Production Technology (Solutions for Text Book Objective & Conventional Practice Questions)

04. Ans: (a) Chapter Sol: Q = 1.6 10-3 m3/sec Metal Casting 1 A = 800 mm2 Q = A  V 01. Ans: (d) 1.6  10-3 = (800 10-6) V VH Sol: Permeability number = V = 2 m/sec = gh2 PAT 2  2  For standard specimen H = D = 5.08 cm h =   = 0.203m 2   P = 5 gm/cm, V=2000 cc, T= 2 min   81.92   08.52000 = 203 mm PN =  1.50 2  5 2  208.5 4 05. Ans: (c)  Sol: Vol. of casting = 2 LD 02. Ans: (c) 4 Sol: Net buoyancy force = Weight of core –  = 2  200150 weight of the liquid 4 which is displaced by core  3534291 mm3

= V.g (  – d ) ht = 200+ 50 = 250 mm  A= A = sprue base area  2  dghd C min 4 400 =  200 mm2  2 2   1130081.918.012.0  1600  4 G.R.= 1:1.5:2 = 193.6N CastingofVolume Pouring time =  VA max.C 03. Ans: (a) 3534291  Volume Sol: Pouring time =  25098102200  VA C max 17671 =  8 Sec 102 6  25098102   175100002200 = 5.34 sec

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06. Ans: (c) 07. Ans: (c)

Sol: The dimension of pouring basin will not Sol: 1 1 affect the pouring time hpb =50 mm 2 2 Let V = maximum velocity of molten metal hs= 200 mm in the gating system, 3 d = dmin = dia. Sprue bottom h = height of sprue = 200 mm castingof.volume A = 650 mm2 Pouring time = P. T  2 5 3 c  VA max Q = flow rate = 6.5 10 mm /s 4 2 353 g = 10 mm/sec  = 25  5 2 105.6 2  Vd V2 =  /1000 Secmm 4 650 3 35 4 V   d/6.2183 2 …… (1) =  h102gh2  pb pb 2  25d 4 hpb = 50 mm = height of molten metal To ensure the laminar flow in the gating in the pouring basin

ht = total height of molten metal above system Re  2000 the bottom of the sprue

= 200 + 50mm For limiting condition Re = 2000 4  dV Vd AVAVAQ 33322  250102 e 2000R  =   =  35 /105.6 smm Vd 2 2000  A3 = 290.7 mm 

2000  9.02000 1800 V    … (2) 08. Ans: (d) d d d Sol: dtop = 225 mm

ht = 250 + 100 = 350 mm From (1) and (2) Volume flow rate Q = 40×106 mm3/sec 6.2183 1800  d 2 d Vbottom = 2= hg t  35098102 6.2183 = 2620 mm/s d  mm21.1 1800 Q = Atop×Vtop=Abottom×Vbottom

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6 2 1040 2 2 D 2 Abottom = =15267.17 mm  SP M SP 6  D  2620  =    M a  a  Cub Cub 6  17.152674 dbottom = = 139.42  140 mm 2 2   2R   2R  =  =    54.1  a   61.1 R  09. Ans: (b) 2  M SP  SP Sol: A2V2 = A3V3   cylM cyl  2  100981022252 2 4  D    2  6  DSp  R2   =       = 1.306 2  D   D   R75.1   b  35098102d  cyl    4  6 

 db = 164.5mm So aspiration will not occur. 12. Ans: 1.205 Sol: Casting – 1 (circular) Common Data for 10 & 11 Diameter = 20mm, length = 50mm Casting -2 (elliptical) 10. Ans: (a) Major/Minor = 2, length = 50mm, 11. Ans: (b) C.S. area of the casting -1 = C.S area of the Sol: 3 castings of spherical, cylindrical and casting -2 cubical  solidifica  1castingoftimetion    Vsp = Vcube  solidifica  2castingoftimetion 

4 2   aR 33  M c1    AV 2c1c  3 =   =   M c2    AV 1c2c  4 a = R 3  = 1.61 R 3  2  2  Vc1 =  hd = 5020 4 4  Vcyl = VSp 3   = 15707.96 mm DH23  R 43  2 Ac1 = 2  dhd 4  33 DR    )HD( 43    2  5020220 1 4  3   16 3 16  D = 3 R    R75.1R 2 3  3  = 3769 mm

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C.S area of cylinder = C.S area of ellipse 13. Ans: 50    axis.minaxis.maj Sol: m = 2 kg, Q = 10 kW  202   4  4 Time taken for removing latent heat  )axis.(min2 2 = 20 – 10 = 10 sec = heatLatent 4 Time  1 Q 2   2 4  Latent heat = time  Q Minor axis =  20   4  2  = 10  10 = 100 kJ Minor axis = 14.14mm 100 Latent heat/kg = = 50 kJ/kg Major axis = 2  minor axis = 28.3mm 2

ba 22 Perimeter = 2 14. Ans: (a) 2 Sol: Circular disc casting Squared disc casting 3.28 C C where a = major axis /2  = 14.14 mm 1 ; 2 2 d  20cm  20a cm 14.14 b = minor axis /2  = 7.07 mm  cm10t ;  cm10t 2  A  Perimeter = 70.24 mm  s   V  1C Surface area of ellipse Freezing ratio (F.R) = X1 =  4.1  A s  = perimeter  length + 2 C.S. area    V  R = 70.2450 + 314  2  A s  2   = 4140 mm = AC2  A   V    s   1C Volume of the ellipse  V  R 4.1 = C.S area  length  A s   A s  3     = 314  50 = 15708 mm = Vc2  V   V  X  2C  2C  4.1 1  A   A   solidifica 1castingoftimetion   s   s    V V solidifica  2castingoftimetion    R   1C 4.1 2  M c1    =  As   As          4.0  M c2       V C 2  V C1  2 2   AV 2c1c    414096.15707  V R =      Volumetric ratio,(V.R) = Y1 =  8.0  AV 1c2c   15708 9.3769  VC

= 1.205  VR = 0.8 V C1

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V 8.0 V Common Data for Q.16 & Q. 17 Now Y  R  C1 2 V V C2 C2 16. Ans: (a) 17. Ans: (a)

  2  Sol: In centrifugal casting 8.0   1020   4  2  = 0.628 Centrifugal force = FC = ma = m r   102020 a = r2

D 15. Ans: (b) 75 g = 2(  N)2 2 Sol: VC = 40 × 30 × 0.3 = 360 cc 2 2 4 VSc = shrinkage volume 75 ×9810 = DN  2 3 =  8.10360 cc  981075 100 Constant = 2 DN   37273 2 2  2 Volume of riser Vr = hd 2 4 Constant = DN  37273   52.05.0 = 2  cc24.5044 D = = 0.51 m = 510 mm 4 2

Vr ≥ 3 Vsc  cc4.328.103V 37273 37273 r N  = 8.55 RPS D 510 Vr ≥ 3 VSc → Satisfied

   Cr 18. Ans: 51.84 mm where 2   m  r = time taken for riser material to solidify R R Sol:    C  mC  C = time taken for casting to solidify  2012080  MM cr m  c  208020120120802  V   V           mc = 7.05  As   As  r casting d V 360 m  givenriserside  R 6  A  403.03.03030402 s m  R  5.1  V  d 4 m     = 0.666 C  A  6 6  s r  d = 51.84 mm 360 =  147.0 2442

 r > C Hence diameter of riser = 4 cm

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19. Ans: 21. Ans: (b) Sol: Given, Sol: Porosity  Permeability test Gating ratio = 1:2:2 Grain size  Sieve analysis (Sprue : Runner : Ingate) Cohesiveness  Shear test Mass, (m) = 30 kg, Strength  UTM Density () = 7.8 g/cc 22. Ans: (d) Solidification time () = 12.6 sec, Sol: Shell mold casting  Phenolic resin as Pouring height (hp) = 250 mm mold material. Sprue height (hs) = 200 mm H = 250 + 200 = 450 mm = 0.45 m Investment casting  Upto 5 kg casting Casting mass Die casting  Only for low melting Choke area =  gH2 point materials Centrifugal casting Axis of rotation is 30    45.081.926.127800 only in horizontal Choke area = 102.73 mm2 = Sprue area (A ) s 23. Ans: (c)  2  73.102d Sol: The main drawbacks of the investment 4 s casting process, which restrict its use to ds = 11.43 mm high quality and small sized castings only. Area of runner = 2  102.73 = 205.46 mm2 Castings weighing from few grams to 5 kg. Area of ingate = 2102.73 = 205.46 mm2

24. Ans: (b) 20. Ans: 0.05 s Sol: For the gating system design is to fill the Sol: Momentum is considered as constant mould in the smallest time. The time for Momentum of water = Momentum of liquid complete filling of mould termed as pouring metal time, is a very important criterion for pressure time pressure time  design. density density Too long a pouring time requires a higher  05.0200  time400 pouring temperatures and too less a pouring  1000 2000 time means turbulent flow in the mould   s05.0time which makes the casting defect prone. There is thus an optimum pouring time for any given casting.

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25. Ans: (c) 30. Ans: (b) Sol: For top riser, D = 2H Sol: For side riser, D = H  Shrinkage cavity  Providing chills  Misrun  Increasing degree of super heat 26. Ans: (a) of molten metal

27. Ans: (c)  Blister  Providing vent holes. Sol:  Rat tail  Adding additives such as saw  Shrinkage cavity  Providing chills dust, cereals hulls, cow dung or horse manure.  Misrun  High pouring temperature

 Sand inclusion  Splash core 31. Ans: (d)  Blister  Providing vent holes Sol: Strainer  Filters the slag

28. Ans: (a) Splash core  Avoids the sand erosion Sol: Skin bob  Separates lighter and heavier  Saw dust or wood powder: They are added impurities to increase the porosity property of Tapered sprue  Avoids sand erosion moulding sand and collapsibility of moulding sand. 32. Ans: (b)  Coal powder: Used for increasing the Sol: refractoriness of the moulding sand.  Skim bob is used for separating impurities  Starch or dextrin: Used for increasing present in molten metal. strength and resistance for deformation of  Riser is acting as a reservoir for supplying moulding sand. the molten metal to the casting cavity to compensate the liquid shrinkages taking 29. Ans: (d) place during solidification. Sol:  Skim bob is used for separating impurities 33. Ans: (c) present in molten metal. Sol: Gating Ratio is the proportion of the cross-  Riser is acting as a reservoir for supplying sectional areas between sprue, runner and the molten metal to the casting cavity to ingate. compensate the liquid shrinkages taking Gating ratio = Ratio of (sprue area: runner place during solidification. area: ingate area)

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34. Ans: (a) Conventional Practice Solutions 35. Ans: (b) 01. 36. Ans: (c) Sol:

37. Ans: (b)

38. Ans: (a)

39. Ans: (a)

40. Ans: (a) Fig: Sand mould sketch for hollow component

40. Ans: (a) 02. Sol: Riser is acting as a reservoir for supplying Sol: Castings are generally cooled from all sides the molten metal to the casting cavity to and, therefore, they are expected to have no compensate the liquid shrinkages taking directional properties. Hence the cast place during solidification. products are isotropic. In forging, when plastic deformation occurs

the metal appears to flow in the solid state along specific directions, which are dependent on the type of processing and the direction of applied force. The crystals or grains of the metal are elongated in the direction of metal flow. Since the grains are elongated in the direction of flow, they would be able to offer more resistance to stresses acting across them. As a result, the mechanically worked metals called wrought products would be able to achieve better

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mechanical strength in specific orientation, 2  2 2   R 2 R  that of the flow direction. t 3 3 cube    The properties are anisotropic mechanical t  4 3  sphere  R  strength (tensile & compressive).  3  In most of engineering applications it 4 2  requires anisotropic material because the   9 3   %35.52 stresses induced in the component in 16 6 different directions are different. 9 2 Ex: In case of pressure vessel, even though 2   2  t  V   hd  uniform internal pressure is applied inside cylinder   cylinder    4  t  V   4 3  the pressure vessel, the stresses induced in sphere  sphere   R   3  the hoop direction is higher than the A = A longitudinal direction. cylinder sphere  2 2  R4dhd 2 03. 4 2  Sol: t  V 22  R4dd 2 2 t cube 100 3d2 = 8R2 t sphere 8 t cylinder d  R 100 3 t sphere 2 2  8 8        2  2 3  R R  t V   t cylinder   3  3 4  cube  cube =   2     t V 4 3 t 4 3 sphere sphere  R  sphere  R       3   3  A = A cube sphere 64 8 1 9 2 2  = 66.67%   R4 9 3 16 16 R2 2   3 04. 2 Sol: (1)  20010070mm   R 3 (2)  20010010mm

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2 2  V   a   2   kt   t  A   6  2   2  2 2 t1  V  t  a  1    2  A1  t  T 2 tT   10100200       10200101001002002  1  CtCT 2  2   70100200   C30Cmm4   1 2   70100702001002002  1  C80Cmm5.6 2 2  200000    C1 = 0.7210  46000  90359.18  2  C2 = 0.0505 1400000  4931.291     0505.01507210.0T = 8.88 mm  82000 

t2 = 0.6485min 07. t2 = 38.91sec Sol: V = 1000cm3 c 5 05. 1000V  cm50 3 s 100 Sol: Q = AV As:Ar:Ag = 1:4:2 3 Vr = 3Vs = 150cm Volume  VA  t cc 2  150hD 4 3750000  gh2A D = 5.758cm t c c D 3750000 m  9598.0 3  3001081.92 r 6 t  V   a 3   t = 3.09 sec m      67.1 c    2   A  a6 c  c

06. mr ≯ mc 2  V  .n Sol:  kt    67.1  A  6

2 D = 10 cm  a 3   kt    2   a6 

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08. around the pattern and is removed from the Sol: Shell molding can produce many types of pattern using built-in ejector pins. castings with close dimensional tolerances Two half-shells are made in this manner and and a good surface finish at low cost. Shell- are bonded or clamped together to form a molding applications include small mold. The thickness of the shell can be mechanical parts requiring high precision, determined accurately by controlling the such as gear housings, cylinder heads, and time that the pattern is in contact with the connecting rods. The process also is used mold. In this way, the shell can be formed widely in producing high-precision molding with the required strength and rigidity to cores. hold the weight of the molten liquid. The shells are light and thin-usually 5 to 10 mm- In this process, a mounted pattern made of a and consequently, their thermal ferrous metal or aluminum is characteristics are different from those for (a) heated to a range of 175° to 370°C, thicker molds. (b) coated with a parting agent (such as silicone), and Shell sand has a much lower permeability (c) clamped to a box or chamber. The box than the sand used for green-sand molding, contains fine sand, mixed with 2.5 to because a sand of much smaller grain size is 4% of a thermosetting resin binder (such used for shell molding. The decomposition as phenol-formaldehyde) that coats the of the shell-sand binder also produces a sand particles. Either the box is rotated high volume of gas. Consequently, unless upside down (shown in below Figure), the molds are vented properly, trapped air or the sand mixture is blown over the and gas can cause serious problems in the pattern, allowing it to form a coating. shell molding of ferrous castings. The high quality of the finished casting can reduce The assembly is then placed in an oven for a cleaning, machining, and other finishing short period of time to complete the curing costs significantly. Complex shapes can be of the resin. In most shell-molding produced with less labor, and the process machines, the oven consists of a metal box can be automated fairly easily. with gas-fired burners that swing over the shell mold to cure it. The shell hardens

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Pattern Coated sand

Coated sand Investment Pattern

Coated Dump box sand (ii). Pattern and dump (iii). Pattern and dump box in (i). Pattern rotated and clamped to dump box box rotated position for the investment

Shell Flask Shells

Excess Sand or coated sand Adhesive Clamps metal beads (vi). Mold placed in flask and (iv). Pattern and shell (v). Mold halves joined together removed from dump box metal poured

Fig : The shell moulding process(Dump box technique)

09. Sol: The hot-chamber process (Fig. 1) involves the use of a piston, which forces a certain volume of metal into the die cavity through a gooseneck and nozzle. Pressures range up to 35 MPa, with an average of about 15 MPa. The metal is held under pressure until it solidifies in the die. To improve die life and to aid in rapid metal cooling (thereby reducing cycle time) dies usually are cooled by circulating water or oil through various passageways in the die block. Low-melting-point alloys (such as zinc, magnesium, tin, and lead) commonly are cast using this process. Cycle times usually range from 200 to 300 shots (individual injections) per hour for zinc, although very small components, such as zipper teeth, can be cast at rates of 18,000 shots per hour.

In the cold-chamber process (Fig. 2), molten metal is poured into the injection cylinder (shot chamber). The chamber is not heated-hence the term cold chamber. The metal is forced into the die cavity at pressures usually ranging from 20 to 70 MPa, although they may be as high as 150 MPa.

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Figure : 1 The hot chamber die casting process.

Figure: 2 Schematic illustration of the cold-chamber die-casting process.

These machines are large compared to the temperatures start at about 600°C for size of the casting, because high forces are aluminum and some magnesium alloys, and required to keep the two halves of the dies increase considerably for copper based and closed under pressure. iron-based alloys. The machines may be horizontal (as in the figure)-or vertical, in which case the shot 10.  63030 chamber is vertical. High-melting-point Sol: m  c  9001801802 alloys of aluminum, magnesium, and copper  D normally are cast using this method,  2.1142.2 although other metals (including ferrous 6 metals) also can be cast. Molten-metal D = 15.428 cm

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14 ESE – Text Book Solutions

11.  It is eliminated by reducing pouring time or Sol: increasing pouring temperature or degree of 1. Dross: The presence of impurities or superheat. foreign particles inside the castings is called as dross. 4. Shrinkage cavity or void: This is due to  A open space or void produced due to non-  Improper separation of impurities present in availability of molten metal for molten metal. compensating liquid shrinkages taking place  It is eliminated by using during solidification (i) providing projections in pouring basin  This is eliminated by directional (ii) using strainer or skim bob solidification, i.e., by providing chills in the (iii) offsetting the axis of the runner & casting process. ingate. 2. Inclusion or Sand inclusion. 5. Shift: The mismatches present between the  Presence of sand particles inside the casting cavities of cope and drag produce the step is called as sand inclusion in the casting is called as shift defect.  This is due to sand erosion taking place It is eliminated by using dowel pins during pouring of molten metal in casting cavity 12.  Sand erosion is due to Sol: Investment Casting: (i) allowing turbulent flow Investment casting (also called "lost wax" (ii) not using the splash core or "precision" casting) is a manufacturing (iii) using top gating system for filling process in which a wax pattern is created, cavities in loose sand moulds gated onto a sprue and repeatedly dipped (iv) using top gating system for cavities into liquid ceramic slurry. Once the ceramic having height more than 200 mm. material hardens, its internal geometry takes 3. Misrun: It is due to non filling of projected the shape of the casting. The wax is melted portion of casting cavity using molten metal out, and molten metal is poured into the is called Misrun. cavity where the wax pattern was. The This is due to metal solidifies within the ceramic mold,  Solidification of molten metal has been and then the metal casting is broken out. started before complete filling of casting cavity.

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1. Tooling and Pattern making : A tool is built 4. De-Waxing and Firing : The molds are to customer provided specifications (A). flash-fired to remove the wax and sprue Cold wax is then injected into the tool to materials and then heated to 1800 and create a wax pattern/prototype (B) that will placed on a sand bed, ready for pouring. hold precise dimensional requirements in the final casting.

5. Casting: Molten metal, up to 3000, is poured into the hollow mold and then cooled. 2. Pattern Assembly : The wax patterns are assembled onto the sprue.

6. Knockout : The ceramic shell is broken off,

and the individual castings are cut away.

3. Dipping and Coating : Successive layers of

ceramic (A) and stucco (B) are applied to

the sprue assembly to form a hard shell.

7. Finishing: Excess metal is removed, surfaces are finished, and castings are heat treated.

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16 ESE – Text Book Solutions

8. Testing and Inspection: Castings undergo through testing and inspection to ensure that they meet dimensional tolerances and specifications.

Die Casting: Die casting is a manufacturing process for producing metal parts by forcing molten metal 9. Packing and Shipping: Castings are under high pressure into a die cavity. These die or securely packaged for shipping to the mold cavities are typically created with hardened customer. tool steel that has been previously machined to the net shape of the die cast parts.

Die inserts machined Release agent Shot sleeve filled Piston forces molten From tool steel Applied to die With molten metal Metal into the die

Piston maintains pressure When sufficiently Individual cast parts are Final machining on molten metal when die cooled, die inserts are trimmed from the operations performed to is filled separated and the casting tree finish part casting tree is removed

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Investment Casting :  Excellent precision, ideal for complex Chapter Welding geometries 2

 Can meet tight tolerance requirements 01. Ans: (a)  Superior surface finish, little additional Sol: V0 = 80 V, IS = 800 A machining required Let for arc welding, V = a + bL  Higher total cost than other casting V0 processes For power source, Vp = V0– I Is  Lower tooling costs For stable V = Vp  Suitable for both ferrous and non-ferrous V VLba  0 I metals 0 Is  Some product size restrictions When L = 5, I = 500 Die Casting 80  a + b × 5 = 80 –  30500  Produces parts with good dimensional 800 tolerance a + 5b = 30  Little secondary machining required when L = 7, I = 460  Ideal for large production runs and high- 80 807ba  34460 volume projects 800  Excellent for producing consistent, By solving, b = 2, a = 20 repeatable parts  V = a + bL = 20 + 2L  High tooling costs 02. Ans: 4860 W, 1.5 mm 13. Sol: For power source, Sol: A1 = A2 (1) (2) I 2 2 Vp = 36 – 4r = 6a 60 2/1 r  3  Va = 2L + 27    r a a  2  At equilibrium conditions

2 2 V = V t  v   4 r3  a P 1   1       3 I t 2  v2   3 a  27 + 2 L = 36 – 60 2 3  4   3  6       = 1.909  3   2  

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18 ESE – Text Book Solutions

I V 40 4  L29L22736 0  60 Is 50 5 I = 60 (9 – 2L) 4 If current is 360 Amps 0 140V  250 5 360 = 60 (9 – 2L) = 140 + 200 = 340 360 9 – 2L =  6 V0 4 0  5V  5340 60 Is    A425 Is 5 4 4 2L = 9 – 6 = 3

3 L =  5.1 04. Ans: 26.7 sec 2 Sol: Rated Power = V I = 50 ×103 If L = 1.5 mm, r r 1050 3 V = 27 + 2 ×1.5 = 27+ 3 = 30 V  I   A2000 r 25 I = 60 (9 – 2 ×1.5) = 360 A D = 50% (rated duty cycle) P = 30 ×360 = 10800 W r If I = 1500 A (desired current) If L = 4 mm, V = 27 + 1.5 ×4 = 33 V d Desired duty cycle, I = 60 (9 – 1.5 ×4) = 180 A 2 2 P = 33 × 180 = 5940 W r DI r  2000  Dd = 2     89.05.0 Change in power = 10800 – 5940 Id  1500  = 4860 W timeonArc Dd = = 0.8930 If the maximum current capacity is 360A, Total welding time the maximum arc length is 1.5mm = 26.7 sec

03. Ans: 425 05. Ans: 27.78 mm/sec Sol: V = 100 + 40 L , Sol: Power = P = 4 + 0.8L – 0.1L2 L = 1 to 2 mm , I = 200 to 250 A For optimum power L = 1, I = 250 dP V  0  0.8 – 0.2L = 0 V = 100 + 401 = 140 V 0  250 dL 0 I s 8.0 L  mm4 L = 2, I = 200 2.0 2 V0 P = 4 + 0.8L – 0.1 L V = 100 + 40  2 = 180 V0  200 2 Is = 4 + 0.8 ×4 – 0.1 × 4 = 5.6 kW V Energy losses = 20% ,  = 80%  5040  0 Is

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19 Production Technology

Area of weld bead (WB) l = 1 m =1000 mm; 1 t = 30 mm, = 2  ACAB 2 d = 4 mm,

= 5 tan 30 × 5 = 14.43 Lt = 450 mm;

5 tan 30 LS = 50 mm, A B 2 A1 = 4  30 = 120 mm 300

0 1 2 60 5 A2 = A3 =  3030tan30 = 259.8 mm 2 Total volume of weld bead C = volume of weld bead + crowning Volume of W.B = 14.43 × 1000 = 1.1  volume of weld bead = 14433 mm3 3 = 1.1 (A1+2A2)1000 = 703560 mm -6 Weight of W.B = 14433 × 10 × 8  Volume /Electrode = 2  LD = 115.5 g 4 e Heat required for melting of W.B   2  )50450(4 = 1600 = 115.5 ×1400 = 161.66 kW 4 66.161 Number of electrodes required Time for welding =  36 Sec  6.58.0 Total beadweldofvolume  1000 Welding speed = Electrode/volume 36 703560   14096.139  78.27 mm/sec 1600 x = 200mm (given) Common data for 06, 07 & 08. 1000 Number of electrodes/pass =  5 200 06. Ans: (d) 07. Ans: (d) 140 Number of passes =  28 5 08. Ans: (c) Total Arc on time

Sol: 1000 30mm (2) (1) (3) =  utesmin28028

30o 30o 100 280 Total weld time =  67.466 minutes 6.0 4mm

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20 ESE – Text Book Solutions

09. Ans: 0.64 mm & 2.1 mm 11. Ans: (b) Sol: Given AC = 10 mm, Sol: Filling rate of weld bead = filled rate by

O1A = O1C = 7 mm, electrode O A = O C = 20 mm  2 2 Area of W.B × Speed = 2  fd

O2 4

r=7  2 D  40002.1 Area of W.B= 4  12.25 mm2 A B C 180

r=20 E Common data for 12 & 13 O1 12. Ans: 2000 J

Sol: H.G = I2 R 

2 -6 5 Height of Bead = BD = O1D – O1B = (10000) ×200×10 ×  J2000 50 = OD– 2  ABAO 2 1 1 13. Ans: 1264 J = 20 – 520 22 Sol: h = 2t – 2 × 0.1 t = 1.8 t = 0.64 mm = 1.8 ×1.5 = 2.7 mm D = t  5.166 = 7.35 mm

Depth of Penetration = BE = O1E–O1B

22 0.1 t = OE12 O A AB

h =  577 22 = 2.10 mm

0.1 t Common Data Q. No 10 and 11

10. Ans: (c)  Sol: I = 200, V = 25, speed = 18 cm /min Vol. of nugget = 2 hD 4 D = 1.2 mm, f = 4 m /min,  = 65%,  2 2  ηIV =  7.235.7 = 114.5 mm Heat input = 4 speed Heat required = Volume × ×heat required /g  600.6520025  = 3  13808105.114 18 = 1264 J = 10.83 kJ / cm

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21 Production Technology

14. Ans: 2.3 & 4.6 MJ 17. Ans: (b)    Sol: Heat dissipated = 360 – 344 = 16 J Sol: RC = 0.85   rn   = Resistivity of metal 18. (i) Ans: (a), (ii) Ans: (b) 3 2 2 Sol: P = 2 kW = 2  10 Watt, 2  V  V (Heat generation)1 = I R =    R =  R  R V = 200 mm/min, L = 300 mm Heat required (HR) = 40 Kcal  10285.0 5 R  = 1.082105 3 C1  02.025 = 4010 4.2 Joule

5 300  10285.0 6 Welding time =  605.1min5.1 R C  = 5.41  10 200 2  02.050

2 = 90sec 5 3 (H.g)1 = = 2310546.04 Heat input = 2 10  90 Joule 10082.1 5 3 2 HR  2.41040 5 HI =  3  9333.0 (H.g)2 = = 4621072.08 HI  90102 1041.5 6 = 93.33%

15. Ans: (c) 19. Ans: (d) Sol: Heat generated = Heat utilized Sol: Heat supplied = Heat utilized I2R = Vol. of nugget × × H. R/g 0.5 J = m (S.H. + L.H) = V (SH+LH) 2 6  .010200I 1 = (a×h)  (Cp (Tm–Tr)+LH)  2   3  1014008000105.1005.0 3 = 0.05 × 10-6 × h × 2700 [896  (933 – 4 303) + 398 × 103] I = 4060 A h = 0.00385 m = 3.85 mm  Common Data Q. 16 & Q. 17 16. Ans: (c) 20. Ans: (c)  Sol: I = 3000 A,  = 0.2, R = 200 Ω Sol: Volume to be melted = 2 2  2)100110( 4 Volume of nugget = 20 mm3  66.3298 mm3 Heat generation = I2R Total heat required = 30002 ×200×10-6 ×0.2 = 360 J = 3298.66 × 10-9 × 64.4 ×106 Heat required =   LHTTcV  rmp = 212.4 Joules -9 3 = 800020×10 ×500(1520 –20)+1400×10 V V 3022 P = VI = V  43.21 = 344 J R R 42

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22 ESE – Text Book Solutions

Total heat required = heat to be generated 23. Ans: 61.53 % 212.4 = Pt Heat required Sol: Thermal efficiency = 100 4.212 pliedsupHeat t =  sec10 43.21 Heat required = 10 × 80 = 800 J 800 thermal = 100 = 61.53 % 21. Ans: (a) 1300 Sol: Frictional force F = Pressure × Area × 24. Ans: 464.758 A  2 = 200  78545.010 Sol: Dd = 100% = 1, Ir = 600A, Dr = 0.6 4 D I2 3 d  r Torque = F  Radius D I2 4 r d 2 3 3 1 600 2 2 Torque = 7854  45.29105 2 d  6.0600I 4 6.0 Id NT2 Power, P =  Id = 464.758A 60000  45.2940002 25. Ans: 17  = 12.33 kW 60000 Sol: Number of electrodes Total metalofvolume deposited  22. Ans: 0.065 sec Volumedeposited onefrom electrode Sol: Given: Total metalofVolume deposited  3  Volume = 80 mm , 2  504503 4 Current (I) = 10000 A, 3 17mm E = 10 J/mm , 19mm 2mm 30o 30o Qlost = Heat lost = 500 J,

R = 0.0002 ohms 2 x 30tan o  Total energy supplied during process 17mm = [(80 × 10) + 500] J x = 9.814 mm Q = 1300J = i2Rt total  1  1300 = (104)2 × 0.0002 × t Area     15.11.1192217814.9  2   t = 0.065 seconds Volume   2   18015.11.1mm85.204

= 46645.30767mm3

 Number of electrodes = 17

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23 Production Technology

26. Ans: 30. Ans: (b) Sol: Given, Butt-welding, Sol: Arc power (Q) = 2.5 kVA = 2.5  103 J  Electro slag welding  Silicon based flux Thickness (t) = 3 mm = 310–3 m  Laser beam welding  Absorptive of the material. V-joint Angle () = 60  Resistance welding  Without use of filler Efficiency (arc) = 0.85 rod 2D – heat transfer :  Soldering  Non fusion welding of metals –5 2 steel = 1.210 m /sec , having melting point < 427C ksteel = 43.6 W/mC  Brazing  Non fusion welding of metals

Assuming TC = 1450C having melting point > 427C  Vb  c 2.0tTk8Q   4  31. Ans: (c) b = width of weld, Sol:  In resistance welding, copper alloys used as b = 2  3  tan30 = 3.464  10–3 m electrode materials must exhibit, high 3 3 3   10464.3V   2.010314506.438105.2  5  electrical and thermal conductivities,   102.14  combined with high strength at elevated

 166.72V2.0647.1 temperatures. For the process to work properly, the Welding speed, V = 20.06 mm/sec  contact resistance must be higher at the

point to be welded than any where else. 27. Ans: (c)

32. Ans: (b) 28. Ans: (c) Sol: At the intersection of inner and outer cone Sol: The cheapest and safest method of the maximum temperature induced is producing oxygen is from atmospheric air. (i) 3260oC in neutral flame O Carburizing flame: 2  95.0to85.0 (ii) 3380oC in oxidizing flame and HC 22 (iii) 3040oC in carburizing flame.

29. Ans: (a)

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24 ESE – Text Book Solutions

33. Ans: (425A) temperature affect the hardness, brittleness, Sol: V = 100 + 40 L , ductility and strength of iron carbon based L = 1 to 2 mm , I = 200 to 250 A metals such as steel and cast iron. Hence L = 1, I = 250 option (c) is correct.

V0 V = 100 + 401 = 140 V0  250 Is 36. Ans: (d) L = 2, I = 200 Sol: Resistance welding can be used for welding of aluminium and its alloys but they require V0 V = 100 + 40  2 = 180 V0  200 Is much higher welding currents because of V high thermal and electrical conductivity  5040  0 compared with steel. Is

V0 40 4  37. Ans: (a) Is 50 5

4 140V  250 0 5 38. Ans: (c) = 140 + 200 = 340 39. Ans: (c) V0 4 0 5V 5340 Is    A425 Is 5 4 4 40. Ans: (c)

Sol: 34. Ans: (b)  Process like laser beam welding and

electron beam welding give a highly 35. Ans: (c) concentrated, limited amount of heat, Sol: resulting in a small HAZ. 1. Preheating slows down the cooling rate of the weld and gives the metal more time to  Electron beam welding is not typically used form a good microstructure, release internal with lead, zinc, cast iron, thermoplastics and stresses and dissipated hydrogen from the most other synthesis. weld. 2. The major purpose of preheating and post 41. Ans: (b) heating of cast iron is to control the rate of temperature change. The level of 42. Ans: (b) temperature and the rate of change of

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25 Production Technology

well as flux-cored electrodes may be used.

Conventional Practice Solutions The guide may be non-consumable (conventional method) or consumable. 01. Electro slag Welding is capable of welding Sol: Electroslag welding (ESW) process and its plates with higher thicknesses ranging from applications are shown in below figure (1). 50 mm to more than 900 mm, and Welding The main difference is that the arc is started is done in one pass. The current required is between the electrode tip and the bottom of about 600 A at 40V - 50 V although higher the part to be welded. Flux is added, which currents are used for thick plates. The travel then melts by the heat of the arc. After the speed of the weld is in the range from 12 to molten slag reaches the tip of the electrode, 36 mm/min and weld quality also good. the arc is extinguished. Heat is produced This process is used for large structural- continuously by the electrical resistance of steel sections, such as heavy machinery, the molten slag. Because the arc is bridges, oil rigs, ships, and nuclear-reactor extinguished, ESW is not strictly an arc- vessels. welding process. Single or multiple solid as

Power source Control panel Regulator Electrode holder Wire reel Tungsten electrode Wire-feed drive Electrode lead Oscillation (optional) Gas Passages Insulating Consumable guide sheath Molten slag tube Inert Work Molten weld pool Welding gas Workpiece Retaining shoe Shielding gas power source supply (ground) lead Water In

Water Out Work Fig 1: Electroslag welding (ESW) Fig 2: Tungsten inert gas welding setup

Gas Tungusten Arc Welding (GTAW) or pool from the atmospheric contamination. The Tungusten Inert Gas Welding (TIG) : shielding gases generally used are argon, helium In this process a non-consumable tungsten or their mixtures. Typical tungsten inert gas electrode is used with an envelope of inert welding setup is shown in Fig.2. shielding gas around it. The shielding gas protects the tungsten electrode and the molten metal weld

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26 ESE – Text Book Solutions

Electrode materials : pool and consequently porous welds. Because of The electrode material may be tungsten, or the use of shielding gases, no fluxes are required tungsten alloy (thoriated tungsten or zirconiated to be used in inert gas shielded arc welding. tungsten). Alloy-tungsten electrodes possess However for thicker sections, it may be desirable higher current carrying capacity, produce a to protect the root side of the joint by providing a steadier arc as compared to pure tungsten flux. The process is generally used for welding electrodes and high resistance to contamination. aluminium, magnesium and stainless steel.

Electric power source Advantages: Both AC and DC power source can be used for  Non-consumable electrodes - It helps to TIG welding. DC is preferred for welding of provide flawless joints because it is not copper, copper alloys, nickel and stainless steel needed to stop for replacing the electrode as whereas DC reverse polarity (DCRP) or AC is in consumable electrode welding. That also used for welding aluminium, magnesium or their contributes to reducing downtime in alloys. DCRP removes oxide film on magnesium production. and aluminium.  No flux is required because inert gas shields molten metal. So no slag and slag inclusion Inert gases : problems. The following inert gases are generally used in  High quality and strong welding achieved by TIG welding: TIG. 1. Argon  Cleaner and more appealing joints. 2. Helium Sometimes they don’t need finishing process. 3. Argon-helium mixtures  They are suitable for welding of very thin 4. Argon-hydrogen mixtures sections.  The versatility of method. They can work Tig Nozzle : with and without filler metal. The nozzle or shield size (the diameter of the  A wide range of metal can be welded. opening of the shroud around the electrode) to be Nonferrous metals like aluminium, copper chosen depends on the shape of the groove to be and dissimilar metal can be welded without welded as well as the required gas flow rate. The any challenge. gas flow rate depends on the position of the weld  Non-corrosive and ductile joints. as well as its size. Too high a gas consumption would give rise to turbulence of the weld metal

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 The minimum amount of flames and spark. 02.

Less distortion due to small heat zone. Sol: Dd = 100% = 1 ,  It can be done in both automatic and manual. Ir = 600A, Dr = 0.6

2 Disadvantages : Dd Ir  2 TIG is a time-consuming process - They are Dr Id

2 slower than any other welding process. Lower 1 600 2 2 2 d  6.0600I filler deposition rate. 6.0 Id

1. More complicated - Highly skilled and Id = 464.758A professional workers are needed to perform TIG welding. 03. 2. Safety issue - Welders, are exposed to high Sol: Commonly used resistance welding intensity of light which can cause eye processes are spot, seam and projection damage. welding which produce lap joints except in 3. High initial cost. case of production of welded tubes by seam 4. It cannot be used in thicker sheets of metal. welding where edges are in butting position. In butt and flash welding, components are in butting position and butt joints are produced.

Electrode force, Fe Fe lw Welding Movable Electrode current, lw Work pieces Single Phase Time, T Line Supply Upper Electrode Fixed Electrodes

Step Down Transformer Fusion Weld Lower Nugget electrode Fig (a): Sketch the set-up for spot welding showing Fig (b): The force / time and current / time diagrams details of power input and electrodes

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The category of resistance welding (RW) The actual temperature rise at the joint covers a number of processes in which the depends on the specific heat and the thermal heat required for welding is produced by conductivity of the metals to be joined. For means of electrical resistance across the two example, metals such as aluminum and components to be joined. These processes copper have high thermal conductivity, so have major advantages, such as not requiring they require high heat concentrations. consumable electrodes, shielding gases, or Similar or dissimilar metals can be joined by flux. The heat generated in resistance welding resistance welding. The magnitude of the is given by the general expression current in resistance-welding operations may H = I2Rt, ------(1) be as high as 100,000 A, but the voltage is where typically only 0.5 to 10 V

H = Heat generated in joules (watt-seconds) Electrodes I = Current (in amperes) Weld nugget R = Resistance (in ohms) Lap joint t = Time of current flow (in seconds). Equation(1) is often modified so that it 1. Force 2. Current 3. Current off, 4. Force represents the actual heat energy available in applied on force on released the weld by including a factor K, which Fig. (a) denotes the energy losses through conduction Electrodes Electrode tip and radiation. The equation then becomes Indentation H = I2RtK,------(2) Weld nugget Sheet separation where the value of K is less than unity.

The total resistance is the sum of the Heat-affected zone

following properties (see Figure 2 below) : Electrode

a. Resistances of the electrodes; Fig. (b)

b. Electrode-workpiece contact resistance; Fig.2(a) Sequence of events in resistance spot welding. (b) c. Resistances of the individual parts to be Cross section of a spot weld, showing the weld welded; nugget and the indentation of the electrode on the d. Contact resistance between the two sheet surfaces. This is one of the most commonly workpieces to be joined faying surfaces. used processes in sheet-metal fabrication and in automotive body assembly.

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The strength of the bond depends on surface produces a number of welds in one pass, roughness and on the cleanliness of the extends electrode life, and is capable of mating surfaces. Oil films, paint, and thick welding metals of different thicknesses, such oxide layers should therefore be removed as a sheet welded over a plate. before welding. The presence of uniform, thin Nuts and bolts can be welded to sheets and plates by this process, with projections that layers of oxide and of other contaminants is are produced by machining or forging. not as critical. Joining a network of rods and wires (such as Applications : The applications of resistance the ones making up metal baskets, grills welding are many. Some of them are: (below fig) oven racks, and shopping carts) Automobile, Auto Ancillary, Auto Electrical, also is considered resistance projection Electrical Industry, Defense & Railways, welding, because of the many small contact Electronics Industry Consumer Durables areas between crossing wires (grids).

Resistance Projection Welding: The main difference as compared to spot welding: In resistance projection welding (RPW), high electrical resistance at the joint is developed by embossing one or more projections (dimples) on one of the surfaces to

be welded (Fig. 3). The projections may be

round or oval for design or strength purposes.

High localized temperatures are generated at

the projections, which are in contact with the flat mating part. The electrodes (typically made of copper-based alloys) are large and flat, and water cooled to keep their temperature low. Weld nuggets similar to those in spot welding are formed as the electrodes exert pressure to soften and compress the projections. Spot-welding Fig:3 (a) Schematic illustration of resistance projection equipment can be used for resistance welding (b) A welded bracket. (c) and (d) Projection projection welding by modifying the electrodes. Although the embossing of the welding of nuts or threaded bosses and studs. (e)

workpieces adds expense, the process Resistance projection-welded grills.

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04. Torch Body Cu Hose Sol: The various fuel gases used for gas welding Convergent O2 nozzle Mixed are Acetylene, Gasoline, Hydrogen, MPS together C2H2 and MAPP gas, Propylene and Fuel Gas, Butane, propane and butane/propane mixes. Flame Rubber Torch or angle Plastic O2 cylinder Hose C2H2 Gas Welding : Cylinder WP i) Oxyacetylene Gas Welding:  In this case heat required for melting of plate is obtained by burning of oxyacetylene gas Oxygen cylinder: R.H Threads, Black colour, mixture 120 KSC,

 Oxygen and acetylene are drawn from their C2H2 Cylinder: L.H. Threads maroon colour/red, respective cylinders through hose pipe into 15 KSC the torch body. Oxyacetylene welding

 Both the gases are mixed together in the torch C2H2 + O2  2CO + H2 + Heat

body so that the mixture is possessing certain 2CO + O2  2CO2 + Heat higher pressure. 1 H2 + O2  H2O + Heat  When this high pressure mixture is passed 2 through the convergent nozzle, the pressure energy gets converted into velocity energy Different Types of Flames used for Welding : and the mixture is coming out from the nozzle O 1. Neutral Flame: 2  ,1 at very high velocity HC 22  If this mixture is given initiation for burning, N = 10 to 15 mm (N = length of inner cone) the continuous flame will be produced. So the Inner cone (Tmax=3260C) heat available in the mixture is used for melting/Joining of work piece.

N

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These two cones can be distinguished based  Because excess supply of oxygen, there is a on their colour. The inner cone will be possibility of free oxygen may present in the yellow or red and outer cone will be light flame blue color  If oxidizing flame is used for Joining of

 Maximum temperature (3260C) of flame is highly reactive metals; the oxidation will take induced at the intersection of inner cone and place, hence it should not be used for Joining outer cones. of highly reactive metals like Al & Mg etc.  Average temperature is (2000-2100)oC. It is  Because of higher average temperature, it can 2 be used for joining of high melting point the of maximum temperature 3 materials

 It is also used for joining brass workpiece. Applications:

 Neutral flame is used for joining and cutting O of all ferrous and non ferrous metals expect Carburizing flame: 2  95.0to85.0 HC brass. 22 Inner cone (T =3040C)  During Joining of Brass, the Zinc present in max

brass will get evaporated.

O 2 2N to 3N 2. Oxidizing Flame:  5.1to15.1 HC 22  Because of short supply of oxygen, the flame Here excess amount of oxygen is present has to travel for a longer distance to N N  The length of Inner cone is about to 3 2 completely burn acetylene i.e length of inner

o Inner cone (Tmax=3380 C) cone is increased to 2N to 3N. o  Maximum temperature is only about 3040 C. o Also Tavg = (1800 – 1900 C)  Because of lower average temperature, high

N/3 to N/2 melting point material can not be joined.  Maximum temperature of flame in oxidizing  This is mostly used for joining of high carbon flame is about 3380oC steels.

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 For connecting O2 cylinder to torch body, the  Maintenance cost is higher because of many flexible hose is made by using copper as a moving parts in it. material.  Arc-blow is severe and difficulty to control.  Voltage drop is relatively higher, it can be  Rubber hose pipes are used for connecting used only at a short distance from the power C2H2 cylinder supply.  If hose pipes are interchanged, then both of  Higher electric energy consumption per kg of the pipes will fail. Hence to avoid the metal deposited (6 to 10 kWh). interchanging of hose pipes, color distinction,  Efficiency of D.C. Welding machine is low size distinction & type of thread distinction only 0.3 to 0.6. will be used.  It has higher operating cost.

05. Hold Time: Sol: Advantages and disadvantages of D.C Nugget welding machine:

Advantages:

 Heat generated is different at the work and the

electrode by changing the polarity.

 D.C. welding machine is suitable for welding Time during which force applied on plate all types of metals by changing the polarity. will be continued to hold until liquid molten  Both coated and bare electrode can be used in metal produced will get solidified. D.C. welding machine.

 It is used for all sorts of work as starting of S. Arc Welding Gas Welding the arc is easier comparatively. No. 1 In the arc welding, In gas welding, fuel  It can be used anywhere with engine driven electricity is used to gases like acetylene, D.C. generator or by rectified A.C. supply. generate heat. hydrogen are used to  The motor in a D.C. Welding has an generate heat. 2. This welding generator This welding generates advantage of high power factor of 0.6 to 0.7. higher temperature than lower temperature than gas welding. The arc welding. The Disadvantages temperature is about temperature is about  D.C. welding machine is two to three times 6000 oC. 3600 oC. 3. This welding generates It gives weaker joint. costlier, larger in size, heavier in weight and stronger joint compare to is more complicated. gas welding.

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4. It gives poor surface This welding gives good I = 540 – 120 L finish surface finish. When I = 300 A 5. In arc welding In gas welding non  300540 consumable electrode is consumable electrode is L  used. used. 120 6. The electrode is A filler rod is used La = 2 mm combined with the filler separately if required. metal. 7. It can be used in welding It can be used in 07. alone. welding, brazing and Sol: V = 20+40L, V0 = 80V, IS = 1000a soldering. 8. There is risk of There is risk of V0 VV 0  I , P = VI explosion due to high explosion due to high Is voltage. pressure. V dP 9. It is mostly used to joint It is mostly use to join 0 VL4020 0  I ,  0 similar material. both similar and Is d different metals. 80 10. The heat is concentrate The heat is distributing 80L4020  I in arc welding. according to the flame. 1000 There is higher loss of 20 + 40L = 80 – 0.08I energy. –60 + 40L = –0.08I 11. It is more efficient. It is less efficient. I = 750 – 500L 12. Speed of welding is Speed of welding is low. high. P = (20 + 40L)(750 – 500L) 13. The initial cost of arc The setup cost of gas = 15000 – 10000L + 30000L – 20000L2 welding is high. welding is low. = 15000–20000L–20000L2

L = 0.5cm 06.

I Sol: Vp = 36  08. 60 Sol: Lw = 300 mm Va = 2La + 27 = La:mm

Hs = 2 kW La I V P V = 200 mm/min 2 300 31 9300 w  2.440 4 60 35 2100 H  kW m t

P = 9300 – 2100  = ? P = 7200 W L 300 t w  60  60300 V 200 300I   2L w 42 a  t = 90 sec

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 Hm = 1.867 kW V = 20+1.5L H 867.1 V = 26V m  %35.93 Hs 2 V0 VV 0  I Is 09. 62 6226  I Sol: V = 20 + 4L 130 L I V I = 75.48A H J 4 mm 550 A 36 V  277.667 6 mm 450 A 44 V  mm

 450550 11. 550I   64 Sol: No. of electrodes  I – 550 = – 50 (L–4) Total metalofvolume deposited   I – 550 = – 50 L + 200 Volumedeposited onefrom electrode  I = 750 – 50 L Total metalofVolume deposited   2 I @ L = 5 mm = 750 – 250 = 500 A 3  05.045.0103 V @ L = 5 mm = 20 + 20 = 40 V 4

 P @ L = 5 mm = 500  40 17 mm 19 mm  P @ L = 5 mm = 20 kW o o 2 mm 30 30 Now: Voltage current characteristics 2  450550 550I   36V x  4436 30tan o  17mm  I – 550 = – 12.5 (V–36) x = 9.814 mm I V  1  1  1000 80 Area     15.11.1192217814.9  2   OCV = 80 V 2 Volume     18015.11.1mm85.204 3 10. = 46645.30767mm No of electrodes = 17 Sol: V0 = 62V; Is = 130A; L = 4mm;  = 0.85 cm15 15 cm 150 mm mm    5.2 min 60 sec 60 sec sec

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12.  w   c 2.0tkT8q  80  4  Sol: 80V  I o 1000 Tc = 1450 C V = 20+4L q = DVI= 10.852.5103 = 2125W P = VI w = 3tan(30o)2 I80 2 –3 i80P  2125 = 843.6310  1450 1000  10464.3v 3  dP  2.0  5   0   102.14  dI 3 I160  1046.31066.9 80  0 40.1  5 1000  102.14 –5 –6 –3 I = 500A 6.72210 = 9.610 + (3.46410 ) 80   = 0.0166m/sec 80V  500 1000 V = 40V 14. 40V = 20+4L Sol:  L = 5 mm 1. Porosity : It is common type. In this defect, air bubbles 13. or gases are present in the weld zone. The Sol: 3 – dimensional heat transfer: distribution of air bubbles in weld zone is 2 w  random. Porosity caused by gases release wkT25.1q c  3 4  during melting of the weld area but trapped 2 – dimensional during solidification, chemical reaction  w  during welding or by contaminants.  c 2.0tkT8q   4  This defect can be minimized by the proper k = thermal conductivity selection or electrode, filler material, improve welding techniques, more attention Tc = Tm–TR t = thickness of sheet to weld area during welding preparation and  = maximum welding speed slower speed to allow gases time to escape. w = width of weld The effect of porosity on performance depends on quality, size and orientation to  = thermal diffusivity –k/sec stresses.

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2. Spatter: Remedy: This defect can only be repaired Metal drop expelled from the weld that stick by grinding down or gouging out and re- to surrounding surface is known as spatter. welding. Spatter can be minimize by correcting the welding condition and should be eliminated 4. Incomplete fusion: by grinding. In this types of welding defect gap is not Causes: totally filled by molten metal. It is due to  Welding current too high. inaccuracy of the welder so pre  Arc is too long. solidification of welding metal.  Incorrect polarity. Causes:  Insufficient gas shielded.  Heat input is too low. Remedies:  Weld pool is too large and running ahead  Reduce welding current and arc length. of the arc.  Use correct polarity according to the  Joint included angle is too low. welding condition.  Electrode and torch angle is incorrect.  Increase torch to plate angle and use  Unfavorable bead position. correct gas shielding. Remedies: 3. Slag inclusions:  Increase welding current and decrease the Slag inclusions are compound such as travel speed. oxides, fluxes and electrode contains  Reduce deposition rate. material that is trapped in the weld zone.  Increase joint include angle. These defect are commonly associated with  Position electrode or plate angle such a undercut, incomplete penetration and lack of way so the plate edges will melt. fusion in weld. Insufficient cleaning  Position bead in such a way that the sharp between multi-pass welds and incorrect edges with other bead or plate are electrode and current can leave slag and avoided. unfused section along the weld joint. Slag inclusion not only reduces cross section area strength of joint but also may serve as initiation point for serious cracking.

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15.

Sol: Given data, Chapter Resistance spot welding (Al) 3 Metal Cutting T = 2 mm, I = 5000 A Common Data for Q. 01 & 02 t = 0.15 sec , R = 75  01. Ans: (a)

dn = 5 mm,

tn = 2.5 mm 02. Ans: (d) usedHeat Sol:  100 Heatgenerated Vf Vs Heat Generated = I2RT –6 = 5000  75  10  0.15 90 

= 0.05625 J Vc  Heat Used = 29 2  J353.1425.2s 4 Vc = 40 m/min; Vf = 20 m/min

353.142 o Vf  100  = 10 ; r  5.0 05625.0 Vc  = 0.039  cosr    tan 1    sinr1    10cos5.0   tan 1    33.28 o   10sin5.01  V V  f cos s sin  20   10cos = 41.5 m/min 33.28sin

03. Ans: 10 Sol: f = 0.25 mm/rev,

t1 = 0.25, i = 10,  = ?

t1 = f cosCS

0.25 = 0.25 cosCS

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cosCS = 1  Cs = 0 Common Data for Q. 08 & 09

 = 90 – CS = 90 08. Ans: (c)

tan b   cossin    itan         09. Ans: (c) tans   sincos   tan

Sol:  = 6 , C  /m1V s tanb = sin tani+costan b = w = 3, d = t1 = 1 mm tanb = sin90 tani + 0 o t2 = 1.5 mm; use 2  90  b = i = 10 t 1 2 r 1  67.0 Common Data for Q.04, 05 & 06 t2 5.1 3

1  6cos67.0  0  tan    62.35 04. Ans: (c) 05. Ans: (b)   6sin67.01  06. Ans: (d) For minimum energy condition use Sol: d = t1 = 2 mm, w = b = 15 mm 2 +   = 90 C  s/m5.0V ,  = 0  62.352690290 0 C  T  30,800F,1200F = 24.76 800   461.076.24tantan  tan1  69.33 0 1200 cf 167.0rvV  60   67.069.33tantan  min/m2.40 60 Power = P = 1200VF  Area of shear plane = As = Ls × b CC 60  bt  31 = 1200 W = 1 =  mm2.5 2 sin  62.35sin Length of shear plane = LS

t 2 = 1  mm4 Common Data for Q. 10 & 11 sin  30sin

10. Ans: (d) 07. Ans: (a)

Sol: For theoretically minimum possible shear 11. Ans: (d) strain to occur Sol:  32D mm,  = 35, K = 0.1 mm,  902 0 1 F = 200 N, V = 10 m/min, 90   690 C C    48o 2 2 L2 = 60 mm, FT = 80 N

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t L 60 60 t 125.0 r 1 2    59.0 r 1  29.0 t 2 L1 D0  32 t 2 43.0

t1 t1 1.0 1  cosr   r t 2  169.0  tan   t 2 r 59.0  sinr1  

 35cos59.0  1 10cos29.0  1 0  tan = 16.73  tan   = 36.15     35sin59.01    10sin29.01 

F 80 1  FT  tanT   tan    F  FC 200  C 

 80  o 1 217   tan 1    tan10   = 32.77  200   517 

o  8.568.2135 Cm = 2  16.73 + 32.77 – 10 = 56.23  .18.56tantan 52 (In general  < 1) 13. Ans: 272 N & 436 W Hence by applying classical friction Sol: S0 = 0.12 mm = t1,

theorem t = 2.0 mm, ta 22  22.0 1  1  Major cutting for, b = pz = Fc   lnln   r 59.0     1tansec.t.S       S0     35 S0 = 0.12,   400 2 2 180 S t = 2 – 0, 5276.0  55.0 t a 22.0 04.1  2 2  83.1 t S 12.0 V 1 0 f  rVVr = 0.59 ×10 = 5.9 m/min cf  = 0 VC Pz = 0.12×2.0×400(1.83sec0–Tan0+1) Vf 9.5 Vs  cos   35cos = 272 N sin  15.36sin V  /m42.8 min Power = p = pVF  f ZCC r

6.52 271Vp  83.1 12. Ans: 56.23 fZ 60 Sol:  = 10, t1 = 0.125, = 436 W Fc = 517 N; FT = 217 N t2 = 0.43; Cm = 2 +  – 

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14. Ans: (d) b c V2  f1   d1        Sol:  = 30 , FT = 800 N, Fc = 1200 N     V1  f 2   d2  F 15.0 C  1  Fs = cos = 2 3.0    11.1 cos  2  F  V11.1V tan (  ) = T 2 1 FC VV % change in speed = 12  %11  800  V1    = tan 1   = 33.69 1200  Productivity is proportional to MRR 1200 % change in productivity Fs =  )69.3330cos(  N23.639 69.33cos MRR  MRR = 2 1 MRR 1 Common Data for Q. 15 & 16  VdfVdf = 111222 = 11 % 15. Ans: (a) Vdf 111 16. Ans: (b) 19. Ans: 49.2 % Sol: D = 100 mm, f = 0.25 mm/sec, Sol: T0 , V0 = original tool life and velocity d = 4 mm If 1  0 1  T5.0TV2.1V 0 V = 90 m/min FC 2  T,V9.0V 20  ? FC = 1500 N FT F n TVTV n FC = N = 1500 N 11  00

FT = F n  T1  V0     T0  V1 Common Data for Q. 17 & 18  V0   1  ln  ln  V 2.1 n   1      263.0 17. Ans: (b) & 18. Ans: (b)  T  l )5.0(n ln 1  Sol: cba  KdfVT    T0  n n a = 0, 3 b = 0, 3, c = 0, 15 00  TVTV 22

f 1 f  1 ,  2dd 1 2 2  V  n  V  263.0 2  0   0   TT 02    T0   = 1.4927T0  V2   V9.0 0  TT 21  60 cba cba % change in tool life  dfTVdfTV 22221111  TT  TT4927.1 = 02  00  4927.0 T0 T0

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20. Ans: (b) V2 80 ln ln Sol: Let Q = no. of parts produced V 47.0 n 1 50  333.0 T 50 41.1 T.C on E.L = T. C on T.L ln 1 ln T 2.12 30 60 2 Q 50080 Q  160 60 60 n n 11  TVTV 33

 16500Q40 Q 1 1 n  V  50 333.0  1     500Q24Q16Q40   TT 13    50   29  V3   60  500  Q =  2183.20 24 23. Ans: 30.8 m/min

Sol: TC = 3min, Tg = 3 min, 21. Ans: (a) Lm = Rs. 0.5/min Sol: n = 0.12, C = 130 Depreciation of tool regrind = Rs 0.5 C1 = 1.1 × 130 = 143, C = 60, n = 0.2 V = V1 = 90 m/min Cg =  33  0.5 + 0.5 = 3.5 1 130  12.0 n n TCVT   min4.21     n L m  90  VOpt = C  .    n1 Cg  1   n 12.0 1111 143  2.0 TCTV     min4.47  2.0 5.0   90   60 . = 30.8 m/min   2.01 5.3  Increased tool life = 47.4 min

Increase in tool life = 47.4 – 21.4 = 26 min 24. Ans: 57.91

C18 C270 Sol: C  , C  , VT 5.0  150 22. Ans: (a) m V t TV 500 Sol: Tool life = T  ,50 TC = k + Cm + Ct 1 10 C18 C270 122 k  T  ,2.12 V TV 2 10 C18 C270 k  1 1  rpm50V , 2  rpm80V V  C  n The feed and depth of are same in both V     V  cases  1   1 n  TVTV n C18 VC270  n  11 22 k  V 1 C n

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TCd fm = 75 mm/min; d = 5 mm; For min TC,  0 dV b = 50 mm; u = 420 MPa;  1   2  1   +  –  = 45; VC270  n    1  C18 n     0  = 0.7 (assumption) V 2 1 C n

 1   2  1  d= t1 VC270  25.0    1 25.0 C18    1 V 2 n C  = tan  = tan-1(0.7) = 34.99 3270 18 V 2   = 45 +  –  1504 V 2 = 45 + 10 – 34.99 = 20.00 4 4 15018 V  FS 3270 u sin  A0  V = 57.91 m/min 2 A0 = t1 × b = 5 × 50 = 250 mm

 A0u 25. Ans: 2.5 & 23 FS  sin  Sol:  =10  250420 =  kN99.306 t1= f.sin = 0.15 sin75 = 0.144 00.20sin 

t1 t2 = 0.36, r  402.0 By using merchant circle t 2 FS FC  cos chip reduction coefficient = t2/t1 cos 1  1099.34cos99.306   5.2K  r  1099.3420cos 1  cosr   99.306  tan   =  99.24cos  = 393.43 kN  sinr1   99.44cos  10cos402.0   tan 1  = 23.18o DN   VC    10sin402.01   601000  20080   837.0 26. Ans: (*)  601000

Sol: Given data: P = W.D = FC × VC z = 15;  = 10; = 393.43  0.837 = 329.30 W N = 200 rpm; D = 80 mm;

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27. Ans: 0.944 Specific cutting energy Sol: T = 60 min  VF = cc =1.49 67 1 c  1000Vbt VA 11.0  min/m70.42 60 Fc =1.49  t1  b  1000 = 1490N 77 Because the power consumption is taking VB =  min/m22.45 60 13.0 place only in the cutting stroke, Under similar conditions with same tool life Velocity of tool in the cutting stroke cutting velocity on material B is greater = length of work or stroke length  RPM than the material A. Hence the of the crank machinability of material ‘B’ is higher than = 200  60 = 12m/min the material ‘A’. Power required = Fc  cutting velocity V 7.42 121490 A  944.0   Watts298 V 22.45 60 B

30. Ans: (*) 28. Ans: 12 Sol: 1. HCS 2. HSS Sol: Given, t1 = 0.2 mm, 3. Stellite 4. Carbides w = 2.5 mm, 5. Ceramics 6. Ceramets Fc = 1177 N, 7. Diamond 8. CBN Ft = 560 N As the cutting is approximated to be 9. UCON 10. Sialon orthogonal. Above list of cutting tool material is in

tani = cos tan b – sin  tan s increasing order with respect to most of

tan 0 = cos tanb – sin tan s condition and property except toughness.

= cos30 tan7 – sin30 tan s If toughness is considered, the correct  s = 12 sequence in decreasing order will be HCS > HSS > Stellite > Diamond > CBN > 29. Ans: 298 UCONS > Sialon > Carbides > Ceramet > Sol: The given problem is the oblique machining Ceramics problem.

Hence, t1.b = f.d = 0.25  4 = 1 mm2 31. Ans: (a)

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32. Ans: (b)

Sol: In machine process, a cutting fluid that Conventional Practice Solutions removes a larger amount of heat. 01. 33. Ans: (d) Sol: Given data: Sol: t2 = 0.4 mm; b = 2.5 mm; f = 0.2 mm/rev; F = 1100 N; c Ft = 290 N, v = 2.5 m/sec;

 = 10; FS = ?;  = ? = t1

t1 2.0 r = chip thickness ratio =  5.0 t2 4.0

FS 34. Ans: (c) F  C – Fsn 35. Ans: (c) F T R F  36. Ans: (a) N

From merchant circle; 37. Ans: (d) F )tan(  T Sol: Machinability of a material cannot be FC quantified directly in the form of some  F   tan 1 T  absolute value; rather it is given in some    F `C  relative form. Hence option (d) is correct.  290   tan10 1  1100  38. Ans: (a)  = 24.76 cosr  10cos5.0  tan  = 1 rsin   10sin5.01   = 28.33 = shear angle F F S  C cos cos

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F 1100 S  09.43cos  76.14cos 

FS = 830.72 N W.D = E = FC×V +  = coefficient of friction = tan  FC FT× feed velocity FT = tan (24.76) = 0.461

02. Sol: Orthogonal cutting:  = 10; The effect of thrust force in machining are t1 = 0.5 mm ; t2 = 1.125 t 5.0 1. It is not influencing the work done or Chip thickness ratio = r = 1  44.0 energy required for machining. t2 125.1 2. The thrust force is increasing wear and 1 cosr     tanangleshear   = 25.13 tear on the tool. Therefore the tool life is  sinr1   reducing. Shear strain = cot + tan (–) 3. The thrust force is varying only due to = cot (25.13) + tan(15.13) variation of depth of cut but it is not = 2.40 influenced by cutting velocity and feed.

4. In thread cutting operation, the thrust 03. force is trying to shear off the threads, Sol: VTn=C therefore the threads will become weak. Where v = cutting velocity in m/min To avoid this problem it is required to T = Tool life in min minimise the thrust force and it is n = Taylor’s exponent depending on possible by minimising the depth of cut. the cutting tool material

04. log(U) Sol: Given data:

ceramic Diamond/CBN z = 15; = 10 ;   carbide N = 200 rpm; D = 80 mm; castalloy Theoretical fm = 75 mm/min; d = 5 mm;

b = 50 mm; u = 420 MPa; HSS

log(T)  +  –  = 45;

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 = 0.4 (assumption) 2. Efficiency of energy utilization is increasing. 3. When friction is reducing, wear and tear d=t1 of the tool is reducing and therefore tool

life will be increasing.

4. Due to reduction of ; B(friction angle)  = Tan   = tan-1(0.4) = 21.80 is reducing so that  (shear angle) is  = 45 + – increasing and due to this MRR is = 45 + 10 – 21.80 = 33.19 increasing F S sin  2 +  –  = 90 (merchant equation) u A 0 90    A 2 F  0u S sin  5. When the friction becomes zero;  250420 Friction angle () becomes zero so that =  kN80.191 19.33sin  2 –  = 90. This is corresponding to 2 the minimum shear strain condition. A0 = t1 × b = 5 × 50 = 250 mm By using merchant circle i.e., as the shear strain is minimum, the amount of energy required for FS FC  cos deforming the chip is minimised. Hence cos the machining becomes more effective 80.191 = 8.11cos  = 265.46 kN and efficient. 99.44cos 6. When lubricant or cutting fluid is used P = W.D = FC × VC = 222.397 for reducing the friction, the heat generated in the machining will be 05. carried away by the cutting fluid so that Sol: The effects of lowering friction at the tool- no heat treatment effects will be taking chip interface:- place. Hence no change in mechanical 1. Due to reduction in friction the energy properties of workpiece will takes place. lost in overcoming the friction will be 7. Lubricant or cutting fluid is forming as a reduced and hence power consumption protective layer over the workpiece for machining operation is reduced. surface and it avoids the atmospheric contamination.

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Disadvantages: 07.  Due to usage of cutting fluid or Sol: Orthogonal cutting: lubricant it may chemically react with  = 10; r = 0.31;

workpiece surface and damages the FC = 1200 N; FT = 650 N; surface of the workpiece. d = 2 mm; f = 0.2 mm/rev;

VC = 200 m/min; 2 +  –  = 90 06. 1 cosr   Sol: Orthogonal cutting:  tan   = 17.88  sinr1   t1 = 0.128 mm; b = 6.36 mm;  F   tan1 T  = 38.44 VC = 2 m/sec;  = 10;    FC  F = 568 N; F = 228 N; C t F F  C sin = 848.47 N t2 = 0.228 mm cos t r = chip thickness ratio = 1 t 2 vf v 128.0 s  561.0 90–  228.0 vc 1 cosr   Shear angle =  tan    sinr1   vf = rVC = 0.31 × 200 = 62 m/min

Work done in friction = F × Vf = 876.75 W 1 )10cos(561.0  = tan    10sin561.01  FC   FS = cos = 942.50 N cos = 31.47

From merchant circle; 08. F Sol: V1 = 100 m/min; T1 = 120 min tan T V = 130 m/min; T = 50 min FC 2 2 n  F   T1  V2  tan 1 T        T2 V1  FC    Taking log on both sides  = 31.87     P = F × V T1 V2 C logn    log   T2   V1  = 568 × 2 = 1.136 kW n = 0.299 V3 = 2.5 × 60 = 150 m/min

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n n V3T3 = V1T1 tool will be acting onto the abrasive 0.299 0.299 150 × (T3) = 100 × (120) particles which inturn produce impact loads

T3 = 30.92 min on to the work piece and due to higher n n V4T4 = V1T1 brittleness and lower toughness the work is

n getting fracturing.  T1  V4 = V1   The chips produced due to this will be  T4  moving along with the abrasive slurry. 299.0 120  = 100  = 112.88 m/min From the above the mechanism by which 80   chip formation taking place is brittle

fracturing 09.  The impact of the hard abrasive grains Sol: fractures the hard and brittle work surface (a) Under similar conditions of machining; the resulting in metal removal in the form of work piece on which discontinuous chips small wear particles, which are carried away are produced will indicate better by slurry. machinability workpiece. This is because the discontinuous are easy to dispose. (b) Present day technology needs to develop Tool vibrates at very some non conventional machining high frequency Abbrassive F = 20 – 30 kHz processes because the available traditional Slurry Amplitude = 50-150 m manufacturing processes can not serve purpose such as (i) New materials with a low machinability Work-piece (ii) Dimensional accuracy requirements (iii) A higher production rate and economy Ultrasonic Machining (USM) The tool material used in USM:  The basic process involves a tool vibrating To withstand for the reaction impact loads with a very high frequency and a continuous acting by the abrasives, the tool must be flow of abrasive slurry in the small gap made by using very soft material the most between the tool and the work surface. commonly used tool material is Cu,  When the tool is vibrating at high frequency, the impact loads produced by the BRASS, and M.S etc.

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The MRR in USM is given by  Number of work pieces machined (in Q  V Z f mass production) Where, Factors Influencing tool life: 1. Properties of work piece material: Q = Volume of MRR, 2. Tool Geometry: V = Volume of material dislodged/impact, 3. Use of Cutting Fluid: Z = No. of particles making impact/cycle, 4. Process Parameters: f = Frequency (i) Cutting speed On simplification of the above equation (ii) Feed

(iii) Depth of cut Application of USM:  Because of uniqueness of process (i) Used to produce holes as small as 0.1 parameters, the researchers tried to mm. establish relationship between process (ii) Used for drilling, grinding, profiling, parameters and tool life. coining etc on materials like SS, glass,  Taylor has assumed that cutting velocity ceramic, carbide quartz, semiconductors is the major parameter influencing the etc. tool life. (iii) Dental doctor uses USM for producing holes in human teeth because it is a Taylor’s tool life equation: VTn = constant = C painless drilling methodology Where V = Cutting velocity in m/min (iv) Also used for piercing of dies and for T = tool life in minutes parting off operations. C = Taylor’s constant

= Cutting velocity for 1 minute tool life 10. n = Taylor’s exponent depending mainly on Sol: Tool Life: The tool life can be defined in cutting tool material three different methods. = 0.05 to 0.1 for H.C steels  The actual machining time between two = 0.1 to 0.2 for H.S.S successive regrinding of a cutting tool is = 0.2 to 0.4  carbides called tool life. It is most commonly expressed in minutes. = 0.4 to 0.6  ceramics  Volume of metal removed (in rough = 0.7 to 0.9  diamond/ CBN machining)

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Modified Taylor’s Tool life equation: machinability of material ‘B’ is higher than V Tn fp dq = C the material ‘A’.

f = Feed mm/rev Drawbacks: This method is applicable only d = depth of cut in mm when more than one material is given but if p, q are constants < 1 only one material is given this method q < p indicates that tool life is more cannot be used. sensitive to the uncut chip thickness than to For maximum production rate (or) the width of cut minimum production time. Effect of tool life on cutting parameters is n  n 1  V > f > d. opt  CV  .    n1 TC 

11.0 11.  11.0 1  A  67V .  min/m91.50 Sol: The machinability of a material is usually opt   11.01 5.1  13.0 defined in terms of four factors:  13.0 1  B  77V .  min/m05.57 1. Surface finish and surface integrity of opt   13.01 5.1  the machined part 2. Tool life 12. 3. Force and power requirements Sol: Orthogonal machining: 4. The level of difficulty in chip control t1 = 0.25 mm; t2 = 0.75 mm; after it is generated b = 2.5 mm;  = 0; Thus, good machinability indicates good FC = 900 N; FT = 400 N surface finish and surface integrity, a long t1 tool life, and low force and power r  33.0 t2 requirements.

As  = 0; FC = N; T = 60 min FT = F; tan  =  = F/N 67 F 400 VA  min/m70.42 T 60 11.0 tan  44.0 FC 900 77 VB =  min/m22.45   = 23.96 60 13.0 FC FS = cos Under similar conditions with same tool life cos cutting velocity on material B is greater 900 =  N37.72939.42cos than the material A. Hence the 96.23cos

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1 cosr   FC  tan   26.18  FS  cos  sinr1   cos 1177 A0 = t1 × b = 0.625 = 72.57cos = 696.08 N 44.25cos FS 2  65.365sin N/mm A0 = t1 × b A 0 = 0.2 × 2.5 = 0.5 mm2 By assuming that machining is taking place FS under minimum energy criteria machining, u sin  = 743.49 MPa A0 the shear stress induced can be taken as ultimate shear stress. 14. 2 0.13 0.6 0.3  = u = 365.65 N/mm Sol: VT f t = C 0.13 0.6 0.3 (40)(60) (0.25) (2) = C 13. C = 36.49

Sol: ASA  7 – S – 6 – 6 – 8 – 30 – 1 V1 = 1.25 × 40 = 50 m/min f1 = 1.25 × 0.25 = 0.3125 mm/rev t1 = 0.2 mm; b = 2.5 mm; t1 = 1.25 × 2 = 2.5 mm i = 0; FC = 1177 N; 1 F = 560 N  C  13.0 T T    min29.2 1 6.0 tfv 3.0  (1) b(ASA) = (ORS) = 7  11 1  1  = 90 – CS = 90 – 30 = 60  C  13.0 T     min75.10 2  3.06.0  tan b   cossin   itan   1 tfv        tanS   sincos  tan 1  C  13.0 T     min37.21 tan S = – cos tani + sin  tan  3  6.0 3.0   1 tvf 1  S = 6.06 1  C  13.0 1 FT  T     min77.35 (2)  =  tan   = 31.50 4  6.0 3.0     tvf 1   FC 

 = tan(32.44) = 0.612 15. (3) Minimum workdone criteria Sol: For maximum production rate; n 2 +  –  = 90  n 1   Cv  90  opt      n1 TC  2 5.0  5.0 1  5.3106.690 = 100  = 33.33 m/min = 28.32   5.0 9 2  

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03. Ans: (b)

Chapter Nfollower Machining Sol: Train value = Gear ratio = 4 NDriver

threadsjobofpitch = 01. Ans: (i) 20 min, (ii) 50 min screwleadofpitch threads L 576  40175.3 127 Sol: Time / cut =   min20  not possible fN 1442.0  406 240 DN  144100 127  201 V    min/m2.45 =  1000 1000 40  206 1 0.75 0.75 75 127 20 VT 75 T    possible V 40 120 333.1  75  =    min96.1 04. Ans: (c)  2.45  Sol:  Plane turning  Taper turning 20 No. of tool changes =  102.91 96.1  Under cutting  Thread cutting

(Because 1 tool is already mounted on W.P) 05. Ans: (d) Total change time / piece = 20 + 10 × 3 N Sol: Gear Ratio = Train value = follower = 50 min N driver 02. Ans: (a) T P = driver  driver Sol: For producing RH threads the direction of Tfollower Pfollower rotation of job and lead screw must be in the Pjob Pspindle N S.L same direction, for this if the designed gear G.R =  P S.L P S.Ls N Spindle train is simple gear train use 1, 3, 5 odd P = pitch number idle gear to get same direction of N Spindle P 6 3 rotation, if the designed gear train is  S.L = = compound gear train use 0, 2, 4,.. even N S.L P job/Spindle  22 2 number of idle gears to get same direction. 06. Ans: (d) In the given problem the designed gear train Sol: With this any change in U will also is a compound gear train, to change the V hand of the thread it requires to change the changes the speed of lead screw, the pitch direction of rotation of job and lead screw of the threads produced depends on the for this use 1, 3, 5… odd number of idle speed of work and speed of lead screw. Us gears. will not affect the speed of the work

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53 Production Technology

07. Ans: (b) 10. Ans: (c) Sol: No. of D.S/min = 10 Sol: Total depth to be removed = 30 – 27 B = 300 min, f = 0.3 mm /stroke = 3 mm B 1 2 Time/cut =  Given, m = = 0.67 f S.Dof.No 3 300 1 feed = 0.5, =  min100 3.0 10 depth = 2 V = 60 m/min 08. Ans: (b) Approach  m50   length wise Sol: L = 2m  min50timeOver  = 50 + 900 + 50 + 50 + 900 + 50 Approach  m5   width wise B = 300 + 5 + 5 = 310  m5timeOver  f = 1 mm/stroke, VC = 1 m /sec, L B 1 Time/cut = M1  M = V f 2 l = 800, B 1 Time per two pieces =  M1 L = 800 + 50 + 50 = 900 f V B = 400 + 5 + 5 = 410 310 2000 =  5.01 = 930 sec 900  2  410 1 1000 Time / cut = 1  60000  3  5.0 930 Time/piece =  sec465 = 20.5 min 2 3 No. of cuts =  cuts25.1 2 09. Ans: (d) Total time = 20.5×2 = 41 mins Sol: Shaping operation

M = 0.6 , L = 500 mm 11. Ans: (b) Double stroke / time = 15 Sol: Time per hole = L/f.N N = time / D.S = 1/15 = 25/(0.25 300) L Average speed, V =  M1 = 1/3 min = 20sec. V Because dia of drill bit was not given, hence 500   6.01 =12000 mm / min  1  AP1 is zero.   15  = 12 m / min

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54 ESE – Text Book Solutions

12. V1000 181000 Nmax =  Sol: D = 15 mm, Vc = 20 m/min, D min  25.6 V1000 201000 V1000 181000 N  = = 425 rpm N =   D  15 min D max  25 N = 425rpm N max 25 r = 16 = 5 f = 0.2 mm /rev N min 25.6 T = 100 min,  45 mml = 1.3195 = 1.32 Time for idle time = 20s Tool change time = 300 s 15. Ans: (d) L  D5.0 Time/hole =   Sol: Hobbing process fN fN No. of teeth = 30 (Not required) 15 45  Module = 3 mm  2  min617.0 Pressure angle = 200 (Not required)  4252.0 Radial depth= Addendum+1m+1.25m = Tm = machining time = 2.25 module = 2.25  3 i) No. of holes produced / drill Radial depth = 6.75 mm 100 = 162 617.0 16. ii) Total time/hole Sol: Part size = 200 × 80 × 60 mm = Tm + idle time + Tool change time D = 100 mm, Z = 12, 20 300 = 617.0  V = 50 m/min, 60 60162 1000V 000 50 = 0.9812 min = 58.87 = 59 sec N = 159 rpm D 100

f0.1mm , AP = OR = 5 mm 13. Ans: (b) t i) With symmetrical milling

14. Ans: (b) 1 22 AP D D w 1   Sol: Given n = 6, Dmax = 25 mm 2 1 Dmin = 6.25 mm =   80100100 22   20mm V = 18 m/min 2

L = 1  ORAPAPl N max r = 1n = 200 + 20 + 5 + 5 = 230 N min

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55 Production Technology

L 18. Ans: (d) Time/cut = t NZf Sol: d = 70 mm , Z = 12 teeth 230 V = 22 m/min   min2.1  121591.0 ft = 0.05 mm/tooth V1000 fm = ftZN, N  ii) If offset = 5mm with asymmetrical milling d

1 2 2  221000 AP1 =   wDD i  fm = 0.05  12  = 60 mm/min 2  7014.3

Where, wi = w+ 2(Of) = 80 + 2 × 5 = 90 19. Ans: (b)

1 22 10 1 4 AP1 =   90100100  Sol: Crank rotation = 1 1  360 2 30 3 3 = 28.2 mm = 480 L = 200 + 28.2 + 5 + 5 = 238.2 CR 480 Job rotation = = = 12 L 40 40 Time/cut = t Nzf 2.238 20. Ans: (b)   min25.1  159121.0 Sol: Given,

Dtool = 15 cm = 150 mm 17. Ans: (b) Feed = 0.08 mm/rev 40 Depth = 0.5 mm = d Sol: Crank rotation = max teethsof.No Length of workpiece, l = 200 mm 40 Cutting Velocity, V = 120 m/min = 28 Total depth to be cut = 2 mm  12  3 = 1  = 1  28  7  9  = 1   21 1 complete revolution and 9 holes in 21 hole circle.

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56 ESE – Text Book Solutions

p = 10 mm/tooth h = 0.075 mm/tooth V = 0.5 m/min O2 O1 Equation for time for broaching operation

A traveltoolofLength d = Linear toolofvelocity B t Length of tool travel = L AP1 = t + Le + AP + OR As (AP + OR) is not given so take it zero

Le = effective length or cutting length l  2026 Depth of cut d = = 3 V1000 1201000 2 N =  = 254.64 rpm D  150 n = no. of teeth = d/h = 3 / 0.075 = 40

Approach = AP1 + O1O2 =  dDd Le = np = 40  10 = 400mm =  5.01505.0 = 8.645 mm Le = 400 mm  Lt Total time/machining Time for broaching = e V = No. of cutsTime/cut  40025 depthTotal 2 = = 8.05 min No. cuts =  = 4 1005.0 cutperdepth 5.0 Time for broaching = 8.05 min L  AP Time/cut = =  fN fN 22. Ans: (c)  645.8200 = = 10.227 min Sol:  25508.0

Total time = 10.227  4 = 40.91 4.5 mm

= 41 min

21. Ans: 8.05 min Sol: Broaching machine d4.5mmtotal  P = 1.5 kW d0f  d1 = 20 mm enlarged to df = 26 mm

t = 25 mm ssS  .080125.0hnd 1

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57 Production Technology

26. Ans: 18 ddrtotalfs dd = 4.5 – 0.1 = 4.4 Sol: The output per annum = 800  52 d 4.4 = 41600 units. n44teethr  r The rejection rate is 20%. h0.1r  The quantity to be produced (including Cutting length = effective length = Le quiredRe output = LLL rejection) = rSf  rejection1( )rate = 44 × 22 + 8 × 20 + 4 × 20 41600  = 52,000 units = 1208mm  )2.01( Total time required for turning 23. Ans: (b) = 52,000  40/60 Sol: Out of all conventional method grinding is = 34666.6 hours one which required largest specific cutting Production time required with 80 per cent energy. efficiency = 34666.6 /0.8 = 43333.3 hours 1) Because of random orientation of Time available per lathe per annum abrasive particles, rubbing energy losses = 48  52 = 2496 hrs will be very high  Number of lathes required 2) Lower penetration of abrasive particle Time )hrs(required 3) Size effect of the larger contact areas = Time available hrs between wheel and work. 33.43333 = = 17.36 = 18 24. Ans: (a) 2496 Sol: Common alignment test for shaper and lathe  No. of lathes required = 18

are (1) Straightness (2) Flatness. 27. Ans: & 28. Ans: Runout is used in lathe. Sol: Hole = 31.75+0.01 mm Parallelism used in shaper. Hole diameter before = 31.24+0.005 mm

p = 15 mm; t = 25 mm 25. Ans: (a) h = 0.025 mm; Sol: The curvature given is the concave Approach = 5 mm curvature hence it increases the stress  DD  42.3175.31 concentration factor therefore it is used for d  if   mm165.0 2 2 supply of lubricating oil to bearing d 165.0 mounting n  7~6.6 h 025.0

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58 ESE – Text Book Solutions

Df = 31.24 + (7 × 0.025 × 2) = 31.59 mm 32. Ans: (c) = 31.29 + (7 × 0.025 × 2) = 31.64 mm i) Lc = n × p = 7 × 15 = 105 mm 33. Ans: (b)

L = t + Lc+ Ap = 25 + 105 + 5 = 135 mm ii) No. of teeth in contact at any point of time 34. Ans: (b) t 25 Sol: Counter boring is internal turning operation = =  267.1 p 15 used for enlarging end of the hole is called Force required = 5000 × 2 = 10 kN as counter boring.

29. Ans: (c) 35. Ans: (b) Sol: Lathe machine  Reducing diameter of 30. Ans: (a) cylindrical job Sol: Milling machine  Key ways on shaft  Half centre: The half centre is similar to Broaching  Rectangular holes an ordinary centre except that a little less than half of the centre has been ground 36. Ans: (b) away. This feature facilitates facing of the bar ends without removal of the centre. 37. Ans: (d)  In the tipped centre, a hard alloy tip is brazed into a steel shank. The hard tip is 38. Ans: (a) wear resistance.  Ball centre is particularly suitable for 39. Ans: (d) taper turning. Sol: Glazing takes place if the wheel is rotated at  Pipe centre is used for supporting pipes, very high speeds and is made with harder shells and hallow end jobs. bonds.

31. Ans: (c) 40. Ans: (c) Sol: In twist drill the helix angle varies from minimum near the center to the maximum towards the periphery and rake is directly proportional to the helix angle.

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59 Production Technology

be used for longer, gradual tapers providing Conventional Practice Solutions the length of taper does not exceed the distance the compound rest will move upon 01. its slide. This method can be used with a Sol: By using taper turning operation in a Lathe high degree of accuracy, but is somewhat machine. limited due to lack of automatic feed and TAPER TURNING the length of taper being restricted to the When the diameter of a piece changes movement of the slide. uniformly from one end to the other, the The compound rest base is graduated in piece is said to be tapered. Taper turning as degrees and can be set at the required angle a machining operation is the gradual for taper turning or boring. With this reduction in diameter from one part of a method, it is necessary to know the included cylindrical workpiece to another part, tapers angle of the taper to be machined. The angle can be either external or internal. If a of the taper with the centerline is one-half workpiece is tapered on the outside, it has the included angle and will be the angle the an external taper; if it is tapered on the compound rest is set for. For example, to inside, it has an internal taper. true up a lathe center which has an included There are three basic methods of turning angle of 60°, the compound rest would be tapers with a lathe. Depending on the set at 30° from parallel to the ways (Figure degree, length, location of the taper 1). (internal or external), and the number of

pieces to be done, the operator will either

use the compound rest, offset the tailstock,

or use the taper attachment. With any of

these methods the cutting edge of the tool

bit must be set exactly on center with the

axis of the workpiece or else the work will

not be truly conical and the rate of taper will Fig 1: Turning of soft center true with the vary with each cut. compound rest.

If there is no degree of angle given for a Compound Rests particular job, then calculate the compound The compound rest is favorable for turning rest setting by finding the taper per inch, or boring short, steep tapers, but it can also

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60 ESE – Text Book Solutions

and then calculate the tangent of the angle  833.0   tan 1     1 41650.0tan (which is the compound rest setting) .  2  For example, the compound rest setting for the workpiece shown in Figure 2 would be 1.000 0.375 calculated with these two formulas: 0.750

d a d Using a handheld calculator, you will see that tan–1(0.41650) = 22.61 L The included angle of the workpiece is Fig: 2 Taper double that of the tangent of angle  dD TPI  (compound rest setting). In this case, the L double of 22° 37’ would equal the included Where, angle of 45°14’. TPI = taper per inch, To machine a taper by this method, the tool D = large diameter, bit is set on center with the workpiece axis. d = small diameter, Turn the compound rest feed handle in a L = length of taper counterclockwise direction to move the a = tan–1 (TPI / 2) compound rest near its rear limit of travel to Where: assure sufficient traverse to complete the a = angle (compound rest setting) in degrees taper. Bring the tool bit into position with tan-1 = the tangent function the workpiece by traversing and cross- TPI = taper per inch feeding the carriage. Lock the carriage to The problem is actually worked out by the lathe bed when the tool bit is in position. substituting numerical values for the letter Cut from right to left, adjusting the depth of variables: cut by moving the cross feed handle and Example: reading the calibrated collar located on the  dD  375.0000.1 TPI   cross feed handle. Feed the tool bit by hand- L 750.0 turning the compound rest feed handle in a 625.0  833.0 clockwise direction 750.0

 TPI   tana 1   2 

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61 Production Technology

02. bonding material are same, it is possible to Sol: Given: vary the grade of a grinding wheel by W = 20 mm; N = 60 rpm varying a pressure applied in the contacting z = 10; d = 5 mm of powder. Metallurgical process used for D = 75 mm;  = 0.5 manufacturing of grinding wheel. i.e., when  = 10;  = 400 N/mm3 large amount of pressure is applied hard

fm = 100 mm/min; wheels are contained and when small  This is the case of redundant data amount of pressure is applied soft wheels are obtained.  Q (Metal Removal Rate) = w.d.fm  cm10cm5.0cm2 It is always desirable to have higher G.R. Q   3 s/cm166.0 60 because in a given machining operation it is Milling power, (from Machinery’s always expected to have higher volume of handbook) material removed from the workpiece and the lower tool wear. P = KpQCW (hp)

where, Kp = Power constant C = Feed factor 04. +0.01 W = tool-wear factor Sol: Hole = 31.80 mm By using Machinery’s handbook Hole diameter before = 31.25+0.005 mm

Kp = 2; C = 1.13; W = 1 p = 15 mm; t = 25 mm

 P = 2  0.166  1.13  1 h = 0.025 mm; FC/teeth = 5000 N P = 0.375 hp Approach = 5 mm

 DD  25.318.31 d  if   mm275.0 03. 2 2 Sol: d 275.0 materialofvolume onremoved workpiece n  11 Grindingratio  h 025.0 Grindingofvolume wearwheel Df = 31.25 + 11 × 0.025 × 2 = 31.8 mm db  R.G   Df = 31.255 + 11 × 0.025 × 2 = 31.805 mm  2 2  DDw 2 4 i f 11 teeth is sufficient and no change of teeth Soft (or) Hard grinding wheel is indicated is required

by using a grade of grinding wheel. Even i) Le = n × p = 11 × 15 = 165 mm

through size of abrasive % of abrasives on L = t + Le + Ap = 25 + 165 + 5 = 195 mm

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62 ESE – Text Book Solutions ii) No. of teeth in contact at any point of time common method of milling, where the t 25 cutting operation is smooth. However, = =  267.1 p 15 the cutter teeth must be sharp, as otherwise the tooth will rub against h=0.025 mm the surface being milled and smear it for some distance before it begins to engage and cut. There may also be a tendency for the cutter to chatter, and

5 mm the workpiece has a tendency to be pulled upward (because of the cutter p=15 mm 195 mm 165 mm rotation direction), thus necessitating proper clamping of the workpiece on 25 mm the table of the machine. h=0.025 mm

In climb milling, also called down milling, cutting starts at the surface of the n=11 workpiece, where the chip is thickest. The Force required = 5000 × 2 = 10 kN advantage of this method is that the direction of rotation of the cutter will push 05. the workpiece downward, thus holding the Sol: Conventional Milling and Climb Milling: workpiece in place, a factor particularly The cutter rotation can be either clockwise important for slender parts. However, or counter-clockwise; this is significant in because of the resulting impact force when the milling operation. a tooth engages the workpiece, this In conventional milling, also called up operation must have a rigid work-holding milling, the maximum chip thickness is at setup, and gear backlash in the table feed the end of the cut as the tooth leaves the mechanism must be eliminated. Climb workpiece surface. Consequently, milling is not suitable for machining of (a) tooth engagement is not a function of workpieces with surface scale, such as workpiece surface characteristics and metals that have been hot worked, forged, (b) contamination or scale (oxide layer) or cast. The scale is hard and abrasive, and on the surface does not adversely thus causes excessive wear and damage to affect tool life. This is the more the cutter teeth, and shortening tool life.

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63 Production Technology

workdone f2 m d f2 d t   FC  1max V Nz D Nnt D C t  0 FC = 952.10 N 1min DN  tt f d VC = t  1max 1min  601000 1avg 2 Nn D t  50010   sec/mm261.0 tA  b 1m avg 601000 T = Torque = 0.6 × F × D f d C = wt   w 1avg = 5712.64 N-mm Nn D t

fw d  2  Q = fND 4 Nn t D 2   5002.010 3 =  = 130.899 mm /sec 4 60 06. P = u  Q Sol: = 1.9  130.899 f P = 248.7 W

NT2 And P  60

 T5002  7.248  60

T = 4.75 N-m i) D = 10 mm, f = 0.2 mm/rev, N = 500 rpm

Specific cutting energy = 1.9 J/mm3 ii) As the grinding process continues the forces Material removal rate = AC × f × N are acting onto the abrasive particles, so the  = 2  NfD 4 abrasive is experiencing wear and tear. Due to this when the wear on the abrasive  2 =   N2.010 4 particle becomes considerable; it is started N rubbing onto the workpiece producing = mm70.15  3 sec/mm83.130 60 rubbing forces. These rubbing forces will be  VF pulling out the blunt abrasive particle so Specific cutting energy = CC MRR that the wear of the grinding wheel will be

FC × VC = 24.85 = workdone = power taking place. As the blunt abrasive particle

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64 ESE – Text Book Solutions

is pulled out, the fresh and new abrasive 07. particle present behind will be started Sol: Effects of vibration: (during turning contacting the workpiece and removing the operation) material from the workpiece. This is also 1. The surface finish produced on the called as “self-sharpening of grinding workpiece is poor. wheel”. 2. The dimensional variations produced on the component is very high. Therefore accuracy is poor. Hence it is required to provide large amount of tolerance on the component dimensions. 3. Due to vibrations, the impact loads are induced on the tool tip and because of lower toughness of carbide tools, the

Out of the four abrasive particles (Al2O3, tool tip may fail abnormally.

SiC, B4C, Diamond) are available, because 4. Presence of vibrations will increase the

of very poor performance in machining B4C forces in machining the tool wear is is not preferable. Due to very high cost the increasing and tool life is decreasing. diamond is not preferable. Hence most Reasons for inducing the vibrations:- commonly used abrasives in grinding wheel 1. High feed and inducing the vibrations.

are Al2O3, SiC. 2. Improper vibration Isolation of Lathe Diamond wheel means the grinding wheel machine. (G.W) is made with diamond abrasive 3. If bed of a machine tool is not made by particles. When diamond is in contact with grey cast iron as a material. the ferrous work piece. i.e., steel because of 4. During fixing the tool, large overhang is presence of atomic attraction, the C-atoms used. are diffusing from the diamond abrasive and 5. The presence of spindle runout in the depositing onto the steel workpiece called lathe machine. as diffusion wear. Therefore when diamond 08. wheels are used for grinding of steel Sol: Finish cut: workpieces, the diffusion wear is high and d = 1 mm; f = 0.1 hence it is not recommended. V = 50 m/min; L = 200 C D = 44 + 2 × 1 = 46 mm

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V1000 MRR = f × d × V = 0.5 × 4 × 24 × 103 N =  rpm95.345 C D = 48 × 103 mm/min L Tfinish cut =  min78.5 Nf 11. Rough cut: Sol: D = 12 mm; t = 50 mm  4652 6 No.of pieces = 100; Tset up = 30 sec No. of cut =  1  32 6 Tdrill set up = 10 sec; Tbringing back = 5 sec V1000  301000 (Peck drilling operation) N    rpm64.183 D  52 Assume VC = 30 m/min (bellow 40 m/min) L 200 HSS Trough cut =   min63.3 Nf  64.1833.0 f = 0.1 mm/rev Total time = T + T = 9.41 min L  CAPt finish rough Time/hole = 

fN f N 09.  125.050 Sol: Cutting stroke = 100 mm × 150 mm  = 0.703 min 301000 1.0  m = 0.7; VC = 20 m/min;  12 d = 3 mm; f = 0.3 mm/stroke Total time/hole = Tm × 60 + 30 + 10 + 5 B L Time/cut =  m1 = 87.22 sec f V Total time = Total time/hole × 100 100 150 =  7.01 = 4.25 min = 8722.30 sec = 145.37 min. 3.0 20000 MRR = f × d × VC 12. 3 = 0.3 × 3 × 20 × 10 Sol: Ist drill : 3 = 18 × 10 mm/min D = 15 mm; t = 40 mm

10. Twithdrawl = 5 sec; VC = 25 m/min; B L f = 0.1; Tap = 10 sec Sol: Time/cut =  m1 f V L  CAPt Tm  1000 1200 f N f N =  75.01 = 175 min 5.0 24000  3.15040   min279.0 (No. of cuts = 10 ,  251000 1.0  Total time = Time/cut × 10 + 1.2 × 9)   15 applicable for multiple cuts Time/hole = Tm × 60 + 5 + 10 = 31.77 sec

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66 ESE – Text Book Solutions

IInd drill: 100 150 10 min =  min/mm5.01 D = 22 mm; 25.0 V t = 30 V = 9 m/sec  30.0t  223.030 T =   min26.1 m  201000 fN 1.0  14.  22 Sol: L = Lw; Time/hole = T × 60 + 10 = 85.88 sec m B = Bw; T.T = 117.65 sec. d = t;

Width of cuter = LC > BW 13. D = diameter; ft = So Sol: Lead Screw -0.2 thread/mm 1 P  mm5 LC 2.0 O2 D O1 Reverse stroke  paul indexing Ratchet by A t 1 teeth BW No. of teeth on ratchet = 20 mm f  LW S.D 1 revolution of ratchet  20 teeth to be indexed L L  CAPL Time/cut =  w  20 D.S or R.S f tN zNf tzNf  1 rev L.S V1000 V1000 N =  C  1 p or Lead D D

 5 mm 2 2 CAP = O1O2 = 1  2AOAO 20 D.S.  5 mm 2  tRR 2 5 .S.D1  fmm25.0 20 =  tRt2 2 B = 100 mm; T = 10 min; m =   )tD(ttR2t 1 m = ; L = 150  tDtL 2 Time/cut = w V1000 C B L t zf  Tm =  m1 D f V

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2 Adoo nn   Ad Chapter 1i

Metal Forming Process 2 5 25  n3.22 5

54.0 01. Ans: (a) y   .MPa592)22.3(315

Sol:  1400 33.0 y 04. Ans: 1.98 MN 1 At maximum load, true strain  Sol: Given: H0 = 4.5 mm 3 H1 = 2.5 mm 33.0  1  H = 2 y  1400   MPa971  3 Droll = 350, Rroller = 175 mm Strip wide = 450 mm = b 02. Ans: (b) Average coefficient of friction = 0.1

Sol: A0p = C.S area of P originally y = 180 MPa st A1p = C.S area of P after 1 reduction RSF = Pavg  projected area = 0.7 A0p 2  L  = y 1   Lb A2p = 0.8 × 0.7 × A0p = 0.56 A0p 3  H4   A   "PinstrainTrue  ln"  op  L = HR =  2175 = 18.7 p  A   2 p   HH  5.25.4 4 = 10  = 3.5  A  2 2  ln op  = 0.58  56.0 A  2   7.181.0   op  1180    7.18450 3   5.34  A0Q = C.S area of Q originally st RSF = 1982.64 kN = 1.98 MN A = C.S area of Q after 1 reduction 1Q = 0.5 A0Q 05. Ans: (a)

 A0Q   1  Sol: Ho = 4, H1 = 3mm, R = 150mm,  ln   ln = 0.693 Q      A1Q   5.0  N = 100 rpm. Velocity of strip at neutral point 03. Ans: (a) = Surface Velocity of rollers DN  100300 Sol: do = 25, di = 5mm    601000  601000  315 54.0 y = 1.57 m/sec

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06. Ans: (a) 07. Ans: (d)

Sol: Ho = 20 mm, Sol: Ho = 16 mm,

b = 100 mm H1 = 10 mm,

H1=18 mm, R = 200 mm R = 250 mm, H BiteofAngle  tan 1  ,rpm10N y  MPa300 R  mm21820H 1016  tan 1  9.9 H 200  = 0.089 R  oflengthL  HRzonendeformatio 08. Ans: (a) Sol: Given rolling process  mm36.222250 Initial thickness H0 = 30 mm 1820 H  19 Final thickness = H = 14 mm 2 1 D = 680 = R = 340 mm 2  L  roller avg F.S.RF y 1Lb   3  H4  y = 200 MPa Thickness at neutral H = 17.2 2   36.22089.0  n 136.22100300   3  194  V H Forward slip = 1 1= n 1 = 795 kN. Vn H1 2.17 = 1 = 0.2285 = 23% avg  ,aFT 14 Where V H Backward slip = 1  0 1 n a = moment arm  L Vn H 0  4.0toL3.0  L 2.17 1 = 42.6%  43% 3 30 Tavg  4.0F L  36.224.010795

= 7110 kN-mm 09. Ans: (b) = 7.11 kN-m Sol: Roll separation distance  NT2  110.7102 P   = 2  R + H1 = 2  300 + 25 av 60 60 = 625 mm  roller/kW44.7

Total Power = 7.44 × 2 = 14.88 kW

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10. Ans: (b) 11. Ans: (a)

Sol: do = 15 mm, Sol: Given wire drawing process

df = 0.1 mm d0 = 6 m, d1 = 5.2 mm drawtheinreduceddia Die angle = 180, diameter land = 4 mm %Reduction= drawbeforedia Coefficient of friction = 0.15  dd Yield dress = 260 MPa  10  drawIst d o  2  A0 = 6 = 136 = 21.237 4 4  dd  21  drawnd2  2  d1 A1 = 2.5 = = 21.237 4 4 a) 3 stages with 80% reduction at each stage Drawing stress = 2  dd 8.0  1o  B    B1   A1   d =    1   o y B   A      0   d1 o  3d2.0 mm B = cot d2 = 0.2. d1 = 0.6 mm 1 1 0 d3 = 0.2 .d2 = 0.12 mm (Error is 20%)  = Die angle = 18=9 2 2

b) 4 stages with 80% reduction in 1st 3 stages  = 9 0 followed by 20% in 4th stage B = 0.15  cot9 = 0.947

d1 = 0.2. d0 = 3 2 = 126.958MPa

d = 0.2. d = 0.6 0 947.0 2 1   947.01   27.21   = 260  1    d3 = 0.2. d2 = 0.12 947.0   270.28        d4 = 0.8. d3 = 0.096 (Error is 4%) = 260(2.056)(0.2375)

c) 5 stages, with 80, 80, 40, 40, 20 etc  L2 R1 Total drawing stress 2 = y + (2y) e d1 = 0.2. d0 = 3 (By considering friction) d2 = 0.2. d1 = 0.6  415.02 d3 = 0.6. d2 = 0.36 = 260 + (130260) e 6.2 d = 0.6. d = 0.0216 4 3 total = 260  81.94 = 178.05 MPa d5 = 0.8. d4 = 0.1728 (Error is 72%) Total drawing load = tA1 From the given multiple choice B, the final = 178.05  21.237 diameter of wire close to 0.1 mm. = 3.781 kN

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Common data for Q 12, 13 & 14 14. Floating mandrel B = 0 12. Ans: (b) , 13. Ans: (c), 14. Ans: (a)   h  2  n 0  Sol: Initial inside diameter of tube    y  h1  d = 52 mm, H = 2.6 0 0  6.2  H = 1.8, D = 50 mm = n  = 0.367 1 1  8.1  2d = 24=12, =0.12

Common data for Q 15. & 16.

2.6mm

1.8 mm 15. Ans: 6 & 16. Ans: 3.4

52mm 50 mm Sol: d0 = 6 mm, df = 1.34 mm Given ideal condition

 = 0.2  = 60

f = 60 MPa

Maximum reduction condition

B2     2   B1  d1   21 = 1  1 =   1   For stationary mandrel B =  B   d   Tan y    0    12.012.0 B = =1.29 B = cot; B = 1.9 )12(Tan B2 B  d1  B 1     B1   H    B1  d    1    0   y2   1    B   H0   B2    d  B  1  1 19.1     29.11   8.1    d0   B1 /y2   1     129.1   6.2   1 =  B1 2/y = 0.64 d 1 1  B2 13. Movable mandrel d0  B1

1 B = cot = (0.12)cot(12) = 0.564  1   1   9.12 d = d  B2   6 564.0 1 0       564.01   8.1     B1    9.11  /  1     519.0 y2  564.0  6.2      d1 = 4.53 ……… (1) stage

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1 Ao Lo = Af Lf B2 d2 = d1 2  B1 A  d  LL o  L  o  1 of A o  d   1  B2 f  f  C =   = 0.756 2   B1   214.12  100    m150  10  nd Dia of wire in 2 stage = 3.424 mm True strain in the drawing process d = d  c 2 1 0 Ad nnoo   0.4 d2 = d1 c = 4.530.756  Ad11 = 3.424 > 1.34 From the graph y at 0.2 , d3 = d2  c  300 MPa = 3.424 0.756 y = 2.589>1.34

d4 = d3c = 1.957>1.34 19. Ans: (b)

d5 = d4  c = 1.4797 > df 20. Ans: (c) d6 = d5c = 1.1186 < df Sol: (Extrusion force) =  A  Hence No. of stages = 6 min 0y  10 102 = 78539.8N Common data for Q 17, 18 4 )F.E( 8.78539 Extrusion force  min  17. Ans: (c) & 18. Ans: (b) ext 4.0 = 196346.5 N Sol: 400 y = 196 Tons

200 21. Ans: (b) Sol: Extrusion constant = K = 250

0.2 0.4 do = 100 mm, df = 50 mm  A o Extrusion Force = o lnKA do = 12.214, Lo = 100m A f d = 10mm, L =? 2 f f  100   2  ln250100   = 2.72 MN. y before  200 MPa , 4  50 

y  MPa400after

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22. Ans: 1 Common data for Q 23 & 24

Sol: Let, d1 = d2 = d

h1 = height of first cylinder 23. Ans: 7068 J & 24. Ans: 0.354 m

h2 = height of second cylinder Sol: do = 100 mm, ho = 50 mm,

Assume h1 < h2 f  mm40h , y  80 MPa Let % reduction in height = 10% h o 50 dd of 100  mm8.111 h f 40 Ist cylinder AF   hh i min y0 f0  1.0 h  0  2  kN318.62880100 4 h0 – hf = 0.1 h0  2 hf = h0 – 0.1 h0 = 0.9 h0 AF  80)8.111( f min yf 4

 kN350.785 A0h0 = Af hf 2 2  FF d0 h0 = df hf F  imin f min  kN834.706 min 2 h h  dd 0  d 0 0f 0 min fo  7068)hh(FD.W J h f h9.0 0  HW2 = 1.054 d0 = 1.054 (d0)1 7068 H   m354.0  10102 3 IInd cylinder

A0h0 = Afhf 2 2 25. Ans: (b) d0 h0 = df hf

h 0  dd 0f 26. Ans: 58% h f st Sol: Area after 1 pass = A1 = (1 – 0.4)A0

h 0 = 0.6 A0  d = 1.054 (d0)2 0 nd h9.0 0 Area after 2 pass = A2 = (1 – 0.3)A1

d d054.1 = 0.7  0.6  A0 = 0.42 A0 Ratio  0 1  0 1  1 d0 2 d054.1 0 2 Overall % reduction = (1 – 0.42)  100 = 58 %

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27. Ans: 7.26 K n  6931.0895 49.0 Y  1   MPa502 Sol: 1n 49.1 0.3= 1.328 O From , the drawing force is  A  Ai  B  f nAYF   C  A  H H  f  0 1 Where,

2   3    502F     6931.0   4 1000   O  

Ho = 10 mm, F = 0.002459 MN 500 0.3 H1 = 7 mm, Power = F  V 1000 B D = 0.002459  0.6 R = = 500 mm 2 = 0.001475 MN m/s C H = 0.001475 MW = 1475 W Angle of bite () = tan 1 R and the actual power will be 35% higher, or  710 Actual power = 1.35  1475 = 1.992 kW = tan 1 = 4.429 500 The die pressure, p = Yf = a

OD where Yf = the flow stress of the material at 328.1  OB the exit of the die. n 0.49 OD = 500cos1.328 = 499.865 Yf = K t = 895  (0.6931) = 748 MPa DC = 500 – OD = 0.1343 mm In this equation,

Thickness of neutral point = At point B  is the drawing stress, d. = 7 + 2  0.1343 = 7.2686 mm Hence, using the actual force, we have F d  28. Ans: A f Sol: The true strain that the material undergoes  002459.035.1 2  41000  in this operation is  32  62  = 470 MPa  n   6931.0 1   2   3  Therefore, the die pressure at the exit is Assume that for this material and condition, p = 748 – 470 = 278 MPa. K = 895 MPa and n = 0.49. Hence,

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29. Ans: 7.687 MPa, 19.7 % rf = 69.098mm

Sol: d0 = 6.25 mm; d1 = 5.60 mm; Forging force 2  = 0; y = 35 N/mm 3   098.6925.02  = 110157.0   B = cot = 0   25.63  B = 29847.44kg = 292.80kN   B1   A   0  1  1    y2       B   A0  0   31. Ans: 20.52 kW By applying L – Hospital rule Sol: d0 = 10 mm;  A  0  AA 10 A1  y2 n  3.0  1   A   1  A 0 A 0  d  d 2  n2  0  13.0  1 y    2  d1  d 0

= 7.687 MPa d1 = 8.36 mm o  AA 2  dd 2 B = cot = 0.1cot(6 ) = .951 % reduction in area = 10  0 1 A d2 B 0 0   B1   A     1  1   = 19.71% y2 B   A      0  

  951.01  951.0 30. Ans: 29.85 tons  240  7.01  951.0  Sol: Initial size = 2525150mm = 141.687 MPa Final size = 6.25100150mm  Drawing load = 141A  d 2  = 0.25; 12 4 1 2 y = 0.7kg/mm   687.141F d 2 As given piece is pressed; height is reduced d 4 1 h0 = 25;  = 141 36.8 2 = 7777.364 = 7.8 kN hf = 6.25 4 A = 25150;  vF 0 P (motor) = d Af = 100150 motor  hr2   5.28.7 Forging force = 1A  f P  fy   95.0  h3 f   P = 20.52 kW (Ac)circular = (Ac)non – circular 2 f 100r  150

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32. Ans: (b) coefficient of friction (i) of about 0.1, Sol: The roll separating force in the rolling can while hot working requires as high 0.6 be reduced by reducing roller diameter, approximately. It is also a fact that hot reducing the friction, heating the material forming or working requires less energy and providing back and front tensions. even for bulk deformation, while cold forming requires more energy even for 33. Ans: (d) lesser degree of deformation. Sol: The pressure exerted over the metal by the roller is not uniform throughout, it is 35. Ans: (d) minimum at both the extremeties L and M Sol: In indirect extrusion, the direction of flow and maximum at a point known as no-slip of the extruded metal is in the opposite point or the point of maximum pressure. At direction of the ram movement. Hence, the this point the surfaces of the metal and the name is given as indirect (or) backward roll move at the same speed. extrusion. The travel speed of the extruded product is Vf Exit point greater than that of the ram. The travel Metal piece O speed of the extruded product is greater than L B Neutral plane S that of the ram. P M h V1 Hn V1 2 t1 M 36. Ans: (d) P S L  O 37. Ans: (d) Roll (Bottom) Sol: Hydrostatic extrusion is similar to direct extrusion. So given assertion is incorrect.

b f 38. Ans: (c)

39. Ans: (c) 34. Ans: (c) Sol: Strain hardening is higher in cold forming, 40. Ans: (b) while it is very low in hot working. Cold Sol: Both statements are correct, but statement forming or working process requires a low (II) is not correct explanation of statement(I).

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41. Ans: (a) 44. Ans: (b) Sol: Hot working is well suited for forming large 42. Ans: (a) parts, since the metal has a low yield Sol: strength and high ductility at elevated  Rolling is used for production in large temperatures. So statement (2) is incorrect. quantities of standard item such as sheets and plates. 45. Ans: (a)  Extrusion more widely used in the Sol: manufacture of solid profiles, hollow  Edgering  Collecting the material at profiles (such as tubes), from (non ferrous localized areas. metals and their alloys, copper, brass and  Upsetting  Rod is deformed so that bronzes etc), but steel and other ferrous diameter increased. alloys can also be successfully processed  Fullering  Distributing the material with the development of molten glass from centre. lubricant.  Trimming  Removing flash from the  Forging is more suited for production of forged part. relatively simple shapes.

 Spinning operation of forming sheet metal

into circular shapes by means of a lathe, forms and hand tools which press and shape the metal about the revolving form.

43. Ans: (d) Sol: Graphite powder  Bright shining surface Air  Burnt surface Soap solutions  Dull surface

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6. Back and front tensions. Conventional Practice Solutions  Recrystalisation temperature (RCT): is the 01. minimum temperature at which formation of Sol: new crystals has been completed.

IMPACT EXTRUSION:  It is a process in which very thin wall tubes 2 =die angle  will be produced from solid rods which are made by very soft materials by using impact d0 Z Z Z3 Z4 d1 2 2 1 load as extrusion load.

ram L 2 to 5 m cover impact load

Zone – 1: Deformation zone: Zone – 2 Entry (or) lubricating zone: Zone – III Sizing zone:  Impact extrusion is mainly used for Zone – IV exit zone producing collapsible tubes like tubes, cosmetic tubes etc. 2 = die angle

d0 = Rod diameter 02. d1 = wire diameter Sol: The process variables in wire drawing operations (i) Bite angle: The angle made by the are deformation zone with respect to the centre 1. Presence of friction at the interface of die and of the rollers is called deformation angle (or) surface of wire. Angle of bite. This depends on the reduction 2. The mechanical properties i.e., yield stress of in thickness and diameter of rollers. the work material. (ii) Neutral point: At the neutral point the 3. Die angle and Die land. relative velocity and slip becomes equal to 4. Length of the sizing zone. zero. 5. Velocity at which the wire is pulled (influencing the power required)

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78 ESE – Text Book Solutions

(iii) Camber: The rolls are often ground so that A 1 1 = 63.21% they are thicker towards the center in such a A 0 way as to exactly offset the deflection that will occur during the process. This extra 04.(i) thickness is called camber. Sol: Simple die: If only operation is performed (iv) Draft refers to reduction in thickness. in one stroke at one stage it is called as a Draft = ho – hf simple die. Compound dies: In these dies, two or more 03. cutting operations may be performed at one Sol: d0 = 6.25 mm; d1 = 5.60 mm; station and at one stroke. The example is 2  = 0; y = 35 N/mm washer is provided by blanking and piercing B = cot = 0 operations simultaneously. These are B accurate and economical in mass production   B1   A   0   1  1    y2     Piercing punch  B   A0  0   By applying L – Hospital rule Punch holder Shedder Blanking die  A   0   y2 n  Job  A1  Scrap strip  d  Stripper  n2  0  y    Blanking punch  d1 

= 7.687 MPa Piercing 2 2 die  AA 10 0  dd 1 % reduction in area =  2 A0 d0 Compound die assembly = 19.71% Limitations: For maximum reduction  Both the methods of reducing the punch

 2  A 0   n1   cannot be used. Hence the force required is   A  y  1  higher A 0 1  718.2e  It is difficult to design and manufacturing the A 1 punch and die combination for producing  AA more than 3 operations. % Reduction in area = 10 A 0

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79 Production Technology

Progressive die: for completing the remaining punching  In progressive die also more than one cutting operation. operation is performed in one stroke but at  Out of two progressive dies are preferable different stages and blanking will be the last because sheet need to be set for every stage operation. at every time.  In progressive die, because the operations are performed at different stages, either the Combination die: provision of shear or staggering of punches  If more than one cutting and forming method will be used for reducing the punch operations are performed in one stroke at force. one stage is called as combination die. For ex: blanking combined with deep Ram drawing operation, blanking combined with bending and punching combined with Blanking punch Pilot Piercing punch drawing etc Scrap Multiple Die:

To produce more than one component per stroke, Die Metal strip more than one die are kept in parallel called as Stop multiple die.

Ex: To produce 10 washers per stroke, then 10 Finished washer compound die or 10 progressive dies are  Same production rate kept in parallel.  Manufacturing is easy.

Disadvantage: 04(ii).  Balancing of force on punch head is difficult Sol: d = 40 mm ; h = 60 mm; Transfer die: r = 2mm ; t = 0.6 mm  In transfer die more than one cutting d 40  20 Neglectingcorner radius operation is performed in one stroke at r 2 different stages. 2 2  mm83.1056040440dh4dD  But here the blanking operation will be the Because nothing is mentioned about the first operation and blank produced in first draw ratio and draw reduction ratio values it stage is traveling from one to another stage is assumed that the draw ratio method with limiting draw ratio = 2.

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80 ESE – Text Book Solutions

D 83.105 forming, the resultant force is acting on the centre d1  4095.52 DR1 2 of the deformation zone. Based on this, to ensure d 915.52 that the strip is pulled by the rollers d 1  404575.26 2 DR 2 2 Rx n = 2 R x

05.

Sol:

entry   /2 Exit  Rx 2

 nipofangle2

06.

Sol: b = 250 mm; H0 = 2.5 mm; Angle of bite : 350 The angle made by the deformation zone w.r.t H1 = 2.0 mm; R  mm175 , 2 center of the roller is called as angle of bite or  05.0 ; y = 180MPa. deformation angle. L2 H n  Tan  cos1DH;  H R H = 2.5–2.0 = 0.5 As angle of bite is increasing reduction in thickness of the strip is increasing. To ensure that  HRL =  mm354.95.0175  354.905.02 the strip is pulled by the rollers, the resultant n   87.1 resolved horizontal component of frictional forces 5.0 must be  0. Based on this maximum value of  At neutral plane; max  Friction givenainangle rollingoperation P  P N lagging N leading Once the rolling operation is started, the location n n     H       H    y 1n  0  1   y 1n  n  1 at which the resultant resolved horizontal  n   H   n   H     n      1   component of force acting is shifting towards the exit and when complete deformation zone is

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81 Production Technology

87.1 n  5.2   H  08.   n 187.1    187.11   1 Sol:  H n   2 

87.1 n  5.2   H  87.0    87.21  n  1  H  2  n   

07. Sol: The tooth paste tubes (also called as collapsible tubes) are manufactured by using impact extrusion process. Impact extrusion is the metal forming process in which the In rolling operation the coefficient of solid rod can be converted into thin wall friction must be as maximum as possible because friction is the mechanism by which tube with the application of impact loads. In the strip is pulled by the roller. Also the this process only very soft materials like reduction per pass in the rolling operation aluminium, copper, brass etc. can be depends upon .

extruded. 2 max  Rh

ram impact 09. load Sol:

(i) The angle made by the deformation zone

with respect to the centre of the rollers is

called deformation angle (or) Angle of bite.

(ii) Percentage reduction in thickness

 AA  Impact extrusion is mainly used for  % Reduction in area (J) = 10 A 0 producing collapsible tubes like tubes,  2 cosmetic tubes etc.  Cross section area of rod A0 = d 0 4

 2  Cross section of the wire A1 = d 4 1

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82 ESE – Text Book Solutions

(iii) Elongation coefficient Draft (or) coefficient 11. A Sol: Initial size = 2525150mm of elongation, K = 0 A1 Final size = 6.25100150mm 2 (iv) The neutral point defined in the deformation  = 0.25; y = 0.7kg/mm zone is dividing the deformation zone into As given piece is pressed; height is reduced

two zones h0 = 25; hf = 6.25

 The zone between the entry and neutral A0 = 25150; Af = 100150 points is called “lagging zone”  hr2  Forging force = 1A  f  The zone between neutral point and exit fy    h3 f  is called “leading zone” (Ac)circular = (Ac)non – circular 2 f  150100r (v) Forward slip: The maximum % slip taking r = 69.098mm place in the leading zone is called as f Forging force “forward slip”. 3   098.6925.02  1  VV V1 = 110157.0  Forward slip =  1   25.63  V V = 29847.44kg = 292.80kN

10.

Sol: H0 = 300 mm, b0 = 600 mm, 12. H = 50 mm; D = 1000 mm Sol: Force Total h = h0–h1 = 15, b = b1–b0 = 5 mm load  = ? Redundant load 50 cos  = 1 – (h/2R) = 1 Ideal load  5002 friction load minimum load   = 18.2 o o 12 48  hh 10 2 And, ductionRe%   67.16 % h 0 In drawing, Now, h0b0l0 = h1b1l1 A bh  600300  n i  1 00   eyd Af  0 bh 11  605250 For maximum reduction,  1  19.1 d = y  0

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83 Production Technology

Ai 14.  n e  1 Af Sol: d0 = 10mm; A  AA A i  718.2 3.0  10 1 1 Af A 0 A 0 A d 2  fi  3678.0 1 13.0  2 Ai d 0 A d = 8.36mm % in reduction = 1 f  %2.63 1 o Ai B = cot = 0.1cot(6 ) = .951 B   B1   A     1  1   13. y2 B   A      0   Sol: t = thickness   951.01  951.0 d = diameter  240  7.01 951.0 Total clearance = c = diametral clearance   Blanking: = 141.687 MPa  Die size = blank size = d Drawing load = 141A  d 2 12 4 1 Punch size = d – c = Die size – c  Punching:  687.141F d 2 d 4 1

2 Fmax  d1     7.0  d 0  Fmax

1  7.010d

d1 = 8.366 mm

Punch size = Hole size = d  Fd = 7.8 kN Die size = Punch size+c = d+c  vF P (motor) = d F dt   max  motor cp  A cp 2  5.28.7 d P  4 95.0 t4    P = 20.52 kW cp d

t4  d min  

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84 ESE – Text Book Solutions

05. Ans: (b)

maxp Kt.F Chapter Sol: F  p 6 Sheet Metal Operations t  IK  25.16.040   14.17 kN  125.16.0

Common data for Q. 1 to 5 maxb ktF Fb   Ikt  25.16.080 01. Ans: (b)   28.34 kN  25.16.0 Sol: For punching operation  .51FFF 42 kN Punch size = Hole size = 12.7 bP

Die size = punch size + clearance Common data for Q. 06, 07 & 08 = 12.7 + 2  0.04 = 12.78 06. Ans: 83.6 N

Sol: 02. Ans: (a) 100 Sol: Die size = Blank size = 25.4mm 30 Punch size = Die size  2(radial clearance) 50mm 0 = 25.4  2(0.04) 45 20

Punch size = 25.32 mm 80mm 20

 508022030100P  28.288 03. Ans: (b) max u  145228.288PtF  6.83 kN Sol: Fmax = Fp max + Fb max

=  80025.17.12  80025.14.25 07. Ans: 66.88 J = 40 +80 = 120 kN Sol: Work done in blanking open

= Fmax.K.t 04. Ans: (c) = 83.61030.42103 Sol: Force required is Max [Fpunch, Fblank] = 66.88 J  force required is Max [40, 80]  force required = 80 kN 08. Ans: 1.98 mm Sol: I = ? F = 24 kN

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Fmax = 83.6 kN 12. Ans: (d)

F(Kt + I) = Fmax  Kt Sol: d = 25 mm, t = 2.5 mm → piercing  KtF  MPa350 I = max Kt u F Diameter clearance  C   24.06.83  =   24.0  = 1.98 mm  tK0064.0  24   3505.20064.0 = 0.3 mm

In piercing 09. Ans: (a) P.S = H.S = 25 mm. 5 Sol: dt5F  dt  max u u  D.S = P.S + C = 25 + 0.3 = 25.3

max dtF u  3505.225 max t4.0d5.1F  u = 68.72 kN. dt4.05.1  u

2 4.05.1   KN3 13. Ans: (a)  Sol: Die size = Blank size = 25 – 0.05

= 24.95 Common Solution for Q. 10 & 11 Punch size = Die size – clearance

= 24.95 – 2  0.06 10. Ans: (a) = 24.83 11. Ans: (b)

Sol: t = 5 mm, L = 200 mm, τu = 100 MPa, Common data for Q. 14 & 15 K = 0.2

W.D = Fmax Kt = L × t × τu × K.t 14. Ans: (b) = 200 × 5 × 100 ×0.2 × 5 before.Dia 10100 3 Sol: Draw Ratio =   J)or(mN100 .Dia after 1000 22.13 Shear provided over a length of  d1 = = 7.34 > 5cm 8.1 20 200 mm  200 = 10 mm 34.7 400  d2 = = 4.08< 5 cm 8.1 F Kt = F (Kt + I) max n = 2 3  52.010100 F   kN1009.9  1052.0

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86 ESE – Text Book Solutions

15. Ans: (a) Common data for 18 & 19

2 18. Ans: 6 Sol: 1  hd4dD 11 2 2 2 2 Sol:  15030430dh4dD 11  dDhd4 1  47.137 2 2 2 2  dD 1  34.722.13 h1   = 4.11 mm 1  6.047.1376.0Dd  3048.82  d4 1  34.74

2  30984.658.048.82d 1 DtP  y 3 7.528.0984.65d  30 =  3155.122.132 4  302.428.07.52d  196238 N = 196.238 kN 5 7.338.02.42d  30 –3 E = P1h1 = 196.2384.1110 278.07.33d  30 = 806.6 kJ 6 n = 6

16. Ans: (b) 19. Ans: 52.7 mm

 dD 1 Sol: 4.0DRR  Sol: d3 = 52.7 mm 1 D

1   12.186.02.304.01Dd 20. Ans: 144.42  75.012.1825.01dd  59.13 d 100 12   Sol:  20to1566.16 r 6 23  75.059.1325.01dd  19.10 2 r d < 12  n = 3 dh4dD  3 2

6 2 251004100  17. Ans: (b) 2 Sol: 1 DtP  y  3.66413522.30 = 138.42 +23 P  42.14432DD mm  1 total 21  2  t2dd 2 4 1 1 21. Ans: (d) 3.641,6   2  2212.1812.18 2 4 22. Ans: (c) n = 65.5 MPa Sol: Number of earing defects produced =2 Where n is an integer So possible option is 64.

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23. Ans: 467 mm 26. Ans: 7.536 kN  Sol: Punching a 10 mm circular hole from 1 mm Sol: 1 18025.015B  180 thickness sheet: = 50.265 mm Punch size = Blank size = 10 mm  Die size = Punch size + 2 C 2 9025.06B  = 10.99 mm 180 C = Clearance = 0t0032.   B2B9220498L 0 21 t = thickness = 1 mm = 466.245 mm where,  = 240 N/mm2

C = 0.0032×1× 240 2mm = 0.0495 mm = 0.05 mm 15 Die size = 10 + 2 × 0.05 = 10.1 mm 92 98 100 Force required = s × d × t

6 mm = 240 ×  × 10 × 1 8 m = 7.536 kN

8 204 8

220m Conventional Practice Solutions 24. Ans: (b) 01. 25. Ans: 3 Sol: 2  dh4dD

2 Sol:  dh4dD 2  mm15010050450 2  mm15010050450  dD 4.0  1  dD D 4.0  1 D 0.4150 = 150 – d1 0.4150 = 150 – d1 d1 = 90mm > 50 d1 = 90mm > 50 d2 = d1(1–0.4) = 54 > 50 d2 = d1(1–0.4) = 54 > 50 d3 = 32.4 < 50 d3 = 32.4 < 50  n = 3  n = 3

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88 ESE – Text Book Solutions

02. Sol: Circular hole: d = 10mm Chapter Rectangular blank: 50200mm 7 Metrology t = 1 mm;

N  240 u mm 2 7.1 Limits, Fits & tolerances Punching : 01. Ans: (a) Punch size = Hole size = 10 mm Sol: For Clearance fit Die size = 10 + 20.05 L- hole > H- shaft Fmax = Ptu = dtu = 101240 = 7.539 kN 02. Ans: (c) 050.0 Sol: Hole = 000.0 mm40 , Blanking : Min. clearance = 0.01 mm, Die size = Blank size = 50200(rectangular) Tolerance on shaft = 0.04 mm , Punch size = Die size –2c Max. clearance of shaft = ? = (50–2c)(200–2c) 0.01 = L.hole – H.shaft = 49.9199.9 0.01 = 40.000 – H.shaft  H.shaft = 40.000 – 0.01 = 39.99mm Fmax = ptu = 2(50+200)1240 = 120 H.shaft – L.shaft = 0.04

L.shaft = 39.99 – 0.04 = 39.95 Max. clearance = H.hole – L.shaft = 40.05 – 39.95 = 0.10 mm

03. Ans: (d)

Sol: Xmax = 50.02 – (37.985 + 9.99) = 2.045

Xmin = 49.98 – (38.015 + 10.01) = 1.955

X = Xmax  Xmin= 0.09 Dimension X = 2 ± 0.045

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04. Ans: (c) 07. (i) Ans: (d)

Sol: D Sol: Let the vertical distance between the holes is ‘y’

2450.05

t t = 0.01 to 0.015mm x

When, t = 0.01 mm 25 y 2500.2 D = 30.01 + 20.01 = 30.03 mm 30

= 30.05 + 2 0.01 = 30.07 mm  2.0  0.0 When, t = 0.015 mm 60 D = 30.01 + 20.015 = 30.04 mm y = 30.05 + 2 0.015 = 30.08 mm sin30 = y = 245sin30  08.0 245   03.0 mm30D ymax = 245maxsin30max

= (245 + 0.05)sin(30 +15/60) = 123.45 05. Ans: (d) 0 ymin = (245 0.05)sin(30 15/60) = 121.55  01.0 Sol: A = 2.25  02.0 (ii) Ans: (c) B = 30.4  0.01 x = 250 – (60 +(30/2) +y +(25/2) ) C = 32.7  0.02 max max min min min min = (250 + 0.2) (60 +15+121.55+12.5) Tmax = Lmax  Amin  Bmin – Cmin = 41.15mm = (118 + 0.08)  (25.2  0.02)  (30.4 xmin = 250min –(60max+(30/2)max + ymax + (25/2)max)  0.01) – (32.7 – 0.02) = (2500.2)  (60.2 + 30.025/2 + 123.45 + = 29.83 = 300.17 25.025/2) T = L  A  B  C min min max max max = 38.625 mm = (118  0.09)  (25.2 + 0.01)  (30.4 Tolerance on X = Xmax – Xmin = 2.525 mm + 0.01) (32.7+ 0.02) = 29.57 08.  43.0 Tmin = 30 Sol: L Hole = BS = 65mm  17.0  43.0 H Hole = BS + Tolerance = 65.05mm  T = 30 (i) Ans: (c)

Allowance = (L.L)  (H.L) 06. Anc: (c) hole Shaft

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90 ESE – Text Book Solutions

 0.09 = 65 – (H.L)shaft 10.

 (H.L)shaft = 65  0.09= 64.91 mm Sol: Let

Tolerance = (HL)shaft  (LL)shaft C = center distance between holes Cmax = max. Outer distance of pins –  0.05 = 64.91 – (LL)shaft sum of min rod holes.  (LL)shaft = 64.86 mm

 09.0 C =100±0.1 Shaft = piston = 65 14.0 14.9  0.025 9.9  0.025 (ii) Ans: (a)

(L.L)hole = 65 mm

(Tolerance)hole = (HL)hole  (LL)hole 15  0.05 10  0.05 x  0.05 = (HL)hole – 65

 (HL)hole = 65.05 mm  9.9   9.14  Xmax = 100max        05.0  2 max  2 max 00.0 Hole = Bore = 65 925.9 925.14 1.100  2 2 (iii) Ans: (b) = 112.525 mm Max Clearance = 65.05 – 64.86  9.9   9.14  Xmin = 100       = 0.19 mm min  2 min  2 min 875.9 875.14 9.99  09. 2 2 Sol: Amax = 15max + 30max = 112.275 mm = 15.06 + 30.1 = 45.16 15  10   XC   Amin = 15min + 30min = 44.84 max max       2 min  2 min  A = 45 ± 0.16. = A ± ∆A  95.14 95.9  Bmax = Amax – 20min 525.112     2 2  = 45.16 – 19.93 = 25.23 mm = 100.075 mm Bmin = Amin – 20max 15  10   = 44.84 – 20.07 = 24.77 mm XC       min min  2 2   B ± ∆ B = 25 ± 0.23.  max  max   05.15 05.10  525.112     2 2 

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= 99.725 mm  075.0 B.S C  100 275.0 F.D=0.02mm

f H. Limit Tol = 0.033mm 11. L. Limit Sol: For the given conditions 875.14 875.9 1.100X  2 2  mm24.233018D = 112.475 mm 3  m3.10010.0D45.0i  05.15 05.10  CX    FD of hole H = 0  2 2  0.41 FD Shaft = 5.5(23.24) = 20m C = 99.925 mm Hole tolerance, IT7 =16i = 20.8m Because C is lying in between the limits, the = 21m = 0.021 mm assembly is possible. Shaft tolerance, IT 8 = 25i

12. Ans: (b) = 32.5m = 33m Sol: Fundamental deviation of hole ‘h’ is zero. = 0.033mm L - hole = basic size =25 mm 13. H - hole = 25 + 0.021 = 25.021 mm  03.0 Sol: Hole = 20 00.0 H - shaft = 25 – 0.02 = 24.98 mm Min. interference = 0.03mm, L - shaft = 24.98 – .033 = 24.947 mm Max. interference = 0.08 mm (i) Ans: (a) 0.03 = L.shaft – H.hole L- hole > H- shaft  Clearance fit L.shaft = 0.03 + 20.03 = 20.06 mm

0.08 = H.shaft – L.hole (ii) Ans: (b) H.shaft = 0.08 + 20.00 = 20.08mm Allowance = difference between max.  08.0  20shaft  06.0 material limits = L.hole – H.shaft = 25.00 – 24.98 = 0.02 mm 14. Sol: (iii) Ans: (b)  02.0 H. Limit H 0.021  021.0 Shaft  25 053.0 , Hole = 25 00.0 B.S L. Limit

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Max clearance = different between 16. Ans: (a) minimum material limits Sol:  mm24.638050D = H.hole – L.shaft i = 1.86 microns = 1.9 microns = (25.021)  (24.947) IT8 = 25i = 47.5 microns = 0.074 mm Tolerance = 0.0475 mm F.D = –5.5 D0.41 = – 5.5 × 63.240.41 (iv) Ans: (a) = 30 Microns = 0.03 mm Size of the GO plug gauge = max. material H. shaft = 60 – F.D = 60 – 0.03 = 59.97 mm limit of hole = L.hole = 25 mm L. shaft = H. shaft – Tolerance = 59.97 – 0.047 = 59.923 mm. (v) Ans: (b) Size of the NOGO plug gauge = min. 17. Ans: (d)

material limit of hole = H.hole = 25.021 mm Sol: Case (i) 25H7 L.L = 25.00 (vi) Ans: (c) U.L = 25.021

Size of the GO ring gauge = max. material Case (2) 25 H8 limit of shaft = H.shaft = 24.98 mm UL = 25.033

Case (3) 25H6, UL - ?

(vii) Ans: (d) (UL)H8 (UL)H7 = (UL)H7  (UL)H6 Size of the NOGO ring gauge = min. 25.03325.021 = 25.021  (25 + x) material limit of shaft = L.shaft = 24.947 x = 0.009

mm  (UL)H6 = 25.009

(viii) Ans: (a) 18. (i) Ans: (a) , (ii) Ans: (a), (iii) Ans: (a), (iv) Ans: (c) 15. Ans: (c) Sol:  2.233018D Sol:

  3.1D001.0D345.0i H 0.025 H. Limit IT8 = 26i = 26 × 1.3 = 33.8 50 L. Limit = 34 m = 0.034 mm

 034.0  000.0  025.0  25H25sizeHole  000.0 8  50Hole

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22. Ans: (d) Sol: To produce an interference fit, L-shaft must p H. Limit 0.042 be greater than H-hole. For this with 0.026 B.S =50 multiple choice D it is possible because L. Limit For D: L-shaft = 20 – 0.02 = 19.98 mm, H-shaft = 20 + 0.02 = 20.02 mm  042.0 L-hole = 20 – 0.035 = 19.965mm,  50Shaft  026.0 H-hole = 20 – 0.03 = 19.97mm, L.hole = B.S = 50 Hence, L-shaft (19.98) > H-hole (19.97) H.hole – L.hole = Tolerance = 0.025 mm

H.hole = L.hole + Tolerance = 50.025 mm 7.2 Angular Measurements Max. interference = difference between max. material limits = H.shaft – L.hole 01. Ans: (a) = 50.042 – 50.00 = 0.042 mm Sol: Sine bar Min. interference = difference between min. Slip gauges material limits = L.shaft - H.hole  = 50.026 – 50.025 = 0.001 mm Given sine bar length = 200 = l

Angle =3256 = 32.085 19. Ans: (c) Slip gauge height = h say

h 20. Ans: (b) sin   Sol: To calculate exactly the data was not given h 085.32sin 0  in the problem. But for shaft “h”, 200 H – Shaft = 25.000  h = 106.235 L – Shaft = less than 25. And h7 → 7 indicates IT 7 not 7 microns. 02. Ans: i-(b), ii-(a) Sol: l = 50 , L = 500 21. Ans: (a) 50  0.08 Sol: GO size = max. material limit of hole 08.0 200  200  32.0 = 20.01 mm 50 NOGO size = min. material limit of hole h = h + 0.32 = 28.87 + 0.32 = 29.19 = 20.05 mm h 19.29 sin  82332 L 200

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03. Ans: (d) 07. Sol: (i) Ans: (b) 04. (i) Ans: (c)  hh sin = 12 h w Sol: sin = L h2  h1 = 100sin30 = 50 h = sin30o 125=62.5 mm  h2 = h1 + 50 = 75

(ii) (ii) Ans: (d) (A) Ans: (a)  25h  0 005.0  )30sin(  h1 h2 d 30tan  = 4.76 005.100  5.62 125   h = 75.0025 mm (B) Ans: (a)  h = 75.0025 + 0.005 = 75.0075 mm  dd 002.0 2 rrdh  12  001.0 12 2 2 08. Ans: (a)  001.0 0  d30tan  = 2  5.62 125 Sol: L = 250 mm, d = 20 mm (C) Ans: (b) h = 100 – (d/2) = 100 – 10 = 90 mm dh = 0.002 90  002.0 0  sin  30tand    = 4 250  5.62 125    = 21.2 deg (D) Ans: (d)

dh =  0.005 09. Ans: 11.556 mm   005.0 0  d30tan     = 10 Sol:  = 2732  5.62 125  o  32  = 27o     60  125 05. Ans: 0.048 mm/m h Sol: Gradient of spirit level = 27.533 h  = Sensitivity specified in mm/m sin = 25 10   1000 = 0.04845 mm/m. 1803600 h = 11.556 mm

06. Ans: (d)

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7.3 Taper Measurement 03. Ans: 112.41 mm Sol: 01. Ans: 19.2 70 mm 30 mm

Sol: h2

2.5 30 h1 h1+r1=O2A+r2+h2 O2 D O2 O1 50 C /2 d1 = 100 m A

15 O1 Diameter = O1C+O1A+O2D

d d 1  2 AOOO 2  2 2 21 2 2  dd 2/sin 12   rrOO =75  ddhh2 1221 2121

1530 O2A = h1 + r1 – r2 – h2 2/sin   15305.522  = 70 + 50  30  25 = 65 15  = 1/6 22  25657550D 15105 = 112.4165 mm   = 19.2

02. Ans: 60 04. Ans: 43.33 mm 5 Sol: tan2/  Sol: 66.8

  = 60

d = 25 O2

42 O2 105=5 35 O1 O1 A

22 O1A =  33.181725 36.345=31.34 D = r + O A + r 36.34 1 = 25 + 18.33 = 43.33 mm 40

50

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05. Ans: 78.074 mm 06. Ans: 1.1

Sol: Sol: d2 – d1 = 10 ; h2 – h1 = 12.138

  dd 12 sin   2  dd)hh(2 1212 A B  = 88.9 12.5mm 12.5mm Error = 90 – 88.9 = 1.1 C M    07. Ans: 38.94 Sol: O X h  D 2  D 2 2  +  = 4550 + 2910 = 75  D  5.37   2 D 2 Sin      2  h  D  Dh2 75 2  = 37.5 –  = 37.5 – 2910 2  D Sin  = 820 2  Dh2 le OBC If D = 0, h = 0 BC D = 1, h = 1 sin 37.5 = OB  1 1 Sin     BC 5.12  2   112 3  OB   533.20 5.37sin 5.37sin     47.19   = 38.94 le OAB  2  OA cos 820 = OB 08. Ans: (d)  OA = OB cos 820 = 20.316 mm  3 Sol: Tan   X = M – (OA + R) 2 54.28 = 110.89 – (20.316 + 12.5)

= 78.074 mm 3  3 2 15.54 + 8 + 5 = 28.54

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  3  7.4 Screw Thread Measurements  Tan 1   = 6 2  54.28  01. Ans: (d)    Taper angle    60  2  Sol: Major diameter = s + (R2 R1) Included angle = 120 = 35.5 + (11.8708 9.3768) = 37.994 mm 09. Ans: (c) 02. Ans: (a) 10 -1 Sol: tan     = tan (1/3)  434.18  Sol: Minor diameter 30 = 30.5 + (15.3768  13.5218) 10mm 10mm Z=40 = 32.355 mm

30mm  03. Ans: (a)

10mm Z=10  p     Sol: best wire diameter, d =  sec  2 2     Z=0  5.3   60  10 – (10/3) =  sec  = 2  2   2  Distance at Z = 0, M = 30.5 + (12.2428  13.3768)  10  = 29.366 mm 0  10230tan10102D    3   p   De = Md tan  = 6.67  2 = 13.33 mm  2 2   5.3  = M2 30tan  1sec  2   1 = 29.366 – 3.010366 = 26.355 mm

With probe diameter compensation 04. Ans: (a) Sol: VED = De  VC Dactual  r sec2334.13   = 13.334 + 2 ×(1 sec 18.435)  cosPVC  P0131.0 1  2 2 = 15.442 mm. P = pitch error

10. Ans: (d) 1, 2  flank angle errors in deg 1 3 1  11667.07 2.0410

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1 3 2 = 9 = 0.15 2.61810 09. Ans: (d) P = 0.004 Sol: Rollers will not used to measure pitch

De = 30.6651 diameter.  = 60 (metric thread)  p     Best size diameter d =  sec  Virtual correction  2   2  VC = (0.004  cos30) + (0.0131   2   60  = sec  3.5(0.11667 + 0.15) )  2   2  VC = 0.01569 = 1.1547 = 1.155

VED = De + VC = 30.6651 + 0.01569 = 30.6807 10. Ans: (d)    Sol: V.C =  P.cos   0131.0 P(1+ 2) 05. Ans: (a)  2  = 0.2 cos30 = 0.346 06. Ans: (d)

  RR 12 Common data Q 11 & 12 Sol: Sin    2  RRMM 1212  8660.04434.1 11. Ans: (a) =  8660.04434.132.2006.22  p     Sol: Best size diameter, d =  sec   = 59.5566 = 593323  2   2   2   60  = sec  =1.155 mm 07. Ans: 16.433 mm  2   2   p   Sol: De = M d tan   2 2  12. Ans: (a) 2  p   M = 14.701 + (1.155+ tan30) =16.433 Sol: Deff = M – d  tan  2  2 2  = 16.455 – 1.155.tan30 = 14.7226 mm 08. Ans: (d) Sol: Lead = pitchno of starts 13. Ans: 1.732 mm lead 3 Sol: The best wire size = (p/2) sec(/2) Pitch = = = 1.5 mm startsofno 2 = (3/2) sec(60/2) = 1.732 mm

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7.5 Surface Finish Measurement (iv) Ans: (d) 2 2 2  h...... hhh 2 Sol: RMS = 1 2 3 n 01. n (i) Ans: (c)  7551212581055 2222222222 R  Sol: Rt = maximum peak – minimum valley s 10 = 42  18 = 24 m = 7.91  m = 8 m

 Ra < Rs < Rz < Rt (ii) Ans: (d) Sol: (v) Ans: (c)  42403535352525232218 Sol: If R value from 18.75 to 37.5 international Mean  a 10 grade of roughness is given by N11. = 30 m

42 40

35 35 35 02. Ans: (c)

A 1 1000 10 12  5 5 5 Sol: Ra =  30 w HM VM

5 12 5 7  480480 1 1000 8 =  = 0.8 8.0 100 15000 25 25 23 22

18 03. Ans: (d)  7551212581055 CLA = R a  05.0 10 Sol: Rt = =50  m 45tan   m = 7.5 m 04. Ans: (c) (iii) Ans: (b) Sol: Sol: Rz 10 point height method  peaksfivelowestofSumpeaksfivehighestofSum  5 40  )2325182225()3542354035( 50 A = 0.105  m 5 = 14.8 m = 15 m 10

2.5

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Am act = 0.105  0.01 2.5 Conventional Practice Solutions = 0.08 A 01. K= m act 3  04.0)5.210( Sol: Basis of a system: (i) Hole base system 08.0 2 1   = 0.8 (ii) Shaft base system 3  04.0105.2 1000

Hole base system : 05. Ans: (a)  In hole based system the hole is made the constant member, provision being made in 06. Ans: (c) the size of the shaft to accommodate the unavoidable variations (tolerances).

 Various fits are obtained by mating with the 07. Ans: (a) tolerances are so disposed as to produce the

desired fit. 08. Ans: 2  In a shaft based system the position is h reversed and shaft becomes the constant Sol: R a  n member.  016416   All the modern systems in the world are hole 32 based systems. 64  m2 32 Shaft base system:  In the shaft based system the above advantage is not there because we need a no. of reamers and other tools required for obtaining different sized holes.  However the argument in support of shaft based system is that the cost of reamers etc can be treated as a negligible one because the reamers come under the category of consumables.

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 Whatever may be the arguments in favour of In the second case the parts are made at specific each of the two systems the fact remains that source (for example in a particular factory) are majority of the limit system in the world are interchangeable, but such parts are not necessary hole based. interchangeable with the parts made any where  However provision is made for shaft based else. system, so that any user industry can use it because of its own preference for the Selective assembly: system.  In certain component designs the maximum The main advantage of the hole base system permissible tolerance as determined by the is that the no. of reamers and other tools functional requirements may be required is very less, accordingly the cost of considerably smaller than that can be such tools is very much reduced. achieved by the manufacturing process it is desired to use. 02.  For example consider two mating Sol: components having an allowance of 0.005 Interchangeable assembly (Interchangeability): mm and tolerance of 0.01 mm. It is one which can be substituted for a similar part manufactured to the same drawing. 10.01 10.00 9.995 The interchangeability is obviously depends upon 9.985 hole shaft 2 factors. (i) The first factor is that it is necessary for the  Tolerance on hole = 10.01 – 10 = 0.01 mm relevant mating parts to be designed  Tolerance on shaft = 9.995 – 9.985 incorporating specified limits of size. = 0.01 mm (ii) The second factor is that the parts must be  Min clearance = 10 – 9.995 = 0.005 manufactured within the specified limits of  Max clearance = 10.01 – 9.985 = 0.025 size, which must be controlled rigidly.  The max possible clearance allowed is 0.025  In practice there are degrees of mm. interchangeability for instance, universal  These clearances are fixed such that the and local interchangeabilities. assembly functions reasonably well and  In the first case it is assumed that similar therefore they should not be altered. parts derived from any source in the  The sizes of the hole and shaft thus required world are interchangeable. are shown in figure.

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(ii) Medium balls are assembled into medium 10.15 10.03 outer and inner rings (or) large inner and 10.02 large outer rings or small inner and small

10.01 10.05 outer rings. 10.00 (iii) Small balls are assembled into small outer 9.995 and large inner ring. Fig.1

9.985 03. Sol: Tolerance:  The tolerance is the amount by which the  After manufacturing the components are given dimension is permitted to vary. sorted out into 3 groups A, B and C each  It is the difference between the high and low grade of external diameter having the same limits of size. relationship to its correspondingly lettered  It is allowed in order to cover the reasonable grade of internal diameters nearly 0.025 imperfection in workmen-ship and the maximum clearance and 0.005 minimum inevitable inaccuracy of the manufacturing clearance process.  Assembly of external diameters with Tolerance internal diameters of the same grade will thus satisfy the theoretical requirement. H-limit  The method of selective assembly has a great advantage in enabling unsuitable L-limit manufacturing methods or the machines which have high process capability. In actual practice it is impossible to make the  Example of selective assembly: component for accurate dimension because  Variations in the properties of the material Ball bearing: A ball bearing essentially consists being machined introduce errors. of an inner ring, outer ring are separated by steel  The production machines themselves have balls, both the types of rings and balls are graded some inherent inaccuracies built into them and when assembled allow the following  It is impossible for an operator to make conditions perfect settings. (i) Large balls are assembled into large outer

ring and small inner ring

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Allowance: Shaft: UL = 19.998 mm Basic size LL = 19.995 mm

t  002.0 hole  005.0 L-hole H-hole shaft  Shaft can be written as 20

H-shaf

L-shaft L-shaft

(ii) Shaft-basis system:  Allowance is defined the difference between

the maximum material limits of the 0.005 mm components. Hence it is taken as the

minimum clearance between the mating 0.002 mm

parts, and is equal to difference between 0.003 mm L.limit of hole and H.limit of shaft. Hole:  It is also equal to max interference between UL = 20.007 mm the mating parts. LL = 20.002 mm From the above diagram  007.0 Min clearance  Hole can be written as 20 002.0 = L.limit of hole – H.limit of shaft Shaft: Max interference UL = 20.000sssssss = H.limit of shaft – L.limit of hole LL = 19.997 mm

 00.0  003.0 04.  Shaft can be written as 20 Sol: (i) Hole-basis system 05. Sol: Methods of measurements of linear 0.005 mm dimensions:

0.002 mm 1. LINE STANDARDS: 0.003 mm When the length being measured is Hole: expressed as the distance between two lines, UL = 20.005 mm then it is called “Line Standard”. LL = 20 Ex: Measuring scales, Imperial standard  005.0 yard, International prototype meter.  Hole can be written as 20  00.0

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104 ESE – Text Book Solutions

Features of line standards:  End standards are highly accurate and are  The scales can be accurately engraved but it is well suited for measurements of close difficult to take full advantage of this tolerances as small as 0.0005 mm accuracy. For example a steel rule can be  These end standards are subject to wear of used to read upto  0.5 mm of true dimension. their measuring faces  A scale is quick and easy to use over a wide  Groups of slip gauges wrung together to range because only one is required. provide a given size therefore frequently  The scale markings are not subjected to wear wringing leads to damage or inaccuracy.  A scale does not posses a built in datum  End standards have a built in datum because which would allow easy scale alignment with their measuring faces are flat and parallel and the axis of measurement. can be +ve ly located on the datum surface  Subjected to parallax error which are a source  The end standards are not subject to parallax of + ve and –ve reading errors. errors, because their use depends on “feel of  Not commitment for close tolerance on length hand” measurement except in combination with Role of Standards: micrometer. The role of standards is to achieve uniform,

consistent and repeatable measurements 2. END STANDARDS: throughout the world. Today our entire industrial When the length being measured is economy is based on the interchangeability of expressed as the distance between two parts and the method of manufacture. parallel faces, then it is called ‘End To achieve this, a measuring system adequate to standard’. define the features to the accuracy required & the Ex: Slip gauges, Gap gauges, Ends of standards of sufficient accuracy to support the micrometer anvils, etc. measuring system are necessary. Features of End standards:

 The end standards are highly accurate and are 06. well suited for measurement of close Sol: Given: Interference assembly tolerances. Nominal diameter = 20 mm  They are time consuming in use and provide Let tolerance of shaft = x only one dimension at a time. Tolerance of hole = 2x  They are not subjected to the parallax effect Minimum interference = 0.03 mm since their use depends on “feel”. Maximum interference = 0.09 mm

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105 Production Technology

Hole: x UL = 30.012 mm y LL = 30  012.0 2x  Hole can be written as 30  00.0

Shaft:

UL = 29.98 mm 2x + y + x = 0.09 LL = 29.972 mm 3x + y = 0.09 ………. (1)  02.0 y = 0.03 ……….. (2)  Shaft can be written as 30 028.0  3x = 0.06 (ii) Shaft-basis system:  x = 0.02 mm

Hole: 1.5x UL = 20.04 mm

LL = 20 mm 0.02 mm

 04.0 x  Hole can be written as 20 00.0 Shaft: Again x = 0.008 mm UL = 20.09 mm Hole: LL = 20+ 0.04 + 0.03 = 20.07 mm UL = 30.032 mm  09.0 LL = 30.02 mm  07.0  Shaft can be written as 20  032.0  02.0  Hole can be written as 30 07. Shaft: Sol: (i) Hole-basis system UL = 30 mm LL = 29.992 mm  00.0  Shaft can be written as 30 008.0

1.5 x

0.02 mm 08.

Sol: Distinguish between allowance and x tolerance (Refer Q.No 3)

Tolerances are two types 1.5 x + 0.02 + x = 0.04 1. Unilateral tolerances:  x = 0.008 mm

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 When the two limit dimensions are only on Conditions for a force fit or an interference fit one side of the nominal size, (either above between mating parts is or below) the tolerances are said to be Interference fit: unilateral.  In this type of fit, the minimum permitted  Normally used in case of mating parts. diameter of the shaft is larger than the Ex: Shaft and bearing, piston & cylinder etc maximum allowable diameter of the hole,  2.0  2.0  2.0 i.e. when H.limit of hole is smaller than Ex:50 2.0 , 50 0.0 , 50 0.0 B.S L.limit of shaft.

 There always the non mating parts will

B.S assembled only with application force,

H-limit

L-limit called interference fit. H-limit L-limit Unilateral  Here allowance is greater than sum of Tolerance tolerances on hole and shaft. 2. Bilateral tolerances: Ex: Bearing bushes fitted into housings  When the two limit dimensions are above

and below nominal size, (i.e., on either side

of the nominal size) the tolerances are said

to be bilateral. Shaft Hole  Normally applied to non mating parts.

Ex: thickness of flywheel, outer diameter of L Hole and L shaft  S > H flange in flange coupling etc L Hole and H. shaft  S > H  2.0 Ex: 50±0.1, 50 2.0 H Hole and L shaft  S > H H Hole and H shaft  S > H Maximum interference (H. shaft L. hole) Bilateral Tolerance = Difference between maximum material The tolerances on a component can be limits of hole and shaft. arrived by considering 2 factors Minimum interference (L. shaft  H hole) 1. Functional requirement A compromise is need among = Difference between minimum material 2. Cost of production this 2 factors limit of hole and shaft. After arriving at a compromised value, the Hole basis system preferred over a shaft basis nearest tolerance should be adapted. system (Refer Q. No 1)

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107 Production Technology

09. (ii). Peak to valley height (Rt or Rmax): It is the Sol: difference between highest peak and deepest (i) Waviness: Waviness is the longer valley. This measurement may not give the wavelength irregularities upon which true characteristic of any surface because roughness is super imposed. Waviness may due to some sudden jerk peak to valley be induced by vibration, hard spots, height value is high in a very small region imperfect turning of a grinding wheel, otherwise the entire surface is quite all right. chatter, heat treatment etc. Average Roughness:  This is also called ‘Centre Line Average’ These are the irregularities arising because of: (CLA) and denoted by ‘R ’. (a) Inaccuracies in the machine tool. a  vvvvvhhhhh 5432154321 Ex: lack of straightness in the guide ways. R a  10 (b) Deformation of work under cutting force )A .. A A (A A A ..  )A 1 1000 R  321 n  (c) Deformation of work due to its own weight. a L M.H VM (d) These are the irregularities caused by A 1 1000 Ra=  vibration of any kind, for example tool L M.H M.V chatter. Where, L = sample length, A = total area.

 1 L

 Roughness: Roughness is the short Ra =   hdL  L 0 wavelength irregularities arising from the V.M. = vertical magnification, production process which comprise H.M. = Horizontal magnification, individual scratch or tool marks such as that h1, h2, h3 … are max peaks (in mm) produced by a single traverse of a planning v1, v2, v3 …are valley’s (in mm). tool across the surface. Such marks contain within them further small irregularities (iii) Redundant data : which are also included in the definition. Hmax = Rt = Maximum peak valley height The reasons are f a) These are the irregularities caused by  S  CcotCtan e machining itself process parameters. 3.0 H   mm048.0 b) These are the irregularities arising from max  10cot30tan rupture of the material during the separation Hmax = 48 microns. of the chip.

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(iv) To improve surface finish the following 8. Use Different Tools to Rough and Finish: methods can be used: 9. Clear the Chips: 1. Increase the speed: Increasing surface feed 10. Check the Tool holding and Work per minute (SFM) reduces built-up edge holding: (BUE). This will prolong tool life and Chatter caused by improper toolholding and reduce the chance that catastrophic tool fixturing, or by a machine tool that is not failure will damage a finished part. rigid, will create nothing but problems. 2. Reduce the Feed: Reducing the inch per Rigid, stable work holding is also key. As, revolution (IPR) will reduce flank wear and the higher the metal removal rate, the more also prolong insert life. important stable work holding becomes.

3. Increase the top Rake Angle: Rake angle H max (v). The approximate value of Ra is is a variable in the insert’s design that can be 4 tailored to achieve the best surface finish. 48  R a  microns12 4. Use a Chip Breaker: Chip breakers can 4 reduce cutting pressures and produce chips that can be evacuated more easily. In 10. materials that produce long, stringy chips, a Sol: Limits: The maximum and minimum chip breaker can help produce smaller chips permissible sizes within which the actual that exit the cutting zone quickly and easily. size of a component lies are called Limits. 5. Use a Large Nose Radius: There is a direct In any manufacturing company, making a relationship between the size of the insert’s component to the exact size is difficult and nose radius and the surface finish produced. costly. While it’s true that a smaller nose radius Tolerance: decreases the pressure on a tool, it also  The tolerance is the amount by which limits the feed rate that can be used. the given dimension is permitted to 6. Use a Wiper: “By using an insert with a vary. Wiper , you can create a smoother surface in  It is the difference between the high and the milling pass.” low limits of size. 7. Use the Correct Technique: Technique  It is allowed in order to cover the also plays a role in achieving fine surface reasonable imperfection in workmen- finishes, and creating a chip that is thick-to- ship and the inevitable inaccuracy of the thin should be the goal. manufacturing process.

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Tolerance For Holes:  Plug gauge will be employed to check the

H-limit limits of holes.  The plug gauge made to the lower limit of the hole is known as a go plug gauge and L-limit this smaller than the L-limit allowed. In actual practice it is impossible to make the  The plug gauge made to the upper limit is component for accurate dimension because known as NOGO or NOTGO plug gauge  Variations in the properties of the material and will not enter any hole which is not being machined introduce errors. larger than the upper limit of the hole  The production machines themselves have allowed. some inherent inaccuracies built into them NOTE: The length of GO gauge should be at  It is impossible for an operator to make least equal to the length or depth of the part perfect settings. to be inspected.

Fit: Fit is defined as the relationship For Shafts: between hole and shaft during assembly.  Ring gauges or Gap gauges are used. The These are of 3 types: gauge made to the upper limit of the shaft is (i) Clearance fit, called GO gauge and will enter any shaft (ii) Interference fit, which is not larger than the H-limit of shaft. (iii) Transition fit.  The gauge made to the lower limit of the 50 H7g6 shaft is called NOGO gauge and it will not Basic size = 50 enter any shaft which is not smaller than the H = fundamental deviation of hole ‘H’ lower limit allowed. 7 = grade of tolerance IT7 for hole. Generalizing both for holes and shafts it can be g = fundamental deviation of shaft ‘g’ clearly stated that Smallest hole 6 = grade of tolerance IT6 for shaft. Max material (L-limit of hole) Limit Gauges Limit of a Checked by Component GO gauge Largest shaft These are the gauges which are used to (H-limit of shaft)

check the limits of a part. Largest hole These are two types. Min material (H-limit) Checked by Limit of a NOGO gauge 1. GO Gauge , 2. NOGO Gauge Component Smallest shaft (L-limit)

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Advantages: 12.  The inspection is rapid Sol: Surface Profilograph:  Even unskilled labor can use them  In this method a drum is wrapped with a  The gauges are usually robust in construction sensitized paper, on which the reflected and so this does not get damaged easily. image of light has been focused so that the mirror is attached to the tracer and which Disadvantages: follows the surface roughness profile.  A close watch on the wear of gauges is a  By measuring profile on the drum we can must. estimate the surface roughness.  No warning possible trouble until the gauges Average Roughness: start rejecting the components, by which time  This is also called ‘Centre Line Average’ a considerable no. of components would have (CLA) and denoted by ‘Ra’. been manufactured.  vvvvvhhhhh R  5432154321 a 10 11. (A A A ..  )A 1 1000 R  321 n  Sol: Accuracy: It refers nearest to true value a L M.H VM For Ex: If an instrument of accuracy  0.01 A 1 1000 Ra=  mm indicates the dimension of part is 20 L M.H M.V mm, it means that true dimension could lie Where, L = sample length, A = total area. between 19.99 and 20.01 mm.  1 L Ra =   hdL Thus saying accuracy of  0.01 mm means  L 0 that measurement can be inaccurate by V.M. = vertical magnification,  0.01 mm or there is uncertainty about the H.M. = Horizontal magnification, true value to the extent of  0.01 mm. h1, h2, h3 … are max peaks (in mm) Precision: It represents the degree of v1, v2, v3 …are valley’s (in mm). repetitiveness or reproducibility. Resolution: It is the ability of the 13. measurement system to detect and faithfully Sol: Internal diameter measurements: indicate small changes in the characteristics 1. Inside Micrometer: of the measurement result. Measurement of internal diameter using Sensitivity: It is the ratio of change in output inside micro meter is similar to that of the of an instrument to the change in input. outside micrometer. But used for

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111 Production Technology

measurement of plane inside diameter only. D = r1+O1A But not for recessed holes 2

D = 2(r1 + O1A) 2. Two balls of dia ‘d’” and slip: 2 2 O1A = 21  2 AOOO D = 2d + S = 2  hrrHrr 2 21  21 

S 5. Pin method: d d (Dynamic method of measurement of inside diameter) le 3. Two balls of d1, d2 and depth gauge OQ = OP+PQ [  OPA] OP = 2  APOP 2 h2 h1 2 2 O2 =  )2/S(L

O Pivoted O1 d2 d A 1

D = r2 + r1 + O1A D 2 2 O1A = O1 O 2 D2A P 2 = (r + r ) (h +r ) (h +r ) B 1 2 1 1 2 2 A S/2 S/2

th 4. 3 balls of diameter d1, 4 ball of diameter d2 Q and depth gauge:

OP  PQ = AP  PB S h  PBAP PQ = d2 OP H O2 S  S S  S = 2 2 = 2 2 O1 2 OP 2 A L  S 2

D d1 2

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Maximum material limit of a component is (c). Master Gauges: checked by GO gauge. Master gauges are also referred as reference  Dimension of GO gauge is equal to largest gauges. These are used only for checking of rod size i.e. 70.05 mm. other gauges. Due to expenditure involved, Minimum material limit of a component is master gauges are seldom used and gauges checked by NO GO gauge are checked by convectional measuring  Dimension of NO GO gauge is equal to instruments like, comparators, slip gauges, smallest rod size i.e. 69.90 mm. optimizers etc.

14. 15. Sol: According to the purpose: Sol: According to the purpose, the gauges may be classified as: SINE BAR : (a) Workshop Gauges,  Rollers of diameter ”d” are used (b) Inspection Gauges, and  The distance between the centers of the (c) Master Gauges (Reference Gauges) rollers is taken as length of sine bar and it is used for specification of sine bar. (a). Workshop Gauges: Holes to make the Workshop gauges are used by the machine gauge lighter in weight operator to check the dimensions of the components as they are being produced. Theses gauges are designed to keep the size of the component near the centre line of the tolerance. Feature:  The upper surface should be flat (b) Inspection Gauges:  Each roller should be round and of same size Inspection gauges are used by the inspectors  The two axes of rollers must be parallel for the final acceptance of manufactured  Centre distance must be exact components when finished. These gauges  Roller axes must parallel to the upper surface have slightly larger tolerance than the  Size of the sine bar is specified with distance workshop gauges so as to accept component between the centers of rollers slightly nearer the tolerance limit than the

workshop gauges.

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Uses:  To set an angle in the horizontal plane or Chapter vertical plane Advanced Machining Methods 8 Numerical Control (NC) Machines  To check/measure unknown angles  Checking of unknown angles of heavy component 01. Ans: (a) Sol: Pitch of lead screw = 5mm Errors in Setting (or) measuring with sine 1 rev = 5mm Bar 1mm = 1/5 rev As “” increases sin does not increase at 200mm = 1/5  200 = 40rev the same rate, hence for higher levels of “” = 40 360 = 14400 deg. even a small error in “h” or “L” will cause greater errors than at lower values of “” 02. Ans: (b) Sol: Pitch of lead screw = 5mm,

Let h = height of slip gauge combination BLU = 0.005mm L = length of sine bar  Distance travelled /pulse  = setting/measuring angle Length of travel = 9mm No.of pulses = L/BLU = 9 / 0.005  Sin  = h/L, = 1800 pulse.

03. Ans: (b)

Sol: For 1 rev of motor 360 are required  360 pulses are required When motor is rotated by 1 rev  lead screw will rotate by 1 rev When Lead screw is rotated by 1 rev 3.6 mm distance is travelled by axis In total For 360 pulses  360 deg of motor  1 rev of motor 1 rev of lead screw  3.6 mm of linear movement of axis

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360 pulses = 3.6mm 07. Ans: (c) 1 pulse = 3.6/360 = 0.01mm = 10 microns Sol: CNC drill table X axis pulses 04. Ans: (b)

Sol: 10V = 100 rpm Pulse Stepper Driver generator motor = 100  5 = 500 mm/min That is for 500mm/min = 10V 1mm /min = 10/500 3000mm/min = 10  3000 / 500=60 V BLU = the distance traveled by the table for one pulse of electrical energy input to the Common Data 05 & 06 motor. 05. Ans: (b) & 06. Ans: (a) Hence 200 pulse = 1 revolution of motor Sol: A, Stepper motor  200 steps / rev = 1 revolution of lead screw = 4mm That is 1 pulse = 4/200 = 1/50 = 0.02mm,  200 pulses /rev hence BLU does not depends on the Pitch = 4 mm, no. of starts = 1, frequency of pulse generator Gear ratio = N0/Ni= 1/4 = U  When frequency of pulse generator is F = 10000 pulses per min doubled, feed rate of table or tool will 200 pulses  1 rev of motor double but BLU remains same.  1/4 rev of lead screw Correct answer is (c) = 1/4  4  1 mm linear distance. = 1mm linear distance 08. Ans: 20 1 pulse = 1/200 = 0.005mm Sol: p = 5 mm = 5 microns = 1 BLU 1000 pulses  1 rev of motor Feed = BLU  pulse /min  1 rev of lead screw = 0.005  10000 = 50mm/min Velocity of table = 6 m/min For changing BLU = 10 microns = 6000 mm/min = 0.01mm = 100 mm/sec  Gear ratio has to be reduced to 1/2 1000 pulses  1 rev of lead screw  5 mm Feed = BLU  pulse /min 5 pulse1  mm005.0  Pulses per min = feed / BLU 1000 = 50/0.01 = 5000 BLU = 0.005 mm

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Table speed = BLU × Rate of Pulses 13. Ans: (d) 100 Sol: Appropriate answer but the correct answer Rate of pulses = 005.0 is = 20000 pulses/sec N05 X5 Y5 = 20000 Hz = 20 kHz N10 G02 X10 Y10 R5 Because in CNC part program we are not 09. Ans: (c) suppose to indicate information about one Sol: axis more than once in one block. 75 14. Ans: 60 centre 55 Sol: In the combined movement, the tool is (50,55) moving for 50mm with a speed of 100mm/min. whereas in the same time tool 50 70 is traveling x-axis by only 30mm. Hence, 10. Ans: (b) For 50mm 100mm/min 100 For 30mm  min/mm6030 11. Ans: (a) 50 Sol: G02 – circular interpolation clockwise 15. Ans: (a) G03 – circular interpolation counter clockwise Sol: Because diameter of milling cutter is 16mm,

the radius is 8mm. the dotted line indicates 12. Ans: (c) cutter center position, which is shifted by 8 Sol: Because the tool has to travel from P1 to P2 mm all around the rectangular slot in clock wise. (–8,58) (108,58) S R

Y P2 = (10, 15) (0,50) (100,50)  Center (15, 15) 

(0,0) (100,0)

P Q  (–8,–8) (108,–8) P1 = (15, 10) If the given shape is rectangular hole, then X the answer is

(8,8), (92,8), (92,42), (8,42), (8,8)

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16. Ans: (a) 18. Ans: (b) Sol: Y Q (4,5) 19. Ans: (b) Sol: Given coordinates (0,0) to (100, 100) 36.9 33.7 R D P 3.2 (1,3) milling C slot (100, 100) O X 22  32PQ = 3.6055 = PC L 100 PD = PC  cos 3.2 = 3.6 10 x co-ordinate of point C = 1 + 3.6 = 4.6

DC = 3.6 sin 3.2 = 0.2 (0, 0) y co-ordinate of point C = 3.0 – 0.2 = 2.8 100

17. Ans: (a) Depth, d = 2 mm, Sol: “P” after translation = (1+2, 3+3, –5 –4) Diameter, D = 10 mm = (3, 6, –9) L = actual distance travel by tool Rotation about z- axis means 2  100100L 2 = 141.42 mm x   00sincos  x disatnce       Time  y  00cossin y speed        z  0100  z 42.141         60  1   1000  1 min/m50   0010   3  = 169.70  170 sec  0001   6       Correct answer is (b)  0100   9      1000   1  20. Ans: 54.166 mm/sec, 10 micron   0060   6 Sol: f = 500 pulse/rev      0003 3 p = 5 mm,         0900   9 N = 650 rpm       1000   1  5650 (i) v = Np = Final point = [–6, 3, –9] 60 v = 54.166 mm/sec

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Now, 1 min = 650 rev 24. Ans: (d) 650 Sol: G90  specifies absolute input dimensions 1 sec = rev 60 G91  Specifies incremental input 650  500f  dimensions 60 G70  Dimensioning in inch units f = 5416.66 pulse/sec G71  Dimensioning in metric units And, v = B.L.U. f

=  66.5416BLU166.54 25. Ans: (c) B.L.U. = 0.01 mm

B.L.U. = 10 microns 26. Ans: (d)

Sol: M00  Program stop, spindle and coolant 21. Ans: 287 off. Sol:  = 0.9 M01  Optional programmable stop. 0.9 = 1 pulse M06  Tool change 360 360   pulses400pulse M10  Clamp 9.0

 1 revolution = 4 mm pitch = 400 pulses 27. Ans: (d)   2.87 mm = 287 pulses

28. Ans: (b) 22. Ans: 100 pulse, 60 mm/min Sol: G90 specifies absolute input dimensions. Sol: Pulse rate = N  pulse/rev

400 15  sec/pulse100 29. Ans: (d) 60 Sol: The track Numbers on paper tape are: Feed rate = 15 rpm  4 mm/rev Track Number Meaning = 60 mm/min 1 to 4 Alphabets

5 Parity check 23. Ans: (b) 6 to 7 Numerals Sol: M03 or M02: This indicates that program 8 End of block has ended. Be sure to specify MO2 or M30

at the end of the program specify this code 30. Ans: (c) in a block by itself. Never include any other words in the same block. 31. Ans: (a)

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32. Ans: (a) feedback to report the amount of vibration Sol: Post processor statements: These statements back to the CNC and counter act against the are applicable to specific machine tools and vibration. If the body of the machine is are used to define machining parameters modeled properly the feedback makes the like feed, speed, coolant on/off, etc. machine stiffer and allows it to run faster, Ex: COOLNT/ON, RAPID, END, smoother and with better accuracy. SPINDLE/500, FEDRAT/2000. 35. Ans: (b) 33. Ans: (a) Sol: Direct Numerical Control distributed Sol: numerical control is common  If a machine is capable of carrying out manufacturing term for networking CNC many number of operations in one machine tools. machine itself is called as General An FMS, (flexible manufacturing system) is Purpose Machine (GPM). i.e., on GPM it only a part of the CIM concept. An FMS is is possible to carryout turning, drilling, a programmable production system milling, grinding etc operations. consisting of a set of automatic work  If CNC is implemented in GPM it is stations mutually connected by material called as machining centre. At general handling systems and governed by a mostly purpose, CNC machine tool capable of hierarchical control system. carrying many number of operations is called as machining centre. 36. Ans: (c)  To carry out the different types of Sol: In incremental programming the present operations the different types of tools are position of tool can be taken as reference required. All these tools will be stored in a for programming to the next position. tool storage unit called as tool magazine, Hence given reason is incorrect. which can be stored maximum number of tools as 32. 37. Ans: (b) Sol: In the given part program, 34. Ans: (a) G71  Dimensioning in metric units. Sol: The deflection on the machine structure is G91  Specifies incremental input due to large acceleration is mostly unknown dimensions. for the CNC controller. It is possible

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38. Ans: (a) The basic elements and operation of a Sol: Type of NC machine Feature typical NC machine in numerical control PTP  Drilling machine and the components basically involved of Contour Control  Making a spherical end data input, data processing and data output. Closed loop control  Feed back system For data input, numerical information is read 4D control  Two tools can work and stored in the tape reader or in computer simultaneously. memory. In data processing, the programs are read into machine control unit for processing. For data output, this information Conventional Practice Solutions is translated into commands, typically

pulsed commands to the motor. The motor 01. moves the table on which the work piece is Sol: Components of the Numerical Control placed to specified positions, through linear System: or rotary movements, by the motors, ball There are three important components of the screw, and others devices. numerical control or NC system. These are: A NC machine can be controlled through 1) Program of instructions two types of circuits, which is open loop and 2) Controller unit, also called as the closed loop. In the open loop system, the machine control unit (MCU) and signals are given to the motor by the 3) Machine tool processor, but the movements and final All these have been shown in the figure destinations of the worktable are not below and also described in the subsequent accurate. The open loop system cannot sections. accurate, but it still can produce the shape

Parts of the Numerical Control Machine: that is required. The closed loop system is equipped with various transducers, sensors, and counters that measure the position of the table accurately. Through feedback control, the position of the worktable is compared Program against the signal. Table movements Machine Control unit (MCU) terminate when the proper coordinates are Machine Tool Basic Components of Numerical Control System reached. For the close loop system normally servomotor is utilized. For open loop system

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normally the stepper motor is utilized. The paths but at different velocities. Because the closed loop system is more complicated and tool cuts as it travels along the path, accurate more expensive than the open loop. control and synchronization of velocities and movements are important. The Control Controller Process contouring system is used on lathes, milling Input signal Output Fig: Open Loop System machines, grinders, welding machinery and machining centers. Control Controller Process Input signal Output

Measuring Element

Fig: Closed Loop System

There are two basic types of control systems in numerical control, point-to-point and contouring. In point-to-point system, also called positioning, each axis of the machine The Coordinate system and Machine is driven separately by ball screw, Motions: depending on the type of operation, at In order for the part programmer to plan the different velocities. The machine moves sequence of position and movements of the initially at maximum velocity in order to cutting tool relative to the workpiece, it is reduce nonproductive time, but decelerates necessary to establish a standard axis system as the tool reaches its numerically defined by which relative position can be specified. position. Thus in an operation such as Using an NC drill press as an example, the drilling or punching, the positioning and drill spindle is fixed in vertical position and cutting take place sequentially. The time the table is moved and controlled relative to required in the operation is minimized for the spindle. However to make things easier efficiency. Point-to-point systems are used for the programmer we adopt the viewpoint mainly in drilling, punching, and straight that the workpiece is stationary while the milling operations. drill is moved relative to it. According, the In the contouring system, also known as the coordinate system of axes is established with continuous path system, positioning and respect to the machine table. Two axis are cutting operations are both along controlled defined as shown in figure.

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+Z direction. For turning operations two axes b +Y are required to command the movement of c the tool relative to the rotational work piece a the arrangement is illustrated in figure. –X +X – X Machine Table – Z + Z –Y –Z Two axes x and y are defined in the plane of + X Fig: The X and Z-axes in NC turning the table. The z axis is defined in the plane perpendicular to the table and the movement Other features of the Location system. The in the z direction is controlled by the vertical purpose of the coordinate system is to motion of the spindle. The positive and provide a means of locating the tool in negative directions of motion of the cutting relation to the workpiece. Depending on the tools are relative to the table along these NC machine the part programmer may have axes. NC drill presses are classified as either several different options available for two axis or the three axis machines, specifying this location. depending on whether or not they have the The programmer must determine the capability to control the z-axis. A numerical position of the tool relative to the origin control machine and similar machine tools (zero point) of the coordinate system. NC use the axis system similar to the drill machines have either of two methods for system. However in addition to the three specifying the zero point. The first linear axes, these machines may posses the possibility is for the machines to have fixed capacity to control one or more rotational zero. In this case the origin is always located axes. These axes are used to specify the at the same position is for the machine table. angle about the x, y and z axes respectively. Usually that position is the southwest corner To distinguish positive from negative of the table and all the tool locations will be angular motions the right hand rules can be defined by positive x and y coordinates. used. Using the right hand with thumb The second and more common feature on pointing in the direction of positive linear modern NC machines allows the machine axis direction the fingers of the hand are operator to set the zero point at any position curled to point the positive rotational on the machinable. This feature is called

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“Floating Point Zero”. The part programmer 03. is the one who decides where the zero point Sol: Advantages of Numerical Control should be located. The decision is based on machine tools : part programming convenience. For  Greater process capability: Greater accuracy example, the work part may be symmetrical and precision is built into the machine and the zero point should be established at resulting in better quality and a high order of the center of symmetry. repeatability.  Higher production rates: Optimum feeds and 02. speeds can be determined for each operation Sol: NC machine operation: with less time spent in non-cutting functions. In this NC machine  Less lead time: Programs can be prepared in G00 : Point to point movement (Rapid less time than conventional jigs and tranverse) templates, less setup time is required. G01 : Linear interpolation  Flexibility: It is easy to change from one part G90 : Programming in absolute coordinates to another or to change part design.

G94 : Specify feed per minute in drilling  Lowe tooling constants: Expensive jigs and templates may not be needed. More general T01 : Tool No. 1 purpose work holders can be used. F150 : Feed rate = 150 mm/ min  Fewer setups per workpiece: More operations S : Spindle speed can be done at each setup of the workpiece. EOB : End of Block  Reduced inventory: The overall inventory M03 : spindle speed clockwise level may be reduced if parts can be run M02 : End of program machine stop economically in smaller quantities. M05 : Spindle stop  Less setup: Operator errors are reduced substantially. The first part of the machine Then program for the operation can be a good part. Follow-on runs of parts N01 G71 G90 G94 x-100 y-0 previously manufactured can readily be z-100 to F200 S3200 EOB duplicated. N02 G00 x 150 y 150 EOB  Less skill required for the operator: Program N03 G01 Z-115 M03 M07 EOB planning in preparing tapes reduces the N04 G00 Z 100 M09 EOB necessity for operator decisions. N05 G00 X-100 y0-0 z100 M05 EOB

N06 M00 EOB

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Limitations of Numerical Control (NC) : Advantages of CNC Systems over There are some problems inherent in Conventional NC Systems : conventional NC which has motivated tool  Because the computer can be readily and builders seek improvements in the basic NC easily reprogrammed, therefore, the system is system. The difficulties occurred by using very flexible. The machine can manufacture a conventional NC machines are following. part followed by other parts of different  NC manufacturing requires training of designs. personnel both for software as well as  More versatility, Editing and debugging hardware. Part programmers are trained to programs, reprogramming; and plotting and write instructions in desired language for the printing part shapes are simpler. machines on the shop floor.  Program to manufacture a component can be  The cost of NC manufacturing setups could easily called. This saves time and eliminates be several times more than for their errors due to tape reading. conventional counter parts. As NC is a  Greater accuracy. complex and sophisticated technology. It also 04. required higher investments for maintenance Sol: The first basic component of NC system is in terms of wages of highly skilled personnel the Controller Unit. This consists of and expensive spares. electronics and hardware that read and  In preparing a punched tape part interpret the program of instructions and programming mistakes are common. The convert it to mechanical actions of the mistakes can be either syntax or numerical machine tool. This works like human brain. errors and it is not uncommon for three or i.e. it takes the input information from input more passes to be required before the NC tape devices such as tape reader system, feed is correct. back device, manual controls etc. , analyze  Paper tape is especially fragile and the data, taking decisions and these susceptibility to wear and tear causes it to be decisions will be implemented through the un-reliable NC component for repeated use on output device such as drive unit. For the shop floor, more durable tape materials. converting low level language to high level Such as ‘Mylar’ and ‘Aluminium Foil’ are language and vice versa an ALU used to overcome this difficulty. How over (Arithmetic Logic Unit) will be used. these materials are relatively expensive. Table speed, V = p rpm = 5  650 = 3250 mm/min = 3.25 m/min

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5 Cellular manufacturing, which is actually an BLU =  mm01.0 500 application of group technology, has been RPM= 650 = (60f)/N = (60f)/500 = 5416 Hz described as a stepping stone to achieving world class manufacturing status. The 05. objective of cellular manufacturing is to Sol: Cellular manufacturing refers to the use of design cells in such a way that some work cells that specialize in the production measure of performance is optimized. This of families of parts or products made in measure of performance could be medium quantities. Parts (and products) in productivity, cycle time, or some other this quantity range are traditionally made in logistics measure. Measures seen in practice batches, and batch production requires include pieces per man hour, unit cost, on- downtime for setup change over and has time delivery, lead time, defect rates, and high inventory carrying costs. Cellular percentage of parts made cell-complete. manufacturing is based on an approach Cellular Manufacturing: called group technology (GT), which Whether part families have been determined minimizes the disadvantages of batch by intuitive grouping, parts classification production by recognizing that although the and coding or production flow analysis, parts are different, they also possess there are advantages in producing those similarities. When these similarities are parts using GT machine cells rather than a exploited in production, operating traditional process-type machine layout. efficiencies are improved. The improvement When the machines are grouped, the term is typically achieved by organizing the cellular manufacturing is used to describe production around manufacturing cells. this work organization. Cellular Each cell is designed to produce one part manufacturing is an application of group family (or a limited number of part technology in which dissimilar machines or families), thereby following the principle of processes have been aggregated into cells, specialization of operations. The cell each of which is dedicated to the production includes special production equipment and of a part or product family, or a limited custom designed tools and fixtures, so that group of families. Typical objectives in the production of the part families can be cellular manufacturing are similar to those optimized. In effect, each cell becomes a group technology. factory within the factory.

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 To shorted manufacturing lead times by close to one another, distinct from other reducing setup, work-part handling, waiting groups. This grouping is called a cell. These times and batch sizes. cells are used to improve many factors in a  To reduce work-in-process inventory. Smaller manufacturing setting by allowing one- batch sizes and shorter lead times reduce piece flow to occur. work-in-process. An example of one-piece flow would be in  To improve quality. This is accomplished by the production of a metallic case part that allowing each cell to specialize in producing a arrives at the factory from the vendor in similar number of different parts. This separate pieces, requiring assembly. First, reduces process variability. the pieces would be moved from storage to  To simplify production scheduling. The the cell, where they would be welded similarity among parts in the family reduces together, then polished, then coated, and the complexity of production scheduling. finally packaged. Instead of scheduling parts through a All of these steps would be completed in a sequence of machines in a process-type shop single cell, so as to minimize various factors layout, the system simply schedules the parts (called non-value-added processes/steps) through the cell. such as time required to transport materials  To reduce setup times. This is accomplished between steps. Some common formats of by using group tooling (cutting tools, jigs, and single cells are: the U-shape (good for fixtures) that have been designed to process communication and quick movement of the part family, rather than part tooling, which workers), the straight line, or the L-shape. is designed for an individual part. This The number of workers inside these reduces the number of individual tools formations depend on current demand and required as well as the time to change tooling can be modulated to increase or decrease between parts. production. For example, if a cell is normally occupied Cell design: by two workers and demand is doubled, Cells are created in a workplace to facilitate four workers should be placed in the cell. flow. This is accomplished by bringing Similarly, if demand halves, one worker together operations (or machines, or people) will occupy the cell. Since cells have a involved in a processing sequence of a variety of differing equipment, it is products natural flow and grouping them

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therefore a requirement that any employee (ii) Pulse rate = N  pulse/rev is skilled at multiple processes. 400 15  sec/pulse100 60 06. Feed rate = 15 rpm  4 mm/rev Sol: FLEXIBLE MANUFACTURING = 60 mm/min SYSTEM (FMS): • FMS is an integrated approach to 08. automating a production operation. Sol: p = 3 mm • The primary characteristic of an FMS is that  = 1.8  200 steps it is a computer controlled manufacturing 3 B.L.U. = 8.1  mm15 system that ties together automated 360 production machines and materials handling Table speed = B.L.U.  frequency equipment. 10100 3 • The FMS is designed to be flexible so that it   f15 60 can fabricate a variety of different products  sec/pulse11.111f of relatively low volumes. v = Np • Flexible manufacturing systems used to get 100 varieties at small batch and at lowest cost. N  rpm33.33 3 Applications: The best application of an FMS is found in 09(i). the production of small sets of products like Sol: Point to point control: those from a mass production. • In point to point control, the machining is done at specific positions, here work is 07. stationary and tool moves from point to Sol:  = 0.9 point. Ex: drilling holes at different positions on a (i) 0.9 = 1 pulse plate. 360 360   pulses400pulse Contour control: 9.0 • In contour control, there are continuous,  1 revolution = 4 mm pitch = 400 pulses simultaneous and co-ordinate motions of the   2.87 mm = 287 pulses tool and the work piece along diff co- ordinate axes. Ex: contours, curved surfaces etc.

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09(ii). is especially useful for contour turning Sol: p = 3 mm operations and close tolerance work. Today, B.L.U. =0.01 mm automatic checkers and bar machines are f = 200 pulse/sec implemented by CNC. v = B.L.U. f = 0.01  200 = 2 mm/sec Point-to-Point Systems: Point-to-point 0.01 mm = 1 pulse systems are those that move the tool or the 1 workpiece from one point to another and 1 mm = pulse 01.0 then the tool performs the required task. 3 Upon completion, the tool (or workpiece) mm3  pulse300 01.0 moves to the next position and the cycle is  Frequency of pulse = /pulse300 rev repeated (Figure). The simplest example for Now, step angle, this type of system is a drilling machine 360 where the workpiece moves.  rev/pulse In this system, the feed rate and the path of 60 the cutting tool (or workpiece) have no  300 significance on the machining process. The   = 1.2 accuracy of positioning depends on the system's resolution in terms of BLU (basic 10. length unit) which is generally between Sol: The modern form of control is computer “0.001 and 0.0001”.

numerical control (CNC), in which the y Next point workpiece machine tool operations are controlled by a 1 3 4 ‘‘program of instructions’’ consisting of 2 Starting point alpha numeric code. CNC provides a more Holes sophisticated and versatile means of control x than mechanical devices. This has led to the Fig: Cutter path between holes in a point-to-point system development of machine tools capable of Continuous Path Systems (Straight cut and more complex machining cycles and part Contouring systems) These systems provide geometries and a higher level of automated continuous path such that the tool can operation than conventional screw machines perform while the axes are moving, enabling and chucking machines. The CNC lathe is the system to generate angular surfaces, an example of these machines in turning. It two-dimensional curves, or three

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dimensional contours . Example is a milling achieved to the control unit. If there is any machine where such tasks are accomplished diff between the input command and the (Figure). Each axis might move actual position achieved, the drive unit is continuously at a different velocity. Velocity actuated by suitable amplifier from the error error is significant in affecting the positions signal. of the cutter (Figure). It is much more Feed back device gives the feed back to the important in circular contour cutting where MCU for exact control of the axis of NC one axis follows sine function while the machines. This will help in obtaining the other follows cosine function. Figure 6 component dimensions within the limits illustrates point-to-point and continuous path specified. for various machines.

Cutter 11(ii). Cutter radius y Sol: G90 - Absolute Programming Correct Error Cutter path M03 - Clock wise Spindle rotation path Incorrect workpiece Machined path 12. surface x (b) Position error caused by the Sol: f = 500 pulse/rev, (a) Continuous path cutting velocity error p = 5 mm, N = 800 rpm 5800 (i) v = Np = 60

v = 66.67 mm/sec

Now, 1 min = 800 rev

800 1 sec = rev 60

800  500f  60

f = 6666.67 pulse/sec

And, v = B.L.U. f 11(i). 5800 800500 Sol: Feed back transducer is provided to check = BLU 60 60 whether the required lengths of tool travel B.L.U. = 0.01 mm has obtained. The feed back transducer B.L.U. = 10 microns sends the information of the actual position

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13. being in error simultaneously is extremely Sol: Error Detection: low. Electrical noise can create errors in data The CRC technique detects errors by communication. In a telephone conversation, performing calculations on the bits. The the receiver can judge whether the message sender appends the results of he calculation was error-free. In case of an error, the to the message. The receiver carries out the receiver responds: could you repeat that same calculation and compares it with the please? In computer communications with appended results to determine whether the no human reasoning capability, the software transmission was error-free. is designed to detect errors. Flexible Manufacturing System (FMS) There are two techniques for error detection: • FMS is an integrated approach to parity check and cyclic redundancy check automating a production operation. (CRC). In the parity technique, a bit, called • The primary characteristic of an FMS is that a parity bit is appended to the character bits it is a computer controlled manufacturing at the left in the most significant position. system that ties together automated The parity bit is either 0 (zero) or 1. Parity production machines and materials handling check may be odd or even. In the odd equipment. method, the parity bit renders the number of • The FMS is designed to be flexible so that it 1’s in the character to the odd. can fabricate a variety of different products In the even parity method, the number of 1’s of relatively low volumes. in the eight-bit group is even parity method. • Flexible manufacturing systems used to get The number of 1’s in the eight-bit group is varieties at small batch and at lowest cost. even. Error detection based on parity is effective Difference between CNC and DNC: when one bit has suffered transmission  In CNC, remote controlling of operation is error. If two bits are in error, the not possible. transmission error will go undetected. In fact  DNC facilitate the remote control. the parity scheme is able to detect only when  CNC is an integral part of the machine. an odd number of errors have occurred. In  DNC is not integral to machines, DNC general, this methods works well, since the computer can locate at a distance from chance of two or more bits out of eight machines.  CNC transferring machining instruction.

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 DNC manage the information distribution to X code: A word that describes a specific the number of machines. position along the X-axis.  CNC computer control one NC machine. X-axis: The linear axis representing motions  Using DNC programmer can control more and positions that travel the longest distance than one NC machine as required. parallel to the worktable.  CNC have low processing power when Y code: A word that describes a specific compared to DNC position along the Y-axis.  DNC have high processing power when Y-axis: The linear axis representing motions compared to CNC and positions that travel the shortest distance  CNC software is to increase the capability of parallel to the worktable. the particular machine tool. Z code: A word that describes a specific  DNC not only control the equipment but also position along the Z-axis. serve as a part of management information Z-axis: The linear axis that represents system. motions and positions perpendicular to the worktable. The Z-axis is always parallel to 14. the spindle. Sol: Part Programming : • CNC part programming language consists of 15. a software package plus the special rules, Sol: conventions and vocabulary words for using (a) Linear interpolation – G01 that language. (b) Circular interpolation CCW rotation – G03 • It is used to communicate part geometry and (c) Dwell – G04 tool information to the computer, so that the (d) Hold / delay – G05 designed part program can be prepared. (e) Thread cutting – G 33, G 34, G 35 N code: A word that acts as the name or title (f) End of program – M02 for a program block. (g) Spindle on CW rotation – M03 S code: A word that determines the speed (h) Tool change – M06 during a cutting operation. (i) Coolant supply No.1 on – M07, M08 T code: A word that determines which (j) Coolant supply off – M09 specific cutting tool will be selected during a tool change

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07. (i) Ans: (a) , (ii) Ans: (c) Chapter Sol: D = 12mm, t = 50mm, R = 40 , NTM, Jigs and Fixtures 9 C = 20 F, Vs = 220V, Vd = 110V

 V  01. Ans (c) 02. Ans: (d) Cycle time = R.C ln  s   t  VV c  ds  03. Ans: (c) –6 220  Sol: In EDM the mechanism of MR is due to = 40 2010 × ln   110 melting and vaporization associated with   6 cavitation and also erosion & cavitation or  55.0sec10554 milli sec spark erosion and cavitation Average power input = W  E   5.0  CV 2  04. Ans: (d) =    d  tc   tc  Sol: The high thermal conductivity of the tool     material will have high electrical = 218 W = 0.218 kW conductivity hence the heat generated with in the tool is low and what ever heat 08. Ans: (b) generated it will be distributed easily Sol: For Rough machining i.e. stock removal the therefore tool melting rate reduces and tool electrolyte should have high electrical wear reduces. Where as due to specific heat conductivity, called passivity electrolyte, of work material, the rise in temp of W.P is where as for finish machining the faster and more amount of MR is possible. electrolyte should have low electrical conductivity called non–passivity 05. Ans: (b) electrolyte will be used. Sol: Given w = 1 + (20.5) = 2 t =5, f = 20 mm/rev 09. Ans: (b) MRR = wtf = 2.5.20 = 200 mm/min Sol: In ECM MRR  gram atomic weight of material 06. Ans: (a) MRR Current density Sol: As the thermal conductivity of tool material 1 is high the heat dissipation from the tool is MRR betweencetandis workandtool taking place and if the specific heat is high, MRR Thermal conduction of electrolyte. it needs large amount of heat for raising the

temps of tool material up to MP.

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10. Ans: (a) 13. Ans: 51.542 Sol: In ECM 1 0.009 L 50 0.009 MRR  gram atomic weight of material Sol: R 0.02   Current density Area Area Area V  Area5.112 1 I Area333.23    betweencetandis workandtool R  009.050 L = 3 + 6 = 9 m = 0.009  Thermal conduction of electrolyte. AI  Area333.2385.55 MRR   ZF 6  965002107860 11. Ans: (b) = 0.98189  Area Sol: I = 5000 A MRR A = 63, Z = 1, F = 96500  sec/mm8590.0 Area AI  635000 MRR   min/mm608590.0 ZF  965001  sec/g264.3 . = 51.542 mm/min

14. Ans: 680 12. Ans: (a) 15. Ans: (c) Sol: A = 55.85, Z = 2, F = 96540 Sol: EDM, ECM and AJM are used for Specific resistance = 2Ω-cm producing straight holes only but in LBM Voltage = 12V by maneuvering or bending laser gun Inter electrode gap = 0.2 mm slightly it is possible perform the Zig – Zag Resistance hole.  IntercetansisRe.Sp electrode gap R  areaSuface 16. Ans: (b) (Both are Correct)  2.0102   01.0 Sol: In EBM Vacuum is provided to avoid the  2020 dispersion of electrons after the magnetic V 12 I  A1200 lense, but this vacuum is giving an addition R 01.0 function of providing efficient shield to the AI 120085.55 MRR  weld bead. ZF  965402

 sec/g3471.0 17. Ans: (d) Sol: Out of all the NTM’s ECM will give large MRR and EBM will give very small MRR.

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18. Ans: (d) 21. (i) Ans: (d), (ii) Ans: 10.6 mm Sol: Relative motion between tool and work Sol: P piece is not necessary.

19. Ans: (c) O1 O2 Sol: A

A B Q

O1 O2 If D = Dmin = 59.9 4 X1 = distance between center of shaft and 3 A 9.59 2 corner of V – block  583.34  34OO 22 = 5 60sin 21 22 1.60 21  xx5OO X 2  698.34 x = 3.5 2 60sin Block of uniform thickness is preferable Error in depth = 2(X – X ) = 0.223 mm 2 1 because of balanced condition.

20. 22. Sol: Resolving the force “F” into Horizontal Sol:  100sinF ………. (1) (a) Fixed rectangular block and movable V –  200100100cosF …… (2) clamp. )1( 100 tan  )2( 200

1 1  O1 O2 Clamping  tan    565.26 · ·  2  100 F   kg6.223 Sin  Taking the moments about vertical axis 30 1003010030100cosxF  20 30.025  x = 10 mm Positional error = 30.025 – 30 = 0.025 mm

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134 ESE – Text Book Solutions

(b) Fixed V – block and movable rectangular 25. Ans: (b) block Sol:  The main drawback of USM process involves relative high tool wear than MRR, uneconomical nature for soft material X2 X1 Clamping machining and frequent tuning of machine. 30 30.025  EBM requires vacuum.  The main disadvantage of ECM process is that only electrical conductive materials can 30 be machined and high specific energy x  34.64 1 60Sin consumption. 025.30 x  34.66 26. Ans: (c) 2 60Sin Sol: (c) Positional error = x2 – x1 = 0.0298 mm  The main drawback of USM process The positional error is mainly depends on involves relative high tool wear than MRR, the fixed element. So when fixed V – block un economical nature for soft material and marble V – block is used, the positional machining and frequent tuning of machine. error is remains same as (b).  EBM requires vacuum. Out of the 3 cases, case (a) is giving lower  The main disadvantage of ECM process is positional error, hence preferable. that only electrical conductive materials can be machined and high specific energy 23. Ans: (b) consumption. Sol: In USM abrasive slurry is used as medium, 27. Ans: (b) EDM The dielectric fluid kerosene is used Sol: A WJM is not used in electronic industry. as medium, ECM conducting electrolyte is Hence statement (3) is incorrect. used as medium and EBM vacuum is used The applications areas, where water jet as medium for avoiding dispersion of machining and abrasive water jet machining electrons. are: paint removal, cleaning, cutting soft

24. Ans: (c) material, cutting frozen meat, leather Sol: Hardening  Sub zero treatment  industry, mass immunization, surgery, Tempering  EDM peening, cutting, pocket milling, drilling, turning.

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135 Production Technology

28. Ans: (c) 32. Ans: (d) Sol: Turbine blades  ECM process. 33. Ans: (c) Cutting of granite  AJM, USM. Sol: Selection factor:

1. Time taken for Manufacturing: welding 29. Ans: (b) < Assembling using standard parts Sol: ECM gives highest MRR - out of all the < Casting. NTM processes. So statement (1) is 2. Damping qualities: Casting > Welding > incorrect. Assembling using standard parts.

30. Ans: (b) 3. Expandability: Assembling using standard Sol: EDM  Fusion & vaporization. parts > welding > casting. ECM  ION displacement 34. Ans: (b) 31. Ans: (c) Sol: The following location principles are to be ensured during pin location of Cartesian co-ordinate system component: 1. Principle of perpendicular planes 2. Principle of minimum locating pins 3. Principle of extreme positions

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