ) 4, 3, 1, 2 (b) 4, 3, 2, 1 (c) 3, 4, 2, 1 (d) 3, 4, 1, 2 Ans. (b) IAS‐2001 Attrition wear Match. List I (Cutting tool materials) with List II M t h Li t I (C tti t l t i l ) ith Li t II (Manufacturing methods) and select the correct answer using the codes given below the Lists: List I List II A. HSS 1. Casting B. Stellite B 2. 2 Forging C. Cemented carbide 3. Rolling D. UCON UCO 4 4. Extrusion 5. Powder metallurgy Codes:A B C D A B C D (a) 3 1 5 2 (b) 2 5 4 3 (c) 3 5 4 2 (d) 2 1 5 3 Ans. (d) Ans (d) The t bonding between th chip and t l material at the hi Th strong b di b t d tool t i l t high temperature is conducive for adhesive wear. The adhesive wear in the rough region is called attrition wear . In the rough region, some parts of the worn surface are still covered b molten chip and the i d by l hi d h irregular attrition wear l ii occurs in this region . The irregular attrition wear is due to the intermittent adhesion during interrupted cutting which makes a periodic attachment and detachment of the work material on the tool surface surface. Therefore, when the seizure between workpiece to tool is broken, the small fragments of tool material are plucked and brought away by the chip. IES‐2005 Consider the following statements: An increase in the C id th f ll i t t t A i i th cobalt content in the straight carbide grades of carbide tools 1. Increases the hardness. 2. D Decreases the hardness. h h d 3. Increases the transverse rupture strength 4. Lowers the transverse rupture strength. Which of the statements given above are correct? (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) 2 and 3 Ans. (d) For-2014 (IES, GATE & PSUs) Page 76 of 78 IES‐1999 Match List‐I with List‐II and select the correct answer M t h Li t I ith Li t II d l t th t using the codes given below the Lists: List I List II (Materials) (Applications) A. Tungsten carbide 1. Abrasive wheels B. Sili B Silicon nitride i id 2. Heating elements H i l C. Aluminium oxide 3. Pipes for conveying liquid metals q D. Silicon carbide 4. Drawing dies Code: A B C D A B C D (a) 3 4 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 4 3 1 2 Ans. (d) Ans (d) IES‐1996 The limit to the maximum hardness of a work Th li it t th i h d f k material which can be machined with HSS tools even at low speeds is set by which one of the t l d i t b hi h f th following tool failure mechanisms? (a) ( ) Attrition (b) Abrasion (c) Diffusion (d) Plastic deformation under compression Plastic deformation under compression. Ans. (a)
- 78. PRODUCTION ANSWERS‐2014 For any doubt send SMS to 9582314327 Page‐1 ‐ A B C A C Page‐2 C D B C D C Page‐6 ‐ D C D B D B ‐ 1597, 0, 429, 1454, 127 1265 B B B D A ‐ D B B B B A A C B D A 828, 1200, 231, 4.021 F291 N457 Fn336 Fs408 β32.49 D ‐ Page‐12 C B B A B C Page‐16 A C B C B ‐ C D B B D C 12 B 0.816 74 A D D A D A D D Page‐8 C&D A C D B B B Page‐7 B Page‐11 Page‐4 Page‐3 B ‐ ‐ Page‐17 B ‐ A B b B B C D C A A B A A A B C D 0.07, 54 A C C D B D A B C D D C A D A ‐ A B A C D A A C C A Page‐26 D C C A D A A Page‐22 D B D D A A Page‐27 C C B D A ‐ A2, B6 C5, D3 B C Page‐31 C A ‐ C ‐ C ‐ D A D B B ‐ B C B C D B A D C B B A C C A A Page‐15 C A B C ‐ D B None C ‐ C B ‐ B ‐ B ‐ B C A C D A C D B A D D C C B A Page‐30 B C C A3, B2, C C1, D5 D C B A C D A A B C A B A B C D A D A 2134 B C B C D B B C C For-2014 (IES, GATE & PSUs) D D D C D A Page‐34 Page‐33 Page 77 of 78 B A C B C D D A C A A B B ‐ Page‐25 Page‐29 A C C(11) A D A B ‐ C Page‐20 A 36.67 2.3, 10.78, 21.42, 35.85 26 195 ‐ C C B A B Page‐10 Page‐24 Page‐28 Page‐32 Gutter A Page‐23 B D B A Page‐19 A Page‐21 A D D Page‐14 Page‐18 B A B A A Page‐9 Page‐13 C C A B C C Page‐5 B B ‐ D C ‐ Page‐35 B C
- 79. Page‐36 539, 467 Centre ‐ 2.04 130 ‐ ‐ Page‐37 3.93 ‐ B 3.63 ‐ ‐ ‐ ‐ 44,25, C 361,231 Page‐41 A C B D D C D D A A A A D C A A C D C B C ‐ A C ‐ B C C B D ‐ B ‐ ‐ C C B A D B B C A B D B B B A A B C C B C C C D C D C A B ‐ ‐ D ‐ D C C A C D B ‐ A ‐ D C ‐ C B D D Extrusion ‐ B A C D A ‐ C ‐ B ‐ ‐ B C ‐ 74, 28 A C A A ‐ B D C D D D C A C B A C A ‐ A D B A C C ‐ A B C D B D B C ‐ A ‐ D ‐ For-2014 (IES, GATE & PSUs) B B B Page 78 of 78 B C A C D D C C C A A A B ‐ A C A C D B B C C ‐ C C D B A C B B ‐ C B B Page‐55 D C C B B B D D C Page‐60 B B B D ‐ D ‐ C B ‐ B B Page‐64 B A A * A C Page‐50 ‐ C C Page‐59 Page‐63 C C ‐ C B 4.2 50,33,4 A C A B B Page‐54 Page‐58 Page‐66 A ‐ C D A Page‐53 B B D Page‐45 Page‐49 Page‐48 Page‐62 C C ‐ Page‐44 Page‐43 Page‐57 Page‐61 ‐ D B A Page‐52 Page‐56 B B A A A B Page‐40 ‐ Page‐47 Page‐51 A D D C D C Page‐39 ‐ Page‐42 Page‐46 B B A Page‐38 D ‐ ‐ Page‐65 C D B