A Textbook of Production Technology PDF

748 A Textbook of Production Technology Vw K = Ratio of return time to cutting time, or ...

748 A Textbook of Production Technology Vw K = Ratio of return time to cutting time, or  Vr 2 3 On an average, K  to 3 4 Now, if Vm = Mean or Average speed, then B 2L Tm   , min. f Vm 1000 2Vw  Vr Vm is given as, Vm  V  V , m / min. w r 3. Drilling :  DN Cutting speed, V  , m / min 1000 Here D = Drill diameter, mm N = rotational speed of the drill, / min  D2 MRR . f.N , mm3 / min 4 D Work x y Fig. II.2. Drilling Operation. L l x y Machining time, Tm   , min. f N f N where l = hole length or depth, mm (Fig. II.2) x = Tool approach, 0.29D (with point angle of 118°) y = Tool over travel, 1 to 2 mm 4. Milling : See Art. 8.5.6., (a) Peripheral Milling : See Fig. 8.62(a), MRR = b.d.F mm3/min where b = cutting edge engagement = width of job d = depth of cut F = table feed, mm/min. Now F = f (mm/rev of cutter) × N = ft × n × N, mm/min. ft = feed per tooth, per rev., mm. n = number of teeth in the cutter lx y Machining time, Tm  , min F l = length of job Appendix-II 749 where, x = cutter approach y = over travel = 1 to 5 mm x is given as : x  d ( D  d ) [See Fig. I.3 (a)]. Cutter D d Job x l y (a) Slab Milling f B x l y (b) Face Milling Fig. II.3. Milling Operaton. (b) Face Milling : For face milling, the tool approach is given as, [Fig. II.3 (b)]  x  0.5 D  D 2  B 2  B is the width of the job. 5. Broaching: The cross - sectional area, Ac, of uncut chip per tooth is Ac = fz b mm2 for a keyway broach = fz.b.n. mm2, for a multiple-Spline”  f z..D, mm2 for a round broach where fz = cut per tooth mm = Difference in height between two successive cutting teeth b = width of broach n = number of splines D = diameter of round broach, mm MRR/pass = MRR/tooth × z 3  MRR/tooth = 1000.Ac.V. mm /min Where z = number of teeth simult aneously in operation. lw  lb Machining time, Tm  V where lw = Length of workpiece lb = Length of broach However, machining time per piece L L T  1000 V 1000 Vr 750 A Textbook of Production Technology where L = Length of stroke, mm Vr = Return velocity of broach. L or T k. 1000 V where k = Co-efficient, taking into account the ratio of working and return stroke speeds = 1.4 to 1.5 6. Grinding : (a) Cylindrical grinding : In centre-type cylindrical grinding, we have two types of grind- ing operations; traverse grinding and plunge-cut grinding. (i) Traverse grinding : In traverse grinding, the depth of cut is the layer of metal removed by the grinding wheel in one traverse stroke parallel to the axis of the job. It is also called “in feed”. So, depth of cut is, Work diameter – final diameter Dw  D f d  2 2 where d = 0.01 to 0.025 mm for rough grinding = 0.005 to 0.015 mm for finish grinding The traverse feed is the longitudinal feed, parallel to the axis of the job, mm/rev for rough grinding, f = (0.3 to 0.5) width of wheel, for Dw < 20 mm = (0.7 to 0.85 width of wheel, for Dw  20 mm For finish grinding, f = (0.2 to 0.4) + width of wheel. Table traverse will be, ft = f.Nw, mm/min where Nw = work speed, rev/min. Peripheral speed of work, Dw N w Vw  , m / min 1000 Peripheral Speed of the Wheel (Cutting speed),  Dg N g Vg  1000 where Dg = grinding wheel diameter, mm Ng = grinding wheel speed, rev/min. When the work and the wheel rotate in opposite direction to each other, the cutting speed can be taken as, V = Vg + Vw Metal Removal Rate : It can be taken approximately as MRR   d Dw f.N w , mm 3 / min Machining Time : L Tm   i  k , min f Nw i = number of passes machining allowance on each side  depth of cut (infeed) k = sparking out factor Appendix-II 751 This factor takes into account the removal of metal even in passes in which there is no in- feed. This occurs due to: the springing of the work, non-uniform wear of the wheel, breaking out of grains and the absence of continuous cutting edge. All this leads to an increase in machining time. k is about 1.2 for rough grinding, and = 1.4 for finish grinding (ii) Plunge- cut Grinding : MRR   Dw.Bg. f p mm3 / min where Bg = wheel width fp = rate of radial feed, mm/min. Machining allowance Machining time, Tm  fp fp = f × Nw, f is in mm/rev. (b) Surface grinding :  Dg N g Cutting speed, V , m / min 1000 MRR = fp.d ft where fp = cross-feed per stroke, mm/stroke ft = table traverse rate, mm/min d = depth of cut per pass. L.B Machining time Tm .i.k. V.1000. f p Where L = Length of workpiece to be ground B = Width of workpiece to be ground V = Traverse Velocity of table, m/min. fp = Crossfeed of grinding wheel, mm/stroke SOLVED EXAMPLES Example II.1: A workpiece of 76 mm diameter is to be machined on a lathe to 68 mm diameter. The total length of the workpiece is 250 mm. The recommended values of the machining elements are : feed = 0.25 mm/rev, a cutting speed of 60 mpm, and a maximum depth of cut of 2.5 mm. How long will it take to finish, the part? Also, determine the metal removal rate. Solution : The machining time is given as, L Tm   i, min. f N where, L=l+x+y x = Tool approach  d cot  Now d = 2.5 mm, Let  = 75° (Cs = 15 to 30°)  x = 2.5 cot 75° = 0.67 mm y = over travel = 2mm (say) f = 0.25 mm/rev. 1000 V 1000  60 N   251 rev./min. D  76 752 A Textbook of Production Technology 76  68 i 2 2  2.5 250  0.67  2  Tm   2 = 8.05 min. 0.25  251 MRR = 1000 f d V = 1000 × 0.25 × 2.5 × 60 = 37.5 × 103 mm3/min. Example II.2 : A part of 25cm in diameter and 50 cm length is to be turned down to 23.5 cm for the entire length. The suggested feed is 1 mm per rev. and the cutting speed is 135 mpm. The maximum allowable depth of cut is 5.0 mm. What are the feed speed, spindle rpm, MRR, and cutting time. Assume the over travel is 12.5mm. 1000 V Solution : Spindle rpm,   D 1000  135   172 rev./min. D feed speed = f × N = 1 × 172 = 172 mm/min. MRR = 1000. f. d. V = 1000 × 1 × 5 × 135 = 67.5 × 104 mm3/min. L Cutting time, Tm  i f N 250  235 Here, i 2 25 and L = 500 + 12.5 = 512.5 mm 512.5  Tm   2  5.96 min 172 Example II.3 : A hole of 25 mm diameter and 62.5 mm depth is to be drilled. The suggested feed is 1.25 mm per rev. and the cutting speed is 60 mpm. What are the feed speed, spindle rpm, and cutting time. Assume the clearance height is 5 mm. Also find the MRR. Solution : Spindle rpm, N  1000 V D 1000  60   764 rev. / min   25 feed speed = f N = 1.25 × 764 = 955 mm/min. L Cutting time  f N Now L = 62.5 + 5 = 67.5 mm. 67.5  Tm   0.0707 min. 955  D2  625 MRR  f N   955  46.8974×104 mm 3 /min. 4 4 Appendix-II 753 Example II.4 : A 50 cm 15 cm surface of cast iron block 50 cm long 15 cm wide 10 cm thick, is to be machined on a shaper. Ram speed is 20 mpm, length of stroke is 510 mm, depth of cut is 4 mm, feed is 1.5 mm/stroke, stock to be removed is 6 mm, side cutting edge angle of the tool is 45°, ratio of time taken in return stroke to the time taken in cutting stroke is 0.5. Determine the metal removal rate and the total cutting time. Solution: Now refer to Art. 8.2.4, LN (1  K ) V 1000 1000  20  N  26. full strokes/min. 510  1.5 MRR = f. d. L. N = 1.5 × 4 × 510 × 26 = 79.56 × 103 mm3/min. Bx y Cutting Time, Tm  f N B = 150 mm x  d cot   4 mm y = 3mm (say) Stock to be removed 6 Now number of passes    2 (say) depth of cut 4 150  4  3  Total cutting time, Tm   2  8.05 min. 1.5  26 Example II.5 : Evaluate the cutting parameters for the slab milling operation for the following data: Diameter of milling cutter = 100 mm, cutter speed = 500 rpm, width of cutter = 100 mm, depth of cut = 5 mm, table feed = 100 mm/min , length of workpiece = 50 cm, width of workpiece = 80 mm, number of teeth in the cutter = 8, Solution : Cutter diameter = 100 mm. Cutter speed = 500 rpm.  DN   100  500 Cutting speed, V   157 m / min 1000 1000 MRR = b. d. F b = widthof job = 80 mm where d = depth of cut = 5 mm F = Table feed = 100 mm/min. lx y Machining time, Tm  F l = 500 mm. Here y = 4 mm (say) x  d ( D  d )  5  95  21.8 mm. 500  21.8  4 Tm   5.258 min. 100 Example II.6 : Determine the machining parameters for the broaching operation from the following date; cutting speed = 12 m/min., return speed of broach = 25 m/min., length of workpiece 754 A Textbook of Production Technology = 200 mm, length of broach = 300 mm, length of stroke = 600 mm, width of workpiece = 10 mm, rise per tooth = 0.10 mm. Solution: MRR = 1000 fz. b. V. fz = cut per tooth = 0.10 mm b = width of contact = 10 mm V = 12 m/min  MRR/tooth = 0.10 × 10 × 1000 × 12 = 12000 mm3/min. lw  lb Machining time, Tm  V lw = 200 mm, lb = 300 mm 200  300  Tm   0.042 min. 12  1000 L L Machining time per piece  1000 V  1000 V r Now L = 600 mm, Vr = 25 m/min. 600 600  Time    0.074 min. 1000  12 1000  25 Example II.7 : Determine the machining parameters for the rough grinding of medium steel shaft from the following date : Work diameter = 38 mm, length of part = 200 mm, Total stock = 0.25 mm, grinding wheel = 50 mm face, depth of cut = 0.025 mm, cutting speed = 15 m/min. Solution : Now Dw = 38 mm  f = feed /rev. = (0.7 to 0.85) × width of wheel Width of wheel = 50 mm  f = 0.8 × 50 = 40 mm. Total stock 0.25 Number of cuts i    10 Depth of cut per pass 0.025 1000 V 1000  15 Wheel speed    125 rev./ min.  Dw  38 L 200  Tm  i  k   10  1.1  0.44 min. f Nw 40 125 Example II.8. A workpiece 200 mm × 300 mm is to be machined on a shaper. Calculate the machining time. Take, Vw = 10 m/min, Vr = 20 m/min, f = 5 mm/full stroke, min clearance of each end = 50 mm. Solution: B = 200 mm L = 300 + 2 × 50 = 400 mm B⎛ L L ⎞  Tm = ⎜  ⎟ f ⎝ Vw  1000 Vr  1000 ⎠ 200 FG 400 400 IJ = 5  H  10  1000 20  1000 K = 2.4 min. Appendix-II 755 Now Average speed, 2  Vw  Vr 2  10  20 40 Vm =   m/min Vw  Vr 10  20 3 B 2L  Tm = f  V  1000 m 200 2  400  3 =  5 40  1000 = 2.4 min. Example II.9. A plain surface, width = 100 mm and length = 320 mm is to be face-milled on a vertical milling machine. The milling allowance is 4 mm. The cutter has 16 teeth and the feed per tooth is 0.25 mm. The spindle speed is 125 rev./min. Calculate the machining time. Diameter of cutter = 160 mm. lxy Solution. Tm = , min. F Where l = length of job = 320 mm x = cutter approach or Added table travel = 0.5 ( D  D 2  B 2 ) = 0.5 (160  160 2  100 2 ) = 17.5 mm Let y = over travel be = 3 mm. F = Table feed per min, mm/min. = Feed per tooth per rev. (mm) × number of teeth in cutter × spindle revolution min. = 0.25 × 16 × 125 = 500 mm/min. 320  17.5  3  Tm = = 0.68 min. 500 Example II.10. The top surface of a slab, 520 mm wide and 4000 mm long is to be planed on a planer. Take cutting speed as 18.8 m/min. and return speed as 75 m/min. Take machining allowance as 10 mm. the tool approach angle is 45°. Calculate the machining time. Cross feed of tool = 3 mm/full stroke. Solution. Refer Art. 8.2.4., the cutting speed or working speed is given as, LN (1  K ) V= , m/min 1000 Where L = Length of table stroke (for planer) = length of job + table overtravel on both sides = 4000 + 325 (say) = 4325 mm. N = Number of full strokes/min Return time Cuttingspeed K=  Cutting time Return speed 18.8 =  0.25 75 1000  18.8  N=  3.5 4325 (1.25) 756 A Textbook of Production Technology Bxy Now, Cutting time, Tm = fN B = Width of job = 520 mm x = Side approach of tool = d cot  = 10. cot 45° = 10 mm y = Side overtravel of tool = 2 to 3 mm, Take 3 mm 520  10  3  Tm = = 51.7 min. 3  3.5 Example II.11. A premachined cylindrical bore 55 mm in diameter and length 62 mm is to be broached on a horizontal broaching machine. Overall length of broach is 570 mm and shank length of broach is 265 mm. Cutting speed is 8 m/min. and return speed is 20 m/min.Take tool overtravel as 50 mm. Determine the broaching time. L Solution. Tm =.k 1000V V 8 k = 1 1  1.4 Vr 20 L = Length of stroke = Length of working part of broach + length of job + tool overtravel = (570 – 265) + 62 + 50 = 417 mm 417  Tm =  1.4 = 0.073 min 1000  8 Example II.12. A shaft of length, 210 mm and 40 mm in diameter is to be longitudinally ground in one pass on a cylindrical grinding machine. The allowance per side is 0.2 mm. Wheel diameter is 600 mm and its width is 63 mm. Cutting speed is 35 m/min. Determine the grinding time. L Solution. Tm =.i. k f. Nw L = Length of table stroke = Length of workpiece + wheel overtravel on each side = 210 + 2 × 0.5 Bg Now, Bg = 63 mm (width of grinding wheel)  L = 273 mm 1000 Vw 1000  35 Nw = Work speed =   Dw   40 = 280 rpm f = traverse feed, mm/rev. = (0.2 to 0.4) × Bg for finish grinding = 0.3 × 63 = 18.9 mm/rev. i = number of passes machiningallowance on each side = depth of cut (infeed) For finish grinding, depth of cut = 0.005 to 0.015 mm = 0.005 mm (say) Appendix-II 757 0.2  i= = 40 0.005 Let k = 1.4 273  Tm =  40  1.4 18.9  280 = 2.89 min. Example II.13. A flat surface of a workpiece is to be ground, workpiece width is 110 mm and length is 280 mm. Machining allowance per side = 0.35 mm. Traverse velocity of table is 16 m/min. Downward feed of the grinding wheel (depth of cut) per pass is 0.015 mm per reverse of the grinding wheel. Cross feed of grinding wheel = 32 mm/table stroke. Determine the total machining time. L.B Solution. Tm = 1000. V. f.i.k p L = 280 mm B = 110 mm V = 16 m/min fp = 32 mm/table stroke i = number of pass 0.35 = 0.015 k = 1.4 (say) 280  110 0.35  Tm =   1.4 1000  16  32 0.015 = 1.965 min. PROBLEMS 1. A 100 mm diameter brass bar of 250 mm length is to be turned to a new diameter of 99.75 mm. (a) Determine the rpm setting for the lathe (b) Determine the cutting time assuming an allowance of 6.25 mm (c) Determine the metal removal rate Take cutting speed = 150 mpm and feed = 0.25 mm/rev. 2. A 19 mm diameter hole is to be drilled in a piece of C-20 steel of thickness 25 mm. Taking cutting speed as 400 rev/min. and feed as 0.50 mm/rev. (a) Calculate the cutting speed in mpm. (b) Determine the machining time assuming an allowance of 12.5 mm. (c) Determine the metal removal rate 3. A 100 mm diameter aluminium bar of 300 mm length is to be turned to a new diameter with a speed of 157.5 mpm, feed of 0.50 mm/rev. and 0.75 mm depth of cut. (a) Determine the rpm to be set on the machine tool. (b) Determine the machining time assuming an allowance of 6.25 mm. (c) Determine the metal removal rate. 4. A slot 6.25 mm wide and 6.25 mm deep is to be milled in one pass in a component. The component is 450 mm long and milled with a 150 mm diameter, 10-tooth cutter. Take cutting speed = 45 mpm and feed = 0.125 mm/tooth (a) Determine the rpm setting. (b) Determine the feed rate in mm/min. (c) Calculate the machining time by taking allowance equal to one cutter diameter. (d) Calculate the metal removal rate. 758 A Textbook of Production Technology 5. A 63.5 mm diameter plain milling cutter having 6 teeth is used to facemill a block of aluminium 18 cm long and 3 cm wide. The spindle speed is 1500 rpm and the feed is 0.125 mm/tooth/rev. (a) Determine the table feed in mm/min. (b) Total table travel. (c) Cutting time. 6. Calculate the time required on the shaper to complete one cut on a plate 600 mm × 900 mm, if the cutting speed is 6 m/min. The return time to cutting time ratio is 1 : 4, and the feed is 2 mm/stroke. The clearance at each end is 75 mm. 7. Determine the machining time for one pass longitudinal turning of a shaft from 70 mm to 64 mm over a length of 200 mm. The speed of lathe spindle is 600 rev./min. Tool feed is 0.4 mm/rev. and plan approach angle of tool is 45°. [Ans. 0.85 min.] 8. Determine the machining time in facing a workpiece of 165 mm diameter on a lathe in one pass. The speed of the spindle is 480 rev/min. and tool feed is 0.3 mm/rev. Machining allowance is 3.5 mm and plan approach angle of tool is 45°. [Ans. 0.61 min] 9. A surface 90 mm wide and 200 mm long is machined on a Shaper. The machining allowance is 2 mm. Take: Cutting speed = 23.8 m/min. Ratio of cutting speed to return speed = 0.8, cross-feed of table = 0.6 mm/full stroke, tool overtravel on each side as 25 mm, plan approach angle of the tool = 45°. [Ans. 2.95 min.] 10. A through hole of diameter 20 mm is to be drilled to a depth of 80 mm on a vertical drilling machine. Take cutting speed as 22.3 m/min. and drill feed as 0.4 mm/rev. [Ans. 0.62 min.] 11. A plain surface 70 mm wide and 600 mm long is to be face milled on a vertical-spindle milling machine. The machining allowance is 3.7 mm, to be removed in one pass. cutter diameter is 110 mm and the cutting speed is 172.7 m/min. Number of teeth on the cutter is 4 and the feed is 0.2 mm/tooth/rev. Find the machining time. [Ans. 1.54 min.] 12. A plain surface 100 mm wide and 320 mm long is to be face milled on a vertical milling machine. Cutter diameter is 160 mm and number of cutter teeth is 16. Cutting speed is 63 m/min. Feed per tooth is 0.25 mm/tooth. Determine the machining time. [Ans. 0.68 min.] 13. The peripheral milling of a plain surface, 75 mm wide and 300 mm long is carried out on the horizontal milling machine. Cutter diameter is 90 mm, cutting speed is 28.3 m/min, Number of cutter teeth is 8 and feed per tooth is 0.2 mm. The machining allowance is 3 mm to be removed in a single pass. Determine the machining time. [Ans. 1.98 min] 14. A plain surface 65 mm wide and 225 mm long is to be machined on a horizontal milling machine. The machining allowance is 1.5 mm. Cutter diameter is 63 mm, cutting speed is 49.5 m/min, number of cutter teeth is 14 and feed per tooth is 0.11 mm. Determine the machining time. [Ans. 0.59 min.] 15. A slot 32 mm wide, 15 mm deep and 250 mm is to be rough machined in a single pass by a disk-type milling cutter on a horizontal milling machine. Cutter diameter is 150 mm, cutting speed is 37.6 m/min, Number of teeth on cutter is 16 and the feed per tooth is 0.078 mm. Determine the machining time. [Ans. 3 min.] 16. A through slot 32 mm wide, 15 mm deep and 300 mm long is to be milled by an end mill cutter on a vertical milling machine. The cutter diameter is 32 mm and it has 6 teeth. Cutting speed is 20 m/min. and the feed per tooth is 0.083 mm/tooth. Determine the machining time. D [Ans. 3.19 min. Hint: Take tool approach = ] 2 17. Determine the metal removal rate (a) Based on mean diamater (b) Based on the difference in initial and final diamter, for the turning operation for the following data: feed = 0.5 mm / rev; Spindle speed = 200 rer./min. Initial diameter = 50 mm Final Diameter = 42 mm. Di  D f 50 – 42 Solution : Depth of cut, d =   4mm 2 2 (a) MRR = d.f. π Dave × N = 4 × 0.5 × π × 46 × 200 ≅ 5.78 × 104 mm3 / min. (b) MRR =  4   502  422  f  N =   4  502  422 × 0.5 × 200 = 5.78 × 104 mm3/min Thus, there is no difference in the two results, It is the personal choice about the method to be used.  APPENDIX III Problems From Competitive Examinations (GATE, IES, IAS) CHAPTER 2 1. Match the terms used in connection with heat-treatment of steel with the microstructural/ physical characteristics: Terms Characteristics (A) Pearlite (P) Extremely hard and brittle phase (B) Martensite (Q) Cementite is finely dispersed in ferrite (C) Austenite (R) Alternate layers of Cementite and Ferrite (D) Eutectoid (S) Can exist only above 723°C (T) Pertaining to state of equilibrium between three solid phases (U) Pertaining to state of equilibrium between one liquid and two solid phase (GATE 1992) (Ans. A – R, B – P, C – S, D– T) 2. The Iron–Carbon diagram and the TTT curves are determined under: (A) equilibrium and non-equilibrium conditions respectively (B) non-equilibrium and equilibrium conditions respectively (C) equi-librium conditions for both (D) non-equilibrium conditions for both [GATE 1996] (Ans. A) 3. On completion of heat treatment, the following structure will have retained Austenite if (A) rate of cooling is greater than the critical cooling rate (B) rate of cooling is less than the critical cooling rate (C) martensite formation starting temperature is above the room temperature (D) martensite formation finish temperature is below the room temperature [GATE 1997] (Ans. C) 4. Cast steel crankshaft surface is hardened by (A) nitriding (B) normalising (C) carburising (D) induction heating [GATE 2000] (Ans. D) 5. The alloying element mainly used to improve the endurance strength of steel materials is: (A) Nickle (B) Vanadium (C) Molybdenum (d) Tungsten [GATE 1997] (Ans. B) 6. Decreasing grain size in a polycrystalline material (A) increases yield strength and corrosion resistance (B) decreases yield strength and corrosion resistance (C) decreases yield strength but increases corrosion resistance (D) increases yield strength but decreases corrosion resistance. [GATE 1998] (Ans. A) 7. Carburised machine components have higher endurance limit, because carburization (A) raises the yield point of the material (B) produces a better surface finish (C) introduces a compressive layer on the surface (D) suppresses any stress concentration produced in the component. [GATE 1992] (Ans. C) 8. Match List I with List II and select the correct answer using the codes given below the lists: List I List II (Mech. properties) (Related to) (A) Malleability 1. Wire drawing (B) Hardness 2. Impact loads (C) Resilience 3. Cold rolling (D) Isotropy 4. Indentation 759 760 A Textbook of Production Technology 5. Direction Codes. A B C D A B C D (a) 4 2 1 3 (b) 3 4 2 5 (c) 5 4 2 3 (d) 3 2 1 5 (Ans. b) (IES 1993) 9. Match List I with List II and select the correct answer using codes given below the lists: List I List II (Steel type) (Product) A. Mild steel 1. Screw driver B. Tool steel 2. Commercial beams C. Medium Carbon steel 3. Crane hooks D. High carbon steel 4. Blanking dies Codes. A B C D A B C D (a) 1 4 3 2 (b) 2 4 1 3 (c) 1 3 4 2 (d) 2 4 3 1 [IES 1993] (Ans. d) 10. Which of the following statements are true of annealing of steels? 1. Steels are heated to 500 to 700°C 2. Cooling is done slowly and steadily 3. Internal stresses are relieved 4. Ductility of steel is increased Select the correct answer using the codes given below: Codes: (A) 2, 3 and 4 (B) 1, 3 and 4 (C) 1, 2 and 4 (D) 1, 2 and 3 [IES 1993] (Ans. A) 11. Duralumin alloy contains aluminium and copper in the ratio of % Al % Cu (A) 94 4 (B) 90 8 (C) 98 10 (D) 86 12 [IES 1993] (Ans. A) 12. Eutectic reaction for iron-carbon system occurs at (A) 600°C (B) 723°C (C) 1147°C (D) 1493°C [IES 1993] (Ans. C) 13. The blade of a power saw is made of (A) boron steel (B) high speed steel (c) stainless steel (D) malleable cast iron [IES 1993] (Ans. B) 14. Quartz is a (A) ferroelectric material (B) ferromagnetic material (C) piezo electric material (D) diamagnetic material [IES 1993] (Ans. C) 15. Match List I with List II and select the correct answer using the codes given below the lists: List I (Material/Part) List II (Techniques) A. Ductile iron 1. Inoculation B. Malleable iron 2. Chilled C. Rail steel joints 3. Annealing D. White cast iron 4. Thermit Welding 5. Isothermal annealing Codes: A B C D A B C D (a) 1 3 4 2 (b) 5 3 2 1 (c) 2 1 4 5 (d) 1 4 2 3 [IAS 1995] (Ans. a) Appendix III 761 16. Consider the following treatments: 1. Normalising 2. Hardening 3. Martempering 4. Cold working Hardness and tensile strength in austenitic steel can be increased by (A) 1, 2 and 3 (B) 1 and 3 (C) 2 and 4 (D) 4 alone [IES 1994] (Ans. D) 17. Killed steels (A) have minimum impurity level (B) have almost zero percentage of phosphorus and sulphur (C) are produced by LD process (D) are free from O2 [IES 1994] (Ans. D) 18. Which of the following pairs are correctly matched: 1. Silicon steels.......... Transformer stampings 2. Duralumin....... Cooking utensils 3. Gun metal........ Bearings Select the correct answer using the codes given below: Codes: (A) 1, 2 and 3 (B) 1 and 2 (C) 1 and 3 (D) 2 and 3 [IES 1994] (Ans. A) 19. Match List I with List II and select the correct answer using the codes given below the lists: List I (Materials) List II (Applications) A. Engineering Ceramics 1. Bearings B. Fibre reinforced plastics 2. Control rods in nuclear reactors C. Synthetic carbon 3. Aerospace industry D. Boron 4. Electrical insulator Codes A B C D A B C D (a) 1 2 3 4 (b) 1 4 3 2 (c) 2 3 1 4 (d) 4 3 1 2 [IES 1994] (Ans. d) 20. Which of the following pairs are correctly matched? 1. Lead screw nut...... Phosphor bronze 2. Piston...... Cast iron 3. Cam........ EN 31 steel 4. Lead screw...... Wrought Iron Select the correct answer using the codes given below:– Codes. (A) 2, 3 and 4 (B) 1, 3 and 4 (C)1, 2 and 4 (D) 1, 2 and 3 [IES 1994] (Ans. D) 21. Babbit lining is used on brass/bronze bearings to (A) increase bearing resistance (B) increase compressive strength (C) provide any friction properties (D) increase wear resistance [IES 1995] (Ans. D) 22. In low carbon steels, presence of small quantities of sulphur improves: (A) Weldability (B) formability (C) machinability (D) hardenability [IES 1995] (Ans. C) 762 A Textbook of Production Technology 23. Match List I with List II and select the correct answer using codes given below the lists: List I (heat treatment) List II (effect on properties) A Annealing 1. Refines grain structure B Nitriding 2. Improves hardness of the whole mass C Martempering 3. Increases surface hardness D Normalising 4. Improves ductility Codes. A B C D A B C D (a) 4 3 2 1 (b) 1 3 4 2 (c) 4 2 1 3 (d) 2 1 3 4 [IES 1995] (Ans. a) 24. Consider the following statements: Addition of Silicon to cast iron 1. promotes graphite nodule formation 2. promotes graphite flake formation 3. increases fluidity of the molten metal 4. improves the ductility of cast iron Of these statements (A) 1 and 4 are correct (B) 2 and 3 are correct (C) 1 and 3 are correct (D) 3 and 4 are correct [IES 1995] (Ans. B) 25. Eutectoid reaction occurs at (A) 600°C (B) 723°C (C) 1147°C (D) 1493°C [IES 1995] (Ans. B) 26. Match List I with List II and select the correct answer using the codes given below the lists: List I (Name of Material) List II (% Carbon Range) A. Hypo-eutectoid steel 1.4.3 – 6.67 B. Hyper-eutectoid steel 2. 2.0 – 4.3 C. Hypo-eutectic cast iron 3. 0.8 – 2.0 D. Hyper-eutectic cast iron 4. 0.008 – 0.8 Codes A B C D A B C D (a) 4 3 2 1 (b) 1 3 2 4 (c) 4 1 2 3 (d) 1 2 3 4 [IES 1995] (Ans. a) 27. Which one of the following constituents is expected in equilibrium cooling of a hypereu- tectoid steel from austenitic state? (A) Ferrite and Pearlite (B) Cementite and Pearlite (C) Ferrite and bainite (D) Cementite and Martensite [IES 1995] (Ans. B) 28. Addition of Magnesium to cast iron increases its (A) hardness (B) ductility and strength in tension (C) Corrosion resistance (D) Creep strength [IES 1995] (Ans. B) 29. Small amount of which one of the following elements/pairs of elements is added to steel to increase machinability (A) Nickle (B) Sulphur and Phosphorous (C) Silicon (D) Manganese and copper [IES 1996] (Ans. B) 30. Which of the following pairs regarding the effect of alloying elements in steel are correctly matched? 1. Molybdenum : Forms abrasion resisting particles 2. Phosphorus: Improves machinability in free cutting steels 3. Cobalt: Contributes to red hardness by hardening ferrite 4. Silicon: Reduces oxidation resistance Select the correct answer using the codes given below: (A) 2, 3, 4 (B) 1, 3 and 4 (C) 1, 2 and 4 (D) 1, 2 and 3 [IES 1996] (Ans. D) Appendix III 763 31. Machine tool guideways are usually hardened by (A) Vacuum hardening (B) Martempering (C) Induction hardening (D) Flame hardening [IES 1996] (Ans. D) 32. 18/8 stainless steel contains: (A) 18% Ni, 8% Cr (B) 18% Cr, 8% Ni (C) 18% Cr, 8% W (D) 18% W, 8% Cr. [IES 1996] (Ans. B) 33. Tin base white metals are used where the bearings are subjected to: (A) large surface wear (B) elevated temperatures (C) high load and pressure (D) high pressure and load [IES 1996] (Ans. D) 34. Alloy steel which is work hardenable and which is used to make the blades of bull dozers bucket wheels excavators and other earth moving equipment contains iron, carbon and (A) Mn (B) Si (C) Cr (D) Mg [IES 1996] (Ans. A) 35. Guide ways of lathe beds are hardened by (A) Carburising (B) Cyaniding (C) nitriding (D) Flame hardening [IES 1997] (Ans. D) 36. A given steel test specimen is studied under metallurgical microscope. Magnification used is 100 X. In that different phases are observed. One of them is Fe3C. The observed phase Fe3C is also known as (A) ferrite (B) cementite (C) austenite (D) martensite [IES 1997] (Ans. B) 37. Match List I (Alloying elements in steel) with List II (Property conferred on steel by the element) and select the correct answer using the codes given below the lists: List I List II A Nickle 1. Corrosion resistance B Chromium 2. Magnetic permeability C Tungsten 3. Heat resistance D Silicon 4. Hardenability Codes: A B C D A B C D (a) 4 1 3 2 (b) 4 1 2 3 (c) 1 4 3 2 (d) 1 4 2 3 [IES 1998] (Ans. a) 38. Match List I (Alloys) with List II (Applications) and select the correct answer using the codes given below the lists: List I List II A Chromel 1. Journal bearing B Babbit alloy 2. Milling Cutter C Nimonic alloy 3. Thermocouple wire D High speed steel 4. Gas turbine blades Codes A B C D A B C D (a) 3 1 4 2 (b) 3 4 1 2 (c) 2 4 1 3 (d) 2 1 4 3 [IES 1998] (Ans. a) 764 A Textbook of Production Technology 39. Match List I with List II and select the correct answer using the codes given below the lists: List I (Materials) List II (Applications) A. Tungsten Carbide 1. Abrasive wheels B. Silicon nitride 2. Heating elements C. Aluminium Oxide 3. Pipes for conveying liquid metals D. Silicon Carbide 4. Drawing dies 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 [IES 1999](Ans. d) 40. Heating the Hypoeutectoid steels 30° above the upper critical temperature line, soaking at that temperature and then cooling slowly to room temperature to form a pearlite and ferrite structure is known as (A) hardening (B) normalising (C) tempering (D) annealing [IES 1999] (Ans. D) 41. In a eutectic system, two elements are completely (A) insoluble in solid and liquid states (B) soluble in liquid state (C) soluble in solid state (D) insoluble in liquid state [IES 1999] (Ans. B) 42. A steel with 0.8% C is called (A) Hypo-eutectoid steel (B) Hyper-eutectoid steel (C) Eutectoid steel (D) None of these (Ans. C) 43. A steel with 0.8% C contains: (A) 100% pearlite (B) 100% austenite (C) ferrite and pearlite (D) pearlite and cementite (Ans. A) 44. The lower critical temperature for all steels is: (A) 700°C (B) 723°C (C) 650°C (D) 910°C (Ans. B) 45. A steel with 0.8% C has (A) One critical point (B) Two critical points (C) No critical point (Ans. A) 46. Cementite consists of: (A) 13% ferrite and 87% pearlite (B) 6.67% C and 93.33% Iron (C) 13% C and 87% ferrite (Ans. B) 47. Pearlite consists of: (A) 87% ferrite and 13% Cementite (B) 87% cementite and 13% ferrite (C) 6.67% C and 93.33% Iron (Ans. A) 48. In the austempering heat treatment process, austenite decomposes into: (A) Sorbite (B) Troostite (C) Bainite (D) Martensite (Ans. C) 49. Do we need tempering operation on a component hardened by austempering process? (Ans. No) 50. The upper critical temperature for steel (A) is constant (B) depends upon the rate of heating (C) varies according to the carbon content in steel (D) none of the above (Ans. C) Appendix III 765 51. The lower critical temperature for steel (A) is constant (B) depends upon the rate of heating (C) varies according to the carbon content in steel (D) None of the above (Ans. A) 52. The beginning of separation of ferrite from solid solution of Austenite is represented by (A) A1 line (B) A3 line (C) Acm line (D) None of the above (Ans. B) 53. The completion of transformation of austenite into ferrite and pearlite is represented by (A) A1 line (B) A3 line (C) Acm line (D) None of the above (Ans. A) 54. The limit of Carbon solubility in austenite is represented by (A) A1 line (B) A3 line (C) Acm line (D) None of the above (Ans. C) 55. Fine grains of austenite (A) decrease hardenability (B) increase hardenability (C) first decrease, then increase hardenability (D) first increase, then decrease hardneability (Ans. A) 56. Dissolved alloying elements in steel (A) decrease hardenability (B) increase hardenability (C) has no effect on hardenability (D) first decrease, then increase hardenability (Ans. B) 57. The temperature at which the first new grain appears is known as (A) Melting temperature (B) Critical temperature (C) Boiling temperature (D) Recrystallisation temperature (Ans. D) 58. The recrystallisation temperature depends upon: (A) grain size (B) type of metal (C) extent of cold deformation (D) annealing time (E) purity of metal (F) all of the above (G) None of the above (Ans. F) 59. The recrystallisation temperature for pure metals is (A) 0.2 Tm (B) 0.3 Tm (C) 0.5 Tm (D) 0.8 Tm Where Tm = melting temperature (Ans. B) 60. The recrystallisation temperature for alloys is approximately: (A) 0.2 Tm (B) 0.3 Tm (C) 0.5 Tm (D) 0.8 Tm (Ans. C) 61. Carbon occurs in steel in the combined state with iron to form the component: (A) Ferrite (B) Cementite (C) Pearlite (D) Bainite (I.MECH, E, L.U., D.U.) (Ans. B) 62. When a steel is heated to above its upper critical temperature, the structure produced is one of: (A) Martensite (B) Austenite (C) Pearlite (D) Sorbite (Ans. B) 63. The predominant structure of a hypereutectoid steel that has been quenched at above its upper critical temperature will be:– (A) Austenite (B) Martensite (C) Troostite (D) Sorbite (B.U., D.U., AIME) (Ans. B) 766 A Textbook of Production Technology 64. Which one of the following structures is predominant in a normalized steel:– (A) Troostite (B) Bain ite (C) Sorbite (D) Martensite (AMIE, UPSC) (Ans. C) 65. When a steel is heated in a furnace and then cooled in air at ordinary temperature, the process is one of (A) Annealing (B) Hardening (C) Normalizing (D) Tempering (AMIE, L.U., P.U.) (Ans. C) CHAPTER 3 1. In a green-sand moulding process, uniform ramming leads to (A) less change of gas porosity (B) uniform flow of molten metal into the mould cavity (C) greater dimensional stability of the casting (D) less sand expansion type of casting defect [GATE 1992] (Ans. C) 2. Match the following moulding casting processes with the product: Moulding/Casting processes Product (A) Slush casting (P) Turbine blades (B) Shell moulding (Q) Machine tool bed (C) Dry sand moulding (R) Cylinder block (D) Centrifugal casting (S) Hollow castings like lamp shades (T) Rain water pipe (U) Cast iron shoe brake [GATE 1992] (Ans. A – S, B – P, C – R, D – T) 3. Centrifugally cast products have (A) large grain structure with high porority (B) fine grain structure with high density (C) fine grain structure with low density (D) segregation of slag towards the outer skin of the casting [GATE 1993] (Ans. B) 4. List I gives a number of processes and List II gives a number of products. Match correct pairs: List I List II (A) Investment casting 1. Turbine rotors (B) Centrifugal casting 2. Turbine blades (C) Die-casting 3. Connecting rods (D) Drop forging 4. Galvanized iron pipe (E) Extrusion 5. Cast iron pipes (F) Shell moulding 6. Carburettor body [GATE 1994] (Ans. A – 2, B – 5, C – 6, D – 3, E – 4, F – 1). 5. Light impurities in the molten metal are prevented from reaching the mould cavity by providing a (A) Strainer (B) Bottom well (C) Skim bob (D) all of the above [GATE 1996] (Ans. C) 6. Chills are used in moulds to (A) achieve directional solidification (B) reduce the possibility of blow holes (C) reduce freezing time (D) Smoothen metal flow for reducing splatter [GATE 1998] [IAS 1994] (Ans. A) 7. Which of the following materials requires the largest shrinkage allowance, while making a pattern for casting? Appendix III 767 (A) Aluminium (B) Brass (C) Cast iron (D) Plain carbon steel [GATE 1999] (Ans. D) 8. Disposable patterns are made of (A) wood (B) rubber (C) metal (D) polystyrene (Ans. D) 9. Match List I with List II and select the correct answer using the codes given below the lists: List I (Equipments) List II (Functions) A. Hot chamber machine 1. Cleaning B. Muller 2. Core making C. Dielectric baker 3. Die casting D. Sand blasting 4. Annealing 5. Mixing Codes: A B C D A B C D (a) 3 5 2 1 (b) 4 2 5 3 (c) 4 2 3 1 (d) 3 5 1 2 [IES 1993] (Ans. a) 10. Which of the following materials can be used for making patterns? 1. Aluminium 2. Wax 3. Mercury 4. Lead Select the correct answer using the codes given below: Codes: (A) 1, 3 and 4 (B) 2, 3 and 4 (C) 1, 2 and 4 (D) 1, 2 and 3 [IES 1994] (Ans. D) 11. Which of the following materials will require the largest size of riser for the same size of casting? (A) Aluminium (B) cast iron (C) steel (D) copper [IES 1995] (Ans. A) 12. Directional solidification in castings can be improved by (A) Chills and chaplets (B) Chills and padding (C) Chaplets and padding (D) Chills, chaplets and padding [IES 1995] (Ans. B) 13. Match List I with List II and select the correct answer taking the help of codes given below the lists: List I (Products) List II (Process of Manufacture) A. Automobile piston in aluminium alloy 1. Pressure die-casting B. Engine crankshaft in spherioidal graphite iron 2. Gravity die-casting C. Carburettor Housing in aluminium alloy 3. Sand casting D. Cast titanium blades 4. Precision investment casting 5. Shell moulding Codes: A B C D A B C D (a) 2 3 1 5 (b) 3 2 1 5 (c) 2 1 3 4 (d) 4 1 2 3 [IES 1995] (Ans. a) 14. Consider the following ingredients used in moulding: 1. Dry silica sand 2. Clay 3. Phenol formaldehyde 4. Sodium silicate Those used for shell mould casting include: (A) 1, 2 and 4 (B) 2, 3 and 4 (C) 1 and 3 (D) 1, 2, 3, 4 [IES 1996] (Ans. C) 15. Which of the following methods are used for obtaining directional solidification for riser design: 768 A Textbook of Production Technology 1. Suitable placement of chills 2. Suitable placement of chaplets 3. Employing padding Select the correct answer: (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 [IES 1996] (Ans. b) 16. Misrun is a casting defect which occurs due to (A) very high pouring temperature of the metal (B) insufficient fluidity of the molten metal (C) absorption of gases by the liquid metal (D) improper alignment of the mould flasks [IES 1996] (Ans. B) 17. Which of the following pairs are correctly matched? 1. Pit moulding........ for large jobs 2. Investment moulding......... Lost wax pro- cess 3. Plaster moulding...... Mould prepared in gypsum (A) 1, 2 and 3 (B) 1 and 2 (C) 1 and 3 (D) 2 and 3 [IES 1996] (Ans. A) 18. Which one of the following pairs is not correctly matched? (A) Aluminium alloy Piston...... Pressure die casting (B) Jewellery........ Lost wax process (C) Large pipes........ Centrifugal casting (D) Large bells...... Loam moulding [IES 1997] (Ans. A) 19. If the melting ratio of a cupola is 10 : 1, then the coke requirement for one ton melt will be (A) 0.1 ton (B) 10 tons (C) 1 ton (D) 11 tons [IES 1997] (Ans. B) 20. Which one of the following are the requirements of an ideal gating system? 1. The molten metal should enter the mould cavity with as high a velocity as possible 2. It should facilitate complete filling of the mould cavity 3. It should be able to prevent the absorption of air or gases from the surroundings of the molten metal while flowing through it Select the correct answer using the codes given below: (A) 1, 2 and 3 (B) 1 and 2 (C) 2 and 3 (D) 1 and 3 [IES 1998] (Ans. C) 21. A spherical drop of molten metal of radius 2 mm was found to solidify in 10 seconds. A similar drop of radius 4 mm will solidify in (A) 14.14 s (B) 20 s (C) 28.30 s (D) 40 s [IES 1998] (Ans. D) FG V IJ 2 (Hint: Solidification for sphere is  H AK ) 22. In solidification of metal during casting, compensation for solid contraction is (A) provided by the oversize pattern (B) achieved by properly placed risers (C) obtained by promoting directional solidification (D) made by providing chills. [IES 1999] (Ans. A) 23. Disk-shaped components are cast by (A) True Centifigural casting (B) Semi-Centrifugal casting (C) Centifuge casting (Ans. B) 24. Consider the following ingredients used in moulding: 1. Dry silica sand 2. Clay 3. Ethyl silicate 4. Phenol formaldehyde Those used for Lost Wax casting method include: (A) 1, 2 and 4 (B) 2, 3 and 4 (C) 1 and 3 (D) 1, 2, 3 and 4 (Ans. C) Appendix III 769 25. Metal patterns require...... (more/less) draft allowance than wooden patterns. (Ans. more) 26. The molten metal is poured from the pouring basin to the gate with the help of a (A) Riser (B) Sprue (C) Runner (D) Core (Ans. B) 27. In cold chamber die-casting process, only non-ferrous alloys with...... (high melting point/ low melting point) are cast. (Ans. High melting point) 28. In the casting of large pipes by true centrifugal casting (A) Core is of sand (B) Core is of metal (C) no core is used (Ans. C) 29. Core prints are provided on patterns (A) to support the core (B) to locate the core in the mould (C) to support as well as locate the core in the mould (Ans. C) 30. Contraction allowance is provided on a pattern: (A) for machining of castings (B) for contraction in metal on cooling (C) for making a good casting (Ans. B) 31. To provide the machining allowance, a pattern should be made...... (larger/smaller) than the size of the finished casting required. (Ans. Larger) 32. An ingate (A) acts as a reservoir for the molten metal (B) delivers molten metal into the mould cavity (C) delivers molten metal from the pouring basin to runner (D) all of the above (Ans. B) 33. A riser (A) acts as a reservoir for the molten metal (B) delivers molten metal into the mould cavity (C) delivers molten metal from the pouring basin to runner (D) feeds the molten metal to the casting in order to compensate for solid shrinkage (Ans. D) 34. The function of a core is: (A) to improve mould surface (B) to form internal cavities in the casting (C) to form a part of a green sand mould (D) None of the above (E) All of the above (Ans. E) 35. When the molten metal flows into the cavity of a metallic mould by gravity, the method of casting is known as: (A) Die-casting (B) Centrifugal casting (C) Permanent mould casting (D) Plaster mould casting (Ans. C) 36. The toys and ornaments of non-ferrous alloys are made by: (A) Die-casting (B) Lost-wax method (C) Permanent mould casting (D) Slush casting (Ans. D) 37. Two castings of the same metal have the same surface area. One casting is in the form of a sphere and the other is a cube. What is the ratio of the solidification time for the sphere to that of a cube. [GATE 1998] FG V IJ 2 Solution. According to Chvorinov’s rule, solidification time is  H AK , where V = Volume of casting A = Surface area of casting Now As = Ac 770 A Textbook of Production Technology FV I Ratio of solidification time = G J 2 HV K s  c F 4 R I 2 G3 JJ 3 = G a GH 3 JK 16 2 R 6 = 9a 6 Where R = Radius of sphere, and a = side of the cube. 38. Risering is not needed when casting Gray cast iron, True/False. (Ans. True) 39. Calculate the ratio of the solidification times of two steel cylindrical risers of sizes 30 cm in diameter by 60 cm height and 60 cm in diameter by 30 cm in height subjected to identical conditions of cooling. [GATE 1992] FG V IJ 2 Solution. According to Chvorinov’s rule solidification time is  H AK (See problem 37)  2 d h V 4  Riser 1. A   d h  2  d2 4 1  900  60 4 13500  6 = 1 2250 30  60   900 2 1  3600  30 V 4 27000    7.5 Riser 2: A 1 3600 60  30   3600 2 F 7.5 I  Ratio of solidification times, = GH 6 JK 2  1.5625 40. An aluminium cube of 10 cm side has to be cast along a cylindrical riser of height equal to its diameter. The riser is not insulated on any surface. If the volume shrinkage of aluminium during solidification is 6%, calculate (i) Shrinkage volume of cube on solidification, and (ii) minimum size of the riser so that it can provide the shrinkage volume. [GATE 1993] Solution: (i) Volume of casting = 10 × 10 × 10 = 1000 cm3 , Shrinkage volume = 6% = 60 cm3 (Depending upon metal, this shrinkage varies from 2.5 to 7.5%) Now from practice, Minimum volume of riser is approximately three times the shrinkage volume, 3  Minimum volume of riser = 3 × 60 = 180 cm Appendix III 771  2  dr  hr  180 4 now hr = dr  From here, dr = 5.2 cm Now for sound casting, the metal in the riser should be the last to cool, that is, the riser should have a longer solidification time than the casting, so FG A IJ  FG A IJ or ⎛ V ⎞  ⎛ V ⎞ H V K H V K ⎜⎝ A ⎟⎠ ⎜⎝ A ⎟⎠ r c r C FG AIJ  6  6 (it is simple to prove this, with d = h ) Now H V K d 5.2 r r r r FG A IJ  6  10  10  3 H V K 10  10  10 5 c FG A IJ FG A IJ As is clear, HV K r is > HV K c , which is not desirable. FG A IJ  FG A IJ  HV K HV K r c  0.6 6 10 or  0.6 ,  1 ,  dr  10. dr dr  Minimum size of riser = 10 cm diameter × 10 cm height   Vr   100  10  785.4 cm3. 4 41. With cylindrical riser, prove that for a longer solidification time, diameter of riser = height of riser. FG V IJ Solution: We know that for longer solidification time, the ratio H AK should be maximum FG A IJ or the ratio, HV K should be minimum.  2 Now, V= d h 4 4V  h=  d2  2 A = d. h  2. d 4 4V  2  A = d.  d d 2 2 4V  2  A=  d d 2 772 A Textbook of Production Technology For A to be minimum for a given V, A / d = 0, and it will be seen that for this, d = h. 4V i.e.   d  0 d2 4V or d3   4V Also,  hd 2   d3 = hd2 or d=h This optimum ratio is true only for a side riser, Fig. 3.22 (a). For a top riser,  2 A  dh  d 4 Doing the analysis as above, it will be seen that d = 2h h 1 or  d 2 42. Shrinkage allowance on pattern is provided to compensate for shrinkage when (a) the temperature of liquid metal drops from pouring to freezing temperature (b) the metal changes from liquid to solid state at freezing temperature (c) the temperature of solid phase drops from freezing to room temperature (d) the temperature of metal drops from pouring to room temperature [GATE 2001] (Ans. c) 43. Two solid workpieces (i) Sphere with radius R, (ii) a cylinder with diameter equal to its height, have to be sand cast. Both workpieces have the same volume. Show that the c ylindrical workpiece will solidify faster than the spherical work piece. [GATE 2001] 4 3 Solution. Volume of sphere = R 3  2 Volume of cylinder = d.h 4  3 = d 4  3 4 3 d  R 4 3 16 3  d3 = R 3 FV I Now solidification time is  G J 2 H AK  2 FG V IJ 4 d.h d Now, H AK cyl   (with d = h)  2 6 dh  2. d 4 Appendix III 773 FG V IJ FG IJ 1 16 1 H AK 3  cyl =. 6 3H K. R = 0.29 R 4 3 FG V IJ R R  3 2   0.33 R Again, H AK sphere 4 R 3 FG V IJ Now, it is known that risers with a higher value of H AK loose heat at a slower rate. Thus, it is clear from above, that a solid casting of cylindrical section will solidify faster than the spherical work piece. 44. The primary purpose of sprue in a casting mould is to (a) Feed the casting at a rate consistent with the rate of solidification (b) Act as a reservoir for molten metal (c) Feed molten metal from the pouring basin to the gate (d) Help feed the casting until all solification takes place (GATE 2002) (Ans. c) 45. In centrifugal casting, the impurities are (a) Uniformly distributed (b) Forced towards the outer surface (c) trapped near the mean radius of the casting (d) collected at the centre of casting [GATE 2002] (Ans. a) 46. Match 4 correct pairs between List I and List II List I List II (A) Sand casting (1) Symmetrical and circular shapes only (B) Plaster mould casting (2) Parts have hardened skin and soft interior (C) Shell mould casting (3) Minimum post casting processing (D) Investment casting (4) Parts have a tendency to warp (5) Parts have soft skin and hard interior (6) Suitable only for non-ferrous metals [GATE 1998] (Ans. A – 2, B – 6, C – 1, D – 3) CHAPTER 4 1. The thickness of the blank needed to produce, by power spinning a missile cone of thick ness 1.5 mm and half cone angle 30°, is (a) 3.00 mm (b) 2.5 mm (c) 2.0 mm (d) 1.55 mm [GATE 1992] (Ans. a) 2. The true strain for low carbon steel bar which is doubled in length by forging is (a) 0.307 (b) 0.5 (c) 0.693 (d) 1.0 [GATE 1992] (Ans. c) 3. The process of hot extrusion is used to produce (a) curtain rods made of aluminium (b) steel pipes for domestic water supply (c) stainless steel tubes used in furniture (d) Large size pipes used in city water mains [GATE 1994] (Ans. a, c) 774 A Textbook of Production Technology 4. Match 4 correct pairs between List I and List II List I List II (a) Rivets for air craft body 1. Forging (b) Carburettor body 2. Cold heading (c) Crank shafts 3. Aluminium-based alloy (d) Nails 4. Pressure die casting 5. Investment casting [GATE 1996] (Ans. a– 3, b – 4, c – 1, d – 2) 5. In a progressive Die (Sheet metal work), the tonnage of the press can be reduced by (a) Grinding the cutting edges sharp. (b) Increasing the hardness of punches. (c) Increasing the hardness of die. (d) Staggering the punches. (Ans. D) 6. List I List II (A) Aluminium brake shoe (1) Deep drawing (B) Plastic water bottle (2) Blow moulding (C) Stainless steel cups (3) sand casting (D) Soft drink can (aluminium) (4) Centrifugal casting (5) Impact extrusion (6) Upset forging [GATE 1998] (Ans. A – 3, B– 2, C – 1, D – 5). 7. In sheet metal work, the cutting force on the tool can be reduced by (A) grinding the cutting edges sharp (B) increasing the hardness of tool (C) providing shear angle on tool (D) increasing the hardness of die [IES 1993] (Ans. C) 8. Tandem drawing of wires and tubes is necessary because (A) it is not possible to reduce in one stage (B) annealing is needed between stages (C) accuracy in dimensions is not possible otherwise (D) surface finish improves after every drawing stage [IES 1993] (Ans. A) 9. In order to get uniform thickness of plate by rolling process, one provides (A) Camber on the rolls (B) offset on the rolls (C) hardening of the rolls (D) antifriction bearings [IES 1993] (Ans. A) 10. A moving mandrel is used in (A) wire drawing (B) tube drawing (C) metal cutting (D) forging [IES 1993] (Ans. B) 11. Which of the following methods can be used for manufacturing 2 metre long seamless metallic tubes? 1. Drawing 2. Extrusion 3. Rolling 4. Spinning [IAS 1994] Select the correct answer using the codes given below: Codes. (A) 1 and 3 (B) 2 and 3 (C) 1, 3 and 4 (D) 2, 3 and 4 (Ans. B) 12. In blanking operation, the clearance provided is (A) 50% on punch and 50% on die (B) on die (C) on punch (D) on die or punch depending upon designers choice [IAS 1995] (Ans. C) 13. Consider the following states of stress: 1. compressive stress in flange 2. tensile stress in the wall 3. tensile stress on the bottom part Appendix III 775 During drawing operation, the states of stress in cup would include (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 [IAS 1995] (Ans. a) 14. The following operations are performed while preparing the billets for extrusion process: 1. Alkaline cleaning 2. Phosphate coating 3. Pickling 4. Lubrication with reactive soap The correct sequence of these operations is (A) 3, 1, 4, 2 (B) 1, 3, 2, 4 (C) 1, 3, 4, 2 (D) 3, 1, 2, 4 [IAS 1995] (Ans. D) 15. Match List I with List II and select the correct answer using the codes given below the lists: List I (Metal forming process) List II (A similar process) A. Blanking 1. Wire drawing B. Coining 2. Piercing C. Extrusion 3. Embossing D. Cup drawing 4. Rolling 5. Bending Codes: A B C D A B C D (a) 2 3 4 1 (b) 2 3 1 4 (c) 3 2 1 5 (d) 2 3 1 5 [IES 1994] (Ans. b) 16. In sheet metal blanking, shear is provided on punches and dies so that (A) press load is reduced (B) good cut edge is obtained (C) warping of sheet is minimised (D) cut blanks are straight [IES 1994] (Ans. A) 17. Which of the following pairs of process and draft are correctly matched? 1. Rolling........ 2 2.Extrusion.........50 3.Forging......4 Select the correct answer using the codes given below: Codes: (A) 1, 2 and 3 (B) 1 and 2 (C) 1 and 3 (D) 2 and 3 [IES 1994] (Ans. A) 18. The mode of deformation of the metal during spinning is (A) bending (B) stretching (C) rolling and stretching (D) bending and stretching [IES 1994] (Ans. D) 19. In drop forging, forging is done by dropping (A) the work piece at high velocity (B) the hammer at high velocity (C) the die with hammer at high velocity (D) a weight on hammer to produce the requisite impact [IES 1994] (Ans. C) 20. Metal extrusion process is generally used for producing (A) uniform solid sections (B) uniform hollow sections (C) uniform solid and hollow sections (D) varying solid and hollow sections [IES 1994] (Ans. C) 21. Which one of the following is an advantage of forging? (A) Good surface finish (B) Low tooling cost (C) close tolerance (D) Improved physical property [IES 1996] (Ans. D) 22. In wire drawing process, the bright shining surface on the wire is obtained if one 776 A Textbook of Production Technology (A) does not use a lubricant (B) uses solid powdery lubricant (C) uses thick paste lubricant (D) uses thin fluid lubricant [IES 1996] (Ans. D) 23. Match List I with List II and select the correct answer List I (Metal forming process) List II (Associated force) A. Wire drawing 1. Shear force B. Extrusion 2. Tensile stress (force) C. Blanking 3. Compressive force D. Bending 4. Spring back force Codes: A B C D A B C D (a) 4 2 1 3 (b) 2 1 3 4 (c) 2 3 1 4 (d) 4 3 2 1 [IES 1996] (Ans. c) 24. In metals subjected to cold working, strain hardening effect is due to (A) slip mechanism (B) twining mechanism (C) dislocation mechanism (D) fracture mechanism [IES 1997] (Ans. A) 25. 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 die upset forging (C) closed die press forging (D) open die progressive forging [IES 1998] (Ans. D) 26. The bending force required for V-bending, U-bending and Edge bending will be in the ratio of (A) 1 : 2 : 0.5 (B) 2 : 1 : 0.5 (C) 1 : 2 : 1 (D) 1 : 1 : 1 [IES 1998] (Ans. A) 27. Match List I with List II and select the correct answer using the codes given below the lists: List I List II A. Drawing