Metal Cutting, Metal Forming & MetrologyQuestions & Answers-2014 (All Questions are in Sequence) IES-1992-2013 (22 Yrs.), GATE-1992-2013 (22 Yrs.), GATE (PI)-2000-2012 (13 Yrs.), IAS-19942011 (18 Yrs.), some PSUs questions and conventional questions are added. Section‐I: Theory of Metal Cutting Chapter-1: Basics of Metal Cutting Chapter-2: Force & Power in Metal Cutting Chapter-3: Tool life, Tool Wear, Economics and Machinability Page-1 Page-7 Page-13 Section‐II: Metal Forming Chapter-4: Cold Working, Recrystalization and Hot Working Chapter-5: Rolling Chapter-6: Forging Chapter-7: Extrusion & Drawing Chapter-8: Sheet Metal Operation Chapter-9: Powder Metallurgy Page-24 Page-27 Page-30 Page-37 Page-43 Page-52 Section‐III: Metrology Chapter-10: Limit, Tolerance & Fits Chapter-11: Measurement of Lines & Surfaces Chapter-12: Miscellaneous of Metrology Section‐IV: Cutting Tool Materials Page-56 Page-61 Page-65 Page‐67 For‐2014 (IES, GATE & PSUs) IES‐2013 Carbide tool is used to machine a 30 mm diameter S‐2001 200 IES For cutting of brass with single‐point cutting tool on a lathe, tool should have ( ) Negative (a) N ti rake k angle l (b) Positive rake angle (c) Zero rake angle (d) Zero side relief angle Theory of Metal Cutting steel shaft at a spindle speed of 1000 revolutions per minute. The cutting speed of the above turning operation is: (a) 1000 rpm B S K Mondal By M d l (b) 1570 m/min (c) 94.2 m/min (d) 47.1 m/min IES‐1995 Single point thread cutting tool should ideally have: a) ) Zero rake b) Positive rake c) Negative rake d) Normal rake GATE‐1995; 2008 Cutting C tti power consumption ti in i turning t i can be b significantly g y reduced by y (a) Increasing rake angle of the tool (b) Increasing the cutting angles of the tool (c) Widening the nose radius of the tool (d) Increasing I i the h clearance l angle l IES‐1993 Assertion (A): For a negative rake tool, the specific cutting pressure is smaller than for a positive rake tool under otherwise identical conditions. Reason (R): The shear strain undergone by the chip in the case of negative rake tool is larger. larger (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 2005 200 IES Assertion (A): Carbide tips are generally given negative rake angle. Reason (R): Carbide tips are made from very hard materials. ( ) 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 S – 2002 IES Assertion (A): Negative rake is usually provided on carbide tipped tools. Reason (R): Carbide tools are weaker in compression. ( ) 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 IES 2011 Which one of the following statement is NOT correct with reference to the purposes and effects of rake angle of a cutting tool? (a) To guide the chip flow direction (b) To reduce the friction between the tool flanks and the machined surface (c) To add keenness or sharpness to the cutting edges. (d) To provide better thermal efficiency. For-2014 (IES, GATE & PSUs) Page 1 of 78 ( ) GATE – 2008 (PI) Brittle materials are machined with tools having zero or negative rake angle because it (a) results in lower cutting force ( ) improves surface finish (b) (c) provides adequate strength to cutting tool (d) results in more accurate dimensions IES 2007 Cast iron with impurities of carbide requires a particular rake angle for efficient cutting with single point tools, what is the value of this rake angle, give reasons for your answer. answer [ 2 marks] S – 1994 99 IAS Consider the following characteristics 1. The cutting edge is normal to the cutting velocity. 2. The Th cutting tti f forces occur i in t two di directions ti only. l 3. The cutting edge is wider than the depth of cut. The characteristics applicable to orthogonal cutting would include (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 1 2 and 3 IES ‐ 2012 During orthogonal cutting, an increase in cutting speed causes (a) An increase in longitudinal cutting force (b) An increase in radial cutting force (c) An increase in tangential cutting force (d) ( ) Cutting g forces to remain unaffected IES‐2006 Which of the following is a single point cutting tool? (a) Hacksaw blade (b) Milling cutter (c) Grinding wheel (d) Parting P ti tool t l IES ‐ 2012 ) Negative g g are preferred p g set‐ Statement ( (I): rake angles on rigid ups for interrupted cutting and difficult‐to machine materials. Statement (II):Negative rake angle directs the chips on to the machined surface ( ) Both (a) B h Statement S (I) and d Statement S (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 (II) ) is not the correct explanation p of true but Statement ( Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true IES‐1995 The the and Th angle l between b t th face f d the th flank fl k of f the th single point cutting tool is known as a) Rake angle b) Clearance angle g c) Lip angle d) Point angle. angle For-2014 (IES, GATE & PSUs) Assertion (A): For drilling cast iron, iron the tool is provided with a point angle smaller than that required for a ductile material. material Reason (R): Smaller point angle results in lower rake k angle. l (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually y 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 2 of 78 IES‐2006 IES‐2002 Consider the following statements: The strength of a single point cutting tool depends upon 1. Rake angle 2. Clearance angle 3. Lip angle Which of these statements are correct? ( ) 1 and (a) d3 (b) 2 and d3 (c) 1 and 2 (d) 1, 2 and 3 The principal and auxiliary orthogonal clearance angles are 10o and 8o. Side relief angle 6.2.4.6. then the side rake angle will be ( ) 5° (a) ° (b) 6° ( ) 8° (c) (d) 10° ° ISRO‐2011 A cutting tool having tool signature as 10. System if the tool nomenclature is 8‐6‐5‐5‐ 10‐15‐2‐mm. 2. 2.IES ‐ 2012 Tool life increase with increase in (a) Cutting speed (b) Nose N radius di (c) Feed (d) Depth of cut IES‐2009 Consider the following statements with respect to the effects of a large nose radius on the tool: 1. 6. 2 and d3 IES‐1995 Consider the following statements about nose radius 1 It improves tool life 1. The inclination angles of the principal and the auxiliary cutting edges are both 0o. 3. 6. Back rake angle 2.7 (b) (c) 1.6. 12.6.3. It improves tool life. 9. GATE & PSUs) Page 3 of 78 . The chip flows in the orthogonal plane. The value of Φ is closest to (a) 00 (b) 450 0 (c) 60 (d) 900 G 20 0 ( ) GATE – 2010 (PI) The tool geometry of a single point right handed turning tool is provided in the orthogonal rake system (ORS). It deteriorates d t i t surface f fi i h finish.3. Nose radius The correct sequence of these tool elements used for correctly specifying the tool geometry is ( ) 1.7 IES‐2009 The following tool signature is specified for a single‐ point cutting tool in American system: 10. Φ is the principal cutting edge angle and its range is 0o ≤ φ ≤ 90o . 8. 1 2 and 3 IES‐1994 Tool geometry of a single point cutting tool is specified by the following elements: 1.7 (d) 1. 20. End relief angle 7. 8.3.5. 2 will have side rake angle (a) 10o (b) 9o (c) 8o (d) 2o GATE‐2008 In a single point turning tool. 8 6.4.5. respectively. The rake angle (in degree) measured on the orthogonal plane is (a) 0 (b) 2 (c) 8 (d) 10 For-2014 (IES.7 (a) ( ) 1. Select the correct answer using g the codes g given below: (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1. 6.4.5. 3 What does the angle 12 represent? (a) Side cutting‐edge angle (b) Side rake angle (c) Back rake angle (d) Side Sid clearance l angle l S‐1993 993 IES In ASA System. Side rake angle 3. Which of the above statements is/are correct? (a) 2 only (b) 3 only ( ) 2 and (c) d 3 only l (d) 1. Side cutting edge angle 5.2. 2. 8. It improves the surface finish. It reduces the cutting force 3. 4. 6 15. The sum of the principal (major) cutting edge angle and the auxiliary (minor) cutting edge angle of the above tool is 90o.2. It increases the possibility of chatter. 5. 3. the side rake angle and orthogonal rake angle are equal. End cutting edge angle 4. 3 0 3 and the rake angle of the tool is 10°.7754×105 5 ( ) 1.e. Plan approach angle 1. 2 2.34 GATE ‐2012 p g to an orthogonal g g Details pertaining metal cutting process are given below. 2. i e uncut thickness is 0.2 and 4 (d) 2.0104×10 (c) (d) 4.0 00 (b) ( ) 0.1781×105 (b) 0. edge 2. A large rake angle means lower strength of the cutting edge.00 (d) GATE 2011 A single – point cutting tool with 12 12° rake angle is used to machine a steel work – piece. 1. W d angle l 4. Tool face B Rake angle B.94 (c) ( ) 50. This allows better access of coolant to the tool work p piece interface Which of the statements given above are correct? (a) 1 and 2 (b) 2 and 3 (c) 2 and 4 For-2014 (d) 3 and 4 (IES. List I with List II and select the correct answer using the codes given below the Lists: List I List II A. GATE & PSUs) IES‐2006 Consider the following statements: 1. Rake R k angle l of f the h cutting i tool. This reduces excessive friction between the tool and d work k piece i 3.31 3 (b) ( ) 0. The depth of cut i. True T f d feed 2. Chip thickness 4.81 0 81 mm.2 and (a) d 3 (b) 1.13 3 (c) 3. l The p parameters which g govern the value of shear angle would include ( ) 1. Cutting g velocity y 3. Chip thickness ratio 04 0.8 mm The shear angle is approximately (a) 22° (b) 26° (c) 56° 5 (d) 76° The following parameters determine the model of continuous chip formation: 1. Clearance angle 3.6 mm R k angle Rake l +10° ° Cutting speed 2. Which of the statements given above is/are correct? (a) ( ) Only y 1 (b) ( ) Only y 2 (c) Both 1 and 2 (d) Neither 1 nor 2 IES‐2004 Match.0 (d) 2.00 (d) 3.3 and 4 IES‐1994 IES ‐ 2009 Minimum shear strain in Mi i h i i g turning g with a cutting g orthogonal tool of zero rake angle is (a) 0.22 ( ) 22. 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 Page 4 of 78 .5 m/s Mean thickness of primary shear zone 25 microns The shear strain rate in s–1 during the process is (a) 0.4. Tool flank C.3 and d4 (c) 1. the chip thickness ratio was obtained as 0.GATE‐2001 During D i orthogonal h l cutting i of f mild ild steel l with ih a 10° rake angle tool.5 5 (c) 1. mm The chip thickness under orthogonal machining condition is 1 8 mm. What is the value l of f the th shear h strain? t i ? (a) ( ) 0. Cutting torque decreases with rake angle.0 IES ‐ 2004 In a machining operation chip thickness ratio is 0. cut.397×105 IES‐2004 Consider the following statements with respect to the relief angle of cutting tool: 1 This affects the direction of chip flow 1. Tool face and flank D Wedge D. C i edge Cutting d 5. This affects tool life 4.4 Undeformed thickness 0.53 (b) 20. The shear angle (in degrees) g ) evaluated from this data is (a) 6. 0 2 0 m/s (b) 2.4 2 4 m/s (c) 1. / depth d th of f cut t = 0. Chip velocity will be 45 m/min. 3 Chip thickness will be 1∙8 mm 3. GATE & PSUs) Page 5 of 78 . Wh t is What i the th velocity l it of f chip hi along l th tool the t l face? (a) 28. Chip velocity will be 80 m/min. Select l the h correct answer using the h code d given b below: l (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 If α is the rake angle of the cutting tool. shear angle 45° and cutting velocity 35 m/min.5 mm. ISRO‐2009 The rake angle of a cutting tool is 15 15°.3 25 3 m/min (d) 23. Then the chip velocity is (a) 2. chip thickness = 0. for turning a work piece of diameter 100 mm at the spindle speed of 480 RPM is (a) 1.0 m/s (d) 1. then the velocity l it of f chip hi sliding lidi along l th shear the h plane l i is given by (a) (c) V cos α cos(φ − α ) V cos α sin(φ − α ) IES‐2001 IES‐2003 An orthogonal cutting operation is being carried out under the following conditions: cutting tti speed d = 2 m/s.5 23 5 m/min IES‐2008 Consider the following statements: In an orthogonal cutting the cutting ratio is found to be 0∙75. tool φ is the shear angle and V is the cutting velocity.5 m/min (b) 27.3 m/min (c) 25.51 (c) 48 (d) 151 For-2014 (IES. mm. The cutting speed is 60 m/min and depth of cut 2∙4 mm. shear angle is the angle between (a) Shear plane and the cutting velocity (b) Shear plane and the rake plane (c) Shear plane and the vertical direction (d) Shear Sh plane l and d the h direction di i of f elongation l i of f crystals l in i the chip (b) (d) V sin φ cos (φ − α ) V sin α sin(φ − α ) IAS‐2002 IAS‐2000 IAS‐1998 The cutting velocity in m/sec.66 m/s IAS‐2003 In orthogonal cutting. Chip thickness will be 3∙2 mm. Which of the following are correct? 1 1. m/min 2.IES‐2003 The angle of inclination of the rake face with respect to the tool base measured in a plane perpendicular to the base and parallel to the width of the tool is called (a) Back rake angle (b) Side rake angle (c) Side cutting edge angle (d) ( ) End cutting g edge g angle g IES‐2004.6 mm.26 (b) 2. 4. is The shear plane angle (in degrees) is (a) ( ) 26. GATE & PSUs) Page 6 of 78 . The chip thickness is 0. (c) Small rake angle. using a cutting tool of rake angle 15o. .8 (d) ( ) 29. using a cutting tool of rake angle 15o. (b) High cutting speed. The chip thickness is 0. Which of the following are correct? C ti Continuous ribbon ibb lik like chip hi i is f formed d when h turning t i 1. the depth of cut is halved and the feed rate is double. A ductile d il material i l Select the correct answer using the code given below: ( ) 1 and (a) d 3 (b) 1 and d 4 (c) 2 and 3 (d) 2 and 4 IAS‐1997 Consider the following machining conditions: BUE will form in (a) Ductile material. (a) Both A and R are individually true and R is the correct explanation l i of fA (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 Ch‐1: Mechanics of Basic Machining Operation Q. ribbon type continuous chips which must be broken into small lengths which otherwise would be difficult to handle and may prove hazardous.8 (b) ( ) 27. the actual chip thickness will be (a) ( ) Doubled (b) ( ) halved (c) Quadrupled (d) Unchanged. Reason (R): High speed turning may produce long. excess metal is removed in the form of chip as in the case of turning on a lathe. (d) Small uncut chip thickness.4 mm and the uncut chip thickness hi k i 0. is The chip velocity (in m/min) is (a) ( )8 (b) ( ) 10 (c) ( ) 12 (d) ( ) 14 4 GATE‐1995 Plain milling of mild steel plate produces (a) Irregular egu a shaped s aped d discontinuous sco t uous c chips ps (b) Regular shaped discontinuous chip (c) Continuous chips without ithout built up edge (d) Joined chips IES 2007 During machining. No 11 12 13 14 15 16 17 18 19 20 Option p D D B C D B B D D B For-2014 (IES. If the chip thickness ratio is unaffected with the changed g cutting g conditions. GATE‐2002 A built is formed b ilt‐up‐edge d i f d while hil machining hi i (a) Ductile materials at high speed (b) Ductile materials at low speed p (c) Brittle materials at high speed (d) Brittle materials at low speed IES‐1997 Assertion (A): For high speed turning of cast iron pistons.2 mm. No 1 2 3 4 5 6 7 8 9 10 Option p C B D C B D B A B B Q. carbide tool bits are provided with chip breakers. At A a l lower cutting i speed d 3.8 7 (c) ( ) 28.4 mm and the uncut chip thickness hi k i 0. A brittle material 4.2 mm.8 9 G ) Common Data S‐2 GATE – 2009 ( (PI) An orthogonal turning operation is carried out at 20 m/min cutting speed.IAS‐1995 In an orthogonal cutting. G ) Common Data S‐1 GATE – 2009 ( (PI) An orthogonal turning operation is carried out at 20 m/min cutting speed. At a higher cutting speed 2. 24 mm/rev and the depth of cut is 2 mm. 8 15.6 (a) 6o (b) 32.48 0 48 mm. N acting at the chip tool interface.56 6 6 (c) 30. width idth of f cut t= 3 mm. coefficient of friction at tool‐chip interface = 0.0 mm depth of cut. shear strength of workpiece material = 460 N/mm2.25 mm. made Tangential component of the cutting force. yield shear stress of work material = 285 2. Mi i Minimum power requirement i t (in (i kW) at t a cutting tti speed d of 150 m/min is (a) 3.56 (a) 6 (b) 26. p g In an orthogonal cutting an HSS tool having the following tool signature in the orthogonal reference system (ORS) ( ) has h been b used: d 0‐10‐7‐7‐10‐75‐1. ) ESE‐2005 Conventional Mild steel machined l is i being b i hi d at a cutting i speed of 200 m/min with a tool rake angle of 10. For the above machining condition determine the values of (i) Friction force. F and normal force.α 2 = 0. mm If the orthogonal rake angle is zero and the principal cutting edge angle is 90 90°. depth of cut = 0. Pz = 1200 N Axial component of the cutting force.4 6 o t ‐2 GATE – 2007 (PI) C Common D Data g g test. (iii) Cutting power consumption in kW. .5 mm Friction angle during machining will be ( ) 22. interface (ii) Yield shears strength of the work material under this machining condition.7. shear strength of workpiece material = 460 N/mm2. the following g In an orthogonal machining observations were made Cutting force 1200 N Thrust force 500 N T l rake Tool k angle l zero Cutting speed 1 m/s Depth of cut 0. coefficient of friction at tool‐chip interface = 0.25 (c) 3. 0. p g In an orthogonal cutting an HSS tool having the following tool signature in the orthogonal reference system (ORS) ( ) has h been b used: d 0‐10‐7‐7‐10‐75‐1. Determine () (i) Cutting force (F ( c) (ii) Radial force (iii) Normal force (N) on tool and (iv) Shear force (Fs ).45 t ‐1 GATE – 2007 (PI) C Common D Data g g test. p y If the average g value of co‐efficient of friction between the tool and the chip is 0.35.2 m/s For-2014(d) 1.6 mm. tool. the shear angle is degree is ( ) 20.5 mm Chip speed along the tool rake face will be ( ) 0. Rake angle = 100. S ‐2003‐ Conventional C i l ESE . 0 (mm) ( ) at speed d of f 400 rpm. GATE & PSUs) GATE‐2007 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool the cutting velocity is 90 m/min. feed of 0.5 0 5 and the shear stress of the work material is 400 N/mm2.65. . Sh Shear plane l angle l (in (i degree) d ) for f minimum i i cutting tti force f is (a) 20. the following observation were made. The chip thickness obtained is 0.1o (c) (d) 67.7. uncut t chip hi thickness thi k = 0. By S K Mondal GATE ‐2008 (PI) Linked S‐1 g g experiment. 10.8 mm Chip thickness 1.8 8o ( ) 57.5 GATE ‐2008 (PI) Linked S‐2 g g experiment.83 (a) 8 m/s / (b) 0.8 mm Chip thickness 1.56 Page 7 of 78 During turning a carbon steel rod of 160 mm diameter by a carbide bid tool l of f geometry.32 mm/rev and 4. m/min The feed is 0.25 mm. Determine (i) shear angle and (ii) Cutting and thrust component of the force. chip thickness ratio = 0.35 (d) 3. 0. the following g In an orthogonal machining observations were made Cutting force 1200 N Thrust force 500 N T l rake Tool k angle l zero Cutting speed 1 m/s Depth of cut 0. 75.6 mm.56 (d) 36.ESE ‐2000 (Conventional) F Force & Power P in i M Metal t lC Cutting tti The the Th following f ll i data d t from f th orthogonal th l cutting tti test t t is available. Given width of cut = 3. mean friction N/ N/mm f i ti co‐efficient ffi i t on tool t l face f = 0. depth of cut = 0. 8.5 (c) 28.8 mm.15 (b) 3. Given width of cut = 3.5 (d) 32.51 mm.5 (b) 24.88 m/s (IES.2 mm respectively.53 m/s / (c) 1. force Px = 800 N Chip thickness (after cut). The width of cut and uncut thickness are 2 mm and 0. GATE – 1995 ‐Conventional While Whil turning t i a C‐15 steel t l rod d of f 160 6 mm diameter di t at t 315 rpm. d The Th side rake angle of the tool was a chosen that the machining hi i operation ti could ld be b approximated i t d to t be b orthogonal cutting. The coefficient of friction at the tool‐chip interface is ( ) 0. t l Th specific machining energy is 2. the following observations were made. / i 0. 90.2 (a) (b) 2 ( ) 200 (c) (d) 2000 Example When the rake angle is zero during orthogonal cutting.000 mm3/min (d) ( ) Can not be calculated with the g given data GATE‐2013 A steel bar 200 mm in diameter is turned at a feed of 0.0 J/mm3. cutting speed 0. IAS‐2003 Main Examination During D i turning t i process with ith 7 ‐ ‐ 6 – 6 – 8 – 30 – 1 (mm) ASA tool the undeformed chip thickness of 2. l it f feed d and d depth d th of f cut t are 120 m/min. 2. What is the rate of metal removal? (a) 1000 mm3/min (b) 60. The rotational speed of the workpiece is 160 rpm.48 mm Draw schematically the Merchant Merchant’s s circle diagram for the cutting force in the present case. The main cutting f force i in N i is (a) 40 (b) 80 (c) 400 (d) 800 C D Question Q i GATE‐2013 (PI) Common Data A disc of 200 mm outer and 80 mm inner diameter is faced of 0.8 mm/rev feed and 1. power the specific cutting energy in J/mm3 is ( ) 0. 80.5 1675 5 GATE‐2007 In turning The I orthogonal th l t i of f medium di carbon b steel. (a) GATE‐2006 Common Data Questions(1) In I an orthogonal th l machining hi i operation: ti Uncut thickness = 0.0 mm and d width idth of f cut t of f 2.46 (c) 0. The facing operation is undertaken at a constant cutting speed of 90 m/min in a CNC lathe.7 mm Thrust force = 200 N Cutting force = 1200 N A Assume M Merchant's h t' th theory.16 mm/rev / b a tool by t l of f geometry t 00.1 mm/rev with a depth of cut of 1 mm.23 (a) (b) 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0. The tangential cutting force and th thrust t force f were 1177 N and d 560 6 N respectively.25 mm/rev with a depth of cut of 4 mm. GATE & PSUs) Page 8 of 78 . The material removal rate in mm3/s is (a) 160 (b) 167. show that 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 T o (d) One Ans.5 mm depth of cut and feed of 0.5 mm were used.95 τs pc = (1 − μ r ) r 1+ r2 p c = specific power of cutting p thickness ratio r = chip μ = coefficient of friction in tool chip interface Where τs is the shear strengrh of the material For-2014 (IES. ti l Calculate: [30 marks] ( ) The (i) h side d rake k angle l (ii) Co‐efficient of friction at the rake face (iii) The dynamic shear strength of the work material IES ‐ 2004 A medium carbon steel workpiece is turned on a lathe at 50 m/min. 100.000 mm3/min (c) 20. The cutting velocity.6 167 6 (c) 1600 (d) 1675. Neglecting the contribution of the feed force towards cutting power.2 mm/rev and 2 mm respectively. Tangential component of the cutting force = 500 N Axial component p of the cutting g force = 200 N Chip thickness = 0.150. The main (tangential) cutting force is 200 N.5 mm depth of cut. 750.85 (d) 0. 0(mm). 65 GATE‐2003 Common Data Questions(1) A cylinder is turned lathe li d i t d on a l th with ith orthogonal th l machining principle. The orthogonal rake angle of the cutting tool in degree is (a) zero (b) 3. GATE & 402. chip thickness ratio ti = 0.3° and 4.4 mm.4 mm. Spindle rotates at 200 rpm. it has been observed that the friction force acting at the chip‐ tool interface is 402.20 mm/rev. 342N (d) 480N. The rate of heat generation (in W) at the primary shear plane is (a) 180. 381N (c) 440N. l are (a) 30. chip thickness ratio ti = 0. The feed velocity is negligibly small with respect to the cutting velocity.25 mm/rev. In the analysis it is found th t the that th shear h angle l is i 27. depth of cut of 4 mm and cutting velocity of 90 m/min.5.18 (b) 0. The feed velocity is negligibly small with respect to the cutting velocity.7 mm Thrust force = 200 N Cutting force = 1200 N A Assume M Merchant's h t' th theory. y.98 GATE‐2008 Common Data Question (1) Orthogonal O th l turning t i is i performed f d on a cylindrical li d i l work k piece with shear strength of 250 MPa.528 mm (c) 0.5 (b) 200. The axial i l feed f d rate t is i 0.2° and 1.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0.71 (d) 0. The axial i l feed f d rate t is i 0. The following conditions diti are used: d cutting tti velocity l it is i 180 8 m/min. The Th orthogonal th l rake k angle l is i 7o.5 N and the friction force is also perpendicular to the cutting velocity vector. 387N (b) 565N. The rake angle is 10°.5.5 PSUs) GATE ‐2010 (PI) Linked S‐2 g g of an engineering g g alloy. The uncut chip thickness is 0.97 (d) 40. / i feed f d is 0.5 mm Cutting speed = 20 m/min Rake angle = 15° Width of cut = 5 mm Chip thickness = 0. The shear plane angle (in ( degree) ) and the shear force respectively are (a) 52: 320 N (b) 52: 400N (c) 28: 400N (d) 28:320N GATE‐2008 Common Data Question (2) Orthogonal O th l turning t i is i performed f d on a cylindrical li d i l work k piece with shear strength of 250 MPa.98 (b) 30.4 mm.75° ° In the above problem. it has In orthogonal turning been observed that the friction force acting at the chip‐ tool interface is 402. interface The main (tangential) cutting force is 1500 N.7 mm Thrust force = 200 N Cutting force = 1200 N A Assume M Merchant's h t' th theory. Apply A l Merchant's theory for analysis.23 (c) 40.58 (c) 5 (d) 7. r Rees e ctively.25 mm per revolution.5 (d) 402.5 (c) 302. it is observed that the main (tangential)cutting force is perpendicular to friction force acting at the chip‐tool interface. Assume that the energy expended during machining is completely converted to heat.2° and 2.75° ° The thickness of the produced chip is (a) 0.20 mm/rev. depth of cut is 3 mm.0 (b) 240.5 Page 9 of 78 Linked Answer Questions GATE‐2013 S‐1 In orthogonal turning of a bar of 100 mm diameter with a feed of 0.0 (c) 360. l ti D th of Depth f cut t is i 0.36 (c) 0.511 mm (b) 0. The ratio of friction force to normal force associated with the chip‐tool interface is 1. The cutting and Thrust forces. In the analysis it is found th t the that th shear h angle l is i 27. The shear force (in N) acting along the primary shear plane is (a) 180.4 04 mm.GATE‐2006 Common Data Questions(2) In I an orthogonal th l machining hi i operation: ti Uncut thickness = 0. The following conditions diti are used: d cutting tti velocity l it is i 180 8 m/min.5 (d) For-2014 (IES.2 0 2 mm and the chip thickness is 0. depth of cut is 3 mm. The cutting velocity is 2 m/s. The rake angle is 10°. The uncut chip thickness is 0.25 mm per revolution.5 N and the friction force is also perpendicular to the cutting velocity vector. 356N GATE ‐2010 (PI) Linked S‐1 In orthogonal turning of an engineering alloy. The Th orthogonal th l rake k angle l is i 7o.16 . a (a) 568N.846 mm GATE‐2003 Common Data Questions(2) A cylinder is turned lathe li d i t d on a l th with ith orthogonal th l machining principle. / i feed f d is 0.3° and 1. respectively. The cutting velocity is 2 m/s. l ti D th of Depth f cut t is i 0.2 mm and the chip thickness is 0. l it The Th ratio ti of f friction f i ti f force t normal to l force f associated with the chip‐tool interface is 1. Apply A l Merchant's theory for analysis. the coefficient of friction at the chip tool interface obtained using Earnest and Merchant theory is (a) 0. Spindle rotates at 200 rpm.818 mm (d) 0. The values of shear angle and shear strain. The percentage of total energy dissipated due to f friction at the h tool l‐chip h interface f is (a) 30% (b) 42% (c) 58% (d) 70% GATE‐2006 Common Data Questions(3) In I an orthogonal th l machining hi i operation: ti Uncut thickness = 0. 5 39 5° (b) 42. depth of cut of 4 mm and cutting velocity of 90 m/min. the thrust force = 600 N and chip shear angle is 30o. Using Merchant's Merchant s shear angle relationship.75 0 75 mm Width of cut = 2.4 N (d) 69.21 150 21 N (b) 18.25 mm Chip thickness = 0.157o . the following data is obtained: Uncut chip thickness = 0.25 mm Chip thickness = 0. The normal force acting at the chip‐tool interface in N is GATE – 2011 (PI) Linked S1 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle. it was observed that the shear angle was 20°.Linked Answer Questions GATE‐2013 S‐2 In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev.66 N GATE – 2011 (PI) Linked S2 During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle. the following data is obtained: Uncut chip thickness = 0.6 969 6 N (c) 479. GATE & PSUs) Page 10 of 78 .25 42 25° (c) 47. respectively.4 1079 4 N (b) 969.γ = friction angle andα = rake angle) GATE‐1997 In using I a typical i l metal l cutting i operation. it is observed that the main (tangential)cutting force is perpendicular to friction force acting at the chip‐tool interface.04 751 04 N (c) 9. are (a) 71. the value of shear angle will be (a) 39. i i a cutting tool of positive rake angle = 10°. an increase in cutting speed causes (a) An increase in longitudinal cutting force (b) An increase in radial cutting force (c) An increase in tangential cutting force (d) ( ) Cutting g forces to remain unaffected IES 2010 The relationship between the shear angle Φ.435 18 435o .218o. 686.5°. the h cutting i force f 900 N. 751.5 mm N Normal l force f = 950 N Thrust force = 475 N The shear angle and shear force. interface The main (tangential) cutting force is 1500 N.75° (d) 50.75 0 75 mm Width of cut = 2.565 71 565o. the friction angle β and cutting rake angle α is given as IES‐2005 Which is Whi h one of f the h following f ll i i the h correct expression for the Merchant's machinability constant? (a) 2φ + γ − α (b) 2φ − γ + α (c) 2φ − γ − α (d) φ + γ − α (Where φ = shear angle. 150. 861. The friction angle g is (a) 45° (b) 30° ( ) 60° (c) ( ) 40° (d) S – 1999 999 IAS In an orthogonal cutting process.5 mm N Normal l force f = 950 N Thrust force = 475 N The ultimate shear stress (in N/mm2) of the work material is (a) 235 (b) 139 (c) 564 (d) 380 (a) 1000 (b) 1500 (c) 20oo (d) 2500 IES ‐ 2012 During orthogonal cutting. Then the chip shear force is (a) 1079.64 N (d) 23.6 N For-2014 (IES. rake angle of the tool is 20° and friction angle is 25.5° IES‐2003 In = I orthogonal h l cutting i test. 2. the cutting force and thrust force were observed to be 1000N and 500 N respectively.6 to 0. the I a machining hi i h percentage of f heat carried away by the chips is typically (a) 5% (b) 25% (c) 50% (d) 75% IES‐1998 In operation. Force exerted by the tool on the chip acting normally to the tool face. Vertical V i l force f which hi h helps h l i holding in h ldi the h tool l in i position. ( ) 1 and (a) d3 (b) 2 and d4 (c) 1 and 4 (d) 2 and 3 GATE‐2007 In carbon I orthogonal th l turning t i of f low l b steel t l pipe i with ith principal cutting edge angle of 90°.5 to 0.4 0 4 to 0. 3. the coefficient of friction in chip‐tool interface will be IES‐1996 Which of the following forces are measured directly by strain gauges or force dynamometers during metal g? cutting 1. GATE & PSUs) Page 11 of 78 . Merchant’s th th ratio the ti of f friction f i ti force to normal force acting on the cutting tool is ( ) 1. 4.4 times the main cutting force (b) 0. E Employing l i M h t’ theory. in that order is (a) 80: 10: 10 (b) 33: 33: 33 (c) 20: 60: 10 (d) 10: 10: 80 S – 2003 IAS As the cutting speed increases (a) More heat is transmitted to the work piece and less heat is transmitted to the tool (b) More heat is carried away by the chip and less heat is t transmitted itt d to t the th tool t l (c) More heat is transmitted to both the chip and the tool (d) ( ) More heat is transmitted to both the work p piece and the tool For-2014 (IES. Horizontal cutting force exerted by the tool on the work piece. tool and work.IES‐2000 In an orthogonal cutting test. Feed force 2 Thrust force 2. 2 (d) 3. 2.64 (a) 2 1 ( b) 2 ( c) 1 ( d) 2 2 IES‐1997 Consider the forces acting on a C id h following f ll i f i finish turning tool: 1. the I metal l cutting i i h approximate i ratio of heat distributed among chip. N The Th shear h angle is 25° and orthogonal rake angle is zero. If the rake angle of tool is zero. 3 (b) 2. 1 IES‐1999 The in single Th radial di l force f i i l ‐point i tool l during d i turning operation varies between (a) 0.6 0 6 times the main cutting force (c) 0.56 (a) ( ) 1.6 times the main cutting force IES‐1995 The tool used Th primary i l force f d in i calculating l l i the total power consumption in machining is the (a) Radial force (b) Tangential force (c) Axial force (d) Frictional force. 3. 2. the main cutting f force i 1000 N and is d the th feed f d force f i 800 is 8 N. 3.25 (b) (c) 0. Frictional resistance of the tool against the chip flow acting along the tool face.80 (d) 0. 1.8 times the main cutting g force (d) 0. IES‐2002 In process.2 to 0. Cutting g force. The correct sequence of the decreasing order of the magnitudes of these forces is (a) 1. 1 (c) 3. single‐point metal cutting. IES 2010 Consider the following statements: In an orthogonal.IES‐2001 Power consumption in is P i i metal l cutting i i mainly due to (a) Tangential component of the force (b) Longitudinal component of the force (c) Normal component p of the force (d) Friction at the metal‐tool interface S – 1995 99 IAS Thrust force will increase with the increase in (a) Side cutting edge angle (b) Tool T l nose radius di (c) Rake angle (d) End cutting edge angle. interface (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A ( ) A is (c) i true but b R is i false f l (d) A is false but R is true For-2014 (IES. 3. (a) Both A and R are individually true and R is the correct explanation of A (b) ( ) Both A and R are individually y true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 2003 IAS The heat generated in metal conveniently be determined by (a) Installing thermocouple on the job (b) Installing thermocouple on the tool (c) Calorimetric set‐up (d) ( ) Using g radiation py pyrometer cutting can IES‐1998 The of pick Th gauge factor f f a resistive i i i k‐up of f cutting force dynamometer is defined as the ratio of (a) Applied strain to the resistance of the wire (b) The proportional change in resistance to the applied strain (c) The resistance to the applied strain (d) Change in resistance to the applied strain Page 12 of 78 IES‐2000 Assertion (A): the A i (A) In I metal l cutting. Reason (R): In electric transducers there is a significant leakage of signal from one axis to the other. Which of these statements are correct? (a) 1 and 3 only (b) 1 and 2 only ( ) 2 and 3 only (c) ( ) 1. Reason (R): Very high temperature is produced at the tool‐chip interface. The tangential g force increases. The longitudinal force drops. as the th side id ‐cutting tti edge d angle l is i increased. GATE & PSUs) . i h normal l laws of sliding friction are not applicable. 2 and 3 (d) IES‐1993 A 'Dynamometer' used the 'D ' is i a device d i d for f h measurement of (a) Chip thickness ratio (b) Forces during metal cutting (c) Wear of the cutting g tool (d) Deflection of the cutting tool IES 2011 The instrument or device used to measure the cutting forces in machining is : ( ) Tachometer (a) T h t (b) Comparator (c) Dynamometer (d) Lactometer S‐2001 200 IAS ) Piezoelectric transducers and preferred p Assertion ( (A): over strain gauge transducers in the dynamometers for measurement of three‐dimensional cutting forces. The Th radial di l force f i increases. such cross error is negligible in the case of piezoelectric transducers. 2. i d 1. 50 (d) 50. front relief face.8 mm. l [ 2 –marks] IES 2010 Flank wear occurs on the (a) Relief face of the tool (b) Rake face (c) Nose of the tool (d) Cutting edge S – 2007 200 IES Flank wear occurs mainly on which of the following? (a) Nose part and top face (b) Cutting edge only (c) Nose part. and side relief face of the cutting tool (d) Face of the cutting tool at a short distance from g edge g the cutting For-2014 (IES.00 (b) 51. State the factors responsible for wear on a t turning i t tool.67 (c) 51. GATE‐2008 (PI) Tool Wear. GATE & PSUs) Page 13 of 78 . Tool Life & Machinability [ 4 – marks] k ] B S K Mondal By M d l For a critical wear land of 1. the wear land (h) has been plotted against machining time (T) as given in i hi i i i i the h following f ll i figure. During machining. the cutting tool life (in minute) is (a) 52.00 IES 2009 Conventional Show Sh crater t wear and d flank fl k wear on a single i l point i t cutting tool.GATE 1992 The of angle h effect ff f rake k angle l on the h mean friction f l in machining can be explained by (A) sliding (Coulomb) model of friction (B) sticking and then sliding model of friction (C) sticking friction (D) ( ) Sliding g and then sticking g model of friction IES‐2004 Assertion (A): The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed as cutting g ratio in orthogonal g cutting g Reason (R): The frictional force is very high due to the occurrence of sticking g friction rather than sliding g friction (a) ( ) Both A and R are individually y true and R is the correct explanation of A (b) ( ) Both A and R are individually y true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true GATE‐1993 The Th effect ff t of f rake k angle l on the th mean friction f i ti angle l in i machining can be explained by ( ) Sliding (coulomb) (a) ( ) model of friction (b) sticking g and then siding g model of friction (c) Sticking friction (d) sliding and then sticking model of friction IAS – 2009 Main Explain ‘sudden‐death mechanism’ of tool failure. Increase in feed of cut at constant cutting 4 g force Select the correct answer using the code given below: (a) 1. Abrasive wears 1. 2 and d 4 For-2014 (IES. 2. Galvanic action B. Reason (R): Measurement of flank wear is simple and more accurate. Temperature of the tool increases drastically Which of the statements given above are correct? (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) 2 and 3 S – 2002 IES Crater wear on tools always starts at some distance from the tool tip because at that point (a) Cutting fluid does not penetrate (b) Normal stress on rake face is maximum (c) Temperature is maximum (d) ( ) Tool strength g is minimum S – 2007 200 IAS Why does crater wear start at some distance from the tool tip? (a) Tool strength is minimum at that region (b) Cutting fluid cannot penetrate that region (c) Tool temperature is maximum in that region (d) ( ) Stress on rake face is maximum at that region g S – 2000 IES Crater wear starts at some distance from the tool tip because (a) Cutting fluid cannot penetrate that region (b) Stress on rake face is maximum at that region (c) Tool strength is minimum at that region (d) ( ) Tool temperature p is maximum at that region g S – 1996 996 IES Notch wear at the outside edge of the depth of cut is due to (a) Abrasive action of the work hardened chip material (b) Oxidation (c) Slip‐stick action of the chip (d) ( ) Chipping. rapid deterioration of tool edge takes place because 1. 3 and (c) d 4 (d) 1.S – 2004 200 IES Consider the following statements: During the third stage of tool‐wear. Plastic deformation 5 5. Chipping of tool edge due to mechanical impact 3. GATE & PSUs) Page 14 of 78 . Diffusion wears 4. Flank wear is only marginal 2. 3 Molecular transfer D. Temperature 3 p of the tool increases g gradually y 4. Gradual wears at tool point 4. Ploughing action C. 2 and 3 (b) 2. Electrolytic y wear 3. Adhesive wears 2. ( ) 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 S – 2008 IES What are the reasons for reduction of tool life in a machining operation? 1 Temperature rise of cutting edge 1. Flank wear is large 3. Metallic bond Code: A B C D A B C D (a) 2 5 1 3 (b) 5 2 1 3 (c) 2 1 3 4 (d) 5 2 3 4 S – 1995 99 IES Crater wear is predominant in (a) Carbon steel tools (b) Tungsten T t carbide bid tools t l (c) High speed steel tools (d) Ceramic tools S – 1994 99 IES Assertion (A): Tool wear is expressed in terms of flank wear rather than crater wear. 3 and 4 ( ) 1. pp g S – 1995 99 IES Match List I with List II and select the correct answer using the codes given below the lists: List I (Wear type) List II (Associated mechanism) A. O id ti Oxidation Which of the above are the causes of tool wear? (a) 2 and 3 (b) 1 and 2 (c) 1. Coolant (a) ( ) 1. High speed steel and ceramic tools. 2. [2 – Marks] Ans. Hot hardness of the tool material. cast alloy and 3. IES‐1996 Chip equivalent is increased by (a) An increases in side‐cutting edge angle of tool (b) An increase in nose radius and side cutting edge angle of tool (c) Increasing the plant area of cut (d) Increasing the depth of cut. determine the shape of the curve between velocity of cutting and life of the tool. 2 and 4 (d) 1 and 3 S – 1999 999 IAS The type of wear that occurs due to the cutting action of the particles in the cutting fluid is referred to as (a) Attritions wear (b) Diffusion Diff i wear (c) Erosive wear (d) Corrosive wear S – 2003 IAS Consider the following statements: Chipping of a cutting tool is due to 1. 2. and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only IFS 2009 With the help of Taylor’s tool life equation. 3. S – 1992 992 IES Tool life is generally specified by (a) Number of pieces machined (b) Volume V l of f metal t l removed d (c) Actual cutting time (d) Any of the above G 200 GATE ‐2004 In a machining operation. 1. Plastic Pl ti deformation d f ti 4. the constants n Taylor’s and C depend upon 1 Work piece material 1. ceramic tools. operation doubling the 1 cutting speed reduces the tool life to 8 th of th original the i i l value. is (a ) 1 8 (b) 1 4 (c ) 1 3 (d ) 1 2 For-2014 (IES. material [10‐Marks] [ ] S 2010 20 0 C i l IES Conventional Draw tool life curves for cast alloy. Which of these statements are correct? (a) 1. Mechanical abrasion 2. 2 and 3 (b) 1 and 3 ( ) 2 and (c) d3 (d) 1 and d2 IES ‐ 2012 In Taylor s tool life equation VTn = C. GATE & PSUs) Page 15 of 78 . Diffusion 3. High speed steel 2. l Th exponent The t n in i Taylor's T l ' n tool life equation VT = C.S – 2002 IAS Consider the following actions: 1. Assume an HSS tool and steel as work material. Tool material 3. High positive rake angle of the tool. Tool T l material t i l being b i too t brittle b ittl 2. 25. C i 3.S – 2000 IES In a tool life test.25 D.521 IES‐2013 A carbide tool(having n = 0.11 0 11 (d) 0. 0.55 (c) 0. the tool life obtained is 10 min at a cutting speed of 100 m/min.08 0 08 to 0.20 to 0.9 ( b ) 3 1 ( c ) 4 1 ( d ) 8 1 S – 2006 IES Which of the following values of index n is associated with carbide tools when Taylor's tool life equation. l f the h value l of f Taylorian l Exponent is (a) 0.262 (b) 0. doubling the cutting speed reduces the tool life to 1/8th of the original.03 0 03 to 0. If the cutting speed is halved.75 (d) 0.25) 0 25) with a mild steel work‐piece was found to give life of 1 hour 21 minutes while cutting at 60 m/min.13 0 13 (c) 0. Taylor's What is the value of n for ceramic tools? ( ) 0. The Taylor's Taylor s tool life index is ( a ) 2 1 S – 1999 999 IES In a single‐point turning operation of steel with a cemented carbide tool.15 to (a) t 0.70 S – 1998 998 IAS g tool material) ) with List ‐ II Match List ‐ I ( (Cutting (Typical value of tool life exponent 'n' in the Taylor's equation V. 0.25 (b) 0.6 to 0.20 0 20 (c) 0. Assuming straight line relationship between cutting speed d and d tool l life. Taylor's tool life exponent is 0.08 0 08 (b) 0. The value of C in Taylor’s tool life equation would be equal to: (a) 200 (b) 180 (c) 150 (d) 100 ISRO‐2011 A 50 mm diameter steel d l rod d was turned d at 284 rpm and d tool failure occurred in 10 minutes.4 to t 0. the tool life is 30 min The value of index (n) in the Taylor min. 0.48 (d) 0.18 8 B. while at 75 m/min cutting speed. V.12 C Ceramic C.8 to 0.423 (d) 0. GATE & PSUs) Page 16 of 78 .Tn = constant is applied? (a) 0∙1 to 0∙15 (b) 0∙2 to 0∙4 ( ) 0.48 to 0. Sintered carbide 4. the tool life will increase by (a) Two times (b) Four times (c) Eight times (d) Sixteen times S – 2008 IES In Taylor s tool life equation is VTn = constant. Cast alloy 2.5 Codes: d A B C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 ( ) 2 (c) 1 4 3 ( ) 1 (d) 2 4 3 ( ) GATE ‐2009 (PI) In an orthogonal machining operation. 0.Tn = C) and select the correct answer using th codes the d given i b below l the th lists: li t List – I List – II A HSS A. Taylor’s s tool life equation (a) 0.21 0 21 (b) 0.323 (c) 0.23 0 23 For-2014 (IES.45 to (c) t 0∙6 (d) 0∙65 6 to t 0∙9 S – 1999 999 IES The approximately variation of the tool life exponent 'n' of cemented carbide tools is (a) 0. minutes The speed was changed to 232 rpm and the tool failed in 60 minutes. 1. 67 6 4 45. Cutting speed Select the correct answer using the codes given below (a) 1.5 42 5 (c) 80. min 2. Nose radius 2.77 9. 2. s tool life exponent (n) is A Taylor Taylor’s 0.94 3. Depth of cut 3. F d Feed The correct sequence of these elements in DECREASING order of their influence on tool life is (a) ( ) 2. Feed 3. The tool life equations are: Carbide tool: VT 1. GATE & PSUs) Page 17 of 78 . Depth D th of f cut t 4.9 142 9 GATE‐2013 Two cutting tools are being compared for a machining operation.76 6 42 42.25 mm/rev and 1 mm depth of cut.6 = 3000 HSS tool: VT 0. Match the graphs for HSS.7 80 7 (d) 142. Depth of cut Select the correct answer using the code given below: (a) 1‐ 2‐ 3 (b) 2‐ 3‐ 1 (c) 3‐ 2‐ 1 (d) 1‐ 3‐ 2 IES 2010 Tool life is affected mainly with (a) Feed (b) Depth of cut (c) Coolant (d) Cutting speed S – 1997 99 IES Consider the following elements: 1. 2 1 (c) 2.3 and B. Feed d rate 2.3 (d) 60. 1 (c) 2.58 8 Tool life. The carbide tool will p provide higher g tool life if the cutting speed in m/min exceeds (a) 15. 3 ISRO‐2012 What is the correct sequence of the following parameters t i in order d of f their th i maximum i t to minimum influence on tool life? 1. without any constraints. Carbide and Ceramic tool materials i l and d select l the h correct answer using the code given below the lists: Code: HSS Carbide Ceramic (a) 1 2 3 (b) 3 2 1 (c) 1 3 2 (d) 3 1 2 G 20 0 GATE ‐2010 For tool A. 1. 3 (d) 4. 6 The Th cutting tti speed d (in (i m/min) above which tool A will have a higher tool life than tool B is (a) 26. 3 1 (d) 3. Speed 2. 1 (b) ( ) 4 4. Cutting speed 3. what is the right sequence to adjust the cutting parameters? 1. 3 2. 2.27 Determine the constants of the Taylor tool life equation VTn = C G GATE ‐2003 A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.87 28. 4 4.45 and constant (K) is 90. 1 2 For-2014 (IES. I.25 mm/rev and depth of cut of 1 mm.7 26 7 (b) 42. Similarly for tool B n = 0.0 (b) 39.4 (c) 49. 3 3.23 23 48 48.4. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0. cut How many components can be produced with one cutting tool at 60 rpm? (a) 29 (b) 31 (c) 37 (d) 42 S – 1994. 2 3 (b) 3. d m/min:49. d K = 60. 1 2. 3 1. 99 200 IES 2007 For increasing the material removal rate in turning.0 Example p The following data was obtained from the tool‐life cutting test: Cutting Speed.74 49 4 49 49.6 = 200 Wh Where V is i the h cutting i speed d in i m/min / i and d T is i the h tool life in min.90 4. 2 3.IES 2010 The above figure shows a typical relationship between tool life and cutting g speed p for different materials. 3 3. S – 2002 IAS Using the Taylor equation VTn = c. the value Taylor's of n = 0. 2.5. S – 2003 IAS The tool life curves for two tools A and B are shown in the figure and they follow the tool life equation VTn = C. 4. 3. Increasing cutting velocity 3. 2 and d 3 S – 1997 99 IAS In the Taylor s tool life equation. Increasing back rake angle up to certain value Which of these statements are correct? (a) 1 and 3 (b) 1 and 2 ( ) 2 and (c) d 3 (d) 1.S – 1992 992 IES Tool life is generally better when (a) Grain size of the metal is large (b) Grain G i size i of f the th metal t l is i small ll (c) Hard constituents are present in the microstructure of the tool material (d) ( ) None of the above 1. Value a ue of o C for o tool too B will be greater g eate than t a that t at for o the t e tool too A. If the tool life is reduced to 45 minutes. . calculate the percentage increase in tool life when the cutting speed is reduced by 50% (n = 0∙5 and c = 400) (a) 300% (b) 400% ( ) 100% (c) % (d) 50% % Which of these statements is/are correct? (a) 1 and 3 (b) 1 and 4 (c) 2 only (d) 4 only S – 2002 IAS Optimum cutting speed for minimum cost (Vc min i ) and optimum cutting speed for maximum production rate (Vr max ) have which one of the following relationships? (a) Vc min = Vr max (b) Vc min > Vr max (c) Vc min < Vr max (d) V2c min = Vr max IES 2010 With increasing cutting velocity. if the cutting speed is halved. Built B ilt ‐ up edge d f formation ti 2. GATE & PSUs) Page 18 of 78 . The tool has a life of 180 minutes at a cutting speed of 18 m/min. Consider the following g statements: Value of n for both the tools is same.25. Value of C for both the tools is same. then the tool life will become (a) Half (b) Two times (c) Eight times (d) ( ) Sixteen times. the total time for machining a component ( ) Decreases (a) D (b) ( ) Increases (c) Remains unaffected (d) First Fi decreases d and d then h increases i S – 2000 IAS Consider the following statements: The tool life is increased by 1. VTn = C. then the cutting speed will be (a) 9 m/min (b) 18 m/min (c) 36 m/min (d) 72 m/min S – 1996 996 IAS The tool life increases with the (a) Increase in side cutting edge angle (b) Decrease D i in side id rake k angle l (c) Decrease in nose radius (d) Decrease in back rake angle S – 1995 99 IAS In a single point turning operation with a cemented carbide and steel combination having a Taylor exponent of 0. Value of C for tool A will be greater than that for the tool B. For-2014 (IES. Find the suitable speed in RPM for turning 300 mm diameter so that tool life is 3 30 min. 00 n = 0. Reason(R): Increases in depth of cut gives rise to relatively small increase in tool temperature. IAS – 2011 Main Determine the optimum speed for achieving maximum production rate in a machining operation. tool life was found to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 What is the p percentage g increase in tool life when the cutting speed is halved? (a) 50% (b) 200% (c) 300% (d) 400% G 999 GATE ‐1999 What is approximate percentage change is the life. v = 40 m/min. The data is as follows : Machining time/job = 6 min.5 and k=1. α. Ta lor's Taylor s equation constants C = 100.0 min i if the cutting speed is doubled. d = depth of cut. is changed from 10 to 7 ? (Hint: Flank wear rate is proportional to cot α (a) 30 % increase (b) 30%. t. A 60 min tool life was obtained using the following cutting condition VT0.5 0 Job handling time = 4 min. of a tool with zero rake angle used i orthogonal in th l cutting tti when h it clearance its l o o angle.0 mm.S – 1995 99 IAS Assertion (A): An increase in depth of cut shortens the tool life.5 and k = 540 (b) n n= 1 and k k=4860 4860 (c) n = ‐1 and k = 0. b d When Wh the th cutting tti speed d was increased i d to 100 m/min.0 The constants in the tool life equation are 60 and 0.0. v = speed. Calculate the effect on tool life if speed. f = 0.2 S ‐2001 200 C i l ESE Conventional In a certain machining operation with a cutting speed of 50 m/min. 999 IAS S ‐2010 20 0 C i l ESE Conventional The following equation for tool life was obtained for HSS tool.25 mm.74 (d) n‐0.50 per min Depreciation of tool regrinds Rs./job Tool l changing h i time i = 9 min. GATE & PSUs) Page 19 of 78 . i [10‐Marks] GATE‐2009 Linked Answer Questions (1) In a machining experiment. ( ) 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 S – 2006 conventional i l IES An HSS tool is used for turning operation operation.15 GATE‐2009 Linked Answer Questions (2) In a machining experiment. tool life of 45 minutes was observed. Estimate the cutting speed for maximum productivity p y if tool change g time is 2 minutes.6d0.3= C. S 2009 Conventional C i l IES Determine the optimum cutting speed for an operation on a Lathe machine using the following information: Tool change time: 3 min T l regrinds Tool i d time: ti 3 min i Machine running cost Rs. The tool life is 1 hr. min Tool life = 90 min.13f0. feed and depth of cut are together increased by 25% and also if they are increased individually by 25%. the tool life decreased to 10 min. where f = feed. 5. when turning is carried at 30 m/min. tool life was found to vary with the cutting speed in the following manner: Cutting speed (m/min) Tool life (minutes) 60 81 90 36 The exponent p (n) ( ) and constant (k) ( ) of the Taylor's y tool life equation are (a) n = 0. / i The Th tool t l life lif will ill be b reduced d d to t 2. S ‐1999. d = 2. 30% decrease (c) 70% increase (d) 70% decrease For-2014 (IES. 2 (c) 3. Curve‐2 B. then after an optimum cutting speed it increases 2. Higher feed rate for the same cutting speed reduces cost of production 4. which one of the following costs remains constant? (a) Machining cost per piece (b) Tool changing cost per piece (c) Tool handling cost per piece (d) ( ) Tool cost p per piece p For-2014 (IES. 2. Minimum cost criterion The correct sequence q in ascending g order of optimum p cutting speed obtained by these approaches is (a) 1. High Hi h metal t l removal l rate t 2. Maximum profit criterion 3. 4 Curve‐4 5. From previous slide Code:A (a) 3 (c) 3 B 2 1 C 4 4 D 5 2 (b) (d) A 4 4 B 1 2 C 3 3 D 2 5 S – 1998 998 IES The variable cost and production rate of a machining process against cutting speed are shown in the given figure. As the cutting speed increases. D.G 200 GATE ‐2005 S – 2007 200 Contd… C d IAS g g economics with A diagram related to machining various cost components is given above. 2. As the cutting speed increases the cost of production also l increases i and d after f a critical i i l value l it i reduces d 3. 2 and 3 (b) 1 and 2 only (c) 2 and 3 only ( ) 3 only (d) S – 2000 IES The magnitude of the cutting speed for maximum profit rate must be (a) In between the speeds for minimum cost and maximum production rate (b) Higher Hi h than th the th speed d for f maximum i production d ti rate t (c) Below the speed for minimum cost (d) Equal to the speed for minimum cost S – 2004 200 IES g statements: Consider the following 1. Tool grinding cost 3. Curve‐l 2. GATE & PSUs) Page 20 of 78 . 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 S – 2002 IES In economics of machining. 3 (b) 1. the range of best cutting speed would be between (a) 1 and 3 (b) 1 and 5 (c) 2 and 4 (d) ( ) 3 and 5 S – 1999 999 IES Consider the following approaches normally applied for the economic analysis of machining: 1 Maximum production rate 1. Curve‐3 D Non‐productive cost 4. Machining cost 1. 2. 3. 1. High cutting tool life 3. Match List I (Cost Element) with List II (Appropriate Curve) and select the correct answer using the code given below the Lists: List I List II (Cost Element) (Appropriate Curve) A. the cost of production initially reduces. 2 IES 2011 The optimum cutting speed is one which should have: 1. Curve‐5 Contd Contd………. Balance the metal removal rate and cutting tool life (a) 1. 1 (d) 3. Tool cost C. For efficient machining. lead and cobalt (c) Aluminium. added to steel.1 to 0. the ratio of the optimum cutting speed for minimum cost and optimum cutting speed for maximum rate of production is always (a) Equal to 1 (b) In I th the range of f 0.6 (d) Greater than 1 IES ‐ 2012 The usual method of defining machinability of a material is by an index based on (a) Hardness of work material (b) Production rate of machined parts (c) Surface finish of machined surfaces (d) ( ) Tool life ti l IES 2011 C Conventional g elements on the Discuss the effects of the following machinability of steels: (i) Aluminium and silicon (ii) Sulphur and Selenium (iii) Lead L d and d Tin Ti (iv) Carbon and Manganese (v) Molybdenum and Vanadium S – 1992 992 IES Ease of machining is primarily judged by (a) Life of cutting tool between sharpening (b) Rigidity Ri idit of f work k ‐piece i (c) Microstructure of tool material (d) Shape and dimensions of work S – 2007. physical (a) h l and d mechanical h l properties and composition of workpiece material. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Page 21 of 78 S – 2009 IES The elements which. Cutting C tti f forces 3. (b) Cutting forces (c) ( ) Type yp of chip p (d) Tool life S – 2003 IES ) The machinability y of steels improves p Assertion ( (A): by adding sulphur to obtain so called 'Free Machining Steels‘. GATE & PSUs) .6 6 t to 1 (c) In the range of 0. Surface finish Which of the above is/are the machinability criterion/criteria? (a) 1. titanium and copper pp For-2014 (IES. Reason (R): Sulphur in steel forms manganese sulphide inclusion which helps to produce thin ribbon like continuous chip. Tool life 2. Sulphur lead and phosphorous (b) Sulphur. production (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1997 99 IAS In turning. 200 2009 IES Consider the following: 1.S – 2007 200 IAS Assertion (A): The optimum cutting speed for the minimum cost of machining may not maximize the profit. Reason (R): The profit also depends on rate of production. lead and copper (d) ( ) Aluminium. 2 and 3 (b) 1 and 3 only (c) 2 and 3 only (d) 2 only [5 Marks] ISRO‐2007 Machinablity depends on ( ) Microstructure. help in chip formation during machining are (a) Sulphur. 4 2 1 1.4.3.2. 1. Power consumption In modern high speed CNC machining with coated carbide tools. The correct sequence of these materials in terms of increasing order of difficulty in machining is (a) 4. Mild steel 3. Better surface finish. 4 S – 1996 996 IES Which of the following machinability? 1 Smaller shear angle 1. 2.3 3 (c) 2. the correct sequence of these criteria in DECREASING order of their importance is (a) ( ) 1.1 (d) 2. 4 (a) 1 and 3 (b) 2 and 4 (c) 1 and 2 (d) 3 and 4 indicate better S – 1996 996 IES Small amounts of which one of the following elements/pairs of elements is added to steel to increase its machinability? (a) Nickel (b) Sulphur and phosphorus ( ) Silicon (c) Sili (d) Manganese M and d copper S – 1995 99 IES In low carbon steels. Grey cast iron. 3 (b) ( ) 2. presence of small quantities sulphur improves (a) Weldability (b) Formability (c) Machinability (d) Hardenability S – 1992 992 IES Machining of titanium is difficult due to (a) High thermal conductivity of titanium (b) Chemical Ch i l reaction ti b between t tool t l and d work k (c) Low tool‐chip contact area (d) None of the above S – 1996 996 IAS Assertion (A): The machinability of a material can be measured as an absolute quantity. 3.2. GATE & PSUs) Page 22 of 78 . 2. 3. 4 (d) 2. 1.1 4 2 3 1 (b) 4. 1. 1. Higher cutting forces 3. Stainless steel 4. 2 Type of chips 3. Reason (R): Machinability index indicates the case with which a material can be machined ( ) 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 IES‐1995 Consider the f following C id th ll i work k materials: t i l 1.3. Tool life 4. 4. 4 4.S – 1998 998 IES Consider the following criteria in evaluating machinability: 1 Surface finish 2. 2. G GATE ‐2009 Friction at the tool‐chip interface can be reduced by ( ) decreasing the rake angle (a) (b) increasing the depth of cut (c) Decreasing the cutting speed (d) increasing the cutting speed IES ‐ 2002 The value of surface roughness 'h' h obtained during the turning operating at a feed 'f' with a round nose tool having radius 'r' r is given as For-2014 (IES. 3 (c) 1. 4 4. Longer tool life 4. Titanium 2. 3. mm/rev Neglecting nose radius of the tool. To achieve h improved surface finish. hydrodynamic lubrication cannot be maintained.036 mm/rev (d) ( ) 0. They are used to turn components under the same machining conditions. the maximum height of surface roughness Hmax produced by a single‐point turning tool is given by (a) ( ) S2/2R (b) S2/4R (c) S2/4R (d) S2/8R IES ‐ 1999 In turning operation. and R = nose radius di in i mm. so Extreme Pressure (EP) additives must be added dd d to the h lubricant. sulfur chlorine. one should (a) decrease nose radius of the cutting tool and increase depth of cut (b) Increase nose radius of the cutting tool (c) Increase feed and decrease nose radius of the cutting tool IAS ‐2009 Main What are extreme‐pressure lubricants? For-2014 (IES. To keep the same level of surface finish. phosphorus.16 mm (b) ( ) 0. these The compounds are activated by the higher temperature resulting from extreme pressure. phosphorus sulfur.187 mm/rev (c) 0. If hp and hQ denote the peak‐ to‐valley heights of surfaces produced by the tools P and Q. pressure As the temperature rises. the nose radius of the tool should be (a) Halved (b) Kept unchanged (c) doubled (d) Made four times GATE ‐ 1997 A cutting tool has a radius of 1.0187 7 mm/rev GATE – 2007 (PI) A tool with Side Cutting Edge angle of and o End Cutting Edge angle of 10 is used for fine turning with a feed of 1 mm/rev. the ratio hp/hQ will be 1993 ISRO‐2008 IES – 1993. one need to improve surface finish without h sacrificing f material l removal l rate.32 mm (d) 0. The feed rate for a theoretical surface roughness of Ra = 5 μ m is (a) 0. For achieving a specific surface finish in single point turning the most important factor to be controlled is (a) Depth of cut (b) Cutting speed ( ) Feed (c) F d (d) Tool T l rake k angle l tan 8o + cot15o tan 8o + cot 30o tan15o + cot7o (c ) tan 30o + cot7o (a) tan15o + cot 8o tan 30o + cot 8o tan7o + cot15o (d ) tan7o + cot 30o (b) IES ‐ 2006 In the selection of optimal cutting conditions.26 mm (c) 0. the maximum (peak to valley) h i h of height f surface f roughness h produced d d will ill be b (a) ( ) 0. chlorine or combination of these. the requirement of surface finish would put a limit on which of the following? (a) The maximum feed (b) The Th maximum i d th of depth f cut t (c) The maximum speed (d) The maximum number of passes GATE ‐2010 (PI) During turning of a low carbon steel bar with TiN coated carbide insert. . GATE & PSUs) (d) Increase depth of cut and Page 23 increase of 78 feed [ 3 – marks] g pressures and rubbing p g action are Where high encountered. EP molecules become reactive and release derivatives such as iron chloride or iron sulfide and forms a solid protective coating. l bi EP lubrication l b i i is i provided id d by b a number of chemical components such as boron.36 0 36 mm/rev (b) 0.8 mm. the feed could be doubled to increase the metal removal rate.48 mm 30o GATE ‐ 2005 Two tools P and Q have signatures 5 5°‐5 5°‐6 6°‐6 6°‐8 8°‐30 30°‐ 0 and 5°‐5°‐7°‐7°‐8°‐15°‐0 (both ASA) respectively.IAS ‐ 1996 Given that S = feed in mm/rev. No Q 12 13 14 15 16 17 18 19 20 21 22 Option p C A A B B B A B A B B Q. Tool Life and Cutting Fluid Q. No 6 7 8 9 Option B A C A GATE‐1995 A test specimen is stressed slightly beyond the Metal Forming B S K Mondal By M d l yield point and then unloaded. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT p of A the correct explanation (c) A is true but R is false (d) A is false but R is true G ‐2003 GATE Cold working of steel is defined as working (a) At its recrystallisation temperature (b) Above Ab it recrystallisation its t lli ti temperature t t (c) Below its recrystallisation temperature (d) At two thirds of the melting temperature of the metal Page 24 of 78 . increase (a) Both Statement (I) and Statement (II) are individually true and d Statement S (II) is i the h correct explanation l i of f Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement ( (I) ) is true but Statement ( (II) ) is false (d) Statement (I) is false but Statement (II) is true For-2014 (IES. R Reason (R) : If they th are worked k d in i cold ld state t t they cannot retain their mechanical properties. No Q 23 24 25 26 27 28 29 30 31 32 33 Option p A C C B B A B A C B C Ch‐4: Economics of Machining Operation Q. Lead Zinc and Tin are always hot worked. GATE & PSUs) IES 2011 Assertion (A): Lead. Statement (II): At higher strain rate and lower temperature the yield strength of structural steel tends to increase.IES ‐ 2001 Dry and compressed air is used as cutting fluid for machining (a) Steel (b) Aluminium (c) Cast iron (d) Brass IES ‐ 2012 The most important function of the cutting fluid is to (a) Provide lubrication (b) Cool C l th the t tool l and d work k piece i (c) Wash away the chips (d) Improve surface finish Ch‐3: Cutting Tools. No 1 2 3 4 5 Option C B A C A Q. No Q 1 2 3 4 5 6 7 8 9 10 11 Option p B A A D D D B A A D C Q. Its yield strength (a) Decreases ( ) Increases (b) (c) Remains same (d) Become equal to UTS IES‐2013 Statement (I): At higher strain rate and lower temperature structural steel tends to become brittle. Stresses are set up in the metal 2. 3 and 4 (d) 1. 1 2 and 4 (c) 2. No heating is required 3. 2. Strength St th is i decreased. 1. in cold working. Less ductility is required 4. Smoother finishes. GATE & PSUs) Page 25 of 78 . 3 and 4 S – 2004 200 IES Assertion (A): Cold working of metals results in increase of strength and hardness Reason (R): Cold working reduces the total number of dislocations per unit volume of the material ( ) 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 S – 2003 IES Cold working produces the following effects: 1. Metal working process is a plastic deformation process. 3 and 4 S – 2009 IES Consider the following characteristics: 1. 4. Surface finish is reduced Which of these statements are correct? (a) 1and 2 (b) 1. Porosity in the metal is largely eliminated. 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only S – 2008 IES Cold forging results in improved quality due to which of the following? 1 Better mechanical properties of the process.G ‐2002. Better surface finish is obtained Which of the statements given above are correct? ( ) 1. Select the correct answer using the code given below: (a) 1. 1 2 and 3 (b) 1. Metal forming decreases harmful effects of impurities and improves mechanical strength. Unbroken grain flow. 2002 ISRO S O‐2012 20 2 GATE Hot rolling of mild steel is carried out (a) At recrystallisation temperature (b) Between B t 100°C °C to t 150°C °C (c) Below recrystallisation temperature (d) Above recrystallisation temperature ISRO‐2010 Materials after cold working are subjected to following process to relieve stresses (a) Hot working (b) Tempering (c) Normalizing (d) Annealing S – 2006 IES Which one of the following is the process to refine the grains of metal after it has been distorted by hammering or cold working? (a) Annealing (b) Softening ( ) Re (c) R ‐crystallizing t lli i (d) Normalizing N li i S – 2004 200 IES Consider the following statements: In comparison to hot working. process 2. 3. High 4 g p pressure. Very intricate shapes can be produced by forging process as compared to casting process. 3. 2 and (a) d 3 (b) 1. 2 and d4 (c) 1 and 3 (d) 2. Grain G i structure t t gets t distorted di t t d 3. Strength and hardness of the metal are decreased 4. strength 2. Higher Hi h forces f are required i d 2. 2 and 3 ( ) 3 and (c) d4 (d) 1 and d4 For-2014 (IES. Which of the statements g given above are correct? (a) 1. Which of the above characteristics of hot working is/are correct? (a) 1 only (b) 3 only (c) 2 and 3 (d) 1 and 3 S – 2008 IES Consider the following statements: 1. Close tolerances cannot be maintained. 1. d d 3. Its hardness and strength increase. li f grain i growth. Reason (R): Cold working is performed below the recrystallisation temperature of the work material. th recrystallisation t lli ti (c) Stress relief. t 2. It is i worked k d below b l room t temperature. It is worked below recrystallisation temperature. Of these correct statements are (a) 1 and 4 (b) 1 and 3 (c) 2 and 3 (d) 2 and 4 S – 2006 IES Assertion (A): In case of hot working of metals. g grain g growth will take place again and spoil the attained structure.S – 2000 IES Assertion (A): To obtain large deformations by cold working intermediate annealing is not required. the temperature at which the process is finally stopped should not be above the recrystallisation y temperature. recrystallisation S – 1996 996 IAS For mild steel. strain hardening effect is due to (a) Slip mechanism (b) Twining mechanism (c) Dislocation mechanism (d) ( ) Fracture mechanism S – 1996 996 IES Consider the following statements: When a metal or alloy is cold worked 1. grain growth (d) Grain growth. ( ) 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 ISRO‐2009 In the metal forming process. GATE & PSUs) Page 26 of 78 . (a) ( ) Both A and R are individually y true and R is the correct explanation of A (b) ( ) Both A and R are individually y true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1992 992 IES Specify the sequence correctly (a) Grain growth. stress relief (b) Stress St relief. Reason (R): Hot working is done above the re‐ crystallization temperature. ( ) 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 For-2014 (IES. 3. stress relief. 4. p . p Reason (R): If the process is stopped above the recrystallisation y temperature. recrystallisation. Its hardness increases but strength does not increase. ( ) 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 S‐2002 IAS Assertion (A): There is good grain refinement in hot working. recrystallisation. the stresses encountered t d are (a) Greater than yield strength but less than ultimate l strength h (b) Less than yield strength of the material (c) Greater than the ultimate strength of the material (d) Less than the elastic limit S – 1997 99 IES In metals subjected to cold working. the hot forging temperature range is (a) 4000C to 6000C (b) 7000C to t 9000C 0 (c) 1000 C to 12000C (d) 13000Cto 15000C S – 2004 200 IAS Assertion (A): Hot working does not produce strain hardening. Reason (R): In hot working physical properties are generally improved. The average flow stress for the plate material is 300 MPa. The power required for the rolling operation in kW is closest to (a) 15.2 15 2 (b) 18.936 (d) 9.936 6 (c) 8.936 (b) 7.4 (d) 45. the of I a rolling lli th state t t of f stress t f the th material undergoing g g deformation is (a) pure compression (b) pure shear (c) compression and shear (d) tension and shear ISRO‐2006 Which of the following processes would produce strongest components? (a) Hot rolling ( ) Extrusion (b) (c) Cold rolling (d) Forging ISRO‐2009 Ring rolling is used (a) To decrease the thickness and increase diameter (b) To increase the thickness of a ring (c) For producing a seamless tube (d) For producing large cylinder S – 2006 IES Which one of the following is a continuous bending process in which opposing rolls are used to produce long sections of formed shapes from coil or strip stock? (a) Stretch forming (b) Roll forming (c) Roll bending (d) Spinning ( ) GATE – 2009 (PI) Anisotropy in rolled components is caused by (a) changes in dimensions (b) scale formation (c) closure of defects (d) grain i orientation i i IFS – 2010 Calculate the neutral plane to roll 250 mm wide annealed copper strip from 2. GATE & PSUs) Page 27 of 78 .S‐2008 IES Which one of the following is correct? Malleability is the property by which a metal or alloy can be plastically deformed by applying (a) Tensile stress (b) Bending stress (c) Shear stress (d) Compressive stress GATE‐2013 In process.0 2 0 mm thickness with 350 mm diameter steel rolls. a 20 mm thick plate with plate width of 100 mm. The roller radius is 250 mm and rotational speed is 10 rpm. 0. is reduced to 18 mm. Take µ = 0 05 and σ’o =180 MPa.2 (c) 30.5 2 5 mm to 2.936 For-2014 (IES. The bite angle in degree will be (a) 5.05 MPa [10‐marks] G ‐2008 GATE In a single pass rolling operation.6 45 G ‐2007 200 GATE The thickness of a metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. the position of of contact does not Diameter of the rolls M t i l of Material f the th rolls ll S – 2001 200 IES g assumptions p Which of the following are correct for cold rolling? 1. The friction coefficient at the work‐roll interface is 0. The material is plastic. g the codes g given below: Select the correct answer using Codes: (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES. The angle subtended by the deformation zone at the roll centre is (in radian) (a) 0.7 mm GATE – 2011 (PI) The thickness of a plate is reduced from 30 mm to 10 mm by successive cold rolling passes using identical rolls of diameter 600 mm. Roll is of diameter 600 mm and it rotates at 100 rpm.1.1.G ‐2004 200 GATE In a rolling process.0 mm (b) 1. the edges are sometimes not straight and flat but are wavy.006 (c) 0. The arc of contact is circular with a radius greater than the radius of the roll. The roll strip contact length will be (a) 5 mm (b) 39 mm (c) 78 mm (d) 120 mm G ‐1998 998 GATE A strip with a cross‐section 150 mm x 4. p 2.031 (d) 0. If the coefficient of friction between the rolls and the work piece is 0.01 0 01 (b) 0.5 mm is being rolled with 20% reduction of area using 450 mm diameter rolls. rollers a strip of width 140 mm and thickness 8 mm undergoes 10% reduction of thickness.22.5 mm (c) (d) 3.02 0 02 (c) 0.600 0 600 G ‐2006 GATE A 4 mm thick sheet is rolled with 300 mm diameter rolls to reduce thickness without any change in its width.5 mm ( ) 2. 3. The minimum possible thickness of the sheet that can be produced in a single pass is (a) 1.03 (d) 0.06 GATE – 2012 Same Q in GATE – 2012 (PI) In a single pass rolling process using 410 mm diameter steel rollers. 01 the minimum number of passes required is ( )3 (a) (b) 4 (c) 6 (d) 7 ( ) GATE‐1992(PI) If factor during rolling of f the h elongation l f d ll f an ingot is 1.062 0 062 (b) 0. 2 and 3 Page 28 of 78 . GATE & PSUs) S – 2002 IES In rolling a strip between two the neutral point in the arc depend on (a) Amount of reduction (b) ( ) Coefficient (c) C ffi i t of f friction f i ti (d) rolls. sheet of 25 mm thickness is rolled to 20 mm thickness. Reason (R): Non‐uniform mechanical properties of the flat material rolled out result in waviness of the edges. mm Assume that there is no change in width. Coefficient of friction is constant over the arc of g the arc of contact and acts in one direction throughout contact. The angle of bite in radians is (a) 0. 1 22 The minimum number of passes needed to produce a section 250 mm x 250 mm from an ingot of 750 mm x 750 mm are (a) 8 ( ) 10 (c) (b) 9 (d) 17 S – 2003 IES ) While rolling g metal sheet in rolling g Assertion ( (A): mill. ( ) 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 For-2014 (IES. Reason (R): The reduction possible in the second pass is less than that in the first pass. pass (a) Both A and R are individually true and R is the correct t explanation l ti of f A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1993 993 IES In order to get uniform thickness of the plate by rolling process.than the entry velocity of the strip and is ……B ….the exit velocity of the strip.5 05 (c) 0. 2000 G 20 0( ) IES GATE‐2010(PI) In the rolling process.10 (c) (d) 94. ( ) less (a) l th / than/greater t less l (b) Greater than/less than S – 2000.20 ( ) GATE‐1990 (PI) While rolling a strip the peripheral velocity of the roll is …. The coefficient of friction for unaided bite should nearly be (a) 0. thickness thi k f a strip t i is i reduced d d from f mm to 3 mm using g3 300 mm diameter rolls rotating g at 100 rpm.S – 2001 200 IES A strip is to be rolled from a thickness of 30 mm to 15 mm using a two‐high mill having rolls of diameter 300 mm.. 993 G 989( ) IES GATE‐1989(PI) The blank diameter used in thread rolling will be (a) Equal to minor diameter of the thread (b) Equal E l to t pitch it h diameter di t of f the th thread th d (c) A little large than the minor diameter of the thread (d) A little larger than the pitch diameter of the thread S – 1992. roll separating force can be decreased by (a) Reducing the roll diameter (b) Increasing the roll diameter (c) Providing back‐up rolls (d) ( ) Increasing g the friction between the rolls and the metal S – 1999 999 IES Assertion (A): In a two high rolling mill there is a limit to the possible reduction in thickness in one pass.A…. one provides (a) Camber on the rolls (b) Offset on the rolls (c) Hardening of the rolls (d) ( ) Antifriction bearings g S – 1993.35 0 35 (b) 0.57 (a) (b) 3.14 ( ) 47.25 (d) 0.07 GATE ‐2008(PI) In of 4 I a rolling lli process.. GATE & PSUs) Page 29 of 78 . The velocity of the strip in (m/s) at the neutral point is ( ) 1. 992 G 992( ) IES GATE‐1992(PI) Thread rolling is restricted to (a) Ferrous materials (b) Ductile D til materials t i l (c) Hard materials (d) None of the above S – 2004 200 IAS Assertion (A): Rolling requires high friction which increases forces and power consumption. lubricants should be used. Reason (R): To prevent damage to the surface of the rolled products. Which of the statements given above are correct? (a) 1 and 2 only (b) 2 and 3 only ( ) 1 and (c) d 3 only l (d) 1. Roof trusses 3. Using large diameter rolls to increase the contact area. Reducing R d i friction f i ti 2. By B small ll working ki roll.S – 2001 200 IAS Consider the following characteristics of rolling process: 1 Shows work hardening effect 1.S. W ldi Welding B. GATE & PSUs) Page 30 of 78 12 . Middle roll rotates by friction B B. Th Three hi h mills high ill 2. 2 and 3 S – 2000 IAS Rolling very thin strips of mild steel requires (a) Large diameter rolls (b) Small S ll diameter di t rolls ll (c) High speed rolling (d) Rolling without a lubricant S – 1998 998 IAS Match List ‐ I (products) with List ‐ II (processes) and select the correct answer using the codes given below the lists: List – I List ‐II A M. one gets (a) One reduction in thickness (b) Two reductions in thickness (c) Three reductions in thickness (d) ( ) Two or three reductions in thickness depending p g upon the setting S – 2007 200 IAS Consider the following statements: Roll forces in rolling can be reduced by 1. Forging C. Rolling Codes:A B C D A B C D ( ) 1 (a) 2 3 4 (b) 4 3 2 1 (c) 1 2 4 3 (d) 4 3 1 2 S – 2007 200 IAS g Match List I with List II and select the correct answer using the code given below the Lists: List I List II ( (Type of Rolling Mill) ) ( (Characteristic) ) A. Casting D. Carburetors 2. A. ll power for rolling is reduced C Four high mills C. No 1 2 3 4 5 6 7 8 9 10 11 Option C B D D A A B D C C B C Forging By S K Mondal For-2014 (IES. Two high non‐reversing mills 1. 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. M S angles l and d channels h l 1. 3 Rolls of equal size are 3. 3. Diameter of working g roll is very small Code:A B C D A B C D ( ) 3 (a) 4 2 1 (b) 2 1 3 4 (c) 2 4 3 1 (d) 3 1 2 4 S – 2003 IAS In one setting of rolls in a 3‐high rolling mill. Gear wheels 4. 2. 2 and d3 Rolling Ch‐14 GATE 2011 The maximum possible draft in cold rolling of sheet increases with the ( ) increase (a) i i coefficient in ffi i t of f friction f i ti (b) decrease in coefficient of friction (c) decrease in roll radius (d) increase in roll velocity Q. Taking 3 g smaller reductions p per p pass to reduce the contact area. Surface finish is not good 3. Cluster mills 4. rotated only in one direction D. 986 1 986 (b) 1. give the main reason for this design aspect. Statement (II): The material is pressed between two surfaces and the compression force applied.693 (d) 1. press forging and upset forging. The percentage change in diameter is (a) 0 (b) 2.4 For-2014 (IES. (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 t true b t Statement but St t t (II) is i not t the th correct t explanation l ti of f Statement (I) (c) Statement (I) ( ) is true but b Statement (II) ( ) is false f l (d) Statement (I) is false but Statement (II) is true IFS‐2011 What advantages does press forging have over drop forging ? Why are pure metals more easily cold worked th alloys than ll ? [5‐marks] IAS‐2011 Main Compare Smith forging. why is parting plane provided in closed die forging? provided.686 1 686 (c) 1.. H t die di steel.5 (c) 0. G 20 0 ( ) GATE ‐2010 (PI) Hot solid forging.307 (b) 0. GATE & PSUs) Page 31 of 78 . ISRO‐2012 The true strain for a low carbon steel bar which is doubled in length by forging is (a) 0.7 (d) 41.0 G ‐2012 20 2 Same Q GATE ‐2012 (PI) GATE A solid lid cylinder li d of f diameter di t 100 mm and d height h i ht 50 mm is forged g between two frictionless flat dies to a height g of 25 mm. [ 2 marks] ] G ‐2007 200 GATE In open‐die forging. [ – Marks] [10 M k ] IES ‐ 2007 Sometimes the parting plane between two forging dies is not a horizontal plane. Mention three points for each.386 (d) 0. gives it a shape. drop forging.( ) GATE‐1989(PI) At the last hammer stroke the excess material from the finishing cavity of a forging die is pushed into……………. a disc of diameter 200 mm and height 60 mm is compressed without any barreling effect.602 GATE‐1992.07 (c) 20. t l used d for f large l lid dies di in i drop d f i should necessarily y have (a) high strength and high copper content (b) high hardness and low hardenability (c) high toughness and low thermal conductivity (d) high hardness and high thermal conductivity IES‐2013 Statement (I): The dies used in the forging process are p made in pair. The true strain is (a) 1. The final diameter of the disc is 400 mm. ti without ith t reducing d i the th strength. y therefore. C b tt body Carburettor b d (A) (B) (C) (D) G ‐1998 998 GATE List I List II Aluminium brake shoe (1) Deep drawing Pl ti water Plastic t bottle b ttl ( ) Blow (2) Bl moulding ldi Stainless steel cups (3) Sand casting Soft drink can (aluminium) (4) Centrifugal casting (5) Impact extrusion (6) Upset U t forging f i S – 2006 IES Assertion (A): Forging dies are provided with taper or draft angles on vertical surfaces. ( ) 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 S – 2005 200 IES Consider the following statements: 1. by which forging can be done easily. p p y IES ‐ 2012 Which of the following statements is correct for forging? (a) Forgeability is property of forging tool. T bi rotors Turbine t (b) Die casting 2. 3. toughness 2.G ‐1994 99 GATE Match 4 correct pairs between List I and List II for the questions List I gives a number of processes and List II gives a number of products List I List II ( ) Investment (a) I t t casting ti 1. Connecting rods (d) Drop forging 4. (a) 1. Turbine blades (c) Centrifugal casting 3. as flaws are always there due to intense working 4 Part are shaped by plastic deformation of material 4. Forged components can be provided with thin sections. IES‐2013 In the forging process: 1. (a) ( ) Both Statement ( (I) ) and Statement ( (II) ) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Page 32 of 78 S – 1993. The metal structure is refined 2. Original unidirectional fibers are distorted. 993 G 99 ( ) IES GATE‐1994(PI) Which one of the following manufacturing processes requires the provision of ‘gutters’? (a) Closed die forging (b) Centrifugal casting (c) Investment casting (d) ( ) Impact p extrusion (c) 1. 2 and 3 (b) 1. 2 and 4For-2014 (IES. t t (c) Certain mechanical properties of the material are influenced by forging. Forging reduces the grain size of the metal. 3 and 4 IES ‐ 2012 ) It is difficult to maintain close tolerance in Statement ( (I): normal forging operation. p poor forging properties. Poor reliability. (d) ( ) Pure metals have g good malleability. Cast iron pipes (f) Shell Sh ll moulding ldi 6 6. t th Which of the statements given above is/are correct? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 S – 1996 996 IES Which one of the following is an advantage of forging? (a) Good surface finish (b) Low tooling cost (c) Close tolerance (d) ( ) Improved p physical p y property. Galvanized iron pipe (e) Extrusion 5. 3 and 4 & PSUs) . (GATE d) 2. easily (b) Forgeability decreases with temperature upto lower critical iti l temperature. (II): ) Forging g g is workable for simple p shapes p Statement ( and has limitation for parts having undercuts. Reason (R): It facilitates complete filling of die cavity and favourable grain flow. which results in a decrease in strength and toughness. 3. Fusion A B C D A B C D (a) 3 2 1 4 (b) 1 2 3 4 ( ) 1 (c) 4 3 2 (d) 3 4 1 2 S – 2003 IES A forging method for reducing the diameter of a bar and in the process making it longer is termed as (a) Fullering (b) Punching (c) Upsetting (d) Extruding S – 2002 IES Consider the following steps involved in hammer forging a connecting rod from bar stock: 1 Blocking 2. 3. 2. 4. 1. 4 S – 2003 IES Consider the following steps in forging a connecting rod from the bar stock: 1 Blocking 2. 1. 4. Finishing 4. 4 (d) 3. C b i Cambering The correct sequence of these operations is (a) 1. 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 S – 2005 200 IES The process of removing the burrs or flash from a forged component in drop forging is called: (a) Swaging (b) Perforating (c) Trimming (d) Fettling IES 2011 Which of the following processes belong to forging operation ? 1. 2 and 5 (b) 4. Reason (R): The gutter helps to restrict the outward flow of metal thereby helping to fill thin ribs and bases in the upper die. 1. 2 Trimming 3. 3 3. 4 (b) 2. 3. 2 Trimming 3. Swaging 3. GATE & PSUs) Page 33 of 78 . 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 S – 2004 200 IES ( y ) with List II Match List I (Different systems) (Associated terminology) and select the correct answer using the codes given below the Lists: List I List II A. The operations involved are: 1 Ground 1. 2 and d3 S – 1999 999 IES Consider the following operations involved in forging a hexagonal bolt from a round bar stock. Welding (a) 1 and 2 only (b) 2 and 3 only ( ) 1 and (c) d 3 only l (b) 1. Riveted J Joints 1. Upsetting 3. Heat treated 4. 3. Fullering F ll i 2. Finishing 4. 2 and 1 (d) 5. 2. Welded joints 2. provision is also to be made in the forging die for additional space called gutter. whose diameter is equal to the bolt diameter: 1. 3. 1 (c) 2. Hot forged on hammers 3. Edging Which of the following is the correct sequence of operations? (a) 1. 3 and 2 (c) 5. 1. 1. 1. 2. Angular movement C Leaf springs C. 2 and 3 only S – 2008 IES The balls of the ball bearings are manufactured from steel rods. Swaging S i 4. Polished 4 What is the correct sequence of the above operations from start? (a) 3‐2‐4‐1 (b) 3‐2‐1‐4 ( ) 2‐3‐1‐4 (c) (d) 2‐3‐4‐1 For-2014 (IES. Nipping pp g B. Fullering D.S – 1997 99 IES ) In drop p forging g g besides the p Assertion ( (A): provision for flash. Fullering 5. Flattening 2. 5. Knuckle joints 4. l it (c) The die with hammer at high velocity (d) a weight on hammer to produce the requisite p impact. Squeezing action C Upset C. Thermoforming D. Press Forging 2. weight which the falling pans weigh. S (I) In I high hi h velocity l i forming f i hi h Statement (I): process. Drop Forging 2. Drop Forging 1. fullering is done to (a) Draw out the material (b) Bend B d the th material t i l (c) Upset the material (d) Extruding the material IES 2011 Consider the following statements : 1. 2. U F i Forging 3. D. For larger work piece of metals that can retain toughness at forging temperature it is preferable to use forge press rather than forge hammer. forging is done by dropping (a) The work piece at high velocity (b) The Th hammer h at t high hi h velocity. Machine Forging 4. Metal is gripped in the dies and pressure is i applied li d on the h heated h d end d B. Calendaring C l d i Code: (a) A B C D (b) A B C D 3 5 1 2 4 5 1 2 (c) ( ) A B C D (d) ( ) A B C D 4 3 3 1 3 1 5 2 . (a) 1 and 2 are correct and 2 is the reason for 1 (b) 1 and 2 are correct and 1 is the reason for 2 (c) 1 and 2 are correct but unrelated (d) 1 only correct S – 2005 200 IES yp of Forging) g g) with List II (Operation) ( p ) Match List I ( (Type and select the correct answer using the code given below the Lists: List I List II A. Large g water tanks 2. Metal M l is i placed l d between b rollers ll and d pushed D Roll Forging 4. Press Forging 3. Done in closed impression dies by hammers in blows D. Plastic sheets 3. weigh (a) Both Statement (I) and Statement (II) are individually t true and d Statement St t t (II) is i the th correct t explanation l ti of f Statement (I) (b) Both h Statement (I) ( ) and d Statement (II) ( ) are individually d d ll true but Statement (II) is not the correct explanation of Statement (I) () (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Page 34 of 78 S – 2009 IES Match List‐I with List‐II and select the correct answer using the the h code d given below b l h Lists: List‐I List‐II (Article) (Processing Method) A. Material is only upset to get the desired shape B. Disposable coffee cups 1. Smith Forging 1. i ht Statement (II): The kinetic energy is the function of mass and velocity. Carried out manually open dies C. high energy can be transferred to metal with relatively small weight. Blow moulding 5. Done in closed impression dies by continuous squeezing force Code: A B C D (a) 2 3 4 1 (b) 4 3 2 1 (c) 2 1 4 3 (d) 4 1 2 3 S – 1998 998 IES Which one of the following processes is most commonly used for the forging of bolt heads of hexagonal shape? (a) Closed die drop forging (b) Open O di upset die t forging f i (c) Close die press forging (d) Open die progressive forging S – 1994. 4 Repeated hammer blows blo s A B C D A B C D ( ) 4 (a) 1 2 3 (b) 3 2 1 4 (c) 4 2 1 3 (d) 3 1 2 4 S – 2008 IES Match List‐I with List‐II and select the correct answer using the code given below the i b l th lists: li t List‐I (Forging Technique) List‐II (Process) A. Rotomoulding B. but Statement is true GATE & (II) PSUs) IES‐2013 IES‐2013 Statement (I): In power forging energy is provided by compressed air or oil pressure or gravity. (a) Both Statement ( (I) ) and Statement ( (II) ) are individually y true and Statement (II) is the correct explanation of () Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement For-2014 (I) is false (IES. Expandable p bead moulding g C.S – 2001 200 IES In the forging operation. Cushion pads 4. Any metal will require some time to undergo complete plastic l ti deformation d f ti particularly ti l l if deforming d f i metal t l has h to fill cavities and corners of small radii. 99 ISRO S O‐2010 20 0 IES In drop forging. Statement (II): The capacity of the hammer is given by the total weight. Lubricated specimens show less surface movement than un‐lubricated ones. M l is Metal i displaced di l d away from f centre. Deep drawing 3. 3. Extrusion 3 4. (a) Both A and R are individually true and R is the p of A correct explanation (b) Both A and R are individually true but R is NOT p of A the correct explanation (c) A is true but R is false (d) A is false but R is true IES‐2013 Consider the following statements pertaining to the open‐die forging of a cylindrical specimen between two flat dies: 1. Barrelling g S. 3 R – 1. Q – 3. Amount of flash depends upon forging force. reducing thickness in middle and increasing length C. R – 4. 2. 2 Q – 3. Rod is pulled through a die D. Pressure a workpiece between two flat dies Codes:A B C D A B C D (a) ( ) 3 2 1 4 (b) ( ) 4 1 2 3 (c) 3 1 2 4 (d) 4 2 1 3 S – 2000 IAS Drop forging is used to produce (a) Small components (b) Large L components t (c) Identical Components in large numbers (d) Medium‐size components S – 1998 998 IAS The forging defect due to hindrance to smooth flow of metal in the component called 'Lap' occurs because (a) The corner radius provided is too large (b) The Th corner radius di provided id d is i too t small ll (c) Draft is not provided (d) The shrinkage allowance is inadequate S – 2002 IAS Consider the following statements related to forging: 1 Flash is excess material added to stock which flows 1. 4. Centre burst R. Upsetting 2. Which of these statements are correct? (a) 1 and 3 For-2014 (b) 1 and 4 (c) GATE 2 and 3& PSUs) (d) 2 and 4 (IES. Lubricated specimens 3 p show more barrelling g than un‐ lubricated ones. Coefficient C ffi i t of f friction f i ti is i constant t t between b t work k piece i and die 2. (B) Fullering (C) Drawing 3. Fullering 3. Match the following ( ) GATE ‐2008 (PI) Group‐2 1. D Drawing i 2. 2 Q – 4. 1 S‐4 Page 35 of 78 . S‐1 (c) P – 2. Closed die forging (b) P – 3. Cold shut (a) P – 2. Lubricated specimens shows less barrelling than un‐ lubricated ones. Flattening 1. Thickness is reduced continuously at different sections along length B B. Flash Fl h helps h l i in filling filli of f thin thi ribs ib and d bosses b in i upper die. 2 and 3 (b) 1 and 2 (c) 1 and 3 (d) 2 and 3 IES 2011 Assertion (A) : Hot tears occur during forging because of inclusions in the blank material Reason (R) : Bonding between the inclusions and the parent material is through physical g and chemical bonding. 3 S‐1 IES ‐ 2012 Assumptions adopted in the analysis of open die forging are 1 Forging force attains maximum value at the middle of 1. Wire drawing 4. g g (D) Swaging Codes:A B C D A B C D (a) ( ) 4 3 2 1 (b) ( ) 2 1 4 3 (c) 4 1 2 3 (d) 2 3 4 1 Click to see file Page 4 – 5 ‐6 S – 2001 200 IAS Match List I (Forging operations) with List II (Descriptions) and the d select l h correct answer using the h codes d given below b l the Lists: List I List II A. 2 and 3 Group ‐1 P . Stress in the vertical (Y‐direction) is zero. R – 1. around parting line. 2. Lubricated specimens show more surface movement than h un‐lubricated l bi d ones. 4. 2. 4 R – 3. S‐2 (d) P – 2.S – 2003 IAS ( g g Operation) p ) with List II ( Match List I (Forging (View of the Forging Operation) and select the correct answer using the codes given below the lists: List‐I List‐II (Forging Operation) (View of the Forging Operation) (A) Edging 1. 3. the operation. (a) ( ) 1 and 2 (b) ( ) 1 and 3 (c) 2 and 3 (d) 1. Q – 4. Which of the above statements are correct? (a) 1. 2. Wrinkling Q. The flow curve equation of the material is given by σ f = 200(0. determine the maximum forging force.05. 178 MPa. Calculate the maximum . forging f Th average yield The i ld stress t of f lead in tension is 7 N/mm2 [10] ti l IES – 2007 C Conventional A cylinder of height 60 mm and diameter 100 mm is forged at room temperature between two flat dies. Take the average yield strength in tension is 7 N/mm2 [Ans.25. Determine maximum forging force.01 + ε ) 0. If the co‐efficient of friction between the job and die is 0. determine the maximum i f i force. 405 MPa] [Hint. Coefficient of friction is 0. Coefficient of friction is 0. First calculate true strain ε and put the value in the equation σ f = 1030ε =σy ] ] For-2014 (IES. First calculate true strain ε and put the value in the equation σ f = 200(0. determine the maximum forging force. GATE & PSUs) Page 36 of 78 . Also find mean die pressure. No 1 2 3 4 5 Option A A A A B Q. [20‐Marks] C ti l IES – 2006 ‐ Conventional A certain disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. 9. Coefficient of friction is 0. There is no sticking.05.2 and flow curve equation is σ f = 1030ε 0.S – 2005 200 C i l IES Conventional A strip 24 mm x 24 t i of f lead l d with ith initial i iti l dimensions di i mm x 150 5 mm is forged g between two flat dies to a final size of 6 mm x 96 mm x 150 mm.25. [ Ans.17 [Ans. pressure in height is upsetted by open die forging to a height of 50 mm. Assume that volume is constant after deformation. [10 – Marks] GATE‐1987 In forging operation the sticking friction condition occurs near the……(Centre/ends) Ch‐15: Forging Q. Find the die load at the end of compression to a height 30 mm. 221 MPa] [Hint. If the coefficient of friction is 0. mean die pressure and maximum pressure.17 height is forged to 40 mm in height.01 + ε ) 0. The yield strength of the work material is given as 120 N/mm2 and the coefficient of friction is 0. 178.24 kN] P ti Problem P bl ‐2 Practice A circular disc of 200 mm in diameter and 100 mm in P ti Problem P bl ‐3 Practice A cylindrical specimen 150 mm in diameter and 100 mm P ti Problem P bl ‐4 Practice A circular disc of 200 mm in diameter and 70 mm in height is compressed between two flat dies to a height of 50 mm. Determine the maximum die pressure. If the coefficient of friction is 0.771 MN. using slab method of analysis.1 and average yield strength in compression is 230 MPa.41 MPa MP MPa . [Ans.05. 46. No 6 7 8 9 Option B C C C P ti Problem P bl ‐1 Practice 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. The average shear yield stress of lead can be taken as 4 N/mm2.26 MN] forging load.41 = σ y 0. p y.21 (c) 6.35 ⎜ 1 + e ⎟ ⎜ ⎟ ⎝ ⎠ ii i ii. . K is the shear yield strength of steel and r is the radial distance of any point (contd .71 (b) 10. are mm).77 6 77 (d) 4. GATE & PSUs) GATE – 2011 (PI) Common Data‐S1 In a multi‐pass drawing operation.39 4 39 Page 37 of 78 .. the coefficient of friction (μ ) is: μ = 0. where p and τ are the normal and shear stresses.. the force (in ) required q for drawing g the bar through g the first die. of friction between the die and steel rod as 0.55 (c) 265. the reduction in cross‐sectional area is 35%.76 (b) 254.Practice Problem ‐5 {GATE‐2010 (PI)} During open die forging process using two flat and parallel dies. During each of these drawing operations. During each of these drawing operations.. respectively.sliding lidi friction f i i prevails. The yield strength of the material is 200 MPa. in the h region i R ss ≤ r ≤ RFN . a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section by drawing it successively through a series of seven dies of decreasing exit diameter. Ignore strain hardening.. ⎛ ⎞ i.) ( RFN − r ) B S K Mondal By M d l IAS‐2010 Main How are metal tooth‐paste tubes made commercially ? Draw the tools configuration with the help of a neat sketch. [30‐Marks] IES 2009 Conventional ( ) GATE‐1994(PI) A moving mandrel is used in (a) Wire drawing (c) Metal Cutting (b) Tube drawing (d) Forging IES – 2011 Conventional A 12. The total true strain applied and the final length (in ). respectively. what would be the draw stress and maximum / N/mm possible reduction ? 2.. a round bar of 10 mm diameter and 100 mm length is reduced in cross‐section by drawing it successively through a series of seven dies of decreasing exit diameter.sticking condition prevails The value of R SS (in mm).45 (d) 278.45 and 345 (c) 3.In the region 0 ≤ r ≤ R SS .15. Ignore strain hardening. Neglecting friction and redundant work. The yield strength of the material is 200 MPa.45 and 8 17 (b) 2.5 12 5 mm diameter rod is to be reduced to 10 mm diameter by drawing in a single pass at a speed of 100 / Assuming g a semi die angle g of 5o and coefficient m/min. 400 N/mm / Take stress of the work material as 4 [15 Marks] For-2014 (IES. Along the die-disc interfaces. (a) 2. calculate: (i) The power required in drawing (ii) ( ) Maximum p possible reduction in diameter of the rod (iii) If the rod is subjected to a back pressure of 50 2 .. is kN) (a) 15.20 Extrusion & Drawing p = 3Ke K H FN and d τ = μ p. is (a) 241. where sticking condition changes to sliding friction.02 3 02 and 2043 (d) 3. a solid lid circular i l steel t l disc di of f initial i iti l radius di (R IN ) 200 mm and d initial i iti l height (H IN ) 50 mm attains a height (H FN ) of 30 mm and radius of R FN . Practice Problem ‐5 {GATE‐2010 (PI)} iii. the reduction in cross‐sectional area is 35%.. il and d R − IN RFN 2μ Contd…….02 3 02 and 3330 GATE – 2011 (PI) Common Data‐S2 In a multi‐pass drawing operation. the p pressure ( (in MPa) ) on the ram will be (a) 416 (b) 624 (c) 700 (d) 832 G ‐1996 996 GATE A wire of 0. N l ti Neglecting f i ti friction and d strain t i hardening. Considering an ideal deformation process (no friction and no redundant work). Take efficiency of motor is 98%.97 kN ( ) 20.1 mm diameter is drawn from a rod of 15 mm diameter.72 MN ( ) 1.36 0 36 (b) 0.36 6 MN GATE – 2009 (PI) Using direct extrusion e trusion process.72 GATE ‐2008 (PI) Linked S‐1 A 10 mm diameter annealed steel wire is drawn through a die at a speed of 0.00 (d) 2.0 14 0 (c) 17.3 m/s. k) extrusion i ratio i 4. . 40% and 80% are available.E l Example Calculate the drawing load required to obtain 30% reduction in area on a 12 mm diameter copper wire.5 (d) 2575. The force required for extrusion is (a) 5. diameter of a steel wire is reduced from 10 mm to 8 mm. Reason (R) : High extrusion speed causes cracks in the material.0 (c) 1287. 20% in a sequence (d) none of the above For-2014 (IES. negligible interface friction and no redundant work. and d average flow fl stress of f material 3 300 MPa. 40%.0 GATE ‐2008 (PI) Linked S‐2 A 10 mm diameter annealed steel wire is drawn through a die at a speed of 0. GATE & PSUs) Page 38 of 78 . 8% 20 0 JWM 2010 ) : Extrusion speed p p Assertion ( (A) depends on work material. 80%. The yield stress of the material is 800 MPa.95 17 95 (d) 28.5 m/s to reduce the diameter by 20%. The yield stress of the material is 800 MPa. The working temperature of 700 700°C C and the extrusion constant is 250 MPa. The ideal force required for drawing (ignoring friction and redundant work) is (a) 4. 200 G 200 ( ) GATE GATE ‐2007 (PI) For rigid perfectly‐plastic work material.63 0 63 (c) 1.40%.5 (b) 357.0 28 0 G ‐2003 GATE A brass billet is to be extruded from its initial diameter of 100 mm to a final diameter of 50 mm.97 8 97 (b) 14.5 m/s to reduce the diameter by 20%. The mean flow stress of the material is 400 MPa.48 kN (b) 8. the theoretically maximum possible reduction in the wire drawing operation is (a) 0. (a) Both A and R are individually true and R is the correct explanation of A (b) Both B h A and d R are individually i di id ll true but b R is i not the h correct explanation of A (c) A is true but R is false (d) ( ) A is false but R is true G ‐2006 GATE In a wire drawing operation. h d i th stress the t required q for drawing g( (in MPa) ) is (a) 178.11 kN (c) (d) 31.36 (c) 6 MN (d) 0.41 kN G ‐2001. the number of stages and reduction at each stage respectively would be (a) 3 stages and 80% reduction for all three stages (b) 4 stages t and d 80% 8 % reduction d ti f first for fi t three th stages t followed by a finishing stage of 20% reduction (c) 5 stages and reduction of 80%. Dies giving reductions of 20%.44 MN (b) 2. process a round billet of 100 mm length and 50 mm diameter is extruded. Th power required The i d for f the th drawing d i process (in (i kW) is i (a) 8. For minimum error in the final size. The following data is given C l l t the Calculate th power of f the th electric l t i motor t if the th drawing speed is 2. 2. The force required on the punch is more in comparison to direct extrusion. The speed of travel of the extruded product is same as that of f the h ram. S – 2007 200 IES Assertion (A): Greater force on the plunger is required in case of direct extrusion than indirect one. (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 p of A (c) A is true but R is false (d) A is false but R is true IES ‐ 2012 Which of the following are correct for an indirect hot extrusion process? 1 Billet remains stationary 1. Reason (R): In case of direct extrusion. ll 3. 2. 2 3 and 4 only S – 1993 993 IES Assertion (A): Direct extrusion requires larger force than indirect extrusion. (b) ( ) Extrusion is used for reducing g the diameter of round bars and tubes by rotating dies which open and close rapidly on the work? (c) Extrusion is used to improve fatigue resistance of the metal by setting up compressive stresses on its surface (d) Extrusion comprises pressing the metal inside a chamber to force it out by high pressure through an orifice which is shaped to provide the desired from of the finished part. ram Which of these Statements are correct? ( ) 1 and (a) d3 (b) 2 and d3 (c) 1 and 4 (d) 2 and 4 For-2014 (IES. thicker walls can be obtained by increasing the forming pressure (b) Extrusion is an ideal process for obtaining rods from metal t l having h i poor density d it (c) As compared to roll forming. There is no friction force between billet and container walls. Extrusion p 4 parts have to be p provided a support. ( ) 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 S – 1994 99 IES Metal extrusion process is generally used for producing (a) Uniform solid sections (b) Uniform hollow sections (c) Uniform solid and hollow sections (d) ( ) Varying y g solid and hollow sections.G ‐1994 99 GATE The process of hot extrusion is used to produce (a) Curtain rods made of aluminium (b) Steel St l pipes/or i / domestic d ti water t supply l (c) Stainless steel tubes used in furniture (d) Large she pipes used in city water mains S – 2007 200 IES g is the correct Which one of the following statement? (a) ( ) Extrusion is used for the manufacture of seamless tubes. 4. steel zinc phosphate coating is used. The ram and the extruded p product travel in the same direction. 4. the direction of the force applied on the plunger and the direction of the movement of the extruded metal are the same. extruding speed is high (d) Impact extrusion is quite similar to Hooker's process g the flow of metal being g in the same direction including S – 1999 999 IES Which one of the following is the correct temperature range for hot extrusion of aluminium? (a) 300‐340 340°C C (b) 350‐400 400°C C (c) 430‐480°C (d) 550‐650°C S – 2000 IES g statements: Consider the following In forward extrusion process 1. 2. pp (a) 1. The ram and the extruded p product travel in the opposite pp direction. The speed of travel of the extruded product is greater than that of the ram. S – 2009 IES Which one of the following statements is correct? (a) In extrusion process. GATE & PSUs) Page 39 of 78 . Reason (R): In indirect extrusion of cold steel. 3. 3 and 4 (b) 1. 2 and 3 only (c) 1. 1 2 and 4 only (d) 2. Reason (R): They lubricate the surface to reduce friction while drawing wires. It is suitable for high‐strength super‐alloys. GATE & PSUs) Page 40 of 78 . p 3.The billet is inserted into the extrusion chamber and pressure is applied by a ram to extrude the billet through the die. The billet is inserted into the extrusion chamber where it is surrounded d d by b a suitable i bl liquid. 2.S – 2009 IES What is the major problem in hot extrusion? (a) Design of punch (b) Design of die ( ) Wear (c) W and d tear t of f die di (d) Wear W of f punch h IES ‐ 2012 Extrusion process can effectively reduce the cost of product through (a) Material saving (b) process time saving (c) Saving in tooling cost (d) ( ) saving g in administrative cost S – 2008. 4. (a) Both A and R are individually true and R is the correct explanation of A (b) Both B h A and d R are individually i di id ll true but b R is i NOT the h correct explanation of A (c) A is true but R is false (d) ( ) A is false but R is true S – 2009 IES Which one of the following stress is involved in the wire drawing process? (a) Compressive (b) Tensile (c) Shear (d) Hydrostatic stress For-2014 (IES. 2. Select the correct answer ans er using the codes given gi en below: belo Codes: ( ) 1 and (a) d3 (b) 1 and d4 (c) 2 and 3 (d) 2 and 4 S – 2006 IES What does hydrostatic pressure in extrusion process improve? (a) Ductility (b) Compressive strength (c) Brittleness (d) Tensile strength ( ) GATE‐1990(PI) Semi brittle materials can be extruded by (a) Impact extrusion (b) Closed cavity extrusion (c) Hydrostatic extrusion (d) Backward B k d extrusion i IES 2010 ) Pickling g and washing g of rolled rods Assertion ( (A): is carried out before wire drawing. 2008 G 989( ) IES GATE‐1989(PI) Which Whi h one of f the th following f ll i methods th d is i used d for f the th manufacture of collapsible p tooth‐p paste tubes? (a) Impact extrusion (c) Deep drawing (b) (d) Direct extrusion Piercing S – 2003 IES The extrusion process (s) used for the production of toothpaste tube is/are 1 Tube extrusion 1. Impact extrusion Select the correct answer using g the codes g given below: Codes: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 3 only S – 2001 200 IES g statements are the salient Which of the following features of hydrostatic extrusion? 1. Forward extrusion 3. li id The Th billet bill is i extruded d d through the die by applying pressure to the liquid. It is suitable for soft and ductile material. (c) uses thick paste lubricant (d) ( ) uses thin film lubricant S – 1994 99 IES Match List I with List II and select the correct answer using the codes given below the Lists: List I (Metal farming process) List II (A similar process) A. Pilots D Sheet metal operations using 4. 5. Direction C d A B Codes:A C D A B C D (a) 4 2 1 3 (b) 3 4 2 5 (c) 5 4 2 3 (d) 3 2 1 5 For-2014 (IES. D. 1 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 3 S – 1999 999 IES Match List‐I with List‐II and select the correct answer using the codes given below the Lists: List‐I List‐II A. 4 D 3 4 (b) (d) Shear force Tensile force Compressive force Spring p g back force A B C D 2 1 3 4 4 3 2 1 A. 4 4. 3. Blanking D. Bending g Codes:A B C (a) 4 2 1 (c) 2 3 1 1. Extrusion C. the bright shining surface on the wire is obtained if one (a) does not use a lubricant (b) uses solid powdery lubricant. B B. Impact loads C. Cold rolling D Isotropy D. D. C. Soap solution B Rolling B. Wire drawing B Hardness B. Blanking Coining C i i Extrusion Cup drawing B 3 2 C 4 1 D 1 5 1. Wire drawing C. Turning D Armature shaft D. Blade 2. 4. 4 4. Wire drawing 3. 4 Crater progressive dies 5. Armature coil wire 3. Indentation 5. 4. GATE & PSUs) S – 2007 200 IES Which metal forming process manufacture of long steel wire? (a) Deep drawing (b) Forging (c) Drawing (d) Extrusion is used for S – 2005 200 IES Which of the following types of stresses is/are involved in the wire‐drawing operation? (a) Tensile only (b) Compressive only (c) A combination of tensile and compressive stresses (d) ( ) A combination of tensile. Resilience 3. Drawing 1.S – 1993 993 IES Tandem drawing of wires and tubes is necessary because (a) It is not possible to reduce at one stage (b) Annealing is needed between stages (c) Accuracy in dimensions is not possible otherwise (d) ( ) Surface finish improves p after every y drawing g stage g S – 2000 IES p ) with List II Match List I (C (Components of a table fan) (Manufacturing processes) and select the correct answer using the codes given below the Lists: List I List II A. Ironing C d A B Code:A C D A B C D (a) 2 5 1 4 (b) 4 1 5 3 (c) 5 2 3 4 (d) 5 2 1 3 S – 1996 996 IES Match List I with List II and select the correct answer List I (Metal/forming process) List II (Associated force) S – 1996 996 IES In wire drawing process. 2 2. compressive p and shear stresses Page 41 of 78 . 2 2. 3. (b) (d) Codes:A ( ) 2 (a) (c) 3 Wire drawing Pi i Piercing Embossing Rolling Bending A B C 2 3 1 2 3 1 D 4 5 S – 1993. Base with stand 1. 2. 993 ISRO S O‐2010 20 0 IES Match List I with List II and select the correct answer using the codes given below the lists: (Mechanical p property) p y) List II ( (Related to) ) List I ( A. Malleability 1. Wire drawing A B. Stamping p g and pressing g B. Camber C. ( ) 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 Page 42 of 78 . 4. Extrusion C. Casting Codes:A B C D A B C D ( ) 3 (a) 1 4 2 (b) 4 1 3 2 (c) 3 2 For-2014 4 1(IES. 2. 2 4 (c) 1. Surfaces having higher 3. Sh peening Shot i C. 4 2 (b) 1. Machine tool beds 3. 2 (d) 3. p The correct sequence of these operations is (a) 3. (d) 4 2 3 1 GATE & PSUs) S – 1997 99 IAS Extrusion force DOES NOT depend upon the (a) Extrusion ratio (b) Type T of f extrusion t i process (c) Material of the die (d) Working temperature S – 2000 IAS Assertion (A): Brittle materials such as grey cast iron cannot be extruded by hydrostatic extrusion. the work material should essentially be (a) Ductile ( ) Hard (c) (b) Tough ( ) Malleable (d) S – 2000 IES Which one of the following lubricants is most suitable for drawing mild steel wires? (a) Sodium stearate (b) Water (c) Lime‐water (d) Kerosene S – 1993 993 IES A moving mandrel is used in (a) Wire drawing (b) Tube T b drawing d i (c) Metal cutting (d) Forging S – 2002 IES Match List I with List II and select the correct answer: List I (Parts) List II (Manufacturing processes) S – 2004 200 IAS Assertion (A): Indirect extrusion operation can be performed either by moving ram or by moving the container. Lubricating 4 g with reactive soap. C ti rods d 1. Collapsible tubes 4. Reason (R): Advantage in indirect extrusion is less quantity of scrap compared to direct extrusion. 1 3. 1 4. W ldi Welding B. Pressure vessels 2. Forging hardness and fatigue strength4. 2.GATE‐1987 For wire drawing operation. 4 A. A and d smooth h tubes b 2. Seamless tubes 1. 1. extrusion billet is uniformly compressed from all sides by the liquid. Forging D. extrusion (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1995 99 IAS The following operations are performed while preparing the billets for extrusion process: 1 Alkaline cleaning 1. 3 2. Pickling 4. Phosphate coating 3. 3 1. Reason(R): In hydrostatic extrusion. Roll forming B Accurate B. Cold forming Codes: A B C A B C (a) 1 4 2 (b) 2 3 1 ( ) 1 (c) 3 2 (d) 2 4 1 S – 2001 200 IAS Match List I (Products) with List II (Suitable processes) and select the correct answer using the codes given below the Lists: List I List II A Connecting A. 3. the rod Assertion ( (A): cross‐section is reduced gradually by drawing it several times in successively reduced diameter dies. D. drawing and Ch‐17: Extrusion Q. Reason (R): Since each drawing reduces ductility of the wire. Also determine the work done if the percentage penetration t ti is i 25 percent t of f material t i l thickness. 4 Spinning 4. No 8 9 10 11 12 13 Option B B A B B A Ch‐16: Drawing Q. if the ultimate shear strength of the material is 450 N/mm2. p.5 5 mm thick metal strip. mm By S K Mondal For-2014 (IES. Extrusion 3 Forming 3. Drawing 2. (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 IES 2011 Match List –I with List –II and select the correct answer using the code given below the lists : IAS 1994 g methods can be used for Which of the following manufacturing 2 metre long seamless metallic tubes? 1. Collapsible p tubes Codes C d A ( ) 2 (a) (c) 2 B 1 4 C 4 1 D 3 3 List –II 1. No 1 2 3 4 5 6 Option A C C B C D E l Example Determine the die and punch sizes for blanking a circular disc of 20‐mm diameter from a sheet whose thickness is 1. E l Example Estimate to E ti t the th blanking bl ki force f t cut t a blank bl k 25 mm wide id and 3 30 mm long g from a 1. Welding 2. 4. so after final drawing the wire is normalized. thi k Sheet Metal Operation p Shear strength of sheet material = 294 MPa Also determine the die and punch sizes for punching a circular hole of 20‐mm diameter from a sheet whose thickness is 1. Select the correct answer using the codes given below Codes: (a) ( ) 1 and 3 (b) ( ) 2 and 3 (c) 1.S – 2002 IAS ) In wire‐drawing g process. GATE & PSUs) Page 43 of 78 .5 mm. Connecting rods B. No 1 2 3 4 5 6 7 Option D C D D B C B Q. p . Pressure vessels C Machine tool beds C.5 1 5 mm. Casting g (b) (d) A 3 3 B 1 4 C 4 1 D 2 2 ( ) GATE‐1991(PI) Seamless long steel tubes are manufactured by rolling. Extrusion 3 Rolling 3. 3 and 4 (d) 2. 3 and 4 List –I A. the maximum crushing stress is 4 times the maximum allowable shearing stress. the diameter of the smallest hole which can be punched is equal to (a) 15 mm (b) 30 mm ( ) 60 (c) 6 mm (d) 120 mm S O‐2008. The cutting blade is 400 mm long and zero‐shear (S = 0) is provided on the edge. Neglect friction. 2008 20 ISRO 2011 With a punch for which the maximum crushing stress is 4 times the maximum shearing stress of th plate. Assuming force vs displacement curve to be trapezoidal. If the allowable crushing stress in the punch is 6 N‐mm‐2. Give suitable shear angle for the punch to bring the work within the capacity it of f a 30T T press. p The ultimate shearing g strength g of the material of the washer is 280 N/mm2. (c) Taking 60% penetration and shear on punch of 1 mm. (b) 1 2 (d) 2 G ‐2010 20 0 Statement S i k d 1 GATE Linked Statement for Linked Answer Questions: In a shear cutting operation.2. 100 mm diameter. q [10‐Marks] [ ] S – 1999 999 IES A hole is to be punched in a 15 mm thick plate having ultimate shear strength of 3N‐mm‐2. (a) ( ) Find the total cutting g force if both p punches act at the same time and no shear is applied to either punch or the die. C l l t in i each h case : (i) Size of punch (ii) Size of die (iii) ( ) Force required.6 6 mm thick. N/ N/mm ) Calculate. 400 G ‐2010 20 0 Statement S i k d 2 GATE Linked Q Statement for Linked Answer Questions: In a shear cutting operation. . For punching the biggest hole. shear stress = 240 2). edge The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0. The ultimate shear strength g of the sheet is 100 MPa and penetration to thickness ratio is 0. cutting g is complete p at 40 per cent penetration of the punch. 5. the l t the th biggest bi t hole h l that th t can be b punched h d in i the plate would be of diameter equal to 1 × Thickness of p plate 4 1 (b) × Thickness of plate 2 (c) Plate thickness (a) ( ) (d) 2 × Plate thickness IES‐2013 A hole of diameter d is to be punched in a plate of thickness t. (b) What will be the cutting force if the punches are staggered. 400 GATE 2011 The shear strength of a sheet metal is 300 MPa. and cutting a rectangular t l blank bl k of f 50 x 200 mm from f a sheet h t of f1 mm thickness (mild steel. a sheet of 5mm thickness is cut along a length of 200 mm. the maximum force (in kN) exerted is (a) 5 (b) 10 Page (c) 44 20 (d) 40 of 78 . With normal clearance on the tools.5 1 5 mm thick sheet is close to (a) 45 kN (b) 70 kN (c) 141 4 kN (d) 3500 kN S S Assuming force vs displacement curve to be rectangular.IAS‐2011 Main For punching a 10 mm circular hole. For the plate material. so that only one punch acts at a time. thi k The Th ultimate lti t shear h stress t i 550 N/mm is N/ E l Example A washer with a 12.7 mm internal hole and an outside diameter of 25. a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long and zero‐shear (S = 0) is provided on the edge.2. GATE & PSUs) A shear of 20 mm (S = 20 mm) is now provided on the blade. the work done (in J) is (a) 100 (b) 200 (c) 250 (d) 300 For-2014 (IES. MPa The blanking force required to produce a blank of 100 mm diameter from a 1. Neglect friction. what will be the cutting g force if both p punches act together. the ratio of diameter of hole to plate thickness should be equal to: (a) 4 (c) 1 1 Example A hole.5 mm thick strip.4 mm is to be made from 1. is to be punched in steel plate 2 . 06 0 06 mm Die allowance 0.00 50 00 and 50. The minimum i i punching hi force f required i d in i kN is i (a) 2. the new blanking force in kN is (a) 3.29 (c) 5. The required radial clearance between the die and the punch is 6% of sheet thickness.12 mm GATE ‐ 2012 Calculate the punch size in mm. The required punch diameter is (a) 19.3 kN G ‐2003 GATE A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The blanking force (in kN) is (a) ( ) 291 9 (b) ( )3 301 (c) ( ) 3 311 (d) ( )3 321 GATE‐2013 (PI) Circular blanks of 10 mm diameter are punched from an aluminium sheet of 2 mm thickness.03 (d) 6. thi k when h ultimate lti t shear h stress of the plate is 500 N/mm2 will be nearly (a) 78 kN (b) 393 kN (c) 98 kN (d) 158 kN G ‐2007 200 GATE The force requirement in a blanking operation of low carbon steel sheet is 5. Size of the blank 25 mm Thickness of the sheet 2 mm Radial clearance between bet een punch and die 0.2 mm.15 50 15 (d) 49.85 and 50. Shear provided on the punch is 2 mm.89 (c) 25.( ) GATE – 2009 (PI) A disk of 200 mm diameter is blanked from a strip of an aluminum alloy of thickness 3.33 S O‐2009 ISRO The force required to punch a 25 mm hole in a mild ild steel t l plate l t 10 mm thick. respectively. GATE & PSUs) Page 45 of 78 .00 (b) 50. are (a) 50.5 d and thickness is reduced to 0.05 mm ( ) 24.30 50 30 (c) 49. Shear strength of the material is 400 N / mm2 and penetration is 40%. The shear strength of aluminium is 80 Mpa.6 kN (b) 37. The thickness of the sheet is ‘t’ t and diameter of the blanked part is ‘d’.17 ( ) GATE‐2008(PI) A blank of 50 mm diameter is to be sheared from a sheet of 2.5 45 (c) 5.0 kN.94 mm ( ) 20. The material shear strength to fracture is 150 MPa. The blanking force during the operation will be (a) 22.88 mm (b) 19.5 mm thickness.0 G ‐2004 200 GATE 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness.0 (d) 8.70 and 50.83 (a) (b) 24.00 50 00 and 50.6 (c) 6 6 kN (d) 94.7 kN ( ) 61. the clearance is provided on (a) The die (b) Both the die and the punch equally (c) The punch (d) ( ) Brittle the p punch nor the die For-2014 (IES.01 (d) 25. if the diameter of the blanked part is increased to 1. The punch and the die clearance is 3%.0 3 0 (b) 4.06 (c) 6 mm (d) 20.00 G ‐2002 GATE In a blanking operation. The punch and die diameters (in mm) for this blanking operation. For the same work material.4 t.57 2 57 (b) 3. for a circular blanking operation for which details are given below. 6 mm and 30 mm (c) 30 mm and 29.00 mm ( ) 50.0 mm thick.G ‐2001 200 GATE The cutting force in punching and blanking operations mainly depends on (a) The modulus of elasticity of metal (b) The shear strength of metal (c) The bulk modulus of metal (d) ( ) The y yield strength g of metal G ‐1996 996 GATE A 50 mm diameter disc is to be punched out from a carbon steel sheet 1.075 mm (d) none of (c) f the th above b S – 1994 99 IES In sheet metal blanking.6 30 6 mm and 29. (c) Warping of sheet is minimized (d) ( ) Cut blanks are straight.8 mm Page 46 of 78 . If 6% clearance is recommended. then the nominal diameters of die and punch are respectively (a) 30. GATE & PSUs) S – 1997 99 IES For 50% penetration of work material. force 2. a punch with single shear equal to thickness will (a) Reduce the punch load to half the value (b) Increase the punch load by half the value (c) Maintain the same punch load (d) ( ) Reduce the p punch load to q quarter load S – 2000 IAS A blank of 30 mm diameter is to be produced out of 10 mm thick sheet on a simple die. Shear increases the life of the punch 4.4 mm (d) ( ) 3 30 mm and 28.4 29 4 mm (b) 30. The diameter of the punch should be (a) 49. Shear Sh i increases th capacity the it of f the th press needed d d 3. g S – 2002 IES Consider the following statements related to piercing and blanking: 1 Shear on the punch reduces the maximum cutting 1.925 mm (b) 50. The total energy needed to make the cut remains provision of shear unaltered due to p Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 S – 1995 99 IAS In blanking operation the clearance provided is (a) 50% on punch and 50% on die (b) On O die di (c) On punch (d) On die or punch depending upon designer’s choice S – 2006 IES In which one of the following is a flywheel generally employed? (a) Lathe (b) Electric motor (c) Punching machine (d) Gearbox S – 2004 200 IES Which one of the following statements is correct? If the size of a flywheel in a punching machine is increased (a) Then the fluctuation of speed and fluctuation of energy will ill both b th decrease d (b) Then the fluctuation of speed will decrease and the fluctuation of energy will increase (c) ( ) Then the fluctuation of speed p will increase and the fluctuation of energy will decrease (d) Then the fluctuation of speed and fluctuation of energy both will increase For-2014 (IES. shear is provided on punches and dies so that (a) Press load is reduced (b) Good cut edge is obtained. the following rule is helpful: (a) Punch size controls hole size die size controls blank size (b) Punch P h size i controls t l both b th hole h l size i and d blank bl k size i (c) Die size controls both hole size and blank size (d) Die size controls hole size. Reason (R): The blank should be of required dimensions. GATE & PSUs) Page 47 of 78 .00 and 35+0. The work done during this shearing operation is (a) 200J (b) 400J (c) 600 J (d) 800 J S – 2002 IAS In deciding the clearance between punch and die in press work in shearing.00 (d) 35+0. The plate thickness is 4 mm and percentage penetration is 25. (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 not t the th correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 2002 IES Which one is not a method of reducing cutting forces to prevent the overloading of press? (a) Providing shear on die (b) Providing shear on punch (c) Increasing die clearance (d) ( ) Stepping pp g p punches S – 2003 IAS Match List I (Press‐part) with List II (Function) and select the correct answer using below the i the h codes d given i b l h lists: li List‐I List‐II (Press‐p ( part) ) (Function) ( ) (A) Punch plate 1. Reason(R): The flywheel stores energy when its speed increase.GATE – 2007 (PI) Circular blanks of 35 mm diameter are punched from a steel sheet of 2 mm thickness. ( ) 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 For-2014 (IES. If the clearance per side between the punch and die is to be kept as 40 microns.080 S – 2007 200 IAS For punching operation the clearance is provided on which one of the following? (a) The punch (b) The die (c) 50% on the punch and 50% on the die (d) ( ) 1/3rd 3 on the p punch and 2/3rd 3 on the die S – 1995 99 IAS Assertion (A): A flywheel is attached to a punching press so as to reduce its speed fluctuations. punch size controls blank size (a) 35+0. 4 Holding g the small p punch in the p proper p position Codes: A B C D A B C D (a) 4 3 2 1 (b) 2 1 4 3 (c) 4 1 2 3 (d) 2 3 4 1 S – 2000 IES Best position of crank for blanking operation in a mechanical press is (a) Top dead centre (b) 20 degrees below top dead centre (c) 20 degrees before bottom dead centre (d) ( ) Bottom dead centre S – 1999 999 IES Assertion (A): In sheet metal blanking operation. clearance must be given to the die.040 and 35‐0.080 (c) 35‐0. Advancing the work‐piece through correct di t distance (C) Stopper 3.040 (b) 35‐0. Ejection of the work‐piece from die cavity (D) ( ) Knockout 4.040 and 35‐0.080 and 35+0. the maximum punch load used in 2 x 105 N. the sizes of punch and di should die h ld respectively i l be b S – 1994 99 IAS In a blanking operation to produce steel washer. Assisting withdrawal of the punch (B) Stripper 2. 5 0 5r (b) Do = d + 2h + 2r (c) Do = d 2 + 2h 2 + 2r For-2014 (IES. Column I Column II P Wrinkling P. S – 2 (b) P – 4. R – 3. The most likely cause and remedy of the phenomenon are. blanks show a tendency to wrinkle up around the periphery (flange). S – 1 (c) P – 2. L Large grain i size i S. Insufficient blank holding force 5. Anisotropy R Stretcher R. Excessive blank holding force (a) P – 6. Yield point elongation Q. ll because b th there i no is blank‐holder (c) Tensile stress in one direction and compressive in the one other direction (d) Compressive in two directions and tensile in the third direction forming? Explain the function of a blank holder.4 mm is to be produced by cup drawing. (A) Buckling due to circumferential compression. How many drawing operations will be necessary if no intervening annealing is done. The number of deductions necessary will be (a) One (b) Two T (c) Three (d) Four E l Example A symmetrical cup of 80 mm diameter and 250 mm height is to be fabricated on a deep drawing die. R – 1.S – 2003 IAS The 'spring spring back back' effect in press working is (a) Elastic recovery of the sheet metal after removal of the load (b) Regaining the original shape of the sheet metal (c) Release of stored energy in the sheet metal (d) ( ) Partial recovery y of the sheet metal S – 1997 99 IES A cup of 10 cm height and 5 cm diameter is to be made from a sheet metal of 2 mm thickness. R – 1. What is drawing ratio and how is the drawing ratio increased ? [10 – marks] [ ] G ‐2003 GATE A shell of 100 mm diameter and 100 mm height with the corner radius of 0. Earing 4. Q – 3 3. Increase blank holder pressure (B) High blank holder pressure and high friction. R – 6. Fine grain size 6. 1 1. S – 4 (d) P – 4. respectively. The required blank diameter is (a) 118 mm (b) 161 mm ( ) 224 mm (c) (d) 312 mm ISRO‐2011 The initial blank diameter required to form a cylindrical li d i l cup of f outside id diameter di 'd‘ and d total height g 'h' having g a corner radius 'r' is obtained using the formula G ‐2006 GATE Match the items in columns I and II. Also find the drawing force IFS ‐ 2009 What is deep drawing process for sheet metal G ‐2008 GATE In the deep drawing of cups. decrease blank holder pressure G ‐1999 999 GATE Identify the stress ‐ state in the FLANCE portion of a PARTIALLYDRAWN CYLINDRICAL CUP when deep ‐ drawing without a blank holder (a) Tensile in all three directions (b) No N stress t i the in th flange fl at t all. GATE & PSUs) (d ) Do = d 2 +Page 4dh − 0. S h strains i 3. Q – 5. S – 2 (a ) Do = d 2 + 4dh − 0. Q – 3. Orange peel 2. Reduce blank holder pressure and apply lubricant (C) High temperature causing increase in circumferential length: Apply coolant to blank (D) Buckling due to circumferential compression.5 r78 48 of . Q – 5 5. To improve 4 p surface finish Select the correct answer using the code given below: (a) 1 and 2 only (b) 1.5 mm and half cone angle 30 30°. small 2. 2. the size of the round blank should be approximately (a) 42 mm (b) 44 mm ( ) 46 (c) 6 mm (d) 48 8 mm S – 2007 200 IAS In drawing operation. the subsequent draws in a deep p drawing g operation p have reduced percentages. 3 and 4 is S – 1997 99 IAS Which one of the following factor promotes the tendency for wrinking in the process of drawing? (a) Increase in the ratio of thickness to blank diameter of work material (b) Decrease D i the in th ratio ti thickness thi k t blank to bl k diameter di t of f work material (c) Decrease in the holding force on the blank (d) ( ) Use of solid lubricants S – 1994 99 IAS Consider the following factors 1. proper lubrication essential for which of the following reasons? 1 To improve die life 1.S – 2008 IES A cylindrical vessel with flat bottom can be deep drawn by (a) Shallow drawing (b) Single action deep drawing (c) Double action deep drawing (d) ( ) Triple p action deep p drawing g S – 1999 999 IES Consider the following statements: Earring in a drawn cup can be due to non‐uniform 1 Speed of the press 1.0 mm (c) (d) 1. 1 2 and 3 (b) 2. 2 3 and 4 (c) 1. The p 4 punch and die alignment g is not p proper. To reduce drawing forces 3.5 mm ( ) 2. by power spinning a missile cone of thickness 1.0 mm (b) 2. the second draw up to 40% reduction and. 2 and 4 S – 1994 99 IES For obtaining a cup of diameter 25 mm and height 15 mm by drawing.5 mm S – 1994 99 IES The mode of deformation of the metal during spinning is (a) Bending (b) Stretching (c) Rolling and stretching (d) ( ) Bending g and stretching. Clearance between the punch and the die is too small. 4 . 3. Clearance between tools 3. GATE & PSUs) Page 49 of 78 . 1 3 and 4 only (c) 3 and 4 only (d) 1. 4. Blank holding 4 g Which of these statements are correct? (a) 1. To reduce temperature 4. 2 3 and 4 (b) 1 and 2 (c) 2 and 4 (d) 1. the third draw of about 3 30% only. y Reason (R): Due to strain hardening. 2. 3 and 4 (d) 1. g For-2014 (IES. The finish at the corners of the die is poor. Material properties 4. (a) ( ) Both A and R are individually y true and R is the correct explanation of A (b) ( ) Both A and R are individually y 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 G ‐1992 992 GATE The thickness of the blank needed to produce. . is (a) 3. (a) 2. The finish at the corners of the punch is poor. 2. 3 and 4 S – 1998 998 IES Assertion (A): The first draw in deep drawing operation can have up to 60% reduction. p The factors responsible for the vertical lines parallel to the axis noticed on the outside of a drawn cylindrical cup would include. the explosive forming has proved to be an g energy g at high g rate and excellent method of utilizing utilizes both the high explosives and low explosives. Reason ( (R): ) The g gas p pressure and rate of detonation can be controlled for both types of explosives. Assertion (A) pulse ‐forming magnetic field produced by eddy currents is used to p create force between coil and workpiece.362 (d) 289 For-2014 (IES.414 1 414 (b) 1.IFS‐2011 Compare metal spinning with press work. GATE & PSUs) Page 50 of 78 .304 1 304 (c) 1. The final thickness of the sheet in mm is (a) 1. (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 t explanation l ti of fA (c) A is true but R is false (d) A is false but R is true IES 2010 ) : In the high g energy gy rate forming g Assertion ( (A) method. .05 and 0.5 mm thick sheet is subject to unequal biaxial stretching and the true strains in the directions of stretching are 0. Reason (R) : It is necessary for the workpiece material to have magnetic properties. [ ‐marks] [2 k ] G ‐2000 GATE A 1.09. [ ‐marks] [2 k ] IES 2011 High energy rate forming process used for forming components from thin metal sheets or deform thin tubes is: (a) Petro‐forming (b) Magnetic pulse forming (c) Explosive p forming g (d) electro‐hydraulic forming 20 0 JWM 2010 ( ) : In magnetic g p g method. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 2007 200 IES Which one of the following metal forming processes is not a high energy rate forming process? (a) Electro‐mechanical forming (b) Roll‐forming (c) Explosive forming (d) ( ) Electro‐hydraulic y forming g S – 2009 IES Which one of the following is a high energy rate forming process? (a) Roll forming (b) Electro‐hydraulic forming (c) Rotary forging (d) ( ) Forward extrusion S – 2005 200 IES Magnetic forming is an example of: (a) Cold forming (b) Hot forming ( ) High (c) Hi h energy rate t forming f i (d) Roll R ll f forming i IFS‐2011 Write four advantages of high velocity forming process. 3‐R. Tension and Compression 5. Die opening p g factor = Die opening 1. Embossing 4. Closed matching dies Codes:A B C D A B C D ( ) 1 (a) 3 4 2 (b) 3 1 4 2 (c) 1 3 2 Page 4 51 (d) 3 1 2 4 of 78 IES 2010 g statements: Consider the following The material properties which principally determine how well a metal may be drawn are 1. Bl ki 1. UTS = 455 MPa.Rate of increase of yield stress relative to progressive p g amounts of cold work.6 mm. 2 and 3 . 3‐Q. Extrusion 3. Flat plates and sheets 5. 3‐S. Solid l d and d hollow h ll parts Codes:A B C D A B C D ( ) 2 (a) 5 3 4 ( ) 1 (b) 2 5 4 (c) 4 1 For-2014 3 2 (d) 4 1 5 2 (IES. 4‐R (c) 1‐P. Wire Drawing 3 Blanking 3. Transfer moulding 5.IES‐1998 g force required q g.33 G ‐2005 200 GATE A 2 mm thick metal sheet is to be bent at an angle of one radian with a bend radius of 100 mm.5 05 (b) 2: 1 : 0. Tension Q Stretch Forming 2. 2‐P. GATE & PSUs) S – 1997 99 IAS Match List‐I (metal forming process) with List‐II (Associated feature) and select the correct answer using the codes given below the Lists: List‐l List‐ II A Blanking A. Forging 2. Coiled stock C. 4‐R Type of f stress P. 3. bending and Edge bending will be in the ratio of (a) 1 : 2 : 0. M ld d luggage l 1. Shear R Tensile and R. Blank thickness. Sh Shear angle l B. compressive S C S. Whi h of Which f the th above b statements t t t is/are i / correct? t? (a) 1 and 2 only (b) 2 and 3 only (c) 1 only (d) 1. U‐ The bending for V‐bending. 2‐Q. Rolling 1. p g = 8t. . 2 Compression R. Packaging containers for liquid 2. Discrete parts B. Tension and Shear C d P Q Codes:P R S P Q R S (a) 2 1 3 4 (b) 3 4 1 5 (c) 5 4 3 1 (d) 3 1 2 4 GATE ‐2012 Same Q in GATE‐2012 (PI) Match the following metal forming processes with their associated stresses in the workpiece. Impact extrusion S. Wide variety of shapes with thin walls ll D. If the stretch factor is 0. 4. Flow forming 2. Blanking 1. Hot rolling R. Coining 3. . Long structural shapes 3. Collapsible tubes 4. Compressive i (b) 1‐S. 4‐Q (d) 1‐P. 4 4. Drawing 4. Ratio of yield stress to ultimate stress. bend length = 1200 mm. Roll forming 3. Metal l forming f i process 1. I j ti moulding Injection ldi Q.5 05 (c) 1: 2 : 1 (d) 1: 1 : 1 E l Example g force for a 45o bend in aluminium Calculate the bending blank. Associated state of stress Processes P. 4‐S G ‐2004 200 GATE Match the following Product Process P Moulded P. 1. C i i Coining (a) P‐1 Q‐4 R‐6 S‐3 (b) P‐4 Q‐5 R‐2 S‐3 (c) P‐1 Q‐5 R‐3 S‐2 (d) P‐5 Q‐1 R‐2 S‐2 S – 1999 999 IAS ( ) with List II ( ) Match List I (Process) (Production of p parts) and select the correct answer using the codes given below the lists: List‐I List‐II A. 2. 2‐P. Rod and Wire C. Mandrel D. the bend allowance is (a) 99 mm (b) 100 mm ( ) 101 mm (c) (d) 102 mm 2mm 1 radian G ‐2007 200 GATE g metal Match the correct combination for following working processes. Rate of work hardening. Tensile Q. Q. Deep D Drawing D i (a) 1‐S. Coining 2. Shear S Deep Drawing S.5. 2‐R. Blow moulding 6 6. 3‐Q. sintering of a component (a) Improves strength and reduces hardness (b) Reduces brittleness and improves strength sintering is required to achieve which property? What i hot is h t iso i ‐static t ti pressing? i ? [ 2 Marks] ] (c) Improves hardness and reduces toughness (d) Reduces R d porosity i and d increases i bi l brittleness IES – 2011 Conventional What is isostatic pressing of metal powders ? What h are its advantage d ? ( ) GATE – 2009 (PI) Which of the following process is used to manufacture products with controlled porosity? (a) Casting ( ) welding (b) (c) formation (d) Powder metallurgy ( ) GATE – 2011 (PI) The binding material used in cemented carbide cutting tti tools t l is i (a) graphite (b) tungsten (c) ( ) nickel (d) cobalt [ 2 Marks] For-2014 (IES. No 1 2 3 4 5 6 7 8 9 Option C B A A D A A A A Q. GATE & PSUs) Page 52 of 78 . but ( ) GATE ‐2010 (PI) In powder metallurgy. aluminum plate is made d anode d (c) High amperage produces powdery deposit of cathode metal on anode (d) ( ) Atomization p process is more suitable for low melting g point metals IES – 2007 Conventional Metal powders are compacted by many methods. No 10 11 12 13 14 15 16 17 Option C C C C D A B D ( ) GATE ‐2011 (PI) Powder Metallurgy gy Which of the following powder production methods th d produces d spongy and d porous particles? ti l ? (a) Atomization (b) Reduction of metal oxides (c) ( ) Electrolytic y deposition p (d) Pulverization By S K Mondal IES ‐ 2012 In electrolysis (a) For making copper powder.Ch‐18: Sheet Metal Forming Q. copper plate is made cathode in electrolyte tank (b) For making aluminum powder. compacting sintering and sizing (b) Blending. Powder metallurgy Codes:A B C D A B C D ( ) 3 (a) 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2 GATE 2011 The operation in which oil is permeated into the pores of a powder metallurgy product is known as ( ) mixing (a) i i (b) sintering (c) impregnation (d) Infiltration S – 1998 998 IES In powder metallurgy. Brake lining 3 Self‐lubricating bearings 3. 2 and 4 (d) 1.IES 2010 Consider the following parts: 1. Powder compacting D. Gears 3. Sand casting 5. Blending compacting. 2 and 3 (b) ( ) 2 only y (c) 2 and 3 only (d) 1 and 2 only IES 2010 Metallic powders can be produced by (a) Atomization (b) Pulverization (c) Electro‐deposition process (d) All of the above S – 2002 IES The rate of production of a powder metallurgy part depends on (a) Flow rate of powder (b) Green strength of compact (c) Apparent density of compact (d) ( ) Compressibility p y of p powder S – 2001 200 IES Match List‐I (Components) with List‐II (Manufacturing Processes) and select the correct answer using the codes given below the lists: List I List II A Car A. blending and sintering (d) ( ) Compacting. Which of these parts are made by powder metallurgy ll technique? h i ? (a) ( ) 1. 2 2. Aluminium gears 3. p g blending. Bakelite gears 2. Filters 4. g sizing g and sintering g S – 1997 99 IES p ) with List‐II ( Match List‐I ( (Gear component) (Preferred method of manufacturing) and select the correct answer using the codes given below the Lists: List‐I List‐II A. 3 and 4 ( (b) ) 2 and 3 (c) 1. sizing and sintering (c) Compacting. Brake linings Select the correct answer using the codes given below: (a) ( ) 1. Engine block 4. Casting C. Automobile transmission gears 4. Grinding wheel 2. Clutch lining 2. the operation carried out to improve the bearing property of a bush is called (a) infiltration (b) impregnation (c) plating (d) heat treatment S – 1997 99 IES Which of the following components can be manufactured by powder metallurgy methods? 1 Carbide tool tips 1. Bearings 3. M hi i Machining B. sizing. compacting. Hobbing B. Gear for clocks 1. . Stamping C. GATE & PSUs) Page 53 of 78 . Sheet metal pressing D. C body b d (metal) ( t l) 1. 3 and 4 S – 1999 999 IES The correct sequence of the given processes in manufacturing by powder metallurgy is (a) Blending. Extrusion Code:A B C D A B C D (a) 2 3 5 1 (b) 5 3 4 2 (c) 5 1 2 3 (d) 2 4 5 3 For-2014 (IES. 2. Porous bearings 3. 3 and 4 (d) 1. journals l Code:A B C D A B C D ( ) 2 (a) 3 4 1 ( ) 3 (b) 2 1 4 (c) 2 3 1 4 (d) 3 2 4 1 S – 2007 200 IES What are the advantages of powder metallurgy? 1. 2. costs 2. 2 and 3 (b) 1. Good Embeddability 3. Low L labour l b cost t 3. the molten metal is forced through a small orifice and broken up p by y a stream of compressed p air. Select the correct answer using the codes given below: (a) Only 1 and 2 (b) Only 3 and 4 (c) Only 1 and 4 (d) Only 1. 4.I. speeds D. Runs well with C. 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only S – 2006 IES Which of the following are the limitations of powder metallurgy? 1 High tooling and equipment costs. Porous B. Wastage of material. High density Which of the above are limitations of powder metallurgy? (a) 1. Available press capacity 4. Extreme purity product 2. 1. Small magnets 4. C. (II): ) Particle shape p in p powder metallurgy gy Statement ( influences the flow characteristic of the powder. 2 and 3 (d) 1 and 2 IES ‐ 2012 ) Parts made by y powder p gy do not Statement ( (I): metallurgy have as good physical properties as parts casted. Parts with intricate shapes 4 p Select the correct answer using the codes given below: Codes: (a) 1. (a) ( ) Both A and R are individually y true and R is the correct explanation of A (b) ( ) Both A and R are individually y true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1997 99 IES Match List‐I with List‐II and select the correct answer using the codes given below the Lists: List‐I List‐II (Bearing materials) (Properties) A. Babbits 1.S – 2001 200 IES Carbide‐tipped cutting tools are manufactured by powder‐ metal technology process and have a composition of (a) Zirconium‐Tungsten (35% ‐65%) (b) Tungsten T t carbide bid ‐Cobalt C b lt (90% ( % ‐ 10%) %) (c) Aluminium oxide‐ Silica (70% ‐ 30%) (d) Nickel‐Chromium‐ Tungsten (30% ‐ 15% ‐ 55%) S – 1999 999 IES Assertion (A): In atomization process of manufacture of metal powder.I. Low equipment cost. Sintered powdered metal 4. 2 and 4 (c) 2. Bronze 2. 2 and 4 S – 2004 200 IES Consider the following factors: 1. It cannot be automated. Reason ( (R): ) The metallic p powder obtained by y atomization process is quite resistant to oxidation. Expensive 4 p metallic p powders. Select the correct answer using the code given below (a) 1. 3. 3 Suitable for high g loads and low C. 1 3 and 4 (b) 2 and 3 (c) 1. Size and shape that can be produced economically 2. (a) ( ) Both Statement ( (I) ) and Statement ( (II) ) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true For-2014 (IES. 3 and 4 Page 54 of 78 . GATE & PSUs) S – 2009 IES Which of the following cutting tool bits are made by powder metallurgy process? (a) Carbon steel tool bits (b) Stellite tool bits (c) Ceramic tool bits (d) HSS tool bits S – 2003 IAS Which of the following are produced by powder metallurgy process? 1 Cemented carbide dies 1. Porosity P it of f the th parts t produced d d 3. Reason (R): In atomization process inert‐gas or water cannot be used as a substitute for compressed air. Q – 1. 3 and 4 (d) 1. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S – 1998 998 IAS Throwaway tungsten manufactured by (a) Forging (c) Powder metallurgy carbide (b) (d) tip tools are Brazing Extrusion S – 1996 996 IAS Which one of the following processes is performed in powder metallurgy to promote self‐lubricating properties in sintered parts? (a) Infiltration (b) Impregnation ( ) Plating (c) Pl ti (d) Graphitization G hiti ti GATE ‐2008 (PI) Match the following Group – 1 P Mulling P. Q – 4. 2 and 4 S – 1997 99 IAS ) C Assertion ( (A): Close dimensional tolerances are NOT possible with isostatic pressing of metal powder in powder metallurgy technique.S – 2003 IAS In parts produced by powder metallurgy process. Reason (R): In the process of isostatic pressing. R – 4. R – 2. Pressing 4 g or compacting p g into the desired shape p Indentify the correct order in which they have to be performed and select the correct answer using the codes given below: (a) 1‐2‐3‐4 (b) 3‐1‐4‐2 (c) 2‐4‐1‐3 (d) Page 4‐3‐2 ‐1 of 78 55 S – 2003 IAS Assertion (A): Atomization method for production of metal powders consists of mechanical disintegration of molten stream into fine particles. (a) . 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 g of the p part S – 2000 IAS Consider the following processes: 1. 2. Flash l h trimming S. (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 p of A (c) A is true but R is false (d) A is false but R is true . p Reason (R): Atomization method is an excellent means of making gp powders from high g temperature p metals. S ‐ 1 (d) P – 4. 2. GATE & PSUs) S – 2004 200 IAS The following are the constituent steps in the process of powder metallurgy: 1 Powder conditioning 1. Processing of f FRP composites 4. Q – 3. Sand casting g ( ) P – 2. Sintering Which of these processes are used for powder preparation in powder metallurgy? (a) 2. R – 3. R – 2. the pressure is equal in all directions which permits uniform density of the metal powder. Sintering 3. Curing g Group ‐2 1 Powder metallurgy 1. Impregnation R. 2 3 and 4 (b) 1. Injection moulding 3. S – 3 S – 2007 200 IAS ) Mechanical disintegration g Assertion ( (A): of a molten metal stream into fine particles by means of a jet of compressed air is known as atomization. Q – 1. S ‐ 3 ( ) P – 4. Q. 1 2 and 3 (c) 1. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true For-2014 (IES. Chemical reduction 4. S – 1 (c) P – 2. Mechanical pulverization 2. Production of metallic powder 4. (b) . Atomization At i ti 3. S – 2007 200 IAS Consider the following basic steps involved in the production of porous bearings: 1 Sintering 1.0001 mm respectively.00 mm (d) 0. Mixing 3.0. 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 +0.0100 GATE ‐ 2004 In an interchangeable assembly. ± 0. By S K Mondal For PSU Tolerances are specified ( ) To obtain (a) b desired d d fits f (b) because it is not possible to manufacture a size exactly (c) ( ) to obtain higher g accuracy y (d) to have proper allowances ISRO‐2010 Expressing a dimension as 25.00 mm. GATE & PSUs) Page 56 of 78 .00 mm and shaft is 50 +0.009 (a) − 0. The maximum possible clearance in the assembly will be ( ) 10 microns (a) i (b) 20 microns (c) 30 microns (d) 60 microns S O‐2010 20 0 ISRO 0.02 0 02 mm (b) 0. Cold‐die‐compaction Which one of the following is the correct sequence of the above steps? Q.016 (d) − 0.02 +0.05 mm is the case of (a) Unilateral tolerance (b) Bilateral tolerance (c) Limiting dimensions (d) All of f the h above b 2010 ISRO‐2012 GATE – 2010.008 (c) − 0. A shaft has a dimension.020 25.01 0 01 mm For-2014 (IES.009. Impregnation 4 p g 5.016 GATE ‐ 1992 Two shafts A and B have their diameters specified as 100 ± 0.000−0.009.02 Dimension of the hole is 50+ mm −0.000 mm. ± 0.0. No 1 2 3 4 Ch‐12: Powder Metallurgy gy Option D B C A Q.000 mm mate with holes of size 25.040 −0. The minimum clearance is (a) 0.φ35−0.025.025 Th respective The ti values l of f fundamental f d t l deviation d i ti and d tolerance are −0. 2. Repressing 4.02 mm (c) -0.1 mm and 0. shafts of size +0.008 (b) − 0. No 5 6 7 8 Option C B D C Limit Tolerance & Fits Limit.025.1 ± 0.3±0. The shafts produced by three different machines A. The 25 +0. is of a unilateral holes and a shafts.02 mm. A milling cutter of 20 mm width is located with reference to the side of the block within ± 0.01 mm.070 0 070 (b) 0. B and C have mean diameters of 19∙99 mm. The maximum clearance in mm between the hole and the shaft is (a) 0. The tolerance on the shaft is 0. with same standard d i ti deviation.015 0 015 +0 GATE ‐2012 Same Q in GATE‐2012 (PI) In an interchangeable assembly. GATE & PSUs) Page 57 of 78 . The maximum offset in mm between the centre lines of the slot and the block is (a) ± 0.09 mm. B d C since i th standard the t d d deviation is same for the three machines (b) Least L t for f machine hi A (c) Least for machine B (d) Least for machine C GATE ‐ 2000 A slot is to be milled centrally on a block with a dimension of 40 ± 0. 0 0 0 mm.070 0 070 (c) ± 0. of nominal diameter 20 mm. with limits. 20∙00 mm and 20. B and C? ( ) Same (a) S f the for th machines hi A Band A.010 mm.01 IAS‐2011 Main An interference assembly.05 (c) 0.10 (d) 0. Engine piston and cylinder (c) Joint between a pulley and shaft transmitting power (d) Joint J i of f lathe l h spindle i dl and d its i bearing b i GATE 2011 A hole is of dimension φ 9 +0.04 −0.001 mm. The mating shaft has a clearance fit with minimum clearance of 0. it is essential that the lower limit of the shaft should be (a) Greater than the upper limit of the hole (b) Lesser than the upper limit of the hole (c) Greater than the lower limit of the hole (d) ( ) Lesser than the lower limit of the hole (a) Joint of cycle axle and its bearing (b) Joint between I.02 0 02 mm. [10‐Marks] corresponding p g shaft is of dimension The resulting assembly has (a) loose running fit (b) close running fit ( ) transition (c) t iti fit (d) interference fit φ 9 +0.01 mm respectively.03 to 0.11 IES 2011 Interference fit joints are provided for: (a) Assembling bush bearing in housing (b) Mounting heavy duty gears on shafts (c) ( ) Mounting gp pulley y on shafts (d) Assembly of flywheels on shafts interference fits? IES‐2013 Which of the following is a joint formed by GATE ‐ 2005 In order to have interference fit.03 25+ +0.C. Wh t will What ill be b the th percentage t rejection j ti f for the shafts produced by machines A. The maximum interference (in microns) in the assembly is ( ) 40 (a) (b) 30 ( ) 20 (c) (d) 10 For-2014 (IES.025 mm. for the two mating parts.04 (b) 0.0 5 0 A hole is specified as 4 0 0 . The manufacturing f t i t l tolerances f the for th holes h l are twice t i that for the shaft.05 mm. shafts of size +0.00 ± 0.04 mm. Permitted interference values are 0.045 GATE ‐ 2007 0 . mm mate with holes of size 0. Determine the sizes.IES ‐ 2005 p y the designer g The tolerance specified by for the diameter of a shaft is 20.020 (d) 0. Cyclic loading 2.IES ‐ 2007 ISRO‐2011 A shaft and hole pair is designated as 50H7d8. Railway carriage wheel and axle 2. An interference fit creates no stress state in the shaft. The amount of interference needed to create a tight joint varies with diameter of the shaft. shaft 2. 2 and 3 IES ‐ 2003 ) with List‐II (Significant ( g Match List‐I ( (Phenomenon) Parameters/Phenomenon) and select the correct answer using the codes given below the Lists: List‐I List‐II (Phenomenon) (Significant Parameters/Phenomenon) / ) A. 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only IES ‐ 2004 Consider the following fits: 1. Interference fit 1. IC engine i cylinder li d and d liner li Which of the above joints is/are the result(s) of interference fit? (a) ( ) 1 only y (b) 2 only (c) Neither 1 nor 2 (d) Both 1 and 2 IES ‐ 2008 Consider the following statements: 1. Interference C. This assembly constitutes (a) Interference fit (b) Transition fit (c) Clearance fit (d) None of the above IES ‐ 2006 Which of the following is an interference fit? (a) Push fit (b) Running R i fit (c) Sliding fit (d) Shrink fit IES ‐ 2009 Consider the following joints: 1. GATE & PSUs) Page 58 of 78 . h ft 3. Induced compressive stress t Codes:A B C D A B C D (a) 3 4 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 4 3 1 2 GATE ‐ 2001 Allowance in limits and fits refers to (a) Maximum clearance between shaft and hole (b) Minimum Mi i clearance l b t between shaft h ft and d hole h l (c) Difference between maximum and minimum size of hole (d) ( ) Difference between maximum and minimum size of shaft GATE ‐ 1998 In the specification of dimensions and fits. Lubricating of bearings 4. The stress state in the hub is similar to a thick‐ walled cylinder with internal pressure. Which of the statements g given above are correct? (a) 1. Ball B ll bearing b i outer t race and d housing h i 3. I. engine cylinder and piston 2. (a) Allowance is equal to bilateral tolerance (b) Allowance All i equal is l to t unilateral il t l tolerance t l (c) Allowance is independent of tolerance (d) Allowance is equal to the difference between p by y the maximum and minimum dimension specified tolerance.C. Notch sensitivity D. For-2014 (IES. Gear meshing 3. Ball bearing inner race and shaft Which of the above fits are based on the interference y system? (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1. Viscosity index B. 24. 2 and 3 (b) 1 and d 2 only l (c) 1 and 3 only (d) 3 only IES ‐ 2002 In the tolerance specification 25 D 6. The fundamental deviation for shaft designation ‘f’ f is ‐5.967 Page 59 of 78 GATE ‐ 2000 A fit is specified as 25H8/e8. shaft (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true S‐2006 Conventional C i l IES Find the limit sizes.5 5 5 D0. Which of the statements given above is/are incorrect? (a) 1. T l Tolerance value l for f lT8 = 25i.5D0.033. Hole H l diameter di t is i 50 mm.41 The values of standard tolerances for grades of IT 8 and d IT 10 are 25i i and d 64i 6 i respectively.970 mm. in μ m= 0. It is a shaft base system. [15‐Marks] IES ‐ 2008 Consider the following statements: A nomenclature φ 50 H8/p8 denotes that 1.046 mm For-2014 (IES. where D is the representative size in mm. Upper Limit = 60.924 mm.001D. i the maximum and minimum limits of dimension for a 25 mm H8 hole (in mm) are (a) 24.016 mm (c) (d) Lower limit = 60.000 mm.954 mm. 2.984 (d) 25. Reason (R): Hole has to be given a larger tolerance band than the mating shaft.45 D1/3 + 0. ‐ve ( ) Zero. If the fundamental deviation for H hole is zero and IT8 = 25 i. Also specify the type of fit that the above pair belongs to.001D.000 (b) 25. ti l Also. Upper Limit = 60. The tolerance value for a nominal diameter of 25 mm in IT8 is 33 microns and fundamental deviation for the shaft is ‐ 40 microns. Given: 100 mm diameter lies in the diameter step range of 80‐120 mm.45 ³√D + 0. 25. i.000 mm ( ) Lower limit = 59. Zero (c) (b) ‐ve. The unit of D is mm.984. i Fundamental deviation for 'f shaft = ‐5. Upper Limit = 59.41 ( ) Lower limit (a) l = 59. 24. & PSUs) GATE – 2008 (PI) Following data are given for calculating limits of dimensions and tolerances for a hole: Tolerance unit i (in µm) = 0. tolerances and allowances for a 100 mm diameter shaft and hole pair designated by F8h10. Fundamental tolerance unit. 24.967 (c) 25. The maximum clearance of the fit in microns is (a) ‐7 (b) 7 (c) 73 (d) 106 . indicate the limits and tolerance on a diagram.970 mm (b) Lower limit = 59.000. GATE Upper Limit = 60. 8 indicates fundamental deviation. 3. the letter D represents (a) Grade of tolerance (b) Upper deviation (c) Lower deviation (d) ( ) Type yp of fit GATE ‐ 2009 pp What are the upper and lower limits of the shaft represented by 60 f8? Use the following data: Diameter 60 lies in the diameter step of 50‐80 mm.IES ‐ 2012 Clearance in a fit is the difference between (a) Maximum hole size and minimum shaft size (b) Minimum Mi i h l size hole i and d maximum i shaft h ft size i (c) Maximum hole size and maximum shaft size (d) Minimum hole size and minimum shaft size ISRO‐2008 Basic shaft and basic hole are those whose upper deviations and lower deviations respectively are (a) +ve. Diameter step is 18‐30 mm.017. +ve ( ) None of the above (d) IES ‐ 2005 Assertion (A): Hole basis system is generally preferred to shaft basis system in tolerance design for getting the required fits. 979 mm (c) 25.005 (c) 25.96 6 (d) 65 6 ±0.00 ± 0.021 mm (b) 25. then the upper (maximum) limit of the bore (in mm) is (a) 25.125 mm. For P = 18.1 0 1 and S = 72. mm When the bore is designated as 25H6. GATE – 2008 (PI) Assuming g a normal distribution for the individual component dimensions. 18 75 ± 0. then the upper (maximum) limit is 25. respectively.05 0 05 A three‐component welded cylindrical assembly is shown below. When the bore is designated as 25H8. B2 and B3 are to be inserted in a channel of width S maintaining a minimum i i gap of f width idth T = 0. GATE & PSUs) Page 60 of 78 .009 (d) 25. 25.000.08.001 (b) 25.The type of fit is (a) Clearance fit (b) Running fit (sliding fit) (c) Push fit (transition fit) (d) ( ) Force fit ( (interference fit) ) IES ‐ 2000 Which one of the following tolerances set on inner diameter and outer diameter respectively of headed jig bush for press fit is correct? (a) G7 h 6 (b) F7 n6 ( ) H 7h (c) h6 (d) F7j6 F j6 ISRO‐2008 Interchangeability can be achieved by (a) Standardization (b) Better process planning (c) Simplification (d) Better B product d planning l i IAS‐2010 main What is the difference between hole basis system and shaft basis system ? Why is hole basis system the more extensive t i in i use ? What are the differences between interchangeability and selective assembly ? [12‐Marks] GATE ‐ 2003 GATE ‐ 1997 Three blocks B1 .033 25 033 mm. The mean length of the three components and their respective tolerances (both in mm) are given in the table below. 24.38 0 38 (c) + 0. (where all dimensions are in mm).013 1996 IES‐2012 GATE – 1996.12.GATE ‐ 2003 The dimensional limits on a shaft of 25h7 are (a) 25. the natural tolerance limits for the length (Y) of the assembly (mm) is ( ) 65 (a) 6 ±2.000.38 0 38 (b) ‐ 0. 25.000.36 6 For-2014 (IES. Q = 25.000 mm and 25.35 + X.021 mm.05 0 05 (d) ‐0.007 mm (d) 25.56 6 (c) ( ) 65 6 ±0.993 mm GATE‐2010 (PI) A small bore is designated as 25H7. The lower (minimum) and upper (maximum) limits of the bore are 25. 0 08. 24.125 28 125 ± 0.000. mm) the tolerance X is (a) + 0. The fit on a hole‐shaft system is specified as H7‐ s6. R = 28.16 6 (b) 65 6 ±1. as shown in Figure. 014 mm and 20. A ring gauge is used to measure (a) Outside diameter but not roundness (b) Roundness R d b t not but t outside t id diameter di t (c) Both outside diameter and roundness (d) Only external threads Measurement of Lines & Surfaces For-2014 (IES.GATE – 2007 (PI) Tolerance on the dimension x in the two component assembly shown below is (All dimensions in mm) ( ) (a) ( ) ±0.025 (a) ( ) ±0.014 mm and 20.0 micron micron.052 (c) 25.020 mm 0 0.080 ( ) 30+ (c) +0..040 mm GATE‐2013 0.084 GATE ‐ 1995 Checking the diameter of a hole using GO‐NO‐GO gauges is an. example of inspection by ….050 b t between 30+ the plating l ti thickness thi k varies i +0.030 mm 0.010 mm diameter are ISRO‐2008 Plug gauges are used to ( ) Measure the (a) h diameter d of f the h workpieces k (b) Measure the diameter of the holes in the workpieces (c) ( ) Check the diameter of the holes in the workpieces (d) Check the length of holes in the workpieces GATE ‐ 2004 GO and NO‐GO plug gages are to be designed for a 0.054 mm electroplated l t l t d in i a shop.054 54 mm (d) 20. tolerance Following ISO system of gage design.070 070 (d) 30+ +0.074 (d) 25.070 (a) 30+ +0.030 3 (a) ±0.040 ( ) ±0.010 mm.01 mm.010 mm and 20. If th between 10 .15 microns. h Thickness Thi k of f the th plating is 30 ±2. sizes of GO and NO‐GO gage will be respectively (a) 20.065 (b) 30+ +0.046 mm (c) ( ) 20.05 hole 200.(variables/attributes) The above statement is ( ) Variables (a) V i bl (b) Attributes (c) Cant say (d) Insufficient data 2006 VS‐2012 GATE – 2006. GATE & PSUs) Page 61 of 78 By S K Mondal .020 Cylindrical pins of 25+ +0.006 mm and 20. the size of the GO gage in mm to inspect the plated components is (a) 25.050 mm (b) 20. Gage tolerances can be taken as 10% of the hole tolerance.042 (b) 25.030 mm 0 080 0. diameter of the hole before plating should be 0. Neglecting gage tolerances.045 (a) GATE ‐2007(PI) The g geometric tolerance that does NOT need a datum for its specification is (a) Concentricity (c) Perpendicularity (b) Runout (d) Flatness GATE – 2007 (PI) Diameter of a hole after plating needs to be controlled 0 050 0. 2011 ISRO‐2009.005 5 mm (c) 0.620 mm (c) 2.02 mm (b) ( ) 0. ISRO‐2008 Standards to be used for reference purposes in laboratories and workshops are termed as (a) Primary standards ( ) Secondary standards (b) (c) Tertiary standards (d) Working standards GATE – 2007 (PI) Which one of the following instruments is a comparator ? ( ) Tool (a) T l Maker’s M k ’ Microscope Mi (b) ( ) GO/NO GO g gauge g (c) Optical Interferometer (d) Dial Di l Gauge G PSU A feeler gauge is used to check the (a) Pitch of the screw (b) Surface roughness (c) Thickness of clearance (d) Flatness Fl of f a surface f For-2014 (IES. over one rotation. In a simple micrometer with screw pitch 0. matching with 24 divisions of main scale (1 main scale divisions = 0. which one of the following would be consistent with the observation? (A) The drill spindle rotational axis is coincident with the drill spindle taper hole axis (B) The drill spindle rotational axis intersects the drill spindle taper hole axis at point P (C) The drill spindle rotational axis is parallel to the drill spindle taper hole axis (D) The drill spindle rotational axis intersects the drill spindle taper hole axis at point Q ISRO‐2010 A master gauge is (a) A new gauge (b) An international reference standard ( ) A standard (c) t d d gauge for f checking h ki accuracy of f gauges used on shop floors (d) A gauge used by experienced technicians A displacement sensor (a dial indicator) measures the lateral displacement of a mandrel mounted on the taper hole inside a drill spindle. Measurements are taken with the sensor placed at two positions P and Q as shown in the figure. The readings are recorded as Rx = maximum deflection minus minimum deflection. The mandrel axis is an extension of the drill spindle taper hole axis and the protruding portion of the mandrel surface is perfectly cylindrical. corresponding to sensor position at X.01 mm (d) 0.5 mm divisions on thimble 50.120 mm (b) ( ) 2.ISRO‐2010 The vernier reading should not be taken at i face its f value l b f before an actual l check h k has h been taken for (a) Zero error (b) Its It calibration lib ti (c) ( ) Flatness of measuring gj jaws (d) Temperature equalization ISRO‐2008 The least count of a metric vernier caliper having 25 divisions on vernier scale.512 5 mm (d) 5. the reading corresponding to 5 divisions on barrel and 12 divisions on thimble is (a) ( ) 2.5 mm) is (a) ( ) 0.005mm and . GATE & PSUs) Page 62 of 78 .012 mm GATE – 2008 S‐1 td from f GATE – 2008 contd… S‐1 If Rp= RQ>0. The calculated taper angle (in degrees) is (a) 21.8 (c) 23. (ii) [10 – marks] ISRO‐2011 A sine bar is specified by (a) Its total length (b) The size of the rollers ( ) The (c) Th centre t distance di t b t between th two the t rollers ll (d) ( ) The distance between rollers and upper pp surface GATE ‐2012 (PI) A sine bar has a length of 250 mm.723 (c) ( ) 2. Explain how (i) shock load is avoided.IAS‐2011 Main Draw a self explanatory sketch of Sigma mechanical comparator. the height g from the surface plate to the centre of a roller is 100 mm.9 ( ) GATE – 2011 (PI) The best wire size (in mm) for measuring effective di diameter t of f a metric t i thread th d (included (i l d d angle l is i 60 6 o) of 20 mm diameter and 2. The effective diameter (in mm) of the thread is (a) 33.1 (b) 22.086 GATE‐2013 A metric t i thread th d of f pitch it h 2 mm and d thread th d angle l 60 6 inspected p for its p pitch diameter using g 3‐wire method. Each roller has a diameter di t of f 20 mm.366 (d) 26. [ ‐marks] [10 k ] of ISRO‐2011 CLA value and RMS values are used for measurement (a) Metal hardness (b) Sharpness of tool edge (c) Surface dimensions (d) Surface roughness For-2014 (IES. GATE & PSUs) Page 63 of 78 . ( ) oscillations of the pointer are damped.000 (c) 1.5 mm pitch using two wire i method th d is i (a) 1.443 (b) 0.532 mm and 15.5 3 5 mm pitch. The micrometer readings over the standard and over the wires are 16.5 mm diameter and a d t two o wires es o of 2 mm d diameter a ete eac each a are e used.6 (d) 68. respectively.366 (b) 30.000 ( ) GATE – 2011 (PI) To measure the effective diameter of an external metric thread (included angle is 60o) of 3. The diameter of the best size wire in mm is (a) 0. p . inspection y ? Define the terms 'roughness g What are they height'.398 mm. During D i t taper angle l measurement of a component.397 IES ‐ 1992 Which grade symbol represents surface rough of broaching? (a) N12 (b) N8 (c) N4 (d) N1 IFS‐2011 What Wh t is i meant t by b interchangeable i t h bl manufacture? f t ? Laser light has unique advantages for inspection.886 (d) 2. a cylindrical standard of 30.866 (b) 1. 'waviness width' and 'lay' in connection with surface irregularities.154 (d) 2.397 (c) 29. IES ‐ 2006 The M and E‐system in metrology are related to measurement of: (a) Screw threads (b) Flatness (c) Angularity (d) Surface finish IES ‐ 2007 What is the dominant direction of the tool marks or scratches in a surface texture having a directional quality, called? (a) Primary texture (b) Secondary texture ( ) Lay (c) L (d) Flaw Fl IES ‐ 2008 What term is used to designate the direction of the predominant surface pattern produced by machining operation? (a) Roughness (b) Lay ( ) Waviness (c) W i (d) Cut C t off ff IES 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) IES ‐ 2008 ISRO‐2010 Surface roughness represented d by b (a) Triangles (b) Circles (c) Squares (d) Rectangles on a drawing is (a) (c) A 4 4 B 2 1 C 1 2 D 3 3 (b) (d) A 3 3 B 2 1 C 1 2 D 4 4 GATE ‐ 1997 List I List II (A) Surface profilometer 1. Calibration (B) Light Section Microscope 2. 2 Form tester (C) Microkater 3. Film thickness measurement (D) Interferometer 4. Centre line average 5 5. Comparator 6. Surface lay measurement C d A B Codes:A C D A B C D (a) 4 1 2 3 (b) 4 3 5 1 (c) 4 2 1 3 (d) 3 1 2 4 For-2014 (IES, GATE & PSUs) ISRO‐2007 Gratings are used in connection with (a) Flatness measurement (b) Roundness measurement (c) Surface texture measurement (d) Linear Li di l displacement measurements GATE ‐ 2003 Two slip gauges of 10 mm width measuring 1.000 1 000 mm and 1.002 mm are kept side by side in contact with each other lengthwise. An optical flat is kept resting on the slip gauges as shown in the figure. Monochromatic light of wavelength 0.0058928 mm is used in the inspection. The total number of straight fringes that can be observed on both slip gauges is (a) 2 (c) 8 (b) 6 (d) 13 Page 64 of 78 GATE – 2011 (PI) Observation of Ob ti f a slip li gauge on a flatness fl t interferometer produced fringe counts numbering 10 and d 14 for f two t readings. di Th second The d reading di is i taken by rotating the set‐up by 180o. Assume that b th faces both f of f the th slip li gauge are flat fl t and d the th wavelength of the radiation is 0.5086 µm. The parallelism ll li error (in (i µm) ) between b t th two the t f faces of f the slip gauge is ( ) 0.2543 (a) (b) 1.172 (c) 0.5086 (d) 0.1272 GATE ‐ 1998 Auto collimator is used to check (a) Roughness (b) Flatness Fl t (c) Angle (d) Automobile balance. Miscellaneous of Metrology By S K Mondal ( ) GATE – 2009 (PI) An autocollimator is used to (a) measure small angular displacements on flat surface ( ) compare known and unknown dimensions (b) (c) measure the flatness error (d) measure roundness error between centers S O‐2010 20 0 ISRO Optical square is (a) Engineer's square having stock and blade set at 90o (b) A constant t t deviation d i ti prism i h i having th angle the l of f deviation between the incident ray and reflected ray, equal l to t 90o (c) A constant deviation prism having the angle of deviation between the incident ray and reflected ray, equal to 45o (d) Used to produce interference fringes IES ‐ 1998 Match List‐I with List‐II and select the correct answer using the codes below the d given i b l h lists: li List‐I List‐II (Measuring ( g Device) ) (Parameter Measured) ( ) A. Diffraction grating 1. Small angular deviations on long flat surfaces B B. Optical flat 2 2. On‐line measurement of moving parts C. Auto collimators 3. Measurement of gear pitch D D. L Laser scan micrometer4. i t S f Surface t t texture using i interferometer i t f t 5. Measurement of very small displacements Code: A B C D A B C D (a) 5 4 2 1 (b) 3 5 1 2 (c) 3 5 4 1 (d) 5 4 1 2 GATE ‐ 1992 y q y Match the instruments with the p physical quantities they measure: Instrument Measurement (A) Pilot‐tube (1) R.P.M. of a shaft g (2) Displacement p (B) McLeod Gauge (C) Planimeter (3) Flow velocity (4) 4 Vacuum (D) LVDT (5) Surface finish (6) ( ) Area Codes:A B C D A B C D (a) ( ) 4 1 2 3 (b) ( ) 3 4 6 2 (c) 4 2 1 3 (d) 3 1 2 4 GATE ‐ 2004 Match the following Feature to be inspected Instrument P Pitch and Angle errors of screw thread 1. 1 Auto Collimator Q Flatness error of a surface plate 2. Optical Interferometer R Alignment Ali error of f a machine hi slide lid way 3. Dividing Di idi Head H d and Dial Gauge S Profile P fil of f a cam 4. Spirit S i i Level L l 5. Sine bar 6. Tool maker's Microscope (a) P‐6 Q‐2 R‐4 S‐6 (b) P‐5 Q‐2 R‐1 S‐6 (c) P‐6 Q‐4 R‐1 S‐3 (d) P‐1 Q‐4 R‐4 S‐2 GATE ‐ 1995 List I (Measuring instruments) (A) Talysurf T l f 1. (B) Telescopic gauge 2. (C) Transfer callipers 3. (D) Autocollimator 4. Codes:A B C D ( ) 4 (a) 1 2 3 (b) (c) 4 2 1 3 (d) List II (Application) T‐slots l t Flatness Internal diameter Roughness A B C D 4 3 1 2 3 1 2 4 For-2014 (IES, GATE & PSUs) Page 65 of 78 GATE ‐ 2010 A taper hole is inspected using a CMM, with a probe of 2 mm diameter. At a height, Z = 10 mm from the bottom, 5 points are touched and a diameter of circle (not compensated for probe size) is obtained as 20 mm. Similarly, a 40 mm diameter is obtained at a height Z = 40 mm. the smaller diameter (in mm) of hole at Z = 0 is (a) 13.334 (b) 15.334 (c) 15.442 (d) 15.542 GATE ‐2008 (PI) An experimental setup is planned to determine the taper of workpiece as shown in the figure. If the two precision rollers have radii 8 mm and 5 mm and the total thickness of slip gauges inserted i d between b the h rollers ll i 15.54 mm, the is h taper angle θ is ( ) 6 degree (a) d (b) 10 degree (c) 11 degree (d) 12 degree g ISRO‐2007 Which Whi h of f the h following f ll i errors are inevitable i i bl in i the h measuring g system y and it would be vain full exercise to avoid them (a) Systematic errors (b) Random R d errors (c) Calibration errors (d) Environmental errors Ch‐13: Metrology gy Q. No 1 2 3 4 5 6 7 8 9 Option C C A C C B C B B Q. No 10 11 12 13 14 15 16 17 Option D A B B D B C B For-2014 (IES, GATE & PSUs) Page 66 of 78 carbides Coated carbides. chief alloying chromium. Page 67 of 78 Contd… . hardness (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Ans. may be present in some p p proportion.9 to 1. This is the major factor in tool life. and wood working Carbon content about 0. and single‐crystal Cutting Tool Materials By S K Mondal proper cutting tool (material and geometry) for a given work material. vanadium. Most of the high speed steels contain tungsten as the y g element. (a) For-2014 (IES. and cost.Introduction Success in metal cutting depends on selection of the Cemented carbides. A wide range g of cutting g tool materials is available with a variety of properties. Reason(R): During machining. dry The hot hardness value is low. l Ceramics. Cermets. etc. GATE & PSUs) High speed steel These steels metals Th t l are used d for f cutting tti t l at t a much h higher cutting speed than ordinary carbon tool steels. ll Contd… natural diamond. machining the tip of the cutting tool is heated to 600/700°C which cause the teal tip to lose its hardness. The high speed steels have the valuable property of retaining their hardness even when heated to red heat. but other elements like cobalt. ( ) Sintered polycrystalline diamond. Sintered d polycrystalline l ll cubic b boron b nitride d (CBN).35% with a hardness ABOUT 62 C Rockwell Maximum cutting speeds about 26 ft/min. Th f i d to mass FIGURE: Improvements in cutting tool materials have reduced machining time. Carbon Steels Limited not suited Li i d tool l life. Whisker reinforced ceramics. performance capabilities. production Can C b formed be f d into i complex l shapes h f for small ll production runs Low L cost Suited to hand tools. ceramics Sialons. Productivity raised by cutting tool materials S – 1997 99 IAS Assertion (A): Cutting tools made of high carbon steel have shorter tool life. Coated d high h h speed d steels. Cast carbides. These include: High carbon Steels and low/medium alloy steels. steels High‐speed steels. Fig. lif Therefore. Cast cobalt b l alloys. This steel contains 20 per cent tungsten. 4 per cent chromium and 2 per cent vanadium. C (d) Cu. planer and shaper g cutters. IAS‐1997 Which of the following processes can be used for production thin. 1% V (d) 18% Cr. 4 per cent chromium. broaches. It is considered to be one of the best of all purpose tool steels. Molybdenum high speed steel This 6 per cent Thi steel t l contains t i 6 per cent t tungsten. chromium and 1 per cent vanadium. Cu Zn Zn. 1% V (b) (c) 18% W. threading g tools. di It has excellent toughness and cutting ability. 2. punches. Sn Ans. TiN . on HSS? 1. Select the correct answer using the codes given below: Codes: (a) 1 and 3 (b) 2 and 3 (c) 2 and 4 (d) 1 and 4 Ans. 4% Ni. V (c) Cr. t t t IES‐2003 The Th correct t sequence of f elements l t of f 18 8‐4‐1 HSS tool is ( ) W. in order t increase to i th cutting the tti efficiency ffi i especially i ll at t high hi h temperatures. heat resistant coating at TiN. lathe. Cr. GATE & PSUs) Page 68 of 78 . The molybdenum high speed steels are better and p than other types yp of steels. cheaper It is particularly used for drilling and tapping operations. milling dies. It is widely used for drills. 4% W. With time the effectiveness and efficiency of HSS IES‐2013 Vanadium in high speed steels: (a) Has a tendency y to p promote decarburization (b) Form very hard carbides and thereby increases the wear resistance of the tool (c) Helps in achieving high hot hardness (d) Has H a tendency d to promote retention i of f Austenite A i Ans (b) Ans. (a) IES 2007 Cutting C tti tool t l material t i l 18 8‐4‐1 HSS has h which hi h one of f the following compositions? ( ) 18% W. For-2014 (IES. etc by Chemical Vapour Deposition (CVD) or Physical Vapour Deposition (PVD) 18‐4‐1 High speed steel This 4 per cent Thi steel t l contains t i 18 8 per cent t tungsten. S Sintering te g u under de reducing educ g at atmosphere. 1% V Ans (a) Ans. . 3. IES‐1993 The Th blade bl d of f a power saw is i made d of f (a) Boron steel (b) High speed steel (c) Stainless steel (d) Malleable cast iron A (b) Ans. hard. (a) (too s) a d t e app cat o range a ge were e e gradually g adua y (tools) and their application enhanced by improving its properties and surface condition through ‐ Refinement of microstructure Addition of large g amount of cobalt and Vanadium to increase hot hardness and wear resistance respectively Manufacture by powder metallurgical process Surface coating g with heat and wear resistive materials like TiC . 2 per cent vanadium and 12 per cent cobalt. Plasma spraying. because cobalt is added from 2 to 15 per cent. Physical vapour deposition. 4% Cr. V (a) (b) Mo. osp e e. Ni. 4% Ni. etc. 1% V (a) ( ) 18% Cr. t t t Super high speed steel This called Thi steel t l is i also l ll d cobalt b lt high hi h speed d steel t l molybdenum. Cr. Chemical vapour deposition with post treatment 4. reamers. 1% V. 2% V. and Ta. Tools of cast cobalt alloys are generally cast to shape and IES 2011 Stellite is a non‐ferrous cast alloy composed of: (a) Cobalt.7% C 3. (a) ( ) Both A and R are individually y 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 of f A (c) A is true but R is false (d) A is false but R is true Ans (a) Ans.IES‐1995 The of Th compositions iti f some of f the th alloy ll steels t l are as under: 1. These tool materials are much harder. nonferrous alloys.15% 4. GATE & PSUs) Page 69 of 78 . Cast cobalt alloys y are hard as cast and cannot be softened or heat treated. C bid hi h are nonferrous f ll l ll d finished f h d to size by b grinding. Reason (R): Chromium and cobalt in High Speed promote martensite formation when the tool is cold worked. Cast cobalt alloys are currently being phased out for cutting‐tool applications because of increasing costs. Reason (R): As a rule. Materials machinable with this tool material include p plain‐ carbon steels. Cast cobalt alloys contain a primary phase of Co‐rich solid solution strengthened by b Cr and W and dispersion hardened by b complex hard. Contd… Other elements added include V. shortages of strategic raw materials (Co. Cutting speed of up to 80‐100 fpm can be used on mild steels. Consequently. B. Although comparable in room‐temperature hardness to high‐ speed d steel l tools. 8% Mo. g titanium. chromium. l cast cobalt b l alloy ll tools l retain i their h i hardness h d to a much higher temperature. 0. 4% Cr. 0. HSS They are more brittle and more expensive and use strategic metals t l (W. 3% Ni. are chemically more stable. techniques Most carbide tools in use today are either straight g carbide ( (WC) ) or multicarbides of W‐Ti or W‐ tungsten Ti‐Ta. T Co) C ) more extensively. The high cost of fabrication is due primarily to the high hardness of the material in the as‐cast condition. tungsten (a) (b) Tungsten. 27% Cr. i h chromium h i ‐tungsten‐ carbon b cast alloys having properties and applications in the intermediate range g between high g ‐speed p steel and cemented carbides. 18% W. depending on the work material to be machined. chromium and tungsten (b) Tungsten. (b) Cast cobalt alloys/Stellite Cast cobalt are cobalt C b l alloys ll b l ‐rich. t i l Contd… For-2014 (IES. W. vanadium (c) Chromium. 8% Ni. 12 Mo 1 W 4 Cr 1 V 3. the heat dissipation is efficient. d They are available only in simple shapes. they can be used at higher g cutting g speeds p (25% ( 5 higher) g ) than HSS tools. Ni. molybdenum. 0. 18 8 W 4 Cr C 1 V 2. and Cr). 0 15% C Which of these relate to that of high speed steel? ( ) 1 and (a) d 3 (b) 1 and d 2 (c) 2 and 3 (d) 2 and 4 Ans. 18 W 8 Cr 1 V The compositions of commonly used high speed steels would include (a) 1 and 2 (b) 2 and 3 (c) 1 and 4 (d) 1 and 3 Ans. vanadium (d) Tungsten. Tungsten chromium. the coolant or cutting fluid preferred is water‐ miscible mineral fatty oil. IAS 1994 Assertion (A): The characteristic feature of High speed Steel is its red hardness. 4% Cr. other superior tool materials at lower cost. 5% Mo. titanium. y tungsten g and chromium (d)Tungsten. 5% Co. vanadium and chromium (c) ( ) Molybdenum. alloy steels. water‐based oils are recommended for high‐speed operations in which high temperatures are generated due to high frictional heat. chromium titanium Ans. sintered (or cemented) carbides because they are manufactured by powder metallurgy techniques.7% C 2. (W Ta. and the development of other. 6 Mo 6 W 4 Cr 1 V 4. and cast iron. such as single‐ point i t tools t l and d saw blades. chromium and vanadium Ans (a) Ans. high stiffness. heat Water being a good coolant. Cemented Carbide Carbides. and lower fi i friction. 6% W. 0.25% C 4 18% Cr. (a) IAS‐2001 Assertion (A): For high‐speed turning of magnesium alloys. Cobalt is the binder. (d) IES‐2000 Percentage P t of f various i alloying ll i elements l t present t in different steel materials are given below: 1. are also called. refractory carbides of W and Cr. bl d b because of f limitations li it ti i the in th casting process and expense involved in the final shaping (grinding). which alloys. have better hot hardness. (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 (c) i true b but R i is f false l (d) A is false but R is true Ans. (b) IES‐1992 The in hi high Steel Th main i alloying ll i elements l t i h speed d St l in i order of increasing proportion are ( ) Vanadium. and d operate at higher hi h cutting i speeds d than h do d HSS. G C l h d T i illi b i i b hi Malleable C. but are not required. hardness > 220 HB HB. steel Turning. grey C. milling. milling.l. P‐50 2. small chips Turning. Turning milling milling. milling. malleable cast iron Steel. shaping at low cutting speeds strength M40 For-2014 (IES. These are used for higher‐speed (> 1000 ft/min) finish machining of steels and some malleable cast irons. low tensile Turning. Steel. (c) S – 1994 99 IAS Assertion (A): Cemented carbide tool tips are produced by powder metallurgy. low cutting speed speed. Ferrous material. roughing cut D. reaming. alloy IES‐1999 Match List‐ M h List Li ‐I (ISO classification l ifi i of f carbide bid tools) l ) with i h Li II (Applications) and select the correct answer using the codes g given below the Lists: List‐I List‐II A. and milling high speed. squares. diamonds. steel castings. steel castings. austenitic steel. large chip section K20 K30 K40 M10 M20 M30 Steel and steel castings Operations requiring high toughness turning. chip section Steel casting. milling. K‐10 3. C l hardness up to 220 Turning milling Turning. spherodized C. alloys scraping containing Si Grey C. boring.l. ferrous and non. GATE & PSUs) Hard grey C.l. of medium or low tensile planning. Non‐ferrous.. and d rounds. brass and light machines. t th therefore th f th bits the bit are often ft brazed b d to t steel t l shanks. d Compressive strength is high compared to tensile strength. milling. finishing cut Code: A B C D A B C D ( ) 4 (a) 3 1 2 ( ) (b) 3 4 2 1 (c) 4 3 2 1 (d) 3 4 1 2 A ( Ans.ferrous material and d non – metals t l like lik Cast C tI Iron. broaching requiring high HB toughness Soft grey C.. primarily applications using predominantly Ni and Mo as a binder. Al. threading. medium cutting speed. Malleable C. reaming under favourable conditions strength steel Soft non-ferrous metals Turning milling etc. medium speed with small chip section Turning. milling. Brass etc. Non‐ferrous. ISO Code P Colour Code Application For machining long chip forming common materials like plain carbon and low alloy steels For machining long or short chip forming ferrous materials like Stainless steel For machining short chipping. Turning. chilled casting. machining high speed Turning.l. casting austentic steel. or used as inserts in holders. Steel. Special alloys are needed to cut steel Contd… IES‐1995 The straight grades of cemented carbide cutting tool materials contain (a) Tungsten carbide only (b) Tungsten carbide and titanium carbide (c) Tungsten carbide and cobalt (d) Tungsten T t carbide bid and d cobalt b lt carbide bid Ans. (a) The standards developed by ISO for grouping of carbide tools and their application ranges are given in Table below.l. medium cutting speed and medium manganese steel. Turning. steel castings. P‐10 1. Ferrous material. planning. planning. cast (a) Both A and R are individually true and R is the correct t explanation l ti of fA (b) Both A and R are individually true but R is not the correct explanation of A (c) ( ) A is true but R is false (d) A is false but R is true Ans.. Steel. Turning. K‐50 4. precision turning and boring. specially in automatic Free cutting steel.l.l. (c) ) Page 70 of 78 . M K Table below shows detail grouping of cemented carbide tools ISO Application group P01 P10 P20 P30 P40 P50 Material Process K01 K10 Steel Steel castings Steel. Al.l. malleable cast iron Steel and steel casting with sand inclusions Precision and finish machining. chip section spherodized C. Reason (R): Carbides cannot be melted and cast. heat medium or large chip section resisting alloys f turning. Low tensile Turning. roughing g g cut B. Steel castings Steel.l. medium cutting speed and medium manganese steel. Cemented carbide tools are available in insert form in many different shapes. milling. Cemented carbide tool materials based on TiC have bee e oped. finishing cut C. triangles. medium speed with small chip section Turning planning Turning. broaching. p a y for o auto industry dust y been de developed. profile strength steel. Turning. l Contd… Speeds up to 300 fpm are common on mild steels Hot hardness properties are very good Coolants C l t and d lubricants l bi t can be b used d to t increase i t l tool life. broaching. alloys with high silicon scraping Grey C. These Th i inserts t may often ft have h negative ti rake k angles. Transformation T f i toughening h i b adding by ddi appropriate i amount of partially or fully stabilised zirconia in Al O powder. Contd… technique – this material is very tough but prone to built‐up‐ edge formation in machining steels Developing SIALON – deriving beneficial effects of Al2O3 and d Si3N4 S Adding carbide like TiC (5 ~ 15%) in Al2O3 powder – to i impart t toughness t h and d thermal th l conductivity d ti it Reinforcing oxide or nitride ceramics by SiC whiskers. materials In view of their ability to withstand high temperatures. sintered. and have higher wear resistance than the other cutting tool materials. p y this novel and inexpensive p tool is still in experimental stage. l Typical products can be machined are brake discs. microstructure. strength toughness and life of the tool and thus productivity spectacularly. Sinterability. Al2O3 d Isostatic and hot isostatic pressing (HIP) – these are very effective but expensive route. case h d hardened d and d hardened h d d steel. if applied should in flooding with copious quantity of fluid. GATE & PSUs) Page 71 of 78 . since ceramics have very poor thermal shock resistance. resistance Else. These can withstand very high temperatures. High Performance ceramics (HPC) IES‐2013 Sialon ceramic is used as: (a) Cutting g tool material (b) Creep resistant (c) Furnace linens IES 2010 Constituents of ceramics are oxides of different materials. cylinder liners and flywheels. acid heated and cooled (d) Ground. they can be used for machining at very high speeds of the order of 10 m/s. p cutting Very high hot hardness properties Often used as inserts in special holders. and the tendency to chipping. fluid to thoroughly wet the entire machining zone. Silicon Nitride (i) Plain (ii) SIALON (iii) Whisker toughened Alumina toughned by (i) Zirconia (ii) SiC whiskers (iii) Metal (Silver (Sil er etc) (d) High strength Ans. palleted and after calcining cooled in oxygen Ans (b) Ans. holders Comparison of important properties of ceramic and tungsten carbide tools Contd… Through last few years remarkable improvements in Introducing nitride ceramic (Si3N4) with proper sintering strength and toughness and hence overall performance of ceramic tools could have been possible by several means which include. which enhanced strength. poor thermal characteristics. Ground washed with acid. Toughening Al2O3 ceramic by adding suitable metal like silver which also impart thermal conductivity and self lubricating gp property. Else it can be machined with no coolant. i which have high hardness. strength and toughness of Al2O3 ceramics were improved to some extent by adding TiO2 and MgO. such as hard castings. i The main problems of ceramic tools are their low refractory materials introduced specifically for high speed machining of difficult to machine materials and cast iron. sintered and p palleted to make ready y ceramics (c) Ground. Contd… Cutting fluid. which are ( ) Cold (a) C ld mixed i d to t make k ceramic i pallets ll t (b) ( ) Ground. They are not suitable for intermittent cutting or for low g speeds. Ceramic C i tools l are used d for f machining hi i work k pieces. (a) For-2014 (IES. .Ceramics Ceramics C i are essentially i ll alumina l i ( Al2O3 ) based b d high hi h It is possible to get mirror finish on cast iron using ceramic i turning. Contd… strength. temperatures are chemically more stable. They can be operated at from two to three times the cutting speeds of tungsten carbide. brake drums. li h smooth and continuous cuts at high speeds.3. H. hard refractory coating of TiC. Multiple coatings are used.IAS‐1996 Match List I with List II and select the correct answer using the codes given below the lists: List I ( (Cutting g tools) ) List II ( (Major j constituent) ) A.4 (c) 3. b Interface coatings are graded to match the properties of the coating and the substrate. GATE & PSUs) Contd… and porosity. Stellite l. The coatings g must be thick enough g to p prolong g tool life but thin enough to prevent brittleness. For-2014 (IES. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) ( ) A is true but R is false (d) A is false but R is true Ans. porosity Naturally.2 1342 Ans (c) Ans. which one of the following is the correct sequence of increasing hardness of the tool materials? (a) Cast alloy y‐HSS‐Ceramic‐Carbide (b) HH‐Cast alloy‐Ceramic‐Carbide (c) HSS‐Cast alloy‐Carbide‐Ceramic (d) Cast alloy‐HSS‐Carbide‐Ceramic Ans. Columbium 5. DCON 4. 2. chemically stable. shock‐resistant carbide that can withstand ith t d high hi h‐temperature t t plastic l ti deformation d f ti and d resist breakage. Tungsten B. cutting A thin. Cobalt C.4. can consistently improve. Chemical vapour p deposition p (CVD) ( ) is the technique q used to coat carbides. tool lif 200 or 300% life % or more.S. Ti i Titanium Codes: A B C D A B C D (a) 5 1 3 4 (b) 2 1 4 3 (c) 2 1 3 4 (d) 2 5 3 4 Ans (c) Ans.1. the coatings must be metallurgically bonded to the h substrate.2.4. used with each layer imparting its own characteristic to the tool. Which h h one of f the h f following ll tool l materials will be suitable for machining the component under d the h specified f d cutting conditions? (a) Sintered carbides (b) Ceramic (c) ( ) HSS (d) ( ) Diamond Ans.4 3124 (d) 1. Contd… . TiN. Reason (R): Ceramics have a high wear resistance and high temperature resistance. High‐speed steel 4. & free of binders industry because coating . d chemically h i ll inert to prevent the tool and the work material from interacting chemically with each other during cutting.2. Contd… Page 72 of 78 The most successful combinations are TiN/TiC/TiCN/TiN and TiN/TiC/ Al2O3 . (d) IAS‐2000 Consider C id the h following f ll i cutting i tool l materials i l used d for f metal‐cutting operation at high speed: 1. Coatings should have a low coefficient of friction so that the chips do not adhere to the rake face. objective The bulk of the tool is a tough. Ceramic C i The correct sequence in increasing order of the range of cutting speeds for optimum use of these materials is (a) 3. In cutting tools. TiN or Al2O3 accomplishes this objective.3.S. i h d and hard.1. 3. material requirements at the surface of the tool l need d to be b abrasion b i resistant. (b) ( ) IES‐1996 A machinist desires to turn a round steel stock of outside diameter 100 mm at 1000 rpm. The depth of cut chosen is 3 mm at a feed rate of 0. Tungsten carbide 2 Cemented titanium carbide 2. The material has tensile strength of 75 kg/mm2. Alumina D.3 mm/rev. (c) ( ) Coated Carbide Tools Coated the d tools l are becoming b h norm in the h metalworking l k The coatings must be fine grained. (b) IES 2007 Which one of the following is not a ceramic? (a) Alumina (b) Porcelain (c) Whisker (d) Pyrosil Ans. Ceramic 3. IAS‐2003 At room temperature.2 (b) 1. IES‐1997 Assertion Ceramic A i (A): (A) C i tools l are used d only l for f light. cuts Modern cermets with rounded cutting edges are suitable for finishing and semi‐finishing of steels at higher speeds. stainless steels but are not suitable for jerky interrupted machining and machining of aluminium and similar materials. gear‐shaper cutters. taps. reground ceramics metal i like lik TiC. carbide even coated carbides in case of light cuts. i d The advantage of TiN‐coated HSS tooling is reduced tool wear. thus allowing individual tools to be g more times. low l thermal h l expansion. zinc and tungsten Ans. (d) S – 2003 IES The correct sequence of cutting tools in the ascending order of their wear resistance is (a) HSS‐Cast non‐ferrous alloy (Stellite)‐Carbide‐ Nitride (b) Cast C t non‐ferrous f alloy ll (St llit )‐HSS‐Carbide (Stellite) C bid ‐ Nitride (c) HSS‐Cast non‐ferrous alloy (Stellite)‐Nitride‐ Carbide (d) Cast non‐ferrous alloy (Stellite)‐Carbide‐Nitride‐ Ans. % Cutting edge sharpness is retained unlike in coated carbide inserts Can machine steels at higher cutting velocity than that used for tungsten carbide. Ni‐Co. IES‐2000 Cermets are (a) Metals for high g temperature p use with ceramic like properties (b) Ceramics with metallic strength and luster (c) Coated tool materials (d) Metal M t l‐ceramic i composites it Ans. (c) best process for coating HSS. and drills. zinc aluminium and magnesium alloys. one such as copper. TiN‐Coated High‐Speed Steel Coated high‐speed steel (HSS) ( ) does not routinely provide as dramatic improvements in cutting speeds as do coated carbides. HSS Therefore. and an assortment of other milling cutters. H d more chemically Harder. g coated by TiN now includes reamers. HSS primarily because it is a relatively low temperature process that does not exceed the tempering point of HSS. h i ll stable t bl and d hence h more wear resistant i t t More brittle and less thermal shock resistant Wt% of f binder bi d metal t l varies i from f 10 to t 20%. Diamonds This is used when good surface finish and dimensional accuracy are desired. bandsaw and circular saw blades. TiN and NaCN (b) TiC and TiN (c) TiN and NaCN (d) TiC and NaCN A (b) Ans. form tools. GATE & PSUs) Page 73 of 78 are the non‐ferrous one. In addition to hobs. spade‐drill blades. materials diamonds are not suitable because of the diffusion of carbon atoms from diamond to the work‐piece Contd… material. copper brass. includes (a) TiC. high heat conductivity. (a) HSS Diamond is the hardest of all the cutting tool materials. Diamond Di d has h the h following f ll i properties: i extreme hardness. On ferrous materials. and a very low co‐efficient of friction. end mills. d i d The work‐materials on which diamonds are successfully employed For-2014 (IES. insert tooling. . Fe etc. Less tool wear results in less stock removal during g tool regrinding. with increases of 10 to 20% being typical. TiC TiN or TiCN and d ‘met’ ‘ ’ from f l (binder) (bi d ) like Ni. broaches. HSS tooling chasers. Contd… Physical vapour deposition (PVD) has proved to be the Cermets These sintered hard inserts are made by combining ‘cer’ from IES 2010 The cutting tool material required to sustain high temperature is (a) High carbon steel alloys (b) Composite of lead and steel (c) Cermet (d) Alloy of steel.IAS‐1999 The coating materials for coated carbide tools. no subsequent heat treatment of the cutting i tool l is i required. brass zinc. It is made by bonding a 0. High wear resistance 3.5 mm) of'fine grain‐ size diamond particles sintered together and metallurgically bonded to a cemented carbide substrate. quality greater toughness. toughness It shows excellent performance in grinding any material of high hardness and strength. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) ( ) A is true but R is false (d) A is false but R is true Ans. the diamond tool is kept flooded with coolant. diamond is preferred because it has 1. strength Contd… For-2014 (IES. carbides Tool life is often greatly improved. (d) S – 1999 999 IAS Assertion (A): During cutting. High 3 g compression p strength g 4. lapping powder and for grinding wheel dressing. finish. . i Oxidation of diamond starts at about 450oC and thereafter it can even crack. Diamond (d) HSS. Contd… improvements over carbides. GATE & PSUs) Page 74 of 78 . toughness and improved wear resistance.5 – 1 mm layer of p y y polycrystalline cubic boron nitride to cobalt based carbide substrate at very high temperature and pressure. (b) ( ) S – 1999 999 IES Consider the following statements: For precision machining of non‐ferrous alloys. Due to their brittle nature. applications If BUE is a problem. HSS (b) Carbide. HSS Di Diamond. honing g boring grinding tools. Low fracture toughness Which of these statements are correct? (a) 1 and 2 (b) 1 and 4 (c) 2 and 3 (d) 3 and 4 Ans. Reason (R): The oxidation of diamond starts at about 4500C ( ) 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 Ans. Polycrystalline diamond (PCD) tools consist of a thin layer (0. Carbide. (a) IES‐1992 Which of the following given the correct order of increasing hot hardness of cutting tool material? (a) Diamond. substrate The main advantages of sintered polycrystalline tools over natural single‐crystal tools are better quality. integrity Positive rake tooling is recommended for the vast majority of diamond tooling applications.( ) GATE – 2009 (PI) Diamond cutting tools are not recommended for machining of ferrous metals due to (a) high tool hardness ( ) high thermal conductivity of work material (b) (c) poor tool toughness (d) chemical affinity of tool material with iron Diamond tools have the applications in single point turning and Diamond tools offer dramatic performance g tools. and surface integrity. and light feeds are used. . d Carbide C bid Ans. as is control over part size.g g wheels. increasing cutting speed and the use of f more positive i i rake k angles l may eliminate li i it. . Diamond. . reamers. (a) Cubic boron nitride/Borazon Next to diamond.5 to 1. Low coefficient of thermal expansion 2. Reason (R): Diamond tools are suitable for high speed machining. carbide. IES‐1995 Assertion Non‐ferrous materials best A i (A): (A) N f i l are b machined with diamond tools. milling g cutters. should be loaded lightly. HSS (c) HSS. resulting from the random orientation of the diamond g grains and the lack of large g cleavage g planes. Reason (R): Diamond tools have very low coefficient of friction. For this reason the diamond tool is kept flooded by the coolant during cutting. cubic boron nitride is the hardest material presently available. (b) ( ) IES‐2001 Assertion Diamond high A i (A): (A) Di d tools l can be b used d at hi h speeds. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) ( ) A is true but R is false (d) A is false but R is true Ans. It remains inert and retains high hardness and fracture oug ess a at e elevated e a ed machining ac g speeds. the diamond tools have poor resistance to shock and so. (a) IES‐2002 Which one of the following is the hardest cutting tool material next only to diamond? (a) Cemented carbides (b) Ceramics (c) Silicon (d) Cubic C bi boron b nitride it id Ans. Nitride D. Diamond 3. cBN tools are also available only y in the form ceramics. (d) IES‐1996 Cubic boron nitride (a) Has a very y high g hardness which is comparable p to that of diamond. i Microfine TiCN particles are uniformly dispersed into the matrix. Consider the following tool materials: 1. Ceramic 4. (c) For heat treatment (d) For none of the above. hard‐chill cast iron. forces tool life and surface finish. GATE & PSUs) Page 75 of 78 .4. and nickel‐ and cobalt‐ based superalloys. l Ans. b d the h coronite based b d tool l is made d of f three h layers. Stellite 4. The only limitation of it is its high cost. Carbide 2. Stellite 2. Correct sequence of these tool materials in increasing order of their ability to retain their hot hardness is ( ) 1. CBN 3. (b) ( ) Coronite Coronite is made basically by combining HSS for strength and IES‐1993 Match List the correct the M t h Li t I with ith List Li t IT and d select l t th t answer using i th codes given below the lists: List ‐ I (Cutting tool Material) List ‐ I I(Major characteristic h t i ti constituent) tit t) A. The operative speed range for cBN when machining grey cast iron is 300 ~400 400 m/min Speed ranges for other materials are as follows: Hard H d cast t iron i ( 400 BHN) : 80 (> 8 – 300 m/min / i Superalloys (> 35 RC) : 80 – 140 m/min Hardened steels (> 45 RC) : 100 – 300 m/min It t is s best to use c cBN too tools s with t a honed o ed o or c chamfered a e ed edge preparation. Molybdenum C. 1. 1. (b) Has a hardness which is slightly more than that of HSS (c) Is used for making cylinder blocks of aircraft engines (d) Is I used d f for making ki optical ti l glasses. (b) toughness and and h d tungsten carbides bid for f heat h d wear resistance.3. Coated carbide tool 4. Carbon y B. Unlike l k a solid l d carbide. UCON Select the correct answer using the codes given below: C d Codes: (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 (d) 2 and 3 Ans. Cobalt Codes: A B C D A B C D (a) 2 1 3 5 (b) 2 5 1 3 (c) 5 2 4 3 (d) 5 4 2 3 Ans. Cemented carbide 2. For-2014 (IES. the th central t l HSS or spring i steel t l core a layer of coronite of thickness around 15% of the tool diameter a thin (2 to 5 μm) PVD coating of TiCN The coronite tools made by b hot extrusion e trusion followed follo ed by b PVD‐ coating of TiN or TiCN outperformed HSS tools in respect of cutting forces. cost Contd… IES‐1994 CBN is less reactive with such materials as hardened steels. Columbium 5 5. 3 Ans. High speed steel 1. 4.2. and surface integrity. of indexable inserts. especially for interrupted cuts. CBN can be used efficiently y and economically y to machine these difficult‐to‐machine materials at higher g removal rate speeds (fivefold) and with a higher (fivefold) than cemented carbide. Ans. (d) IAS‐1998 Which of the following tool materials have cobalt as a constituent element? 1. Borazon. Cermet 3. 4 (d) 2.2. (a) IES‐1994 Cubic boron nitride is used (a) As lining g material in induction furnace (b) For making optical quality glass.4 (a) (b) 1.3 (c) 2. 3. Like . finish. and with superior accuracy. 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. The adhesive wear in the rough region is called attrition wear . IAS‐2001 Match. 2 Ans. 1. when the seizure between workpiece to tool is broken. 2. 1. Which of the statements given above are correct? (a) 1 and 3 (b) 2 and 4 (c) 1 and 4 (d) 2 and 3 Ans. Silicon carbide 4. Attrition wear The bonding the Th strong t b di between b t th chip hi and d tool t l material t i l at t IES‐1996 The to th the maximum hardness of Th limit li it t i h d f a work k material which can be machined with HSS tools even at t low l speeds d is i set t by b which hi h one of f the th following tool failure mechanisms? ( ) Attrition (a) (b) Abrasion (c) Diffusion (d) Plastic deformation under compression. 4 Extrusion 5. 2 (a) (b) 4. Cemented carbide 3. (b) IES‐1999 Match List the correct M t h List Li t‐I with ith Li t‐II and d select l t th t answer using the codes given below the Lists: List I List II (Materials) (Applications) A. compression Ans. UCO UCON 4. 4. Cemented carbide 3. 2. Rolling D. (b) IES‐2000 Consider the following tool materials: 1. 2. Decreases D the h hardness. 1 (d) 3. Sili Silicon nitride i id 2. List (Cutting List M t h Li t I (C tti tool t l materials) t i l ) with ith Li t II (Manufacturing methods) and select the correct answer using the codes given below the Lists: List I List II A. surface Therefore. Casting B B. IES‐2005 Consider An increase in the C id the th following f ll i statements: t t t A i i th cobalt content in the straight carbide grades of carbide tools 1. 4. 1 (c) 3. Increases the hardness. 3. the small fragments of tool material are plucked and brought away by the chip. Lowers the transverse rupture strength. some parts of the worn surface are still covered d by b molten l chip hi and d the h irregular i l attrition ii wear occurs in this region . (a) high temperature is conducive for adhesive wear. Tungsten carbide 1. Diamond The correct sequence of these materials in decreasing order of their cutting speed is ( ) 4. Stellite 2 2. Increases the transverse rupture strength 4. GATE & PSUs) Page 76 of 78 . In the rough region. Forging C. HSS 1. Ceramics 4. Pipes for conveying liquid q metals D. h d 3. Aluminium oxide 3. HSS 2. Abrasive wheels B B. (d) For-2014 (IES. Drawing dies Code: A B C D A B C D (a) 3 4 1 2 (b) 4 3 2 1 (c) 3 4 2 1 (d) 4 3 1 2 Ans (d) Ans. 3.IES‐2003 Which one of the following is not a synthetic abrasive material? (a) Silicon Carbide (b) Aluminium Oxide (c) Titanium Nitride (d) Cubic Boron Nitride Ans. H i elements Heating l C. Powder metallurgy Codes:A B C D A B C D (a) 3 1 5 2 (b) 2 5 4 3 (c) 3 5 4 2 (d) 2 1 5 3 Ans (d) Ans. B2. B6 C5.42. 35. 429. 21. 4.78.49 D ‐ Page‐3 C B D C B ‐ C D B A Page‐4 B D C D B C C A B A A ‐ A D D B Page‐5 B A B D Page‐6 C&D Page‐7 1597. 1200. GATE & PSUs) Page 77 of 78 . D3 B C A C D B D B C D D C C D A A B C C C B C D A D B B D D D D D A A D A C A D D C D C C B A D C ‐ A C C Page‐21 B A C Page‐22 D B D Page‐23 None C ‐ C B ‐ A C C(11) A D A A A Page‐24 Page‐25 Page‐26 A D A Page‐27 C C A Page‐28 Page‐29 Page‐30 C B D B A ‐ A B B B B C Page‐31 C ‐ C Page‐32 A3. 231.85 26 195 ‐ C C B ‐ B A B B B C B B ‐ B C B A C C C D A D B A Page‐20 B A A ‐ A D C C Gutter C B D C D A A C C B A ‐ D 0. C C1. 127 1265 Page‐8 12 B 0. D5 D C B A Page‐33 Page‐34 Page‐35 ‐ C D A 2134 B For-2014 (IES. 0.3. 54 A D C C A D A ‐ A2.021 C C A B D A B b D A D D B A D C B B A C Page‐11 A C D B B A A C B B C A B C A Page‐12 A C B B ‐ ‐ A Page‐13 B ‐ A A B B C A B D A A Page‐14 C B C C A A C A B Page‐15 C ‐ D ‐ B ‐ C B Page‐16 D Page‐17 A Page‐18 Page‐19 A 36.816 74 Page‐9 Page‐10 B B B D A ‐ D B B B A 828.PRODUCTION ANSWERS‐2014 For any doubt send SMS to 9582314327 Page‐1 ‐ A B B C A C C D B B C D C ‐ Page‐2 ‐ D C D B D B B B F291 N457 Fn336 Fs408 β32. 10.67 2.07. 1454. GATE & PSUs) Page 78 of 78 .63 ‐ ‐ ‐ ‐ 44.4 A C A B B Page‐45 * A C B B ‐ B B B D ‐ D ‐ B B B A C C B B D D C D ‐ ‐ C D B ‐ B C ‐ 74.93 ‐ B 3.Page‐36 539.04 130 ‐ ‐ Page‐41 A C B B B A A D D B B A ‐ D B A ‐ D D C D D A A C D C B D ‐ D C D B ‐ ‐ A A A C B C ‐ B ‐ C A C A ‐ D Page‐42 C D C B D D A C D B B D C A ‐ A C C B ‐ A ‐ C C D D A D B ‐ A B Page‐43 B ‐ C A A D C A C D B C C ‐ Extrusion ‐ Page‐44 C B 4. 467 Centre Page‐37 3.33.231 ‐ B D ‐ B B Page‐38 A B A B A B A C C Page‐39 D B C B C C B D C Page‐40 C D C A A B ‐ 2. 28 A ‐ B D C B A D B C A ‐ D C D A B B B B ‐ A D B B Page‐46 D C A Page‐47 A A B Page‐48 Page‐49 B C A C C C C A C C C ‐ C D D A A A C C D B A A ‐ C C D C C B B B ‐ C B Page‐50 Page‐51 ‐ A C Page‐52 ‐ ‐ D Page‐53 Page‐54 Page‐55 Page‐56 ‐ B C Page‐57 C ‐ C Page‐58 Page‐59 Page‐60 Page‐61 ‐ C C Page‐62 C C ‐ Page‐63 Page‐64 Page‐65 Page‐66 ‐ For-2014 (IES. C 361.25.2 50.