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IES Officer (Railway); GATE Percentile 99.96 IES Officer (Railway); GATE Percentile 99.96 Teaching Experience (6 Years); NTPC Ltd (5 Years Teaching Experience (6 Years); NTPC Ltd (5 Years) E-mail: mail: [email protected] [email protected]
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Page 1: Production Q a 2013 S K Mondal Mobile Version

IES Officer (Railway); GATE Percentile 99.96IES Officer (Railway); GATE Percentile 99.96

Teaching Experience (6 Years); NTPC Ltd (5 YearsTeaching Experience (6 Years); NTPC Ltd (5 Years))EE--mail: mail: [email protected][email protected]

Page 2: Production Q a 2013 S K Mondal Mobile Version

Theory of Metal Cutting

By  S K Mondal

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IAS 2009 mainIAS 2009 mainName four independent variables and threedependent variables in metal cuttingdependent variables in metal cutting.

[ 5 marks]

d d bl d blIndependent Variables Dependent Variables

•Starting materials  •Force or power requirementsg

(tool/work)

l

p q

•Maximum temperature in 

•Tool geometry

•Cutting Velocity

cutting

•Surface finish

•Lubrication

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

Ans. (c)

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IES 1995IES‐1995

Single point thread cutting tool should ideally have:

a) Zero rakeb) Positive rake)c) Negative raked) Normal raked) Normal rakeAns. (a) 

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GATE 1995 2008GATE‐1995; 2008Cutting power consumption in turning can be g p p g

significantly reduced by                                                   

( )  I i   k   l   f  h   l (a)  Increasing rake angle of the tool 

(b)  Increasing the cutting angles of the tool( ) g g g

(c)  Widening the nose radius of the tool   

(d)  Increasing the clearance angle

A  ( )Ans. (a)

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IES 1993IES‐1993Assertion (A): For a negative rake tool, the specific

i i ll h f i i kcutting pressure is smaller than for a positive raketool under otherwise identical conditions.R (R) Th h t i d b th hiReason (R): The shear strain undergone by the chipin the case of negative rake tool is larger.(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) 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 Ans. (d)

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IES 2005IES – 2005Assertion (A): Carbide tips are generally given

i k lnegative rake angle.Reason (R): Carbide tips are made from very hard

t i lmaterials.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (b)(d) A is false but R is true Ans. (b)

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IES 2002IES – 2002Assertion (A): Negative rake is usually provided on

bid i d lcarbide tipped tools.Reason (R): Carbide tools are weaker in

icompression.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (c)(d) A is false but R is true Ans. (c)

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IES 2011IES 2011Which one of the following statement is NOT correctwith reference to the purposes and effects of rake angleof a cutting tool?( ) d h h fl d(a) To guide the chip flow direction(b) To reduce the friction between the tool flanks andh hi d fthe machined surface(c) To add keenness or sharpness to the cutting edges.(d) d b h l ff(d) To provide better thermal efficiency.Ans. (b)

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IES ‐ 2012S 0Statement (I): Negative rake angles are preferred on rigid set‐ups for interrupted cutting and difficult‐to machine materialsmaterials.Statement (II):Negative rake angle directs the chips on to the machined surfacethe machined surface(a) Both Statement (I) and Statement (II) areindividually true and Statement (II) is the correct

( )explanation of Statement (I)(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is not the correctindividually true but Statement (II) is not the correctexplanation of Statement (I) Ans. (b)(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

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GATE 2008 (PI)GATE – 2008 (PI)Brittle materials are machined with tools having zero orBrittle materials are machined with tools having zero or

negative rake angle because it

(a) results in lower cutting force

(b) improves surface finish(b) improves surface finish

(c) provides adequate strength to cutting tool

(d) results in more accurate dimensions

Ans (c)Ans. (c)

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IAS 1994IAS – 1994Consider the following characteristics1. The cutting edge is normal to the cutting velocity.2. 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 3Ans. (d)

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IES 2006IES‐2006Which of the following is a single point cutting Which of the following is a single point cutting tool?(a) Hacksaw blade( )(b) Milling cutter(c) Grinding wheel(c) Grinding wheel(d) Parting toolAns  (d)Ans. (d)

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IES 1995IES‐1995The angle between the face and the flank of thegsingle point cutting tool is known as

a) Rake angleb) Clearance anglec) Lip angle) p gd) Point angle.Ans. (c)( )

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S 2006IES‐2006Assertion (A): For drilling cast iron, the tool isprovided with a point angle smaller than thatrequired for a ductile material.

( ) ll l l lReason (R): Smaller point angle results in lowerrake angle.( ) B h A d R i di id ll d R i h(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(c) A is true but R is false(d) A is false but R is true Ans. (c)

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IES 2002IES‐2002Consider the following statements:The strength of a single point cutting tool dependsupon1. Rake angle2. Clearance angle3. Lip angleWhich of these statements are correct?(a) 1 and 3 (b) 2 and 3(c) 1 and 2 (d) 1, 2 and 3( ) ( ) , 3Ans. (d)

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IES ‐ 2012IES ‐ 2012Tool life increase with increase in(a) Cutting speed (b) Nose radius (c) Feed (d) Depth of cutAns. (b)

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IES 2009IES‐2009Consider the following statements with respectto the effects of a large nose radius on the tool:1. It deteriorates surface finish.2. It increases the possibility of chatter.3. It improves tool life.Which of the above statements is/are correct?(a) 2 only (b) 3 onlyy y(c) 2 and 3 only (d) 1, 2 and 3Ans. (c)( )

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IES 1995IES‐1995Consider the following statements about nose

diradius1. It improves tool life2. It reduces the cutting force3. It improves the surface finish.Select the correct answer using the codes given below:(a) 1 and 2 (b) 2 and 3(c) 1 and 3 (d) 1, 2 and 3Ans. (c)

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IES‐1994IES‐1994Tool geometry of a single point cutting tool is specified bythe following elements:the following elements:1. Back rake angle2. Side rake angle3. End cutting edge angle4. Side cutting edge angle

d l f l5. Side relief angle6. End relief angle7 Nose radius7. Nose radiusThe correct sequence of these tool elements used forcorrectly specifying the tool geometry isy p y g g y(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 Ans. (b)

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IES 2009IES‐2009The following tool signature is specified for a single‐point cutting tool in American system:10, 12, 8, 6, 15, 20, 3What does the angle 12 represent?(a) Side cutting‐edge angle(b) Side rake angle(c) Back rake angleg(d) Side clearance angleAns. (b)( )

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

Ans. (b)

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ISRO‐2011A cutting tool having tool signature as 10, 9, 6, 6,

8 8 2 will have side rake angle8, 8, 2 will have side rake angle

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

Ans. (b)

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GATE 2008GATE‐2008In a single point turning tool, the side rake angle

d h l k l l Φ i hand orthogonal rake angle are equal. Φ is theprincipal cutting edge angle and its range is

Th hi fl i th th l l0 90o oφ≤ ≤ . The chip flows in the orthogonal plane.The value of Φ is closest to(a) 00 (b) 450

0 90o oφ≤ ≤

(a) 00 (b) 450

(c) 600 (d) 900

A (d) i i l tti d l iAns. (d) principal cutting edge angle is = 90 ‐cs

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IAS ‐ 2009 Main

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GATE 2001GATE‐2001During orthogonal cutting of mild steel withg g ga 10° rake angle tool, the chip thickness ratiowas obtained as 0.4. The shear angle (in4 g (degrees) evaluated from this data is(a) 6 53  (b) 20 22 (a) 6.53  (b) 20.22 (c) 22.94  (d) 50.00    A  ( )Ans. (c)

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GATE 2011GATE 2011A 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 chipthi k d th l hi i diti ithickness under orthogonal machining condition is1.8 mm. The shear angle is approximately(a) 22°(a) 22(b) 26°( ) 6°(c) 56°(d) 76°A (b)Ans. (b)

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IES 1994IES‐1994The following parameters determine the

d l f i hi f imodel of continuous chip formation:1. True feed2. Cutting velocity3 Chip thickness3. Chip thickness4. Rake angle of the cutting tool.Th hi h h l f hThe parameters which govern the value of shearangle would include(a) 1,2 and 3 (b) 1,3 and 4(c) 1,2 and 4 (d) 2,3 and 4 Ans. (b)

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IES 2009Minimum shear strain in orthogonal

IES ‐ 2009g

turning with a cutting tool of zerorake angle is

( )(a) 0.0(b) 0.5(c) 1.0(d) 2.0Ans. (d)

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I hi i i hi

IES ‐ 2004In a machining operation chipthickness ratio is 0.3 and the rakeangle of the tool is 10°. What is thevalue of the shear strain?(a) 0.31 (b) 0.13( ) (d)(c) 3.00 (d) 3.34Ans. (d)( )

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GATE ‐2012GATE ‐2012Details pertaining to an orthogonal metal cuttingprocess are given belowprocess are given below.

Chip thickness ratio 0.4U d f d thi k 6Undeformed thickness 0.6 mmRake angle +10°C i d /Cutting speed 2.5 m/sMean thickness of primary shear zone 25 microns

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

(c) 1.0104×105 (d) 4.397×105 Ans. (c)

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GATE 2007GATE‐2007In orthogonal turning of a low carbon steel barof diameter 150 mm with uncoated carbidetool, the cutting velocity is 90 m/min. The feedi / d th d th f t iis 0.24 mm/rev and the depth of cut is 2 mm.The chip thickness obtained is 0.48 mm. If theorthogonal rake angle is zero and the principalorthogonal rake angle is zero and the principalcutting edge angle is 90°, the shear angle isdegree isg(a) 20.56 (b) 26.56(c) 30.56 (d) 36.56 Ans. (b)( ) 3 5 ( ) 3 5 ( )

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IES 2004IES‐2004Consider the following statements with respect to h   li f  l   f  i   l                                             the relief angle of cutting tool:                                            1.  This affects the direction of chip flow2.  This reduces excessive friction between the tool and work piece  Thi   ff   l lif3.  This affects tool life

4.  This allows better access of coolant to the tool k  i  i t fwork piece interface

Which of the statements given above are correct?( )   d  (b)   d (a) 1 and 2 (b) 2 and 3(c) 2 and 4 (d) 3 and 4 Ans. (b) 

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IES 2006IES‐2006Consider the following statements:1. A large rake angle means lower strength of the cutting edge.2. Cutting torque decreases with rake angle.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

Ans. (c) 

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IES 2004IES‐2004Match. List I with List II and select the correct answer 

i  th   d   i  b l  th  Li tusing the codes given below the Lists:List I List II

A Plan approach angle 1 Tool faceA. Plan approach angle 1. Tool faceB. Rake angle 2. Tool flankC Clearance angle 3 Tool face and flankC. Clearance angle 3. Tool face and flankD. Wedge angle 4. Cutting edge

5 Tool nose      [Ans  (c)]5. Tool nose      [Ans. (c)]A  B  C D  A B C D

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

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IES 2003IES‐2003The angle of inclination of the rake face with

h l b d i lrespect to the tool base measured in a planeperpendicular to the base and parallel to the widthof the tool is calledof the tool is called(a) Back rake angle(b) Side rake angle(b) Side rake angle(c) Side cutting edge angle(d) E d tti d l(d) End cutting edge angleAns. (b)

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IES 2004 ISRO 2009IES‐2004, ISRO‐2009The rake angle of a cutting tool is 15°, shearangle 45° and cutting velocity 35 m/min.What is the velocity of chip along the toolface?(a) 28.5 m/min (b) 27.3 m/min( ) 5 ( ) 7 3(c) 25.3 m/min (d) 23.5 m/minAns (a)Ans. (a)

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IES 2008IES‐2008Consider 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 mm   Which of the following are correct?mm.  Which of the following are correct?1. Chip velocity will be 45 m/min.2. Chip velocity will be 80 m/min.2. Chip velocity will be 80 m/min.3. Chip thickness will be 1∙8 mm.4. Chip thickness will be 3∙2 mm.4. Chip thickness will be 3 2 mm.Select the correct answer using the code given below:(a) 1 and 3 (b) 1 and 4( ) 3 ( ) 4(c) 2 and 3 (d) 2 and 4 Ans. (b)

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IES 2001IES‐2001If α is the rake angle of the cutting tool, is theφshear angle and V is the cutting velocity, then thevelocity of chip sliding along the shear plane isgiven by

(a) (b)coscos( )V α

φ α− ( )sin

cos −V φ

φ α

(c) (d) Ans. (a)cossin( )−V α

φ αsin

sin( )V α

φ α−

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IES 2003IES‐2003An orthogonal cutting operation is beingcarried out under the following conditions:cutting 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(c) 1.0 m/s (d) 1.66 m/s Ans. (d)(c) 1.0 m/s (d) 1.66 m/s Ans. (d)

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

Ans  (a)Ans. (a)

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IAS 2002IAS‐2002

A  ( )Ans. (a)

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IAS 2000IAS‐2000

Ans  (d)Ans. (d)

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IAS 1998IAS‐1998The cutting velocity in m/sec, for turning a work piecef di t t th i dl d f 8 RPM iof diameter 100 mm at the spindle speed of 480 RPM is

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

Ans (b)Ans. (b)

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IAS 1995IAS‐1995In an orthogonal cutting, the depth of cut is halved andth f d t i d bl If th hi thi k ti ithe feed rate is double. If the chip thickness ratio isunaffected with the changed cutting conditions, theactual chip thickness will bep(a) Doubled (b) halved(c) Quadrupled (d) Unchanged.( ) Q p ( ) g

Ans. (b)

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GATE 2009 (PI) Common Data S 1GATE – 2009 (PI) Common Data S‐1An 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 chip15 . The chip thickness is 0.4 mm and the uncut chip

thickness is 0.2 mm.

The shear plane angle (in degrees) is

( ) 6 8 (b) 8 ( ) 8 8 (d) 8(a) 26.8 (b) 27.8 (c) 28.8 (d) 29.8

Ans. (c)( )

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GATE 2009 (PI) Common Data S 2GATE – 2009 (PI) Common Data S‐2An 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 chip15 . The chip thickness is 0.4 mm and the uncut chip

thickness is 0.2 mm.

The chip velocity (in m/min) is

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

Ans. (b)( )

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GATE 1995GATE‐1995Plain milling of mild steel plate produces Plain milling of mild steel plate produces (a) Irregular shaped discontinuous chips(b) Regular shaped discontinuous chip(b) Regular shaped discontinuous chip(c) Continuous chips without built up edge(d) J i d  hi(d) Joined chips

A  (b)Ans. (b)

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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 of chip as in the case of turning on a lathe. Which of the following are correct?Continuous ribbon like chip is formed when turningp g1. At a higher cutting speed2. At a lower cutting speedg p3. A brittle material4. A ductile materialSelect the correct answer using the code given below:(a) 1 and 3 (b) 1 and 4(c) 2 and 3 (d) 2 and 4 Ans. (b)

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

Ans  (a)Ans. (a)

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GATE 2002GATE‐2002A built‐up‐edge is formed while machining              

(a) Ductile materials at high speed

(b) Ductile materials at low speed

(c) Brittle materials at high speed(c) tt e ate a s at g speed

(d) Brittle materials at low speed

Ans  (b)Ans. (b)

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IES 1997IES‐1997Assertion (A): For high speed turning of cast ironi t bid t l bit id d ith hipistons, carbide tool bits are provided with chip

breakers.Reason (R): High speed turning may produce long,ibb i hi hi h b b kribbon type continuous chips which must be brokeninto small lengths which otherwise would bedifficult to handle and may prove hazardous.(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 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(d) A is false but R is true Ans (d)(d) A is false but R is true Ans. (d)

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Ch 1 M h i f B i M hi i O tiCh‐1: Mechanics of Basic Machining OperationQ. No Option Q. No Option

1 C 11 D

2 B 12 D

3 D 13 B

4 C 14 C

5 B 15 D

6 D 16 B

7 B 17 B

8 A 18 D

9 B 19 D

10 B 20 B

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Force & Power in Metal Cuttingg

By  S K Mondal

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ESE 2000 (C ti l)ESE ‐2000 (Conventional)The following data from the orthogonal cutting testg g gis available. Rake angle = 100, chip thickness ratio =0.35, uncut chip thickness = 0.51 mm, width of cut =

i ld h f k i l3 mm, yield shear stress of work material = 285N/mm2, mean friction co‐efficient on tool force =0 65 Determine0.65, Determine

(i) Cutting force (Fc)(ii) Radial force(ii) Radial force(iii) Normal force (N) on tool and(i ) Sh f h l (F )(iv) Shear force on the tool (Fs )

Ans. 1597 N, 0 N, 1453.8 N, 1265 N

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ESE 2005 C ti lESE‐2005 ConventionalMild steel is being machined at a cuttingg gspeed 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 averagevalue of co‐efficient of friction between thetool and the chip is 0.5 and the shear stress ofthework material is 400 N/mm2,4 / ,

Determine (i) shear angle and(ii)Cutting and thrust component of the(ii)Cutting and thrust component of theforce. [Ans. 429 N , 127 N]

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GATE ‐2008 (PI) Linked S‐1GATE  2008 (PI) Linked S 1In an orthogonal cutting experiment, an HSS tool having

the following tool signature in the orthogonal reference

system (ORS) has been used: 0‐10‐7‐7‐10‐75‐1. Given

width of cut = 3.6 mm; shear strength of workpiece

material = 460 N/mm2; depth of cut = 0 25 mm;material = 460 N/mm2; depth of cut = 0.25 mm;

coefficient of friction at tool‐chip interface = 0.7.

Shear plane angle (in degree) for minimum cutting force

isis

(a) 20.5 (b) 24.5 (c) 28.5 (d) 32.5 [Ans. (d)]

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GATE ‐2008 (PI) Linked S‐2GATE  2008 (PI) Linked S 2In an orthogonal cutting experiment, an HSS tool having

the following tool signature in the orthogonal reference

system (ORS) has been used: 0‐10‐7‐7‐10‐75‐1. Given

width of cut = 3.6 mm; shear strength of workpiece

material = 460 N/mm2; depth of cut = 0 25 mm;material = 460 N/mm2; depth of cut = 0.25 mm;

coefficient of friction at tool‐chip interface = 0.7.

Minimum power requirement (in kW) at a cutting speed

of 150 m/min isof 150 m/min is

(a) 3.15 (b) 3.25 (c) 3.35 (d) 3.45 [Ans. (b)]

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GATE – 2007 (PI) Common Data‐1GATE – 2007 (PI) Common Data‐1In an orthogonal machining test, the followingobservations were madeobservations were madeCutting force 1200 NTh t f NThrust force 500 NTool rake angle zeroC i d /Cutting speed 1 m/sDepth of cut 0.8 mmChip thickness 1.5 mmFriction angle during machining will be(a) 22.6o (b) 32.8o (c) 57.1o (d) 67.4o [Ans. (a)]

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GATE – 2007 (PI) Common Data‐2GATE – 2007 (PI) Common Data‐2In an orthogonal machining test, the followingobservations were madeobservations were madeCutting force 1200 NTh t f NThrust force 500 NTool rake angle zeroC i d /Cutting speed 1 m/sDepth of cut 0.8 mmChip thickness 1.5 mmChip speed along the tool rake face will be(a) 0.83 m/s (b) 0.53 m/s(c) 1.2 m/s (d) 1.88 m/s [Ans. (b)]

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ESE‐2003‐ ConventionalS 003 o e t o aDuring turning a carbon steel rod of 160 mm diameter by acarbide tool of geometry; 0, 0, 10, 8, 15, 75, 0 (mm) at speed of

f d f / d d th f t th400 rpm, feed of 0.32 mm/rev and 4.0 mm depth of cut, thefollowing observation weremade.

Tangential component of the cutting force, Pz = 1200 Na ge t a co po e t o t e cutt g o ce, z 00Axial component of the cutting force, Px = 800 NChip thickness (after cut), =2 0.8mm.α

For the abovemachining condition determine the values of(i) Friction force, F and normal force, N acting at the chip tooli t finterface.(ii) Yield shears strength of the work material under thismachining condition.g(iii) Cutting power consumption in kW.Ans. 828 N, 1200 N, 231.15 Mpa, 4.021 kW

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GATE 1995 C ti lGATE – 1995 ‐ConventionalWhile turning a C‐15 steel rod of 160 mm diameter atg 5315 rpm, 2.5 mm depth of cut and feed of 0.16mm/rev by a tool of geometry 00, 100, 80, 90,150, 750,( ) h f ll i b i d0(mm), the following observations weremade.

Tangential component of the cutting force = 500 NAxial component of the cutting force = 200 NChip thickness = 0.48 mm

Draw schematically the Merchant’s circle diagramfor the cutting force in the present case.Ans. F = 291 N, N = 457.67 N, Fn = 355.78 N, Fs = 408.31N, Friction angle = 32.49o

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IAS‐2003 Main ExaminationIAS 2003 Main ExaminationDuring turning process with 7 ‐ ? ‐ 6 – 6 – 8 – 30 – 1(mm) ASA tool the undeformed chip thickness of(mm) ASA tool the undeformed chip thickness of2.0 mm and width of cut of 2.5 mm were used. Theside rake angle of the tool was a chosen that thegmachining operation could be approximated to beorthogonal cutting. The tangential cutting force andh f N d 6 N i lthrust force were 1177 N and 560 N respectively.Calculate: [30 marks](i) Th id k l(i) The side rake angle(ii) Co‐efficient of friction at the rake face( ) h d h h f h k l(iii) The dynamic shear strength of the work materialAns. 12o, 0.82, 74.43 MPa

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IES ‐ 2004IES ‐ 2004A medium carbon steel workpiece is turned on al h / i i d 8 / f dlathe at 50 m/min. cutting speed 0.8 mm/rev feedand 1.5 mm depth of cut. What is the rate of metalremoval?removal?(a) 1000 mm3/min(b) 60 000 mm3/min(b) 60,000 mm3/min(c) 20,000 mm3/min(d) C t b l l t d ith th i d t(d) Can not be calculated with the given dataAns. (b)

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GATE 2007GATE‐2007In orthogonal turning of medium carbon steel. The g gspecific machining energy is 2.0 J/mm3. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 

/   d      i l   h   i   i  mm/rev and 2 mm respectively. The main cutting force in N is( )    (b) 8  (a) 40  (b) 80 (c) 400  (d) 800

(d)Ans. (d)

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

Ans. (a)( )

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i ( )GATE‐2006 Common Data Questions(1)In an orthogonal machining operation:g g pUncut thickness = 0.5 mm Cutting speed = 20 m/min  Rake angle = 15°g p / g 5Width of cut = 5 mm  Chip thickness = 0.7 mmThrust force = 200 N  Cutting force = 1200 NThrust force   200 N  Cutting force   1200 NAssume Merchant's theory.The coefficient of friction at the tool‐chip interface is   The coefficient of friction at the tool‐chip interface is   (a) 0.23  (b) 0.46 (c) 0 85  (d) 0 95 Ans. (b)(c) 0.85  (d) 0.95 Ans. (b)

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i ( )GATE‐2006 Common Data Questions(2)In an orthogonal machining operation:g g pUncut thickness = 0.5 mm Cutting speed = 20 m/min  Rake angle = 15°g p / g 5Width of cut = 5 mm  Chip thickness = 0.7 mmThrust force = 200 N  Cutting force = 1200 NThrust force   200 N  Cutting force   1200 NAssume Merchant's theory.The percentage of total energy dissipated due to The percentage of total energy dissipated due to friction at the tool‐chip interface is 

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

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i ( )GATE‐2006 Common Data Questions(3)In an orthogonal machining operation:g g pUncut thickness = 0.5 mm Cutting speed = 20 m/min  Rake angle = 15°g p / g 5Width of cut = 5 mm  Chip thickness = 0.7 mmThrust force = 200 N  Cutting force = 1200 NThrust force   200 N  Cutting force   1200 NAssume Merchant's theory.The values of shear angle and shear strain  The values of shear angle and shear strain, respectively, are                  

(a) 30 3° and 1 98  (b) 30 3° and 4 23 (a) 30.3 and 1.98  (b) 30.3 and 4.23 (c) 40.2° and 2.97  (d) 40.2° and 1.65        Ans. (d)

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i ( )GATE‐2003 Common Data Questions(1)A cylinder is turned on a lathe with orthogonaly gmachining principle. Spindle rotates at 200 rpm. Theaxial feed rate is 0.25 mm per revolution. Depth of cut is

h k l i h l i i i f d0.4 mm. The rake angle is 10°. In the analysis it is foundthat the shear angle is 27.75°

Th  thi k   f th   d d  hi  iThe thickness of the produced chip is(a) 0.511 mm  (b) 0.528 mm ( ) (d)(c) 0.818 mm (d) 0.846 mmAns. (a)

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i ( )GATE‐2003 Common Data Questions(2)A cylinder is turned on a lathe with orthogonaly gmachining principle. Spindle rotates at 200 rpm. Theaxial feed rate is 0.25 mm per revolution. Depth of cut is

h k l i h l i i i f d0.4 mm. The rake angle is 10°. In the analysis it is foundthat the shear angle is 27.75°I  th   b   bl  th   ffi i t  f f i ti   t In the above problem, the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is    Merchant theory is    (a) 0.18  (b) 0.36 (c) 0 71  (d) 0 98 Ans  (d)(c) 0.71  (d) 0.98 Ans. (d)

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i ( )GATE‐2008 Common Data Question (1)Orthogonal turning is performed on a cylindrical workg g p ypiece with shear strength of 250 MPa. The followingconditions are used: cutting velocity is 180 m/min. feedi / d h f i hi hi kis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessratio = 0.5. The orthogonal rake angle is 7o. ApplyMerchant's theory for analysisMerchant s theory for analysis.The shear plane angle (in degree) and the shear force respectively are force respectively are (a) 52: 320 N (b) 52: 400N     (c) 28: 400N     (d) 28:320N  Ans  (d)(c) 28: 400N     (d) 28:320N  Ans. (d)

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i ( )GATE‐2008 Common Data Question (2)Orthogonal turning is performed on a cylindrical workg g p ypiece with shear strength of 250 MPa. The followingconditions are used: cutting velocity is 180 m/min. feedi / d h f i hi hi kis 0.20 mm/rev. depth of cut is 3 mm. chip thicknessratio = 0.5. The orthogonal rake angle is 7o. ApplyMerchant's theory for analysisMerchant s theory for analysis.The cutting and thrust forces, respectively, are              (a)  68N  38 N        (b)  6 N  38 N      (a) 568N; 387N        (b) 565N; 381N      (c) 440N; 342N (d) 480N; 356N [Ans. (b)] 

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GATE 2010 (PI) Li k d S 1GATE ‐2010 (PI) Linked S‐1In orthogonal turning of an engineering alloy, it hasb b d h h f i i f i h hibeen 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 feedperpendicular to the cutting velocity vector. The feedvelocity is negligibly small with respect to the cuttingvelocity. The ratio of friction force to normal forceyassociated 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.The shear force (in N) acting along the primary shearl iplane is

(a) 180.0 (b) 240.0 (c) 360.5 (d) 402.5 Ans. (a)

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GATE 2010 (PI) Li k d S 2GATE ‐2010 (PI) Linked S‐2In orthogonal turning of an engineering alloy, it hasbeen observed that the friction force acting at the chipbeen 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 feedp p g yvelocity 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.Assume 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 Ans. (d)

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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 datawith a cutting tool of zero rake angle, the following datais obtained:Uncut chip thickness = 0.25 mmUncut chip thickness 0.25 mmChip thickness = 0.75 mmWidth of cut = 2 5 mmWidth of cut = 2.5 mmNormal force = 950 NThrust force = 475 N [Ans (b)]Thrust force = 475 N [Ans. (b)]The shear angle and shear force, respectively, are(a) 71 565o 150 21 N (b) 18 435o 751 04 N(a) 71.565o, 150.21 N (b) 18.435o , 751.04 N(c) 9.218o, 861.64 N (d) 23.157o , 686.66 N

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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 datawith a cutting tool of zero rake angle, the following datais obtained:

Uncut chip thickness = 0.25 mmUncut chip thickness 0.25 mmChip thickness = 0.75 mmWidth of cut = 2 5 mmWidth of cut = 2.5 mmNormal force = 950 NThrust force = 475 NThrust force = 475 N

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

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IES ‐ 2012IES ‐ 2012During orthogonal cutting, an increase in cutting speed causes(a) An increase in longitudinal cutting force( )(b) An increase in radial cutting force(c) An increase in tangential cutting force(d) Cutting forces to remain unaffected

Ans. (d)

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IES 2010IES 2010The relationship between the shear angle Φ,h f i i l β d i k lthe friction angle β and cutting rake angle αis given as

Ans. (b) 

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IES 2005IES‐2005Which one of the following is the correctgexpression for the Merchant's machinabilityconstant?(a)(b)

2φ γ α+ −2φ(b)

(c)2φ γ α− +

2φ γ α− −

(d)(Where = shear angle, = friction angle

φ γ α+ −

φ γ( g , gand = rake angle) Ans. (a)

γα

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GATE 1997GATE‐1997In a typical metal cutting operation, using a yp g p , gcutting tool of positive rake  angle = 10°, it was observed that the shear angle was 20°. gThe friction angle is        (a) 45° (b) 30°(a) 45 (b) 30(c) 60° (d) 40°

Ans. (c)

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IAS 1999IAS – 1999In an orthogonal cutting process, rake angle of the

l i ° d f i i l i ° U itool is 20° and friction angle is 25.5°. UsingMerchant's shear angle relationship, the value ofshear anglewill beshear anglewill be(a) 39.5° (b) 42.25°(c) 47 75° (d) 50 5°(c) 47.75 (d) 50.5

A (b)Ans. (b)

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IES 2003IES‐2003In orthogonal cutting test, the cutting force =g g , g900 N, the thrust force = 600 N and chipshear angle is 30o. Then the chip shear force isg 3 p(a) 1079.4 N (b) 969.6 N(c) 479 4 N (d) 69 6 N(c) 479.4 N (d) 69.6 N

Ans. (c)

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IES 2000IES‐2000In an orthogonal cutting test the cutting force andIn an orthogonal cutting test, the cutting force and

thrust force were observed to be 1000N and 500 N

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

coefficient of friction in chip‐tool interface will becoefficient of friction in chip tool interface will be

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

Ans. (a)

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IES 1996IES‐1996Which of the following forces are measured directly by

f d d lstrain gauges or force dynamometers during metalcutting ?1 Force exerted by the tool on the chip acting normally to1. Force exerted by the tool on the chip acting normally tothe tool face.2. Horizontal cutting force exerted by the tool on the workipiece.

3. Frictional resistance of the tool against the chip flowacting along the tool faceacting along the tool face.4. Vertical force which helps in holding the tool inposition.(a) 1 and 3 (b) 2 and 4(c) 1 and 4 (d) 2 and 3 Ans. (b)

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GATE 2007GATE‐2007In orthogonal turning of low carbon steel pipe withg g p pprincipal cutting edge angle of 90°, themain cuttingforce is 1000 N and the feed force is 800 N. The shear

l i d h l k l iangle is 25° and orthogonal rake angle is zero.Employing Merchant’s theory, the ratio of frictionforce to normal force acting on the cutting tool isforce to normal force acting on the cutting tool is(a) 1.56 (b) 1.25(c) 0 80 (d) 0 64(c) 0.80 (d) 0.64

A ( )Ans. (c)

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IES 1997IES‐1997Consider the following forces acting on ag gfinish turning tool:1 Feed force1. Feed force2. Thrust force

C i f3. Cutting force.The correct sequence of the decreasing order ofthe magnitudes of these forces is(a) 1, 2, 3 (b) 2, 3, 1( ) , , 3 ( ) , 3,(c) 3, 1, 2 (d) 3, 2, 1 [Ans. (c)]

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IES 1999IES‐1999The radial force in single‐point tool duringg p gturning operation varies between(a) 0 2 to 0 4 times the main cutting force(a) 0.2 to 0.4 times the main cutting force(b) 0.4 to 0.6 times the main cutting force( ) i h i i f(c) 0.6 to 0.8 times the main cutting force(d) 0.5 to 0.6 times the main cutting force

Ans (a)Ans. (a)

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IES 1995IES‐1995The primary tool force used in calculatingp y gthe total power consumption in machining isthe(a) Radial force (b) Tangential force(c) Axial force (d) Frictional force(c) Axial force (d) Frictional force.

Ans. (b)

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IES 2002IES‐2002In a machining process, the percentage ofg p , p gheat carried away by the chips is typically(a) 5% (b) 25%(a) 5% (b) 25%(c) 50% (d) 75%

Ans. (d)

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IES 1998IES‐1998In metal cutting operation, the approximateg p , ppratio of heat distributed among chip, tooland work, in that order is,(a) 80: 10: 10 (b) 33: 33: 33(c) 20: 60: 10 (d) 10: 10: 80(c) 20: 60: 10 (d) 10: 10: 80

Ans. (a)

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IAS 2003IAS – 2003As the cutting speed increases(a) More heat is transmitted to the work piece and less heat is transmitted to the tool( )(b) More heat is carried away by the chip and less heat is transmitted to the tool( ) M  h  i   i d   b h  h   hi   d  h  (c) More heat is transmitted to both the chip and the tool(d) M  h t i  t itt d t  b th th   k  i   d (d) More heat is transmitted to both the work piece and the tool

Ans. (b)

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IES 2001IES‐2001Power consumption in metal cutting isp gmainly due to(a) Tangential component of the force(a) Tangential component of the force(b) Longitudinal component of the force( ) N l f h f(c) Normal component of the force(d) Friction at the metal‐tool interface

Ans (a)Ans. (a)

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

Ans. (a)

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IES 2010IES 2010Consider the following statements:In an orthogonal, single‐point metal cutting,as the side‐cutting edge angle is increased,

1. The tangential force increases.2 The longitudinal force drops2. The longitudinal force drops.3. The radial force increases.Whi h f h ?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 Ans. (c)

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IES 1993IES‐1993A 'Dynamometer' is a device used for theymeasurement of(a) Chip thickness ratio(a) Chip thickness ratio(b) Forces during metal cutting( ) W f h i l(c) Wear of the cutting tool(d) Deflection of the cutting tool

Ans (b)Ans. (b)

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

Ans. (c)

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IAS 2001IAS‐2001Assertion (A): Piezoelectric transducers and preferredover strain gauge transducers in the dynamometers forover 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 piezoelectrictransducers.(a) Both A and R are individually true and R is the correctexplanation of A(b) h d d d ll b 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(c) A is true but R is false(d) A is false but R is true Ans. (c)

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IAS 2003IAS – 2003The heat generated in metal cutting can

i l b d i d bconveniently be determined by(a) Installing thermocouple on the job( )(b) Installing thermocouple on the tool(c) Calorimetric set‐up(d) Using radiation pyrometer

Ans. (c)

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IES 1998IES‐1998The gauge factor of a resistive pick‐up ofg g p pcutting force dynamometer is defined as theratio of(a) Applied strain to the resistance of the wire(b) The proportional change in resistance to the(b) The proportional change in resistance to theapplied strain( ) Th i h li d i(c) The resistance to the applied strain(d) Change in resistance to the applied strainAns. (b)

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IES 2000IES‐2000Assertion (A): In metal cutting, the normal( ) g,laws of sliding friction are not applicable.Reason (R): Very high temperature isReason (R): Very high temperature isproduced at the tool‐chip interface.(a) Both A and R are individually true and R is(a) Both A and R are individually true and R isthe correct explanation of A(b) B h A d R i di id ll b R i(b) Both A and R are individually true but R isnot the correct explanation of A(c) A is true but R is false(d) A is false but R is true Ans. (a)

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GATE 1992GATE 1992The effect of rake angle on the mean friction angleThe effect of rake angle on the mean friction anglein machining can be explained by(A) sliding (Coulomb) model of friction( ) g ( )(B) sticking and then sliding model of friction(C) sticking friction(C) sticking friction(D) Sliding and then sticking model of friction

Ans. (b)

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IES 2004IES‐2004Assertion (A): The ratio of uncut chip thickness to

t l hi thi k i l l th d iactual chip thickness is always less than one and istermed as cutting ratio in orthogonal cuttingReason (R): The frictional force is very high due to theReason (R): The frictional force is very high due to theoccurrence of sticking friction rather than slidingfriction(a) Both A and R are individually true and R is the correctexplanation of A(b) h d d d ll b 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(c) A is true but R is false(d) A is false but R is true Ans. (b)

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GATE 1993GATE‐1993The effect of rake angle on themean friction angle ing gmachining can be explained by(a) Sliding (coulomb) model of friction(b) sticking and then siding model of friction(c) Sticking friction( ) g(d) sliding and then sticking model of friction

Ans. (b)Ans. (b)

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

By  S K Mondal

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

[ 4 – marks]

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IES 2009 C ti lIES 2009 ConventionalShow crater wear and flank wear on a single point g pcutting tool. State the factors responsible for wear on a turning tool.

[ 2 –marks]

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

Ans. (a)

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IES 2007IES – 2007Flank wear occurs mainly on which of the f ll i ?following?(a) Nose part and top face( )(b) Cutting edge only(c) Nose part, front relief face, and side relief face of the 

i   lcutting tool(d) Face of the cutting tool at a short distance from th   tti   dthe cutting edgeAns. (c)

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IES 2004IES – 2004Consider the following statements:During the third stage of tool‐wear, rapiddeterioration of tool edge takes place because1. Flank wear is only marginal2. Flank wear is large3. Temperature of the tool increases gradually4. Temperature of the tool increases drasticallyWhich of the statements given above are correct?(a) 1 and 3 (b) 2 and 4(c) 1 and 4 (d) 2 and 3 Ans. (b)

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IES 2002IES – 2002Crater wear on tools always starts at some distance f   h   l  i  b     h   ifrom the tool tip because at that point(a) Cutting fluid does not penetrate( )(b) Normal stress on rake face is maximum    (c) Temperature is maximum(d) Tool strength is minimum

Ans. (c)

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IAS 2007IAS – 2007Why does crater wear start at some distance from h   l  i ?the tool tip?(a) Tool strength is minimum at that region( )(b) Cutting fluid cannot penetrate that region(c) Tool temperature is maximum in that region(d) Stress on rake face is maximum at that region

Ans. (c)

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IES 2000IES – 2000Crater wear starts at some distance from the tool tip bbecause(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

Ans. (d)

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IES 1996IES – 1996Notch wear at the outside edge of the depth of cut is d  due to(a) Abrasive action of the work hardened chip material( )(b) Oxidation(c) Slip‐stick action of the chip (d) Chipping.

Ans. (b)

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IES 1995IES – 1995Match List I with List II and select the correct 

  i   h   d   i  b l   h  lianswer using the codes given below the lists:List I (Wear type)  List II (Associated mechanism) A. Abrasive wears  1. Galvanic actionB. Adhesive wears  2. Ploughing actionC. Electrolytic wear  3. Molecular transferD. Diffusion wears 4. Plastic deformation

5. Metallic bondCode: A B C D A B C D

[Ans. (a)]

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

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IES 1995IES – 1995Crater wear is predominant in(a) Carbon steel tools (b) Tungsten carbide tools(c) High speed steel tools (d) Ceramic tools

Ans. (a) 

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IES 1994IES – 1994Assertion (A): Tool wear is expressed in terms of fl k    h   h    flank wear rather than crater wear.Reason (R): Measurement of flank wear is simple 

d    tand more accurate.(a) Both A and R are individually true and R is the correct explanation of Acorrect explanation of A(b) Both A and R are individually true but R is not the correct explanation of A correct explanation of A (c) A is true but R is false(d) A is false but R is true [Ans  (c) ](d) A is false but R is true [Ans. (c) ]

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IES 2008IES – 2008What are the reasons for reduction of tool life in a 

hi i   i ?machining operation?1. 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 forceSelect the correct answer using the code given below:(a) 1, 2 and 3 (b) 2, 3 and 4(c) 1, 3 and 4 (d) 1, 2 and 4 [Ans. (a)]

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IAS 2002IAS – 2002Consider the following actions:1. Mechanical abrasion 2. Diffusion3. 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

Ans. (c)

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IAS 1999IAS – 1999The type of wear that occurs due to the cutting

i f h i l i h i fl id iaction of the particles in the cutting fluid isreferred to as( ) Att iti(a) Attritions wear(b) Diffusion wear( ) E i(c) Erosive wear(d) Corrosive wear

Ans. (c)

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IAS 2003IAS – 2003Consider the following statements:

Chipping of a cutting tool is due to1. Tool material being too brittle2. Hot hardness of the tool material.3. High positive rake angle of the tool.Which of these statements are correct?(a) 1, 2 and 3 (b) 1 and 33 3(c) 2 and 3 (d) 1 and 2Ans. (b)( )

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IES ‐ 2012IES ‐ 2012In Taylor’s tool life equation VTn = C, the constants n 

d C d d and C depend upon1. Work piece material2. Tool material3. Coolant(a) 1, 2, and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 onlyAns. (a)

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IFS 2009With the help of Taylor’s tool life equation,

determine the shape of the cur e bet een elocitdetermine the shape of the curve between velocity

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

and steel as work material.

[ M k ][10‐Marks]

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IES 1996IES‐1996Chip equivalent is increased by(a) An increases in side‐cutting edge angle of tool(b) An increase in nose radius and side cutting(b) An increase in nose radius and side cuttingedge angle of tool(c) Increasing the plant area of cut(c) Increasing the plant area of cut(d) Increasing the depth of cut.Ans. (b)

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

Ans. (d)

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GATE 2004GATE‐2004In a machining operation, doubling the

1cutting speed reduces the tool life to th ofthe original value. The exponent n in Taylor's

18

tool life equation VTn = C, is1 1 1 1( ) ( ) ( ) ( )a b c d( ) ( ) ( ) ( )8 4 3 2

a b c d

A  ( )Ans. (c)

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IES 2000IES – 2000In a tool life test, doubling the cutting speedd h l lif /8 h f h i i l Threduces the tool life to 1/8th of the original. The

Taylor's tool life index is

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

Ans. (b)

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IES 1999IES – 1999In a single‐point turning operation of steel with a 

d  bid   l  T l '   l lif    i  cemented carbide tool, Taylor's tool life exponent is 0.25. If the cutting speed is halved, the tool life will increase byincrease by(a) Two times  (b) Four times(c) Eight times (d) Sixteen times(c) Eight times (d) Sixteen times

A  (d)Ans. (d)

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IES 2008IES – 2008In Taylor's tool life equation is VTn = constant.What is the value of n for ceramic tools?(a) 0.15 to 0.25 (b) 0.4 to 0.55(c) 0.6 to 0.75 (d) 0.8 to 0.9

Ans. (c)

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IES 2006IES – 2006Which of the following values of index n is

i d i h bid l h T l ' l lifassociated with carbide tools when Taylor's tool lifeequation, V.Tn = constant is applied?( ) t (b) t(a) 0∙1 to 0∙15 (b) 0∙2 to 0∙4(c) 0.45 to 0∙6 (d) 0∙65 to 0∙9

Ans. (b)

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IES 1999IES – 1999The approximately variation of the tool life

' ' f d bid l iexponent 'n' of cemented carbide tools is(a) 0.03 to 0.08 (b) 0.08 to 0.20( ) ( )(c) 0.20 to 0.48 (d) 0.48 to 0.70

Ans. (c)

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IAS 1998IAS – 1998Match List ‐ I (Cutting tool material) with List ‐ II (Typical value of tool life exponent 'n' in the Taylor's (Typical value of tool life exponent  n  in the Taylor s equation V.Tn = C) and select the correct answer using the codes given below the lists:List – I List – IIA. HSS 1. 0.18B Cast alloy 2 0 12B. Cast alloy 2. 0.12C. Ceramic 3. 0.25D Sintered carbide 4 0 5

Ans. (d)D. Sintered carbide 4. 0.5

Codes: A B C D A B C D(a)  1 2 3 4 (b)  2 1 3 43 4 3 4(c)  2 1 4 3 (d)  1 2 4 3

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GATE ‐2009 (PI)In an orthogonal machining operation, the tool life

obtained is 0 min at a cutting speed of 00 m/minobtained 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 life

equationequation

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

Ans. (a)

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ISRO 2011ISRO‐2011A 50 mm diameter steel rod was turned at 284 rpm andA 50 mm diameter steel rod was turned at 284 rpm and

tool failure occurred in 10 minutes. The speed was

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

Assuming straight line relationship between cuttingg g p g

speed and tool life, the value of Taylorian Exponent is

(a) 0.21 (b) 0.13 (c) 0.11 (d) 0.23

Ans (c)Ans. (c)

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IES 2010IES 2010The above figure shows a typicall i hi b l lif drelationship between tool life and

cutting speed for differentmaterials. Match the graphs forg pHSS, Carbide and Ceramic toolmaterials and select the correctanswer using the code givenanswer using the code givenbelow the lists:

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

Ans. (a) 

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GATE‐2008 (PI)G 008 ( )During machining, the wear land (h) has been plottedagainst machining time (T) as given in the followingfigure.

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

Ans. (b)

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GATE 2010GATE‐2010For tool A, Taylor’s tool life exponent (n) is0.45 and constant (K) is 90. Similarly for toolB, 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

Ans. (a)

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GATE 2003GATE‐2003A batch of 10 cutting tools could produce 500

t hil ki t ithcomponents while working at 50 rpm with atool feed of 0.25 mm/rev and depth of cut of 1mm A similar batch of 10 tools of the samemm. A similar batch of 10 tools of the samespecification could produce 122 componentswhile working at 80 rpm with a feed of 0.25while working at 80 rpm with a feed of 0.25mm/rev and 1 mm depth of cut. How manycomponents can be produced with onep pcutting tool at 60 rpm?(a) 29 (b) 319 3(c) 37 (d) 42 Ans. (a)

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IES 1994 2007IES – 1994, 2007For increasing the material removal rate in turning, i h     i   h  i   h   i h    without any constraints, what is the right sequence 

to adjust the cutting parameters?S d F d D th  f  t1. Speed 2. Feed 3. Depth of cut

S l   h       i   h   d   i  b lSelect the correct answer using the code given below:(a) 1‐ 2‐ 3 (b) 2‐ 3‐ 1( ) ( )(c) 3‐ 2‐ 1 (d) 1‐ 3‐ 2Ans. (c) [read question again, common error (a)]

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

Ans. (d)

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IES 1997IES – 1997Consider the following elements:1. Nose radius 2. Cutting speed3. 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 (c) 2,4, 1, 3  (d) 4, 2, 1, 3

Ans. (a)

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ISRO‐2012What is the correct sequence of the followingparameters in order of their maximum to minimumi fl l lif ?influence on tool life?1. Feed rate2. Depth of cut3. Cutting speedSelect the correct answer using the codes given below(a) 1, 2, 3 (b) 3, 2, 1 (c) 2, 3, 1 (d) 3, 1, 2

Ans. (d)

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IES 1992IES – 1992Tool life is generally better  when(a) Grain size of the metal is large(b) Grain size of the metal is small(c) Hard constituents are present in the microstructure of the tool material(d) None of the above

Ans. (a)

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IAS 2003IAS – 2003The tool life curves for two tools A and B are shown in th  fi   d th  f ll  th  t l lif   ti  VTn  C  the figure and they follow the tool life equation VTn = C. Consider the following statements:1. Value of n for both the tools is same.1. 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.

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

Ans. (a)

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IAS 2002IAS – 2002Using the Taylor equation VTn = c, calculate the 

 i  i   l lif   h   h   i  percentage increase in tool life when the cutting speed is reduced by 50% (n = 0∙5 and c = 400)( ) % (b) %(a) 300% (b) 400%(c) 100% (d) 50%

( )Ans. (a)

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IAS 2002IAS – 2002Optimum cutting speed for minimum cost (Vc min )

d i i d f iand optimum cutting speed for maximumproduction rate (Vr max ) have which one of thefollowing relationships?following relationships?(a) Vcmin = Vrmax (b) Vcmin > Vrmax

(c) V < V (d) V2 V(c) Vcmin < Vrmax (d) V2c min = Vrmax

Ans. (c)

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IES 2010IES 2010With increasing cutting velocity, the totali f hi itime formachining a component

(a) Decreases(b) Increases(c) Remains unaffected(c) Remains unaffected(d) First decreases and then increases

Ans. (d)

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IAS 2000IAS – 2000Consider the following statements:The tool life is increased by1. Built ‐up edge formation2. Increasing cutting velocity3. Increasing back rake angle up to certain valueWhich of these statements are correct?(a) 1 and 3 (b) 1 and 23(c) 2 and 3 (d) 1, 2 and 3

Ans. (a)

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IAS 1997IAS – 1997In the Taylor's tool life equation, VTn = C, the valuef Th l h lif f 8 iof n = 0.5. The tool has a life of 180 minutes at a

cutting speed of 18 m/min. If the tool life is reducedto 45 minutes then the cutting speed will beto 45 minutes, then the cutting speed will be(a) 9 m/min (b) 18 m/min(c) 36 m/min (d) 72 m/min(c) 36 m/min (d) 72 m/min

A ( )Ans. (c)

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IAS 1996IAS – 1996The tool life increases with the(a) Increase in side cutting edge angle(b) Decrease in side rake angle(c) Decrease in nose radius(d) Decrease in back rake angle

Ans. (a)

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IAS 1995IAS – 1995In a single point turning operation with a cemented 

bid   d  l  bi i  h i    T l  carbide and steel combination having a Taylor exponent of 0.25, if the cutting speed is halved, then the tool life will becomethe tool life will become(a) Half (b) Two times(b) Two times(c) Eight times(d) Si t  ti(d) Sixteen times.

A  (d)Ans. (d)

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IAS 1995IAS – 1995Assertion (A): An increase in depth of cut shortensh l lifthe tool life.Reason(R): Increases in depth of cut gives rise tol ti l ll i i t l t trelatively small increase in tool temperature.

(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 thecorrect explanation of Acorrect 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)]

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IES 2006 conventionalIES – 2006 conventionalAn HSS tool is used for turning operation. Thetool life is 1 hr. when turning is carried at 30m/min. The tool life will be reduced to 2.0 min ifthe cutting speed is doubled. Find the suitablespeed in RPM for turning 300 mm diameter sothat tool life is 30 min.

Ans (36 66 rpm)Ans. (36.66 rpm)

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ESE 1999; IAS 2010 ConventionalESE‐1999; IAS ‐2010 ConventionalThe following equation for tool life was obtained for HSS

l A 6 i l lif b i d i h f ll itool. A 60 min tool life was obtained using the followingcutting condition VT0.13f0.6d0.3= C. v = 40 m/min, f = 0.25mm d = 2 0 mm Calculate the effect on tool life ifmm, d = 2.0 mm. Calculate the effect on tool life ifspeed, feed and depth of cut are together increased by25% and also if they are increased individually by 25%;5 y y y 5 ;where f = feed, d = depth of cut, v = speed.

Ans. 2.3 min, 10.78 min, 21.42 min, 35.85 min

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IES 2009 ConventionalIES 2009 ConventionalDetermine the optimum cutting speed for an

i L h hi i h f ll ioperation on a Lathe machine using the followinginformation:T l h ti iTool change time: 3 minTool regrinds time: 3 minM hi i R iMachine running cost Re.0.50 perminDepreciation of tool regrinds Rs. 5.0The constants in the tool life equation are 60 and0.2A 6 / iAns. 26 m/min

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ESE 2001 ConventionalESE‐2001 ConventionalIn a certain machining operation with a cuttingspeed of 50 m/min, tool life of 45 minutes wasobserved. When the cutting speed was increasedto 100 m/min, the tool life decreased to 10 min.Estimate the cutting speed for maximumproductivity if tool change time is 2 minutes.

Ans. 195 m/min

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GATE 2009 Li k d A Q i (1)GATE‐2009 Linked Answer Questions (1) In a machining experiment, tool life was found to vary i h  h   i   d i   h  f ll i  with the cutting speed in the following manner:

Cutting speed (m/min) Tool life (minutes)60 8190 36The exponent (n) and constant (k) of the Taylor's tool life equation are( ) ( )(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=1.15Ans. (a)

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GATE 2009 Li k d A Q i (2)GATE‐2009 Linked Answer Questions (2) In a machining experiment, tool life was found to vary i h  h   i   d i   h  f ll i  with the cutting speed in the following manner:

Cutting speed (m/min) Tool life (minutes)60 8190 36What is the percentage increase in tool life when the cutting speed is halved?( ) ( )(a) 50% (b) 200%(c) 300%  (d) 400%      Ans. (c)

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

Ans. (b)

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GATE 2005GATE‐2005

Ans. (a)

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IAS 2007 ContdIAS – 2007             Contd…A diagram related to machining economics withvarious cost components is given above Match List Ivarious cost components is given above. Match List I(Cost Element) with List II (Appropriate Curve) andselect the correct answer using the code given belowg gthe Lists:

List I  List II( ) ( )(Cost Element) (Appropriate Curve) A. Machining cost  1. Curve‐lB T l    CB. Tool cost  2. Curve‐2C. Tool grinding cost  3. Curve‐3D N d ti   t  CD. Non‐productive cost  4. Curve‐4

5. Curve‐5

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Contd………. From previous slideContd………. From previous slide

A  (b)

C d A B  C  D A  B  C  D

Ans. (b)

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

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IES 1998IES – 1998The variable cost and production rate of a

hi i i i d hmachining process against cutting speed are shownin the given figure. For efficient machining, therange of best cutting speed would be betweenrange of best cutting speed would be between(a) 1 and 3(b) 1 and 5(b) 1 and 5(c) 2 and 4(d) d(d) 3 and 5

Ans. (c)

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IES 1999IES – 1999Consider the following approaches normally

li d f h i l i f hi iapplied for the economic analysis of machining:1. Maximumproduction rate2. Maximumprofit criterion3. Minimum cost criterionThe correct sequence in ascending order of optimumcutting speed obtained by these approaches is( ) ( )(a) 1, 2, 3 (b) 1, 3, 2(c) 3, 2, 1 (d) 3, 1, 2Ans. (c)

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IES 2011IES 2011The optimum cutting speed is one whichshould have:

1. Highmetal removal rate2. High cutting tool life3. Balance the metal removal rate andcutting tool life

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

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IES 2000IES – 2000The magnitude of the cutting speed for maximum

fi bprofit ratemust be(a) In between the speeds for minimum cost and

i d ti tmaximum production rate(b) Higher than the speed for maximum production rate( ) B l h d f i i(c) Below the speed for minimum cost(d) Equal to the speed for minimum cost

Ans. (a)

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IES 2004IES – 2004Consider the following statements:

A th tti d i th t f d ti1. 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 productionalso increases and after a critical value it reduces3 Higher feed rate for the same cutting speed reduces cost3. Higher feed rate for the same cutting speed reduces costof production4. Higher feed rate for the same cutting speed increases the4 g g pcost of productionWhich of the statements given above is/are correct?( ) d (b) d(a) 1 and 3 (b) 2 and 3(c) 1 and 4 (d) 3 only [Ans. (a)]

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

Ans. (c)

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IAS 2007IAS – 2007Assertion (A): The optimum cutting speed for thei i f hi i i i hminimum cost of machining may not maximize the

profit.R (R) Th fit l d d t fReason (R): The profit also depends on rate ofproduction.(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) 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 [Ans. (a) ]

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

Ans. (b)

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IES ‐ 2012IES ‐ 2012The usual method of defining machinability of a

i l i b i d b dmaterial is by an index based on(a) Hardness of work material( )(b) Production rate of machined parts(c) Surface finish of machined surfaces(d) Tool life

Ans. (d)

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IES 2011 ConventionalIES 2011 ConventionalDiscuss the effects of the following elements on themachinability of steels:machinability of steels:(i) Aluminium and silicon(ii) S l h d S l i(ii) Sulphur and Selenium(iii) Lead and Tin(i ) C b d M(iv) Carbon and Manganese(v) Molybdenum and Vanadium [5 Marks]

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IES 1992IES – 1992Ease of machining is primarily judged by(a) Life of cutting tool between sharpening(b) Rigidity of work ‐piece(c) Microstructure of tool material(d) Shape and dimensions of work

Ans. (a)

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IES 2007 2009IES – 2007, 2009Consider the following:1. Tool life2. 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

Ans. (a) 

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

Ans. (a)( )

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IES 2003IES – 2003Assertion (A): The machinability of steels improvesby adding sulphur to obtain so called 'Freeby adding sulphur to obtain so called FreeMachining Steels‘.Reason (R): Sulphur in steel forms manganeseReason (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 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 Ans. (c)

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IES 2009IES – 2009The elements which, added to steel, help in chipf i d i hi iformation during machining are(a) Sulphur, lead and phosphorous( )(b) Sulphur, lead and cobalt(c) Aluminium, lead and copper(d) Aluminium, titanium and copper

Ans. (a)

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IES 1998IES – 1998Consider the following criteria in evaluating 

hi bilimachinability:1. Surface finish 2. Type of chips3. Tool life 4. Power consumptionIn modern high speed CNC machining with coated 

bid   l   h       f  h   i i  carbide tools, the correct sequence of these criteria in DECREASING order of their importance is( )         (b)        (a) 1, 2, 4, 3  (b) 2, 1, 4, 3 (c) 1, 2, 3, 4  (d) 2, 1, 3, 4A  ( )Ans. (c)

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IES 1996IES – 1996Which of the following indicate better

hi bili ?machinability?1. Smaller shear angle2. Higher cutting forces3. Longer tool life4. Better surface finish.(a) 1 and 3 (b) 2 and 4(c) 1 and 2 (d) 3 and 4

Ans. (d)

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IES 1996IES – 1996Small amounts of which one of the followingl / i f l i dd d lelements/pairs of elements is added to steel to

increase its machinability?( ) Ni k l (b) S l h d h h(a) Nickel (b) Sulphur and phosphorus(c) Silicon (d) Manganese and copper

Ans. (b)

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IES 1995IES – 1995In low carbon steels, presence of small quantities 

l h isulphur improves(a) Weldability (b) Formability( ) ( )(c) Machinability (d) Hardenability

Ans. (c)

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IES 1992IES – 1992Machining of titanium is difficult due to(a) High thermal conductivity of titanium(b) Chemical reaction between tool and work(c) Low tool‐chip contact area(d) None of the above

Ans. (b)

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IAS 1996IAS – 1996Assertion (A): The machinability of a material can b   d      b l   ibe measured as an absolute quantity.Reason (R): Machinability index indicates the case ith  hi h    t i l   b   hi dwith which a material can be machined

(a) Both A and R are individually true and R is the correct explanation of Acorrect explanation of A(b) Both A and R are individually true but R is not the correct explanation of A 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 trueAns. (d)

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GATE 2009GATE‐2009Friction at the tool‐chip interface can be reduced by

(a) decreasing the rake angle ( ) g g(b) increasing the depth of cut(c) Decreasing the cutting speed (c) Decreasing the cutting speed (d) increasing the cutting speed

Ans. (d)

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IES ‐ 2002IES ‐ 2002The value of surface roughness 'h' obtained duringh i i f d 'f' i h dthe turning operating at a feed 'f' with a round nosetool having radius 'r' is given as

Ans. (b)

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IAS ‐ 1996IAS ‐ 1996Given thatS = feed in mm/rev. andR = 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(d) S2/8RAns. (d)

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IES ‐ 1999IES ‐ 1999In turning operation, the feed could be doubled toi h l l T k hincrease the metal removal rate. To keep the samelevel of surface finish, the nose radius of the toolshould beshould be(a) Halved (b) Kept unchanged(c) doubled (d) Made four times(c) doubled (d) Made four timesAns. (d)

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GATE ‐ 1997GATE ‐ 1997A cutting tool has a radius of 1.8 mm. The feed rate f     h i l  f   h   f R          ifor a theoretical surface roughness of Ra = 5     m is(a) 0.36 mm/rev( )

μ

(b) 0.187 mm/rev(c) 0.036 mm/rev(d) 0.0187 mm/rev

Ans. (none) 

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GATE – 2007 (PI)GATE – 2007 (PI)A tool with Side Cutting Edge angle of 30o andE d C i Ed l f i d f fiEnd Cutting Edge angle of 10o is used for fineturning with a feed of 1 mm/rev. Neglecting nosedi f h l h i ( k ll )radius of the tool, the maximum (peak to valley)

height of surface roughness produced will be(a) 0.16 mm (b) 0.26 mm(c) 0.32 mm (d) 0.48 mm( ) 3 ( ) 4Ans. (a)

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GATE ‐ 2005GATE ‐ 2005Two tools P and Q have signatures 5°‐5°‐6°‐6°‐8°‐30°‐0 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively.5 5 7 7 5 ( ) p yThey are used to turn components under the samemachining conditions. If hpand hQ denote the peak‐p Qto‐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 o

Ans. (b)

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IES – 1993 ISRO‐2008IES – 1993, ISRO‐2008For achieving a specific surface finish in single point

i h i f b ll dturning the most important factor to be controlledis( ) D th f t (b) C tti d(a) Depth of cut (b) Cutting speed(c) Feed (d) Tool rake angle

Ans. (c)

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IES ‐ 2006IES ‐ 2006In the selection of optimal cutting conditions, the

i f f fi i h ld li irequirement of surface finish would put a limit onwhich of the following?( ) Th i f d(a) The maximum feed(b) The maximum depth of cut( ) Th i d(c) The maximum speed(d) The maximum number of passes

Ans. (a)

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GATE ‐2010 (PI)GATE ‐2010 (PI)During turning of a low carbon steel bar with TiN coatedcarbide insert, one need to improve surface finishwithout sacrificing material removal rate. To achieveimproved surface finish, one should

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

(b) Increase nose radius of the cutting tool(b) Increase nose radius of the cutting tool

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

(d) Increase depth of cut and increase feed [Ans. (b) ]

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

[ 3 – marks][ 3  marks]Where high pressures and rubbing action areencountered, hydrodynamic lubrication cannot bey ymaintained; so Extreme Pressure (EP) additives must beadded to the lubricant. EP lubrication is provided by anumber of chemical components such as boronnumber of chemical components such as boron,phosphorus, sulfur, chlorine, or combination of these.The compounds are activated by the higher temperatureresulting from extreme pressure. As the temperaturerises, EP molecules become reactive and releasederivatives such as iron chloride or iron sulfide andderivatives such as iron chloride or iron sulfide andforms a solid protective coating.

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IES ‐ 2001IES ‐ 2001Dry and compressed air is used as cutting fluid for

hi imachining(a) Steel (b) Aluminium( ) ( )(c) Cast iron (d) Brass

Ans. (c)

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

Ans. (b)

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Ch 3 C tti T l T l Lif d C tti Fl idCh‐3: Cutting Tools, Tool Life and Cutting FluidQ. No Option Q. No Option Q. No Option1 B 12 C 23 A1 B 12 C 23 A2 A 13 A 24 C3 A 14 A 25 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 D 20 A 31 C10 D 21 B 32 B10 D 21 B 32 B11 A 22 B 33 C

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Ch 4: Economics of Machining OperationCh‐4: Economics of Machining OperationQ. No Option Q. No Option1 C 6 B2 B 7 A3 A 8 C3 A 8 C4 C 9 A5 A5 A

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M t l F iMetal Forming

d lBy  S K Mondal

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IES 2011IES 2011Assertion (A): Lead, Zinc and Tin are always hotworked.Reason (R) : If they are worked in cold stateh h h lthey cannot retain theirmechanical properties.(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 NOTth t l ti f Athe correct explanation of A(c) A is true but R is false(d) A i f l b R i A (b)(d) A is false but R is true Ans. (b)

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GATE 2003GATE‐2003 Cold working of steel is defined as working(a) At its recrystallisation temperature(b) Above its recrystallisation temperature(c) Below its recrystallisation temperature(d) At two thirds of the melting temperature of themetal

Ans. (c)

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GATE 2002 ISRO 2012GATE‐2002, ISRO‐2012Hot rolling of mild steel is carried out(a) At recrystallisation temperature(b) Between 100°C to 150°C(c) Below recrystallisation temperature(d) Above recrystallisation temperature

Ans. (d)

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

following process to relieve stresses

(a) Hot working

(b) T i(b) Tempering

(c) Normalizing(c) Normalizing

(d) Annealing Ans. (d)g

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IES 2006IES – 2006Which one of the following is the process to refineh i f l f i h b di d bthe grains of metal after it has been distorted byhammering or cold working?( ) A li (b) S ft i(a) Annealing (b) Softening(c) Re‐crystallizing (d) Normalizing

( )Ans. (c)

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IES 2004IES – 2004Consider the following statements:In comparison to hot working, in cold working,1. Higher forces are required2. No heating is required3. Less ductility is required4. Better surface finish is obtainedWhich of the statements given above are correct?g(a) 1, 2 and 3 (b) 1, 2 and 4(c) 1 and 3 (d) 2, 3 and 4( ) 3 ( ) , 3 4Ans. (b)

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IES 2009IES – 2009Consider the following characteristics:1. Porosity in the metal is largely eliminated.2. 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 3Ans. (d)

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IES 2008IES – 2008Consider the following statements:1. Metal forming decreases harmful effects ofimpurities and improves mechanical strength.2. Metal working process is a plastic deformationprocess.

V i i h b d d b f i3. Very intricate shapes can be produced by forgingprocess as compared to casting process.Whi h f th t t t i b t?Which of the statements given above are correct?(a) 1, 2 and 3 (b) 1 and 2 only( ) d l (d) d l A (b)(c) 2 and 3 only (d) 1 and 3 only Ans. (b)

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IES 2008IES – 2008Cold forging results in improved quality due tohi h f h f ll i ?which of the following?

1. Better mechanical properties of the process.2. Unbroken grain flow.3. Smoother finishes.4. High pressure.Select the correct answer using the code given below:(a) 1, 2 and 3 (b) 1, 2 and 4(c) 2, 3 and 4 (d) 1, 3 and 4 Ans. (a)

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IES 2004IES – 2004Assertion (A): Cold working of metals results ini f h d h dincrease of strength and hardnessReason (R): Cold working reduces the total numberf di l ti it l f th t i lof dislocations per unit volume of thematerial

(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (c)(d) A is false but R is true Ans. (c)

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IES 2003IES – 2003Cold working produces the following effects:1. Stresses are set up in the metal2. Grain structure gets distorted3. Strength and hardness of the metal are decreased4. Surface finish is reducedWhich of these statements are correct?(a) 1 and 2 (b) 1, 2 and 33(c) 3 and 4 (d) 1 and 4Ans. (a)( )

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IES 2000IES – 2000Assertion (A): To obtain large deformations by cold

ki i di li i i dworking intermediate annealing is not required.Reason (R): Cold working is performed below the

t lli ti t t f th k t i lrecrystallisation temperature of thework material.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (d)(d) A is false but R is true Ans. (d)

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ISRO‐2009In the metal forming process, the stressesencountered are(a) Greater than yield strength but less thanultimate strength(b) Less than yield strength of the material(c) Greater than the ultimate strength of the

i lmaterial(d) Less than the elastic limitAns. (a)

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IES 1997IES – 1997In metals subjected to cold working, strainh d i ff i dhardening effect is due to(a) Slip mechanism( )(b) Twining mechanism(c) Dislocation mechanism(d) Fracture mechanism

Ans. (c)

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IES 1996IES – 1996Consider the following statements:When a metal or alloy is cold worked1. 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 increase.Of these correct statements are(a) 1 and 4  (b) 1 and 3 (c) 2 and 3  (d) 2 and 4 Ans. (c)

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IES 2006IES – 2006Assertion (A): In case of hot working of metals, thet t t hi h th i fi ll t dtemperature at which the process is finally stoppedshould not be above the recrystallisation temperature.Reason (R): If the process is stopped above theReason (R): If the process is stopped above therecrystallisation temperature, grain growth will takeplace again and spoil the attained structure.(a) Both A and R are individually true and R is the correctexplanation of A(b) h d d d ll b 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(c) A is true but R is false(d) A is false but R is true Ans. (d)

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IES 1992IES – 1992Specify the sequence correctly(a) Grain growth, recrystallisation, stress relief(b) Stress relief, grain growth, recrystallisation(c) Stress relief, recrystallisation, grain growth(d) Grain growth, stress relief, recrystallisation

Ans. (c)

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

Ans. (c)

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IAS 2004IAS – 2004Assertion (A): Hot working does not produce strainh d ihardening.Reason (R): Hot working is done above the re‐

t lli ti t tcrystallization temperature.(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 thecorrect explanation of Acorrect 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)

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IAS 2002IAS‐2002Assertion (A): There is good grain refinement in hot

kiworking.Reason (R): In hot working physical properties are

ll i dgenerally improved.(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 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 trueAns. (b) Ulta hai. Assertion reasonme hona chaihe.

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IES 2008IES‐2008Which one of the following is correct?Malleability is the property by which a metal oralloy can be plastically deformed by applying( ) ( )(a) Tensile stress (b) Bending stress(c) Shear stress (d) Compressive stress

Ans. (d)

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R lliRolling

By  S K Mondal

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ISRO‐2006Which of the following processes would produce

strongest components?strongest components?

(a) Hot rolling

(b) Extrusion

(c) Cold rolling

(d) Forging Ans (c)(d) Forging Ans. (c)

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

Ans. (a)( )

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IES 2006IES – 2006Which one of the following is a continuous bending

i hi h i ll d dprocess in which opposing rolls are used to producelong sections of formed shapes from coil or stripstock?stock?(a) Stretch forming (b) Roll forming(c) Roll bending (d) Spinning(c) Roll bending (d) Spinning

A ( )Ans. (c)

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GATE – 2009 (PI)Anisotropy in rolled components is caused by

( ) h d(a) changes in dimensions

(b) scale formation(b) sca e o a o

(c) closure of defects

(d) grain orientation

A (d)Ans. (d)

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GATE 2008GATE‐2008In a single pass rolling operation, a 20 mm thickl i h l id h f i d d 8plate with plate width of 100 mm, is reduced to 18

mm. The roller radius is 250 mm and rotationalspeed is 10 rpm The average flow stress for the platespeed is 10 rpm. The average flow stress for the platematerial is 300 MPa. The power required for therolling operation in kW is closest tog p(a) 15.2(b) 18.2(b) 18.2(c) 30.4(d) 45 6 Ans (a)(d) 45.6 Ans. (a)

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GATE 2007GATE‐2007The thickness of a metallic sheet is reduced from ani i i l l f 6 fi l l f iinitial value of 16 mm to a final value of 10 mm inone single pass rolling with a pair of cylindricalrollers each of diameter of 400 mm The bite anglerollers each of diameter of 400 mm. The bite anglein degreewill be(a) 5 936(a) 5.936(b) 7.936(c) 8 936(c) 8.936(d) 9.936 Ans. (d)

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GATE 2004GATE‐2004In a rolling process, sheet of 25 mm thickness is

ll d hi k R ll i f di 6rolled to 20 mm thickness. Roll is of diameter 600mm and it rotates at 100 rpm. The roll strip contactlength will belength will be(a) 5 mm (b) 39 mm(c) 78 mm (d) 120 mm(c) 78 mm (d) 120 mm

A (b)Ans. (b)

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GATE 1998GATE‐1998A strip with a cross‐section 150 mm x 4.5 mm isb i ll d i h % d i f ibeing rolled with 20% reduction of area using 450mm diameter rolls. The angle subtended by thedeformation zone at the roll centre is (in radian)deformation zone at the roll centre is (in radian)(a) 0.01 (b) 0.02(c) 0 03 (d) 0 06(c) 0.03 (d) 0.06

A (d) it i i diAns. (d) it is in radian.

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GATE 2012GATE – 2012 Same Q in GATE – 2012 (PI)

In a single pass rolling process using 410 mmIn a single pass rolling process using 410 mm

diameter steel rollers, a strip of width 140 mm and

thickness 8 mm undergoes 10% reduction of

thickness. The angle of bite in radians isg

(a) 0.006 (b) 0.031

(c) 0.062 (d) 0.600

Ans (c)Ans. (c)

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GATE 2006GATE‐2006A 4 mm thick sheet is rolled with 300 mm diameter

ll d hi k i h h i irolls to reduce thickness without any charge in itswidth. The friction coefficient at the work‐rollinterface is 0 1 The minimum possible thickness ofinterface is 0.1. The minimum possible thickness ofthe sheet that can be produced in a single pass is(a) 1 0 mm (b) 1 5 mm(a) 1.0 mm (b) 1.5 mm(c) 2.5 mm (d) 3.7 mm

Ans. (c) find hmin

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GATE 2011 (PI)GATE – 2011 (PI)The thickness of a plate is reduced from 30 mm to10 mm by successive cold rolling passes usingidentical rolls of diameter 600 mm. Assume thatth i h i idth If th ffi i t fthere is no change in width. If the coefficient offriction between the rolls and the work piece is 0.1,the minimum number of passes required isthe minimum number of passes required is(a) 3(b) 4(b) 4(c) 6(d) 7 Ans (d)(d) 7 Ans. (d)

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IES 2003IES – 2003Assertion (A): While rolling metal sheet in rollingmill the edges are sometimes not straight and flatmill, the edges are sometimes not straight and flatbut are wavy.Reason (R): Non‐uniform mechanical properties ofReason (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 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 Ans. (c)

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IES 2002IES – 2002In rolling a strip between two rolls, the position ofh l i i h f dthe neutral point in the arc of contact does notdepend on( ) A t f d ti (b) Di t f th ll(a) Amount of reduction (b) Diameter of the rolls(c) Coefficient of friction (d) Material of the rolls

Ans. (d)

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IES 2001IES – 2001Which of the following assumptions are correct forcold rolling?cold rolling?

1. The material is plastic.2 The arc of contact is circular with a radius greater than2. 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 of3

contact and acts in one direction throughout the arc ofcontact.

S l h i h d i b lSelect the correct answer using the codes given below:Codes:( ) d (b) d(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 1, 2 and 3 Ans. (a)

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IES 2001IES – 2001A strip is to be rolled from a thickness of 30 mm to

i hi h ill h i ll f15 mm using a two‐high mill having rolls ofdiameter 300 mm. The coefficient of friction forunaided bite should nearly beunaided bite should nearly be(a) 0.35 (b) 0.5(c) 0 25 (d) 0 07(c) 0.25 (d) 0.07

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GATE 2008(PI)GATE ‐2008(PI)In a rolling process, thickness of a strip is reduced from 4g p p 4

mm to 3 mm using 300 mm diameter rolls rotating at 100

Th l i f h i i ( / ) h lrpm. The velocity of the strip in (m/s) at the neutral

point is

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

Ans (a)Ans. (a)

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IES 2000 GATE 2010(PI)IES – 2000, GATE‐2010(PI)In the rolling process, roll separating force can bed d bdecreased by(a) Reducing the roll diameter( )(b) Increasing the roll diameter(c) Providing back‐up rolls(d) Increasing the friction between the rolls and themetal

Ans. (a)

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IES 1999IES – 1999Assertion (A): In a two high rolling mill there is ali i h ibl d i i hi k ilimit to the possible reduction in thickness in onepass.R (R) Th d ti ibl i th dReason (R): The reduction possible in the secondpass is less than that in the first pass.(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) 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 Ans. (b)

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IES 1993IES – 1993In order to get uniform thickness of the plate by

lli idrolling process, one provides(a) Camber on the rolls( )(b) Offset on the rolls(c) Hardening of the rolls(d) Antifriction bearings

Ans. (a)

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IES 1993IES – 1993The blank diameter used in thread rolling will be(a) Equal to minor diameter of the thread(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

Ans. (d)

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IES 1992IES – 1992Thread rolling is restricted to(a) Ferrous materials(b) Ductile materials(c) Hard materials(d) None of the above

Ans. (b)

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IAS 2004IAS – 2004Assertion (A): Rolling requires high friction whichi f d iincreases forces and power consumption.Reason (R): To prevent damage to the surface of the

ll d d t l b i t h ld b drolled products, lubricants should be used.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (c)(d) A is false but R is true Ans. (c)

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IAS 2001IAS – 2001Consider the following characteristics of rollingprocess:1. 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(c) 2 and 3 (d) 1, 2 and 3 Ans. (c)

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IAS 2000IAS – 2000Rolling very thin strips of mild steel requires(a) Large diameter rolls(b) Small diameter rolls(c) High speed rolling(d) Rolling without a lubricant Ans. (b)

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IAS 1998IAS – 1998Match List ‐ I (products) with List ‐ II (processes)

d l h i h d iand select the correct answer using the codes givenbelow the lists:

Li t I Li t IIList – I List ‐IIA. M.S. angles and channels 1. WeldingB C b F iB. Carburetors 2. ForgingC. Roof trusses 3. CastingD. Gear wheels 4. RollingCodes:A B C D A B C D

[Ans. (d)]

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

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IAS 2007IAS – 2007Match List I with List II and select the correct answer usingthe code given below the Lists:the code given below the Lists:

List I List II(Type of Rolling Mill) (Characteristic)A. Two high non‐reversing mills 1. Middle roll rotates by frictionB. Three high mills 2. By small working roll, power

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

rotated only in one directionD Cl t ill Di t f ki ll iD. Cluster mills 4. Diameter of working roll is

very smallCode:A B C D A B C D

Ans. (d)

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

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

Ans. (b)

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IAS 2007IAS – 2007Consider the following statements:Roll forces in rolling can be reduced by1. Reducing friction2. Using large diameter rolls to increase the contactarea.3. Taking smaller reductions per pass to reduce thecontact area.Which of the statements given above are correct?(a) 1 and 2 only (b) 2 and 3 only(c) 1 and 3 only (d) 1, 2 and 3 Ans. (c)

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

(a) increase in coefficient of friction(b) decrease in coefficient of friction(c) decrease in roll radius(d) increase in roll velocity

Ans. (a)

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[For IES Conventional Only]

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A ti i R lliAssumptions in Rolling1. Rolls are straight, rigid cylinders.2. Strip is wide compared with its thickness, so that no

widening of strip occurs (plane strain conditions).3. The arc of contact is circular with a radius greater than

the radius of the roll.Th i l i i id f l l i ( i ld4. The material is rigid perfectly plastic (constant yieldstrength).Th ffi i t f f i ti i t t th t l5. The co‐efficient of friction is constant over the tool‐work interface.

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Stress Equilibrium of an Element in Rolling

Considering the thickness of the element perpendicular to theConsidering the thickness of the element perpendicular to theplane of paper to be unity, We get equilibrium equation in x‐direction as,

x

- σ h + (σ +dσ ) (h + dh) - 2pR dθ sin θ + 2 τ R dθ cos θ = 0

x x x

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xFor sliding friction, τ = μp Simplifying and neglecting second order terms, sin and cos 1, we getθ θ θ≅ =

( ), , g

2 ( )xd hpR

θ μθ

= ∓

'0 0

23x

d

p

θ

σ σ σ− = =

( ) ( )'0

3

2d h p pRd

σ θ μθ⎡ ⎤− =⎣ ⎦ ∓( ) ( )

( )

0

'0 1 2

dd ph pR

θ

σ θ μ

⎣ ⎦

⎡ ⎤⎛ ⎞− =⎢ ⎥⎜ ⎟ ∓( )0 '

0

'

1 2h pRd

d p p

σ θ μθ σ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎛ ⎞

( ) ( )'d⎛ ⎞'0 ' '

0 0

d p phd

σθ σ σ⎛ ⎞

+ −⎜ ⎟⎝ ⎠

( ) ( )'01 2d h pR

dσ θ μ

θ⎛ ⎞

=⎜ ⎟⎝ ⎠

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'0

'

Due to cold rolling, increases as h decreases,

thus nearly a constant and itsderivative zeroh

σ

σ

( )0

'0

thus nearly a constant and itsderivative zero.

/ 2

hd p R

σ

σ( )( )

0

'0

2/

p Rdp h

θ θ μσ

= ∓

( )( )

2

'

2 1 cos

/ 2

f fh h R h R

d p R

θ θ

σ

= + − ≈ +

( )( ) ( )0

2'0

/ 2/ f

d p R dh Rp

σθ μ θ

θσ=

+∓

( )

( )Integrating both side

2R dθ θ 2Rμ∫ ∫( )'

02ln /f

R dph

θ θσ =+ 2 2

2 ( )f

R d I II sayR h R

μ θθ θ

=+∫ ∫∓ ∓

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2f

2Rθdθ 2Rθdθ 2θdθ hI lnh h / R Rh Rθ

⎛ ⎞= = = = ⎜ ⎟+ ⎝ ⎠∫ ∫ ∫2fhNow h / R θ

R= +

d hor 2θdθ R

⎛ ⎞ =⎜ ⎟⎝ ⎠

2f

2RμII dθh Rθ

=+∫

2f

2μ dθh / R θ

=+∫

1

f f

R R2μ .tan .θh h

− ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠f f⎝ ⎠

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( ) − ⎛ ⎞⎛ ⎞∴ = +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

' 10

f f

h R Rln p / σ ln 2μ .tan .θ lnCR h h

∓⎝ ⎠ ⎝ ⎠

⎛ ⎞∴ = ⎜ ⎟⎝ ⎠

f f

' μH0

hp Cσ eR

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

1

R

R Rwhere H 2 .tan .θh h⎜ ⎟

⎝ ⎠=

f fh hNow at entry ,θ αH H H ith θ l d b i b ti= ∝

=0

'

Hence H H with θ replaced by in above equationAt exit θ 0

= '0Therefor p σ

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−⎛ ⎞= ⎜ ⎟

⎝ ⎠oμH' o

0hIn theentryzone, p C.σ eR⎝ ⎠

= oμH

RRand C .eh

( )−= 0

o

μ H H'

hhp σ e ( )= 0

0

p σ . eh

In the exit zone⎛ ⎞

= ⎜ ⎟' μH0

In the exit zonehp σ .e⎜ ⎟

⎝ ⎠0

f

p σ .eh

At theneutral po int aboveequationsAt theneutral po int aboveequationswill givesameresults

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( )− =0 n nμ H H μHn nh h. e . eh h

( )−= 0 n

0 f

μ H 2Ho

h hhor eh

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

f

0

h

h1 1or H H ln= −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞

0n 0

f

or H H ln2 μ h

R R− ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

1

f f

R RFrom H 2 .tan .θh h

⎛ ⎞∴ = ⎜ ⎟⎜ ⎟

⎝ ⎠

f f nn

h h Hθ .tan .R R 2

( )

⎜ ⎟⎝ ⎠

= + −n f n

R R 2

and h h 2R 1 cosθ

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bIf back tension σ is thereat Entry,

( ) ( )−′= − 0μ H Ho b

hp σ σ . eh0

f

hIf front tension σ is thereat Exit,

( )′= − μHo f

f

hp σ σ . ehfh

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IFS – 2010 Calculate the neutral plane to roll 250 mm wideannealed copper strip from 2.5 mm to 2.0 mmthickness with 350 mm diameter steel rolls Take µthickness with 350 mm diameter steel rolls. Take µ =0.05 and σ’o =180 MPa.

[10‐marks][10 marks]

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Q. No OptionRolling Ch‐14

1 C2 B3 D4 D5 A6 A

B7 B8 D9 C9 C10 C11 B11 B12 C

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F iForgingg g

By  S K MondalBy  S K Mondal

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GATE 2010 (PI)GATE ‐2010 (PI)Hot die steel, used for large solid dies in drop forging,g p g g

should necessarily have

(a) high strength and high copper content

(b) high hardness and low hardenability(b) high hardness and low hardenability

(c) high toughness and low thermal conductivity

(d) high hardness and high thermal conductivity

Ans. (c)

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IFS‐2011What advantages does press forging have over drop

forging ? Why are pure metals more easily cold workedforging ? Why are pure metals more easily cold worked

than alloys ?

[5‐marks]

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IAS‐2011 MainCompare Smith forging, drop forging, press

forging and upset forging Mention three pointsforging and upset forging. Mention three points

for each.

[10 – Marks]

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IES ‐ 2007Sometimes the parting plane between two forging

dies is not a hori ontal plane gi e the main reasondies is not a horizontal plane, give the main reason

for this design aspect, why is parting plane

provided, in closed die forging?

[    k ][ 2 marks]

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GATE 2007GATE‐2007In open‐die forging, a disc of diameter 200 mm andh i h 6 i d i h b liheight 60 mm is compressed without any barrelingeffect. The final diameter of the disc is 400 mm. Thetrue strain istrue strain is(a) 1.986 (b) 1.686(c) 1 386 (d) 0 602(c) 1.386 (d) 0.602

A ( )Ans. (c)

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GATE 1992 ISRO 2012GATE‐1992, ISRO‐2012The true strain for a low carbon steel bar which is The true strain for a low carbon steel bar which is doubled in length by forging is      (a) 0.307( ) 3 7(b) 0.5(c) 0.693(c) 0.693(d)  1.0

Ans. (c)

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GATE 2012 S Q GATE 2012 (PI)GATE‐2012 Same Q GATE ‐2012 (PI)

A solid cylinder of diameter 100 mm and height 50y g 5

mm is forged between two frictionless flat dies to a

h i h f Th h iheight of 25 mm. The percentage change in

diameter is

(a) 0  (b) 2.07  (c) 20.7  (d) 41.4

Ans  (d) Ans. (d) 

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IES 2006IES – 2006Assertion (A): Forging dies are provided with taper

d f l i l for draft angles on vertical surfaces.Reason (R): It facilitates complete filling of die

it d f bl i flcavity and favourable grain flow.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (c)(d) A is false but R is true Ans. (c)

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IES 2005IES – 2005Consider the following statements:

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

2. Forged components can be provided with thinsections, without reducing the strength.

Whi h f h i b i / ?Which of the statements given above is/are correct?(a) Only 1 (b) Only 2( ) ( )(c) Both 1 and 2 (d) Neither 1 nor 2Ans. (b)

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IES 1996IES – 1996Which one of the following is an advantage off i ?forging?(a) Good surface finish( )(b) Low tooling cost(c) Close tolerance(d) Improved physical property

Ans. (d)

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IES ‐ 2012IES ‐ 2012Which of the following statements is correct for forging?(a) Forgeability is property of forging tool, by whichforging can be done easily.( )(b) Forgeability decreases with temperature upto lowercritical temperature.( ) C i h i l i f h i l(c) Certain mechanical properties of the material areinfluenced by forging.(d) P t l h d ll bilit th f(d) Pure metals have good malleability, therefore, poorforging properties.Ans (c)Ans. (c)

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IES ‐ 2012St t t (I) It i diffi lt t i t i l t l iStatement (I): It is difficult to maintain close tolerance innormal forging operation.Statement (II): Forging is workable for simple shapesStatement (II): Forging is workable for simple shapesand 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) B th St t t (I) d St t t (II)(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is not the correctexplanation of Statement (I)p ( )(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true( ) ( ) ( )Ans. (b)

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IES 1993IES – 1993Which one of the following manufacturing

i h i i f ‘ ’?processes requires the provision of ‘gutters’?(a) Closed die forging( )(b) Centrifugal casting(c) Investment casting(d) Impact extrusion

Ans. (a)

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IES 1997IES – 1997Assertion (A): In drop forging besides the provisionfor flash provision is also to be made in the forgingfor flash, provision is also to be made in the forgingdie for additional space called gutter.Reason (R): The gutter helps to restrict the outwardReason (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 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 Ans. (c)

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IES 2003IES – 2003A forging method for reducing the diameter of a bar

d i h ki i l i dand in the process making it longer is termed as(a) Fullering (b) Punching( ) ( )(c) Upsetting (d) Extruding

Ans. (a)

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IES 2002IES – 2002Consider the following steps involved in hammerf i i d f b kforging a connecting rod from bar stock:1. Blocking 2. Trimming3. Finishing 4. Fullering 5. EdgingWhich of the following is the correct sequence of

i ?operations?(a) 1, 4, 3, 2 and 5( )(b) 4, 5, 1, 3 and 2(c) 5, 4, 3, 2 and 1(d) 5, 1, 4, 2 and 3Ans. (b)

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IES 2003IES – 2003Consider the following steps in forging a connecting

d f h b krod from the bar stock:1. Blocking 2. Trimming3. Finishing 4. EdgingSelect the correct sequence of these operations using the

d i b lcodes given below:Codes:( ) ( )(a) 1‐2‐3‐4 (b) 2‐3‐4‐1(c) 3‐4‐1‐2 (d) 4‐1‐3‐2Ans. (d)

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IES 2005IES – 2005The process of removing the burrs or flash from af d i d f i i ll dforged component in drop forging is called:(a) Swaging (b) Perforating( ) ( )(c) Trimming (d) Fettling

Ans. (c)

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

1. Fullering2. Swaging3. Welding

(a) 1 and 2 only(b) 2 and 3 onlyy(c) 1 and 3 only(b) 1, 2 and 3 only Ans. (a)( ) , 3 y ( )

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IES 2008IES – 2008The balls of the ball bearings are manufacturedf l d Th i i l dfrom steel rods. The operations involved are:1. Ground2. Hot forged on hammers3. Heat treated4. PolishedWhat is the correct sequence of the aboveoperations from start?(a) 3‐2‐4‐1 (b) 3‐2‐1‐4(c) 2‐3‐1‐4 (d) 2‐3‐4‐1Ans. (None) Correct sequence is 2 – 1 – 3 ‐ 4

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IES 2001IES – 2001In the forging operation, fullering is done to   (a) Draw out the material (b) Bend the material(c) Upset the material(d) Extruding the material

Ans. (a)

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IES 2011IES 2011Consider the following statements :1. Any metal will require some time to undergo completeplastic deformation particularly if deforming metal has

f ll d f ll dto fill cavities and corners of small radii.2. For larger work piece of metals that can retain

h f i i i f bltoughness at forging temperature it is preferable to useforge press rather than forge hammer.( ) d t d i th f(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( ) d b l d(c) 1 and 2 are correct but unrelated(d) 1 only correct Ans. (b)

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

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

pressure is applied on the heated endB Press Forging 2 Squeezing actionB. Press Forging 2. Squeezing actionC. Upset Forging 3. Metal is placed between rollers and

pushedpD. Roll Forging 4. Repeated hammer blows

A B C D A B C DAns. (c)

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

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IES – 2008Match List‐I with List‐II and select the correct answer usinggthe code given below the lists:

List‐I (Forging Technique)

List‐II (Process) Technique)

A. Smith Forging 1. Material is only upset to get the desired shape

B. Drop Forging 2. Carried out manually open diesC. Press Forging  3. Done in closed impression dies by hammers

in blowsin blowsD. Machine Forging

4. Done in closed impression dies bycontinuous squeezing force

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

Ans. (a)

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IES 1998IES – 1998Which one of the following processes is most

l d f h f i f b l h d fcommonly used for the forging of bolt heads ofhexagonal shape?( ) Cl d di d f i(a) Closed die drop forging(b) Open die upset forging( ) Cl di f i(c) Close die press forging(d) Open die progressive forging

( )Ans. (c)

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IES 1994 ISRO 2010IES – 1994, ISRO‐2010In drop forging, forging is done by dropping(a) The work piece at high velocity(b) The hammer at high velocity.(c) The die with hammer at high velocity(d) a weight on hammer to produce the requisiteimpact.

Ans. (c)

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IAS 2003IAS – 2003Match List I (Forging Operation) with List II (View of theForging Operation) and select the correct answer using theForging 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 1. 2(B) Fullering(C) Drawing(D) S i

2.

3.4(D) Swaging

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

4.

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

Ans. (a)

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IAS 2001IAS – 2001Match List I (Forging operations) with List II (Descriptions)and select the correct answer using the codes given belowg gthe Lists:

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

different sections along lengthB. Drawing 2. Metal is displaced away from centre,

reducing thickness in middle andgincreasing length

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

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

Ans. (b)

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

Ans. (a)

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IAS 1998IAS – 1998The forging defect due to hindrance to smooth flowf l i h ll d 'L 'of metal in the component called 'Lap' occurs

because( ) Th di id d i t l(a) The corner radius provided is too large(b) The corner radius provided is too small( ) D f i id d(c) Draft is not provided(d) The shrinkage allowance is inadequate

Ans. (b)

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IAS 2002IAS – 2002Consider the following statements related to f iforging:1. Flash is excess material added to stock which flows 

d  ti  liaround parting line.2. Flash helps in filling of thin ribs and bosses in upper diedie.3. Amount of flash depends upon forging force.Whi h  f th   b   t t t     t?Which of the above statements are correct?(a) 1, 2 and 3 (b) 1 and 2( )   d  (d)   d (c) 1 and 3 (d) 2 and 3Ans. (b)

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IES 2011IES 2011Assertion (A) : Hot tears occur during forgingb f i l i i h bl k i lbecause of inclusions in the blank materialReason (R) : Bonding between the inclusionsand the parent material is through physicaland the parent material is through physicaland chemical bonding.(a) Both A and R are individually true and R is the( ) ycorrect explanation of A(b) Both A and R are individually true but R is NOTh l i f Athe correct explanation of A(c) A is true but R is false(d) A i f l b t R i t A ( )(d) A is false but R is true Ans. (c)

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GATE ‐2008 (PI)Match the followingMatch the following

Group ‐1 Group‐2  i kli   iP . Wrinkling 1. Upsetting

Q. Centre burst 2. Deep drawingR. Barrelling 3. ExtrusionS. 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 

Ans. (c)

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IES 2005 ConventionalIES – 2005 ConventionalA strip of lead with initial dimensions 24 mm x 24p 4 4

mm x 150 mm is forged between two flat dies to a

fi l i f 6 6 If hfinal size of 6 mm x 96 mm x 150 mm. If the

coefficient of friction is 0.25, determine the

maximum forging force. The average yield stress of

l d i t i i N/ 2lead in tension is 7 N/mm2

[10]

Will be discussed in class

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IES – 2007 ConventionalIES – 2007 ConventionalA cylinder of height 60 mm and diameter 100 mm isf d b fl di Fi dforged at room temperature between two flat dies. Findthe die load at the end of compression to a height 30mm using slab method of analysis The yield strength ofmm, using slab method of analysis. The yield strength ofthe work material is given as 120 N/mm2 and thecoefficient of friction is 0.05. Assume that volume is5constant after deformation. There is no sticking. Alsofind mean die pressure. [20‐Marks]

Will be discussed in classWill be discussed in class

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IES – 2006 ConventionalIES – 2006 ‐ ConventionalA certain disc of lead of radius 150 mm and thickness 50

i d d hi k f b dimm is reduced to a thickness of 25 mm by open dieforging. If the co‐efficient of friction between the job anddie is 0 25 determine the maximum forging force Thedie is 0.25, determine the maximum forging force. Theaverage shear yield stress of lead can be taken as 4N/mm2. [10 – Marks]/ [ ]

Will be discussed in classWill be discussed in class

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Practice Problem 1Practice Problem‐1A 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 96 mm 0 mm If the coefficient of friction is6 mm x 96 mm x 150 mm. If the coefficient of friction is

0.05, determine the maximum forging force. Take the

average yield strength in tension is 7 N/mm2

[Ans. 178.24 kN][ 7 4 ]

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Practice Problem 2Practice Problem‐2A circular disc of 200 mm in diameter and 100 mm in

height is compressed between two flat dies to a height of

0 mm Coefficient of friction is 0 and a erage ield50 mm. Coefficient of friction is 0.1 and average yield

strength in compression is 230 MPa. Determine the

maximum die pressure.

[Ans. 405 MPa][ 4 5 ]

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Practice Problem 3Practice Problem‐3A 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 curvemm. Coefficient of friction is 0.2 and flow curve

equation is MPa . Calculate the maximum17.01030εσ =f

forging force.

[Ans. 46.26 MN][ 4 ]

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

]170the equation ]yf σεσ == 17.01030

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Practice Problem 4Practice Problem‐4A 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 materialfriction is 0.05. The flow curve equation of the material

is given by . Determine maximumMPa)01.0(200 41.0εσ +=f

forging load, mean die pressure and maximum pressure.

[ Ans. 9.771 MN, 178 MPa, 221 MPa][ 9 77 , 7 , ]

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

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

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Practice Problem ‐5 {GATE‐2010 (PI)}{ ( )}During open die forging process using two flat and parallel dies, a solid circular steel disc of initial radius (R ) 200 mm and initial IN

height (H )50 mm attains a height (H )of 30 mm and radius of R .Al

N

IN FN FN

ong the die-disc interfaces.

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

FN

RReμ μ

−⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

( )

,

2

ii. in the region R sliding friction prevails, andss FN

R

r Rμ

⎝ ⎠≤ ≤

( ) 3 ,

FNFN

R rHp Ke and pμ

τ 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

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Contd…….

Practice Problem ‐5 {GATE‐2010 (PI)}

SS

iii.In the region 0 r R ,sticking condition prevailsThe value of R (in mm), where sticking condition changes to sliding

SS≤ ≤

friction, is(a) 241.76 (b) 254.55 (c) 265.45 (d) 278.20

Ans. (b)( )

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Ch 15: ForgingCh‐15: ForgingQ. No Option Q. No Option1 A 6 B2 A 7 C3 A 8 C3 A 8 C4 A 9 C5 B5 B

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Extrusion & Drawingg

d lBy  S K Mondal

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IAS 2010 MainIAS‐2010 MainHow are metal tooth‐paste tubes madeHow are metal tooth‐paste tubes made

commercially ? Draw the tools configuration with

the help of a neat sketch.

[30 Marks][30‐Marks]

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IES 2009 C ti lIES 2009 Conventional

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IES 2011 C ti lIES – 2011 ConventionalA 12.5 mm diameter rod is to be reduced to 10 mmdi b d i i i l d fdiameter by drawing in a single pass at a speed of 100m/min. Assuming a semi die angle of 5o and coefficientof friction between the die and steel rod as 0.15,o ct o betwee t e d e a d stee od as 0. 5,calculate:(i) The power required in drawing(ii) Maximum possible reduction in diameter of the rod(iii) If the rod is subjected to a back pressure of 50N/ 2 h ld b h d d iN/mm2 , what would be the draw stress and maximumpossible reduction ?Take stress of the work material as 400 N/mm2Take stress of the work material as 400 N/mm .Will be discussed in the class                           [15 Marks] 

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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‐sectiondiameter and 100 mm length is reduced in cross sectionby drawing it successively through a series of seven diesof decreasing exit diameter. During each of theseg gdrawing operations, the reduction in cross‐sectional areais 35%. The yield strength of the material is 200 MPa.I i h d iIgnore strain hardening.The total true strain applied and the final length (in

) ti lmm), respectively, are(a) 2.45 and 8 17 (b) 2.45 and 345( ) d (d) d(c) 3.02 and 2043 (d) 3.02 and 3330Ans. (c)

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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‐sectiondiameter and 100 mm length is reduced in cross sectionby drawing it successively through a series of seven diesof decreasing exit diameter. During each of theseg gdrawing operations, the reduction in cross‐sectional areais 35%. The yield strength of the material is 200 MPa.I i h d iIgnore strain hardening.Neglecting friction and redundant work, the force (in kN)  i d f  d i  th  b  th h th  fi t di  ikN) required for drawing the bar through the first die, is(a) 15.71  (b) 10.21 ( )    (d) (c) 6.77  (d) 4.39Ans. (d)

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JWM 2010JWM 2010Assertion (A) : Extrusion speed depends on workmaterialmaterial.Reason (R) : High extrusion speed causes cracks inthematerialthematerial.(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 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 trueAns. (a)

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GATE 2006GATE‐2006In a wire drawing operation, diameter of a steel wirei d d f 8 Th flis reduced from 10 mm to 8 mm. The mean flowstress of the material is 400 MPa. The ideal forcerequired for drawing (ignoring friction andrequired for drawing (ignoring friction andredundantwork) is(a) 4 48 kN (b) 8 97 kN(a) 4.48 kN (b) 8.97 kN(c) 20.11 kN (d) 31.41 kNAns (b)Ans. (b)

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GATE 2001 GATE 2007 (PI)GATE‐2001, GATE ‐2007 (PI)For rigid perfectly‐plastic work material, negligiblei f f i i d d d k hinterface friction and no redundant work, thetheoretically maximum possible reduction in thewire drawing operation iswire drawing operation is(a) 0.36 (b) 0.63(c) 1 00 (d) 2 72(c) 1.00 (d) 2.72Ans. (b)

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GATE ‐2008 (PI) Linked S‐1G 008 ( ) ed SA 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.y

Neglecting friction and strain hardening, the stress

required for drawing (in MPa) is

(a) 178 5 (b) 357 0 (c) 1287 5 (d) 2575 0(a) 178.5 (b) 357.0 (c) 1287.5 (d) 2575.0

Ans. (b)

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GATE ‐2008 (PI) Linked S‐2G 008 ( ) ed SA 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.y

The power required for the drawing process (in kW) is

(a) 8.97 (b) 14.0 (c) 17.95 (d) 28.0

Ans (a)Ans.(a)

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GATE 2003GATE‐2003A brass billet is to be extruded from its initialdi f fi l di fdiameter of 100 mm to a final diameter of 50 mm.The working temperature of 700°C and theextrusion constant is 250 MPa The force requiredextrusion constant is 250 MPa. The force requiredfor extrusion is(a) 5 44 MN (b) 2 72 MN(a) 5.44 MN (b) 2.72 MN(c) 1.36 MN (d) 0.36 MN

Ans. (b)

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

mm length and 50 mm diameter is extruded.

Considering an ideal deformation process (noConsidering an ideal deformation process (no

friction and no redundant work), extrusion ratio 4,

and average flow stress of material 300 MPa, the

pressure (in MPa) on the ramwill bep ( )

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

Ans. (a)

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GATE 1996GATE‐1996A wire of 0.1 mm diameter is drawn from a rod of 15

di Di i i d i f % %mm diameter. Dies giving reductions of 20%, 40%and 80% are available. For minimum error in thefinal size the number of stages and reduction atfinal size, the number of stages and reduction ateach stage respectively would be(a) 3 stages and 80% reduction for all three stages(a) 3 stages and 80% reduction for all three stages(b) 4 stages and 80% reduction for first three stagesfollowed by a finishing stage of 20% reductionfollowed by a finishing stage of 20% reduction(c) 5 stages and reduction of 80%, 80%.40%, 40%, 20%in a sequenceq(d) none of the above Ans. (b)

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

Ans. (a)

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IES 2007IES – 2007Which one of the following is the correctstatement?statement?(a) Extrusion is used for the manufacture of seamlesstubes.tubes.(b) Extrusion is used for reducing the diameter of roundbars 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 surfacemetal by setting up compressive stresses on its surface(d) Extrusion comprises pressing the metal inside achamber to force it out by high pressure through anchamber to force it out by high pressure through anorifice which is shaped to provide the desired from of thefinished part. Ans. (d)

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IES 2007IES – 2007Assertion (A): Greater force on the plunger is requiredi f di t t i th i di tin case of direct extrusion than indirect one.Reason (R): In case of direct extrusion, the direction ofthe force applied on the plunger and 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 correct( ) yexplanation of A(b) Both A and R are individually true but R is not the

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

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IES ‐ 2012IES ‐ 2012Which of the following are correct for an indirect hot

i ?extrusion process?1. Billet remains stationary2. There is no friction force between billet and containerwalls.

Th f i d h h i i3. The force required on the punch is more incomparison to direct extrusion.

E t i t h t b id d t4. Extrusion parts have to be provided a support.(a) 1, 2, 3 and 4 (b) 1, 2 and 3 only( ) d l (d) d l(c) 1, 2 and 4 only (d) 2, 3 and 4 onlyAns. (c)

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IES 1993IES – 1993Assertion (A): Direct extrusion requires larger forceh i di ithan indirect extrusion.Reason (R): In indirect extrusion of cold steel, zinch h t ti i dphosphate coating is used.

(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (b)(d) A is false but R is true Ans.(b)

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IES 1994IES – 1994Metal extrusion process is generally used for

d iproducing(a) Uniform solid sections( )(b) Uniform hollow sections(c) Uniform solid and hollow sections(d) Varying solid and hollow sections.Ans. (c)

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IES 2009IES – 2009Which one of the following statements is correct?(a) In extrusion process, thicker walls can be obtainedby increasing the forming pressure( )(b) Extrusion is an ideal process for obtaining rods frommetal having poor density( ) A d ll f i di d i hi h(c) As compared to roll forming, extruding speed is high(d) Impact extrusion is quite similar to Hooker's processi l di th fl f t l b i i th di tiincluding the flow of metal being in the same directionAns. (c)

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IES 1999IES – 1999Which one of the following is the correct

f h i f l i i ?temperature range for hot extrusion of aluminium?(a) 300‐340°C (b) 350‐400°C( ) ( )(c) 430‐480°C (d) 550‐650°CAns. (c)

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IES 2000IES – 2000Consider the following statements:I f d t iIn forward extrusion process

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

2. The ram and the extruded product travel in the oppositedirection.h d f l f h d d d i h3. The speed of travel of the extruded product is same as that

of the ram.4 The speed of travel of the extruded product is greater than4. The speed of travel of the extruded product is greater than

that of the ram.Which of these Statements are correct?(a) 1 and 3 (b) 2 and 3(c) 1 and 4 (d) 2 and 4 Ans. (c)

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

Ans. (b)

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IES ‐ 2012IES ‐ 2012Extrusion process can effectively reduce the cost of

d h hproduct through(a) Material saving( )(b) process time saving(c) Saving in tooling cost(d) saving in administrative cost

Ans. (c)

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IES 2008IES – 2008Which one of the following methods is used for the

f f ll ibl h b ?manufacture of collapsible tooth‐paste tubes?(a) Impact extrusion (b) Direct extrusion( ) ( )(c) Deep drawing (d) Piercing

Ans. (a)

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IES 2003IES – 2003The extrusion process (s) used for the production of

h b i /toothpaste tube is/are1. Tube extrusion2. Forward extrusion3. Impact extrusionSelect the correct answer using the codes given below:Codes:(a) 1 only (b) 1 and 2(c) 2 and 3 (d) 3 onlyAns. (d)

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IES 2001IES – 2001Which of the following statements are the salientfeatures of hydrostatic extrusion?features of hydrostatic extrusion?

1. It is suitable for soft and ductile material.2. It is suitable for high‐strength super‐alloys.g g p y3.The billet is inserted into the extrusion chamber and pressureis applied by a ram to extrude the billet through the die.h bill i i d i h i h b h i i4. The billet is inserted into the extrusion chamber where it is

surrounded by a suitable liquid. The billet is extrudedthrough the die by applying pressure to the liquid.g y pp y g p qSelect the correct answer using the codes given below:Codes:(a) 1 and 3 (b) 1 and 4(c) 2 and 3 (d) 2 and 4 Ans. (d)

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

Ans. (a)

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IES 2010IES 2010Assertion (A): Pickling and washing of rolled rodsis carried out beforewire drawingis carried out beforewire drawing.Reason (R): They lubricate the surface to reducefriction while drawing wiresfriction while drawing wires.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (c)(d) A is false but R is true Ans. (c)

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IES 2009IES – 2009Which one of the following stress is involved in thei d i ?wire drawing process?

(a) Compressive (b) Tensile( ) ( )(c) Shear (d) Hydrostatic stress

Ans. (b)

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IES 1993IES – 1993Tandem drawing of wires and tubes is necessarybbecause(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

Ans. (a)

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IES 2000IES – 2000Match List I (Components of a table fan) with List II(Manufacturing processes) and select the correct(Manufacturing processes) and select the correctanswer using the codes given below the Lists:

List I List IIList I List IIA. Base with stand 1. Stamping and

pressingB. Blade 2. Wire drawingC. Armature coil wire 3. TurningD. Armature shaft 4. CastingCodes:A B C D A B C D( ) ( )

Ans. (d)

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

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

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

progressive dies 5. Ironingprogressive dies 5. IroningCode:A B C D A B C D(a) 2 5 1 4 (b) 4 1 5 3( ) 5 4 ( ) 4 5 3(c) 5 2 3 4 (d) 5 2 1 3

Ans. (d)  

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IES 1996IES – 1996Match List I with List II and select the correct answerList I (Metal/forming process) List II (Associated force)

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

Ans. (c)

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IES 1996IES – 1996In wire drawing process, the bright shining surface

h i i b i d ifon thewire is obtained if one(a) does not use a lubricant( )(b) uses solid powdery lubricant.(c) uses thick paste lubricant(d) uses thin film lubricant

Ans. (d)

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IES 1994IES – 1994Match List I with List II and select the correct answer 

i   h   d   i  b l   h  Liusing the codes given below the Lists:List I (Metal farming process) List II (A similar process)  

A Bl ki   Wi  d iA. Blanking  1. Wire drawingB. Coining  2. PiercingC E i E b iC. Extrusion 3. EmbossingD. Cup drawing  4. Rolling

5. BendingCodes:A B  C  D A  B  C  D

Ans. (d)

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

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IES 1993 ISRO 2010IES – 1993, ISRO‐2010Match List I with List II and select the correctanswer using the codes given below the lists:answer using the codes given below the lists:List I (Mechanical property) List II (Related to)A Malleability 1 Wire drawingA. Malleability 1. Wire drawingB. Hardness 2. Impact loadsC Resilience 3 Cold rollingC. Resilience 3. Cold rollingD. Isotropy 4. Indentation

5. Direction Ans  (b)5. DirectionCodes:A B C D A B C D(a) 4 2 1 3 (b) 3 4 2 5

Ans. (b)

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

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IES 2007IES – 2007Which metal forming process is used for

f f l l i ?manufacture of long steel wire?(a) Deep drawing (b) Forging( ) ( )(c) Drawing (d) Extrusion

Ans. (c) Wire Drawing

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IES 2005IES – 2005Which of the following types of stresses is/arei l d i h i d i i ?involved in thewire‐drawing operation?(a) Tensile only( )(b) Compressive only(c) A combination of tensile and compressive stresses(d) A combination of tensile, compressive and shearstresses

Ans. (a)

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IES 2000IES – 2000Which one of the following lubricants is most

i bl f d i ild l i ?suitable for drawing mild steel wires?(a) Sodium stearate (b) Water( ) ( )(c) Lime‐water (d) Kerosene

Ans. (c)

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

Ans. (b)

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

A S l t b R ll f iA. Seamless tubes 1. Roll formingB. Accurate and smooth tubes 2. Shot peeningC S f h i hi h F iC. Surfaces having higher 3. Forging

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

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

Ans. (a)

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IAS 2004IAS – 2004Assertion (A): Indirect extrusion operation can be

f d i h b i b i hperformed either by moving ram or by moving thecontainer.R (R) Ad t i i di t t i i lReason (R): Advantage in indirect extrusion is lessquantity of scrap compared to direct extrusion.(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) 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 Ans. (d)

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IAS 1995IAS – 1995The following operations are performed while

i h bill f ipreparing the billets for extrusion process:1. Alkaline cleaning2. Phosphate coating3. Pickling4. Lubricating with reactive soap.The correct sequence of these operations is(a) 3, 1, 4, 2 (b) 1, 3, 2, 4(c) 1, 3. 4, 2 (d) 3, 1, 2, 4Ans. (d)

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IAS 2001IAS – 2001Match List I (Products) with List II (Suitable

) d l h i hprocesses) and select the correct answer using thecodes given below the Lists:

Li t I Li t IIList I List IIA. Connecting rods 1. WeldingB P l E iB. Pressure vessels 2. ExtrusionC. Machine tool beds 3. Forging

( )D. Collapsible tubes 4. Casting Ans. (a)Codes:A B C D A B C D(a) 3 1 4 2 (b) 4 1 3 2(c) 3 2 4 1 (d) 4 2 3 1

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IAS 1997IAS – 1997Extrusion force DOES NOTdepend upon the(a) Extrusion ratio(b) Type of extrusion process(c) Material of the die(d) Working temperature

Ans. (c)

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IAS 2000IAS – 2000Assertion (A): Brittle materials such as grey casti b d d b h d i iiron cannot be extruded by hydrostatic extrusion.Reason(R): In hydrostatic extrusion, billet is

if l d f ll id b th li iduniformly compressed from all sides by the liquid.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (d)(d) A is false but R is true Ans. (d)

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IAS 2002IAS – 2002Assertion (A): In wire‐drawing process, the rodcross section is reduced gradually by drawing itcross‐section is reduced gradually by drawing itseveral times in successively reduced diameter dies.Reason (R): Since each drawing reduces ductility ofReason (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 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 Ans. (b)

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

List –I  List –II

A  Connecting rods 1  WeldingA. Connecting rods 1. Welding

B. Pressure vessels 2. Extrusion

C. Machine tool beds 3. Forming

D  Collapsible tubes 4  Casting               Ans  (b)Codes

A B C D A B C D

D. Collapsible tubes 4. Casting               Ans. (b)

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

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IAS 1994IAS 1994Which of the following methods can be used formanufacturing 2 metre long seamless metallicmanufacturing 2 metre long seamless metallictubes?1 Drawing 2 Extrusion1. Drawing 2. Extrusion3. Rolling 4. SpinningSelect the correct answer using the codes given belowSelect the correct answer using the codes given belowCodes:( ) d (b) d(a) 1 and 3 (b) 2 and 3(c) 1, 3 and 4 (d) 2, 3 and 4A (b)Ans. (b)

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Wire Drawing Analysis (Home Work)g y ( )

Theequilibrium equation in x-direction will bedx⎛ ⎞( ) ( )2 2 cos 2

cosx x x xdxd r dr r rσ σ π σ π τ α πα

⎛ ⎞+ + − + ⎜ ⎟⎝ ⎠

⎛ ⎞sin 2 0cosx

dxP rα πα

⎛ ⎞+ =⎜ ⎟⎝ ⎠

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22 2 2 tan 0x x x xor rdr d r r dx P rdxσ σ τ α+ + + =2Dividing by r dr and taking dx/dr = cot we get

22dα

( ) 22 cot 0x xx x

d Pdr r rσ τσ α+ + + =

Vertical component of due to small half die angles and that of can be neglected

x xP Pτ

≅angles and that of can be neglected.Thefore,

xτ two principal stresses are andx xPσ −

Both Tresca's and Von-Mises criteria willgivePσ σ+ =

( )andx x o

x x o x

PP

σ στ μ μ σ σ

+

= = −

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( )22 cot 0o xx od μ σ σσ σ α−

+ + =cot 0

Let cotdr r r

B

α

μ α

+ +

=

( )2 1xx o

d B Bd

μσ σ σ= − +⎡ ⎤⎣ ⎦( )

2x

dr rdor drσ

⎣ ⎦

=⎡ ⎤( )1

Integrating both sidex o

or drrB Bσ σ− +⎡ ⎤⎣ ⎦

( ) ( )

Integrating both side1ln 1 2lnB B rCσ σ− + × =⎡ ⎤⎣ ⎦( ) ( )ln 1 2ln

{Cis integration cont.}

x oB B rCB

σ σ+ ×⎡ ⎤⎣ ⎦

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( ) ( )21 Bx oor B B rCσ σ− + =

1

. ,so x bBC at r r σ σ= =

( )1

21 Bb oB B

Cr

σ σ− +⎡ ⎤⎣ ⎦∴ =

( ) 2 21

1

o

B Bo

r

B r rσ ⎡ ⎤+ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟

( ) 1 .ox b

o o

orB r r

σ σ⎢ ⎥= − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

( ) 2 21

( ) 1 .B B

f fod b

r rBDrawing stress

B r rσ

σ σ⎡ ⎤+ ⎛ ⎞ ⎛ ⎞⎢ ⎥∴ = − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦o oB r r⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

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Extrusion Analysis (Home Work)y ( )

s

For a round bar both wiredrawing and extrusion willgivesame equation except B C

( ) ( )2

same equation except B.C

1 Bx oB B rCσ σ∴ − + =( ) ( )

1

. , 0 (at exit stress is zero)sf xB C at r r σ= =

( )1

21 BoB

Crσ− +⎡ ⎤⎣ ⎦∴ =

( )2

1

f

B

r

B rσ ⎡ ⎤⎛ ⎞+ ⎢ ⎥⎜ ⎟( )1

1ox

f

B rorB r

σσ

⎛ ⎞+ ⎢ ⎥= − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

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( )2

1

o

B

at r r

B

=

⎡ ⎤⎛ ⎞( )11o o

xof

B rB r

σσ

⎡ ⎤⎛ ⎞+ ⎢ ⎥= − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦2

Extrusion ratio for round baro oA rR

⎝ ⎠⎣ ⎦

⎛ ⎞= = ⎜ ⎟⎜ ⎟Extrusion ratio, for round bar

f f

RA r

h

= = ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞

for flat stocko

f

hh

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

( ) 211o B

xo

BR

σ+

⎡ ⎤= −⎣ ⎦B

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If effect of container friction is consideredram pressure required by container friction

uniforminterfaceshear stress betweenfp

τ

=

= uniforminterfaceshear stress betweenbillet and container wall

iτ =

20 0

2. 2 if i f

o

Lp r r L or prτπ π τ= =

tTotal Extrusion Pressure(P ) xo fpσ∴ = +2d E t i L da 2

0nd Extrusion Load .tp rπ=

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Ch 17: ExtrusionCh‐17: ExtrusionQ. No Option Q. No Option1 D 8 B2 C 9 B3 D 10 A3 D 10 A4 D 11 B5 B 12 B5 B 12 B6 C 13 A7 B7 B

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Ch 16: DrawingCh‐16: Drawing

Q. No Option1 A2 C3 C4 B5 C6 D6 D

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Sheet Metal OperationSheet Metal Operation

By  S K MondalBy  S K Mondal

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ExampleExampleDetermine the die and punch sizes for blanking a circulardi f di t f h t h thi k idisc of 20‐mm diameter from a sheet whose thickness is 1.5mm.

Shear strength of sheet material = 294 MPa

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

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ExampleExampleEstimate the blanking force to cut a blank 25 mm wideg 5

and 30 mm long from a 1.5 mm thick metal strip, if the

l i h h f h i l i N/ultimate shear strength of the material is 450 N/mm2.

Also determine the work done if the percentage

penetration is 25 percent of material thickness.

Ans. 74.25 kN, 27.84 NmAns. 74.25 kN, 27.84 Nm

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IAS‐2011 MainFor punching a 10 mm circular hole, and cutting arectangular blank of 50 x 200 mm from a sheet of 1

hi k ( ild l hmm thickness (mild steel, shear stress = 240N/mm2), Calculate, in each case :(i) Si f h(i) Size of punch(ii) Size of die(iii) F i d [ M k ](iii) Force required. [10‐Marks]

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IES 1999IES – 1999A hole is to be punched in a 15 mm thick plateh i l i h h f N If hhaving ultimate shear strength of 3N‐mm‐2. If theallowable crushing stress in the punch is 6 N‐mm‐2,the diameter of the smallest hole which can bethe diameter of the smallest hole which can bepunched is equal to(a) 15 mm (b) 30 mm(a) 15 mm (b) 30 mm(c) 60 mm (d) 120 mm

Ans. (b)

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ISRO 2008 2011ISRO‐2008, 2011With a punch for which the maximum crushingstress is 4 times the maximum shearing stress ofthe plate, the biggest hole that can be punched inthe plate would be of diameter equal to

11(a) Thickness of plate41

×

1(b) Thickness of plate2

( ) Pl t thi k

×

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

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E lExampleA hole 100 mm diameter is to be punched in steel plateA hole, 100 mm diameter, is to be punched in steel plate

5.6 mm thick. The ultimate shear stress is 550 N/mm2 .

With normal clearance on the tools, cutting is complete

at 40 per cent penetration of the punch. Give suitable4 p p p

shear angle for the punch to bring the work within the

it f Tcapacity of a 30T press.

Ans. 4.2o

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ExampleExampleA washer with a 12.7 mm internal hole and an outsidediameter of 25 4 mm is to be made from 1 5 mm thickdiameter of 25.4 mm is to be made from 1.5 mm thickstrip. The ultimate shearing strength of the material ofthe washer is 280 N/mm2.(a) Find the total cutting force if both punches act atthe 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.( ) k % d h h f(c) Taking 60% penetration and shear on punch of 1mm, what will be the cutting force if both punches acttogether.gAns. 50.25 kN, 33.51 kN, F = 45.24 kN

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GATE 2010 Statement Linked 1GATE‐2010 Statement Linked 1Statement 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 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.

400400

Assuming force vs displacement curve to be rectangular,h k d (i J) i

S

the work done (in J) is(a) 100 (b) 200 (c) 250 (d) 300 [Ans. (a)]

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GATE 2010 Statement Linked 2GATE‐2010 Statement Linked 2Statement for Linked Answer Questions:In a shear cutting operation a sheet of 5mm thicknessIn 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 edge.The ultimate shear strength of the sheet is 100 MPa andThe ultimate shear strength of the sheet is 100 MPa andpenetration to thickness ratio is 0.2. Neglect friction.

400

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 [Ans. (b)]

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GATE 2011GATE 2011The 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 to( ) k(a) 45 kN(b) 70 kN(c) 141 kN(d) 3500 kN

Ans. (c)

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

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

A (b)Ans. (b)

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

mild steel plate 10 mm thick, when ultimate shear

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

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

Ans (b)Ans. (b)

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GATE 2007GATE‐2007The force requirement in a blanking operation ofl b l h i kN Th hi k flow carbon steel sheet is 5.0 kN. The thickness ofthe sheet is ‘t’ and diameter of the blanked part is‘d’ For the same work material if the diameter ofd. For the same work material, if the diameter ofthe blanked part is increased to 1.5 d and thicknessis reduced to 0.4 t, the new blanking force in kN is4 , g(a) 3.0 (b) 4.5(c) 5.0 (d) 8.0(c) 5.0 (d) 8.0Ans. (a)

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GATE 2004GATE‐200410 mm diameter holes are to be punched in a steelh f hi k Sh h f hsheet of 3 mm thickness. Shear strength of thematerial is 400 N / mm2 and penetration is 40%.Shear provided on the punch is 2 mm The blankingShear provided on the punch is 2 mm. The blankingforce during the operation will be(a) 22 6 kN (b) 37 7 kN(a) 22.6 kN (b) 37.7 kN(c) 61.6 kN (d) 94.3 kN

Ans. (a)

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GATE 2003GATE‐2003A metal disc of 20 mm diameter is to be punchedf h f hi k Th h d hfrom a sheet of 2 mm thickness. The punch and thedie clearance is 3%. The required punch diameter is( ) 88 (b)(a) 19.88 mm (b) 19.94 mm(c) 20.06 mm (d) 20.12 mm

Ans. (a)

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GATE ‐ 2012Calculate the punch size in mm, for a circular blankingoperation for which details are given below.Size of the blank 25 mmThickness of the sheet 2 mmRadial clearance between punch and die 0.06 mmDie allowance 0.05 mm

(a) 24.83 (b) 24.89(c) 25.01 (d) 25.17( ) 5 ( ) 5 7

Ans. (a)

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GATE‐2008(PI)A blank of 50 mm diameter is to be sheared from a

sheet of 2.5 mm thickness. The required radialsheet of 2.5 mm thickness. The required radial

clearance between the die and the punch is 6% of

sheet thickness The punch and die diameters (in mm)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

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

Ans. (c)

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

Ans. (c)

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GATE 2001GATE‐2001The cutting force in punching and blanking

i i l d doperations mainly depends on(a) The modulus of elasticity of metal( )(b) The shear strength of metal(c) The bulk modulus of metal(d) The yield strength of metal

Ans. (b)

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GATE 1996GATE‐1996A 50 mm diameter disc is to be punched out from a

b l h hi k Th di fcarbon steel sheet 1.0 mm thick. The diameter ofthe punch should be( ) (b)(a) 49.925 mm (b) 50.00 mm(c) 50.075 mm (d) none of the above

Ans. (d)

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IES 1994IES – 1994In sheet metal blanking, shear is provided on

h d di hpunches and dies so that(a) Press load is reduced( )(b) Good cut edge is obtained.(c) Warping of sheet is minimized(d) Cut blanks are straight

Ans. (a)

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IES 2002IES – 2002Consider the following statements related toi i d bl kipiercing and blanking:

1. Shear on the punch reduces the maximum cuttingfforce2. Shear increases the capacity of the press needed

Sh i h lif f h h3. Shear increases the life of the punch4. The total energy needed to make the cut remains

lt d d t i i f hunaltered due to provision of shearWhich of these statements are correct?( ) d (b) d(a) 1 and 2 (b) 1 and 4(c) 2 and 3 (d) 3 and 4 Ans. (b)

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

Ans. (c)

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IES 2006IES – 2006In which one of the following is a flywheel generally

l d?employed?(a) Lathe (b) Electric motor( ) ( )(c) Punching machine (d) Gearbox

Ans. (c)

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IES 2004IES – 2004Which one of the following statements is correct?If the size of a flywheel in a punching machine isincreased( )(a) Then the fluctuation of speed and fluctuation ofenergy will both decrease(b) Th h fl i f d ill d d h(b) Then the fluctuation of speed will decrease and thefluctuation of energy will increase( ) Th th fl t ti f d ill i d th(c) Then the fluctuation of speed will increase and thefluctuation 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 Ans. (a)

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IES 1997IES – 1997For 50% penetration of work material, a punch withi l h l hi k illsingle shear equal to thickness will(a) Reduce the punch load to half the value( )(b) Increase the punch load by half the value(c) Maintain the same punch load(d) Reduce the punch load to quarter load

Ans. (a)

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IAS 2000IAS – 2000A blank of 30 mm diameter is to be produced out of

hi k h i l di If 6% l i10 mm thick sheet on a simple die. If 6% clearance isrecommended, then the nominal diameters of dieand punch are respectivelyand punch are respectively(a) 30.6 mm and 29.4 mm(b) 30 6 mm and 30 mm(b) 30.6 mm and 30 mm(c) 30 mm and 29.4 mm(d) d 8 8(d) 30 mm and 28.8 mm

A (d)Ans. (d)

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GATE – 2007 (PI)GATE – 2007 (PI)Circular blanks of 35 mm diameter are punchedf l h f hi k If hfrom a steel sheet of 2 mm thickness. If theclearance per side between the punch and die is

b k i h i f h dto be kept as 40 microns, the sizes of punch anddie should respectively be

(a) 35+0.00 and 35+0.040 (b) 35‐0.040 and 35‐0.080

(c) 35‐0.080 and 35+0.00 (d) 35+0.040 and 35‐0.080(c) 35 and 35 (d) 35 4 and 35

Ans. (c)

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IAS 1994IAS – 1994In a blanking operation to produce steel washer, the

i h l d d i N Th lmaximum punch load used in 2 x 105 N. The platethickness is 4 mm and percentage penetration is 25.Thework done during this shearing operation isThework done during this shearing operation is(a) 200J (b) 400J(c) 600 J (d) 800 J(c) 600 J (d) 800 J

A ( )Ans. (a)

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IAS 2002IAS – 2002In deciding the clearance between punch and die in

k i h i h f ll i l i h l f lpress work in shearing, the following rule is helpful:(a) Punch size controls hole size die size controls blankisize(b) Punch size controls both hole size and blank size( ) Di i l b h h l i d bl k i(c) Die size controls both hole size and blank size(d) Die size controls hole size, punch size controls blankisize

A ( )Ans. (a)

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IAS 2007IAS – 2007For punching operation the clearance is provided

hi h f h f ll i ?onwhich one of the following?(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

Ans. (b)

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IAS 1995IAS – 1995Assertion (A): A flywheel is attached to a punching

d i d fl ipress so as to reduce its speed fluctuations.Reason(R): The flywheel stores energy when itsspeed increasespeed increase.(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 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 trueAns. (a)

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IES 2002IES – 2002Which one is not a method of reducing cuttingf h l di f ?forces to prevent the overloading of press?(a) Providing shear on die( )(b) Providing shear on punch(c) Increasing die clearance(d) Stepping punches

Ans. (c)

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IAS 2003IAS – 2003Match List I (Press‐part) with List II (Function) and select thecorrect answer using the codes given below the lists:

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

distance(C) Stopper 3. Ejection of the work‐piece from die cavity( ) pp 3 j p y(D) Knockout 4. Holding the small punch in the proper

positionCodes:A B C D A B C D

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

Ans. (c)

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IES 1999IES – 1999Assertion (A): In sheet metal blanking operation,l b i h diclearancemust be given to the die.

Reason (R): The blank should be of requireddi idimensions.(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 thecorrect explanation of Acorrect explanation of A(c) A is true but R is false(d) A is false but R is true Ans (d)(d) A is false but R is true Ans. (d)

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IAS 2003IAS – 2003The 'spring back' effect in press working is(a) Elastic recovery of the sheet metal after removal ofthe 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

Ans. (a)

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IES 1997IES – 1997A cup of 10 cm height and 5 cm diameter is to be

d f h l f hi k Thmade from a sheet metal of 2 mm thickness. Thenumber of deductions necessarywill be( ) O(a) One(b) Two( ) Th(c) Three(d) Four

Ans. (c)

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IFS ‐ 2009What is deep drawing process for sheet metal

forming? E plain the function of a blank holderforming? Explain the function of a blank holder.

What is drawing ratio and how is the drawing ratio

increased ?

[   k ][10 – marks]

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GATE 2008GATE‐2008In the deep drawing of cups, blanks show a tendency to

i kl d th i h (fl )wrinkle up around the periphery (flange).The most likely cause and remedy of the phenomenon are,respectivelyrespectively,(A) Buckling due to circumferential compression; Increaseblank holder pressurep(B) High blank holder pressure and high friction; Reduceblank holder pressure and apply lubricant(C) High temperature causing increase in circumferentiallength: Apply coolant to blank(D) B kli d i f i l i d(D) Buckling due to circumferential compression; decreaseblank holder pressure [Ans. (a)]

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GATE 1999GATE‐1999Identify the stress ‐ state in the FLANCE portion of aPARTIALLYDRAWN CYLINDRICAL CUP h dPARTIALLYDRAWN CYLINDRICAL CUP when deep ‐drawing without a blank holder( ) T il i ll th di ti(a) Tensile in all three directions(b) No stress in the flange at all, because there is noblank holderblank‐holder(c) Tensile stress in one direction and compressive inthe one other directionthe one other direction(d) Compressive in two directions and tensile in thethird directionthird directionAns. (b)

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GATE 2003GATE‐2003A shell of 100 mm diameter and 100 mm height withh di f i b d d bthe corner radius of 0.4 mm is to be produced bycup drawing. The required blank diameter is( ) 8 (b) 6(a) 118 mm (b) 161 mm(c) 224 mm (d) 312 mm

Ans. (c)

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ISRO‐2011The initial blank diameter required to forma cylindrical cup of outside diameter 'd‘ andy ptotal height 'h' having a corner radius 'r' isobtained using the formulag

2( ) 4 0.5( ) 2 2

oa D d dh rb D d h

= + −

2 2

( ) 2 2

( ) 2 2o

o

b D d h r

c D d h r

= + +

= + +2

( )

( ) 4 0.5o

od D d dh r= + −Ans. (d)

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

Column I Column IIP. Wrinkling 1. Yield point elongationQ O l A iQ. Orange peel 2. AnisotropyR. Stretcher strains 3. Large grain sizeS E i I ffi i t bl k h ldiS. Earing 4. Insufficient blank holding

force5 Fine grain size5. Fine grain size6. Excessive blank holding force

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

Ans. (d)

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IES 2008IES – 2008A cylindrical vessel with flat bottom can be deepd bdrawn by(a) Shallow drawing( )(b) Single action deep drawing(c) Double action deep drawing(d) Triple action deep drawing

Ans. (c)

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

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IES 1994IES – 1994For obtaining a cup of diameter 25 mm and height 15

b d i h i f h d bl k h ldmm by drawing, the size of the round blank shouldbe approximately( ) (b)(a) 42 mm (b) 44 mm(c) 46 mm (d) 48 mm

Ans. (c)

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IAS 2007IAS – 2007In drawing operation, proper lubrication is

i l f hi h f h f ll i ?essential forwhich of the following reasons?1. To improve die life2. To reduce drawing forces3. To reduce temperature4. To improve surface finishSelect the correct answer using the code given below:(a) 1 and 2 only (b) 1, 3 and 4 only(c) 3 and 4 only (d) 1, 2, 3 and 4Ans. (d)

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IAS 1997IAS – 1997Which one of the following factor promotes the

d f i ki i h f d i ?tendency forwrinking in the process of drawing?(a) Increase in the ratio of thickness to blank diameterf k t i lof work material

(b) Decrease in the ratio thickness to blank diameter ofwork materialwork material(c) Decrease in the holding force on the blank(d) U f lid l b i t(d) Use of solid lubricants

A ( )Ans. (c)

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IAS 1994IAS – 1994Consider the following factors1. Clearance between the punch and the die is toosmall.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.The factors responsible for the vertical lines parallel tothe axis noticed on the outside of a drawn cylindrical cupwould include.( ) d (b) d(a) 2, 3 and 4 (b) 1 and 2(c) 2 and 4 (d) 1, 3 and 4 Ans. (d)

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IES 1998IES – 1998Assertion (A): The first draw in deep drawing operation

h t 6 % d ti th d d tcan have up to 60% reduction, the second draw up to40% reduction and, the third draw of about 30% only.Reason (R): Due to strain hardening the subsequentReason (R): Due to strain hardening, the subsequentdraws in a deep drawing operation have reducedpercentages.(a) Both A and R are individually true and R is the correctexplanation of A(b) h d d d ll b 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(c) A is true but R is false(d) A is false but R is true Ans. (a)

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GATE 1992GATE‐1992The thickness of the blank needed to produce, by 

  i i     i il     f  hi k      power spinning a missile cone of thickness 1.5 mm and half cone angle 30°, is( )     (b)    (a) 3.0 mm  (b) 2.5 mm (c) 2.0 mm  (d) 1.5 mm

Ans. (a)

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IES 1994IES – 1994The mode of deformation of the metal during

i i ispinning is(a) Bending( )(b) Stretching(c) Rolling and stretching(d) Bending and stretching.

Ans. (d)

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IFS‐2011Compare metal spinning with press work.

[2‐marks]

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

Ans. (b)

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JWM 2010JWM 2010Assertion (A) : In magnetic pulse‐forming method,magnetic field produced by eddy currents is used tomagnetic field produced by eddy currents is used tocreate force between coil and workpiece.Reason (R) : It is necessary for the workpieceReason (R) : It is necessary for the workpiecematerial to havemagnetic properties.(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) 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( )(d) A is false but R is true Ans. (a)

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IES 2010IES 2010Assertion (A) : In the high energy rate formingmethod the explosive forming has proved to be anmethod, the explosive forming has proved to be anexcellent method of utilizing energy at high rate andutilizes both the high explosives and low explosives.g p pReason (R): The gas pressure and rate of detonationcan be controlled for both types of explosives.(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 thecorrect 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. (c)

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IES 2007IES – 2007Which one of the following metal forming

i hi h f i ?processes is not a high energy rate forming process?(a) Electro‐mechanical forming( )(b) Roll‐forming(c) Explosive forming(d) Electro‐hydraulic forming

Ans. (b)

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

Ans. (b)

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IES 2005IES – 2005Magnetic forming is an example of:(a) Cold forming (b) Hot forming(c) High energy rate forming (d) Roll forming

Ans. (c)

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

[2‐marks]

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GATE 2000GATE‐2000A 1.5 mm thick sheet is subject to unequal biaxial

hi d h i i h di i fstretching and the true strains in the directions ofstretching are 0.05 and 0.09. The final thickness ofthe sheet in mm isthe sheet in mm is(a) 1.414 (b) 1.304(c) 1 362 (d) 289(c) 1.362 (d) 289

A (b)Ans. (b)

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

A ( )Ans. (a)

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GATE 2005GATE‐2005A 2 mm thick metal sheet is to be bent at an angle of

di i h b d di f If hone radian with a bend radius of 100 mm. If thestretch factor is 0.5, the bend allowance is( ) (b)(a) 99 mm (b) 100 mm(c) 101 mm (d) 102 mm

2mm

1 radian

Ans. (c)

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GATE 2007GATE‐2007Match the correct combination for following metalworking processesworking processes.Processes Associated state of stressP Blanking 1 TensionP. Blanking 1. TensionQ. Stretch Forming 2. CompressionR Coining 3 ShearR. Coining 3. ShearS. Deep Drawing 4. Tension and Compression

5. Tension and Shear5. Tension and ShearCodes:P Q R S P Q R S(a) 2 1 3 4 (b) 3 4 1 5( ) 3 4 ( ) 3 4 5(c) 5 4 3 1 (d) 3 1 2 4

Ans. (d)

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GATE ‐2012 Same Q in GATE‐2012 (PI)Match the following metal forming processes with theirassociated stresses in the workpiece.Metal forming process Type of stress1. Coining  P. Tensile2. Wire Drawing  Q. Shearg Q3. Blanking  R. Tensile and 

compressive

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

Ans. (a)

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GATE 2004GATE‐2004Match the following

Product ProcessP. Moulded luggage 1. Injection mouldingQ. Packaging containers for liquid 2. Hot rollingR. Long structural shapes 3. Impact extrusionS. Collapsible tubes 4. Transfer moulding

5. Blow moulding5 g6. Coining

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

Ans. (b)

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

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

wallsD Drawing 4 Flat plates and sheetsD. Drawing 4. Flat plates and sheets

5. Solid and hollow parts [Ans. (d)]Codes: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

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IAS 1997IAS – 1997Match List‐I (metal forming process) with List‐II(A i d f ) d l h(Associated feature) and select the correct answerusing the codes given below the Lists:

Li t l Li t IIList‐l List‐ IIA. Blanking 1. Shear angleB Fl f i C il d kB. Flow forming 2. Coiled stockC. Roll forming 3. Mandrel [Ans. (c)]D. Embossing 4. Closed matching dies

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

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IES 2010IES 2010Consider the following statements:The material properties which principallydetermine 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.3. Rate of work hardening.gWhich of the above statements is/are correct?(a) 1 and 2 only (b) 2 and 3 only(a) 1 and 2 only (b) 2 and 3 only(c) 1 only (d) 1, 2 and 3 Ans. (d)

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Ch 18: Sheet Metal FormingCh‐18: Sheet Metal FormingQ. No Option Q. No Option1 C 10 C2 B 11 C3 A 12 C3 A 12 C4 A 13 C5 D 14 D5 D 14 D6 A 15 A7 A 16 B7 A 16 B8 A 17 D9 A

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Powder MetallurgyPowder Metallurgy

By  S K Mondal

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GATE ‐2011 (PI)Which of the following powder productionmethods produces spongy and porous

i l ?particles?(a) Atomization(b) Reduction of metal oxides(c) Electrolytic deposition(d) Pulverization

Ans. (b)

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IES ‐ 2012IES ‐ 2012In electrolysis(a) For making copper powder, copper plate is madecathode in electrolyte tank( )(b) For making aluminum powder, aluminum plate ismade anode( ) Hi h d d d i f h d(c) High amperage produces powdery deposit of cathodemetal on anode(d) At i ti i it bl f l lti(d) Atomization process is more suitable for low meltingpoint metalsAns (b)Ans. (b)

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

sintering is required to achieve which property? Whatsintering is required to achieve which property? What

is hot iso‐static pressing?

[ 2 Marks]

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

( ) h d d h d(a) Improves strength and reduces hardness

(b) Reduces brittleness and improves strength(b) educes b e ess a d p o es s e g

(c) Improves hardness and reduces toughness

(d) Reduces porosity and increases brittleness

A (b)Ans. (b)

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IES – 2011 ConventionalWhat is isostatic pressing of metal powders ?What are its advantage ?

[ 2 Marks]

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

manufacture products ith controlled porosit ?manufacture products with controlled porosity?

(a) Casting

(b) welding

(c) formation

(d) Powder metallurgy Ans (d)(d) Powder metallurgy Ans. (d)

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GATE – 2011 (PI)The binding material used in cementedcarbide cutting tools is(a) graphite(b) tungsten(c) nickel(d) cobalt

Ans. (d)( )

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IES 2010IES 2010Consider the following parts:1. Grinding wheel2. Brake liningg3. Self‐lubricating bearingsWhich of these parts are made by powderWhich of these parts are made by powdermetallurgy technique?( ) d (b) l(a) 1, 2 and 3 (b) 2 only(c) 2 and 3 only (d) 1 and 2 onlyAns. (c)

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

Ans. (d)

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IES 2002IES – 2002The rate of production of a powder metallurgy partd ddepends on(a) Flow rate of powder( )(b) Green strength of compact(c) Apparent density of compact(d) Compressibility of powder

Ans. (c)

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IES 2001IES – 2001Match List‐I (Components) with List‐II(M f i P ) d l h(Manufacturing Processes) and select the correctanswer using the codes given below the lists:

Li t I Li t IIList I List IIA. Car body (metal) 1. MachiningB Cl h li i C iB. Clutch lining 2. CastingC. Gears 3. Sheet metal pressingD. Engine block 4. Powder metallurgy

Codes:A B C D A B C DAns. (d)

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

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

Ans. (c)( )

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IES 1998IES – 1998In powder metallurgy, the operation carried out toi h b i f b h i ll dimprove the bearing property of a bush is called(a) infiltration (b) impregnation( ) ( )(c) plating (d) heat treatment

Ans. (b)

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IES 1997IES – 1997Which of the following components can be

f d b d ll h d ?manufactured by powdermetallurgymethods?1. 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(c) 1, 2 and 4 (d) 1, 2, 3 and 4

Ans. (d)

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IES 1999IES – 1999The correct sequence of the given processes in

f i b d ll imanufacturing by powdermetallurgy is(a) Blending, compacting, sintering and sizing( )(b) Blending, compacting, sizing and sintering(c) Compacting, sizing, blending and sintering(d) Compacting, blending, sizing and sintering

Ans. (a)

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IES 2001IES – 2001Carbide‐tipped cutting tools are manufactured by

d l h l d hpowder‐ metal technology process and have acomposition of( ) Zi i T t ( % 6 %)(a) Zirconium‐Tungsten (35% ‐65%)(b) Tungsten carbide‐Cobalt (90% ‐ 10%)( ) Al i i id Sili ( % %)(c) Aluminium oxide‐ Silica (70% ‐ 30%)(d) Nickel‐Chromium‐ Tungsten (30% ‐ 15% ‐ 55%)

Ans. (b)

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IES 1999IES – 1999Assertion (A): In atomization process of manufacture of

t l d th lt t l i f d th hmetal powder, the molten metal is forced through asmall orifice and broken up by a stream of compressedair.Reason (R): The metallic powder obtained byatomization process is quite resistant to oxidation.(a) Both A and R are individually true and R is the correctexplanation of A(b) h d d d ll b 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(c) A is true but R is false(d) A is false but R is true Ans. (c)

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IES 2007IES – 2007What are the advantages of powdermetallurgy?1. Extreme purity product2. 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(c) 2 and 3 only (d) 1 and 3 only3 y 3 y

Ans. (b)

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IES 2006IES – 2006Which of the following are the limitations ofpowdermetallurgy?powdermetallurgy?1. High tooling and equipment costs.2 Wastage of material2. Wastage of material.3. It cannot be automated.4 Expensive metallic powders4. Expensive metallic powders.Select 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

Ans. (c)

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IES 2004IES – 2004Consider the following factors:1. Size and shape that can be produced economically2. Porosity of the parts produced3. Available press capacity4. High densityWhich of the above are limitations of powdermetallurgy?(a) 1, 3 and 4 (b) 2 and 3(c) 1, 2 and 3 (d) 1 and 2Ans. (a)

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IES ‐ 2012St t t (I) P t d b d t ll dStatement (I): Parts made by powder metallurgy donot have as good physical properties as parts casted.Statement (II): Particle shape in powder metallurgyStatement (II): Particle shape in powder metallurgyinfluences 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) B th St t t (I) d St t t (II)(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is not the correctexplanation of Statement (I)p ( )(c) Statement (I) is true but Statement (II) is false(d) Statement (I) is false but Statement (II) is true( ) ( ) ( )Ans. (b)

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IES 2009IES – 2009Which of the following cutting tool bits are made by

d ll ?powdermetallurgy process?(a) Carbon steel tool bits (b) Stellite tool bits( ) ( )(c) Ceramic tool bits (d) HSS tool bits

Ans. (c)

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IAS 2003IAS – 2003Which of the following are produced by powder

ll ?metallurgy process?1. Cemented carbide dies2. Porous bearings3. Small magnets4. Parts with intricate shapesSelect the correct answer using the codes given below:Codes:(a) 1, 2 and 3 (b) 1, 2 and 4(c) 2, 3 and 4 (d) 1, 3 and 4 Ans. (a)

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IAS 2003IAS – 2003In parts produced by powder metallurgy process,

i i i dpre‐sintering is done to(a) Increase the toughness of the component( )(b) Increase the density of the component(c) Facilitate bonding of non‐metallic particles(d) Facilitate machining of the part

Ans. (d)

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IAS 2000IAS – 2000Consider the following processes:

M h i l l i i1. Mechanical pulverization2. Atomization

Ch i l d ti3. 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

Ans. (b)

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IAS 1997IAS – 1997Assertion (A): Close dimensional tolerances areNOT possible with isostatic pressing of metalNOT possible with isostatic pressing of metalpowder in powdermetallurgy technique.Reason (R): In the process of isostatic pressing, theReason (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 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 Ans. (d)

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IAS 1998IAS – 1998Throwaway tungsten carbide tip tools are

f d bmanufactured by(a) Forging (b) Brazing( ) ( )(c) Powder metallurgy (d) Extrusion

Ans. (c)

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IAS 1996IAS – 1996Which one of the following processes is performedi d ll lf l b i iin powder metallurgy to promote self‐lubricatingproperties in sintered parts?( ) I filt ti (b) I ti(a) Infiltration (b) Impregnation(c) Plating (d) Graphitization

Ans. (b)

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

G   G  Group – 1 Group ‐2P. Mulling 1. Powder metallurgyQ  I i  I j i ldiQ. Impregnation 2. Injection mouldingR. Flash trimming 3. Processing of FRP composites

( ) P   Q   R   S  (b) P   Q   R   S 

S. Curing 4. Sand casting

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

Ans. (d)

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IAS 2007IAS – 2007Assertion (A): Mechanical disintegration of amolten metal stream into fine particles by means ofmolten metal stream into fine particles by means ofa jet of compressed air is known as atomization.Reason (R): In atomization process inert‐gas orReason (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 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 Ans. (c)

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IAS 2004IAS – 2004The following are the constituent steps in the

f d llprocess of powdermetallurgy:1. Powder conditioning2. Sintering3. Production of metallic powder4. Pressing or compacting into the desired shapeIndentify the correct order in which they have to beperformed and select the correct answer using the codesgiven below:( ) (b)(a) 1‐2‐3‐4 (b) 3‐1‐4‐2(c) 2‐4‐1‐3 (d) 4‐3‐2‐1 Ans. (b)

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IAS 2003IAS – 2003Assertion (A): Atomization method for production of

t l d i t f h i l di i t ti fmetal powders consists of mechanical disintegration ofmolten stream into fine particles.Reason (R): Atomization method is an excellent meansReason (R): Atomization method is an excellent meansof making powders from high temperaturemetals.(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

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

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IAS 2007IAS – 2007Consider the following basic steps involved in the

d i f b iproduction of porous bearings:1. Sintering2. Mixing3. Repressing4. Impregnation5. Cold‐die‐compactionWhich one of the following is the correct sequence of theabove steps? Ans. (b)

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C ti l Q tiConventional Questions1. Explain why metal powders are blended. Describe what

h d i i i [IES M k ]happens during sintering. [IES‐2010, 2 Marks]

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C ti l Q tiConventional Questions1. Discuss the terms fineness and particle size

distribution in powder metallurgy [IES 2010 2distribution in powder metallurgy. [IES‐2010, 2Marks]

Ans.Ans.Fineness: Is the diameter of spherical shaped particleand mean diameter of non‐spherical shaped particle.

Particle size distribution: Geometric standarddeviation (a measure for the bredth or width of a(distribution), is the ratio of particle size diameterstaken at 84.1 and 50% of the cumulative undersizedweight plot respectively and mean mass diameterweight plot, respectively and mean mass diameterdefine the particle size distribution.

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C ti l Q tiConventional QuestionsEnumerate the steps involved in “powder metallurgy”

Di h N h i l dprocess. Discuss these steps. Name the materials usedin “powder metallurgy”. What are the limitations ofpowder metallurgy? [IES‐2005 10 Marks]powder metallurgy? [IES‐2005, 10 Marks]

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Ch‐12: Powder Metallurgy

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

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Limit, Tolerance & Fits,

By  S K Mondal

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

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

Ans. (b) ( )

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

( ) l l l(a) Unilateral tolerance

(b) Bilateral tolerance(b) a e a o e a ce

(c) Limiting dimensions

(d) All of the above

A (b)Ans. (b)

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GATE – 2010 ISRO‐2012GATE – 2010, ISRO‐20120 009−

A shaft has a dimension,The respective values of fundamental deviation and

0.0090.02535φ −

tolerance are(a) 0.025, 0.008 (b) 0.025,0.016− ± −(c) 0.009, 0.008 (d) 0.009,0.016− ± −

Ans. (d)

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GATE ‐ 1992GATE ‐ 1992Two shafts A and B have their diameters specified as 

      d         i l100 ± 0.1 mm and 0.1  ± 0.0001 mm respectively.Which of the following statements is/are true?( )(a) Tolerance in the dimension is greater in shaft A(b) The relative error in the dimension is greater in shaft AA(c) Tolerance in the dimension is greater in shaft B( )(d) The relative error in the dimension is same for shaft A and shaft B

Ans. (a)

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

0.040+ 0.020+

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

0.010025.000− 0.00025.000−

will be(a) 10 microns(b) i(b) 20 microns(c) 30 microns( )(d) 60 microns

Ans. (c)

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ISRO 2010ISRO‐20100.02 0.020.00 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(a) 0.02 mm (b) 0.00 mm (c) -0.02 mm (d) 0.01 mm

Ans. (c)

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GATE ‐ 2007GATE ‐ 2007A hole is specified as mm. The matingh f h l fi i h i i l f

0 . 0 5 00 . 0 0 04 0

shaft has a clearance fit with minimum clearance of0.01 mm. The tolerance on the shaft is 0.04 mm. Themaximum clearance in mm between the hole andmaximum clearance in mm between the hole andthe shaft is(a) 0 04(a) 0.04(b) 0.05(c) 0 10(c) 0.10(d) 0.11

Ans. (c)

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IES ‐ 2005IES ‐ 2005The tolerance specified by the designer for thediameter of a shaft is 20 00 ± 0 025 mm The shaftsdiameter 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 and9 9920.01 mm respectively, with same standarddeviation. What will be the percentage rejection forthe shafts produced bymachines A B and C?the shafts produced bymachines A, B and C?(a) Same for the machines A, Band C since the standard

deviation is same for the three machines(b) Least for machine A(c) Least for machine B(d) Least for machine C Ans. (c)

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GATE ‐ 2000GATE ‐ 2000A slot is to be milled centrally on a block with adi i f A illi fdimension of 40 ± 0.05 mm. A milling cutter of 20mm width is located with reference to the side ofthe block within ± 0 02 mm The maximum offset inthe block within ± 0.02 mm. The maximum offset inmm between the centre lines of the slot and theblock is(a) ± 0.070 (b) 0.070(c) ± 0.020 (d) 0.045(c) ± 0.020 (d) 0.045

Ans (c)Ans. (c)

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GATE – 2007 (PI)GATE – 2007 (PI)

Diameter of a hole after plating needs to be controlled0.0500.010

Diameter of a hole after plating needs to be controlledbetween 30 mm. If the plating thickness varies +

+

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

0.070.030

plating should be(a) 30+

+0 0.065

0.020mm (b) 30 mm ++

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

Ans. (d)

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IES 2011IES 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(b) Mounting heavy duty gears on shafts(c) Mounting pulley on shafts(d) A bl f fl h l h ft(d) Assembly of flywheels on shafts

A ( )Ans. (a)

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GATE ‐ 2005GATE ‐ 2005In order to have interference fit, it is essential that

the lower limit of the shaft should be

(a) Greater than the upper limit of the hole

(b) Lesser than the upper limit of the hole(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

Ans. (a)

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GATE 2011GATE 20110.015+

A hole is of dimension mm. The0.010+

09φ +

corresponding shaft is of dimension mm.The resulting assembly has

0.0019φ +

(a) loose running fit(b) close running fitg(c) transition fit(d) interference fit( )Ans. (c)

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GATE ‐2012 Same Q in GATE‐2012 (PI)

0 04+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 assemblyis(a) 40 (b) 30 (c) 20 (d) 10

Ans. (c)

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IAS‐2011 MainAn interference assembly, of nominal diameter 20

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

manufacturing 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,3 9

for the two mating parts.

[ M k ][10‐Marks]

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IES ‐ 2007IES ‐ 2007

Ans. (a)

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ISRO‐2011A shaft and hole pair is designated as 50H7d8. This 

assembly constitutesassembly constitutes

(a) Interference fit 

(b) Transition fit

(c) Clearance fit ( )

(d) None of the above

( )Ans. (c)

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

Ans. (d)

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IES ‐ 2009IES ‐ 2009Consider the following joints:

R il i h l d l1. Railway carriage wheel and axle2. IC engine cylinder and linerWhi h f th b j i t i / th lt( ) fWhich of the above joints is/are the result(s) ofinterference fit?(a) 1 only(a) 1 only(b) 2 only(c) Neither 1 nor 2(c) Neither 1 nor 2(d) Both 1 and 2

Ans. (d)

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IES ‐ 2008IES ‐ 2008Consider the following statements:1. The amount of interference needed to create a tightjoint varies with diameter of the shaft.2. An interference fit creates no stress state in theshaft.

Th i h h b i i il hi k3. The stress state in the hub is similar to a thick‐walled cylinder with internal pressure.Whi h f th t t t i b t?Which of the statements given above are correct?(a) 1, 2 and 3 (b) 1 and 2 only( ) d l (d) d l(c) 2 and 3 only (d) 1 and 3 onlyAns. (d)

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IES ‐ 2004IES ‐ 2004Consider the following fits:1. I.C. engine cylinder and piston2. Ball bearing outer race and housing3. Ball bearing inner race and shaftWhich of the above fits are based on the interferencesystem?(a) 1 and 2(b) 2 and 3(c) 1 and 3(d) 1, 2 and 3 Ans. (b)

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

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

stress [Ans. (b)]Codes:A B C D A B C D( ) (b)(a) 3 4 1 2 (b) 4 3 2 1(c) 3 4 2 1 (d) 4 3 1 2

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GATE ‐ 2001GATE ‐ 2001 Allowance in limits and fits refers to(a) Maximum clearance between shaft and hole(b) Minimum clearance between shaft and hole(c) Difference between maximum and minimum size ofhole(d) Difference between maximum and minimum size ofshaft

Ans. (b)

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GATE ‐ 1998GATE ‐ 1998In the specification of dimensions and fits,(a) Allowance is equal to bilateral tolerance(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 thetolerance.

Ans. (c)

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IES ‐ 2012IES ‐ 2012Clearance in a fit is the difference between

(a) Maximum hole size and minimum shaft size

(b) Minimum hole size and maximum shaft size

(c) Maximum hole size and maximum shaft size(c) Maximum hole size and maximum shaft size

(d) Minimum hole size and minimum shaft size

Ans. (b)

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ISRO‐2008Basic shaft and basic hole are those whose upper

de iations and lo er de iations respecti el aredeviations and lower deviations respectively are

(a) +ve, ‐ve (b) ‐ve, +ve

(c) Zero, Zero (d) None of the above

Ans (c)Ans. (c)

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IES ‐ 2005IES ‐ 2005Assertion (A): Hole basis system is generally

f d h f b i i l d ipreferred to shaft basis system in tolerance designfor getting the required fits.R (R) H l h t b i l t lReason (R): Hole has to be given a larger toleranceband than themating shaft.(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) 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 Ans. (c)

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IES 2006 ConventionalIES‐2006 ConventionalFind the limit sizes, tolerances and allowances for a

di h f d h l i d i d b100 mm diameter shaft and hole pair designated byF8h10. Also specify the type of fit that the above pairbelongs tobelongs to.Given: 100 mm diameter lies in the diameter steprange of 80‐120 mm The fundamental deviation forrange of 80 120 mm. The fundamental deviation forshaft designation ‘f’ is ‐5.5 D0.41

The values of standard tolerances for grades of IT 8The values of standard tolerances for grades of IT 8and IT 10 are 25i and 64i respectively.Also, indicate the limits and tolerance on a diagram., g

Will be discussed in class [15‐Marks]

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IES ‐ 2008IES ‐ 2008Consider the following statements:A nomenclature 50 H8/p8 denotes that1. Hole diameter is 50 mm.

φ

2. It is a shaft base system.3. 8 indicates fundamental deviation.Which of the statements given above is/are incorrect?(a) 1, 2 and 33(b) 1 and 2 only(c) 1 and 3 only( ) 3 y(d) 3 only Ans. (a)

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

Ans. (c)

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GATE ‐ 2009GATE ‐ 2009What are the upper and lower limits of the shaftrepresented by 60 f ?represented by 60 f8?Use the following data:Diameter 60 lies in the diameter step of 50‐80 mm.p 5Fundamental tolerance unit,i, in m= 0.45 D1/3 + 0.001D, where D is therepresentative size in mm;

μrepresentative size in mm;Tolerance value for lT8 = 25i. Ans. (a)Fundamental deviation for 'f shaft = ‐5.5D0.41Fundamental deviation for f shaft = 5.5D(a) Lower limit = 59.924 mm, Upper Limit = 59.970 mm(b) Lower limit = 59.954 mm, Upper Limit = 60.000 mm59 954 pp(c) Lower limit = 59.970 mm, Upper Limit = 60.016 mm(d) Lower limit = 60.000 mm, Upper Limit = 60.046 mm

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GATE 2008 (PI)GATE – 2008 (PI)Following data are given for calculating limits of

dimensions and tolerances for a hole: Tolerance unit i (in

µm) 0 4 ³√D 0 00 D The unit of D is mm Diameterµm) = 0.45 ³√D + 0.001D. The unit of D is mm. Diameter

step is 18‐30 mm. If the fundamental deviation for H

hole is zero and IT8 = 25 i, the maximum and minimum

limits of dimension for a 25 mmH hole (in mm) arelimits of dimension for a 25 mmH8 hole (in mm) are

(a) 24.984, 24.967 (b) 25.017, 24.984

(c) 25.033, 25.000 (d) 25.000, 24.967 Ans. (c)

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GATE ‐ 2000GATE ‐ 2000A fit is specified as 25H8/e8. The tolerance value for

i l di f i IT8 i ia nominal diameter of 25 mm in IT8 is 33 micronsand fundamental deviation for the shaft is ‐ 40microns The maximum clearance of the fit inmicrons. The maximum clearance of the fit inmicrons is(a) ‐7(a) 7(b) 7(c) 73(c) 73(d) 106

Ans. (d)

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GATE – 1996 IES‐2012GATE – 1996, IES‐2012The fit on a hole‐shaft system is specified as H7‐6 Th f fi is6.The type of fit is(a) Clearance fit( ) ( )(b) Running fit (sliding fit)(c) Push fit (transition fit)(d) Force fit (interference fit)

Ans. (d)

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

Ans. (b)

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

(minimum) and upper (maximum) limits of the bore

are 25.000 mm and 25.021 mm, respectively. When the

bore is designated as 25H8, then the upper (maximum)

limit is 25.033 mm. When the bore is designated as

25H6 then the upper (maximum) limit of the bore (in25H6, then the upper (maximum) limit of the bore (in

mm) is

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

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

shaft basis system ? Why is hole basis system the moreshaft basis system ? Why is hole basis system the more

extensive in use ?

What are the differences between interchangeability

and selective assembly ?y

[12‐Marks]

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GATE ‐ 2003GATE ‐ 2003 

[Ans. (b)] 

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GATE ‐ 1997GATE ‐ 1997Three blocks B1 , B2 and B3 areto be inserted in a channel ofto be inserted in a channel ofwidth S maintaining aminimum gap of width T =

h i Fi0.125 mm, as shown in Figure.For P = 18. 75 ± 0.08;Q = 25 00 ± 0 12;Q = 25.00 ± 0.12;R = 28.125 ± 0.1 andS = 72.35 + X, (where all7 35 , (dimensions are in mm), thetolerance X is

( )    8 (b)  8 ( )    (d) (a) + 0.38 (b) ‐ 0.38 (c) + 0.05 (d) ‐0.05Ans. (d)

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IES ‐ 2000IES ‐ 2000Which one of the following tolerances set on innerdi d di i l f h d ddiameter and outer diameter respectively of headedjig bush for press fit is correct?( ) G h 6 (b) F 6(a) G7 h 6 (b) F7 n6(c) H 7h 6 (d) F7j6

Ans. (b)

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ISRO‐2008Plug gauges are used to(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(c) Check the diameter of the holes in theworkpieces(d) Check the length of holes in the workpieces(d) Check the length of holes in the workpieces

Ans. (c)( )

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GATE ‐ 2004GATE ‐ 2004GO and NO‐GO plug gages are to be designed for a

0 05hole mm. Gage tolerances can be taken as 10%of the hole tolerance. Following ISO system of gaged i i f GO d NO GO ill b

0.050.0120

design, sizes of GO and NO‐GO gage will berespectively(a) 20 010 mm and 20 050 mm(a) 20.010 mm and 20.050 mm(b) 20.014 mm and 20.046 mm( ) 6 d(c) 20.006 mm and 20.054 mm(d) 20.014 mm and 20.054 mmA (b)Ans. (b)

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GATE ‐ 1995GATE ‐ 1995Checking the diameter of a hole using GO‐NO‐GO

i l f i i bgauges is an, example of inspection by…..(variables/attributes)Th b t t t iThe above statement is(a) Variables(b) A ib(b) Attributes(c) Cant say( )(d) Insufficient data

Ans. (b)

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

Ans. (c)

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Measurement of Lines & Surfaces

By  S K Mondal

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ISRO‐2010The vernier reading should not be taken at itsface value before an actual check has beenk ftaken for

(a) Zero error(b) Its calibration(c) Flatness of measuring jaws(d) Temperature equalization

Ans. (a)

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ISRO‐2008The least count of a metric vernier caliper

ha ing 2 di isions on ernier scale matchinghaving 25 divisions on vernier scale, matching

with 24 divisions of main scale (1 main scale

divisions = 0.5 mm) is

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

(c) 0.02 mm (d) 0.005 mm( ) ( ) 5

Ans. (c)

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

mm and di isions on thimble 0 the readingmm and divisions on thimble 50, the reading

corresponding to 5 divisions on barrel and 12

divisions on thimble is

( ) 6 (b)(a) 2.620 mm (b) 2.512 mm

(c) 2.120 mm (d) 5.012 mm( ) ( ) 5

Ans. (a)

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

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

ll l t th d ill i dl t h lparallel to the drill spindle taper holeaxis(D) The drill spindle rotational axisi h d ill i dl h lintersects the drill spindle taper holeaxis at point Q

Ans. (c)

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ISRO 2010ISRO‐2010Amaster gauge isAmaster gauge is

(a) A new gauge

(b) An international reference standard

( ) A d d f h ki f d(c) A standard gauge for checking accuracy of gauges used

on shop floors

(d) A gauge used by experienced technicians

Ans. (c)

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ISRO‐2008Standards to be used for reference purposes in

laboratories and orkshops are termed aslaboratories and workshops are termed as

(a) Primary standards

(b) Secondary standards

(c) Tertiary standards

(d) Working standards Ans (d)(d) Working standards Ans. (d)

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PSUA feeler gauge is used to check the

( ) h f h(a) Pitch of the screw

(b) Surface roughness(b) Su ace oug ess

(c) Thickness of clearance

(d) Flatness of a surface

A  ( )Ans. (c)

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ISRO‐2011A sine bar is specified by

( ) l l h(a) Its total length

(b) The size of the rollers(b) e s e o e o e s

(c) The centre distance between the two rollers

(d) The distance between rollers and upper surface

A ( )Ans. (c)

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

has a diameter of 20 mm. During taper angle

measurement of a component, the height

from the surface plate to the centre of a rollerp

is 100 mm.

( )The calculated taper angle (in degrees) is

(a) 21.1 (b) 22.8 (c) 23.6 (d) 68.9( ) ( ) ( ) 3 ( ) 9

Ans. (a)

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GATE – 2011 (PI)The best wire size (in mm) for measuringeffective diameter of a metric thread(i l d d l i 6 ) f di d(included angle is 60o) of 20 mm diameter and2.5 mmpitch using twowiremethod is( ) (b)(a) 1.443 (b) 0.723(c) 2.886 (d) 2.086

Ans. (a)

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IES ‐ 1992IES ‐ 1992Which grade symbol represents surface rough ofb hi ?broaching?(a) N12 (b) N8

( ) ( )(c) N4 (d) N1

Ans. (b)

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IFS 2011IFS‐2011What is meant by interchangeable manufacture?y g

Laser light has unique advantages for inspection.

What are they ? Define the terms 'roughness

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

with surface irregularities.

[10‐marks]

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ISRO‐2011CLA value and RMS values are used for 

measurement ofmeasurement of

(a) Metal hardness 

(b) Sharpness of tool edge

(c) Surface dimensions (c) Surface dimensions 

(d) Surface roughness

Ans. (d)

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IES ‐ 2006IES ‐ 2006The M and E‐system in metrology are related toy gy

measurement of:

(a) Screw threads (b) Flatness

(c) Angularity (d) Surface finish(c) Angularity (d) Surface finish

Ans. (d)

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IES ‐ 2007IES ‐ 2007What is the dominant direction of the tool marks or

h i f h i di i lscratches in a surface texture having a directionalquality, called?( ) P i t t (b) S d t t(a) Primary texture (b) Secondary texture(c) Lay (d) Flaw

Ans. (c)

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IES ‐ 2008IES ‐ 2008What term is used to designate the direction of theg

predominant surface pattern produced by

hi i i ?machining operation?

(a) Roughness (b) Lay( ) g ( ) y

(c) Waviness (d) Cut off

Ans. (b)

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IES 2010IES 2010Match List I with List II and select the correct answer

i th d i b l th li tusing the code given below the lists:List I List II

(Symbols for direction of lay) (Surface texture)( y y) ( )

A B  C  D  A  B  C  D

[Ans. (b)]

A 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

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IES ‐ 2008IES ‐ 2008 

Ans. (c)( )

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ISRO‐2010Surface roughness on a drawing is represented

byby

(a) Triangles

(b) Circles

( ) S(c) Squares

(d) Rectangles

Ans. (a)

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GATE ‐ 1997GATE ‐ 1997 List I List II

(A) S f fil C lib i(A) Surface profilometer 1. Calibration(B) Light Section Microscope 2. Form tester(C) Mi k t Fil thi k(C) Microkater 3. Film thickness

measurement(D) Interferometer 4 Centre line average(D) Interferometer 4. Centre line average

5. Comparator6. Surface lay measurement6. Surface lay measurement

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

Ans. (b) 

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GATE ‐ 2003GATE ‐ 2003Two slip gauges of 10 mm width measuring 1.000 mm

d k id b id i i h hand 1.002 mm are kept side by side in contact with eachother lengthwise. An optical flat is kept resting on theslip gauges as shown in the figure Monochromatic lightslip gauges as shown in the figure. Monochromatic lightof wavelength 0.0058928 mm is used in the inspection.The total number of straight fringes that can be observedg gon both slip gauges is

(a)  2 (b) 6(a)  2 (b) 6(c) 8 (d) 13

Ans. (a)

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Miscellaneous of Metrology

By  S K Mondal

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

Ans. (c)

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GATE – 2009 (PI)An autocollimator is used to

( ) ll l d l fl f(a) measure small angular displacements on flat surface

(b) compare known and unknown dimensions(b) co pa e o a d u o d e s o s

(c) measure the flatness error

(d) measure roundness error between centers

A ( )Ans. (c)

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

(b) A constant deviation prism having the angle ofdeviation between the incident ray and reflected ray,equal to 90o

( ) A d i i i h i h l f(c) A constant deviation prism having the angle ofdeviation between the incident ray and reflected ray,equal to 45oequal to 45

(d) Used to produce interference fringes

Ans. (b)

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IES ‐ 1998IES ‐ 1998Match List‐I with List‐II and select the correct answer using thecodes given below the lists:

List‐I List‐II(Measuring Device) (ParameterMeasured)A. Diffraction grating 1. Small angular deviations on longg g g g

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

partsC A lli M f i hC. Auto collimators 3. Measurement of gear pitchD. Laser scan micrometer4. Surface texture using interferometer

5. Measurement of very smalldi l tdisplacements

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

Ans. (d)

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GATE ‐ 1992GATE ‐ 1992Match the instruments with the physical quantities theymeasure:measure:Instrument Measurement(A) Pilot‐tube (1) R.P.M. of a shaft( ) ( )(B) McLeod Gauge (2) Displacement(C) Planimeter (3) Flow velocity(D) LVDT (4) Vacuum

(5) Surface finish(6) A [A (b)](6) Area [Ans. (b)]

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

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GATE ‐ 2004GATE ‐ 2004Match the followingFeature to be inspected InstrumentP Pitch and Angle errors of screw thread 1. Auto CollimatorQ Fl f f l O i l I fQ Flatness error of a surface plate 2. Optical InterferometerR Alignment error of a machine slide way 3. Dividing Head

and Dial Gaugeand Dial GaugeS Profile of a cam 4. Spirit Level

5 Sine bar5. Sine bar6. Tool maker's Microscope

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

Ans. (b)

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GATE ‐ 1995GATE ‐ 1995List I List II

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

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

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GATE ‐ 2010GATE ‐ 2010A taper hole is inspected using a CMM, with a probef di A h i h Z f hof 2 mm diameter. At a height, Z = 10 mm from the

bottom, 5 points are touched and a diameter ofcircle (not compensated for probe size) is obtainedcircle (not compensated for probe size) is obtainedas 20 mm. Similarly, a 40 mm diameter is obtainedat a height Z = 40 mm. the smaller diameter (in mm)g 4 ( )of hole at Z = 0 is(a) 13.3343 334(b) 15.334(c) 15.442( ) 5 44(d) 15.542

Ans. (a)

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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 rollersworkpiece as shown in the figure. If the two precision rollershave radii 8 mm and 5 mm and the total thickness of slipgauges inserted between the rollers is 15.54 mm, the taper

l θ iangle θ is(a) 6 degree(b) d(b) 10 degree(c) 11 degree(d) d(d) 12 degree

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IES ‐ 2000IES ‐ 2000A screw thread specified by M 20 x 2.5 C as per BIS h d   thread system means(a) Metric thread of 20 mm nominal diameter and 2.5 

  it h h i    t lmm pitch having coarse tolerance(b) Metric thread of 20 mm root diameter and 2.5 mm pitch having coarse tolerancepitch having coarse tolerance(c) Metric thread of fine class having 20 mm root diameter and 2 5 mm pitchdiameter and 2.5 mm pitch(d) Metric thread of 20 mm shank diameter and 2.5 mm thread depth with coarse tolerancethread depth with coarse toleranceAns. (c)

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Ch‐13: MetrologyQ. No Option Q. No Option1 C 10 D2 C 11 A3 A 12 B3 A 12 B4 C 13 B

C D5 C 14 D6 B 15 B7 C 16 C8 B 17 B9 B