Metal Cutting, Metal Forming & MetrologyQuestions & Answers-2016 (All Questions are in Sequence) IES-1992-2015 (24 Yrs.), GATE-1992-2015 (24 Yrs.), GATE (PI)-2000-2015 (16 Yrs.), IAS-19942011 (18 Yrs.), some PSUs questions and conventional questions IES, IAS, IFS are added. Section‐I: Theory of Metal Cutting Chapter-1: Basics of Metal Cutting Chapter-2: Analysis of Metal Cutting Chapter-3: Tool life, Tool Wear, Economics and Machinability Section‐II: Metrology Answer & Explanations Page-145 Page- 149 Page-158 Questions Answer & Explanations Page-36 Page-48 Page-61 Chapter-4: Cold Working, Recrystalization and Hot Working Chapter-5: Rolling Chapter-6: Forging Chapter-7: Extrusion & Drawing Chapter-8: Sheet Metal Operation Chapter-9: Powder Metallurgy Section‐IV: Cutting Tool Materials Page-2 Page-10 Page-18 Chapter-10: Limit, Tolerance & Fits Chapter-11: Measurement of Lines & Surfaces Chapter-12: Miscellaneous of Metrology Section‐III: Metal Forming Questions Page-166 Page-171 Page-173 Questions Answer & Explanations Page-66 Page-71 Page-82 Page-93 Page-107 Page-126 Page‐135 Page-174 Page-175 Page-177 Page-182 Page-186 Page-190 Page-191 For‐2016 (IES, GATE & PSUs) For-2016 (IES,GATE & PSUs) Page 1 of 210 Rev.0 Theory of Metal Cutting p Whyy even a batteryy operated pencil sharpener p p cannot be accepted as a machine tool? y Ans. In spite of having all other major features of machine tools, the sharpener is of low value. By S K Mondal IAS 2009 main y Name four independent variables and three dependent variables in metal cutting. [ 5 marks] Independent Variables Dependent Variables •Starting materials •Force or power requirements (tool/work) •Maximum temperature in •Tool geometry T l cutting i •Cutting Velocity •Surface finish •Lubrication 1 2 Speed feed Speed, feed, and depth of cut • Feed & Depth of cut IES‐2013 3 IES‐2001 Carbide tool is used to machine a 30 mm diameter For cutting of brass with single‐point cutting tool steel t l shaft h ft att a spindle i dl speed d off 1000 revolutions l ti per on a lathe, tool should have minute The cutting speed of the above turning minute. operation p is: (a) Negative rake angle (b) Positive rake angle k l (a) 1000 rpm ( ) Zero rake angle (c) Z k l (b) 1570 m/min (d) Zero side relief angle Z id li f l (c) 94.2 m/min Cutting speed, feed, and depth of cut for a turning operation 4 IES‐1995 (d) 47.1 m/min 5 IES‐1993 GATE‐1995; 2008 Single point thread cutting tool should ideally Cutting power consumption in turning can be have: significantly reduced by f l d db (a) Increasing rake angle of the tool a) Zero rake (b) Increasing the cutting angles of the tool b) Positive rake (c) Widening the nose radius of the tool c) Negative rake (d) Increasing the clearance angle d) Normal rake l k For-2016 (IES,GATE & PSUs) 7 Page 2 of 210 6 8 Assertion (A): A i (A) For F a negative i rake k tool, l the h specific ifi cutting pressure is smaller than for a positive rake tool under otherwise identical conditions. Reason (R): The shear strain undergone by the chip in the case of negative rake tool is larger. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Rev.0 9 IES‐2015 IES – 2005 Assertion (A): are generally A i (A) Carbide C bid tips i ll given i negative rake angle. Reason (R): Carbide tips are made from very hard materials. materials (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true Statement (I) : The ceramic tools used in machining of material have highly brittle tool tips. Statement (II) : Ceramic tools can be used on hard‐to hard to machine work material. ( ) Both (a) B th statement t t t (I) and d (II) are individually i di id ll true t and d statement (II) is the correct explanation of statement (I) (b) Both statement (I) and statement(II) are individually true but statement(II) ( ) is not the correct explanation p of statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but statement (II) is true 10 IES – 2002 Assertion (A): A i (A) Negative N i rake k is i usually ll provided id d on carbide tipped tools. Reason (R): Carbide tools are weaker in compression. compression (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true 11 12 For IES Only IES 2011 Which one of the following statement is NOT correct IES 2007 Conventional GATE – GATE – 2008 (PI) 2008 (PI) Cast iron with impurities of carbide requires a with h reference f to the h purposes and d effects ff off rake k angle l B ittl materials Brittle t i l are machined hi d with ith tools t l particular ti l rake k angle l for f efficient ffi i t cutting tti with ith single i l of a cutting tool? having zero or negative rake angle because it point tools, what is the value of this rake angle, give (a) To guide the chip flow direction (a) results in lower cutting force reasons for your answer. (b) To reduce the friction between the tool flanks and ((b)) improves p surface finish [ 2 marks] Answer: Free carbides in castings reduce their machinability and cause tool chipping or fracture, necessitating tools with the machined surface ((c)) p provides adequate q strength g to cutting g tool (c) To add keenness or sharpness to the cutting edges. hi h toughness. high t h Z Zero rake k tool t l is i perfect f t for f this thi purpose. (d) results in more accurate dimensions (d) To provide better thermal efficiency. 13 y Differentiate between orthogonal and oblique cutting. [4 – marks] 16 Consider the following characteristics C id h f ll i h i i g g g y 1. 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. Th i d i id h h d h f pp g g The characteristics applicable to orthogonal cutting would include ( ) 1 and 2 (a) d (b) 1 and 3 d (c) 2 and 3 (d) 1, 2 and 3 Page 3 of 210 15 IES ‐ 2014 IAS – 1994 IES‐2014 Conventional IES‐2014 Conventional For-2016 (IES,GATE & PSUs) 14 Which Whi h one off the h following f ll i statements is i correct about b an oblique cutting? (a) Direction of chip flow velocity is normal to the cutting edge of the tool (b) Only two components of cutting forces act on the tooll (c) cutt cutting g edge o of tthee too tool iss inclined c ed at aan acute aangle ge to the direction of tool feed (d) Cutting C tti edge d clears l th width the idth off the th workpiece k i 17 Rev.0 18 IES ‐ 2012 IES ‐ 2012 IES‐2006 During orthogonal cutting, an increase in cutting speed D i h l i i i i d 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 Which of the following is a single point cutting tool? (a) Hacksaw blade (b) Milling cutter (c) Grinding wheel (d) Parting tool 19 20 21 IES‐2015 IES‐2003 The off the with Th angle l off inclination i li i h rake k face f ih respect to the tool base measured in a plane perpendicular to the base and parallel to the width of the tool is called (a) Back rake angle (b) Side S d rake k angle l (c) S Side de cutt cutting g edge aangle ge (d) End cutting edge angle IAS – 1996 GATE(PI)‐1990 The purpose of providing side rake angle in the cutting The diameter and rotational speed of a job are 100 mm and tool is 500 rpm respectively. The high spot (Chatter marks) are (a) avoid work from rubbing against tool f found d at a spacing off 30 deg d on the h job b surface. f The h chatter h (b) Control chip flow flo frequency is (c) Strengthen tool edge ((a)) 5 Hz ((d)) 5500 Hz Page 4 of 210 24 IES 2010 Consider C id the th following f ll i statements: t t t In an orthogonal, single single‐point point metal cutting, as the side‐cutting edge angle is increased, 1. The tangential force increases. 2 The longitudinal force drops. 2. drops 33. The radial force increases. Which of these statements are correct? (a) 1 and 3 only (b) 1 and 2 only ( ) 2 and (c) d 3 only l (d) 1, 2 and d3 Thrust force will increase with the increase in Th f ill i i h h i i ((a)) Side cutting edge angle g g g (b) Tool nose radius ( ) Rake angle (c) R k l ( ) (d) End cutting edge angle. g g g 25 ((c)) 100 Hz 23 IAS – 1995 The tool life increases with the Th l lif i i h h ((a)) Increase in side cutting edge angle g g g (b) Decrease in side rake angle ( ) Decrease in nose radius (c) D i di ((d)) Decrease in back rake angle g ((b)) 12 Hz (d) Break chips p 22 For-2016 (IES,GATE & PSUs) Statement (I): Negative rake angles are preferred on rigid set‐ St t t (I) N ti k l f d i id t ups for interrupted cutting and difficult‐to machine materials. materials Statement (II):Negative rake angle directs the chips on to the machined surface (a) Both Statement (I) and Statement (II) are individually t true and d Statement St t t (II) is i the th correctt explanation l ti off Statement (I) (b) Both B h Statement S (I) and d Statement S (II) are individually i di id ll true but Statement (II) is not the correct explanation of St t Statement t (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true 26 Rev.0 27 5.7 For-2016 (IES. if the tool nomenclature is 8‐6‐5‐5‐ 10‐15‐2‐mm. 12.5. (a) Both A and R are individually true and R is the correct explanation p of A (b) Both A and R are individually true but R is not the correctt explanation l ti off A (c) A is true but R is false (d) A is false but R is true 29 Consider the following statements with respect to the effects of a large g nose radius on the tool: 1. End d cutting edge d angle l 4. 2.2. Select the correct answer using the codes given below: ( ) 1 and (a) d2 (b) 2 and d3 (c) 1 and 3 (d) 1.6. It improves tool life 2 It reduces the cutting force 2.0 36 .5.3. End relief angle 7. 3 Wh does What d the h angle l 12 represent?? ((a)) Side cutting‐edge g g angle g (b) Side rake angle ( ) Back (c) k rake k angle l (d) Side de cclearance ea a ce aangle ge Page 5 of 210 33 35 In ASA System. then the side rake angle g will be (a) 5° (b) 6° (c) 8° (d) 10° Rev.3. Nose radius The correct sequence of these tool elements used for correctly specifying the tool geometry is ( ) 1.. 6.2. Lip angle Which hi h off these h statements are correct? (a) 1 aand d3 ((b)) 2 aand d3 (c) 1 and 2 (d) 1.IES‐1995 IES‐2002 IES‐2006 The angle between the face and the flank of the single point cutting tool is known as a) Rake angle b) Clearance angle c) Lip angle d) Point angle. Reason (R): Smaller point angle results in lower rake angle. Side cutting edge angle 5.GATE 30 IES‐1995 IES‐2009 Tool life increase with increase in T l lif i i h i i ( ) (a) Cutting speed g p (b) Nose radius ( ) F d (c) Feed ( ) p (d) Depth of cut Consider the following statements: The strength of a single point cutting tool depends upon 1. Clearance angle 3. 4. It I increases i the h possibility ibili off chatter. Side rake angle 3. 2 and 3 32 IES‐2009 IES‐1993 The following tool signature is specified for a single‐ point cutting p g tool in American system: y 10.6. 2. 15.6. 8.&6.4. 20.PSUs) 3. It improves the surface finish. 2 and 3 Consider the statements about nose C id h following f ll i b radius 1. Side relief angle 6. It improves p tool life.7 (a) 6 (b) 1. the tool is provided with a p p point angle g smaller than that required for a ductile material. 5 . 3. Back rake angle 2.3.4. h 33.7 (d) 1. Which of the above statements is/are correct? ( ) 2 only (a) l (b) 3 only l (c) 2 aand d3o onlyy (d) 1. Rake R k angle l g 2.2. 28 IES ‐ 2012 Assertion (A): For drilling cast iron.4.5. 2 aand d3 31 IES‐1994 Tool geometry of a single point cutting tool is specified by the following elements: 1 1. It deteriorates surface finish.7 6 34 (c) 1. uncut thickness is 0.8 mm.22 (c) 22.34 IES‐2014 Conventional IES‐2014 Conventional operation is ……………….53 (b) (b) 20.e.5 mm. The value of Φ is closest to (a) 00 (b) 450 (c) 600 (d) 900 41 The parameters determine the Th following f ll i t d t i th p formation: model of continuous chip 1. Cutting velocity 3 Chip thickness 3. 4. The chip thickness under orthogonal machining condition is 1.2 (a) 0. I a single i l point i turning i l the h side id rake k angle l and orthogonal rake angle are equal.00 IAS IAS ‐ 2009 Main 2009 Main In tool.31 (b) 0.GATE‐2008 ISRO‐2011 A cutting tool having tool signature as 10. What thickness of the chip produced is 0. The chip flows in the orthogonal plane.4. True feed 2.0 45 . The shear strain produced during the (c) 3. 6 8. The depth p of cut. it is found that the is 0. Determine shear h angle l also l if the h rake k angle l is i 10o During g p pure orthogonal g turning g operation p of a In a machining g operation p chip p thickness ratio hollow cylindrical pipe. The parameters which govern the value of shear angle l would ld include l d (a) 1. find the cutting ratio.3 and 4 42 GATE‐2014 IES ‐ 2004 A bar of 70 mm diameter is being cut orthogonally and is reduced to 68 mm by a cutting tool. Φ is the principal cutting edge angle and its range is plane 0o ≤ φ ≤ 90o .94 (d) 50.13 [10 Marks] mm/rev. Hint: length of uncut chip = πD For-2016 (IES.2 1 2 and 3 (b) 1. 10 9. 8 g 2 will have side rake angle (a) 10o (b) 9o (c) 8o (d) 2o 37 GATE‐2001 38 A single – point cutting tool with 12° rake angle is used to machine a steel work – p piece. The shear angle (in degrees) evaluated from this data is ( ) 6 (a) 6. 6 6.9 mm. 8 8. tool In case mean length of the chip is 68.GATE & PSUs) 43 Page 6 of 210 44 Rev. 9 6. tool the chip thickness ratio was obtained as 0. i. The shear angle is approximately ( ) 22°° (a) ((b)) 26° (c) 56° (d) 76° 40 39 IES‐1994 GATE 2011 During orthogonal cutting of mild steel with a 10 10° rake angle tool. Rake angle 4 g of the cutting g tool.3 and the rake angle of the tool is 10°.3 1 3 and 4 (c) 1.2 and 4 (d) 2. The feed is the value of the shear strain? given to the zero degree rake angle tool is 0.81 mm.00 (d) 3. 5 ( ) 1. Cl l 3. Chip thickness will be 1∙8 mm.GATE & PSUs) 52 Details pertaining to an orthogonal metal cutting process are given below. This allows better access of coolant to the tool work piece interface Which of the statements given above are correct? h h f h b (a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4 49 (b) angle l theoretically minimum possible shear strain to 46 (a) 28.3 f for the h IES‐2006 Consider the following statements with respect to C id h f ll i i h the relief angle of cutting tool: 1. Plan approach angle 1. Select the correct answer using the code given below: (a) 1 and 3 (b) 1 and 4 ( ) 2 and 3 (c) d (d) 2 and 4 d 53 Page 7 of 210 51 IES ‐ 2014 In = I an orthogonal h l turning i process.7754×105 (c) 1.0 54 . List I with List II and select the correct answer M h Li I i h Li II d l h using the codes given below the Lists: List I List II A.1781×105 (b) 0.5 For-2016 (IES. The cutting speed is 60 m/min and depth of cut 2∙4 mm.0 20 A single point cutting tool with 120 rake angle is used for orthogonal machining of a ductile material. feed = 0.IES ‐ 2009 Minimum shear strain in orthogonal th l turning t i with ith a cutting tti tool of zero rake angle g is (a) 0. T l f Tool face and flank d fl k D.0 (c) (d) 2. then the cutting ratio will be (a) 2. This affects tool life 4 This allows better access of coolant to the tool 4. Chip velocity will be 80 m/min 3. Chip velocity will be 45 m/min. Wedge angle g g 4. Tool face B.4 Undeformed thickness 0 6 mm 0. What h is the h velocity l of chip along the tool face? (c) 25. shear angle 45° 27 3 m/min 27. 2 Chip velocity will be 80 m/min.0 (b) 0. the h chip hi thickness hi k 0.5 m/s Mean thickness of primary shear zone 25 microns The shear strain rate in s–1 during the process is (a) 0.32 mm.397×10 4 397×105 47 48 IES‐2004 Consider the following statements: C id h f ll i g g g 1.8 18 Rev.5 28 5 m/min GATE 2012 GATE ‐2012 GATE (PI)‐1990 Consider the following statements: C id h f ll i In an orthogonal cutting the cutting ratio is found to be 0∙75. A large rake angle means lower strength of the cutting edge. Chip thickness ratio 0.2 (c) 1. 2. Chip thickness will be 3∙2 mm. This affects the direction of chip flow 2 This reduces excessive friction between the tool 2. Rake angle 2. Tool nose A B C D A B C D (a) 1 4 2 5 (b) 4 1 3 2 (c) 4 1 2 3 (d) 1 4 3 5 50 IES‐2008 The rake angle of a cutting tool is 15°.3 25 3 m/min (d) 23 5 m/min 23. 4.6 Rake angle +10° Cutting speed 2.2 mm/rev.0104×10 1 0104×105 (d) 4. 4 Cutting edge g g 5. Which of the following are correct? g 1. i l Th The shear h plane l IES‐2004 IES‐2004.6 26 (b) 3.6 (d) 1. This reduces excessive friction between the tool and work piece 3. 2 Cutting torque decreases with rake angle. Cutting torque decreases with rake angle Which of the statements given above is/are correct? (a) Only 1 (b) Only 2 ( ) Both 1 and 2 (c) B th d (d) Neither 1 nor 2 N ith Match. Tool flank C Clearance angle C. 2. ISRO‐2009 occur (a) 51 (b) 45 (c) 30 (d) None of these and d cutting velocity l 35 m/min. Then Th th chip the hi velocity y is (a) 2.26 1 26 58 IAS‐1995 In the I an orthogonal h l cutting.2 mm. chip hi thickness thi k = 0.4 4 mm and the uncut chip p thickness is 0.0 63 . d the h actual chip thickness will be (a) Doubled (b) halved (c) Quadrupled (d) Unchanged.4 m/s (c) 1. thickness is 0. If the chip thickness ratio is unaffected ff d with h the h changed h d cutting conditions. for turning a work piece off diameter di t 100 mm att the th spindle i dl speed d off 480 8 RPM is i (a) 1. Unchanged For-2016 (IES. l φ is i the h shear angle and V is the cutting velocity. speed using a cutting tool of rake angle m/min cutting speed.8 (c) 28.51 (c) 48 (d) 151 59 60 GATE – 2009 (PI) Common Data S‐1 GATE – 2009 (PI) Common Data S‐2 An operation is A orthogonal h l turning i i i carried i d out at 20 An operation is A orthogonal h l turning i i i carried i d out at 20 m/min cutting speed.6 6 mm.IES‐2001 If α is i the h rake k angle l off the h cutting i tool.0 m/s (b) 2.8 (a) 8 (b) 27. speed using a cutting tool of rake angle 155o. depth of cut = 0. The shear plane angle (in degrees) is The chip velocity (in m/min) is (a) 26. then the velocity of chip sliding along the shear plane is given i b by (a) ( ) (c) V cos α cos(φ − α ) V cos α sin(φ − α ) (b) (d) V sin φ cos (φ − α ) IES‐2003 IAS‐2003 An operation is A orthogonal h l cutting i i i being b i carried out under the following conditions: cutting speed = 2 m/s.0 m/s (d) 1.5 mm.8 Page 8 of 210 62 (b) 10 (c) 12 (d) 14 Rev. The chip p thickness is 0.66 m/s V sin α sin(φ − α ) 55 In orthogonal cutting.GATE & PSUs) 61 (b) 2 51 2. i h depth d h off cut is i halved h l d and d the feed rate is double.8 (d) 29. The chip p thickness is 0. shear angle is the angle between ( ) Shear plane and the cutting velocity (a) (b) Shear plane and the rake plane Sh l d th k l (c) Shear plane and the vertical direction ((d)) Shear plane and the direction of elongation of crystals in p g y the chip 56 IAS‐2002 57 IAS‐2000 IAS‐1998 The cutting velocity in m/sec.2 mm.4 4 mm and the uncut chip p 155o. low cutting speed and Consider the following machining conditions: BUE will A built‐up‐edge is formed while machining f form in i insufficient cooling help produce (a) Ductile material.1 Rev. a a e a g e. (b) Ductile materials at low speed Small uncut chip thickness. A brittle material 4.88 4 B 14 C 24 C 34 5 D 15 C 25 A 828.90 2 B 12 D 22 C 32 2. Reason (R): High speed turning may produce long. Ductile material (a) continuous chips in ductile materials (a) Ductile materials at high speed (b) (c) S Small rake angle.7.10 3 D 13 B 23 D 33 18 88 18. Option O ti Q No N O ti 1 C 11 B 21 A 31 2. ( ) (c) Brittle materials at high speed (c)continuous chips with built built‐up up edges in ductile (d) B i l (d) Brittle materials at low speed i l l d materials (d) discontinuous chips in brittle materials 67 68 69 WORKBOOK GATE‐2009 IES‐1997 Friction at the tool‐chip interface can be reduced by (a) decreasing the rake angle (b) increasing the depth of cut ( ) Decreasing the cutting speed (c) (d) increasing i i the h cutting i speed d For-2016 (IES. 1200.75 9 B 19 C 28 B 10 B 20 B 30 0. 231 4 231. At a lower cutting speed 3.0 72 . 4. (a) Both A and R are individually true and R is the correct explanation p of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 9 of 210 71 Q. (d) (b) discontinuous chips in ductile materials High cutting speed High cutting speed. Option Q. excess metal is removed in the form During machining excess metal is removed in the form of chip as in the case of turning on a lathe. At a higher cutting speed 2. 127 8 C 18 D 28 D 36 324. A ductile material 4 Select the correct answer using the code given below: (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 Plain milling of mild steel plate produces (a) Irregular shaped discontinuous chips (b) Regular shaped discontinuous chip (c) Continuous chips without built up edge (d) Joined chips 64 Wh t are the What th conditions diti th t would that ld allow ll a continuous chip to be formed in metal cutting? [4 Marks] 65 IES‐2015 66 GATE‐2002 IAS‐1997 Coarse feed . Option Q. S a u cut c p t c ess.IES 2007 GATE‐1995 IES‐2015 Conventional IES‐2015 Conventional During machining. 36 7 429 429.GATE & PSUs) 70 Ch‐1: Mechanics of Basic Machining Operation Assertion (A): For high speed turning of cast iron pistons. Which of the following are correct? Continuous ribbon like chip is formed when turning 1 1. low rake angle. carbide tool bits are provided with chip b k breakers. ribbon type continuous chips which must be broken into small lengths which otherwise would be difficult to handle and may yp prove hazardous.21 21 6 D 16 B 26 B 35 7 A 17 B 27 WRONG 36. Q No N O ti Q No N O ti Q No N Option Q. depth of cut of 4 mm with a feed of 0. The Th cutting i velocity l i is i 2 m/s. The feed velocity l it is i negligibly li ibl small ll with ith respectt to t the th cutting tti velocity. rake angle of the used in an orthogonal machining process.5°.2 mm and the chip thickness is 0.5 Linked Answer Questions GATE‐2013 S‐2 In orthogonal turning of a bar of 100 mm diameter In orthogonal turning of a bar of 100 mm diameter with a feed of 0. mean friction co‐efficient on tool face = 0. Merchant relationship the value of N If the coefficient of friction between N. / Assume that the energy gy expended p during g machining g is completely converted to heat.16 y The normal force acting at the chip‐tool interface in N is (a) 1000 (b) 1500 76 GATE 2015 GATE-2015 In off an engineering alloy.5 75 (c) 20oo (d) 2500 77 78 IAS – 1999 IES‐2014 Conventional A single point cutting tool with 0° rake angle is Show schematically the Merchant’s force circle in In an orthogonal cutting process. y The orthogonal rake angle of the cutting tool in degree is (a) zero (b) 3. Rake angle = 100.5 N and the friction force is also perpendicular p p to the cutting g velocityy vector. The main (tangential) cutting force is 1500 N.0 (c) 360. The uncut chip hi thickness thi k i 0. The ratio of friction force to normal force associated i t d with ith the th chip‐tool hi t l interface i t f i 1. then the power consumption(in also the assumptions p made while arriving g at the ((a)) 39 39. The feed velocity is negligibly small with respect to the cutting velocity The ratio of friction force to normal force velocity. depth of cut of 4 mm and d cutting tti velocity l it off 90 m/min.25° 4 5 kW) for the machining g operation p is _____ final equations.5 N and is d the th friction f i ti f force i also is l perpendicular to the cutting velocity vector. width of cut = 3 mm. The cutting velocity is 2 m/s. The shear force (in N) acting along the primary shear plane is (a) 180. The rate of heat generation (in W) at the p g primaryy shear p plane is (a) 180. At a orthogonal cutting. 74 73 Linked Answer Questions GATE‐2013 S‐1 GATE ‐2010 (PI) Linked S‐2 In off an engineering alloy.25 mm/rev. bet een the tool properties and cutting process parameters. to friction force acting at the chip‐tool interface. mm yield shear stress of work material = 285 N/mm2.35. it has I orthogonal th l turning t i i i ll h been observed that the friction force acting at the chip‐ t l interface tool i t f i 402.51 mm.5 (b) 200.5 (c) 302.7. Using cutting speed of 180 m/min. The is Th uncutt chip thickness is 0. Derive the equations for shear tooll is 20° and d friction f angle l is 25. chip thickness ratio = 0. 6 D Determine i ((i)) Cutting g force ((Fc) (ii) Radial force (iii) Normal N l force f (N) on tooll and d ((iv)) Shear force ((Fs )).2 mm and is d the th chip hi thickness thi k i 0. associated with the chip‐tool interface is 1.5° 5 ((b)) 42. it I orthogonal h l turning i i i ll i has h been observed that the friction force acting at the chip‐ tool interface is 402.4 is mm. the thrust force is 490 and d frictional f i i l forces f i terms off the in h material i l Merchant'ss shear angle relationship.58 (c) 5 (d) 7. / i it is i observed b d that th t and d cutting tti velocity l it off 90 m/min.ESE ‐2000 (Conventional) Analysis of Metal Cutting B S K M d l By S K Mondal GATE ‐2010 (PI) Linked S‐1 The following data from the orthogonal cutting test is available. ((c)) 47 47.5 (d) 402.65.5° 5 [15‐Marks] For-2016 (IES.5 (d) 402.0 81 .25 mm/rev. / i it is i observed b d that th t the main (tangential)cutting force is perpendicular the main (tangential)cutting force is perpendicular to friction force acting at the chip‐tool interface.75° 75 ((d)) 550. The main (tangential) cutting force is 1500 N.4 mm.0 (b) 240. parameters State shear angle will be and the chip is 0. uncut chip thickness = 0.GATE & PSUs) 79 Page 10 of 210 80 Rev. 7. The friction angle is ( ) 45° (a) (b) 30° ( ) 60° (c) 6 ° (d) 40°° GATE 2008 (PI) Li k d S 1 GATE ‐2008 (PI) Linked S‐1 ESE‐2005 Conventional Mild steel is being machined at a cutting speed of 200 m/min with a tool rake angle of 10.GATE‐1997 In a typical metal cutting operation. co efficient of friction between the tool and the chip is 0.6 mm.15 (b) 3. Reduction in friction angle decreases cutting force R Reduction R.35 (d) 3.5 (b) 24.γ = friction angle andα = rake angle) Page 11 of 210 89 In orthogonal cutting test. shear strength of workpiece 2. coefficient of friction at tool tool‐chip chip interface = 0. experiment an HSS tool having 83 (a) 20. shear strength of workpiece material i l = 460 6 N/ N/mm2.6 N ( ) 479. Reduction in friction angle g increases cutting g force Q.25 mm. depth material i l = 460 6 N/mm N/ d h off cut = 0. Φ g β and cutting g rake angle g α the friction angle is given as For-2016 (IES. the thrust force = 600 N and chip shear angle is 30o.45 85 IES 2010 The the angle Th relationship l ti hi between b t th shear h l Φ.6 Rev. 07 Determine (i) shear angle and Shear p plane angle g ((in degree) g ) for minimum cutting g force ((ii)) Cutting g and thrust component p of the force.GATE & PSUs) 88 86 87 IES‐2005 IES‐2003 Which one of the following is the correct expression for the Merchant Merchant'ss machinability constant? (a) 2φ + γ − α (b) 2φ − γ + α (c) 2φ − γ − α ((d)) φ + γ − α (Where φ = shear angle.5 and the 2. experiment an HSS tool having the following g tool signature g in the orthogonal g reference system (ORS) has been used: 0‐10‐7‐7‐10‐75‐1. Given width of cut = 3. The width of cut and uncut thickness are 2 mm and 0. is 82 GATE 2008 (PI) Li k d S 2 GATE ‐2008 (PI) Linked S‐2 In an orthogonal cutting experiment.2 mm respectively If the average value of co‐efficient respectively. depth d h off cut = 0.0 90 . 07 Minimum p power requirement q ((in kW)) at a cutting g speed p (c) 28. Then the chip shear force is ( ) 1079. R d ti in i friction f i ti angle l increases i chip hi thickness thi k g decreases chip p thickness S.25 (c) 3.7. the cutting force = 900 N. using a cutting tool of positive rake angle = 10°. Reduction in friction angle (a) P and R (b) P and S ( ) Q and (c) dR (d) Q and dS Better surface finish is obtained with a large rake angle because (a) the area of shear plane decreases resulting in the decrease in shear force and cutting g force (b) the tool becomes thinner and the cutting force is reduced d d (c) less heat is accumulated in the cutting g zone (d) the friction between the chip and the tool is less of 150 m/min is (a) 3.4 N (c) (d) 6 6N 69. it was observed that the shear angle was 20°. coefficient of friction at tool tool‐chip chip interface = 0.25 mm.5 (d) 32.5 GATE‐2014 In an orthogonal cutting experiment.5 84 GATE‐2014 Which pair of following statements is correct for orthogonal cutting using a single‐point cutting t l? tool? P. Given width of cut = 3.4 N (a) (b) 969. shear h stress off the h work k material i l is i 400 N/mm N/ the following g tool signature g in the orthogonal g reference system (ORS) has been used: 0‐10‐7‐7‐10‐75‐1.6 mm. Then the shear force will be approximately ( given sin 11.8 Chip thickness 1.435 435 .04 4N (c) 9.85 (d) 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0.66 N 95 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle. 686.83 0 83 m/s (b) 0. 861. the following data i obtained: is bt i d p thickness = 0.IES ‐ 2014 GATE – 2007 (PI) Common Data‐1 ( ) IES‐2000 11.8 32 8o (c) 57.21 5 N ((b)) 18.23 (b) 0 46 (b) 0.7 mm Thrust force 200 N Thrust force = 200 N Cutting force 1200 N Cutting force = 1200 N Assume Merchant's theory. the following data i obtained: is bt i d p thickness = 0. f i l i N d N i l Determine (i) The coefficient of friction between the tool and chip.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0. width of cut = 5 mm.2) (a) 650 N (b) 720 N (c) 620 N (d) 680 N In an orthogonal cutting test.GATE & PSUs) 92 G 20 ( ) i k d S GATE – 2011 (PI) Linked S1 GATE – 2007 (PI) Common Data‐2 ( ) 97 In an orthogonal machining test.5 mm 950 N Normal force = 95 Thrust force = 475 N Th shear The h angle l and d shear h f force.31o = 0. respectively.2 mm. the coefficient of friction in chip‐tool chip tool interface will ill be 1 (a) 2 ( b) 2 ( c) 1 ( d) 2 2 91 In an orthogonal machining test.64 N (d) 23. 150.8 Chip thickness 1.25 mm Uncut chip Chip thickness = 0.7 mm Thrust force 200 N Thrust force = 200 N Cutting force 1200 N Cutting force = 1200 N Assume Merchant's theory. I an orthogonal h l cutting i operation i shear h l = cutting force = 900 N and thrust force = 810 N.53 0 53 m/s 94 (c) 1. ti l are o o ((a)) 771. test the following observations were made Cutting force 1200 N Thrust force 500 N Tool rake angle zero Cutting speed 1 m/s Depth of cut 0 8 mm 0.0 99 . the cutting force and thrust force were observed to be 1000N and 500 N respectively.5 mm Chip speed along the tool rake face will be (a) 0. 75 751.6 22 6o (b) 32.2 m/s (d) 1.88 m/s IFS 2012 IFS‐2012 An orthogonal machining operation is being carried out under the following conditions : depth of cut = 0.95 Page 12 of 210 98 96 In an orthogonal machining operation: Uncut thickness 0 5 mm Uncut thickness = 0.4 67 4o 93 G 20 ( ) i k d S2 GATE – 2011 (PI) Linked S2 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle.5 mm Friction angle during machining will be (a) 22. test the following observations were made Cutting force 1200 N Thrust force 500 N Tool rake angle zero Cutting speed 1 m/s Depth of cut 0 8 mm 0.218o.1 mm. rake angle = 10o The force components along and normal to the direction of cutting velocity are 500 N and 200 N respectively.75 mm Width off cutt = 2.46 (c) 0. If the rake angle of tool is zero.5 mm 950 N Normal force = 95 Thrust force = 475 N 2) of the work Th lti t h t The ultimate shear stress (in N/mm (i N/ ) f th k material is (a) 235 (b) 139 (c) 564 (d) 380 GATE‐2006 Common Data Questions(1) GATE‐2006 Common Data Questions(2) In an orthogonal machining operation: Uncut thickness 0 5 mm Uncut thickness = 0. The percentage of total energy dissipated due to friction at the tool chip interface is friction at the tool‐chip interface is (a) 30% (b) 42% (c) 58% (d) 70% Rev.25 mm Uncut chip Chip thickness = 0. (ii) Ultimate shear stress of the workpiece material [10] (ii) Ultimate shear stress of the workpiece material. [10] For-2016 (IES.75 mm Width off cutt = 2.565 5 5 .31o In angle . The coefficient of friction at the tool‐chip interface is (a) 0 23 (a) 0.157o . chip thickness = 0.1 57 1o (d) 67. 100.4 4 times the main cutting g force (b) 0.4 to 0. p . 8.2 to 0. 0. feed eed o of 0. The feed is 0.6 to 0.0 108 . 3 (b) 2.8 Power consumption in metal cutting is mainly due to ((a)) Tangential g component p of the force (b) Longitudinal component of the force (c) Normal component of the force (d) Friction F i ti att the th metal‐tool t l t l interface i t f 101 IES‐1999 Consider the following forces acting on a finish turning tool: 1. 80. 100 IES‐1997 ESE‐2003‐ Conventional For the above machining condition determine the values of (i) Friction force.0 mm dept depth o of cut. 0(mm) the following observations were made. made Tangential component of the cutting force = 500 N Axial component of the cutting force = 200 N Chip thickness = 0. 2. (ii) Yield shears strength of the work material under this machining condition.6 6 times ti th main the i cutting tti force f 103 During turning a carbon steel rod of 160 mm diameter by a carbide tool of geometry.2 mm/rev and the depth of cut is 0.GATE & PSUs) IES‐2001 A straight turning operation is carried out using a single point cutting tool on an AISI 1020 steel rod. (after t) α 2 = 0. 2.3° and 1. 104 GATE – 1995 ‐Conventional IAS‐2003 Main Examination While turning a C‐15 steel rod of 160 mm diameter at 315 rpm. Cutting force.48 0 48 mm Draw schematically the Merchant’s circle diagram for the cutting force in the present case.IES‐1995 GATE‐2006 Common Data Questions(3) In an orthogonal machining operation: Uncut thickness 0 5 mm Uncut thickness = 0. Th correctt sequence off the The th decreasing d i order d off the magnitudes g of these forces is (a) 1. respectively are respectively. Tangential component of the cutting force. 2 (d) 3.5 mm were used. 10. 1 (c) 3. Calculate: [30 marks] (i) The side rake angle (ii) Co‐efficient of friction at the rake face (iii) The dynamic shear strength of the work material Rev. 1 For-2016 (IES. 0 (mm) at speed of 400 rpm. i Th tangential The i l cutting i force f and d thrust force were 1177 N and 560 N respectively. Feed force 2.97 (d) 40. tthe e following observation were made.7 mm Thrust force 200 N Thrust force = 200 N Cutting force 1200 N Cutting force = 1200 N Assume Merchant's theory. 0.23 (c) 40.2° and 2. 750.150. 0(mm).5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0. F and normal force. 1. 90. Chi thickness Chip thi k ( ft cut). 15.0 mm and width of cut of 2. N acting at the chip tool interface. 106 102 GATE‐2014 The radial force in single‐point tool during turning operation varies between ((a)) 0. 3.5 to t 0. Thrust force 3. mm. (iii) Cutting power consumption in kW. The values of shear angle and shear strain.8 times the main cutting force (d) 0. Page 13 of 210 105 107 During turning process with 7 ‐ ‐ 6 – 6 – 8 – 30 – 1 (mm) ASA tool the undeformed chip thickness of 2.32 mm/rev / ev a and d 4.5 mm The tool has a side cutting edge angle of 60o.3 0. 75. Pz = 1200 N Axial component of the cutting force.6 times the main cutting force (c) 0.5 mm depth of cut and feed of 0.16 mm/rev by a tool of geometry 00.3° and 4. are (a) 30. Px = 800 N 0 8 mm mm.2° and 1.65 The primary tool force used in calculating the total power consumption in machining is the (a) Radial force (b) Tangential force (c) Axial force (d) Frictional force. The uncut chip thickness (in mm) is ……….98 (b) 30. 2. The side rake angle of the tool was a chosen that the machining operation could be approximated to be orthogonal h l cutting. 528 mm (c) 0.25 mm per revolution.24 0 24 mm/rev and the depth of cut is 2 mm.71 (d) 0.80 For-2016 (IES. The shear angle is 25° 2 ° and orthogonal rake angle is zero. the shear angle is d degree i is (a) 20.20 0 20 mm/rev. Apply M h ' theory Merchant's h f analysis.19 9 110 111 GATE‐2003 Common Data Questions(1) GATE‐2003 Common Data Questions(2) A cylinder is turned on a lathe with orthogonal machining principle. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°.In orthogonal turning of a low carbon steel bar of diameter 150 5 mm with uncoated carbide tool. rpm The axial feed rate is 0.75° In the above problem. In the analysis it is found that the shear angle is 27. feed is 0.36 (c) 0. principle Spindle rotates at 200 rpm. the coefficient of friction at the chip tool interface obtained using Earnest and p g Merchant theory is (a) 0 18 (a) 0.56 An A orthogonal h l turning i operation i is i carried i d out under d the following conditions: rake angle = 55°.5 ((b)) 26. the cutting velocity is 90 m/min.64 4 GATE‐2008 Common Data Question (1) Orthogonal turning of mild steel tube with a tool b) ° ´ and b)22°20´ d 3. tc is found to be 3 mm . Apply M h ' theory Merchant's h f analysis. 0. The feed is 0. feed is 0. The values of shear angle and shear strain is a) 28°20´ and 2. mm/rev depth of cut is 3 mm. Depth of cut is 0 4 mm. axial feed = 0.5.511 mm (b) 0 528 mm (b) 0. p .846 mm 112 Orthogonal turning is performed on a cylindrical work piece with shear strength of 250 MPa. 342N N N (d) (d) 480N. The cutting and Thrust forces.19 d)37°20´ 37 and 55. spindle rotational speed p = 4 400 rpm.56 . 381N ( ) (c) 440N. mm chip thickness ratio = 0.25 mm per revolution.GATE & PSUs) 115 out at a feed of 0. ero Employing Merchant Merchant’ss theory.56 5 GATE 2015 GATE-2015 GATE 2015 GATE-2015 GATE‐2007 A cylinder is turned on a lathe with orthogonal machining principle.0 117 . mm The chip thickness obtained is 0.V V = b t V = fdV b V fdV Where Ac = cross‐section area of uncut chip (mm2) V = cutting speed = π DN . Depth of cut is 0 4 mm. The shear angle (in degrees) in GATE‐2007 In orthogonal turning of low carbon steel pipe with principal cutting edge angle of 90°. If the thickness of chip produced is 0. mm chip thickness ratio = 0.255 ((c)) 0. for l i g . The orthogonal rake angle is 7o. mm / min Rev.14 mm/rev.818 mm (d) 0.4 4 m/min / and radial depth of cut = 5 mm. 0.75° The thickness of the produced chip is (a) 0 511 mm (a) 0. 387N (b) 565N. rpm The axial feed rate is 0.4 mm The rake angle is 10 10°. the ratio of friction force to normal force acting g on the cutting g tool is ((b)) 1. 356N N N Page 14 of 210 116 114 Metal Removal Rate (MRR) M l Metal removal rate (MRR) = A l (MRR) Ac.5 (c) 30.20 0 20 mm/rev. principle Spindle rotates at 200 rpm.18 (b) 0 36 (b) 0. mm/rev depth of cut is 3 mm.56 (d) 36.28 mm.98 113 GATE‐2008 Common Data Question (2) Orthogonal turning is performed on a cylindrical work piece with shear strength of 250 MPa.4 mm The rake angle is 10 10°. In the analysis it is found that the shear angle is 27. respectively.53 this turning process is _____ 109 ((a)) 1. The chip thickness. MPa The following conditions are used: cutting velocity is 180 m/min.19 4 19 ((d)) 0.5. The orthogonal rake angle is 7o.56 . the main cutting force is 1000 N and the feed force is 800 N. are (a) 568N. for l i p g ( g ) The shear plane angle (in degree) and the shear force respectively are (a) 52: 320 N (b) 52: 400N (c) 28: 400N (d) 28:320N rake angle of 10° is carried c) 24 24°20´ 20 and 4. p y. MPa The following conditions are used: cutting velocity is 180 m/min.48 mm. Fc.8 mm/rev feed and 1. The specific machining energy is 2.2 mm/rev and 2 mm respectively. feed and depth of cut are 120 m/min. 0. divided by the cross section area of the undeformed chip gives the nominal cutting stress or the specific cutting pressure.IES ‐ 2004 GATE‐2013 A medium carbon steell workpiece is di b k i i turned d on a lathe at 50 m/min.GATE & PSUs) 124 Example When the rake angle is zero during orthogonal cutting. cutting speed 0. The facing operation is undertaken at a constant cutting speed p of 9 90 m/min / in a CNC lathe.000 0. Neglecting the contribution of the feed force towards cutting power.5 mm depth of cut.1 mm/rev with a depth of cut of 1 mm. m/s d / 60 118 119 GATE(PI)‐1991 Specific Energy Consumption e= ( GATE‐2007 Amount of energy consumption per unit volume of Fc Power (W ) = 3 MRR mm / s 1000 fd metal removal is maximum in ) (a) Turning (b) Milling (c) Reaming (d) Grinding 120 In orthogonal turning of medium carbon steel. The rotational speed of the workpiece is 160 rpm. The material removal rate in mm3/s is (a) 160 (b) 167.0 J/mm3.0 126 . the specific cutting energy in J/mm3 is (a) 0. respectively The main cutting force in N is (a) 40 (b) 80 (c) 400 (d) 800 Sometimes it is also known as specific power consumption.6 (c) 1600 (d) 1675.5 c Where Fc = cutting force (in N) πDN V tti V = cutting speed = . pressure pc pc = pc = specific p ppower of cutting g r = chip thickness ratio μ = coefficient of friction in tool chip interface Page 15 of 210 123 125 Fc Fc = bt fd Rev.25 mm/rev with a depth of cut of 4 mm. show that τs pc = (1 − μ r ) r 1+ r2 Wh τs is Where i the h shear h strengrhh off the h material i l Specific Cutting Pressure Specific Cutting Pressure The cutting force.000 mm3//min (d) Can not be calculated with the given data Power Consumed During Cutting A steell bar is b 200 mm in i diameter di i turned d at a feed f d off 0. The cutting velocity. consumption 121 122 GATE‐2013 (PI) Common Data Question A disc diameter is di off 200 mm outer and d 80 8 mm inner i di i faced of 0.2 (b) 2 (c) 200 (d) 2000 For-2016 (IES.000 mm3/min / (c) 20. What is the rate of metal removal? (a) 1000 mm3/min (b) 60. The main (tangential) cutting force is 200 N. applicable ) Very y high g temperature p is Reason ((R): produced at the tool‐chip interface.1 mm/rev / feed rate. ( ) Both (a) h A and d R are individually d d ll true and d R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES‐2002 133 Assertion (A): to A i (A) The Th ratio i off uncut chip hi thickness hi k actual chip thickness is always less than one and is termed d as cutting ratio in orthogonal h l cutting Reason ((R): ) The frictional force is very y high g due to the occurrence of sticking friction rather than sliding friction (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation l i off A (c) A is true but R is false (d) A is false but R is true 131 Page 16 of 210 132 IES‐1998 In a machining process.GATE & PSUs) 129 134 In metal cutting operation. the normal laws of sliding friction are not applicable.0 135 .Friction in Metal Cutting l GATE‐2014 GATE 1992 The force acting on a tooll during the Th main i cutting i f i d i h turning (orthogonal cutting) operation of a metal is 400 N. The specific p cutting pressure is The effect of rake angle on the mean friction angle in machining can be explained by (A) sliding (Coulomb) model of friction (B) sticking and then sliding model of friction ((C)) sticking g friction (D) Sliding and then sticking model of friction (a) 1000 (b) 2000 (c) 3000 (d) 4000 127 GATE‐1993 128 IES‐2000 The effect of rake angle on the mean friction angle in machining can be explained by (a) Sliding (coulomb) model of friction (b) sticking and then siding model of friction (c) Sticking friction (d) sliding and then sticking model of friction 130 H Di ib i i M l C i Heat Distribution in Metal Cutting IES‐2004 Assertion (A): In metal cutting. the percentage of heat carried away by the chips is typically ((a)) 55% ((b)) 25% 5 (c) 50% (d) 75% For-2016 (IES. chip tool and work. the approximate ratio of heat distributed among chip. in that order is (a) 80: 10: 10 (b) 33: 33: 33 ( ) 20: 60: 10 (d) 10: 10: 80 (c) Rev. The turning was performed using 2 mm depth p of cut and 0. (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) Page 2 and 143 173 of 210 Strain S i gauge type Or piezoelectric type St i gauge type Strain t d dynamometers t are inexpensive i i but b t less l accurate and consistent. 3. whereas.0 144 .375 0.5 0. forces y For Orthogonal Cutting use 2D dynamometer y For Oblique q Cutting g use 33D dynamometer y IES‐1993 For-2016 (IES.2 y Tool‐work thermocouple b 0. the piezoelectric type are highly accurate. Force exerted byy the tool on the chip the tool face.GATE & PSUs) The instrument or device used to measure the cutting forces in machining is : g (a) Tachometer (b) Comparator (b) C ( ) y (c) Dynamometer (d) Lactometer 140 141 Types of Dynamometers IES‐1996 A 'Dynamometer' is a device used for the measurement of ((a)) Chip p thickness ratio (b) Forces during metal cutting (c) Wear of the cutting tool (d) Deflection D fl ti off the th cutting tti tool t l 142 138 IES 2011 Dynamometer 139 y Infra ray detection method Which of the following forces are measured directly by strain gauges or force dynamometers during metal cutting ? p acting g normallyy to 1.For IES Only IAS – 2003 Mean Temperature in Turning As the cutting speed increases A h i d i ((a)) More heat is transmitted to the work piece and less p heat is transmitted to the tool (b) More heat is carried away by the chip and less heat is transmitted to the tool (c) More heat is transmitted to both the chip and the too tool (d) More heat is transmitted to both the work piece and th t l the tool Determination of cutting heat & temperature E Experimentally i t ll HEAT y Calorimetric method Mean temperature ∞ V a f b TEMPERATURE y Decolourising g agent g Tool Material HSS Carbide a 0. Frictional resistance of the tool against the chip flow acting along the tool face. position. reliable and consistent but very expensive p for high g material cost and stringent g construction. Rev. Horizontal cutting piece. 4 Vertical force which helps in holding the tool in 4.125 y Moving thermocouple technique M i h l h i y Embedded thermocouple technique p q y Using compound tool y Indirectly from Hardness and structural transformation di l f d d l f i y Photo‐cell technique q 136 IAS – 2003 The heat generated metall Th h d in i conveniently be determined by (a) Installing thermocouple on the job (b) Installing thermocouple on the tool (c) Calorimetric set‐up (d) Using radiation pyrometer cutting i can 137 y Dynamometers are used for measuring Cutting forces. g force exerted byy the tool on the work 2. GATE & PSUs) the cutting edge 151 Page 18 of 210 distance from 152 Rev. front relief face.Strain Gauge Dynamometers IAS‐2001 IES‐1998 The changes the Th strain. brittle fracture. 150 149 IES 2010 IES – 2007 Flank wear occurs on the Flank wear occurs mainly on which of the ( ) Relief face of the tool (a) f ll i ? following? (b) Rake R k face f (a) Nose part and top face ( ) Nose (c) N off the th tool t l (b) Cutting edge only IAS – IAS – 2009 Main 2009 Main y Explain ‘sudden‐death mechanism’ of tool failure.0 153 . the cutting y 2. [ 4 – marks] ((c)) Nose p part.0 for conductive gauges) The change in resistance of the gauges connected in a Wh t t Wheatstone b id produces bridge d voltage lt output t t ΔV. of the strain gauges which are firmly pasted on the surface of the tool‐holding beam as ΔR ΔR = Gε R where. these processes will dominate and when tool failure will occur. Tool Life & Tool Wear. Sudden‐death: Failures leading to premature end of the tool y The sudden‐death type of tool failure is difficult to predict. R. Slow‐death: The g gradual or p progressive g wearing g away of rake face (crater wear) or flank (flank wear) of g tool or both. Tool Life & Machinability Machinabilityy By S K Mondal 148 Tool failure is two types y 1. Tool Wear. and side relief face of the (d) Cutting edge cutting tool (d) Face of the cutting tool at a short For-2016 (IES. Explain sudden death mechanism of tool failure. ΔV through th h a strain measuring bridge (SMB) The gauge factor of a resistive pick‐up of cutting force dynamometer is defined as the ratio of (a) Applied strain to the resistance of the wire (b) The h proportionall change h in resistance to the h applied strain (c) The resistance to the applied strain (d) Change in resistance to the applied strain 145 Assertion (A): and A ti (A) Piezoelectric Pi l t i transducers t d d preferred f d over strain gauge transducers in the dynamometers for measurement of three‐dimensional three dimensional cutting forces. fatigue fracture or edge pp g However it is difficult to p predict which of chipping. transducers (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correctt explanation l ti off A (c) A is true but R is false (d) A is false but R is true 146 147 Tool Failure For PSU & IES For PSU & IES In strain gauge dynamometers the use of how many active gauge makes the dynamometers more effective (a) Four (b) Three (c) Two (d) One . i ε induced i d d by b the h force f h h electrical l i l resistance. forces Reason (R): In electric transducers there is a significant leakage of signal from one axis to the other. other such cross error is negligible in the case of piezoelectric transducers. Tool failure mechanisms include plastic deformation. G = gauge factor (around 2. GATE & PSUs) 160 Page 19 of 210 161 Rev. Secondary y wear The region where the wear progresses at a uniform rate. y Shearing off the micro welds between tool and work material. l y At low speed p flank wear p predominates. chip breakage and variable dimensions of cut (d) Chip breakage. For-2016 (IES. cutting g fluid and chip p breakage g (b) Variable thermal stresses. Flank Wear: (Wear land) Cutting tooll is the C i i much h harder h d than h h work‐piece. Reason (R): Measurement of flank wear is simple and more accurate.. k i Yet the tool wears out during the tool‐work interaction. y Flank wear is usually the most common determinant of tool life. breakage variable thermal stresses and cutting fluid Assertion (A): Tool wear is expressed in terms of A i (A) T l i d i f flank wear rather than crater wear. material y Abrasion by fragments of built‐up‐edge ploughing against i the h clearance l f face off the h tool.0 162 . life 158 159 Flank Wear: (Wear land) Stages Flank Wear: (Wear land) Primary wear y Flank Fl k Wear W occurs in i three h stages off varying i wear rates The region Th i where h the h sharp h cutting i edge d i quickly is i kl broken down and a finite wear land is established. y If MRR increased flank wear increased. chip breakage and variable dimensions of cut (c) Abrasive friction. because (a) extra hardness is imparted to the work work‐piece piece due to coolant used (b) oxide d layers l on the h work‐piece k surface f impart extra hardness to it (c) extra hardness is imparted to the work‐piece due to severe rate of strain (d) vibration is induced in the machine tool 157 Effect y Flank Fl k wear directly di l affect ff the h component dimensions di i produced.l Tool Wear IES ‐ 2014 IES – 1994 The failure off a tooll is Th fatigue f i f il i due d to ((a)) abrasive friction. and more accurate (a) Both A and R are individually true and R is the correct explanation of A l f (b) Both A and R are individually true but R is not ot a d R a e d v dua y t ue but R s ot tthe e correct explanation of A ( ) A is true but R is false (c) A i t b t R i f l (d) A is false but R is true 154 155 156 GATE‐2014 Flank Wear: (Wear land) Reason y Abrasion Ab i b hard by h d particles i l and d inclusions i l i i the in h work k piece. GATE 2008 (PI) GATE‐2008 (PI) IES – 2004 Flank Wear: (Wear land) Tertiary wear The region Th i where h wear progresses at a gradually d ll increasing rate. y In the tertiary region the wear of the cutting tool has become sensitive to increased tool temperature due to high wear land. y Re‐grinding R i di i recommended is d d before b f they h enter this hi region. Consider C id the h following f ll i statements: g the third stage g of tool‐wear,, rapid p During deterioration of tool edge takes place because 1 Flank wear is only marginal 1. 2. Flank wear is large 3. Temperature of the tool increases gradually 4. Temperature T t off the th tool t l increases i d ti ll drastically Which of the statements g given above are correct? (a) 1 and 3 (b) 2 and 4 ( ) 1 and (c) d4 (d) 2 and d3 163 IFS‐2012 Explain the mechanism of flank wear of a cutting tool. l Plot l a flank fl k wear rate curve and d indicate d the h region of tool failure. failure 164 u g machining, ac g, tthee wea a d ((h)) has as bee otted During wear land been p plotted against machining time (T) as given in the following figure. figure For a critical wear land of 1.8 mm, the cutting tool life (in ( minute) is (a) 52.00 (b) 51.67 (c) 51.50 (d) 50.00 165 Tool life criteria ISO Crater wear (A certain width of flank wear (VB) is the most common (A i id h f fl k (VB) i h criterion) y Uniform wear: 0.3 mm averaged over all past y Localized wear: 0.5 mm on any individual past Localized wear: 0 5 mm on any individual past y More common in ductile materials which produce [10 Marks] continuous chip. h y Crater C wear occurs on the h rake k face. f y At very high hi h speed d crater t wear predominates d i t y For crater wear temperature is main culprit and tool defuse into the chip material & tool temperature is maximum at some distance from the tool tip. 166 Crater wear Contd….. y Crater depth exhibits linear increase with time. y It increases with MRR. MRR 167 168 IES – 2002 IAS – 2007 Crater wear on tools always starts at some distance C l l di from the tool tip because at that point (a) Cutting fluid does not penetrate (b) Normal stress on rake face is maximum (c) Temperature is maximum (d) Tool strength is minimum Why does crater wear start at some distance from Wh d di f 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 y Crater wear has little or no influence on cutting forces, forces work piece tolerance or surface finish. For-2016 (IES,GATE & PSUs) 169 Page 20 of 210 170 Rev.0 171 IES – 2000 IES – 1995 Crater wear starts at some distance from the tool tip C di f h l i because (a) Cutting fluid cannot penetrate that region (b) Stress on rake face is maximum at that region (c) Tool strength is minimum at that region (d) Tool temperature is maximum at that region IES 2009 Conventional Crater wear is predominant in C i d i i ((a)) Carbon steel tools (b) Tungsten carbide tools ( ) High speed steel tools (c) Hi h d l l ((d)) Ceramic tools 172 Show crater wear and flank wear on a single point cutting tool State the factors responsible for wear cutting tool. State the factors responsible for wear on a turning tool. [ [ 2 –marks] k ] 173 IAS – 2002 Wear Mechanism IES – 1995 Consider C id the h following f ll i actions: i 1. Mechanical abrasion 2. Diffusion 3. Plastic deformation 4. Oxidation Whi h off the Which h above b are the h causes off tooll wear?? ((a)) 2 and 3 ((b)) 1 and 2 (c) 1, 2 and 4 (d) 1 and 3 1. Abrasion wear 2. Adhesion wear 3 Diffusion wear 3. 4. Chemical or oxidation wear 175 IAS – 1999 The to the Th type off wear that h occurs due d h cutting i action of the particles in the cutting fluid is referred to as (a) Attritions wear (b) Diffusion wear (c) Erosive wear (d) Corrosive wear 176 Why chipping off or fine cracks developed at the cutting edge d l d h d 178 Consider C id the h following f ll i statements: pp g of a cutting g tool is due to Chipping 1. Tool material being too brittle 2. Hot H hardness h d off the h tooll material. i l 33. High g p positive rake angle g of the tool. Which of these statements are correct? ( ) 1, 2 and (a) d 3 (b) 1 and d3 (c) 2 and 3 (d) 1 and 2 y Tool material is too brittle y Weak design of tool, such as high positive rake angle y As a result of crack that is already in the tool Page 21 of 210 Match List I with List II and select the correct M h Li I i h Li II d l h answer using the codes given below the lists: List I (Wear type) List II (Associated mechanism) A Abrasive wears A. 1 1. Galvanic action B. Adhesive wears 2. Ploughing action C. Electrolytic wear 3. Molecular transfer D Diffusion wears D. Diff i 4. Pl ti d f Plastic deformation ti 5. Metallic bond Code: A B C D A B C D ( ) 2 (a) 5 1 3 (b) 5 2 1 3 3 4 (d) 5 2 3 4177 (c) 2 1 IAS – 2003 y Excessive E i static i or shock h k loading l di off the h tool. l For-2016 (IES,GATE & PSUs) 174 179 Rev.0 180 Notch Wear IES – 1996 y Notch wear on the trailing edge is to a great extent an oxidation id ti wear mechanism h i occurring i where h th cutting the tti edge leaves the machined workpiece material in the feed direction. y But abrasion and adhesion wear in a combined effect can Notch wear at the outside edge of the depth of cut is N h h id d f h d h f i due to (a) Abrasive action of the work hardened chip material (b) Oxidation (c) Slip‐stick action of the chip (d) Chipping. List the important properties of cutting tool materials and explain why each is important. t i l d l i h hi i t t y Hardness at high temperatures ‐ this provides longer life of the cutting tool and allows higher cutting speeds. y Toughness ‐ to provide the structural strength needed to resist impacts and cutting forces y Wear resistance ‐ to prolong usage before replacement doesn’t chemically react ‐ another wear factor contribute to the formation of one or several notches. y Formable/manufacturable ‐ can be manufactured in a useful geometry 181 182 Why are ceramics normally provided as inserts for tools, and not as entire tools? i f l d i l? IES‐2014 Conventional IES‐2014 Conventional y Describe the characteristics of tool materials. [4‐marks] Ceramics are brittle materials and cannot provide the C i b i l i l d id h structural strength required for a tool. 183 IES 2015 IES‐2015 g statements are be correct for Which of the following temperature rise in metal cutting operation? 1. It adversely affects the properties of tool material 2. It provides better accuracy during machining 3. It I causes dimensional di i l changes h i the in h work‐piece k i and d affects the accuracy of machining 4. It can distort the accuracyy of machine tool itself. 4 184 IES – 1992 Tool Life Criteria Tool life criteria can be defined as a predetermined numerical value of any type of tool deterioration which can be measured. Some of 185 the ways y Actual cutting time to failure. (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 only (d) 1, 3 and 4 186 IAS – 2012 Main Tool life is generally specified by T l lif i ll ifi d b ((a)) Number of pieces machined p (b) Volume of metal removed ( ) Actual cutting time (c) A l i i ((d)) Any of the above y Define “tool life” and list down four methods for quantitative measurement off tooll life. lf [M k ] [Marks ‐12] y Volume of metal removed. Volume of metal removed y Number of parts produced. p p y Cutting speed for a given time y Length of work machined. For-2016 (IES,GATE & PSUs) 187 Page 22 of 210 188 Rev.0 189 IES ‐ 2012 Values of Exponent ‘n’ Taylor’s Tool Life Equation based on Flank Wear Causes y Sliding of the tool along the machined surface y Temperature rise VT n = C Where, V = cutting speed (m/min) T Time (min) T = Time (min) n = exponent depends on tool material C = constant based on tool and work material and cutting 190 condition. n = 0.08 to 0.2 for HSS tool = 0.1 to 0.15 for Cast Alloys = 0.2 to 0.4 for f carbide bid tooll [IAS‐1999; [IAS 1999; IES IES‐2006] 2006] = 0.55 to 0.77 for ceramic tool [NTPC‐2003] In Taylor’s tool life equation VT I T l ’ l lif i VTn = C, the constants n C h and C depend upon 1. Work piece material 2 Tool material 2. Tool material 3. Coolant (a) 1, 2, and 3 (b) d l (b) 1 and 2 only (c) 2 and 3 only y (d) 1 and 3 only 191 IES – 2008 192 IES – 2006 In Taylor's tool life equation is VT I T l ' l lif i i VTn = constant. What is the value of n for ceramic tools? (a) 0.15 to 0.25 (b) 0.4 to 0.55 ( ) 0.6 to 0.75 (c) 6 (d) 0.8 to 0.9 IES – 1999 Which values off index n is Whi h off the h following f ll i l i d i associated with carbide tools when Taylor's tool life equation, V.Tn = constant is applied? (a) 0∙1 to 0∙15 (b) 0∙2 to 0∙4 (c) 0.45 to 0∙6 (d) 0∙65 to 0∙9 193 194 IAS – 1998 Match List ‐ M t h Li t I (Cutting tool material) with List ‐ I (C tti t l t i l) ith Li t II (Typical value of tool life exponent 'n' in the Taylor's equation V Tn = C) and select the correct answer using equation V.T C) and select the correct answer using the codes given below the lists: List I List – List II List – A. HSS 1. 0.18 B. Cast alloy 2. 0.12 33. 0.255 C. Ceramic D. Sintered carbide 4. 0.5 Codes: A B C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 (c) 2 1 4 3 (d) 1 2 4 3 196 For-2016 (IES,GATE & PSUs) The off the Th approximately i l variation i i h tooll life lif exponent 'n' of cemented carbide tools is (a) 0.03 to 0.08 (b) 0.08 to 0.20 (c) 0.20 0 20 to 0.48 0 48 (d) 0.48 0 48 to 0.70 0 70 195 ISRO‐2011 GATE ‐2009 (PI) GATE ‐2009 (PI) A 50 mm diameter steel rod was turned at 284 rpm and I an orthogonal In th l machining hi i operation, ti th tool the t l life lif tool failure occurred in 10 minutes. The speed was obtained is 10 min at a cutting speed of 100 m/min, m/min changed h d to 232 rpm and d the h tooll failed f il d in i 60 6 minutes. i while at 75 m/min cutting g speed, p the tool life is 330 Assuming straight line relationship between cutting min. The value of index (n) in the Taylor’s tool life speed p and tool life,, the value of Taylorian y Exponent p is equation (a) 0.262 (a) 0.21 (b) 0.323 (c) 0.423 Page 23 of 210 (b) 0.133 (c) 0.11 (d) 0.233 (d) 0.521 197 Rev.0 198 5 and C= 300 for the cutting speed and the tool original. (b) 180 ((c)) 336 m/min ((d)) 772 m/min (c) 150 (d) 100 For-2016 (IES.GATE‐2004. the value An HSS tool is used for turning operation. when cutting speed is reduced by 25% . The h tooll has h a life l f off 180 minutes at a 1 hr.GATE & PSUs) 205 Page 24 of 210 206 Rev. then the cutting speed will be doubled. IES‐2000 IES – 1999. VTn = C.25. doubling the IES‐2015 cutting speed reduces the tool life to 1/16th of the Using the Taylor equation VTn = c. if the cutting speed is halved. is 1 (a) 8 1 (b) 4 1 (c ) 3 1 (d ) 2 IAS – 1995 In a single‐point turning operation of steel with a In a single point turning operation with a cemented cemented d carbide b d tool. Taylor’s tool life index (n) for this tool‐ percentage increase in tooll life l f when h the h cutting life relation. The tool life is work‐piece k i was found f d to t give i lif off 1 hour life h 21 off n = 0.25. Find the suitable speed in RPM for turning 300 (a) 200 ((a)) 9 m/min / ((b)) 18 m/min / mm diameter so that tool life is 30 min. m/min If the tool life is reduced will be reduced to 2. 0. then increase by the tool life will become ((a)) Two times ((b)) Four times ((c)) Eight g times ((d)) Sixteen times (a) Half (b) Two times (c) Eight times (d) Sixteen times 199 200 GATE-2015 GATE 2015 201 IAS – 2002 Under certain cutting conditions. if workpiece combination will be _______ speed is reduced by 50% (n = 0∙5 0 5 and c = 400) the h tooll life lif is i increased i d by b (a) 300% (b) 400% a) 100% (c) 100% (d) 50% b)95% )95 c) 78% d)50% 202 203 IES‐2013 IAS – 1997 204 IES – 2006 conventional A carbide tool(having n = 0.0 207 . The h tooll life lf minutes while cutting at 60 m/min.5. calculate the If n = 0. ISRO‐2013 In operation. doubling the I a machining hi i i d bli h cutting i 1 speed reduces the tool life to 8 th of the original value. h when h turning is carried d at 30 m/min.. m/min The value of C cutting speed of 18 m/min.0 2 0 min if the cutting speed is in Taylor’s y tool life equation q would be equal q to: to 45 minutes.25) with a mild steel In the Taylor's tool life equation. The exponent n in Taylor's tool life equation VTn = C. l Taylor's l ' tooll life l f exponent is carbide bid and d steel t l combination bi ti h i having a Taylor T l 0 25 If the cutting speed is halved. halved the tool life will exponent of 0. Similarly l l for f tooll B. m/min:49.9 4 9 210 The following data was obtained from the tool‐life cutting test: Cutting Speed.77 Where V is the cutting g speed p in m/min and T is the ((b)) 4 42.27 tool life in min.87 28. angle α.25 0 25 mm/rev and 1 mm depth of ((Hint: Flank wear rate is p proportional p to cot α)) cut. decrease one cutting tool at 60 rpm? (c) 70% increase (d) 70% decrease (a) 29 (b) (IES.55 ((c)) 80.15 In a machining experiment.58 Tool life. Also estimate the cutting speed for a tool life of 80 min. The carbide tool will provide higher Determine the constants of the Taylor tool life equation VTn = C C tool life if the cutting speed in m/min exceeds (a) 15.5 m/s.74 (d) n‐0. 60 The cutting speed (in m/min) above which tool Carbide tool: VT 1. T(in min) T l lif T(i i ) 100 130 120 50 Derive Taylor's tool life equation for this operation and estimate the tool life at a speed of 2.76 42.90 4. How many y components p can be p produced with ((a)) 330 % increase ((b)) 330%.GATE 31 (c)&37PSUs) (d) 42 For-2016 214 Extended or Modified Taylor’s equation GATE‐1999 g tools could p A batch of 10 cutting produce 500 213 Page 25 of 210 215 i.GATE‐2009 Linked Answer Questions (1) IFS 2013 IFS‐2013 GATE‐2009 Linked Answer Questions (2) In a machining experiment. the tool life was found to vary with the cutting speed in the following manner : C i Cutting speed.5 and k=1.e Cutting speed has the greater effect followed by feed g p g y and depth of cut respectively. Taylor’s tool life exponent (n) is 0. V (in m/min) d V (i / i ) Tool life.0 216 .0 211 212 GATE 2003 GATE‐2003 components while working at 50 rpm with a tool feed of 0.45 and Two cutting tools are being compared for a constant (K) ( ) is 90. tool life was found to vary I hi i i l lif f d with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 The exponent (n) and constant (k) of the Taylor's tool life equation are (a) n = 0. 208 209 Example p GATE‐2013 GATE‐2010 For tool A. tool life was found to vary I hi i i l lif f d with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 What is the percentage increase in tool life when the cutting speed is halved? (a) 50% (b) 200% (c) 300% (d) 400% In a metal cutting experiment.25 mm/rev and depth of cut of 1 mm. off a tooll with h zero rake k angle l used d in orthogonal h l cutting when its clearance angle.23 48.4 (c) 49.5 and k = 540 (b) n= 1 and k=4860 (c) n = ‐1 and k = 0. ti Th tool The t l life lif equations ti are: = 60.6 = 200 ((a)) 26. t. α is changed from 10o to 7o? rpm with a feed of 0. Rev.3 (d) 60.67 45.77 ((d)) 142.94 3.74 49. A similar batch of 10 tools of the same specification could ld produce d 122 components t while hil working ki att 80 8 What is approximate percentage change is the life.3 and dK machining hi i operation.77 9.6 = 3000 A will have a higher tool life than tool B is HSS tool: VT 0.0 (b) 39. min 2. n = 0. 2 3. 3. 3 and 4 (c) 1. 1. 2. 2 3 (b) 3.GATE & PSUs) 1. Reason(R): Increases in depth of cut gives rise to relatively small increase in tool temperature.0 225 . 3 and 4 1 3 and 4 (d) 1.6 0 6d0. what h is the h right h sequence to adjust the cutting parameters? 2 2. 3 218 IES – 1994. 1 2. Temperature rise of cutting edge 2 Chipping of tool edge due to mechanical impact 2. 2 and 3 (b) 2. l A 60 min tooll life l f was obtained b d using the h following f ll With the th help h l off Taylor’s T l ’ tool t l life lif equation. 25% and also if they are increased individually by 25%.0 mm. 3. Ceramic Page 26 of 210 224 Rev. HSS 223 2. m/min f = 0. Cutting speed 3. 3 2. 2 1 (c) 2. I. Calculate the effect on tool life if of cutting g and life of the tool. 3 2 4 1 3 (d) 4 2 I 3 4. Feed rate 2. ti 0 13f0. Depth of cut Select the correct answer using the code given below: (a) 11‐ 22‐ 3 (b) 22‐ 33‐ 1 ((c)) 33‐ 2‐ 1 ((d)) 1‐ 33‐ 2 1. IAS ‐2010 Conventional 222 IFS 2009 IFS 2009 The following equation for tool life was obtained for HSS tool.4. 3. v = speed. Feed The correct sequence of these elements in DECREASING order of their influence on tool life is (a) 2. Increase in feed of cut at constant cutting force S l t th Select the correct answer using the code given t i th d i below: (a) 1. feed and depth of cut are together increased by and steel as work material.IES 2010 Tool life is affected mainly with (b) Depth D h off cut ( ) Coolant (c) C l t (d) Cutting speed 217 IES – 2008 For increasing the material removal rate in turning.3 0 3= C. Gradual wears at tool point 4.25 0 25 determine the shape of the curve between velocity mm. 1 2 ( ) Feed (a) 1 1. Nose radius 2. 1 2 4 3 1 (b) 4 2 3 1 4. cutting condition VT0. without h any constraints. d = depth of cut. 3 1 (d) 3. [10‐Marks] where f = feed. temperature (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true 221 Tool Life Curve l f ESE‐1999.13 C v = 40 m/min. Select the correct answer using the codes given below (a) 1. 1 (c) 2. 2 and 4 1 2 and 4 220 219 Assertion (A): in A i (A) An A increase i i depth d h off cut shortens h the tool life. 2. Depth of cut 4. Carbide 3. For-2016 (IES. Depth of cut 3 Cutting speed 3. IES – 1997 ISRO‐2012 IAS – 1995 What are the reasons for reduction of tool life in a Wh h f d i f l lif i machining operation? 1. 3 1. Feed 3 3. Assume an HSS tool speed. 4. d = 2. 2007 Speed Consider the following elements: What is the correct sequence of the following parameters in order of their maximum to minimum influence on tool life? 1. Annealed A l d 186 86 97 3 2. As Cast A C 207 60 40 Tool life Tests Tool life Tests y Conventional test: Using empirical formula g p y Accelerated test: Estimate the tool life quickly E t Extrapolating of steady wear rate l ti f t d t High speed test‐will take less time Variable speed test Multi pass turning Taper turning Draw a figure showing variation of tool life with cutting speed and the effect of workpiece hardness and microstructure. Match the graphs for HSS Carbide HSS. 2. Which of these statements is/are correct? Whi h f th t t t i / t? (a) 1 and 3 (b) 1 and 4 (c) 2 only (d) 4 only 227 Effect of Rake angle on tool life 228 Effect of Clearance angle on tool life If clearance angle increased it reduces flank wear but weaken the cutting edge.GATE & PSUs) Value of n for both the tools is same. Hardness(HB) Ferrite Pearlite 1. Value of C for both the tools is same.IES 2010 Conventional y Draw tool life curves for cast alloy. C bid and d Ceramic C i tooll materials and select the correct answer using i th code the d given i below the lists: C d HSS Code: SS Carbide C bid C Ceramic i (a) 1 2 3 (b) 3 2 1 (c) 1 3 2 (d) 3 1 2 The tool life curves for two tools A and B are shown in Th l lif f l A d B h i the figure and they follow the tool life equation VTn = C.0 234 . Answer IES 2013 Conventional 2013 C ti l IES – For IES Only You are asked to turn ductile cast iron with various microstructure and hardness as shown in the following table. Value of C for tool A will be greater than that for the tool B Value of C for tool A will be greater than that for the tool B. High speed steel and D l lif f ll Hi h d l d ceramic tools. 232 Page 27 of 210 233 Refer: B. High speed steel 2. Value of C for tool B will be greater than that for the tool A. microstructure For-2016 (IES. 229 230 231 IES – 2013 Con. cast alloy and 3. edge so best compromise is 80 for 0 HSS and 5 for carbide tool. ceramic tools. 4. Consider the following statements: d h f ll 1. [2 – Marks] Ans. Annealed 170 100 ‐ 4.L Juneja+Nitin Seth Rev. HSS (min) 30 m/min < Cast alloyy < Carbide Effect of work piece on tool life < Cemented carbide 150 m/min < Cermets y With hard micro‐constituents in the matrix gives poor < Ceramics or sintered oxide (max) 600 m/min tool life. 226 Cutting speed used for different tool materials IAS – 2003 IES 2010 The above figure shows a typical relationship between tool life and cutting speed for different materials. 3 3. As Cast 2655 20% 80% 3. y With larger grain size tool life is better. Ans 1. IES ‐ 2014 Chip Equivalent In main I accelerated l d tooll life lif tests, the h three h i types off quick and less costly tool life testing are (a) Extrapolation on the basis of steady wear; conventional measurement of flank and crater wear; comparative performance against tool chipping (b) Measurement off abrasive b wear; multi l –pass turning; conventional measurement of diffusion wear (c) Extrapolating on the basis of steady wear, multi‐pass turning; taper turning (d) comparative performance against tool chipping; taper turning; measurement of abrasive wear ChipEquivalent(q) = • The SCEA alters the length of the engaged cutting Engaged cutting edge length Plan area of cut edge without affecting the area of cut. As a result, the chip equivalent changed. When the SCEA is increased, y It I is i used d for f controlling lli the h tooll temperature. the h chip h equivalent l is increased, d without h significantly f l changing h i the th cutting tti forces. f • Increase I i nose radius in di also l increases i th value the l off the th chip equivalent and improve tool life. life 235 236 237 239 240 E i f t l tti Economics of metal cutting IES‐1996 Chip is Chi equivalent i l i increased i d by b (a) An increases in side‐cutting edge angle of tool (b) An increase in nose radius and side cutting g edge angle of tool ( ) Increasing (c) I i the th plant l t area off cutt (d) Increasing the depth of cut. cut 238 l Formula Vo Ton = C Optimum tool life for minimum cost ⎛ C ⎞⎛ 1− n ⎞ To = ⎜ Tc + t ⎟ ⎜ ⎟ Cm ⎠ ⎝ n ⎠ ⎝ C ⎛ 1− n ⎞ = t ⎜ ⎟ Cm ⎝ n ⎠ if Tc , Ct & Cm given Tooling cost (Ct) = tool regrind cost + tool depreciation per service/ replacement Machining cost (Cm) = labour ) labour cost + over head cost per min if Ct & Cm g given Optimum p tool life for Maximum Productivity y (minimum production time) ⎛ 1− n ⎞ To = Tc ⎜ ⎟ (IES,GATE & PSUs) n ⎠ ⎝For-2016 IES 2009 Conventional g g Units:Tc – min (Tool changing time) Ct – Rs./ servicing or replacement (Tooling cost) Cm – Rs/min (Machining cost) V – m/min (Cutting speed) 241 Page 28 of 210 242 Determine the cutting speed an D i h optimum i i d for f operation on a Lathe machine using the following information: Tool change time: 3 min Tool regrinds time: 3 min Machine running cost Rs.0.50 per min Depreciation of tool regrinds Rs. Rs 5.0 50 The constants in the tool life equation are 60 and 0.2 Rev.0 243 GATE‐2014 ESE‐2001 Conventional If the Taylor’s tool life exponent n is 0.2, and the tooll changing h time is 1.5 min, then h the h tooll life l f (in ( min) for maximum production rate is ………………. In I a certain i machining hi i operation i with i h a cutting i speed of 50 m/min, tool life of 45 minutes was observed. When the cutting speed was increased t 100 m/min, to / i the th tool t l life lif decreased d d to t 10 min. i g speed p for maximum Estimate the cutting productivity if tool change time is 2 minutes. 244 GATE‐2005 The optimum cutting speed is one which should have: 1. High metal removal rate 2. High Hi h cutting i tooll life lif 33. Balance the metal removal rate and cutting g tool life (a) 1, 1 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 3 only l For-2016 (IES,GATE & PSUs) 250 A diagram related economics di l t d to t machining hi i i with ith various cost components is given above. Match List I (C t Element) (Cost El t) with ith List Li t II (Appropriate (A i t Curve) C ) and d select the correct answer using the code given below th Lists: the Li t List I List II (Cost Element) (Appropriate Curve) A Machining cost A. 1 1. Curve‐l B. Tool cost 2. Curve‐2 C. Tool grinding cost l d 3. Curve‐3 D. Non‐productive cost p 4. 4 Curve‐4 4 5. Curve‐5 248 Vmax.production > Vmax.profit > Vmin. cost Page 29 of 210 246 C d Contd………. F From previous i slide lid Minimum Cost Vs Production Rate IES 2011 Determine the optimum p speed p for achieving g maximum production rate in a machining p The data is as follows : operation. Machining time/job = 6 min. T l life Tool lif = 90 min. i Taylor's equation constants C = 100, n = 0.5 Job handling time = 4 min./job Tool changing time = 9 min. min [10‐Marks] 245 IAS – 2007 Contd… 247 IAS – IAS – 2011 Main 2011 Main 251 Code:A ( ) 3 (a) (c) 3 B 2 1 C 4 4 D 5 2 ((b)) (d) A 4 4 B 1 2 C 3 3 D 2 5 249 IES – 1999 Consider the following approaches normally C id h f ll i h ll applied for the economic analysis of machining: 1. Maximum production rate 2 Maximum profit criterion 2. 3. Minimum cost criterion The correct sequence in ascending order of optimum cutting speed obtained by these approaches is (a) 1, 2, 3 (b) 1, 3, 2 (c) 3, 2, 1 (d) 3, 1, 2 Rev.0 252 IES – 1998 IAS – 2002 The rate off a Th variable i bl cost and d production d i machining process against cutting speed are shown in the given figure. For efficient machining, the range g of best cutting g speed p would be between (a) 1 and 3 (b) 1 and d5 (c) 2 aand d4 (d) 3 and 5 Optimum cutting speed cost (Vc min ) O i i d for f minimum i i and optimum cutting speed for maximum production rate (Vr max ) have which one of the following g relationships? p (a) Vc min = Vr max (b) Vc min > Vr max ( ) Vc min < Vr max (c) (d) V2c min = Vr max 253 IES – 2000 In turning, the ratio of the optimum cutting speed I i h i f h i i d for minimum cost and optimum cutting speed for maximum rate of production is always (a) Equal to 1 (b) In the range of 0.6 to 1 (c) In the range of 0.1 to 0.6 (d) Greater than 1 254 255 IES – 2004 The speed Th magnitude i d off the h cutting i d for f maximum i profit rate must be (a) In between the speeds for minimum cost and maximum production rate (b) Higher than the speed for maximum production rate (c) Below the speed for minimum cost (d) Equal to the speed for minimum cost 256 IES – 2002 Consider C id the th following f ll i statements: t t t 1. As the cutting speed increases, the cost of production i i i ll reduces, initially d then h after f an optimum i cutting i speed d it i increases 2. As A the h cutting i speed d increases i the h cost off production d i also increases and after a critical value it reduces 3. Higher feed rate for the same cutting speed reduces cost of production 4. Higher feed rate for the same cutting speed increases the cost of production Which of the statements given above is/are correct? ((a)) 1 and 3 ((b)) 2 and 3 (c) 1 and 4 (d) 3 only 257 IES 2010 IAS – 2007 Assertion (A): cutting speed A i (A) The Th optimum i i d for f the h minimum cost of machining may not maximize the profit. Reason (R): The profit also depends on rate of production. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2016 (IES,GATE & PSUs) IAS – 1997 259 With increasing cutting velocity, i i tti l it the th total t t l time for machining g a component p (a) Decreases (b) Increases (c) Remains unaffected ((d)) First decreases and then increases Page 30 of 210 260 In economics of machining, which one of the I i f hi i hi h f h 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 258 Machinability‐Definition Machinability M hi bili can be b tentatively i l defined d fi d as ‘ability ‘ bili off being machined’ and more reasonably as ‘ease of machining’. Such h ease off machining h or machining h characters h of any y tool‐work p pair is to be jjudged g by: y y Tool wear or tool life y Magnitude of the cutting forces y Surface finish y Magnitude g of cutting g temperature p y Chip forms. Rev.0 261 For IES Only Free Cutting steels y Addition of lead in low carbon re‐sulphurised steels and also in aluminium, aluminium copper and their alloys help reduce their τs. The dispersed lead particles act as discontinuity and solid lubricants and thus improve machinability by reducing friction, cutting forces and temperature, tool wear and d BUE formation. f i y It contains less than 0.35% 35 lead byy weight g . y A free cutting steel contains C C‐0.07%, % Si‐0.03%, Si % Mn‐0.9%, M % P‐0.04%, P % S‐0.22%, S % Pb‐0.15% Pb % IES ‐ 2012 Machinability Index Or Machinability Rating The machinability index KM is defined by KM = V60 6 /V60R 6 R Where V60 is the cutting speed for the target material that h ensures tooll life lif off 60 6 min, i V60R is i the h same for f the h reference material. If KM > 1, the machinability of the target material is better that this of the reference material, material and vice versa 262 The machinability off a Th usuall method h d off defining d fi i hi bili material is by an index based on (a) Hardness of work material (b) Production rate of machined parts (c) Surface finish of machined surfaces (d) Tool life 263 264 For IES Only IAS – 1996 Machinability of Steel Assertion (A): The machinability of a material can A i (A) Th hi bili f i l be measured as an absolute quantity. Reason (R): Machinability index indicates the case with which a material can be machined (a) Both A and R are individually true and R is the correct explanation of A l f (b) Both A and R are individually true but R is not ot a d R a e d v dua y t ue but R s ot tthe e correct explanation of A ( ) A is true but R is false (c) A i t b t R i f l (d) A is false but R is true y Mainly and machinability M i l sulfur lf d lead l d improve i hi bili off steel. y Resulfurized steel: Sulfur is added to steel only if th there i sufficient is ffi i t manganese in i it. it Sulfur S lf f forms manganese sulfide which exists as an isolated phase and act as internal lubrication and chip breaker. y If insufficient manganese is there a low melting iron sulfide will formed around the austenite grain boundary. Such steel is very weak and brittle. y Tellurium and selenium is similar to sulfur. sulfur dispersed fine particle and act as solid lubricants. At high speed lead melts and acting as a liquid lubricants. As lead is toxin and p pollutant,, lead free steel is p produced using Bismuth and Tin. y Rephosphorized steel: Phosphorus strengthens the ferrite, causing increased hardness, result in better chip f formation and d surface f f finish. h y Calcium Calcium‐Deoxidized Deoxidized steel: Oxide flakes of calcium silicates are formed. Reduce friction, tool temp, crater wear specially at high speed. speed 266 For IES Only y Stainless Steel: S i l S l Difficult Diffi l to machine hi due d to abrasion. b i y Aluminum and Silicon in steel: Reduce machinability y due to aluminum oxide and silicates formation, which are hard and abrasive. abrasive y Carbon and manganese in steel: Reduce machinability h b l due d to more carbide. bd y Nickel, c e,C Chromium, o u , molybdenum, o ybde u , a and d va vanadium ad u in steel: Reduce machinability due to improved property. y Effect Eff t off boron b i is negligible. li ibl O Oxygen i improve machinability. Nitrogen and Hydrogen reduce machinability. For-2016 (IES,GATE & PSUs) 268 Machinability of Steel contd… y Leaded off L d d steel: l Lead L d is i insoluble i l bl and d takes k the h form f 265 Machinability of Steel contd… For IES Only 267 For IES Only IES 2011 Conventionall Role of microstructure on Machinability y Discuss the effects of the following elements on the machinability of steels: (i) Aluminium and silicon (ii) Sulphur and Selenium (iii) Lead and Tin (iv) Carbon and Manganese (v) Molybdenum and Vanadium For IES Only [5 Marks] Coarse microstructure leads to lesser value of τ C i l d l l f s. Therefore, τs can be desirably reduced by y Proper heat treatment like annealing of steels P h lik li f l y Controlled addition of materials like sulphur p ((S), lead ), (Pb), Tellerium etc leading to free cutting of soft ductile metals and alloys. metals and alloys y Brittle materials are relatively more machinable. Page 31 of 210 269 Rev.0 270 Increasing back rake angle up to certain value g g p Which of these statements are correct? ( ) 1 and 3 (a) d (b) 1 and 2 d (c) 2 and 3 (d) 1. and again too large clearance reduces the tool strength g and tool life hence machinability.GATE & PSUs) 277 (c) 2 and 3 only (b) 1 and 3 only (d) 322 only Page of 210 (d) Tool life 278 Rev. Microstructure physical and mechanical ( ) Microstructure of tool material (c) 3 3. g . hmax y The Th variation i ti in i the th cutting tti edge d angles l does d nott affect ff t cutting force or specific energy requirement for cutting. C tti f Cutting forces (a) Microstructure. Surface finish properties and composition of workpiece material. hmax Inadequate clearance angle reduces tool life and surface finish by tool – work rubbing. 271 Consider the following statements: C id h f ll i y The tool life is increased by 1. cutting y Increase in SCEA and reduction in ECEA improves surface finish sizeably in continuous chip formation hence Machinability. Tool life M hi bilit depends Machinability d d on (b) Rigidity of work ‐piece 2. 2 and 3 272 For IES Only Effects of Cutting Edge angle(s) on machinability IAS – 2000 273 For IES Only Effects of clearance angle on machinability For IES Only Effects of Nose Radius on machinability Proper tool nose radiusing improves machinability to some extent through y increase in tool life by increasing mechanical strength and d reducing d i temperature at the h tooll tip i y reduction of surface roughness. 2 and 3 For-2016 (IES.0 279 . Built ‐up edge formation 2. (d) Shape and dimensions of work Sh d di i f k Which of the above is/are the machinability ((b)) Cutting g forces criterion/criteria? ((c)) Type yp of chip p (a) 1. 2009 ISRO‐2007 Consider the following: Ease of machining is primarily judged by f2 = 8R 8R (a) Life of cutting tool between sharpening 1.For IES Only IES – 1992 Tool life is generally better when T l lif i ll b h ((a)) Grain size of the metal is large g (b) Grain size of the metal is small ( ) Hard constituents are present in the microstructure (c) H d i i h i of the tool material (d) None of the above ff k angle(s) l ( ) on Effects off tooll rake machinability y As Rake angle increases machinability increases. 274 275 IES – 1992 276 IES – 2007. Increasing cutting velocity I i i l i 33. y But too much increase in rake weakens the cutting edge. Surface finish 2. Better surface finish. Type of chips 3 Tool life 3. 4 4.GATE & PSUs) 286 Page 33 of 210 287 Rev.IES – 2003 IES – 2009 Assertion (A): A ti (A) The Th machinability hi bilit off steels t l improves i by adding sulphur to obtain so called 'Free M hi i Steels‘. Aluminium lead and copper (d) Aluminium. 4. (a) Both A and R are individually true and R is the correct co ect eexplanation p a at o o of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A is false but R is true (d) The elements which. causing IES – 2013 Conventional Why does titanium have poor machinability? ( ) Low tool‐chip contact area (c) (d) None of the above N f h b y Almost all tool materials tend to react chemically y with titanium. 3 ( ) 1. Smaller shear angle increase its machinability? ( ) Weldability (a) W ld bili (b) F Formability bili 2.0 288 . presence of small quantities machinability? hi bilit ? elements/pairs l off elements l is added dd d to steell to sulphur l h improves 1 1. 2. 1. Power consumption In modern high speed CNC machining with coated carbide tools. 4. Higher cutting forces (a) Nickel (b) Sulphur and phosphorus ( ) Machinability (c) M hi bilit (d) H d Hardenability bilit 33. 3. Aluminium titanium and copper 280 of the following Consider the following criteria in evaluating C id h f ll i i i i l i machinability: 1. 2. 1. 4 (c) (d) 2. the correct sequence of these criteria in DECREASING order of their importance is CR S G o de o t e po ta ce s (a) 1. 4 281 IES – 1996 Which IES – 1998 282 IES – 1996 indicate better IES – 1995 Small amounts of which one of the following In low carbon steels. S l h lead l d and d cobalt b lt (c) Aluminium. Longer g tool life (c) Silicon (d) Manganese and copper 4. 3. For-2016 (IES. 3 (b) 2. conductivity g high g temperature p rise and BUE. Machining St l ‘ Reason (R): Sulphur in steel forms manganese sulphide inclusion which helps to produce thin ribbon like continuous chip. chipping (b) Chemical reaction between tool and work y Titanium and its alloys have poor thermal conductivity. added to steel. help in chip f formation d during machining h are ( ) Sulphur. (a) S l h lead l d and d phosphorous h h (b) Sulphur. (a) 1 and 3 (b) 2 and 4 (c) 1 and 2 (d) 3 and 4 283 284 285 For IES Only For IES Only IES – 1992 Machinability of Titanium y Titanium is very reactive and the chips tend to weld to Machining of titanium is difficult due to the h tooll tip leading l d to premature tooll failure f l d to edge due d (a) High thermal conductivity of titanium chipping. 1 (b) 4. To keep k the h same μ f for a theoretical surface roughness of is 5 m is h l f h f level of surface finish.268 mm/rev (a) 68 / should be (b) 0.IES ‐ 2002 Surface Roughness IES ‐1995 Consider the following work materials: 1 Titanium 1. 4.26 mm (c) 0. finish the nose radius of the tool ( ) 0. If hp and hQ denote the p peak‐ to‐valley heights of surfaces produced by the tools P and Q. 1. 4. 1.0 297 . 3 y Ideal Surface ( Zero nose radius) f tan SCEA + cot ECEA h f and (Ra) = = 4 4 ( tan SCEA + cot ECEA ) Peak to valley roughness (h) = The roughness 'h' obtained Th value l off surface f h b i d during d i the turning operating at a feed 'f' with a round nose tool having radius 'r' is given as y Practical Surface ( with nose radius = R) h= f2 8R Ra = and f2 18 3R Change g in feed ((f)) is more important p than a change g in nose radius (R) and depth of cut has no effect on surface roughness. 289 290 IAS ‐ 1996 IES ‐ 1999 Given that Gi h / and S = feed in mm/rev.8 mm. 2 2. Grey cast iron. Stainless steel 4. 3. R = nose radius in mm.32 mm (d) 0. the feed could be doubled to A cutting tool has a radius of 1.16 mm 291 296 For achieving a specific surface finish in single point turning the h most important factor f to be b controlled ll d is (a) Depth of cut (b) Cutting speed (c) Feed (d) Tool rake angle Rev. 2. the h maximum i h i h off surface height f roughness h Hmax produced by a single‐point turning tool is given by (a) S2/2R (b) S2/4R / R (c) S2/4R (d) S2/8R GATE ‐ 1997 In turning operation.48&mm For-2016 (IES. Mild steel 3. ISRO‐2008 Two tools signatures 5°‐5°‐6°‐6°‐8°‐30°‐ T l P and d Q have h i ° ° 6° 6° 8° ° 0 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively. 1 (d) 2.GATE PSUs) 294 GATE ‐ 2005 30o turning with a feed of 1 mm/rev. The feed rate increase the h metall removall rate. tool the maximum (peak to valley) height of surface roughness produced will be 295 IES – 1993. mm/rev Neglecting nose (a) 0. 2. 3.187 mm/rev 8 / ((a)) Halved ((b)) Kept p unchanged g (c) 0. Q the ratio hp/hQ will be o o o o tan 8 + cot15 tan15 + cot 8 (b) o o tan 8 + cot 30 tan 30o + cot 8o tan15o + cot7o tan7o + cot15o (c ) ( d ) o tan 30o + cot7 tan7o + cot 30o Page 34 of 210 (a) (b) 0.0187 mm/rev 0 0187 mm/rev 292 293 GATE – 2007 (PI) ( ) A tool Edge t l with ith Side Sid Cutting C tti Ed angle l off End Cutting Edge angle of 10o and d is used for fine radius of the tool. 3 (c) 2.036 mm/rev 0 036 mm/rev ((c)) doubled ((d)) Made four times (d) 0. They are used to turn components under the same machining g conditions. The correct sequence of these materials in terms of increasing order of difficulty in machining is (a) 4. 46 3 4 5 C A B 8 9 10 6. Soluble oil for high speed machining and grinding y Brass: Machined dry or straight mineral oil with or without ih EPA EPA. Q. Q. Q. chlorine. In finishing pass. ddi i 304 Dry fluid D and d compressed d air i is i used d as cutting i fl id for f machining (a) Steel (b) Aluminium (c) Cast iron (d) Brass 302 303 WORKBOOK WORKBOOK Ch‐2: Cutting Tools. Tool Life and Cutting Fluid Ch‐3: Economics of Machining Operation Q. i EP molecules l l b become reactive i and d release l derivatives such as iron chloride or iron sulfide and f forms a solid lid protective i coating.67 10 D 20 A 30 C Page 35 of 210 300 IES ‐ 2001 [ 3 – marks] Wh Where hi h pressures and high d rubbing bbi action i are encountered. Q No N 1 2 Option O ti C B Q. sulfur. roughness g on the work surface can be reduced byy (a) reducing the spindle speed (b) increasing the h spindle dl speed d (c) reducing educ g tthee feed eed o of too tool (d) increasing the feed of tool In the selection of optimal cutting conditions. i IES ‐ 2012 The most important function of the cutting fluid is to Th i f i f h i fl id i ( ) (a) Provide lubrication (b) Cool the tool and work piece ( ) W h (c) Wash away the chips h hi ( ) p (d) Improve surface finish (d) Increase depth of cut and increase feed IAS 2009 Main IAS ‐2009 Main pressure lubricants? y What are extreme What are extreme‐pressure lubricants? 301 For-2016 (IES.0 195 02 0.25 9 A 19 B 29 B 39 36.2 T1: 26 m/min Rev. one should 305 Q. y Cast Iron: Machined dry or compressed air. i i d so Extreme E P Pressure (EP) additives ddi i must be b added to the lubricant.0 306 .IES ‐ 2006 GATE‐2014 (PI) ( ) GATE ‐2010 (PI) ( ) A spindle i dl speed d off 300 rpm and d a feed f d off 0. EP lubrication is provided by a number b off chemical h i l components such h as boron. As the temperature rises. Q No N Option O ti Q No N Option O ti Q No N Option O ti Q No N Option O ti B 1 B 11 C 21 2 A 12 C 22 B 32 B 3 A 13 A 23 A 33 A 4 D 14 A 24 C 34 B 31 C A 5 D 15 B 25 C 35 6 D 16 B 26 A 36 C 7 B 17 B 27 C 37 B 8 A 18 A 28 B 38 0. l bi t reducing d i the th friction f i ti ff t att the tool‐chip interface and the work‐blank regions. To achieve improved surface finish. or combination of these. y Aluminium: Machined dry y or kerosene oil mixed with mineral oil or soluble oil y Stainless steel and Heat resistant alloy: High performance soluble oil or neat oil with high concentration i with i h chlorinated hl i d EP additive. one need to improve surface finish without sacrificing material removal rate. Q No N 6 7 Option O ti A 56. the requirement off surface f f finish h would ld put a limit l on which of the following? (a) The maximum feed (b) The maximum depth of cut (c) The maximum speed (a) decrease nose radius of the cutting tool and increase depth of cut (b) Increase nose radius of the cutting tool (c) Increase feed and decrease nose radius of the cutting tool ((d)) The maximum number of p passes 298 299 Cutting fluid Cutting fluid y The cutting fluid acts primarily as a coolant and secondly effects dl as a lubricant.GATE & PSUs) During turning of a low carbon steel bar with TiN coated carbide insert.3 mm/revolution are chosen for longitudinal turning operation on an engine lathe. hydrodynamic lubrication cannot be maintained. b phosphorus. Th compounds The d are activated i d by b the h higher hi h temperature resulting from extreme pressure. o e a ce & ts Limit. +0 18 e. tolerances)) are applied.10 e.00 25 00 mm Upper Limit = 25..GATE & PSUs) 3 7 e.10 Bilateral Limits occur when the maximum limit is above and the minimum limit is below the basic size.04 mm L Lower Li it = 24.00 25 00 mm Upper Limit = 25.. e.08 mm 8 For PSU For PSU Tolerances are specified (a) To obtain desired fits (b) because it is not possible to manufacture a size exactlyy (c) to obtain higher accuracy (d) to have proper allowances h ll Rev.10 mm For-2016 (IES. ¾The tolerance may be unilateral or bilateral. etc. -0. whereas the smallest size is known as lower or minimum limit. force. The largest permissible size for a dimension is called upper or high or maximum limit.96 Limit 6 mm Page 36Tolerance of 210 = 0...20 Basic Size = 25. sketch k h the h basic b i size. y Actual size: Actual measured dimension of the part.10 mm Tolerance = 0.18 For Unilateral Limits. IAS 2014 (Main) IAS 2014 (Main) a ssize: e: S ed in tthee d aw g.08 mm 0 ‐0. otherwise it will interfere with the interchangeability of the mating parts. variation ((i. Ø25 +0.e.g.0 9 . and tolerance on a shaft and hole assembly. temperature.90 mm Lower Limit = 24. i deviation. 5 6 Terminology Contd Terminology Contd..00 mm Upper Limit = 24. t t. 10 e g Ø25 -00. Ø25 +0.10 0. such as dimensions. 4 Terminology Contd.20 . A According di to the h ISO system. Ø25 0 +0.80 mm Tolerance = 0. y No Nominal Sizee o of a pa partt spec specified drawing.Terminology Metrology y The Th science i off measurement. pp Basic dimension is theoretical dimension.g.. Tolerance & Fits y The purpose of this discipline is to establish means of determining physical quantities.g.10 Basic Size = 25.. y Tolerance ¾The difference between the upper limit and lower limit. e o ogy y Limits of sizes: There are two extreme permissible sizes for a dimension of the part.. The difference between the basic size and the actual size should not exceed a certain limit. y Basic size: Size of a part to which all limits of It is used for general identification purpose.. By S K Mondal 2 1 Terminology Terminology C td Terminology Contd. Unilateral Limits occurs when both maximum limit and minimum limit are either above or below the basic size. Ø25 ±0. ¾It is the maximum permissible variation in a dimension.g. Limits a case may occur when one of the limits coincides with the basic size.04 Basic Size = 25.18 mm Lower Limit = 25. bl shafts h f off size i 25.0001 mm respectively.009.01 mm respectively..040 −0. ± 0.070 (b) 0.025...025 mm. d C have mean diameters of 19∙99 mm. which is nearest one y Upper U d i i deviation: I the Is h algebraic l b i difference diff b between (d) All of the above y Mean deviation: Is the arithmetical mean of upper pp and lower deviations.016 (d) − 0. the maximum size and the basic size..000 +0. Terminology C td Terminology Contd. g line corresponding p g to the basic y Zero line: A straight ISRO‐2010 E Expressing i a dimension di i as 25. y Deviation: Is the algebraic difference between a size (actual. (a) Unilateral tolerance an actuall size i and d the h corresponding di basic b i size. 20∙00 mm and 20.016 IES ‐ 2005 GATE ‐ 2004 Two shafts A and B have their diameters specified as T h f A d B h h i di ifi d 100 ± 0. possible clearance in the assembly y The maximum p will be ( ) 10 microns (a) i (b) 20 microns i (c) 30 microns (d) 60 microns 14 GATE ‐ 2000 The specified for the Th tolerance t l ifi d by b the th designer d i f th diameter of a shaft is 20.02 mm. The shafts produced d d by b three th diff different t machines hi A B and A.070 (c) ± 0.008 y Fundamental deviation: This is the deviation. ± 0. etc.0 18 . φ35 −0.008 (a) (c) − 0..000 +0.1 ± 0.020 −0. Band C since the standard deviation is same for the three machines (b) Least for machine A (c) Least for machine B (d) Least L t for f machine hi C For-2016 (IES. i ((c)) Limiting g dimensions to zero line for either a hole or shaft.045 Page 37 of 210 15 17 Fits: (assembly condition between “Hole” & “Shaft”) Hole – A feature engulfing a component Shaft – A feature being engulfed by a component p Rev. size 10 GATE – 2010.020 (d) 0.025.) and the corresponding basic size. A milling cutter of 20 mm width is located with reference to the side of the block within ± 0. What will be the percentage rejection for th shafts the h ft produced d d by b machines hi A B and A.0100 mm mate with holes of size 25. Which of the following statements is/are true? (a) Tolerance in the dimension is greater in shaft A (b) The relative error in the dimension is greater in shaft A (c) Tolerance in the dimension is greater in shaft B (d) The relative error in the dimension is same for shaft A and shaft B d h f 13 16 In I an interchangeable i h bl assembly.1 mm and 0. max. The deviations are measured from this line. either the upper or the lower deviation. d C? (a) Same for the machines A. ti l with ith same standard t d d deviation.GATE & PSUs) 12 GATE ‐ 1992 −0.0.05 mm.009. The maximum offset in mm between the centre lines of the slot and the block is (a) ± 0.Terminology C td Terminology Contd. ISRO‐2012 11 The respective values of fundamental deviation and tolerance are (b) ( ) − 0.009 mm A shaft has a dimension. y Actual deviation: Is the algebraic difference between ((b)) Bilateral tolerance y Lower deviation: Is the algebraic difference between Fit A slot l is i to be b milled ill d centrally ll on a block bl k with ih a dimension of 40 ± 0.00 ± 0.025 0 025 ( ) − 0...000 mm.05 mm is i the th case of size. the minimum size and the basic size.3 ±0.0. The limits of both hole and shaft using hole basis system will be a) low limit of hole = 25 mm.LL of shaft Min.99 mm and low limit of shaft = 25 mm Holes of diameter 25 (c) Joint between a pulley and shaft transmitting power (d) Assembly of flywheels on shafts (d) Joint of lathe spindle and its bearing Page 38 of 210 26 Rev.026 mm.0 27 .02 mm and a minimum clearance of 0. 22 Max I Shaft Tolerance zones never meet but crosses each other Min I +0. upper limit of shaft = 24. C = LL of hole . 0. upper limit of shaft = 24. 25 For-2016 (IES. press fit.11 The clearance fits may be slide fit.01 mm The tolerance on the shaft is 0. upper limit of shaft = 24. mm The Shaft maximum clearance in mm between the hole and the shaft is Max.UL of shaft Min I = UL of hole . The hole tolerance is to be 1.008 mm 0 008 mm mm.0 5 0 0 .LL of shaft Min. 2. high limit of hole = 25.C. heavy drive fit and The interference fits may be shrink fit heavy drive fit and light drive fit.10 (d) 0.040 20 IES‐2015 A hole and a shaft have a basic size of 25 mm and are to have a clearance fit with a maximum clearance of 0. 23 ((a)) 1 onlyy ((b)) Both 1 and 2 (c) 2 only (d) Neither 1 nor 2 24 IES‐2013 IES 2011 Which of the following is a joint formed by Interference fit joints are provided for: i t f interference fit ? fits? (a) Assembling bush bearing in housing (a) Joint of cycle axle and its bearing (b) Joint between I. The mating shaft h f has h a clearance l f with fit h minimum clearance l off 0 01 mm. I C Engine piston and cylinder (c) Mounting pulley on shafts The interference fits may be shrink fit.005 a) 0.GATE & PSUs) mm are assembled −0. In transition fit. pulley on shaft). C = UL of hole .05 (c) 0. +0.006 mm. The difference between hole size and shaft size is called Max I allowance.006 mm. I = LL of hole . high Ch limit of hole = 25 mm. I = LL of hole .048 mm b) 0. small positive or negative clearance b between the h shaft h f and d hole h l member b is i employable l bl Max. high limit of hole = 25.99 mm d) low limit of hole = 25. slack running fit and loose running fit. upper limit of shaft = 25 mm and low limit of shaft = 24.76 mm c) low limit of hole = 24 mm.04 0 04 mm. high limit of hole = 25.01 mm. 21 IES 2015 IES‐2015 Hole Consider the following statements Max C Tolerance zones always overlap Shaft In case of assemblyy of mating gp parts 1.020 interchangeably with the pins of diameter 25 Interference Fits Hole +0.99 mm and low limit of shaft = 24‐ 986 mm b) low limit of hole = 25 mm. wringing fit (Gear.04 (b) 0.LL of shaft Max.UL of shaft Which of the above statements is/are correct? The transition fits may be tight fit and push fit.8 mm and low limit of shaft = 24. C = UL of hole .006 mm. running Th l fit b lid fit lidi fit i 19 fit.015 mm c) 0 0.005 mm 005 mm d) 0.Clearance Fits A hole is specified as 4 0 Hole Max C Min C GATE 2015 GATE-2015 GATE ‐ 2007 T l Tolerance zones never meet 0 . easy sliding fit.008 The minimum clearance in the assembly will be Transition Fits (b) Mounting heavy duty gears on shafts Max.5 times the shaft tolerance.UL of shaft (a) 0.0 0 0 mm. 03 to 0. An interference assembly. o o a d a ete 20 0 mm.001 mm. (a) Both Statement (I) and Statement (II) are individuallyy true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement St t t (I) is i false f l but b t Statement St t t (II) is i true t 28 GATE 2011 29 +0 mm The mm.02 25+−0. the i with ith limits.01 mm. shaft Permitted interference values are 0. the outer diameter of the inner cylinder will be more than the inner diameter of the hollow outer cylinder St t Statement‐II: t II These Th fit are recommended fits d d for f two t parts frequently dismantled and assembled. li it for f the th two t mating ti parts.09 mm. Statement (II) : The amount of clearance obtained from the assemblyy of hole and shaft resulting g in interference fit is called positive clearance.015 A hole is of dimension φ 9 IES‐2015 Statement (I) : In fit. it is essential that the h lower l l limit off the h shaft h f should h ld be b ( ) Greater (a) G than h the h upper limit li i off the h hole h l (b) Lesser L th the than th upper limit li it off the th hole h l (c) Greater than the lower limit of the hole (d) Lesser than the lower limit of the hole Statement‐I: In interference fit.IES ‐ 2014 IES GATE ‐ 2005 In order to have interference fit.03 0.. The maximum interference (in microns) in the assembly is (a) 4 40 ((b)) 330 (c) 20 (d) 10 31 30 IAS 2011 Main IAS‐2011 Main GATE ‐2012 Same Q in GATE‐2012 (PI) +0. corresponding shaft is of dimension The resulting g assemblyy has (a) loose running fit (b) close l running i fit fi (c) transition a s o fit (d) interference fit φ9 In an interchangeable assembly.0 36 . assembly shafts of size +0. [10‐Marks] Hint: Use unilateral hole basis system. mm mate with holes of size 0 03 25++0.010 +0. The manufacturing tolerances for the holes are twice that for the shaft. (a) Push fit This assembly constitutes y (b) Running fit (a) Interference fit ( ) Sliding fit (c) (b) Transition fit (d) Shrink Sh i k fit fi (c) Clearance fit (d) None of the above For-2016 (IES. off S I interference i f fi the h outer diameter di the shaft is greater than the inner diameter of the hole. D t Determine i th sizes. of nominal diameter is of a unilateral holes and a shafts.GATE & PSUs) 34 Page 39 of 210 35 Rev. (a) Both statement (I) and (II) are individually true and statement (II) is the correct explanation of statement (I) (b) Both B th statement t t t (I) and d statement(II) t t t(II) are individually i di id ll true but statement(II) is not the correct explanation of statement (I) () (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but statement (II) is true te e e ce asse b y.04 0. 32 IES ‐ 2007 33 IES ‐ 2006 ISRO‐2011 Which of the following is an interference fit? A shaft and hole pair is designated as 50H7d8 A shaft and hole pair is designated as 50H7d8. The stress state in the hub is similar to a thick‐ walled a ed cy cylinder de with t internal te a p pressure. Wh t th diff What are the different types of fits possible with t t f fit ibl ith reference to mechanical systems? [4 Marks] 40 41 GATE ‐ 2001 GATE ‐ 1998 42 IES ‐ 2012 Allowance in limits and fits refers to In the specification of dimensions and fits. 2 and d3 y It is Minimum clearance or maximum interference.C. The amount of interference needed to create a tight joint varies with diameter of the shaft. The allowance mayy be positive or negative.0 45 . shaft.GATE & PSUs) 43 Page 40 of 210 44 Rev. I. Ball B ll bearing b i inner i race and d shaft h f Which of the above fits are based on the interference system? ( ) 1 and (a) d2 (b) 2 and 3 (c) 1 and 3 (d) 1. Clearance in a fit is the difference between (a) Maximum clearance between shaft and hole (a) Allowance is equal to bilateral tolerance (a) Maximum hole size and minimum shaft size (b) Minimum clearance between shaft and hole (b) Allowance is equal to unilateral tolerance (b) Minimum hole size and maximum shaft size ( ) Difference between maximum and minimum size of (c) ( ) Allowance is independent of tolerance (c) ( ) Maximum hole size and maximum shaft size (c) hole (d) Allowance All i equall to the is h difference diff b between (d) Minimum Mi i h l size hole i and d minimum i i shaft h f size i (d) Difference between maximum and minimum size of maximum and minimum dimension specified by the shaft tolerance.IES ‐ 2009 IES 2015 IES‐2015 IES ‐ 2008 Consider C id the h following f ll i joints: j i g wheel and axle 1. 2 and (a) d3 (b) 1 and d 2 only l (c) 2 and 3 only (d) 1 and 3 only 37 In fit a shaft I an interference i f fi between b h f and d a hub. Ball bearing outer race and housing 3. Which of the statements given above are correct? ( ) 1. h b the h state of stress in the shaft due to interference fit is a)) onlyy compressive p radial stress b)a tensile radial stress and a compressive tangential stress c) a tensile tangential stress and a compressive radial stress d)a compressive tangential stress and a compressive radial stress 38 39 Allowance IES ‐ 2004 Consider C id the h following f ll i fits: fi 1. It is IES‐2015 Conventional IES‐2015 Conventional the intentional difference between the basic dimensions of the mating gp parts. IC engine cylinder and liner Whi h off the Which h above b j i joints i / is/are the h result(s) l ( ) off interference fit? (a) 1 only (b) 2 only l (c) Neither 1 nor 2 (d) Both 1 and 2 Consider C id the h following f ll i statements: g 1. For-2016 (IES. 3. essu e. Railwayy carriage 2. 2 An interference fit creates no stress state in the 2. engine g cylinder y and p piston 2. +ve ((c)) Zero.GATE & PSUs) 52 Page 41 of 210 53 Assertion (A): A i (A) Hole H l basis b i system is i generally ll preferred to shaft basis system in tolerance design for getting the required fits. y If shaft basis system is used considerable no of reamers and other precision tools are required for producing different classes of holes for one class of shaft for obtaining different fits which increases IES ‐ 2005 ISRO‐2008 y Holes can be finished by y tools like reamers. Zero ((d)) None of the above cost of production.00 0.e. y The basic size of the hole is taken as the lower limit of y Basic size of the shaft is taken Upper limit for the shaft ( size of the hole ( Maximum metal condition). y .Hole Basis System IES – 2012 Conventionall ISRO‐2010 Explain the difference between tolerance and Dimension of the hole is 50 allowance.02 0 02 mm Hole basis system y The hole is kept as a constant member (i. h stands for dimensions of shafts whose lower deviation is zero. (i e when the upper deviation of the shaft is zero) y Different fits are obtained by varying the hole size then the limit system is said to be on a shaft basis.. drills.00 +0. and their sizes are not adjustable.. Shaft basis system: y The actual size of hole is always more than basic size or equal to basic size si e but never ne er less than Basic size.00 mm (d) 0. ve. Reason (R): Hole has to be given a larger tolerance band than the mating shaft.. y Lower limit of the shaft and two limits of hole are selected to give the desired fit. H stands for dimensions of holes y For shaft basis system. The shaft sizes can be easily obtained by external machining.0 54 . +0. y It is economical For-2016 (IES.02 mm −0. and shaft is 50 +0. si e 51 For IES Only Why Hole Basis Systems are Preferred? Why Hole Basis Systems are Preferred? broaches. ‐ve ve (b) ‐ve.e.02 0 02 mm. whose upper deviation is zero. when the (b) 0.01 0 01 mm 46 lower deviation of the hole is zero)) y Different fits are obtained by varying the shaft size then the limit system is said to be on a hole basis. 50 y Actual size of shaft is always less than basic size or equal to basic size si e but never ne er more than basic size. si e 49 y When the shaft is kept as a constant member (i. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Rev. B i shaft Basic h ft and d basic b i hole h l are those th whose h upper deviations and lower deviations respectively are (a) +ve.02 mm (c) -0. y . Maximum metal condition) y The higher limit of size of the hole and two limits of size for the shaft are then selected to give desired fits. basis 48 47 Shaft Basis system y For hole basis system.00 The minimum clearance is (a) 0. 8 + 0. J.6)(ITn -IT6) 61 Which Whi h one off the h following f ll i is i not correct in i hole h l basis b i system of fits? (a) The hole size is kept constant. f.6)(ITn -IT6) Page 42 of 210 = 1000i 57 62 work Rev. zb. fit = 640i 10(1. g. U V. fg. h. D. j. P. V X. T is the tolerance (in µm) Standard Tolerance unit or Fundamental tolerance unit i = 0. ZB ZC.3 + 0. y IT01 to IT4 ‐ For production of gauges. (d) The high limit of the size of the hole and the two limits off size i off the th shaft h ft are selected l t d to t give i desired d i d fit. za. provided T = K ×i Where. e. js. R S. ZC Shaft : a. the concepts of What is the difference between hole basis system and h l basis hole b and d shaft h f basis b in terms off assembly bl fit f shaft basis system y ? Whyy is hole basis system y the more specifications Which of the two is preferred and specifications. (IS‐919) But ISO 286 specify 20 grades upto IT18 y There are 25 (IS 919) and 28 (ISO 286) types of fundamental deviations.02D =a Upto and including (mm) ( ) ‐ ‐ 3 3 ‐ 6 6 ‐ 10 6 10 ‐ 18 18 8 ‐ 30 30 ‐ 50 550 ‐ 80 80 ‐ 120 120 ‐ 180 180 ‐ 250 250 ‐ 315 315 ‐ 400 For-2016 (IES. r. cd.6)(ITn -IT6) = 100i IT15 10(1.IFS ‐ 2013 IAS‐2010 main IAS‐2010 main IES ‐ 2005 Explain. x. (c) The actual size of a hole that is within the tolerance limits ts aalways ays less ess tthan a tthee bas basicc ssize. FG. extensive in use ? why? [ [8 –Marks] ] [8‐Marks] 55 µ µm y Limits and fits comprises 18 grades of fundamental • It is defined graphically by the magnitude of the Tolerance Zone tolerance and by its position in relation to the zero line. X Y. 55 20 Basic Size 58 Diameter Steps Diameter Steps Above (mm) ( ) tolerances for both shaft and hole.008D 0. u. with the help of sketches. E. casting general cutting 10(1.0 63 . z. in mm)) [For IT6 to IT16] K = is a constant Grades of Tolerance y It is an indication of the level of accuracy.6)(ITn -IT6) = 400i IT16 y IT12 to IT14 – For Sheet metal working or press working y IT15 to IT16 – For processes like casting. CD. (b) The basic size of the hole is taken as the low limit of size of the hole. y.6)(ITn -IT6) = 10i IT10 IT6) 10(1.6)(ITn -IT6) = 160i IT9 10(1 6)((ITn -IT6)) 10(1.. N.6) = 40i IT13 10(1. B. s. v. F.6)(ITn -IT6) = 250i IT2 ar r = 101/5 IT6 6 10(1. EF. measuring i instruments i y IT5 IT to t IT 7 ‐ For F fits fit in i precision i i engineering i i applications li ti y IT8 to IT11 – For General Engineering 10(1. These are called standard tolerances.6) ( )(ITn -IT6) = 16i IT11 10(1. G.012D 0. IT0 and IT1 to IT16. designated as IT01. e. M. c.45 3 D + 0. R.GATE 400 ‐ 500 & PSUs) 56 Limits and Fits Limits and Fits Tolerance Zone 10(1. b. deviations Hole: A. plug gauges. y A unilateral hole basis system y is recommended but if necessary a unilateral or bilateral shaft basis system may 59 also be used IT01 Value of the Tolerance IT0 IT1 IT3 3 ar2 IT4 ar3 IT5 5 ar4 = 7i IT7 IT8 0. n. zc. JS. Z ZA. K. ef. Y Z. T U. d.6) 10(1 6)(ITn -IT6) = 64i IT14 Tolerance Designation (IS) Tolerance on a shaft or a hole can be calculated by using table provided. S T. m. p.5 + 0. k. ZA ZB. C. t.001D in μ m D = D1D2 (D1 and D2 are the nominal sizes marking the beginning and the end of a range of sizes.6) 0( 6)(IT n -IT6) = 25i IT12 10(1. H. H I. It I is i a shaft h f base b system.3D ) for D ≤ 120 mm − 3.oversized shafts Letters a to g . IT7 III. y For example. Which of the statements given above is/are incorrect? ( ) 1.6 0 63 D + ( IT 7 − IT 6) n p 0 34 + 5D 0.41 g 2 5D 0. y A fit is designated by its basic size followed by symbols representing the limits of each of its two components.5D IT 8 + 3. 5 GATE 2014 GATE‐2014 Q III III 0 85D ) for D ≤ 160 mm − (140 + 0. h stands for a dimension whose upper deviation refers to the basic size. Hole Tolerance Grade S S. 2 and (a) d3 (b) 1 and 2 only (c) 1 and 3 only ( ) 3 only (d) Page 43 of 210 69 Find tolerances and for Fi d the h limit li i sizes. IT8 II.34 + IT 7 + (0 − 5) s Geometric mean of values of ei for p and s IT 8 + 1 to 4 for D ≤ 50 mm t u IT 7 + 0.41 f − 5. Shaft Tolerance Grade G d (a) (c) 65 deviation d i ti refers f t the to th basic b i size.oversized holes Letters P to ZC . 33. mm The fundamental deviation for shaft designation ‘f’ is ‐5.8 D for D > 160 mm − 52D 0. indicate the limits and tolerance on a diagram. i The Th hole h l H for f which hi h the lower deviation is zero is called a basic hole.63 0 63D IT 7 + D v x y IT 7 + 1. Shaft Type Q. Given: 100 mm diameter lies in the diameter step range of 80‐120 80 120 mm.4 D for D > 50 mm IT 7 + 0. for shafts.5D 0. example 100 H6/g5 means basic size is 100 mm and the tolerance grade for the hole is 6 and for the shaft is 5.44 e − 11D 0. 8 indicates fundamental deviation. y Similarly. H stands for a dimension whose lower g y 5 g p For the given assembly: 25 H7/g8. the hole being quoted first.5D for D > 120 mm c EI = ES – IT ES = EI + IT Fundamental Deviation a R IV IV S II I (b) (d) P I II For-2016 (IES.2 for D ≤ 40 mm − (95 + 0.6 D IT 7 + 2D z za zb IT 7 + 2.r. match Group A with Group B P I II b Fundamental Deviation − k4 to t k7 m r es = upper deviation of shaft pp ei = lower deviation of shaft ES = upper deviation of hole EI= lower deviation of hole − (265 + 1. 2.Fundamental Deviation is chosen to locate the tolerance zone w. i l d allowances ll f a 100 mm diameter shaft and hole pair designated by F8h10.t.5 h 0 66 y For hole.5 D0. The shaft h for which the upper deviation is zero is called a basic shaft.25D IT 7 + 1. Also. Also specify the type of fit that the above pair belongs g to. g IV Sh ft T l IV. [ M k ] [15‐Marks] 71 Rev.GATE & PSUs) Q IV IV R III III S II I 70 Basic size F d Fundamental Deviation t l D i ti Hole Tolerance Zone Shaft Tolerance Zone IT# 68 IES ‐ 2008 IES‐2006 Conventional Consider C id the h following f ll i statements: A nomenclature φ 550 H8/p8 /p denotes that 1.15D IT 9 + 4 D zc IT 10 + 5D 67 Group A p Group B p P.41 − 2.undersized holes ei = es – IT es = ei + IT y For Hole F H l Shafts are designated by small letter: Letters m to zc .undersized shafts H is used for holes and h is used for shafts whose fundamental deviation is zero 64 Shaft j5 to j8 + 0. Hole Type R.8 0 8D ) f D > 40 mm for d − 16 D 0.85 −1. Hole diameter is 50 mm.41 The values of standard tolerances for grades of IT 8 and IT 10 are 25i and 64i respectively.0 72 . the zero line Shaft Calculation for Upper and Lower Deviation y For Shaft Holes are designated by capital letter: Letters A to G . The tolerance unit is. [ M k ] [10 Marks] 73 74 GATE ‐ 2000 GATE – 2008 (PI) GATE ‐ 2003 Following data are given for calculating limits of A fit is specified as 25H8/e8. 24. where D is the representative size in mm. ti l When Wh the th bore is designated as 25H8.000. respectively. 24.0 81 .046 mm 75 78 GATE – 1996.The type of fit is (a) Clearance fit (b) Running fit (sliding fit) (c) Push fit (transition fit) (d) Force fit (interference fit) A small bore is designated as 25H7.GATE & PSUs) 79 Page 44 of 210 80 Rev.45 D1/3 + 0. in μ m= 0.000 mm. then the upper (maximum) limit is 25. 0 001D The unit of D is mm. Upper Limit = 60.000 (d) 25. mm Upper Limit = 60.954 mm. 5 . 25.41 (a) Lower limit = 59.984. 24.000. The maximum clearance of the fit in hole is zero and IT8 = 25 i.45 0 45 ³√D √D + 0.013 For-2016 (IES.033. i i.033 mm. D I the h tolerance l ifi i 6 the h letter l represents (a) Grade of tolerance (b) Upper deviation (c) Lower deviation (d) Type of fit The tolerance grade for number 8 quality is 25i and for 9 quality is 40i.979 mm ( ) 25. When the bore is designated as 25H6.000 mm and d 25.007 mm (c) ((d)) 25. F d Fundamental l tolerance l unit. H8d The diameter steps are 18 mm and 30 mm.021 5 mm (b) 25.967 (b) 25.009 (d) 25.44. the maximum and minimum microns is limits of dimension for a 25 mm H8 hole (in mm) are (a) ‐7 (b) 7 (a) 24.000.001 (b) 25.000.45 3 D + 0.993 4 993 mm 77 Recommended Selection of Fits GATE‐2010 (PI) What limits Wh t are the th upper and d lower l li it off the th shaft h ft represented by 60 f8? U the Use h following f ll i data: d Diameter 60 lies in the diameter step of 50‐80 mm. 5 .967 76 The Th dimensional di i l limits li i on a shaft h f off 25h7 h are ((a)) 25.016 60 0 6 mm (d) Lower limit = 60.IES ‐ 2002 IES‐2015 Conventional Determine the fundamental deviation and tolerances and the li it f i f h l d h ft i i th fit limits of size for hole and shaft pair in the fit: 25 mm H8d9.001D.000.017. The lower (minimum) and upper (maximum) limits of the bore are 25.970 9 9 0 mm. The fundamental deviation for d shaft is given as ‐16D0.5D0. If the fundamental deviation for H microns.001D GATE ‐ 2009 In specification 25 D 6. i = −0. The tolerance value for di dimensions i and d tolerances t l f a hole: for h l Tolerance T l unit it i (in (i a nominall diameter d off 25 mm in IT8 is 33 microns µm) = 0.970 59 970 mm (b) Lower limit = 59.000 mm (c) Lower Lo er limit = 59.924 59 924 mm. 5 Tolerance value for lT8 = 25i. then the upper (maximum) limit of the bore (in mm)) is ( ) 25. 25. 25.021 mm. 24.001D. Upper Limit = 60. mm Upper Limit = 59. IES‐2012 The Th fit fi on a hole‐shaft h l h f system is i specified ifi d as H7‐ H s6.984 (c) 73 (d) 106 (c) 25. Fundamental deviation for 'f shaft = ‐5. mm Diameter and fundamental deviation for the shaft is ‐ 40 step p is 18‐30 3 mm.005 (c) (a) ( ) 25. 24. y All the parts (hole & shaft) produced are measured and graded into a range of dimensions within the tolerance groups.05 ) 5 90 . y Reduces d the h cost off production d y No.of group = For IES Only y This facilitates to select at random from a large number of parts for an assembly and results in a considerable saving in the cost of production. Process capability Tolerance desired 82 ISRO‐2008 quite closely related to standardization and is realized through standardization.05 ) 5 Rev. is y Interchangeability.38 ) 3 Page 45 of 210 89 ((b) ‐ ) 0.35 + X. reduce assembly time.00 ± 0.e. the tolerance l X is y A design engineer keeps one section of the part blank (without tolerance) so that production engineer can d dump all ll the h tolerances l on that h section i which hi h becomes b most inaccurate dimension of the part. (where all dimensions are in mm).0 ((d) ‐0. part y Position of sink can be changing the reference point. B2 and d B3 are to be inserted in a channel of width S maintaining a minimum gap of width T = 0 125 mm. point y Tolerance for the sink is the cumulative sum of all the tolerances and only like minded tolerances can be added i. 0.08. R = 28. easy 83 84 IES 2010 Conventional IES 2010 Conventional IAS‐2010 main IAS‐2010 main What is meant by interchangeable manufacture ? I t h Interchangeability bilit can be b achieved hi d by b What are the differences between interchangeability and selective assemblyy ? (a) Standardization [4‐Marks] (b) Better process planning (c) Simplification (d) Better product planning 85 86 GATE ‐ 2003 Tolerance Sink GATE ‐ 1997 Three blocks Th bl k B1 .38 3 ((c) + 0. all parts of equivalent size will be equally fit for operating in machines and mechanisms and the mating parts will give the required fitting. either equally bilateral or equally unilateral.125 ± 0. replacement and repair becomes very easy. Figure For P = 18.125 mm as shown in Figure. y If the variation of items are within certain limits.12.GATE & PSUs) 87 88 ((a) + 0. Q = 25.1 and S = 72.For IES Only IES ‐ 2000 Interchangeability Selective Assembly Which tolerances set on inner Whi h one off the h following f ll i l i diameter and outer diameter respectively of headed jig bush for press fit is correct? (a) G7 h 6 (b) F7 n6 (c) H 7h 6 (d) F7j6 g y a maintainabilityy design g factor. 75 ± 0. For-2016 (IES. diameter of the hole before plating l i should h ld be b the dimension L4 in mm.002 mm Rev.012 ±0 016 ±0.000 0.00±0.010 mm diameter are electroplated in a shop.084 25 084 94 Allocation of manufacturing tolerances ll i f f i l y Unilateral system: gauge tolerance zone lies t l li entirely within the work tolerance zone.07 0. Neglecting gage tolerances.004 98 Taking example as above: ∴ Dimension of 'GO' Plug gauge = 24.016 c) 2.030 mm (d) 30++0. Fig.080 0. Ring and snap gauges 95 96 • Bilateral system: in this Example system.070 0. holes The GO plug gauge is Cylindrical pins of 25++0.02 25 02 mm Lower limit of hole = 24.01 L2 = L3 = 10.GATE & PSUs) 97 +0.052 25 052 (c) 25 25.GATE 2007(PI) GATE ‐2007(PI) GATE 2015 GATE-2015 In the assembly shown below.98 mm −0.0−0.0 ±0. If the plating thickness varies between 10 .002 0 002 mm +0. Gap or Ring gauge: used for gauging the shaft and male l components. ((a)) Concentricityy ((b)) Runout ((c)) Perpendicularity p y ((d)) Flatness between 30++0.98 24 98 mm Work tolerance = 0. Size of the hole to be checked 25 ± 0.15 microns.0 micron.010 mm.02 mm Page 46 of 210 −0.002 Dimension of 'NOT GO' Plug gauge = 25.000 0 000 +0.065 (b) 30++0. a) 2.00±0. y work tolerance zone becomes smaller by the sum of the gauge tolerance.002 −0. hole y Snap. Hi Higher h limit li it off hole h l = 25. Thickness of the plating is 30 ±2.020 0.008 b) 2.020 91 0 (a) 30++0. the size of the GO gage in mm tto iinspectt the th plated l t d components t iis (a) 25 25.02 99 . the GO and NO GO gauge tolerance zones are bisected by the high and low limits off the work tolerance zone.00±0.98 +0.074 074 (d) 25.002 +0.00 ) ±0 020 d) d) 2. Plug gauge ISRO‐2008 Plug gauges are used to (a) Measure the diameter of the workpieces (b) Measure the diameter of the holes in the workpieces p (c) Check the diameter of the holes in the workpieces (d) Check the length of holes in the workpieces Fig. The Th Go G snap gauge is i off a size i corresponding to the high (maximum) limit of the shaft. the part dimensions are: L 1 = 22.020 mm (c) 30++0. tolerance the size of the low limit of the hole while the NOT GO plug gauge corresponds to the high limit of the hole.04 mm ∴ Gauge tolerance = 10% of work tolerance = 0.042 042 (b) 25.040 mm 92 93 Limit Gauges Limit Gauges GATE‐2013 y Plug gauge: used to check the holes.004 ∴ Dimension of 'GO' Plug gauge = 24.004 mm For-2016 (IES.000 Dimension of 'NOT GO' Plug gauge = 25. while hil the h NOT GO gauge corresponds d to the h low l (minimum limit).030 mm 0.02 mm H Here.005 mm Assuming normal distribution of part dimensions Assuming normal distribution of part dimensions.050 0.0 10 0 GATE – 2007 (PI) ( ) The geometric tolerance that does NOT need a datum mm Diameter of a hole after plating needs to be controlled for its specification is ±0. 054 mm (d) 20. VS‐2012 A ring i gauge is i used d to measure ((a)) Outside diameter but not roundness (b) Roundness but not outside diameter ( ) Both (c) B h outside id diameter di and d roundness d ((d)) Onlyy external threads This principle states that the GO gauge should always be so designed d d that h it will ll cover the h maximum metall IES 2010 Conventional IES 2010 Conventional Discuss a Go gauge.000 101 100 GATE 2015 GATE-2015 GATE ‐ 2004 GO and d NO‐GO NO GO plug l gages are to be b designed d i d for f a 0.• Taking example of above: ∴Wear Allowance = 5% of work tolerance = 0.0 4 mm aand d 20. the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system is (a ) 24.003 −0.006 (d ) 24 24.98 24 98 + 0.GATE & PSUs) 106 Page 47 of 210 107 Rev. condition (MMC) of as many dimensions as possible in the same limit gauges. y The size of go plug gauge is reduced while that of go snap gauge increases. Following ISO system of gage design sizes of GO and NO design.006 mm and 20. example of inspection by …. whereas a NOT GO gauges to cover the minimum metal condition of one dimension only.015 mm.046 mm (c) 20. Nominal size of GO plug gauge = 24. Wear allowance is usually taken as 5% of the work tolerance.0 108 .02 Dimension of 'NOT GO' Plug +0..003 +0.015 +0.000 −0.010 mm and (a) d 20.050 mm (b) 20.03 −0.985 985 +0.014 0.000 −0.(variables/attributes) The above statement is (a) Variables (b) Attributes (c) Cant say (d) Insufficient data 104 105 For IES Only T l ’ Pi i l Taylor’s Principle GATE – 2006.014 mm and d 20. For-2016 (IES.003 −0. NO‐GO GO gage will be respectively ( ) 20.002 0 002 mm +0.004 ∴ Dimension Di i off 'GO' Plug Pl gauge = 24. i y To increase service life of gauges wear allowance is g g added to the go gauge in the direction opposite to wear. Considering 10% of work tolerance to be the g gauge g tolerance and no wear condition.982 24 982 mm −0.046 0.002 mm g g y y Wear allowance: GO gauges which constantly rub against the surface of the parts in the inspection are subjected to wear and loose their initial size.03 (b) 25. y Wear allowance is applied to a nominal diameter W ll i li d i l di before gauge tolerance is applied.054 mm GATE ‐ 2014 GATE ‐ 1995 Which one of the following statements is TRUE? a) The ‘GO’ GO gage controls the upper limit of a hole b)The ‘NO GO’ gage controls the lower limit of a shaft c) The ‘GO’ gage controls the lower limit of a hole d)The ‘NO GO’ gage controls the upper limit of a hole 103 102 Checking the off a hole GO‐NO‐GO Ch ki h diameter di h l using i GO NO GO gauges is an. Gage of the hole tolerance. Discuss a ‘Go’ gauge.004 0 004 mm A GO‐NOGO gauge is GO NOGO plug l i to be b designed d i d for f measuring a hole of nominal diameter 25 mm with a hole tolerance of ± 0.000 g ggauge g = 25.985 (c) 24 24.01 0 01 mm.985 985 +0.05 g tolerances can be taken as 10% hole 200. ( R10 : 1. y These numbers are called preferred numbers having common ratios as..58. all follows a definite pattern or series.12 and 40 10 1. l Precision data have small dispersion p ( spread p or scatter ) but may be far from the true value. y These are named as Renard series.. machine tool speed and feed.0004D and Deviation = 0 Medium force fit : 1/3 1/3 Tolerance = 0.1. 25 5 10 10..P. or both. The expected ability for a system to discriminate between two settings..6. A measurement can be accurate but not precise. e ca Sta da d ssoc at o o e a ce Syste American Standard Association Tolerance System 1.06 y Depending on the common ratio.1.1. deviations y It has been observed that if the sizes are put in the form of g geometric p progression. 1 0 1 06 1 12 ( ) ) 1000 1000...... settings Smaller the bias more accurate the data. g .. Medium force fit M di f fi that is permitted by the following classes of fit 3. 5 1000... 40 1000 1000. rejected 109 formed..For IES Only Why is a unilateral tolerance Why is a unilateral tolerance preferred over bilateral tolerance ? preferred over bilateral tolerance ? y This system is preferred for Interchangeable manufacturing.0.25.6.0005 0 0005D − 0 0.0006D 0 0006D and Deviation De iation = 0. y It is easy and simple to determine deviations... 40 100 100. 0 1 6 2.1..GATE & PSUs) 115 By S K Mondal 116 Page 48 of 210 y (true) value of the quantity being measured..26 1 26 :1. 100 etc. R20 and R40 110 111 For IES Only Preferred Number Contd..) 10 10. 20 10 ≈ 1. series i are given i b l below.0. Medium fit 7.. these are R5 . 1000 20 IES‐2013 conventional IES‐2013 conventional 4. 100 40 20 10 10.12.4..0.. 1000 10 20 R 40 : 1. R10 .0. manufacturing Preferred Number Preferred Number IES 2010(Conv) IAS‐2014 (Main) IES 2010(Conv) IAS‐2014 (Main) y A designed product needs standardization. engine power.1. Tight fit as p per ANSI class 4 : Snug g fit and class 7 :Medium force fit. y Typical values of the common ratio for four basic G. A measurementt system t i called is ll d valid lid if it is i both b th accurate t and precise. then wide ranges g are covered with a definite sequence.58 58 :1 :1. :1 0 1 25 1 6 ( R 20 : 1. 10 10 ≈ 1.. Also mention applications.0006 0006 D Measurement of Lines & Surfaces y y Snug fit is applicable where no shake is permissible S fi i li bl h h k i i ibl y Medium force fit is applicable for shrink fit on cast iron y y For-2016 (IES.. Free fit 8..0 .26. precise precise but not accurate. :1 0 1 12 1 4 ( R5 : 11.06. 6. neither. 117 Rev. y Motor speed. Why is a unilateral tolerance preferred over bilateral tolerance ? y This also helps in interchangeability of products. 10.1.1..12. y It helps p standardize the GO g gauge g end y Helpful for operator because he has to machine the upper limit of the shaft and the lower limit of the hole knowing 5 fully well that still some margin is left for machining before 10 ≈ 1. 100. C td Preferred Number …. Loose fit ) 112 113 114 Accuracy & Precision Accuracy & Precision Snug fit y Accuracy ‐ The ability of a measurement to match the actual 1/3 Tolerance = 0. ratio four basic series are the part is rejected. Wringing fit 5 Snug fit 5.5. y Many other derived series are formed by multiplying or dividing the basic series by 10. 5 100 100. 10 10 100. 100 10 1000.06 1 06 :1. Heavy force shrunk fit W it the Write th amountt off allowance ll and d tolerance t l 2.... P i i Precision ‐ The Th precision i i off an instrument i i di indicates i its ability to reproduce a certain reading with a given accuracy ‘OR’ it i is i the h degree d off agreement between b repeated d results..12 1 12 :1. Reliability of measurement Reliability of measurement Repeatability y It is the ability of a measuring system to reproduce output readings when the same input is applied to it consecutively. Eg. t bilit y It is a quantitative characteristic which implies confidence in the measured results depending on whether or not the frequency q y distribution characteristics of their deviations from the true values of the corresponding p g q quantities are known. instruments These may be reduced or eliminated byy correct choice of instruments. Primary Standard) Page 49 of 210 120 Errors y Systematic S i errors or fixed fi d errors (Bias): (Bi ) Due D to faulty f l or improperly calibrated instruments. y The vernier scale is that a certain number n of divisions on the vernier scale is equal in length to a different number (usually one less) of main main‐scale scale divisions. It is usuall achieved usually achie ed by b means of a direct comparison against measurement standards or certified reference materials. 122 Which of the following errors are inevitable in the measuring system and d it would ld be b vain full f ll exercise to avoid them (a) Systematic errors (b) Random errors (c) Calibration errors (d) Environmental errors 124 Some off the used the S h instruments i d for f h linear li measurements are: y Rules (Scale) y Vernier y Micrometer (Most widely used.0 . y Calibration determines the performance characteristics of an instrument. y Least count = S − V = 1 S n 126 Rev. i l under d the h same conditions. Errors stemming from environmental variations. divisions nV = (n −1)S where h n = number b off divisions d on the h vernier scale l V = The length g of one division on the vernier scale and S = Length of the smallest main‐scale division y Least count is applied to the smallest value that can be read directly by use of a vernier scale. system or reference material. Which of these targets represents accurate shooting? Precise shooting? h ti ? Reliable R li bl shooting? h ti ? 118 Calibration usually by adjusting it to match or conform to a dependably known value or act of checking. Due E E i f i l i i D 123 to Insufficient sensitivity of measuring system Vernier Caliper Linear measurements ISRO‐2007 For-2016 (IES. secondary Standard) y Height gauge y Bore B gauge y Dial indicator y Slip gauges or gauge blocks (Most accurate. Th variation The i ti range is i referred f d to t as repeatability. and mostly. It is the probability that the results will be predicted. mostly a sticker is provided for the instrument. These cannot be eliminated. alters the results as shown. cycle a process does not stop at the same location. y A calibration certificate is issued and. y It is very widely used in industries. words the gauge or measuring instrument should be 10 times as accurate as the characteristic to be measured. It denotes the maximum changes in an input signal g that will not initiate a response p on the output. Dose this show h which hi h shooting h ti was the th mostt reliable? 125 y A vernier scale is an auxiliary scale that slides along the main scale. Eg. di i and d in i the h same direction. such as wind. In other words. 121 measuring instruments y Resolution: It is the smallest change of the measured quantity tit which hi h changes h th indication the i di ti off a measuring i instruments y Sensitivity: The smallest change in the value of the measured variable to which the instrument respond is sensitivity. Errors of technique q etc. y Imperfections in mechanical systems can mean that during a Mechanical cycle.GATE & PSUs) 119 y Drift: It is a slow change of a metrological characteristics of a y It or correcting off a measuring device I is i the h setting i i i d i A change in one variable. or move through the same spot each ti time. p y Rule of 10 or Ten‐to one rule: That the discrimination (resolutions) of the measuring instrument should divide the tolerance of the characteristic to be measured into ten parts. g calibration errors. y Random errors: Random errors are due to non‐specific cause like natural disturbances that may occur during the experiment. 01 mm (c) 0. Page 50 of 210 132 pp cat o s o d a d cato c ude: Applications of dial indicator include: y centering workpices to machine tool spindles y offsetting lathe tail stocks y aligning a vice on a milling machine y checking dimensions Principle of a dial indicator 134 Rev.01 mm to 0.005mm Vernier Caliper 127 128 M t i Mi t Metric Micrometer 129 ISRO‐2009. the reading y It consists a main scale and a thimble corresponding p g to 5 divisions on barrel and 12 Method of Measurement divisions on thimble is Step‐I: Find the whole number of mm in the barrel Step‐I: Find the reading of barrel and multiply by 0.012 mm 131 y Dial indicator: Converts a linear displacement into a radial movement to measure over a small ll range off movement for f the h plunger.0 135 .005 mm (b) 0.5 in common use. sizes For-2016 (IES.512 mm (c) 2.001 mm.5 mm) is ((c)) Flatness of measuring g jjaws ((d)) Temperature p equalization q (a) 0.02 mm (d) 0.01 Step‐III: Add the value in Step‐I and Step‐II 130 y Bore Gauge: used for measuring bores of different g g from small‐to‐large g sizes. matching (b) Its calibration with 24 4 divisions of main scale ((1 main scale divisions = 0. It is one of the most accurate mechanical devices I a simple In i l micrometer i t with ith screw pitch it h 0.620 mm (b) 2.ISRO‐2010 ISRO‐2008 The should Th vernier i reading di h ld not be b taken k at its i face f value before an actual check has been taken for Th least The l t countt off a metric t i vernier i caliper li (a) Zero error having 25 divisions on vernier scale. sizes ranging y Provided with various extension arms that can be added for different sizes. 2011 ISRO‐2009 2011 y A micrometer allows a measurement of the size of a body. mm and divisions on thimble 50. mm y It is possible to use the dial indicator as a comparator by mounting g it on a stand at anyy suitable height.120 mm (d) 5.GATE & PSUs) Micrometer 133 (a) 2. y The typical least count that can be obtained with suitable gearing dial indicators is 0. 020 1 020 _______ 35. 140 To make up a Slip Gauge pile to 41. >0 which one of the following would be consistent with the observation? (A) The drill spindle rotational axis is coincident with the drill spindle p taper p hole axis (B) The drill spindle rotational axis i intersects the h drill d ill spindle i dl taper hole h l axis at point P (C) The Th drill d ill spindle i dl rotational t ti l axis i is i parallel to the drill spindle taper hole axis (D) The drill spindle rotational axis intersects the drill spindle taper hole axis at point Q 136 GATE – 2014(PI) S‐2 ( ) This whether Thi test inspects i h h the h ((a)) spindle p vertical feed axis is p perpendicular p to the base plate (b) axis of symmetry of the cylindrical spindle is perpendicular to the base plate (c) axis of symmetry. t y A typical yp set consisting g of 88 p pieces for metric units is shown in. g This is so that theyy will be the are 2 mm wear g only ones that will wear down. y To T build b ild any given i di dimension. g identifyy a set of blocks.GATE & PSUs) 142 138 y Come in sets with different number of pieces in a given y These are small Th ll blocks bl k off alloy ll steel. The mandrel axis is an extension of the drill spindle p taper p hole axis and the protruding portion of the mandrel surface is perfectly cylindrical Measurements are taken with the sensor cylindrical.100 1 100 _______ 34.005 1 00 _______ 36.125 -4. 141 41. y Take away the thickness of the two wear gauges.000 4 000 _______ 30. The measuring surfaces f off the th gauge blocks bl k are finished fi i h d to t a very high hi h degree of finish.100 -1. y Not to be used for regular and continuous measurement measurement. y Generally G ll the h top and d bottom b Sli Gauges Slip G i the in h pile il gauges. placed at two positions P and Q as shown in the figure.000 -4.. y Rectangular blocks with thickness representing the dimension of the block. y Decide what height g y you want to set up. Th readings The di are recorded d d as Rx = maximum i d fl ti deflection minus minimum deflection. y Are hardened and finished to size. turn starting with the lowest. For-2016 (IES. and it is much cheaper to replace l two gauges than h a whole h l set. GATE – 2008 contd… from S‐2 d f If Rp= RQ>0.125mm.0 144 .125 -1.. corresponding to sensor position at X. The cross‐section of the block iss usua usuallyy 332 mm x 9 mm.125 mm To make up a Slip Gauge pile to 41. over one rotation. p in this case 41. and then use the gauges in the set to remove each place of decimal in turn. A dial indicator is fixed to the cylindrical spindle and d the th spindle i dl is i rotated to make the indicator d touch h the h base p plate at different points 137 Slip Gauges or Gauge blocks y Used in the manufacturing g shops p as length g standards. i it is i necessary to t put together.000 Page 51 of 210 143 Rev. flatness and accuracy.000 30 000 _______ 0.GATE – 2008 S‐1 A displacement sensor (a di l ( dial di l indicator) i di ) measures the h lateral displacement of a mandrel mounted on the taper hole inside a drill spindle. which are to be p y Number of blocks used should always be the smallest. the rotational axis and the vertical feed eed aaxiss o of tthee sp spindle d e aaree aall co coincident c de t (d) spindle rotational axis is perpendicular to the base plate l t The test Th alignment li “Spindle square with base plate” is applied to the radial drilling g machine.000 -30.000 ______ 37.120 -1. l 139 y A Slip pile Sli Gauge G il is i sett up with ith the th use off simple i l GATE – 2014(PI) S‐1 ( ) sett to t suit it the th requirements i t off measurements.125 mm maths. 005 1.500 0. a number of principles are used such as mechanical. di i y During g the measurement. pneumatic and electrical.001 0. mm Increment mm ‐ 0.490 0 500 to 9.010 0 500 0.500 10. Principle of a comparator Feeler Gauge 149 Limit Gauges PSU A f l A feeler gauge is used to check the i d t h k th (a) Pitch of the screw (b) Surface roughness For-2016 (IES.GATE & PSUs) 151 150 Gauge For Measuring Snap Gauge External Dimensions Plug Gauge g g Internal Dimensions Taper Plug Gauge Taper hole Ring Gauge External Diameter (c) Thickness of clearance G G Gap Gauge G d G Gaps and Grooves (d) Flatness of a surface Radius Gauge Gauging radius Thread pitch Gauge p g External Thread Page 52 of 210 152 Rev.A M t i li t (88 Pi ) A Metric slip gauge set (88 Pieces) Slip gauges size or range. electrical 148 comparator t ? (a) Tool Maker Maker’ss Microscope (b) GO/NO GO gauge (c) Optical Interferometer (d) Dial Gauge Fig.001 to 1. a comparator p is able to g give the deviation of the dimension from the set dimension.010 to 1.500 to 9 500 10 to 100 Increment. which is quick and more convenient for checking h ki large l number b off identical id ti l dimensions. optical. y To magnify the deviation. y Highly reliable.0 153 .009 1. mm 1. y Cannot measure absolute dimension but can only compare two dimensions.000 Number of Pieces 1 9 49 19 10 ISRO‐2010 A master gauge is (a) A new gauge (b) An international te at o a reference e e e ce sta standard da d (c) A standard gauge for checking accuracy of gauges used on shop floors (d) A gauge used by experienced technicians ISRO‐2008 St d d to Standards t be b used d for f reference f purposes in i laboratories and workshops are termed as (a) Primary standards ((b)) Secondaryy standards ((c)) Tertiaryy standards (d) Working standards 145 146 147 Comparators GATE – 2007 (PI) ( ) y Comparator is another form of linear measuring Which one of the following instruments is a method. Optical amplification A d O i l lifi i y 50 /1 x 2 plunger. Sigma Mechanical Comparator 154 155 156 Optical Comparators Mechanical Comparators Mechanical Comparators y These devices use a plunger to rotate a mirror. needle The assembly provides a frictionless movement with a resistant pressure provided by the springs. while the other is fixed. and simply by the virtue of distance. f The h most common angular l measuring y The float height is essentially proportional to the air tools are: that h escapes from f the h gauge head h d y Bevel protractor y Master M t gauges are used d to t find fi d calibration lib ti points i t on y Sine bar the scales y The input 159 pressure is regulated to allow magnification adjustment For-2016 (IES. y The Eden‐Rolt Reed system y uses a pointer attached to the end of two reeds.GATE & PSUs) 160 Page 53 of 210 161 Rev. then image is tilted by an angle δθ then image y will be tilted by 2 x δθ. because if mirror one reed d moves relative l ti to t the th other. y And. th y is tilted by an angle δθ. One reed is pushed by a y In this system. wrapped d band b d wrapped d about b a driving d d drum to turn a pointer needle.0 162 .Mechanical Comparators Mechanical Comparators Sigma Mechanical Comparator y The Mikrokator principle The Sigma Mechanical Comparator uses a partially greatly magnifies any d i ti deviation i size in i so that th t even small deviations produce d l large d fl deflections off the p pointer over the scale. As y It is multiplied by 2. the small rotation of the mirror can be converted to a significant g translation with little friction. Thus overall the pointer that they are commonly y magnification of this system attached to will deflect. Mechanical amplification y = 20 /1 . y = 2 x (20/1) ( 50/1 =2000 units) 2 x (20/1) ( 50/1 2000 units) 157 Pneumatic Comparators 158 Angular Measurement Pneumatic Comparators y Flow type: This involves the measurement of angles of tapers and similar l surfaces. A light Th d i l i li h beam is reflected off that mirror. the angle can be calculated using the sine discrimination of one degree. the various part of sine bar are hardened before grinding & lapping. Sine bars cannot be used for conveniently for y Threads are normally specified by the major diameter. can be used. the metric V‐thread shown in Fig y 2. (c) The centre distance between the two rollers Th calculated The l l t d taper t angle l (in (i degrees) d ) is i (d) The distance between rollers and upper surface ( ) 21. & 300 mm. 100 mm. possible to measure the angle. y Screw pitch gauge y Pitch measuring machine y Thread form For-2016 (IES. 2 Misalignment of workpiece with sine bar may y Major diameter y Micrometer y Pitch diameter y Floating Carriage micrometer y Wire method (Three wire and two wire) y Pitch sometimes introduce considerable errors.6 (c) 6 (d) 68. g formula. 200 mm. the sine bar will tilt to a specific on the h workpiece k i and d then h by b rotating i the h rule.e. the most common thread encountered is the metric V thread shown in Fig. During taper angle (a) Its total length s in θ = H L measurement of a component. i. the height from the (b) The size of the rollers surface f plate l to the h centre off a roller ll is i 100 mm. a sine bar y Is part of the machinist's combination square. Each roller has A sine i bar b is i specified ifi d by b a diameter of 20 mm. angle It will typically have a y Knowing the height differential of the two rollers in alignment with the workpiece .GATE & PSUs) 169 Page 54 of 210 170 y Optical projector Rev. y A sine bar is specified by the distance between the centre of the two rollers.1 (a) 166 Thread Measurements y 1.8 8 168 y The parameters that are normally measured are: y Though there are a large variety of threads used in g g y engineering. Th d ll ifi d b h j di adjustment problems.9 68 167 Dis‐advantages measuring angles l more than h 60o because b off slip l gauge (b) 22. l it i is i angle. y Basically a sine bar is a bar of known length. A Bevel Protractor 163 164 165 ( ) GATE ‐2012 (PI) ISRO‐2011 A sine bar has a length of 250 mm. length When gauge blocks y The flat base of the protractor helps in setting it firmly are placed under one end. problems ( ) 23.0 171 .Sine Bar Sine Bar Bevel Protractor y When a reference for a non‐square angle is required. 5156 p y Best wire size D s = Micrometer reading over the plug screw gauge with the wire P = Pitch value p α sec 2 2 For ISO metric thread.GATE & PSUs) 174 y Pitch Pit h Diameter Di t or Eff Effective ti Di Dia.532 mm and 15. Two-Wire Method y Distance W over the outer edge α⎞ p α ⎛ W = D p + d ⎜1 + cosec ⎟ − cot 2⎠ 2 2 ⎝ For ISO metric thread.866 (b) 1. two on one side and one on other side p α ⎞ ⎛α ⎞ ⎛ cot ⎜ ⎟ − d ⎜ cosec − 1⎟ 2 2 ⎠ ⎝2⎠ ⎝ T = Dimensions under the wire = D + ( Dm − Ds ) =T + D=D Diameter i t off master t or standard t d d cylinder li d Dm = Micrometer reading over standard cylinder with two wire W = D + 3d − 1.5774 p d= y Best wire size d= 175 A metric thread of pitch 2 mm and thread angle 60 The best wire size (in mm) for measuring effective diameter of a metric thread (included angle is 60o) of 20 mm diameter and 2.398 mm.55 mm diameter and two wires of 2 mm diameter each are used.6496 p p = pitch of thread .366 (d) 26. The micrometer readings over the standard and over the wires are 16. and α = thread angle 173 inspected d for f its pitch h diameter d using 3‐wire method The diameter of the best size wire in mm is method.397 Rev.5 mm pitch using two wire method is (a) 1. The effective diameter (in mm) of the thread is (a) 33.886 (d) 2. α = 60 and D p = D − 0.5 mm pitch. a cylindrical p y standard of 330.0 180 ..086 178 α p sec 2 2 176 GATE‐2013 GATE – GATE – 2011 (PI) 2011 (PI) For-2016 (IES. α = 60 d = 0. respectively.443 (b) 0.154 Page 55 of 210 (d) 2. (a) 0.723 0 723 (c) 2.Three-Wire Method The Three-Wire Method of Measuring Threads y Three wires of equal diameter placed in thread.366 33 366 (b) 30.397 30 397 (c) 29.000 (c) 1. two on one side and one on other side y Standard micrometer used to measure distance over wires (M) y Different sizes and pitches of threads require diff different t sizes i off wires i D p = pitch diameter or Effective diameter 172 Dp = T + P y Two wires of equal diameter placed in thread.000 179 177 GATE – GATE – 2011 (PI) 2011 (PI) To measure the effective diameter of an external metric thread (included angle is 60o) of 3.. 181 182 183 185 186 g g y Roughness height: is the p parameter with which generally the surface finish is indicated. of 210 honing.0 . y The greater the width. performs Measurement of Surfaces fS f y Surface texture can be difficult to analyse q quantitatively. Perpendicular lay: Lay perpendicular to the Surface. For-2016 (IES. width the smoother is the surface and thus is more desirable. width y Determined by the height of the waviness and its width. Lay Contd. 189 Rev. Circular lay: Approximately circular relative to the center. but most Real surfaces are rarely so flat or smooth but most y Two surfaces may y be entirelyy different. Surface is produced d d by b grinding. Radial lay: Approximately radial relative to the center of the nominal surface. ways surface f quality. Surface is produced d d b by shaping. y Lay L di direction: i i the is h direction di i off the h predominant d i surface pattern produced on the workpiece by the tool marks. It is specified either as arithmetic average value or the root mean square value. i di lapping. or smooth. g y Roughness width cut‐off: is the maximum width of the surface that is included in the calculation of the roughness height. the same CLA (Ra) value.. l and the better is performs. which constitutes the predominant pattern of the roughness. Surface is produced by 188 Page 56knurling. l the h longer l a product d generally ll lasts. y Flaw: are surface irregularities that are present which are random and therefore will not be considered. yyet still p provide commonly a combination of the two. h i planning etc. Diagram Symbol Description Multidirectional lay: Lay multidirectional. Surface is produced b shaping by h i and d planning l i Crossed lay: Lay angular in y y g both directions. but the better the ways.Surfaces y Surface geometry can be quantified a few different y No surface is perfectly smooth. super finishing.GATE & PSUs) 187 Diagram Symbol Description Parallel lay: Lay parallel to the Surface. S f Surface i produced is d d by b facing. y y Real surfaces are rarely so flat. 184 Lay y Waviness: refers to those surface irregularities that have a greater spacing than that of roughness width. y Roughness R h width: idth is i the th distance di t parallel ll l to t the th nominal part surface within which the peaks and valleys. 100”. 0.030”.” 0.025 N1 ∇∇ ∇∇∇ ∇∇∇∇ 192 IFS‐2011 y Waviness width ‐ the distance between peaks or What is meant by interchangeable manufacture? valleys y y Roughness width cutoff ‐ a value greater than the maximum roughness width that is the largest separation of surface irregularities included in the measurements. [2 marks] 190 IES ‐ 1992 Which rough Whi h grade d symbol b l represents surface f h off broaching? (a) N12 (b) N8 (c) N4 (d) N1 191 deviation denoted as Ra.4 N5 0. 0. 5.3 N9 32 3. 2.2 N4 0. The average of the five highest peak and five deepst valleys ll i the in th sample.5 N11 N10 ∇ 6. Centre line average (CLA) or arithmetic mean Roughness Grade Number y Waviness height ‐ the distance from a peak to a valley 193 Evaluation of Surface Roughness Roughness Ra (μm) 50 195 Determination of Mean Line y M‐System: After plotting the characteristic of any surface a horizontal line is drawn by joining two points.003”. irregularities [10 marks] [10‐marks] 194 Determination of Mean Line 1. profile For-2016 (IES.05 N2 0.GATE & PSUs) 196 Page 57 of 210 197 Rev. What are they ? Define the terms 'roughness h i ht' 'waviness height'. ( ” 0. ' i width' idth' and d 'lay' 'l ' in i connection ti with surface irregularities. points This line is shifts up and down in such a way that 50% area is above the line and 50% area is below the line 3. 4.IES‐2012 Representation of Surface Roughness In texture define I connection i with i h surface f d fi ((a)) waviness (b) flaws. l 5 The average or leveling depth of the profile.0 198 . Typical T i l values l are (0. Maximum peak to valley roughness (hmax) y E‐System: (Envelop System) A sphere of 25 mm diameter is rolled over the surface and the locus of its centre is being traced out called envelope.6 N7 0.1 N3 0 05 0.300”) y Lay ‐ the direction the roughness pattern should follow y Stylus travel is perpendicular to the lay specified. This is called mean envelope l and d the h system off datum d i called is ll d E‐ E system.8 N6 04 0. Laser light has unique advantages for inspection. This envelope is shifted in downward direction till the area above the line is equal to the area below the line.2 N8 1. (c) l List three defects found on surfaces. and ( ) lay.010”. Root mean square value (Rg) : rms value Roughness Symbol N12 ‐ 25 2 12. Surface roughness parameters 199 IES ‐ 2006 200 201 IES ‐ 2007 IES ‐ 2008 The M and E‐system in metrology are related to What is the dominant direction of the tool marks or What term is used to designate the direction of the measurement of: f scratches h in a surface f texture having h a directional d l predominant d machining operation? surface f pattern produced d d ( ) Screw (a) S threads h d (b) Fl Flatness quality called? quality.Arithmetical Average: y Measured are added M d for f a specified ifi d area and d the h figures fi dd d together and the total is then divided by the number of measurements taken to obtain the mean or arithmetical average g ((AA). ) y It is also sometimes called the centre line average or CLA value. This in expression form is g p RRMS = 1 N ISRO‐2011 CLA value and RMS values are used for measurement ∑y of 2 i (a) Metal hardness (b) Sharpness of tool edge L Ra = 1 1 y ( x ) dx ≅ L ∫0 N ∑y (c) Surface dimensions i (d) Surface roughness Fig.0 207 .GATE 3 (d) & PSUs) 3 1 C 1 2 D 4 4205 Page 58 of 210 206 Rev. ( ) Angularity (c) A l it (d) S f Surface fi i h finish (a) Primary texture (b) Secondary texture (a) Roughness (b) Lay (c) Lay Flaw (c) Waviness (d) Cut off (d) 202 203 204 IES ‐ 2008 IES 2010 b by ISRO‐2010 Match List I with List II and select the correct answer using the code given below the lists: List I List II (Symbols for direction of lay) (Surface texture) Surface roughness on a drawing is represented by (a) Triangles (b) Circles (c) Squares (d) Rectangles (a) (c) A 4 4 B C D A B 2 1 3 (b) 3 2 For-2016 1 2 (IES. value This in equation form is given by y The other parameter that is used sometimes is the root mean square value of the deviation in place of the arithmetic average . processes y Interferometry ‐ uses light wave interference patterns (discussed later) y One method of note is the finger nail assessment of roughness and touch method. 208 209 210 212 213 Stylus Equipment y uses a stylus that in l h tracks k small ll changes h i surface f height. Surface lay measurement Codes:A B C D A B C D (a) 4 1 2 3 (b) 4 3 5 1 ( ) 4 (c) 2 1 3 (d) 3 1 2 4 Rev. 4 Centre line average a erage 5. y A profilometer can measure small surface variations in vertical stylus displacement as a function of position. in order to quantify its roughness. Observation Methods should h ld b be made d against i t matched t h d identical processes.GATE & PSUs) GATE ‐ 1997 Contact profilometers 214 y A diamond stylus is moved vertically in contact with a A di d l i d i ll i i h sample and then moved laterally across the sample for a specified distance and specified contact force. y One O example l off this h is the h Brown & Sharpe Sh S f Surfcom unit. y Vertical resolution is usually in the nanometre level. y The relative motion between the skid and the stylus is measured with a magnetic circuit and induction coils. For-2016 (IES. y To give the human tester a reference for what they are y Observation and touch ‐ the human finger g is veryy irregularities What are their causes? irregularities. hi commercial i l sets off standards d d are available.Methods of measuring Surface Roughness h d f i S f h IAS – IAS – 2013 Main 2013 Main There are a number of useful techniques for measuring F a machined For hi d surface.0 216 . Form tester (C) Microkater 3 3. Page 59 of 210 215 List List Li t I Li t II (A) Surface profilometer 1. 211 Profilometer y Measuring instrument used to measure a surface's profile. y Human perception is highly relative. il bl y Comparison C i y stylus based equipment ‐ very common Explain any three of them. y The radius of diamond stylus ranges from 20 h d fd d l f nanometres to 25 μm. Comparator 6. Calibration (B) Light Section Microscope 2. f show h macro‐ and d micro‐ i surface roughness: What are the various measures of surface finish? perceptive to surface roughness touching. and a skid that follows large changes in surface height. though h h lateral l l resolution l is usually ll poorer. Film thickness measurement (D) Interferometer 4. Non‐contact Profilometers Optical Flats Advantages of optical Profilometers y An optical profilometer is a non‐contact method for providing much of the same information as a stylus y Because the non‐contact profilometer does not touch the h surface f the h scan speeds d are dictated d d by b the h light l h reflected from the surface and the speed of the based profilometer. map 217 218 219 220 221 222 For IES Only For IES Only IAS – 2012 Main y When the fringes are perfectly straight and same fringe width for dark and bright band we conclude that the surface is perfectly flat.GATE & PSUs) 223 Explain how flatness of a surface is measured with an optical flat. holography. aces. in a similar way to the contour lines on a map. flat y For convex surface the fringes curve around the point of contact. interference y The spacing between the fringes is smaller where the gap is changing more rapidly. [ [12 marks] k ] λ 2 Page 60 of 210 224 Rev. y When a flat surface of another optic p is p placed on the optical flat.0 225 . y For concave surface the fringes curve away from the point of contact. sensor) confocal microscopy and digital acquisition electronics. rapidly indicating a departure from flatness in one of the two surfaces. y Optical p profilometers do not touch the surface and p therefore cannot be damaged by surface wear or careless operators. both id y Used with a monochromatic light to determine the flatness of other optical surfaces by interference. l d such h as laser l ti triangulation l ti (triangulation sensor). The distance of air gap between two successive fringes is given by = nλ Distance of air ggap p of interference fringe g of n th order is = 2 For-2016 (IES. interference fringes are seen due to te e e ce in tthee ttinyy gap bet between ee tthee ttwo o su surfaces. g p y y Optical‐grade fused structures O ti l d clear l f d quartz t or glass l t t lapped and polished to be extremely flat on one or b th sides. y There are many different techniques which are currently tl being b i employed. transducer 229 Cli t Clinometer 230 horizontal. and measuring the deflection of the returned image against a scale. fl t name the th types t off surfaces.0058928 mm is used in the inspection. height t 232 Miscellaneous of Metrology By S K Mondal 231 A t lli t Autocollimator Clinometer ca instrument st u e t for o non‐contact o co tact measurement easu e e t o y An opt optical of y An A optical i l device d i for f measuring i elevation l i angles l above b For-2016 (IES.0 234 . and not the entire system.GATE & PSUs) 228 Talysurf Two slip 1. detector y A visual autocollimator can measure angles as small as 0. g Monochromatic light g of wavelength 0.For IES Only Optical flat as a comparator Optical flat as a comparator IES – 2012 Conventionall are supplied for two completely different surfaces using nλ l 2 optical ti l flat. f and d draw d if Wh l = separation of edges Where Δh = Write in short about optical flat. line y The clinometer can read easily and accurately angles of elevation that would be very difficult to measure in any other simple and inexpensive way. The total number of straight fringes that can be observed on both slip gauges is (a) 2 (c) 8 227 small angles or small angular tilts of a reflecting surface.002 mm are kept side by side in contact with each other lengthwise. either visually or by means of an electronic detector. 226 GATE ‐ 2003 (b) 6 (d) 13 y It I is i based b d upon measuring i the h generated d noise i due d to dry friction of a metallic blade which travels over the surface under consideration.5 arcsecond. Page 61 of 210 233 Rev. roughness and independent of the measured specimen size and material. and the prototype transducer. Fringe patterns for two completely different types of surfaces. y An autocollimator works by projecting an image onto a target mirror. while an electronic autocollimator can be up p to 100 times more accurate. y Compass clinometers are fundamentally just magnetic compasses held with their plane vertical so that a plummet or its equivalent can point to the elevation of the sight line. l y The specimen surface roughness was measured by a widely used commercial instrument (Talysurf 10). y Used U d to t align li components t and d measure deflections d fl ti i in optical or mechanical systems. y If the frictional force is made small enough to excite the blade. Two fringe patterns n = number of fringes / cm Δh = The difference of height between gauges λ = wevlength of monochomatic light required Fig. then the noise will ill be proportional to surface roughness. y A fairly common use of a clinometer is to measure the h i ht off trees.000 mm T li gauges off 10 mm width id h measuring i and 1. An optical flat is kept resting on the slip pg gauges g as shown in the figure. y Measurement of forms. y Measurement of gap between rollers.5 cm deep. y Applications : y Measurement of outer dia. rollers Rev. And roundness of cylinder. at a high temperature. ray equal to 90o (c) A constant deviation prism having the angle of deviation dev at o bet between ee tthee incident c de t ray ay aand d reflected e ected ray. y Servo S autocollimators t lli t are specialized i li d compactt forms f of electronic autocollimators that are used in high speed servo feedback loops for stable platform applications. g .GATE & PSUs) An Optical Square 241 239 y about 5 cm in diameter and 12. One mirror(horizon glass) is half silvered and other(index glass) is wholly silvered.0 243 . Angle between the first incident ray and the last reflected ray is 90o Used to find out the foot of the perpendicular from a given point i t to t a line. equal to 45o (d) Used U d to t produce d i t f interference fi fringes y The LSM features a high scanning rate which allows Page 62 of 210 242 inspection p of small workpiece p even if theyy are fragile. pp 235 GATE – GATE – 2009 (PI) 2009 (PI) 236 237 Optical Square GATE – GATE – 2014 y An A Optical O ti l square consists i t off a small ll cylindrical li d i l metal t l box. Auto collimator is A lli i used d to check h k ((a)) Roughness g (b) Flatness ( ) Angle (c) A l ((d)) Automobile balance. t d d for f monitoring it i angular l movement over long periods of time and for checking angular position repeatability in mechanical systems. 240 Two mirrors may be replaced by two prisms. y Electronic and digital autocollimators are used as angle l measurementt standards. sheets y Measurement of spacing if IC chips. in which two mirrors are placed at an angle of 45o to each other and at right angles to the plane of the instrument. b A autocollimator An t lli t is i used d to t Th flatness The fl t off a machine hi b d can be bed b (a) measure small angular displacements on flat measured using y (a) Vernier calipers y surface ((b)) compare p known and unknown dimensions ((b)) Auto collimator ((c)) measure the flatness error ((c)) Height g g gauge g y (d) measure roundness error between centers (d) Tool maker’s microscope y 238 For-2016 (IES. Th optical The ti l square belongs b l t a reflecting to fl ti instruments i t t which hi h measure angles by reflection. ay.y Visual autocollimators are used for lining up laser rod GATE ‐ 1998 Autocollimator ends the parallelism off optical d and d checking h ki th face f ll li ti l windows and wedges. y Measurement of thickness of film and sheets. li Used to set out right angles at a given point on a line in the fi ld field. ISRO‐2010 Laser Scanning Micrometer Optical O i l square is i ((a)) Engineer's g square q having g stock and blade set at 9 90o (b) A constant deviation prism having the angle of deviation between the incident ray and reflected ray. in motion or vibrating. P.5 . a common type yp of electromechanical transducer that can convert the rectilinear motion of an object to which it is coupled mechanically into a corresponding electrical signal. Measurement of gear pitch D. one can pressure using g Boyle's y Law equation.GATE & PSUs) 249 251 . y g y 247 LVDT 248 LVDT GATE ‐ 1992 Page 63 of 210 Match M t h the th instruments i t t with ith the th physical h i l quantities titi they th measure: I Instrument M Measurement (A) Pilot‐tube (1) R. Generally McLeod gauge is used for calibration lib ti purpose. y LVDT linear li position i i sensors are readily dil available il bl that h can measure movements as small as a few millionths of an inch up to several inches. "Compression of known pressure g gas to higher g pressure and p volume of low p measuring resulting volume & pressure. gauge for the reason that compression will cause condensation .01 micron i t 50 mm Hg to H can be b measured. On‐line measurement of moving p parts C.IES ‐ 1998 Match List‐I List I with List‐II List II and select the correct answer using the codes given below the lists: List‐I List‐II (Measuring Device) (Parameter Measured) A. ). Small angular deviations on long flat surfaces B. y A pressure from f 0. surface by tracing the boundary of the area.0 y Acronym for Linear Variable Differential Transformer. Optical flat 2. Diffraction grating 1. but are also capable of measuring easu g pos positions o s up to o ±20 inches c es ((±0.M. Surface texture using interferometer 5. Auto collimators 3. For-2016 (IES.5 m)." q calculate initial p y Pressure of gases containing vapours cannot normally measured with a McLeod gauge. y A rotary variable differential transformer (RVDT) i a type is t off electrical l t i l transformer t f used d for f measuring i 250 angular displacement. Measurement off very small ll displacements Code: A B C D A B C D (a) 5 4 2 1 (b) 3 5 1 2 ((c)) 3 5 4 1 ((d)) 5 4 1 2 244 GATE‐2014 Which one of the following instruments is widely used to check and calibrate geometric features of machine tools during their assembly? (a) Ultrasonic probe (b) Coordinate Measuring Machine (CMM) ( ) Laser interferometer (c) f (d) Vernier calipers 245 McLeod gauge 246 Planimeter y Used d to measure vacuum by b application l off the h y A device used for measuring the area of any plane principle of Boyle's law. of a shaft (B) McLeod Gauge (2) Displacement ((C)) Planimeter (3) Flow velocityy (D) LVDT (4) Vacuum (5) Surface finish (6) Area C d A Codes:A B C D A B C D (a) 4 1 2 3 (b) 3 4 6 2 (c) 4 2 1 3 (d) 3 1 2 4 252 Rev. Laser scan micrometer4. y Works on the principle. worms and similar components. y Elements El t off external t l thread th d forms f off screw plug l gauges. a 40 mm diameter is obtained at a height Z = 40 mm. maker'ss Microscope (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 An essential part of engineering inspection. y The highly sensitive machine measures parts down to the fraction of an inch. 5 points are touched and a diameter of circle ((not compensated p for p probe size)) is obtained as 20 mm. e. punches and dies. CMM with i h a probe b of 2 mm diameter. Z = 10 mm from the bottom. mm the smaller diameter (in mm) of hole at Z = 0 is ( ) 13. 254 GATE ‐ 1995 List Li I ((Measuring g instruments)) (A) Talysurf 1. A taper hole h l is i inspected i d using i a CMM. (D) Autocollimator 4. Similarly. Sine bar 6 Tool maker 6. a CMM contains many highly sensitive air bearings on which the measuring arm floats.334 (c) 15. y Shallow bores and recesses. tools plate and template gauges. ((C)) Transfer callipers p 33. 2 Optical Interferometer R Alignment error of a machine slide way 3. Advantages. 253 Telescopic Gauges Telescopic Gauges easu e a bo e s ssize.GATE & PSUs) 255 259 Page 64 of 210 260 Rev.GATE ‐ 2004 Tool Maker’s Microscope Match M h the h following f ll i Feature to be inspected Instrument P Pitch and Angle errors of screw thread 1. y Specifically. (B) Telescopic Tl i gauge 2. y They Th are a direct di t equivalent i l t off inside i id callipers lli and d require the operator to develop the correct feel to obtain repeatable results. At a height. measurement and calibration in metrology gy labs. y Costly C l y fixturing g is critical y requires a very good tolerance model For-2016 (IES.0 261 . Spirit p Level 5. Auto Collimator Q Flatness error of a surface plate 2.334 (a) (b) 15.542 (d) y can automate inspection process y less prone to careless errors l l y allows direct feedback into computer system p y Disadvantages. Codes:A d B C D (a) 4 1 2 3 (b) (c) 4 2 1 3 (d) List Li II ((Application) pp ) T‐slots Fl Flatness Internal diameter Roughness A B C D 4 3 1 2 3 1 2 4 256 C di t M i M hi Coordinate Measuring Machine (CMM) y An instrument that locates point coordinates on three st u e t t at ocates po t coo d ates o t ee dimensional structures mainly used for quality control applications applications.442 ( ) 15. 257 258 GATE ‐ 2010 g . Dividing Head and d Dial Di l Gauge G S Profile of a cam 4. y Measurement of g glass g graticules and other surface marked parts. annular grooved and threaded h d d hobs h b etc. by ttransferring a se g tthee y Used to measure bore's internal dimension to a remote measuring tool. Hence is used to the following: y Examination of form tools. taps. D BCD B D -0. Th The diameter (D.C.0 . N None can tteach h you. ----- Rev. If the two precision rollers have radii 8 mm and 5 mm and the total thickness of slip gauges inserted between the rollers is 15.55 mm.41 Ref Note Ref. No 1 2 3 4 5 6 7 8 9 10 Ch‐13: Ch 13: Metrology Metrology Option A C C C C B C B B D Q. No 21 22 23 24 25 26 264 You have to grow from the inside out. the taper angle θ is (a) 6 degree (b) 10 degree (c) 11 degree (d) 12 degree GATE 2014 GATE ‐2014 The diameter of a recessed ring was measured by using two spherical h i l balls b ll off diameter di d2 = 60 6 mm and d d1 = 40 mm as shown in the figure. There is no other teacher But your own soul.55 mm and H1 = 20. ‐Swami Vivekananda Swami Vivekananda 265 For-2016 (IES. i mm)) off the in th ring gauge is …………. The distance H2 = 35.GATE & PSUs) 266 Page 65 of 210 Option B. H1 H2 d1 Diameter C H B A R Recessed Ring D 262 d2 Diameter 263 Q.54 mm. No 11 12 13 14 15 16 17 18 19 20 Option A D D C C B D B D A Q.WorkBook GATE 2008 (PI) GATE ‐2008 (PI) An experimental setup is planned to determine the taper of workpiece as shown in the figure. grain fragmentation. or larger Billet: is hot rolled from a bloom and is square. For-2016 (IES.5 in. Page 66 of 210 8 Rev. unloaded Its yield strength grain distortion or fragmentation does not take (a) Decreases place. I i the h first fi solid lid form f l Bloom: is the p product of first breakdown of ingot g has square q cross section 6 x 6 in. i h cross‐section i off workpiece does not change—the material is only subjected to shape changes. or more wide and 1. state there is no A test t t specimen i i stressed is t d slightly li htl beyond b d the th recrystalization y of g grains and thus recoveryy from y The Th ratio i cross‐section i area/volume / l i very high. 600 mm y For F mostt operations. ti h t or warm working hot ki y Strip is the product with a thickness < 5 mm and width conditions are preferred although some operations are carried out at room temperature. greater resistance ((c)) Remains same t this to thi action ti results lt in i increased i d hardness h d and d (d) Become equal to UTS strength i. Sl b is Slab: i the th hot h t rolled ll d ingot i t or bloom bl rectangular t l cross section 10 in.GATE & PSUs) GATE‐1995 y When metal is formed in cold state. or more thick. is hi h 7 6 yield point and then unloaded. Mill product Ingot y The cross‐section of workpiece p changes g without atoms and lattice distortion. in on a side or larger. y y y Forging: The workpiece is compressed between two g Metal Forming opposing dies so that the die shapes are imparted to the work. deformation y Due to slip. y Drawing: The diameter of a wire or bar is reduced by By S K Mondal pulling it through a die opening (bar drawing) or a series off die di openings i ( i drawing) (wire d i ) 1 Terminology Plastic Deformation y Plate is the product with thickness > 5 mm Billet slab 3 y These processes involve large amount of plastic deformation. y The ratio cross‐section area/volume is small. volume change. y They are always performed as cold working operations. y Sheet metalworking operations are performed on thin (less than 5 mm) sheets. strain hardening. movement of y Sheet is the product with thickness < 5 mm and width > Bloom Bulk Deformation Processes y Deformation beyond elastic limits. square 1. strips or coils of metal by means off a set off tools l called ll d punch h and d die di on machine hi tools l called stamping presses. < 600 6 mm 4 5 Sheet‐Forming Processes Strain Hardening Strain Hardening y In metall working the I sheet h ki operations.Terminology Four Important forming techniques are: Four Important forming techniques are: g The pprocess of pplasticallyy deformingg metal byy y Rolling: passing it between rolls. y Extrusion: The work material is forced to flow through a die opening taking its shape y y Semi‐finished product Ingot: is off steel.e. ((b)) Increases y As grain deformation proceeds.0 9 .5 1 5 in. Ductility and strain hardening limit the amount of forming 4. g y Incorporation of impurity atoms and insoluble second phase y Lead. Better surface finish 2. p essential to be so strong. Grain flow during deformation can cause desirable directional properties in product that can be done 4. has already received. closer tolerances 1.. paint y For Pure metal Rx temp. (Kelvin) y Rx temp. it may occur in all polycrystalline materials. driving force for this process is reduction in grain boundary energy. approximately equi‐axed grains to replace all the deformed crystals. soft steel. fracture g Cold Working y A malleable material should be plastic but it is not y In practical applications. The higher the cold work. Working below recrystalization g y temp. = 0. temp (Kelvin). y Rx temp. temp = 0. i lower l will ill be b the h Rx R temp 11 Grain growth h Malleability ll b l y Grain growth follows complete crystallization if the materials y Malleability is the property of a material whereby it can left at elevated temperatures.3 x Melting temp. p 12 b shaped be h d when h cold ld by b hammering h or rolling. Better accuracy. wrought iron. of Iron is 450oC and for steels around 1000°C y Finer Fi i the is h initial i i i l grain i size. 55. p of lead and Tin is below room temp. grain growth is not desirable. copper and aluminium are y Grain growth is very strongly dependent on temperature. ductility whereas working below are cold‐working process. of Cadmium and Zinc is room temp. p . y Rx temp. In some operations. ll ll l ll l y Rx temp. 13 Advantages of Cold Working d f ld k 14 15 Disadvantages of Cold Working d f ld k Equipment of higher forces and power required 1. y For Alloy Rx temp.IES‐2013 Statement (I): At higher strain rate and lower temperature structural steel tends to become brittle.) y “The temperature at which “Th minimum i i hi h the h completed l d recrystallisation of a cold worked metal occurs within a specified period of approximately one hour”. Page 67 of 210 17 Rev. y It I involves i l replacement l off cold‐worked ld k d structure by b a new set of strain‐free. (Kelvin). hot‐working g p process 10 recrystallization. y If working g above Rx temp. temp varies between 1/3 to ½ melting paint. metal must be annealed to allow further deformation 5 No heating of work required (less total energy) 5. Contd. the lower would ld be b the h Rx R temp. particles are effective in retarding grain growth. Statement (II): At higher strain rate and lower temperature p the yyield strength g of structural steel tends to increase. y Rx temp. temp decreases strength and increases ductility. Surfaces of starting work piece must be free of scale and S f f t ti k i t b f f l d dirt 3. Some metals are simply not ductile enough to be cold py g For-2016 (IES. ( ) Both (a) B th Statement St t t (I) and d Statement St t t (II) are individually i di id ll true and Statement (II) is the correct explanation of Statement (I) () (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Recrystallisation Temperature (Rx temp. p y Rx temp.5 0 5 x Melting temp. ll y Grain growth does not need to be preceded by recovery and y A malleable ll bl material i l is i capable bl off undergoing d i plastic l i y In contrary to recovery and recrystallization.0 18 . temp depends on the amount of cold work a material deformation without fracture. Strain hardening increases strength and hardness 3. 2. some materials in order of diminishing malleability.GATE & PSUs) 16 worked. p 19 Micro‐Structural Changes in a Hot Mi St t l Ch i H t Working Process (Rolling) Working Process (Rolling) 20 Annealing g •Annealing relieves the stresses from cold working – three stages: recovery. loads increase material ductility. p . 1. physical properties of the cold‐worked material i l are restored d without ih any observable b bl change h i in microstructure. forming it reduces loads. refined 3.GATE & PSUs) GATE‐2003 IES 2011 y During hot forming. sacrificing g die life for p product q quality. non uniform deformation and cracking of the surface. y The Th dies di or tooling li must be b heated h d to the h workpiece k i temperature. 23 25 Assertion (A): Lead. low residual stresses and uniform metal For-2016 (IES. good surface finish and tighter dimensional tolerances. Working above recrystalization g y temp. it produce less scaling and decarburization. recovery recrystallization and grain growth. y For temp. 2. It is useful for forging of details with intricate shapes. percentage elongation. 24 26 Cold C ld working ki off steell is i defined d fi d as working ki ((a)) At its recrystallisation y temperature p (b) Above its recrystallisation temperature ( ) Below (c) B l its i recrystallisation lli i temperature ((d)) At two thirds of the melting g temperature p of the metal Rev. growth •During recovery. Zinc and Tin are always hot worked. l 4. y Warm forming f i i a precision is i i f i forging operation i carried i d 22 Isothermal Forming h l interior and the variations in strength can result in non‐ interior. properties (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A ( ) A is true but (c) b R is false f l (d) A is false but R is true Page 68 of 210 out at a temperature range between 550–950°C.0 27 . 2. The impurities like slag are squeezed into fibers and di ib d throughout distributed h h the h metal. The porosity of the metal is largely eliminated. g . 2 It produces poor surface finish. y y Close tolerances. y Compared to cold forming. finish due to the rapid oxidation and scale formation on the metal surface. l tolerance l cannot be maintained. The mechanical p 4 properties p such as toughness.‐sensitive materials deformation is performed under isothermal conditions. Due D to the h poor surface f fi i h close finish. and resistance to shock and vibration are improved due to the refinement of grains. y Compared to hot forming. 2 The grain structure of the metal is refined. 3. 21 W F i Warm Forming y Deformation produced at temperatures intermediate to hot and cold forming is known as warm forming.Hot Working Advantages of Hot Working Dis‐advantages of Hot Working 1. percentage reduction in area. Reason (R) : If they are worked in cold state they cannot retain their mechanical properties. with desirable grain flow. better dimensional precision and smoother surfaces. cooler surfaces surround a hotter flow. It requires expensive tools. process. Strength is decreased. 4. Metal forming impurities and improves mechanical strength. g. 3. 3 and 4 IES – 2008 34 Consider C id the h following f ll i statements: g decreases harmful effects of 1. ocess. 3. 3 and 4 (d) 1. 2 Metal working process is a plastic deformation 2. in cold working. Higher forces are required 2. Grain structure gets distorted 3. 3 and 4 For-2016 (IES. i i d g is/are / Which of the above characteristics of hot working correct? ( ) 1 only (a) l (b) 3 only l (c) 2 and 3 (d) 1 and 3 31 30 IES – 2003 Assertion (A): off metals A i (A) Cold C ld working ki l results l in i increase of strength and hardness Reason (R): Cold working reduces the total number of dislocations per unit volume of the material (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true Page 69 of 210 33 35 Cold the C ld working ki produces d h following f ll i effects: ff p in the metal 1. S l t the Select th correctt answer using i the th code d given i b l below: (a) 1. 2 and 3 (b) 1. Less ductilityy is required q 4. Better mechanical properties of the process. Stresses are set up 2. Smoother finishes. g. 2 and d3 (c) 3 and 4 (d) 1 and 4 Rev.GATE & PSUs) IES – 2008 Consider C id the h following f ll i characteristics: h i i g y eliminated. In comparison 1. 2 and 4 (c) 1 and 3 (d) 2. High pressure. flow 3. 2 and 3 (b) 1. 2 Unbroken grain flow. Better surface finish is obtained Which h h off the h statements given above b are correct? (a) 1. Close Cl tolerances l cannot be b maintained. 2. 2 and (a) d3 (b) 1 and d 2 only l (c) 2 and 3 only (d) 1 and 3 only 32 IES – 2004 Cold results to C ld forging f i l in i improved i d quality li due d which of the following? 1. Which of the statements given above are correct? ( ) 1. No N heating h i is i required i d 33. Porosityy in the metal is largely 2. Surface finish is reduced 4 Which of these statements are correct? ( ) 1and (a) d2 (b) 1.0 36 . ISRO‐2012 IES – 2006 Materials are subjected M t i l after ft cold ld working ki bj t d to t Hot H rolling lli off mild ild steell is i carried i d out ((a)) At recrystallisation y temperature p (b) Between 100°C to 150°C ( ) Below (c) B l recrystallisation lli i temperature ((d)) Above recrystallisation y temperature p follo ing process to relieve following relie e stresses ( ) Hot (a) H working ki Which is Whi h one off the h following f ll i i the h process to refine fi the grains of metal after it has been distorted by hammering or cold working? (a) Annealing (b) Softening (c) Re‐crystallizing (d) Normalizing (b) Tempering (c) Normalizing (d) Annealing 28 IES – 2004 29 IES – 2009 Consider C id the h following f ll i statements: p to hot working. Strength S h and d hardness h d off the h metall are decreased d d 4. 1.ISRO 2010 ISRO‐2010 GATE‐2002. Very intricate shapes can be produced by forging process p ocess as co compared pa ed to cast casting gp process. 2 and 4 (c) 2. l the h hot h forging f i temperature range is i ((a)) 4 4000C to 6000C (b) 7000C to 9000C ( ) 10000C to 12000C (c) ((d)) 1300 3 0Cto 1500 5 0C Assertion (A): Hot working does not produce strain A i (A) H ki d d i hardening. recrystallisation ( ) Stress (c) S relief. lli i grain i growth h ((d)) Grain g growth.IES – 2000 Assertion (A): deformations by A i (A) To T obtain b i large l d f i b cold ld working intermediate annealing is not required. g 4. stress relief (b) Stress relief. A i (A) In I case off hot h working ki l the h temperature at which the process is finally stopped should h ld not be b above b the h recrystallisation ll temperature..GATE & PSUs) 39 IES – 1992 Assertion (A): off metals. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation l i off A (c) A is true but R is false (d) A is false but R is true Consider the following statements: C id h f ll i y When a metal or alloy is cold worked 1. grain growth. It is worked below room temperature. Reason (R): Hot working is done above the re‐ crystallization temperature. Reason ((R): ) If the p process is stopped pp above the recrystallisation temperature. Reason (R): Cold working is performed below the recrystallisation temperature of the work material. Reason (R): In hot working physical properties are generally improved. stress relief. 33. 2. recrystallisation y 41 IAS – 1996 For-2016 (IES. recrystallisation. crystallization temperature (a) Both A and R are individually true and R is the correct explanation of A l f (b) Both A and R are individually true but R is not ot a d R a e d v dua y t ue but R s ot tthe e correct explanation of A ( ) A is true but R is false (c) A i t b t R i f l (d) A is false but R is true 43 Specify S if the h sequence correctly l ((a)) Grain g growth. improved (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true Rev. y . grain growth will take place again p g and spoil p the attained structure. the stresses encountered are ( ) Greater (a) G than h yield i ld strength h but b l less than h ultimate strength (b) Less than yield strength of the material (c) Greater than the ultimate strength of the material (d) Less than the elastic limit 37 IES – 1996 IES – 1997 ISRO‐2009 38 IES – 2006 40 IAS – 2004 For F mild ild steel. It is worked below recrystallisation I i k d b l lli i temperature.0 45 . li f recrystallisation.. Its hardness increases but strength does not increase. material (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true In the metal forming process. strain I l bj d ld ki i hardening effect is due to (a) Slip mechanism (b) Twining mechanism (c) Dislocation mechanism (d) Fracture mechanism Page 70 of 210 42 IAS‐2002 44 Assertion is in A i (A): (A) There Th i good d grain i refinement fi i hot h working. Of these correct statements are (a) 1 and 4 (b) 1 and 3 ( ) 2 and 3 (c) d (d) 2 and 4 d In metals subjected to cold working. Its hardness and strength increase.. ll g Rolling y Most M widely id l used. 2 and 3 Rev.0 54 . rod. l y Friction F i ti b t between th rolls the ll and d the th metal t l surface f produces high compressive stress. t y Breakdown of ingots into blooms and billets is done by hot‐rolling.IES‐2008 Rolling y Definition: The process of plastically deforming metal Which one of the following is correct? Malleability is the property by which a metal or alloy ll can be b plastically l i ll deformed d f d by b applying l i ( ) Tensile (a) T il stress t (b) B di stress Bending t (c) Shear stress (d) Compressive stress b passing it between by b rolls. rail. bar. stress y Hot Hot‐working working (unless mentioned cold rolling. sheet. Shows work hardening effect 2 Surface finish is not good 2. For-2016 (IES. g This is followed byy further hot‐rolling g into plate. 3. falls to about (50 to 100°C) above the recrystallization temp. y Surface S f quality lit and d final fi l dimensions di i are less l accurate. Heavy reduction in areas can be obtained Which of these characteristics are associated with hot rolling? (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1.) By S K Mondal 46 y Metal will undergo g bi‐axial compression. y Page 71 of 210 53 IAS – 2001 Consider the characteristics off rolling C id h following f ll i h i i lli process: 1.GATE & PSUs) 52 Hot rolling is an effective way to reduce grain size in metals for improved p strength g and ductility. y Hot rolling is terminated when the temp. pipe. y Results fine grained structure. d high hi h production d i and d close l tolerance. the state of stress of the materiall undergoing d d f deformation is ( ) pure compression (a) i (b) pure shear h (c) compression and shear (d) tension and shear 49 Ch Change in grains structure in Hot‐rolling i i t t i H t lli Hot Rolling y Done above the recrystallization temp. p 47 48 50 51 GATE‐2013 In a rolling process. 0 63 . with good surface fi i h and finish d increased i d mechanical h i l strength h with i h close l produce strongest components? product dimensions. y Upper roll being adjustable to control the degree of curvature. l bending. where plates. ring rolling. dimensions (a) Hot rolling ((b)) Extrusion y Performed on four four‐high high or cluster cluster‐type type rolling mills. Ring rolling is used (a) To decrease the thickness and increase diameter (b) To T increase i th thickness the thi k off a ring i (c) For producing a seamless tube (d) For producing large cylinder y As the rolls squeeze and rotate.Cold Rolling ISRO‐2006 y Done below the recrystallization temp. t For-2016 (IES. 58 Sheet rolling 59 60 Roll Forming Roll Bending y In sheet rolling we are only attempting to reduce the y A continuous form of three‐point bending is roll cross section thickness h k off a material. foil etc. mills ((Due to high g force and p power)) ((c)) Cold rolling g (d) Forging 55 56 57 Ring Rolling ISRO‐2009 y Ring rolls are used for tube rolling. strip. i y Shaped Sh d rolls ll can be b used d to t produce d a wide id variety i t off cross‐section cross section profiles.GATE & PSUs) 61 Page 72 of 210 62 Rev.. Whi h off the Which th following f ll i processes would ld y Products are sheet. profiles y Ring rolls are made of spheroidized graphite bainitic and pearlitic matrix or alloy cast steel base. and rolled shapes can be bent to a desired curvature on forming rolls. sheets. the wall thickness is reduced d d and d the h diameter di off the h ring i increases. p bl k blank. and more wear‐resistant surface than cut threads.002 i h) than inch) h the h pitch i h diameter of the thread. thread y Restricted to ductile materials. y Used to produce threads in substantial quantities. materials y No metal is removed. 67 68 69 IES – 1992. y Blank Bl k diameter di i little is li l larger l ( (0. such h as aluminium l f l foil. GATE‐1992(PI) Thread rolling is restricted to (a) Ferrous materials (b) Ductile materials ( ) Hard materials (c) (d) None of the above N f h b For-2016 (IES. Thread rolling contd….IES – 2006 Shape rolling Which one of the following is a continuous bending process in which h h opposing rolls ll are used d to produce d long sections of formed shapes from coil or strip stock? ((a)) Stretch forming g ((b)) Roll forming g ((c)) Roll bending g ((d)) Spinning p g 64 65 Pack rolling 66 Thread rolling y Pack rolling involves hot rolling multiple sheets of materiall at once. harder. y Major diameter is always greater than the diameter of the y This is a cold‐forming process in which the threads are y Improved I d productivity d i i f formed d by b rolling lli a thread h d blank bl k between b h d hardened d dies di y Aluminum sheets (aluminum foil) y Matte. g smoother. g greater strength.0 72 .GATE & PSUs) 70 Page 73 of 210 71 Rev. h b h side bright d – foil‐to‐roll f l ll contact due d to high h h contact stresses with polished rolls that cause the metal to flow radially into the desired y A thin surface oxide film prevents their welding shape. satin side – foil‐to‐foil contact y Shiny. one gets ( ) One (a) O reduction d i in i thickness hi k (b) Two T reductions d ti i thickness in thi k (c) Three reductions in thickness (d) Two or three reductions in thickness depending upon p the setting g Rev. integrity ( ) A little large than the minor diameter of the thread (c) y Employed for high productivity and high quality. rolls which help in guiding the metal. y Because of the angle at which the roll meets the metal..IES – 1993. y These rolls have a central cylindrical portion with the sides tapering slightly. rolling y High accuracy and surface integrity. power for rolling is reduced C. Gear rolling between three gear roll tools 76 77 78 IAS – 2007 For-2016 (IES. T high hi h non‐reversing i mills ill 1. Fig. y The billet or round stock is rolled between two rolls.GATE & PSUs) 79 Match M t h List Li t I with ith List Li t II and d select l t the th correctt answer using i the code given below the Lists: List I List II (Type of Rolling Mill) (Characteristic) A Two A. an additional axial advance. y This cross‐rolling g action makes the metal friable at the centre which is then easily pierced and given a cylindrical shape by the central central‐piercing piercing mandrel. 75 y It is a variation of rolling called roll piercing. By small working roll.0 81 . it gets in addition to a rotary motion. Cluster mills 4. quality (costly (d) A little li l larger l than h the h pitch i h diameter di off the h thread h d machine)) y Larger size gears are formed by hot rolling and then 73 74 Roll piercing ll finished by machining. slightly There are two small side rolls. Three high mills 2. GATE‐1989(PI) IES – 2013 Conventional Manufacture of gears by rolling The blank diameter used in thread rolling will be Write two advantages of thread rolling and explain y The straight and helical teeth of disc or rod type external (a) Equal to minor diameter of the thread with figure two‐die cylindrical machine. both of them rotating in the same direction with their axes at an angle of 4. hf d l d l h steell gears off small ll to medium d d diameter and d module d l are [ M k ] [ 5 Marks] (b) Equal to pitch diameter of the thread generated by cold rolling. Rolls of equal size are rotated only in one direction D.5 degree. Four high mills 3. which brings the metal into the rolls. Diameter of working roll is very small Code:A B C D A B C D (a) 3 4 2 1 (b) 2 1 3 4 (c) 2 4 3 1 (d) 3 1 2 4 80 Page 74 of 210 IAS – 2003 In one setting of rolls in a 3‐high rolling mill.5 to 6. Middle Middl roll ll rotates t t by b friction f i ti B. ll Therefore. Reason (R): To prevent damage to the surface of the rolled products. il low‐ l (b) Offset Off t on the th rolls ll (c) Hardening of the rolls viscosity lubricants. p y Camber can be used to correct the roll deflection (at only one value of the roll force). y As each pair of planetary rolls ceases to have contact with the work piece.0 90 . buckle Edge breaks Non uniform Non‐uniform deformation Alligatoring Page 75 of 210 89 GATE – GATE – 2009 (PI) 2009 (PI) A i t Anisotropy i rolled in ll d components t is i caused d by b (a) changes in dimensions (b) scale formation (c) closure of defects (d) grain orientation Rev. and a pair of planishing rolls on the exit to improve the surface finish. rust. il emulsions l i and d fatty f acids id are used. products lubricants should be used. used (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true For-2016 (IES. y Each planetary roll gives an almost constant reduction to the (a) Large diameter rolls slab as it sweeps out a circular path between the backing rolls and the slab. mill y The operation requires feed rolls to introduce the slab into the mill. lubricants such as mineral oils. d y Cold C ld rolling lli l bi lubricants t are water‐soluble t l bl oils.. (b) Small diameter rolls ( ) High speed rolling (c) (d) Rolling without a lubricant R lli i h l b i 82 83 IES – 1993 Camber 84 Lubrication for Rolling In order to get uniform thickness of the plate by y Hot rolling of ferrous metals is done without a lubricant. 85 86 87 Defects in Rolling IAS – 2004 Defects Assertion (A): requires high which A i (A) Rolling R lli i hi h friction f i i hi h increases forces and power consumption. d i y The overall reduction is the summation of a series of small reductions d ti b each by h pair i off rolls. rolling ll process. Inclusions and scratches. oils emulsions. emulsions (d) Antifriction bearings paraffin and fattyy acids. another pair of rolls makes contact and repeat that h reduction. impurities p in the pits. one provides d y Hot rolling of non‐ferrous metals a wide variety of ( ) Camber (a) C b on the h rolls ll compounded d d oils.IAS – 2000 Planetary mill y Consist of a pair of heavy backing rolls surrounded by a large Rolling very thin strips of mild steel requires number of planetary rolls. centre Due to roll bending g edges g elongates more and buckle. cracks materials Wavy edges Strip is thinner along g its edges than at its centre. Th f th planetary the l t mill ill can reduce a slab directly to strip in one pass through the mill.GATE & PSUs) 88 What is Cause Surface Defects Scale. 936 For-2016 (IES.033 ((d)) 0. Formula in Rolling but are wavy. the reduction in blooming g mills is about 100 [10 marks] related to roll friction? mm and in slabbing mills.01 ((b)) 0.hf =2(R. y Reason (R) : Non uniform mechanical properties of the flat material rolled out result in waviness of the edges.031 3 ((a)) 0. 91 Draft 92 93 IAS‐2012 Main Angle of bite: y Total T l reduction d i or “draft” “d f ” taken k in i rolling.Rcos α) =D(1.0 99 .Geometry of Rolling Process IES‐2003 A Assertion ti (A): (A) While Whil rolling lli metal t l sheet h t in i rolling lli mill the edges are sometimes not straight and flat mill. The bite angle thickness The angle of bite in radians is thickness. rolls The angle subtended by the rollers each of diameter of 400 mm.600 ((c)) 0. deformation zone at the roll centre is (in radian) in degree will be ((a)) 0. and (2) Angle of nip: Δh=h0 .5 mm is initiall value l off 16 mm to a final f l value l off 10 mm in diameter steel rollers.062 (d) 0.02 (c) 0.06 (a) 5.cos α) angle l off bite b during d rolling ll operation? How are they h y Usually.936 (c) 8.936 (b) 7.GATE & PSUs) 97 Page 76 of 210 98 Rev. y. a strip of width 140 mm and b being rolled ll d with h 20% % reduction d off area using 450 one single pass rolling with a pair of cylindrical thickness h k 8 mm undergoes d 10% % reduction d off mm diameter rolls. about 50 to 60 mm.936 (d) 9. 94 GATE‐2007 95 96 GATE‐1998 GATE – 2012 Same Q in GATE – 2012 (PI) The thickness of a metallic sheet is reduced from an In a single pass rolling process using 410 mm A strip with a cross‐section 150 mm x 4.006 ((b)) 0. lli What is the significance of (1) angle of nip. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Rev. sheet of 25 mm thickness is L = R α rolled ll d to 20 mm thickness. T (c) S.1. rpm The roll strip contact [ α must be in radian] length will be ((a)) 5 mm ((b)) 39 mm ((c)) 778 mm ((d)) 120 mm 100 Maximum Draft Possible 101 sheet increases with the 2 =μ R (a) increase in coefficient of friction (b) decrease d in coefficient ff off friction f ( ) decrease (c) d i roll in ll radius di (d) increase i i roll in ll velocity l it 103 GATE 2015 GATE-2015 GATE 2015 GATE-2015 500 mm . process the maximum possible draft. h t mainly i l depends on which pair of the following parameters? P: Strain Q: Strength of the work material R: Roll diameter S: Roll velocity T: Coefficient of friction between roll and work (a) Q.0 108 . Reason (R): The reduction possible in the second pass is less than that in the first pass.GATE & PSUs) 102 GATE 2014 GATE 2011 The maximum possible draft in cold rolling of ( Δh )max μ ≥ tan α 107 Assertion (A): mill is A i (A) In I a two high hi h rolling lli ill there h i a limit to the possible reduction in thickness in one pass. reduced to less than 20 mm in one pass. h k Roll ll is off diameter d 600 mm and it rotates at 100 rpm. plate is _______ If µ = 0. T (d) P. draft defined as the difference between the initial and th final the fi l thickness thi k off the th metal t l sheet. the maximum angle subtended by the deformation zone at the centre of the roll (bite angle l in degrees) d ) is _____ Page 77 of 210 105 IES – 1999 In operation . S (b) R. the I a slab l b rolling lli i h maximum i 106 In a rolling process. if a 25 mm thick thi k plate l t cannott be b thickness reduction(∆h)max is given by ∆hmax = µ µ²R R.Roll strip contact length Roll strip contact length GATE‐2004 For Unaided entry y Roll strip contact length R ll i l h In a rolling process. R 104 In a rolling operation using rolls of diameter For-2016 (IES. the where R is the radius of the roll and µ is the co‐ coefficient of friction between the roll and the efficient of friction between the roll and the sheet. i t the th work k material t i l has h rough h surface finish and on the other side. the n= Δhrequired q GATE‐1990 (PI) GATE‐1990 (PI) Whil rolling While lli a strip t i the th peripheral i h l velocity l it off the roll is …. what is the minimum coefficient of friction required for unaided rolling to be possible? (a) 0. by using rolls of 800 mm diameter.255 ((d)) 0.0 mm (b) 1.158 (c) 0. i t Vf = final or output velocity Vr − Vo × 100% Vr A 80 with 8 mm thick hi k steell plate l i h 400 mm width id h is i rolled to 40 mm thickness in 4 passes with equal reduction in each pass.GATE & PSUs) Vr Velocity of the strip > Velocity of the roll 114 IES 2014 IES ‐ The point Th i where h roll ll velocity l i equals l work velocity is known as the no‐slip V0 = input velocity point i t or the th neutral t l point.55 ((c)) 0.077 113 Neutral Point and Neutral Plane V f − Vr R roll radius R = roll radius ho = back height hf =output or final fi l thickness α = angle of bite N N neutral point or no N‐N = neutral point or no‐ slip point To the left of the Neutral Point: Velocity of the strip < Velocity of the roll To the right of the Neutral Point: Forward slip = ×100% 115 For-2016 (IES. (a) less than/greater less (b) Greater than/less than Page 78 of 210 116 In the process of metal rolling operation.than A than the entry velocity of the strip p and is ……B ….Minimum Possible Thickness (h f min ) GATE‐2006 Number of pass needed A 4 mm thick sheet is rolled with 300 mm diameter ho − h f min = μ 2 R rolls ll to reduce d thickness h k without h any change h in its width The friction coefficient at the work‐roll width.. the work material is in compression p (b) On one side of this point.1.0 . Assume that there is no diameter 300 mm. Assuming g the p plane‐strain deformation. but on either side of neutral point.. along the arc off contact in i the h roll ll gap there h is i a point i called ll d the h neutral point. side it has velocity lesser than that of the roll ( ) On (c) O one side id off this thi point.5 mm (d) 3..7 mm 109 IES – 2001 A strip is to be rolled from a thickness of 30 mm to mm by successive cold rolling passes using identical 15 mm using a two‐high h h mill ll having h rolls ll off rolls of diameter 600 mm.1..111 (b) 0.the exit velocity y of the strip.5 mm (c) 2.223 (d) 0.35 35 ((b)) 0. mm The coefficient of friction for change in width. the material width increases 117 Rev.316 (d) 7 112 Backward slip = 111 GATE‐2014(PI) The off a plate is 30 mm to 10 Th thickness hi k l i reduced d d from f minimum number of passes required is Δhmax 110 G 20 ( ) GATE – 2011 (PI) between the rolls and the work piece is 0. If the coefficient of friction unaided bite should nearly be (a) 3 (b) 4 (c) 6 ((a)) 0. the work material is in tension and on the other side. the work material has very fine fi finish fi i h (d) At this p point there is no increase in material width.A…. work roll interface is 0. because (a) On one side of this point. the work material has velocity greater than that of the roll and on the other side. The minimum possible thickness of the sheet that can be produced in a single pass is (a) 1. 5 Lp for o hot ot rolling o g from an ingot of 750 mm x 750 mm are For-2016 (IES. thickness of a strip is reduced In rolling a strip between two rolls.20 (d) disadvantageous after the neutral point 118 119 C ti it E Continuity Equation ti Elongation Factor or Elongation Co‐efficient GATE‐2014 y Generally rolling increases the A mild steel plate has to be rolled in one pass such work width from an initial value of bo to a final one of bf and this is called spreading. Nm N 1000 Total power for two roller ( P ) = 2Rev.57 57 ((b)) 33. N ingot is 1.e. y The Th inlet i l and d outlet l volume l rates of material flow must be the h same. inW T Torque per roller ll (T ) = F × 126 .IES – 2002 GATE ‐2008(PI) Selected Questions In a rolling process. the exit velocityy ((in m/min) / ) hobovo = hf bfvf 121 f n − pass for 123 RollSeparating p g Force ( F ) = σ o × L p × b . mm GATE‐1992(PI) (c) 10 L1 Ao = Lo A1 122 Force. mm If the plate widens byy 2% during g rolling. mm 2 If the elongation factor during rolling of an (d) 12 En = for single pass Projected length ( L p ) = R sin α = RΔh . Torque and Power (b) 9 E= is …………… where vo and vf are the entering g and exiting velocities of the work. MPa ] needed d d to produce d a section i 250 mm x 250 mm Arm length e gt ( a in mm ) = 0. g. The minimum number of passes [σ o in N / mm 2 i.0 T ω .14 4 ((c)) 47 47. hi k with i h the h entrance speed d off 10 m/min / i and roll diameter of 500 mm.GATE & PSUs) Ln Ao = Lo An Projected j Area ( Ap ) = Lp × b . the position of The effect of friction on the rolling mill is f from 4 mm to 3 mm using 300 mm diameter d rolls ll the h neutrall point in the h arc off contact does d not (a) always bad since it retards exit of reduced metal rotating at 100 rpm. that h is.22. (a) 8 120 = 0. i that the final plate thickness is 2/3rd of the initial thickness.10 ((d)) 94 94.45 Lp for cold rolling 124 Page 79 of 210 Will b be discussed in class 125 a . rpm The velocity of the strip in depend on (b) always good since it drags metal into the gap between (m/s) at the neutral point is (a) Amount of reduction (b) Diameter of the rolls th rolls the ll (c) Coefficient of friction (d) Material of the rolls (c) advantageous before the neutral point ((a)) 1. Taking smaller reductions per pass to reduce the contact area. Using Ui l large di diameter rolls ll to increase i the h contact area. The power required for the rolling operation in kW is closest to (a) 15. Reducing friction 2.0 135 . 5.2 (b) 18. t sin i θ ≅ θ and d cos θ = 11. 3 The arc of contact is circular with a radius greater than 3.GATE‐2008 IES – 2000.4 30 4 (d) 45.2 (c) 30. 3. Rolls R ll are straight. contact and acts in one direction throughout the arc of contact. The material is plastic.. Strip p is wide compared p with its thickness. li d 2. The material is rigid perfectly plastic (constant yield st e gt ).2pR dθ sin θ σ x h + (σ x +dσ + 2 τ x R dθ cos θ = 0 133 For-2016 (IES. thus σ 0' h nearly a constant and itsderivative zero. 3 Coefficient of friction is constant over the arc of 3. unity We get equilibrium equation in -x‐direction as. strength). is reduced to 18 mm. x ) (h + dh) . [For IES Conventional Only] the radius of the roll. d (σ x h ) = 2 pR (θ ∓ μ ) dθ 2 p −σ x = σ 0 = σ 0' 3 d ⎡ h ( p − σ 0' ) ⎤ = 2 pR (θ ∓ μ ) ⎦ dθ ⎣ ⎞⎤ d ⎡ ' ⎛ p ⎢σ 0 h ⎜ ' − 1 ⎟ ⎥ = 2 pR (θ ∓ μ ) dθ ⎣ ⎝σ0 ⎠⎦ d ⎛ p σ 0' h ⎜ d θ ⎝ σ 0' Which assumptions are correctt for Whi h off the th following f ll i ti f cold rolling? 1. The roller radius is 250 mm and rotational speed p is 10 rpm. p The average g flow stress for the p plate material is 300 MPa. i h rigid i id cylinders. 2 and 3 132 d ( p / σ 0' ) 2R dθ = (θ ∓ μ ) p / σ 0' h h = h f + 2 R (1 − cos θ ) ≈ h f + Rθ 2 d ( p / σ 0' ) ( p /σ ) ' 0 134 = 2R (θ ∓ μ ) dθ h f + Rθ 2 Integrating both side 2 Rθ dθ ln ( p / σ 0' ) = ∫ ∓ h f + Rθ 2 ∫h 2 Rμ dθ = I ∓ II ( say ) 2 f + Rθ Rev. 1 2 and 3 128 129 IES – 2001 Assumptions in Rolling 1. σ 0' increases as h decreases. τ x = μp Simplifying and neglecting secondd order d terms. so that no widening of strip occurs (plane strain conditions). contact Select the correct answer using the codes given below: Codes: d ((a)) 1 and 2 ((b)) 1 and 3 (c) 2 and 3 (d) 1. a 20 mm thick I a single i l pass rolling lli i hi k plate with plate width of 100 mm. we gett ⎞ ⎛ p ⎞ d (σ 0' h ) = 2 pR (θ ∓ μ ) ⎟ + ⎜ ' − 1⎟ σ d θ ⎠ ⎝ 0 Page ⎠ 80 of 210 Due to cold rolling. The arc of contact is circular with a radius g greater than the radius of the roll. 2. 4. 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. i t f 130 Stress Equilibrium of an Element in Rolling Considering the thickness of the element perpendicular to the plane of paper to be unity.6 IAS – 2007 In the rolling process. GATE‐2010(PI) In operation.GATE & PSUs) 131 For sliding friction. The co‐efficient of friction is constant over the tool‐ work k interface. roll separating force can be d decreased d by b ( ) Reducing (a) R d i the h roll ll diameter di (b) Increasing I i the th roll ll diameter di t (c) Providing back‐up back up rolls (d) Increasing the friction between the rolls and the metal 127 Consider C id the h following f ll i statements: g can be reduced byy Roll forces in rolling 1. σ '0 ⎜ o ⎟ e− μHo ⎝R⎠ R μHo and C = . M.tan −1 hf 140 IAS – 1998 Calculate the neutral plane to roll 250 mm wide annealed copper pp strip p from 2. [10‐marks] For-2016 (IES.S. angles and channels 1. e ( 0 n ) = n . eμ Hn h0 hf If back tension σ b is there at Entry.0 144 .1414 5.e ho ⎛h⎞ p = C σ '0 ⎜ ⎟ e∓ μH ⎝R⎠ where H = 2 2Rμ R II = ∫ dθ h f + Rθ2 R . p = C. No 1 2 3 4 5 6 7 8 9 10 Rolling Ch‐14 Option C B D D A A B B C C Q. ⎞⎤ ⎟⎥ ⎠ ⎥⎦ ⎛ R ⎞ .θ ⎟ ⎜ ⎟ hf ⎝ hf ⎠ h μ H −H .GATE & PSUs) What is friction hill ? What is "friction hill" ? h .05 and σ’o =180 MPa.71 Rev. Forging C Roof trusses C. Carburetors 2. ⎜ R 2 ⎟⎟ R ⎝ ⎠ and h n = h f + 2R (1 − cos θn ) From H = 2 137 Match M h List Li ‐ I (products) ( d ) with i h List Li ‐ II (processes) ( ) and select the correct answer using the codes given below the lists: List – I List ‐II II A.θ = α Hence H = H0 with θ replaced by ∝ in above equation 2μ dθ =∫ h f / R + θ2 = 2μ ⎛h⎞ ln p / σ '0 = ln ⎜ ⎟ ∓ 2μ ⎝R⎠ I th In the exit it zone ⎛ h ⎞ p = σ '0 ⎜ ⎟ . Entry p = ( σ ′o − σ b ) ho μ H − 2H or = e ( 0 n) hf or Hn = ⎛ h0 1⎡ 1 ⎢H0 − ln ⎜ 2 ⎢⎣ μ ⎝ hf p = ( σ ′o − σ f ) R .θ ⎟ + ln C ⎜⎜ ⎟ ⎝ hf ⎠ ⎛h ⎞ In the entry y zone.tan −1 ⎜ .θ ⎟ ⎜⎜ ⎟ ⎝ hf ⎠ ⎛ h f Hn ⎞ hf ∴ θn = .θ ⎟ ⎜⎜ ⎟ ⎝ hf ⎠ p = σ '0 Now at entry .0 mm thickness with 350 mm diameter steel rolls. No 11 12 13 14 15 16 17 18 19 20 Option C C B C B C B 14 7 14. e ( 0 ) h0 will give same results 136 hn h μ H −H .I= 2Rθdθ = 2 f + Rθ ∫h ∫ Now h / R = or 2Rθdθ = h 2θdθ ∫h/R ⎛h⎞ = ln ⎜ ⎟ ⎝R⎠ ∴ hf + θ2 R ∴ d ⎛h⎞ = 2θ θ\ dθ ⎜⎝ R ⎟⎠ ( ) R . Take µ = 0.tan −1 hf ⎛ R ⎞ . Welding B. Rolling Codes: A B C D A B C D (a) 1 2 3 4(b) 4 3 2 1 of 2104 (c) 1 2 Page 4 813(d) 3 1 2143 141 WorkBook Q. e ( 0 ) h0 If front tension σ f is there at Exit. 3 3. Gear wheels 4.tan ⎜ .7 0.eμH ⎝ hf ⎠ At the neutral po int above equations At exit θ = 0 Therefor p = σ '0 ⎛ R ⎞ R . Casting D..tan −1 hf ⎛ R ⎞ .55 mm to 2. eμ H hf 139 IFS – IFS – 2010 2010 142 138 IAS ‐2012 Main h μ H −H . d d y Mechanical properties and reliability of the materials increases due to improve in crystal structure. which increases strength and toughness. Part are shaped by plastic deformation of material p yp y Forging operations are limited to simple shapes and has limitations for parts having undercuts. pair Statement (II): The material is pressed between two surfaces and the compression force applied. h hammer and d shape h i is y In closed die forging. Original unidirectional fibers are distorted (b) Low L tooling t li costt 3 Poor reliability as flaws are always there due to intense 3.Forging Closed Die forging g g g g p g p y Forging process is a metal working process byy which metals or alloys are plastically deformed to the desired shapes by a compressive force applied with the help of a pair of dies. For-2016 (IES. y In open die forging. y Forging reduces the grain size of the metal. Poor reliability.GATE & PSUs) 151 Page 82 of 210 152 (a) 1. 149 Disadvantages of Forging d f 150 IES‐2013 IES – 1996 y Costly C l Which one of the following is an advantage of In the forging process: I h f i y Poor dimensional accuracy and surface finish. y Discrete shape off product can be Di h d b produced. finish forging? 1. gives it a shape. Forging By S K Mondal y 146 145 IES‐2013 Statement (I): The dies used in the forging process are made in pair. 2. 3 and 4 153 . as flaws are always there due to intense (c) Close tolerance working g (d) Improved physical property. such that the best mechanical properties p p can be obtained based on the specific p application. p (a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement ((II)) is not the correct explanation p of Statement (I) (c) Statement (I) is true but Statement (II) is false 148 (d) Statement (I) is false but Statement (II) is true 147 Open and Closed die forging d l dd f Advantages of Forging d f y Depending upon complexity off the is D di l i h part forging f i i carried out as open die forging and closed die forging. the desired configuration is obtained b i d by b squeezing i the h workpiece k i b between two shaped and closed dies. 2 and 3 (b) 1. re‐entrant surfaces etc surfaces. 2 and 4 Rev. property 4 4.0 (d) 2. toughness y Fatigue and creep strength increases. it is possible to closely control the grain flow in the specific direction. y Because B off the th manipulative i l ti ability bilit off the th forging f i process. 3 and 4 (c) 1. the metal is compressed by repeated blows by bl b a mechanical h i l manipulated manually. The metal structure is refined Th t l t t i fi d ( ) Good (a) G d surface f fi i h finish 2 Original unidirectional fibers are distorted. y In forging favorable grain orientation of metal is obtained that strengthen the component but forging distorts the previously created uni‐directional fibre. easily (b) Forgeability decreases with temperature upto lower criticall temperature. ll bilit therefore. (c) Ce Certain ta mechanical ec a ca p properties ope t es o of tthee material ate a aaree influenced by forging. bili y Forgeability increases with temperature. by which forging can be done easily. l y Internal surfaces require more draft than external surfaces. y Adequate draft should be provided‐at least 3o for aluminum l i and d 5 to 7o for f steel. Contd… y A flash for blows from the fl h acts as a cushion hi f impact i bl f h finishing impression and also helps to restrict the outward flow of metal. l the h externall surfaces f are likely lik l to be separated. g 2. y The amount of flash depends on the forging size and may vary from 10 to 50 per cent. Which of the statements given above is/are correct? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 154 Forgeability bl Which processes induce more Whi h off the h following f ll i i d stress in the metal? (a) Hot rolling (b) Forging (c) Swaging ( ) Turning (d) 156 IES ‐ 2012 y The off a metall can be Th forgeability f bili b defined d fi d as its i capability to undergo deformation by forging without cracking. 157 y The Th draft d f provided id d on the h sides id for f withdrawal i hd l off the h forging. (d) Pure P metals t l have h good d malleability. y Upsetting test and Hot‐twist test are used to determine f forgeability. temperature Draft f Which statements is Whi h off the h following f ll i i correct for f forging? (a) Forgeability is property of forging tool. y Metal M l which hi h can be b formed f d easily il without ih cracking. th f poor forging properties. flow (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true For-2016 (IES. During cooling. Forged components can be provided with thin sections. y The Th forging f i l d can be load b decreased d d by b increasing i i the h flash thickness. ki with ih low force has good forgeability. without reducing the strength. thus helping in filling of thin ribs and bosses in the upper pp die. Statement (II): Forging is workable for simple shapes and has limitation for parts having undercuts. Page 83 of 210 161 Rev. Forging reduces the grain size of the metal. 158 159 Flash l h IES – 2006 Assertion (A): dies A i (A) Forging F i di are provided id d with i h taper or draft angles on vertical surfaces. Reason (R): It facilitates complete filling of die cavity and favourable grain flow. which results in a decrease in strength g and toughness. forging tends to shrink towards i centre and its d as a result.IES 2005 IES – IES ‐ 2012 Consider the following statements: 1.GATE & PSUs) ISRO‐2013 Statement (I): It is difficult to maintain close tolerance in normal forging operation.0 162 . whereas the internal surfaces tend to cling to the die more strongly. undercuts (a) Both Statement (I) and Statement (II) are i di id ll true and individually d Statement S (II) is i the h correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individuallyy true but Statement ((II)) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true 155 160 Flash l h The Th excess metall added dd d to the h stock k to ensure complete l filling of the die cavity in the finishing impression is called Flash. Flash is excess material added to stock which flows around parting line. (a) Both A and R are individually true and R is the correct co ect eexplanation p a at o o of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A is false but R is true (d) 167 IES‐2015 For-2016 (IES. Gutter G tt is i provided id d to t ensure complete l t closing l i off the th die. provision is also to be made in the forging di for die f additional dditi l space called ll d gutter. Amount of flash depends upon forging force. Gathers the material as required in the final forging. fl h provision i i should h ld be b made d in i the h 170 Fullering or swaging Edging or rolling Reducing cross section and making it longer. material ( ) Closed (a) Cl d die di forging f i (b) Centrifugal C t if l casting ti (c) Investment casting (d) Impact extrusion 166 of flash gutter and flash land around the parts to be At th l t h At the last hammer stroke the excess material from t k th t i l f the finishing cavity of a forging die is pushed into……………. This is called g gutter. the finishing impression Blocking Semi‐finishing impression.. GATE‐1994(PI) Contd…. 2 and 3 ((d)) 2 and 3 onlyy Page 84 of 210 168 Sequential steps involved in closed die forging Which of the following statements apply to provision GATE‐1989(PI) Assertion (A): forging A ti (A) In I drop d f i besides b id the th provision i i for flash. y Without a gutter.IES ‐ 2014 Gutter IAS – 2002 In thin off material I hot h die di forging. which indicates the quality of the forging (d) Cavity. 164 165 IES – 1993. which fills up hot gases (b) Flash. 169 forged? 1. around parting line 2. not allowing ll the h dies d to close l completely. Flash the width of it is an indicator of the pressure developed in the cavity (c) Coining. 3. f i hi layer l i l all ll around d the forging is (a) Gutter space.GATE & PSUs) y In I addition ddi i to the h flash. extra material. 171 Rev. Cavity which is filled with hot impurities in the material Consider the following statements related to C id h f ll i l d forging: 1.0 Trimming or cut off Removal of flash present around forging . about 20%‐25% of the volume of forging. 3. ou t o as depe ds upo o g g o ce. Which of the above statements are correct? ( ) 1. Flash land and Gutter provided to the die. 2 and 3 (a) d (b) 1 and 2 d (c) 1 and 3 (d) 2 and 3 163 Gutter die for additional space so that any excess metal can flow and help in the complete closing of the die. Small cavities are provided which are directly outside the die impression.. l l Which one of the following IES – 1997 manufacturing processes requires the h provision off ‘gutters’? ‘ ’ y Gutter G d h and depth d width id h should h ld be b sufficient ffi i to accommodate the extra. (a) 1 and 2 only (b) 1 and 3 only ((c)) 1. Preform shape. Bending Required for those parts which have a bent shape p Drawing or cogging Like fullering but c/s of only one end is reduced Flattening Flatten the stock so that it fits properly into the finishing impression. tt Reason (R): The gutter helps to restrict the outward flow of metal thereby helping to fill thin ribs and bases in the upper die. Flash helps in filling of thin ribs and bosses in upper die. 2 The volume of flash land and flash gutter should be 2. Imparts to the f i it’ forging it’s general but not exact or final shape. a flash may become excessively thick. l b t t t fi l h Finishing Final impression. fullering is done to commonly l used d for f the h forging f off bolt b l heads h d off (a) Draw out the material hexagonal shape? (b) Bend the material (a) Closed die drop forging ( ) Upset the material (c) (b) Open die upset forging (d) Extruding the material E di h i l (c) Close die press forging ((d)) Open p die p progressive g forging g g 172 IES – 2003 and d in the h process making k it longer l is termed d as (b) P Punching hi ( ) Upsetting (c) U tti (d) E t di Extruding IES – 2005 Which of the following processes belong to forging operation p ? 1. 3. 4 Edging Select the correct sequence of these operations using the codes given below: Codes: (a) 1‐2‐3‐4 (b) 2‐3‐4‐1 (c) 3‐4‐1‐2 (d) 4‐1‐3‐2 Page 85 of 210 177 179 Match List I (Forging operations) with List II (Descriptions) and select the correct answer using the codes given below the Lists: List I List II A. Thickness is reduced continuously at diff different sections i along l l length h B.Example l IES – 1998 IES – 2001 Which one of the following processes is most In the forging operation. 4 Fullering 5 Edging 5. 3. Drawing 2. 1 4. 2 and 3 only 175 IES – 2002 178 The process of removing the burrs or flash from a f forged d component in drop d f forging is called: ll d ( ) Swaging (a) S i (b) ( ) Trimming (c) Ti i (d) P f Perforating i F ttli Fettling 176 IES – 2003 Consider steps involved C id the h following f ll i i l d in i hammer h forging a connecting rod from bar stock: 1.0 180 . Blocking 2. Wire drawing 4. 5 1. Which of the following is the correct sequence of operations? (a) 1. Welding g (a) 1 and 2 only (b) 2 and d 3 only l (c) 1 aand d3o onlyy (b) 1. 3 and 2 (c) 5.GATE & PSUs) 174 IES 2011 A forging method for reducing the diameter of a bar ( ) Fullering (a) F ll i 173 IAS – 2001 Consider C id the h following f ll i steps in i forging f i a connecting i rod from the bar stock: 1. 1. 4 3. Flattening 1. 3. Fullering g 33. Pressure a workpiece between two flat dies Codes:A B C D A B C D (a) 3 2 1 4 (b) 4 1 2 3 ( ) 3 (c) 1 2 4 ( ) (d) 4 2 1 3 Rev. 2 and 1 (d) 5. 5. Swaging S i 33. 1 4. Trimming 3 Finishing 4. 4 2 and 3 For-2016 (IES. Fullering 2. Metal is displaced away from centre. 4. Rod is p pulled through g a die D. reducing thickness in middle and increasing length C. 3 2 and 5 (b) 4. Trimming 3 Finishing 4. Blocking 2. 2. 4.0 189 .GATE & PSUs) the upper half is fixed to the ram. Advantages of Press Forging over Drop Forging y Press forging is faster than drop forging y Alignment of the two die halves can be more easily y Structural St t l quality lit off the th product d t is i superior i to t drop d forging. h while hl Page 86 of 210 188 186 IES 2011 IES 2011 Consider the following statements : 1.IES‐2012 Conventionall In I forging f i define d fi the h terms ((i)) Edging g g (ii) Fullering and d (iii) Flash Fl h 181 IAS – 2003 Drop Forging Match Operation) off the M h List Li I (Forging (F i O i ) with i h List Li II (View (Vi h Forging Operation) and select the correct answer using the codes d given i b l below th lists: the li t List‐I List‐II (Forging Operation) (View of the Forging Operation) g g (A) Edging 1 1. the complete cavity is formed. y Metal is squeezed gradually by a hydraulic or mechanical h lf off the half h die d is fixed f d to the h anvill off the h machine. For larger work piece of metals that can retain toughness h at forging f i temperature it i is i preferable f bl to use forge press rather than forge hammer. 183 Drop forging D f i is i used d to produce d ((a)) Small components p (b) Large components ( ) Identical (c) Id i l Components C i large in l numbers b ((d)) Medium‐size components p 185 maintained i i d than h with i h hammering. hence the dimensional accuracy is much better than d drop f i forging. in quick succession so that the metal spreads and completely fills the die cavity. forging y With ejectors in the top and bottom dies. Any metal will require some time to undergo complete p plastic deformation p p particularly y if deforming metal has to fill cavities and corners of small radii. When the two die halves close. it is possible to handle reduced die drafts. (B) Fullering (C) Drawing 3. The lower IES – 1994. (D) Swaging C d A B Codes:A 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 1 182 Click to see file Page 4 – 5 ‐6 y The drop forging die consists of two halves. ( ) The (c) Th die di with i h hammer h at high hi h velocity l i ((d)) a weight g on hammer to p produce the requisite q impact. ram The heated stock is IAS – 2000 In forging. 2. h i For-2016 (IES. forging by I drop d f i f i is i done d b dropping d i ((a)) The work p piece at high g velocityy (b) The hammer at high velocity. y Drop forging is used to produce small components. 187 kept in the lower die while the ram delivers four to five blows on the metal. (a) 1 and 2 are correct and 2 is the reason for 1 (b) 1 and 2 are correct and 1 is the reason for 2 (c) 1 and 2 are correct but unrelated (d) 1 only correct Rev. ISRO‐2010 184 Press Forging g g press and component is produced in a single closing of die. ball 193 Smith Forging h IES – 2005 y Blacksmith Bl k i h uses this hi forging f i method h d y Quality of the product depends on the skill of the operator. y When the rolls are in the open position. the h materiall is only l upset forging g g ? Whyy are p pure metals more easilyy cold worked to get the desired shape. Squeezing action p Forging g g 33. hammers in blows D. D F i Forging 1.GATE & PSUs) 194 196 Match off Forging) M t h List Li t I (Type (T F i ) with ith List Li t II (Operation) (O ti ) and select the correct answer using the code given below the Lists: List I List II A Drop A. roll down the stock. Press Forging 3 Done in closed impression dies by 3. y Not used in industry. Done in closed impression dies by continuous squeezing q g force Code: A (a) 2 (c) 2 B 3 1 C 4 4 D 1 3 (b) (d) A 4 4 B C 3 2 1 2 Rev. stock The stock is transferred to a second set of grooves. The rolls turn again and so on until the piece is finished. Repeated p hammer blows A B C D A B C D (a) 4 1 2 3 (b) 3 2 1 4 (c) 4 2 1 3 (d) 3 1 2 4197 Page 87 of 210 195 IES – 2008 IES Match List‐I with List‐II and select the correct answer using the code given below the lists: List‐I (Forging Technique) List‐II (Process) A Smith Forging A. Metal is p placed between rollers and C. cavities [5‐marks] 190 191 Roll Forging ll 192 Roll Forging ll Contd…. M l is Metal i gripped i d in i the h dies di and d pressure is applied on the heated end B.0 D 1 3 198 . Skew Rolling y Skew rolling produces metal ball is advanced d d up to a stop.Carried out manually open dies C Press Forging C. As the h rolls ll rotate. Press Forging 2. Upset pushed D. Drop Forging 2. Roll Forging g g 4 4. in machine h f forging. Material is only upset to get the desired shape B. Machine Forging 4. they h grip and d y Round stock is fed continuously to two p y designed g specially opposing rolls. For-2016 (IES. h y Employs E l split li dies di that h contain i multiple l i l positions ii or than alloys ? cavities. Smith Forging 11.IFS‐2011 Machine Forging h Upset Forging y Unlike the drop or press forging where the material is y Increasing the diameter of a material by compressing its What advantages does press forging have over drop d drawn out. y Metal is forged by each of the grooves in the rolls and emerges from the end as a metal ball. the heated stock y A rapid id process. shape l length. y Cold Shut or fold: A small crack at the corners and at right g angles g to y For cold forging: g g mineral oil and soaps. C Compare S ith forging. Cause: improper y In hot forging. Al. y Ram strokes short (due to high acceleration) y Productivity high. Heat treated 4. 199 Statement (I): In high velocity forming process. rocket component. it is applied to the&workpiece. y For hot forging: graphite. overall production cost low y Used for Alloy steel. wear. f ll d due d to improper y Die Shift: Due to Misalignment of the two die halves or thermal barrier. barrier design of die making the two halves of the forging to be of improper shape. press forging and upset forging. y Improper Grain Flow: This is caused by the improper design of the die. (a) Both Statement (I) and Statement (II) are individually true and Statement ((II)) is the correct explanation p of Statement (I) (b) Both B th Statement St t t (I) and d Statement St t t (II) are individually i di id ll true but Statement (II) is not the correct explanation of Statement (I) () (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true 200 IFS‐2011 Write four advantages of high velocity forming process. 205 Page 88 of 210 206 Rev. Polished Wh t is What i the th correctt sequence off the th above b operations from start? (a) 3‐2‐4‐1 (b) 3‐2‐1‐4 (c) 2‐3‐1‐4 (d) 2‐3‐4‐1 203 Lubrication for Forging b f Forging Defects f 204 Forging Defects f Contd…. y Lubricants influence: friction. tolerance for each. die which makes the flow of metal not flowing the final intended directions. to fabricate one piece complex components of smaller size like valve. ) y Cost and size of machine low. k completely l l filled.GATE PSUs) improper cleaning of the stock. y As K. y The process deforms using veryy high velocities. the lubricant is applied to the dies. Smith f i d drop f i forging. provided on the movements of rams and dies. high energy is delivered to the metal with relativelyy small weights g ((ram and die). d non‐sticking. [2‐marks] 201 For IES Only Flashless forging y The work material is completely surrounded by the die IAS‐2011 Main IAS‐2011 Main cavity during compression and no flash is formed. deforming forces y Unfilled Sections: Die cavity is not y Scale Irregular on the due S l Pits: Pi I l depressions d i h surface f d to and d flow fl off materiall in die‐cavities. Mg. titanium.0 207 . high gy can be transferred to metal with relativelyy small energy weight. but in design of the die cold forging.For IES Only IES‐2013 Hi h V l it F i (HVF) High Velocity Forming (HVF) ep ocess de o s metals eta s by us g ve g ve oc t es. Ground 2 Hot forged on hammers 2. 3. forging Mention three points y Most important requirement in flashless forging is that the work volume must equal the space in the die cavity to a very close tolerance. [10 – Marks] 202 IES – 2008 The are manufactured Th balls b ll off the h ball b ll bearings b i f d from steel rods.E ∞ V2. graphite MoS2 and sometimes molten g glass. The operations involved are: 1. Statement (II): The kinetic energy is the function of mass and velocity. y Flakes: l k Internall ruptures caused d by b the h improper cooling. p the forged surface. For-2016 (IES. 0 216 . (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A ( ) A is true but (c) b R is false f l ((d)) A is false but R is true 209 210 B lli Barrelling GATE ‐2008 (PI) GATE ‐2008 (PI) IES ‐ IES ‐ 2007 Match the following Group ‐1 P Wrinkling P . Q (c) P 2 Q – 3. Cold shut S Sometimes ti th parting the ti plane l b t between t two f i forging Group‐2 1 Upsetting 1. Which of these statements are correct? (a) 1 and 3For-2016 (b) 1 and 4 (c) 2 & and 3 (d) 2 and 4214 (IES. Upsetting 2. S 1 S‐4 4 (b) P – 3. flow y Hot tears and thermal cracking: These are surface cracks occurring g due to non‐uniform cooling g from the The due to smooth Th forging f i defect d f d to hindrance hi d h flow fl of metal in the component called 'Lap' occurs because (a) The corner radius provided is too large (b) The corner radius provided is too small (c) Draft is not provided (d) The shrinkage allowance is inadequate forging stage or during heat treatment. Q.GATE PSUs) Die Materials Should have l h ld h GATE ‐2010 (PI) y Good hardness. y Forging Laps: These are folds of metal squeezed together h during d f forging. Barrelling S C ld h t S. Deep drawing 3. Q – 3. S‐1 (c) P – 2. p whyy is p parting g p plane provided. toughness and ductility at low and Good hardness toughness and ductility at low and Hot die steel. R 4 R – 3. 2. used for large solid dies in drop forging. R – 1. They h have h irregular l contours and occur at right angles to the direction of metal flow. 208 IES 2011 Assertion (A) ( ) : Hot tears occur during d forging f because of inclusions in the blank material Reason (R) : Bonding between the inclusions parent material is through g p physical y and the p and chemical bonding. S 3 S‐1 1 211 212 Inhomogeneous deformation with barreling of the workpiece 213 IES‐2013 Consider C id the h following f ll i statements pertaining i i to the h open‐die forging of a cylindrical specimen between two flat dies: 1 Lubricated specimens show more surface movement 1. Lubricated specimens show more barrelling than un‐ lubricated ones. R 3 R – 1. than un‐lubricated ones. Q – 4. Lubricated b d specimens show h l less surface f movement than un‐lubricated ones. Lubricated specimens shows less barrelling than un‐ lubricated ones. plane give the main reason for this design g aspect.Forging Defects f IAS – 1998 Contd…. 3. Centre burst R. S‐2 (d) P – 2. elevated temperatures p should h ld necessarily l have h y Adequate fatigue resistance ( ) high (a) hi h strength h and d high hi h copper content y Sufficient hardenability y Low thermal conductivity (b) high hi h hardness h d and d low l hardenability h d bilit y Amenability to weld repair A bili ld i (c) high toughness and low thermal conductivity y Good machinability (d) high hardness and high thermal conductivity Material: Cr‐Mo‐V‐alloyed tool steel and Cr‐Ni‐Mo‐ y alloyed tool steel. in closed die forging? [ 2 marks] (a) P – 2. Extrusion 4. Closed die forging Cl d di f i dies is not a horizontal plane. R – 4. Q (d) P 2 Q – 4. ones 4. Page 89 of 210 215 Rev. it ) The capacity p y of the hammer is g given byy Statement ((II): the total weight.For IES Only For IES Only IES 2013 IES‐2013 Statement (I): In power forging energy is provided by compressed d air i or oil il pressure or gravity.602 σ = Kε n Where K is strength coefficient and n is strain‐hardening (or work‐hardening) work hardening) exponent and at UTS.686 ( ) 1.693 0 693 (d) 1.0 225 . which the falling pans weigh.986 (b) 1. (a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of S Statement (I) ((b)) Both Statement ((I)) and Statement ((II)) are individuallyy true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true 217 218 True Stress & True Strain GATE‐2014 True stress (σ T ) = σ (1 + ε ) ⎛ L⎞ ⎛A True strain (ε T ) = ln(1 ( + ε ) = ln ⎜ ⎟ = ln ⎜ o ⎝ A ⎝ Lo ⎠ GATE‐1992. GATE‐2007 by the flow curve: In a disc 200 mm and I open‐die di forging. The true strain is (a) 1.GATE & PSUs) 223 Page 90 of 210 224 Rev.0 221 222 Strain Hardening & Flow Stress region the material behaviour is expressed y In the plastic region. For-2016 (IES. VS‐2013 The relationship between true strain (εT ) and ⎞ ⎛ do ⎞ ⎟ = 2 ln ⎜ d ⎟ ⎠ ⎝ ⎠ engineering strain (ε ( E ) in a uniaxiall tension test is ( ) E = ln(1 + ε (a) ε l ( T ) (b) E = ln(1 ‐ (b) ε l ( εT ) ( ) εT = ln(1 + ε (c) l ( E ) (d) T = ln(1 ‐ (d) ε l ( εE ) 220 219 The true strain for a low carbon steel bar which is doubled in length by forging is (a) 0.386 (c) (d) 0. stress but on an average value over the stress – strain curve from the beginning of strain to the final (maximum) value that occurs during deformation.5 (c) 0.307 (b) 0. f i di off diameter di d height 60 mm is compressed without any barreling effect. ISRO‐2012. The final diameter of the disc is 400 mm. Average g flow stress (σ o ) = K ε nf 1+ n Here εf is the maximum strain value during deformation. UTS ε = n Average Flow Stress y Average (mean) flow stress is not on the basis of instantaneous flow stress. and the variation of stress field along g yy‐ direction is negligible.25. operation 2. A strip of lead with initial dimensions 24 mm x 24 mm x 150 mm is forged f d between b two flat fl dies d to a final size of 6 mm x 96 mm x 150 mm.35 (d) σ = 775 ε0. mm If the coefficient of friction is 0. h d i then the true stress‐true strain relation (stress in MPa) in the plastic deformation range is: m 1 dh v Platen Velocity ε= = = h dt h Instantaneous height The strain hardening exponent n of stainless steall SS 304 with h distinct d yield ld and d UTS values l undergoing plastic deformation is a) n<0 0 30 (a) σ = 540 ε0.30 (b) σ = 775 ε0. The average yield stress of lead in tension is 7 N/mm2 [10] Page 91 of 210 233 Rev. 25 mm The percentage change in diameter is 25 mm.07 (c) 20 7 (c) 20. The percentage change in diameter is (a) 0 (b) 2 07 (b) 2. forging A solid cylinder of diameter 100 mm and height 50 mm y Coefficient of friction is constant between workpiece and is forged between two frictionless flat dies to a height of f db f l fl d h h f dies (platens).0 234 . problem is plain strain type.35 b)n=0 c) 0<n<1 d)n=1 226 227 228 For IES Only Assumption GATE‐2012 Same Q GATE ‐2012 (PI) y Forging force is maximum at the end of the forging.4 dimensions. 2 and d3 For-2016 (IES. % If the h material i l obeys b power law l off hardening.. St (a) 1 and 2 (b) 1 and 3 ( ) 2 and (c) d3 (d) 1.St i Strain rate effects t ff t y Strain rate effect (hot Working) σ o = Cε GATE 2015 GATE-2015 GATE‐2006 The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%.30 (c) σ = 540 ε0. e o. Forging force attains maximum value at the middle of the operation. Coefficient of friction is constant between work piece and d die d Stress ess in tthee ve vertical t ca ((Y‐direction) d ect o ) iss zero. type y The entire workpiece p is in the p plastic state during g the 229 230 For IES Only IES ‐ 2012 Rectangular Bar Forging l 232 231 For IES Only For IES Only IES – 2005 Conventional Assumptions adopted A i d d in i the h analysis l i off open die di forging f i are 1.GATE & PSUs) process. IES Conventional Only y Length is much more than width.7 For IES Only y Thickness Thi k off the h workpiece k i i small is ll compared d with i h other h (d) 41 4 (d) 41.. determine the maximum forging force. 2. Also fi d mean die find di pressure. 405 MPa] For-2016 (IES.For IES Only For IES Only IES – 2007 Conventionall Axi‐symmetric Forging IES – 2006 ‐ Conventionall A cylinder 100 mm is li d off height h i h 60 6 mm and d diameter di i forged at room temperature between two flat dies.26 MN] [ Ans.05. mm Coefficient of friction is 0. determine the maximum forging g g force. mean die pressure and maximum pressure. 46.25. MPa . Coefficient of friction is 0. Determine the equation is maximum die pressure.41 MPa . Fi First calculate l l true strain i ε and d put the h value l in i the equation 242 σ f = 200(0.2 and flow curve friction is 0. There is no sticking.1 0 1 and average yield mm. 221 MPa] the equation 241 is given by forging force. 178. Determine maximum forging load. 0 05 Assume that volume is constant after deformation.771 MN.24 kN] 239 For IES Only 240 For IES Only For IES Only Practice Problem‐2 bl Practice Problem‐3 bl Practice Problem‐4 bl y A circular disc of 200 mm in diameter and 100 mm in y A cylindrical specimen 150 mm in diameter and 100 mm y A circular disc of 200 mm in diameter and 70 mm in h i ht is height i compressed d between b t t flat two fl t dies di to t a height h i ht off in height is upsetted by open die forging to a height of 50 height is forged to 40 mm in height. y The yyield strength g of the work material is given as 120 N/mm2 and the coefficient of friction is 0. 5 determine the maximum forging g g force. Fi First calculate l l true strain i ε and d put the h value l in i [Ans. f i i l disc di is i gradually d ll compressed between two flat platens. Calculate the maximum σ f = 200(0. 9. 5.0 243 .01 + ε ) 0.17 =σy ] Page 92 of 210 [Hi [Hint. using g slab method of analysis.GATE & PSUs) σ f = 1030ε 0. The average shear yield stress of lead can be taken as 4 N/mm2. [Ans. 5 The flow curve equation q of the material strength g in compression p is 230 3 MPa. 178 MPa. [Hi [Hint.41 = σ y ] Rev. i h i di t that indicates th t (a) there is no sticking friction anywhere on the flat face of the disc (b) st sticking c g friction ct o aand d ssliding d g friction ct o co co‐exist e st o on tthee flat face of the disc (c) the flat face of the disc is frictionless (d) there is only sticking friction on the flat face of the disc di 238 y A strip of metal with initial dimensions 24 mm x 24 mm x 150 mm is i forged f d between b t t flat two fl t dies di to t a final fi l size i off 6 mm x 96 mm x 150 mm. from the center of the disc towards its periphery.17 σ f = 1030ε 0. Take the average yield strength in tension is 7 N/mm2 [Ans. [10 – Marks] 236 237 For IES Only Practice Problem‐1 bl GATE‐2014(PI) GATE‐1987 I f i In forging operation the sticking friction condition ti th ti ki f i ti diti occurs near the (Centre/ends) occurs near the……(Centre/ends) In a circular I an open die di forging. The exponential ti l decay d off normall stress t on the th flat fl t face f of the disc. If the co‐efficient of friction between the job and die is 0.05. height Coefficient of 50 mm. [ M k ] [20‐Marks] 235 For IES Only A certain 50 i disc di off lead l d off radius di 150 mm and d thickness hi k mm is reduced to a thickness of 25 mm by open die forging.05..01 + ε ) 0. Find the die load at the end of compression to a height 30 mm. mm If the coefficient of friction is 0. where p and τ are the normal and shear stresses.. iii. (high yield strengths. respectively. in the region R ss ≤ r ≤ RFN . Q No N 1 2 Option O ti A A Q.. ti Al derive Also d i tensile t il and d shear h yield i ld stress relationships for their approaches.20 [10 Marks] 2μ ( RFN − r ) H FN p = 3Ke and τ = μ p. K iis the th shear h yield i ld strength t th off steel t l andd r is i th the radial di l distance di t 244 of any point (contd . Discuss Tresca and Von Mises yield criterion for metal a solid circular steel disc of initial radius (R IN ) 200 mm and initial height (H IN ) 50 mm attains a height (H FN ) of 30 mm and radius of R FN .. the coefficient of friction (μ ) is: μ = 0.76 R − IN ⎞ ⎛ i. and nickel‐based alloys are For-2016 (IES. friction. where sticking condition changes to sliding Along the die-disc interfaces.35 ⎜ 1 + e RFN ⎟ ⎜ ⎟ ⎝ ⎠ ii. 250 249 248 y Steels. y Metal will undergo tri‐axial compression. welding with suitably bl shaped h d die d to form f a product d with h reduced d d but b wall)..sticking condition prevails The value of R SS (in mm). section lubricants . and alloys of these metals are commonly extruded. aluminum.sliding friction prevails.41 3. is ((a)) 241.. (b) Extrusion is used for reducing the diameter of round g dies which open p and close bars and tubes byy rotating rapidly on the work? (c) Extrusion is used to improve fatigue resistance of the metal by setting up compressive stresses on its surface (d) Extrusion E t i comprises i pressing i th metal the t l inside i id a chamber to force it out by high pressure through an orifice ifi which hi h is i shaped h d to t provide id the th desired d i d from f off the th finished part.) W kB k WorkBook 245 Extrusion & Drawing Ch‐15: Forging Q. magnesium. copper. Use phosphate‐based and molten glass constant cross section. 252 Rev. stainless steels. Q No N 11 12 Option O ti C B 3 4 5 6 7 8 9 10 A A B B C C C D 13 14 15 16 17 18 19 C C B A 75 7.. approaches Which of these criterion is more realistic? Whyy ? ((d)) 278.45 f forming i operations.0 . IFS‐2012 Practice Problem ‐5 {GATE‐2010 (PI)} During open die forging process using two flat and parallel dies.93 MN Extrusion y The extrusion process is like squeezing toothpaste out of a tube. employed y Lead. By S K Mondal 247 y Metal is compressed and forced to flow through a difficult to extrude.GATE & PSUs) 246 Page 93 of 210 251 IES – 2007 Which following is correctt Whi h one off the th f ll i i the th statement? (a) Extrusion is used for the manufacture of seamless tubes.In the region 0 ≤ r ≤ R SS .For IES Only Practice Problem 5 {GATE 2010 (PI)} Practice Problem ‐5 {GATE‐2010 (PI)} For IES Only For IES Only Contd Contd……..55 ((c)) 265. and (b) ( ) 254. compression y Hot extrusion is commonly employed. Extrusion Ratio DRDO‐2008 y Ratio of the cross‐sectional area of the billet to the cross‐ sectionall area off the h product. The Th process off hot h extrusion i is i used d to produce d ((a)) Curtain rods made of aluminium (b) Steel pipes/or domestic water supply ( ) Stainless (c) S i l steell tubes b used d in i furniture f i ((d)) Large g shape p p pipes p used in cityy water mains y Medium and small batch production. y Conversion from one product to another requires only a single die change y Good dimensional precision. ll off sections i and d pipes i off complex l configuration. the percentage reduction in the h cross‐sectionall area off the h billet b ll after f the h y about b 40: 1 for f hot h extrusion i off steell extrusion will be y 400: 1 for f aluminium l i i (a) 98% y Any cross‐sectional shape can be extruded from the nonferrous f metals. 253 IES ‐ 2012 254 Which Whi h one off the h following f ll i statements is i correct?? ((a)) In extrusion p process. For-2016 (IES. d Advantages of Extrusion d f If the extrusion ratio is 20.. d 257 IES – 1994 y Working of poorly plastic and non ferrous metals and y Manufacture M f Limitation of Extrusion f IES – 2009 Extrusion process can effectively the E i ff i l reduce d h cost off product through (a) Material saving (b) process time saving (c) Saving in tooling cost (d) saving in administrative cost 255 258 GATE‐1994 Metal process is M l extrusion i i generally ll used d for f producing (a) Uniform solid sections (b) Uniform hollow sections (c) Uniform solid and hollow sections (d) Varying solid and hollow sections. configuration y Cross section must be uniform for the entire length of the h product. thicker walls can be obtained by increasing the forming pressure (b) Extrusion is an ideal process for obtaining rods from metal having poor density (c) As compared to roll forming. production y Manufacture of parts of high dimensional accuracy. extruding speed is high (d) Impact extrusion is quite similar to Hooker Hooker'ss process including the flow of metal being in the same direction 256 Application l alloys.0 261 .GATE & PSUs) 259 Page 94 of 210 260 Rev. t l y Many shapes (than rolling) (b) 95% (c) 20% (d) 5% y No od draft at y Huge reduction in cross section. 2. l i y Design D i off die di is i a problem. used 265 IES – 1993 Assertion extrusion larger force A i (A): (A) Direct Di i requires i l f than indirect extrusion. y Since no relative motion. The speed of travel of the extruded product is same as that of the ram.0 270 . y Used U d to produce d curtain i rods d made d off aluminum. used (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true For-2016 (IES. The speed of travel of the extruded product is greater than that of the ram. t ti confined billet. friction between the billet and the chamber is eliminated. The ram and the extruded product travel in the same direction. 3.IAS – 2012 main IES – 1999 E t i Extrusion Hot Which following is correct Whi h one off the h f ll i i the h temperature range for hot extrusion of aluminium? (a) 300‐340°C (b) 350‐400°C (c) 430‐480°C 430 480°C (d) 550‐650°C 550 650°C Cold Direct Indirect Forward Hydrostatic Classify the process of extrusion with the help of sketches. Rev. k h Backward Cold Extrusion Forging Impact Extrusion 263 262 Hot Extrusion Process Direct Extrusion IES – 2009 y The temperature range for hot extrusion of aluminum is 430‐480°C 264 y A solid the billet lid ram drives di h entire i bill to and d through h h a What in Wh is i the h major j problem bl i hot h extrusion? i ? ((a)) Design g of p punch ((b)) Design g of die (c) Wear and tear of die (d) Wear of punch stationary die and must provide additional power to overcome the h frictional f l resistance between b the h surface f off the h moving billet and the confining chamber. zinc phosphate coating is used. Reason (R): In indirect extrusion of cold steel. 4. bl y Either direct or indirect method used.GATE & PSUs) 268 266 267 IES – 2000 Indirect Extrusion d Consider C id the th following f ll i statements: t t t In forward extrusion process 1. The ram and the extruded product travel in the opposite direction. Which of these Statements are correct? (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) Page 2 and 269 954of 210 y A hollow ram drives the h ll di th die di back b k through th h a stationary. walls. 3. the direction of the force applied pp on the p plunger g and the direction of the movement of the extruded metal are the same. Reason (R): Advantage in indirect extrusion is less quantity of scrap compared to direct extrusion. Impact extrusion Select the correct answer using the codes given below: C d Codes: (b) 1 and 2 (a) 1 onlyy (c) 2 and 3 (d) 3 only Rev. The force required on the punch is more in comparison co pa so to d direct ect eextrusion. 4. (a) Both A and R are individually true and R is the correct explanation of A (b) Both B th A and d R are individually i di id ll true t b t R is but i nott the th correct explanation of A (c) A is true but R is false ((d)) A is false but R is true 278 The Th extrusion i process (s) ( ) used d for f the h production d i off toothpaste tube is/are 1. medications and other creams. 275 IES – 2008. t us o . articles y Now‐a‐days also been used for forming mild steel parts. 271 IAS – 2004 Assertion (A): extrusion operation can be A i (A) Indirect I di i i b performed either by moving ram or by moving the container. Billet remains stationary 2 There is no friction force between billet and container 2. 2 and d 3 only l (c) 1. Extrusion parts have to be provided a support. ll for shielding electronic components and larger cans for y Used U d for f making ki collapsible ll ibl tubes. 3. 3 and 4 only y Required force is lower (25 to 30% less) y Low process waste.Indirect Extrusion d Contd… IES ‐ 2012 IES – 2007 Which are correct for hot Whi h off the h following f ll i f an indirect i di h extrusion process? 1. 277 For-2016 & to PSUs) 273 Cold Extrusion ld 274 Impact Extrusion Assertion (A): force on the is A i (A) Greater G f h plunger l i required i d in case of direct extrusion than indirect one. 2. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true 272 Backward cold extrusion k d ld y Used with low‐strength metals such as lead. GATE‐1989(PI) manufacture f off collapsible ll bl tooth‐paste h tubes? b ( ) Impact (a) I extrusion i (b) Di Direct extrusion i ( ) Deep (c) D d drawing i (d) Pi i Piercing Page 96 of 210 276 IES – 2003 Which one of the following methods is used for the y The extruded parts are stripped by the use of a stripper plate. 3 and (a) d4 (b) 1. 2 and 4 only (d) 2. Reason (R): In case of direct extrusion. Tube extrusion 2 Forward extrusion 2. ( ) 1. y The metal is extruded through the gap between the and d aluminum l to produce d collapsible ll bl tubes b f for punch h and d die d opposite to the h punch h movement. because they (IES.GATE tend to stick the punch.0 279 . t b cans for f liquids li id and d food and beverages. toothpaste. toothpaste medications. small "cans" cans y For F softer f materials i l such h as aluminium l i i and d its i alloys. tin. similar articles. zinc. a stripper will ll strip it from f the h punch. bl k a hollow h ll slug l can also l be b used. y High‐pressure fluid applies the force to the workpiece through h h a die. billet is uniformly compressed from all sides by the liquid.GATE & PSUs) 282 Hydrostatic Extrusion d Contd…. liquid (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true y Extrusion of nuclear reactor fuel rod E t i f l t f l d y Cladding of metals y Making wires for less ductile materials Page 97 of 210 285 287 Rev. t i b t the but th fluid fl id pressure surrounding the billet prevents upsetting. d l y A flat blank of specified diameter and thickness is placed in a y y y y suitable i bl die di and d is i forced f d through h h the h opening i off the h die di with ih the punch when h the h punch h starts downward d d movement. y Another type of cold extrusion process. h Small copper tubes and cartridge cases are extruded by this method. and the pressurized fluid acts as a lubricant between the billet and the die. molybdenum. highly plastic. stainless steel. Reason(R): In hydrostatic extrusion. di y It i forward is f d extrusion. y Relatively brittle materials like cast iron. Crack formation is suppressed leading to a phenomenon known suppressed. [ [30‐Marks] k ] 280 281 Hooker Method k h d Hydrostatic Extrusion d y The Th ram/punch / h has h a shoulder h ld and d acts t as a mandrel. Pressure P i is exerted by the shoulder of the punch. I place In l off a flat fl solid lid blank. tungsten and various inter inter‐metallic metallic compounds can be plastically deformed without fracture and materials with limited ductility become fracture. y Temperature is the sink T i limited li i d since i h fluid fl id acts as a heat h i k and the common fluids (light hydrocarbons and oils) burn or decomposes at moderately low temperatures. the metal being forced t flow to fl th through h the th restricted t i t d annular l space between b t th the punch and the opening in the bottom of the die. y The metal deformation is performed in a high high‐ compression environment. d If the tube sticks to the punch on its upward stroke. 283 Hydrostatic Extrusion d Contd…. 284 IAS – 2000 Application Assertion (A): A i (A) Brittle B i l materials i l such h as grey cast iron cannot be extruded by hydrostatic extrusion.0 288 . upsetting y Billet Billet‐chamber chamber friction is eliminated.IES ‐ 2014 Hooker Method k h d IAS‐2010 Main A toothpaste tube h b can be b produced d d by b ((a)) Solid forward extrusion (b) Solid backward extrusion ( ) Hollow (c) H ll backward b k d extrusion i ((d)) Hollow forward extrusion How are metal tooth‐paste tubes made commercially ? Draw the tools configuration with the help of a neat sketch. 286 For-2016 (IES. kno n as pressure‐induced ductility. IES – 2006 What hydrostatic Wh does d h d i pressure in i extrusion i process improve? (a) Ductility (b) Compressive strength (c) Brittleness (d) Tensile strength IES – 2001 GATE‐1990(PI) S i brittle Semi b ittl materials t i l can be b extruded t d d by b (a) Impact extrusion (b) Closed cavity extrusion (c) Hydrostatic extrusion (d) Backward extrusion 289 290 Which statements Whi h off the th following f ll i t t t are the th salient li t features of hydrostatic extrusion? 1. k Process variables in Extrusion Process variables in Extrusion Explain the processes of extrusion given below. i l 2. It is suitable for high‐strength super‐alloys. through the die by applying pressure to the liquid. Temperature and Metallurgy: Variations in temperature during extrusion seem to influence flow behaviour in number of ways. d it may lead l d to t melting lti off alloy ll constituents (a) Both Statement (I) and Statement (II) are individuallyy true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement St t t (I) is i false f l but b t Statement St t t (II) is i true t For-2016 (IES. Most effective lubricant is a phosphate conversion coating on the h workpiece. critical especially with steels. Page 98 of 210 296 JWM 2010 Assertion (A) : Extrusion speed depends on work material. because the material starts melting g or cracking if it is leaves the die with too high temperature. di Tendency T d i increases with ih increasing die angle and amount of impurities. It is known that the extrusion pressure mayy be lowered if either the temperature p p of the billet or the velocity of the stem is increased. Reason (R) : High extrusion speed causes cracks in the material. because of the possibility of sticking (seizure) bet een the workpiece between orkpiece and the tooling if the lubrication breaks down. Tendency decrease with increasing extrusion ratio and friction. stainless steels and high temperature metals and alloys. 3. y Bamboo defects at low temperature due to sticking of metals in die land. the initial temperature of billet should be high. 4. and that there are certain limitations. It I is i suitable i bl for f soft f and d ductile d il material. p y Surface crack due to high high high friction etc. The billet is inserted into the extrusion chamber where it is q The billet is extruded surrounded byy a suitable liquid. As indicated. g the codes g given below: Select the correct answer using Codes: (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 291 For IES Only Lubrication for Extrusion b f IES 2009 Conventional y For glass is lubricant with F hot h extrusion i l i an excellent ll l bi ih steels. y Centre Burst or Chevron defect are attributed to a state of hydrostatic tensile stress at the centreline in the d f deformation i zone in i the h die.0 297 . p g speed. (iv) Impact Extrusion 292 293 For IES Only IES ‐ 2014 IES Statement‐I: For high extrusion pressure. like funnel called pipe. high Statement‐II: As the speed of hot extrusion is i increased. flow patterns may be changed g considerablyy byy rendering g the temperature p distribution in the container.The billet is inserted into the extrusion chamber and pressure is applied by a ram to extrude the billet through the die. extrusion lubrication is critical. y Pipe defects or tail pipe or fishtailing. (a) Both A and R are individually true and R is the correctt explanation l ti off A (b) Both A and R are individuallyy true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Rev. Experimental p studies of flow 2. during extrusion surface oxides and impurities p are driven towards the centre of the billet. y For cold extrusion. Indicate one typical product made through each of these processes: (i) Direct Di t Extrusion E t i (ii) Indirect Extrusion (iii) Hydrostatic Extrusion 1.GATE & PSUs) 295 294 For IES Only Extrusion Defects Extrusion Defects g temperature. )) so that it freelyy enters the die orifice and sticks out behind the die. diameter.GATE‐2014 With respect to metal working. coiled y The wire is subjected to tension only. only But when it is in contact with dies then a combination of tensile. working match Group A with Group B Group A Group B P: Defect in extrusion I: alligatoring Q: Defect in rolling II: scab R: Product of skew III: fish tail rolling lli S: Product of rolling IV: seamless tube through cluster mill V: thin sheet with tight tolerance VI: semi‐finished balls of ball bearing IAS – 2012 main Enumerate the conditions under which central burst may occur.GATE & PSUs) portion only. IES – 2007 y Wire getting continuously wound on the reel. y For fine wire. compressive and shear stresses Page 99 of 210 305 GATE‐1987 F i d For wire drawing operation. compressive and shear stresses will be there in that 301 IES – 2009 302 303 IES – 2005 Which Whi h one off the h following f ll i stress is i involved i l d in i the h wire drawing process? (a) Compressive (b) Tensile (c) Shear (d) Hydrostatic stress For-2016 (IES. drawing the end of the rod or wire to be drawn is p pointed ((byy swaging g g etc. di receiving i i successive i reductions d i i in Which process Whi h metall forming f i manufacture of long steel wire? (a) Deep drawing (b) Forging (c) Drawing (d) Extrusion is i used d for f diameter before being coiled. 300 Wire Drawing Contd…. 304 Which types off stresses is/are Whi h off the h following f ll i i / involved in the wire‐drawing operation? (a) Tensile only (b) Compressive only (c) A combination of tensile and compressive stresses (d) A combination of tensile. d y Same S process as bar b drawing d i except that h it i involves i l smaller‐diameter material.0 306 . the work material i ti th k t i l should essentially be (a) Ductile (b) Tough ((c) Hard ) ((d) Malleable ) Rev. the material may be passed through a number b off dies. Where does a 'pipe' occur ? h d ' ' 298 (a) (c) P II III Q III I Wire Drawing Contd…. material y At the start of wire drawing. R VI IV S V VI (b) (d) P III I Q I II R VI V S V VI 299 Wire Drawing y A cold working process to obtain wires from rods of b bigger d diameters through h h a die. lubricant (c) uses thick paste lubricant (d) uses thin film lubricant Bundle Drawing Bundle Drawing In this process. the tube may be annealed and straightened. 3. 2 (d) 3.Cleaning and Lubrication in wire Drawing y Cleaning is done to remove scale and rust by acid pickling. y The tubes are also first pointed and then entered through the die where the point is gripped in a similar way a as the bar drawing dra ing and pulled through in the form desired along a straight line. y Phosphating: A thin film of Mn. cross‐section of round d wire i is i reduced d d by b pulling lli it through th h a die di g p produces wires that Statement‐II: Bundle drawing are polygonal in cross‐section rather than round (a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation l i off Statement S (I) ((b)) Both Statement ((I)) and Statement ((II)) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true For-2016 (IES. y Sulling: The wire is coated with a thin coat of ferrous hydroxide which when combined with lime acts as filler for the lubricant.GATE & PSUs) 313 Wire Drawing Die Rod and Tube Drawing d d b y Rod R d drawing d i is i similar i il to wire i drawing d i except for f the h fact f y Die materials: tool steels or tungsten carbides or polycrystalline diamond wire) 314 Page (for 100 fine of 210 that the dies are bigger because of the rod size being larger than the wire. 3. l l 310 311 312 For IES Only IES ‐ 2014 IES Statement‐I: In drawing process. the wires are separated from each other by a suitable material. 4. Th correctt sequence off these The th operations ti i is (a) 3. the surface I wire i drawing d i h bright b i h shining hi i f on the wire is obtained if one (a) does not use a lubricant (b) uses solid powdery lubricant. Alkaline cleaning 2 Phosphate coating 2. The cross‐section of the wires is somewhat h polygonal. 4. y The practice of drawing tubes without the help of an i t internal l mandrel d l is i called ll d tube t b sinking. 4 (c) 1. 4 In process. 3. 2. 1. To prevent sticking. 2. Lubricating with reactive soap. IES 2010 IES – 2000 Assertion (A): Pickling and washing of rolled rods is carried out before wire drawing. process many wires (as much as several thousand)) are drawn simultaneouslyy as a bundle. electrolytic y coating g of copper is used to reduce friction. Fe or Zn phosphate is applied on the wire. 1. y When the final size is obtained. i ki Rev. (a) Both A and R are individually true and R is the correctt explanation l ti off A (b) Both A and R are individuallyy true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 307 Which lubricants is Whi h one off the h following f ll i l b i i most suitable for drawing mild steel wires? (a) Sodium stearate (b) Water (c) Lime‐water Lime water (d) Kerosene 308 309 For IES Only IAS – 1995 IES – 1996 The operations are performed Th following f ll i i f d while hil preparing the billets for extrusion process: 1. Reason (R): They lubricate the surface to reduce friction while drawing g wires. wire y Electrolytic y coating: g For veryy thin wires. 2 (b) 1. Pickling 4.0 315 . Cleaning is done to remove scale and rust by acid pickling y Lubrication boxes precede the individual dies to help reduce friction drag and prevent wear of the dies.. Stamping and p pressing g B.0 . 323 Match M t h List Li t I (Components (C t off a table t bl fan) f ) with ith List Li t II (Manufacturing processes) and select the correct answer using i the th codes d given i b l below th Lists: the Li t List I List II A. g The term swaging g g is also applied pp to (c) Swaging processes where material is forced into a confining die to (d) Wire Wi Drawing D i 319 320 IES – 1993 IES‐2015 a) Round bars and tubes b)Billets b)Bill c) Dies d)Rectangular ) g blocks 322 Page 101 of 210 321 IES – 2000 Tandem drawing off wires and T d d i i d tubes b is i necessary because (a) It is not possible to reduce at one stage (b) Annealing is needed between stages (c) Accuracy in dimensions is not possible otherwise (d) Surface finish improves after every drawing stage Rotary swaging is a process for shaping For-2016 (IES. l causing the metal to flow inward and assume the shape (a) Hydrostatic Extrusion of the die. Wire drawing C Armature C. Base with stand 1. T i Turning D. GATE‐1994(PI). h y Repeated R d blows bl are delivered d li d from f various i angles. Blade 2. Casting Codes:A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 2 3 4 1 (d) 4 1 2 3324 Rev. Armature shaft 4. (b) Tube drawing y It is cold working. A t coil il wire i 3. 2014(PI) IES 1993.Rod and Tube Drawing d d b Contd… IES‐1993. GATE 1994(PI) 2014(PI) A moving mandrel is used in Tube Sinking Fixed Plug Drawing Floating plug Drawing Moving Mandrel (a) Wire drawing (b) Tube drawing ((c)) Metal Cutting g ((d)) Forging g g Back 316 317 Swaging or kneading k d IAS‐2006 318 Swaging or kneading k d Contd… y The hammering of a rod or tube to reduce its diameter Which one of the following processes necessarily requires q mandrel of requisite q diameter to form the internal hole? where h the h die d itself lf acts as the h hammer.GATE & PSUs) reduce its diameter. 3 3. Direction Codes:A B C D A B C D (a) 4 2 1 3 (b) 3 4 2 5 ( ) 5 (c) 4 2 3 (d) 3 2 1 5 P Q R S IES – 1994 332 g Match List –I with List –II and select the correct answer using the code given below the lists : List I List –I List II List –II A Connecting rods A. Collapsible tubes 4.R -3 . Blanking Coining Extrusion Cup drawing Codes:A (a) 2 (c) 3 B 3 2 C 4 1 List I (Parts) 331 List II (Manufacturing processes) A.R R -1 1 . Drawing g 1. 4. Shot peening C. 3 3.GATE & PSUs) List I (Metal farming process) List II (A similar process) A. Indentation 5.Q -4 .Q Q -3 3 .IES – 1999 IES – 1996 Match M t h List‐I Li t I with ith List‐II Li t II and d select l t the th correctt answer using the codes given below the Lists: List‐I List‐II A. Cold rolling D. Impact loads C Resilience C.R R -22 . Bending C d A B Codes:A C (a) 4 2 1 (c) 2 3 1 1.R -2 . Malleabilityy 1.S S -2 2 b ) P -4 4 . Extrusion C. D D. D 3 4 (b) (d) Shear force T il force Tensile f Compressive p force Spring back force A B C D 2 1 3 4 4 3 2 1 325 326 IES – 1993. 5. Seamless tubes A 1 Roll forming 1.Q -4 . Rolling 2. A B. Ironing Code:A B C D A B C D (a) 2 5 1 4 (b) 4 1 5 3 ( ) 5 (c) 2 3 4 (d) 5 2 1 3 Match List I with List II and select the correct answer List I (Metal/forming process) List II (Associated force) A. Welding B. Surfaces having higher 3. Connecting rods 1. Forming D. di Reason (R): Since each drawing reduces ductility of the wire.Q Q -3 3 . Connecting rods 1 Welding 1. strength4 Cold forming Codes: A B C A B C (a) 1 4 2 (b) 2 3 1 (c) 1 3 2 (d) 2 4 1 329 D 1 5 (b) (d) Wire drawing Piercing Embossing Rolling Bending A B C 2 3 1 2 3 1 D 4 5327 Match I (Products) II (Suitable M h List Li (P d ) with i h List Li (S i bl processes) and select the correct answer using the codes given below the Lists: List I List II A. Welding B. Soap p solution B. 3. so after final drawing the wire is normalized. Crater progressive dies 5. C. Extrusion C Machine tool beds C. Casting 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 1 330 IES 2011 IES 2011 IAS – 2002 Assertion (A): process. 3. Hardness 2.0 1 D 2 2333 . Wire drawing g B. dra ing 3 3. Sheet metal operations using 4. Pressure vessels 2. B. 2.S -1 For-2016 (IES. 4 4. Casting Codes A (a) 2 (c) 2 B 1 4 C 4 1 D 3 3 (b) (d) A 3 3 B C 1 4 4 Rev. E t i g C. Blanking D. ISRO‐2010 GATE 2015 GATE-2015 g p Match the following products with p preferred manufacturing processes: 1 2 3 4 Process Blow molding Extrusion F i Forging Rolling g a ) P -4 4 . Pilots D.S S -1 1 c ) P -2 . Machine tool beds 3.S -1 d ) P -3 . Wire drawing B Extrusion B. 2. Collapsible tubes 4. Forging D. Forging hardness and fatigue strength4. IAS – 2001 Match the M h List Li I with i h List Li II and d select l h correct answer: 328 Product Rails Engine Crankshaft Al i i Aluminium Ch Channels l PET water bottles Match List I with List II and select the correct answer M h Li I i h Li II d l h using the codes given below the Lists: IES – 2002 Match M t h List Li t I with ith List Li t II and d select l t the th correctt answer using the codes given below the lists: List I (Mechanical property) List II (Related to) A. Isotropy 4. Camber C Wire drawing C. Accurate and smooth tubes 2. the A ti (A) In I wire‐drawing i d i th rod d cross‐section is reduced gradually by drawing it severall times ti i successively in i l reduced d d diameter di t dies. (a) Both A and R are individually true and R is the correct co ect eexplanation p a at o o of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A is false but R is true (d) Page 102 of 210 11. Pressure vessels 2. a round billet of 100 d diameter off 100 mm to a final f l diameter d off 50 mm. 3 and 4 (c) ( ) 2.GATE & PSUs) ⎞ ⎟⎟ ⎠ ⎞ ⎟⎟ ⎠ 340 σE = ⎛A P = σ o ln ⎜ o ⎜A A0 ⎝ f deformation.36 MN (a) 416 Page 103 of 210 341 (b) 624 (c) 700 Rev. lli 2 Extrusion drawing and 3 3. t t deformation. Spinning 334 335 y Approximate A i method h d (Uniform (U if friction) “work – formula” How are the seamless tubes produced ? ⎛A P = Aoσ o ln ⎜ o ⎜A ⎝ f ⎞ ⎛d πd2 ⎟⎟ = 2 × o × σ o × ln ⎜⎜ o 4 ⎠ ⎝ df Extrusion Stress deformation. Rolling S Seamless l l long steel t l tubes t b are manufactured f t d by b rolling. The force required friction and no redundant work).Seamless tube Manufacturing Seamless tube Manufacturing IAS 1994 GATE‐1991(PI) 1 Rolling 1. Extrusion 1. d f i no ⎞ π d o2 × σ o × ln ( R ) ⎟⎟ = 4 ⎠ y For real conditions ⎛A P = KAo ln ⎜ o ⎜A ⎝ f ⎞ ⎛d πd × K × ln ⎜ o ⎟⎟ = 2 × ⎜ 4 ⎠ ⎝ df ⎞ ⎟⎟ ⎠ ⎛A P = Af σ o ln ⎜ o ⎜A ⎝ f Drawing Stress ⎞ ⎛d π d 2f × σ o × ln ⎜ o ⎟⎟ = 2 × ⎜d 4 ⎠ ⎝ f ⎛A P σd = = σ o ln ⎜ o ⎜A Af ⎝ f ⎞ ⎛d ⎟⎟ = 2 × σ o × ln ⎜⎜ o ⎠ ⎝ df For-2016 (IES. 3 and 4 (d) 339 GATE‐2003 GATE – 2009 (PI) A brass billet is to be extruded from its initial Using direct extrusion process. for extrusion is and average flow stress of material 300 MPa. extrusion ratio 4. the (a) 5. Rolling 4.44 MN (b) 2. t t 338 Force required in Wire or Tube drawing no method h d (Uniform (U if friction) “work – formula” σE = K = extrusion t i constant. d f i ⎞ ⎛d ⎟⎟ = 2 × σ o × ln ⎜⎜ o ⎠ ⎝ df no ⎞ ⎟⎟ = σ o × ln ( R ) ⎠ ⎛A P = K ln ⎜ o ⎜ A0 ⎝ Af ⎞ ⎛d ⎟⎟ = 2 × K × ln ⎜⎜ o ⎠ ⎝ df ⎞ ⎟⎟ ⎠ K = extrusion t i constant. d d The working temperature of 700°C 700 C and the Considering an ideal deformation process (no extrusion constant is 250 MPa.0 (d) 832 342 . deformation y Approximate A i y For real conditions 2 o 337 method (Uniform friction) “work – formula” 336 Extrusion Load IES‐2012 Conventional IES‐2012 Conventional y Approximate Which of the following methods can be used for manufacturing 2 meter long seamless metallic tubes? b g 2. Spinning Select l the h correct answer using the h codes d given below b l Codes: (a) 1 and 3 (b) 2 and 3 ( ) 1. Tube Drawing g 4.72 MN pressure (in MPa) on the ram will be (c) 1.36 MN (d) 0. mm length l h and d 50 mm diameter d is extruded. Drawing 3. 40% and 80% are available. diameter of a steel wire is reduced d d from f 10 mm to 8 mm.0 344 345 IES ‐ 2014 GATE‐2001. the number of stages g and reduction at each stage respectively would be (a) 3 stages and 80% reduction for all three stages (b) 4 stages and 80% reduction for first three stages followed by a finishing stage of 20% reduction (c) 5 stages and reduction of 80%.97 kN (c) 20. GATE ‐2007 (PI) (i) What kind of products are manufactured by wire For rigid perfectly‐plastic work material.36 3 ((b)) 0.40%. the f ti fractional l reduction d ti ( ti off change (ratio h i cross – in σd = σ o (1 + B ) ⎡ B 2B 2B ⎛r ⎞ ⎤ ⎛r ⎞ ⎢1 − ⎜ f ⎟ ⎥ + ⎜ f ⎟ .5 (b) 357. the stress The power required for the drawing process (in kW) required for drawing (in MPa) is is (a) 178. The yield stress of the material is diameter by 20%. The h mean flow fl stress of the material is 400 MPa.24 (b) 0.58 (c) 0. 80% 40% 40%.0 (a) 8. k the h (ii) How H much h force f will ill approximately i l be b required i d theoretically maximum possible reduction in the to draw a wire from 1.σ b ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎝ ro ⎠ sectional area to initial cross cross‐sectional sectional area) in the first stage g is 0. For minimum error in the final size.72 7 In wire‐drawing operation.3.0 343 IES‐2014 Conventional GATE 2008 (PI) Linked S 2 GATE ‐2008 (PI) Linked S‐2 (b) 14.48 kN (b) 8.97 (c) 1287. 40% 20% in a sequence ( ) none of the above (d) For-2016 (IES. The overall fractional reduction is _____ (a) 0.0 f Page 104 of 210 350 Rev.GATE & PSUs) 348 349 Wire Drawing Wire Drawing In a two stage wire drawing operation . 80% 80%.95 (d) 28.5 (d) 2575.GATE 2008 (PI) Linked S 1 GATE ‐2008 (PI) Linked S‐1 GATE‐2006 In a wire drawing operation.0 10 wire drawing operation is mm diameter wire if the average g y yield strength g of ((a)) 0.0 351 . 800 MPa.55 m/s to reduce the diameter by 20%. the maximum I i d i i h i reduction per pass for perfectly plastic material in ideal condition is (a) 68 % (b) 63 % (c) 58 % (d) 50% [10 Marks] 346 347 GATE 2015 GATE-2015 GATE‐1996 A wire is from a rod i off 0.55 m/s to reduce the through g a die at a speed p of 0.0 (c) 17.1 mm diameter di i drawn d f d off 15 mm diameter. Neglecting friction and strain hardening..41 kN A 10 mm diameter annealed steel wire is drawn A 10 mm diameter annealed steel wire is drawn through g a die at a speed p of 0. negligible drawing process? interface f f friction and d no redundant d d work. 4 The fractional reduction in the second stage is 0.00 ((d)) 2. MPa The ideal force required for drawing (ignoring friction and redundant work) is (a) 4.5 1 5 mm diameter steel wire to 1. The yield stress of the material is 800 MPa.60 (d) 1.633 the work material is 300 MPa? ((c)) 1.11 kN (d) 31. Dies giving reductions of 20%.4. multi‐pass operation a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section b drawing by d i it successively i l through th h a series i off seven dies di of decreasing exit diameter. are (a) 2. is (a) 15. The true strain rate 356 D= Ao − Af Ao = Do2 − D 2f Page 105 of 210 357 IAS – 2006 In and I drawing d i operation i if Di = initial i i i l diameter di d Do = Outgoing diameter. During each of these d drawing operations. multi‐pass operation a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section b drawing by d i it successively i l through th h a series i off seven dies di of decreasing exit diameter.. then what is the degree of drawing equal to? (a) Do2 ⎛A ThereforeTrue strain ( ε ) = ln ⎜ o ⎜A ⎝ f 358 Extrusion DOES NOT depend E i force f d d upon the h ((a)) Extrusion ratio (b) Type of extrusion process ( ) Material (c) M i l off the h die di ((d)) Working g temperature p at the end of one minute is ………. Ignore strain hardening. The yyield strength g of the material is 200 MPa.77 (d) 4. The Degree of Drawing or Reduction factor in Cross Sectional area in Cross Sectional area Write W i the h process variables i bl in i wire i drawing. Friction 354 GATE – 2014 355 IES – 2012 In a multi pass drawing operation.15. σ b σo = σ o (1 + B ) ⎡ B 2B ⎛r ⎞ ⎤ ⎢1 − ⎜ f ⎟ ⎥ ⎢⎣ ⎝ ro ⎠ ⎥⎦ 352 A metal rod of initial length IAS – 1997 is subjected to a d drawing process. Assuming a semi die angle of 5o and coefficient of friction between the die and steel rod as 0.71 (b) 10.39 For-2016 (IES. The h length l h off the h rod d at any instant is given by the expression. Neglecting friction and redundant work the force (in Neglecting friction and redundant work.02 and 2043 (d) 3.Maximum Reduction per pass A 12.45 and 8 17 (b) 2. the force (in kN) required for drawing the bar through the first die. σ b σo = σ o (1 + B ) ⎡ 2B 2B ⎛ rf ⎞ ⎤ ⎛ rf ⎞ ⎢1 − ⎜ ⎟ ⎥ + ⎜ ⎟ . what would be the draw stress and maximum possible ibl reduction d ti ? Take stress of the work material as 400 N/mm2 .5 mm diameter d rod d is to be b reduced d d to 10 mm diameter by drawing in a single pass at a speed of 100 m/min. The total true strain applied and the final length (in mm). the h reduction d in cross‐sectionall area is 35 35%. The yyield strength g of the material is 200 MPa. d i Ans.45 and 345 (c) 3. expression L(t) = Lo(1 + t2) where t is the time in minutes.σ b ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎝ ro ⎠ B Without back stress.21 (c) 6. calculate: ((i)) The p power required q in drawing g (ii) Maximum possible reduction in diameter of the rod (iii) If the rod is subjected to a back pressure of 50 N/mm2 . respectively. Reduction in cross sectional area 2.0 360 . During each of these d drawing operations. the h reduction d in cross‐sectionall area is 35 35%. [15 Marks] With back stress.GATE & PSUs) G 20 ( ) C S GATE – 2011 (PI) Common Data‐S1 IES – 2011 Conventional Di − Do Di (b) Do − Di Do (c ) Di2 − Do2 Di2 (d ) Di2 − Do2 Di2 ⎞ ⎛ 1 ⎞ ⎟⎟ = ln ⎜ ⎟ ⎝ 1− D ⎠ ⎠ 359 Rev. 1. Ignore strain hardening. Die Di angle l 33.02 and 3330 353 G 20 ( ) C S2 GATE – 2011 (PI) Common Data‐S2 In a multi pass drawing operation. π r0 2 ⎞ ⎟⎟ ⎠ 2B ⎤ ⎥ ⎥ ⎦ 366 Ch‐17: Extrusion Q. [8‐Marks] 1 2B 2B 2B σ o (1 + B ) ⎡ ⎛ r ⎞ ⎤ ⎛ r ⎞ ⎢1 − ⎜ ⎟ ⎥ + ⎜ ⎟ .π r0 2 = 2π r0τ i L or p f = σ o (1 + B ) ⎡ 365 τ i = uniform interface shear stress between 2 σ xo = ⎛A P = σ o ln ⎜ o ⎜A Af ⎝ f p f = ram pressure required by container friction A ⎛r ⎞ Extrusion ratio. dies determine the drawing stress and the reduction in area when the 2. Also yield i ld stress t f for aluminium l i i i 35 N/mm is N/ Al calculate the tangential stress at the exit.σ b ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎝ ro ⎠ B 2B 364 at r = ro σ xo = B ⎛r ⎢1 − ⎜ o ⎢ ⎜r ⎣ ⎝ f ⎞ ⎟⎟ ⎠ 2B B ∴ Bσ x − (1 + B ) σ o = ( rC ) B.σ b or σ x = B r r ⎣⎢ ⎝ o ⎠ ⎦⎥ ⎝ o ⎠ σ o (1 + B ) ⎡ 362 IFS 2013 IFS‐2013 B. σ x = 0 or σ x = τ x = μ Px = μ (σ o − σ x ) at Exit τ = μ P = μ (σ o − σ d ) = 0 ⎞ for flat stock ⎟⎟ ⎠ 2B (at exit stress is zero) ⎛ ⎢1 − ⎜ r ⎢ ⎝⎜ rf ⎣ B Workbook billet and container wall p f . Neglecting friction between the rod and the dies. 361 σ o (1 + B ) ⎡ ⎛r ⎞ ⎤ ⎛r ⎞ ⎢1 − ⎜ f ⎟ ⎥ + ⎜ f ⎟ . Q No N 8 9 Option O ti B B 3 4 5 6 7 D D B C B 10 11 12 13 14 C C B B 416 ⎡⎣1 − R 2 B ⎤⎦ For-2016 (IES.C at r = ro .60 mm diameter.255 mm d diameter.0 369 . The equilibrium equation in x-direction will be 2 or σ x 2rdr + dσ x r 2 + 2rτ x dx + Px 2rdx tan α = 0 2τ i L ro ∴ Total Extrusion Pressure(Pt ) = σ xo + p f and Extrusion Load = pt . iss d aw into to An a aluminium 6. drawn a wire 5.. 6. a ete .Wire Drawing Analysis (Home Work) Wire Drawing Analysis (Home Work) (σ x + dσ x ) π ( r + dr ) dx ⎞ ⎛ − σ xπ r + τ x cos α ⎜ 2π r ⎟ cos α ⎠ ⎝ dx ⎞ ⎛ + Px sin α ⎜ 2π r ⎟=0 cos α ⎠ ⎝ Thefore two principal stresses are σ x and − Px Thefore.} Extrusion Analysis (Home Work) Extrusion Analysis (Home Work) u u rod. Shear Stress σd = 363 ⎡ − (1 + B ) σ o ⎦⎤ 2 B ∴C = ⎣ rf Hint: Drawing Stress 2B 1 = 2 ln ( rC ) B {Cis integration cont. Q No N 1 2 Option O ti D C Q.GATE & PSUs) 367 Page 106 of 210 368 Rev.Cs If effect of container friction is considered ⎤ ⎥ ⎥ ⎦ ⎛h =⎜ o ⎜h ⎝ f For a round bar both wire drawingg and extrusion will ggive 1 ⎞ ⎛r ⎞ ⎟⎟ = 2 × σ o × ln ⎜⎜ o ⎟⎟ ⎠ ⎝ rf ⎠ For Tangential Stress i.C s at r = rf . R = o = ⎜ o ⎟ for round bar Af ⎜⎝ rf ⎟⎠ σ o (1 + B ) same equation except B. od.e.σ x = σ b ∴ Drawing stress (σ d ) = ln ⎡⎣ Bσ x − (1 + B ) σ o ⎤⎦ × 2B s ⎡ Bσ b − (1 + B ) σ o ⎤⎦ ∴C = ⎣ ro and τ x = μ Px = μ (σ o − σ x ) dσ x 2σ o 2 μ (σ o − σ x ) + + cotα = 0 dr r r Let μ cotα = B dσ x 2 = ⎡⎣ Bσ x − (1 + B ) σ o ⎤⎦ dr r dσ x 2 or = dr ⎡⎣ Bσ x − (1 + B ) σ o ⎤⎦ r Integrating both side Both Tresca's and Von-Mises criteria will g give σ x + Px = σ o 2 or Bσ x − (1 + B ) σ o = ( rC ) Dividing by r 2 dr and taking dx/dr = cotα we get dσ x 2 2τ + (σ x + Px ) + x cotα = 0 dr r r Vertical i l component off Px ≅ Px due d to small ll half h lf die di angles and that of τ x can be neglected neglected. 0 378 . 373 Clearance (VIMP) Clearance (VIMP) Clearance Clearance Contd Contd…. ‐ office furniture By S K Mondal y 370 ‐ frames for home appliances 371 372 374 375 Piercing (Punching) and Blanking Piercing (Punching) and Blanking ( h ) d l k y Piercing and blanking are shearing operations. strength. ‐ automotive bodies and frames. ‘clearance’ y Punching Punch = size of hole Die = punch size +2 clearance y clearance is determined with g following equation C = 0. y Die opening must be larger than punch and known as y The clearance. y Note: In pu punching c g cclearance ea a ce iss p provided ov ded o on Die e In Blanking is & provided For-2016clearance (IES. y In blanking.0032t τ Remember: In punching punch is correct size. 2C 377 Rev. the piece being punched out becomes the h workpiece k i and d any major j burrs b or undesirable d i bl features should be left on the remaining strip.e.GATE PSUs) on punch 376 y Total clearance between punch and die size will be Blanking Punching Page 107 of 210 twice these ‘C’ i. y Where τ is the shear strength of the material in y Blanking N/mm2(MPa) Die = size Di i off product d Punch = Die size ‐2 clearance y Remember: In blanking die size will be correct. strip y In piercing (Punching). p y Both done on some form of mechanical press.0 Sheet Metal y Product has light weight and versatile shape as compared to forging/casting Sheet Metal Operation y Most commonly used – low carbon steel sheet (cost.Workbook Ch‐16: Drawing Q. formability) y Aluminium and titanium for aircraft and aerospace y Sheet metal has become a significant material for. No 1 2 3 4 Option A B C B 5 6 7 8 9 A B B B 1. (Punching) the punch punch‐out out is the scrap and the remaining g strip p is the workpiece. Also determine the work done if the percentage [Where C is a constant and equal to 1.5 mm thick h k metall strip. steel shear stress = 240 N/mm2). respectively. [10‐Marks] Page 108 of 210 384 Minimum Diameter of Piercing f IAS‐2011 Main IAS‐2011 Main A rectangular hole of size 100 mm × 50 mm is to be made on a 5 mm thick sheet of steel having ultimate tensile strength and shear strength of 500 MPa and 300 MPa.075 given i th then C = 0. 3% The required punch diameter is Shear strength of sheet material = 294 MPa C = 0. The hole is made b punching by hi process.75 depending upon the profile] th fil ] The punching force for holes which are smaller than the stock thi k thickness may be estimated as follows: b ti t d f ll Fmax = penetration is 25 percent of material thickness.Example l Clearance in % Clearance in % y If the th allowance ll f the for th material t i l is i a = 0. and cutting a rectangular blank of 50 x 200 mm from a sheet of 1 mm thickness (mild steel.5 mm. Neglecting N l i the h effect ff off clearance.01 0 01 x thickness of the sheet A metal disc of 20 mm diameter is to be punched (a) 19.88 19 88 mm (b) 19 94 mm 19. the punching force (in kN) is (a) 300 (b) 450 (c) 600 (d) 750 For-2016 (IES. in each case : (i) Size of punch (ii) Size of die (iii) Force required. π dtσ 3 d t 382 GATE‐2014 385 For punching a 10 mm circular hole.12 381 Example l Estimate the blanking force to cut a blank 25 mm wide Press capacity will be = Fmax ×C and d 30 mm long l f from a 1. = Strength of punch.0 387 . h k The h punch h and d the h die clearance is 3%.94 (c) 20.5 mm.1 to 1.06 20 06 mm (d) 20 12 mm 20. y If clearance is 1% given then Also determine the die and punch sizes for punching a circular hole of 20‐mm diameter from a sheet whose thickness is 1. 4 Rev. 379 380 Punching Force and Blanking Force h d l k Capacity of Press for Punching and Blanking Fm ax = Ltτ f from a sheet h off 2 mm thickness. Calculate.t Piercing pressure.075 0 075 x thickness of the sheet GATE‐2003 Determine for a circular D i the h die di and d punch h sizes i f blanking bl ki i l disc of 20‐mm diameter from a sheet whose thickness is 1. iff the h ultimate shear strength of the material is 450 N/mm2.GATE & PSUs) 383 386 π σc × d 2 τs πd. Iff the h allowable crushing stress in the punch is 6 N N‐mm mm‐22. F the th plate l t material. For-2016 (IES.IES ‐ 2014 IES – 1999 A hole is to be punched in a 15 mm thick plate h having ultimate l shear h strength h off 3N‐mm‐2.0 396 . using punching force of 44 kN. 100 mm diameter. The ultimate shear stress is 550 N/mm2 . h Where p is percentage penetration required for rupture Ideal power in press ( P inW ) = E×N 60 [Where N = actual number of stroke per minute] y It distribute the cutting action over a period of time. With normal clearance on the tools. mm S = shear on the punch or die. The maximum value of t is (a) 0. shear is ground on the face of (Unit:Fmax a in kN and t in mm othrwise use Fmax a in N and t in m) the h die d or punch. the (a) 1 4 (c) 1 1 (b) 2 (d) 2 391 total work done remains same. plate would be of diameter equal to 1 (a) × Thickness of plate 4 1 (b) × Thickness of plate 2 (c) Plate thickness (d) 2 × Plate thickness 390 Energy and Power for Punching and Blanking Shear on Punch h h Id l Energy Ideal E (E in i J) = maximum i force f x punch h travell = Fmax × ( p × t ) y To reduce shearing force. ess. For p punching g the biggest gg hole.5 mm (b) 10 mm (c) 1 mm (d) 2 mm 388 IES‐2013 A hole of diameter d is to be punched in a plate of thi k thickness t For t. is to be punched in steel plate 5.GATE & PSUs) 394 Page 109 of 210 395 Rev. 2011 389 With a punch for which the maximum crushing stress is 4 times the h maximum shearing h stress off the h plate the biggest hole that can be punched in the plate. 392 Force required with shear on Punch F × pt F = max S 393 Example y A hole.1 to 1.75 depending upon the profile E×N A Actual l power in i press ( P iinW W)= 60 ×η WhereE is actual energy and η is efficiency of the press crushing stress is 4 times the maximum allowable shearing g stress.6 mm thick. t i l the th maximum i ISRO‐2008. h Give Gi suitable i bl Where p = penetration of punch as a fraction S shear on the punch or die. mm shear angle for the punch to bring the work within the capacity capac ty o of a 30 30T p press. the diameter of the smallest hole which can be punched is equal to (a) 15 mm (b) 30 mm (c) 60 mm (d) 120 mm A hole 35 mm is h l off diameter di i to be b punched h d in i a sheet metal of thickness t and ultimate shear strength 400 MPa. ratio of diameter of hole to plate thickness should Actual Energy ( E in J ) = Fmax × ( p × t ) × C y Shear only reduces the maximum force to be applied but be equal to: Where C is a constant and equal to 1. cutting is complete at 40 per cent penetration i off the h punch. mm The cutting blade is 400 mm long and zero‐shear (S = 0) is provided on the edge.2. y Nibbling Nibbli ‐ a single i l punch h is i moved d up and d down d rapidly. (b) What will be the cutting force if the punches are staggered. The h ultimate l shearing h strength h off the h materiall off the washer is 280 N/mm2. Neglect p g friction. idl y Used from low‐strength U d to blank bl k shapes h f l h materials. Statement for Linked Answer Questions: In a shear cutting operation. y Notching: Metal pieces are cut from the edge of a sheet. what will be the cutting force if both punches act together. ( ) Taking (c) T ki 6 % penetration 60% i and d shear h on punch h off 1 mm. GATE‐2010 Statement Linked 2 Statement for Questions: S f Linked Li k d Answer A Q i In a shear cutting operation. bl k 400 401 Dinking k y Lancing – A hole is partially cut and then one side is bent down to form a sort of tab or louver. no scrap. d l the h maximum force f ( kN) (in k ) exerted d is 399 (a) 5 (b) 10 (c) 20 (d) 40 y Slitting ‐ moving rollers trace out complex paths during y Trimming ‐ Cutting unwanted excess material from the periphery of a previously formed component. Assuming g force vs displacement curve to be trapezoidal. g The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0. y The shank of a die is either struck with a hammer or mallet or the entire die is driven downward byy some form of mechanical press. No metal removal.0 405 . strip t i or blank. The cutting blade is 400 mm long g and zero‐shear ((S = 0)) is p provided on the edge. Neglect friction.Example l GATE‐2010 Statement Linked 1 A washer h with i h a 12. y Squeezing ‐ Metal is caused to flow to all portions of a die cavity under the action of compressive forces. The final result is blanks that have extremelyy close tolerances.7 mm internal i l hole h l and d an outside id diameter of 25.GATE & PSUs) 403 Page 110 of 210 404 Rev.5 mm thick strip. fiber. a sheet of 5 mm thickness is cut along a length of 200 mm. the work done ((in J) is (a) 100 (b) 200 (c) 250 (d) 300 398 A shear of 20 mm (S = 20 mm) is now provided on the blade. y Perforating: Multiple holes which are very small and y Shaving h ‐ Accurate dimensions d off the h part are obtained b d by b removing a thin strip of metal along the edges.4 mm is to be made from 1. h 397 Fine Blanking l k Fine Blanking ‐ dies small Fi Bl ki di are designed d i d that h have h ll clearances and pressure pads that hold the material while it is sheared. a sheet of 5mm thickness is cut along a length of 200 mm. 402 y Steel Rules ‐ soft materials are cut with a steel strip shaped h d so that h the h edge d is the h pattern to be b cut. material This allows a simple p die to cut complex p slots. (a) Find the total cutting force if both punches act at the same time and no shear is applied pp to either p punch or the die. or cloth. For-2016 (IES. so that only one punch acts at a time. edges close together are cut in flat work material. The ultimate shear strength g of the sheet is 100 MPa and penetration to thickness ratio is 0. each time cutting off a small amount of material. cutting (like a can opener).2. 400 400 S S Assuming force vs displacement curve to be rectangular. i l such h as rubber. IAS – 2003 The in Th 'spring ' i back' b k' effect ff i press working ki is i ((a)) Elastic recoveryy of the sheet metal after removal of the load (b) Regaining the original shape of the sheet metal (c) Release of stored energy in the sheet metal (d) Partial recovery of the sheet metal pressure is released. spring back. replace mounting mounting. released there is an elastic recovery and the total deformation will g get reduced a little. p especially for attached to it. the h wear occurring on the h press bed b d is high.0 414 . For-2016 (IES.. ti when h th the strength. t th Higher Hi h the th yield i ld y At the th end d off a metal t l working ki operation. wear y Used to locate and hold the punch h in position.. 406 407 ISR0– 2013 408 Punching Press h Punch and Die material Spring back in metal forming depends on (a) Modulus of Elasticity y Commonly used – tool steel y For high production ‐ carbides (b) Load Applied (c) Strain Rate (d) None N off these h 409 410 Bolster plate l l 411 Bolster plate l l Contd.Elastic recovery or spring back l b k Elastic recovery or spring back l b k Contd. y This Thi is i a useful f l way off y Relatively cheap and easy to replace.. This y To compensate this. d f deformation.. Punch plate h l y When many dies are to run in the same press at different times. h h The h bolster plate is incorporated to take this wear. y It depends d d on the th yield i ld strength. greater spring back. the cold deformation be carried beyond the desired limit by an amount equal to the phenomenon is called as "spring back".GATE & PSUs) 412 Page 111 of 210 413 Rev. y Total deformation = elastic deformation + plastic y More important in cold working. y Attached to the press bed and the die shoe is then small punches. Knockout k y The stripper removes the stock from the punch after a piercing or blanking operation.03 For-2016 (IES.GATE & PSUs) (d) 6. mm t = stock thickness.5 mm with the cut at the edge or near a preceding cut = 0. mm k hi k K = stripping constant. The blanking g force required q to p produce a blank of 100 mm diameter from a 1. when ultimate shear (b) 3. ld 418 419 GATE‐2013 (PI) ( ) GATE – GATE – 2009 (PI) 2009 (PI) 417 420 ISRO‐2009 Circular blanks of 10 mm diameter are punched The force required to punch a 25 mm hole in a A disk di k off 200 mm diameter di t is i blanked bl k d from f a strip ti f from an aluminium l i i sheet h t off 2 mm thickness. for freeing a work piece from a die. kN stripping force kN L = perimeter of cut.Stripper Stripper Contd.0 423 .57 (a) 78 kN (b) 393 kN (c) 98 kN (d) 158 kN (a) 291 (b) 301 (c) 311 mild steel plate 10 mm thick..33 421 Page 112 of 210 422 Rev. The minimum punching force required in kN is stress of the plate is 500 N/mm2 will be nearly blanking force (in kN) is (a) 2. thi k Th The of an aluminum alloy of thickness 3.0103 for low‐ carbon steels thinner than 1. Ps = KLt Where Ps = stripping force.0207 for low‐ carbon steels above 1.5‐mm thickness = 0. mm The shear strength of aluminium is 80 Mpa.0241 for harder materials f h d l 415 416 Pitman Dowel pin l GATE 2011 The shear strength of a sheet metal is 300 MPa.29 (d) 321 (c) 5.5 mm thick sheet is close to (a) 45 kN (b) 70 kN (c) 141 kN (d) 3500 kN y It is a connecting rod which is used to transmit motion f from the h main drive d shaft h f to the h press slide. The material shear strength g to fracture is 150 5 MPa. y Knockout is usually K k i a mechanism. h i ll connected d to and d operated by the press ram.0145 for same materials but for other cuts f i l b f h = 0... = 0.2 3 2 mm. Shear on the punch reduces the maximum cutting force 2.83 24 83 (b) 24.0 (b) 4.00 427 GATE‐1996 IES – 1994 A 50 mm diameter disc a di di is i to be b punched h d out from f carbon steel sheet 1.17 425 GATE‐2002 GATE‐2008(PI) GATE ‐ GATE ‐ 2012 426 GATE‐2001 In operation.. Shear increases the life of the punch 4 The total energy needed to make the cut remains 4.0 kN.7 3 kN (c) 61. The required radial clearance between the die and the punch is 6% of sheet thickness.6 22 6 kN (b) 37. t the new blanking force in kN is (a) 3. if the diameter of the blanked part is increased to 1.15 (c) 49. Shear p provided on the p punch is 2 mm.00 50 00 mm (c) 50.89 24 89 (c) 25. The blanking g force during the operation will be (a) 22.3 kN 424 A blank of 50 mm diameter is to be sheared from a sheet of 2.00 (d) 49. respectively are (a) 50.GATE‐2007 GATE‐2004 The requirement in operation off Th force f i i a blanking bl ki i low carbon steel sheet is 5.0 . Shear increases the capacity of the press needed 3.70 and 50. The punch and die diameters (in mm) Calculate the p punch size in mm.30 (b) 50.925 49 925 mm (b) 50.06 mm Die allowance 0. the is I a blanking bl ki i h clearance l i provided id d on (a) The die (b) Both the die and the punch equally (c) The punch (d) Brittle the punch nor the die The force in punching and Th cutting i f i hi d blanking bl ki operations mainly depends on (a) The modulus of elasticity of metal (b) The shear strength of metal (c) The bulk modulus of metal (d) The yield strength of metal for this blanking operation.5 d and thickness is reduced to 0. Page 113 of 210 429 431 Consider the statements related C id h following f ll i l d to piercing and blanking: 1. operation respectively. shear is I sheet h bl ki h i provided id d on punches and dies so that (a) Press load is reduced (b) Good cut edge is obtained.6 kN (d) 94. The diameter of the punch should be (a) 49.5 (c) 5.075 mm (d) none of the above For-2016 (IES. for a circular blanking operation for which details are given below. obtained (c) Warping of sheet is minimized (d) Cut blanks are straight.00 and 50.4 0 4 t.01 (d) 25. The thickness of the sheet is ‘t’ and diameter of the blanked part is ‘d’.0 (d) 8. Size of the blank 25 mm Thi k Thickness off the th sheet h t 2 mm Radial clearance between punch and die 0.0 10 mm diameter holes di h l are to be b punched h d in i a steell sheet of 3 mm thickness.. Shear strength of the material is 400 N / mm2 and penetration is 40%.00 and 50.5 mm thickness. For the same work material.GATE & PSUs) 428 430 IES – 2002 In metall blanking.05 mm (a) 24.85 and 50.0 mm thick. unaltered due to provision of shear Which h h off these h statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 432 Rev. If the clearance per side between the punch and die is to be kept as 40 microns.00 and 35+0. then the nominal diameters of die and p punch are respectively p y (a) 30.GATE & PSUs) IES – 2004 IAS – 2007 In between punch I deciding d idi the h clearance l b h and d die di in i press work in shearing.040 (b) 35‐0.00 (d) 35+0. size punch size controls blank size Page 114 of 210 438 440 For operation the is F punching hi i h clearance l i provided id d on which 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 Rev. The work done during g this shearing g operation p is (a) 200J (b) 400J ( ) 600 J (c) (d) 800 J 435 GATE – 2007 (PI) ( ) A blank is bl k off 30 mm diameter di i to be b produced d d out off 10 mm thick sheet on a simple die.IAS – 1995 IES – 2006 In provided I blanking bl ki operation i the h clearance l id d is i ((a)) 550% on p punch and 550% on die (b) On die ( ) On (c) O punch h ((d)) On die or p punch depending p g upon p designer’s g choice In I which hi h one off the h following f ll i is i a flywheel fl h l generally ll employed? (a) Lathe (b) Electric motor (c) Punching machine (d) Gearbox 433 IES – 1997 439 Circular Ci l blanks bl k off 35 mm diameter di t are punched h d from a steel sheet of 2 mm thickness.6 mm and d 30 mm (c) 30 mm aand d 29.080 437 IAS – 2002 In steell washer.4 mm (d) 30 mm and 28. the following rule is helpful: (a) Punch size controls hole size die size controls blank size (b) Punch size controls both hole size and blank size (c) Die size controls both hole size and blank size (d) Die size controls hole size.080 and 35+0. microns the sizes of punch and die should respectively be (a) 35+0.8 mm 436 IAS – 1994 Which Whi h one off the h following f ll i statements is i correct?? If the size of a flywheel y in a p punching g machine is increased (a) Then the fluctuation of speed and fluctuation of energy will both decrease (b) Then the fluctuation of speed will decrease and the fluctuation uctuat o o of eenergy e gy will increase c ease (c) Then the fluctuation of speed will increase and the fl t ti off energy will fluctuation ill decrease d p and fluctuation of (d) Then the fluctuation of speed energy both will increase 434 IAS – 2000 For F 50% % penetration i off work k material.4 mm (b) 30. If 6% clearance is recommended.040 and 35‐0.6 mm and 29.040 and 35‐0.080 (c) 35‐0. i l a punch h with ih single shear equal to thickness will (a) Reduce the punch load to half the value (b) Increase the punch load by half the value (c) Maintain the same punch load (d) Reduce the punch load to quarter load For-2016 (IES.0 441 . The plate thickness is 4 mm and percentage penetration is 25. I a blanking bl ki operation i to produce d h the h maximum punch load used in 2 x 105 N.4 9. the metal wide varietyy of shapes. thickness very little or not at all. Reason(R): The flywheel stores energy when its speed p increase. Reason (R): The blank should be of required dimensions. l with a depth more than several times the thickness of y Cold C ld drawing d i uses relatively l i l thin hi metal.. Assisting withdrawal of the punch (B) Stripper S i 2. Ejection of the work‐piece work piece from die cavity (D) Knockout 4. dimensions (a) Both A and R are individually true and R is the correct explanation l off A (b) Both ot A aand d R aaree individually d v dua y ttrue ue but R iss not ot tthee correct explanation of A ( ) A is (c) i true t b t R is but i false f l (d) A is false but R is true 445 Drawing 446 447 Drawing Drawing y Drawing is a plastic deformation process in which a flat y Hot drawing is used for thick‐walled parts of simple sheet h or plate l is formed f d into a three‐dimensional h d l part geometries.IAS – 1995 IES – 2002. For-2016 (IES. all and produces parts in a y As a p punch descends into a mating g die.0 450 . (a) Both A and R are individually true and R is the correct explanation e planation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Which cutting Whi h one is i not a method h d off reducing d i i forces to prevent the overloading of press? (a) Providing shear on die (b) Providing shear on punch (c) Increasing die clearance (d) Stepping punches 442 IES – 2000 IAS – 2003 Match List I (Press‐part) (Press part) with List II (Function) and select the correct answer using the codes given below the lists: List‐I List‐II (Press‐part) (Function) (A) Punch plate 1. A i (A) In I sheet h bl ki i clearance must be given to the die. l changes h the h the metal. Holding the small punch in the proper position Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 ( ) 4 (c) 1 2 3 (d) 2 3 4 1 443 444 IES – 1999 Best position off crank operation in B ii k for f blanking bl ki i i a mechanical press is (a) Top dead centre (b) 20 degrees below top dead centre (c) 20 degrees before bottom dead centre (d) Bottom dead centre Assertion (A): metall blanking operation. Ad Advancing i the h work‐piece k i through h h correct distance (C) Stopper 3 3.GATE & PSUs) 448 Page 115 of 210 449 Rev. thinning h takes k place. p assumes the desired configuration. GATE(PI)‐2003 Assertion (A): A i (A) A flywheel fl h l is i attached h d to a punching hi press so as to reduce its speed fluctuations. the side walls increase in thickness as the height is increased. The Th maximum i generallyy to be one‐third of the drawing g limit is g force. Calculate the blank size for the (ii) Decide whether it can be drawn in a single draw. y Due to the radial flow of material. of 50 mm diameter and 100 mm height. is to be A symmetrical cup of circular cross section with d drawn f from l low carbon b steell sheet. y Draw D Cl Clearance Punch diameter = Die opening diameter – 2. drawing [10 Marks] [10 marks] For-2016 (IES.7 r ) when d < 10r For obtaining a cup of diameter 25 mm and height 15 A shell of 100 mm diameter and 100 mm height with mm by b drawing. drawing The required blank diameter is (a) 42 mm (b) 44 mm (a) 118 mm (b) 161 mm (c) 46 mm (d) 48 mm (c) 224 mm (d) 312 mm 451 452 The initial blank diameter required to form a cylindrical cup of outside diameter 'd‘ and totall height h h 'h' having h a corner radius d ' ' is 'r' obtained using the formula (a ) Do = d 2 + 4dh − 0. draw if drawn cup.0 459 . y A cylindrical vessel with flat bottom can be deep drawn by double action deep drawing.GATE & PSUs) 453 y Drawing Force ISRO‐2011 (d ) Do = d 2 + 4dh − 0. Rev. g y Easy with ductile materials.Blank Size Blank Size D = d 2 + 4dh IES – 1994 When d > 20r D = d 2 + 4dh − 0. h Neglecting l the h d diameter 40 mm and d height h h 60 mm with h a corner influence of thickness and corner radii: radius of 2 mm is to be obtained in C20 steel of 0.6 06 (i) Calculate the blank diameter mm thickness.5 25t 454 455 IAS – 2013 Main 456 Deep drawing d IFS‐2013 A cup.5r when15r ≤ d ≤ 20r D= ( d − 2r ) 2 + 4d ( h − r ) + 2π r ( d − 0.5r (b) Do = d + 2h + 2r (c) Do = d 2 + 2h 2 + 2r ⎡D ⎤ P = π dtσ ⎢ − C ⎥ ⎣d ⎦ y Blank Holding Force Blank holding force required depends on the wrinkling i kli t d tendency off the th cup. single step? y Drawing when cup height is more than half the diameter is termed deep p drawing. Will it be possible to draw the cup in maximum reduction p permitted is 4 40%.5r GATE‐2003 457 Page 116 of 210 458 y Deep drawing ‐ is a combination of drawing and stretching.4 mm is to be b produced d d by b be approximately cup drawing. d the h size off the h round d blank bl k should h ld the h corner radius d off 0. g (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation l i off A (c) A is true but R is false (d) A is false but R is true 464 465 For IES Only Die Design IFS ‐ IFS ‐ 2009 Progressive dies Perform two or more operations simultaneously in a single stroke component is k off a punch h press. l Reason ((R): ) Due to strain hardening.GATE & PSUs) 466 Page 117 of 210 467 Rev. drawing y In wall of the cup: simple tension y This ratio depends upon material. operation [10 – marks] For-2016 (IES. d the h third h d draw d off about b 30% % only. a compound die consists of the necessary sets off punches h and d dies. y Progressive dies y What Wh t is i deep d d drawing i process for f sheet h t metal t l forming? Explain the function of a blank holder.6 to 2. h k The h d ⎞ ⎛ Reduction = ⎜ 1 − ⎟ × 100% = 50% D⎠ ⎝ Th b rule: Thumb l First draw:Reduction = 50 % Second draw:Reduction = 30 % Third draw:Reduction = 25 % Fourth draw:Reduction = 16 % Fifth draw:Reduction = 13 % number of deductions necessary will be (a) One (b) Two (c) Three ((d)) Four 463 462 Assertion (A): draw in drawing A i (A) The Th first fi d i deep d d i operation i can have up to 60% reduction. (LDR) after f which hi h the h punch will pierce a hole in the blank instead of drawing. p y Limiting drawing ratio (LDR) is 1. To do more than one set of operations. etc. D/d.0 468 . in a single stroke of the ram. the subsequent q draws in a deep drawing operation have reduced p percentages.IES – 2008 A cylindrical can be li d i l vessell with i h flat fl bottom b b deep d drawn by (a) Shallow drawing (b) Single action deep drawing (c) Double action deep drawing (d) Triple action deep drawing Deep Drawability bl Stresses on Deep Drawing Stresses on Deep Drawing y The ratio of the maximum blank diameter to the y In flange of blank: d diameter off the h cup drawn d .3 460 461 IES – 1997 Limiting Drawing Ratio (LDR) y The average reduction in deep drawing IES – 1998 A cup of 10 cm height and 5 cm diameter is to be d = 0.5 05 D made d from f a sheet h metall off 2 mm thickness. so that h a complete l obtained for each stroke. holder g ratio and how is the drawing g ratio What is drawing For IES Only Compound dies All the necessary operations are carried out at a single station. i. material amount of friction uni‐axial uni axial present. g. d Bi‐axial tension and compression y There Th i a limiting is li i i drawing d i ratio i (LDR).e. di y Compound dies y Combination dies increased ? Combination C bi i dies di A combination die is same as that of a compound die with th main the i difference diff th t here that h non‐cutting tti operations ti such h as bending and forming are also included as part of the operation. the second draw up to 40% % reduction d and. k ears or lobes l b tend d to occur develop on the flange. 3. Progressive piercing and blanking die for making a simple washer. To reduce drawing forces. 472 In drawing operation. and the corners (if any). 4. For-2016 (IES. [ 6 Marks] [ 6 – M k ] 2. To improve die life 2 To reduce drawing forces 2. bottom. because of the anisotropy induced by the rolling operation. d IAS – 2007 y In drawing operation. may cause a thinning of the walls and a fracture at the flange. proper lubrication is essential for 1. proper lubrication I d i i l b i i essential for which of the following reasons? 1. To improve die life.GATE & PSUs) Flange Wrinkle Wall Wrinkle 475 Page 118 of 210 476 Rev. 2. 3 and 4 onlyy (a) 1 and 2 onlyy (c) 3 and 4 only (d) 1. To reduce temperature 4. which may also extend to the wall of the cup. To reduce temperature.0 477 . To improve surface finish S l t the Select th correctt answer using i the th code d given i b l below: (b) 1. 3 and 4 473 is i 474 Defects in Drawing ‐ f wrinkle kl Defects in Drawing ‐ f Fracture Defects in Drawing ‐earing f y An blank A insufficient i ffi i bl k holder h ld pressure causes wrinkles i kl to y Further. Lubrication b compound and progressive dies. making a simple washer For IES Only IAS‐1996 Compound die performs (a) Two or more operations at one station in one stroke (b) Two or more operations at different stations in one stroke t k (c) high frequency sound wave (d) High frequency eddy current Back 469 Back 470 471 For IES Only IFS‐2013 y Differentiate among the simple. F h too much h off a blank bl k holder h ld pressure and d friction fi i y While Whil drawing d i a rolled ll d stock. 3. Part is blanked (a) and subsequently pierced (b) The blanking punch contains the die for piercing.For IES Only For IES Only Method for making a simple washer in a compound piercing and blanking die. To improve surface finish. S – 1 (c) P – 2. p y. Q – 3.0 . Clearance between tools 3. For-2016 (IES. S – 4 (d) P – 4. Large grain size S.GATE & PSUs) products from metal stock that has a coarse grain size. Speed of the press p p 2. S – 2 486 Rev. 478 St t h t i (lik L d Lines) Stretcher strains (like Luders Li ) y Caused due to C d by b plastic l ti deformation d f ti d t inhomogeneous i h 479 y These lines can criss‐cross the surface of the workpiece p and lubrication may be visibly objectionable. Q – 5. respectively. Fine grain size 6 6. because there is no bl k h ld blank‐holder (c) Tensile e s e st stress ess in o onee d direction ect o aand d co compressive p ess ve in the one other direction (d) Compressive C i in i two t di ti directions and d tensile t il in i the th third direction Page 119 of 210 483 485 Match in I and M h the h items i i columns l d II. This defect is known as miss strike. blanks bl k show h d wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are. 481 IAS – 1997 484 In drawing a tendency to I the h deep d d i off cups. R – 3. Q – 3. 4 Defects in Drawing – f miss strike k Defects in Drawing – f Orange peel l y Due off the D to the h misplacement i l h stock. Increase blank holder pressure (B) High blank holder pressure and high friction. 2 and 3 (a) d (b) d (b) 2. S – 2 (b) P – 4. R – 1. k unsymmetrical i l y A surface roughening (defect) f h i (d f ) encountered d in i forming f i flanges may result. Material properties M i l i 4. R – 1. Excessive blank holding force (a) P – 6. 3 and 4 ((c) 1. y Low carbon steel and aluminium shows more stretcher strains. insufficient yielding. 2 2. Q – 5. Anisotropy R.IES‐1999 Consider the following statements: E i in a drawn cup can be due non‐uniform Earing i d b d if 1. II Column I Column II P. Reduce bl k holder blank h ld pressure and d apply l lubricant l bi t (C) High temperature causing increase in circumferential length: Apply coolant to blank ((D)) Buckling g due to circumferential compression. (A) Buckling due to circumferential compression. overly large grains usually the result of annealing at too high a temperature. Wrinkling 1. Yield point elongation Q Orange peel Q. p . Insufficient blank holding force 5. Earing 4. y It is due to uneven flow or to the appearance of the GATE‐2006 Identify Id if the h stress ‐ state in i the h FLANCE portion i off a PARTIALLYDRAWN CYLINDRICAL CUP when deep ‐ drawing without a blank holder (a) Tensile in all three directions (b) No stress in the flange at all. 3 and 4 ) . R – 6. Stretcher strains 3. Blank holding 4 g Which of these statements are correct? ( ) 1. decrease blank holder pressure 482 GATE‐1999 Which factor promotes the Whi h one off the h following f ll i f h tendency for wrinking in the process of drawing? (a) Increase in the ratio of thickness to blank diameter of work material (b) Decrease in the ratio thickness to blank diameter of work k materiall (c) Decrease ec ease in tthee holding o d g force o ce o on tthee b blank a (d) Use of solid lubricants 480 GATE‐2008 Surface scratches Surface scratches y Die or punch not having a smooth surface. 2 and 4 ) .3 4 ((d) 1. IES – 1999 Spinning IAS – 1994 Consider the statements: Earring in C id h following f ll i E i i a drawn cup can be due to non‐uniform 1.5 mm (c) Rolling and stretching (d) Bending and stretching. Th factors The f t responsible ibl for f the th vertical ti l lines li parallel ll l to t the axis noticed on the outside of a drawn cylindrical cup would ld include. The finish at the corners of the die is poor. The punch and die alignment is not proper. A mandrel d l (or ( die di for f internal i l pieces) i ) is i placed l d on a rotating axis (like a turning center). pushing the h blank bl k against the h mandrel. 4. Material properties 4. l d (a) 2. y Localized pressure is applied through a simple round‐ended wooden or metal tool or small roller.GATE & PSUs) 493 Page 120 of 210 494 Rev.0 mm 2 0 mm (d) 1 5 mm 1. 2 and 3 (b) 2. 2. t d and d the th partt is i removed d and d trimmed. 3 and 4 488 489 Spinning Spinning y Spinning S i i i a cold‐forming is ld f i operation ti i which in hi h a rotating disk of sheet metal is shaped over a male form. 2 The finish at the corners of the punch is poor. Thee pa partt co conforms o s to tthee sshape ape o of tthee mandrel a d e ((with t some springback). Clearance between the p small. 490 491 492 GATE‐1992 tc = tb sinα IES – 1994 The thickness of the blank needed to produce. 3 and 4 (c) 1. 2. 3 A roller is pushed against the material near the 3. by The mode of deformation of the metal during power spinning a missile cone of thickness 1.0 mm 3 0 mm (b) 2 5 mm 2. 3 and 4 (d) 1. 3 and 4 (b) 1 and 2 (c) 2 and 4 (d) 1.5 mm l f h k spinning is and half cone angle 30°. or mandrel. 3. 2 and 4 487 Consider C id the h following f ll i factors f punch and the die is too 1. center of rotation.5 mm (b) Stretching St t hi (c) 2. 5. d l 4.0 495 . stretching For-2016 (IES. Blank holding Whi h off these Which th statements t t t are correct? t? (a) 1. is and half cone angle 30 is ( ) Bending (a) B di (a) 3. and slowly moved outwards. poor 3. Speed of the press 2 Clearance between tools 2. which traverses the entire surface of the part 1. The Th process is i stopped. A blank or tube is held to the face of the mandrel. 0 a die or mating workpiece.GATE & PSUs) 502 Page 121 of 210 503 around d a cylinder.For IES Only For IES Only HERF IFS‐2011 y High Energy Rate Forming. y HERF makes it possible to form large work pieces and difficult‐to‐form metals with less‐expensive equipment and tooling required. y No N springback i b k 496 497 For IES Only Underwater explosions. f forming i can be b utilised tili d to t form f a wide id variety i t off metals. 498 For IES Only For IES Only U d Underwater explosions l i U d Underwater Explosions E l i y A shock wave in the fluid medium (normally water ) is Electro‐magnetic Electro magnetic (the use of rapidly formed magnetic fields). y A large capacitor bank is discharged. y A capacitor bank is charged through the charging circuit. y Used for bulging operations in small parts. materials Internal combustion of gaseous g mixtures. l d resulting lti i a spark in k within the electrode g gap p to discharge g the capacitors. aircraft industries and automobile related components. t l the th discharge induces a secondary current in the workpiece. p surge through h h a coiled l d conductor. . t l from f [2‐marks] g gy g( ) High Energy Rate Forming(HERF) aluminum to high strength alloys. HERF generated d by b detonating d an explosive l charge. y Employed Pneumatic‐ P i mechanical means in Aerospace. For-2016 (IES. b tl a switch it h is i closed. y Applied a large amount of energy in a very sort time interval. y TNT and d dynamite d i for f higher hi h energy and d gun powder d for f lower energy is used. d y If the coil has been placed within a conductive cylinder. used y Used for parts of thick materials. li d or adjacent dj t the th flat fl t sheet h t off metal. producing a current subsequently. 499 500 For IES Only 501 For IES Only For IES Only Electro‐hydraulic Forming l h d l Electromagnetic or Magnetic Pulse Forming y An operation using electric discharge in the form of y Based B d on the h principle i i l that h the h electromagnetic l i field fi ld off sparks to generate a shock wave in a fluid is called electrohydrulic forming. causing it to be repelled from the coil and conformed to 504 Rev. an induced current always opposes the electromagnetic field of the inducing current. also known as HERF or explosive Compare metal spinning with press work. h Underwater spark discharge (electro discharge (electro‐ hydraulic). y Energy level and peak pressure is lower than underwater explosions but easier and safer. y Why might less springback be observed in HERF? (b) Electro‐hydraulic El t h d li forming f i [5 marks] (c) Rotary forging (d) Forward extrusion For-2016 (IES. method magnetic field produced by eddy currents is used to create force f b between coil il and d workpiece. y This process is most effective for relatively thin materials (0. or to permanently assemble component parts.25 mm thick). ( ) Both (a) B th A and d R are individually i di id ll true t and d R is i the th correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 508 IES – 2007 Assertion (A) : In the high energy rate forming method. (c) Explosive forming (d) electro electro‐hydraulic hydraulic forming 505 506 507 IES 2010 JWM 2010 Assertion (A) : In magnetic pulse‐forming pulse forming method. y The workpiece must be electrically conductive but need not be magnetic.GATE & PSUs) 511 Page 122 of 210 512 Rev.0 513 .25 to 1. Reason (R): The gas pressure and rate of detonation yp of explosives.For IES Only For IES Only IES 2011 Electromagnetic or Magnetic Pulse Forming y The Th process is i very rapid id and d is i used d primarily i il to expand d or contract tubing. k i y for the workpiece p Reason ((R)) : It is necessary material to have magnetic properties. magnetic High energy rate forming process used for forming components from thin metal sheets or d f deform thin hi tubes b is: i ( ) Petro‐forming (a) P t f i (b) Magnetic pulse forming y Short life of the coil is the major problem. p can be controlled for both types (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true 509 IES – 2009 IES – 2005 Which one of the following metal forming processes is not a high h h energy rate forming f process? ( ) Electro‐mechanical (a) El h i l forming f i (b) Roll‐forming R ll f i (c) Explosive forming (d) Electro‐hydraulic Electro hydraulic forming 510 IES‐2013 Conventional IES‐2013 Conventional Which one of the following is a high energy rate Magnetic forming is an example of: forming process? (a) Cold forming (b) Hot forming ( ) Roll (a) R ll forming f i (c) High energy rate forming (d) Roll forming y Name at least four methods by which high energy release rates are obtained. the explosive forming has proved to be an excellent ll method h d off utilizing ili i energy at high hi h rate and d utilizes both the high explosives and low explosives. 521 Rev.. and the workpiece conforms very closely to the shape of the tool. 517 518 519 For IES Only Ironing Contd.. Ironing Force Embossing b y Neglecting off the N l i the h friction f i i and d shape h h die.. the form blocks can often b made be d off wood. di the h ironing i i y It is a very shallow drawing operation where the depth of force can be estimated using the following equation.5 mm thick is to unequall biaxial hi k sheet h i subject bj bi i l stretching and the true strains in the directions of stretching are 0. less than those normally encountered in bending or o g..414 (b) 1. y Popular in the aircraft industry and is frequently used to f form aluminum l and d stainless l steell y Low‐carbon L b steell can be b stretch h formed f d to produce d l large panels for the automotive and truck industry.0 522 . 514 Stretch Forming h Contd. g y Examples p of ironed p products include brass cartridge g cases and the thin‐walled beverage can. Stretch Forming h Contd.. lengthened while the thickness of the base remains unchanged. y There is very little springback...GATE & PSUs) 520 Page 123 of 210 unchanged.. forming.09.. thickness y The walls are thinned and lengthened. tool y Because the forces are so low.304 ( ) 1...Stretch Forming h Stretch Forming h Contd. d low‐melting‐point l lti i t metal..362 (c) (d) 289 y The process of thinning the walls of a drawn cylinder by passing it between b a punch h and d die d whose h separation is less than the original wall thickness. the forces on the form block are far tensile stretching. the h draw d is limited l d to one to three h times the h thickness h k off the metal. t l or even plastic. The final thickness of the sheet in mm is (a) 1. industry 515 516 Ironing GATE‐2000 A 1.. y A sheet of metal is gripped by two or more sets of jaws that stretch it and wrap it around a single form block. metal and the material thickness remains largely ⎛t ⎞ F = π dt ttσ av ln l ⎜ o⎟ ⎝ tt ⎠ For-2016 (IES.....05 and 0.. y Because most of the deformation is induced by the g. y Produce large sheet or limited P d l h t metal t l parts t in i low l li it d quantities. Lb = α(R+kt) α(R+kt) where R = bend radius k k = constant (stretch factor) ( h f ) For R > 2t k = 0.5 ((c)) 1: 2 : 1 ((d)) 1: 1 : 1 Edge-Bending For U or channel bending force required is double than V bending For U or channel bending force required is double than V – For edge bending it will be about one‐half that for V ‐ bending 526 527 528 For IES Only Example l Spanking k GATE‐2005 y Calculate the bending force for a 45o bend in aluminium blank. bi‐axial compression and bi‐axial tension is there.40 0. MP Die Di opening i factor f = 1.67 0. di i mm K = die-opening factor . UTS = 455 MPa. h y Bending parts depends upon material properties at the location of the bend.6 mm. coins medals and similar articles. For-2016 (IES.GATE & PSUs) 529 Page 124 of 210 530 Rev.67 W > = 16t 1. Di opening Die i = 8t.5 (b) 2: 1 : 0.33 A 2 mm thick is at an angle hi k metall sheet h i to be b bent b l off one radian with a bend radius of 100 mm. volume shape. the bend allowance is (a) 99 mm (b) 100 mm (c) 101 mm (d) 102 mm 2mm y During bending.20 2. bend length = 1200 mm. 1.33 t = thickness of material hi k f i l α = bend angle ( g (in radian) 523 524 525 For IES Only Bending Force Bending Force Klσ ut t 2 Bending F= y The strain on the outermost fibers of the bend is ε= 1 2R +11 + t IES 1998 IES‐1998 The bending force required for V‐bending. h b d y At bend.Bending Coining Bending y After basic shearing operation. (can be used followin table) Condition V-Bending U-Bending W < 16t 1. mm w = width idth off die-opening.5. mm of 2 σ ut = Ultimate Ulti t tensile t il strength. any bulging developed earlier will be 1 radian completely presses or “spanked” out.6 (a) 1 : 2 : 0. t th MPa MP (N/mm (N/ ) t = blank thickness.5 For R < 2t k = 0. U‐ w b di and bending d Edge Ed bending b di will ill be b in i the th ratio ti Where l =Bend length = width of the stock.0 531 . Blank thickness. surfaces punch and die are completely p y closed on so that when the p the blank. y Coining is used for making coins.33 2. we can bend a part to give it some y Coining is essentially a cold‐forging operation except for the h fact f that h the h flow fl off the h metall occurs only l at the h top layers and not the entire volume. the area of the sheet under the punch h a tendency has d to flow fl and d form f a bulge b l on the h outer surface. Bend allowance. If the stretch factor is 0. surface y The lower die should be provided with mating surfaces. Impact p extrusion R. H rolling Hot lli g structural shapes p 33. Flow forming 2.9 9 mm (ii) 50 x 200 mm (iii) 120 KN Rev. Ratio R ti off yield i ld stress t t ultimate to lti t stress. Shear R.Rate of increase of yyield stress relative to progressive amounts of cold work. 3 3. Tension and Compression 5. Closed matching dies Codes:A B C D A B C D (a) 1 3 4 2 (b) 3 1 4 2 For-2016 (c) 1 3 2 (IES.51 KN ((iii)) 45. Shear angle B. (ii)B 7 A 17 8 A 18 B 9 A 19 B 10 C 20 C Option O ti D 51. Rod and Wire C. 3‐S. Coiled stock C Roll forming C. 4‐Q (d) 1‐P. Collapsible tubes 4. 3 3. t 2. h d i Which of the above statements is/are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1 only (d) 1. Extrusion 3. Option Q. Shear S.9 49 9 x 199 199. Coining (a) P‐1 Q‐4 R‐6 S‐3 (b) P‐4 Q‐5 R‐2 S‐3 ( ) P‐1 Q‐5 R‐3 S‐2 (c) ( ) P‐5 Q‐1 R‐2 S‐2 (d) Match off parts) M t h List Li t I (Process) (P ) with ith List Li t II (Production (P d ti t ) and select the correct answer using the codes given below the lists: List‐I List‐II A Rolling A. 3‐Q.0 540 . Wire Drawing 3. Wide variety of shapes with thin walls D. Processes Associated state of stress P. Forging 2. Deep Drawing 4 p g (a) 1‐S. Blow l moulding ld 6. Drawing 4. Deep Drawing 4. Tension and Shear Codes:P Q R S P Q R S (a) 2 1 3 4 (b) 3 4 1 5 ( ) 5 (c) 4 3 1 (d) 3 1 2 4 532 GATE ‐2012 Same Q in GATE‐2012 (PI) Match the following metal forming processes with their associated stresses in the workpiece. 2 and 3 Page 125 of 210 539 Q. 3‐Q. Stretch Forming 2. Tensile and compressive i S. Blanking g 1.1 1 mm (iii) 7. 2‐Q. 3‐R. 4‐R ( ) 1‐P. Embossing 4. 3. R lli 1. P k i containers i f liquid for li id 2. Q No N O ti Q No N O ti Q No N 1 A 11 C 21 2 C 12 C 22 3 B 13 C 23 4 A 14 D 5 A 15 A 6 D 16 A 24 (i)A.GATE‐2007 Match for metal M t h the th correctt combination bi ti f following f ll i t l working processes. Long S. p Metal forming process 1. Option Q. Blanking 4. Injection moulding Q Packaging Q.GATE 4 (d)& PSUs) 3 1 2 4538 534 536 WorkBook IES 2010 Ch‐18: Sheet Metal Forming Consider the following follo ing statements: statements The material p properties p which p principally p y determine how well a metal may be drawn are 1. Coining 2 Wire Drawing 2. 2‐P. 4‐R (c) Type of stress P. 2‐R.23 mm. Tensile Q Shear Q. Mandrel D.24 5 KN For 10 mm hole (i) 10 mm (ii) 10 10. Moulded luggage 1. Compressive p (b) 1‐S. Blanking 1. Di Discrete parts B. Solid and hollow parts 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 2537 535 IAS – 1997 Match process)) with M h List‐I Li I (metal ( l forming f i i h List‐II Li II (Associated feature) and select the correct answer using the codes given below the Lists: List‐ll List List‐ II List A. Transfer moulding 5. Compression R Coining R. Rate R off work k hardening. 2‐P.27 KN (ii) 33.54 KN For 50 x 200 mm (i) 49. Flat plates and sheets 5 5. Tension Q. 5 (i) 50. 4‐S 533 GATE‐2004 IAS – 1999 Match M h the h following f ll i Product Process P. the molten metal is forced through a small ll orifice f and d broken b k up by b a stream off compressed d air. (a) Both A and R are individually true and R is the correct co ect eexplanation p a at o o of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A is false but R is true (d) 544 Assertion off A i (A): (A) In I atomization i i process off manufacture f metal powder. Al. l Reason ((R): ) Atomization method is an excellent means of making powders from high temperature metals..Manufacturing of Powder Manufacturing of Powder Atomization using a gas stream Powder Metallurgy Powder Metallurgy y Powder metallurgy is the name given to the Powder Metallurgy process byy which fine p p powdered materials are blended. Cu. low melting point materials brass. Pb etc. Reason (R): The metallic powder obtained by atomization p process is q quite resistant to oxidation. t i ti Reason (R): In atomization process inert‐gas or water cannot be used as a substitute for compressed air. Ni and 547 Grinding g This metallic powder is nothing but the unburnt tiny chips formed during the process of grinding. tungsten. y The h irregular i l sponge‐like lik particles i l are soft. molybdenum. For-2016 (IES. Molten metal is f forced d through th h a small orifice and is disintegrated by a jet of compressed air. By S K Mondal 542 541 IAS – 2003 IES – 1999 Assertion (A): off a A ti (A) Mechanical M h i l disintegration di i t ti molten metal stream into fine particles by means of a jet j t off compressed d air i is i known k as atomization. materials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both B th A and d R are individually i di id ll true t b t R is but i nott the th correct explanation of A (c) A is true but R is false ((d)) A is false but R is true Cobalt. Zn. air inert gas or water jet It is used for jet. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation l i off A (c) A is true but R is false (d) A is false but R is true 545 Manufacturing of Powder Manufacturing of Powder 546 Manufacturing of Powder Manufacturing of Powder Reduction GATE ‐2011 (PI) GATE ‐2011 (PI) y Metal oxides are turned to pure metal powder when exposed to below melting point gases results in a product of cake of sponge metal. Tn.GATE & PSUs) 543 IAS – 2007 Assertion (A): method off A i (A) Atomization A i i h d for f production d i metal powders consists of mechanical disintegration of molten l stream into fine f particles. pressed into a desired shape (compacted). (a) Atomization compressible and give compacts of good pre‐sinter compressible.0 549 . f readily dil Whi h off Which th the f ll i following powder d production d ti methods produces spongy and porous particles? ((b)) Reduction of metal oxides (“green”) g strength g ((c)) Electrolytic y deposition p y Used for iron. and then heated (sintered) in a controlled ll d atmosphere h to bond b d the h contacting surfaces of the particles and establish the desired p p properties. brass bronze. (d) Pulverization Page 126 of 210 548 Rev. 557 Rev. ferro‐silicon are produces 550 551 IES ‐ 2012 Manufacturing of Powder In I electrolysis l l i ((a)) For making g copper pp p powder. y For making copper powder. materials Condensation – Metals are boiled to produce metal vapours and then condensed to obtain metal powders. Mg. Used for Zn. Mg Cd. 552 GATE‐2014 (PI) Granulations ‐ as metals are cooled they are stirred rapidly Machining ‐ coarse powders such as magnesium Milling g ‐ crushers and rollers to break down metals. Atomization 3. 3 and 4 (b) 1.. Mechanical pulverization 2. y The cost of manufacturing is high. 3 and 4 (d) 1. Cd 553 555 IES 2010 IAS – 2000 Consider C id the h following f ll i processes: p 1. h d dried d i d and d pulverized to the desired g p grain size. testing A standard sieves with mesh size varies between (100) and (325) are used to determine particle ti l size i and d particle ti l size i distribution di t ib ti off powder d in i a certain range. Chemical Ch i l reduction d i 4. The cathode plated are taken out and powder d is i scrapped d off. powder aluminum plate is made anode (c) High amperage produces powdery deposit of cathode metal eta o on aanode ode (d) Atomization process is more suitable for low melting point i t metals t l y Used for iron. 2 and 4 For-2016 (IES.Only for IES Manufacturing of Powder Manufacturing of Powder Manufacturing of Powder Manufacturing of Powder IES‐2013 Conventionall Comminution C i i Electrolytic Deposition Explain the terms comminution and reduction used in y Granular G l material.0 558 . 2 and 3 (c) 1. copper plates are placed as [ 2 marks] y Process is used for production of very fine powders such as are required for injection moulding . Brittle materials such as inter inter‐metallic metallic compounds. whereas the aluminium plates l t are placed l d in i the th electrolyte l t l t to t actt as cathode.. copper pp p plate is made cathode in electrolyte tank (b) For making aluminum powder. the copper pp g gets deposited p When DC current is p on cathode. t i l which hi h may be b coarsely l atomized t i d p powder. Sintering 4 g Which of these processes are used for powder preparation ti in i powder d metallurgy? t ll ? (a) 2. copper. silver anode in the tank of electrolyte. th d passed. y Particle size distribution: refers to amount of each particle size in the powder and have a great effect in determining flowability. can be determined either by pouring the powder through a sieve or by microscopic testing. compounds ferro ferro‐alloys alloys ‐ ferro ferro‐ chromium. apparent density and final porosity of product. ff The Th powder d is i washed. Shooting ‐ drops of molten metal are dropped in water. . Used for brittle materials. is fed in a stream of g gas under p pressure through g a venturi and is cooled and thereby embrittled by the adiabatic di b i expansion i off the h gas before b f i i i impinging on a g on which the g granules shatters target y Process is similar to electroplating. used Which methods Whi h one off the h following f ll i h d is i NOT used d for producing metal powders? (a) Atomization (b) Compaction (c) Machining and grinding (d) Electrolysis for low melting point materials. electroplating powder metallurgy.GATE & PSUs) 554 556 Metallic can be M t lli powders d b produced d d by b (a) Atomization (b) Pulverization (c) Electro‐deposition process (d) All off the h above b Page 127 of 210 Characteristics Ch i i off metall powder: d y Fineness: refers to p particle size of p powder. Brittleness reduces.IES – 1999 Conventional Questions y Discuss the terms fineness and size Di h fi d particle i l i distribution in powder metallurgy. respectively and mean mass diameter define the particle size distribution. Blending compacting. y Heat to 0. [IES‐2010. t y Lubricants such as graphite or stearic acid improve the flow Why lubricants are used to mix the metal powders? [ 2 marks] characteristics and compressibility at the expense of reduced strength. is the ratio of particle size diameters taken at 84. sintering and sizing (b) Blending. Thermoplastics or a water water‐soluble soluble methylcellulose binder is used. y Binders produce the reverse effect of lubricants. blending and sintering (d) Compacting.0 567 . ~70% density y Part shrinks in size y May be done cold or warm (higher density) y Density increases. recrystalization P ti l bi d t th diff i t li ti product complexity) and grain growth takes place. Porosity St th i B ittl d P it decreases. diffusion. g For-2016 (IES.GATE & PSUs) 565 Page 128 of 210 566 Rev. g g p y Still very porous. ).75*T melt y Particles bind together. sizing. 2 Marks] Ans. compacting sizing and sintering (c) Compacting. y Most lubricants or binders are not wanted in the final product and are removed ( volatilized or burned off) 562 563 C ti Compacting Compacting 564 Sintering y Powder is pressed into a “green compact” y Controlled atmosphere: no oxygen y 40 to 1650 MPa pressure (Depends on materials. compacting. The processes in Th correct sequence off the h given i i manufacturing by powder metallurgy is (a) Blending. Fineness: Is the diameter of spherical shaped particle and mean diameter of non‐spherical shaped particle. Toughness increases. up to 95% y Strength increases. sizing and sintering Particle size distribution: Geometric standard deviation ((a measure for the bredth or width of a distribution).1 and 50% of the cumulative undersized weight plot. 559 560 Blending l d 561 IES‐2013 Conventionall y Blending can be either Bl di or mixing i i operations ti b done d ith dry d or wet. blending. simultaneously y No lubricant is needed yU Used in the p production of billets of super‐alloys. etc. (Due to high y The Th flexible fl ibl mould ld is i then h pressurized i d by b means off temperature) high‐pressure water or oil. t ll sintering i t i off a componentt (a) Improves strength and reduces hardness (b) Reduces brittleness and improves strength The off a powder part Th rate off production d i d metallurgy ll depends on (a) Flow rate of powder (b) Green strength of compact (c) Apparent density of compact (d) Compressibility of powder IES – IES – 2007 Conventional 2007 Conventional y Metal powders are compacted by many methods. where the integrity y High and uniform density can be achieved 571 572 IAS – 1997 For-2016 (IES. (a) Both A and R are individually true and R is the correct co ect eexplanation p a at o o of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false ( ) A is false but R is true (d) 574 Page 129 of 210 575 of the materials is a prime consideration 573 IES – IES – 2011 Conventional 2011 Conventional y What is isostatic Wh t i i t ti pressing of metal powders ? i f t l d ? y What are its advantage ? [ 2 Marks] Rev. p y . the pressure is equal in all directions which permits uniform density of the metal powder. rubber bb or some other h elastomer l materiall y The flexible mould is made of sheet metal. Reason (R): In the process of isostatic pressing. methods g is required q to achieve which but sintering property? What is hot iso‐static pressing? [ 2 Marks] (c) Improves hardness and reduces toughness (d) Reduces porosity and increases brittleness 568 Cold Isostatic ld Pressing (CIP) ( ) y The powder is contained in a flexible mould made of 569 570 H t I t ti Pressing (HIP) P i (HIP) Hot Isostatic Cold Isostatic Pressingg y Is carried out at high g temperature p and p pressure using ga gas such as argon. high‐ g speed steels.0 576 .IES – 2002 GATE ‐2010 (PI) GATE ‐2010 (PI) I powder In d metallurgy.GATE & PSUs) and d Assertion (A): dimensional are A ti (A) Close Cl di i l tolerances t l NOT possible with isostatic pressing of metal powder d in i powder d metallurgy t ll t h i technique. oil (same pressure in all y Compaction C i directions)) sintering i i are completed l d simultaneously. ceramics. titanium. it is deposited p into a cooler For IES Only elevated temperature. y Complex shapes that are impossible with conventional compaction.For IES Only S Spray Deposition D iti y Spray deposition S d iti is i a shape‐generation h ti process. impregnation y A part surface can be infiltrated with a low melting point metal to increase density. y Achieve density above 99%. (a) Atomiser (b) Spray chamber with inert atmosphere (c) Mould for producing preforms. y High production rate.GATE & PSUs) 581 583 582 Production of magnets d f y For parts. ti mechanical h i l properties ti same as wrought ht product d t 577 y Sheet metal for electrical and electronic components and for coins can be made by this process. 578 579 For IES Only l Explosive Compaction y High Energy Rate Forming (HERF) or Explosive Forming of the metal powders at rather higher velocities 3500 m/s than h that h off the h usuall speed d off compaction i during d i the h ordinary die compacting. heat treating and machining operations can also b used. sintered. be d y Al‐Ni‐Fe is used for permanent magnets p g Page 130 of 210 584 y Sintering is done in a wire coil to align the magnetic poles of the material y H2 is used to rapidly cool the part (to maintain magnetic alignment) y Total shrinkage is approximately 3‐7% (for accurate parts an extra sintering step may be added before magnetic alignment) li t) y The sintering temperature is 600°C in H g p 2 Rev. y After the metal is atomised. ductility and impact resistance. y Plating. y The powders are put in a container and are forced by a ram between two rotating rolls. y Basic components of a spray deposition process y The powder‐polymer mixture is then injected into split dies. a sintering F high hi h tolerance l i i part is i put back b k into i y 50:50 Fe‐Al alloys is used for magnetic parts F Al ll i d f i a die and repressed. In general this makes the part more accurate with a better surface finish. preforms preheated to remove the binder and. fine grain structure. Another method uses vacuum acuum first. as a polymer or a wax‐based binder. For IES Only Metal Injection Moulding Metal Injection Moulding R ll C Roll Compaction ti y Fine metal powders are blended with an organic binder such y Powders are compacted by passing between two rolls rotating in opposite direction.5 m/s. finally. p . y Volumetric shrinkage during sintering is very high. y A part has many voids that can be impregnated. impregnated One method is to use an oil bath. process y The rolling g p processes can be carried out at room or at y Good dimensional accuracy. y Good mechanical properties. first then impregnation. molten component/particle wetting..0 585 . rolls and is compacted into a continuous strip at speeds of up to 0. hardness. capillary action of the liquid 580 ISRO 2013 ISRO ‐2013 Features of PM products f d Following is a process used to form powder metal to shape h ( ) Sintering (a) Si i (b) Explosive E l i Compacting C ti (c) Isostatic Molding (d) All of these For-2016 (IES. from the lower MP component. mayy exist y Alloying may take place at the particle‐particle interface y Molten M l component may surround d the h particle i l that h has h not melted y High compact density can be quickly attained y Higher sintered strength y Important variables: y More uniform density distribution y Nature of alloy. y Higher green densities For IES Only Liquid Phase Sintering y During sintering a liquid phase. preformed mould. strength. such as cobalt ( 5 to 12%).g. 2 Wastage of material. Size and shape 2. Extreme p purityy p product 2. 3 and 4 (b) 2 and 3 (c) 1. (a) Both Statement (I) and Statement (II) are individuallyy true and Statement ((II)) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false ( ) Statement (I) (d) ( ) is false but Statement (II) ( ) is true 590 Consider C id the h following f ll i factors: f p that can be p produced economicallyy 1. y Pressure or flow regulators.0 594 . and compression of the Part size is limited by the press and compression of the powder used. y Small S ll gears.GATE & PSUs) GATE – GATE – 2009 (PI) 2009 (PI) y Physical properties can be controlled 593 y Oil impregnated bearings made from either iron or Oil‐impregnated copper alloys for home appliance and automotive applications li ti y P/M filters can be made with p pores of almost anyy size. S l t the Select th correctt answer using i the th codes d given i b l below: (b) Onlyy 3 and 4 (a) Onlyy 1 and 2 (c) Only 1 and 4 (d) Only 1. g the code g given below Select the correct answer using (a) 1. 589 IES – 2006 592 Statement do St t t (I): (I) Parts P t made d by b powder d metallurgy t ll d nott have as good physical properties as parts casted. 2 and 3 (b) 1 and 2 only ( ) 2 and (c) d 3 only l (d) 1 and d 3 only l For-2016 (IES. y Good tolerances and surface finish G d l d f fi i h y Highly complex shapes made quickly g y p p q y y Can produce porous parts and hard to manufacture materials (e. y Sharp corners and varying thickness can be hard to p oduce produce y Non‐moldable features are impossible to produce. 2 and 4 588 IES ‐ 2012 y Metal powders deteriorate quickly when stored M l d d i i kl h d What off powder Wh are the h advantages d d metallurgy? ll ? 1. t y Products where the combined p properties p of two or more metals (or both metals and nonmetals) are desired. High 4 g densityy Which of the above are limitations of powder metallurgy? t ll ? (a) 1. 2 and 3 (d) 1 and 2 Page 131 of 210 591 A li ti Applications IES – 2004 Which are the off Whi h off the h following f ll i h limitations li i i powder metallurgy? 1. Low L equipment i cost.Advantages d Advantages d Contd…. b lt ( t %) f ll d b li id h i t i Rev. cams etc. y Cemented carbides are produced by the cold‐ Cemented carbides are produced by the cold compaction of tungsten carbide powder in a binder. Low labour cost 3. Porosity of the parts produced 3. material 3. High tooling and equipment costs. 2. Available A il bl press capacity i 4. 4. cemented oxides) materials (e g cemented oxides) y Pores in the metal can be filled with other materials/metals y Surfaces can have high wear resistance y Porosity can be controlled y Low waste y Automation is easy 586 y Variation from part to part is low Whi h off the Which th following f ll i process is i used d to t y Hard to machine metals can be used easily H d t hi t l b d il manufacture products with controlled porosity? y No molten metals (a) Casting y No need for many/any finishing operations ((b)) welding g y Permits high volume production of complex shapes g p p p ((c)) formation y Allows non‐traditional alloy combinations (d) Powder metallurgy y Good control of final density 587 Disadvantages d IES – 2007 improperly y Fixed and setup costs are high y Part size is limited by the press. It cannot be automated. Statement (II): Particle shape in powder metallurgy influences the flow characteristic of the powder. Expensive metallic powders. followed by liquid‐phase sintering. In metal powder. g 4. 4 4. 2 and 3 (b) 1. Grinding wheel 2.0 . Self‐lubricating bearings Whi h off these Which h parts are made d by b powder d gy technique? q metallurgy (a) 1. 2. Bearings 3 Filters 3. Small magnets 4. 2 and (c) d 4 (d) 1. 3 and 4 Rev.IES 2010 IES‐2015 Conventional IES‐2015 Conventional Consider C id the th following f ll i parts: t 1. 3 and (c) d 4 (d) 1. 3 and 4 (b) 2 and 3 ( ) 1. Parts with intricate shapes S l t the Select th correctt answer using i the th codes d given i b l below: Codes: (a) 1. Brake linings Select the correct answer using the codes given below: (a) 1. gy p [10 Marks] 595 IES – 2009 Which of the following cutting tool bits are made by (b) S lli tooll bits Stellite bi ( ) Ceramic (c) C i tool t l bits bit (d) HSS tool t l bits bit Throwaway tungsten Th manufactured by (a) Forging (c) Powder metallurgy carbide bid (b) (d) tools l are 597 IAS – 2003 GATE – GATE – 2011 (PI) 2011 (PI) 598 tip i Brazing Extrusion 596 The binding material used in cemented carbide cutting tools is ( ) graphite (a) hi ((b)) tungsten g (c) nickel (d) cobalt b l powder d metallurgy ll process? ( ) Carbon (a) C b steell tooll bits bi IAS – 1998 Which are produced Whi h off the h following f ll i d d by b powder d metallurgy process? 1. 2 and 4 ( ) 2. 3 and d4 For-2016 (IES. Carbide tool tips 2. Brake lining 3. The powder heated in die or mould at high temperature is then pressed and compacted to get desired shape and strength. 3 and 4 only (d) 1. In sintering the metal powder is gradually heated resulting in coherent bond. Refractory materials made of tungsten can be manufactured easily. 2 and (a) d 3 only l (b) 1. 3 and d4 599 600 IES‐2015 IES – 1997 IES – 2001 Which components can be Whi h off the h following f ll i b manufactured by powder metallurgy methods? 1. easily 2. 3. Give examples of each. bond Which of the above statements are correct? ( ) 1. 2 and d 4 only l 603 (c) 2.GATE & PSUs) 601 Carbide‐tipped tools C bid i d cutting i l are manufactured f d by b powder‐ metal technology process and have a composition of (a) Zirconium Zirconium‐Tungsten Tungsten (35% ‐65%) 65%) (b) Tungsten carbide‐Cobalt (90% ‐ 10%) (c) Aluminium oxide‐ Silica (70% ‐ 30%) (d) Nickel‐Chromium‐ Tungsten (30% ‐ 15% ‐ 55%) Page 132 of 210 602 Consider the following statements regarding powder metallurgy : 1. control of grain size results in relatively l i l much h uniform if structure 3. 2. 2 and 3 (b) 2 only (c) 2 and 3 only (d) 1 and 2 only Classify the products that are commonly produced by p powder metallurgy. Cemented carbide dies 2 Porous bearings 2. l b y The e co components po e ts ca can abso absorb b bet between ee 12% % aand d 30% o oil by volume. the pre pre‐sintering sintering operation which is done before sintering operation. ll h operation i i d out to improve the bearing property of a bush is called (a) infiltration (b) impregnation (c) plating (d) heat treatment (b) sintering (c) impregnation (d) Infiltration For-2016 (IES.0 612 . y It is i being b i used d on P/M self‐lubricating lf l b i ti b bearing i components since the late 1920's. I parts produced d d by b powder d metallurgy ll pre‐sintering is done to (a) Increase the toughness of the component (b) Increase the density of the component (c) Facilitate bonding of non‐metallic particles (d) Facilitate machining of the part 604 Infiltration fl alloy ll liquid l d improve the h mechanical h l properties. the oil is released l d slowly l l to provide d the h necessary lubrication. p y The p process is used q quite extensivelyy with ferrous p parts using copper as an infiltrate but to avoid erosion. y Further F h improvement i i achieved is hi d by b re‐sintering. i i 605 Impregnation y Component is dipped into a low melting‐temperature y Repressing is performed to increase the density and 606 Oil‐impregnated Porous Bronze Bearings y Impregnation I i is i similar i il to infiltration i fil i y PM component p is kept p in an oil bath. The oil p penetrates into the voids by capillary forces and remains there. the strength g of the component. of copper containing iron and manganese is often used. In process. 607 608 IAS – 1996 GATE 2011 The operation in which oil is permeated into the pores of a powder metallurgy product is known as ( ) mixing (a) i i 609 IES – 1998 Which processes is Whi h one off the h following f ll i i performed f d in powder metallurgy to promote self‐lubricating properties in sintered parts? (a) Infiltration (b) Impregnation (c) Plating (d) Graphitization In the carried I powder d metallurgy. During the actual service conditions. y The Th liquid li id would ld flow fl i into the h voids id simply i l by b capillary ill action thereby decreasing the porosity and improving action.Pre ‐ Sintering Repressing IAS – 2003 y If a part made by PM needs some machining. it will be rather h very difficult d ff l iff the h materiall is very hard h d and d strong These machining operations are made easier by strong.GATE & PSUs) 610 Page 133 of 210 611 Rev. an alloy y The oil is used for lubrication of the component when necessary. 0 621 . stresses (b) Low melting point metal is filled in the pores of a sintered d powder d metallurgy ll product d (c) Liquid qu d o oil o or g grease ease iss filled ed in tthee po pores es o of a ssintered te ed powder metallurgy product (d) During D i sintering i t i operation ti off powder d metallurgy. Powder conditioning 2 Sintering 2. Q – 1. 3 3. Injection 3. Q – 1. Sintering 2 Mixing 2. rapid cooling is performed to avoid thermal stresses. Impregnation R. S ‐ 3 617 618 WorkBook Conventional Questions Enumerate the E h steps involved i l d in i “powder “ d metallurgy” ll ” process. t ll rapid heating is performed to avoid sudden produce of high internal pressure due to volatilization of lubricant IAS – 2004 Consider basic C id the h following f ll i b i steps involved i l d in i the h production of porous bearings: 1. R – 3.GATE & PSUs) Conventional Questions Matc t e o ow g Match the following Group – 1 P. Curing The are the steps in Th following f ll i h constituent i i the h process of powder metallurgy: 1.IES ‐ 2014 IAS – 2007 The in metallurgy Th process off impregnation i i i powder d ll technique is best described by which of the following? (a) After sintering operation of powder metallurgy. R – 2. Engine block 4. Cold‐die‐compaction C ld di ti g is the correct sequence q of the Which one of the following above steps? 613 614 GATE 2008 (PI) GATE ‐2008 (PI) IES – 2001 Match List‐I (Components) with List‐II M h Li I (C ) ih Li II (Manufacturing Processes) and select the correct answer using the codes given below the lists: List I List II A. 5. Sheet metal pressing D. Car body (metal) 1. S ‐ 1 (d) P – 4. Q – 3. S – 1 (c) P – 2. 10 Marks]] For-2016 (IES. Powder metallurgy 2 Injection moulding 2. Sand casting (a) P – 4. Name the materials used in “powder metallurgy”. Pressing or compacting into the desired shape I d tif the Indentify th correctt order d in i which hi h they th have h t be to b performed and select the correct answer using the codes given below: b l (a) 11‐2‐3‐4 234 (b) 33‐1‐4‐2 142 (c) 2‐4‐1‐3 (d) 4‐3‐2‐1 615 619 Ch‐19: Powder Metallurgy Ch 19: Powder Metallurgy Q N Q. Mulling Q Impregnation Q. S – 3 y Explain why metal powders are blended. Describe what h happens h d during sintering. Processing of FRP composites 4. Q – 4. Clutch lining 2. Production of metallic powder 4. Repressing 4. R – 2. 3. 3. No 1 2 3 4 5 6 Option O ti D B A C B C Q N Q. [IES‐2010. No 7 8 9 10 11 Page 134 of 210 Option O ti C D D C B 620 Rev. [ 2 Marks] k ] (b) P – 2. Impregnation 5. Discuss these steps. Casting C Gears C. Flash trimming S. Machining B. R – 4. Powder metallurgy Codes:A B C D A B C D (a) 3 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2616 Group ‐2 1. What are the limitations of powder metallurgy? p gy [[IES‐2005. GATE & PSUs) 5 7 6 y With time the effectiveness and efficiency of HSS y These steels are used for cutting metals at a much higher cutting speed than ordinary carbon tool steels. but other elements like cobalt. chromium vanadium. steels y The high speed steels have the valuable property of retaining i i their h i hardness h d even when h heated h d to red d heat. p .9 to 1. alloys y Cast carbides. Productivity raised by cutting tool materials 8 Contd… (tools) and their application range were gradually (t l ) d th i li ti d ll enhanced by improving its properties and surface condition through ‐ diti th h y Refinement of microstructure y Addition of large amount of cobalt and Vanadium to increase hot hardness and wear resistance respectively y Manufacture by powder metallurgical process y Surface coating with heat and wear resistive materials like TiC t i l lik TiC . Therefore. the tip of the cutting tool is heated to 600/700 600/700°C C which cause the tool tip to lose its hardness. y Can be formed into complex shapes for small production runs y Low cost y Suited to hand tools. and wood working y Carbon content about 0.0 9 . y Cermets. y Ceramics. y These include: y High carbon Steels and low/medium alloy steels. y Cast cobalt alloys.35% with a hardness FIGURE: Improvements in cutting tool materials have reduced machining time. and single‐crystal natural diamond.. Page 135 of 210 Fig. dry and used upto 250oC y The hot hardness value is low. TiN TiN . y A wide range of cutting tool materials is available with a varietyy of p properties. y Success in metal cutting depends on selection of the Cutt g oo ate a s Cutting Tool Materials By S K Mondal proper cutting tool (material and geometry) for a given work material. y Coated high speed steels. ( ) Both (a) h A and d R are individually d d ll true and d R is the h correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2016 (IES. etc may be present in some proportion. Sintered polycrystalline cubic boron nitride (CBN) y Sintered polycrystalline diamond. y High‐speed steels. life 4 IAS – 1997 High speed steel Assertion (A): tools A i (A) Cutting C i l made d off high hi h carbon b steel have shorter tool life. p . y Sintered polycrystalline cubic boron nitride (CBN). p performance capabilities. l di d 2 Contd… 1 3 Carbon Steels Carbon Steels y Limited tool life. h y Most of the high g speed p steels contain tungsten g as the chief alloying element. vanadium etc. not suited to mass production.Introduction y Cemented carbides. etc by Chemical Vapour t b Ch i l V Deposition (CVD) or Physical Vapour Deposition (PVD) Rev. ABOUT 62 C Rockwell y Maximum cutting speeds about 8 m/min. chromium. Reason(R): During machining. This is the major factor in tool life. y Sialons. Cermets y Whisker reinforced ceramics. y Coated carbides. and cost. 16 12 IES‐1993 Cutting tool material 18‐4‐1 HSS has which one of the following compositions? (a) 18% W. hard.IES‐2013 Vanadium in high speed steels: ( ) Has a tendency to promote decarburization (a) (b) Form F very hard h d carbides bid and d thereby h b increases i the h wear resistance of the tool (c) Helps in achieving high hot hardness (d) Has a tendency to promote retention of Austenite IAS‐1997 18‐4‐1 High speed steel Which of the following processes can be used for production thin. Sn chromium and 1 per cent vanadium. vanadium y It is considered to be one of the best of all purpose tool Page 136 of 210 15 IES‐1995 17 The compositions of some of the alloy steels are as under: 1. 18 W 8 Cr 1 V The compositions of commonly used high speed steels would include (a) 1 and 2 (b) 2 and 3 (c) 1 and 4 (d) 1 and 3 Rev. dies. 4 per cent 10 IES‐2003 The blade of a power saw is made of (a) Boron steel (b) High speed steel (c) Stainless steel (d) Malleable cast iron 14 Molybdenum high speed steel Super high speed steel y This steel contains 6 per cent tungsten. 1% V (d) 18% Cr. p 1.. bili y The molybdenum y high g speed p steels are better and cheaper than other types of steels. 4. y It is particularly used for drilling and tapping operations. Plasma spraying. 3 Chemical vapour deposition with post treatment 3.GATE & PSUs) steels. V (c) Cr. heat resistant coating at TiN. 3. 6 per cent y This steel is also called cobalt high speed steel molybdenum. 6 Mo 6 W 4 Cr 1 V 4. Cr. p planer and shaper p tools. S l t th Select the correct answer using the codes given below: t i th d i b l Codes: (a) 1 and 3 (b) 2 and 3 ((c)) 2 and 4 4 ((d)) 1 and 4 4 y This steel contains 18 per cent tungsten. broaches. 1% V (b) 18% Cr. 4% Ni.. lathe. cent in order to increase the cutting efficiency especially at high temperatures. Ni. 4% W. etc 11 IES 2007 The correct sequence of elements of 18‐4‐1 HSS tool is (a) W. because cobalt is added from 2 to 15 per cent. Cr. on HSS? y p deposition.0 18 . C Cr Ni C (d) Cu. milling cutters. V (b) Mo. 4% Cr. 4 per cent chromium and 2 per cent molybdenum vanadium. punches etc. Sintering under reducing atmosphere. threading dies punches. y It I has h excellent ll toughness h and d cutting i ability. Physical vapour 2. temperatures y This steel contains 20 per cent tungsten. Zn. 18 W 4 Cr 1 V 2 12 Mo 1 W 4 Cr 1 V 2. 4% Ni. reamers. 2 per cent vanadium and 12 per cent cobalt. l y It is widely y used for drills. 1% V 13 For-2016 (IES. 1% V (c) 18% W. 4 per cent chromium. chromium. nonferrous alloys. The high cost of fabrication is due primarily to the high hardness of the material in the as as‐cast cast condition. Water being a good coolant the heat dissipation is efficient coolant. Reason (R): Chromium and cobalt in High Speed promote martensite formation when the tool is cold p worked. y Tools of cast cobalt alloys are generally cast to shape and IAS‐2001 Assertion (A): For high‐speed turning of magnesium alloys. 0. temperature Consequently. shapes such as single‐ single point tools and saw blades. 8% Ni. 4% Cr. 5% Mo. chromium‐tungsten‐ carbon Compare HSS and ceramic tools with regard to their application in high speed machining. and the development of other. 0. 5% Co. y Cast cobalt alloys are currently being phased out for cutting‐tool cutting tool applications because of increasing costs. vanadium and chromium (c) Molybdenum.GATE & PSUs) 21 Cast cobalt alloys/Stellite IAS – IAS – 2013 Main 2013 Main y Cast cobalt alloys are cobalt‐rich. 0. costs shortages of strategic raw materials (Co. 6% W. titanium. 0 25% C 4. 24 Contd… 23 IES 2011 Stellite is a non‐ferrous cast alloy composed of: (a) Cobalt. 25 For-2016 (IES. B. titanium. W. tungsten and chromium (d)Tungsten molybdenum.7% C 3 27% Cr. (d)Tungsten.IES‐2000 IES‐1992 Percentage of various alloying elements present in different steel materials are given below: 1. and Ta. water‐based oils are recommended f hi h for high‐speed operations in which high temperatures are d ti i hi h hi h t t generated due to high frictional heat.25% C 3.15% C Which of these relate to that of high speed steel? (a) 1 and 3 (b) 1 and 2 (c) 2 and 3 (d) 2 and 4 The main alloying elements in high speed Steel in order of increasing proportion are (a) Vanadium. refractory carbides of W and Cr. the heat dissipation is efficient. 18% W. pp g p g y y y y 22 y Other elements added include V. chromium. y They are available only in simple shapes. steels Cast cobalt alloys are hard as cast and cannot be softened or heat treated. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct e p a at o o explanation of A (c) A is true but R is false (d) A is false but R is true cast alloys y having g p properties p and applications pp in the intermediate range between high‐speed steel and cemented carbides. Although comparable in room‐temperature hardness to high‐ speed steel tools. 3% Ni. 5% Mo. y Materials machinable with this tool material include plain‐ carbon steels. 3% Ni. molybdenum chromium and vanadium Page 137 of 210 26 IAS – IAS – 2013 Main 2013 Main What are the desirable properties while selecting a tool material for metal‐cutting applications? g pp Rev. Cutting speed of up to 80‐100 80 100 fpm can be used on mild steels. Consequently they can be used at higher cutting speeds (25% higher) than HSS tools. ((a)) Both A and R are individually true and R is the correct y explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true 20 finished to size by grinding.0 27 . 4% Cr. cast cobalt alloy tools retain their hardness to a much higher temperature. the coolant or cutting fluid preferred is water‐ miscible mineral fatty oil. and cast iron. vanadium Chromium titanium vanadium (d) Tungsten. treated Cast cobalt alloys contain a primary phase of Co‐rich solid solution strengthened by Cr and W and dispersion hardened by complex hard. 0. 8% Mo.7% C 2. because of limitations in the casting process and expense involved in the final shaping (grinding). Cobalt chromium and tungsten (b) Tungsten. alloy steels. 2% V. 18% Cr. and Cr). Ni. i ibl i l f il Reason (R): As a rule. superior tool materials at lower cost. titanium 19 IAS 1994 Assertion (A): The characteristic feature of High speed Steel is its red hardness. 27% Cr. tungsten (b) Tungsten. 1% V. vanadium (c) Chromium. Turning milling medium cutting speed and medium manganese steel. austentic steel.GATE & PSUs) 34 life but are not required. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct co ect eexplanation p a at o o of A (c) A is true but R is false (d) A is false f l but b R is true 32 ISO Application group Application y Coolants and lubricants can be used to increase tool IAS – 1994 The straight grades of cemented carbide cutting tool materials contain (a) Tungsten carbide only (b) Tungsten carbide and titanium carbide (c) Tungsten carbide and cobalt (d) Tungsten carbide and cobalt carbide ISO Code y Hot hardness p properties p are veryy g good 29 Contd… IES‐1995 The standards developed by ISO for grouping of carbide tools pp g g and their application ranges are given in Table below.l. Malleable C. small chips Turning. These tool materials are much harder. y Speeds are common on mild S d up to 300 fpm f ild steels l Material M t i l 33 K01 Process P K10 P01 Steel.. medium speed with small chip section i P40 Steel and steel casting with ith sand d iinclusions l i Turning. For-2016 (IES. steel castings. milling. boring. y These inserts may often have negative rake angles. Steel austenitic steel. scraping Al alloys with high silicon Grey C. brass and light machines. diamonds. malleable ll bl cast iiron Turning. or used as inserts in holders. high speed K20 P10 Steel Steel castings Steel. threading. Steel castings Precision and finish machining. milling. chip section spherodized C. Turning. Steel steel castings Steel. y Compressive strength is high compared to tensile strength.l. therefore the bits are often brazed to steel shanks. alloy 36 Rev. strength steel Soft non-ferrous metals Turning milling etc. y y y y sintered ((or cemented)) carbides because theyy are manufactured by powder metallurgy techniques.l. squares. high stiffness. chipping ferrous and non. grey C. Iron Brass etc. and lower friction. steel Turning milling. Steel.l.l. broaching. i y Cemented carbide tools are available in insert form in many different shapes. large chip section ti P50 Steel and steel castings Operations requiring high toughness turning. Turning threading and milling high speed speed. medium speed with small chip section P30 Steel.l. precision turning and boring.Cemented Carbide y Cemented carbide tool materials based on TiC have y Carbides.l. castings. extensively been developed. harder are chemically more stable. Turning. low cutting speed. Al.ferrous material and non – metals like Cast Iron.. and rounds. C l Low tensile Turning reaming under favourable conditions Turning. medium cutting speed and medium manganese steel steel. Al. milling.0 . hardness > 220 HB. Reason (R): Carbides cannot be melted and cast. are also called. off medium di or llow ttensile il planning. required y Special alloys are needed to cut steel 30 Contd… Assertion (A): are A i (A) Cemented C d carbide bid tooll tips i produced by powder metallurgy. Turning. chip section Steel casting. Th These are used d for f hi h higher‐speed d (> ( 1000 ft/min) finish machining of steels and some malleable castt irons. b d l d primarily i il f for auto t i d t industry applications using predominantly Ni and Mo as a bi d binder. malleable cast iron Turning. specially in automatic strength steel. milling. alloys scraping containing Si Grey C. They are more brittle and more expensive and use strategic metals (W.. Most carbide tools in use today are either straight tungsten carbide (WC) or multicarbides of W‐Ti or W‐ p g on the work material to be machined. have better hot hardness. Ti‐Ta. milling. heat medium or large chip section resisting alloys Free cutting steel steel. and operate at higher cutting speeds than do HSS. angles 28 Contd… 31 P M K Colour Code Table below shows detail grouping of cemented carbide tools For machining long chip forming common materials like plain carbon and low alloy steels For machining long or short chip forming g ferrous materials like Stainless steel For machining short chipping. triangles. l i shaping h i att llow cutting tti speeds d strength Page 138 of 210 35 K40 M10 M20 M30 M40 Hard grey C. milling. reaming. milling planning.l. depending Cobalt is the binder. planning medium cutting speed. Ta Co) more extensively. broaching. requiring high HB toughness Soft grey C. life. planning. (W Ta. milling. hardness up to 220 Turning. chilled casting. K30 P20 Steel. Turning.. speed spherodized C. turning. low tensile Turning profile turning Turning.l. which are nonferrous alloys. Malleable C. Turning. steel castings. y They are not suitable for intermittent cutting or for low cutting speeds. are chemically more stable. Sinterability microstructure microstructure. roughing cut B. g C. 41 Contd… 42 Contd… IES‐2013 High Performance ceramics (HPC) Sialon ceramic is used as: ( ) (a) Cutting tool material (b) C (b) Creep resistant i ( ) F (c) Furnace linens li Silicon Nitride Sili Nit id (i) Plain (ii) SIALON (iii) Whisker toughened (d) High strength Alumina toughned Al i t h d by b (i) Zirconia (ii) SiC whiskers (iii) Metal (Silver etc) Page 139 of 210 44 Rev. 38 Contd… 37 b e to get mirror o finish s o o us g y Itt iss poss possible on cast iron using ceramic turning. y In I view i off their h i ability bili to withstand ih d high hi h temperatures. y Typical products can be machined are brake discs. y Very V hi h hot high h t hardness h d properties ti y Often used as inserts in special p holders. drums cylinder liners and flywheels. they can be used for machining at very high speeds of the h order d off 10 m/s. Ferrous material. which have high hardness. Ferrous material. it can be machined with no coolant. l l Toughening Al2O3 ceramic by adding suitable metal like silver which also impart thermal conductivity and self lubricating property. roughing cut 5 4. / y They can be operated at from two to three times the y p cutting speeds of tungsten carbide. finishing cut e ous a e a . coolant y Ceramic tools are used for machining work pieces. poor thermal characteristics. and the tendency to chipping. case hardened and hardened steel. stable and have higher wear resistance than the other cutting tool materials. toughness and life of the tool and thus productivity d i i spectacularly. to thoroughly wet the entire machining zone.. this novel and inexpensive tool is still i experimental in i l stage. Else. P‐10 1. y Sinterability. if applied pp g with y Cutting should in flooding copious quantity of fluid. P‐50 5 2. K‐50 Code: A B C D A B C D (a) 4 3 1 2 (b) 3 4 2 1 (c) 4 3 2 1 (d) 3 4 1 2 y Ceramics are essentially alumina ( Al2O3 ) based high refractory materials introduced specifically for high speed machining of difficult to machine materials and cast iron. powder y Isostatic and hot isostatic pressing (HIP) – these are very effective ff i but b expensive i route. y Through last few years remarkable improvements in strength and toughness and hence overall performance of ceramic tools could have been p possible byy several means which include. brake drums. 40 Contd… g fluid. K‐10 3. M O y Transformation toughening g g byy adding g appropriate pp p amount of partially or fully stabilised zirconia in Al2O3 powder.0 45 . Non‐ferrous. strength and toughness of Al2O3 ceramics were improved to some extent by b adding ddi TiO2 TiO and d MgO. flywheels For-2016 (IES. finishing cut . h k which h h enhanced strength. Non‐ferrous. such as hard castings. since ceramics have very poor thermal shock resistance. iron y These can withstand very high temperatures.GATE & PSUs) 43 Comparison of important properties of ceramic and tungsten carbide tools 39 y Introducing nitride ceramic (Si3N4) with proper sintering y y y y technique – this material is very tough but prone to built‐up‐ built up edge formation in machining steels Developing SIALON – deriving beneficial effects of Al2O3 and Si3N4 Addi Adding carbide bid like lik TiC (5 ( ~ 15%) %) in i Al2O3 Al O powder d – to t impart toughness and thermal conductivity Reinforcing f oxide d or nitride d ceramics by b SiC whiskers.IES‐1999 Ceramics Match List‐I (ISO classification of carbide tools) with List‐ II (Applications) and select the correct answer using the pp g codes given below the Lists: List‐I List‐II A. y The Th main i problems bl off ceramic i tools t l are their th i low l strength. s g cu D. can consistently improve. Columbium 55. 2. which are (a) Cold mixed to make ceramic pallets (b) Ground. palleted and after calcining cooled l d in i oxygen IAS‐1996 46 IES‐1996 IES‐1997 Match List I with List II and select the correct answer using the codes given below the lists: List I (Cutting tools) ( l ) List II (Major constituent) ( ) A. improve tool life 200 or 300% or more. shock‐resistant carbide that can withstand high high‐temperature temperature plastic deformation and resist breakage. TiN. sintered. chemically stable. sintered and palleted to make ready ceramics (c) Ground.2 Which one of the following is not a ceramic? ( ) Alumina (a) Al i ((b)) Porcelain (c) Whisker (d) Pyrosil P il 49 50 IAS‐2003 Coated Carbide Tools At room temperature.1.. Cobalt C. Ceramic 3. y The coatings must be thick enough to prolong tool life but thin enough to prevent brittleness.3.3 h f h i f d f mm/rev. hard. brittleness y Coatings should have a low coefficient of friction so that the chips do not adhere to the rake face. Ceramic The correct sequence in increasing order of the range of cutting speeds for optimum use of these materials is (a) 3. heated and cooled (d) Ground. Tungsten carbide 2. The d depth of cut chosen is 3 mm at a feed rate of 0.4. j y The bulk of the tool is a tough. DCON 4. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true A i f l b R i 52 51 g must be fine g y The coatings grained. High‐speed steel Hi h d l 4.IES 2010 Constituents are oxides off C tit t off ceramics i id different materials.4 (d) 1. 54 Rev. and chemically i inert to prevent the h tooll and d the h work k material i l from f interacting chemically with each other during cutting. t l material t i l requirements i t att the th surface f off the th tool need to be abrasion resistant.4. Naturally the coatings must be metallurgically bonded to the substrate. H. g p smooth and continuous cuts at high speeds. used with each layer imparting its own characteristic to the tool. 53 Page 140 of 210 Contd… and porosity. Which one of the following tool materials will be suitable for machining the component under the specified cutting p p g conditions? (a) Sintered carbides (b) Ceramic (c) HSS (d) Diamond IAS‐2000 Consider the following cutting tool materials used for g p metal‐cutting operation at high speed: 1 1. which one of the following is the correct sequence of increasing hardness of the tool materials? (a) Cast alloy‐HSS‐Ceramic‐Carbide (b) HH HH‐Cast alloy‐Ceramic‐Carbide Cast alloy Ceramic Carbide (c) HSS‐Cast alloy‐Carbide‐Ceramic (d) Cast alloy‐HSS‐Carbide‐Ceramic y Coated tools are becoming the norm in the metalworking For-2016 (IES. Alumina D. hard refractory coating of TiC. Titanium Codes: A B C D A B C D ((a)) 5 1 33 4 ((b)) 2 1 4 3 (c) 2 1 3 4 (d) 2 5 3 4 47 48 IES 2007 A machinist desires to turn a round steel stock of outside diameter 100 mm at 1000 rpm The outside diameter 100 mm at 1000 rpm. y Naturally.1. The material has tensile strength of 75 kg/mm2.S. Stellite l. Tungsten B.2. Reason (R): Ceramics have a high wear resistance and high temperature resistance. & free of binders industry because coating .2 (b) 1.S. or Al2O3 accomplishes p this objective.0 Contd… . y Multiple coatings are used.4 (c) 3. y A thin. y In I cutting tti tools.GATE & PSUs) Assertion (A): Ceramic tools are used only for light. washed with acid.3. Cemented titanium carbide 3. y Interface coatings are graded to match the properties of the coating g and the substrate.2. Ni‐Co. taps.. y Less tool wear results in less stock removal during tool regrinding. chasers spade chasers.. y Therefore. Fee etc. TiN and NaCN (b) TiC and TiN (c) TiN and NaCN (d) TiC and NaCN 55 Contd… TiN‐Coated High‐Speed Steel 56 y p p y Physical vapour deposition (PVD) has p proved to be the y Coated high‐speed steel (HSS) does not routinely provide as dramatic improvements in cutting speeds as do coated carbides. TiN or TiCN and ‘met’ from metal (binder) likee Ni.IAS‐1999 y The most successful combinations are TiN/TiC/TiCN/TiN and TiN/TiC/ Al2O3 . mills. broaches bandsaw and circular saw blades. Modern cermets with rounded cutting edges are suitable for finishing and semi‐finishing of steels at higher speeds. g gear‐shaper p cutters. The coating materials for coated carbide tools. 58 Contd… IES 2010 The tool Th cutting tti t l material t i l required i d to t sustain high temperature is ((a)) High g carbon steel alloys y (b) Composite of lead and steel (c) Cermet (d) Alloy of steel.GATE & PSUs) 61 59 C t Cermets cer from y These sintered hard inserts are made by combining ‘cer’ y y y y y y ceramics like TiC. end mills and an assortment of other milling cutters.. Co. i l y In addition to hobs. more chemically stable and hence more wear resistant More brittle and less thermal shock resistant Wt% of binder metal varies from 10 to 20%. cutters best process for coating HSS. primarily because it is a relatively low temperature process that does not exceed the tempering point of HSS. Harder. Cutting edge sharpness is retained unlike in coated carbide inserts Can machine steels at higher cutting velocity than that used for tungsten carbide. insert tooling. blades broaches. and drills.. with increases of 10 to 20% being typical. form tools. spade‐drill drill blades. stainless steels but are not suitable for jerky interrupted machining and 60 machining of aluminium and similar materials. even coated carbides in case of light cuts. IES – 2003 IES‐2000 Cermets are ( ) Metals for high temperature use with ceramic like (a) M l f hi h i h i lik properties (b) Ceramics with metallic strength and luster (c) Coated tool materials (d) Metal‐ceramic composites Page 141 of 210 57 62 The tools Th correct sequence off cutting i l in i the h ascending order of their wear resistance is (a) HSS‐Cast non‐ferrous alloy (Stellite)‐Carbide‐ Nitride (b) Cast non‐ferrous alloy (Stellite)‐HSS‐Carbide‐ Nitride d (c) HSS‐Cast SS Cast non‐ferrous o e ous aalloy oy (Ste (Stellite)‐Nitride‐ te) t de Carbide (d) Cast C t non‐ferrous f alloy ll (St llit ) C bid Nit id (Stellite)‐Carbide‐Nitride‐ HSS Rev. Th f no subsequent b t heat h t treatment t t t off the th cutting tool is required. i di thus h allowing ll i i di id l tools individual l to be b reground more times. y The advantage of TiN‐coated HSS tooling is reduced tool wear. zinc and tungsten For-2016 (IES. includes (a) TiC. y Chemical h l vapour deposition d ( (CVD) ) is the h technique h used to coat carbides. HSS tooling coated by TiN now includes reamers.0 63 . increasing cutting speed and the use of more positive rake angles may eliminate it. Reason (R): The oxidation of diamond starts at about 4500C (a) Both A and R are individually true and R is the correct explanation of A l f (b) Both A and R are individually true but R is not ot a d R a e d v dua y t ue but R s ot tthe e correct explanation of A ( ) A is true but R is false (c) A i t b t R i f l (d) A is false but R is true Rev. i d li h f d d Consider the following statements: C id h f ll i p g y. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true A i f l b R i 68 IAS – 1999 IES‐1992 Which of the following given the correct order of increasing hot hardness of cutting tool material? (a) Diamond. and light feeds are used. such as copper.5 to 1. hardness Di Diamond d cutting tti t l are nott recommended tools d d for f y low thermal expansion. y This is used when good surface finish and dimensional accuracy are desired. Carbide. h l honing h i tools. p g y y Oxidation of diamond starts at about 450oC and thereafter it can even crack For this reason the thereafter it can even crack. finish. y Due D to their h i brittle b i l nature. as is control over part size. HSS (c) HSS. diamonds are not suitable because of the diffusion of carbon atoms from diamond to the work‐piece 64 Contd… material. grinding i di wheels. High compression strength 4. and improved wear resistance. y Positive rake tooling is recommended for the vast majority of diamond tooling applications.Di d Diamonds y Diamond tools have the applications in single point turning and GATE – GATE – 2009 (PI) 2009 (PI) g tool materials. y If BUE is a problem. resulting from the random orientation of the diamond grains and the lack of large cleavage planes. HSS (b) Carbide. aluminium and magnesium alloys. Diamond HSS carbide Diamond (d) HSS. Carbide Page 142 of 210 69 71 Assertion (A): During cutting. machining of ferrous metals due to y high hi h heat h conductivity. t l milling illi cutters. the h diamond di d tools l have h poor resistance to shock and so. 2. d i i and d y a very low co‐efficient of friction. ((c)) p poor tool toughness g (d) chemical affinity of tool material with iron improvements over carbides. Low fracture toughness L f t t h Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 ( ) 2 and 3 (c) d (d) 3 and 4 d For-2016 (IES. For precision machining of non‐ferrous alloys.. y Diamond is the hardest of all the cutting y Diamond has the following properties: y extreme hardness. pp brass. l Assertion (A): Diamond tools can be used at high p speeds. should be loaded lightly. Tool life is often greatly improved. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true A i f l b R i 67 IES – 1999 66 Contd… 65 IES‐1995 y Diamond tools offer dramatic performance boring b i tools.5 mm) of'fine grain‐ size diamond particles sintered together h and d metallurgically ll ll bonded b d d to a cemented d carbide bd substrate. and p p surface integrity.0 72 . greater toughness. Diamond. High wear resistance 3. y On ferrous materials. zinc. lapping powder and for grinding wheel dressing. For this reason the diamond tool is kept flooded by the coolant during cutting. diamond is preferred because it has 1 Low coefficient of thermal expansion 1. tt reamers. (a) high tool hardness ((b)) high g thermal conductivityy of work material y The work‐materials on which diamonds are successfully employed are the non‐ferrous one. Diamond.GATE & PSUs) 70 IES‐2001 Assertion (A): Non‐ferrous materials are best machined with diamond tools. Reason (R): Diamond tools are suitable for high speed machining. y The main advantages of sintered polycrystalline tools over natural single‐crystal tools are better quality. Reason (R): Diamond tools have very low coefficient of friction. carbide. y Polycrystalline diamond (PCD) tools consist of a thin layer (0. the diamond tool is A i (A) D i i h di d l i kept flooded with coolant. GATE & PSUs) 77 80 toughness and tungsten carbides for heat and wear resistance.4 (b) 1. y It shows excellent performance in grinding any material of high g hardness and strength. g p g 76 IES‐1994 y Coronite is made basically by combining HSS for strength and Which of the following tool materials have cobalt as a constituent element? 1. (b) Has a hardness which is slightly more than that of HSS (c) Is used for making cylinder blocks of aircraft engines i ((d)) Is used for making optical glasses. hard‐chill cast iron. y CBN can be used efficiently and economically to machine these difficult‐to‐machine materials at higher g speeds (fivefold) and with a higher removal rate (fivefold) than cemented carbide and with superior (fivefold) than cemented carbide. 4 Borazon. 74 Contd… 75 IES‐1994 IES‐2002 IES‐1996 Consider the following tool materials: 1. y the central HSS or spring steel core y a layer of coronite of thickness around 15% of the tool diameter y a thin (2 to 5 μm) PVD coating of TiCN y The coronite tools made by y hot extrusion followed byy PVD‐ coating of TiN or TiCN outperformed HSS tools in respect of g forces.5 – 1 mm layer of polycrystalline cubic boron nitride to cobalt based carbide substrate at veryy high g temperature p and pressure. 4. 73 Contd… steels.2. F f h b For-2016 (IES. tool life and surface finish. g p q yg (c) For heat treatment (d) For none of the above. and nickel‐ steels hard chill cast iron and nickel and cobalt‐ and cobalt based superalloys. g grey castt iron is i i 300 ~400 m/min / i y CBN is less reactive with such materials as hardened y Speed p ranges g for other materials are as follows: y Hard cast iron (> 400 BHN) : 80 – 300 m/min y Superalloys S ll ( 35 RC) : 80 (> 8 – 140 m/min / i y Hardened steels ((> 45 RC) : 100 – 3 300 m/min y It is best to use cBN tools with a honed or chamfered edge preparation. Stellite 4. and with superior accuracy. y Microfine TiCN particles are uniformly dispersed into the matrix.4. 3 Which one of the following is the hardest cutting tool material next only to diamond? (a) Cemented carbides (b) Ceramics (c) Silicon (d) Cubic boron nitride Cubic boron nitride ( ) Has a very high hardness which is comparable to (a) H hi h h d hi h i bl that of diamond. available y It is made by bonding a 0.0 81 . Cemented carbide 2. 1. y It remains inert and retains high hardness and fracture toughness at elevated machining speeds. cuts Like ceramics. UCON Select the correct answer using the codes given below: Codes: (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 (d) 2 and 3 79 Page 143 of 210 78 C it Coronite IAS‐1998 Cubic boron nitride is used ( ) As lining material in induction furnace (a) A li i i l i i d i f ((b)) For making optical quality glass. i y The only y limitation of it is its high g cost. Carbide C bid 2. the coronite based tool is made of three layers. 4 (d) 2. CBN 3.3 (c) 2. Correct sequence of these tool materials in increasing order of their ability to retain their hot hardness is (a) 1. cBN tools are also available only in the form off indexable i d bl inserts.Cubic boron nitride/Borazon y The operative speed range for cBN when machining y Next to diamond. finish.2. C Cermet 33. 3. cutting Rev. Ceramic 4. 1. preparation especially for interrupted cuts.3. matrix y Unlike a solid carbide. cubic boron nitride is the hardest material presently available. and surface integrity. IES‐1993 Match List I with List IT and select the correct answer using the codes given below the lists: Li I (Cutting tool Material) List ‐ I (C i l M i l) List ‐ Li I I(Major I I(M j characteristic constituent) A.GATE & PSUs) 84 88 Page 144 of 210 89 Rev. 1 (c) 3. Increases the transverse rupture strength I h h 4. 2 (b) 4. High speed steel 1. HSS 2. 1. Increases the hardness 2. 1 1. y The adhesive wear in the rough region is called attrition wear . 2. Extrusion 5 5. Drawing dies Code: A B C D A B C D (a) 3 4 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 4 3 1 2 Match. Forging C Cemented carbide C. 1. Decreases the hardness. 4 Diamond The correct sequence of these materials in decreasing order of their cutting speed is (a) 4. 2. For-2016 (IES. h 87 IES‐2005 Consider the following statements: An increase in the cobalt content in the straight carbide grades g g of carbide tools 1 Increases the hardness. y Therefore. Ceramics 4. 3. Lowers the transverse rupture strength. 3 3. Nitride D. 4 4. UCON 4. Ab i h l Abrasive wheels B. 2 82 83 Match List‐I with List‐II and select the correct answer using the codes given below the Lists: List I List II (Materials) (Applications) A Tungsten carbide A. 3. Powder metallurgy Codes:A B C D A B C D (a) 3 1 5 2 (b) 2 5 4 3 (c) 3 5 4 2 (d) 2 1 5 3 85 86 IES‐1999 IAS‐2001 IES‐1996 Attrition wear y The strong bonding between the chip and tool material at high temperature is conducive for adhesive wear.0 . Al i i oxide id 3. Silicon nitride 2. bid 1. List I (Cutting tool materials) with List II (Manufacturing methods) and select the correct answer using the codes given below the Lists: i th d i b l th Li t List I List II A HSS A. 3. Cobalt C d A Codes: B C D A B C D (a) 2 1 3 5 (b) 2 5 1 3 (c) 5 2 4 3 (d) 5 4 2 3 IES‐2003 IES‐2000 Which one of the following is not a synthetic abrasive material? (a) Silicon Carbide (b) Aluminium Oxide (c) Titanium Nitride (d) Cubic Boron Nitride Consider the following tool materials: 1. some parts of the worn surface are still covered by molten chip and the irregular attrition wear occurs in this region . Molybdenum C. 1 (d) 3. Casting B. Coated carbide tool 4. Diamond 3. Heating elements C Aluminium C. Pi f Pipes for conveying i liquid metals D Silicon carbide D. y The irregular attrition wear is due to the intermittent adhesion during interrupted cutting which makes a periodic attachment and detachment of the work material on the tool surface. 4 p g Which of the statements given above are correct? (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) 2 and 3 The limit to the maximum hardness of a work material which can be machined with HSS tools even at low speeds is set by which one of the following tool failure mechanisms? (a) Attrition (b) Abrasion (c) Diffusion (d) Plastic deformation under compression. when the seizure between workpiece to tool is broken. Rolling D. the small fragments of tool material are plucked and d brought b h away by b the h chip. Stellite 2. 1. Carbon B. C Cemented carbide d bid 33. y In the rough region. Stellite 2. 4. 4. Columbium 5. m / min = Ans.4 Slide No. In turning positive back rake angle takes the chips away from the machined surface. and cutting force will be slightly reduce so we take as cutting forces will not be affected by the cutting velocity. As velocity increases power consumption will increase and temperature will increase.2 m / min Ans.17 Ans. otherwise. Hardness and brittleness is different property IES-2015 Page No. (a) It is true form-cutting procedure. (b) Both are correct. An obvious disadvantage of this method is that the absence of side and back rake results in poor cutting (except on cast iron or brass).2 Speed (V ) = IES-2001 SlideNo. IES-2011 Page No.4 Slide No.0 .2008 Page No.12 Ans. (b)The rake angle does not have any effect on flank but clearance anglehas to reduce the friction between the tool flank and the machined surface. GATE-1995. and the top of the tool must be horizontal and be set exactly in line with the axis of rotation of the work. but if you think deeply there is no reason for increasing the force. 3 Slide No. IES-2006 Page No.(c)Carbide tools are stronger in compression. (c) For cutting brass recommended rake angle is -5 to +5 IES-1995 Page No.4 Slide No. Common sense will say force will increase. Negative rake is used as carbides are brittle not due to hardness.9 specific pressure.8 Ans. (a) Increasing rake angle reduces the cutting force on the tool and thus power consumption is reduced.GATE & PSUs) Page 145 of 210 Rev. (d)All other tools are multi-point.20 Ans.7 Ans. (c) IES-2012 Page No. 3 Slide No. IES-2012 Page No. But in actual practice it will reduce due to high temperature.10 Ans.(d) When cutting velocity is increased. Cutting force feed×depth of cut IES-2005 Page No.11 Ans.3 Slide No. (b) For-2016 (IES. (c) Brittle workpiece materials are hard and needs stronger tool.2 Slide No. the resulting thread profile will not be correct.14 Ans.22 Ans.e. same material same hardness.2 Slide No. GATE-2008(PI) Page No. (d) Negative rake angle increases the cutting force i. (b) IES-2015 Page No. (d) IES-2014 Page No. Tools having zero or negative rake angle provides adequate strength to cutting tool due to large lip angle.4 Slide No.4 Slide No.3 Slide No.2 degree. 3 Slide No.6 1000 m / min = 94.13 Ans. IES-1993 Page No.3 Slide No. Specific pressure = Ans.19 Ans. same tool same sharpness.(c) π × 30 ×1000 Slide No. IAS-1994 Page No. no rake should be ground on the tool. Therefore the force will remain unaffected.Ch-1 Basics of Metal Cutting: Answers with Explanations IES-2013 Page No.18 Ans.(b)Carbide tips are generally given negative rake angle it is very hard and very brittle material. We have used negative rake angles for different purpose but not for the direction of chip.5 π DN 1000 Page No. Whereas negative back rake angle directs the chip on to the machined surface.23 Ans.2 Slide No. IES-2003 Page No. it will lead to increase in power and temperature.2 Slide No.(b) IES-2002 Page No.21 Ans. The surface finish on steel usually will be poor. (d)Strength of a single point cutting tool depends on lip angle but lip angle also depends on rake and clearance angle.81 = = 0.(d) We may use principal cutting edge angle or approach angle = 90 - CS .then α S = α .8 r cos α 0. When. There is an improvement in surface finish and permissible cutting speed as nose radius is increases from zero value.31 Ans. IES-2012 Page No. 6 φ = tan −1 IES-1994 Page No.36 Ans.944 1 − r sin α 1 − 0.(c) r cos α 0.4 cos10 = tan −1 = 22.GATE(PI)-1990 Page No.33 Ans.4sin10 GATE-2011Page No.37 Ans.35 Ans. (a) Fy = Ft cos λ = Ft sin CS IES-2010 Page No. But this question is not for Orthogonal Cutting it should be turning.27 Ans.5 Slide No.6 Slide No.28 Ans.4 Slide No. (c) IES-2006 Page No.(c) Smaller point angle results in higher rake angle. GATE-2001Page No. (b) GATE-2008 Page No.40 Ans.GATE & PSUs) Ans.45 tc 1.45sin12o Slide No. (b)The second item is the side rake angle. Thus 6° is IES-1993 Page No. ISRO-2011 Page No.30 Ans. 6 Slide No.5 Slide No. (b) IES-2009 Page No.90o 1 − r sin α 1 − 0.6 φ = tan −1 Slide No. (b) r = t 0. 6 Slide No.34 Ans. (a) IAS-1995 Page No. (c) It will increase tool cutting force.5 Slide No.5 Slide No.5 Slide No.4 Slide No. Side cutting edge angle is not principal cutting edge angle.29 Ans.5 theside rake angle. (b) Slide No.26 Ans.0 . IES-1994 Page No. IES-1995 Page No.5 Slide No. But too large a nose radius will induce chatter. (c) No of chattering per cycle 360/30 = 12 No of cycle per second = 500 /60 Therefore chattering frequency is 12 x 500/60 = 100 Hz IAS-1996 Page No. (c)Large nose radius improves tool life.41 Ans.25 Ans.5 Slide No.42 For-2016 (IES. tangential force. (b) Page 146 of 210 Rev. (c) Fy ↑= Ft cos λ = Ft sin CS ↑ (radial force) Fx ↓= Ft sin λ = Ft cos CS ↑ (axial force)and SCEA has no influence on cutting force i.4 Slide No.5 Slide No.32 Ans.38 Ans.e. A sharp point on the end of a tool is highly stressed.4 Slide No. short lived and leaves a groove in the path of cut. (b) IES-2009 Page No. Don’t confuse with side cutting edge angle.45cos12o = tan −1 = 25. IES-1995 Page No.24 Ans. IES-2002 Page No. principal cutting edge angle =90. But best choice is (b) IES-2006 Page No.4.7 Slide No. φ = tan −1 VS V = sin ( 90 − φ + α ) sin ( 90 − α ) VS 2.(d) r = 0.94 1 − r sin α 1 − 0.GATE-2014 Page No.GATE & PSUs) Page 147 of 210 Rev.4sin10 from the velocity triangle.52 Ans.44 Ans.2 × sin 90o = (for turning) = = 0.ISRO-2009 Page No.6 Slide No.5 = sin ( 90 − 22. α = 10 0.9 IES-2004 Page No. (c)Cutting torque decreases with increase in rake angle.80 − 0)o = 2.31o 1 − r sin α 1 − 0. IES-2004 Page No.6 r= Slide No.4 cos10 = tan −1 = 22.0 t f sin λ 0. (c) IES-2004.80o + tan(21.5 r cos α r cos 0 = tan −1 = tan −1 r = tan −1 0. shear strain (ε ) = cot φ + tan φ Slide No. (d)as rake angle is zero.46 GATE-1990(PI)Page No.3cos10 r cos α φ = tan −1 = tan −1 = 17.31o + tan (17.49 Ans.7 Slide No.7 Slide No.0 .7 Ans.80o φ = tan −1 Shear strain(ε )= cot φ + tan (φ − α ) = cot 21.4 1 − r sin α 1 − r sin 0 φ = 21.8 to 3. (b) actually 2.526 m / s = = 1.34 o IES-2009 Page No. 7 Slide No.577m / min sin ( 90 + 15 − 45 ) For-2016 (IES.50 Ans.4 tc tc 0. (a) VC V = sin φ sin ( 90 + α − φ ) VC = 35 × sin 45 = 28.94 + 10 ) sin ( 90 − 10 ) VS = 2.51 Ans.3sin10 shear strain (ε ) = cot φ + tan (φ − α ) = cot17.526m / s Shear strain rate(ε i ) = VS 2. 2.48 Ans.3.47 Ans. (c) r = 0.31 − 10 ) = 3. α = 10 r cos α 0. 7 Slide No.3 and 4 are correct.7 Slide No. (a) ε = cot φ + tan (φ − 12 ) dε = − cos ec 2φ + sec2 (φ − 12 ) = 0 gives φ = 51o dφ GATE-2012 Page No.45 Ans.0104 × 105 / s tS 25 × 10−6 m IES-2004 Page No. 58 Ans.59 Ans. (b) IES-2015 Slide No. uncut chip thickness is also halved and hence chip thickness will be halved.7 Slide No.75 VC sinφ = = r = 0. V = 60 m / min t 2.4 r = ⇒ tC = = 3.6 IAS-2003 Page No.63 Ans. α = 15° ∴r = Slide No.5cos15° = 0.67 Ans.02° tc 1 − r sin α 1 − 0.65 Ans.common data S-2) Page No.68 Ans.8 Slide No.8 Slide No.IES-2008 Page No.2 mm. (a) Most of the students get confused in this question.5 mm . (a) Page No.55 Ans.57 Ans.9 IAS-1997Page No.8 V = 20 m / min.75 × 60 = 45 m / min IES-2014 Page No.54 Ans. (d) VC t 0.6 2 × 0.2/0.8 Slide No.53 Ans.2 mm tC 0.5 =r= = V tc 0.75 V sin(90 + α − φ ) VC = 0. (b) Cutting ratio means chip thickness ratio. (c) Slide No.64 Ans.8 t = 0.5 = 1.66 m / s 0.7 Slide No.625 But examiner has given reciprocal value = 1.61 Ans.9 Slide No.5 orVC = 10 m / min V GATE-1995 Page No.8 Slide No. IES-2003 Page No.GATE & PSUs) Page 148 of 210 Rev. (d) IAS-1998Page No. or VC = Ans. Slide No. φ = tan −1 = tan −1 = 29.5.56 t = 0. (a) IAS-2000 Page No.8 Slide No. r = 0.60 Ans. tc= 0.6 IES-2001 Page No. (b) V= π DN 1000 × 60 m/s = π ×100 × 480 1000 × 60 m / s = 2. (b) It is orthogonal cutting means depth of cut equal to uncut chip thickness.0 . Velocity of chip sliding along the shear plane is shear velocity (Vs) and velocity of chip along rake face is chip velocity (Vc ).9 Slide No.51 m / s IAS-1995 Page No.32 mm.5sin15° Nearest option is ( c ) GATE-2009 (PI. cutting ratio = chip thickness ratio = t / tc = 0.8 Slide No.32 = 0. Ductile material with multipoint cutter will produce regular shaped discontinuous chip.9 For-2016 (IES. As depth of cut halved. GATE-2009 (PI -common data S-1) Page No. tc = 0.4 mm. IES-2007Page No. (a) IAS-2002 Page No.6 mm. (b) VC = r = 0. (c)f = f sinλ = 0.62 Ans.2 sin 90 = 0.2 mm .75.8 Slide No. tc = 0.(c) t r cos α 0. (b)It is multi-point cutter and mild steel is ductile material. .335.9 N Anss.023° Froom Merchannt Circle: ∴ Fs = R cos (φ + β − α ) ⇒R= FS = 1735. τ y = 285 N / mm m 2 .2 mm.5. This T micro-weld have to break du ue to rela ative motion between chiip and tool.35 × cos10 = = 0..9 N Anss. (d)Low cu utting speed d means long g chip tool co ontact time. annd = μ = 1( given) ∴ N = 402. ∴ β = tan −1 μ = tan −1 0.565° tan β = μ = 1.51 mm bt FS = τ y = 285 N / mm m2 sin 20. μ = 0. r = = = 0. tc = 0.4 r cos α r coss 0 = r = 0.3 N (ii) Radial forcee (Fy ) =0 [Th his force is present p in turning t but it is orthogonnal cutting] c β = 1455. ∵ Friction forcce is perpend dicular to thhe cuttting velocityy vector thaat means α = 0° F F = 402. So chip breaker is not n needed. (b b)Low cuttin ng speed m means long chip c tool con ntact timee.(∵α = 0°) = 1 − r sinn α 1 − r siin 0 or φ = tan −1 r = tan −1 0.75 Ans.7 N tan φ = μ = tan β .71 IES S-1997Page No.1152° sin φ = 12265. Slide No. And long contact c timee will sufficient to form bond b betwee en chip and tool.86 N GA ATE-2010 (PI) linke ed S-1 Pa age No.9 Slide No. b=3mm m.7 7N Notte : Shear fo orce on tool face f is fricttion force ( F ) = R sin β = 945.5 N .10 Slide No o. r = 0.35sin10 φ = tan −1 0.51 mm m . (d)Cast iron mea ans brittle material and will form disccontinuous chip..5 = 26.GATE & PSUs) Page 149 of 210 Rev.69 Ans.2 N (iii)) Normal forrce on tool (N): N = R cos (iv))Shear forcee (Fs ):1265. t = 0. And long g contact tim me will suffficient to foorm bond beetween chip p and tool..4mm.65 = 33..10 Slide No.152° 3 mm × 0. Ch h-2: An nalysis of Mettal Cuttting: Answer A rs with h Expla anation ns ESE E-2000(co onvention nal) Pag ge No. Slide No.0 ..655 r cos α 0.2 t = 0. It will increease co-efficient of frictioon. β = 45° tan φ = For-2016 (IES. α = 10°.74 Ans.5 tc 0.3669 1 − r sinn α 1 − 0.70 GAT TE-2009 Pa age No.36669 = 20.6N cos (φ + β − α ) (i) Cutting C force (Fc ): Fc = R cos ( β − α ) = 1597.GAT TE-2002 Pa age No.55 N N t 0. ∴ β = tan −1 μ = tan −1 0. Ft will form a rectangle.11 Using Merchant Analysis: φ = 45° + 2 Page No.22 N In FS and FC triangle: Fs =Rcos ( β − α + φ ) = 569. (c) β 2 10° β ° − ⇒ β = 60° 20° = 45° + 2 2 ESE-2005(conventional) Page No.10 α 2 − Slide No. FC .11 μ = 0. N.10 Slide No.2361 m / s = 402.78 From merchant circle. using Merchant Analysis: φ = 45° + α 2 − β 2 or φ = 45° + 5° − 26.2361 m s ∴ Heat generation at the primary shear zone will be because of shear velocity and shear force Heat =FS × VS = 180. (b) Fc = N = 1500 N Ft = F GATE-2015 IAS-1999 φ =45° + α β − 2 ° 20 25.10 Page No.0 .565 ) N = 180.GATE & PSUs) Page 150 of 210 Rev. Ans.5 = 26.5.82 Ans.1 Ans. ∵α = 0.76 m .5W GATE-2013 linked queS-1 Page No.79 Slide No. 2.25° 2 2 GATE-1997 Page No.81 Ans.5 N sin 45° = 569.0 N GATE-2010 (PI) linked S-2 Page No.77 Ans.From merchant circle: F = R sin β or R = 402.83 Ans.0 N × 2. (b) Slide No. V 2 = S sin 90 sin ( 90 − 26.5° φ = 45° + − = 42. applying sine rule.717° 2 For-2016 (IES. V =2 s From the velocity triangle . (a) From Merchant Circle if cutting force ( FC ) is perpendicular to the friction force ( F ) then the rake angle will be zero GATE-2013 linked queS-2 Page No.10 Slide No.565° Slide No.10 Slide No.22 × cos ( 45 − 0 + 26.565 ) VS = 2. VS V = sin {90 − (φ − α )} sin ( 90 − α ) Ans. then F.565° = 36. ((b) f Fs : Calculaating shear force 3.bt 2 mm × 0.2 25 KW 60 GA ATE-2014 Pag ge No.4N For-2016 (IES.99° − = 32.11 Slide No.11 Slide No.11 Slide No.7717° Froom merchant circle: FS = τ y Fs = R cos (φ + β -α ) N = R cos(36. φ ↑= 45° + − 2 2 Power= =FC iV = 12999.565° − 10 0° ) = 127.6 mm × 0.5° Using Merchant M Circle: C In trianngle formed by Fs . ((d) Forr minimum cutting c forcee we have too use mercha ant Theory α = 10° . μ = 0.52 7 N sin φ sin 322.6 cos(26.11 Slide No..87 Ans.11 Slide No. (b) Merchant M Analysis IES S-2005 Page No. (d) α β↓ Using Merchant Analysiis. Fc =Rcoos ( β − α ) = 1433. GA ATE-2014 Pag ge No..61N sin φ sin 36.52 N = R ( N ) × cos(32. Fn annd R.85 Ans. φ ↑= 45° + − 2 2 As shear anglee increases area a of the shear plane e decreased.86 Ans.99° − 10 1 ° ) = 1299.5 N × As shear anglee increases cutting force will decrease and lengtth of shear p plane decrea ased results chip thicckness decreease. Here option IES S-2010 Pa age No.7 × coos(34.90 Ans.8W ≈ 3.7 = tan β or β = tan −1 μ = 34. (a) IES S-2003 Page No..5° φ = 45° + 2 2 GA ATE-2008 (PI) Link ked S-2 Page No. Fc annd R.22 mm = 400 N / mm m2 = 2677. Fs = τ y × c (φ + β − α ) Fs = R cos or 770.7 N In trianngle formed by Ft .89 Ans.61 or R = 447.565° − 10 1 ° ) = 429. surrface finish..88 Ans.611N GA ATE-2008 (PI) Link ked S-1 Page No.002 N Ft = 447.6sin(226.11 b = 3. it results ccutting force e hence imp prove n (b) is also correct c but best b one is op ption (a).5 N 1500 m / s = 32448..99° − 10° ) ⇒ R = 1433.717° + 26.GATE & PSUs) Page 151 of 210 Rev.5655° − 10° ) 2 or 267. (a) α β↓ Ussing Merchaant Analysiss.6 N andd Fc = R cos ( β -α ) FC = 447.5° + 34..25 0 mm bt = 4660 ( N / mm 2 ) = 770.99° M A Analysis Using Merchant φ = 45° + α β − 2 2 10° 34.11 Slide No.0 .(c) Fs = Fc cos φ − Ft sin φ = 900cos 9 30 − 600sin 30 = 479.84 Ans.6 mm Slide No. 435° = 751. Ft = 475 N r= Ans.25 mm.12 Slide No.75 r cos α r cos 0 = = r = 0.12 Slide No.33333 tc 0. (a) F = Fc sin α + Ft cos α N = Fc cos α − Ft sin α ∵ α = 0.31° = 723 N But we have to calculate without using calculator sin11.31° = 1 − sin 2 11. α = 0°.94 Ans.GATE & PSUs) Page 152 of 210 Rev. N = 950 N .25 = = 0.98 900 cos11. (d) bt We know. so F= F = 500 N t N = Fc = 1000 N 500 1 F = = N 1000 2 GATE-2007 (PI) Linked S-1 Page No.04 N GATE-2011 (PI) linked S-2 Page No.435° ⇒ τ y = 379.8 mm.6° 1200 GATE-2007 (PI) Linked S-2 Page No.5 mm.12 Slide No. tc = 0.8 Vc Vc = = = tc 1.12 FC = 1200 N .25 mm 751. t = depth of cut = 0.5 mm. Ft = 500 N .2 given ∴ cos 11.93 Ans.95 t = 0.12 Slide No. (a) Using the relations: F = FC sin α + Ft cos α = 1200sin 0 + 500 cos 0 = 500 N N = FC cos α − Ft sin α = 1200 cos 0 − 500sin 0 = 1200 N F 500 = N 1200 500 ∴ β = tan −1 = 22.75 mm.31° = 900 × 0. V =1m/s μ = tan β = r= Slide No. For-2016 (IES.98 − 810 × 0.12 Orthogonal machining.2 = 720 N IES-2000 Page No.90 N mm 2 IFS-2012 Page No.96 Ans.31° − 810sin11. N = Fc = 950 N tan φ = Fs = Fc cos φ − Ft sin φ = 950 cos18. b = 2.04 N = τ y ( N / mm 2 ) sin18.92 Ans.53 m / s GATE-2011 (PI) linked S-1 Page No. .435° To calculate shear force. α = 0° μ = Slide No.12 Slide No. (b) Fs = Fc cos ϕ − Ft sin ϕ = 900 cos11. t c =1.0 .97 Ans.31° − 810sin11.31° = 1 − 0. as α = 0.33333 1 − r sin α 1 − r sin 0 φ ⇒ 18.5 mm × 0.435° − 475sin18.31° = 0. (b) t 0.22 = 0. (b) t 0.IES-2014 Page No.91 Ans.5 V 1 ⇒ Vc = 0. sin φ 2. FS = τ y . tc = 0.12 V = 20m / min.7mm.455 ≈ 0.2 mm 0.98 Ans.5 = 20 m / min 0.77 N × 60 Total power = Fc × V = 1200 N × For-2016 (IES.79 = = 0.7142 tc 0.GATE & PSUs) Page 153 of 210 Rev.d = t = 0.95W Frictional power=F × VC = 503.(a) Vc 0.5° 2 μ = tan β = tan 24. Fc = 1200 N .12 α = 15.334° − 200sin 28.3 Alternatively Using Merchant Theory: ∴μ = φ = 45° + α β − 2 2 15° β = 45° + − 40.334° τ y = 327.0 .5sin10 Using relation.46 N 1107.77 = = 0.62 N 457. t = 0. φ = tan −1 Fs = Fc cos φ − Ft sin φ = 500 cos 28.7 orVC = 14.46 GATE-2006 common data Q-2 Page No.79 N N = Fc cos α − Ft sin α = 500 cos10° − 200sin10° = 457.5° = 0.67 N F 283.7142sin15° F = Fc sin α + Ft cos α =1200sin15° + 200 cos15° = 503.286 m / min 20 m / s = 400W 60 14.67 To calculate shear strength.99 Ans.334° = 345.5 tc 0. Fc = 1200 N .3 N F 503.24° 1 − r sin α 1 − 0.1 mm = = 0.18 N ⎛ bt ⎞ Shear force(FS ) = shear strength(τ y ) × shear area ⎜ ⎟ ⎝ sin φ ⎠ 5 × 0. Ft = 200 N ( given) Vc t =r= V tc or Slide No.2mm. Fc = 500 N .7142 cos15° = tan −1 = 40. Ft = 200 N .65 N / mm 2 GATE-2006 common data Q-1 Page No.286 m / s = 119.77 N φ = tan −1 N = Fc cos α − Ft sin α = 1200 cos15° − 200sin15° = 1107.1 bt Fs = τ y ⇒ 345. (b) t 0.5 = = 0.334° 1 − r sin α 1 − 0.24° = 24.1 mm.5mm.18 = τ y × sin φ sin 28. α = 10° To calculate co-efficient of friction: Using relations: F = Fc sin α + Ft cos α = 500sin10° + 200 cos10° = 283. μ= r= t 0. Ft = 200 N r= Slide No.7 r cos α 0.456 ≈ 0. tc = 0. b = 5 mm.5cos10 r cos α = tan −1 = 28. (c) IES-1999 Page No.Percentage of total energy dissipated as frictional power is F × VC 119.32197 tc r cosα = 0.2sin 30° = 0. (a) GATE-2014 Page No.106 Ans. t = f sin λ = 0.f = 0.48 mm tanφ = For-2016 (IES.22 N N = Fc cos α − Ft sin α = 1200 cos 0 − 828.76 N φ = tan −1 bt sin φ F 820.351 m / s 60 60 Power consumption(P)=Fc × V = 1200 × 3.988% ≈ 30% 400 Fc × V GATE-2006 common data Q-3 Page No.13 Slide No.102 Ans. (b) Tangential force accounts for 99% of the power consumption.16 mm/rev or φ =18. r = = = 0.13 = 820.105 Ans.13 Slide No.13 Slide No. IES-2001 Page No.104 Ans.24 + tan(40. α = 0.2mm / rev.103 Ans.5650 t c = 0. sin λ sin 75° t 0.100 Ans. 0.021kW GATE-1995(conventional) Page No. tc = 0.1411× 0.22sin 0 = 1200 N (ii ) ∵ it is a turning operation. d = 4mm. speed = 400rpm ∵ it is a turning operation. Fc = 1200 N Fx 800 = = 828.76 or τ y = s = = 230.13° 1 − r sin α 1 − r sin 0 Fs = Fc cos φ − Ft sin φ = 1200 cos 21. (a) IES-1997 Page No.GATE & PSUs) Page 154 of 210 Rev.08 to 0.351W = 4.38638 tc 0.13 Slide No.38638 = 21.13 Slide No. dia ( D) = 160mm.13 Slide No.101 Ans.93 N / mm 2 bt 4.3091 mm Now .13 Slide No.22 cos 0 = 828.33586 1 − r sinα f = 0.8 mm r cos α r cos 0 = tan −1 = tan −1 r = tan −1 0. t = f sin λ = 0.13 Slide No.0 .1411 mm.3091 mm d 4 b= = = 4.24 − 15 ) = 1.12 t = f sin λ λ = 90° − CS = 90° − 60° = 30° t = 0.107 Fs = τ y × Given : α =100 λ = 750 Ans.95 ⇒ ×100% = × 100% = 29.15455mm r= t = 0.13 − 828.3091 sin φ sin 21.22sin 21.32sin 75° = 0.653 IES-1995 Page No.1 mm ESE-2003(conventional) Page No.13 π DN π × 0.8mm. (d) ° ° ° ε = cot φ + tan(φ − α ) = cot 40.22 N sin λ sin 75° (i) Using the force relations Ft = F=FC sin α + Ft cos α = 1200sin 0 + 828.160 × 400 (iii ) V = = = 3. GATE & PSUs) Slide No..214° N 1009.34 N / mm 2 2 × 2.111 Page 155 of 210 An ns. (a) Rev.288° − 560 sin 32.87 = 74.14 Slide No.288° 2 2 2 2 Usiing force rellation: Fs = Fc cos φ − Ft sin φ = 1177 1 cos 32.0 .14 Pag ge No..6 N F 824.68N Ft = N = Fc cos α − Ft sin α = 500 cos10 − 207sin10 = 456.79° 39.42N Fs = Fc cos φ − Ft sin φ = 500 cos18.816 = 39.288 sinn φ Pag ge No.110 0 Slide No.44 N N = FC cos α − Ft sin α = 1177 1 cos13. CS = 300.108 Ans.56N F 29 90.63 3667 N 45 56.(i ) tan α b = tann α cos λ + sin λ tan i t 7° = tan α cos 60° + sin 60° tan 0° or tan or α = 13.. t = 2mm.88 An ns..56 β = tan −1 μ = 32.484 o μ= Fn = Fc sin n φ + Ft cos φ = 500 sin18.5 565 − 207 sin 18.14 For-2016 (IES..∴ λ = 90° − 30° = 60° Slide No.444 μ= = = 0..5 = 26...68 = = 0.565 = 355.79° Froom (i ) α s = 12° 1 Usiing force relaations F = FC sin α + Ft cos α = 1177 sin13.79 = 8224. α = 0° t = f sinn λ = f sin 90 9 ° = 0...48 mm φ = tan −1 GA ATE-2015 GA ATE-2015 r cos 0 r cos α = tann −1 r = tan −1 0.79 + 560 coss13.24 mm = = 0.87 N Sheear strength (τ y ) = Fs 6695.6 Usiing Merchannt Theory α β 3.5 565 + 207 coss18. b = 2..5 tc 0.08 N IAS S-2003(ma ain exam m) Page No.5mm. Fc = 1177 1 N ..816 ⇒ β = tan −1 μ = tan −1 0.288° = 695.448 mm r= An ns.55 sin 32. Ft = 560 N Usiing relationss: tan α = tan α s sin λ + taan α b cos λ or tan t α = tan α s sin 60° + tan 7° cos 60°. (b) t 0.24 mm m tc = 0.2214° 13 φ = 45°+ − = 45° + − = 32..13 N Givven : α b = 7..109 9 bt = GA ATE-2007 λ = 90° .Fc = 500N N Fx 200 = = 207 N sin λ sin75 F = Fc sin α + Ft cos α = 500sin10 + 207 cos10 = 290.79 = 100 09.779 − 560sinn13.565 = 408 8. 18.566° = tan −1 1 − r sin α 1 − r sin 0 Pag ge No. 5° = 0.25 mm r cos α r cos10° tan φ = = 1 − r sin α 1 − r sin10° r cos10° tan 27..75° t = f sin λ = 0.5. (d) β 2 10° β − ⇒ β = 44. (d) Slide No..75° = 45° + GATE-2008(common data)S-1 τ y = 250 MPa..α = 0 . r = 0.20 mm / rev.14 using Merchant Analysis.09 N ≈ 320 N sin φ sin 27..(λ = 90).113 Ans.14 Slide No. (b) f = 0. d = b & t = f t = f sin λ ⎥⎦ Fs = τ y bt 0.51138 mm GATE-2003(common data)S-2Page No.25 r = 0.GATE & PSUs) Page 156 of 210 Rev.14 7 β 7° β ⇒ 28° = 45° + − = 41° 2 2 2 2 Using Merchant Circle: φ = 45° + − Fs = R cos (φ + β − α ) 320 = R cos(28° + 41° − 7°) ⇒ R = 681.62 N For-2016 (IES.5cos 7° = tan −1 = 27. Page No.8 N 1000 GATE-2003(common data)S-1 Page No.4 mm.116 Ans. φ =45°+ α 2 − Slide No.(c) Fx 800 = = 800 N sin λ sin 90° Using relations: We know.114 Ans..25 mm / rev. d = 3 mm.GATE-2007 Page No.5° 2 2 μ = tan β = tan 44..9826 27.14 Slide No.5sin 7° We know.4888 = = tc tc μ= Slide No.20 × 3 = 250 × = 321.85° ≈ 28° 1 − r sin α 1 − 0. Ft = F = Fc sin α + Ft cos α = 1000sin 0 + 800 cos 0 = 800 N N = Fc cos α − Ft sin α = 1000 cos 0 − 800sin 0 = 1000 N F 800 = = 0. d = 0. φ = tan −1 d = b sin λ ⎤ .85° GATE-2008(common data)S-2 Using Merchant Theory: Page No.115 Ans.V = 180 m / min. φ = 25 . α = 10°.75° = 1 − r sin10° t 0.25sin 90° = 0..14 f = 0.0 ..112 ° ° ° λ = 90 . φ = 27. (a) ⇒ tc = 0. Fc = 1000 N Ans. α = 7° r cos α 0. 16 Slide No.1998 Page No. (c) IES-1993 Page No.147 Ans.1× 1 GATE-2014 Page No.16 Slide No. All the four gauges in each bridge are active gauges.17 Slide No.122 Ans. (d) The energy consumption per unit volume of material removal. total of which adds up to four gauges.1996 Page No.136 Ans.15 Slide No. 4 for main cutting force. (b) IES-2000 Page No.62 cos(41 − 7) = 565.18 Slide No.17 Slide No.(d) π DN π × 200 ×160 MRR = fdV = fd × mm / s = 0. and the bridge fully compensates for temperature changes.(b) ) = 50 × 103 × 0.16 N IES-2004 Page No. two or 2.18 Slide No.131 Ans.GATE & PSUs) Page 157 of 210 Rev. (b) R IAS.124 Ans.129 Ans.16 Slide No.1 GATE-1992 Page No.5mm3 / s 60 60 GATE(PI)-1991 Page No.8 × 1. (a)For higher sensitivity. (c) IES-2011 Page No.146 Ans.134 Ans. (b) IES-2002 Page No.0 . e= Fc Power (W ) = 3 MRR mm / s 1000 fd ( ) Fc ⇒ Fc = 800 N 1000 × 0.16 Slide No.143 Ans.17 Slide No.17 Slide No.141 Ans.127 Ans. (a) IES-2004 Page No.62sin(41 − 7) = 381.17 Slide No.132 Ans.25mm × 4 mm × mm / s = 1675.16 Slide No. while two others are for compressive strain. commonly known as specific energy. (b) GATE-1993 Page No.09 N Ft = R × sin ( β − α ) = 681. (b) ΔR = Gε IES.2001 Page No. (b) IES.142 Ans.16 Slide No.135 Ans.130 Ans.15 MRR = Vfd ( Slide No. (d) chip : work piece : tool = 80 : 10 : 10 IES-1998 Page No. (b) FC FC 400 = = = 2000 N / mm 2 Specific pressure = bt fd 2 × 0.15 Slide No.119 Ans.123 Ans.2 × 2 GATE-2013(PI) common data question Page No.Fc = R × cos ( β − α ) = 681. (b) IAS-2003 Page No. For-2016 (IES.16 Slide No.0 = gauges are used for tensile strain.5mm3 / min = 60000mm3 / min GATE-2013 Page No.15 Slide No.139 Ans. (b) FC 200 J / mm3 = = 2 J / mm3 Specific cutting energy = 1000 fd 1000 × 0. 4 for Radial force and 4 for feed force.18 Slide No. (c) FOR PSU & IES Page No. (a) IAS-2003 Page No.118 Ans. For 3-D lathe dynamometer total 12 strain gauge is needed.15 Slide No. (d) GATE-2007 Page No.148 Ans. (d) IES-2012 Page No.20 Slide No.80 → for Ceramics.186 Ans. (b) Think only the parameters which can produce cyclic stress on tool material.180 Ans.(c) Flank wear directly affect the component dimensions.172 Ans.(c) IES-1995 Page No.22 Slide No.23 Slide No.S.173 Ans.156 Ans.21 Slide No.(b) Solving using straight line equation: y − y1 = y2 − y1 i( x − x1 ) x2 − x1 1.176 Ans.20 → for H.23 Slide No.GATE & PSUs) Page 158 of 210 Rev.158 Ans.23 Slide No.21 Slide No. Slide No.V1 = π D × 284 m / min 1000 π D × 232 N 2 = 232rpm. IES-2006 Page No.23 Slide No.192 Ans.(c) N1 = 284rpm.(a) IAS-1999 Page No.22 Slide No. (c) In Taylor’s tool life equation is n = 0.08 – 0.197 Ans.22 Slide No.S. IAS-2003 Page No.154 Ans.19 Slide No. T2 = 60 min.21 Slide No.18 Slide No.152 Ans. IES-1994 Page No.(d) IES-1992 Page No.8 − 0.23 Slide No.20 Slide No.60 – 0.(b) IES-1999 Page No.(d) GATE-2009(PI) Page No. n = 0.V2 = m / min 1000 Using Taylor's Tool Life Equation.VT n = C V1T1n = V2T2 n For-2016 (IES.(b) IES-2015 Page No.2616 ( can be solved using solve function on calculator ) ISRO-2011 Page No. And high carbon tool steel withstands least temperature 200oC.21 Slide No.21 Slide No.666 IES-2002 Page No. n = 0.19 Slide No.164 Ans.(a) IES-2008 Page No.151 Ans.20 Slide No.(a) Using Taylor's Tool Life Equation VT n = C V1T1n = V2T2 n or 100 × 10n = 75 × 30n or n = 0.8 ( x − 10 ) 60 − 10 x = 51.(c) IES-2014 Page No.(d) IES-1995 Page No.177 Ans.195 Ans.8 = 2 − 0. IAS-2007 Page No.(b) IES-1996 Page No.(c)For crater wear temperature is main culprit and tool defuse into the chip material & tool temperature is maximum at some distance from the tool tip that so why crater wear start at some distance from tip.0 .(c) IES-2004 Page No.19 Slide No.(b) GATE-2008(PI) Page No.165 Ans.178 Ans.23 D = 50mm.171 Ans.193 Ans.20 Slide No.(c) IES-2000 Page No.194 Ans.20 – 0.23 Slide No.(c) IAS-1998 Page No. T1 = 10 min.60 → for Carbides.Ch‐3 Tool Life: Answers with Explanation IES-2010 Page No.182 Ans.198 Ans.(c)Chemical reaction between abrasive and workpiece material at elevated temperature and in the presence of grinding fluid.196 Ans.(a) IES-2007 Page No.21 Slide No.18 Slide No. GATE-2014 Page No.(a)Crater wear occurs due to temperature mainly.170 Ans.188 Ans. IAS-2002 Page No. 5 ⎝2⎠ or T2 = 20. T2 = 2 min.V2 = ? Putting in equation: or 18 ×1800.V2 = V1 2 V1 × T 0.202 Ans.24 Slide No.204 Ans.5 T1 = 4T1 %change = T2 − T1 4T − T ×100% = 1 1 × 100% = 300% T1 T1 IES-2015 IES-2013 Page No.25 = C = 180 Using Taylor's Tool Life Equation.5.201 Ans.0 .207 Ans. VT = C Ans. VT n = C V1T1n = V2T2 n ⎛V ⎞ or V1T10. Given: V1 = 30 m/min. n = 0.25.(a) Using Taylor's Tool Life Equation.VT n = C V1T1n = V2T2 n V1 ⎤ ⎡ ⎢Where.199 Using Taylor's Tool Life Equation.25. T1 = 1 hr = 60 min.ISRO-2013 Page No. T2 = 45 min.24 Page No.24 Slide No.24 Slide No.V = 60m / min.(c) Using Taylor's Tool Life Equation.25 ⎠ T1 = 16T1 IAS-1995 Page No.24 Slide No.(d) Using Taylor's Tool Life Equation.5 = ⎜ 1 ⎟ T20.25 = ⎜ 1 ⎟ T2 0.25 Slide No.V2 = 2 ⎥ ⎣ ⎦ ⎛V ⎞ or V1T10.24 Page No. VT n = C V1T1n = V2T2 n or V1 × T10.or π D × 284 ×10n = π D × 232 1000 or n = 0. V2 = 2V1.5 = V2 × 450.24 Slide No. VT n = C V1T1n = V2T2 n n = 0.200 Ans.5 or V2 = 36m / min IES-2006 (conventional) Page No.1128 1000 GATE-2004.205 Ans.24 Slide No. Taylor tool life equation gives For-2016 (IES. T3 = 30 min.25 = GATE-2015 IAS-2002 or n = 0. IES-2000 × 60n Page No.25 ⎝2⎠ ⎛ 1 ⎞ ⎜ ⎟ or T2 = 2⎝ 0.(b) IAS-1997 Page No.203 Ans.(c) Slide No.(d) Using Taylor's Tool Life Equation.5.25 or T2 = 16T1 2 2 Page No. T = 1 hr 21min = 81min 60 × 810.206 Ans.0. V1 = 18 m / min.GATE & PSUs) Page 159 of 210 Rev.24 Slide No. T1 = 180 min.(c) n V1T1n = V2T2 n n 1 ⎛T ⎞ or V × T n = 2V × ⎜ ⎟ or 8n = 2 or n = 8 3 ⎝ ⎠ IES-1999. VT n = C n = 0. according to the given question V = V1 60 = = 30m / min.204 or n = ⎛T ⎞ ⎛ 60 ⎞ ln ⎜ ⎟ In ⎜ 1 ⎟ T ⎝ 2 ⎠ ⎝ 2⎠ Now for = T3 = 30 min.25 Slide No.25 Slide No.209 Ans.(a) V1 T1n = V2 T2n 60 × 81n = 90 × 36n n 90 ⎛ 81 ⎞ or ⎜ ⎟ = = 1.66 rpm πd π×0.5 − 1 = 300% Slide No. For-2016 (IES.55 m/min V3 34.3 GATE-2009 Linked Q-1 VTn = C Page No.5 60 ⎝ 36 ⎠ In1.210 Ans.GATE & PSUs) Page 160 of 210 Rev. 2 2 1/n ⎛C⎞ T1 = ⎜ ⎟ ⎝ V1 ⎠ 1/n ⎛ C⎞ T2 = ⎜ ⎟ ⎝ V2 ⎠ 1/n T2 − T1 ⎛ V1 ⎞ =⎜ ⎟ T1 ⎝ V2 ⎠ IFS-2013 −1 = ( 2) Page No.25 1/0.208 Ans. V3 = ? Here V1 T1n = V3 T3n n ⎛T ⎞ V1 T1n ⎛ 60 ⎞ = V1 x ⎜ 1 ⎟ = 30 × ⎜ ⎟ n T T3 ⎝ 30 ⎠ ⎝ 3⎠ V3 = πdN or V3 = or N = 0.204 = 34.VTn = C or V1 T1n = V2 T2n n ⎛ T1 ⎞ ⎛ V2 ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ T2 ⎠ ⎝ V1 ⎠ taking log on both side we get or ⎛T ⎞ ⎛V ⎞ n ln ⎜ 1 ⎟ = ln ⎜ 2 ⎟ ⎝ T2 ⎠ ⎝ V1 ⎠ ⎛V ⎞ ⎛ 2V ⎞ ln ⎜ 2 ⎟ ln ⎜ 1 ⎟ ⎝ V1 ⎠ = ⎝ V1 ⎠ = 0.(c) Now.5 ⎛ 81 ⎞ In ⎜ ⎟ ⎝ 36 ⎠ C =60 × 810.5 n= ⇒ n = 0.0 .5 = 540 = K GATE-2009 Linked Q-2 Page No.5 = 90 × 360.55 = =36. K 2 = 60 Using Taylor's Tool Life Equation:VT n =C let cutting speed is x m/min x×TA 0.49 Tool life when cutting speed is 2.45.45 ⎛ 60 ⎞ = 60 ⇒ TB = ⎜ ⎟ ⎝ x ⎠ 0.3 = 420. This can be solved using regression analysis: Using Taylor's Tool Life Equation:VT n = C taking log on both sides: log V + n log T = log C or log V = log C − n log T For-2016 (IES.06 min Velocity when tool life is 80 min V × 800.0 .3 = 420. K1 = 90 for HSS → n2 = 0.49 or T = 31.389 m/min EXAMPLE Page No.Using Taylor's Tool Life Equation: V1T1n = V2T2 n or 100 × 120n = 130 × 50n or n = 0.211 Ans.5 m/s = 2.25 Slide No.60 ⎜ x ⎟ >⎜ x ⎟ ⎝ ⎠ ⎝ ⎠ Solve for x using calculator.6.6.6 =3000 ⇒ Tc = ⎜ ⎟ ⎝ x ⎠ and x×TH 0.5 × 60 m/min ( 2. K1 = 3000 for B → n2 = 0. 213 Ans.6 ⎛ 200 ⎞ = 200 ⇒ TH = ⎜ ⎟ ⎝ x ⎠ 1 0.25 for A → Slide No.(b) n1 = 0.GATE & PSUs) Page 161 of 210 Rev.7 m/min GATE-2013 Page No.25 Slide No. K 2 = 200 Using Taylor's Tool Life Equation:VT n =C let cutting speed is x m/min ⎛ 3000 ⎞ x×Tc1.(a) for Carbide → n1 = 1. x = 26.30 for TA >TB 1 1 ⎛ 90 ⎞ 0.6 1 1.45 and x×TB ⎛ 90 ⎞ =90 ⇒ TA = ⎜ ⎟ ⎝ x ⎠ 1 0.3 = 420.49 or V = 112.30 >⎜ ⎟ ⎜ x ⎟ ⎝ ⎠ ⎝ x ⎠ Solve for x using calculator.45 ⎛ 60 ⎞ 0.6 ⎛ 200 ⎞ 0.3.3 1 0.94 m / min GATE-2010 Page No.6 for Tc >TH 1 1 ⎛ 3000 ⎞ 1.212 Ans.3 NowC = 100 ×1200. x = 39.5 × 60 ) T 0.2997 ≈ 0. log49.214 Ans.25 × 50 ⎠ 0.74 X1= log2. B = −n T V X=logT Y=logV 2.47 then press DT(M+) it will display n =1 then press AC x2 .25mm / rev Let t2 be the time to produce 1 component in 2 nd case.77 Y3= log48.77 48. x = log T .2 × ⎟ 0.e. log42.07 From y = A + Bx.23 then press DT(M+) it will display n =2 then press AC x3 .94 Y1= log49.2components × t2 = 12. 1 cutting tool will produce 50 components we know : V = π DNm / min For the 1st case: N = 50rpm.0 . A = log C ⇒ C = 10 A = 54 B = − n = −0.76 28.23 X2= log3. 1 cutting tool will produce 12.67 then press DT(M+) it will display n =3 then press AC x4 .in Casio Calculator: y = A + Bx y = log V .t 3 = min fN L Tool life(T3 ) = ( x)components × t2 = x × min f × 60 For-2016 (IES.2 components For the 2nd case: N = 80rpm.58 X5= log28.25 Slide No.27 42.76 then press DT(M+) it will display n =4 then press AC x5 .25mm / rev L Let t3 be the time to produce 1 component in 3rd case.87 45.25 For 3rd case : N = 60rpm.94.47 3.e.27 Y5= log42.27.25mm / rev Let t1 be the time to produce 1 component in 1st case.67 X3= log4.94 49.y2 i.23 4.y5 i.87 Y4= log45. log4.07 = 54 GATE-2003 Page No.90.732 Again press the right arrow 2 times then B(2) press 2 and = it will give B = -0.67 9.90 49.58 then press DT(M+) it will display n =5 then press AC After entering all values then press shift then S-VAR(on number 2).GATE & PSUs) Page 162 of 210 Rev.07 ⇒ n = 0. log45. Press mode – 2 times-then press ‘2’ (Reg-2) Then select the type – (lin-1) Then start entering the values as below.y4 i.07 ∴ equation becomes:VT n = C ⇒ VT 0.y3 i. log3.then press the right arrow 2 times then A (1) press 1 then = it will give A = 1.76 X4= log9.2499 ≈ 0. log28.t1 = Tool life(T1 ) = 50components × t1 = 50 × L min fN L min f × 50 10 cutting tools produce 122 components Therefore.e.25 × 80 ⎟⎠ ⎝ ⎝ or n = 0. log9.90 Y2= log49.58 Now on calculator.2 × n n L L ⎛ ⎞ ⎛ ⎞ or π D × 50 × ⎜ 50 × = π D × 80 × ⎜ 12. f = 0.t 2 = L min fN L min f × 80 Using Taylor's Tool Life Equation:VT n = C ⇒ V1T1n = V2T2 n Tool life(T2 ) = 12. log49. A = log C .y1 i. log2.e. log48.77. f = 0.87.On comparing with straight line equation.(a) 10 cutting tools produce 500 components Therefore. x1 .e. f = 0. then tool life Now when f is increased by 25% New V.VT 0.13 f 0.6 × 20.99 T = 2.3 = C V = 40m / min. (a)Both A and R are true and R is the correct explanation of A ESE-1991.50.f and d are:V = 40 + 0.f and d are:V = 40m / min.(b) ∵ flank wear ∝ cot α ∴ tool life ∝ tan α % change in life = tan α 2 − tan α1 tan 7 − tan10 × 100% = × 100% = −30% tan α1 tan10 ∴ % change in life = 30% decrease IES-2010 Page No. f and d are increased by 25% New V.25 × 0.25 × 2 = 2.25 + 0. d = 2mm putting in given equation: VT 0.215 Ans.31250.13 f 0.49 or 40 × T 0.IAS-2010( conventional) Page No. rest parameters remain same.99 or 50 × T 0. f = 0.GATE & PSUs) Page 163 of 210 Rev.25 × 40 = 50 m / min f = 0.2007 Page No.25 Slide No.779 min When only feed is increased by 25% .13 × 0.6 d 0. then For-2016 (IES.3 = 34.5mm putting in given equation:VT 0. then Now when V is increased by 25% New V.25 × 50 ⎟⎠ ⎝ 0.219 Ans. given.3 = 36.25 L ⎛ ⎞ = π D × 60 × ⎜ x × 0.3125mm / rev d = 2 + 0.26 Slide No. feed & depth of cut are together increased by 25%.25 × 0.926 ≈ 29components GATE-1999 Page No.6 d 0.49 When speed. T = 60 min.220 Ans. rest parameters remain same.(a) IES-1994.(c) We know that cutting speed has the greatest effect on tool life followed by feed and depth of cut respectively. V1T10. d = 2mm putting in given equation:VT 0.1 respectively.2.25 = 0.25 = V3T30.25 × 40 = 50m / min. f = 0.3 = 34.0 .6 × 20.3 = 36. tool life will be Now when V.29 min When only speed is increased by 25%.41min When only depth of cut is increased by 25%.31250.26 Slide No.(d) ISRO-2012 Page No.250.13 × 0.3 f 0.f and d are:V = 40 + 0. For maximizing tool life we will adjust 3.26 Slide No.6 d 0.(a) IAS-1995 Page No.222 Ans. IES-2008 Page No.218 Ans.6 d 0.25 L ⎛ ⎞ or π D × 50 × ⎜ 50 × 0. rest parameters remain same.217 Ans.3 = C ⇒ C = 36.now.25mm / rev.25mm / rev.3125mm / rev.26 Slide No.49 50 × T 0.26 Slide No.223 Ans.3 = 36.(d) IES-1997 Page No.6 × 20.3 = 34.99 or T = 10.3 × 0.49 or T = 21.13 × 0.25 × 60 ⎟⎠ ⎝ 0.25 + 0.26 Slide No.6 × 2. f = 0.25 0.25 = 0.250.221 Ans.13 f 0.26 Slide No.25 ⎛ x ⎞ or 50 = 60 × ⎜ ⎟ ⎝ 60 ⎠ or x = 28. d = 2mm Putting in equation : 40 × 600. 0 = Rs.6 d 0.2 ⎟⎠ ⎝ Using Taylor's Tool Life Equation:VoTon = C Vo ( 64 ) GATE-2014 0.2 = 60 orVo = 26. T1 = 45 min. n = 0.5 ⎠ Using Taylor's Tool Life Equation:VoTon = C . C = 100 Using optimum tool life equation for max productivity: ⎛ 1− n ⎞ ⎛ 1 − 0. n = 0.(c) For-2016 (IES. n = 0.5 / regrind Putting in equation: 6.46.50 ⎟⎠ ⎜⎝ 0.29 Page No.25 × 2 = 2.252 Ans.235 Slide No. T2 = 10 min.28 IES-1996 Page No.3 = 36. Cm = Rs.5. 5.50 / min× 3min + Rs.49 or 40 × T 0.f and d are:V = 40m / min.244 Ans. Tc = 2 min Using Taylor's Tool Life Equation:VT n = C ⇒ V1T1n = V2T2 n 50 × 45n = 100 ×10n or n = 0.29 Page No.246 Ans.5 = 100 orVo = 33. d = 2 + 0. Vo × 90.250.250 Ans.50.(a) Slide No.340. C = 60 Using the equation for optimum tool life for minimum cost C ⎞ ⎛ 1− n ⎞ ⎛ To = ⎜ Tc + t ⎟ ⎜ Cm ⎠ ⎝ n ⎟⎠ ⎝ Ct = Cm × Tr + Depriciation cost Ct = Rs.6.0.(a) Ans.5 ⎞ ⎛ 1 − 0.0 . Ans.(b) Slide No.9 to 6. Tc = 3min.5 ⎞ To = Tc ⎜ or To = 9 ⎜ ⎟ ⎟ = 9 min ⎝ n ⎠ ⎝ 0.46 = 288 orVo = 195m / min IAS-2011(main) Page No.5.243 Ans.49 or T = 35.V2 = 100m / min.Now when d is increased by 25% New V.46 ⎟ = 2.6 × 2.46 ⎞ ⎛ 1− n ⎞ To = Tc ⎜ ⎟ = 2 ⎜ 0.0.2 Slide No.5 min. Tool regrind time(Tr ) = 3min.29 Slide No.50 / min Depriciation cost = Rs.28 IES-2009(conventional) Slide No.238 Page No.227 Slide No.34 n ⎝ ⎠ ⎝ ⎠ Again Using Taylor's Tool Life Equation:VoTon = C Vo × 2.1min Using optimum tool life equation for maximum productivity: ⎛ 1− n ⎞ ⎛ 1 − 0.28 Ans.245 V1 = 50m / min.33 m / min GATE-2005 IAS-2007 IES-2011 IES-1999 Page No.27 IAS-2003 Page No.3 = 36.13 f 0.(a) Ans.2 ⎠ ESE-2001(conventional) Page No.(d) Slide No.2 ⎞ ⎛ = 64 min To = ⎜ 3 + 0. Tc = 9 min.248-249 Ans. C = 50 × 450. f = 0.0.GATE & PSUs) Page 164 of 210 Rev.29 Tc = 1.29 Slide No.5 ⎜ ⎟ ⎟ = 6 min ⎝ n ⎠ ⎝ 0.29 Page No.5.228 Slide No.(b) Slide No.247 Ans.25mm / rev.5mm putting in given equation:VT 0.46 = 288 Using equation of optimum tool life: ⎛ 1 − 0.13 × 0.2 ⎞ To = Tc ⎜ or To = 1.2.11 m / min Page No.844 min IES-2010 Page No.27 IES-2014 Page No.29 Slide No. 259 Ans.(d) smaller shear angle means higher force.269 Ans.30 Slide No.32 Slide No.(d)Machinability is a comparative measure not absolute. Sulfur (0.33 Slide No.31 Slide No.285 Ans.273 Ans. lower prodeuction rate.282 Ans.e. bismuth. which has low shear strength. IES-2011(conventional) Page No.(a)Machinability: Machinability can be tentatively defined as ‘ability of being machined’ and more reasonably as ‘ease of machining’.1 is not possible.30 Slide No.278 Ans. Lead and Phosphorous are added to steel which when added to Manganese forms Manganese sulphide etc. higher cost per piece as it reduces tool life.258 Ans.32 Slide No. chip Increases breaker 4. IES-2000 Page No.6%) to form soft manganese sulfide inclusions. Effect of elements on machinability of steels: S.277 Ans. Lead & Tin Internal Lubrication.e. IES-1992 Page No.(a) IES-2007. Titanium’s work-hardening characteristics are such that titanium alloys demonstrate a complete absence of “built-up edge”. while the maximum production rate criteria will result higher cutting speed i. Carbon Carbide former Decreases 5. vanadium Strong carbide former Decreases IES-1992 Page No. Because of the lack of a stationary mass of metal (BUE) ahead of the cutting tool.256 Ans.(a) large grain means soft workpiece material. Molybedenum.284 Ans.2009 Page No.283 Ans. We have to select best option that so why chosen (a) IES-2003 Page No.30 Slide No. coupled with the friction developed by For-2016 (IES.265 Ans. IES-2002 Page No.260 Ans. Almost all tool materials tend to react chemically with titanium. selenium.255 Ans.NO ELEMENTS Cause MACHINABILITY 1.271 Ans. IAS-1997 Page No. The inclusions also provide a built-in lubricant that prevents formation of a built-up edge on the cutting tool and imparts an altered cutting geometry.0 .30 Slide No.33 Slide No.(a) Sulphur.(a) To improve MRR = fdv i. (b) Titanium is very reactive and the chips tend to weld to the tool tip leading to premature tool failure due to edge chipping.33 Slide No.279 Ans.(c) IES-1992 Page No. IAS-2000 Page No.7 to 1.32 Slide No.253 Ans.254 Ans.287 Ans. Sulphur& selenium Internal lubrication.33 Slide No. dimensional accuracy and surface finish is prime factor. IES-1996 Page No.33 Slide No.(b) IES-1995 Page No.257 Ans.(c) The minimum cost criterion will give a lower cutting speed i. but increase in velocity will reduce the tool drastically so will increase cost more than feed.e.GATE & PSUs) Page 165 of 210 Rev.264 Ans. a high shearing angle is formed. These act as discontinuities in the structure which serve as sites to form broken chips.(a)Built up edge protects the cutting edge of the tool from wear.30 Slide No. Such ease of machining or machining characters of any tool-work pair is to be judged by: • Magnitude of the cutting forces • Tool wear or tool life • Surface finish • Magnitude of cutting temperature • Chip forms ISRO-2007 Page No.(b) it is less than one but very close to each other so 0.(a) IES-2004 Page No.280 Ans. IES-1996 Page No.(a)At optimum cutting speed for the minimum cost of machining gives low production rate.32 Slide No. tellurium. lead. Sulfur. So tool life increased but it changes the geometry of the cutting.(a)But All of the above are machinability criteria.(d) IAS-1996 Page No.30 Slide No. Aluminium& silicon Hard oxide former Decrease 2.30 Slide No. IES-1998 Page No.31 Slide No.(c) IAS-2002 Page No. This causes a thin chip to contact a relatively small area on the cutting tool face and results in high loads per unit area. IES-2009 Page No. and phosphorus have all been added to enhance machinability. chip Increases breaker 3.30 Slide No.31 Slide No.33 Slide No.08 to 0.33%) combines with manganese (0.IES-1998 Page No.(c)Free-machining steels are basically carbon steels that have been modified by an alloy addition to enhance machinability.(d) After some time cutting speed will be so that tool changing time will be significant.33 Slide No.281 Ans.(c) IAS-2007 Page No. These high forces. IES-2010 Page No. productivity we can increase velocity or feed.32 Slide No. IES-2012 Page No.(c) It is CNC machine. 304 Ans.34 Slide No.975 mm Fundamental Deviation = basic size-nearer limit = 35-34.16mm tan SCEA + cot ECEA tan 30 + cot10 Slide No.35 Slide No. used for finish machining. GATE-2010(PI) Page No. (a) Titanium high cost and need 10 times much energy than steel to produce.8 × 10−3 −6 GATE-2007(PI) Page No.34 Slide No. properties between steel and aluminium. Slow cutting results in formation of built-up edge.0999=0.37 Ans. Here MRR remains same therefore feed remains same only nose radius can be changed.34.294 Ans.1-99.37 GATE-2010.14 Ans. (b) f tan SCEA + cot ECEA f f hP = and hQ = tan 30 + cot 8 tan15 + cot 8 ( tan15 + cot 8) h ∴ P = hQ ( tan 30 + cot 8 ) Using formula:h = IES-1993. (d) IES-1999 Page No.34 Slide No. 10 Page No. IES-2006 Page No.68 × 10−4 m / rev = 0.296 Ans. (b) Ch-4: Limit.9 Slide No. (b) For increasing surface finish means reduce roughness we have to increase nose radius and reduce feed.293 Ans.(a) f = 1mm / rev.(b) Slide No. especially as titanium has a tendency to gall and weld to the tool surface.GATE & PSUs) Page 166 of 210 Rev.34 Slide No.35 Slide No. (c) Surface roughness is directly dependent on square of feed.03 mm = 30 microns For-2016 (IES. GATE-1997 Page No. and pressure.291 Ans.025 = 34.Light weight.991 .8mm = 1.297 Ans.298 Ans.289 Ans. means that tool life can be short.the chip as it passes over the cutting area.0002 So.37 Slide No. and hc remains the same.2 Tolerance of shaft B=0. So f has maximum effect.35 Slide No.(a) Tolerance of shaft A=100.34 f 1 = = 0.15 Ans. SCEA = 30°.ISRO-2008 Page No. IES-1995 Page No. (b) IAS-1996 Page No.303 Ans.299 Ans.ISRO-2012 Slide No.8 × 10−3 m f2 f2 we know : h = or 5 ×10−6 = or f = 2. (c) IES-2012 Page No.34 Slide No. (a) refer previous question GATE-2014(PI) Page No. R = 1.36 ISRO-2010 Page No. Tolerance & Fit: Answers with Explanations For PSU PageNo.1001-0.295 Ans.35 Slide No. tolerance of shaft A > tolerance of shaft B GATE-2004 Page No.9=0. but after certain speed the finish remains same.34 f2 f2 f2 4f = 1 or = ⇒ R1 = 4 R 8 R 8 R1 8 R 8 R1 Slide No.(b) Ans.975 = 0.292 Ans. ECEA = 10° Using formula: h = GATE-2005 Page No.0 .(d) upper limit = 35-0.016 mm GATE-1992 Page No. but its effect is small at speeds.300 Ans. Rake angle has noticeable effect at slow speeds. (a) h = 5μ m = 5 ×10 m. IES-2001 Page No.009 mm Tolerance = upper limit-lower limit = 34.009 = 34.13 Ans.34 Slide No.(c) Maximum clearance = Higher limit of hole – lower limit of shaft = 25.991= 0.990 = 0. result in a great increase in heat on a very localized portion of the cutting tool. corrosion resistant. strong. (d) we know : hc = f2 8R When f1 = 2 f .35 Slide No.268mm / rev 8R 8 ×1. IES-2002 Page No.020-24.991mm lower limit = 35-0. All this heat (which the titanium is slow to conduct away).37 Slide No. 27 Ans.(c) max clearance = upper limit of hole . minimum rejection will be of the batch having mean diameter 20mm GATE-2000 Page No.1 mm GATE-2015 Page No.50-39. GATE-2007 Page No.26 Ans.(c) Allowance is the minimum clearance between shaft and hole.38 Slide No.39 Page No.(a) IES-2013 Page No.005 mm = 0.28 Ans.38 Slide No.IES-2005 Page No.GATE & PSUs) Ans.(a) GATE-2005 Page No.(c) Reference will be taken from side therefore only tolerance on toll position will affect our product. IES-2011 Page No.30 For-2016 (IES.20 Ans.(b)This being an clearance fit .95 = 0.22 Ans.38 Slide No.(a) IES-2014 IES-2015 Page No.0 .39 Slide No.38 Slide No.38 Slide No.015 mm IES-2015 Page No.020-25.37 Slide No.39 Slide No.lower limit of shaft = 40.(a) IES-2015 Page No.17 Ans.(c) Since basic size is 20mm so. (c) Page 167 of 210 Rev.24 Ans.29 Slide No.38 Slide No.21 Ans. so minimum clearance will be Min C= LL of hole –UL of Shaft Min C = 25.16 Ans. (c) Ans.37 Slide No. 04+0.(c) GATE -2012 Same Q in GATE-2012 (PI) Page No.03mm.43 Ans.02 mm ∴ size of hole: lower limit = 20 mm upper limit =20+2x = 20+2 × 0.09 mm. Basic size = 20 mm Refering the figure:2 x + 0. (a) IES-2004 Page No.(c) Ans. (a) Smax=50.39 Slide No.(d) IES-2009 Page No.(b) GATE-2001 Page No.40 Slide No. Using formula of minimum clearance Min clearance = upper limit of shaft .04) – (25 + 0.40 Ans.(b) GATE-1998 Page No.54 Slide No.0 . Using unilateral hole base system.36 Ans.02 mm Here Minimum clearance is negative i.40 Slide No.40 Slide No. Max clearance = 0.35 Ans.47 Ans.03 + x = 0.39 Slide No.e.37 Ans.53 Slide No.39 Ans.39 Slide No.(c) It is transition fit.(c) Ans.(c) IES-2006 Page No. Hmin=50.57 For-2016 (IES.39 Slide No.04 mm size of shaft:Lower limit = 20.(c) IES-2012 Page No.03= 20.09 mm IES-2007 Page No.41 Slide No.lower limit of hole = -0.09 or x = 0.(a) For clearance fit Maximum metal condition of shaft will be smaller than minimum metal condition of the hole.07 mm upper limit = 20.GATE-2011 Page No.40 Slide No.40 Slide No.07+ x =20.(d) IES-2008 Page No. ISRO-2008 IES-2005 IES-2005 Page No.45 Ans. Min clearance = 0.39 Slide No.02=20. maximum inteference occur.07+0.42 Slide No.40 Slide No.000.34 Ans.(b) ISRO-2010 Page No.41 Page No. IES-2015 Page No.02) mm = 20 µm IAS-2011(main) Page No.40 Slide No.41 Page No.005 so Smax<Hmin ISRO-2011 Page No.02=20.(c) Maximum Interference = Maximum size of shaft – Minimum size of hole = (25 + 0.39 Slide No.33 Ans.GATE & PSUs) Ans.31 Ans.32 Ans.44 Ans.(d) An interference fit creates stress state in the shaft therefore “2” is wrong.(c) Page 168 of 210 Rev.38 Ans. 30776 × 10−3 mm For-2016 (IES.41 = −36 μ m o hole = +36 3 μm ∴ Fundamentall deviation of 1 i = 0.936mm LL of sh haft = UL .052 2 2 = 24..tolerance t = 24.(c))Without callculating wee can choose option (c) ass fundamental deviattion is zero therefore t LL L = Basic size = 25.455 D 3 + 0.43 3 Slide No.45 × 3 D + 0.001 186 = 0.5 D 0.246]0.0 .76 Ans.72 Ans.17 μ m = 139 μ m Alloowance = min n clearance = 36 µm S-2015(conv ventional) IES Page No.033mm m Toleran nce of shaft = 40i = 52μm = 0.75 N Anss.(a) All stattements are wrong.246)] = 1.45(63.44 Slide No. And ex S-2006(conv ventional) Page No.70 N Anss.0 064mm = 24.444 = −64 μm = −0.D = Dmin ×Dmax = 50 × 80 = 63.23 mm 1 i = 0.000 mm m But proper calculattion is D = D1 × D2 = 18 1 × 30 = 23.(c) GAT TE-2009 Page No.GATE & PSUs) Page 169 of 210 Rev.43 3 Slide No.44 4 Slide No.8 859 μm = 0.033mm UL of sh haft = BS +F FD = 25. Therefore Geomettric mean diameter.45 0 D 3 + 0.97mm m Funndamental deviation d off shaft = − 5. = −5.052mm m Fundam mental devia ation of shafft d = -16D0.45D1/3 +0. is basic size.74 N Anss. 50 m mm is not ho ole diameterr it xaminer ask k INCORREC CT not corre ect options.936-0.884 mm m IES S-2002 Page No.73 Ans Geomettric mean diameter(D) = 18 × 30 = 23.001D = 1. D = D1 × D2 = 80 ×1200 = 97.5[6 63.GAT TE-2014 Page No.1711μ m IT 8 = 25i = 25 × 2.064mm m Fundam mental devia ation of holee H = zero LL of hoole = BS =25 5mm UL of hole h = BS + Tolerance T = 25. Therefore T alll options arre incorrect.71 N Anss.44 4 Slide No.001D) + μm m 1 + 0.41 = −0.030115mm GAT TE-2008(PII) Pag ge No.41 = −5.001(63 = [0 0. IES Bassic size =1000 mm.3074 μ m Toleran nce of hole = 25i = 33μm = 0..246)1/3 3.5D D0.(d) IES S-2008 Page No.0464 46 mm Fundam mental devia ation for 'f'sh haft.(a) 60 mm diameter liees in the dia ameter step of o 50-80mm m..001D = 2.43 N Slid de No.00 0186 mm Foor IT8 = 25i = 25 × 0.001 × D = 1.44 N Slid de No.17 μ m = 54 μ m IT 10 1 = 64i = 644 × 2.238mm Standarrd tolerancee unit (i) = 0.3076μ m = 1.0.246 mm m Fundam mental tolerance unit ( i ) = (0. 021=24.78 Ans.45 3 D + 0.44 Slide No. at least one section must be there that treated as sink.(d) Fundamental deviation of all the bore is zero. i = 0.021 mm ∵ it is a shaft base system: Upper limit = basic size=25.(b) ISRO-2008 Page No.44 Slide No.013 mm Therefore Upper Limit = 25.96 – 0.34 ×10−3 mm For IT 7 = 16i = 16 ×1. and tolerance of sink will be cumulative sum of all tolerances.995=2.04 = 24.032mm ∵ hole base system so.45 Slide No.013 mm for IT6 GATE-1996.79 Ans.(d) Remember H7 with p6.01 w = 9.92 Ans.13 Page No.45 Slide No.08 + 0.02mm R = 13. For IT7.46 For-2016 (IES.0. 89 Ans.033 = 24.927 mm Therefore Maximum clearance = Higher limit of hole – lower limit of shaft = 25.81 Ans. Tolerance = 25i = 0.021 mm For IT8.(a) It requires centre.978 mm GATE-2010(PI) Page No.45 Slide No.01-9. IES-2000 Page No.033 = 25.00 mm Lower limit = Upper limit − tolerance = 25.82 Ans.GATE & PSUs) Slide No. s6: Interference fit H7 with k6.106 mm = 106 microns GATE-2003 Page No.(d) LL of L4 = LL of L1 –UL of L2 –UL of L3 = 21.0326mm upper limit of hole = basic size + tolerance = 25+0. 91 Ans.005=1.99mm Tolerance are probabilities and not the absolute value on any part.77 Ans.45 Slide No.005-10.033-0.0 .(a) GATE-2003 Page No. so P =Q+W+R Considering dimension ⇒ 35 = 12 + W + 13.98mm UL of L4 = UL of L1 –LL of L2 –LL of L3 = 22.34 ×10−3 = 0. n6: Transition fit All other fits are clearance fit.34μ m = 1. lower limit = basic size=25mm GATE-2000 Page No.(b) ∵ diametric steps are not given we take given dia as the basic diameter only. ∴ GATE-1997 Tolerance = 0.033 mm Shaft: Higher limit = basic size – fundamental deviation = 25 – 0.021 mm – (0.927 = 0.00+0.03 = 0.99-10.033 mm For IT6.02 = 13.46 Slide No. Page 170 of 210 Rev.995-9.08mm Q = 12.02mm GATE-2007(PI) Page No.00 .033 – 24.For IT 8 = 25i = 0.0 +−0.44 Slide No.96 mm Lower limit = Higher limit – tolerance = 24.IES-2012 Page No.(d) Hole: Lower limit = Basic size = 25 mm Higher limit = lower limit + tolerance = 25 + 0.0326=25.04 0.(b) P = 35+0. 90 Ans.(d) S = P +Q + R +T GATE-2015 or T = S − P − Q − R or Tmin = Smin − Pmax − Qmax − Rmax Page No.44 Slide No.03 Now all have same bilateral tolerance.001D = 1. Tolerance = 10i = 16i – 6i = 0.021)x(6/9) mm =0.85 Ans.02 + 0. Tolerance = 16i = 0.01 ± 0. 150 Ans. min thickness = 2 × (10 × 10−3 ) mm = 0.GATE-2007(PI) Page No.014 mm Dimension of ‘NOT GO’ gauge = 20.104 Ans.084 mm ISRO-2008 GATE-2014 Page No.50 Slide No.07 mm Minimum limit will correspond to max thickness.010 mm Max thickness of plating =2 × 0.105 Ans.05 – 0.01 + 0.62 mm GATE-2008 Page No.5/25 = 0.032 = 25.004 = 20. ISRO-2008 Page No.04 mm GATE-2013 Page No.03mm Min limit = min size of hole + max thickness = 30.5/50) x 12 = 2.(c) Ch-5 Measurement of Lines & Surface: Answers with Explanations ISRO-2010 Page No.02mm Max limit = max size of hole + min thickness = 30.Guage = 24.010+0.146 Ans.94 Ans(c) Upper limit of pin = 25.147 Ans.050+0.GATE & PSUs) Page 171 of 210 Rev.(d) Lower limit of hole = 25-0.136.47 Slide No.51 Slide No.52 Slide No.056 mm Minimum size will correspond to max thickness Size of GO-Guage = Lower limit of pin + Max thickness of plating Size of GO-Guage = 25.128 Ans.93 Ans.01 mm Work tolerance = 20. Dimension of ‘GO’ gauge = 20.(d) For-2016 (IES.so.01 = 0.02=30.47 Slide No.05 – 20.102 Ans.46 Slide No.05 mm Lower limit of hole = 20.of div in thimble 0.137 Ans.(a) Total dimension= pitch × No.0 .064 mm Min thickness of plating =2 × 0.50 Slide No. In workshop it has no use.(c)Least count = 0.(c) Slide No.04 mm Gauge tolerance = 10% of work tolerance = 0.(c) A measuring device of a standard size that is used to calibrate other measuring instruments.015=24.52 Slide No.46 Slide No.(b) GATE – 2006.958 mm lower limit of GO-Guage = 24.02 mm ISRO-2009. 2011 Page No. Engineering metrology ISRO-2008 Page No. (d) ISRO-2010 Page No.03=30.51 Slide No. Maximum limit will correspond to min thickness.96 Ans. max thickness = 2 × (15 × 10−3 ) mm = 0.of div + Pitch × reading No.106 Ans.46 Page No.004 = 20.47 Slide No.020 mm Lower limit of pin = 25.50 Slide No.015) = 0.138.5x5 + (0.139 Ans.028=0.(a)The vernier reading should not be taken at its face value before an actual check has been taken for zero.004 mm Therefore.(b) Higher limit of hole = 20.47 Slide No. VS-2012 Page No.so.129 Ans.(d) Primary standards are used for calibration only.(c) If there is axial intersection Rp must not equal to RQ GATE-2014(PI) Page No.003 mm GATE-2004 Page No.132 Ans.103 Ans.(b &c) GATE-1995 Page No. GATE-2007(PI) Page No.52 Slide No.9850+0.985 mm Work tolerance = 10% of guage tolerance = 10% of (2 × 0.020+ 2x0.(d) Before plating the hole size will be bigger .032=0.003mm Upper limit of GO.47 Slide No.046 mm GATE-2015 Page No. Refersildes for theory Ans.152 Ans.1 ⎠ ⎝ ⎠ Page No.(d) IES-2006 Page No.(b)Lay – directional of predominant surface texture produced by machining operation is called Lay. Cannot measure absolute dimension but can only compare two dimensions.443 mm 2 2 2 ⎝ 2 ⎠ Slide No.58 Slide No.58 Slide No.55 Slide No.532-15.201 Ans.168 Ans.(c) p α 2 ⎛ 60 ⎞ Best Wire Size: d = sec = sec ⎜ ⎟ = 1.59 Page No.205 Ans.5 ⎛ 60 ⎞ sec = sec ⎜ ⎟ = 1.134=29.(c) ISRO-2010 IAS-2013(main) GATE-1997 IAS-2012(main) Page No.204 Ans. a comparator is able to give the deviation of the dimension from the set dimension.52 Slide No.58 Slide No.Refer slides for theory ISRO-2011 Page No.167 Ans.207 Slide No.58 Slide No.179 Ans.180 Ans.58 Page No.0 .54 Slide No.398=1.57 Slide No.208 Slide No.57 Slide No.203 Ans.206 Ans.(a) L = 250 mm.54 Slide No.190 Ans.57 Slide No.58 Slide No.(b) IES-2008 Page No.( Diameter d = 20 mm or r = 10 mm) or H = 90 mm ⎛H ⎝L θ = sin −1 ⎜ GATE-2011(PI) 90 ⎞ ⎞ −1 ⎛ o ⎟ = sin ⎜ 250 ⎟ = 21. H + r = 100 mm .5-1.Refer sildes for theory Page 172 of 210 Rev.55 Slide No.55 Best Wire Size : d = GATE-2013 Page No.(d) IES-2007 Page No.216 SlideNo.(a) Ans.202 Ans.(c)Lay direction: is the direction of the predominant surface pattern produced on the workpiece by the tool marks.134mm Diameter of cylindrical standard = 30.59 Page No.(a) p α 2.60 For-2016 (IES.(b) IFS-2011 Page No.225 Ans.GATE & PSUs) Slide No.366mm IES-2012 Page No.(b) Ans.5mm ∴ Effective diameter= 30.Refer slides for theory IES-1992 Page No.1547 mm 2 2 2 ⎝ 2 ⎠ GATE-2011(PI) Page No.178 Ans.(c) A sine bar is specified by the distance between the centre of the two rollers GATE-2012(PI) Page No. (Rest all the options will give reading of the dimension measured it will not compare) PSU Page No.(c) Difference between the readings of micrometers= 16.During the measurement. IES-2010 Page No.(c)A feeler gauge is used to check theThickness of clearance ISRO-2011 Page No.58 Slide No.193 Ans.195 Ans. IES-2008 Page No. 54 = 6. 257 Ans. 242 Ans.245 Ans.61 (1. (a) GATE-2014 Page No. (a) 10 30 θ = 18.238Ans.261 Ans.237 Ans.434 = x 10 x = 3.63 Slide No. Angle between the first incident ray and the last reflected ray is 90o. 255 Ans. 239 Ans.229 Ans.64 Slide No.002 − 1.62 Slide No.000) ×10−1 cm = Slide No.357 ≈ 2 fringes Ch-6 Miscellaneous of Metrology: Answers with Explanations GATE-1998 Page No. 252 Ans.GATE_2003 nλl Δh = 2 Page No.334 ∴ diameter at z = 0 is (20 − 2 x) ∴ diameter = (20 − 2 × 3. (c)Laser interferometeris widely used to check and calibrate geometric features of machine tools during their assembly GATE-1992 Page No. 262 Ans.65 Slide No. (d) 3 C2 A = 5 + 15.54 + 8 28. IES-1998 Page No.0058928 × 10−1 cm × 10−1 cm 2 n = 0.334) = 13. 246 Ans.336 GATE-2008(PI) tan θ 2 θ 2 = Page No.006 θ = 12. ISRO-2010 Page No.62 Slide No.678 / cm So for both fringes=2 × n = 1.63 Slide No.(a) n × 0.62 Slide No.62 Slide No. (d) GATE-2014 Page No.63 Slide No. (b)In optical square two mirrors are placed at an angle of 45o to each other and at right angles to the plane of the instrument. (b) GATE-1995 Page No.434 Also. (c)Autocollimator isan optical instrument for non-contact measurement of small angles or small angular tilts of a reflecting surface GATE-2009(PI) Page No. (b) GATE-2010 Page No.64 Slide No.Two mirrors may be replaced by two prisms.001 For-2016 (IES.0 . (b) GATE-2004 Page No. (b)Autocollimator is also measure flatness.64 Slide No.GATE & PSUs) Page 173 of 210 Rev. x tan θ = 10 tan θ = tan18. 3 r 30+smaller radius r D = 30 + 43.6 69 Slide No. m reliieve interna al stresses. (d) Annea aling requireed.55 H − 40. (a) IES S-2000 Page No. an n energy lev vel will reached when the old graiin structuree (which is coarse due to previous cold working g) starts disintegrating.9 N Anss. 69 soften material.. (b) IES S-2008 Page No. (d) Streng gth increases due to gra ain refinement.6 67 Slide No. 69 6 Slide No. N 26 Anss. 70 7 Slide No. Page No. IES S-2009 Page No. GAT TE-2003 Page No. N 29 Anss. N 10 Anss.GATE & PSUs) Page 174 of 210 Rev.6 68 Slide No. (d)Ann nealing i is used to indu uce ducttility. f This phenomen non is know wn as “recrysstallisation” IES S-2004 Page No. N 38 Anss. diislocation density incre eases.32 N Anss. IES S-2006 Page No. 70 7 Slide No. (b)If the specimen s is stressed slig ghtly beyond the yield point p and unloaded then n the phenoomena of strain s harde ening takes place as a result of which strength increasees. metal is simply not ductile enou IES S-2004 Page No. N 27 Anss. 70 7 Slide No.55 = 55 55 − 30(radius) = 25 = CB B CB 25 tan θ = = (30 + 200) 50 θ = 30 R cos 30 = 50 R = 43. IES S-2003 Page No.55.65 5 H1 = 20. refine the e structure by makin ng it hom mogeneous. (d) ISR RO-2010 6 Slide No. N 34 Anss. 69 6 Slide No.55 Slide No. N 69 Slid de No. (a) IES S-1997 Page No. (c) menon wheree ductile meetals become stronger and a harder when they are deform med plasticallly is • Phenom called sttrain harden ning or work k hardening. (c)If work king below Rx R temp then n it is cold-w working proccess GAT TE-2002. For-2016 (IES.55+60=95 5.55 = 955.30 ∴ D = bigger radius+43. And thus t their iinteraction with each other o resultin ng in increasse in yield sttress. N 31 Anss.55mm H1 + 20(radiuss) = 40. 28 Ans. N 36 Anss. (a) Adv vantages of Cold Formin ng vs. (c)When a metal is heated & d deformed un nder mechan nical force. closer tolerances ace finish • Better surfa dening increa ases strengtth and hardn ness • Strain hard d deforrmation can cause desirrable directioonal propertties in produ uct • Grain flow during uired (less total energy)) • No heating of work requ Diss-advantagess of Cold Forrming • Equipment of higher forces and pow wer required d • Surfaces of starting worrk piece must be free off scale and dirt nd strain harrdening limiit the amoun nt of forming that can b be done • Ductility an o allow furth her deformattion • In some opeerations. N 33 Anss. a portion n of the defo ormation en nergy becomess stored within the matterial in thee form of add ditional disllocations an nd increased grain boun ndary surface area. • During plastic defoormation. stress free) with red duced grain size Starts forming. 68 6 Slide No. IES S-2013 Page No. metal must be annealed to ugh to be colld worked • In other casses. Simultaneoously. 69 6 Slide No. H 2 = 35. an en ntirely new grain structture (equi-a axed.55 − 40.263 N Anss.0 .30 N Anss.330mm Ch-7 Metal Form ming: Answe A ers wiith Ex xplanations GAT TE-1995 Page No.35 N Anss.300 + 20 = 93. and improvee cold workiing propertiees. (b) If cold d worked it will w improve mechanicall properties.ISRO-2012 Page No. N 37 Anss. (b) IES S-20011 Page No. Hot Working: W • Better accurracy. 69 6 Slide No. 69 6 Slide No.Normalizzation is an n annealing process in w which a me etal is cooleed in air afterr heating. IES S-2008 Page No. 69 6 Slide No. 66 6 Slide No. N 39 Anss. (c) Durin ng deformation. (b) For cold working metal m should d have high ductility. H=3 35. ISR RO-2009 Page No.GAT TE-2014 Page No. 73 Slide No. rolling is accomplished progressively in many steps. IES-2008 Page No. (a) IAS-2002 Page No. wrought iron.65 Ans. GATE -2007 Page No. In addition.81 Ans.72 Slide No. because of cold working. and rolled shapes can be bent to a desired curvature on forming rolls. sheet and strip are rolled between rolls having a smooth.(b) IAS-2000 Page No. 71 Slide No.76 Slide No. 40 Ans.75 Slide No.74 Slide No. • Grain growth does not need to be preceded by recovery and recrystallization. where plates.(c)Rolling means hot working it will not show work hardening ISRO-2006 Page No. 70 Slide No. fatigue. one provides camber on the rolls to take care of unavoidable tool bending. 41 Ans. Plate. 45 Ans.(b)Since brittle materials cannot handle plastic deformation. soft steel. and wear properties are improved through strain hardening. (b)Ulta hai. • During recovery.49 Ans. 70 Slide No. IES-1993 Page No.R=200 mm∴Δh = 6 mm. When an accurate class of fit is desired. (d) Should be above the recrystallisation temperature. Ch-8 Rolling: Answers with Explanations GATE -2013 Page No.72 Slide No.D = 400 mm. ISRO-2009 Page No.76 Slide No.57 Ans. IAS-2004 Page No. (c) Wavy edges due to roller deflection. If it is desired to have the body of a bolt larger than the outside diameter of the rolled thread.86 Ans. GATE-1992(PI) Page No.54 Ans.IES-1996 Page No.71 Slide No.73 Slide No. Because of the limitations in the equipment and workability of the metal.(c) Rolling means hot rolling where no lubricant is used. (d)Malleability. IES-2006 Page No.75 Slide No.GATE & PSUs) Page 175 of 210 Rev.(d) ho = 16 mm . GATE-1989(PI) Page No.91 Ans. Strength. but Cold rolling (cold working) is mentioned therefore answer will be (c) because forging means hot working.002 inch larger than the thread-pitch diameter. GATE -2009(PI) Page No. recrystallisation temp is of the order of 10000C IAS-2004 Page No. Lead. harder.(b)Rolling with smaller diagram rolls requires lower force. 70 Slide No. the diameter of the blank is made about 0. Refer slides IAS-2007 Page No. it may occur in all polycrystalline materials. sheets. 70 Slide No. and a smoother.(a)In order to get uniform thickness of the plate by rolling process.75 Slide No. copper and aluminium are some materials in order of diminishing malleability. 70 Slide No.74 Slide No.80 Ans.75 Slide No. • Grain growth follows complete crystallization if the material is left at elevated temperatures.74 Slide No. • One obvious characteristic of a rolled thread is that its major diameter always is greater than the diameter of the blank.60 Ans. the threads have greater strength than cut threads. 46 Ans. resulting in substantial savings. Δh = D(1 − cos α ) 6 = 400(1 − cos α ) ⇒ α = 9.(c) A continuous form of three-point bending is roll bending. 44 Ans.88 Ans. slightly cambered (convex) or concave working surface. Assertion reason me hona chahiye.97 Ans.(d) Due to directional granule deformation.936° For-2016 (IES. (c) • Annealing relieves the stresses from cold working – three stages: recovery. recrystallization and grain growth. cylindrical. and more wear-resistant surface is obtained. IES-1992 Page No. 70 Slide No.(d) • Thread rolling is used to produce threads in substantial quantities.It is a special case of ductility which permits materials to be rolled or hammered into thin sheets.74 Ans.(a) IES-2006 Page No. IES – 1993. IES – 1992. h f = 10 mm .0 . IES-2003 Page No. IAS-1996 Page No.72 Ans.90 Ans. A malleable material should be plastic but it is not essential to be so strong. Because no metal is removed in the form of chips. IES-2013(conventional) Page No.73 Ans. physical properties of the cold-worked material are restored without any observable change in microstructure.(d) IAS -2003 Page No. (c)Cold working increases the strength and hardness of the material due to strain hardening. This is a cold-forming process operation in which the threads are formed by rolling a thread blank between hardened dies that cause the metal to flow radially into the desired shape.82 Ans. less material is required. the blank for the thread is made smaller than the body. 42 Ans.71 Slide No.(c)Bi axial compression and frictional force between roller and workpiece produces shear stress IAS-2001 Page No. 43 Ans.74 Slide No. Cylindrical rollers would result in production of plate with convex surface.(c) You may confused with Forging. (c)For Mild Steel. 104 Ans.min = μ 2 R = 0.116 Slide No.calculation Δh = μ 2 R or μ = GATE-2014(PI) GATE-1990(PI) IES-2014 GATE-2008(PI) Page No.118 Ans.117 Slide No.79 Slide No.76 Slide No.114 Slide No.1414 GATE -2015 Page No.(c & d) Slide No.2 = 0.77 Slide No.(d) For strip rolling sheet rolling width remains same.min = 2.402° = 0.(d) Slide No. Ans. D = 410 mm.78 Page No. Δh = D(1 − cos α ) or 0.77 Slide No.(a)Maximum possible draft.58o = 3.79 Selected Questions Page No.110 Ans.12 × 150 mm = 1.105 Ans.(b) Ans.(b) Roll strip contact length.(d) R = ( 0. 30 mm -10 mm = 20 mm 20 ∴ Number of pass needed = ≈7 3 Page No.77 Slide No. Initial thickness (h1) = 4.(a) The velocity at neutral point is equal to the velocity of roller.129 = 38.8 x 4.8 m/min The inlet and outlet volume rates of material flow must be the same. GATE-2006 Page No. Therefore in actual practice the reduction in second pass is less than in the first pass.(c) ( Δh )max = ho − h f . Δhmax = μ 2 R GATE -2011 Page No.78 ( Δh )max = μ 2 Slide No.1) × 300 mm = 3 mm 2 Therefore we cannot reduce more than 3 mm in a single pass but we have IES-2001 to reduce total. h f = 15 mm Δh 15 1 1 = = ≈ ≈ 0.77 Slide No. (b) Ans.108 Ans. we have to use above apporx.(c) Δh = 10% of 8 mm = 0.02b o .300m × 100rpm V= = = 1.122 Ans.57 m / s 60 60 IES-2002 Page No. as there is no slip occur π DN π × 0.5 − 3.76 Slide No.8 mm ⎤⎦ We know that.8 = 410 (1 − cos α ) or α = 3.99 Ans. Final thickness (h2 = 0.(b) Actually metal will get hardened in every pass due to strain hardening.107 Ans.79 Slide No.71 IES-1999 Page No.5 or h f .78 Page No.6 to 14.0.77 Slide No.106 Ans. L = R α Δh = D(1 − cos α ) or ( 25 − 20 ) = 600 (1 − cos α ) or α = 7.1) = 7.something In IESobjective exam calculators are not allowed.6 mm Δh = D (1 − cos θ ) or ( 4.119 Ans.5 mm GATE – 2011 (PI) Page No. ho = 30 mm. b = 1.5 mm.62° = 0.101 Ans. h f = 8 x (1 – 0.(b) GATE -2015 Page No. vo = 10m / min 3 o f For-2016 (IES.78 Slide No.062 rad GATE -1998 Page No. h o bo vo = h f b f v f hf = 2 h .(a) R = 150 mm.6 ) = 450 (1 − cosθ ) or θ = 3.0 . that is.112 Ans.GATE – 2012 Same Q in GATE – 2012 (PI) Page No.8 mm ⎡⎣ alternative : ho − h f = 8 − 7.3something R 150 10 3. Δhmax GATE -2014 Page No.129 rad Therefore L = R α = 300 x 0.58 × π 180 rad = 0.77 Slide No. As width constant therefore 20% reduction in area means 20% reduction in thickness also.min = 1.5 = 3.76 mm ≈ 39 mm = μ2R Maximum possible draft.98 ho = 8 mm .14.(b) same as above Ans.79 GATE-2014 Page No. 5.78 Slide No.120 Ans.113 Ans.78 Page No.063radian GATE -2004 Page No.GATE & PSUs) Page 176 of 210 Rev.2 mm.5 mm or 4 − h f . 83 Slide No.84 Slide No.250m Pr ojected Length (L p ) = RΔh = 0..002 = 0. IES-2001 Page No. 178 Ans.143 Ans.163 Ans.e.250 × 0.82 Slide No.(a)The roll-separating force which separates the two rolls apart can be obtained by multiplying the average roll pressure with the total contact area. which is a function of the contact length. 153 Ans.84 Slide No.8 × 10 × 0.(a) IES-2011 Page No.80 Slide No..168 Ans.132 Ans.81 Slide No.706 m / min 3 o Page No. 152 Ans.84 Slide No.82 Slide No.80 Slide No.5 0.02236m Arm length (a) = λ RΔh = 0.. GATE-1989(PI) Page No.(c) IES-2002 Page No. 158 Ans.. Smaller contact lengths means lesser friction forces acting.85 Slide No.(b) IES-2012 Page No.8 kN Torque(T) = F × a ( 3 ) [Force F on both roller] = 670.7 KW 60 60 IES – 2000. smaller rolls would have less contact length than larger rolls for the same reduction.(c) IES-2006 Page No.142 Ans.5 × 103 × = 15. IFS-2010 Page No.22n = or 11.128 Ans. 177 Ans. 160 Ans.002m R = 250 mm = 0. 154 Ans. Gutter IES-2015 Page No.(c) Ao = 1.(a) IES-2003 Page No.85 Slide No. is wrong.(d) Total Power for both roller (P) = 2 × T × ω = 2 × T × ( ) Ch-9 Forging: Answers with Explanations IES-2013 Page No. 175 Ans.002 = 0. 164 Ans.1 = 670. IAS-2007 Page No..01118m ( ) Force(F) = Pr essure × Pr ojected area = σ o × (L p × b) = 300 × 106 × 0. 155 Ans.79 Elongation factor = E = En = Slide No. Thus.80 Slide No. point3.(b) IES-2012 Page No.124 Ans.h o b o × 10 = GATE-1992(PI) 2 h × 1.84 Slide No. 148 Ans.80 Slide No. means (a).02b o v f ⇒ v f = 14.83 Slide No.(a)Closed die forging requires the provision of gutters to provide space for excess material and ensure complete closure of die and defect free forged part.(c)Use small dia rolls to reduce Roll force.01118 = 7.(a) IES-1996 Page No.( given) A1 Ao 750 × 750 or 1.. 173 Ans. This in turn.85 Slide No. The average roll pressure can be decreased by reducing the maximum pressure. 156 Ans..02236 × 0. 169 Ans.81 Slide No. (b) and (d) wrong.(a) IES-2005 Page No.82 Slide No..GATE & PSUs) Page 177 of 210 Rev. 170 Ans.(b)Amount of flash depends on the forging size not on forging force.(c) The provision of gutters to provide space for excess material and ensure complete closure of die and defect free forged part. GATE-2010(PI) Page No.85 Slide No.(c) The draft provided on the sides for withdrawal of the forging.83 Slide No.83 Slide No. since. 174 Ans. by reducing the contact length. Use analysis of rolling (Home work portion) IAS-1998 Page No.(c) IES-1998 Page No. it is possible to decrease the roll-separating force.83 Slide No.. IES-2005 Page No.(c) Forging components poses high reliability i.(a)Coefficient of friction is constant over the arc of contact and But does not acts in one direction throughout the arc of contact. GATE-1994(PI) Page No.(a) Δh = 20mm −18 mm = 2mm = 0.04 close to (c) An 250 × 250 GATE-2008 Page No.0 .129 Ans. (b) IAS-2002 Page No.(b) If undercut is present it is not moldable means can’t be withdrawn from die. 176 Ans. IES – 1993. ISRO-2013 Page No..84 Slide No.(d) IES-2013 Page No. IES-1997 Page No.250 × 0.5kNm 2π N 2π × 10 = 2 × 7.85 Slide No. can be achieved by reducing the roll diameter.(b) For-2016 (IES.84 Slide No.22. IES-2014 Page No.85 Slide No.167 Ans.127 Ans.(c) IES-2001 Page No. 87 Slide No.202 Ans.85 Slide No. 2.It can be located such that the line will surround the largest projected area of the piece. y Similar to drop forging.(c) IES-2007 Page No. y Drop forging is used to produce small components.87 Slide No. K.88 Slide No.(a) IES-2013 Page No. 5. IFS-2011 Page No. ISRO-2010 Page No. Smith Forging • Blacksmith uses this forging method • Quality of the product depends on the skill of the operator.89 Slide No. 180 Ans. much shallower angles are preferred.86 Slide No.E ∞ V2.87 Slide No.(b) IES-2011 Page No.85 Slide No. IES-2011 Page No. Advantages of High Velocity Forming: 1. The lower half of the die is fixed to the anvil of the machine. the complete cavity is formed. • Upset forging generally employs split dies that contain multiple positions or cavities.86 Slide No. IES-2013 Page No.86 Slide No. Drop Forging y The drop forging die consists of two halves. Cost and size of machine low. • Not used in industry. 185 Ans.(a) The term swaging is also applied to processes where material is forced into a confining die to reduce its diameter. in general.(d) IAS-2001 Page No. EDGING: Preform shape. Press Forging y Metal is squeezed gradually by a hydraulic or mechanical press and component is produced in a single closing of die. 181 Ans.89 Slide No. the complete cavity is formed. 211 Ans. in quick succession so that the metal spreads and completely fills the die cavity. 190 Ans.(None) Correct sequence is 2 – 1 – 3 .88 Slide No.(b) IFS-2011 Page No. Ram strokes short (due to high acceleration) IAS-2011(main) Page No. The angle of the surface at the parting line from the primary parting plane should not exceed 75o. y Most commonly used for the forging of bolt heads of hexagonal shape is close die press forging. When the two die halves close.86 Slide No.86 Slide No. IES – 1994.(c) IES-2008 Page No. while the upper half is fixed to the ram. 214 Ans. Upset Forging • Upset forging involves increasing the diameter of a material by compressing its length.Select the parting line so that no undercut are in either die impression at the time of ejection of workpiece. 3.89 Slide No. press forging is also done in closed impression dies with the exception that the force is a continuous squeezing type applied by the hydraulic presses.(a) Due to low toughness.E ∞ V2.(c)The drop forging die consists of two halves. When the two die halves close.212 Ans. 210 Ans.201 Ans. Productivity high. 198 Ans.IES-2003 Page No. The heated stock is kept in the lower die while the ram delivers four to five blows on the metal.(a) As K.GATE & PSUs) Page 178 of 210 Rev. IES-2008 Page No.0 . high energy is delivered to the metal with relatively small weights (ram and die). 209 Ans. IAS-2000 Page No. FLASH: The excess metal added to the stock to ensure complete filling of the die cavity in the finishing impression is called Flash. 179 Ans. hence the dimensional accuracy is much better than drop forging. 197 Ans.(c) Bonding between the inclusions and the parent material is through physical bonding no chemical bonding possible. A shapes having straight or tapered reduced sections may be forged with the aid of rolls. Gathers the material as required in the final forging. 182 Ans.4 IAS-1998 Page No. high energy is delivered to the metal with relatively small weights (ram and die).88 Slide No. in quick succession so that the metal spreads and completely fills the die cavity.89 Slide No. FULLERING: Reducing cross section and making it longer. The lower half of the die is fixed to the anvil of the machine. The heated stock is kept in the lower die while the ram delivers four to five blows on the metal.88 Slide No. IAS-2003 Page No. • Parts can be upset forged both hot and cold on special high-speed machines where the workpiece is rapidly moved from station to station.(b) IES-2012 Conventional Page No. GATE-2008(PI) Page No.89 Slide No. 186 Ans. The mating surfaces of the two halves of the die define a parting line around the edges of the forging as they come together. 204 Ans. overall production cost low 4. 200 Ans. 189 Ans.(b) For-2016 (IES. while the upper half is fixed to the ram. Refer slides IES-2005 Page No. d 2 − d1 × 100% = 41. IES S-2013 Page No.(b) GAT TE-2014 Page No.232 N Anss.(c)Forging g force attaiins maximu um value att the end off the For-2016 (IES. 222 2 Ans. N 228 Anss. N 227 Anss.(b) Page No. N 219 Anss.90 0 Slide No..90 0 L True strain (∈T ) = ∫ Lo Pag ge No.90 0 Slide No.(c) ⎛ L ⎞⎟ ⎛D ⎞ ⎛A ⎞ ⎛ 200 ⎞⎟ dx = ln ⎜⎜⎜ ⎟⎟ = ln ⎜⎜ o ⎟⎟⎟ = 2ln ⎜⎜ o ⎟⎟⎟ = 2ln ⎜⎜ ⎟ = −1.91 1 operatioon. N 223 Anss.42% d1 Slide No.(c))Low therm mal conductiv vity because low heat loss from woorkpiece. GAT TE-2006 Page No. N 221 Anss.(c)) ⎞ ⎟ = ln 2 = 0.693 ⎠ Slide No.0 .(c) TE-2012 Sa ame Q GATE -2012 (PII) GAT Pag ge No.42 mm = 100 × h2 25 Percenttage change in diameterr = IES S-2012 Page No.91 1 Slide No.91 Slide No.Ao Lo = AL L Ao (π / 4 ) d o 2 = = Lo A (π / 4 ) d 2 ln A d L = ln o = 2 lnn o Lo A d GAT TE-1992.(c) Engineeering strainn or Conventional Strainn(ε E ) = elongation origginal lengthh elongaation us length i instantaneo If suppose x is the length.GAT TE-2010(PII) Pag ge No.91 1 TE-2015 GAT Slide No. IS SRO-2012. dx is the elonggation which h is infinitelly small True Sttrain(ε T ) = εT = ∫ L Lo dx L = ln x Lo ΔL L Lo + ΔL = = 1+ = 1+ εE Lo Lo Lo as ε T = ln (1 + ε E ) ∵ volum me change will w not be there t so.(d)) Volumee of materiaal will remainn same due to t incompressibility π d12 4 × h1 = d 2 = d1 π d 22 4 × h2 h1 50 = 1411. VS-2013 V ⎛ L ⎝ Lo ε T = ln ⎜ GAT TE-2007 ⎞ ⎛ 2 L0 ⎟ = ln ⎜ ⎠ ⎝ Lo Page No. 229 2 Ans.386 ⎜⎝ 400 ⎠⎟ ⎟ ⎟ ⎜ ⎜ x ⎝A⎠ ⎝D⎠ ⎝⎜ Lo ⎠⎟ negativ ve sign indica ates compresssive strain.GATE & PSUs) Page 179 of 210 Rev.89 Slide No.90 Slide No. 215 2 Ans. 25 mm 2 × 0.5) e 6 ⋅ 150 ⋅ dx ⎨14 + 6 ⎭ 0 ⎩ 39.25 ⎝ 2 × 0.1 mm τ y = 4 N/ mm2 (Shear yield stress) = K By Tresca Theory.25 2 39.1 − 25 1 ⎛ ⎞ ln ⎜ ⎟ = 177.25 ⎞ ⎟(212.92 Slide No. Solution: R1 = 150 mm.25 (48 − x ) 6 ⋅ 150 .IES – 2005 Conventional Page No. 236 Ans.1 ⎨⎜ ⎧ ⎫ ⎛2×4⎞ ⎪⎝ ∫0 ⎨⎩16 + ⎜⎝ 25 ⎟⎠ (177.25 ⎟⎠ ⎝ 2μ ⎠ xS L 2μ (L − x ) 2K ⎫ ( x s − x )⎬ B. 237 Ans.25 ⎝ 2 × 0.68 − x )⎬ ⋅ 150 ⋅ dx +2 ∫ (2 × 3. 2L = 96 mm. d1 = 100 mm.R ⎤ ⎡ R 1 eh ⎢ ⎥ = 2.1 − r ) ⎬ 25 ⎠ ⎪⎭ 177. dx + 2 × ∫ 2K e h B . h = 25 mm.93 MN (Tresca’s Theory) Ftotal = .68 mm ln ⎜ ⎟ = 48 − ln ⎜ 2μ 2 × 0. h1 = 60 mm. dx h ⎭ xS Ftotal = 2 × ∫ ⎧⎨Ps + 0 ⎩ Applying Von-Mises theory or σ0 = 4.25 ⎠ For-2016 (IES.68 ∫ (2 × 4.25 ⎝ 3 × 0. dx + 2 × 39.5 ⎧ ⎫ (39.05 F =2× ∫ or π d12 h1 = π R 2 h 4 or or 1002 × 60 = R 2 × 30 4 R = 70.04) e = 510 kN + 29.234 Ans.5 N / mm2 . μ = 0.4 212. h1 = 50 mm.4 − r )⎬⎭ . R = ?. μ= 0.4 ∫ (8) × e ⎩ = 3.25 (48 − x ) 2 × 3.04 N / mm2 3 K = 16.68 442 kN + 25 kN = 467kN (Tresca ' s) IES – 2007 Conventional Page No.16 N / mm2 Ps = μ K= 39.68 or 48 2 × 0.5 = 3. Solution: h = 6 mm.GATE & PSUs) Page 180 of 210 Rev.4 mm to 212.4 mm 2 × 0. Solution: Given.04 MN − − + F = 2π σ 0 0 2 2 ⎥ ⎢ ⎛ −2μ ⎞ ⎛ −2μ ⎞ ⎛ −2μ ⎞ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ h ⎠ ⎦⎥ ⎣⎢ ⎝ h ⎠ ⎝ h ⎠ IES – 2006 Conventional Page No. dx σo K 3.0 .25 σ0 = 2 K = 2 × 4 = 8 N / mm2 ⎫⎪ ⎪⎧⎛ 2×0.68 48 2×0. R s = 212.4 mm → sticking ⎤ ⎢177. 2 πr dr + 177.91 Slide No.25 xs = L − h 6 1 ⎛ 1 ⎞ ⎛ ⎞ = 39. 2 πr dr Von Miscs Theory.7 mm 2μ . h = 30 mm σ0 = 120 N/ mm2 and μ = 0. K = 2×0.25 πR12 h1 = π R 2 h R = 212.68 − x )⎬⎭ ⋅ 150 .25 ⎠ ⎡0 mm to 177.1 − 25 1 ⎛ ⎞ ln ⎜ ⎟ = 170. Ps = = = 14 N / mm2 μ 0.10 kN = 539 kN (Von − Mises) F = 2× Applying Tresca’s Theory.16 + 6 (39. R s = 212.1mm → sliding ⎥ ⎣ ⎦ Ps = K 4 = = 16 N / mm2 μ 0.25 ⎧ ⎫ ∫0 ⎨⎩16.92 Slide No. h1 = 100 mm.155 mm xs = L − Since K = 4.92 Slide No.25 ∫ 4 3.25 − r )⎬⎭ ⋅ 2 πr dr + 170.21 − ln ⎜ = −261.25 mm → sticking ⎤ ⎢170.⎡0 mm to 170.1 ⎝ 3 × 0.1 ⎟⎠ According to Tresca Rs = R − h ⎛ 1 ⎞ 50 1 ⎛ ⎞ ln ⎜ = 141.6 MN (Von Misces) GATE-1987 Page No.25 212.1 ⎟⎠ ∵ Rs came out to be negative so only sliding friction takes place.92 Slide No. 238 Ans. Ftotal = 2×0.25 (212.04 × 150 ⎫⎪ ⎪⎧⎛ 2×0.1. σ Y = 230 MPa = σ O ∵ Volume before forging = Volume after forging π d12 h1 = π R 2 h or π × 2002 ×100 = π R 2 × 50 ⇒ R = 141.2.04 N/mm2 xs came negative so there will be no sticking only sliding will take place. P = Pmax Pmax = 230 × e 2×0.1 mm ⎟ 2μ ⎝ 2μ ⎠ 2 × 0. 242 Ans. (a) Practice Problem -1 Page No. 240 Ans. 239 Ans.1 ⎝ 3 × 0.66 mm Page 181 of 210 Rev. 241 Ans.98 kN 0 Practice Problem -2 Page No.05 h ⎛ 1 ⎞ ln 2μ ⎜⎝ 2μ ⎟⎠ x s = − 90. B = 150 mm. e = 3. 2μ P = σ oe h ( R−r ) at r = 0.25 σ0 = K 3 = 4 3 N / mm2 RS ∫ Ftotal = 0 2K ⎧ ⎫ (R s − r ) ⎬ ⋅ 2 πr dr + ⎨Ps + h ⎩ ⎭ R ∫σ 2μ 0 eh (R − r ) ⋅ 2 πr dr RS 170. d1 = 150 mm.21 − ln ⎜ = −297. L 2μ F = 4 KB ∫ e h (L − x ) dx 0 48 = 4 × 4.05 ⎞ ⎨⎜ ⎟ (48 − x ) ⎬ 6 ⎠ ⎭⎪ ⎪⎝ ∫ e⎩ dx = 177. h = 6 mm.1 2×4 ⎧ ⎫ ∫0 ⎨⎩16 + 25 (170.0 . μ = 0.92 Slide No.1mm → sliding ⎥ ⎣ ⎦ Ps = K 4 = = 16 N / mm2 μ 0.421 mm 4 4 According to Von-Mises h ⎛ 1 ⎞ 50 1 ⎛ ⎞ ln ⎜ = 141.1 − r ) 25 ⋅ 2 πr dr Given: 2L = 96 mm.94 MPa Page No. Center GATE-2014(PI) Page No.92 Slide No.1 mm ⎟ 2 μ ⎝ 3μ ⎠ 2 × 0. μ = 0. h1 = 100 mm. h = 50 mm. μ = 0.25 mm to 212. L = 48 mm.GATE & PSUs) ×1502 × 100 = π R 2 × 50 ⇒ R = 106.21) 50 Practice Problem -3 = 404. Rs = R − The formula for pressure we get after the slab method of analysis of forging.92 Slide No. d1 = 200 mm.1 (141. ∵ Volume before forging = Volume after forging π 4 d12 h1 = π R 2 h or π 4 For-2016 (IES. h = 50 mm. 41 ∵ Volume before forging = Volume after forging π d12 h1 = π R 2 h or π × 2002 × 70 = π R 2 × 40 ⇒ R = 132.05.2 mm 2002 × 50 = RFN 200 − ⎛ ⎞ and μ = 0.2 ⎠ Von Miscs Theory. d1 = 200 mm. On one end of the container.01 + ε )0.51 ⎝ 3 × 0. Practice Problem -4 Page No.246 Ans.93 Slide No.0 .66 − 50 ⎛ 1 ⎞ ln ⎜ ⎟ =-7. Refer forging analysis Ch-10 Extrusion and Drawing: Answers with Explanations IES-2007 Page No. 244-245 Ans.51 ⎠ Page No.93 Slide No.(c) Advantages: 1. (d) The equipment consists of a cylinder or container into which the heated metal billet is loaded. 252 Ans.41 σ f = 200(0.78 = σ o Now use Tresca’s theory Von‐Mises Theory Practice Problem -5 {GATE-2010 (PI)} Page No. and acquiring the shape of the opening.92 Slide No. π R H IN = π R H FN 2 IN or 2 FN 2 × 30 ⇒ RFN = 258. h = 40 mm.6930. μ = 0.35 ⎜1 + e 258.2 ⎟ = 0.28 mm 4 h 40 True strain ( ε ) = ln = ln = −0.94 Slide No.693 h1 100 Flow stress (σ o ) = σ f = 1030ε 0. the die plate with the necessary opening is fixed. From the other end. Saving in tooling cost All are correct but only a die change can change the product therefore (c) is most appropriate.55 mm ⎟ = 258.01 + ε )0. The extruded metal is then carried by the metal-handling system as it comes out of the die.41 = 158. 254 Ans.51 ⎝ ⎠ Now at Rss Shear stress in sticking K = shear stress in sliding ( μ Pss ) or K = μ 3Ke 2μ ( RFN − Rss ) H FN ⎛ 1 ⎞ 2μ or ln ⎜ ( RFN − Rss ) ⎟= ⎝ 3μ ⎠ H FN ⎛ 1 ⎞ H or FN ln ⎜ ⎟ = RFN − Rss 2μ ⎝ 3μ ⎠ or Rss = RFN − IFS-2012 H FN ⎛ 1 ⎞ 30 1 ⎛ ⎞ ln ⎜ ln ⎜ = 254.5596 h1 70 4 σ f = 200(0. σ f = 200(0. 256 Ans. 243 Ans.2 ⎝ 2 × 0. a plunger or ram compresses the metal billet against the container walls and the die plate.93 Slide No.2 − ⎟ 2μ ⎝ 3μ ⎠ 2 × 0. Process time saving 3. Material saving 2.True strain ( ε ) = ln h 50 = ln = −0.17 = 1030 × 0. For-2016 (IES. thus forcing it to flow through the die opening.01 + 0. h1 = 70 mm.74 MPa By Tresca Theory.87mm 2 × 0.(b) IES-2012 Page No. R s = 106.5596)0.94 Slide No. DRDO-2008 Page No.17 = 967.GATE & PSUs) Page 182 of 210 Rev. For sketches refer slides. 303 Ans.298 Ans. (b) Both A and R are true but R is not correct explanation of A. the initial temperature of billet should be low. (d) For-2016 (IES. GATE-1989(PI) Page No. 272 Ans. 318 Ans.280 Ans. 274 Ans.(c) IAS-1995 Page No.(c) Cleaning is done to remove scale and rust by acid pickling. (d) IES-1999 Page No. IES-2007 Page No. IES-2014 Page No.94 Slide No.98 Slide No.96 Slide No. (b) IES-1993 Page No. (a)Impact Extrusion is used for manufacture of collapsible toothpaste tubes IES-2003 Page No. And materials with limited ductility become highly plastic. (b) IES-2007 Page No. GATE-1994 Page No.99 Slide No.100 Slide No. Refer slide GATE-2014 Page No. 308 Ans. 100 Slide No.95 Slide No. 263 Ans.(c)As diameter decreases therefore for same mass flow rate the speed of travel of the extruded product must be greater than that of the ram.(c) IES-2001 Page No. IES-2012 Page No. (d) IES-2014 Page No.309 Ans. IES-2005 Page No. 281 Ans.99 Slide No.98 Slide No.95 Slide No.99 Slide No. 101 Slide No. phosphate coating.101 Slide No. IES-2000 Page No. IAS-2004 Page No.98 Slide No. 98 Slide No. For indirect extrusion. 288 Ans.99 Slide No.295 Ans.98 Slide No.(c) IES-1994 Page No. Lubrication boxes precede the individual dies to help reduce friction drag and prevent wear of the dies. 311 Ans. (b) IES-1993. 97 Slide No. and lubricating with reactive soap.96 Slide No. 306 Ans.GATE & PSUs) Page 183 of 210 Rev. (d) IES-1996 Page No. GATE-1990(PI) Page No. (a) It is pressure induced ductility. IES – 2008.100 Slide No. alkaline cleaning.0 . GATE-1994(PI) Page No. 323 Ans.102 Slide No. 305 Ans. (c) IAS-2010(main) Page No. 319 Ans.(c) IES-2009 Page No. (a)Tandem drawing of wires and tubes is necessary because it is not possible to reduce at one stage. electroplating.(c) The force required on the punch is less in comparison to direct extrusion.(c) IES-1994 Page No.100 Slide No. 322 Ans. IES-2000 Page No.291 Ans.(c)Metal extrusion process is generally used for producing constant solid and hollow sections over any length.101 Slide No.95 Slide No. Extrusion of nuclear fuel reactor fuel rod (iv) Impact Extrusion-Collapsible tubes for toothpastes. 273 Ans.102 Slide No.IES-2009 Page No.96 Slide No. (d)Hydrostatic extrusion suppresses crack formation by pressure induced ductility. 325 Ans.101 Slide No. 297 Ans.94 Slide No.278 Ans.100 Slide No. friction with the chamber opposes forward motion of the billet.94 Slide No. (b)The wire is subjected to tension only.289 Ans.97 Slide No. 257 Ans. (b)In direct extrusion. (b) IAS-2006 Page No.99 Slide No. since there is no relative motion. 310 Ans. Relative brittle materials can be plastically deformed without fracture.313 Ans.102 Slide No. (d) IES-2014 Page No. 299 Ans. Refer slides IAS-2000 Page No. 326 Ans. It is done by sulling. 264 Ans. (a) IES-2010 Page No. 290 Ans. 304 Ans.96 Slide No. 324 Ans. (d)The correct sequence for preparing a billet for extrusion process is pickling. 261 Ans. IES-1996 Page No. (d) For high extrusion pressure.(c) IAS-2012(main) Page No. IES-2006 Page No. compressive and shear stresses will be there in that portion only. JWM-2010 Page No. 266 Ans. Zinc phosphate coating is used to prevent metal contact. (d) IES-2009(conventional) Page No. IES-2000 Page No. (a) IES-1999 Page No. 327 Ans. (a) IES-1993 Page No.95 Slide No.99 Slide No. (c) IES-2015 Page No. (i)Direct Extrusion-curtain rods (ii) Indirect Extrusion(iii) Hydrostatic Extrusion-Cladding of metals. (a) IAS-2012(main) Page No. Refer slides IES-2009 Page No.In direct extrusion. 268Ans. friction with the chamber opposes forward motion of the billet. (a) GATE-1987 Page No.269 Ans. there is no friction.97 Slide No. creams etc. (d) Only ram movement is there. But when it is in contact with dies then a combination of tensile.96 Slide No. 279 Ans.101 Slide No. 260 Ans. phosphating. 293 Ans.95 Slide No.98 Slide No. 0 . For-2016 (IES. (b) Extrusion constant ( k ) = 250MPa Initial area ( A o ) = π do 2 and Final area ( A f ) = 4 Force required for extrusion: ⎛A P = kA0 ln ⎜ o ⎜A ⎝ f GATE-2009(PI) πdf 2 4 ⎞ ⎛ π / 4 × 0. Extrusion IAS-1994 Page No.104 Slide No.104 Slide No.IES-1993. 332 Ans.102 Slide No. hardness to indentation.8 ) × (1 − 0. (b) ⎛ Ao ⎞ σ d = σ o ln ⎜ ⎟ For Maximum reduction. (b) Case(a ) : 3 − stage reduction final dia =15 × (1 − 0.4 ) × (1 − 0.12 mm ∴ error = 0. Ignore friction and redundant work means ⎛r ⎞ π × 82 ⎛ 5 ⎞ Ideal Force = 2σ 0 A f ln ⎜ o ⎟ = 2 × 400 × ln ⎜ ⎟ = 8.342 Ans. 329 Ans. (a) IAS-2001 Page No.2) × 10 = 8 mm ⎛A Stress (σ d ) = σ o ln ⎜ o ⎜A ⎝ f ⎞ ⎛d ⎟⎟ = 2 × σ o × ln ⎜⎜ o ⎠ ⎝ df GATE -2008 (PI) Linked S-2 Power = Drawing force × Velocity ⎞ ⎛ 10 ⎞ ⎟⎟ = 2 × 800 × ln ⎜ ⎟ = 357 MPa ⎝ 8⎠ ⎠ Page No. resilience to impact loads.353 Ans. (b) d o = 10 mm. 103 Slide No. 347 Ans.004mm (c) 5 − stage reduction final dia =15 × (1 − 0. 345 Ans. 333 Ans. (b)since malleability is related to cold rolling. IES-2002 Page No.103 Slide No. Refer slide GATE-2003 Page No. (b) IES-2011 Page No. 341 Ans.336 Ans.102 Slide No.104 Slide No.2) × d o = (1 − 0.103 Slide No.8 ) × (1 − 0.12 ⎞ π 2 = 2. d f = (1 − 0.8 ) = 0. 104 Slide No. 102 Slide No. 328 Ans.8 ) × (1 − 0. 343 Ans.104 Slide No.2 ) = 0. GATE -2007 (PI) Page No.97 KW GATE-2001.102 Slide No.72219 MN ⎟⎟ = 250 × × 0. 105 Slide No.8 ) × (1 − 0. (b) Given : Do = 10mm.8 ) × (1 − 0. 330 Ans. 337 Ans.0728mm GATE-2015 Page No.5W = 8.096mm ∴ error = 0.σ 0 = 400 MPa.102 Slide No.103 Slide No. 331 Ans. (b) GATE-1996 Page No.8 ) × (1 − 0. and isotropy to direction. ISRO-2010 Page No.GATE & PSUs) Page 184 of 210 Rev. (b) IAS-2002 Page No. IES-2012(conventional) Page No.σ d = σ o ⎝ Af ⎠ ⎛A ⎞ A A − Af ⎛ 1⎞ σ o = σ o ln ⎜ o ⎟ or o = e = 2. (b) IES-2011(conventional) Page No. 335 Ans. (a) ⎛A Pressure (σ ) = σ o ln ⎜ o ⎜A ⎝ f ⎞ ⎟⎟ = σ o ln r = 300 ln 4 = 416 MPa ⎠ GATE-2006 Page No. Df = 8mm.1 ln ⎜ 2 ⎟ 4 ⎝ π / 4 × 0. (a) GATE-2015 Page No.1728mm∴ error = 0.348 Ans. (b) Extrusion and skew rolling produce seamless metallic tubes. (a) = Stress (σ d ) × area ( Af ) ×Velocity = 357 × π × 82 4 × 0. 349 Ans.8 ) × (1 − 0.102 Slide No.05 ⎠ ⎠ Page No. 350 Ans.97 kN 4 ⎝4⎠ ⎝ rf ⎠ GATE -2008 (PI) Linked S-1 Page No.2 ) = 0.104 Slide No.02mm (b) 4 − stage reduction final dia =15 × (1 − 0.344 Ans.104 Slide No.103 Slide No. (b) GATE-1991(PI) Page No.71828 ∴ o × 100 = ⎜1 − ⎟ × 100 = 63% A A A e⎠ ⎝ f o ⎝ f⎠ IES-2014 Page No.4 ) × (1 − 0. 7145 ⎤ ⎛ rf min ⎞ 400(1 + 1. μ = 0.7145 × 50 = 361.25 ⎠ d − d f min Max possible % reduction in diameter = o × 100% = 23..54 ⎞ True strain = ln ⎜ ⎟ = ln ⎜ o ⎟ = ln ⎜ ⎟ = 3.7145 ⎤ ⎛ rf min ⎞ ⎤ 400(1 + 1.26MPa For maximum possible reduction.7145 ⎤ 400(1 + 1.7145) ⎡ ⎛ rf min ⎞ ⎢1 − ⎜ ⎥ ⎢ ⎥ or rf min = 4.7145 ⎤ ⎛ 5 ⎞ 400(1 + 1. d f = 10mm.653 × 102 N 4 π 100 m / s = 44..V = 100m / min.3% do ro If the rod is subjected to a back pressure of 50 N/mm2 σd = σ o (1 + B ) ⎡ B 2B 2B ⎛ r f ⎞ ⎤ ⎛ rf ⎞ ⎢1 − ⎜ ⎟ ⎥ + ⎜ ⎟ . 354 Ans.85 ⎠ ⎝ A⎠ ⎝ Lo ⎠ and Ao Lo = A7 L7 or 78.54 mm 2 = 51 mm 2 After second pass area ( A 2 ) = (1 − 0.35 ) Ao = (1 − 0. 2 After 7th pass area ( A 7 ) = (1 − 0.GATE & PSUs) Page 185 of 210 Rev..329 kW Power = P × V = 338.σ o = σ d σo = σ o (1 + B ) ⎡ B 2B 2B ⎛ rf min ⎞ ⎤ ⎛ rf min ⎞ ⎢1 − ⎜ ⎟ .7145 ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎢⎣ ⎝ ro ⎠ ⎥⎦ d − d f min ro − rf min Max possible reduction in dia = o ×100% = ×100% = 25.15cot 5 = 1.85 mm 2 7 7 ⎛ L⎞ ⎛A ⎞ ⎛ 78..54 ×100 = 3. σ o = 400 MPa B = μ cot α = 0. 105 Slide No.. α = 5°.25 ⎠ 2×1.0 .67mm or = − σo = 400 1 ⎟ ⎜ ⎟ B 1.7145 ⎣⎢ ⎝ 6.54 mm 2 = 3.35 ) Ao = (1 − 0.25 ⎠ ⎥⎦ π Force ( P ) = 338.σ b ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎝ ro ⎠ 2×1.5% do Max possible % reduction in area = GATE – 2011 (PI) Common Data-S1 Initial area ( A o ) = πd 2 o = π ×10 Ao − Af min Ao × 100% = 41.d o = 12.7145) ⎡ ⎛ 5 ⎞ σd = 1 − ⎢ ⎜ ⎥ = 338.54 mm 2 4 4 After first pass area ( A1 ) = (1 − 0.15.35 ) × 78.35) Ao and then .7145 σ o (1 + B ) ⎡ 2B ⎛ rf ⎞ ⎤ ⎢1 − ⎜ ⎟ ⎥ σd = B ⎢⎣ ⎝ ro ⎠ ⎥⎦ 2×1.78 mm ⎟ ⎟ 1..85 × L7 ⇒ L7 = 2040 mm For-2016 (IES.02 ⎝ 3.25 ⎠ ⎦⎥ ⎝ 6.653 × 102 × 4 60 Maximum possible reduction.7145 ⎢⎣ ⎝ 6.7145) ⎡ ⎛ rf min ⎞ 400 = ⎢1 − ⎜ ⎥+⎜ × 50 ⇒ rf min = 4.σ b ⎟ ⎥+⎜ ⎢⎣ ⎝ ro ⎠ ⎥⎦ ⎝ ro ⎠ 2×1.653MPa ⎟ 1.7145) ⎡ ⎛ 5 ⎞ σd = ⎢1 − ⎜ ⎥+⎜ ⎟ ⎟ 1.5mm.35 ) × 78.(c) 2 mm 2 = 78.7145 2×1.25 ⎠ ⎥⎦ ⎝ 6.5% Page No.7145 ⎢⎣ ⎝ 6.σ o = σ d σ o (1 + B ) ⎡ 2B 2×1.. 108 Slide No.GATE & PSUs) Page 186 of 210 Rev.106 Slide No. Refer slides Ans.GATE – 2011 (PI) Common Data-S-2 ⎛A P = σ o Af ln ⎜ o ⎜A ⎝ f GATE-2014 Page No. 356 Ans.5 × 294 = 0. GATE-2014 Page No.τ = 240MPa (i) Punch size = size of hole = 10 mm For-2016 (IES.105 Page No.0032 × t × √τ Shear strength of annealed C20 steel = 294 MPa Hence. Hint given on slide Ch-11 Sheet Metal Operation: Answers with Explanations Example Page No.1 True strain at any instant t ( ε T ) = ∫ ( t dL dL 2tdt =∫ =∫ = ln ⎡⎣1 + t 2 ⎤⎦ 2 2 L L0 1 + t 0 1+ t ( ) ( ) ) as.88 mm Example Page No. 380 Ans. C = 0.9 to 1. 384 Ans. 365 Ans. Punch size= size of hole = 20 mm Die size = Punch size +2 C = 20 +2 × 0. t = 1 mm.40 KN 51 A ⎝ ⎠ 1 ⎝ ⎠ ⎠ Page No. The clearance to be provided is given by. (c) Ans. 355 Ans.(a) It is blanking operation Therefore Diameter of die = metal disc diameter = 20 mm 3% clearance (c) = 0.381 Ans. = 0.1646 mm GATE-2003 Page No.0 2 1 + 12 dt 1+ t ( ) Page No. 105 Slide No. 358 Slide No.54 ⎞ ⎟⎟ = σ o A1 ln ⎜ o ⎟ = 200 × 51× ln ⎜ ⎟ N = 4.108 Slide No. 357 Slide No. Die size = blank size = 20 mm Punch size = blank size – 2 C = 20 – 2 × 0.0823 = 19.0032 ×1. C = 0.06 = 19. 385 Ans.108 Slide No. 360 Slide No.0823 mm Since it is a blanking operation.0 .108 Slide No.06 mm on both side of the die (of sheet thickness) Therefore Diameter of punch = 20 – 2c = 20 – 2 x 0.105 Page No. 386 Ans.∴ dL = L0 2tdt εT = IAS-1997 IES-2012 IAS-2006 IFS-2013 dεT 2t 2 ×1 = = = 1.105 Page No. L = L0 1 + t 2 .0823 = 20. (d) ⎞ ⎛A ⎞ ⎛ 78.(b) Punching Force(F) = Ltτ F = 2(a + b)tτ = 2(100 + 50) × 5 × 300 = 450 kN IAS-2011(main) Page No.108 Slide No.83 mm Now when it is punching operation. For punching operation 10 mm circular hole d = 10 mm.105 Slide No.(c) Ans. (c) π dtτ fs 3 = 4 × 20 × = 30 mm fc 6 Fmax × pt 968 × 0.2 t Fmax = Ltτ = 200 × 5 × 100 = 100 kN Work Done = Fmax × ( pt ) = 100 × (0.5 mm.GATE & PSUs) Page 187 of 210 Rev.109 Slide No.6 ×1.90 mm Width will be = 50 -2C =50 .90 mm ( ii ) Die size = correct size Length will be = 200 mm Width will be = 50 mm (iii ) F = 2(a + b)tτ = 2(200 + 50) × 1× 240 = 120 kN IES-1999 Page No.5 × 280 ) × 0. t = 1. F= (π d1tτ + π d 2tτ ) × pt = (π 25.81 kN =294 kN F= EXAMPLE Ans.110 Slide No.40 × 5.2 × 5) = 100 J For-2016 (IES. d1 = 25.109 Slide No.7 mm.109 Slide No.6 x 0. 396 Ans.5 × 280 = 33.(c)same as previous question Page No. L = 200 mm.109 Slide No. The cutting force if both punches act together.7 ×1.36/100 or θ = 4. 391 Ans. 398 Ans.4 ×1.09914 mm (iii) Punching Force(F) = Ltτ = π dtτ = π ×10 × 1× 240 N = 7.6 N = 968 kN Work requires to shear hole = 968 x 5.0032 ×1× 240)= 10. F = π d1tτ + π d 2tτ = π × 25.4 = 2168 J Force with shear = 30 T = 30 x 9. 397 Ans.225kN S GATE-2010 Statement Linked 1 1 Page No.(b) min dia =4t IES-2014 Page No.( ii ) Die size = Punch size + 2 C = 10 + 2(0.(a) p t = 5 mm.4 mm.0032t τ )=10 + 2(0. Maximum force without shear = 550 x 100 x π x 5.4 × 1.τ = 100 MPa.54 KN For blanking 50×200 mm rectangular blank (i) Punch size = size -2C Length will be = 200-2C =200 . so that only one punch acts at a time: Fmax = π d outsidetτ = π × 25.2(0.36 mm s s Angle of shear. (c) ISRO-2008.2o Page No.0032 ×1× 240) =199.5 × 280 + π × 12.7 × 1. d 2 = 12.τ = 280 N / mm 2 Total cutting force when both punches act at the same time and no shear is applied to either the die or punch. tanθ = 7.388 Ans.110 Slide No. = 0.109 Slide No.515kN Taking 60% penetration and shear on punch of 1 mm.0 .6 or 294 = ⇒ s = 7.5 × 280 = 50.0032 ×1× 240) = 49.271 kN The cutting force if the punches are staggered.4 × 1. 390 IES-2013 EXAMPLE ≤ σc × πd2 4 or d ≥ 4tτ σc = 4tτ =t 4τ Page No.5 × 280 + π ×12.5 = 45.2(0.389 Ans. 2011 Page No. 7 × 0.4 GATE-2004 Page No.(c) Blanking Force(F) = Ltτ = π dtτ = π ×100 × 1.(b) GATE-1996 Page No.6 kN S 2 GATE_2012 Page No. 434 Ans.GATE & PSUs) Page 188 of 210 Rev. 431 Ans.08 mm It is blanking operation : Punch size = 35 − 0.5 mm = 0.5 × 0.25 × 4 = 200 J [2 × 105 N = 200kN ] Page No.371kN GATE-2009(PI) Page No.114 Slide No.428 Ans. Due to insufficient data we cant calculate.4t × τ F1 5 1 π dtτ = or = or F2 = 3 F2 π ×1.114 Slide No.(a) Page No. Therefore.112 Slide No.(c) IES-2004 Page No.113 Slide No. IES-1994 Page No. 440 Ans.114 Slide No.88 – 0.111 Slide No.(c) IES-2006 Page No.113 Slide No.4t × τ F2 1.(d) It is blanking operation so clearance must be provided on punch.(b) GATE-2013(PI) Page No.113 Slide No.15 mm = 49.(c) C = 40 microns = 0. final punch size will be = 24.(a) Punch size without allowance = Die size – 2 x radial clearance = 25 – 2 x 0. 408 Ans.06×2.e. Die size = blank size = 30 mm Punch size = blank size – 2C = 30 -2 x 0.113 Slide No.(c) GATE-2001 Page No.112 Slide No.112 Slide No. 433 Ans. GATE-2011 Page No. Fmax ( pt ) S 100 F= = 10kN 10 F= IAS-2003 ISRO-2013 back.113 Slide No. 2C = 0.(b) For 400mm length shear is 20mm.7 kN F= Fmax × pt 37.113 Slide No.(b) IAS-1995 Page No. 420 Ans. therefore for 200mm length it becomes10mm.(d) Clearance only on punch for Blanking operation.111 Slide No. 437 Ans. 409 Ans.2×0.(c) C = 6% of t = 0.114 Slide No.(b) GATE-2007 Page No.026kN ISRO-2009 Page No.2C = 50 .5 × 300 = 141. 438 Ans.2 × 150 = 301.0 . 422 Ans.112 Slide No.(a) The blanking force (Fmax ) =π dtτ = π × 10 × 3 × 400 = 37.5d × 0.425 Ans.(a) Higher the modulus of elasticity higher will be the spring Page No.040 mm.113 Slide No.88 mm We need another gap (die allowance ) i. 423 Ans. 426 Ans.00 mm GATE-2002 Page No.421 Ans.113 Slide No. Only 200 mm length is effective.114 Slide No.080 mm and Die size = 35 mm IAS-1994 IAS-2002 Page No.(a) In punching usable part is sheet so punch size is For-2016 (IES.110 Slide No.592kN Blanking Force(F) = Ltτ = π dtτ = π ×10 × 2 × 80 = 5.40 × 3 = = 22. 435 Ans.(c) Blanking Force(F) = Ltτ = π dtτ = π × 200 × 3.GATE-2010 Statement Linked 2 Page No.(a) IAS-2000 Page No. 436 Ans.8 mm GATE-2007(PI) Page No.69kN Blanking Force(F) = Ltτ = π dtτ F1 = 5 = π dtτ and F2 = π × 1.(a) Blanking Force(F) = Ltτ = π dtτ = π × 25 ×10 × 500 = 392.(a) Work done = Fmax × pt = 200 kN × 0.114 Slide No. 430 Ans.06 x t = 30 – 2 x 0. 439 Ans.70 mm Die size = 50.05 = 24. 429 Ans.432 Ans. 424 Ans.83 mm GATE-2008(PI) Page No.114 Slide No. 427 Ans. 399 Ans.(a) IES-2002 Page No.5d × 0.(a) IES-1997 Page No.06 x 10 = 28.15 mm Punch size = die size .113 Slide No.06 = 24.114 Slide No. 117 Slide No.116 Slide No.4 ISRO-2011 Page No.118 Slide No.415mm Second draw 30% reduction. Refer slides for theory Ans. 444 Ans. it can't be draw in a single draw. IAS-1995 Page No.(b)In punching useable part is punched sheet so size of hole must be accurate i.(a)Both A and R are true and R is the correct explanation of A IES-2002 Page No.(c) IAS-2003 Page No.D = r 0.e. Reduction = 1 − IFS-2013 Page No.(a)An insufficient blank holder pressure causes wrinkles to Page 189 of 210 Rev. d = 40 mm. 472 IAS-2007 Page No.119 Slide No.5. 483 For-2016 (IES. h = 60 mm.(c) A cylindrical vessel with flat bottom can be deep drawn by double action deep drawing IES-1997 Page No. 1st Reduction. (b) Ans.44mm (possible) It is not possible to draw the cup in single step.(d) In drawing operation.(d) In blanking operation clearance is always given on the punch . d3 = 3. d1 = 7. 453 d 100 Here = = 250 For d ≥ 20r .115 Slide No.. D ⎝ D⎠ Thumb rule: First draw:Reduction = 50 % Second draw:Reduction = 30 % Third draw:Reduction = 25% Fourth draw:Reduction =16 % Fifth draw:Reduction = 13% IFS-2009 Page No. (a) Ans.5 cm Second draw 30% reduction. To improve die life.(c) D = d + 4dh = 15 cm First draw 50% reduction. 445 Ans.25 cm Third draw 25% reduction. BlankDia ( D ) = d 2 + 4dh = 502 + 4 × 50 × 100 = 150 mm d d . 454 IAS-2013(mains) Page No. 460 Ans. 442Ans.471 IFS-2013 Page No.(c) IES-2000 Page No.116 Slide No..(b) IES-1999 Page No.5D=52. To reduce drawing forces.0 . 466 IAS-1996 Page No. IES-2008 Page No. 443 Ans. 474 1. d2 = 5.118 Slide No. 452 Ans.115 Slide No.117 Slide No. d1 = 0. 446 Ans.115 Slide No. To reduce temperature. IES-1999 Page No. 478 GATE-2008 Page No.GATE & PSUs) Ans.Refer slides for theory Ans.83mm First draw 50% reduction. 4.Correct and clearance on die. 458 Ans. size of punch must be accurate. proper lubrication is essential for Ans.117 Slide No. h = 15mm. Die size is always the exact dimension IES-1994 Page No. 2.116 Ans.we have to use double step.119 Slide No.94 cm possible IES-1998 Page No.114 Slide No. 3.(c) d 2 + 4dh = 1002 + 4 × 100 × 100 = 224 mm Ans.6d1 = 31.115 Slide No. h = 100 mm. 457 Ans. r = 2 mm D = d 2 + 4dh = 402 + 4 × 40 × 60 = 105.(d) 2 d ⎛ d⎞ Reduction = ⎜1 − ⎟ × 100% = 50% = 0.4 = 1 − ⇒ d = 90 mm D 150 So.115 Slide No.(a) Slide No.(c) d = 25mm.117 Slide No.We know ( D ) = d 2 + 4dh = 252 + 4 × 25 ×15 = 46 mm GATE-2003 Page No.116 Slide No.118 Slide No. To improve surface finish.116 Slide No. d = 50 mm. d 2 = 0. 464 Ans. In blanking usable part is punched out circular part so die size is correct and clearance on punch IAS-2007 Page No. 0. Clearance have be given on Die only. 465 Ans. 441 Ans. 124 Slide No.(b) IAS-1994 Page No.124 Slide No.555 Ans. (b)Sintering used for making bond IES-2010 Page No. IES-1999 Page No.127 Slide No. (b) GATE -2014 (PI) Page No.(a) tc = 1. IAS-2000 Page No.119 Slide No. 485 Ans. 486 Ans.(d) Ans.GATE & PSUs) Page 190 of 210 Rev. 484 Ans.120 Slide No.5 × 2) = 101 mm GATE-2007 Page No. Refer slides for theory GATE-2000 Page No.125 Slide No. 510 Ans.(d)Mode of deformation of metal during spinning is bending and stretching.(b) Option (b) Magnetic pulse forming and (d) Eletro-hydraulic formingboth are High Energy Rate Forming (HERF).(c) IES-2007 Page No.557 Ans.120 Slide No.556 Ans. 537 IAS-1997 Page No. 518 Ans. 494 Ans.127 Slide No.09 Page No.5.05. 508 Ans.119 Slide No.545 Ans.125 Slide No. t = 2mm Lb = α (R + kt) = 1(100 + 0.(b) IES-2005 Page No. (c)It is for low melting point temperature metals.127 Slide No.(d) Page No. 536 IAS-1999 Page No. (v) Electro-magnetic (the use of rapidly formed magnetic fields) IES-2009 Page No.(d) IES-1999 Page No. Refer slide IES-2012 Page No.125 Slide No. 511 Ans. ε 2 = 0. (c)An oxide film is formed in the case of air atomization and that film can be avoided by using an inert gas. 530 Ans. 528 Ans.5 = tb sin 30 ⇒ tb = 3 mm IES-1994 Page No. 488 Ans. 509 Ans. 539 Ans.126 Slide No.125 Slide No.123 Slide No.535 Ans.(a) Ans. JWM-2010 Page No.(c) Ans.122 Slide No. 487 Ans.(b) High-Energy-Rate-Forming is metal forming through the application of large amount of energy in a very sort time interval. R = 100mm. k = 0.(b)It is without a blank holder.(b) Ans.126 Slide No. α = 30° now ( tc ) = tb sin α . 507 Ans.05 × e0.(d) Ch-12 Powder Metallurgy: Answers with Explanations IAS-2003 Page No.develop on the flange. 512 Ans. High energy-release rate can be obtained by five distinct methods: (i) Underwater explosions. IAS-2007 Page No. IAS-1997 Page No.122 Slide No. (iv)Internal combustion of gaseous mixtures. (d) For-2016 (IES. 495 Ans. 513 Ans. But Question is "usedfor forming components form thin metal sheets or deform thin tubes"it is done by Magnetic pulse forming only. which may also extend to the wall of the cup. ε1 = 0.122 Slide No. GATE-2006 Page No.127 Slide No.120 Slide No. (b) Compaction is used for making product.(c) IES-2013(conventional) Page No. Refer slides IES-2011 Page No. GATE-2011(PI) Page No.544 Ans.(c) GATE-1999 Page No.122 Slide No.125 Slide No.5mm.5mm.122 Slide No. 534 GATE -2012 Same Q in GATE-2012 (PI) GATE-2004 Page No. 538 IES-2010 Page No. (iii)Pneumatic-mechanical means.551 Ans.(c) α = 1 radian. IFS-2011 Page No. (c)In atomization process inert gas or water may be used as a substitute for compressed air.120 Slide No.127 Slide No.(c) IES-2010 Page No.125 Slide No. or 1. so no stress. 496 Ans.5 = 1.126 Slide No.(b) For bi-axial strectching of sheets: ε1 = ln Final thickness = li1 l and ε 2 = ln i 2 lo1 lo 2 initial thickness (t ) e ε1 × e ε 2 t = 1.119 Slide No.0 .121 Slide No. (b)In reduction Metal oxides are turned to pure metal powder when exposed to below melting point gases results in a product of cake of sponge metal. (ii) Underwater spark discharge (electro-hydraulic).122 Slide No.548 Ans.553 Ans.(a) Page No. IES-2013(conventional) Page No.546 Ans.304 mm e0.09 ∴Final thickness = IES-1998 GATE-2005 1.126 Slide No.122 Slide No.(d) GATE-1992 Page No. 4 per cent chromium and 1 per cent vanadium IES-2007 Page No.137 Slide No.(a)Carbon steel tools have Limited tool life.133 Slide No.132 Slide No.136 Slide No.128 Slide No. 4 per cent chromium and 1 per cent vanadium • Molybdenum high speed steel contains 6 per cent tungsten.599 Ans. Maximum cutting speeds about 8 m/min.601 Ans. (c) IES-2009 Page No. IES-2011(conventional) Page No.132 Slide No. (a) IES-2015(conventional) Page No. (d) GATE-2008(PI) Page No. dry and used upto 250oC IES-2013 Page No.619 Ans.137 Slide No.133 Slide No.(a)Coating if TiC and TiN on HSS is done by by Chemical Vapour Deposition (CVD) or Physical Vapour Deposition (PVD) IES-2003 Page No.131 Slide No.613 Ans.136 Slide No.129 Slide No.615 Ans. (d) GATE-2011 Page No.132 Slide No.614 Ans.591 Ans.(b) Addition of large amount of cobalt and Vanadium to increase hot hardness and wear resistance respectively IAS-1997 Page No. (b) IES-2006 Page No.131 Slide No.132 Slide No.(a) IES-1993 Page No.(b)The blade of a power saw is made of high speed steel.612 Ans. (d) IAS-2003 Page No.130 Slide No.0 .583 Ans.132 Slide No.GATE & PSUs) Page 191 of 210 Rev.564 Ans.617 Ans.593 Ans.136 Slide No.It may be automated though it is difficult for automation.134 Slide No.136 Slide No.15 Ans.610 Ans.596 Ans.132 Slide No.132 Slide No.602 Ans. GATE-2010(PI) Page No.10 Ans. (b) IES-2014 Page No. (d) Conventional Question. (c)No wastage of material.11 Ans. (d)A is false. 6 per cent molybdenum.611 Ans. Slide No. 595 Ans.597 Ans. Refer slide Ch-xx Tool Materials: Answers with Explanations IAS-1997 Page No.(a) IAS-2001 Page No.contains 18 per cent tungsten.129 Slide No.7 Ans.(a)18-4-1 High speed steel. Refer slide Conventional Question. (b) IES-2001 Page No. IES-2012 Page No. (c) IAS-1996 Page No. (c) GATE-2011(PI) Page No132.14 Ans. (d) IES-2007 Page No. (b& c)Best choice will be ( c) GATE-2009(PI) Page No.IES-1999 Page No. (c) IAS-2007 Page No. (b) IAS-2003 Page No.598 Ans.620 Ans.605 Ans.128 Slide No.21 Ans.134 Slide No.18 Ans.136 Slide No.134 Slide No.133 Slide No.131 Slide No. this is not true IES-2004 Page No.570 Ans.IES-2005 Page No.588 Ans. (a) IES-1997 Page No.131 Slide No. IES-2000 Page No. IES-1995 Page No.137 Slide No.132 Slide No. (b) IAS-2004 Page No. (b)Due to formation of bonding brittleness reduces.576 Ans. Refer slide IAS-1997 Page No.616 Ans.contains 18 per cent tungsten.26 Ans.569 Ans.133 Slide No. (b) IES-1998 Page No.592 Ans.131 Slide No.134 Slide No.19 Ans. (d) IES-2001 Page No.137 Slide No.IES-2010 Page No. (c) IAS-1998 Page No.(b) IES-1992 Page No. Refer slides IES-2010 Page No. 4 per cent chromium and 2 per cent vanadium.561 Ans.129 Slide No. (b)Disadvantage of PM is relatively high die cost.134 Slide No. Closed dimensional tolerances are possible with iso-static pressing of metal power in powder metallurgy technique.137 Slide No.(a) IAS-1994 Page No.134 Slide No.600 Ans.134 Slide No. Refer Slide ISRO-2013 Page No.568 Ans.129 Slide No. (a) IES-2013(conventional) Page No.589 Ans.136 Slide No. (b) IES-2015 Page No.13 Ans. IES-2002 Page No. 603 Ans. Lubricants such as graphite or stearic acid improve the flow characteristics and compressibility at the expense of reduced strength.575 Ans. (c) IES-2007(conventional) Page No.(d) • 18-4-1 High speed steel.129 Slide No.20 Ans.(a) For-2016 (IES.22 Ans.(b) IES-2011 Page No.135 Slide No. IAS-1998 Page No.89 Ans.138 Slide No.(d) For-2016 (IES.143 Slide No.142 Slide No.GATE & PSUs) Page 192 of 210 Rev.52 Ans.140 Slide No.84 Ans.(b)Nonferrous materials are best to work with diamond because ferrous materials have affinity towards diamond and diffusion of carbon atoms takes place.140 Slide No. Hence the abrasive used in grinding wheels is generally manufactured from the aluminium ore.79 Ans.” .76 Ans.(c) IES-2013 Page No. bauxite. Thus C matches with 1.(b) IES-1999 Page No. IES-2000 Page No. Thus A matches with 2.(a)“Oxidation of diamond starts at about 450oC and thereafter it can even crack.(a) IES-2002 Page No.(b)Ceramic tools are used only for light.88 Ans. sintered and palleted to make ready ceramics IES-1996 Page No. Thus D matches with 3 IES-2003 Page No. Coated carbide tools are treated by nitriding. Silicon carbide (SiC) Silicon carbide is made from silica sand and coke with small amounts of common salt.142 Slide No.(c) H.144 Slide No.33 Ans.144 Slide No.(a) IES-2010 Page No. IES-2001 Page No.32 Ans. For this reason the diamond tool is kept flooded by the coolant during cutting.86 Ans.140 Slide No. IES-1999 Page No.142 Slide No.62 Ans. Thus B matches with 5.15m / min Cutting speed in this case is 314 m / min.(b) V= IES-2007 IAS-2000 IAS-2003 IAS-1999 IES-2010 IES-2000 π D(mm) N 1000 m / min = π ×100 × 1000 1000 = 314.(a)Hardness of CBN is comparable to diamond IES-1994 Page No.138 Slide No.Book B L Juneja and Nitin seth page 88 IES-1994 Page No.(a) IES-2005 Page No. The major constituent of diamond is carbon.47 Ans. and is also called corundum and emery.144 Slide No.0 .S.45 Ans.140 Slide No.(d) IES-1996 Page No.144 Slide No. Page No. at which ceramic is suited. which areGround.(a) IES-1999 Page No.72 Ans.(a) GATE-2009(PI) Page No.69 Ans.144 Slide No. diamonds are not suitable because of the diffusion of carbon atoms from diamond to the work-piece material.(c) IES-1994 Page No.48 Ans.IES-1995 Page No.78 Ans. the natural abrasives generally have impurities and.(d)WC is used for drawing dies. and light feeds are used.(a) IES-1992 Page No.139 Slide No.63 Ans.141 Slide No.This is because of low strength of ceramics IES-1996 Page No.142 Slide No.(b) Page No.(d) IES-1996 Page No. However.68 Ans.143 Slide No. and silicon carbide for heating elements.49 Ans.70 Ans.(b)Cutting speed of diamond is very high but small feed rate with low depth of cut.143 Slide No.(c) IAS-1999 Page No.(b)This is one of the natural abrasives found.S < Cast alloy < Carbide < Cemented carbide < Cermets < ceramics Page No.37 Ans.144 Slide No. IAS-2001 Page No. their performance is inconsistent. Degarmo and Kalpakjian both book written this.(b) IES-1993 Page No.140 Slide No.141 Slide No.143 Slide No.142 Slide No. Cr & V. IES-1995 Page No.(c) Page No.(d)Cermets are Metal-ceramic composites IES-2003 Page No.(c) Page No.142 Slide No.71 Ans.51 Ans. Non ferrous alloys (stellites) are high in cobalt.77 Ans.65 Ans.(d) Page No.140 Slide No.140 Slide No.(d)On ferrous materials.50 Ans. in addition to W.83 Ans.61 Ans.139 Slide No.85 Ans.57 Ans.(b) High speed steel.(c) IES-1997 Page No.82 Ans.141 Slide No. smooth and continuous cuts at high speeds.143 Slide No. as a result. Al2O3 for abrasive wheels.(d)None of the uses is true for CBN. silicone nitride for pipes to carry liquid metal.(b)Constituents of ceramics are oxides of different materials.141 Slide No.144 Slide No. has Mo as the most influencing constituent.80 Ans.46 Ans. For-2016 (IES.GATE & PSUs) Page 193 of 210 Rev. σult – Ultim mate tensile stress here neck formation starts. f we need flow stress s and fllow stress iss not constaant and depends on stress of the workpiecee.e in between σ y and σult σ y – Yield Stress σ o – For forging. [ ∵ Ao Lo = AL ] It is valid v till th he neck form mation. Flow Stress s When a m material deforms plasticallly strain harrdening occurs. Forging occurs in plasstic zone i. (σT ) = W Where lload = σ (1 + ε) Instanta aneous area a σ annd ε is thee engineering stress and a engineeering strain n respective ely. Trrue strain L ⎛L⎞ ⎛A ⎞ ⎛d ⎞ Elongation E dxx =∫ = ln ⎜⎜⎜ ⎟⎟⎟ = ln (1 + ε) = ln ⎜⎜ o ⎟⎟⎟ = 2 ln ⎜⎜ o ⎟⎟⎟ (εT ) = ⎜⎝ A ⎠ ⎜⎝ d ⎠ ⎜⎝ Lo ⎠⎟ Instan ntaneouslen nght L x o orr engineerin ng strain ( ε ) = e εT -1 The volume of the specimen s iss assumed to t be consta ant during plastic defformation.Ana alysis s of Forg ging True stress and Trrue Strrain Th he true strress is defiined as thee ratio of th he load to the cross section are ea at any in nstant.0 . Geometry should bee taken at end e of forgiing P B dx τx x (σx+d σx) h x =0 2L L P x τx a which th he material does not move m in any y direction. x = 0 .3 Here σ o iss flow stresss but it is truee stress and ε T is truestraain.GATE & PSUs) Page 194 of 210 Rev. σo = 1000 0 (∈T )0. width won’t increease.g.0 . and (2) Axi‐sym mmetric forgging 1. At the en nd of the forging. (b) Withou ut strain hard dening: σ o = σ y We will an nalyze only o open die forging using sllab method of analysis ffor (1) Rectan ngular Bar forging. f forrce will be maximum m because oof the area involved between the die and d the workp piece is ma aximum. Rec ctangu ular Bar Forgiing Before forging After fforging (lenggth ↑ height ↓) B B h h1 2L 2L1 • • • Here we are using plane p strain n condition n i.Equation o of Flow Curvve (a) With sttrain hardening σo = K(∈T )n e. is the point at Take an element dxx at a distance of x (en nlarged view w): For-2016 (IES.e. B h − 2τx . eories of plasticity.Fig. Soo on upper side force= = P × area = (P × B× dx) d & similarly on low wer side = (P P B dx ) • As metal is moving ou utwards soo friction force f will act a in oppoosite directtion.0 . this frriction force is shear force and d will causse shear sttress on th he surface equal to (ττx B dx ) in lo ower and upper u surface. Upper die will U w give preessure on upper u surface and lower surface will get pre essure by loower die. thereefore net reesultant forrce in any direction n is zero. τx and x so we reduce condition n. • On left sidee (stress x area) = Force (σ x × Bh ) and on other side force e will be B (σ x + dσ x ) Bh • . h − 2τx dx = 0 dσ x 2τ x − =0 h dx (1 1) r it in nto two va ariables by applying Here theere are thrree variablees σx . At the en nd of forgin ng the systeem must bee in equilibrrium. Von-M Mises Theo ory: (σ1 − σ2 )2 + ( σ2 − σ3 )2 + ( σ3 − σ1 )2 = 2 σ02 2. B dx = 0 or or d σx B h − 2τx B dx = 0 d σx . ∴ Σ Fx = 0.GATE & PSUs) Page 195 of 210 Rev. FBD F of Elements Now elem ment will loook like a slab s and hence its nam me slab metthod of ana alysis. Tresca’s Theory y: σ1 − σ3 = σ0 For-2016 (IES. For a ducctile materrial there arre two the 1. Gives ( σx + d σx ) B h − σx . ν r = ⎟ as volume change nott occur.(2)) [where K = 0 = flow shear strress] 3 or or or or or resca’s th heory : From Tr σ1 − σ3 = σ0 or or σ x + P = σ0 σ x + P = 2 K …. σ2 = Therefore.0 . From Vo on-Mises theory: t 1 (σ − P ) 2 x 2 2 1 ⎧ ⎫ ⎧1 ⎫ 2 2 ⎨σ x − ( σ x − P) ⎬ + ⎨ ( σ x − P) + P ⎬ + ( − P − σ x ) = 2σ0 2 ⎭ ⎭ ⎩ ⎩2 ( σ x + P)2 ( σ x + P)2 + ( σ x + P)2 = 2σ20 + 4 4 3 2 ( σ x + P)) = 2 σ20 2 4 ( σ x + P)2 = σ20 3 2 ( σ x + P) = σ0 3 σ σx + P = 2 K ….GATE & PSUs) Page 196 of 210 Rev.(2)) [where K = σ0 = flow w shear strress] 2 Differenttiating equation (2) For-2016 (IES.(∵ Plane stra ain conditioon) ∈2 = 0 As σ2 νσ1 νσ3 − − =0 E E E σ2 = ν( σ1 + σ3 ) orr σ2 = ν( σx − P) 1⎞ 2⎠ ⎛ ⎝ Note: In theories off plasticity ⎜ Poisson's ratio. dσ x d P + =0 dx dxx dσx −d P = dx dxx or . PB dx τx = μ P uation (1) and a (3) From equ or or or d σ x 2τ x − =0 h dx d P 2μ P − =0 − x h dx dP 2μ ∫ P = − h ∫ dxx 2μ ln P = − x +C h .(3) Condition-1: dering sliding fric ction all over the e surface ( τx Consid = μP ) PBd dx τ × Bdx F = μN N or dF = μdN μ τx B dx = μ ..L h ln ( 2 K ) = − t values of o C in equ uation (4) Putting the 2μ 2μ x + lln ( 2 K ) + ....L h h ⎛ P ⎞ 2μ (L ln ⎜ L − x) ⎟= ⎝ 2K ⎠ h ln P = − or or ⎛ P ⎞ 2μ (L ln ⎜ L − x) ⎟= ⎝ 2K ⎠ h For-2016 (IES.GATE & PSUs) Page 197 of 210 Rev. σ x = 0 (because ( no o force is applied soo no stresss on that Boundary condition surface) and σx + P = 2K gives P = 2 K or or 2μ L+C h 2μ C = ln ( 2 K ) + . at x = L..0 ..(4) ns... Pmin = 2K K eh 2μ (0) = 2K Elementa al force.. e h .. dx ) ⎜⎜∵ ∫ gives half 0 ⎝ 0 ⎞ ⎟⎟ ⎠ . From equ uation (1) and a (3) or or or τ dσ x −2 x =0 h dx P 2K dP − − =0 dx x h −2K ∫ d P = h ∫ dxx 2K P=− x +C h For-2016 (IES. dx 0 Condition-2: Consid dering sticking frriction all over the surfac ce ( τx = τy = K ) Shear faiilure will occur at each and everry point.GATE & PSUs) .dx 2μ dF = 2K e h L (L − x ) 2μ Integratiing..dx .0 ... B. Pmax = 2K Ke At x = L... dF F = P.B.P = 2K .∫ e h (L − x ) L ⎛ L h portion F so for 2L w we use 2∫ . (5) (Pressure ( d distribution n equation) 2μ (L) h At x = 0. B . (L − x ) 0 L 2μ F = 4 KB .(6) Page 198 of 210 Rev. F = 2 × ∫ (2 K ... e or 2μ (L − x ) h . Boundary condition ns..GATE & PSUs) Page 199 of 210 Rev. dx or K+ dF = ⎨2K ⎧ ⎩ x=0 x=L 2K ⎫ (L − x ) ⎬ B . dx x (L h ⎭ 0⎩ Condition-3: Consid dering sticking an nd slidin ng both model m off friction (∵ Temperatu T re is same throughout body) For Slidiing Region: dσx 2τx − =0 dx h For-2016 (IES. Pmin = 2 K ….L . (7) {Presssure distriibution equ uation} 2K . dx h ⎭ L 2K ⎧ ⎫ F = 2 × ∫ ⎨2K + L − x ) ⎬ B ... Pmax = 2K K+ x = L.L h 2h 2k Elementa al force.L h 2k + 2k .. 2K = − or i equation n (6) Putting in P=− 2K 2K 2K + . σ x = 0 (because ( no o force is applied soo no stresss on that surface) and σx + P = 2K gives P = 2 K 2K L +C h 2K C = 2K + . dF = P . B .x +2 h h P = 2K + At 2K 2 (L − x ) h x = 0. at x = L.0 .L h So. .or or or d P 2μ P − =0 dx h dP 2μ ∫ P = − h ∫ dx 2μ ln P = − x +C h − ..0 . (5) .. at x = L.... xs h Ps = − 2K 2K x + Ps + xs h h 2K P = Ps + ( xs − x ) .....L h ln ( 2 K ) = − Putting the values of C in equation (4) 2μ 2μ .GATE & PSUs) Page 200 of 210 Rev.. e For Sticking Region: or or or At or or 2μ (L − x ) h dσ x τ −2 x =0 dx h dP 2K − − =0 dx h −2 K ∫ d P = h ∫ dx 2K P=− x +C h ... σ x = 0 (because no force is applied so no stress on that surface) and σx + P = 2 K gives P = 2 K or or 2μ L+C h 2μ C = ln ( 2 K ) + . P = Ps 2K xs + C h 2K C = Ps + ....(4) Boundary conditions.......(6) x = x s .L x + ln ( 2 K ) + h h ⎛ P ⎞ 2μ ln ⎜ (L − x ) ⎟= ⎝ 2K ⎠ h ln P = − or or or ⎛ P ⎞ 2μ ln ⎜ (L − x ) ⎟= ⎝ 2K ⎠ h P = 2K ..(8) h P=− For-2016 (IES..... For-2016 (IES. dx + 2 ∫ ⎨2K e h = 2 ∫ ⎨Ps + ⎬ B .25 2 × ⎝ ⎠ ⎝ ⎠ 0 to 36. B dx d Xs x 2μ s (L − x ) ⎫ ⎧⎪ 2K ⎧ ⎫ ⎪ ( x s − x ) ⎬ B .(10) (in any ques stion first w we find thiss xs ) n we can deecide the coondition of friction. h = 10 mm & μ = 0.13 mm stick king and 36 6.. Shear strresses are same τx = K τx = μ Ps τx = K = μ Ps Ps = At (if consid dering stick king) (if consid dering sliding) K μ .GATE & PSUs) Page 201 of 210 Rev.13 mm to 50 mm slid ding will ta ake place.0 . P = Ps 2μ Ps = 2 K e h (L − x s ) 2μ (L − xs ) K = 2K e h μ or 2μ (L − xs ) 1 =eh 2μ or ⎛ 1 ⎞ 2μ (L − x s ) ln ⎜ ⎟ = ⎝ 2μ ⎠ h or h ⎛ 1 ⎞ .. B . dx + 2 0 L ∫P Sliding .... f Using this equation Example 1: L = 50 mm. dx h ⎭ ⎪ 0 ⎩ xs ⎩ ⎭⎪ L To find Ps and x s s for booth sticking g and slidin ng At x = x s . ln ⎜ ⎟ = L − x s 2μ ⎝ 2μ ⎠ xs = L − h ⎛ 1 ⎞ ln ⎜ ⎟ 2μ ⎝ 2μ ⎠ .13mm 2μ μ 2 × 0.FTotal = FSticking + FSliding =2 Xs ∫ PSticking .25 2 0.(9) x = xs ..25 xs = L − h 10 1 ⎛ 1 ⎞ ⎛ ⎞ ln ln n ⎜ ⎟ = 50 − n⎜ ⎟ = 36. Example 2: L = 50 mm, h = 10 mm & μ = 0.08 xs = L − h 10 1 ⎛ 1 ⎞ ⎛ ⎞ ln ⎜ ⎟ = 50 − ln ⎜ ⎟ = − 64.53 mm ( absurd value ) 2μ 2 2 0.08 2 0.08 μ × × ⎝ ⎠ ⎝ ⎠ (∵ x cannot be –ve) i.e only sliding no sticking occur. Example 3: L = 50 mm, h = 10 mm & μ = 0.65 xs = L − h 10 1 ⎛ 1 ⎞ ⎛ ⎞ ln ⎜ ⎟ = 50 − ln ⎜ ⎟ = 52.01 mm 2μ 2 × 0.65 ⎝ 2 × 0.65 ⎠ ⎝ 2μ ⎠ Only sticking no sliding NOTE: If μ > 0.5 then only sticking, In hot forging ( μ ) is larger if μ > 0.5 only sticking condition will occur. IES – 2005 Conventional A strip of lead with initial dimensions 24 mm x 24 mm x 150 mm is forged between two flat dies to a final size of 6 mm x 96 mm x 150 mm. If the coefficient of friction is 0.25, determine the maximum forging force. The average yield stress of lead in tension is 7 N/mm2 Solution: h = 6 mm, 2L = 96 mm, μ = 0.25 xs = L − h 6 1 ⎛ 1 ⎞ ⎛ ⎞ ln ⎜ ⎟ = 48 − ln ⎜ ⎟ = 39.68 mm 2μ 2 μ 2 × 0.25 2 × 0.25 ⎝ ⎠ ⎝ ⎠ xS Ftotal = 2 × ∫ ⎧⎨Ps + 0 ⎩ L 2μ (L − x ) 2K ⎫ ( x s − x ) ⎬ B. dx + 2 × ∫ 2 K e h B . dx h ⎭ xS Applying Von-Mises theory K = σ0 = 4.04 N / mm2 3 K Ps = = 16.16 N / mm2 μ or 39.68 or 48 2 × 0.25 ⎧ (39.68 − x )⎫⎬ ⋅ 150 . dx + 2 × ∫ (2 × 4.04) e ⎨16.16 + 6 ⎭ 0 ⎩ 39.68 = 510 kN + 29.10 kN = 539 kN (Von − Mises) F = 2× ∫ Applying Tresca’s Theory, K = ⋅ 150 . dx σo K 3.5 = 3.5 N / mm2 ; Ps = = = 14 N / mm2 2 μ 0.25 39.68 48 2 × 3.5 ⎧ ⎫ (39.68 − x ) ⎬ ⋅ 150 ⋅ dx +2 ∫ (2 × 3.5) e ⎨14 + 6 ⎭ 0 ⎩ 39.68 442 kN + 25 kN = 467kN (Tresca ' s) F = 2× 2×0.25 (48 − x ) 6 ∫ 2×0.25 (48 − x ) 6 ⋅ 150 ⋅ dx Practice Problem-1 A strip of metal with initial dimensions 24 mm x 24 mm x 150 mm is forged between two flat dies to a final size of 6 mm x 96 mm x 150 mm. If the coefficient of friction is 0.05, determine the maximum forging force. Take the average yield strength in tension is 7 N/mm2 Given: 2L = 96 mm; L = 48 mm; h = 6 mm; B = 150 mm; μ = 0.05 h ⎛ 1 ⎞ ln ⎜ ⎟ 2μ ⎝ 2μ ⎠ x s = − 90.155 mm xs = L − K = 4.04 N/mm2 Since xs came negative so there will be no sticking only sliding will take place. For-2016 (IES,GATE & PSUs) Page 202 of 210 Rev.0 L 2μ F = 4 KB ∫ e h (L − x ) dx 0 48 = 4 × 4.04 × 150 ∫e ⎧⎪⎛ 2×0.05 ⎞ ⎫⎪ ⎨⎜ ⎟(48 − x ) ⎬ ⎠ ⎩⎪⎝ 6 ⎭⎪ dx = 177.98 kN 0 Axi – Symmetrical Forging (Open Die): Using cylindrical co-ordinate system (r, θ, z ) and Using Slab Method of analysis d1 R h1 h At the start of forging At end of forging Volume before forging = Volume after forging π 2 d1 × h1 = π R 2 h 4 At an angle θ, we take an dθ element at a radius r we take dr element. σσ + dσr dr dθ σ θ + dσ θ r σr σθ θ For axi-symmetrical forging dσ θ will be zero. For-2016 (IES,GATE & PSUs) Page 203 of 210 Rev.0 Net resultant force in radially y outward direction d is 0. ( σr + d σr ) (r + dr ) dθ . h − ( σr . r dθ . h ) − 2 τr . r dθ . dr − 2 σ θ dr h . ssin dθ =0 2 σθ dr.h cos dθ 2 dθ 2 σθ dr σθ dr.h cos dθ 2 σθ drh σθ dr.h sin dθ 2 dθ 2 σθ dr.h sin dθ 2 dθ ⎛ ⎞ gets can ncelled ∴ the ey are oppossite ⎟ ⎜ coss ⎝ ⎠ 2 For Axi--Symmetr ry forging ∈r = ∈θ σr = σ θ i.e. From aboove equatioon, ( σr + d σr ) ( r + dr ) dθ . h − ( σr . r dθ . h ) − 2 τr . r dθ . dr − 2 σr dr h . or dθ =0 2 d θ dθ ⎞ ⎛ ≈ ⎜ U sin g : σ θ = σr ; sin ⎟ ⎝ 2 2 ⎠ (σ σr + d σr ) (r + dr ) . h − ( σr . rh) − 2τr . r dr − σr . dr d .h =0 For-2016 (IES,GATE & PSUs) Page 204 of 210 Rev.0 or ( σr . r . h + dσr . rh + dσr . drh + σr drr . h) − σr . rh h − 2τr . r drr − σr dr. h = 0 or d σr . r h = 2τr . r dr or d σ r 2 τr − =0 d dr h ...(1) For ductiile materia al there are two theoriies of plastiicity 1. Tresca’s Theory y: σ1 − σ3 = σ0 ...(2) σ +P=σ r 0 or 2. Von Miscs M Theo ory: (σ σ1 − σ2 )2 + ( σ2 − σ3 )2 + ( σ3 − σ1 )2 = 2 σ20 or ( σr − σr )2 + ( σr + P)2 + ( − P − σr )2 = 2 σ20 or 2 ( σr + P)2 = 2 σ20 ...(2) σ +P=σ r 0 or On differrentiating; d σr d P + =0 dr dr d σr dP =− dr dr ...(3) Condittion 1: Consid dering sliding s f friction all over r the sur rface τr = μ P or d P 2τr − =0 dr h dP μP − −2. =0 dr h or d P −2 μ P = dr h From (1) and (3); dP 2μ =−∫ . dr P h 2μ ln P = − .r +C h ∫ or At or − r = R; σr = 0 ...(4) (becausse on this su urface therre will be noo force) and d σr + P = σ 0 ; P = σ 0 ln σ0 = − For-2016 (IES,GATE & PSUs) 2μ . R +C h Page 205 of 210 Rev.0 ⎛ 2μ ⎞ ⎢ ⎜− ⎟ ⎜ ⎝ ⎣ ⎝ h ⎠ R ⎞⎤ e ⎥ d ⎟⎟ ⎥ dr ⎛ 2μ ⎞ ⎜ − ⎟ ⎟⎥ ⎝ h ⎠ ⎠⎦0 2μ (R − r ) h R orr orr 2μ 2μ (R − r ) (R − r ) ⎤ ⎡ eh r. 2πr .e h (R − r ) . e h F = 2πσ0 ⎢ − ∫ ⎜⎜1 .GATE & PSUs) Page 206 of 210 Rev. Pmax = σ0 e h r = R. dr dr 0 orr 2μ ⎡ (R − r ) ⎛ ⎢r .C = ln σ0 + or 2μr 2 μR + ln σ0 + h h P 2μ = (R − r ) ln h σ0 uation (4) From equ ln P = − or 2μ P = σ0 . drr 2μ ∫ d F = ∫ σ0 . Pmin = σ0 r = 0 means a point ding force For find Elementa al force (dF F) dr r R dA = 2π πr dr dF = d F = P .0 .R r = 0. (R − r ) . e h R orr 2μ F = 2πσ0 ∫ r.. (5) Pressure distribution n .eh ⎥ ⎢ F = 2πσ0 − 2 ⎥ ⎢ ⎛ 2μ ⎞ ⎛ 2μ ⎞ ⎢ ⎜− ⎟ ⎜− ⎟ ⎥ ⎢⎣ ⎝ h ⎠ ⎝ h ⎠ ⎥⎦ 0 2μ R ⎡ ⎤ h e R 1 ⎢ ⎥ − F = 2π σ 0 −0+ 2 ⎥ 2 ⎢ ⎛ −2μ ⎞ − μ − 2 2 μ ⎛ ⎞ ⎛ ⎞ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⎢⎣ ⎝ h ⎠ ⎝ h ⎠ ⎝ h ⎠ ⎦⎥ For-2016 (IES. 2πr . e h or At At Here 2 μR h 2μ (R − r ) ……… ………. 7 mm ⎡ R 1 F = 2π σ 0 ⎢ − −0+ 2 ⎢ ⎛ −2μ ⎞ 2 − μ ⎛ ⎞ ⎢⎜ ⎟ ⎜ ⎟ ⎣⎢ ⎝ h ⎠ ⎝ h ⎠ Mean Die pressure = ⎤ ⎥ = 2. h1 = 60 mm.04 × 106 = ≈ 130 MPa Total Area π × 70.(6) (because on this surface there will be no force) and σr + P = σ0 . using slab method of analysis.R Total force 2. There is no sticking.72 GATE – 2014 (PI) In an open die forging. indicates that (a) there is no sticking friction anywhere on the flat face of the disc (b) sticking friction and sliding friction co-exist on the flat face of the disc (c) the flat face of the disc is frictionless (d) there is only sticking friction on the flat face of the disc Answer: (a) Condition -2: Considering sticking friction all over the surface τr = K From (1) equation (3) or d σr 2 τr − =0 dr h d P 2K − − =0 dr h or ∫dP = − ∫ or P=− At r = R. from the center of the disc towards its periphery. [20-Marks] Solution: Given. Find the die load at the end of compression to a height 30 mm. Also find mean die pressure.0 ..IES – 2007 Conventional A cylinder of height 60 mm and diameter 100 mm is forged at room temperature between two flat dies.GATE & PSUs) Page 207 of 210 Rev. Assume that volume is constant after deformation. dr h 2K . σr = 0 σ0 = − 2K . a circular disc is gradually compressed between two flat platens.05.R+C h For-2016 (IES. The exponential decay of normal stress on the flat face of the disc.04 MN 2 ⎥ ⎛ −2μ ⎞ ⎥ ⎜ ⎟ ⎝ h ⎠ ⎦⎥ 2μ eh . h = 30 mm σ0 = 120 N/ mm 2 and μ = 0. d1 = 100 mm.05 or or or π d12 h1 = π R2 h 4 1002 × 60 = R2 × 30 4 R = 70. The yield strength of the work material is given as 120 N/mm2 and the coefficient of friction is 0.r +C h . P = σ0 2K .. C = σ0 + or 2K 2K . r + σ0 + R h h 2 2K P = σ0 + . (R − r ) h From (6) P=− or At 2K R h r = 0; Pmax = σ0 + ...(7) Pressure D Distribution n linear 2K .R h r = R; Pmin = σ0 For find ding force: r R 2πrdr d F = P . 2 πr dr d or F = ∫ P . 2 πr dr d R or 2K K ⎡ ⎤ F = ∫ ⎢ σ0 + . (R − r ) ⎥ 2π r dr h ⎦ 0⎣ For-2016 (IES,GATE & PSUs) Page 208 of 210 Rev.0 Condition 3: When there is sticking and sliding both frictions occur St ic ki ng d Sl i i ng Sliding Ps Sticking r = Rs For sliding region pressure distribution is same as we derived in previous condition same boundary condition same differential equation. 2μ P = σ0 . e h (R − r ) For sticking region: Using equation (6). P=− 2K .r +C h Boundary condition at r = R; or P = Ps 2K . Rs + C h 2K C = Ps + . Rs h Ps = − or Putting in equation (6) 2K 2K (r ) + Ps + Rs h h 2K P = Ps + . (R s − r ) h P=− or ...(8) FTotal = Fsticking + Fsliding Ftotal = Rs R ∫P sticking ⋅ 2 πr dr + 0 Ftotal = Rs ∫ 0 ∫P sliding ⋅ 2 πr dr Rs 2K ⎡ ⎤ ⎢⎣ Ps + h (R s − r )⎥⎦ ⋅ 2 πr dr + R 2μ ∫ σ0 . e h (R − r ) . 2 πr dr Rs To find Ps and Rs τr = μ Ps = K or At Ps = r = Rs ; ......(9) P = Ps 2μ Ps = σ0 e h or K μ (R − Rs ) 2μ (R − R s ) K = σ0 e h μ ⎛ K ln ⎜ ⎝ μ σ0 ⎞ 2μ (R − R s ) ⎟= ⎠ h For-2016 (IES,GATE & PSUs) Page 209 of 210 Rev.0 Rs = R − h ⎛ K ⎞ ln ⎜ ⎟ 2μ ⎝ μ σ0 ⎠ or According to Tresca’s theory K= σ0 2 Rs = R − K 1 = σ0 2 or h ⎛ 1 ⎞ ln ⎜ ⎟ 2μ ⎝ 2μ ⎠ ...(10) According to Von-Miscs Theory K= σ0 or 3 Rs = R − K 1 = σ0 3 h ⎛ 1 ⎞ ln ⎜ ⎟ 2μ ⎝ 3 μ⎠ ...(11) IES – 2006 – Conventional A certain disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. If the co-efficient of friction between the job and die is 0.25, determine the maximum forging force. The average shear yield stress of lead can be taken as 4 N/mm2 [10 – Marks] Solution: R1 = 150 mm, h1 = 50 mm, R = ?, h = 25 mm, μ= 0.25 πR12h1 = π R 2 h R = 212.1 mm τ y = 4 N/ mm2 (Shear yield stress) = K By Tresca Theory; R s = 212.1 − 25 1 ⎛ ⎞ ln ⎜ ⎟ = 177.4 mm 2 × 0.25 ⎝ 2 × 0.25 ⎠ ⎡0 mm to 177.4 mm → sticking ⎤ ⎢177.4 mm to 212.1mm → sliding ⎥ ⎣ ⎦ Ps = K 4 = = 16 N / mm2 μ 0.25 σ0 = 2 K = 2 × 4 = 8 N / mm2 ⎫⎪ ⎪⎧⎛ 2×0.25 ⎞ ⎟(212.1 − r ) ⎬ 25 ⎠ ⎪⎭ 177.4 212.1 ⎨⎜ ⎧ ⎫ ⎛2×4⎞ Ftotal = ∫ ⎨16 + ⎜ (177.4 − r ) ⎬ . 2 πr dr + ∫ (8) × e ⎪⎩⎝ ⎟ ⎝ 25 ⎠ ⎭ 0 ⎩ 177.4 = 3.93 MN (Tresca’s Theory) . 2 πr dr Von Miscs Theory; R s = 212.1 − 25 1 ⎛ ⎞ ln ⎜ ⎟ = 170.25 mm 2 × 0.25 ⎝ 3 × 0.25 ⎠ ⎡0 mm to 170.25 mm → sticking ⎤ ⎢170.25 mm to 212.1mm → sliding ⎥ ⎣ ⎦ Ps = K 4 = = 16 N / mm2 μ 0.25 σ0 = K 3 = 4 3 N / mm2 Ftotal = RS ∫ 0 2K ⎧ ⎫ (R s − r ) ⎬ ⋅ 2 πr dr + ⎨Ps + h ⎩ ⎭ R ∫ 2μ σ0 e h (R − r ) ⋅ 2 πr dr RS 170.25 212.1 2×0.25 (212.1 − r ) 2×4 ⎧ ⎫ 25 16 + (170.25 − ) ⋅ 2 π + 4 3. ⋅ 2 πr dr r r dr e ⎨ ⎬ ∫0 ⎩ ∫ 25 ⎭ 170.25 = 3.6 MN (Von Misces) Ftotal = For-2016 (IES,GATE & PSUs) Page 210 of 210 Rev.0 GATE -2016: Mechanical Following Topics are very important for GATE Examination Grade-A Subjects Mathematics Production SOM TOM Thermodynamics Industrial Engineering Grade-B Subjects (2008-17%, 2002-22%, 1996-19%) (2008-13%, 2002-14%, 1996-15%) (2008-7%, 2002-9%, 1996-9%) (2008-11%, 2002-6%, 1996-6%) (2008-9%, 2002-14%, 1996-8%) (2008-9%, 2002-2%, 1996-5%) (2008-6%, 2002-6%, 1996-6%) (2008-5%, 2002-5%, 1996-5%) ………….less than 5 revisions required RAC Power plant Engineering Mechanics Grade-D Subjects (2008-16%, 2002-8%, 1996-7%) ………….more than 5 revisions required Design Heat Transfer Fluid Mechanics & Machines Grade-C Subjects ………….more than 10 revisions required (2008-3%, 2002-3%, 1996-4%) (2008-1%, 2002-7%, 1996-8%) (2008-2%, 2002-0%, 1996-4%) ………….1 or 2 revisions are sufficient IC Engine Engineering Materials (2008-1%, 2002-4%, 1996-4%) (2008-0%, 2002-0%, 1996-0%) Detailed Topics 1. Mathematics a) Probability and Statistics b) Eigen Values and Eigen Vectors c) Differential Equations By: S K Mondal (GATE 99.96 Percentile) E-Mail:
[email protected] Page 1 force. Multiple Integrals Complex Variables Matrix Algebra Fourier Series Partial Derivatives 2.in Page 2 . Special welding theory Casting: allowances. mass calculation. e) Critical Speed of Shafts By: S K Mondal (GATE 99.co. Pouring time calculations. milling. Gradient. EDM theory. tool life Rolling calculations Wire drawing and Extrusion Calculations Sheet metal operations. TOM a) b) c) d) Mechanism (VIMP) Linear Vibration Analysis of Mechanical Systems (VIMP) Gear train (VIMP) Flywheel (Coefficient of Fluctuation of speed. Riser Design. end conditions) Strain Energy Method (Castigliano’s theory) Theories of failure Thin cylinder Riveted and Welded Joint Spring 4. comparison of all NTMM NC/CNC Machine. clearance. Resistance welding calculations. Sprue Design. shear calculations Lathe. Special Castings.96 Percentile) E-Mail: swapan_mondal_01@yahoo. power. BLU calculations. Mohr Circle(VIMP) Stress and Strain (VIMP) Bending Moment and Shear Force Diagram (VIMP) Deflection of Beam Torsion Theories of Column (Euler method. 3. tolerance. shaping cutting time calculations Grinding and finishing ECM MRR.d) e) f) g) h) i) j) Transform Theory Numerical Methods Calculus. drilling. Coefficient of Fluctuation of energy). SOM a) b) c) d) e) f) g) h) i) j) k) Principal Stress and Principal Strain. feed calculations. Production a) b) c) d) e) f) g) h) i) j) k) l) Theory of metal cutting. fit Jig & Fixture. Casting Defects. forces. upto M & G code Limit. 3-2-1 principle Welding: V-I Characteristics calculations. NTU) Radiation (The Stefan-Boltzmann Law.in Page 3 .5. Shape Factor Algebra.96 Percentile) E-Mail: swapan_mondal_01@yahoo. Brake 8. Sommerfeld Number Fluctuating Load Consideration for Design Clutch . Load-life Relationship Sliding contact bearing.co. Availability Pure Substance (VIMP) Gases and Gas mixture Thermodynamics relations 6. Thermal Boundary Layer Compressible Flow By: S K Mondal (GATE 99. theriblings symbol) Break even Analysis The Scheduling Problem and Johnson’s Rule Assembly line balancing 7. Industrial Engineering a) b) c) d) e) f) g) h) i) EOQ Models (VIMP) PERT & CPM (VIMP) Queuing Model (VIMP) Forecasting (VIMP) LPP Work Study & Work Measurement (standard time calculations. Fluid Mechanics & Fluid Machines a) b) c) d) e) f) g) Properties of fluid Pressure measurement. Thermodynamics a) b) c) d) e) f) Basic Concepts Application of First law Entropy. manometers Fluid kinematics (VIMP) Bernoulli’s Equation Venturimeter Boundary Layer. Heat Transfer a) b) c) d) e) Conduction Critical Thickness of Insulation Unsteady Conduction (Lumped Parameter Analysis) Heat Exchangers (LMTD. Modes of Lubrication. Heat Exchange between Nonblack Bodies) 9. Design a) b) c) d) Rolling Contact Bearings. Power plant a) Steam Cycle (VIMP) b) Gas Cycle c) Compressor 12. IC Engine a) Gas Power Cycles (All) (VIMP) b) IC Engine Performances c) Octane number. cetane number 14.in Page 4 . TTT diagram.co. RAC a) Heat engine. heat pump. Engineering materials a) Iron-carbon Equilibrium diagram.96 Percentile) E-Mail:
[email protected]) Hydraulic Turbine i) Centrifugal Pump 10. b) Heat treatment. refrigerator (VIMP) b) Vapour Compression Systems c) Psychrometry (VIMP) 11. c) Crystal structure & crystal defects By: S K Mondal (GATE 99. Engineering Mechanics a) Equilibrium b) Truss 13. General Guidelines Please follow the step by step procedure given below for preparing GATE where only objective type questions are asked. For regular eye on your progress Paste the topic list in front of study table. tick mark the topics completed by you (class note + all questions from question set) on daily basis.IAS) SOM (Class note + My Question set GATE. GATE papers are set by Professors of IITs and they check student’s fundamentals of the subject. One tick when topic is completed once. So you don’t need to remember the questions and answers but you must remember the funda behind it. The questions are not repeated but the theory (Funda) which is needed to solve the question remains same. I found that in all competitive examinations similar type of questions are asked. enormous calculations etc and our University examinations are also conventional type.IAS) By: S K Mondal (GATE 99.co. We have to prepare for Objective Questions. For Students’ who are attending coaching Attend all classes.96 Percentile) E-Mail: swapan_mondal_01@yahoo. They are similar but not the same. Mathematics (Class note + My Question set) Manufacturing Science (Class note + My Question set GATE. You know that in the engineering books are not made for objective type questions. so that formulae can be remembered on the long run. The theory involves rigorous derivations. Topic list should contain as many tick marks as many times revision for a particular topic is completed. and so on. Meticulously read class note From the question set solve all previous year asked questions which are given in topic wise sequence. then second tick after the second revision. It is recommended to solve my question set on your own and then cross check with my explanations. Once one subject gets completed it has to be revised frequently (once in a month or so).IES. So we must be prepared with fundamentals. That’s why funda is repeated.IES.in Page 5 . IES.IAS) Engineering Mechanics ( Book: D P Sharma or Bansal ) Thermodynamics {Book Rajput + P K Nag Unsolved all (My solutions available)} Fluid Mechanics & Machines (Book Bansal or Rajput or my book + Only 20 Years GATE Questions.IAS } SOM (Book Sadhu Singh or B. My set) IC Engine (Book V Ganeshan + Only 20 Years GATE Questions.IAS) Industrial Engineering (Class note + Question set GATE. My set) Power plant (Book PK Nag + Only 20 Years GATE Questions. Mathematics (Grewal or Dash + My Question set) Manufacturing Science { Book P.IES. TOM (Class note + Question set GATE. My set) RAC (Book Rajput or CP Arora+ Only 20 Years GATE Questions.96 Percentile) E-Mail:
[email protected] Sharma (Vol 1& 2) + My Question set GATE. My set) Heat Transfer (Book Rajput or Domkundwar + Only 20 Years GATE Questions.IES.IES.in Page 6 . My set) Engineering Materials (NPTEL IISc material from net + Only 20 Years GATE Questions. Practice all solved examples from the books of respective subject as listed below of ONLY the topics mentioned in the topic list It is recommended to solve my question set on your own and cross check with my explanations.IAS) Industrial Engineering (Ravi Shankar or Kanti Swarup or my book + My Question set GATE.IES.IES.C Punmia or only My Book + My Question set GATE.IAS) TOM ( Book GhoshMallick + My Question set GATE. My set) By: S K Mondal (GATE 99. My set) Design (Book Bhandari or Khurmi + Only 20 Years GATE Questions.IAS) Engineering Mechanics ( Book: D P Sharma or Bansal or Khurmi) Thermodynamics {Class note + P K Nag Unsolved all (My solutions available)} Fluid Mechanics & Machines (Class note + Only 20 Years GATE Questions) Heat Transfer (Class note + Only 20 Years GATE Questions) Design (Class note + Only 20 Years GATE Questions) RAC (Class note + Only 20 Years GATE Questions) Power plant (Class note + Only 20 Years GATE Questions) IC Engine (Class note + Only 20 Years GATE Questions) Engineering Materials (Class note + Only 20 Years GATE Questions) For self preparing students Meticulously read book only of the topics mentioned in topic list.co.