production engg. question set with answers

For‐2014 (IES, GATE & PSUs)

MetalCutting,MetalForming&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-1994-2011 (18 Yrs.), some PSUs questions and conventional questions are added. Section‐I:TheoryofMetalCutting

Chapter-1: Basics of Metal Cutting Page-1 Chapter-2: Force & Power in Metal Cutting Page-7 Chapter-3: Tool life, Tool Wear, Economics and Machinability Page-13

Section‐II:MetalFormingChapter-4: Cold Working, Recrystalization and Hot Working Page-24 Chapter-5: Rolling Page-27 Chapter-6: Forging Page-30 Chapter-7: Extrusion & Drawing Page-37 Chapter-8: Sheet Metal Operation Page-43 Chapter-9: Powder Metallurgy Page-52

Section‐III:MetrologyChapter-10: Limit, Tolerance & Fits Page-56 Chapter-11: Measurement of Lines & Surfaces Page-61 Chapter-12: Miscellaneous of Metrology Page-65

Section‐IV:CuttingToolMaterials Page‐67

Theory of Metal CuttingTheory of Metal Cutting

B S K M d lBy S K Mondal

IES 2013IES‐2013Carbide 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

S 200IES‐2001For cutting of brass with single point cutting toolFor cutting of brass with single‐point cutting toolon a lathe, tool should have( ) N ti k l(a) Negative rake angle(b) Positive rake angle(c) Zero rake angle (d) Zero side relief angle(d) Zero side relief angle

IES‐1995

Single point thread cutting tool should ideally have:

a) Zero rake)b) Positive rakec) Negative rakec) Negative raked) Normal rake

GATE‐1995; 2008C tti ti i t i b Cutting power consumption in turning can be

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) I i h l l(d) Increasing the clearance angle

IES‐1993Assertion (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

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

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

IES 2011Which 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.

For-2014 (IES, GATE & PSUs) Page 1 of 78

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

IES 2007Cast iron with impurities of carbide requires aparticular rake angle for efficient cutting with singleparticular rake angle for efficient cutting with singlepoint tools, what is the value of this rake angle, givereasons for your answerreasons for your answer.

[ 2 marks]

S 99IAS – 1994Consider the following characteristicsConsider the following characteristics1. The cutting edge is normal to the cutting velocity.

Th tti f i t di ti l2. The cutting forces occur in two directions only.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(c) 2 and 3 (d) 1, 2 and 3

IES 2012IES ‐ 2012During orthogonal cutting, an increase in cutting speed During orthogonal cutting, an increase in cutting speed causes(a) An increase in longitudinal cutting force(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‐2006Which of the following is a single point cutting tool?(a) Hacksaw blade(b) Milling cutter(b) Milling cutter(c) Grinding wheel(d) P ti t l(d) Parting tool

IES 2012IES ‐ 2012Statement (I): Negative rake angles are preferred on rigid set‐( ) g g p gups for interrupted cutting and difficult‐to machine materials.Statement (II):Negative rake angle directs the chips on to the machined surface( ) 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

IES‐1995Th 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.

IES‐2006Assertion (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

IES‐2002Consider 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


IES 2012IES ‐ 2012Tool life increase with increase inTool life increase with increase in(a) Cutting speed (b) N di (b) Nose radius (c) Feed (d) Depth of cut

IES‐2009Consider 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

IES 1995IES‐1995Consider 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

IES 1994IES‐1994Tool 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

IES 2009IES‐2009The following tool signature is specified for a singleThe following tool signature is specified for a single‐point cutting tool in American system:

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

S 993IES‐1993In ASA System if the tool nomenclature is 8 6 5 5In ASA System, if the tool nomenclature is 8‐6‐5‐5‐10‐15‐2‐mm, then the side rake angle will be( ) ° (b) 6° ( ) 8° (d) °(a) 5° (b) 6° (c) 8° (d) 10°

ISRO‐2011A 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

GATE‐2008In a single point turning tool, the side rake angleIn a single point turning tool, the side rake angleand orthogonal rake angle are equal. Φ 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 Φ is closest to0 90o oφ≤ ≤The value of Φ is closest to(a) 00 (b) 450

(c) 600 (d) 900

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


GATE‐2001D 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

GATE 2011A 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) 56°5(d) 76°

IES‐1994The 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

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

IES ‐ 2004In a machining operation chipthickness ratio is 0 3 and the rakethickness ratio is 0.3 and the rakeangle of the tool is 10°. What is the

l f th h t i ?value of the shear strain?(a) 0.31 (b) 0.13( ) 3 ( ) 3(c) 3.00 (d) 3.34

GATE 2012GATE ‐2012Details 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 +10°Cutting speed 2.5 m/sMean thickness of primary shear zone 25 microns

The shear strain rate in s–1 during the process isThe shear strain rate in s during the process is(a) 0.1781×105 (b) 0.7754×105

( ) 5 (d) 5(c) 1.0104×105 (d) 4.397×105

IES‐2004Consider the following statements with respect to Consider the following statements with respect to the relief angle of cutting tool: 1 This affects the direction of chip flow1. This affects the direction of chip flow2. This reduces excessive friction between the tool

d k iand work piece3. This affects tool life4. This allows better access of coolant to the tool work piece interfacepWhich of the statements given above are correct?(a) 1 and 2 (b) 2 and 3(a) 1 and 2 (b) 2 and 3(c) 2 and 4 (d) 3 and 4

IES‐2006Consider the following statements:Consider the following statements:1. A large rake angle means lower strength of the cutting edgecutting edge.2. Cutting torque decreases with rake angle.Which of the statements given above is/are correct?(a) Only 1 (b) Only 2( ) y ( ) y(c) Both 1 and 2 (d) Neither 1 nor 2

IES‐2004Match. List I with List II and select the correct answer Match. List I with List II and select the correct answer using the codes given below the Lists:

List I List IIList I List IIA. Plan approach angle 1. Tool faceB Rake angle 2 Tool flankB. Rake angle 2. Tool flankC. Clearance angle 3. Tool face and flankD W d l C i dD. Wedge angle 4. Cutting edge

5. Tool noseA 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


IES‐2003The 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

IES‐2004, ISRO‐2009The rake angle of a cutting tool is 15° shearThe rake angle of a cutting tool is 15 , shearangle 45° and cutting velocity 35 m/min.Wh t i th l it f hi l th t lWhat is the velocity of chip along the toolface?(a) 28.5 m/min (b) 27.3 m/min(c) 25 3 m/min (d) 23 5 m/min(c) 25.3 m/min (d) 23.5 m/min

IES‐2008Consider the following statements: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/min1. Chip velocity will be 45 m/min.2. Chip velocity will be 80 m/min.3 Chip thickness will be 1 8 mm3. Chip thickness will be 1∙8 mm.4. Chip thickness will be 3∙2 mm.

l h h d b lSelect the correct answer using the code given below:(a) 1 and 3 (b) 1 and 4(c) 2 and 3 (d) 2 and 4

IES‐2001If α is the rake angle of the cutting tool is theφIf α 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

φ

velocity of chip sliding along the shear plane isgiven by

(a) (b)cosV α sinV φ(a) (b)cos( )φ α− ( )cos −

φφ α

(c) (d)cossin( )−V α

φ αsin

sin( )V α

φ α−

IES‐2003An 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

IAS‐2003In orthogonal cutting, shear angle is the angle betweenIn orthogonal cutting, shear angle is the angle between(a) Shear plane and the cutting velocity(b) Shear plane and the rake plane(b) Shear plane and the rake plane(c) Shear plane and the vertical direction(d) Sh l d h di i f l i f l i (d) Shear plane and the direction of elongation of crystals in the chip

IAS‐2002 IAS‐2000 IAS‐1998The cutting velocity in m/sec, for turning a work pieceThe cutting velocity in m/sec, for turning a work pieceof diameter 100 mmat the spindle speed of 480 RPM is(a) 1.26 (b) 2.51(a) 1.26 (b) 2.51(c) 48 (d) 151


IAS‐1995In 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.

G 2009 ( )GATE – 2009 (PI) Common Data S‐1An 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

G 2009 ( )GATE – 2009 (PI) Common Data S‐2An 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

GATE‐1995Plain milling of mild steel plate produces (a) Irregular shaped discontinuous chips(a) egu a s aped d sco t uous c ps(b) Regular shaped discontinuous chip(c) Continuous chips ithout built up edge(c) Continuous chips without built up edge(d) Joined chips

IES 2007IES 2007During 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?C ti ibb lik hi i f d h t iContinuous ribbon like chip is formed when turning1. At a higher cutting speed

A l i d2. At a lower cutting speed3. A brittle material

A d il i l4. A ductile materialSelect the correct answer using the code given below:( ) d (b) d(a) 1 and 3 (b) 1 and 4(c) 2 and 3 (d) 2 and 4

IAS‐1997Consider the following machining conditions: BUE will Consider the following machining conditions: BUE will form in(a) Ductile material. (b) High cutting speed.(a) Ductile material. (b) High cutting speed.(c) Small rake angle. (d) Small uncut chip thickness.

GATE‐2002A b ilt d i f d hil hi i A built‐up‐edge is formed while machining

(a) Ductile materials at high speed(a) Ductile materials at high speed

(b) Ductile materials at low speedp

(c) Brittle materials at high speed

(d) Brittle materials at low speed

IES‐1997Assertion (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

Ch‐1: Mechanics of Basic Machining OperationQ. 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 BFor-2014 (IES, GATE & PSUs) Page 6 of 78

S K Mondal

New Stamp

F & P i M t l C ttiForce & Power in Metal Cutting

By S K Mondal

ESE ‐2000 (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 co‐efficient 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 ).

ESE‐2005 ConventionalMild l i b i hi d iMild steel is being machined at a cuttingspeed of 200 m/min with a tool rake angle of10. The width of cut and uncut thickness are 2mm and 0.2 mm respectively. If the averagep y gvalue of co‐efficient of friction between thetool and the chip is 0 5 and the shear stress oftool and the chip is 0.5 and the shear stress oftheworkmaterial is 400 N/mm2,

Determine (i) shear angle and(ii)Cutting and thrust component of the(ii)Cutting and thrust component of theforce.

GATE 2008 (PI) Linked S 1GATE ‐2008 (PI) Linked S‐1In 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: 0‐10‐7‐7‐10‐75‐1. Given

width 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 tool‐chip 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

GATE 2008 (PI) Linked S 2GATE ‐2008 (PI) Linked S‐2In 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: 0‐10‐7‐7‐10‐75‐1. Given

width 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 tool‐chip 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

GATE 2007 (PI) C D t 1GATE – 2007 (PI) Common Data‐1In 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

GATE 2007 (PI) C D t 2GATE – 2007 (PI) Common Data‐2In 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

GATE‐2007In 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

S 2003 C i lESE‐2003‐ 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.α

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.


GATE – 1995 ‐ConventionalWhil t i C t l d f 6 di t tWhile turning a C‐15 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 Merchant’s circle diagramDraw schematically the Merchant s circle diagramfor the cutting force in the present case.

IAS‐2003 Main ExaminationD 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) Co‐efficient of friction at the rake face(iii) The dynamic shear strength of the work material

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

GATE 2013GATE‐2013A 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

GATE‐2007I th l t i f di b t l Th In orthogonal turning of medium carbon steel. The specific machining energy is 2.0 J/mm3. The cutting

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

GATE 2013 (PI) C D Q iGATE‐2013 (PI) Common Data QuestionA 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

ExampleWhen the rake angle is zero during orthogonal cutting, show that

( )2

11

s r rp r

μτ −=

+

s

1Where τ is theshear strengrh of the material

cp r+

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

For PSU & IESIn strain gauge dynamometers the use of howmany active gauge makes the dynamometers moremany active gauge makes the dynamometers moreeffective(a) Four(a) Four(b) Three(c) T o(c) Two(d) One

Ans. (a)

GATE‐2006 Common Data Questions(1)I th l hi i tiIn an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NA M h t' thAssume Merchant's theory.The coefficient of friction at the tool‐chip interface is ( ) (b)(a) 0.23 (b) 0.46 (c) 0.85 (d) 0.95


GATE‐2006 Common Data Questions(2)I th l hi i tiIn an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NA M h t' thAssume Merchant's theory.The percentage of total energy dissipated due to f h l h ffriction at the tool‐chip interface is

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

GATE‐2006 Common Data Questions(3)I th l hi i tiIn an orthogonal machining operation:Uncut thickness = 0.5 mm Cutting speed = 20 m/min Rake angle = 15°Width of cut = 5 mm Chip thickness = 0.7 mmWidth of cut 5 mm Chip thickness 0.7 mmThrust force = 200 N Cutting force = 1200 NA M h t' thAssume Merchant's theory.The values of shear angle and shear strain,

lrespectively, are (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‐2003 Common Data Questions(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°

The thickness of the produced chip is(a) 0.511 mm (b) 0.528 mm (c) 0.818 mm (d) 0.846 mm(c) 0.818 mm (d) 0.846 mm

GATE‐2003 Common Data Questions(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 27.75°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(c) 0.71 (d) 0.98

GATE‐2008 Common Data Question (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.( )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 (c) 28: 400N (d) 28:320N

GATE‐2008 Common Data Question (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.The cutting and Thrust forces, r Rees e ctively, are (a) 568N; 387N (b) 565N; 381N (c) 440N; 342N (d) 480N; 356N

GATE ‐2010 (PI) Linked S‐1In orthogonal turning of an engineering alloy, it hasIn orthogonal turning of an engineering alloy, it hasbeen observed that the friction force acting at the chip‐tool 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 chip‐tool 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

GATE ‐2010 (PI) Linked S‐2In orthogonal turning of an engineering alloy, it hasg g g g y,been observed that the friction force acting at the chip‐tool 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 chip‐tool 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

Linked Answer Questions GATE‐2013 S‐1In 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 chip‐tool interface.

Themain (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) 7.16For-2014 (IES, GATE & PSUs) Page 9 of 78

S K Mondal

Rectangle

Linked Answer Questions GATE‐2013 S‐2In 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 chip‐tool interface.

Themain (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

GATE 2011 (PI) Linked S1GATE – 2011 (PI) Linked S1During 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

GATE 2011 (PI) Linked S2GATE – 2011 (PI) Linked S2During 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

The ultimate shear stress (in N/mm2) of the work material is(a) 235 (b) 139 (c) 564 (d) 380

IES 2012IES ‐ 2012During orthogonal cutting, an increase in cutting speed During orthogonal cutting, an increase in cutting speed causes(a) An increase in longitudinal cutting force(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 2010IES 2010The relationship between the shear angle Φ,The relationship between the shear angle Φ,the friction angle β and cutting rake angle αis given asis given as

IES‐2005Whi h f h f ll i i hWhich one of the following is the correctexpression for the Merchant's machinabilityconstant?(a) 2φ γ α+(a)(b)

2φ γ α+ −2φ γ α− +

(c)(d)

2φ γ α− −

φ γ α+ −(d)(Where = shear angle, = friction angle

φ γ α+

φ γand = rake angle)α

GATE‐1997I i l l i i i In a typical metal cutting operation, using a 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°

S 999IAS – 1999In an orthogonal cutting process, rake angle of theIn an orthogonal cutting process, rake angle of thetool is 20° and friction angle is 25.5°. UsingMerchant's shear angle relationship, the value ofMerchant s shear angle relationship, the value ofshear angle will be(a) 39 5° (b) 42 25°(a) 39.5 (b) 42.25(c) 47.75° (d) 50.5°

IES‐2003I h l i h i fIn orthogonal cutting test, the cutting force =900 N, the thrust force = 600 N and chipshear angle is 30o. Then the chip shear force is(a) 1079 4 N (b) 969 6 N(a) 1079.4 N (b) 969.6 N(c) 479.4 N (d) 69.6 N


IES‐2000In 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 chip‐tool interface will be

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

IES‐1996Which 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

GATE‐2007I th l t i f l b t l i ithIn orthogonal turning of low carbon steel pipe withprincipal cutting edge angle of 90°, themain cuttingf i N d th f d f i 8 N Th hforce is 1000 N and the feed force is 800 N. The shearangle is 25° and orthogonal rake angle is zero.E l i M h t’ th th ti f f i tiEmploying Merchant’s theory, the ratio of frictionforce to normal force acting on the cutting tool is( ) ( )(a) 1.56 (b) 1.25(c) 0.80 (d) 0.64

IES‐1997C 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

IES‐1999Th di l f i i l i l d iThe radial force in single‐point 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

IES‐1995Th i l f d i l l iThe primary tool force used in calculatingthe total power consumption in machining isthe(a) Radial force (b) Tangential force(a) Radial force (b) Tangential force(c) Axial force (d) Frictional force.

IES‐2002I 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%

IES‐1998I 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

S 2003IAS – 2003As the cutting speed increasesAs the cutting speed increases(a) More heat is transmitted to the work piece and less heat is transmitted to the toolheat is transmitted to the tool(b) More heat is carried away by the chip and less heat is t itt d t th t ltransmitted to the tool(c) More heat is transmitted to both the chip and the tool(d) More heat is transmitted to both the work piece and ( ) pthe tool


IES‐2001P 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 metal‐tool interface

S 99IAS – 1995Thrust force will increase with the increase inThrust force will increase with the increase in(a) Side cutting edge angle(b) T l di (b) Tool nose radius (c) Rake angle(d) End cutting edge angle.

IES 2010IES 2010Consider the following statements:Consider the following statements:In an orthogonal, single‐point metal cutting,

th id tti d l i i das the side‐cutting 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

IES‐1993A '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

IES 2011The instrument or device used to measure the cutting The instrument or device used to measure the cutting forces in machining is :( ) T h t(a) Tachometer(b) Comparator(c) Dynamometer(d) Lactometer(d) Lactometer

S 200IAS‐2001Assertion (A): Piezoelectric transducers and preferred( ) pover strain gauge transducers in the dynamometers formeasurement of three‐dimensional 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

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 set‐up(d) Using radiation pyrometer( ) g py

IES‐1998Th f f i i i k fThe gauge factor of a resistive pick‐up 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

IES‐2000A 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 tool‐chip 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 trueFor-2014 (IES, GATE & PSUs) Page 12 of 78

GATE 1992h 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

IES‐2004Assertion (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

GATE‐1993Th 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

Tool Wear, Tool Life & Tool Wear, Tool Life & MachinabilityMachinability


IAS – 2009 MainExplain ‘sudden‐death mechanism’ of tool failure.

[ k ][ 4 – marks]

GATE 2008 (PI)GATE‐2008 (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

IES 2009 ConventionalSh t d fl k i l i t Show crater wear and flank wear on a single point cutting tool. State the factors responsible for wear t i t lon a turning tool.

[ 2 –marks]

IES 2010IES 2010Flank wear occurs on theFlank wear occurs on the(a) Relief face of the tool(b) Rake face(c) Nose of the tool(c) Nose of the tool(d) Cutting edge

S 200IES – 2007Flank wear occurs mainly on which of the Flank wear occurs mainly on which of the following?(a) Nose part and top face(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 the cutting edgeg g


S 200IES – 2004Consider the following statements:Consider the following statements:During the third stage of tool‐wear, 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

S 2002IES – 2002Crater wear on tools always starts at some distance Crater wear on tools always starts at some distance from the tool tip because at that point(a) Cutting fluid does not penetrate(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 200IAS – 2007Why does crater wear start at some distance from Why does crater wear start at some distance from the tool tip?(a) Tool strength is minimum at that region(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 2000IES – 2000Crater wear starts at some distance from the tool tip Crater wear starts at some distance from the tool tip because(a) Cutting fluid cannot penetrate that region (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 996IES – 1996Notch wear at the outside edge of the depth of cut is Notch wear at the outside edge of the depth of cut is due to(a) Abrasive action of the work hardened chip material(a) Abrasive action of the work hardened chip material(b) Oxidation(c) Slip‐stick action of the chip (d) Chipping.( ) pp g

S 99IES – 1995Match List I with List II and select the correct 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) List I (Wear type) List II (Associated mechanism) A. Abrasive wears 1. Galvanic actionB. Adhesive wears 2. Ploughing actionC. Electrolytic wear 3. Molecular transfery 3D. Diffusion wears 4. Plastic deformation

5 Metallic bond5. Metallic bondCode: 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 99IES – 1995Crater wear is predominant inCrater wear is predominant in(a) Carbon steel tools (b) T t bid t l(b) Tungsten carbide tools(c) High speed steel tools (d) Ceramic tools

S 99IES – 1994Assertion (A): Tool wear is expressed in terms of Assertion (A): Tool wear is expressed in terms of flank wear rather than crater wear.Reason (R): Measurement of flank wear is simple Reason (R): Measurement of flank wear is simple and more accurate.( ) 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 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(d) A is false but R is true

S 2008IES – 2008What are the reasons for reduction of tool life in a What are the reasons for reduction of tool life in a machining operation?1 Temperature rise of cutting edge1. Temperature rise of cutting edge2. Chipping of tool edge due to mechanical impact3. Gradual wears at tool point4. Increase in feed of cut at constant cutting force4 gSelect the correct answer using the code given below:below:(a) 1, 2 and 3 (b) 2, 3 and 4( ) d (d) d (c) 1, 3 and 4 (d) 1, 2 and 4


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

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

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

IES 2012IES ‐ 2012In Taylor’s tool life equation VTn = C, the constants n In Taylor s tool life equation VT C, the constants n and C depend upon1 Work piece material1. Work piece material2. Tool material3. Coolant(a) 1, 2, and 3 ( ) 3(b) 1 and 2 only (c) 2 and 3 only (c) 2 and 3 only (d) 1 and 3 only

IFS 2009With the help of Taylor’s 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.

[10‐Marks][ ]

S 20 0 C i lIES 2010 ConventionalDraw tool life curves for cast alloy, High speed steel and Draw tool life curves for cast alloy, High speed steel and ceramic tools. [2 – Marks]

Ans.

1. High speed steel 2. cast alloy and 3. ceramic tools.

IES‐1996Chip equivalent is increased byChip equivalent is increased by(a) An increases in side‐cutting 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.

S 992IES – 1992Tool life is generally specified byTool life is generally specified by(a) Number of pieces machined(b) V l f t l d(b) Volume of metal removed(c) Actual cutting time(d) Any of the above

G 200GATE‐2004In a machining operation doubling theIn a machining operation, doubling thecutting speed reduces the tool life to th ofth i i l l Th t i T l '

18

the original value. The exponent n in Taylor'stool life equation VTn = C, is

1 1 1 1( ) ( ) ( ) ( )8 4 3 2

a b c d


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

S 999IES – 1999In 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 0.25. If the cutting speed is halved, the tool life will increase by(a) Two times (b) Four times(a) Two times (b) Four times(c) Eight times (d) Sixteen times

S 2008IES – 2008In Taylor's tool life equation is VTn = constant.In Taylor s tool life equation is VT = constant.What is the value of n for ceramic tools?( ) t (b) t (a) 0.15 to 0.25 (b) 0.4 to 0.55(c) 0.6 to 0.75 (d) 0.8 to 0.9

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) 0∙1 to 0∙15 (b) 0∙2 to 0∙4( ) t 6 (d) 6 t(c) 0.45 to 0∙6 (d) 0∙65 to 0∙9

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

S 998IAS – 1998Match List ‐ I (Cutting tool material) with List ‐ II ( g )(Typical value of tool life exponent 'n' in the Taylor's equation V.Tn = C) and select the correct answer using th d i b l th li tthe codes given below the lists:List – I List – IIA HSS 8A. HSS 1. 0.18B. Cast alloy 2. 0.12C C iC. Ceramic 3. 0.25D. Sintered carbide 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

GATE ‐2009 (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 Taylor’s 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

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

work‐piece was found to give life of 1 hour 21

minutes while cutting at 60 m/min. The value of C

in Taylor’s tool life equationwould be equal to:

( )(a) 200

(b) 180(b) 180

(c) 150

(d) 100

ISRO‐2011d 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


IES 2010The 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

G 20 0GATE‐2010For tool A Taylor’s tool life exponent (n) isFor tool A, Taylor s tool life exponent (n) is0.45 and constant (K) is 90. Similarly for toolB d K 6 Th tti d (iB, n = 0.3 and K = 60. The cutting speed (inm/min) above which tool A will have a highertool life than tool B is(a) 26 7 (b) 42 5 (c) 80 7 (d) 142 9(a) 26.7 (b) 42.5 (c) 80.7 (d) 142.9

GATE 2013GATE‐2013Two 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) 60.0

ExamplepThe following data was obtained from the tool‐life cutting test:cutting test:

d 49 4 49 23 48 6 4 6 42 8Cutting Speed, m/min:49.74 49.23 48.67 45.76 42.58Tool life, min 2.94 3.90 4.77 9.87 28.27

Determine the constants of the Taylor tool life equation Determine the constants of the Taylor tool life equation VTn = C

G 2003GATE‐2003A batch of 10 cutting tools could produce 500A batch of 10 cutting tools could produce 500components while working at 50 rpm with atool feed of 0.25 mm/rev and depth of cut of 1tool feed of 0.25 mm/rev and depth of cut of 1mm. A similar batch of 10 tools of the samespecification could produce 122 componentsspecification could produce 122 componentswhile working at 80 rpm with a feed of 0.25mm/rev and 1 mm depth of cut How manymm/rev and 1 mm depth of cut. How manycomponents can be produced with onecutting tool at 60 rpm?cutting tool at 60 rpm?(a) 29 (b) 31(c) 37 (d) 42

S 99 200IES – 1994, 2007For increasing the material removal rate in turning, For increasing the material removal rate in turning, without any constraints, what is the right sequence to adjust the cutting parameters?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 (d) 1‐ 3‐ 2(c) 3 2 1 (d) 1 3 2

IES 2010IES 2010Tool life is affected mainly withTool life is affected mainly with(a) Feed(b) Depth of cut(c) Coolant(c) Coolant(d) Cutting speed

S 99IES – 1997Consider the following elements:Consider the following elements:1. Nose radius 2. Cutting speed

D th f t F d3. Depth of cut 4. FeedThe correct sequence of these elements in DECREASING order of their influence on tool life is(a) 2, 4, 3, 1 (b) 4, 2, 3, 1 ( ) 4 3 ( ) 4 3(c) 2,4, 1, 3 (d) 4, 2, I, 3

ISRO‐2012What 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


S 992IES – 1992Tool life is generally better whenTool life is generally better when(a) Grain size of the metal is large(b) G i i f th t l i ll(b) Grain size of the metal is small(c) Hard constituents are present in the microstructure of the tool material(d) None of the above( )

S 2003IAS – 2003The tool life curves for two tools A and B are shown in 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:g1. Value of n for both the tools is same.2. Value of C for both the tools is same.3. Value of C for tool A will be greater than that for the tool B.4. Value of C for tool B will be greater than that for the tool A.4. a ue o C o too be g eate t a t at o t e too .

Which of these statements is/are correct?(a) 1 and 3 (b) 1 and 4(a) 1 and 3 (b) 1 and 4(c) 2 only (d) 4 only

S 2002IAS – 2002Using the Taylor equation VTn = c, calculate the Using the Taylor equation VT = 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%( ) % (d) %(c) 100% (d) 50%

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

IES 2010IES 2010With 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

S 2000IAS – 2000Consider the following statements:Consider the following statements:The tool life is increased by

B ilt d f ti1. Built ‐up edge formation2. Increasing cutting velocity3. Increasing back rake angle up to certain valueWhich of these statements are correct?Which of these statements are correct?(a) 1 and 3 (b) 1 and 2( ) d (d) d (c) 2 and 3 (d) 1, 2 and 3

S 99IAS – 1997In the Taylor's tool life equation, VTn = C, the valueIn the Taylor s tool life equation, VT = C, the valueof n = 0.5. The tool has a life of 180 minutes at acutting speed of 18 m/min. If the tool life is reducedcutting speed of 18 m/min. If the tool life is reducedto 45 minutes, then the cutting speed will be(a) 9 m/min (b) 18 m/min(a) 9 m/min (b) 18 m/min(c) 36 m/min (d) 72 m/min

S 996IAS – 1996The tool life increases with theThe tool life increases with the(a) Increase in side cutting edge angle(b) D i id k l(b) Decrease in side rake angle(c) Decrease in nose radius(d) Decrease in back rake angle

S 99IAS – 1995In a single point turning operation with a cemented 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 exponent of 0.25, if the cutting speed is halved, then the tool life will become(a) Half (a) Half (b) Two times(c) Eight times(d) Sixteen times.( )


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 2006 i lIES – 2006 conventionalAn HSS tool is used for turning operation TheAn HSS tool is used for turning operation. Thetool life is 1 hr. when turning is carried at 30/ i Th t l lif ill b d d t i ifm/min. The tool life will be reduced to 2.0 min if

the cutting speed is doubled. Find the suitablespeed in RPM for turning 300 mm diameter sothat tool life is 30 min.3

S 999 S 20 0 C i lESE‐1999; IAS ‐2010 ConventionalThe following equation for tool life was obtained for HSSThe following equation for tool life was obtained for HSStool. A 60 min tool life was obtained using the followingcutting condition VT0.13f0.6d0.3= C. v = 40 m/min, f = 0.25cutting condition VT f d C. v 40 m/min, f 0.25mm, d = 2.0 mm. Calculate the effect on tool life ifspeed, feed and depth of cut are together increased byspeed, feed and depth of cut are together increased by25% and also if they are increased individually by 25%;where f = feed, d = depth of cut, v = speed.where f feed, d depth of cut, v speed.

S 2009 C i lIES 2009 ConventionalDetermine 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

S 200 C i lESE‐2001 ConventionalIn 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

IAS – 2011 MainDetermine 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.[10‐Marks]

GATE‐2009 Linked Answer Questions (1) In a machining experiment, tool life was found to vary 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)Cutting speed (m/min) Tool life (minutes)60 8190 36The exponent (n) and constant (k) of the Taylor's p ( ) ( ) ytool life equation are(a) n = 0.5 and k = 540 (b) n= 1 and k=4860 (a) n 0.5 and k 540 (b) n 1 and k 4860 (c) n = ‐1 and k = 0.74 (d) n‐0.5 and k=1.15

GATE‐2009 Linked Answer Questions (2) In a machining experiment, tool life was found to vary 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)Cutting speed (m/min) Tool life (minutes)60 8190 36What is the percentage increase in tool life when p gthe cutting speed is halved?(a) 50% (b) 200%(a) 50% (b) 200%(c) 300% (d) 400%

G 999GATE‐1999What is approximate percentage change isWhat is approximate percentage change isthe life, t, of a tool with zero rake angle usedi th l tti h it lin orthogonal cutting when its clearanceangle, α, is changed from 10o to 7o?(Hint: Flank wear rate is proportional to cot α(a) 30 % increase (b) 30% decrease(a) 30 % increase (b) 30%, decrease(c) 70% increase (d) 70% decrease


G 200GATE‐2005 S 200 C dIAS – 2007 Contd…A 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:

List I List II(Cost Element) (Appropriate Curve) A. Machining cost 1. Curve‐lB. Tool cost 2. Curve‐2C. Tool grinding cost 3. Curve‐3D Non‐productive cost 4 Curve‐4D. Non productive cost 4. Curve 4

5. Curve‐5

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

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

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

IES 2011The 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

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

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

S 2002IES – 2002In economics of machining, which one of the In economics of machining, which one of the following costs remains constant? (a) Machining cost per piece(a) Machining cost per piece(b) Tool changing cost per piece(c) Tool handling cost per piece(d) Tool cost per piece( ) 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


S 99IAS – 1997In turning, the ratio of the optimum cutting speed In turning, the ratio of the optimum cutting speed for minimum cost and optimum cutting speed for maximum rate of production is alwaysmaximum rate of production is always(a) Equal to 1 (b) I th f 6 t (b) In the range of 0.6 to 1(c) In the range of 0.1 to 0.6 (d) Greater than 1

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( )

IES 2011 C ti lIES 2011 ConventionalDiscuss the effects of the following elements on thegmachinability of steels:(i) Aluminium and silicon(i) Aluminium and silicon(ii) Sulphur and Selenium(iii) L d d Ti(iii) Lead and Tin(iv) Carbon and Manganese(v) Molybdenum and Vanadium [5 Marks]

S 992IES – 1992Ease of machining is primarily judged byEase of machining is primarily judged by(a) Life of cutting tool between sharpening(b) Ri idit f k i(b) Rigidity of work ‐piece(c) Microstructure of tool material(d) Shape and dimensions of work

S 200 2009IES – 2007, 2009Consider the following:Consider the following:1. Tool life

C tti f2. Cutting forces3. Surface finishWhich 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(c) 2 and 3 only (d) 2 only

ISRO‐2007Machinablity depends on( ) h l d h l(a) Microstructure, physical and mechanicalproperties and composition of workpiece material.(b) Cutting forces(c) Type of chip( ) yp p(d) Tool life

S 2003IES – 2003Assertion (A): The machinability of steels improves( ) y pby adding sulphur to obtain so called 'FreeMachining Steels‘.Reason (R): Sulphur in steel forms manganesesulphide inclusion which helps to produce thinribbon like continuous chip.(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

S 2009IES – 2009The elements which, added to steel, help in chipThe elements which, added to steel, help in chipformation during machining are(a) Sulphur lead and phosphorous(a) Sulphur, lead and phosphorous(b) Sulphur, lead and cobalt(c) Aluminium, lead and copper(d) Aluminium, titanium and copper( ) pp


S 998IES – 1998Consider the following criteria in evaluating Consider the following criteria in evaluating machinability:1 Surface finish 2 Type of chips1. Surface finish 2. Type of chips3. Tool life 4. Power consumptionIn 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 (b) 2, 1, 4, 3 ( ) 4 3 ( ) 4 3(c) 1, 2, 3, 4 (d) 2, 1, 3, 4

S 996IES – 1996Which of the following indicate betterWhich of the following indicate bettermachinability?1 Smaller shear angle1. Smaller shear angle2. Higher cutting forces3. Longer tool life4. Better surface finish.4(a) 1 and 3 (b) 2 and 4(c) 1 and 2 (d) 3 and 4(c) 1 and 2 (d) 3 and 4

S 996IES – 1996Small amounts of which one of the followingSmall amounts of which one of the followingelements/pairs of elements is added to steel toincrease its machinability?increase its machinability?(a) Nickel (b) Sulphur and phosphorus( ) Sili (d) M d(c) Silicon (d) Manganese and copper

S 99IES – 1995In low carbon steels, presence of small quantities In low carbon steels, presence of small quantities sulphur improves(a) Weldability (b) Formability(a) Weldability (b) Formability(c) Machinability (d) Hardenability

S 992IES – 1992Machining of titanium is difficult due toMachining of titanium is difficult due to(a) High thermal conductivity of titanium(b) Ch i l ti b t t l d k(b) Chemical reaction between tool and work(c) Low tool‐chip contact area(d) None of the above

S 996IAS – 1996Assertion (A): The machinability of a material can Assertion (A): The machinability of a material can be measured as an absolute quantity.Reason (R): Machinability index indicates the case Reason (R): Machinability index indicates the case with which a material can be machined( ) 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 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(d) A is false but R is true

IES‐1995C id th f ll i k t i lConsider the following work materials:1. Titanium 2. Mild steel 3. Stainless steel 4. Grey cast iron.The correct sequence of these materials in terms of The correct sequence of these materials in terms of increasing order of difficulty in machining is(a) 4 2 3 1 (b) 4 2 1 3 (a) 4,2,3,1 (b) 4,2, 1,3 (c) 2,4,3,1 (d) 2, 4, 1, 3.

G 2009GATE‐2009Friction 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(b) increasing the depth of cut(c) Decreasing the cutting speed (d) increasing the cutting speed

IES 2002IES ‐ 2002The value of surface roughness 'h' obtained duringThe value of surface roughness h obtained duringthe turning operating at a feed 'f' with a round nosetool having radius 'r' is given astool having radius r is given as


IAS 1996IAS ‐ 1996Given thatGiven thatS = feed in mm/rev. andR di iR = nose radius in mm,the maximum height of surface roughness Hmaxproduced by a single‐point turning tool is given by(a) S2/2R( )(b) S2/4R(c) S2/4R(c) S2/4R(d) S2/8R

IES 1999IES ‐ 1999In turning operation, the feed could be doubled toIn turning operation, the feed could be doubled toincrease the metal removal rate. To keep the samelevel of surface finish, the nose radius of the toollevel of surface finish, the nose radius of the toolshould be(a) Halved (b) Kept unchanged(a) Halved (b) Kept unchanged(c) doubled (d) Made four times

GATE 1997GATE ‐ 1997A cutting tool has a radius of 1.8 mm. The feed rate A cutting tool has a radius of 1.8 mm. The feed rate for a theoretical surface roughness of Ra = 5 m is(a) 0 36 mm/rev

μ(a) 0.36 mm/rev(b) 0.187 mm/rev(c) 0.036 mm/rev(d) 0.0187 mm/rev( ) 7

GATE 2007 (PI)GATE – 2007 (PI)A tool with Side Cutting Edge angle of 30o andA tool with Side Cutting Edge angle of 30 andEnd Cutting Edge angle of 10o is used for fineturning with a feed of 1 mm/rev Neglecting noseturning with a feed of 1 mm/rev. Neglecting noseradius of the tool, the maximum (peak to valley)h i h f f h d d ill bheight of surface roughness produced will be(a) 0.16 mm (b) 0.26 mm( ) ( )(c) 0.32 mm (d) 0.48 mm

GATE 2005GATE ‐ 2005Two tools P and Q have signatures 5°‐5°‐6°‐6°‐8°‐30°‐Two tools P and Q have signatures 5 5 6 6 8 300 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively.They are used to turn components under the sameThey are used to turn components under the samemachining conditions. If hpand hQ denote the peak‐to‐valley heights of surfaces produced by the tools Pto valley heights of surfaces produced by the tools Pand Q, the ratio hp/hQ will be

+ ++ +

tan8 cot15 tan15 cot8( ) ( )tan8 cot30 tan30 cot8

o o o o

o o o oa b+ ++ +

tan8 cot30 tan30 cot8tan15 cot7 tan7 cot15( ) ( )

o o o o

c d+ +

( ) ( )tan30 cot7 tan7 cot30o o o oc d

IES 1993 ISRO 2008IES – 1993, ISRO‐2008For achieving a specific surface finish in single pointFor achieving a specific surface finish in single pointturning the most important factor to be controlledisis(a) Depth of cut (b) Cutting speed( ) F d (d) T l k l(c) Feed (d) Tool rake angle

IES 2006IES ‐ 2006In the selection of optimal cutting conditions, theIn the selection of optimal cutting conditions, therequirement of surface finish would put a limit onwhich of the following?which of the following?(a) The maximum feed(b) Th i d th f t(b) The maximum depth of cut(c) The maximum speed(d) The maximum number of passes

GATE 2010 (PI)GATE ‐2010 (PI)During turning of a low carbon steel bar with TiN coatedDuring turning of a low carbon steel bar with TiN coatedcarbide insert, one need to improve surface finish

h f l l hwithout sacrificing material removal rate. To achieveimproved surface finish, one should

(a) decrease nose radius of the cutting tool and increasedepth of cutdepth of cut

(b) Increase nose radius of the cutting tool

(c) Increase feed and decrease nose radius of the cuttingtooltool

(d) Increase depth of cut and increase feed

IAS 2009 MainIAS ‐2009 MainWhat are extreme‐pressure lubricants?

[ 3 – marks]Where high pressures and rubbing action areg p gencountered, hydrodynamic lubrication cannot bemaintained; so Extreme Pressure (EP) additives must bedd d h l b i EP l b i i i id d badded to the lubricant. EP lubrication is provided by anumber of chemical components such as boron,phosphorus sulfur chlorine or combination of thesephosphorus, sulfur, chlorine, or combination of these.The compounds are activated by the higher temperatureresulting from extreme pressure As the temperatureresulting from extreme pressure. As the temperaturerises, EP molecules become reactive and releasederivatives such as iron chloride or iron sulfide andforms a solid protective coating.


IES 2001IES ‐ 2001Dry and compressed air is used as cutting fluid forDry and compressed air is used as cutting fluid formachining(a) Steel (b) Aluminium(a) Steel (b) Aluminium(c) Cast iron (d) Brass

IES 2012IES ‐ 2012The most important function of the cutting fluid is toThe most important function of the cutting fluid is to(a) Provide lubrication (b) C l th t l d k i(b) Cool the tool and work piece(c) Wash away the chips (d) Improve surface finish

Ch‐3: Cutting Tools, Tool Life and Cutting FluidQ. No Option Q. No Option Q. No OptionQ p Q p Q p1 B 12 C 23 A2 A 13 A 24 C3 A 14 A 25 C4 D 15 B 26 B5 D 16 B 27 B6 D 17 B 28 A7 B 18 A 29 B8 A 19 B 30 A9 A 20 A 31 C10 D 21 B 32 B11 C 22 B 33 C

Ch‐4: Economics of Machining OperationQ. No Option Q. No OptionQ. No Option Q. No Option1 C 6 B2 B 7 A2 B 7 A3 A 8 C4 C 9 A4 C 9 A5 A

Metal FormingMetal Forming


GATE‐1995A test specimen is stressed slightly beyond the

yield point and then unloaded. Its yield strength

(a) Decreases

( )(b) Increases

(c) Remains same(c) Remains same

(d) Become equal to UTS(d) Become equal to UTS

IES‐2013Statement (I): At higher strain rate and lowerStatement (I): At higher strain rate and lowertemperature structural steel tends to become brittle.Statement (II): At higher strain rate and lowerStatement (II): At higher strain rate and lowertemperature the yield strength of structural steel tendsto increaseto increase.

(a) Both Statement (I) and Statement (II) are individuallyd S (II) i h l i ftrue and Statement (II) is the correct explanation of

Statement (I)(b) Both Statement (I) and Statement (II) are individuallytrue but Statement (II) is not the correct explanation ofStatement (I)

(c) Statement (I) is true but Statement (II) is false( ) ( ) ( )(d) Statement (I) is false but Statement (II) is true

IES 2011Assertion (A): Lead Zinc and Tin are always hotAssertion (A): Lead, Zinc and Tin are always hotworked.R (R) If th k d i ld t tReason (R) : If they are worked in cold statethey cannot retain theirmechanical properties.(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 NOTthe correct explanation of Ap(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

G 2003GATE‐2003 Cold working of steel is defined as workingCold working of steel is defined as working(a) At its recrystallisation temperature(b) Ab it t lli ti t t(b) Above its recrystallisation temperature(c) Below its recrystallisation temperature(d) At two thirds of the melting temperature of themetal


S K Mondal

New Stamp

S K Mondal

New Stamp

G 2002 S O 20 2GATE‐2002, ISRO‐2012Hot rolling of mild steel is carried outHot rolling of mild steel is carried out(a) At recrystallisation temperature(b) B t °C t °C(b) Between 100°C to 150°C(c) Below recrystallisation temperature(d) Above recrystallisation temperature

ISRO 2010ISRO‐2010Materials after cold working are subjected toMaterials after cold working are subjected to

following process to relieve stresses

(a) Hot working

(b) Tempering

(c) Normalizing

(d) Annealing

S 2006IES – 2006Which one of the following is the process to refineWhich one of the following is the process to refinethe grains of metal after it has been distorted byhammering or cold working?hammering or cold working?(a) Annealing (b) Softening( ) R t lli i (d) N li i(c) Re‐crystallizing (d) Normalizing

S 200IES – 2004Consider the following statements:Consider the following statements:In comparison to hot working, in cold working,

Hi h f i d1. Higher forces are required2. No heating is required3. Less ductility is required4. Better surface finish is obtained4. Better surface finish is obtainedWhich of the statements given above are correct?( ) d (b) d(a) 1, 2 and 3 (b) 1, 2 and 4(c) 1 and 3 (d) 2, 3 and 4

S 2009IES – 2009Consider the following characteristics:Consider the following characteristics:1. Porosity in the metal is largely eliminated.

St th i d d2. Strength is decreased.3. Close tolerances cannot be maintained.Which of the above characteristics of hot working is/arecorrect?(a) 1 only (b) 3 only(c) 2 and 3 (d) 1 and 3(c) 2 and 3 (d) 1 and 3

S 2008IES – 2008Consider the following statements:Consider the following statements:1. Metal forming decreases harmful effects ofimpurities and improves mechanical strengthimpurities and improves mechanical strength.2. Metal working process is a plastic deformationprocess.3. Very intricate shapes can be produced by forgingprocess 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(c) 2 and 3 only (d) 1 and 3 only

S 2008IES – 2008Cold forging results in improved quality due toCold forging results in improved quality due towhich of the following?1 Better mechanical properties of the process1. Better mechanical properties of the process.2. Unbroken grain flow.3. Smoother finishes.4. High pressure.4 g pSelect the correct answer using the code given below:(a) 1 2 and 3 (b) 1 2 and 4(a) 1, 2 and 3 (b) 1, 2 and 4(c) 2, 3 and 4 (d) 1, 3 and 4

S 200IES – 2004Assertion (A): Cold working of metals results inAssertion (A): Cold working of metals results inincrease of strength and hardnessReason (R): Cold working reduces the total numberReason (R): Cold working reduces the total numberof dislocations per unit volume of thematerial( ) 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 2003IES – 2003Cold working produces the following effects:Cold working produces the following effects:1. Stresses are set up in the metal

G i t t t di t t d2. Grain structure gets distorted3. Strength and hardness of the metal are decreased4. Surface finish is reducedWhich of these statements are correct?Which of these statements are correct?(a) 1and 2 (b) 1, 2 and 3( ) d (d) d(c) 3 and 4 (d) 1 and 4


S 2000IES – 2000Assertion (A): To obtain large deformations by coldAssertion (A): To obtain large deformations by coldworking intermediate annealing is not required.Reason (R): Cold working is performed below theReason (R): Cold working is performed below therecrystallisation temperature of thework material.( ) 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

ISRO‐2009In the metal forming process, the stresses

t dencountered are(a) Greater than yield strength but less thanl hultimate strength

(b) Less than yield strength of the material(c) Greater than the ultimate strength of thematerial(d) Less than the elastic limit

S 99IES – 1997In metals subjected to cold working, strain In metals subjected to cold working, strain hardening effect is due to(a) Slip mechanism(a) Slip mechanism(b) Twining mechanism(c) Dislocation mechanism(d) Fracture mechanism( )

S 996IES – 1996Consider the following statements:Consider the following statements:When a metal or alloy is cold worked

It i k d b l t t1. It is worked below room temperature.2. It is worked below recrystallisation temperature.3. Its hardness and strength increase.4. Its hardness increases but strength does not 4. Its hardness increases but strength does not increase.Of these correct statements areOf these correct statements are(a) 1 and 4 (b) 1 and 3 (c) 2 and 3 (d) 2 and 4

S 2006IES – 2006Assertion (A): In case of hot working of metals, theAssertion (A): In case of hot working of metals, thetemperature at which the process is finally stoppedshould not be above the recrystallisation temperature.y pReason (R): If the process is stopped above therecrystallisation temperature, grain growth will takey p , g gplace again and spoil the attained structure.(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

S 992IES – 1992Specify the sequence correctlySpecify the sequence correctly(a) Grain growth, recrystallisation, stress relief(b) St li f i th t lli ti(b) Stress relief, grain growth, recrystallisation(c) Stress relief, recrystallisation, grain growth(d) Grain growth, stress relief, recrystallisation

S 996IAS – 1996Formild steel, the hot forging temperature range isFormild steel, the hot forging temperature range is(a) 4000C to 6000C(b) 0C t 0C(b) 7000C to 9000C(c) 10000C to 12000C(d) 13000Cto 15000C

S 200IAS – 2004Assertion (A): Hot working does not produce strain Assertion (A): Hot working does not produce strain hardening.Reason (R): Hot working is done above the reReason (R): Hot working is done above the re‐crystallization 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 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(d) A is false but R is true

S 2002IAS‐2002Assertion (A): There is good grain refinement in hotAssertion (A): There is good grain refinement in hotworking.Reason (R): In hot working physical properties areReason (R): In hot working physical properties aregenerally improved.( ) 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 2008IES‐2008Which one of the following is correct?Which one of the following is correct?Malleability is the property by which a metal oralloy can be plastically deformed by applyingalloy can be plastically deformed by applying(a) Tensile stress (b) Bending stress(c) Shear stress (d) Compressive stress

GATE 2013GATE‐2013I lli th t t f t f thIn a rolling process, the state of stress of the

material undergoing deformation isg g

(a) pure compression

(b) pure shear

(c) compression and shear

(d) tension and shear

ISRO‐2006Which of the following processes would produce

strongest components?

(a) Hot rolling

( )(b) Extrusion

(c) Cold rolling(c) Cold rolling

(d) Forging(d) Forging

ISRO‐2009Ring rolling is used(a) To decrease the thickness and increasediameterdiameter(b) To increase the thickness of a ring(c) For producing a seamless tube(d) For producing large cylinder(d) For producing large cylinder

S 2006IES – 2006Which one of the following is a continuous bendingWhich one of the following is a continuous bendingprocess in which opposing rolls are used to producelong sections of formed shapes from coil or striplong sections of formed shapes from coil or stripstock?(a) Stretch forming (b) Roll forming(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) i i i(d) grain orientation

IFS – 2010 Calculate the neutral plane to roll 250 mm wideannealed copper strip from 2 5 mm to 2 0 mmannealed copper strip from 2.5 mm to 2.0 mmthickness with 350 mm diameter steel rolls. Take µ =0 05 and σ’ =180 MPa0.05 and σ o =180 MPa.

[10‐marks]

G 2008GATE‐2008In a single pass rolling operation, a 20 mm thickIn a single pass rolling operation, a 20 mm thickplate with plate width of 100 mm, is reduced to 18mm. The roller radius is 250 mm and rotationalmm. The roller radius is 250 mm and rotationalspeed is 10 rpm. The average flow stress for the platematerial is 300 MPa. The power required for thematerial is 300 MPa. The power required for therolling operation in kW is closest to(a) 15 2(a) 15.2(b) 18.2(c) 30.4(d) 45.645

G 200GATE‐2007The thickness of a metallic sheet is reduced from anThe thickness of a metallic sheet is reduced from aninitial value of 16 mm to a final value of 10 mm inone single pass rolling with a pair of cylindricalone single pass rolling with a pair of cylindricalrollers each of diameter of 400 mm. The bite anglein degreewill bein degreewill be(a) 5.936(b) 6(b) 7.936(c) 8.936(d) 9.936


G 200GATE‐2004In a rolling process, sheet of 25 mm thickness isIn a rolling process, sheet of 25 mm thickness isrolled to 20 mm thickness. Roll is of diameter 600mm and it rotates at 100 rpm. The roll strip contactmm and it rotates at 100 rpm. The roll strip contactlength will be(a) 5 mm (b) 39 mm(a) 5 mm (b) 39 mm(c) 78 mm (d) 120 mm

G 998GATE‐1998A strip with a cross‐section 150 mm x 4.5 mm isA strip with a cross section 150 mm x 4.5 mm isbeing rolled with 20% reduction of area using 450mm diameter rolls. The angle subtended by themm diameter rolls. The angle subtended by thedeformation zone at the roll centre is (in radian)(a) 0 01 (b) 0 02(a) 0.01 (b) 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 anddiameter steel rollers, a strip of width 140 mm and

thickness 8 mm undergoes 10% reduction of

thickness. The angle of bite in radians is

(a) 0.006 (b) 0.031

(c) 0 062 (d) 0 600(c) 0.062 (d) 0.600

G 2006GATE‐2006A 4 mm thick sheet is rolled with 300 mm diameterA 4 mm thick sheet is rolled with 300 mm diameterrolls to reduce thickness without any change in itswidth. The friction coefficient at the work‐rollwidth. The friction coefficient at the work rollinterface is 0.1. The minimum possible thickness ofthe sheet that can be produced in a single pass isthe sheet that can be produced in a single pass is(a) 1.0 mm (b) 1.5 mm( ) (d)(c) 2.5 mm (d) 3.7 mm

GATE – 2011 (PI)The thickness of a plate is reduced from 30 mm toThe thickness of a plate is reduced from 30 mm to10 mm by successive cold rolling passes usingidentical rolls of diameter 600 mm Assume thatidentical rolls of diameter 600 mm. Assume thatthere is no change in width. If the coefficient offriction between the rolls and the work piece is 0 1friction between the rolls and the work piece is 0.1,the minimum number of passes required is( )(a) 3(b) 4(c) 6(d) 7(d) 7

( )GATE‐1992(PI)f h l f d ll fIf the elongation factor during rolling of an ingot

is 1 22 The minimum number of passes needed tois 1.22. The minimum number of passes needed to

produce a section 250 mm x 250 mm from an ingot

of 750 mm x 750 mm are

(a) 8 (b) 9

( ) (d)(c) 10 (d) 17

S 2003IES – 2003Assertion (A): While rolling metal sheet in rolling( ) g gmill, the edges are sometimes not straight and flatbut are wavy.Reason (R): Non‐uniform mechanical properties ofthe flat material rolled out result in waviness of theedges.(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

S 2002IES – 2002In rolling a strip between two rolls, the position ofIn rolling a strip between two rolls, the position ofthe neutral point in the arc of contact does notdepend ondepend on(a) Amount of reduction (b) Diameter of the rolls( ) C ffi i t f f i ti (d) M t i l f th ll(c) Coefficient of friction (d) Material of the rolls

S 200IES – 2001Which of the following assumptions are correct forg pcold rolling?

1. The material is plastic.p2. 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

contact and acts in one direction throughout the arc ofgcontact.

Select the correct answer using the codes given below:g gCodes:(a) 1 and 2 (b) 1 and 3(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 1, 2 and 3


S 200IES – 2001A strip is to be rolled from a thickness of 30 mm toA strip is to be rolled from a thickness of 30 mm to15 mm using a two‐high mill having rolls ofdiameter 300 mm. The coefficient of friction fordiameter 300 mm. The coefficient of friction forunaided bite should nearly be(a) 0 35 (b) 0 5(a) 0.35 (b) 0.5(c) 0.25 (d) 0.07

GATE ‐2008(PI)I lli thi k f t i i d d fIn a rolling process, thickness of a strip is reduced from 4

mm to 3 mm using 300 mm diameter rolls rotating at 1003 g 3 g

rpm. The velocity of the strip in (m/s) at the neutral

point is

( ) (b) ( ) (d)(a) 1.57 (b) 3.14 (c) 47.10 (d) 94.20

GATE‐1990 (PI)( )While rolling a strip the peripheral velocity of the

roll is ….A…..than the entry velocity of the strip

and is ……B …..the exit velocity of the strip.

( ) l th / t l(a) less than/greater less

(b) Greater than/less than(b) Greater than/less than

S 2000 G 20 0( )IES – 2000, GATE‐2010(PI)In the rolling process, roll separating force can beIn the rolling process, roll separating force can bedecreased by(a) Reducing the roll diameter(a) Reducing the roll diameter(b) Increasing the roll diameter(c) Providing back‐up rolls(d) Increasing the friction between the rolls and the( ) gmetal

S 999IES – 1999Assertion (A): In a two high rolling mill there is a Assertion (A): In a two high rolling mill there is a limit to the possible reduction in thickness in one pass.pass.Reason (R): The reduction possible in the second pass is less than that in the first passpass is less than that in the first pass.(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 the correct explanation of A (c) A is true but R is false(d) A is false but R is true

S 993IES – 1993In order to get uniform thickness of the plate byIn order to get uniform thickness of the plate byrolling process, one provides(a) Camber on the rolls(a) Camber on the rolls(b) Offset on the rolls(c) Hardening of the rolls(d) Antifriction bearings( ) g

S 993 G 989( )IES – 1993, GATE‐1989(PI)The blank diameter used in thread rolling will beThe blank diameter used in thread rolling will be(a) Equal to minor diameter of the thread(b) E l t it h di t f th th d(b) Equal to pitch diameter of the thread(c) A little large than the minor diameter of the thread(d) A little larger than the pitch diameter of the thread

S 992 G 992( )IES – 1992, GATE‐1992(PI)Thread rolling is restricted toThread rolling is restricted to(a) Ferrous materials(b) D til t i l(b) Ductile materials(c) Hard materials(d) None of the above

S 200IAS – 2004Assertion (A): Rolling requires high friction whichAssertion (A): Rolling requires high friction whichincreases forces and power consumption.Reason (R): To prevent damage to the surface of theReason (R): To prevent damage to the surface of therolled products, lubricants should be used.( ) 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 200IAS – 2001Consider the following characteristics of rollingConsider the following characteristics of rollingprocess:1 Shows work hardening effect1. Shows work hardening effect2. Surface finish is not good3. Heavy reduction in areas can be obtainedWhich of these characteristics are associated with hotrolling?(a) 1 and 2 (b) 1 and 3(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 1, 2 and 3

S 2000IAS – 2000Rolling very thin strips of mild steel requiresRolling very thin strips of mild steel requires(a) Large diameter rolls(b) S ll di t ll(b) Small diameter rolls(c) High speed rolling(d) Rolling without a lubricant

S 998IAS – 1998Match List ‐ I (products) with List ‐ II (processes)Match List I (products) with List II (processes)and select the correct answer using the codes givenbelow the lists:below the lists:

List – I List ‐IIA M S l d h l W ldiA. M.S. angles and channels 1. WeldingB. Carburetors 2. ForgingC. Roof trusses 3. CastingD. Gear wheels 4. RollingD. Gear wheels 4. RollingCodes:A B C D A B C D( ) (b)(a) 1 2 3 4 (b) 4 3 2 1(c) 1 2 4 3 (d) 4 3 1 2

S 200IAS – 2007Match List I with List II and select the correct answer usinggthe 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 frictionB Th hi h ill B ll ki llB. Three high mills 2. By small working roll, power

for rolling is reducedC Four high mills 3 Rolls of equal size areC. Four high mills 3. Rolls of equal size are

rotated only in one directionD. Cluster mills 4. Diameter of working roll isg

very smallCode:A B C D A B C D( ) (b)(a) 3 4 2 1 (b) 2 1 3 4(c) 2 4 3 1 (d) 3 1 2 4

S 2003IAS – 2003In one setting of rolls in a 3‐high rolling mill, oneIn one setting of rolls in a 3 high rolling mill, onegets(a) One reduction in thickness(a) One reduction in thickness(b) Two reductions in thickness(c) Three reductions in thickness(d) Two or three reductions in thickness depending( ) p gupon the setting

S 200IAS – 2007Consider the following statements:Consider the following statements:Roll forces in rolling can be reduced by

R d i f i ti1. Reducing friction2. Using large diameter rolls to increase the contactarea.3. Taking smaller reductions per pass to reduce the3 g p pcontact area.Which of the statements given above are correct?Which of the statements given above are correct?(a) 1 and 2 only (b) 2 and 3 only( ) d l (d) d(c) 1 and 3 only (d) 1, 2 and 3

GATE 2011The maximum possible draft in cold rolling of sheetThe maximum possible draft in cold rolling of sheetincreases with the

( ) i i ffi i t f f i ti(a) increase in coefficient of friction(b) decrease in coefficient of friction(c) decrease in roll radius(d) increase in roll velocity(d) increase in roll velocity

Q. No OptionRolling Ch‐14Q. No Option1 C2 B2 B3 D4 D4 D5 A6 A6 A7 B8 D9 C10 C11 B12 C

ForgingForging

By S K MondalBy S K MondalFor-2014 (IES, GATE & PSUs) Page 30 of 78

S K Mondal

New Stamp

GATE‐1989(PI)( )At the last hammer stroke the excess material from

the finishing cavity of a forging die is pushed

into……………..

G 20 0 ( )GATE ‐2010 (PI)H t di t l d f l lid di i d f iHot die steel, used for large solid dies in drop forging,

should necessarily havey

(a) high strength and high copper content

(b) high hardness and low hardenability

(c) high toughness and low thermal conductivity

(d) high hardness and high thermal conductivity

IES‐2013Statement (I): The dies used in the forging process aremade in pair.pStatement (II): The material is pressed between twosurfaces and the compression force applied, gives it asurfaces and the compression force applied, gives it ashape.

(a) Both Statement (I) and Statement (II) are individually(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 individuallyt b t St t t (II) i t th t l ti ftrue but Statement (II) is not the correct explanation ofStatement (I)

( ) ( ) b ( ) f l(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true

IFS‐2011What advantages does press forging have over drop

forging ? Why are pure metals more easily cold worked

th ll ?than alloys ?

[5‐marks][5 marks]

IAS‐2011 MainCompare Smith forging, drop forging, press

forging and upset forging. Mention three points

for each.

[ M k ][10 – Marks]

IES ‐ 2007Sometimes the parting plane between two forging

dies is not a horizontal plane, give the main reason

for this design aspect, why is parting plane

provided in closed die forging?provided, in closed die forging?

[ 2 marks][ ]

G 200GATE‐2007In open‐die forging, a disc of diameter 200 mm andIn open die forging, a disc of diameter 200 mm andheight 60 mm is compressed without any barrelingeffect. The final diameter of the disc is 400 mm. Theeffect. The final diameter of the disc is 400 mm. Thetrue strain is(a) 1 986 (b) 1 686(a) 1.986 (b) 1.686(c) 1.386 (d) 0.602

GATE‐1992, ISRO‐2012The true strain for a low carbon steel bar which is doubled in length by forging is (a) 0.307(b) 0.5(b) 0.5(c) 0.693(d) (d) 1.0

G 20 2GATE‐2012 Same Q GATE ‐2012 (PI)

A lid li d f di t d h i ht A solid cylinder of diameter 100 mm and height 50 mm

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


G 99GATE‐1994Match 4 correct pairs between List I and List II forMatch 4 correct pairs between List I and List II forthe questions List I gives a number of processes andList II gives a number of productsList II gives a number of products

List I List II( ) I t t ti T bi t(a) Investment casting 1. Turbine rotors(b) Die casting 2. Turbine blades(c) Centrifugal casting 3. Connecting rods(d) Drop forging 4. Galvanized iron pipe(d) Drop forging 4. Galvanized iron pipe(e) Extrusion 5. Cast iron pipes(f) Sh ll ldi 6 C b tt b d(f) Shell moulding 6. Carburettor body

G 998GATE‐1998List I List IIList I List II

(A) Aluminium brake shoe (1) Deep drawing(B) Pl ti t b ttl ( ) Bl ldi(B) Plastic water bottle (2) Blow moulding(C) Stainless steel cups (3) Sand casting(D) Soft drink can (aluminium)

(4) Centrifugal casting(4) Centrifugal casting(5) Impact extrusion(6) U t f i(6) Upset forging

S 2006IES – 2006Assertion (A): Forging dies are provided with taperAssertion (A): Forging dies are provided with taperor draft angles on vertical surfaces.Reason (R): It facilitates complete filling of dieReason (R): It facilitates complete filling of diecavity and favourable grain flow.( ) 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 200IES – 2005Consider the following statements:Consider the following statements:

1. Forging reduces the grain size of the metal, whichresults in a decrease in strength and toughnessresults in a decrease in strength and toughness.

2. Forged components can be provided with thinti ith t d i th t thsections, without reducing the strength.

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(c) Both 1 and 2 (d) Neither 1 nor 2

S 996IES – 1996Which one of the following is an advantage ofWhich one of the following is an advantage offorging?(a) Good surface finish(a) Good surface finish(b) Low tooling cost(c) Close tolerance(d) Improved physical property.( ) p p y p p y

IES 2012IES ‐ 2012Which of the following statements is correct for forging?Which of the following statements is correct for forging?(a) Forgeability is property of forging tool, by whichforging can be done easilyforging can be done easily.(b) Forgeability decreases with temperature upto loweriti l t tcritical temperature.

(c) Certain mechanical properties of the material areinfluenced by forging.(d) Pure metals have good malleability, therefore, poor( ) g y pforging properties.

IES 2013IES‐2013In the forging process: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 material4. Part are shaped by plastic deformation of material

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

(c) 1, 2 and 4 (d) 2, 3 and 4

IES 2012IES ‐ 2012Statement (I): It is difficult to maintain close tolerance in( )normal forging operation.Statement (II): Forging is workable for simple shapes( ) g g p pand has limitation for parts having undercuts.(a) Both Statement (I) and Statement (II) are( ) ( ) ( )individually true and Statement (II) is the correctexplanation of Statement (I)(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is not the correctexplanation of Statement (I)(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true

S 993 G 99 ( )IES – 1993, GATE‐1994(PI)Which one of the following manufacturingWhich one of the following manufacturingprocesses requires the provision of ‘gutters’?(a) Closed die forging(a) Closed die forging(b) Centrifugal casting(c) Investment casting(d) Impact extrusion( ) p


S K Mondal

New Stamp

S K Mondal

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S 99IES – 1997Assertion (A): In drop forging besides the provision( ) p g g pfor flash, provision is also to be made in the forgingdie for additional space called gutter.Reason (R): The gutter helps to restrict the outwardflow of metal thereby helping to fill thin ribs andbases in the upper die.(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

S 200IES – 2004Match List I (Different systems) with List II( y )(Associated terminology) and select the correctanswer using the codes given below the Lists:

List I List IIA. Riveted Joints 1. NippingJ pp gB. Welded joints 2. Angular movementC Leaf springs 3 FulleringC. Leaf springs 3. FulleringD. Knuckle joints 4. Fusion

A B C D A B C DA B C D A B C D(a) 3 2 1 4 (b) 1 2 3 4( ) (d)(c) 1 4 3 2 (d) 3 4 1 2

S 2003IES – 2003A forging method for reducing the diameter of a barA forging method for reducing the diameter of a barand in the process making it longer is termed as(a) Fullering (b) Punching(a) Fullering (b) Punching(c) Upsetting (d) Extruding

S 2002IES – 2002Consider the following steps involved in hammerConsider the following steps involved in hammerforging a connecting rod from bar stock:1 Blocking 2 Trimming1. Blocking 2. Trimming3. Finishing 4. Fullering 5. EdgingWhich of the following is the correct sequence ofoperations?(a) 1, 4, 3, 2 and 5(b) 4, 5, 1, 3 and 2(b) 4, 5, 1, 3 and 2(c) 5, 4, 3, 2 and 1(d) d(d) 5, 1, 4, 2 and 3

S 999IES – 1999Consider the following operations involved inConsider the following operations involved inforging a hexagonal bolt from a round bar stock,whose diameter is equal to the bolt diameter:whose diameter is equal to the bolt diameter:1. Flattening 2. Upsetting

S i C b i3. Swaging 4. CamberingThe 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(c) 2, 1, 3, 4 (d) 3, 2, 1, 4

S 2003IES – 2003Consider the following steps in forging a connectingConsider the following steps in forging a connectingrod from the bar stock:1 Blocking 2 Trimming1. Blocking 2. Trimming3. Finishing 4. EdgingSelect the correct sequence of these operations using thecodes given below:Codes:(a) 1‐2‐3‐4 (b) 2‐3‐4‐1(a) 1 2 3 4 (b) 2 3 4 1(c) 3‐4‐1‐2 (d) 4‐1‐3‐2

S 200IES – 2005The process of removing the burrs or flash from aThe process of removing the burrs or flash from aforged component in drop forging is called:(a) Swaging (b) Perforating(a) Swaging (b) Perforating(c) Trimming (d) Fettling

IES 2011Which of the following processes belong to forgingWhich of the following processes belong to forgingoperation ?

F ll i1. Fullering2. Swaging3. Welding

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

S 2008IES – 2008The balls of the ball bearings are manufacturedThe balls of the ball bearings are manufacturedfrom steel rods. The operations involved are:1 Ground1. Ground2. Hot forged on hammers3. Heat treated4. Polished4What is the correct sequence of the aboveoperations from start?operations from start?(a) 3‐2‐4‐1 (b) 3‐2‐1‐4( ) (d)(c) 2‐3‐1‐4 (d) 2‐3‐4‐1


S 200IES – 2001In the forging operation, fullering is done to In the forging operation, fullering is done to (a) Draw out the material (b) B d th t i l(b) Bend the material(c) Upset the material(d) Extruding the material

IES 2011Consider the following statements :Consider the following statements :1. Any metal will require some time to undergo completel ti d f ti ti l l if d f i t l hplastic deformation particularly if deforming metal has

to fill cavities and corners of small radii.2. For larger work piece of metals that can retaintoughness at forging temperature it is preferable to useforge 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(c) 1 and 2 are correct but unrelated(d) 1 only correct

S 200IES – 2005Match List I (Type of Forging) with List II (Operation)( yp g g) ( p )and select the correct answer using the code givenbelow the Lists:

List I List IIA. Drop Forging 1. Metal is gripped in the dies and

i li d h h d dpressure is applied on the heated endB. Press Forging 2. Squeezing actionC U F i M l i l d b ll dC. Upset Forging 3. Metal is placed between rollers and

pushedD Roll Forging 4 Repeated hammer blo sD. Roll Forging 4. Repeated hammer blows

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

S 2008IES – 2008Match List‐I with List‐II and select the correct answer using the codei b l th li tgiven below the lists:

List‐I (Forging Technique) List‐II (Process)A. Smith Forging 1. Material is only upset to get the desired shapeB. Drop Forging 2. Carried out manually open diesC. Press Forging 3. Done in closed impression dies by hammers in

blowsD. Machine Forging 4. Done in closed impression dies by continuous

squeezing forceCode: A B C DCode: C

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

S 998IES – 1998Which one of the following processes is mostWhich one of the following processes is mostcommonly used for the forging of bolt heads ofhexagonal shape?hexagonal shape?(a) Closed die drop forging(b) O di t f i(b) Open die upset forging(c) Close die press forging(d) Open die progressive forging

S 99 S O 20 0IES – 1994, ISRO‐2010In drop forging, forging is done by droppingIn drop forging, forging is done by dropping(a) The work piece at high velocity(b) Th h t hi h l it(b) The hammer at high velocity.(c) The die with hammer at high velocity(d) a weight on hammer to produce the requisiteimpact.p

IES‐2013S (I) I hi h l i f i hi hStatement (I): In high velocity forming process, highenergy can be transferred to metal with relatively small

i htweight.Statement (II): The kinetic energy is the function ofmass and velocity.

(a) Both Statement (I) and Statement (II) are individually( ) ( ) ( ) ytrue and Statement (II) is the correct explanation ofStatement (I)( )

(b) Both Statement (I) and Statement (II) are individuallytrue but Statement (II) is not the correct explanation oftrue but Statement (II) is not the correct explanation ofStatement (I)

(c) Statement (I) is true but Statement (II) is false(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true

IES‐2013Statement (I): In power forging energy is provided bycompressed air or oil pressure or gravity.compressed air or oil pressure or gravity.Statement (II): The capacity of the hammer is given bythe total weight which the falling pans weighthe total weight, which the falling pans weigh.

(a) Both Statement (I) and Statement (II) are individuallyt d St t t (II) i th t l ti ftrue and Statement (II) is the correct explanation ofStatement (I)

(b) h ( ) d ( ) d d ll(b) Both Statement (I) and Statement (II) are individuallytrue 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

S 2009IES – 2009Match List‐I with List‐II and select the correct answer usingh d b l hthe code given below the Lists:

List‐I List‐II(Article) (Processing Method)(Article) (Processing Method)

A. Disposable coffee cups 1. RotomouldingB. Large water tanks 2. Expandable bead mouldingg p gC. Plastic sheets 3. ThermoformingD. Cushion pads 4. Blow moulding

C l d i5. CalendaringCode:(a) A B C D (b) A B C D(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 2For-2014 (IES, GATE & PSUs) Page 34 of 78

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S 2003IAS – 2003Match List I (Forging Operation) with List II (View of the( g g p ) (Forging Operation) and select the correct answer using thecodes given below the lists:

List‐I List‐II(Forging Operation) (View of the Forging Operation)(A) Edging(B) Fullering

1. 2.

(C) Drawing(D) Swaging

3. 4.g g

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

Click to see file Page 4 – 5 ‐6

S 200IAS – 2001Match List I (Forging operations) with List II (Descriptions)

d l h h d b land select the correct answer using the codes given belowthe Lists:

List I List IIList I List IIA. Flattening 1. Thickness is reduced continuously at

different sections along lengthB D i M l i di l d fB. Drawing 2. Metal is displaced away from centre,

reducing thickness in middle andincreasing length

C. Fullering 3. Rod is pulled through a dieD. Wire drawing4. Pressure a workpiece between two flat

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

S 2000IAS – 2000Drop forging is used to produceDrop forging is used to produce(a) Small components(b) L t(b) Large components(c) Identical Components in large numbers(d) Medium‐size components

S 998IAS – 1998The forging defect due to hindrance to smooth flowThe forging defect due to hindrance to smooth flowof metal in the component called 'Lap' occursbecausebecause(a) The corner radius provided is too large(b) Th di id d i t ll(b) The corner radius provided is too small(c) Draft is not provided(d) The shrinkage allowance is inadequate

S 2002IAS – 2002Consider the following statements related to Consider the following statements related to forging:1 Flash is excess material added to stock which flows 1. Flash is excess material added to stock which flows around parting line.

Fl h h l i filli f thi ib d b i 2. Flash helps in filling of thin ribs and bosses in upper 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(c) 1 and 3 (d) 2 and 3

IES 2011Assertion (A) : Hot tears occur during forgingAssertion (A) : Hot tears occur during forgingbecause of inclusions in the blank materialReason (R) : Bonding between the inclusionsReason (R) : Bonding between the inclusionsand the parent material is through physicaland chemical bonding.g(a) Both A and R are individually true and R is thecorrect explanation of Ap(b) Both A and R are individually true but R is NOTthe correct explanation of Ap(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

IES‐2013Consider the following statements pertaining to theConsider the following statements pertaining to theopen‐die forging of a cylindrical specimen betweentwo flat dies:two flat dies:1. Lubricated specimens show more surface movementh l b i dthan un‐lubricated ones.2. Lubricated specimens show less surface movementthan un‐lubricated ones.3. Lubricated specimens show more barrelling than un‐3 p glubricated ones.4. Lubricated specimens shows less barrelling than un‐4. Lubricated specimens shows less barrelling than unlubricated ones.Which of these statements are correct?Which of these statements are correct?(a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4

GATE ‐2008 (PI)( )Match the following

Group ‐1 Group‐2Group ‐1 Group‐2P . Wrinkling 1. UpsettingQ. Centre burst 2. Deep drawingR. Barrelling 3. Extrusiong 3S. Cold shut 4. Closed die forging

(a) P – 2, Q – 3, R – 4, S‐1 (b) P – 3, Q – 4, R – 1, S‐2 (c) P 2 Q 3 R 1 S 4 (d) P 2 Q 4 R 3 S 1 (c) P – 2, Q – 3, R – 1, S‐4 (d) P – 2, Q – 4, R – 3, S‐1

IES 2012IES ‐ 2012Assumptions adopted in the analysis of open die forgingAssumptions adopted in the analysis of open die forgingare1 Forging force attains maximum value at the middle of1. Forging force attains maximum value at the middle ofthe operation.

C ffi i t f f i ti i t t b t k i2. Coefficient of friction is constant between work pieceand die2. Stress in the vertical (Y‐direction) is zero.(a) 1 and 2 (b) 1 and 3( ) ( ) 3(c) 2 and 3 (d) 1, 2 and 3


S 200 C i lIES – 2005 ConventionalA t i f l d ith i iti l di iA strip of lead with initial dimensions 24 mm x 24

mm x 150 mm is forged between two flat dies to a5 g

final size of 6 mm x 96 mm x 150 mm. If the

coefficient of friction is 0.25, determine the

i f i f Th i ld t fmaximum forging force. The average yield stress of

lead in tension is 7 N/mm27

[10]

IES 2007 C ti lIES – 2007 ConventionalA cylinder of height 60 mm and diameter 100 mm isA cylinder of height 60 mm and diameter 100 mm isforged at room temperature between two flat dies. Findthe die load at the end of compression to a height 30the die load at the end of compression to a height 30mm, using slab method of analysis. The yield strength ofthe work material is given as 120 N/mm2 and thethe work material is given as 120 N/mm and thecoefficient of friction is 0.05. Assume that volume isconstant after deformation. There is no sticking. Alsoconstant after deformation. There is no sticking. Alsofind mean die pressure. [20‐Marks]

IES 2006 C ti lIES – 2006 ‐ ConventionalA certain disc of lead of radius 150 mm and thickness 50A certain disc of lead of radius 150 mm and thickness 50mm is reduced to a thickness of 25 mm by open dieforging. If the co‐efficient of friction between the job andforging. If the co efficient of friction between the job anddie is 0.25, determine the maximum forging force. Theaverage shear yield stress of lead can be taken as 4average shear yield stress of lead can be taken as 4N/mm2. [10 – Marks]

GATE‐1987In forging operation the sticking friction condition

occurs near the……(Centre/ends)

Ch‐15: ForgingQ. No Option Q. No OptionQ. No Option Q. No Option1 A 6 B2 A 7 C2 A 7 C3 A 8 C4 A 9 C4 A 9 C5 B

P ti P bl 1Practice Problem‐1A strip of metal with initial dimensions 24 mm x 24 mmA 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/mm2average yield strength in tension is 7 N/mm

[Ans. 178.24 kN]

P ti P bl 2Practice Problem‐2A circular disc of 200 mm in diameter and 100 mm inA circular disc of 200 mm in diameter and 100 mm in

height is compressed between two flat dies to a height of

50 mm. Coefficient of friction is 0.1 and average yield

strength in compression is 230 MPa. Determine the

maximum die pressuremaximum die pressure.

[Ans. 405 MPa]

P ti P bl 3Practice Problem‐3A cylindrical specimen 150 mm in diameter and 100 mmA cylindrical specimen 150 mm in diameter and 100 mm

in height is upsetted by open die forging to a height of 50

mm. Coefficient of friction is 0.2 and flow curve

equation is MPa Calculate the maximum17.01030equation is MPa . Calculate the maximum

forging force.

17.01030εσ =f

[Ans. 46.26 MN]

[Hint. First calculate true strain ε and put the value in

the equation ]σεσ == 17.01030the equation ]yf σεσ ==1030

P ti P bl 4Practice Problem‐4A circular disc of 200 mm in diameter and 70 mm inA circular disc of 200 mm in diameter and 70 mm in

height is forged to 40 mm in height. Coefficient of

friction is 0.05. The flow curve equation of the material

is given by Determine maximumMP)010(200 41.0+is given by . Determine maximum

forging load, mean die pressure and maximum pressure.

MPa)01.0(200 41.0εσ +=f

[ Ans. 9.771 MN, 178 MPa, 221 MPa]

[Hint. First calculate true strain ε and put the value in

the equation ]410)010(200the equation ]y41.0)01.0(200 σεσ =+=f


S K Mondal

New Stamp

Practice Problem ‐5 {GATE‐2010 (PI)}Practice Problem 5 {GATE 2010 (PI)}During open die forging process using two flat and parallel dies,

lid i l t l di f i iti l di (R ) 200 d i iti la solid circular steel disc of initial radius (R ) 200 mm and initial height (H )50 mm attains a height (H )of 30 mm and radius of R .

IN

IN FN FN

Along the die-disc interfaces.INR

R−⎛ ⎞

⎜ ⎟ i. the coefficient of friction ( ) is: = 0.35 1

ii i h i R lidi f i i il d

FNRe

R

μ μ +⎜ ⎟⎜ ⎟⎝ ⎠

≤ ≤

( )

,

2

ii. in the region R sliding friction prevails, and

3FN

FN

ss FN

R rH

r R

K dμ

−

≤ ≤

3 ,

FNHp Ke and pτ μ= = where p and are the normal and shear stresses, respectively; τ

K is the shear yield strength of steel and r is the radial distance of any point ( ........)contd

Contd…….

Practice Problem ‐5 {GATE‐2010 (PI)}

iii.In the region 0 r R ,sticking condition prevailsSS≤ ≤

SSThe value of R (in mm), where sticking condition changes to slidingfriction, is(a) 241.76 (b) 254.55 (c) 265.45 (d) 278.20

Extrusion & DrawingExtrusion & Drawing


IAS‐2010 MainHow are metal tooth‐paste tubes made

commercially ? Draw the tools configuration withcommercially ? Draw the tools configuration with

the help of a neat sketch.

[30‐Marks]

IES 2009 Conventional GATE‐1994(PI)( )Amoving mandrel is used in

(a) Wire drawing (b) Tube drawing

(c) Metal Cutting (d) Forging

IES – 2011 ConventionalA 12 5 mm diameter rod is to be reduced to 10 mmA 12.5 mm diameter rod is to be reduced to 10 mmdiameter by drawing in a single pass at a speed of 100m/min. Assuming a semi die angle of 5o and coefficient/ g g 5of 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 50N/mm2 , what would be the draw stress and maximum/ ,possible reduction ?Take stress of the work material as 400 N/mm2 .4 /

[15 Marks]

GATE 2011 (PI) Common Data S1GATE – 2011 (PI) Common Data‐S1In a multi‐pass drawing operation, a round bar of 10 mmdiameter and 100 mm length is reduced in cross‐sectionby drawing it successively through a series of seven diesof decreasing exit diameter. During each of thesedrawing operations, the reduction in cross‐sectional areais 35%. The yield strength of the material is 200 MPa.Ignore strain hardening.The total true strain applied and the final length (inmm), respectively, are), p y,(a) 2.45 and 8 17 (b) 2.45 and 345(c) 3 02 and 2043 (d) 3 02 and 3330(c) 3.02 and 2043 (d) 3.02 and 3330

GATE 2011 (PI) Common Data S2GATE – 2011 (PI) Common Data‐S2In a multi‐pass drawing operation, a round bar of 10 mmdiameter and 100 mm length is reduced in cross‐sectionby drawing it successively through a series of seven diesof decreasing exit diameter. During each of thesedrawing operations, the reduction in cross‐sectional areais 35%. The yield strength of the material is 200 MPa.Ignore strain hardening.Neglecting friction and redundant work, the force (in kN) required for drawing the bar through the first die, is) q g g ,(a) 15.71 (b) 10.21 (c) 6 77 (d) 4 39(c) 6.77 (d) 4.39


E lExampleCalculate the drawing load required to obtain 30%Calculate the drawing load required to obtain 30%reduction in area on a 12 mm diameter copper wire.The following data is givenThe following data is given

C l l t th f th l t i t if thCalculate the power of the electric motor if thedrawing speed is 2.3 m/s. Take efficiency of motor is8%98%.

20 0JWM 2010Assertion (A) : Extrusion speed depends on work( ) p pmaterial.Reason (R) : High extrusion speed causes cracks inReason (R) : High extrusion speed causes cracks inthematerial.(a) Both A and R are individually true and R is the(a) Both A and R are individually true and R is thecorrect explanation of A(b) B h A d R i di id ll b R i h(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( )

G 2006GATE‐2006In a wire drawing operation, diameter of a steel wireIn a wire drawing operation, diameter of a steel wireis reduced from 10 mm to 8 mm. The mean flowstress of the material is 400 MPa. The ideal forcestress of the material is 400 MPa. The ideal forcerequired for drawing (ignoring friction andredundantwork) isredundantwork) is(a) 4.48 kN (b) 8.97 kN( ) kN (d) kN(c) 20.11 kN (d) 31.41 kN

G 200 G 200 ( )GATE‐2001, GATE ‐2007 (PI)For rigid perfectly‐plastic work material, negligibleFor rigid perfectly plastic work material, negligibleinterface friction and no redundant work, thetheoretically maximum possible reduction in thetheoretically maximum possible reduction in thewire drawing operation is(a) 0 36 (b) 0 63(a) 0.36 (b) 0.63(c) 1.00 (d) 2.72

GATE ‐2008 (PI) Linked S‐1GATE 2008 (PI) Linked S 1A 10 mm diameter annealed steel wire is drawn through

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.

N l ti f i ti d t i h d i th tNeglecting friction and strain hardening, the stress

required for drawing (in MPa) isq g ( )

(a) 178.5 (b) 357.0 (c) 1287.5 (d) 2575.0

GATE ‐2008 (PI) Linked S‐2GATE 2008 (PI) Linked S 2A 10 mm diameter annealed steel wire is drawn through

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.

Th i d f th d i (i kW) iThe power required for the drawing process (in kW) is

(a) 8 97 (b) 14 0 (c) 17 95 (d) 28 0(a) 8.97 (b) 14.0 (c) 17.95 (d) 28.0

G 2003GATE‐2003A brass billet is to be extruded from its initialA brass billet is to be extruded from its initialdiameter of 100 mm to a final diameter of 50 mm.The working temperature of 700°C and theThe working temperature of 700 C and theextrusion constant is 250 MPa. The force requiredfor extrusion isfor extrusion is(a) 5.44 MN (b) 2.72 MN( ) 6 MN (d) 6 MN(c) 1.36 MN (d) 0.36 MN

GATE – 2009 (PI)Using direct e trusion process a round billet of 100 mmUsing direct extrusion process, a round billet of 100 mm

length and 50 mm diameter is extruded. Considering an

ideal deformation process (no friction and no redundant

k) i i d fl fwork), extrusion ratio 4, and average flow stress of

material 300 MPa, the pressure (in MPa) on the ram will3 , p ( )

be

(a) 416 (b) 624 (c) 700 (d) 832

G 996GATE‐1996A wire of 0.1 mm diameter is drawn from a rod of 15A wire of 0.1 mm diameter is drawn from a rod of 15mm diameter. Dies giving reductions of 20%, 40%and 80% are available. For minimum error in theand 80% are available. For minimum error in thefinal size, the number of stages and reduction ateach stage respectively would beeach stage respectively would be(a) 3 stages and 80% reduction for all three stages(b) t d 8 % d ti f fi t th t(b) 4 stages and 80% reduction for first three stagesfollowed 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


G 99GATE‐1994The process of hot extrusion is used to produceThe process of hot extrusion is used to produce(a) Curtain rods made of aluminium(b) St l i / d ti t l(b) Steel pipes/or domestic water supply(c) Stainless steel tubes used in furniture(d) Large she pipes used in city water mains

S 200IES – 2007Which one of the following is the correctgstatement?(a) Extrusion is used for the manufacture of seamless( )tubes.(b) Extrusion is used for reducing the diameter of round( ) gbars and tubes by rotating dies which open and closerapidly on the work?(c) Extrusion is used to improve fatigue resistance of themetal by setting up compressive stresses on its surface(d) Extrusion comprises pressing the metal inside achamber to force it out by high pressure through anorifice which is shaped to provide the desired from of thefinished part.

S 200IES – 2007Assertion (A): Greater force on the plunger is requiredAssertion (A): Greater force on the plunger is requiredin case of direct extrusion than indirect one.Reason (R): In case of direct extrusion, the direction ofReason (R): In case of direct extrusion, the direction ofthe force applied on the plunger and the direction ofthemovement of the extruded metal are the same.(a) Both A and R are individually true and R is the correctexplanation of Ap(b) Both A and R are individually true but R is not thecorrect explanation of Ap(c) A is true but R is false(d) A is false but R is true(d) A is false but R is true

IES 2012IES ‐ 2012Which of the following are correct for an indirect hotWhich of the following are correct for an indirect hotextrusion process?1 Billet remains stationary1. Billet remains stationary2. There is no friction force between billet and container

llwalls.3. The force required on the punch is more incomparison to direct extrusion.4. Extrusion parts have to be provided a support.4 p p pp(a) 1, 2, 3 and 4 (b) 1, 2 and 3 only(c) 1 2 and 4 only (d) 2 3 and 4 only(c) 1, 2 and 4 only (d) 2, 3 and 4 only

S 993IES – 1993Assertion (A): Direct extrusion requires larger forceAssertion (A): Direct extrusion requires larger forcethan indirect extrusion.Reason (R): In indirect extrusion of cold steel zincReason (R): In indirect extrusion of cold steel, zincphosphate coating is used.( ) 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 99IES – 1994Metal extrusion process is generally used forMetal extrusion process is generally used forproducing(a) Uniform solid sections(a) Uniform solid sections(b) Uniform hollow sections(c) Uniform solid and hollow sections(d) Varying solid and hollow sections.( ) y g

S 2009IES – 2009Which one of the following statements is correct?Which one of the following statements is correct?(a) In extrusion process, thicker walls can be obtainedby increasing the forming pressureby increasing the forming pressure(b) Extrusion is an ideal process for obtaining rods from

t l h i d itmetal having poor density(c) As compared to roll forming, extruding speed is high(d) Impact extrusion is quite similar to Hooker's processincluding the flow of metal being in the same directiong g

S 999IES – 1999Which one of the following is the correctWhich one of the following is the correcttemperature range for hot extrusion of aluminium?(a) 300 340°C (b) 350 400°C(a) 300‐340 C (b) 350‐400 C(c) 430‐480°C (d) 550‐650°C

S 2000IES – 2000Consider the following statements:gIn forward extrusion process

1. The ram and the extruded product travel in the samepdirection.

2. The ram and the extruded product travel in the oppositep ppdirection.

3. The speed of travel of the extruded product is same as thatf hof the ram.

4. The speed of travel of the extruded product is greater thanthat of the ramthat of the ram.

Which of these Statements are correct?( ) d (b) d(a) 1 and 3 (b) 2 and 3(c) 1 and 4 (d) 2 and 4For-2014 (IES, GATE & PSUs) Page 39 of 78

S 2009IES – 2009What is themajor problem in hot extrusion?What is themajor problem in hot extrusion?(a) Design of punch (b) Design of die( ) W d t f di (d) W f h(c) Wear and tear of die (d) Wear of punch

IES 2012IES ‐ 2012Extrusion process can effectively reduce the cost ofExtrusion process can effectively reduce the cost ofproduct through(a) Material saving(a) Material saving(b) process time saving(c) Saving in tooling cost(d) saving in administrative cost( ) g

S 2008 G 989( )IES – 2008, GATE‐1989(PI)Whi h f th f ll i th d i d f thWhich one of the following methods is used for the

manufacture of collapsible tooth‐paste tubes?p p

(a) Impact extrusion (b) Direct extrusion

(c) Deep drawing (d) Piercing

S 2003IES – 2003The extrusion process (s) used for the production ofThe extrusion process (s) used for the production oftoothpaste tube is/are1 Tube extrusion1. Tube extrusion2. Forward extrusion3. Impact extrusionSelect the correct answer using the codes given below:g gCodes:(a) 1 only (b) 1 and 2(a) 1 only (b) 1 and 2(c) 2 and 3 (d) 3 only

S 200IES – 2001Which of the following statements are the salientgfeatures of hydrostatic extrusion?

1. It is suitable for soft and ductile material.2. It is suitable for high‐strength super‐alloys.3.The billet is inserted into the extrusion chamber and pressurepis applied by a ram to extrude the billet through the die.

4. The billet is inserted into the extrusion chamber where it isd d b i bl li id Th bill i d dsurrounded by a suitable liquid. The billet is extruded

through the die by applying pressure to the liquid.Select the correct ans er using the codes gi en beloSelect the correct answer using the codes given below:Codes:( ) d (b) d(a) 1 and 3 (b) 1 and 4(c) 2 and 3 (d) 2 and 4

S 2006IES – 2006What does hydrostatic pressure in extrusion processWhat does hydrostatic pressure in extrusion processimprove?(a) Ductility (b) Compressive strength(a) Ductility (b) Compressive strength(c) Brittleness (d) Tensile strength

GATE‐1990(PI)( )Semi brittle materials can be extruded by

(a) Impact extrusion

(b) Closed cavity extrusion

(c) Hydrostatic extrusion

(d) B k d i(d) Backward extrusion

IES 2010IES 2010Assertion (A): Pickling and washing of rolled rods( ) g gis carried out beforewire drawing.Reason (R): They lubricate the surface to reduceReason (R): They lubricate the surface to reducefrictionwhile drawing wires.(a) Both A and R are individually true and R is the(a) Both A and R are individually true and R is thecorrect explanation of A(b) B h A d R i di id ll b R i NOT h(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( )

S 2009IES – 2009Which one of the following stress is involved in theWhich one of the following stress is involved in thewire drawing process?(a) Compressive (b) Tensile(a) Compressive (b) Tensile(c) Shear (d) Hydrostatic stress


S 993IES – 1993Tandem drawing of wires and tubes is necessaryTandem drawing of wires and tubes is necessarybecause(a) It is not possible to reduce at one stage(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 2000IES – 2000Match List I (Components of a table fan) with List II(C p )(Manufacturing processes) and select the correctanswer using the codes given below the Lists:

List I List IIA. Base with stand 1. Stamping andp g

pressingB. Blade 2. Wire drawinggC. Armature coil wire 3. TurningD Armature shaft 4 CastingD. Armature shaft 4. CastingCodes:A B C D A B C D(a) 4 3 2 1 (b) 2 1 4 3(a) 4 3 2 1 (b) 2 1 4 3(c) 2 3 4 1 (d) 4 1 2 3

S 999IES – 1999Match List‐I with List‐II and select the correctanswer using the codes given below the Lists:

List‐I List‐IIA. Drawing 1. Soap solutionB Rolling 2 CamberB. Rolling 2. CamberC. Wire drawing 3. PilotsD Sheet metal operations using 4 CraterD. Sheet metal operations using 4. Crater

progressive dies 5. IroningC d A B C D A B C DCode: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 996IES – 1996Match List I with List II and select the correct answerMatch List I with List II and select the correct answerList I (Metal/forming process) List II (Associated force)

A Wire drawing 1 Shear forceA. Wire drawing 1. Shear forceB. Extrusion 2. Tensile forceC. Blanking 3. Compressive forceD. Bending 4. Spring back forceg 4 p gCodes:A B C D A B C D(a) 4 2 1 3 (b) 2 1 3 4(a) 4 2 1 3 (b) 2 1 3 4(c) 2 3 1 4 (d) 4 3 2 1

S 996IES – 1996In wire drawing process, the bright shining surfaceIn wire drawing process, the bright shining surfaceon thewire is obtained if one(a) does not use a lubricant(a) does not use a lubricant(b) uses solid powdery lubricant.(c) uses thick paste lubricant(d) uses thin film lubricant( )

S 99IES – 1994Match List I with List II and select the correct answer 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) List I (Metal farming process) List II (A similar process)

A. Blanking 1. Wire drawingB C i i Pi iB. Coining 2. PiercingC. Extrusion 3. EmbossingD. Cup drawing 4. Rolling

5. Bending5. BendingCodes:A B C D A B C D( ) (b) (a) 2 3 4 1 (b) 2 3 1 4(c) 3 2 1 5 (d) 2 3 1 5

S 993 S O 20 0IES – 1993, ISRO‐2010Match List I with List II and select the correctanswer using the codes given below the lists:List I (Mechanical property) List II (Related to)( p p y) ( )A. Malleability 1. Wire drawingB Hardness 2 Impact loadsB. Hardness 2. Impact loadsC. Resilience 3. Cold rollingD Isotropy 4 IndentationD. Isotropy 4. Indentation

5. DirectionC d A B C D A B C DCodes: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

S 200IES – 2007Which metal forming process is used forWhich metal forming process is used formanufacture of long steel wire?(a) Deep drawing (b) Forging(a) Deep drawing (b) Forging(c) Drawing (d) Extrusion

S 200IES – 2005Which of the following types of stresses is/areWhich of the following types of stresses is/areinvolved in thewire‐drawing operation?(a) Tensile only(a) Tensile only(b) Compressive only(c) A combination of tensile and compressive stresses(d) A combination of tensile, compressive and shear( ) pstresses


GATE‐1987For wire drawing operation, the work material

should essentially be

(a) Ductile (b) Tough

( ) ( )(c) Hard (d) Malleable

S 2000IES – 2000Which one of the following lubricants is mostWhich one of the following lubricants is mostsuitable for drawing mild steel wires?(a) Sodium stearate (b) Water(a) Sodium stearate (b) Water(c) Lime‐water (d) Kerosene

S 993IES – 1993Amoving mandrel is used inAmoving mandrel is used in(a) Wire drawing(b) T b d i(b) Tube drawing(c) Metal cutting(d) Forging

S 2002IES – 2002Match List I with List II and select the correctMatch List I with List II and select the correctanswer:List I (Parts) List II (Manufacturing processes)List I (Parts) List II (Manufacturing processes)

A. Seamless tubes 1. Roll formingB A d h b Sh iB. Accurate and smooth tubes 2. Shot peeningC. Surfaces having higher 3. Forging

hardness and fatigue strength4. Cold formingCodes: A B C A B CCodes: A B C A B C

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

S 200IAS – 2004Assertion (A): Indirect extrusion operation can beAssertion (A): Indirect extrusion operation can beperformed either by moving ram or by moving thecontainer.container.Reason (R): Advantage in indirect extrusion is lessquantity of scrap compared to direct extrusionquantity of scrap compared to direct extrusion.(a) Both A and R are individually true and R is the


S 99IAS – 1995The following operations are performed whileThe following operations are performed whilepreparing the billets for extrusion process:1 Alkaline cleaning1. Alkaline cleaning2. Phosphate coating3. Pickling4. Lubricating with reactive soap.4 g pThe correct sequence of these operations is(a) 3 1 4 2 (b) 1 3 2 4(a) 3, 1, 4, 2 (b) 1, 3, 2, 4(c) 1, 3. 4, 2 (d) 3, 1, 2, 4

S 200IAS – 2001Match List I (Products) with List II (SuitableMatch List I (Products) with List II (Suitableprocesses) and select the correct answer using thecodes given below the Lists:codes given below the Lists:

List I List IIA C ti d W ldiA. Connecting rods 1. WeldingB. Pressure vessels 2. ExtrusionC. Machine tool beds 3. ForgingD. Collapsible tubes 4. CastingD. Collapsible tubes 4. Casting

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

S 99IAS – 1997Extrusion force DOES NOT depend upon theExtrusion force DOES NOT depend upon the(a) Extrusion ratio(b) T f t i(b) Type of extrusion process(c) Material of the die(d) Working temperature

S 2000IAS – 2000Assertion (A): Brittle materials such as grey castAssertion (A): Brittle materials such as grey castiron cannot be extruded by hydrostatic extrusion.Reason(R): In hydrostatic extrusion billet isReason(R): In hydrostatic extrusion, billet isuniformly compressed from all sides by the liquid.( ) 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 2002IAS – 2002Assertion (A): In wire‐drawing process, the rod( ) g p ,cross‐section is reduced gradually by drawing itseveral times in successively reduced diameter dies.Reason (R): Since each drawing reduces ductility ofthe wire, so after final drawing the wire isnormalized.(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

IES 2011IES 2011Match List –I with List –II and select the correct answer using the code given below the lists :the code given below the lists :

List –I List –II

A. Connecting rods 1. Welding

B. Pressure vessels 2. Extrusion

C Machine tool beds 3 Forming

C d

C. Machine tool beds 3. Forming

D. Collapsible tubes 4. CastingCodes

A B C D A B C D( ) (b)

p g

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

IAS 1994IAS 1994Which of the following methods can be used forgmanufacturing 2 metre long seamless metallictubes?1. Drawing 2. Extrusion3 Rolling 4 Spinning3. Rolling 4. SpinningSelect the correct answer using the codes given belowCodes:(a) 1 and 3 (b) 2 and 3( ) 3 ( ) 3(c) 1, 3 and 4 (d) 2, 3 and 4

GATE‐1991(PI)( )Seamless long steel tubes are manufactured by

rolling, drawing and

Ch‐17: ExtrusionQ. No Option Q. No OptionQ. No Option Q. No Option1 D 8 B2 C 9 B2 C 9 B3 D 10 A4 D 11 B4 D 11 B5 B 12 B6 C 13 A6 C 13 A7 B

Ch‐16: Drawing

Q. No Option1 A1 A2 C3 C4 B5 C6 D

Sheet Metal Operationp

By S K MondalBy S K Mondal

E lExampleDetermine the die and punch sizes for blanking a circularDetermine the die and punch sizes for blanking a circulardisc of 20‐mm diameter from a sheet whose thickness is 1.5mm.

Shear strength of sheet material = 294 MPaShear strength of sheet material = 294 MPa

Also determine the die and punch sizes for punching aAlso determine the die and punch sizes for punching acircular hole of 20‐mm diameter from a sheet whosethickness is 1 5 mmthickness is 1.5 mm.

E lExampleE ti t th bl ki f t t bl k idEstimate the blanking force to cut a blank 25 mm wide

and 30 mm long from a 1.5 mm thick metal strip, if the3 g 5 p,

ultimate shear strength of the material is 450 N/mm2.

Also determine the work done if the percentage

t ti i t f t i l thi kpenetration is 25 percent of material thickness.


S K Mondal

New Stamp

S K Mondal

New Stamp

IAS‐2011 MainFor punching a 10 mm circular hole, and cutting a

t l bl k f f h t frectangular blank of 50 x 200 mm from a sheet of 1mm thickness (mild steel, shear stress = 240N/ 2) C l l t i hN/mm2), Calculate, in each case :(i) Size of punch(ii) Size of die(iii) Force required. [10‐Marks]( ) q [ ]

S 999IES – 1999A hole is to be punched in a 15 mm thick plateA hole is to be punched in a 15 mm thick platehaving ultimate shear strength of 3N‐mm‐2. If theallowable crushing stress in the punch is 6 N‐mm‐2,allowable crushing stress in the punch is 6 N mm ,the diameter of the smallest hole which can bepunched is equal topunched is equal to(a) 15 mm (b) 30 mm( ) 6 (d)(c) 60 mm (d) 120 mm

S O 2008 20ISRO‐2008, 2011With a punch for which the maximum crushingWith a punch for which the maximum crushingstress is 4 times the maximum shearing stress ofth l t th bi t h l th t b h d ithe plate, the biggest hole that can be punched inthe plate would be of diameter equal to

1(a) Thickness of plate4×( ) p

41(b) Thickness of plate×(b) Thickness of plate2

(c) Plate thickness

×

(c) Plate thickness(d) 2 Plate thickness×

IES 2013IES‐2013A hole of diameter d is to be punched in a plate ofA 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:of diameter of hole to plate thickness should be equal to:

(a) (b)14

1( ) ( )

(c) 1 (d) 2

4 2

ExampleA hole, 100 mm diameter, is to be punched in steel plate

6 thi k Th lti t h t i N/ 25.6 mm thick. The ultimate shear stress is 550 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

it f Tcapacity of a 30T press.

E lExampleA washer with a 12.7 mm internal hole and an outsideA washer with a 12.7 mm internal hole and an outsidediameter of 25.4 mm is to be made from 1.5 mm thickstrip. The ultimate shearing strength of the material ofp g gthewasher is 280 N/mm2.(a) Find the total cutting force if both punches act at( ) g pthe same time and no shear is applied to either punchor the die.(b) What will be the cutting force if the punches arestaggered, so that only one punch acts at a time.(c) Taking 60% penetration and shear on punch of 1mm, what will be the cutting force if both punches actg ptogether.

G 20 0 S i k dGATE‐2010 Statement Linked 1Statement for Linked Answer Questions:Statement for Linked Answer Questions:In a shear cutting operation, a sheet of 5 mm thicknessis cut along a length of 200 mm. The cutting blade is 400is cut along a length of 200 mm. The cutting blade is 400mm long and zero‐shear (S = 0) is provided on the edge.The ultimate shear strength of the sheet is 100 MPa andgpenetration to thickness ratio is 0.2. Neglect friction.

400

S

Assuming force vs displacement curve to be rectangular,thework done (in J) isthework done (in J) is(a) 100 (b) 200 (c) 250 (d) 300

G 20 0 S i k d 2GATE‐2010 Statement Linked 2Statement for Linked Answer Questions:QIn a shear cutting operation, a sheet of 5mm thicknessis cut along a length of 200 mm. The cutting blade is 400mm long and zero shear (S = 0) is provided on the edgemm long and zero‐shear (S = 0) is provided on the edge.The ultimate shear strength of the sheet is 100 MPa andpenetration to thickness ratio is 0.2. Neglect friction.

400

S

A shear of 20 mm (S = 20 mm) is now provided on theblade. Assuming force vs displacement curve to be

S

blade. Assuming force vs displacement curve to betrapezoidal, the maximum force (in kN) exerted is(a) 5 (b) 10 (c) 20 (d) 40

GATE 2011The shear strength of a sheet metal is 300 MPa TheThe shear strength of a sheet metal is 300 MPa. Theblanking force required to produce a blank of 100mm diameter from a 1 5 mm thick sheet is close tomm diameter from a 1.5 mm thick sheet is close to(a) 45 kN(b) 70 kN(c) 141 kN4(d) 3500 kN


GATE – 2009 (PI)( )A disk of 200 mm diameter is blanked from a strip

of an aluminum alloy of thickness 3.2 mm. The

material shear strength to fracture is 150 MPa. The

blanking force (in kN) isblanking force (in kN) is

(a) 291 (b) 301 (c) 311 (d) 321( ) 9 ( ) 3 ( ) 3 ( ) 3

GATE 2013 (PI)GATE‐2013 (PI)Circular blanks of 10 mm diameter are punchedCircular blanks of 10 mm diameter are punched

from an aluminium sheet of 2 mm thickness. The

shear strength of aluminium is 80 Mpa. The

i i hi f i d i kN iminimum punching force required in kN is

(a) 2 57(a) 2.57

(b) 3.29

(c) 5.03

(d) 6.33

S O 2009ISRO‐2009The force required to punch a 25 mm hole in a

ild t l l t thi k h lti t hmild steel plate 10 mm thick, when ultimate shear

stress of the plate is 500 N/mm2 will be nearlystress of the plate is 500 N/mm will be nearly

(a) 78 kN (b) 393 kN (c) 98 kN (d) 158 kN(a) 78 kN (b) 393 kN (c) 98 kN (d) 158 kN

G 200GATE‐2007The force requirement in a blanking operation ofThe force requirement in a blanking operation oflow carbon steel sheet is 5.0 kN. The thickness ofthe sheet is ‘t’ and diameter of the blanked part isthe sheet is t and diameter of the blanked part is‘d’. For the same work material, if the diameter ofthe blanked part is increased to 1.5 d and thicknessthe blanked part is increased to 1.5 d and thicknessis reduced to 0.4 t, the new blanking force in kN is(a) 3 0 (b) 4 5(a) 3.0 (b) 4.5(c) 5.0 (d) 8.0

G 200GATE‐200410 mm diameter holes are to be punched in a steel10 mm diameter holes are to be punched in a steelsheet of 3 mm thickness. Shear strength of thematerial is 400 N / mm2 and penetration is 40%.material is 400 N / mm and penetration is 40%.Shear provided on the punch is 2 mm. The blankingforce during the operationwill beforce during the operationwill be(a) 22.6 kN (b) 37.7 kN( ) 6 6 kN (d) kN(c) 61.6 kN (d) 94.3 kN

G 2003GATE‐2003A metal disc of 20 mm diameter is to be punchedA metal disc of 20 mm diameter is to be punchedfrom a sheet of 2 mm thickness. The punch and thedie clearance is 3%. The required punch diameter isdie clearance is 3%. The required punch diameter is(a) 19.88 mm (b) 19.94 mm( ) 6 (d)(c) 20.06 mm (d) 20.12 mm

GATE ‐ 2012Calculate the punch size in mm, for a circularblanking operation for which details are givenblanking operation for which details are givenbelow.Size of the blank 25 mmSize of the blank 25 mmThickness of the sheet 2 mmRadial clearance bet een punch and die 0 06 mmRadial clearance between punch and die 0.06 mmDie allowance 0.05 mm

( ) (b)(a) 24.83 (b) 24.89(c) 25.01 (d) 25.17

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% ofclearance 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 (b) 50 00 and 50 15(a) 50.00 and 50.30 (b) 50.00 and 50.15

(c) 49.70 and 50.00 (d) 49.85 and 50.00

G 2002GATE‐2002In a blanking operation, the clearance is providedIn a blanking operation, the clearance is providedon(a) The die(a) The die(b) Both the die and the punch equally(c) The punch(d) Brittle the punch nor the die( ) p


G 200GATE‐2001The cutting force in punching and blankingThe cutting force in punching and blankingoperationsmainly depends on(a) The modulus of elasticity of metal(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

G 996GATE‐1996A 50 mm diameter disc is to be punched out from aA 50 mm diameter disc is to be punched out from acarbon steel sheet 1.0 mm thick. The diameter ofthe punch should bethe punch should be(a) 49.925 mm (b) 50.00 mm( ) (d) f th b(c) 50.075 mm (d) none of the above

S 99IES – 1994In sheet metal blanking, shear is provided onIn sheet metal blanking, shear is provided onpunches and dies so that(a) Press load is reduced(a) Press load is reduced(b) Good cut edge is obtained.(c) Warping of sheet is minimized(d) Cut blanks are straight.( ) g

S 2002IES – 2002Consider the following statements related toConsider the following statements related topiercing and blanking:1 Shear on the punch reduces the maximum cutting1. Shear on the punch reduces the maximum cuttingforce

Sh i th it f th d d2. Shear increases the capacity of the press needed3. Shear increases the life of the punch4. The total energy needed to make the cut remainsunaltered due to provision of shearpWhich of these statements are correct?(a) 1 and 2 (b) 1 and 4(a) 1 and 2 (b) 1 and 4(c) 2 and 3 (d) 3 and 4

S 99IAS – 1995In blanking operation the clearance provided isIn blanking operation the clearance provided is(a) 50% on punch and 50% on die(b) O di(b) On die(c) On punch(d) On die or punch depending upon designer’s choice

S 2006IES – 2006In which one of the following is a flywheel generallyIn which one of the following is a flywheel generallyemployed?(a) Lathe (b) Electric motor(a) Lathe (b) Electric motor(c) Punching machine (d) Gearbox

S 200IES – 2004Which one of the following statements is correct?Which one of the following statements is correct?If the size of a flywheel in a punching machine isincreasedincreased(a) Then the fluctuation of speed and fluctuation of

ill b th denergy will both decrease(b) Then the fluctuation of speed will decrease and thefluctuation of energy will increase(c) Then the fluctuation of speed will increase and the( ) pfluctuation of energy will decrease(d) Then the fluctuation of speed and fluctuation of(d) Then the fluctuation of speed and fluctuation ofenergy both will increase

S 99IES – 1997For 50% penetration of work material, a punch withFor 50% penetration of work material, a punch withsingle shear equal to thickness will(a) Reduce the punch load to half the value(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

S 2000IAS – 2000A blank of 30 mm diameter is to be produced out ofA blank of 30 mm diameter is to be produced out of10 mm thick sheet on a simple die. If 6% clearance isrecommended, then the nominal diameters of dierecommended, then the nominal diameters of dieand punch are respectively(a) 30 6 mm and 29 4 mm(a) 30.6 mm and 29.4 mm(b) 30.6 mm and 30 mm(c) 30 mm and 29.4 mm(d) 30 mm and 28.8 mm( ) 3


GATE 2007 (PI)GATE – 2007 (PI)Circular blanks of 35 mm diameter are punchedCircular blanks of 35 mm diameter are punchedfrom a steel sheet of 2 mm thickness. If theclearance per side between the punch and die isclearance per side between the punch and die isto be kept as 40 microns, the sizes of punch anddi h ld i l bdie should respectively be(a) 35+0.00 and 35+0.040 (b) 35‐0.040 and 35‐0.080(a) 35 and 35 (b) 35 and 35(c) 35‐0.080 and 35+0.00 (d) 35+0.040 and 35‐0.080

S 99IAS – 1994In a blanking operation to produce steel washer, theIn a blanking operation to produce steel washer, themaximum punch load used in 2 x 105 N. The platethickness is 4 mm and percentage penetration is 25.thickness is 4 mm and percentage penetration is 25.Thework done during this shearing operation is(a) 200J (b) 400J(a) 200J (b) 400J(c) 600 J (d) 800 J

S 2002IAS – 2002In deciding the clearance between punch and die inIn deciding the clearance between punch and die inpress work in shearing, the following rule is helpful:(a) Punch size controls hole size die size controls blank(a) Punch size controls hole size die size controls blanksize(b) P h i t l b th h l i d bl k i(b) Punch size controls both hole size and blank size(c) Die size controls both hole size and blank size(d) Die size controls hole size, punch size controls blanksize

S 200IAS – 2007For punching operation the clearance is providedFor punching operation the clearance is providedonwhich one of the following?(a) The punch(a) The punch(b) The die(c) 50% on the punch and 50% on the die(d) 1/3rd on the punch and 2/3rd on the die( ) 3 p 3

S 99IAS – 1995Assertion (A): A flywheel is attached to a punchingAssertion (A): A flywheel is attached to a punchingpress so as to reduce its speed fluctuations.Reason(R): The flywheel stores energy when itsReason(R): The flywheel stores energy when itsspeed increase.(a) Both A and R are individually true and R is the(a) Both A and R are individually true and R is thecorrect explanation of A(b) B th A d R i di id ll t b t R i t th(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

S 2002IES – 2002Which one is not a method of reducing cuttingWhich one is not a method of reducing cuttingforces to prevent the overloading of press?(a) Providing shear on die(a) Providing shear on die(b) Providing shear on punch(c) Increasing die clearance(d) Stepping punches( ) pp g p

S 2003IAS – 2003Match List I (Press‐part) with List II (Function) and select the

i h d i b l h licorrect answer using the codes given below the lists:List‐I List‐II

(Press‐part) (Function)( p ) ( )(A) Punch plate 1. Assisting withdrawal of the punch(B) Stripper 2. Advancing the work‐piece through correct

di tdistance(C) Stopper 3. Ejection of the work‐piece from die cavity(D) Knockout 4. Holding the small punch in the proper( ) 4 g p p p

positionCodes:A B C D A B C D

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

S 2000IES – 2000Best position of crank for blanking operation in aBest position of crank for blanking operation in amechanical press is(a) Top dead centre(a) Top dead centre(b) 20 degrees below top dead centre(c) 20 degrees before bottom dead centre(d) Bottom dead centre( )

S 999IES – 1999Assertion (A): In sheet metal blanking operation,Assertion (A): In sheet metal blanking operation,clearancemust be given to the die.Reason (R): The blank should be of requiredReason (R): The blank should be of requireddimensions.( ) 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 2003IAS – 2003The 'spring back' effect in press working isThe spring back effect in press working is(a) Elastic recovery of the sheet metal after removal ofthe loadthe 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

S 99IES – 1997A cup of 10 cm height and 5 cm diameter is to beA cup of 10 cm height and 5 cm diameter is to bemade from a sheet metal of 2 mm thickness. Thenumber of deductions necessary will benumber of deductions necessary will be(a) One(b) T(b) Two(c) Three(d) Four

E lExampleA symmetrical cup of 80 mm diameter and 250 mmA symmetrical cup of 80 mm diameter and 250 mmheight is to be fabricated on a deep drawing die.How many drawing operations will be necessary ifHow many drawing operations will be necessary ifno intervening annealing is done.Also find the drawing forceAlso find the drawing force

IFS ‐ 2009What 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 ?increased ?

[10 – marks][ ]

G 2008GATE‐2008In the deep drawing of cups, blanks show a tendency toIn the deep drawing of cups, blanks show a tendency towrinkle up around the periphery (flange).The most likely cause and remedy of the phenomenon are,The most likely cause and remedy of the phenomenon are,respectively,(A) Buckling due to circumferential compression; Increase(A) Buckling due to circumferential compression; Increaseblank holder pressure(B) High blank holder pressure and high friction; Reduce(B) High blank holder pressure and high friction; Reduceblank holder pressure and apply lubricant(C) High temperature causing increase in circumferential(C) High temperature causing increase in circumferentiallength: Apply coolant to blank(D) Buckling due to circumferential compression; decrease(D) Buckling due to circumferential compression; decreaseblank holder pressure

G 999GATE‐1999Identify the stress ‐ state in the FLANCE portion of aIdentify the stress state in the FLANCE portion of aPARTIALLYDRAWN CYLINDRICAL CUP when deep ‐drawing without a blank holderdrawing without a blank holder(a) Tensile in all three directions(b) N t i th fl t ll b th i(b) No stress in the flange at all, because there is noblank‐holder(c) Tensile stress in one direction and compressive inthe one other direction(d) Compressive in two directions and tensile in thethird direction

G 2003GATE‐2003A shell of 100 mm diameter and 100 mm height withA shell of 100 mm diameter and 100 mm height withthe corner radius of 0.4 mm is to be produced bycup drawing. The required blank diameter iscup drawing. The required blank diameter is(a) 118 mm (b) 161 mm( ) (d)(c) 224 mm (d) 312 mm

ISRO‐2011The initial blank diameter required to form

li d i l f id di 'd‘ da cylindrical cup of outside diameter 'd‘ andtotal height 'h' having a corner radius 'r' isg gobtained using the formula

2( ) 4 0 5D d dh2( ) 4 0.5( ) 2 2

oa D d dh rb D d h r

= + −

= + +2 2

( ) 2 2

( ) 2 2o

o

b D d h rc D d h r

+ +

= + +2( ) 4 0.5od D d dh r= + −

G 2006GATE‐2006Match the items in columns I and II.Match the items in columns I and II.

Column I Column IIP Wrinkling 1 Yield point elongationP. Wrinkling 1. Yield point elongationQ. Orange peel 2. AnisotropyR S h i L i iR. Stretcher strains 3. Large grain sizeS. Earing 4. Insufficient blank holding

fforce5. Fine grain size6. Excessive blank holding force

(a) P – 6, Q – 3, R – 1, S – 2 (b) P – 4, Q – 5, R – 6, S – 13 5(c) P – 2, Q – 5, R – 3, S – 4 (d) P – 4, Q – 3, R – 1, S – 2


S 2008IES – 2008A cylindrical vessel with flat bottom can be deepA cylindrical vessel with flat bottom can be deepdrawn by(a) Shallow drawing(a) Shallow drawing(b) Single action deep drawing(c) Double action deep drawing(d) Triple action deep drawing( ) p p g

S 999IES – 1999Consider the following statements: Earring in aConsider the following statements: Earring in adrawn cup can be due to non‐uniform1 Speed of the press1. Speed of the press2. Clearance between tools3. Material properties4. Blank holding4 gWhich of these statements are correct?(a) 1 2 and 3 (b) 2 3 and 4(a) 1, 2 and 3 (b) 2, 3 and 4(c) 1, 3 and 4 (d) 1, 2 and 4

S 99IES – 1994For obtaining a cup of diameter 25 mm and height 15For obtaining a cup of diameter 25 mm and height 15mm by drawing, the size of the round blank shouldbe approximatelybe approximately(a) 42 mm (b) 44 mm( ) 6 (d) 8(c) 46 mm (d) 48 mm

S 200IAS – 2007In drawing operation, proper lubrication isIn drawing operation, proper lubrication isessential forwhich of the following reasons?1 To improve die life1. To improve die life2. To reduce drawing forces3. To reduce temperature4. To improve surface finish4 pSelect the correct answer using the code given below:(a) 1 and 2 only (b) 1 3 and 4 only(a) 1 and 2 only (b) 1, 3 and 4 only(c) 3 and 4 only (d) 1, 2, 3 and 4

S 99IAS – 1997Which one of the following factor promotes theWhich one of the following factor promotes thetendency forwrinking in the process of drawing?(a) Increase in the ratio of thickness to blank diameter(a) Increase in the ratio of thickness to blank diameterof work material(b) D i th ti thi k t bl k di t f(b) Decrease in the ratio thickness to blank diameter ofwork material(c) Decrease in the holding force on the blank(d) Use of solid lubricants( )

S 99IAS – 1994Consider the following factorsConsider the following factors1. Clearance between the punch and the die is toosmallsmall.2. The finish at the corners of the punch is poor.3. The finish at the corners of the die is poor.4. The punch and die alignment is not proper.4 p g p pThe factors responsible for the vertical lines parallel tothe axis noticed on the outside of a drawn cylindrical cupthe axis noticed on the outside of a drawn cylindrical cupwould include.(a) 2 3 and 4 (b) 1 and 2(a) 2, 3 and 4 (b) 1 and 2(c) 2 and 4 (d) 1, 3 and 4

S 998IES – 1998Assertion (A): The first draw in deep drawing operationAssertion (A): The first draw in deep drawing operationcan have up to 60% reduction, the second draw up to40% reduction and, the third draw of about 30% only.4 , 3 yReason (R): Due to strain hardening, the subsequentdraws in a deep drawing operation have reducedp g ppercentages.(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

G 992GATE‐1992The thickness of the blank needed to produce, by The thickness of the blank needed to produce, by power spinning a missile cone of thickness 1.5 mm and half cone angle 30°, isand half cone angle 30 , is(a) 3.0 mm (b) 2.5 mm ( ) (d) (c) 2.0 mm (d) 1.5 mm

S 99IES – 1994The mode of deformation of the metal duringThe mode of deformation of the metal duringspinning is(a) Bending(a) Bending(b) Stretching(c) Rolling and stretching(d) Bending and stretching.( ) g g


IFS‐2011Compare metal spinning with press work.

[ k ][2‐marks]

IES 2011High energy rate forming process used for formingHigh energy rate forming process used for formingcomponents from thin metal sheets or deform thintubes is:tubes is:(a) Petro‐forming(b) Magnetic pulse forming(c) Explosive formingp g(d) electro‐hydraulic forming

20 0JWM 2010Assertion (A) : In magnetic pulse‐forming method,( ) g p g ,magnetic field produced by eddy currents is used tocreate force between coil and workpiece.pReason (R) : It is necessary for the workpiecematerial to havemagnetic properties.material to havemagnetic properties.(a) Both A and R are individually true and R is thecorrect explanation of Acorrect explanation of A(b) Both A and R are individually true but R is NOT the

t l ti f Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true

IES 2010IES 2010Assertion (A) : In the high energy rate forming( ) g gy gmethod, the explosive forming has proved to be anexcellent method of utilizing energy at high rate andg g gutilizes both the high explosives and low explosives.Reason (R): The gas pressure and rate of detonation( ) g pcan be controlled for both types of explosives.(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(b) Both A and R are individually true but R is NOT thecorrect 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

S 200IES – 2007Which one of the following metal formingWhich one of the following metal formingprocesses is not a high energy rate forming process?(a) Electro mechanical forming(a) Electro‐mechanical forming(b) Roll‐forming(c) Explosive forming(d) Electro‐hydraulic forming( ) y g

S 2009IES – 2009Which one of the following is a high energy rateWhich one of the following is a high energy rateforming process?(a) Roll forming(a) Roll forming(b) Electro‐hydraulic forming(c) Rotary forging(d) Forward extrusion( )

S 200IES – 2005Magnetic forming is an example of:Magnetic forming is an example of:(a) Cold forming (b) Hot forming( ) Hi h t f i (d) R ll f i(c) High energy rate forming (d) Roll forming

IFS‐2011Write four advantages of high velocity forming process.

[ k ][2‐marks]

G 2000GATE‐2000A 1.5 mm thick sheet is subject to unequal biaxialA 1.5 mm thick sheet is subject to unequal biaxialstretching and the true strains in the directions ofstretching are 0.05 and 0.09. The final thickness ofstretching are 0.05 and 0.09. The final thickness ofthe sheet in mm is(a) 1 414 (b) 1 304(a) 1.414 (b) 1.304(c) 1.362 (d) 289


IES 1998IES‐1998The bending force required for V‐bending, U‐g q g,bending and Edge bending will be in the ratio of(a) 1 : 2 : 0 5 (b) 2: 1 : 0 5(a) 1 : 2 : 0.5 (b) 2: 1 : 0.5(c) 1: 2 : 1 (d) 1: 1 : 1

E lExampleCalculate the bending force for a 45o bend in aluminiumg 45blank. Blank thickness, 1.6 mm, bend length = 1200 mm,Die opening = 8t, UTS = 455 MPa, Die opening factor =p g , 455 , p g1.33

G 200GATE‐2005A 2 mm thick metal sheet is to be bent at an angle ofA 2 mm thick metal sheet is to be bent at an angle ofone radian with a bend radius of 100 mm. If thestretch factor is 0.5, the bend allowance isstretch factor is 0.5, the bend allowance is(a) 99 mm (b) 100 mm( ) (d)(c) 101 mm (d) 102 mm

2mm

1 radian

G 200GATE‐2007Match the correct combination for following metalgworking processes.Processes Associated state of stressP. Blanking 1. TensionQ Stretch Forming 2 CompressionQ. Stretch Forming 2. CompressionR. Coining 3. ShearS Deep Drawing 4 Tension and CompressionS. Deep Drawing 4. Tension and Compression

5. Tension and ShearC d P Q R S P Q R SCodes: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 theirMatch the following metal forming processes with theirassociated stresses in the workpiece.

l f i f Metal forming process Type of stress1. Coining P. Tensile2. Wire Drawing Q. Shear3 Blanking R Tensile and 3. Blanking R. Tensile and

compressive D D i S C i

(a) 1‐S, 2‐P, 3‐Q, 4‐R (b) 1‐S, 2‐P, 3‐R, 4‐Q4. Deep Drawing S. Compressive

(c) 1‐P, 2‐Q, 3‐S, 4‐R (d) 1‐P, 2‐R, 3‐Q, 4‐S

G 200GATE‐2004Match the followingMatch the following

Product ProcessP M ld d l I j ti ldiP. Moulded luggage 1. Injection mouldingQ. Packaging containers for liquid 2. Hot rollingR. Long structural shapes 3. Impact extrusionS. Collapsible tubes 4. Transfer mouldingS. Collapsible tubes 4. Transfer moulding

5. Blow moulding6 C i i6. Coining

(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

S 999IAS – 1999Match List I (Process) with List II (Production of parts)( ) ( p )and select the correct answer using the codes givenbelow the lists:

List‐I List‐IIA. Rolling 1. Discrete partsB. Forging 2. Rod andWireC. Extrusion 3. Wide variety of shapes with thin

llwallsD. Drawing 4. Flat plates and sheets

l d d h ll5. Solid and hollow partsCodes:A B C D A B C D( ) ( )(a) 2 5 3 4 (b) 1 2 5 4(c) 4 1 3 2 (d) 4 1 5 2

S 99IAS – 1997Match List‐I (metal forming process) with List‐IIMatch List I (metal forming process) with List II(Associated feature) and select the correct answerusing the codes given below the Lists:using the codes given below the Lists:

List‐l List‐ IIA Bl ki Sh lA. Blanking 1. Shear angleB. Flow forming 2. Coiled stockC. Roll forming 3. MandrelD. Embossing 4. Closed matching diesD. Embossing 4. Closed matching dies


IES 2010IES 2010Consider the following statements:gThe material properties which principallydetermine howwell a metal may be drawn aredetermine howwell a metal may be drawn are1. Ratio of yield stress to ultimate stress.2.Rate of increase of yield stress relative toprogressive amounts of cold work.p g3. Rate of work hardening.Whi h f th b t t t i / t?Which of the above statements is/are correct?(a) 1 and 2 only (b) 2 and 3 only(c) 1 only (d) 1, 2 and 3


Ch‐18: Sheet Metal FormingQ. No Option Q. No OptionQ. No Option Q. No Option1 C 10 C2 B 11 C2 B 11 C3 A 12 C4 A 13 C4 A 13 C5 D 14 D6 A 15 A6 A 15 A7 A 16 B8 A 17 D9 A

Powder Metallurgygy

By S K Mondal

GATE ‐2011 (PI)( )Which of the following powder production

th d d d ti l ?methods produces spongy and porous particles?(a) Atomization(b) Reduction of metal oxides(c) Electrolytic deposition( ) y p(d) Pulverization

IES 2012IES ‐ 2012In electrolysisIn electrolysis(a) For making copper powder, copper plate is madecathode in electrolyte tankcathode in electrolyte tank(b) For making aluminum powder, aluminum plate is

d dmade anode(c) High amperage produces powdery deposit of cathodemetal on anode(d) Atomization process is more suitable for low melting( ) p gpoint metals

IES – 2007 ConventionalMetal powders are compacted by many methods, but

sintering is required to achieve which property? What

i h t i t ti i ?is hot iso‐static pressing?

[ 2 Marks][ ]

GATE ‐2010 (PI)( )In powder metallurgy, sintering of a component

(a) Improves strength and reduces hardness

(b) Reduces brittleness and improves strength

(c) Improves hardness and reduces toughness

(d) R d i d i b i l(d) Reduces porosity and increases brittleness

IES – 2011 ConventionalWhat is isostatic pressing of metal powders ?

h dWhat are its advantage ?[ 2 Marks]

GATE – 2009 (PI)( )Which of the following process is used to

manufacture products with controlled porosity?

(a) Casting

( )(b) welding

(c) formation(c) formation

(d) Powder metallurgy(d) Powder metallurgy

GATE – 2011 (PI)( )The binding material used in cemented carbide

tti t l icutting tools is(a) graphite(b) tungsten(c) nickel( )(d) cobalt


S K Mondal

New Stamp

IES 2010IES 2010Consider the following parts:Consider the following parts:1. Grinding wheel2. Brake lining3 Self‐lubricating bearings3. Self lubricating bearingsWhich of these parts are made by powder

ll h i ?metallurgy technique?(a) 1, 2 and 3 (b) 2 only( ) , 3 ( ) y(c) 2 and 3 only (d) 1 and 2 only

IES 2010IES 2010Metallic powders can be produced byMetallic powders can be produced by(a) Atomization(b) Pulverization(c) Electro‐deposition process(c) Electro deposition process(d) All of the above

S 2002IES – 2002The rate of production of a powder metallurgy partThe rate of production of a powder metallurgy partdepends on(a) Flow rate of powder(a) Flow rate of powder(b) Green strength of compact(c) Apparent density of compact(d) Compressibility of powder( ) p y p

S 200IES – 2001Match List‐I (Components) with List‐IIMatch List I (Components) with List II(Manufacturing Processes) and select the correctanswer using the codes given below the lists:answer using the codes given below the lists:

List I List IIA C b d ( t l) M hi iA. Car body (metal) 1. MachiningB. Clutch lining 2. CastingC. Gears 3. Sheet metal pressingD. Engine block 4. Powder metallurgyD. Engine block 4. Powder metallurgy


GATE 2011The operation in which oil is permeated into theThe operation in which oil is permeated into thepores of a powder metallurgy product is known as( ) i i(a) mixing(b) sintering(c) impregnation(d) Infiltration(d) Infiltration

S 998IES – 1998In powder metallurgy, the operation carried out toIn powder metallurgy, the operation carried out toimprove the bearing property of a bush is called(a) infiltration (b) impregnation(a) infiltration (b) impregnation(c) plating (d) heat treatment

S 99IES – 1997Which of the following components can beWhich of the following components can bemanufactured by powdermetallurgymethods?1 Carbide tool tips 2 Bearings1. Carbide tool tips 2. Bearings3. Filters 4. Brake liningsSelect the correct answer using the codes given below:(a) 1, 3 and 4 (b) 2 and 3( ) 3 4 ( ) 3(c) 1, 2 and 4 (d) 1, 2, 3 and 4

S 999IES – 1999The correct sequence of the given processes inThe correct sequence of the given processes inmanufacturing by powdermetallurgy is(a) Blending compacting sintering and sizing(a) Blending, compacting, sintering and sizing(b) Blending, compacting, sizing and sintering(c) Compacting, sizing, blending and sintering(d) Compacting, blending, sizing and sintering( ) p g g g g

S 99IES – 1997Match List‐I (Gear component) with List‐II (Preferred( p ) (method of manufacturing) and select the correctanswer using the codes given below the Lists:

List‐I List‐IIA. Gear for clocks 1. HobbingB. Bakelite gears 2. StampingC. Aluminium gears 3. Powder compactingD. Automobile transmission gears 4. Sand casting

5. ExtrusionCode: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


S K Mondal

New Stamp

S 200IES – 2001Carbide‐tipped cutting tools are manufactured byCarbide tipped cutting tools are manufactured bypowder‐ metal technology process and have acomposition ofcomposition of(a) Zirconium‐Tungsten (35% ‐65%)(b) T t bid C b lt ( % %)(b) Tungsten carbide‐Cobalt (90% ‐ 10%)(c) Aluminium oxide‐ Silica (70% ‐ 30%)(d) Nickel‐Chromium‐ Tungsten (30% ‐ 15% ‐ 55%)

S 999IES – 1999Assertion (A): In atomization process of manufacture ofAssertion (A): In atomization process of manufacture ofmetal powder, the molten metal is forced through asmall orifice and broken up by a stream of compressedp y pair.Reason (R): The metallic powder obtained by( ) p yatomization process is quite resistant to oxidation.(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

S 99IES – 1997Match List‐I with List‐II and select the correct answerusing the codes given below the Lists:

List‐I List‐II(Bearing materials) (Properties)A. Babbits 1. PorousB. Bronze 2. Good EmbeddabilityC. C.I. 3. Suitable for high loads and low3 g

speedsD. Sintered powdered metal 4. Runs well with C.I.

ljournalsCode: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 200IES – 2007What are the advantages of powdermetallurgy?What are the advantages of powdermetallurgy?1. Extreme purity product

L l b t2. Low labour cost3. Low equipment cost.Select the correct answer using the code given below(a) 1, 2 and 3 (b) 1 and 2 only(a) 1, 2 and 3 (b) 1 and 2 only(c) 2 and 3 only (d) 1 and 3 only

S 2006IES – 2006Which of the following are the limitations ofWhich of the following are the limitations ofpowdermetallurgy?1 High tooling and equipment costs1. High tooling and equipment costs.2. Wastage of material.3. It cannot be automated.4. Expensive metallic powders.4 p pSelect the correct answer using the codes given below:(a) Only 1 and 2 (b) Only 3 and 4(a) Only 1 and 2 (b) Only 3 and 4(c) Only 1 and 4 (d) Only 1, 2 and 4

S 200IES – 2004Consider the following factors:Consider the following factors:1. Size and shape that can be produced economically

P it f th t d d2. Porosity of the parts produced3. Available press capacity4. High densityWhich of the above are limitations of powderWhich of the above are limitations of powdermetallurgy?(a) 1 3 and 4 (b) 2 and 3(a) 1, 3 and 4 (b) 2 and 3(c) 1, 2 and 3 (d) 1 and 2

IES 2012IES ‐ 2012Statement (I): Parts made by powder metallurgy do not( ) y p gyhave as good physical properties as parts casted.Statement (II): Particle shape in powder metallurgy( ) p p gyinfluences the flow characteristic of the powder.(a) Both Statement (I) and Statement (II) are( ) ( ) ( )individually true and Statement (II) is the correctexplanation of Statement (I)(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is not the correctexplanation of Statement (I)(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true

S 2009IES – 2009Which of the following cutting tool bits are made byWhich of the following cutting tool bits are made bypowdermetallurgy process?(a) Carbon steel tool bits (b) Stellite tool bits(a) Carbon steel tool bits (b) Stellite tool bits(c) Ceramic tool bits (d) HSS tool bits

S 2003IAS – 2003Which of the following are produced by powderWhich of the following are produced by powdermetallurgy process?

1 Cemented carbide dies1. Cemented carbide dies2. Porous bearings3. Small magnets4. Parts with intricate shapes4 pSelect the correct answer using the codes given below:Codes:Codes:(a) 1, 2 and 3 (b) 1, 2 and 4(c) 2, 3 and 4 (d) 1, 3 and 4


S 2003IAS – 2003In parts produced by powder metallurgy process,In parts produced by powder metallurgy process,pre‐sintering is done to(a) Increase the toughness of the component(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 2000IAS – 2000Consider the following processes:Consider the following processes:1. Mechanical pulverization

At i ti2. Atomization3. Chemical reduction4. SinteringWhich of these processes are used for powderWhich of these processes are used for powderpreparation in powder metallurgy?(a) 2 3 and 4 (b) 1 2 and 3(a) 2, 3 and 4 (b) 1, 2 and 3(c) 1, 3 and 4 (d) 1, 2 and 4

S 99IAS – 1997Assertion (A): Close dimensional tolerances are( ) CNOT possible with isostatic pressing of metalpowder in powdermetallurgy technique.Reason (R): In the process of isostatic pressing, thepressure is equal in all directions which permitsuniformdensity of themetal powder.(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

S 998IAS – 1998Throwaway tungsten carbide tip tools areThrowaway tungsten carbide tip tools aremanufactured by(a) Forging (b) Brazing(a) Forging (b) Brazing(c) Powder metallurgy (d) Extrusion

S 996IAS – 1996Which one of the following processes is performedWhich one of the following processes is performedin powder metallurgy to promote self‐lubricatingproperties in sintered parts?properties in sintered parts?(a) Infiltration (b) Impregnation( ) Pl ti (d) G hiti ti(c) Plating (d) Graphitization

GATE 2008 (PI)GATE ‐2008 (PI)Match the following

Group – 1 Group ‐2P Mulling 1 Powder metallurgyP. Mulling 1. Powder metallurgyQ. Impregnation 2. Injection moulding

l h fR. Flash trimming 3. Processing of FRP compositesS. Curing 4. Sand casting

(a) P – 4, Q – 3, R – 2, S – 1 (b) P – 2, Q – 4, R – 3, S ‐ 1

g g

( ) 4, Q 3, , ( ) , Q 4, 3,(c) P – 2, Q – 1, R – 4, S – 3 (d) P – 4, Q – 1, R – 2, S ‐ 3

S 200IAS – 2007Assertion (A): Mechanical disintegration of a( ) gmolten metal stream into fine particles by means ofa jet of compressed air is known as atomization.Reason (R): In atomization process inert‐gas orwater cannot be used as a substitute for compressedair.(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

S 200IAS – 2004The following are the constituent steps in theThe following are the constituent steps in theprocess of powdermetallurgy:1 Powder conditioning1. Powder conditioning2. Sintering3. Production of metallic powder4. Pressing or compacting into the desired shape4 g p g pIndentify the correct order in which they have to beperformed and select the correct answer using the codesperformed and select the correct answer using the codesgiven below:(a) 1 2 3 4 (b) 3 1 4 2(a) 1‐2‐3‐4 (b) 3‐1‐4‐2(c) 2‐4‐1‐3 (d) 4‐3‐2‐1

S 2003IAS – 2003Assertion (A): Atomization method for production ofAssertion (A): Atomization method for production ofmetal powders consists of mechanical disintegration ofmolten stream into fine particles.pReason (R): Atomization method is an excellent meansof making powders from high temperaturemetals.g p g p(a) Both A and R are individually true and R is the correctexplanation of Ap(b) Both A and R are individually true but R is not thecorrect explanation of Ap(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 200IAS – 2007Consider the following basic steps involved in theConsider the following basic steps involved in theproduction of porous bearings:1 Sintering1. Sintering2. Mixing3. Repressing4. Impregnation4 p g5. Cold‐die‐compactionWhich one of the following is the correct sequence of theWhich one of the following is the correct sequence of theabove steps?

Ch‐12: Powder Metallurgygy

Q. No Option Q. No Option1 D 5 C2 B 6 B3 C 7 D4 A 8 C

Limit Tolerance & FitsLimit, Tolerance & Fits

By S K Mondal

For PSUTolerances are specified ( ) b d d f(a) To obtain desired fits(b) because it is not possible to manufacture a size

exactly(c) to obtain higher accuracy( ) g y(d) to have proper allowances

ISRO‐2010Expressing a dimension as 25.3±0.05 mm is the case of

(a) Unilateral tolerance

(b) Bilateral tolerance

(c) Limiting dimensions

(d) All f h b(d) All of the above

GATE 2010 ISRO 2012GATE – 2010, ISRO‐2012

A shaft has a dimension,Th ti l f f d t l d i ti d

0.0090.02535φ−−

The respective values of fundamental deviation andtolerance are(a) 0.025, 0.008 (b) 0.025,0.016(c) 0.009, 0.008 (d) 0.009,0.016

− ± −− ± −(c) 0.009, 0.008 (d) 0.009,0.016±

GATE 1992GATE ‐ 1992Two shafts A and B have their diameters specified as 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?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 (d) The relative error in the dimension is same for shaft A and shaft B

GATE 2004GATE ‐ 2004 In an interchangeable assembly, shafts of sizeIn an interchangeable assembly, shafts of size

mm mate with holes of size mm.The maximum possible clearance in the assembly

0.0400.010025.000+−

0.0200.00025.000+−

The maximum possible clearance in the assemblywill be( ) i(a) 10 microns(b) 20 microns(c) 30 microns(d) 60 microns(d) 60 microns

S O 20 0ISRO‐20100.02 0.02Dimension of the hole is 50 mm and shaft is 50 mm+ +0.00 0.00Dimension of the hole is 50 mm and shaft is 50 mm.

The minimum clearance is− +

(a) 0.02 mm (b) 0.00 mm (c) 0 02 mm (d) 0 01 mm(c) -0.02 mm (d) 0.01 mm


S K Mondal

New Stamp

IES 2005IES ‐ 2005The tolerance specified by the designer for thep y gdiameter of a shaft is 20.00 ± 0.025 mm. The shaftsproduced by three different machines A, B and Chave mean diameters of 19∙99 mm, 20∙00 mm and20.01 mm respectively, with same standardd i ti Wh t ill b th t j ti fdeviation. What will be the percentage rejection forthe shafts produced bymachines A, B and C?( ) S f th hi A B d C i th t d d(a) Same for the machines A, Band C since the standard

deviation is same for the three machines(b) L t f hi A(b) Least for machine A(c) Least for machine B(d) Least for machine C

GATE 2000GATE ‐ 2000A slot is to be milled centrally on a block with aA slot is to be milled centrally on a block with adimension of 40 ± 0.05 mm. A milling cutter of 20mm width is located with reference to the side ofmm width is located with reference to the side ofthe block within ± 0.02 mm. The maximum offset inmm between the centre lines of the slot and themm between the centre lines of the slot and theblock is(a) ± 0 070 (b) 0 070(a) ± 0.070 (b) 0.070(c) ± 0.020 (d) 0.045

GATE 2007GATE ‐ 2007A hole is specified as mm. The mating

0 . 0 5 00 . 0 0 04 0A hole is specified as mm. The mating

shaft has a clearance fit with minimum clearance of0.01 mm. The tolerance on the shaft is 0.04 mm. The

4 0

0.01 mm. The tolerance on the shaft is 0.04 mm. Themaximum clearance in mm between the hole andthe shaft isthe shaft is(a) 0.04(b)(b) 0.05(c) 0.10(d) 0.11

IES 2011Interference fit joints are provided for:Interference fit joints are provided for:(a) Assembling bush bearing in housing(b) Mounting heavy duty gears on shafts(c) Mounting pulley on shafts( ) g p y(d) Assembly of flywheels on shafts

IES 2013IES‐2013Which of the following is a joint formed byWhich of the following is a joint formed by

interference fits?

(a) Joint of cycle axle and its bearing

(b) Joint between I.C. Engine piston and cylinder

(c) Joint between a pulley and shaft transmitting power

(d) J i f l h i dl d i b i(d) Joint of lathe spindle and its bearing

GATE 2005GATE ‐ 2005In order to have interference fit, it is essential thatIn order to have interference fit, it is essential thatthe lower limit of the shaft should be(a) Greater than the upper limit of the hole(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( )

GATE 20110 015

A hole is of dimension mm. The0.015

09φ++

corresponding shaft is of dimension mm.0.0100.0019φ

++p g

The resulting assembly has(a) loose running fit

φ

(a) loose running fit(b) close running fit( ) t iti fit(c) transition fit(d) interference fit

GATE ‐2012 Same Q in GATE‐2012 (PI)

In an interchangeable assembly, shafts of size 0.040.0125+−

mmmate with holes of size mm.0.030 0225++

The maximum interference (in microns) in the assembly

0.0225+

The maximum interference (in microns) in the assemblyis( ) (b) ( ) (d)(a) 40 (b) 30 (c) 20 (d) 10

IAS‐2011 MainAn interference assembly, of nominal diameter 20

mm, is of a unilateral holes and a shafts. The

f t i t l f th h l t imanufacturing tolerances for the holes are 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]


IES 2007IES ‐ 2007ISRO‐2011

A shaft and hole pair is designated as 50H7d8. This

assembly constitutes

(a) Interference fit

(b) Transition fit(b) Transition fit

(c) Clearance fit

(d) None of the above

IES 2006IES ‐ 2006Which of the following is an interference fit?Which of the following is an interference fit?(a) Push fit(b) R i fit(b) Running fit(c) Sliding fit(d) Shrink fit

IES 2009IES ‐ 2009Consider the following joints:Consider the following joints:1. Railway carriage wheel and axle

IC i li d d li2. IC engine cylinder and linerWhich of the above joints is/are the result(s) ofinterference fit?(a) 1 only( ) y(b) 2 only(c) Neither 1 nor 2(c) Neither 1 nor 2(d) Both 1 and 2

IES 2008IES ‐ 2008Consider the following statements:Consider the following statements:1. The amount of interference needed to create a tightjoint varies with diameter of the shaftjoint varies with diameter of the shaft.2. An interference fit creates no stress state in theh ftshaft.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(c) 2 and 3 only (d) 1 and 3 only

IES 2004IES ‐ 2004Consider the following fits:Consider the following fits:1. I.C. engine cylinder and piston

B ll b i t d h i2. Ball bearing outer race and housing3. Ball bearing inner race and shaftWhich of the above fits are based on the interferencesystem?y(a) 1 and 2(b) 2 and 3(b) 2 and 3(c) 1 and 3(d) 1, 2 and 3

IES 2003IES ‐ 2003Match List‐I (Phenomenon) with List‐II (Significant( ) ( gParameters/Phenomenon) and select the correctanswer using the codes given below the Lists:

List I List IIList‐I List‐II(Phenomenon) (SignificantParameters/Phenomenon)/ )A. Interference fit 1. Viscosity indexB. Cyclic loading 2. InterferenceC. Gear meshing 3. Notch sensitivityD. Lubricating of bearings 4. Induced compressive

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

GATE 2001GATE ‐ 2001 Allowance in limits and fits refers toAllowance in limits and fits refers to(a) Maximum clearance between shaft and hole(b) Mi i l b t h ft d h l(b) Minimum clearance between shaft and hole(c) Difference between maximum and minimum size ofhole(d) Difference between maximum and minimum size of( )shaft

GATE 1998GATE ‐ 1998In the specification of dimensions and fits,In the specification of dimensions and fits,(a) Allowance is equal to bilateral tolerance(b) All i l t il t l t l(b) Allowance is equal to unilateral tolerance(c) Allowance is independent of tolerance(d) Allowance is equal to the difference betweenmaximum and minimum dimension specified by thep ytolerance.


IES 2012IES ‐ 2012Clearance in a fit is the difference betweenClearance in a fit is the difference between(a) Maximum hole size and minimum shaft size(b) Mi i h l i d i h ft i(b) Minimum hole size and maximum shaft size(c) Maximum hole size and maximum shaft size(d) Minimum hole size and minimum shaft size

ISRO‐2008Basic 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 2005IES ‐ 2005Assertion (A): Hole basis system is generallyAssertion (A): Hole basis system is generallypreferred to shaft basis system in tolerance designfor getting the required fits.for getting the required fits.Reason (R): Hole has to be given a larger toleranceband than themating shaftband than themating shaft.(a) Both A and R are individually true and R is the


S 2006 C i lIES‐2006 ConventionalFind the limit sizes, tolerances and allowances for aFind the limit sizes, tolerances and allowances for a100 mm diameter shaft and hole pair designated byF8h . Also specify the type of fit that the above pairF8h10. Also specify the type of fit that the above pairbelongs to.Given: 100 mm diameter lies in the diameter stepGiven: 100 mm diameter lies in the diameter steprange of 80‐120 mm. The fundamental deviation forshaft designation ‘f’ is 5 5 D0.41shaft designation f is ‐5.5 D0.41

The values of standard tolerances for grades of IT 8d IT i d 6 i ti land IT 10 are 25i and 64i respectively.

Also, indicate the limits and tolerance on a diagram.[15‐Marks]

IES 2008IES ‐ 2008Consider the following statements:Consider the following statements:A nomenclature 50 H8/p8 denotes that

H l di t iφ

1. Hole diameter is 50 mm.2. It is a shaft base system.3. 8 indicates fundamental deviation.Which of the statements given above is/are incorrect?Which of the statements given above is/are incorrect?(a) 1, 2 and 3(b) d l(b) 1 and 2 only(c) 1 and 3 only(d) 3 only

IES 2002IES ‐ 2002In the tolerance specification 25 D 6, the letter DIn the tolerance specification 25 D 6, the letter Drepresents(a) Grade of tolerance(a) Grade of tolerance(b) Upper deviation(c) Lower deviation(d) Type of fit( ) yp

GATE 2009GATE ‐ 2009What are the upper and lower limits of the shaftpprepresented by 60 f8?Use the following data:Diameter 60 lies in the diameter step of 50‐80 mm.Fundamental tolerance unit,i, in m= 0.45 D1/3 + 0.001D, where D is therepresentative size in mm;T l l f lT8 i

μ

Tolerance value for lT8 = 25i.Fundamental deviation for 'f shaft = ‐5.5D0.41

( ) l(a) Lower 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, Upper Limit = 60.046 mm

GATE 2008 (PI)GATE – 2008 (PI)Following data are given for calculating limits ofFollowing 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 minimumhole is zero and IT8 = 25 i, the maximum and minimum

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

GATE 2000GATE ‐ 2000A fit is specified as 25H8/e8. The tolerance value forA fit is specified as 25H8/e8. The tolerance value fora nominal diameter of 25 mm in IT8 is 33 micronsand fundamental deviation for the shaft is ‐ 40and fundamental deviation for the shaft is 40microns. The maximum clearance of the fit inmicrons ismicrons is(a) ‐7(b)(b) 7(c) 73(d) 106


GATE 2003GATE ‐ 2003The dimensional limits on a shaft of 25h7 areThe dimensional limits on a shaft of 25h7 are(a) 25.000, 25.021 mm(b)(b) 25.000, 24.979 mm(c) 25.000, 25.007 mm(d) 25.000, 24.993 mm

GATE‐2010 (PI)A small bore is designated as 25H7. The lower

(minimum) and upper (maximum) limits of the bore(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 aslimit is 25.033 mm. When the bore is designated as

25H6, then the upper (maximum) limit of the bore (in

mm) is

(a) 25.001 (b) 25.005 (c) 25.009 (d) 25.013

GATE 1996 IES 2012GATE – 1996, IES‐2012The fit on a hole‐shaft system is specified as H7‐The fit on a hole shaft system is specified as H7s6.The type of fit is(a) Clearance fit(a) Clearance fit(b) Running fit (sliding fit)(c) Push fit (transition fit)(d) Force fit (interference fit)( ) ( )

IES 2000IES ‐ 2000Which one of the following tolerances set on innerWhich one of the following tolerances set on innerdiameter and outer diameter respectively of headedjig bush for press fit is correct?jig bush for press fit is correct?(a) G7 h 6 (b) F7 n6( ) H h 6 (d) F j6(c) H 7h 6 (d) F7j6

ISRO‐2008Interchangeability can be achieved by

(a) Standardization

(b) Better process planning

(c) Simplification

(d) B d l i(d) Better product planning

IAS‐2010 mainWhat is the difference between hole basis system and

shaft basis system ? Why is hole basis system the more

t i i ?extensive in use ?

What are the differences between interchangeabilityWhat are the differences between interchangeability

and selective assembly ?

[12‐Marks]

GATE 2003GATE ‐ 2003 GATE 1997GATE ‐ 1997Three blocks B1 , B2 and B3 are1 , 2 3to be inserted in a channel ofwidth S maintaining ai i f idth Tminimum gap of width T =

0.125 mm, as shown in Figure.For P 18 75 ± 0 08;For P = 18. 75 ± 0.08;Q = 25.00 ± 0.12;R 28 125 ± 0 1 andR = 28.125 ± 0.1 andS = 72.35 + X, (where alldimensions are in mm) thedimensions are in mm), thetolerance X is

(a) + 0 38 (b) ‐ 0 38 (c) + 0 05 (d) ‐0 05(a) + 0.38 (b) 0.38 (c) + 0.05 (d) 0.05

GATE 2008 (PI)GATE – 2008 (PI)A three‐component welded cylindrical assembly is shownbelow. The mean length of the three components and theirbelow. The mean length of the three components and theirrespective tolerances (both in mm) are given in the tablebelow.

Assuming a normal distribution for the individualgcomponent dimensions, the natural tolerance limits for thelength (Y) of the assembly (mm) is( ) 6 6 (b) 6 6 ( ) 6 6 (d) 6 6(a) 65 ±2.16 (b) 65 ±1.56 (c) 65 ±0.96 (d) 65 ±0.36


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GATE 2007 (PI)GATE – 2007 (PI)Tolerance on the dimension x in the twoTolerance on the dimension x in the twocomponent assembly shown below is

(All dimensions in mm)( )

( )(a) ±0.025(a) ±0.030( ) 3(a) ±0.040( )(a) ±0.045

GATE ‐2007(PI)GATE 2007(PI)The geometric tolerance that does NOT need a datumg

for its specification is

(a) Concentricity (b) Runout

(c) Perpendicularity (d) Flatness

GATE 2007 (PI)GATE – 2007 (PI)

0 050

Diameter of a hole after plating needs to be controlledb t 30 If th l ti thi k i+0.050

0.010between 30 mm. If the plating thickness varies between 10 - 15 microns, diameter of the hole before

++

between 10 15 microns, diameter of the hole beforeplating should be

0.070.030(a) 30++

0 0.0650.020

0 080 0 070

mm (b) 30 mm

( ) 30 (d) 30

++

+ +0.080 0.0700.030 0.040(c) 30 mm (d) 30 mm + ++ +

GATE 2013GATE‐20130.0200.010Cylindrical pins of 25 mm diameter are

l t l t d i h Thi k f th

++

2.0

electroplated in a shop. Thickness of the plating is 30 micron Neglecting gage ±plating is 30 micron. Neglecting gage 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

ISRO‐2008Plug gauges are used to( ) h d f h k(a) Measure the diameter of the workpieces(b) Measure the diameter of the holes in theworkpieces(c) Check the diameter of the holes in the( )workpieces(d) Check the length of holes in the workpieces(d) Check the length of holes in the workpieces

GATE 2004GATE ‐ 2004GO and NO‐GO plug gages are to be designed for aGO and NO GO plug gages are to be designed for ahole mm. Gage tolerances can be taken as 10%of the hole tolerance Following ISO system of gage

0.050.0120

of the hole tolerance. Following ISO system of gagedesign, sizes of GO and NO‐GO gage will berespectivelyrespectively(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 1995GATE ‐ 1995Checking the diameter of a hole using GO‐NO‐GOChecking the diameter of a hole using GO NO GOgauges is an, example of inspection by…..(variables/attributes)…..(variables/attributes)The above statement is( ) V i bl(a) Variables(b) Attributes(c) Cant say(d) Insufficient data(d) Insufficient data

GATE 2006 VS 2012GATE – 2006, VS‐2012A ring gauge is used tomeasureA ring gauge is used tomeasure(a) Outside diameter but not roundness(b) R d b t t t id di t(b) Roundness but not outside diameter(c) Both outside diameter and roundness(d) Only external threads

Measurement of Lines & SurfacesMeasurement of Lines & Surfaces

By S K MondalFor-2014 (IES, GATE & PSUs) Page 61 of 78

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ISRO‐2010The vernier reading should not be taken ati f l b f l h k hits face value before an actual check hasbeen taken for(a) Zero error(b) It lib ti(b) Its calibration(c) Flatness of measuring jaws( ) g j(d) Temperature equalization

ISRO‐2008The least count of a metric vernier caliper having

25 divisions on vernier scale, matching with 24

divisions of main scale (1 main scale divisions = 0.5

mm) ismm) is

(a) 0.005 mm (b) 0.01 mm( ) 5 ( )

(c) 0.02 mm (d) 0.005mm

ISRO‐2009, 2011,In a simple micrometer with screw pitch 0.5 mm

and divisions on thimble 50, the reading

corresponding to 5 divisions on barrel and 12

divisions on thimble isdivisions on thimble is

(a) 2.620 mm (b) 2.512 mm( ) ( ) 5

(c) 2.120 mm (d) 5.012 mm

GATE 2008 S 1GATE – 2008 S‐1 A displacement sensor (a dial indicator) measures theA displacement sensor (a dial indicator) measures thelateral displacement of a mandrel mounted on the taperhole inside a drill spindle. The mandrel axis is anhole inside a drill spindle. The mandrel axis is anextension of the drill spindle taper hole axis and theprotruding portion of the mandrel surface is perfectlyprotruding portion of the mandrel surface is perfectlycylindrical. Measurements are taken with the sensorplaced at two positions P and Q as shown in the figure.placed at two positions P and Q as shown in the figure.The readings are recorded as Rx = maximum deflectionminus minimum deflection, corresponding to sensorminus minimum deflection, corresponding to sensorposition at X, over one rotation.

GATE 2008 td f S 1GATE – 2008 contd… from S‐1 If Rp= RQ>0, which one of thep Qfollowing would be consistent with theobservation?(A) The drill spindle rotational axis is(A) The drill spindle rotational axis iscoincident with the drill spindle taperhole axis(B) The drill spindle rotational axisintersects the drill spindle taper holeaxis at point Paxis at point P(C) The drill spindle rotational axis isparallel to the drill spindle taper holeaxis(D) The drill spindle rotational axisintersects the drill spindle taper holeintersects the drill spindle taper holeaxis at point Q

ISRO‐2010A master gauge isA master gauge is(a) A new gauge(b) An international reference standard( ) A t d d f h ki f(c) A standard gauge for checking accuracy ofgauges used on shop floors

(d) A gauge used by experienced technicians

ISRO‐2008Standards to be used for reference purposes in

laboratories and workshops are termed as

(a) Primary standards

( )(b) Secondary standards

(c) Tertiary standards(c) Tertiary standards

(d) Working standards(d) Working standards

GATE 2007 (PI)GATE – 2007 (PI)Which one of the following instruments is aWhich one of the following instruments is acomparator ?( ) T l M k ’ Mi(a) Tool Maker’s Microscope(b) GO/NO GO gauge( ) g g(c) Optical Interferometer(d) Di l G(d) Dial Gauge

PSUA feeler gauge is used to check the

(a) Pitch of the screw

(b) Surface roughness

(c) Thickness of clearance

(d) Fl f f(d) Flatness of a surface


IAS‐2011 MainDraw a self explanatory sketch of Sigma

mechanical comparator. Explain how

(i) shock load is avoided,

( )(ii) oscillations of the pointer are damped.

[10 marks][10 – marks]

ISRO‐2011A sine bar is specified by

(a) Its total length

(b) The size of the rollers

( ) Th t di t b t th t ll(c) The centre distance between the two rollers

(d) The distance between rollers and upper surface( ) pp

GATE ‐2012 (PI)A sine bar has a length of 250 mm. Each roller has

di t f D i t la diameter of 20 mm. During taper angle

measurement of a component, the height from thep , g

surface plate to the centre of a roller is 100 mm.

The calculated taper angle (in degrees) is

(a) 21.1 (b) 22.8 (c) 23.6 (d) 68.9

GATE – 2011 (PI)( )The best wire size (in mm) for measuring effectivedi t f t i th d (i l d d l i 6 o)diameter of a metric thread (included angle is 60o)of 20 mm diameter and 2.5 mm pitch using twoi th d iwire method is

(a) 1.443(b) 0.723(c) 2.886( )(d) 2.086

GATE 2013GATE‐2013A t i th d f it h d th d l 6A metric thread of pitch 2 mm and thread angle 60

inspected for its pitch diameter using 3‐wirep p g 3

method. The diameter of the best size wire in mm is

(a) 0.866 (b) 1.000 (c) 1.154 (d) 2.000

GATE – 2011 (PI)( )To measure the effective diameter of an externalmetric thread (included angle is 60o) of 3 5 mmmetric thread (included angle is 60 ) of 3.5 mmpitch, a cylindrical standard of 30.5 mm diameterand two wires of 2 mm diameter each are used.a d t o es o d a ete eac a e used.The micrometer readings over the standard andover the wires are 16.532 mm and 15.398 mm,respectively. The effective diameter (in mm) of thethread is(a) 33.366 (b) 30.397(c) 29.366 (d) 26.397

IES 1992IES ‐ 1992Which grade symbol represents surface rough ofWhich grade symbol represents surface rough ofbroaching?(a) N (b) N(a) N12 (b) N8

(c) N4 (d) N1

IFS‐2011Wh t i t b i t h bl f t ?What is meant by interchangeable manufacture?

Laser light has unique advantages for inspectionLaser light has unique advantages for inspection.

What are they ? Define the terms 'roughnessy g

height', 'waviness width' and 'lay' in connection

with surface irregularities.

[ k ][10‐marks]

ISRO‐2011CLA value and RMS values are used for measurement

of

(a) Metal hardness

(b) Sharpness of tool edge(b) Sharpness of tool edge

(c) Surface dimensions

(d) Surface roughness


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IES 2006IES ‐ 2006The M and E‐system in metrology are related toThe M and E system in metrology are related tomeasurement of:(a) Screw threads (b) Flatness(a) Screw threads (b) Flatness(c) Angularity (d) Surface finish

IES 2007IES ‐ 2007What is the dominant direction of the tool marks orWhat is the dominant direction of the tool marks orscratches in a surface texture having a directionalquality, called?quality, called?(a) Primary texture (b) Secondary texture( ) L (d) Fl(c) Lay (d) Flaw

IES 2008IES ‐ 2008What term is used to designate the direction of theWhat term is used to designate the direction of thepredominant surface pattern produced bymachining operation?machining operation?(a) Roughness (b) Lay( ) W i (d) C t ff(c) Waviness (d) Cut off

IES 2010Match List I with List II and select the correct answerMatch List I with List II and select the correct answerusing the code given below the lists:List I List II

(Symbols for direction of lay) (Surface texture)

A B C D A B C DA B C D A B C D(a) 4 2 1 3 (b) 3 2 1 4(c) 4 1 2 3 (d) 3 1 2 4

IES 2008IES ‐ 2008 ISRO‐2010

Surface roughness on a drawing isd brepresented by

(a) Triangles(a) Triangles(b) Circles(c) Squares(d) Rectangles(d) Rectangles

GATE 1997GATE ‐ 1997 List I List II

(A) Surface profilometer 1. Calibration(B) Light Section Microscope 2 Form tester(B) Light Section Microscope 2. Form tester(C) Microkater 3. Film thickness

measurementmeasurement(D) Interferometer 4. Centre line average

5 Comparator5. Comparator6. Surface lay measurement

C d A B C D A B C DCodes: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

ISRO‐2007Gratings are used in connection with

(a) Flatness measurement

(b) Roundness measurement

(c) Surface texture measurement

(d) Li di l(d) Linear displacement measurements

GATE 2003GATE ‐ 2003Two slip gauges of 10 mm width measuring 1 000 mmTwo slip gauges of 10 mm width measuring 1.000 mmand 1.002 mm are kept side by side in contact with eachother lengthwise. An optical flat is kept resting on theother lengthwise. An optical flat is kept resting on theslip gauges as shown in the figure. Monochromatic lightof wavelength 0.0058928 mm is used in the inspection.of wavelength 0.0058928 mm is used in the inspection.The total number of straight fringes that can be observedon both slip gauges ison both slip gauges is

(a) 2 (b) 6(c) 8 (d) 13


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GATE – 2011 (PI)Ob ti f li fl tObservation of a slip gauge on a flatnessinterferometer produced fringe counts numbering

d f t di Th d di i10 and 14 for two readings. The second reading istaken by rotating the set‐up by 180o. Assume thatb th f f th li fl t d thboth faces of the slip gauge are flat and thewavelength of the radiation is 0.5086 µm. The

ll li (i ) b t th t f fparallelism error (in µm) between the two faces ofthe slip gauge is( ) (b)(a) 0.2543 (b) 1.172(c) 0.5086 (d) 0.1272

Miscellaneous of MetrologyMiscellaneous of Metrology

By S K Mondal

GATE 1998GATE ‐ 1998 Auto collimator is used to checkAuto collimator is used to check(a) Roughness(b) Fl t(b) Flatness(c) Angle(d) Automobile balance.

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(c) measure the flatness error

(d) measure roundness error between centers(d) measure roundness error between centers

S O 20 0ISRO‐2010Optical square isOptical square is(a) Engineer's square having stock and blade set at 90o

(b) A t t d i ti i h i th l f(b) A constant deviation prism having the angle ofdeviation between the incident ray and reflected ray,

l t oequal to 90o

(c) A constant deviation prism having the angle ofdeviation between the incident ray and reflected ray,equal to 45o

(d) Used to produce interference fringes

IES 1998IES ‐ 1998Match List‐I with List‐II and select the correct answer using the

d i b l h licodes given below the lists:List‐I List‐II

(Measuring Device) (ParameterMeasured)( g ) ( )A. Diffraction grating 1. Small angular deviations on long

flat surfacesB Optical flat 2 On‐line measurement of movingB. Optical flat 2. On‐line measurement of moving

partsC. Auto collimators 3. Measurement of gear pitchD L i t S f t t i i t f tD. Laser scan micrometer4. Surface texture using interferometer

5. Measurement of very smalldisplacements

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(c) 3 5 4 1 (d) 5 4 1 2

GATE 1992GATE ‐ 1992Match the instruments with the physical quantities theyp y q ymeasure:Instrument Measurement(A) Pilot‐tube (1) R.P.M. of a shaft(B) McLeod Gauge (2) Displacementg p(C) Planimeter (3) Flow velocity(D) LVDT (4) Vacuum4

(5) Surface finish(6) Area( )

Codes:A B C D A B C D(a) 4 1 2 3 (b) 3 4 6 2( ) 4 3 ( ) 3 4(c) 4 2 1 3 (d) 3 1 2 4

GATE 2004GATE ‐ 2004Match the followingMatch the followingFeature to be inspected InstrumentP Pitch and Angle errors of screw thread 1 Auto CollimatorP Pitch and Angle errors of screw thread 1. Auto CollimatorQ Flatness error of a surface plate 2. Optical InterferometerR Ali f hi lid Di idi H dR Alignment error of a machine slide way 3. Dividing Head

and Dial GaugeS P fil f S i i L lS Profile of a cam 4. Spirit Level

5. Sine bar6. Tool maker's Microscope

(a) P‐6 Q‐2 R‐4 S‐6 (b) P‐5 Q‐2 R‐1 S‐65(c) P‐6 Q‐4 R‐1 S‐3 (d) P‐1 Q‐4 R‐4 S‐2

GATE 1995GATE ‐ 1995List I List IIList I List II

(Measuring instruments) (Application)(A) T l f T l t(A) Talysurf 1. T‐slots(B) Telescopic gauge 2. Flatness(C) Transfer callipers 3. Internal diameter(D) Autocollimator 4. Roughness(D) Autocollimator 4. Roughness



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GATE 2010GATE ‐ 2010A taper hole is inspected using a CMM, with a probeA taper hole is inspected using a CMM, with a probeof 2 mm diameter. At a height, Z = 10 mm from thebottom, 5 points are touched and a diameter ofbottom, 5 points are touched and a diameter ofcircle (not compensated for probe size) is obtainedas 20 mm. Similarly, a 40 mm diameter is obtainedas 20 mm. Similarly, a 40 mm diameter is obtainedat a height Z = 40 mm. the smaller diameter (in mm)of hole at Z = 0 isof hole at Z 0 is(a) 13.334(b)(b) 15.334(c) 15.442(d) 15.542

GATE 2008 (PI)GATE ‐2008 (PI)An experimental setup is planned to determine the taper ofworkpiece as shown in the figure. If the two precision rollershave radii 8 mm and 5 mm and the total thickness of slip

i d b h ll i hgauges inserted between the rollers is 15.54 mm, the taperangle θ is( ) 6 d(a) 6 degree(b) 10 degree(c) 11 degree(d) 12 degreeg

ISRO‐2007Whi h f h f ll i i i bl i hWhich of the following errors are inevitable in the

measuring system and it would be vain fullg y

exercise to avoid them

(a) Systematic errors

(b) R d(b) Random errors

(c) Calibration errors(c) Calibration errors

(d) Environmental errors

Ch‐13: MetrologygyQ. No Option Q. No Option

C D1 C 10 D2 C 11 A3 A 12 B4 C 13 B4 C 13 B5 C 14 D6 B 15 B7 C 16 C7 C 16 C8 B 17 B9 B


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Cutting Tool MaterialsCutting Tool Materials

By S K Mondal

IntroductionSuccess in metal cutting depends on selection of theproper cutting tool (material and geometry) for a givenwork material.A wide range of cutting tool materials is available withg ga variety of properties, performance capabilities, andcost.These include:

High carbon Steels and low/medium alloy steelsHigh carbon Steels and low/medium alloy steels,High‐speed steels,

b l llCast cobalt alloys,Contd…

Cemented carbides,Cast carbides Cast carbides, Coated carbides,

d h h d lCoated high speed steels, Ceramics, Cermets, Whisker reinforced ceramics Whisker reinforced ceramics, Sialons,

d l ll b b d ( )Sintered polycrystalline cubic boron nitride (CBN), Sintered polycrystalline diamond, and single‐crystal natural diamond.

FIGURE: Improvements in cutting tool materials have reduced machining time.

Carbon SteelsLi i d l lif Th f i dLimited tool life. Therefore, not suited to massproductionC b f d i l h f llCan be formed into complex shapes for smallproduction runsLLow costSuited to hand tools, and wood workingCarbon content about 0.9 to 1.35% with a hardnessABOUT 62 C RockwellMaximum cutting speeds about 26 ft/min. dryThe hot hardness value is low. This is the major factorin tool life.

Fig. Productivity raised by cutting tool materials

S 99IAS – 1997Assertion (A): Cutting tools made of high carbonAssertion (A): Cutting tools made of high carbonsteel have shorter tool life.Reason(R): During machining the tip of the cuttingReason(R): During machining, the tip of the cuttingtool is heated to 600/700°C which cause the teal tipto lose its hardnessto lose its hardness.(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 Ans. (a)

High speed steelTh t l d f tti t l t hThese steels are used for cutting metals at a muchhigher cutting speed than ordinary carbon tool steels.The high speed steels have the valuable property ofretaining their hardness even when heated to red heat.Most of the high speed steels contain tungsten as thechief alloying element, but other elements like cobalt,y gchromium, vanadium, etc. may be present in someproportion.p p

Contd…For-2014 (IES, GATE & PSUs) Page 67 of 78

With time the effectiveness and efficiency of HSS (tools) and their application range were gradually (too s) a d t e app cat o a ge e e g adua yenhanced by improving its properties and surface condition through ‐

Refinement of microstructureAddition of large amount of cobalt and Vanadium to gincrease hot hardness and wear resistance respectivelyManufacture by powder metallurgical processSurface coating with heat and wear resistive gmaterials like TiC , TiN , etc by Chemical Vapour Deposition (CVD) or Physical Vapour Deposition (PVD)

IES 2013IES‐2013Vanadium in high speed steels: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 d i f A i(d) Has a tendency to promote retention of Austenite

Ans (b)Ans. (b)

IAS‐1997Which of the following processes can be used for Which of the following processes can be used for production thin, hard, heat resistant coating at TiN, on HSS?1. Physical vapour deposition.2. Sintering under reducing atmosphere.. S te g u de educ g at osp e e.3. Chemical vapour deposition with post treatment4. Plasma spraying.4. Plasma spraying.Select the correct answer using the codes given below:Codes:Codes:(a) 1 and 3 (b) 2 and 3(c) 2 and 4 (d) 1 and 4 Ans. (a)(c) 2 and 4 (d) 1 and 4 Ans. (a)

18‐4‐1 High speed steel Thi t l t i 8 t t t tThis steel contains 18 per cent tungsten, 4 per centchromium and 1 per cent vanadium.It is considered to be one of the best of all purpose toolsteels.It is widely used for drills, lathe, planer and shapertools, milling cutters, reamers, broaches, threadingg gdies, punches, etc.

IES‐2003Th t f l t f 8 HSS The correct sequence of elements of 18‐4‐1 HSS tool is( )(a) W, Cr, V (b) Mo, Cr, V(c) Cr, Ni, C(d) Cu Zn Sn(d) Cu, Zn, Sn

Ans. (a)

IES 2007C tti t l t i l 8 HSS h hi h f Cutting tool material 18‐4‐1 HSS has which one of 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)

IES‐1993Th bl d f i d fThe blade of a power saw is made of(a) Boron steel(b) High speed steel(c) Stainless steel(c) Stainless steel(d) Malleable cast iron

A (b)Ans. (b)

Molybdenum high speed steelThi t l t i 6 t t t 6 tThis steel contains 6 per cent tungsten, 6 per centmolybdenum, 4 per cent chromium and 2 per cent

divanadium.It has excellent toughness and cutting ability.The molybdenum high speed steels are better andcheaper than other types of steels.p ypIt is particularly used for drilling and tappingoperations.operations.

Super high speed steelThi t l i l ll d b lt hi h d t lThis steel is also called cobalt high speed steelbecause cobalt is added from 2 to 15 per cent, in ordert i th tti ffi i i ll t hi hto increase the cutting efficiency especially at hightemperatures.This steel contains 20 per cent tungsten, 4 per centchromium, 2 per cent vanadium and 12 per cent cobalt.


IES‐1995Th iti f f th ll t l The compositions of some of the alloy steels are as under:

8 W C V1. 18 W 4 Cr 1 V2. 12 Mo 1 W 4 Cr 1 V3. 6 Mo 6 W 4 Cr 1 V4. 18 W 8 Cr 1 VThe 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)(c) 1 and 4 (d) 1 and 3 Ans. (d)

IES‐2000P t f i ll i l t t Percentage of various alloying elements present in different steel materials are given below:1. 18% W; 4% Cr; 1% V; 5% Co; 0.7% C2. 8% Mo; 4% Cr; 2% V; 6% W; 0.7% C3. 27% Cr; 3% Ni; 5% Mo; 0.25% C4 18% Cr; 8% Ni; 0 15% C4. 18% Cr; 8% Ni; 0.15% CWhich of these relate to that of high speed steel?( ) d (b) d(a) 1 and 3 (b) 1 and 2 (c) 2 and 3 (d) 2 and 4 Ans. (b)

IES‐1992Th i ll i l t i hi h d St l i The main alloying elements in high speed Steel in order of increasing proportion are( )(a) Vanadium, chromium, tungsten(b) Tungsten, titanium, vanadiumg(c) Chromium, titanium, vanadium(d) Tungsten chromium titanium(d) Tungsten, chromium, titanium

Ans. (a)

IAS‐2001Assertion (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

l i f A explanation of A (c) A is true but R is false(d) A is false but R is true Ans (a)(d) A is false but R is true Ans. (a)

IAS 1994Assertion (A): The characteristic feature of High Assertion (A): The characteristic feature of High speed Steel is its red hardness.Reason (R): Chromium and cobalt in High Speed 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 ( ) A i b R i f l(c) A is true but R is false(d) A is false but R is true Ans. (b)

Cast cobalt alloys/StelliteC b l ll b l i h h i bCast cobalt alloys are cobalt‐rich, chromium‐tungsten‐ carboncast alloys having properties and applications in theintermediate range between high‐speed steel and cementedg g pcarbides.Although comparable in room‐temperature hardness to high‐

d l l b l ll l i h i h dspeed steel tools, cast cobalt alloy tools retain their hardness toa much higher temperature. Consequently, they can be used athigher 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 oryheat treated.Cast cobalt alloys contain a primary phase of Co‐rich solidsolution strengthened b Cr and W and dispersion hardened bsolution strengthened by Cr and W and dispersion hardened bycomplex 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 andf h d b dfinished to size by grinding.They are available only in simple shapes, such as single‐

i t t l d bl d b f li it ti i thpoint tools and saw blades, because of limitations in thecasting process and expense involved in the final shaping(grinding). The high cost of fabrication is due primarily to(grinding). The high cost of fabrication is due primarily tothe high hardness of the material in the as‐cast condition.Materials machinable with this tool material include plain‐pcarbon steels, alloy steels, nonferrous alloys, and cast iron.Cast cobalt alloys are currently being phased out forcutting‐tool applications because of increasing costs,shortages of strategic raw materials (Co, W, and Cr), andthe development of other superior tool materials at lowerthe development of other, superior tool materials at lowercost.

IES 2011Stellite is a non ferrous cast alloy composed of: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 vanadiumAns (a)Ans. (a)

Cemented CarbideC bid hi h f ll l ll dCarbides, which are nonferrous alloys, are also called,sintered (or cemented) carbides because they aremanufactured by powder metallurgy techniquesmanufactured by powder metallurgy techniques.Most carbide tools in use today are either straighttungsten carbide (WC) or multicarbides of W‐Ti or W‐g ( )Ti‐Ta, depending on the work material to be machined.Cobalt is the binder.These tool materials are much harder, are chemically morestable, have better hot hardness, high stiffness, and lowerf i i d hi h i d h d HSSfriction, and operate at higher cutting speeds than do HSS.They are more brittle and more expensive and use strategic

t l (W T C ) t i lmetals (W, Ta, Co) more extensively.


Cemented carbide tool materials based on TiC havebeen developed, primarily for auto industrybee de e oped, p a y o auto dust yapplications using predominantly Ni and Mo as abinder. These are used for higher‐speed (> 1000ft/min) finish machining of steels and some malleablecast irons.Cemented carbide tools are available in insert form inmany different shapes; squares, triangles, diamonds,

d dand rounds.Compressive strength is high compared to tensilet th th f th bit ft b d t t lstrength, therefore the bits are often brazed to steelshanks, or used as inserts in holders.Th i t ft h ti k lThese inserts may often have negative rake angles.

Contd…

Speeds up to 300 fpm are common on mild steelsSpeeds up to 300 fpm are common on mild steelsHot hardness properties are very goodC l t d l b i t b d t i t lCoolants and lubricants can be used to increase toollife, but are not required.Special alloys are needed to cut steel

Contd…

IES‐1995The straight grades of cemented carbide cutting tool materials contain(a) Tungsten carbide only(b) Tungsten carbide and titanium carbide(b) Tungsten carbide and titanium carbide(c) Tungsten carbide and cobalt(d) T t bid d b lt bid(d) Tungsten carbide and cobalt carbide

Ans. (c)

S 99IAS – 1994Assertion (A): Cemented carbide tool tips areAssertion (A): Cemented carbide tool tips areproduced by powdermetallurgy.Reason (R): Carbides cannot bemelted and castReason (R): Carbides cannot bemelted and cast.(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 Ans. (a)

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 nonferrous material

d t l lik C t I and non – metals like Cast Iron, Brass etc.

Table below shows detail grouping of cemented carbide tools

ISO Material ProcessApplicationgroup

P01 Steel Steel castings Precision and finish machining high speedP01 Steel, Steel castings Precision and finish machining, high speed

P10 Steel, Steel castings Turning, threading, and milling high speed, small chipssmall chips

P20 Steel, steel castings, malleable cast iron

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 Turning planning low cutting speed large chipP40 Steel and steel casting with sand inclusions

Turning, planning, low cutting speed, large chip section

P50 Steel and steel castings Operations requiring high toughness turning, of medium or low tensile strength

planning, shaping at low cutting speeds

K01 Hard grey C.l., chilled casting, Al. alloys with high silicon

Turning, precision turning and boring, milling, scraping

K10 G C l h d 220 HB T i illi b i i b hiK10 Grey C.l. hardness > 220 HB. Malleable C.l., Al. alloys

containing Si

Turning, milling, boring, reaming, broaching, scraping

K20 Grey C l hardness up to 220 Turning milling broaching requiring highK20 Grey C.l. hardness up to 220 HB

Turning, milling, broaching, requiring high toughness

K30 Soft grey C.l. Low tensile strength steel

Turning, reaming under favourable conditions

K40 Soft non-ferrous metals Turning milling etc.M10 Steel, steel castings,

manganese steel, grey C.l.Turning, milling, medium cutting speed and medium

chip sectionM20 Steel casting austentic steel Turning milling medium cutting speed and mediumM20 Steel casting, austentic steel,

manganese steel, spherodized C.l., Malleable

C.l.

Turning, milling, medium cutting speed and medium chip section

M30 Steel, austenitic steel, spherodized C.l. heat

resisting alloys

Turning, milling, planning, medium cutting speed, medium or large chip section

fM40 Free cutting steel, low tensile strength steel, brass and light

alloy

Turning, profile turning, specially in automatic machines.

IES‐1999M h Li I (ISO l ifi i f bid l ) i h LiMatch List‐I (ISO classification of carbide tools) with List‐II (Applications) and select the correct answer using the codes given below the Lists:gList‐I List‐IIA. P‐10 1. Non‐ferrous, roughing cutg gB. P‐50 2. Non‐ferrous, finishing cutC. K‐10 3. Ferrous material, roughing cutD. 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 2A ( )Ans. (c)


CeramicsC i i ll l i ( ) b d hi hCeramics are essentially alumina ( ) based highrefractory materials introduced specifically for highspeed machining of difficult to machine materials and

2 3Al O

speed machining of difficult to machine materials andcast iron.These can withstand very high temperatures areThese can withstand very high temperatures, arechemically more stable, and have higher wearresistance than the other cutting tool materialsresistance than the other cutting tool materials.In view of their ability to withstand high temperatures,they can be used for machining at very high speeds ofthey can be used for machining at very high speeds ofthe order of 10 m/s.They can be operated at from two to three times the They can be operated at from two to three times the cutting speeds of tungsten carbide.

Contd… Comparison of important properties of ceramic and tungsten carbide tools

It is possible to get mirror finish on cast iron usingi iceramic turning.

The main problems of ceramic tools are their lowstrength, poor thermal characteristics, and thetendency to chipping.They are not suitable for intermittent cutting or for lowcutting speeds.g pVery high hot hardness propertiesOften used as inserts in special holdersOften used as inserts in special holders.

Contd…

Through last few years remarkable improvements instrength and toughness and hence overall performancestrength and toughness and hence overall performanceof ceramic tools could have been possible by severalmeans which include;means which include;

Sinterability, microstructure, strength andtoughness of Al2O3 ceramics were improved totoughness of Al2O3 ceramics were improved tosome extent by adding TiO2 and MgO,T f i h i b ddi iTransformation toughening by adding appropriateamount of partially or fully stabilised zirconia inAl O dAl2O3 powder,Isostatic and hot isostatic pressing (HIP) – these arevery effective but expensive route.

Contd…

Introducing nitride ceramic (Si3N4) with proper sinteringtechnique – this material is very tough but prone to built‐up‐edge formation in machining steelsDeveloping SIALON – deriving beneficial effects of Al2O3

d Sand Si3N4Adding carbide like TiC (5 ~ 15%) in Al2O3 powder – toi t t h d th l d ti itimpart toughness and thermal conductivityReinforcing oxide or nitride ceramics by SiC whiskers, whichenhanced strength toughness and life of the tool and thusenhanced strength, toughness and life of the tool and thusproductivity spectacularly.Toughening Al2O3 ceramic by adding suitable metal likeToughening Al2O3 ceramic by adding suitable metal likesilver which also impart thermal conductivity and selflubricating property; this novel and inexpensive tool is stillg p p y pin experimental stage.

Contd…

Cutting fluid, if applied should in flooding withcopious quantity of fluid to thoroughly wet the entirecopious quantity of fluid, to thoroughly wet the entiremachining zone, since ceramics have very poorthermal shock resistance Else it can be machinedthermal shock resistance. Else, it can be machinedwith no coolant.C i l d f hi i k iCeramic tools are used for machining work pieces,which have high hardness, such as hard castings, caseh d d d h d d lhardened and hardened steel.Typical products can be machined are brake discs,brake drums, cylinder liners and flywheels.

High Performance ceramics (HPC)

Silicon Nitride Alumina toughned by(i) Plain (i) Zirconia(i) Plain (i) Zirconia(ii) SIALON (ii) SiC whiskers(iii) Whisker toughened (iii) Metal (Sil er etc)(iii) Whisker toughened (iii) Metal (Silver etc)

IES 2013IES‐2013Sialon ceramic is used as:Sialon ceramic is used as:

(a) Cutting tool material( ) g

(b) Creep resistant

(c) Furnace linens

(d) High strength

Ans. (a)

IES 2010IES 2010Constituents of ceramics are oxides ofConstituents of ceramics are oxides ofdifferent materials, which are( ) C ld i d t k i ll t(a) Cold mixed to make ceramic pallets(b) Ground, sintered and palleted to make ready( ) , p yceramics(c) Ground washed with acid heated and cooled(c) Ground, washed with acid, heated and cooled(d) Ground, sintered, palleted and after calciningcooled in oxygen

Ans (b)Ans. (b)


IAS‐1996Match List I with List II and select the correct answer Match List I with List II and select the correct answer using the codes given below the lists:List I (Cutting tools) List II (Major constituent)( g ) ( j )A. Stellite l. TungstenB. H.S.S. 2. CobaltC. Ceramic 3. AluminaD. DCON 4. Columbium

Ti i5. TitaniumCodes: A B C D A B C D(a) 3 4 (b) 2 4 3(a) 5 1 3 4 (b) 2 1 4 3(c) 2 1 3 4 (d) 2 5 3 4

Ans (c)Ans. (c)

IES‐1997A i (A) C i l d l f li h Assertion (A): Ceramic tools are used only for light, smooth and continuous cuts at high speeds.Reason (R): Ceramics have a high wear resistance and Reason (R): Ceramics have a high wear resistance and high temperature resistance.(a) Both A and R are individually true and R is the (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 (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‐1996A 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

h h f h f ll lmm/rev. Which one of the following tool materials will be suitable for machining the

d h f dcomponent under the specified cutting conditions?(a) Sintered carbides (b) Ceramic(c) HSS (d) Diamond( ) ( )

Ans. (b)

IES 2007Which one of the following is not a ceramic?(a) Alumina(b) Porcelain(c) Whisker(c) Whisker(d) Pyrosil

(d)Ans. (d)

IAS‐2000C id h f ll i i l i l d f Consider the following cutting tool materials used for metal‐cutting operation at high speed:high speed:1. Tungsten carbide 2 Cemented titanium carbide2. Cemented titanium carbide3. High‐speed steel

C i4. CeramicThe correct sequence in increasing order of the range of cutting speeds for optimum use of these materials iscutting speeds for optimum use of these materials is(a) 3,1,4,2 (b) 1,3,2,4(c) 3 1 2 4 (d) 1 3 4 2 Ans (c)(c) 3,1,2,4 (d) 1,3,4,2 Ans. (c)

IAS‐2003At room temperature, which one of the following is the correct sequence of increasing hardness of the tool materials?(a) Cast alloy‐HSS‐Ceramic‐Carbidey(b) HH‐Cast alloy‐Ceramic‐Carbide(c) HSS‐Cast alloy‐Carbide‐Ceramic(c) HSS‐Cast alloy‐Carbide‐Ceramic(d) Cast alloy‐HSS‐Carbide‐Ceramic

( )Ans. (c)

Coated Carbide Toolsd l b h h l kCoated tools are becoming the norm in the metalworking

industry because coating , can consistently improve, toollif %life 200 or 300% or more.In cutting tools, material requirements at the surface of the

l d b b i i h d d h i lltool need to be abrasion resistant, hard, and chemicallyinert to prevent the tool and the work material frominteracting chemically with each other during cuttinginteracting chemically with each other during cutting.A thin, chemically stable, hard refractory coating of TiC,TiN or accomplishes this objectiveAl OTiN, or accomplishes this objective.The bulk of the tool is a tough, shock‐resistant carbide that

ith t d hi h t t l ti d f ti d

2 3Al O

can withstand high‐temperature plastic deformation andresist breakage.

Contd…

The coatings must be fine grained, & free of bindersand porosityand porosity.Naturally, the coatings must be metallurgically bonded

h bto the substrate.Interface coatings are graded to match the propertiesof the coating and the substrate.The coatings must be thick enough to prolong tool lifeg g p gbut thin enough to prevent brittleness.Coatings should have a low coefficient of friction soCoatings should have a low coefficient of friction sothat the chips do not adhere to the rake face.Multiple coatings are used with each layer impartingMultiple coatings are used, with each layer impartingits own characteristic to the tool.

Contd…

The most successful combinations areTiN/TiC/TiCN/TiN and TiN/TiC/ .Chemical vapour deposition (CVD) is the technique

2 3Al Op p ( ) q

used to coat carbides.


IAS‐1999The coating materials for coated carbide tools, includes(a) TiC, TiN and NaCN (b) TiC and TiN(c) TiN and NaCN (d) TiC and NaCN(c) TiN and NaCN (d) TiC and NaCN

A (b)Ans. (b)

TiN‐Coated High‐Speed Steel( )Coated high‐speed steel (HSS) does not routinely

provide as dramatic improvements in cutting speeds asdo coated carbides, with increases of 10 to 20% beingtypical.In addition to hobs, gear‐shaper cutters, and drills,HSS tooling coated by TiN now includes reamers, taps,gchasers, spade‐drill blades, broaches, bandsaw andcircular saw blades, insert tooling, form tools, endmills, and an assortment of other milling cutters.

Contd…

Physical vapour deposition (PVD) has proved to be thebest process for coating HSS primarily because it is abest process for coating HSS, primarily because it is arelatively low temperature process that does notexceed the tempering point of HSSexceed the tempering point of HSS.Therefore, no subsequent heat treatment of the

i l i i dcutting tool is required.The advantage of TiN‐coated HSS tooling is reducedtool wear.Less tool wear results in less stock removal during toolgregrinding, thus allowing individual tools to bereground more times.g

CermetsCermetsThese sintered hard inserts are made by combining ‘cer’ from

i lik TiC TiN TiCN d ‘ ’ f l (bi d )ceramics like TiC, TiN or TiCN and ‘met’ from metal (binder)like Ni, Ni‐Co, Fe etc.H d h i ll t bl d h i t tHarder, more chemically stable and hence more wear resistantMore brittle and less thermal shock resistantWt% f bi d t l i f t %Wt% of binder metal varies from 10 to 20%.Cutting edge sharpness is retained unlike in coated carbideinsertsinsertsCan machine steels at higher cutting velocity than that used fortungsten carbide even coated carbides in case of light cutstungsten carbide, even coated carbides in case of light cuts.Modern cermets with rounded cutting edges are suitable forfinishing and semi‐finishing of steels at higher speeds, stainlessfinishing and semi finishing of steels at higher speeds, stainlesssteels but are not suitable for jerky interrupted machining andmachining of aluminium and similar materials.

IES 2010IES 2010The cutting tool material required toThe cutting tool material required tosustain high temperature is(a) High carbon steel alloys(b) Composite of lead and steel(b) Composite of lead and steel(c) Cermet(c) Cermet(d) Alloy of steel, zinc and tungsten

Ans. (c)

IES‐2000Cermets are(a) Metals for high temperature use with ceramic like g p

properties(b) Ceramics with metallic strength and luster(b) Ceramics with metallic strength and luster(c) Coated tool materials(d) M t l i it(d) Metal‐ceramic composites

Ans. (d)

S 2003IES – 2003The correct sequence of cutting tools in theThe correct sequence of cutting tools in theascending order of their wear resistance is(a) HSS Cast non ferrous alloy (Stellite) Carbide(a) HSS‐Cast non‐ferrous alloy (Stellite)‐Carbide‐Nitride(b) C t f ll (St llit ) HSS C bid(b) Cast non‐ferrous alloy (Stellite)‐HSS‐Carbide‐Nitride(c) HSS‐Cast non‐ferrous alloy (Stellite)‐Nitride‐Carbide(d) Cast non‐ferrous alloy (Stellite)‐Carbide‐Nitride‐HSS Ans. (a)

DiamondsDiamondsDiamond is the hardest of all the cutting tool materials.Di d h h f ll i iDiamond has the following properties:

extreme hardness,l h llow thermal expansion,high heat conductivity, anda very low co‐efficient of friction.

This is used when good surface finish and dimensional accuracyd i dare desired.

The work‐materials on which diamonds are successfully employedare the non ferrous one such as copper brass zinc aluminiumare the non‐ferrous one, such as copper, brass, zinc, aluminiumand magnesium alloys.On ferrous materials diamonds are not suitable because of theOn ferrous materials, diamonds are not suitable because of thediffusion of carbon atoms from diamond to the work‐piecematerial. Contd…For-2014 (IES, GATE & PSUs) Page 73 of 78

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(c) poor tool toughness

(d) chemical affinity of tool material with iron(d) chemical affinity of tool material with iron

Diamond tools have the applications in single point turning andboring tools, milling cutters, reamers, grinding wheels, honingg , g , , g g , gtools, lapping powder and for grinding wheel dressing.Due to their brittle nature, the diamond tools have poorresistance to shock and so, should be loaded lightly.Polycrystalline diamond (PCD) tools consist of a thin layer (0.5to 1.5 mm) of'fine grain‐ size diamond particles sinteredtogether and metallurgically bonded to a cemented carbidesubstratesubstrate.The main advantages of sintered polycrystalline tools overnatural single‐crystal tools are better quality greater toughnessnatural single crystal tools are better quality, greater toughness,and improved wear resistance, resulting from the randomorientation of the diamond grains and the lack of large cleavageg g gplanes.

Contd…

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 integritysurface integrity.Positive rake tooling is recommended for the vast majority of diamond tooling applicationsmajority of diamond tooling applications.If BUE is a problem, increasing cutting speed and the

f i i k l li i i use of more positive rake angles may eliminate it. 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.

IES‐1995A i (A) N f i l b Assertion (A): Non‐ferrous materials are best machined with diamond tools. Reason (R): Diamond tools are suitable for high speed Reason (R): Diamond tools are suitable for high speed machining.(a) Both A and R are individually true and R is the (a) Both A and R are individually true and R is the



Ans. (b)( )

IES‐2001A i (A) Di d l b d hi h Assertion (A): Diamond tools can be used at high speeds.Reason (R): Diamond tools have very low coefficient Reason (R): Diamond tools have very low coefficient of friction.(a) Both A and R are individually true and R is the (a) Both A and R are individually true and R is the



Ans. (b)( )

S 999IES – 1999Consider the following statements:Consider the following statements:For precision machining of non‐ferrous alloys, diamond is preferred because it hasis preferred because it has1. Low coefficient of thermal expansion 2. High wear resistance3. High compression strength3 g p g4. Low fracture toughnessWhich of these statements are correct?Which of these statements are correct?(a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 Ans. (a)

IES‐1992Which of the following given the correct order of increasing hot hardness of cutting tool material?(a) Diamond, Carbide, HSS(b) Carbide, Diamond, HSS(b) Carbide, Diamond, HSS(c) HSS, carbide, Diamond(d) HSS Di d C bid(d) HSS, Diamond, Carbide

Ans. (d)

S 999IAS – 1999Assertion (A): During cutting, the diamond tool is Assertion (A): During cutting, the diamond tool is kept flooded with coolant.Reason (R): The oxidation of diamond starts at Reason (R): The oxidation of diamond starts at about 4500C( ) 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 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. (a)(d) A is false but R is true Ans. (a)

Cubic boron nitride/BorazonNext to diamond, cubic boron nitride is the hardestmaterial presently available.It is made by bonding a 0.5 – 1 mm layer ofpolycrystalline cubic boron nitride to cobalt basedp y ycarbide substrate at very high temperature andpressure.It remains inert and retains high hardness and fracturetoughness at elevated machining speeds.oug ess a e e a ed ac g speeds.It shows excellent performance in grinding anymaterial of high hardness and strengthmaterial of high hardness and strength.


The operative speed range for cBN when machininggrey cast iron is 300 ~400 m/mingrey cast iron is 300 400 m/minSpeed ranges for other materials are as follows:

H d t i ( BHN) 8 / iHard cast iron (> 400 BHN) : 80 – 300 m/minSuperalloys (> 35 RC) : 80 – 140 m/minHardened steels (> 45 RC) : 100 – 300 m/min

It is best to use cBN tools with a honed or chamferedt s best to use c too s t a o ed o c a e ededge preparation, especially for interrupted cuts. Likeceramics, cBN tools are also available only in the form, yof indexable inserts.The only limitation of it is its high costThe only limitation of it is its high cost.

Contd…

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 ymachine these difficult‐to‐machine materials at higher speeds (fivefold) and with a higher removal rate g(fivefold) than cemented carbide, and with superior accuracy, finish, and surface integrity.

IES‐1994Consider the following tool materials:1. Carbide 2. Cermet3. Ceramic 4. Borazon.Correct sequence of these tool materials in increasing Correct sequence of these tool materials in increasing order of their ability to retain their hot hardness is( ) (b)(a) 1,2,3,4 (b) 1,2,4,3(c) 2, 1, 3, 4 (d) 2, 1, 4, 3

Ans. (a)

IES‐2002Which one of the following is the hardest cutting tool material next only to diamond?(a) Cemented carbides(b) Ceramics(b) Ceramics(c) Silicon (d) C bi b it id(d) Cubic boron nitride

Ans. (d)

IES‐1996Cubic 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 (b) Has a hardness which is slightly more than that of

HSS(c) Is used for making cylinder blocks of aircraft (c) Is used for making cylinder blocks of aircraft

engines(d) I d f ki ti l l(d) Is used for making optical glasses.

Ans. (a)

IES‐1994Cubic boron nitride is used(a) As lining material in induction furnaceg(b) For making optical quality glass.(c) For heat treatment(c) For heat treatment(d) For none of the above.

(d)Ans. (d)

IAS‐1998Which of the following tool materials have cobalt as a constituent element?1. Cemented carbide 2. CBN3. Stellite 4. UCON3. Stellite 4. UCONSelect the correct answer using the codes given below:C dCodes:(a) 1 and 2 (b) 1 and 3(c) 1 and 4 (d) 2 and 3

Ans. (b)( )

CoroniteCoroniteCoronite is made basically by combining HSS for strength and

h d bid f h d itoughness and tungsten carbides for heat and wear resistance.Microfine TiCN particles are uniformly dispersed into the matrix.

l k l d b d h b d l d f hUnlike a solid carbide, the coronite based tool is made of three layers;

th t l HSS i t l the central HSS or spring steel corea layer of coronite of thickness around 15% of the tool diameterdiametera thin (2 to 5 μm) PVD coating of TiCN

The coronite tools made b hot e trusion follo ed b PVDThe coronite tools made by hot extrusion followed by PVD‐coating of TiN or TiCN outperformed HSS tools in respect ofcutting forces tool life and surface finishcutting forces, tool life and surface finish.

IES‐1993M t h Li t I ith Li t IT d l t th t i th Match List I with List IT and select the correct answer using the codes given below the lists:List ‐ I (Cutting tool Material) List ‐ I I(Major

h t i ti tit t)characteristic constituent)A. High speed steel 1. CarbonB. Stellite 2. MolybdenumyC. Diamond 3. NitrideD. Coated carbide tool 4. Columbium

5 Cobalt5. CobaltCodes: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)


IES‐2003Which one of the following is not a synthetic abrasive material?(a) Silicon Carbide (b) Aluminium Oxide(c) Titanium Nitride (d) Cubic Boron Nitride(c) Titanium Nitride (d) Cubic Boron Nitride

Ans. (b)

IES‐2000Consider the following tool materials:1. HSS 2. Cemented carbide 3. Ceramics 4. DiamondThe correct sequence of these materials in decreasing The correct sequence of these materials in decreasing order of their cutting speed is( ) (b) (a) 4, 3, 1, 2 (b) 4, 3, 2, 1(c) 3, 4, 2, 1 (d) 3, 4, 1, 2

Ans. (b)

IES‐1999M t h Li t I ith Li t II d l t th t Match List‐I with List‐II and select the correct answer using the codes given below the Lists:List I List II(Materials) (Applications)A. Tungsten carbide 1. Abrasive wheelsB Sili i id H i lB. Silicon nitride 2. Heating elementsC. Aluminium oxide 3. Pipes for conveying

liquid metalsqD. 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)

IAS‐2001M t h Li t I (C tti t l t i l ) ith Li t II Match. List I (Cutting tool materials) with List II (Manufacturing methods) and select the correct answer using the codes given below the Lists:List I List IIA. HSS 1. CastingB Stellite 2 ForgingB. Stellite 2. ForgingC. Cemented carbide 3. RollingD. UCON 4. ExtrusionUCO 4

5. Powder metallurgyCodes: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)

Attrition wearTh t b di b t th hi d t l t i l tThe strong bonding between the chip and tool material athigh temperature is conducive for adhesive wear.The adhesive wear in the rough region is called attritionThe adhesive wear in the rough region is called attritionwear .In the rough region, some parts of the worn surface are still

d b l hi d h i l i icovered by molten chip and the irregular attrition wearoccurs in this region .The irregular attrition wear is due to the intermittentThe irregular attrition wear is due to the intermittentadhesion during interrupted cutting which makes aperiodic attachment and detachment of the work materialon the tool surfaceon the tool surface.Therefore, when the seizure between workpiece to tool isbroken, the small fragments of tool material are pluckedbroken, the small fragments of tool material are pluckedand brought away by the chip.

IES‐1996Th li it t th i h d f k The limit to the maximum hardness of a work material which can be machined with HSS tools

t l d i t b hi h f th even at low speeds is set by which one of the following tool failure mechanisms?( )(a) Attrition(b) Abrasion(c) Diffusion(d) Plastic deformation under compression(d) Plastic deformation under compression.

Ans. (a)

IES‐2005C id th f ll i t t t A i i th Consider the following statements: An increase in the cobalt content in the straight carbide grades of carbide toolscarbide tools1. Increases the hardness.

D h h d2. Decreases the hardness.3. Increases the transverse rupture strength4. 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)(c) 1 and 4 (d) 2 and 3 Ans. (d)


‐ C C C ‐ D B C C C B B B A BA A D D D B B B B D D C A D AB C B C C D B D ‐ B C C A D B

B C&D B ‐ 1597, 0, 1454, 1265

429, 127

F291 N457 Fn336 Fs408 β32.49

12 0.816 74

B A D A ‐ B D

B B A D B A D D B D D B D B AB D ‐ B B 828,

1200, 231, 4.021

‐ A b A D A C B C

A B C A A C B B B B C C C C BC A B B C C ‐ ‐ B D B A A ‐ ‐D A B C B A ‐ A C A C A B D C

B D C A A B A A A A 36.67 2.3, 10.78, 21.42, 35.85

A ‐ B

B C D 0.07, 54 A C C D A 26 195 ‐ C C D

A B C D A D C A D A C B A A C

A B D C D B D D None C B ‐ D D C‐ A A C B D A B C ‐ ‐ B B D BA C A A D B A B ‐ B B C A C A

D A C D A C B D C A A B C B DC D C A C D C D C(11) A B A D B CC A B ‐ A D C D A D B C A ‐ ‐

Gutter C A A2, B6 C5, D3

A3, B2, C1, D5

C C D A A B C A B A

‐ ‐ ‐ B D C B D D A C C B B CC C D C B A C A 2134 B B D B C C

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PRODUCTION ANSWERS‐2014 For any doubt send SMS to 9582314327


539, 467

2.04 130

3.93 3.63

‐ B ‐ ‐ A B A D B B C A

Centre ‐ ‐ ‐ ‐ B B B A C B C D D A‐ ‐ ‐ 44,25,

361,231C D B A B C C C C C B

A D D A C B B B B ‐ B C B * BC D D A D D Extrusion ‐ ‐ C 4.2 50,33,4 A A AB C A A C D ‐ ‐ 74, 28 A B C A C C

B D A C A A A C ‐ C B C ‐ B CB C C B A C ‐ A B D C D C B BA A D C B D C A D A A D C ‐ B

A ‐ C ‐ ‐ B C D C B C A D B DD A B B ‐ B D C B B C A C B DD C D ‐ D D D A A B C A C B C

B ‐ ‐ C C C A C D B C C B D DB B D A ‐ A D D B ‐ A C B ‐ ‐A C C C C ‐ B B C A C D B D ‐

‐ ‐ D A C A ‐ C A D C B ‐ ‐ CD C B ‐ C C A C ‐ B C A C B DB C ‐ D ‐ C B ‐ D B ‐ A B B B

A ‐ ‐‐







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production engg. question set with answers

Education

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true page

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