2015 Metal Cutting Metal Forming Metrology

For -2015 Rev1

Metal Cutting, Metal Forming & Metrology

Theory, Questions & Answers (All Questions are in Sequence)

IES-1992-2014 (23 Yrs.), GATE-1992-2014 (23 Yrs.), GATE (PI)-2000-2014 (15 Yrs.), IAS-1994-2011

(18 Yrs.), some PSUs questions and conventional questions are added.

Section-I: Theory of Metal Cutting Theory & Questions Answer &Explanation

Chapter-1: Basics of Metal Cutting Page-2 Page-137

Chapter-2: Force & Power in Metal Cutting Page-9 Page-141

Chapter-3: Tool life, Tool Wear, Economics and Machinability Page-16 Page-151

Section-II: Metrology

Chapter-4: Limit, Tolerance & Fits Page-34 Page-160

Chapter-5: Measurement of Lines & Surfaces Page-44 Page-165

Chapter-6: Miscellaneous of Metrology Page-55 Page-166

Section-III: Metal Forming

Chapter-7: Cold Working, Recrystalization and Hot Working Page-60 Page-167 Chapter-8: Rolling Page-65 Page-168 Chapter-9: Forging Page-75 Page-171 Chapter-10: Extrusion & Drawing Page-86 Page-176 Chapter-11: Sheet Metal Operation Page-99 Page-179 Chapter-12: Powder Metallurgy Page-117 Page-184

Section-IV: Cutting Tool Materials Page-127 Page-185

Section -V: Forging Analysis -- Page-188

For-2015 (IES, GATE & PSUs) Page 1 of 205 Rev.1

TheoryofMetalCuttingTheoryofMetalCutting

B SKM d lBySKMondal

1

IAS 2009 iIAS2009mainy Name four independent variables and three dependentp pvariables in metal cutting. [ 5 marks]

I d d tV i bl D d tV i blIndependentVariables DependentVariables

Startingmaterials ForceorpowerrequirementsStartingmaterials(tool/work)

ForceorpowerrequirementsMaximumtemperaturein

Toolgeometry cuttingCuttingVelocityLubrication

SurfacefinishLubrication

Feed&Depth ofcut 2

GATE 2014GATE2014Better surface finish is obtained with a large rakegangle because(a) the area of shear plane decreases resulting in the(a) the area of shear plane decreases resulting in thedecrease in shear force and cutting force(b) the tool becomes thinner and the cutting force is(b) the tool becomes thinner and the cutting force isreduced( ) l h i l d i h i(c) less heat is accumulated in the cutting zone(d) the friction between the chip and the tool is less

3

IES 2013IES2013Carbide tool is used to machine a 30 mm diameterCarbide tool is used to machine a 30 mm diameter

steel shaft at a spindle speed of 1000 revolutions per

minute. The cutting speed of the above turning

operation is:

( )(a) 1000 rpm

(b) 1570 m/min(b) 1570 m/min

(c) 94.2 m/min

(d) 47.1 m/min 4

S 200IES2001For cutting of brass with single point cutting toolFor cutting of brass with singlepoint cutting toolon a lathe, tool should have( ) N ti k l(a) Negativerakeangle(b) Positiverakeangle(c) Zerorakeangle(d) Zerosidereliefangle(d) Zerosidereliefangle

5

IES1995Singlepointthreadcuttingtoolshouldideallyhave:

a) Zerorake)b) Positiverakec) Negativerakec) Negativeraked) Normalrake

6

GATE1995;2008C tti ti i t i b Cuttingpowerconsumptioninturningcanbesignificantlyreducedbyg y y(a)Increasingrakeangleofthetool

(b)Increasingthecuttinganglesofthetool

(c)Wideningthenoseradiusofthetool

(d)I i h l l(d)Increasingtheclearanceangle

7

IES1993Assertion (A): For a negative rake tool, the specificAssertion (A): For a negative rake tool, the specificcutting pressure is smaller than for a positive raketool under otherwise identical conditions.tool under otherwise identical conditions.Reason (R): The shear strain undergone by the chipin the case of negative rake tool is largerin the case of negative rake tool is larger.(a) Both A and R are individually true and R is the

t l ti f Acorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true

8

S 200IES 2005Assertion (A): Carbide tips are generally givenAssertion (A): Carbide tips are generally givennegative rake angle.Reason (R): Carbide tips are made from very hardReason (R): Carbide tips are made from very hardmaterials.( ) B th A d R i di id ll t d R i th(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

9For-2015 (IES, GATE & PSUs) Page 2 of 205 Rev.1

S 2002IES 2002Assertion (A): Negative rake is usually provided onAssertion (A): Negative rake is usually provided oncarbide tipped tools.Reason (R): Carbide tools are weaker inReason (R): Carbide tools are weaker incompression.( ) B th A d R i di id ll t d R i th(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

10

IES2011Which one of the following statement is NOT correctWhich one of the following statement is NOT correctwith reference to the purposes and effects of rake angleof a cutting tool?of a cutting tool?(a) To guide the chip flow direction(b) To reduce the friction between the tool flanks andthe machined surface(c) To add keenness or sharpness to the cutting edges.(d) To provide better thermal efficiency.(d) To provide better thermal efficiency.

11

GATE 2008(PI)( )Brittle materials are machined with tools

having zero or negative rake angle because it

(a) results in lower cutting force

( )(b) improves surface finish

(c) provides adequate strength to cutting tool(c) provides adequate strength to cutting tool

(d) results in more accurate dimensions(d) results in more accurate dimensions

12

l

ForIESOnly

IES2007ConventionalCast iron with impurities of carbide requires aCast 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

f [ k ]reasons for your answer. [ 2 marks]

Answer: Free carbides in castings reduce their machinabilityAnswer: Free carbides in castings reduce their machinability

and cause tool chipping or fracture, necessitating tools with

high toughness. Zero rake tool is perfect for this purpose.

13

S 99IAS 1994ConsiderthefollowingcharacteristicsConsiderthefollowingcharacteristics1. Thecuttingedgeisnormaltothecuttingvelocity.

Th tti f i t di ti l2. Thecuttingforcesoccurintwodirectionsonly.3. Thecuttingedgeiswiderthanthedepthofcut.Thecharacteristicsapplicabletoorthogonalcuttingwouldinclude(a) 1and2 (b) 1and3(c) 2and3 (d) 1 2and3(c) 2and3 (d) 1,2and3

14

IES 2014IES 2014Which one of the following statements is correct aboutWhich one of the following statements is correct aboutan oblique cutting?(a) Direction of chip flow velocity is normal to the(a) Direction of chip flow velocity is normal to thecutting edge of the tool(b) O l t t f tti f t th(b) Only two components of cutting forces act on thetool(c) cutting edge of the tool is inclined at an acute angleto the direction of tool feed(d) Cutting edge clears the width of the workpiece

15

IES 2012IES 2012Duringorthogonalcutting,anincreaseincuttingspeedDuringorthogonalcutting,anincreaseincuttingspeedcauses(a)Anincreaseinlongitudinalcuttingforce(a)Anincreaseinlongitudinalcuttingforce(b)Anincreaseinradialcuttingforce(c)Anincreaseintangentialcuttingforce(d)Cuttingforcestoremainunaffected( ) g

16

IES2006Whichofthefollowingisasinglepointcuttingtool?(a) Hacksawblade(b) Millingcutter(b) Millingcutter(c) Grindingwheel(d) P ti t l(d) Partingtool

IES 2012IES 2012Statement(I):Negativerakeanglesarepreferredonrigidset( ) g g p gupsforinterruptedcuttinganddifficulttomachinematerials.Statement(II):Negativerakeangledirectsthechipsontothemachinedsurface( ) B h S (I) d S (II) i di id ll(a) Both Statement (I) and Statement (II) are individuallytrue and Statement (II) is the correct explanation ofStatement (I)Statement (I)(b) Both Statement (I) and Statement (II) are individuallytrue but Statement (II) is not the correct explanation of( ) pStatement (I)(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true


IES2003The angle of inclination of the rake face withThe angle of inclination of the rake face withrespect to the tool base measured in a planeperpendicular to the base and parallel to the widthperpendicular to the base and parallel to the widthof the tool is called(a) Back rake angle(a) Back rake angle(b) Side rake angle(c) Side cutting edge angle(d) End cutting edge angle( ) g g g

19

GATE(PI)1990The diameter and rotational speed of a job are 100 mm and

500 rpm respectively The high spot (Chatter marks) are500 rpm respectively. The high spot (Chatter marks) are

found at a spacing of 30 deg on the job surface. The chatter

frequency is

(a) 5 Hz (b) 12 Hz (c) 100 Hz (d) 500 Hz

20

S 996IAS 1996ThetoollifeincreaseswiththeThetoollifeincreaseswiththe(a) Increaseinsidecuttingedgeangle(b) D i id k l(b) Decreaseinsiderakeangle(c) Decreaseinnoseradius(d) Decreaseinbackrakeangle

21

S 99IAS 1995ThrustforcewillincreasewiththeincreaseinThrustforcewillincreasewiththeincreasein(a) Sidecuttingedgeangle(b)T l di (b)Toolnoseradius(c) Rakeangle(d)Endcuttingedgeangle.

22

IES 2010IES2010Consider the following statements:Consider the following statements:In an orthogonal, singlepoint metal cutting,

th id tti d l i i das the sidecutting edge angle is increased,1. The tangential force increases.g2. The longitudinal force drops.Th di l f i3. The radial force 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

23

IES1995Th l b t th f d th fl k f thThe angle between the face and the flank of thesingle point cutting tool is known as

a) Rake angleb) Clearance anglegc) Lip angled) Point angled) Point angle.

24

IES2006Assertion (A): For drilling cast iron the tool isAssertion (A): For drilling cast iron, the tool isprovided with a point angle smaller than thatrequired for a ductile materialrequired for a ductile material.Reason (R): Smaller point angle results in lowerk lrake angle.

(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not theycorrect explanation of A(c) A is true but R is false(c) A is true but R is false(d) A is false but R is true

25

IES2002Consider the following statements:Consider the following statements:The strength of a single point cutting tool dependsupon1. Rake angle2. Clearance angle3. Lip angle3. Lip angleWhich of these statements are correct?( ) d (b) d(a) 1 and 3 (b) 2 and 3(c) 1 and 2 (d) 1, 2 and 3

26

IES 2012IES 2012ToollifeincreasewithincreaseinToollifeincreasewithincreasein(a)Cuttingspeed(b)N di (b)Noseradius(c)Feed(d)Depthofcut


IES2009Consider the following statements with respectConsider the following statements with respectto the effects of a large nose radius on the tool:

It d t i t f fi i h1. It deteriorates surface finish.2. It increases the possibility of chatter.3. It improves tool life.Which of the above statements is/are correct?Which of the above statements is/are correct?(a) 2 only (b) 3 only( ) d l (d) d(c) 2 and 3 only (d) 1, 2 and 3

28

IES 1995IES1995Consider the following statements about noseConsider the following statements about noseradius1 It improves tool life1. It improves tool life2. It reduces the cutting force3. 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(c) 1 and 3 (d) 1, 2 and 3

29

IES 1994IES1994Tool geometry of a single point cutting tool is specified bythe following elements:1. Back rake angle2. Side rake angle3. End cutting edge angle4. Side cutting edge angle5. Side relief angle6. End relief angle7. Nose radiusThe correct sequence of these tool elements used forcorrectly 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 30

IES 2009IES2009The following tool signature is specified for a singleThe following tool signature is specified for a singlepoint cutting tool in American system:

8 610, 12, 8, 6, 15, 20, 3What does the angle 12 represent?(a) Side cuttingedge angle(b) Side rake angle(b) Side rake angle(c) Back rake angle(d) Sid l l(d) Side clearance angle

31

S 993IES1993In ASA System if the tool nomenclature is 8 6 5 5In ASA System, if the tool nomenclature is 865510152mm, then the side rake angle will be( ) (b) 6 ( ) 8 (d) (a) 5 (b) 6 (c) 8 (d) 10

32

ISRO2011A cutting tool having tool signature as 10, 9, 6, 6, 8, 8,

2 will have side rake angle

(a) 10o (b) 9o (c) 8o (d) 2o

33

GATE2008In a single point turning tool, the side rake angleIn a single point turning tool, the side rake angleand orthogonal rake angle are equal. is theprincipal cutting edge angle and its range isprincipal cutting edge angle and its range is

. The chip flows in the orthogonal plane.The value of is closest to0 90o o The value of is closest to(a) 00 (b) 450

(c) 600 (d) 900

34

G 20 0 ( )GATE 2010(PI)The tool geometry of a single point right handed turningThe tool geometry of a single point right handed turningtool is provided in the orthogonal rake system (ORS).The sum of the principal (major) cutting edge angle andThe sum of the principal (major) cutting edge angle andthe auxiliary (minor) cutting edge angle of the above toolis 90o The inclination angles of the principal and theis 90o. The inclination angles of the principal and theauxiliary cutting edges are both 0o. The principal andauxiliary orthogonal clearance angles are 10o and 8oauxiliary orthogonal clearance angles are 10 and 8 ,respectively. The rake angle (in degree) measured on theorthogonal plane isorthogonal plane is(a) 0 (b) 2 (c) 8 (d) 10

35

IAS 2009Main


GATE2001D i h l i f ild l i hDuring orthogonal cutting of mild steel witha 10 rake angle tool, the chip thickness ratiowas obtained as 0.4. The shear angle (indegrees) evaluated from this data isg )(a)6.53 (b)20.22( ) ( )(c)22.94 (d)50.00

37

GATE2011A single point cutting tool with 12 rake angle isA single point cutting tool with 12 rake angle isused to machine a steel work piece. The depth ofcut i e uncut thickness is 0 81 mm The chipcut, i.e. uncut thickness is 0.81 mm. The chipthickness under orthogonal machining condition is1 8 mm The shear angle is approximately1.8 mm. The shear angle is approximately(a) 22(b) 26(c) 565(d) 76

38

IES1994The following parameters determine theThe following parameters determine themodel of continuous chip formation:

T f d1. True feed2. Cutting velocityg y3. Chip thickness

R k l f h i l4. Rake angle of the cutting tool.The parameters which govern the value of shearp gangle would include( ) d (b) d(a) 1,2 and 3 (b) 1,3 and 4(c) 1,2 and 4 (d) 2,3 and 4 39

GATE2014During pure orthogonal turning operation of a

hollow cylindrical pipe, it is found that the

thickness of the chip produced is 0 5 mm The feedthickness of the chip produced is 0.5 mm. The feed

given to the zero degree rake angle tool is 0.2g g g

mm/rev. The shear strain produced during the

operation is .

40

IES 2004In a machining operation chip thickness ratio

is 0.3 and the rake angle of the tool is 10. What

is the value of the shear strain?is the value of the shear strain?

(a) 0.31 (b) 0.13(a) 0.3 (b) 0. 3

(c) 3.00 (d) 3.34

41

Mi i h i i

IES 2009Minimum shear strain inorthogonal turning with a cuttingg g gtool of zero rake angle is

(a) 0 0(a) 0.0(b) 0.5( ) 5(c) 1.0(d)(d) 2.0

42

GATE(PI)1990A single point cutting tool with 120 rake angle is

used for orthogonal machining of a ductileused for orthogonal machining of a ductile

material. The shear plane angle for the

theoretically minimum possible shear strain to

occur

(a) 51 (b) 45(a) 51 (b) 45

(c) 30 (d) None of these( ) 3 ( )

43

GATE 2012GATE2012Details pertaining to an orthogonal metal cuttingp g g gprocess are given below.

Chip thickness ratio 0 4Chip thickness ratio 0.4Undeformed thickness 0.6 mmR k l Rake angle +10Cutting speed 2.5 m/sMean thickness of primary shear zone 25 microns

The shear strain rate in s1 during the process isThe shear strain rate in s during the process is(a) 0.1781105 (b) 0.7754105

( ) 5 (d) 5(c) 1.0104105 (d) 4.397105

44

IES2004ConsiderthefollowingstatementswithrespecttoConsiderthefollowingstatementswithrespecttothereliefangleofcuttingtool:1 Thisaffectsthedirectionofchipflow1.Thisaffectsthedirectionofchipflow2.Thisreducesexcessivefrictionbetweenthetool

d k iandworkpiece3.Thisaffectstoollife4.ThisallowsbetteraccessofcoolanttothetoolworkpieceinterfacepWhichofthestatementsgivenabovearecorrect?(a) 1and2 (b) 2and3(a) 1and2 (b) 2and3(c) 2and4 (d) 3and4 45For-2015 (IES, GATE & PSUs) Page 6 of 205 Rev.1

IES2006Considerthefollowingstatements:Considerthefollowingstatements:1. Alargerakeanglemeanslowerstrengthofthecuttingedgecuttingedge.2. Cuttingtorquedecreaseswithrakeangle.Whichofthestatementsgivenaboveis/arecorrect?(a) Only1 (b) Only2( ) y ( ) y(c) Both1and2 (d) Neither1nor2

46

IES2004Match.ListIwithListIIandselectthecorrectanswerMatch.ListIwithListIIandselectthecorrectanswerusingthecodesgivenbelowtheLists:

ListI ListIIListI ListIIA. Planapproachangle 1. ToolfaceB Rakeangle 2 ToolflankB. Rakeangle 2. ToolflankC. Clearanceangle 3. ToolfaceandflankD W d l C i dD. Wedgeangle 4. Cuttingedge

5. ToolnoseA B C D A B C D

(a) 1 4 2 5 (b) 4 1 3 25 3(c) 4 1 2 3 (d) 1 4 3 5

47

IES2004,ISRO2009Th k l f tti t l i h l The rake angle of a cutting tool is 15, shear angle 45

and cutting velocity 35 m/min. What is the velocityg y 35 / y

of chip along the tool face?

(a) 28.5 m/min (b) 27.3 m/min

(c) 25.3 m/min (d) 23.5 m/min

48

IES2008Considerthefollowingstatements:Considerthefollowingstatements:Inanorthogonalcuttingthecuttingratioisfoundtobe075.Thecuttingspeedis60m/minanddepthofcut240 75.Thecuttingspeedis60m/minanddepthofcut2 4mm.Whichofthefollowingarecorrect?1 Chipvelocitywillbe45m/min1. Chipvelocitywillbe45m/min.2. Chipvelocitywillbe80m/min.3 Chipthicknesswillbe1 8mm3. Chipthicknesswillbe18mm.4. Chipthicknesswillbe32mm.

l h h d b lSelectthecorrectanswerusingthecodegivenbelow:(a) 1and3 (b) 1and4(c) 2and3 (d) 2and4

49

IES 2014IES 2014In an orthogonal turning process, the chip thickness =In an orthogonal turning process, the chip thickness0.32 mm, feed = 0.2 mm/rev. then the cutting ratio willbebe(a) 2.6(b)(b) 3.2(c) 1.6(d) 1.8

50

IES2001If is the rake angle of the cutting tool is theIf is the rake angle of the cutting tool, is theshear angle and V is the cutting velocity, then thel it f hi lidi l th h l i

velocity of chip sliding along the shear plane isgiven by

(a) (b)cosV sinV (a) (b)cos( ) ( )cos

(c) (d)cossin( )V

sin

sin( )V

51

IES2003An orthogonal cutting operation is beingAn orthogonal cutting operation is beingcarried out under the following conditions:

tti d / d th f tcutting speed = 2 m/s, depth of cut = 0.5 mm,chip thickness = 0.6 mm. Then the chipvelocity is(a) 2 0 m/s (b) 2 4 m/s(a) 2.0 m/s (b) 2.4 m/s(c) 1.0 m/s (d) 1.66 m/s

52

IAS2003Inorthogonalcutting shearangleistheanglebetweenInorthogonalcutting,shearangleistheanglebetween

(a) Shearplaneandthecuttingvelocity( ) p g y

(b) Shearplaneandtherakeplane

(c) Shearplaneandtheverticaldirection

(d) Shearplaneandthedirectionofelongationofcrystalsinthechipthechip

53

IAS2002


IAS2000

55

IAS1998The cutting velocity in m/sec for turning a work pieceThe cutting velocity in m/sec, for turning a work piece

of diameter 100 mmat the spindle speed of 480 RPM is

(a) 1.26 (b) 2.51 (c) 48 (d) 151

56

IAS1995In an orthogonal cutting, the depth of cut is halved andIn an orthogonal cutting, the depth of cut is halved andthe feed rate is double. If the chip thickness ratio isunaffected with the changed cutting conditions, theg g ,actual chip thickness will be(a) Doubled (b) halved( ) ( )(c) Quadrupled (d) Unchanged.

57

G 2009 ( )GATE 2009(PI)CommonDataS1An orthogonal turning operation is carried out at 20An orthogonal turning operation is carried out at 20

m/min cutting speed, using a cutting tool of rake angle

15o. The chip thickness is 0.4 mm and the uncut chip

hi k ithickness is 0.2 mm.

The shear plane angle (in degrees) isThe shear plane angle (in degrees) is

(a) 26.8 (b) 27.8 (c) 28.8 (d) 29.8( ) ( ) 7 ( ) ( ) 9

58

G 2009 ( )GATE 2009(PI)CommonDataS2An orthogonal turning operation is carried out at 20An orthogonal turning operation is carried out at 20

m/min cutting speed, using a cutting tool of rake angle

15o. The chip thickness is 0.4 mm and the uncut chip

hi k ithickness is 0.2 mm.

The chip velocity (in m/min) isThe chip velocity (in m/min) is

(a) 8 (b) 10 (c) 12 (d) 14( ) ( ) ( ) ( ) 4

59

GATE1995Plainmillingofmildsteelplateproduces(a)Irregularshapeddiscontinuouschips(a) egu a s aped d sco t uous c ps(b)Regularshapeddiscontinuouschip(c)Continuouschips ithoutbuiltupedge(c)Continuouschipswithoutbuiltupedge(d)Joinedchips

60

IES 2007IES2007Duringmachining,excessmetalisremovedintheformofchipasinthecaseofturningonalathe.Whichofthefollowingarecorrect?C ti ibb lik hi i f d h t iContinuousribbonlikechipisformedwhenturning1. Atahighercuttingspeed

A l i d2. Atalowercuttingspeed3. Abrittlematerial

Ad il i l4. AductilematerialSelectthecorrectanswerusingthecodegivenbelow:( ) d (b) d(a) 1and3 (b) 1and4(c) 2and3 (d) 2and4

61

IAS1997Considerthefollowingmachiningconditions BUEwillConsiderthefollowingmachiningconditions:BUEwillformin

(a) Ductilematerial. (b) Highcuttingspeed.

(c) Smallrakeangle. (d) Smalluncutchipthickness.

62

GATE2002Ab ilt d i f d hil hi i Abuiltupedgeisformedwhilemachining

(a)Ductilematerialsathighspeed(a)Ductilematerialsathighspeed

(b)Ductilematerialsatlowspeedp

(c)Brittlematerialsathighspeed

(d)Brittlematerialsatlowspeed


G 2009GATE2009F i ti t th t l hi i t f b d d bFriction at the toolchip interface can be reduced by

(a) decreasing the rake angle(a) decreasing the rake angle

(b) increasing the depth of cutg p

(c) Decreasing the cutting speed

(d) increasing the cutting speed

64

IES1997Assertion (A): For high speed turning of cast ironAssertion (A): For high speed turning of cast ironpistons, carbide tool bits are provided with chipbreakers.Reason (R): High speed turning may produce long,ribbon type continuous chips which must be brokeninto small lengths which otherwise would beinto small lengths which otherwise would bedifficult to handle and may prove hazardous.(a) Both A and R are individually true and R is the

l i f Acorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

65

WorkbookCh1:MechanicsofBasicMachiningOperation

Q. No Option Q. No Optionp p

1 C 11 D

2 B 12 D2 B 12 D

3 D 13 B

4 C 14 C

5 B 15 D

6 D 16 B

7 B 17 B7 B 17 B

8 A 18 D

9 B 19 D9 B 19 D

10 B 20 B 66

A l i f M t l C ttiAnalysis of Metal Cutting

BySKMondal67

ESE2000(Conventional)Th f ll i d t f th th l tti t tThe following data from the orthogonal cutting testis available. Rake angle = 100, chip thickness ratio =

t hi thi k idth f t0.35, uncut chip thickness = 0.51 mm, width of cut =3 mm, yield shear stress of work material = 285N/ 2 f i ti ffi i t t l fN/mm2, mean friction coefficient on tool face =0.65, Determine

( ) ( )(i) Cutting force (Fc)(ii) Radial force(iii) Normal force (N) on tool and(iv) Shear force (F )(iv) Shear force (Fs ).

68

GATE2010(PI)LinkedS1In orthogonal turning of an engineering alloy, it hasIn orthogonal turning of an engineering alloy, it hasbeen observed that the friction force acting at the chiptool interface is 402.5 N and the friction force is alsotool interface is 402.5 N and the friction force is alsoperpendicular to the cutting velocity vector. The feedvelocity is negligibly small with respect to the cuttingvelocity is negligibly small with respect to the cuttingvelocity. The ratio of friction force to normal forceassociated with the chiptool interface is 1. The uncutassociated with the chip tool interface is 1. The uncutchip thickness is 0.2 mm and the chip thickness is 0.4mm. The cutting velocity is 2 m/s.mm. The cutting velocity is 2 m/s.The shear force (in N) acting along the primary shearplane isplane is(a) 180.0 (b) 240.0 (c) 360.5 (d) 402.5 69

GATE2010(PI)LinkedS2In orthogonal turning of an engineering alloy, it hasg g g g y,been observed that the friction force acting at the chiptool interface is 402.5 N and the friction force is alsoperpendicular to the cutting velocity vector. The feedvelocity is negligibly small with respect to the cuttingl it Th ti f f i ti f t l fvelocity. The ratio of friction force to normal force

associated with the chiptool interface is 1. The uncutchip thickness is 0 2 mm and the chip thickness is 0 4chip thickness is 0.2 mm and the chip thickness is 0.4mm. The cutting velocity is 2 m/s.Assume that the energy expended during machining isAssume that the energy expended during machining iscompletely converted to heat. The rate of heatgeneration (inW) at the primary shear plane isgeneration (inW) at the primary shear plane is(a) 180.5 (b) 200.5 (c) 302.5 (d) 402.5

70

LinkedAnswerQuestionsGATE2013S1In orthogonal turning of a bar of 100 mm diameter

with a feed of 0.25 mm/rev, depth of cut of 4 mmwith 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 interfaceto friction force acting at the chiptool interface.Themain (tangential) cutting force is 1500 N.

y The orthogonal rake angle of the cutting tool in degree is(a) zero (b) 3.58 (c) 5 (d) 7.16

71

LinkedAnswerQuestionsGATE2013S2In orthogonal turning of a bar of 100 mm diameter

with a feed of 0.25 mm/rev, depth of cut of 4 mmwith 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 interfaceto friction force acting at the chiptool interface.Themain (tangential) cutting force is 1500 N.

y The normal force acting at the chiptool interface in N is(a) 1000 (b) 1500 (c) 20oo (d) 2500


GATE 2014GATE2014Which pair of following statements is correct fororthogonal cutting using a singlepoint cuttingtool?P. Reduction in friction angle increases cutting forceQ Reduction in friction angle decreases cutting forceQ. Reduction in friction angle decreases cutting forceR. Reduction in friction angle increases chip thicknessS R d i i f i i l d hi hi kS. Reduction in friction angle decreases chip thickness(a) P and R (b) P and S(c) Q and R (d) Q and S

73

S 999IAS 1999I th l tti k l f thIn an orthogonal cutting process, rake angle of the

tool is 20 and friction angle is 25.5. Usingg 5 5 g

Merchant's shear angle relationship, the value of

shear angle will be

( ) (b)(a) 39.5 (b) 42.25

(c) 47 75 (d) 50 5(c) 47.75 (d) 50.5

74

GATE1997In a typical metal cutting operation, using a cutting

tool of positive rake angle = 10 it was observedtool of positive rake angle = 10 , it was observed

that the shear angle was 20. The friction angle is

(a) 45 (b) 30

(c) 60 (d) 40

75

ESE2005ConventionalMild t l i b i hi d t tti d fMild steel is being machined at a cutting speed of200 m/min with a tool rake angle of 10. The width of

t d t thi k dcut and uncut thickness are 2 mm and 0.2 mmrespectively. If the average value of coefficient off i ti b t th t l d th hi i d thfriction between the tool and the chip is 0.5 and theshear stress of thework material is 400 N/mm2,

Determine

(i) h l d(i) shear angle and

(ii) Cutting and thrust component of the force.(ii) Cutting and thrust component of the force.

76

GATE 2008 (PI) Linked S 1GATE2008(PI)LinkedS1In an orthogonal cutting experiment, an HSS tool havingg g p g

the following tool signature in the orthogonal reference

( ) h b dsystem (ORS) has been used: 0107710751. Givenwidth of cut = 3.6 mm; shear strength of workpiecewidth of cut 3.6 mm; shear strength of workpiece

material = 460 N/mm2; depth of cut = 0.25 mm;

coefficient of friction at toolchip interface = 0.7.

Sh l l (i d ) f i i tti fShear plane angle (in degree) for minimum cutting force

is

(a) 20.5 (b) 24.5 (c) 28.5 (d) 32.5 77

GATE 2008 (PI) Linked S 2GATE2008(PI)LinkedS2In an orthogonal cutting experiment, an HSS tool havingg g p g

the following tool signature in the orthogonal reference

( ) h b dsystem (ORS) has been used: 0107710751. Givenwidth of cut = 3.6 mm; shear strength of workpiecewidth of cut 3.6 mm; shear strength of workpiece

material = 460 N/mm2; depth of cut = 0.25 mm;

coefficient of friction at toolchip interface = 0.7.

Mi i i t (i kW) t tti dMinimum power requirement (in kW) at a cutting speed

of 150 m/min is

(a) 3.15 (b) 3.25 (c) 3.35 (d) 3.45 78

IES 2010IES2010The relationship between the shear angle ,The relationship between the shear angle ,the friction angle and cutting rake angle is given asis given as

79

IES2005Whi h f h f ll i i hWhich one of the following is the correctexpression for the Merchant's machinabilityconstant?(a) 2 +(a)(b)

2 + 2 +

(c)(d)

2 + (d)

(Where = shear angle, = friction angle

+

and = rake angle)80

IES2003In orthogonal cutting test, the cutting force = 900 N,

the thrust force = 600 N and chip shear angle is 30othe thrust force = 600 N and chip shear angle is 30o.

Then the chip shear force is

(a) 1079.4 N (b) 969.6 N

(c) 479.4 N (d) 69.6 N


IES 2014IES 2014In an orthogonal cutting operation shear angle = 11 31o ,In an orthogonal cutting operation shear angle 11.31 ,cutting force = 900 N and thrust force = 810 N. Then theshear force will be approximately ( given sin 11.31o = 0.2)shear force will be approximately ( given sin 11.31 0.2)(a) 650 N(b) N(b) 720 N(c) 620 N(d) 680 N

82

IES2000In an orthogonal cutting test, the cutting force and

thrust force were observed to be 1000N and 500 Nthrust force were observed to be 1000N and 500 N

respectively. If the rake angle of tool is zero, the

coefficient of friction in chiptool interface will be

( ) ( ) ( ) ( )1 1a b 2 c d 2 2 22 2

83

GATE 2007 (PI) C D t 1GATE 2007(PI)CommonData1In an orthogonal machining test, the followingg g , gobservations were madeCutting force 1200 NCutting force 1200 NThrust force 500 NT l k lTool rake angle zeroCutting speed 1 m/sDepth of cut 0.8 mmChip thickness 1.5 mmChip thickness 1.5 mmFriction angle during machining will be( ) 6o (b) 8o ( ) o (d) 6 o(a) 22.6o (b) 32.8o (c) 57.1o (d) 67.4o

84

GATE 2007 (PI) C D t 2GATE 2007(PI)CommonData2In an orthogonal machining test, the followingg g , gobservations were madeCutting force 1200 NCutting force 1200 NThrust force 500 NT l k lTool rake angle zeroCutting speed 1 m/sDepth of cut 0.8 mmChip thickness 1.5 mmChip thickness 1.5 mmChip speed along the tool rake face will be( ) 8 / (b) /(a) 0.83 m/s (b) 0.53 m/s(c) 1.2 m/s (d) 1.88 m/s 85

GATE 2011 (PI) Linked S1GATE 2011(PI)LinkedS1During orthogonal machining of a mild steel specimenwith a cutting tool of zero rake angle, the following datais obtained:Uncut chip thickness = 0.25 mmChip thickness = 0 75 mmChip thickness = 0.75 mmWidth of cut = 2.5 mmN l f NNormal force = 950 NThrust force = 475 NThe shear angle and shear force, respectively, are(a) 71 565o, 150 21 N (b) 18 435o , 751 04 N(a) 71.565 , 150.21 N (b) 18.435 , 751.04 N(c) 9.218o, 861.64 N (d) 23.157o , 686.66 N

86

GATE 2011 (PI) Linked S2GATE 2011(PI)LinkedS2During orthogonal machining of a mild steel specimenwith a cutting tool of zero rake angle, the following datais obtained:

Uncut chip thickness = 0.25 mmChip thickness = 0 75 mmChip thickness = 0.75 mmWidth of cut = 2.5 mmN l f NNormal force = 950 NThrust force = 475 N

Theultimateshearstress(inN/mm2)oftheworkmaterialis(a)235 (b)139 (c)564 (d)380

87

IFS2012IFS2012An orthogonal machining operation is being carried outunder the following conditions :under the following conditions :depth of cut = 0.1 mm,h h kchip thickness = 0.2 mm,width of cut = 5 mm,rake angle = 10o

TheforcecomponentsalongandnormaltothedirectionTheforcecomponentsalongandnormaltothedirectionofcuttingvelocityare500Nand200Nrespectively.DetermineDetermine(i)Thecoefficientoffrictionbetweenthetoolandchip.(ii)Ul i h f h k i i l [ ](ii)Ultimateshearstressoftheworkpiecematerial.[10]

88

GATE2006CommonDataQuestions(1)I th l hi i tiInanorthogonalmachiningoperation:Uncutthickness=0.5mmCuttingspeed=20m/min Rakeangle=15Widthofcut=5mm Chipthickness=0.7mmWidthofcut 5mm Chipthickness 0.7mmThrustforce=200N Cuttingforce=1200NA M h t' thAssumeMerchant'stheory.Thecoefficientoffrictionatthetoolchipinterfaceis( ) (b)(a)0.23 (b)0.46(c)0.85 (d)0.95

89

GATE2006CommonDataQuestions(2)I th l hi i tiInanorthogonalmachiningoperation:Uncutthickness=0.5mmCuttingspeed=20m/min Rakeangle=15Widthofcut=5mm Chipthickness=0.7mmWidthofcut 5mm Chipthickness 0.7mmThrustforce=200N Cuttingforce=1200NA M h t' thAssumeMerchant'stheory.Thepercentageoftotalenergydissipatedduetof h l h ffrictionatthetoolchipinterfaceis

(a)30% (b)42%(c)58% (d)70%


GATE2006CommonDataQuestions(3)I th l hi i tiInanorthogonalmachiningoperation:Uncutthickness=0.5mmCuttingspeed=20m/min Rakeangle=15Widthofcut=5mm Chipthickness=0.7mmWidthofcut 5mm Chipthickness 0.7mmThrustforce=200N Cuttingforce=1200NA M h t' thAssumeMerchant'stheory.Thevaluesofshearangleandshearstrain,

lrespectively,are(a)30.3 and1.98 (b)30.3 and4.23(c)40.2 and2.97 (d)40.2 and1.65

91

IES1995The primary tool force used in calculating the total

power consumption in machining is thepower consumption in machining is the

(a) Radial force (b) Tangential force( ) ( ) g

(c) Axial force (d) Frictional force.

92

IES2001P i i l i iPower consumption in metal cutting ismainly due to(a) Tangential component of the force(b) Longitudinal component of the force(b) Longitudinal component of the force(c) Normal component of the forcep(d) Friction at the metaltool interface

93

IES1997C id h f ll i f iConsider the following forces acting on afinish turning tool:1. Feed force2 Thrust force2. Thrust force3. Cutting force.gThe correct sequence of the decreasing order ofthe magnitudes of these forces isthe magnitudes of these forces is(a) 1, 2, 3 (b) 2, 3, 1(c) 3, 1, 2 (d) 3, 2, 1

94

IES1999Th di l f i i l i l d iThe radial force in singlepoint tool duringturning 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(b) 0.4 to 0.6 times the main cutting force(c) 0.6 to 0.8 times the main cutting forceg(d) 0.5 to 0.6 times the main cutting force

95

GATE2014A straight turning operation is carried out using a

single point cutting tool on an AISI 1020 steel rodsingle point cutting tool on an AISI 1020 steel rod.

The feed is 0.2 mm/rev and the depth of cut is 0.5

mm. The tool has a side cutting edge angle of 60o.

The uncut chip thickness (in mm) is .

96

S 2003 C i lESE2003 ConventionalDuring turning a carbon steel rod of 160 mm diameter by a

bid l f 8 ( ) d fcarbide tool of geometry; 0, 0, 10, 8, 15, 75, 0 (mm) at speed of400 rpm, feed of 0.32 mm/rev and 4.0 mm depth of cut, thefollowing observationweremadefollowing observationweremade.

Tangential component of the cutting force, Pz = 1200 NAxial component of the cutting force P = 800 NAxial component of the cutting force, Px = 800 NChip thickness (after cut),

For the abovemachining condition determine the values of=2 0.8mm.

For the abovemachining condition determine the values of(i) Friction force, F and normal force, N acting at the chip toolinterfaceinterface.(ii) Yield shears strength of the work material under thismachining condition.machining condition.(iii) Cutting power consumption in kW.

97

GATE 1995ConventionalWhil t i C t l d f 6 di t tWhile turning a C15 steel rod of 160 mm diameter at315 rpm, 2.5 mm depth of cut and feed of 0.16

/ b t l f t 0 0 80 0 0 0mm/rev by a tool of geometry 00, 100, 80, 90,150, 750,0(mm), the following observationsweremade.Tangential component of the cutting force = 500 N

Axial component of the cutting force = 200 Np gChip thickness = 0.48 mm

Draw schematically the Merchants circle diagramDraw schematically the Merchant s circle diagramfor the cutting force in the present case.

98

IAS2003MainExaminationD i t i ith 6 6 8During turning process with 7 6 6 8 30 1(mm) ASA tool the undeformed chip thickness of

d idth f t f d Th2.0 mm and width of cut of 2.5 mm were used. Theside rake angle of the tool was a chosen that the

hi i ti ld b i t d t bmachining operation could be approximated to beorthogonal cutting. The tangential cutting force andth t f N d 6 N ti lthrust force were 1177 N and 560 N respectively.Calculate: [30 marks]( ) h d k l(i) The side rake angle(ii) Coefficient of friction at the rake face(iii) The dynamic shear strength of the work material


GATE2007In orthogonal turning of a low carbon steel barIn orthogonal turning of a low carbon steel barof diameter 150 mm with uncoated carbidetool the cutting velocity is 90 m/min The feedtool, the cutting velocity is 90 m/min. The feedis 0.24 mm/rev and the depth of cut is 2 mm.The chip thickness obtained is 0 48 mm If theThe chip thickness obtained is 0.48 mm. If theorthogonal rake angle is zero and the principalcutting edge angle is 90 the shear angle iscutting edge angle is 90 , the shear angle isdegree is( ) 6 (b) 6 6(a) 20.56 (b) 26.56(c) 30.56 (d) 36.56

100

GATE2007In orthogonal turning of low carbon steel pipe with

principal cutting edge angle of 90 themain cuttingprincipal cutting edge angle of 90 , themain cutting

force is 1000 N and the feed force is 800 N. The shear

angle is 25 and orthogonal rake angle is zero.

Employing Merchants theory, the ratio of friction

force to normal force acting on the cutting tool isforce to normal force acting on the cutting tool is

(a) 1 56 (b) 1 25 (c) 0 80 (d) 0 64(a) 1.56 (b) 1.25 (c) 0.80 (d) 0.64

101

GATE2003CommonDataQuestions(1)A li d i t d l th ith th lA cylinder is turned on a lathe with orthogonalmachining principle. Spindle rotates at 200 rpm. Thei l f d t i l ti D th f t iaxial feed rate is 0.25 mm per revolution. Depth of cut is

0.4 mm. The rake angle is 10. In the analysis it is foundth t th h l i that the shear angle is 27.75

Thethicknessoftheproducedchipis(a)0.511mm (b)0.528mm(c)0.818mm (d)0.846mm(c)0.818mm (d)0.846mm

102

GATE2003CommonDataQuestions(2)A li d i t d l th ith th lA cylinder is turned on a lathe with orthogonalmachining principle. Spindle rotates at 200 rpm. Thei l f d t i l ti D th f t iaxial feed rate is 0.25 mm per revolution. Depth of cut is

0.4 mm. The rake angle is 10. In the analysis it is foundth t th h l i that the shear angle is 27.75Intheaboveproblem,thecoefficientoffrictionatthechiptoolinterfaceobtainedusingEarnestandMerchanttheoryis(a)0.18 (b)0.36(c)0.71 (d)0.98(c)0.71 (d)0.98

103

GATE2008CommonDataQuestion(1)O th l t i i f d li d i l kOrthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following

diti d tti l it i 8 / i f dconditions are used: cutting velocity is 180 m/min. feedis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessti Th th l k l i o A lratio = 0.5. The orthogonal rake angle is 7o. Apply

Merchant's theory for analysis.( )Theshearplaneangle(indegree)andtheshear

forcerespectivelyare(a)52:320N (b)52:400N(c)28:400N (d)28:320N(c)28:400N (d)28:320N

104

GATE2008CommonDataQuestion(2)O th l t i i f d li d i l kOrthogonal turning is performed on a cylindrical workpiece with shear strength of 250 MPa. The following

diti d tti l it i 8 / i f dconditions are used: cutting velocity is 180 m/min. feedis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessti Th th l k l i o A lratio = 0.5. The orthogonal rake angle is 7o. Apply

Merchant's theory for analysis.ThecuttingandThrustforces,respectively,are(a)568N;387N (b)565N;381N(c)440N;342N (d)480N;356N

105

IES 2004IES 2004A medium carbon steel workpiece is turned on aA medium carbon steel workpiece is turned on alathe at 50 m/min. cutting speed 0.8 mm/rev feedand 1.5 mm depth of cut. What is the rate of metaland 1.5 mm depth of cut. What is the rate of metalremoval?(a) 1000 mm3/min(a) 1000 mm3/min(b) 60,000 mm3/min(c) 20,000 mm3/min(d) Can not be calculated with the given data( ) g

106

GATE 2013GATE2013A steel bar 200 mm in diameter is turned at a feed ofA steel bar 200 mm in diameter is turned at a feed of0.25 mm/rev with a depth of cut of 4 mm. Therotational speed of the workpiece is 160 rpm. Therotational speed of the workpiece is 160 rpm. Thematerial removal rate in mm3/s is(a) 160 (b) 167 6 (c) 1600 (d) 1675 5(a) 160 (b) 167.6 (c) 1600 (d) 1675.5

107

GATE(PI)1991Amount of energy consumption per unit volume of

metal removal is maximum inmetal removal is maximum in

(a) Turning (b) Milling( ) g ( ) g

(c) Reaming (d) Grinding


GATE2007I th l t i f di b t l ThIn orthogonal turning of medium carbon steel. Thespecific machining energy is 2.0 J/mm3. The cutting

l it f d d d th f t / ivelocity, feed and depth of cut are 120 m/min, 0.2mm/rev and 2 mm respectively. The main cuttingf i N iforce in N is(a) 40 (b) 80(c) 400 (d) 800

109

GATE 2013 (PI) C D Q iGATE2013(PI)CommonDataQuestionA disc of 200 mm outer and 80 mm inner diameter isA disc of 200 mm outer and 80 mm inner diameter isfaced of 0.1 mm/rev with a depth of cut of 1 mm. Thefacing operation is undertaken at a constant cuttingfacing operation is undertaken at a constant cuttingspeed of 90 m/min in a CNC lathe. The main(tangential) cutting force is 200 N.(tangential) cutting force is 200 N.Neglecting the contribution of the feed forcetowards cutting power the specific cutting energytowards cutting power, the specific cutting energyin J/mm3 is( ) (b) ( ) (d)(a) 0.2 (b) 2 (c) 200 (d) 2000

110

ExampleWhentherakeangleiszeroduringorthogonalcutting,showthat

( )2

11

s r rp r

= +s

1Where is theshear strengrh of the material

cp r+

cp = specific power of cuttingr = chip thickness ratiop = coefficient of friction in toolchip interface

111

GATE2014The main cutting force acting on a tool during theThe main cutting force acting on a tool during theturning (orthogonal cutting) operation of a metal is400 N. The turning was performed using 2 mm400 N. The turning was performed using 2 mmdepth of cut and 0.1 mm/rev feed rate. The specificcutting pressure iscutting pressure is

(a) 1000

(b) 2000

(c) 3000

(d) 4000(d) 4000

112

GATE1992h ff f k l h f lThe effect of rake angle on the mean friction angle in

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

113

GATE1993Th ff t f k l th f i ti l iThe effect of rake angle on themean friction angle inmachining can be explained by( ) ( )(a) Sliding (coulomb) model of friction(b) sticking and then siding model of frictiong g(c) Sticking friction(d) sliding and then sticking model of friction(d) sliding and then sticking model of friction

114

IES2000A i (A) I l i h lAssertion (A): In metal cutting, the normallaws of sliding friction are not applicable.Reason (R): Very high temperature isproduced at the toolchip interfaceproduced at the tool chip interface.(a) Both A and R are individually true and R isthe correct explanation of A(b) Both A and R are individually true but R is(b) Both A and R are individually true but R isnot the correct explanation of A( ) A i b R i f l(c) A is true but R is false(d) A is false but R is true 115

IES2004Assertion (A): The ratio of uncut chip thickness toAssertion (A): The ratio of uncut chip thickness toactual chip thickness is always less than one and istermed as cutting ratio in orthogonal cuttingg g gReason (R): The frictional force is very high due to theoccurrence of sticking friction rather than slidingg gfriction(a) Both A and R are individually true and R is the correct( ) yexplanation of A(b) Both A and R are individually true but R is not the( ) ycorrect explanation of A(c) A is true but R is false(c) A is true but R is false(d) A is false but R is true

116

IES2002I hi i h fIn a machining process, the percentage ofheat carried away by the chips is typically(a) 5% (b) 25%(c) 50% (d) 75%(c) 50% (d) 75%


IES1998I l i i h iIn metal cutting operation, the approximateratio of heat distributed among chip, tooland work, in that order is(a) 80: 10: 10 (b) 33: 33: 33(a) 80: 10: 10 (b) 33: 33: 33(c) 20: 60: 10 (d) 10: 10: 80

118

S 2003IAS 2003AsthecuttingspeedincreasesAsthecuttingspeedincreases(a) Moreheatistransmittedtotheworkpieceandlessheatistransmittedtothetoolheatistransmittedtothetool(b) Moreheatiscarriedawaybythechipandlessheatist itt dt th t ltransmittedtothetool(c) Moreheatistransmittedtoboththechipandthetool(d) Moreheatistransmittedtoboththeworkpieceand( ) pthetool

119

S 2003IAS 2003The heat generated in metal cutting canThe heat generated in metal cutting canconveniently be determined by(a) Installing thermocouple on the job(a) Installing thermocouple on the job(b) Installing thermocouple on the tool(c) Calorimetric setup(d) Using radiation pyrometer( ) g py

120

IES2011TheinstrumentordeviceusedtomeasurethecuttingTheinstrumentordeviceusedtomeasurethecuttingforcesinmachiningis:( )T h t(a)Tachometer(b)Comparator(c)Dynamometer(d)Lactometer(d)Lactometer

121

IES1993A 'D ' i d i d f hA 'Dynamometer' is a device used for themeasurement of(a) Chip thickness ratio(b) Forces during metal cutting(b) Forces during metal cutting(c) Wear of the cutting toolg(d) Deflection of the cutting tool

122

IES1996Which of the following forces are measured directly byWhich of the following forces are measured directly bystrain gauges or force dynamometers during metalcutting ?g1. Force exerted by the tool on the chip acting normally tothe tool face.2. Horizontal cutting force exerted by the tool on the workpiece.3. Frictional resistance of the tool against the chip flowacting along the tool face.

V i l f hi h h l i h ldi h l i4. Vertical force which helps in holding the tool inposition.( ) d (b) d(a) 1 and 3 (b) 2 and 4(c) 1 and 4 (d) 2 and 3 123

IES1998Th f f i i i k fThe gauge factor of a resistive pickup ofcutting force dynamometer is defined as theratio of(a) Applied strain to the resistance of the wire(a) Applied strain to the resistance of the wire(b) The proportional change in resistance to theapplied strain(c) The resistance to the applied strain(c) The resistance to the applied strain(d) Change in resistance to the applied strain

124

S 200IAS2001Assertion (A): Piezoelectric transducers and preferred( ) pover strain gauge transducers in the dynamometers formeasurement of threedimensional cutting forces.Reason (R): In electric transducers there is a significantleakage of signal from one axis to the other, such crosserror is negligible in the case of piezoelectricerror is negligible in the case of piezoelectrictransducers.(a) Both A and R are individually true and R is the correct(a) Both A and R are individually true and R is the correctexplanation of A(b) Both A and R are individually true but R is not the( ) ycorrect explanation of A(c) A is true but R is false(d) A is false but R is true

125

ForPSU&IESIn strain gauge dynamometers the use of how

ti k th d tmany active gauge makes the dynamometers moreeffective( )(a) Four(b) Three(c) Two(d) One(d) One


ToolWear,ToolLife&ToolWear,ToolLife&MachinabilityMachinability

B SKM d lBySKMondal127

ToolFailureTool failure is two typesTool failure is two typesy 1. Slowdeath: The gradual or progressive wearingaway of rake face (crater wear) or flank (flank wear) ofaway of rake face (crater wear) or flank (flank wear) ofthe cutting tool or both.y 2.Suddendeath:Failuresleadingtoprematureend. Sudde deat : a u es ead g to p e atu e e dofthetooly The suddendeath type of tool failure is difficult toyppredict. Tool failure mechanisms include plasticdeformation, brittle fracture, fatigue fracture or edgechipping However it is difficult to predict which ofchipping. However it is difficult to predict which ofthese processes will dominate and when tool failurewill occur.will occur.

128

IES 2010IES2010Flank wear occurs on theFlank wear occurs on the

(a) Relief face of the tool( )

(b) Rake face

(c) Nose of the tool

(d) Cutting edge

129

S 200IES 2007Flank wear occurs mainly on which of theFlank wear occurs mainly on which of the

following?

(a) Nose part and top face

(b) Cutting edge only

( ) N f li f f d id li f f f h(c) Nose part, front relief face, and side relief face of the

cutting toolg

(d) Face of the cutting tool at a short distance from

the cutting edge130

IAS 2009Mainy Explainsuddendeathmechanismoftoolfailure.

[ k ][4 marks]

131

IES 2014IES 2014The fatigue failure of a tool is due toThe fatigue failure of a tool is due to(a) abrasive friction, cutting fluid and chip breakage(b) V i bl th l t hi b k d i bl(b) Variable thermal stresses, chip breakage and variabledimensions of cut(c) Abrasive friction, chip breakage and variabledimensions of cut(d) Chip breakage, variable thermal stresses and cuttingfluid

132

ToolWear( ) Fl kW(a) FlankWear

(b) CraterWear(b) CraterWear

(c) Chipping off of the cutting edgepp g g g

133

Tool WearToolWear

134

S 99IES 1994Assertion(A):ToolwearisexpressedintermsofAssertion(A):Toolwearisexpressedintermsofflankwearratherthancraterwear.Reason(R):MeasurementofflankwearissimpleReason(R):Measurementofflankwearissimpleandmoreaccurate.( ) B thA dR i di id ll t dRi th (a) BothAandRareindividuallytrueandRisthecorrectexplanationofA(b) BothAandRareindividuallytruebutRisnot thecorrectexplanationofA(c) AistruebutRisfalse(d) AisfalsebutRistrue(d) AisfalsebutRistrue


FlankWear:(Wearland)ReasonReasony Abrasion by hard particles and inclusions in the worky ppiece.y Shearing off the micro welds between tool and workShearing off the micro welds between tool and workmaterial.y Abrasion by fragments of built up edge ploughingy Abrasion by fragments of builtupedge ploughingagainst the clearance face of the tool.y At l d fl k d i ty At low speed flank wear predominates.y If MRR increased flank wear increased.

136

G 20GATE2014Cutting tool is much harder than the workpiece.Cutting tool is much harder than the work piece.Yet the tool wears out during the toolworkinteraction, becauseinteraction, because(a) extra hardness is imparted to the workpiece due tocoolant usedcoolant used(b) oxide layers on the workpiece surface impart extrah d t ithardness to it(c) extra hardness is imparted to the workpiece due tosevere rate of strain(d) vibration is induced in the machine tool

137

FlankWear:(Wearland)EffectEffecty Flank wear directly affect the component dimensionsy pproduced.y Flank wear is usually the most common determinant ofFlank wear is usually the most common determinant oftool life.

138

FlankWear:(Wearland)StagesStagesy FlankWear occurs in three stages of varying wear ratesg y g

139

FlankWear:(Wearland)PrimarywearPrimarywearThe region where the sharp cutting edge is quicklyg p g g q ybroken down and a finite wear land is established.

Secondar earSecondarywearThe region where the wear progresses at a uniform rate.g p g

140

FlankWear:(Wearland)Tertiary wearTertiary wearThe region where wear progresses at a graduallyg p g g yincreasing rate.y In the tertiary region the wear of the cutting tool hasIn the tertiary region the wear of the cutting tool hasbecome sensitive to increased tool temperature due tohigh wear land.high wear land.y Regrinding is recommended before they enter thisregionregion.

141

S 200IES 2004Consider the following statements:Consider the following statements:During the third stage of toolwear, rapiddeterioration of tool edge takes place becausedeterioration of tool edge takes place because1. Flank wear is only marginal2. Flank wear is large3. Temperature of the tool increases gradually3 p g y4. Temperature of the tool increases drasticallyWhich of the statements given above are correct?Which of the statements given above are correct?(a) 1 and 3 (b) 2 and 4(c) 1 and 4 (d) 2 and 3

142

GATE 2008 (PI)GATE2008(PI)During machining, the wear land (h) has been plotted

i hi i i (T) i i h f ll iagainst machining time (T) as given in the followingfigure.

For a critical wear land of 1.8 mm, the cutting tool life (inminute) isminute) is(a) 52.00 (b) 51.67 (c) 51.50 (d) 50.00

143

IFS2012E l i th h i f fl k f ttiExplain the mechanism of flank wear of a cutting

tool. Plot a flank wear rate curve and indicate the

region of tool failure.

[10 Marks]


ToollifecriteriaISO(Acertainwidthofflankwear(VB)isthemostcommon(Acertainwidthofflankwear(VB)isthemostcommoncriterion)y Uniformwear:0 3mmaveragedoverallpasty Uniformwear:0.3mmaveragedoverallpasty Localizedwear:0.5mmonanyindividualpast

145

CraterwearyM i d til t i l hi h dyMore common in ductile materials which producecontinuous chip.p

y Crater wear occurs on the rake face.

y At very high speed crater wear predominates

y For crater wear temperature is main culprit and toold f i h hi i l & l idefuse into the chip material & tool temperature is

maximum at some distance from the tool tip.a u at so e d sta ce o t e too t p.

146

CraterwearContd..y Crater depth exhibits linear increase with timey Crater depth exhibits linear increase with time.y It increases with MRR.

y Crater wear has little or no influence on cutting forces,work piece tolerance or surface finish.

147

S 2002IES 2002CraterwearontoolsalwaysstartsatsomedistanceCraterwearontoolsalwaysstartsatsomedistancefromthetooltipbecauseatthatpoint(a) Cuttingfluiddoesnotpenetrate(a) Cuttingfluiddoesnotpenetrate(b) Normalstressonrakefaceismaximum(c) Temperatureismaximum(d) Toolstrengthisminimum( ) g

148

S 200IAS 2007WhydoescraterwearstartatsomedistancefromWhydoescraterwearstartatsomedistancefromthetooltip?(a) Toolstrengthisminimumatthatregion(a) Toolstrengthisminimumatthatregion(b) Cuttingfluidcannotpenetratethatregion(c) Tooltemperatureismaximuminthatregion(d) Stressonrakefaceismaximumatthatregion( ) g

149

S 2000IES 2000CraterwearstartsatsomedistancefromthetooltipCraterwearstartsatsomedistancefromthetooltipbecause(a) Cuttingfluidcannotpenetratethatregion(a) Cuttingfluidcannotpenetratethatregion(b) Stressonrakefaceismaximumatthatregion(c) Toolstrengthisminimumatthatregion(d) Tooltemperatureismaximumatthatregion( ) p g

150

S 99IES 1995CraterwearispredominantinCraterwearispredominantin(a) Carbonsteeltools(b) T t bid t l(b) Tungstencarbidetools(c) Highspeedsteeltools(d) Ceramictools

151

IES2009ConventionalSh t dfl k i l i tShowcraterwearandflankwearonasinglepointcuttingtool.Statethefactorsresponsibleforwear

t i t lonaturningtool.[2marks]

152

WearMechanism1. Abrasionwear

2. Adhesionwear

3. Diffusionwear

4. Chemicaloroxidationwear


S 2002IAS 2002Consider the following actions:Consider the following actions:1. Mechanical abrasion 2. Diffusion

Pl ti d f ti O id ti3. Plastic deformation 4. OxidationWhich 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(c) 1, 2 and 4 (d) 1 and 3

154

S 99IES 1995MatchListIwithListIIandselectthecorrectMatchListIwithListIIandselectthecorrectanswerusingthecodesgivenbelowthelists:ListI(Weartype) ListII(Associatedmechanism)ListI(Weartype) ListII(Associatedmechanism)A. Abrasivewears 1. GalvanicactionB. Adhesivewears 2. Ploughing actionC. Electrolyticwear 3. Moleculartransfery 3D. Diffusionwears 4. Plasticdeformation

5 Metallicbond5. MetallicbondCode: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 4155

S 999IAS 1999The type of wear that occurs due to the cuttingThe type of wear that occurs due to the cuttingaction of the particles in the cutting fluid isreferred to asreferred to as(a) Attritions wear(b) Diff i(b) Diffusion wear(c) Erosive wear(d) Corrosive wear

156

h hi i ff fi kWhychippingofforfinecracksdeveloped at the cutting edgedevelopedatthecuttingedge

y T l t i l i t b ittly Tool material is too brittle

k d f l h h h k lyWeak design of tool, such as high positive rake angle

y As a result of crack that is already in the tool

y Excessive static or shock loading of the tool.

157

S 2003IAS 2003Consider the following statements:Consider the following statements:

Chipping of a cutting tool is due toT l t i l b i t b ittl1. Tool material being too brittle

2. Hot hardness of the tool material.3. High positive rake angle of the tool.Which of these statements are correct?Which of these statements are correct?(a) 1, 2 and 3 (b) 1 and 3( ) d (d) d(c) 2 and 3 (d) 1 and 2

158

hNotchWeary Notch wear on the trailing edge is to a great extent any Notch wear on the trailing edge is to a great extent anoxidation wear mechanism occurring where the cutting

edge leaves the machined workpiece material in the feed

direction.

y B t b i d dh i i bi d ff ty But abrasion and adhesion wear in a combined effect cancontribute to the formation of one or several notches.

159

S 996IES 1996NotchwearattheoutsideedgeofthedepthofcutisNotchwearattheoutsideedgeofthedepthofcutisdueto(a) Abrasiveactionoftheworkhardenedchipmaterial(a) Abrasiveactionoftheworkhardenedchipmaterial(b) Oxidation(c) Slipstickactionofthechip(d) Chipping.( ) pp g

160

Listtheimportantpropertiesofcuttingtoolmaterials and explain why each is importantmaterialsandexplainwhyeachisimportant.

yHardness at high temperatures this provides longerlife of the cutting tool and allows higher cutting speeds.y Toughness to provide the structural strength neededToughness to provide the structural strength neededto resist impacts and cutting forcesyW i t t l b f l tyWear resistance to prolong usage before replacementdoesnt chemically react another wear factory Formable/manufacturable can be manufactured in auseful geometryuse u geo et y

161

Whyareceramicsnormallyprovidedasinserts for tools and not as entire tools?insertsfortools,andnotasentiretools?Ceramicsarebrittlematerialsandcannotprovidethepstructuralstrengthrequiredforatool.


163

ToolLifeCriteriaT l lif it i b d fi d d t i dTool life criteria can be defined as a predeterminednumerical value of any type of tool deterioration which

b dcan be measured.

Some of thewaysSome of thewaysyActualcuttingtimetofailure.yVolumeofmetalremoved.yNumberofpartsproducedyNumberofpartsproduced.yCuttingspeedforagiventimeyLengthofworkmachined.

164

S 992IES 1992ToollifeisgenerallyspecifiedbyToollifeisgenerallyspecifiedby(a) Numberofpiecesmachined(b) V l f t l d(b) Volumeofmetalremoved(c) Actualcuttingtime(d) Anyoftheabove

165

S 20 2 iIAS 2012MainD fi t l lif d li t d f th d fDefine tool life and list down four methods for

quantitative measurement of tool life.q

[Marks12]

166

TaylorsToolLifeEquationbasedonFlankWearCausesy Slidingofthetoolalongthemachinedsurfacey Temperaturerise

nVT C=Where,V=cuttingspeed(m/min)

VT CT=Time(min)n=exponentdependsontoolmaterialC=constantbasedontoolandworkmaterialandcuttingcondition. 167

ValuesofExponentn8 f HSS ln = 0.08 to 0.2 for HSS tool

= 0 1 to 0 15 for Cast Alloys= 0.1 to 0.15 for Cast Alloys= 0.2 to 0.4 for carbide tool

[IAS1999; IES2006]t f i t l= 0.5 to 0.7 for ceramic tool

[NTPC2003][ 3]

168

IES 2012IES 2012InTaylorstoollifeequationVTn =C,theconstantsnInTaylor stoollifeequationVT C,theconstantsnandCdependupon1 Workpiecematerial1.Workpiecematerial2.Toolmaterial3.Coolant(a)1,2,and3( ) 3(b)1and2only(c)2and3only(c)2and3only(d)1and3only

169

S 2008IES 2008InTaylor'stoollifeequationisVTn =constant.InTaylor stoollifeequationisVT =constant.Whatisthevalueofnforceramictools?( ) t (b) t (a) 0.15to0.25 (b) 0.4to0.55(c) 0.6to0.75 (d) 0.8to0.9

170

S 2006IES 2006Which of the following values of index n isWhich of the following values of index n isassociated with carbide tools when Taylor's tool lifeequation, V.Tn = constant is applied?equation, V.T constant is applied?(a) 01 to 015 (b) 02 to 04( ) t 6 (d) 6 t(c) 0.45 to 06 (d) 065 to 09


S 999IES 1999The approximately variation of the tool lifeThe approximately variation of the tool lifeexponent 'n' of cemented carbide tools is(a) 0 03 to 0 08 (b) 0 08 to 0 20(a) 0.03 to 0.08 (b) 0.08 to 0.20(c) 0.20 to 0.48 (d) 0.48 to 0.70

172

S 998IAS 1998MatchList I(Cuttingtoolmaterial)withList II( g )(Typicalvalueoftoollifeexponent'n'intheTaylor'sequationV.Tn =C)andselectthecorrectanswerusingth d i b l th li tthecodesgivenbelowthelists:List I List IIA HSS 8A. HSS 1. 0.18B. Castalloy 2. 0.12C C iC. Ceramic 3. 0.25D. Sinteredcarbide 4. 0.5dCodes:A B 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

173

GATE2009(PI)( )In an orthogonal machining operation, the tool life

obtained is 10 min at a cutting speed of 100 m/min,

while at 75 m/min cutting speed, the tool life is 30

min The value of index (n) in the Taylors tool lifemin. The value of index (n) in the Taylor s tool life

equation

(a) 0.262 (b) 0.323 (c) 0.423 (d) 0.521

174

ISRO2011d l d d dA 50 mm diameter steel rod was turned at 284 rpm and

tool failure occurred in 10 minutes The speed wastool failure occurred in 10 minutes. The speed was

changed to 232 rpm and the tool failed in 60 minutes.

Assuming straight line relationship between cutting

d d l l f h l f lspeed and tool life, the value of Taylorian Exponent is

(a) 0 21 (b) 0 13 (c) 0 11 (d) 0 23(a) 0.21 (b) 0.13 (c) 0.11 (d) 0.23

175

G 200GATE2004In a machining operation, doubling the cuttingIn a machining operation, doubling the cuttingspeed reduces the tool life to the of the originalvalue. The exponent n in Taylor's tool life equation

18value. The exponent n in Taylor s tool life equation

VTn = C, is1 1 1 11 1 1 1( ) ( ) ( ) ( )8 4 3 2

a b c d

176

S 2000IES 2000In a tool life test, doubling the cutting speedIn a tool life test, doubling the cutting speedreduces the tool life to 1/8th of the original. TheTaylor's tool life index isTaylor s tool life index is

( ) ( ) ( ) ( )1 1 1 1a b c d 2 3 4 8( ) ( ) ( ) ( )2 3 4 8

177

S 999 S O 20 3IES 1999,ISRO2013I i l i t t i ti f t l ithIn a singlepoint turning operation of steel with acemented carbide tool, Taylor's tool life exponent is, y p

0.25. If the cutting speed is halved, the tool life will

increase by

( ) (b)(a) Two times (b) Four times

(c) Eight times (d) Sixteen times(c) Eight times (d) Sixteen times

178

S 2002IAS 2002U i th T l ti VTn l l t thUsing the Taylor equation VTn = c, calculate the

percentage increase in tool life when the cuttingp g g

speed is reduced by 50% (n = 05 and c = 400)

(a) 300% (b) 400%

(c) 100% (d) 50%

179

S 99IAS 1995In a single point turning operation with a cementedIn 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

h l lif ill bthe tool life will become

(a) Half(a) Half

(b) Two times

(c) Eight times

(d) Sixteen times180For-2015 (IES, GATE & PSUs) Page 21 of 205 Rev.1

IES 2013IES2013A carbide tool(having n = 0 25) with a mild steelA carbide tool(having n = 0.25) with a mild steel

workpiece was found to give life of 1 hour 21minutes while cutting at 60 m/min. The value of C

in Taylors tool life equationwould be equal to:

( )(a) 200

(b) 180(b) 180

(c) 150

(d) 100 181

S 99IAS 1997I th T l ' t l lif ti VTn C th lIn the Taylor's tool life equation, VTn = C, the value

of n = 0.5. The tool has a life of 180 minutes at a5

cutting speed of 18 m/min. If the tool life is reduced

to 45 minutes, then the cutting speed will be

( ) (b)(a) 9 m/min (b) 18 m/min

(c) 36 m/min (d) 72 m/min(c) 36 m/min (d) 72 m/min

182

S 2006 i lIES 2006conventionalA HSS t l i d f t i ti Th t l lif iAn HSS tool is used for turning operation. The tool life is

1 hr. when turning is carried at 30 m/min. The tool lifeg 3 /

will be reduced to 2.0 min if the cutting speed is

doubled. Find the suitable speed in RPM for turning 300

di t th t t l lif i imm diameter so that tool life is 30 min.

183

GATE2009LinkedAnswerQuestions(1)Inamachiningexperiment,toollifewasfoundtovaryInamachiningexperiment,toollifewasfoundtovarywiththecuttingspeedinthefollowingmanner:Cuttingspeed(m/min) Toollife(minutes)Cuttingspeed(m/min) Toollife(minutes)60 8190 36Theexponent(n)andconstant(k)oftheTaylor'sp ( ) ( ) ytoollifeequationare(a)n=0.5andk=540 (b)n=1andk=4860(a)n 0.5andk 540 (b)n 1andk 4860(c)n=1andk=0.74 (d)n0.5andk=1.15

184

GATE2009LinkedAnswerQuestions(2)Inamachiningexperiment,toollifewasfoundtovaryInamachiningexperiment,toollifewasfoundtovarywiththecuttingspeedinthefollowingmanner:Cuttingspeed(m/min) Toollife(minutes)Cuttingspeed(m/min) Toollife(minutes)60 8190 36Whatisthepercentageincreaseintoollifewhenp gthecuttingspeedishalved?(a)50% (b)200%(a)50% (b)200%(c)300% (d)400%

185

IFS2013In a metal cutting experiment, the tool life was

found to vary with the cutting speed in the

f ll ifollowing manner :Cuttingspeed,V(inm/min) Toollife,T(inmin)

100 120

Derive Taylor's tool life equation for this operation

130 50

y q p

and estimate the tool life at a speed of 2.5 m/s. Also

estimate the cutting speed for a tool life of 80 min.186

G 20 0GATE2010F t l A T l t l lif t ( ) i dFor tool A, Taylors tool life exponent (n) is 0.45 and

constant (K) is 90. Similarly for tool B, n = 0.3 and K( ) 9 y , 3

= 60. The cutting speed (in m/min) above which tool

A will have a higher tool life than tool B is

( ) (b) ( ) (d)(a) 26.7 (b) 42.5 (c) 80.7 (d) 142.9

187

GATE 2013GATE2013Two cutting tools are being compared for aTwo 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

Wh V i h i d i / i d T i hWhere V is the cutting speed in m/min and T is the

tool life in min. The carbide tool will provide higherp g

tool life if the cutting speed in m/min exceeds

(a) 15.0 (b) 39.4 (c) 49.3 (d) 60.0188

ExamplepThefollowingdatawasobtainedfromthetoollifecuttingtest:cuttingtest:

d 49 4 49 23 48 6 4 6 42 8CuttingSpeed,m/min:49.74 49.23 48.67 45.76 42.58Toollife,min 2.94 3.90 4.77 9.87 28.27

DeterminetheconstantsoftheTaylortoollifeequationDeterminetheconstantsoftheTaylortoollifeequationVTn =C


GATE2003GATE2003A 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 Afeed of 0.25 mm/rev and depth of cut of 1 mm. A

similar batch of 10 tools of the same specificationp

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 components can be produced withcut. How many components can be produced with

one cutting tool at 60 rpm?g p

(a) 29 (b) 31 (c) 37 (d) 42 190

G 999GATE1999Wh t i i t t h i th lif tWhat is approximate percentage change is the life, t,

of a tool with zero rake angle used in orthogonalg g

cutting when its clearance angle, , is changed from10o to 7o?

( l k l )(Hint: Flank wear rate is proportional to cot )

(a) 30 % increase (b) 30% decrease(a) 30 % increase (b) 30%, decrease

(c) 70% increase (d) 70% decrease(c) 70% c ease (d) 70% dec ease

191

E t d d M difi d T l tiExtendedorModifiedTaylorsequation

i e Cuttingspeedhasthegreatereffectfollowedbyfeedi.e Cuttingspeedhasthegreatereffectfollowedbyfeedanddepthofcutrespectively.

192

IES 2010IES2010Tool life is affected mainly withTool life is affected mainly with

(a) Feed( )

(b) Depth of cut

(c) Coolant

(d) Cutting speed

193

ISRO2012What is the correct sequence of the following

t i d f th i i tparameters in order of their maximum tominimum influence on tool life?

d1. Feed rate2. Depth of cut3. Cutting speedSelect the correct answer using the codes givenSelect the correct answer using the codes givenbelow(a) 1 2 3 (b) 3 2 1 (c) 2 3 1 (d) 3 1 2(a) 1, 2, 3 (b) 3, 2, 1 (c) 2, 3, 1 (d) 3, 1, 2

194

S 99IES 1997C id th f ll i l tConsiderthefollowingelements:

1 Noseradius 2 Cuttingspeed1. Noseradius 2. Cuttingspeed

3. Depthofcut 4. Feed3 p

ThecorrectsequenceoftheseelementsinDECREASINGorderoftheirinfluenceontoollifeis

(a) 2,4,3,1 (b) 4,2,3,1

( ) (d) I (c) 2,4,1,3 (d) 4,2,I,3195

S 99 200IES 1994,2007F i i th t i l l t i t iFor increasing the material removal rate in turning,

without any constraints, what is the right sequencey , g q

to adjust the cutting parameters?

1. Speed 2. Feed 3. Depth of cut

Select the correct answer using the code given below:

( ) (b)(a) 1 2 3 (b) 2 3 1

(c) 3 2 1 (d) 1 3 2(c) 3 2 1 (d) 1 3 2196

S 2008IES 2008WhatarethereasonsforreductionoftoollifeinaWhatarethereasonsforreductionoftoollifeinamachiningoperation?1 Temperatureriseofcuttingedge1. Temperatureriseofcuttingedge2. Chippingoftooledgeduetomechanicalimpact3. Gradualwearsattoolpoint4. Increaseinfeedofcutatconstantcuttingforce4 gSelectthecorrectanswerusingthecodegivenbelow:below:(a) 1,2and3 (b) 2,3and4( ) d (d) d(c) 1,3and4 (d) 1,2and4

197

S 99IAS 1995Assertion (A): An increase in depth of cut shortensAssertion (A): An increase in depth of cut shortensthe tool life.Reason(R): Increases in depth of cut gives rise toReason(R): Increases in depth of cut gives rise torelatively small increase in tool temperature.( ) B th A d R i di id ll t d R i th(a) Both A and R are individually true and R is thecorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true


S 999 S 20 0 C i lESE1999;IAS2010ConventionalTh f ll i ti f t l lif bt i d f HSSThe following equation for tool life was obtained for HSS

tool. A 60 min tool life was obtained using the followingg g

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

d f d d d th f t t th i d bspeed, feed and depth of cut are together increased by

25% and also if they are increased individually by 25%;5 y y y 5 ;

where f = feed, d = depth of cut, v = speed.

199

T l Lif CToolLifeCurve

1.HSS 2.Carbide 3.Ceramic200

IFS2009With the help of Taylors tool life equation,

determine the shape of the curve between velocity

of cutting and life of the tool. Assume an HSS tool

and steel as work materialand steel as work material.

[10Marks][ ]

201

S 20 0 C i lIES2010Conventionaly Drawtoollifecurvesforcastalloy,HighspeedsteelandDrawtoollifecurvesforcastalloy,Highspeedsteelandceramictools. [2 Marks]

Ans.

1.Highspeedsteel 2.castalloyand3.ceramictools.202

IES2010The above figure shows a typicalThe above figure shows a typicalrelationship between tool life andcutting speed for differentg pmaterials. Match the graphs forHSS, Carbide and Ceramic tool

i l d l hmaterials and select the correctanswer using the code givenbelow the lists:below the lists:

Code: HSS Carbide Ceramic(a) 1 2 3(a) 1 2 3(b) 3 2 1(c) 1 3 2(c) 1 3 2(d) 3 1 2

203

S 2003IAS 2003ThetoollifecurvesfortwotoolsAandBareshowninThetoollifecurvesfortwotoolsAandBareshowninthefigureandtheyfollowthetoollifeequationVTn =C.Considerthefollowingstatements:g1. Valueofnforboththetoolsissame.2. ValueofCforboththetoolsissame.3. ValueofCfortoolAwillbegreaterthanthatforthetoolB.4. ValueofCfortoolBwillbegreaterthanthatforthetoolA.4. a ue o C o too be g eate t a t at o t e too .

Whichofthesestatementsis/arecorrect?(a) 1and3 (b) 1and4(a) 1and3 (b) 1and4(c) 2only (d) 4only

204

Cuttingspeedusedfordifferentl ltoolmaterials

HSS (min) 30 m/min < Cast alloy < Carbide

< Cemented carbide 150 m/min < Cermets

< Ceramics or sintered oxide (max) 600 m/min< Ceramics or sintered oxide (max) 600 m/min

205

EffectofRakeangleontoollife

206

EffectofClearanceangleontoollifeIf clearance angle increased it reduces flank wear butweaken the cutting edge, so best compromise is 80 forHSS and 50 for carbide tool.

Effect of work piece on tool lifeEffectofworkpieceontoollifeyWith hard microconstituents in the matrix gives poor

l liftool life.yWith larger grain size tool life is better.


IES 2013 ConventionalIES 2013ConventionalYou are asked to turn ductile cast iron with variousmicrostructure and hardness as shown in themicrostructure and hardness as shown in thefollowing table.

H d (HB) F it P litHardness(HB) Ferrite Pearlite1.Annealed 186 97 32.AsCast 265 20% 80%3.Annealed 170 100

D fi h i i i f l lif i h

3.Annealed 170 1004.AsCast 207 60 40Draw a figure showing variation of tool life withcutting speed and the effect of workpiece hardness

d iand microstructure.208

IES 2013Con.Answer

209

ForIESOnly

ToollifeTestsy Conventionaltest:Usingempiricalformulay A l t dt t E ti t th t llif i kly Acceleratedtest:Estimatethetoollifequickly

ExtrapolatingofsteadywearrateHighspeedtestwilltakelesstimeVariablespeedtestMultipassturningTaperturningp g

Refer:B.LJuneja+Nitin Seth

IES 2014IES 2014In accelerated tool life tests, the three main types ofIn accelerated tool life tests, the three main types ofquick and less costly tool life testing are(a) Extrapolation on the basis of steady wear;(a) Extrapolation on the basis of steady wear;conventional measurement of flank and crater wear;comparative performance against tool chippingcomparative performance against tool chipping(b) Measurement of abrasive wear; multi pass turning;

ti l t f diff iconventional measurement of diffusion wear(c) Extrapolating on the basis of steady wear, multipassturning; taper turning(d) comparative performance against tool chipping;p p g pp gtaper turning; measurement of abrasive wear

211

ChipEquivalentE d tti d l thEngaged cutting edgelengthChipEquivalent(q)

Planareaof cut=

y It is used for controlling the tool temperature.

212

The SCEA alters the length of the engaged cutting

d i h ff i h f A l hedge without affecting the area of cut. As a result, the

chip equivalent changed When the SCEA is increased,chip equivalent changed. When the SCEA is increased,

the chip equivalent is increased, without significantly

changing the cutting forces.

Increase in nose radius also increases the value of the

chip equivalent and improve tool life.

213

IES1996Chip equivalent is increased byChip equivalent is increased by(a) An increases in sidecutting edge angle of tool(b) An increase in nose radius and side cuttingedge angle of tooledge angle of tool(c) Increasing the plant area of cut(d) Increasing the depth of cut.

214

Economics of metal cuttingEconomicsofmetalcutting

215 216For-2015 (IES, GATE & PSUs) Page 25 of 205 Rev.1

FormulaFormulanV T C=no oV T C

O ti t l lif f i i tOptimum tool life for minimum cost

1T T if T & giventC n C C = + T T if T , & given

1

o c c t mm

C CC n

C n

= + 1 if & given

Optimum tool life for Maximum Productivity

tt m

m

C n C CC n

= Optimum tool life for Maximum Productivity(minimum production time)

1T To cn

n = 217

Units:Tc min(Toolchangingtime)C / i i l ( li Ct Rs./servicingorreplacement(Toolingcost)Cm Rs/min(Machiningcost)V m/min(Cuttingspeed)/ ( g p )

Toolingcost(C ) toolregrindcostToolingcost(Ct)=toolregrindcost+tooldepreciationperservice/replacement

Machiningcost(Cm)=labour cost+overheadcostpermin

218

S 2009 C i lIES2009ConventionalDetermine the optimum cutting speed for anDetermine the optimum cutting speed for anoperation on a Lathe machine using the followinginformation:information:Tool change time: 3 minT l i d ti iTool regrinds time: 3 minMachine running cost Rs.0.50 perminDepreciation of tool regrinds Rs. 5.0The constants in the tool life equation are 60 andThe constants in the tool life equation are 60 and0.2

219

G 20GATE2014If th T l t l lif t i d thIf the Taylors tool life exponent n is 0.2, and the

tool changing time is 1.5 min, then the tool life (ing g 5 , (

min) formaximum production rate is .

220

S 200 C i lESE2001ConventionalIn a certain machining operation with a cuttingIn a certain machining operation with a cuttingspeed of 50 m/min, tool life of 45 minutes wasb d Wh th tti d i dobserved. When the cutting speed was increased

to 100 m/min, the tool life decreased to 10 min.Estimate the cutting speed for maximumproductivity if tool change time is 2 minutes.p y g

221

IAS 2011MainDetermine the optimum speed for achievingmaximum production rate in a machiningmaximum production rate in a machiningoperation. The data is as follows :Machining time/job = 6 minMachining time/job = 6 min.Tool life = 90 min.Ta lor's equation constants C 00 n 0Taylor s equation constants C = 100, n = 0.5Job handling time = 4 min./job

l h i i iTool changing time = 9 min.[10Marks]

222

G 200GATE2005

223

S 200 C dIAS 2007ContdA diagram related to machining economics withg gvarious cost components is given above. Match List I(Cost Element) with List II (Appropriate Curve) andselect the correct answer using the code given belowthe Lists:

ListI ListII(CostElement) (AppropriateCurve)A. Machiningcost 1. CurvelB. Toolcost 2. Curve2C. Toolgrindingcost 3. Curve3D Nonproductivecost 4 Curve4D. Non productivecost 4. Curve 4

5. Curve5224

Contd From previous slideContd. From previous slide

Code:A B C D A B C D(a) 3 2 4 5 (b) 4 1 3 2(c) 3 1 4 2 (d) 4 2 3 5(c) 3 1 4 2 (d) 4 2 3 5


IES2011The optimum cutting speed is one which shouldThe optimum cutting speed is one which shouldhave:

Hi h t l l t1. High metal removal rate2. High cutting tool life3. Balance the metal removal rate and cutting

tool life(a) 1, 2 and 3(b) 1 and 2 only(b) 1 and 2 only(c) 2 and 3 only( )(d) 3 only

226

MinimumCostVsProductionRate

max.production max.profit min. costV >V >V

227

S 999IES 1999Consider the following approaches normallyConsider the following approaches normallyapplied for the economic analysis of machining:1 Maximumproduction rate1. Maximumproduction rate2. Maximumprofit criterion3. Minimum cost criterionThe correct sequence in ascending order of optimumq g pcutting speed obtained by these approaches is(a) 1, 2, 3 (b) 1, 3, 2(a) 1, 2, 3 (b) 1, 3, 2(c) 3, 2, 1 (d) 3, 1, 2

228

S 998IES 1998The variable cost and production rate of aThe variable cost and production rate of amachining process against cutting speed are shownin the given figure. For efficient machining, thein the given figure. For efficient machining, therange of best cutting speed would be between(a) 1 and 3(a) 1 and 3(b) 1 and 5(c) 2 and 4(d) 3 and 5( ) 3 5

229

S 2002IAS 2002Optimum cutting speed for minimum cost (V i )Optimum cutting speed for minimum cost (Vc min )and optimum cutting speed for maximumproduction rate (V ) have which one of theproduction rate (Vr max ) have which one of thefollowing relationships?(a) V = V (b) V > V(a) Vcmin = Vrmax (b) Vcmin > Vrmax(c) Vcmin < Vrmax (d) V2c min = Vrmax

230

S 99IAS 1997Inturning,theratiooftheoptimumcuttingspeedInturning,theratiooftheoptimumcuttingspeedforminimumcostandoptimumcuttingspeedformaximumrateofproductionisalwaysmaximumrateofproductionisalways(a) Equalto1(b) I th f 6t (b) Intherangeof0.6to1(c) Intherangeof0.1to0.6(d) Greaterthan1

231

S 2000IES 2000The magnitude of the cutting speed for maximumThe magnitude of the cutting speed for maximumprofit ratemust be(a) In between the speeds for minimum cost and(a) In between the speeds for minimum cost andmaximum production rate(b) Hi h th th d f i d ti t(b) Higher than the speed for maximum production rate(c) Below the speed for minimum cost(d) Equal to the speed for minimum cost

232

S 200IES 2004Consider the following statements:g1. As the cutting speed increases, the cost of productioninitially reduces, then after an optimum cutting speed itincreases2. As the cutting speed increases the cost of productionl i d f i i l l i dalso increases and after a critical value it reduces3. Higher feed rate for the same cutting speed reduces costof productionof production4. Higher feed rate for the same cutting speed increases thecost of productioncost of productionWhich of the statements given above is/are correct?(a) 1 and 3 (b) 2 and 3(a) 1 and 3 (b) 2 and 3(c) 1 and 4 (d) 3 only

233

S 2002IES 2002Ineconomicsofmachining,whichoneoftheIneconomicsofmachining,whichoneofthefollowingcostsremainsconstant?(a) Machiningcostperpiece(a) Machiningcostperpiece(b) Toolchangingcostperpiece(c) Toolhandlingcostperpiece(d) Toolcostperpiece( ) p p


S 200IAS 2007Assertion (A): The optimum cutting speed for theAssertion (A): The optimum cutting speed for theminimum cost of machining may not maximize theprofit.profit.Reason (R): The profit also depends on rate ofproductionproduction.(a) Both A and R are individually true and R is the

t l ti f Acorrect explanation of A(b) Both A and R are individually true but R is not thecorrect explanation of A(c) A is true but R is false(d) A is false but R is true

235

IES 2010IES2010With increasing cutting velocity, the totalWith increasing cutting velocity, the totaltime formachining a component

( ) D(a) Decreases(b) Increases( )(c) Remains unaffected(d) Fi d d h i(d) First decreases and then increases

236

MachinabilityDefinitionMachinability can be tentatively defined as ability ofMachinability can be tentatively defined as ability ofbeing machined and more reasonably as ease ofmachining.machining.

Such ease of machining or machining charactersof any toolwork pair is to be judged by:yTool wear or tool lifeTool wear or tool lifeyMagnitude of the cutting forcesy Surface finishyMagnitude of cutting temperatureyMagnitude of cutting temperatureyChip forms. 237

ForIESOnly

FreeCuttingsteelsAdditi f l d i l b l h i d t l dy Addition of lead in low carbon resulphurised steels andalso in aluminium, copper and their alloys help reduceth i Th di d l d ti l t di ti ittheir s. The dispersed lead particles act as discontinuityand solid lubricants and thus improve machinability byd i f i ti tti f d t t t lreducing friction, cutting forces and temperature, tool

wear and BUE formation.y It contains less than 0.35% lead by weight .y A free cutting steel containsgC0.07%, Si0.03%, Mn0.9%, P0.04%, S0.22%, Pb0.15%

238

MachinabilityIndexO M hi bilit R tiOrMachinabilityRating

The machinability index KM is defined byThe machinability index KM is defined byKM = V60/V60R

Wh V i th tti d f th t t t i lWhere V60 is the cutting speed for the target materialthat ensures tool life of 60 min, V60R is the same for thef t i lreference material.

If KM > 1, the machinability of the target material isbetter that this of the reference material, and vice versa

239

IES 2012IES 2012The usual method of defining machinability of aThe usual method of defining machinability of amaterial is by an index based on(a) Hardness of work material(a) Hardness of work material(b) Production rate of machined parts(c) Surface finish of machined surfaces(d) Tool life( )

240

S 996IAS 1996Assertion(A):ThemachinabilityofamaterialcanAssertion(A):Themachinabilityofamaterialcanbemeasuredasanabsolutequantity.Reason(R):MachinabilityindexindicatesthecaseReason(R):Machinabilityindexindicatesthecasewithwhichamaterialcanbemachined( ) B thA dR i di id ll t dRi th (a) BothAandRareindividuallytrueandRisthecorrectexplanationofA(b) BothAandRareindividuallytruebutRisnot thecorrectexplanationofA(c) AistruebutRisfalse(d) AisfalsebutRistrue(d) AisfalsebutRistrue

241

ForIESOnly

MachinabilityofSteelyMainly sulfur and lead improve machinability ofMainly sulfur and lead improve machinability ofsteel.R lf i d t l S lf i dd d t t l l ifyResulfurized steel: Sulfur is added to steel only ifthere is sufficient manganese in it. Sulfur forms

lf d h h l d hmanganese sulfide which exists as an isolated phaseand act as internal lubrication and chip breaker.y If insufficient manganese is there a low melting ironsulfide will formed around the austenite grainsulfide will formed around the austenite grainboundary. Such steel is very weak and brittle.y Tellurium and s

Documents

2015 Metal Cutting Metal Forming Metrology