Production-technology.pdf

Production Technology Solutions for Volume – I _ Classroom Practice Questions 04.

Chapter‐ 1

Ans: (a)

Sol: Q = 1.6 10-3 m3/sec

Metal Casting

A = 800 mm2 Q=AV

01.

Ans: (d)

1.6  10-3 = (800 10-6) V

Sol: Permeability number =

VH PAT

V = 2 m/sec =

2

  2  = 0.203m h =   2  9.81 

For standard specimen H = D = 5.08 cm P = 5 gm/cm2, V=2000 cc, T= 2 min PN =

02.

2000  5.08  50.1 2  2 5   5.08  2 4

= 203 mm 05.

Ans: (c)

Sol: Vol. of casting =

Ans: (c)

Sol: Net buoyancy force = Weight of core –

=

weight of the liquid = V.g (  – d )



 2  d h  g    d  4

AC = Amin = sprue base area

 2  0.12   0.18  9.81  11300  1600  4

=

400  200 mm2 2

G.R.= 1:1.5:2 Pouring time =

Ans: (a)



Volume Sol: Pouring time = A C  Vmax 

  150 2  200 4

ht = 200+ 50 = 250 mm

= 193.6N 03.

 2 D L 4

 3534291 mm3

which is displaced by core



2gh

2  10

6

=

Volume of Casting A C.  Vmax 3534291

200  2  9810  250 17671 2  9810  250

 8 Sec

200  2  10000  175

= 5.34 sec ACE Engineering Publications

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:2: 06.

ME – GATE_ Vol – I_Solutions 07.

Ans: (c)

Sol: The dimension of pouring basin will not

Sol:

affect the pouring time

Ans: (c) 1

1 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 Pouring time = P. T  

V

A2 = 650 mm2

volume. of casting A c  Vmax

353  2 d V 4

Q = flow rate = 6.5 105 mm3/s g = 104 mm/sec2

= 25

6.5  105  1000 mm 2 / Sec V2 = 650

35 3

= 2gh pb  2  10 4  h pb

 2183.6 / d 2 …… (1)  2 d  25 4 To ensure the laminar flow in the gating

hpb = 50 mm = height of molten metal

system Re  2000

ht = total height of molten metal above

in the pouring basin the bottom of the sprue = 200 + 50mm

For limiting condition Re = 2000

Q  A2 V2  A3V3  A3 2  10 4  250

 V d Vd = R e  2000     2000  V

= 6.5  105 mm 3 / s

Vd 

 A3 = 290.7 mm2

2000 2000 0.9 1800   … (2) d d d

Ans: (d)

Sol: dtop = 225 mm

ht = 250 + 100 = 350 mm

From (1) and (2)

Volume flow rate Q = 40×106 mm3/sec

2183.6 1800  d d2 d

08.

Vbottom = 2183.6  1.21mm 1800

= 2620 mm/s Q = Atop×Vtop=Abottom×Vbottom Abottom =

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2  g  ht = 2  9810  350

40  10 6 =15267.17 mm2 2620

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:3: 4  15267.17

dbottom =



Production Technology 2

2

 6 2

2

09. Ans: (b)

 2R   2R  =   1.54  =  a   1.61R 

Sol: A2V2 = A3V3   2252  2  9810  100 4    d 2b  2  9810  350 4



2

 SP  M SP   D 6  D  =    a   Cub  M Cub   a

=139.42 mm

 SP  M SP     cyl  M cyl  2

2 2 D    D  2R  Sp   6    = 1.306 =    D   1.75 R   D  cyl    6

db = 164.5mm

So aspiration will not occur. 12.

Common Data for 10 & 11

2

Ans: 1.205

Sol: Casting – 1 (circular) 10. Ans: (a)

Diameter = 20mm, length = 50mm

11. Ans: (b)

Casting -2 (elliptical)

Sol: 3 castings of spherical, cylindrical and

Major/Minor = 2, length = 50mm, C.S. area of the casting -1 = C.S area of the

cubical

casting -2

Vsp = Vcube 4 3 R  a 3 3 a =R

3

 solidification time of casting  1   solidification time of casting  2   

4  = 1.61 R 3

Vcyl = VSp

2

 V  A c2  M  =  c1  =  c1   M c2   Vc 2  A c1     Vc1 =  d 2  h =  20 2  50 4 4 



 D 2 H   R3 4 3

= 15707.96 mm3



 D 3   R 3 (D  H) 4 3 1

D=

3

16 3  16  3 R    R  1.75R 3 3

 Ac1 = 2   d 2  dh 4     20 2  2    20  50 4  = 3769 mm2

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:4:

C.S area of cylinder = C.S area of ellipse  2  4  20

13.

= 20 – 10 = 10 sec Time  1 2

Latent heat Q

Latent heat = time  Q = 10  10 = 100 kJ

Minor axis = 14.14mm

Latent heat/kg =

Major axis = 2  minor axis = 28.3mm a 2  b2 2

Perimeter = 2

14.

28.3 where a = major axis /2  = 14.14 mm 2 b = minor axis /2 

14.14 = 7.07 mm 2

Perimeter = 70.24 mm Surface area of ellipse = perimeter  length + 2 C.S. area = 70.2450 + 314  2 = 4140 mm2 = AC2 Volume of the ellipse = C.S area  length = 314  50 = 15708 mm3 = Vc2  solidifica tion time of casting  1   solidifica tion time of casting  2   

M  =  c1   M c2 

2

2

 V  Ac2  15707.96  4140  =  c1     15708  3769.9   Vc 2  A c1  = 1.205 ACE Engineering Publications

Q = 10 kW

Time taken for removing latent heat

  2  (min .axis) 2 4

 4  Minor axis =   20 2    2  4

Ans: 50

Sol: m = 2 kg,

   maj.axis  min .axis   4 =

ME – GATE_ Vol – I_Solutions

2

100 = 50 kJ/kg 2

Ans: (a)

Sol: Circular disc casting Squared disc casting C1 ; d  20cm

C2 a  20cm

t  10cm ;

t  10cm

 As     V  C1  1.4 Freezing ratio (F.R) = X1 =  As     V R  As     As   V  C1    1 .4  V R  As   As       V  C2  V  C2 X1    1.4  As   As       V R  V  C1 1.4    As  A       s   0.4    V   V  C1 C2   Volumetric ratio,(V.R) = Y1 =

VR  0 .8 VC

 VR = 0.8 VC1 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata

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Now Y2 

0.8VC1 VR  VC 2 VC2

  0.8  20 2  10  4  = 0.628  20  20  10 15.

Production Technology Common Data for Q.16 & Q. 17 16. Ans: (a)

17. Ans: (a)

Sol: In centrifugal casting

Centrifugal force = FC = ma = m r 2 a = r2

Ans: (b)

75 g =

Sol: VC = 40 × 30 × 0.3 = 360 cc

VSc = shrinkage volume =

75 ×9810 = N 2 D 

3  360  10.8 cc 100

Volume of riser Vr = =

4 2 2

Constant = N 2 D 

75  9810  37273 2 2

 2 d h 4

Constant = N 2 D  37273

 2  4  4  50.24 cc 4

D=

0.5  0.52 = 0.51 m = 510 mm 2

N

37273 37273  = 8.55 RPS D 510

Vr ≥ 3 Vsc  Vr  3  10.8  32.4cc Vr ≥ 3 VSc → Satisfied

r C

18.

where r = time taken for riser material to solidify C = time taken for casting to solidify Mr  Mc

Sol:

Ans: 51.84 mm

R  m R    C  m C  mc 

V V        A s  r  A s  casting V 360  As 240  30  30  0.3  0.3  40  V d 4      = 0.666  As r 6 6 360 =  0.147 2442 

D (2 N)2 2

2

80  120  20 280  120  120  20   80  20

mc = 7.05 mR  

d  side riser given 6

mR  1.5 mC

 d = 51.84 mm

r > C

Hence diameter of riser = 4 cm ACE Engineering Publications

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:6: 20.

ME – GATE_ Vol – I_Solutions

Ans: 0.05 s

Sol: Momentum is considered as constant

Chapter‐ 2

Momentum of water = Momentum of liquid

Welding

metal pressure  time pressure  time  density density 200  0.05 400  time  1000 2000 

time  0.05 s

01.

Ans: (a)

Sol: V0 = 80 V,

IS = 800 A

Let for arc welding V = a+bL For power source, Vp = V0–

V0 I Is

For stable V = Vp  a  b L  V0 

V0 I Is

When L = 5, I = 500  a + b × 5 = 80 –

80  500  30 800

a + 5b = 30 when L = 7, I = 460 a  b  7  80 

80  460  34 800

By solving, b = 2, a = 20  V = a + bL = 20 + 2L 02. Ans: 4860 W, 1.5 mm Sol: For power source,

Vp = 36 –

I 60

Va = 2L + 27 At equilibrium conditions Va = VP 27 + 2 L = 36 –

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I 60

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Production Technology

V0 40 4   I s 50 5

I  36  27  2L  9  2L 60

I = 60 (9 – 2L)

4 V0  140   250 5

If current is 360 Amps 360 = 60 (9 – 2L) 9 – 2L =

= 140 + 200 = 340

360 6 60

V0 4 V  5 340  5   Is  0   425 A Is 5 4 4

2L = 9 – 6 = 3 L=

3  1 .5 2

04.

Ans: 26.7 sec

Sol: Rated Power = Vr Ir = 50 ×103

If L = 1.5 mm,

 Ir 

V = 27 + 2 ×1.5 = 27+ 3 = 30 V I = 60 (9 – 2 ×1.5) = 360 A

Dr = 50% (rated duty cycle)

P = 30 ×360 = 10800 W

If Id = 1500 A (desired current)

If L = 4 mm, V = 27 + 1.5 ×4 = 33 V

Desired duty cycle,

I = 60 (9 – 1.5 ×4) = 180 A

2

I 2 D  2000  Dd = r 2 r     0.5  0.89 Id  1500 

P = 33 × 180 = 5940 W Change in power = 10800 – 5940 = 4860 W

Dd =

If the maximum current capacity is 360A,

Ans: 425

05.

Sol: V = 100 + 40 L ,

Arc on time = 0.8930 Total welding time

= 26.7 sec

the maximum arc length is 1.5mm 03.

50  10 3  2000 A 25

Ans: 27.78 mm/sec

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  0  0.8 – 0.2L = 0 dL

V = 100 + 401 = 140  V0 

V0  250 Is

L = 2, I = 200 V V = 100 + 40  2 = 180  V0  0  200 Is  40  50  ACE Engineering Publications

V0 Is

L

0.8  4 mm 0 .2

P = 4 + 0.8L – 0.1 L2 = 4 + 0.8 ×4 – 0.1 × 42 = 5.6 kW Energy losses = 20% ,  = 80%

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:8:

Area of weld bead (WB)

ME – GATE_ Vol – I_Solutions

l = 1 m =1000 mm;

t = 30 mm,

1 = 2   AB  AC 2

d = 4 mm,

= 5 tan 30 × 5 = 14.43

Lt = 450 mm; LS = 50 mm,

5 tan 30 A B 300

A1 = 4  30 = 120 mm2

600

A2 = A3 =

5

1  30 tan 30  30 = 259.8 mm2 2

Total volume of weld bead

C

= volume of weld bead + crowning

Volume of W.B = 14.43 × 1000 3

= 14433 mm

-6

Weight of W.B = 14433 × 10 × 8 = 115.5 g Heat required for melting of W.B = 115.5 ×1400 = 161.66 kW Time for welding =

161.66  36 Sec 0.8  5.6

1000 Welding speed = 36

 27.78 mm/sec

= 1.1  volume of weld bead = 1.1 (A1+2A2)1000 = 703560 mm3 Volume /Electrode = 

  D2  Le 4

 2  4  (450  50) = 1600 4

No of electrodes required 

Total volume of weld bead volume / Electrode



703560  139.96  140 1600

x = 200mm (given) Common data for 06, 07 & 08. 06.

Ans: (d)

08.

Ans: (c)

07. Ans: (d)

No of electrodes/pass = No of passes =

1000 5 200

140  28 5

Total Arc on time

Sol: 30mm

(2) (1) (3) 30o 30o

=

1000  28  280 minutes 100

Total weld time =

280  466.67 minutes 0.6

4mm

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:9: 09.

Ans: 0.64 mm & 2.1 mm

Production Technology 11.

Sol: Given AC = 10 mm,

Ans: (b)

Sol: Filling rate of weld bead = filled rate by

O1A = O1C = 7 mm,

electrode

O2A = O2C = 20 mm

Area of W.B × Speed =

O2

A

B

r=20

E

4

d2  f

  1.2 2  4000 4  25.12 mm2 Area of W.B= 180

r=7

D



C

Common data for 12 & 13

O1

12.

Ans: 2000 J

Sol: H.G = I2 R 

= (10000)2×200×10-6×

Height of Bead = BD = O1D – O1B = O1D–

O1 A2  AB 2

= 20 – 20  5 2

2

13.

5  2000 J 50

Ans: 1264 J

Sol: h = 2t – 2 × 0.1 t = 1.8 t

= 1.8 ×1.5 = 2.7 mm

= 0.64 mm

D = 6 t  6 1.5 = 7.35 mm Depth of Penetration = BE = O1E–O1B =  O1 E  

 O2 A    AB  2

= 7  7 2  52

2

= 2.10 mm

0.1 t h

0.1 t

Common Data Q. No 10 and 11 10.

Ans: (c)

Sol: I = 200,

V = 25,

speed = 18 cm /min

D = 1.2 mm, f = 4 m /min,  = 65%, Heat input = 

VIη speed 25  200  0.65  60 18

Vol. of nugget = =

 4

D2h

 7.352  2.7 = 114.5 mm2 4

Heat required = Volume × ×heat required /g = 114.5  10 3  8  1380 = 1264 J

= 10.83 kJ / cm ACE Engineering Publications

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: 10 : 14.

Ans: 2.3 & 4.6 MJ    Sol: RC = 0.85    nr 

17.

Ans: (b)

Sol: Heat dissipated = 360 – 344 = 16 J

 = Resistivity of metal

18.

R C1 

0.85  2  10 5 = 1.082105 25    0.02

R C2 

0.85  2  10 5 = 5.41  106 50    0.02

(i) Ans: (a), (ii) Ans: (b)

Sol: P = 2 kW = 2  103 Watt,

2

V2 V (Heat generation)1 = I2R =    R = R R

(H.g)1 =

ME – GATE_ Vol – I_Solutions

V = 200 mm/min, L = 300 mm Heat required (HR) = 40 Kcal = 401034.2 Joule Welding time =

300  1.5 min  1.5  60 200 = 90sec

52 = 2310546.04 1.082  10 5

Heat input = 2 103  90 Joule

52 (H.g)2 = = 4621072.08 5.41 10 6

HI =

HR 40  10 3  4.2   0.9333 HI 2  10 3  90 = 93.33%

15.

Ans: (c)

19.

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)

I 2  200  10 6  0.1

= (a×h)  (Cp (Tm–Tr)+LH)

 2  0.005  1.5  10 3  8000  1400  103 4

= 0.05 × 10-6 × h × 2700 [896  (933 – 303) + 398 × 103]

I = 4060 A

 h = 0.00385 m = 3.85 mm

Common Data Q. 16 & Q. 17 16.

20.

Ans: (c)

Sol: I = 3000 A,

Ans: (d)

Ans: (c)

 Sol: Volume to be melted = (110 2  100 2 )  2 4

 = 0.2, R = 200 Ω

Volume of nugget = 20 mm3

 3298.66 mm3

2

Heat generation = I R = 30002 ×200×10-6 ×0.2 = 360 J



Heat required = V c p Tm  Tr   LH -9



= 800020×10 ×500(1520 –20)+1400×10 = 344 J ACE Engineering Publications

Total heat required = 3298.66 × 10-9 × 64.4 ×106 = 212.4 Joules

3

P = VI = V 

V V 2 30 2    21.43 42 R R

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: 11 :

Total heat required = heat to be generated 212.4 = Pt t=

21.

212.4  10 sec 21.43

23. Ans: 61.53 % Sol: Thermal efficiency =

Ans: (a)

= 200 

 4

 10  0.5  7854 2

Heat required  100 Heat supplied

Heat required = 10 × 80 = 800 J thermal =

Sol: Frictional force F = Pressure × Area ×

800  100 = 61.53 % 1300

24. Ans: 464.758 A Sol: Dd = 100% = 1, Ir = 600A, Dr = 0.6

3 Torque = F   Radius 4

D d I 2r  D r I d2

3 Torque = 7854   5  10 3  29.45 4

1 600 2  2  I d2  600 2  0.6 0.6 Id

Power, P = 

22.

Production Technology

2NT 60000 2  4000  29.45 = 12.33 kW 60000

 Id = 464.758A 25.

Ans: 17

Sol: Number of electrodes



Ans: 0.065 sec

Sol: Given:



3

Volume = 80 mm , Current (I) = 10000 A, E = 10 J/mm3, Qlost = Heat lost = 500 J, R = 0.0002 ohms Total energy supplied during process = [(80 × 10) + 500] J Qtotal = 1300J = i2Rt 1300 = (104)2 × 0.0002 × t  t = 0.065 seconds

Total volume of metal deposited Volume deposited from one electrode Total Volume of metal deposited  2 3  450  50  4

 

17mm 30o

2mm

x tan 30  17 mm

30o

19mm

2

o

x = 9.814 mm 1  Area    9.814  17  2  2  19  1.1  1.15 2  Volume  204.85mm 2  1.1  1.15  180 = 46645.30767mm3  Number of electrodes = 17

ACE Engineering Publications

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: 12 :

ME – GATE_ Vol – I_Solutions

cosCS = 1  Cs = 0

Chapter‐ 3

 = 90 – CS = 90

Metal Cutting

 tan  b   sin  cos    tan i   tan     cos  sin    tan      s 

Common Data for Q. 01 & 02 01.

Ans: (a)

02.

Ans: (d)

tanb = sin tani+costan tanb = sin90 tani + 0  b = i = 10 Common Data for Q.04, 05 & 06

Sol: Vf

04. Ans: (c)

Vs

06.



90

Ans: (d)

Sol: d = t1 = 2 mm,

Vc

VC  0.5m / s,

Vc = 40 m/min; Vf = 20 m/min  = 10o;

r

Vf  0 .5 Vc

=0

FC  1200,

FT  800,   300

    tan 1

800  33.690 1200

Power = P = FC  VC  1200 

 0.5 cos10  o  tan 1    28.33 1 0 . 5 sin 10   

60 60

= 1200 W Length of shear plane = LS

V Vs  f  cos  sin 

=

20  cos10 = 41.5 m/min sin 28.33 07.

t1 2   4mm sin  sin 30

Ans: (a)

Sol: For theoretically minimum possible shear

03. Ans: 10

strain to occur

Sol: f = 0.25 mm/rev,

t1 = 0.25,

w = b = 15 mm

  tan   tan 33.69  0.67

 r cos     tan 1    1  r sin  



05. Ans: (b)

i = 10,  = ?

t1 = f cosCS



0.25 = 0.25 cosCS ACE Engineering Publications

2    90 90   90  6   48 o 2 2

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: 13 :

Production Technology

Common Data for Q. 08 & 09 08. 09.

Ans: (c) Ans: (c)

Sol:  = 6 ,

VC  1 m / s

b = w = 3,

r

t1 L 2 60 60     0.59 t 2 L1 D 0   32

r

t1 t 0.1  t2  1   0.169 t2 r 0.59

 0.59 cos 35   = 36.150   tan 1   1  0.59 sin 35 

d = t1 = 1 mm use 2      90 o

t2 = 1.5 mm;

r

tan     

1 2 t1    0.67 t2 1.5 3

FT 80  FC 200

 80      tan 1    200 

 0.67 cos 6    35.62 0   tan 1   1  0.67 sin 6 

 35  21.8  56.8o

For minimum energy condition use

  tan   tan 56.8  1.52

2 +   = 90

(In general  < 1)

  90    2  90  6  2  35.62

Hence

  tan   tan 24.76  0.461  40.2 m / min Area of shear plane = As = Ls × b



t b 1 3  5.2 mm 2 = = 1 sin  sin 35.62

Vf VC

Common Data for Q. 10 & 11

11.

Ans: (d)

friction

0.5276  0.55 1.04  r  V f  rVc = 0.59 ×10 = 5.9 m/min

Vs 

Vf 5.9 cos    cos 35 sin  sin 36.15  8.42 m / min

Sol: D 0  32 mm,

FC = 200 N,

classical

1  1  ln  ln  r 0.59           35  2 2 180

Vf  rv c  0.67 1  60

Ans: (d)

applying

theorem

= 24.76

10.

by

 = 35, K1 = 0.1 mm, VC = 10 m/min,

L2 = 60 mm, ACE Engineering Publications

FT = 80 N

12.

Ans: 56.23

Sol:  = 10,

t1 = 0.125,

Fc = 517 N;

FT = 217 N

t2 = 0.43;

Cm = 2 +  – 

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: 14 : r

14.

t 1 0.125   0.29 t2 0.43

tan (  ) =

Cm = 2  16.73 + 32.77 – 10 = 56.23 Ans: 272 N & 436 W

Sol: S0 = 0.12 mm = t1, a 2  t 2  0.22

Fs =

15.

Ans: (a)

16.

Ans: (b)

Sol: D = 100 mm, f = 0.25 mm/sec,

d = 4 mm V = 90 m/min

 S0 .t.S  sec   tan   1

FC = 1500 N

 S  400

FT

FC = N = 1500

FC

F

N

FT = F

t = 2 – 0,



1200  cos(30  33.69)  639.23N cos 33.69

Common Data for Q. 15 & 16

Major cutting for, b = pz = Fc S0 = 0.12,

FT FC

 800      = tan 1   = 33.69 1200  

  

 217   10o  tan 1   = 32.77  517 

t = 2.0 mm,

FC cos      cos   

Fs =

 0.29 cos10   tan 1   = 16.73  1  0.29 sin 10 

13.

Ans: (d)

Sol:  = 30, FT = 800 N, Fc = 1200 N

 r cos     tan 1    1  r sin  

F     tan 1  T  FC

ME – GATE_ Vol – I_Solutions

t2 a2 0.22    1.83 t1 S0 0.12

Common Data for Q. 17 & 18

=0 Pz = 0.12×2.0×400(1.83sec0–Tan0+1)

Power = p = FC  VC  p Z 

Vf r

52.6  p Z  Vf    271 1.83 60

ACE Engineering Publications

Ans: (b)

&

18. Ans: (b)

Sol: VT a f b d c  K

= 272 N

= 436 W

17.

a = 0, 3 f2 

f1 , 2

b = 0, 3,

c = 0, 15

d 2  2d

T1  T2  60

V1T1a f1b d1c  V2T2a f 2b d 2c

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: 15 :

V2  f1    V1  f 2 

b

 d1     d2 

1 = 2 0.3   2 V2 1.11 V1

Production Technology 20.

c

Ans: (b)

Sol: Let Q = no. of parts produced

T.C on E.L = T. C on T.L

0.15

 1.11

30 60  Q  80  500   Q  160 60 60

40Q  500  16Q

V V % change in speed = 2 1  11% V1

40 Q  16Q  24Q  500

Productivity is proportional to MRR

 Q=

% change in productivity =

MRR 2  MRR 1 MRR 1

21.

f d V f d V = 2 2 2 1 1 1 = 11 % f 1d 1 V1

C1 = 1.1 × 130 = 143, V = V1 = 90 m/min 1

 130  0.12 VT n  C  T    21.4 min   90 

T0 , V0 = original tool life and velocity If V1  1.2V0

T1  0.5 T0

V2  0.9V0 ,

 143  V T C T    90 

T2  ?

1

V T  V0 T n 1 1

Ans: (a)

Sol: n = 0.12, C = 130

19. Ans: 49.2 % Sol:

n 0

 T1  V    0 V1  T0  V  ln  0  ln  1  V 1.2  n  1   0.263  T1  ln (0.5) ln    T0 

1n

V  T2  T0  0   V2 

n

 V   T0  0   0.9V0 

1 0.263

22.

= 1.4927T0

T2  T0 1.4927T0  T0   0.4927 T0 T0

1

0.12

 47.4 min

Ans: (a)

Sol: Tool life = T1 

T2  V1  50rpm ,

% change in tool life

ACE Engineering Publications

1

Increase in tool life = 47.4 – 21.4 = 26 min

V0 T0n  V2 T2n 1

1

Increased tool life = 47.4 min

n

=

500  20.83  21 24

500  50, 10 122  12.2, 10 V2  80 rpm

The feed and depth of are same in both cases V1T1n  V2 T2n

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: 16 : V2 80 ln V1 50  0.47  0.333  n  50 T 1.41 ln ln 1 12.2 T2

 18C  V2

V T  V3 T

n 3

V   T3  T1  1   V3 

1

n

1

 50  0.333  29  50   60 

270C V

1   2  

270C V  n

1

Ans: 30.8 m/min

Sol: TC = 3min,

  1 2    0.25 

Lm = Rs. 0.5/min

18  150 4 V  270  3 4

Depreciation of tool regrind = Rs 0.5 C = 60, n = 0.2

 V = 57.91 m/min

Cg = 3  3  0.5 + 0.5 = 3.5 n

 0.2 0.5   60  . 1  0.2 3.5  24.

25.

t1= f.sin = 0.15 sin75 = 0.144

0.2

= 30.8 m/min

t2 = 0.36, r 

t1  0.402 t2

chip reduction coefficient = t2/t1

18C 270C Sol: C m  , Ct  , VT 0.5  150 V TV TC = k + Cm + Ct

k

18C 270C  V TV

k

18C  V



1  K  2.5 r  r cos     tan 1    1  r sin    0.402 cos10   = 23.18o  tan 1   1  0.402 sin 10 

270C 1

 C n V  V

18C 270C V  1 V Cn

ACE Engineering Publications

Ans: 2.5 & 23

Sol:  =10

Ans: 57.91

k

 1    1  0.25   18C 1 V2 n

18 270  3  V2  2 4 V 150

Tg = 3 min,

 n Lm  VOpt = C  .  1  n C g 

1     1 n  0

Cn

C

23.

dTC  0 dV

For min TC,

ln

n 1 1

ME – GATE_ Vol – I_Solutions

1   1  n 

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: 17 : 27. Sol:

Production Technology

Ans: 0.944

Chapter‐ 4

T = 60 min VA 

67

600.11

Machining

 42.70 m / min

01.

77  45.22 m / min VB = 600.13

Sol: Time / cut =

Under similar conditions with same tool life

V

cutting velocity on material B is greater than

the

material

A.

Ans: (i) 20 min, (ii) 50 min

Hence

the

L 576   20 min fN 0.2  144

DN  100 144   45.2 m / min 1000 1000

 75  VT 0.75  75  T    V

machinability of material ‘B’ is higher than the material ‘A’.

1.333

 75  =   45.2 

VA 42.7   0.944 VB 45.22

 1.96 min

No. of tool changes = 28. Ans: 12 Sol: Given,

20  1  9.2  10 1.96

(Because 1 tool is already mounted on W.P) t1 = 0.2 mm,

Total change time / piece = 20 + 10 × 3

w = 2.5 mm, Fc = 1177 N, Ft = 560 N As the cutting is approximated to be orthogonal. tani = cos tan b – sin  tan s tan 0 = cos tanb – sin tan s = cos30 tan7 – sin30 tan s 

1 0.75

s = 12

ACE Engineering Publications

= 50 min 02.

Ans: (a)

Sol: For producing RH threads the direction of rotation of job and lead screw must be in the same direction, for this if the designed gear train is simple gear train use 1, 3, 5 odd number idle gear to get same direction of rotation, if the designed gear train is compound gear train use 0, 2, 4,.. even number of idle gears to get same direction. In the given problem the designed gear train is a compound gear train, to change the hand of the thread it requires to change the direction of rotation of job and lead screw for this use 1, 3, 5… odd number of idle gears.

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: 18 : 03.

Ans: (b)

07.

Sol: Train value = Gear ratio =



3.175  40 127  not possible  6  40 240

=

127 1  20  40 6  20



127 20  possible  40 120

=

f = 1 mm/stroke, M=

4. Thread cutting

VC = 1 m /sec,

1 2

Time per two pieces =

= P job PL.S

N follower N driver



=

Tdriver P  driver Tfollower Pfollower



Pspindle PLs.S



N L.S N Spindle

Time/piece =

09.

B 1  1  M  f V

310 2000 1  0.5 = 930 sec  1 1000 930  465 sec 2

Ans: (d)

Sol: Shaping operation

P = pitch N L.S

Ans: (b)

= 50 + 900 + 50 + 50 + 900 + 50

2. Taper turning

Sol: Gear Ratio = Train value =

N Spindle

300 1    100 min 0.3 10

Sol: L = 2m

Ans: (d)

G.R =

B 1  f No. of D.S

Time/cut =

08.

f = 0.3 mm /stroke

B = 300 + 5 + 5 = 310

3. under cutting

06.

B = 300 min,

Ans: (c)

Sol: 1. Plane turning

05.

Ans: (b)

Sol: No. of D.S/min = 10

pitch of job threads pitch of lead screw threads

=

04.

N follower N Driver

ME – GATE_ Vol – I_Solutions

PL.S PSpindle / job

=

6 3 = 2 2 2

Ans: (d)

M = 0.6 ,

Double stroke / time = 15 N = time / D.S = 1/15 Average speed, V =

Sol: With this any change in UV will also

changes the speed of lead screw, the pitch of the threads produced depends on the speed of work and speed of lead screw. Us will not affect the speed of the work ACE Engineering Publications

L = 500 mm



L 1  M  V

500 1  0.6 =12000 mm / min 1    15 

= 12 m / min

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: 19 : 10.

Ans: (c)

Production Technology 12.

Sol: Total depth to be removed = 30 – 27

Sol: D = 15 mm, Vc = 20 m/min,

= 3 mm

N

2 = 0.67 3

Given, m =

1000 V 1000  20 = = 425 rpm  D  15

N = 425rpm

feed = 0.5,

f = 0.2 mm /rev

depth = 2 V = 60 m/min

Time for idle time = 20s

Approach  50 m   length wise Over time  50 min 

Tool change time = 300 s Time/hole =

Approach  5 m   width wise Over time  5 m  Time/cut =

L   0.5D  fN fN

15 2  0.617 min  0.2  425 45 

L 1  M   B f V

= Tm = machining time

l = 800, i)

L = 800 + 50 + 50 = 900

No. of holes produced / drill

B = 400 + 5 + 5 = 410 Time / cut =

900  1  60000 

= 2  410  3  0.5

ii)

No. of cuts =

3  1.5  2cuts 2

Sol: Time per hole = L/f.N

= 25/(0.25 300) = 1/3 min = 20sec.

Total time/hole

= 0.617 

13.

Ans: (b)

14.

Ans: (b)

Sol: Given n = 6,

Dmax = 25 mm Dmin = 6.25 mm

Because dia of drill bit was not given, hence

V = 18 m/min

AP1 is zero. r=

ACE Engineering Publications

20 300  60 162  60

= 0.9812 min = 58.87 = 59 sec

Total time = 20.5×2 = 41 mins Ans: (b)

100  162 0.617

= Tm + idle time + Tool change time

= 20.5 min

11.

l  45 mm

T = 100 min,

n 1

N max N min

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: 20 :

Nmax =

1000V 1000  18    6.25 D min

Time/cut =

Nmin =

1000V 1000  18    25 D max



r=

N max = N min

6 1

25 6.25

5

ii)

AP1 =

No. of teeth = 30 (Not required) Module = 3 mm





1 100  100 2  90 2 2

L = 200 + 28.2 + 5 + 5 = 238.2

Radial depth= Addendum+1m+1.25m

Time/cut =

= 2.25 module = 2.25  3 Radial depth = 6.75 mm

 16. Sol: Part size = 200 × 80 × 60 mm

D = 100 mm,

17.

Z = 12,

238.2  1.25 min 0.1 12 159

Ans: (b)

1000 V 000  50   159 rpm D  100

=

f t  0.1 mm , AP = OR = 5 mm 1 D  D2  w 2 2



100 2  80 2

  20 mm

L = l  AP1  AP  OR

40 No. of teeths 40 28

3  12  = 1  = 1 7  28 

With symmetrical milling

 1 = 100  2

L f t Nz

Sol: Crank rotation =

V = 50 m/min,



= 28.2 mm

Pressure angle = 200 (Not required)

AP1 



1 D  D 2  w i2 2

= 80 + 2 × 5 = 90

Sol: Hobbing process

i)

230  1.2 min 0.1159 12

Where, wi = w+ 2(Of)

Ans: (d)

N=

L f t NZ

If offset = 5mm with asymmetrical milling AP1 =

= 1.3195 = 1.32 15.

ME – GATE_ Vol – I_Solutions

9 = 1   21  1 complete revolution and 9 holes in 21 hole circle.

= 200 + 20 + 5 + 5 = 230 ACE Engineering Publications

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: 21 : 18.

Production Technology

Ans: (d)

Sol: d = 70 mm ,

Z = 12 teeth O1

V = 22 m/min ft = 0.05 mm/tooth

A

1000 V fm = ftZN, N  d

B

t

l

Ans: (b)

Sol: Crank rotation = 1

10 1 4  1   360 30 3 3

N=

= 480 Job rotation =

20.

d

AP1

1000  22 = 60 mm/min fm = 0.05  12  3.14  70 19.

O2

1000V 1000  120  = 254.64 rpm   150 D

Approach = AP1 + O1O2 =

CR 480 = = 12 40 40

d D  d 

0.5150  0.5 = 8.645 mm

=

Total time/machining = No. of cutsTime/cut

Ans: (b)

Sol: Given,

No. cuts =

Total depth 2 =4  depth per cut 0.5

Time/cut =

L   AP = fN fN

Dtool = 15 cm = 150 mm Feed = 0.08 mm/rev Depthmax = 0.5 mm = d Length of workpiece, l = 200 mm

200  8.645 = 10.227 min 0.08  255

=

Cutting Velocity, V = 120 m/min

Total time = 10.227  4 = 40.91

Total depth to be cut = 2 mm

= 41 min 21.

Ans: 8.05 min

Sol: Broaching machine

P = 1.5 kW d1 = 20 mm enlarged to df = 26 mm t = 25 mm p = 10 mm/tooth h = 0.075 mm/tooth ACE Engineering Publications

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: 22 :

V = 0.5 m/min

nr 

Equation for time for broaching operation =

ME – GATE_ Vol – I_Solutions

Length of tool travel Linear velocity of tool

d r 4.4   44 teeth h r 0.1

Cutting length = effective length = Le = L r  LS  L f

Length of tool travel = L

= 44 × 22 + 8 × 20 + 4 × 20

= t + Le + AP + OR

= 1208mm

As (AP + OR) is not given so take it zero Le = effective length or cutting length Depth of cut d =

26  20 =3 2

23.

Ans: (b)

Sol: Out of all conventional method grinding is

one which required largest specific cutting

n = no. of teeth = d/h = 3 / 0.075 = 40

energy.

Le = np = 40  10 = 400mm

1) Because of random orientation of

Le = 400 mm

abrasive particles, rubbing energy losses

Time for broaching = =

t  Le V

will be very high 2) Lower penetration of abrasive particle

25  400 = 8.05 min 0.5  100

3) Size effect of the larger contact areas between wheel and work.

Time for broaching = 8.05 min 24. Ans: (a) 22. Ans: (c)

Sol: Common alignment test for shaper and lathe

Sol:

are (1) Straightness (2) Flatness. Runout is used in lathe.

4.5 mm

Parallelism used in shaper. 25.

Ans: (a)

Sol: The

d total  4.5 mm

curvature

given

is

the

concave

curvature hence it increases the stress

df  0

concentration factor therefore it is used for

d S  n s  h s  0.0125  8  0.1

supply

d r  d total   d f  d s 

of

lubricating

oil

to

bearing

mounting

= 4.5 – 0.1 = 4.4 ACE Engineering Publications

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: 23 : 26.

Production Technology

Ans: 18

Sol: The output per annum = 800  52

Chapter‐ 5

= 41600 units.

Metal Forming Process

The rejection rate is 20%.  The quantity to be produced (including rejection) = 

Re quiredoutput (1  rejection rate)

01.

Ans: (a)

Sol:  y  1400 0.33

41600 = 52,000 units (1  0.2)

At maximum load, true strain 

Total time required for turning

1  y  1400  3

= 52,000  40/60

1 3

0.33

 971 MPa

= 34666.6 hours Production time required with 80 per cent

02.

efficiency = 34666.6 /0.8 = 43333.3 hours

Sol: A0p = C.S area of P originally

Time available per lathe per annum

A1p = C.S area of P after 1st reduction

= 48  52 = 2496 hrs

= 0.7 A0p

 Number of lathes required =

Time required (hrs) Time available hrs 

=

43333.33 = 17.36 = 18 2496

Ans: (b)

A2p = 0.8 × 0.7 × A0p = 0.56 A0p  Aop  p  True strain in " P"  ln A  2p  Aop  ln  0.56 A op 

 No. of lathes required = 18

   

  = 0.58  

A0Q = C.S area of Q originally A1Q = C.S area of Q after 1st reduction = 0.5 A0Q  A0Q Q  ln A  1Q 03.

   ln 1  = 0.693   0.5  

Ans: (a)

Sol: do = 25,

di = 5mm

 y  315 0.54 ACE Engineering Publications

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: 24 :

d  A   n o   n  o  A1  di 

2

2

 25    n    3.22  5   y  315 (3.22) 0.54  592 MPa.

06. Ans: (a) Sol: Ho = 20 mm,

b = 100 mm H1=18 mm, R = 250 mm, N  10 rpm,

H0 = 4.5 mm

H = 0.089 R



H1 = 2.5 mm

L  length of deformation zone  RH

H = 2 Droll = 350, Rroller = 175 mm Strip wide = 450 mm = b Average coefficient of friction = 0.1 y = 180 MPa RSF = Pavg  projected area =

2  L    y 1  bL 3  4H  RH = 175  2 = 18.7

L= 4= 

H 0  H1 4.5  2.5  = 3.5 2 2  0.1  18.7   1801    450  18.7 4  3.5  3 

2

RSF = 1982.64 kN = 1.98 MN

 250  2  22.36 mm H

20  18  19 2

Favg  R.S.F  

 L   y b  L 1    4H  3

2

 0.089  22.36   300  100  22.36 1   4  19 3 

2

= 795 kN. T  Favg  a , Where a = moment arm   L

 0.3L to 0.4  L T  Favg  0.4L  795  10 3  0.4  22.36

05. Ans: (a) Sol: Ho = 4,

 y  300 MPa

H  20  18  2mm

04. Ans: 1.98 MN Sol: Given:

ME – GATE_ Vol – I_Solutions

= 7110 kN-mm

H1 = 3mm, R = 150mm,

= 7.11 kN-m

N = 100 rpm. Velocity of strip at neutral point = Surface Velocity of rollers 

  300  100 DN  1000  60 1000  60

Pav 

2 NT 2  10  7.110  60 60  7.44 kW / roller

Total Power = 7.44 × 2 = 14.88 kW

= 1.57 m/sec ACE Engineering Publications

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: 25 : 07.

Production Technology 10.

Ans: (d)

Ans: (b)

Sol: Ho = 16 mm,

Sol: do = 15 mm,

H1 = 10 mm,

df = 0.1 mm

R = 200 mm Angle of Bite    tan 1  tan 1

08.

%Reduction=

H R

16  10  9 .9 200

Ans: (a)

a)

Sol: Given rolling process



d 0  d1  Ist draw do



d1  d 2  2nd draw d1

3 stages with 80% reduction at each stage 0 .8 

Initial thickness H0 = 30 mm

dia reduced in the draw dia before draw

d o  d1 do

Final thickness = H1 = 14 mm

d1  0.2 d o  3mm

Droller = 680 = R = 340 mm

d2 = 0.2. d1 = 0.6mm

y = 200 MPa

d3 = 0.2 .d2 = 0.12mm

Thickness at neutral Hn = 17.2 Forward slip = =

V1 H 1= n 1 Vn H1

b)

d1 = 0.2. d0 = 3 d2 = 0.2. d1 = 0.6 d3 = 0.2. d2 = 0.12

V H Backward slip = 1  0  1  n Vn H0

17.2 = 42.6%  43% 30

4 stages with 80% reduction in 1st 3 stages followed by 20% in 4th stage

17.2  1 = 0.2285 = 23% 14

 1

(Error is 20%)

d4 = 0.8. d3 = 0.096 c)

(Error is 4%)

5 stages, with 80, 80, 40, 40, 20 etc d1 = 0.2. d0 = 3 d2 = 0.2. d1 = 0.6

09.

Ans: (b)

d3 = 0.6. d2 = 0.36

Sol: Roll separation distance

= 2  R + H1 = 2  300 + 25 = 625 mm

d4 = 0.6. d3 = 0.0216 d5 = 0.8. d4 = 0.1728

(Error is 72%)

From the given multiple choice B, the final diameter of wire close to 0.1 mm.

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: 26 : 11. Ans: (a)

ME – GATE_ Vol – I_Solutions

Common data for Q 12, 13 & 14

Sol: Given wire drawing process

d0 = 6 m,

d1 = 5.2 mm

12.

0

Die angle = 18 , diameter land = 4 mm

Ans: (b) ,

13. Ans: (c),

14. Ans: (a)

Sol: Initial inside diameter of tube

Coefficient of friction = 0.15

d0 = 52 mm,

H0 = 2.6

Yield dress = 260 MPa

H1 = 1.8,

D1 = 50 mm

A0 =

 2  6 = 136 = 21.237 4 4

A1 =

 2  5.2 = = 21.237 4 4

2d = 24=12, =0.12 2.6mm 1.8 mm

Drawing stress = 2  1  B   A 1 = y   1  B   A 0

  

B

52mm

   

50 mm

B = cot =

1 1 Die angle =  18 =90 2 2

=9

For stationary mandrel B =

B = 0.15  cot90 = 0.947 2 = 126.958MPa

B=

0  1  0.947   21.27     1 = 260     0.947   28.270 

0.947

   

1  1.29    1.8  2 / y    1    1.129    2.6 

1.19

2 L R1

(By considering friction) = 260 + (130260) e

Total drawing load = tA1 = 178.05  21.237 = 3.781 kN ACE Engineering Publications

  

2/y = 0.64

20.154 2.6

total = 260  81.94 = 178.05 MPa

0.12  0.12 =1.29 Tan (12)

B 1  B    H1   1   2   y     B    H 0  

= 260(2.056)(0.2375) Total drawing stress 2 = y + (2y) e

1   2 Tan

13.

Movable mandrel B = cot = (0.12)cot(12) = 0.564 0.564 1  0.564    1.8   2 / y      0.519 1    0.564    2.6  

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: 27 : 14.

Production Technology

Floating mandrel B=0

d2 = d1 2 B 1 1 B

h  2  n  0  y  h1 

 1  2B C=   = 0.756 1 B 

1

 2.6  = n   = 0.367  1.8 

Dia of wire in 2nd stage = 3.424 mm d1 = d0  c

Common data for Q 15. & 16.

d2 = d1 c = 4.530.756 = 3.424 > 1.34

15.

Ans: 6

&

Sol: d0 = 6 mm,

16. Ans: 3.4

d3 = d2  c

df = 1.34 mm

= 3.424 0.756

Given ideal condition

= 2.589>1.34

 = 60

 = 0.2

d4 = d3c = 1.957>1.34

f = 60 MPa

d5 = d4  c = 1.4797 > df

Maximum reduction condition 2 y

d6 = d5c = 1.1186 < df

2B  1  B   d1   =11=   1  B   d 0  

B = cot; d  B  1   1  1 B  d0   d1     d0 

2B

 Hence No. of stages = 6 Common data for Q 17, 18

B = 1.9 2B

17. Sol:

B 1 1 B =

1 1 B

Ans: (c) &

400 y 200 0.2

d1 2 B 1  d0 1 B 1

 1  21.9 1    6 d1 = d 0  2 B    1  1.9   1 B 

d1 = 4.53 ……… (1) stage

ACE Engineering Publications

18. Ans: (b)

0.4 

do = 12.214,

Lo = 100m

df = 10mm,

Lf =?

 y before  200 MPa ,  y after  400 MPa

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: 28 :

Ao Lo = Af Lf

22. Ans: 1

d A L f  L o  o  L o  o Af  df

  

Sol: Let, d1 = d2 = d

2

h1 = height of first cylinder h2 = height of second cylinder

2

 12.214   100     150m  10 

Assume h1 < h2

True strain in the drawing process 2

d  A     n o   n  o   0.4 A1  d1 

From the graph

 y at   0.2 ,  y  300 MPa

19.

Ans: (b)

20.

Ans: (c)

ME – GATE_ Vol – I_Solutions

Let % reduction in height = 10% Ist cylinder

h0  hf  0.1 h0 h0 – hf = 0.1 h0 hf = h0 – 0.1 h0 = 0.9 h0 A0h0 = Af hf d02h0 = df2 hf

Sol: (Extrusion force)min =  y  A 0

  10  10 2 = 78539.8N 4

Extrusion force 

(E.F) min 78539.8   ext 0.4

= 196346.5 N = 196 Tons

df  d0

= 1.054 d0 = 1.054 (d0)1 IInd cylinder

A0h0 = Afhf d02h0 = df2 hf df  d0

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

do = 100 mm,

df = 50 mm

Extrusion Force = A o K ln

Ao Af

h0 h0  d0 hf 0 .9 h 0

 d0

Ratio 

h0 hf h0 = 1.054 (d0)2 0 .9 h 0

d 0 1 d 0 2



1.054d 0 1 1 1.054d 0 2

2

  100   100 2  250 ln  = 2.72 MN. 4  50 

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: 29 : Common data for Q 23 & 24

Production Technology 27.

Ans: 7.26

Sol: 23.

0.3= 1.328

Ans: 7068 J & 24. Ans: 0.354 m

Sol: do = 100 mm, h f  40 mm , df  do

ho = 50 mm,  y  80 MPa



A B H0

O C

H1

ho 50  100  111.8mm hf 40

Fi min  A 0   y 

O

  1002  80  628.318 kN 4

Ff min  A f   y 

 (111.8) 2  80 4

 785.350 kN

Fmin

F  Ff min  i min  706.834 kN 2

Ho = 10 mm,

500 0.3

H1 = 7 mm, R=

1000 = 500 mm 2

Angle of bite () = tan

7068 H  0.354 m 2  10  103

1

= tan 1

W.D  Fmin  (h o  h f )  7068J  2 W  H

B

1.328 

H R

D C

10  7 = 4.429 500

OD OB

OD = 500cos1.328 = 499.865 DC = 500 – OD = 0.1343 mm

25.

Ans: (b)

26.

Ans: 58%

Thickness of neutral point = At point B = 7 + 2  0.1343 = 7.2686 mm

Sol: Area after 1st pass = A1 = (1 – 0.4)A0

= 0.6 A0 Area after 2nd pass = A2 = (1 – 0.3)A1 = 0.7  0.6  A0 = 0.42 A0 Overall % reduction = (1 – 0.42)  100 = 58 %

ACE Engineering Publications

29. Ans: 7.687 MPa, 19.7 % Sol: d0 = 6.25 mm;

 = 0;

d1 = 5.60 mm; y = 35 N/mm2

B = cot = 0 B  1  B   A1   0 2   y   1     B   A 0   0

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: 30 :

ME – GATE_ Vol – I_Solutions

By applying L – Hospital rule

31.

A  2   y n  0  A1

Sol: d0 = 10 mm;

  

Ans: 20.52 kW

0 .3 

d   y  2n  0  d1

  

0.3  1 

= 7.687 MPa A 0  A1 d 02  d12  % reduction in area = A0 d 02 = 19.71% 30.

A 0  A1 A  1 1 A0 A0

d1 = 8.36 mm B = cot = 0.1cot(6o) = .951  1  B   A 1 2  y   1  B   A 0

Ans: 29.85 tons



  

B

   

 1  0.951  0.951  240  1  0.7   0.951 

Sol: Initial size = 2525150mm

Final size = 6.25100150mm  = 0.25;

d 12 d 02



= 141.687 MPa Drawing load =  2  A 1  141 

2

y = 0.7kg/mm

As given piece is pressed; height is reduced h0 = 25;

 Fd  141.687 d12 4

hf = 6.25

= 141 

A0 = 25150; Af = 100150

P (motor) =

 2hrf  Forging force =  y A f 1   3h f  

(Ac)circular = (Ac)non – circular rf2  100  150

 2 d 1  4

P 

 8.36 2  = 7777.364 = 7.8 kN 4 Fd  v motor

7.8  2.5 0.95

P = 20.52 kW

rf = 69.098mm Forging force  2  0.25  69.098  = 0.7  15  10 3 1   3  6.25  = 29847.44kg = 292.80kN

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: 31 :

Production Technology 05.

Chapter‐ 6

Fp max .Kt

Sol: Fp 

Sheet Metal Operation



Common data for Q. 1 to 5 01.

Ans: (b)

Kt  I 40  0.6  1.25  17.14 kN 0.6  1.25  1

Fb 

Ans: (b)



Sol: For punching operation

Fb max kt kt  I

80  0.6  1.25  34.28 kN 0.6  1.25

F  FP  Fb  51.42 kN

Punch size = Hole size = 12.7 Die size = punch size + clearance = 12.7 + 2  0.04 = 12.78

Common data for Q. 06, 07 & 08 06. Ans: 83.6 N Sol:

02.

Ans: (a)

100

Sol: Die size = Blank size = 25.4mm

Punch size = Die size  2(radial clearance)

30

50mm 450

= 25.4  2(0.04) Punch size = 25.32 mm 03.

80mm

20

20

P  100  30  20 2  80  50  288.28

Ans: (b)

Fmax  Pt u  288.28  2  145  83.6 kN

Sol: Fmax = Fp max + Fb max

=   12.7  1.25  800    25.4  1.25  800 = 40 +80 = 120 kN

07.

Ans: 66.88 J

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 ACE Engineering Publications

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: 32 :

Fmax = 83.6 kN

12.

F(Kt + I) = Fmax  Kt

Sol: d = 25 mm, t = 2.5 mm → piercing

I=

Diameter clearance  C  0.0064 K t  0.0064  2.5 350 = 0.3 mm In piercing

Ans: (a)

Sol: Fmax

Ans: (d)

 u  350 MPa

Fmax  Kt Kt F

  83.6  0.4  2  0.4  2  = 1.98 mm = 24   09.

ME – GATE_ Vol – I_Solutions

P.S = H.S = 25 mm.

5  5  dt  u  dt u  

D.S = P.S + C = 25 + 0.3 = 25.3 Fmax  dt  u    25  2.5  350

Fmax    1.5d  0.4t   u

= 68.72 kN.

   1.5  0.4  dt u    1.5  0.4 

2  3 KN 

13.

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 = 24.83

10. Ans: (a) 11.

Ans: (a)

Ans: (b)

Sol: t = 5 mm, L = 200 mm, τu = 100 MPa,

K = 0.2 W.D = Fmax Kt = L × t × τu × K.t = 200 × 5 × 100 ×0.2 × 5 

Common data for Q. 14 & 15 14.

Ans: (b)

Sol: Draw Ratio =

100  10 3  100 N  m (or ) J only 1000

 d1 =

13.22 = 7.34 > 5cm 1.8

 d2 =

7.34 = 4.08<5 cm 1.8

Shear provided over a length of 20 200 mm   200 = 10 mm 400

n=2

Fmax Kt = F (Kt + I) 100  10 3  0.2  5  9.09  10 kN F 0.2  5  10

Dia.before Dia.after

15.

Ans: (a)

Sol: D  d 12  4d 1 h 1 ACE Engineering Publications

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: 33 :

Production Technology

4d 1 h 1  D 2  d 12

d 2  82.48  0.8  65.984  30

D 2  d 12 13.22 2  7.34 2 h1   = 4.11 mm 4  d1 4  7.34

d 3  65.984  0.8  52.7  30 d 4  52.7  0.8  42.2  30

P1  Dt y

d 5  42.2  0.8  33.7  30

=   132.22  1.5  315

d 6  33.7  0.8  27  30

 196238 N = 196.238 kN

n=6

–3

E = P1h1 = 196.2384.1110 = 806.6 kJ 19. 16.

Ans: (b)

Sol: DRR 1  0.4 

Ans: 52.7 mm

Sol: d3 = 52.7 mm D  d1 D

d 1  D1  0.4   30.2  0.6  18.12 d 2  d 1 1  0.25  18.120.75  13.59

20.

Ans: 144.42

Sol:

d 100   16.66  15 to 20 6 r

d 3  d 2 1  0.25  13.590.75  10.19

D  d 2  4dh 

d3 < 12  n = 3

r 2

 100 2  4  100  25  17.

Ans: (b)

= 138.42 +23

Sol: P1  Dt y    30.2  2  35  6641.3  21 



P1

 2 2 d 1  d 1  2 t  4







6,641.3

 2 18.12 2  18.12  2  2 4 = 65.5 MPa



6 2

D total  D  2  3  144.42 mm 21.

Ans: (d)

22.

Ans: (c)

Sol: Number of earing defects produced =2n

Where n is an integer So possible option is 64.

Common data for 18 & 19 18. Ans: 6

23.

Sol: D  d 2  4dh  30 2  4  30  150

Sol: B1  15  0.5  2  180 

 137.47 d 1  D  0.6  137.47  0.6  82.48  30 ACE Engineering Publications

Ans: 467 mm

 180

= 50.265 mm

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: 34 :

B 2  6  0.5  2  90 

 = 10.99 mm 180

C = Clearance = 0.0032 t  t = thickness = 1 mm

L 0  98  204  92  B1  2B 2

where,  = 240 N/mm2

= 466.245 mm

C = 0.0032×1× 240 = 0.0495 mm = 0.05 mm

2mm

Die size = 10 + 2 × 0.05 = 10.1 mm

15

Force required = s × d × t

92 98

100 6 mm

= 240 ×  × 10 × 1 = 7.536 kN

8m

204

8

ME – GATE_ Vol – I_Solutions

8

220m

24.

Ans: (b)

25. Ans: 3 Sol: D  d 2  4dh  50 2  4  50  100  150mm 0 .4 

D  d1 D

0.4150 = 150 – d1 d1 = 90mm > 50 d2 = d1(1–0.4) = 54 > 50 d3 = 32.4 < 50

n=3 26.

Ans: 7.536 kN

Sol: Punching a 10 mm circular hole from 1 mm

thickness sheet: Punch size = Blank size = 10 mm Die size = Punch size + 2 C ACE Engineering Publications

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: 35 :

Production Technology 04. Sol:

Chapter‐ 7

Ans: (c) D

Metrology t t = 0.01 to 0.015mm

7.1 Limits, Fits & tolerances

When, t = 0.01 mm 01.

D = 30.01 + 20.01 = 30.03 mm

Ans: (a)

Sol: For Clearance fit

= 30.05 + 2 0.01 = 30.07 mm

L- hole > H- shaft

When, t = 0.015 mm D = 30.01 + 20.015 = 30.04 mm

02.

Ans: (c)

Sol: Hole = 40

= 30.05 + 2 0.015 = 30.08 mm 0.050 0.000

0.08

mm ,

D  30 0.03 mm

Min. clearance = 0.01 mm, Tolerance on shaft = 0.04 mm , Max. clearance of shaft = ? 0.01 = L.hole – H.shaft

05.

Ans: (d) 0.01

Sol: A = 25.2 0.02

0.01 = 40.000 – H.shaft

B = 30.4  0.01

 H.shaft = 40.000 – 0.01 = 39.99mm

C = 32.7  0.02

H.shaft – L.shaft = 0.04 L.shaft = 39.99 – 0.04 = 39.95

Tmax = Lmax  Amin  Bmin – Cmin = (118 + 0.08)  (25.2  0.02)  (30.4

 0.01) – (32.7 – 0.02)

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

= 29.83 = 300.17 Tmin = Lmin  Amax  Bmax  Cmax

03.

Ans: (d)

= (118  0.09)  (25.2 + 0.01)  (30.4

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

+ 0.01) (32.7+ 0.02) = 29.57 Tmin = 30 0.43 0.17

 T = 30 0.43

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: 36 : 06.

(i)

Anc: (c)

ME – GATE_ Vol – I_Solutions Ans: (c)

Allowance = (L.L)hole  (H.L)Shaft 07. (i) Ans: (d)



Sol: Let the vertical distance between the holes

 (H.L)shaft = 65  0.09= 64.91 mm

is ‘y’

Tolerance = (HL)shaft  (LL)shaft



2450.05

25

y

30

0.05 = 64.91 – (LL)shaft

 (LL)shaft = 64.86 mm

x

60

0.09 = 65 – (H.L)shaft

Shaft = piston = 65

0.09  0.14

2500.2

(ii)

0.2  0.0

Ans: (a)

(L.L)hole = 65 mm (Tolerance)hole = (HL)hole  (LL)hole

sin30 =

y y = 245sin30 245



0.05 = (HL)hole – 65

 (HL)hole = 65.05 mm

ymax = 245maxsin30max

Hole = Bore = 65

= (245 + 0.05)sin(30 +15/60) = 123.45

0.05 0.00

0

ymin = (245 0.05)sin(30 15/60) = 121.55 (iii) Ans: (b) (ii)

Max Clearance = 65.05 – 64.86

Ans: (c)

= 0.19 mm

xmax = 250max– (60min+(30/2)min+ymin+(25/2)min) = (250 + 0.2) (60 +15+121.55+12.5) 09.

= 41.15mm xmin = 250min –(60max+(30/2)max + ymax + (25/2)max) = (2500.2)  (60.2 + 30.025/2 + 123.45 +

Sol: Amax = 15max + 30max

= 15.06 + 30.1 = 45.16 Amin = 15min + 30min = 44.84

25.025/2)

A = 45 ± 0.16. = A ± ∆A

= 38.625 mm Tolerance on X = Xmax – Xmin = 2.525 mm

Bmax = Amax – 20min = 45.16 – 19.93 = 25.23 mm Bmin = Amin – 20max

08. Sol: L Hole = BS = 65mm

H Hole = BS + Tolerance = 65.05mm ACE Engineering Publications

= 44.84 – 20.07 = 24.77 mm

 B ± ∆ B = 25 ± 0.23.

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Production Technology

10.

 C  100

Sol: Let

C = center distance between holes Cmax = max. Outer distance of pins – sum of min rod holes.

11. Sol: For the given conditions

X  100.1 

C =100±0.1

14.875 9.875  2 2

= 112.475 mm

9.9  0.025

14.9  0.025

0.075  0.275

 15.05 10.05  C  X   2   2 15  0.05 x

Xmax = 100 max

 9 .9   14.9       2  max  2  max

9.925 14.925  100.1   2 2 = 112.525 mm Xmin = 100 min

 99.9 

C = 99.925 mm

10  0.05

 9 .9   14.9       2  min  2  min

Because C is lying in between the limits, the assembly is possible. 12.

Ans: (b)

Sol: Fundamental deviation of hole ‘h’ is zero. 13. 0.03

Sol: Hole = 20 0.00

Min. interference = 0.03mm,

9.875 14.875  2 2

Max. interference = 0.08 mm 0.03 = L.shaft – H.hole

= 112.275 mm

L.shaft = 0.03 + 20.03 = 20.06 mm

 15   10   C max  X max         2  min  2  min 

0.08 = H.shaft – L.hole H.shaft = 0.08 + 20.00 = 20.08mm

 14.95 9.95   112.525     2   2

0.08

shaft  20  0.06

= 100.075 mm  15   10   C min  X min         2  max  2  max 

 15.05 10.05   112.525     2   2

14. Sol: H. Limit

H

0.021 B.S

L. Limit

= 99.725 mm ACE Engineering Publications

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: 38 :

ME – GATE_ Vol – I_Solutions

= H.hole – L.shaft B.S F.D=0.02mm f

H. Limit

= (25.021)  (24.947) = 0.074 mm

Tol = 0.033mm

L. Limit

(iv) Ans: (a) Size of the GO plug gauge = max. material limit of hole = L.hole = 25 mm

D  18  30  23.24mm i  0.453 D  0.0010  1.3m FD of hole H = 0

(v)

Size of the NOGO plug gauge = min.

FD Shaft = 5.5(23.24)0.41 = 20m

material limit of hole = H.hole = 25.021 mm

Hole tolerance, IT7 =16i = 20.8m = 21m = 0.021 mm Shaft tolerance, IT 8 = 25i

(vi) Ans: (c) Size of the GO ring gauge = max. material

= 32.5m = 33m

limit of shaft = H.shaft = 24.98 mm

= 0.033mm L - hole = basic size =25 mm H - hole = 25 + 0.021 = 25.021 mm

(vii) Ans: (d) Size of the NOGO ring gauge = min.

H - shaft = 25 – 0.02 = 24.98 mm

material limit of shaft = L.shaft = 24.947

L - shaft = 24.98 – .033 = 24.947 mm (i)

Ans: (a)

L- hole > H- shaft  Clearance fit (ii)

Ans: (b)

mm (viii) Ans: (a)

Ans: (b)

15.

Ans: (c)

Allowance = difference between max.

Sol: D  18  30  23.2

material limits = L.hole – H.shaft

i  0.45 3 D  0.001 D  1.3

= 25.00 – 24.98 = 0.02 mm

IT8 = 26i = 26 × 1.3 = 33.8 = 34 m = 0.034 mm

(iii) Ans: (b)

0.034

0.02

Shaft  250.053 , Hole = 25

0.021  0.00

Hole size  25H 8  25  0.000

Max clearance = different between minimum material limits ACE Engineering Publications

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: 39 : 16.

Production Technology

Ans: (a)

Hole  50

Sol: D  50  80  63.24 mm

i = 1.86 microns = 1.9 microns

H. Limit

IT8 = 25i = 47.5 microns

0.025  0.000

0.042

p 0.026

Tolerance = 0.0475 mm

B.S =50

L. Limit

F.D = –5.5 D0.41 = – 5.5 × 63.240.41 = 30 Microns = 0.03 mm H. shaft = 60 – F.D = 60 – 0.03 = 59.97 mm

0.042

Shaft  50 0.026

L. shaft = H. shaft – Tolerance

L.hole = B.S = 50

= 59.97 – 0.047 = 59.923 mm. 17.

H.hole – L.hole = Tolerance = 0.025 mm H.hole = L.hole + Tolerance = 50.025 mm

Ans: (d)

Max. interference = difference between

Sol: Case (i) 25H7

max. material limits = H.shaft – L.hole

L.L = 25.00

= 50.042 – 50.00 = 0.042 mm

U.L = 25.021

Min. interference = difference between min.

Case (2) 25 H8

material limits = L.shaft - H.hole

UL = 25.033

= 50.026 – 50.025 = 0.001 mm

Case (3) 25H6, UL - ? (UL)H8 (UL)H7 = (UL)H7  (UL)H6 25.03325.021 = 25.021  (25 + x) x = 0.009



19.

Ans: (c)

20. Ans: (b)

(UL)H6 = 25.009

Sol: To calculate exactly the data was not given

in the problem. But for shaft “h”,

18. (i) Ans: (a) , (ii) Ans: (a),

H – Shaft = 25.000

(iii) Ans: (a), (iv) Ans: (c)

L – Shaft = less than 25. And h7 → 7 indicates IT 7 not 7 microns.

Sol: H

H. Limit

21. Ans: (a)

0.025 50

L. Limit

Sol: GO size = max. material limit of hole

= 20.01 mm NOGO size = min. material limit of hole = 20.05 mm

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: 40 : 22.

Ans: (d)

ME – GATE_ Vol – I_Solutions

03.

Ans: (d)

04.

(i) Ans: (c)

Sol: To produce an interference fit, L-shaft must

be greater than H-hole. For this with multiple choice D it is possible because For D:

L-shaft = 20 – 0.02 = 19.98 mm,

Sol: sin  =

h L

h = sin30o  125=62.5 mm

H-shaft = 20 + 0.02 = 20.02 mm L-hole = 20 – 0.035 = 19.965mm, H-hole = 20 – 0.03 = 19.97mm, Hence, L-shaft (19.98) > H-hole (19.97)

(ii) (A) Ans: (a)

0.005   0 d  tan 30  = 4.76  62.5 125 

7.2 Angular Measurements

(B) Ans: (a) 01. Sol:

Ans: (a)

dh  r2  r1 

Sine bar

Slip gauges

 0.001 0  d  tan 30  = 2  62.5 125 



Given sine bar length = 200 = l

(C) Ans: (b)

 

Angle =325 6 = 32.085

dh = 0.002

Slip gauge height = h say

0   0.002 d  tan 30   = 4  62.5 125 

sin  

h 



(D) Ans: (d)



sin 32.085 0 

h 200

dh =  0.005 0    0.005 d  tan 30   = 10 125   62.5

 h = 106.235 02.

d 2  d1 0.002   0.001 2 2

Ans: i-(b), ii-(a)

Sol: l = 50 , L = 500

05.

Sol: Gradient of spirit level

50  0.08

= Sensitivity specified in mm/m

0.08  0.32 200  200  50



h = h + 0.32 = 28.87 + 0.32 = 29.19 Sin  

h 29.19   82332 L 200

ACE Engineering Publications

Ans: 0.048 mm/m

06.

10    1000 = 0.04845 mm/m. 3600 180

Ans: (d)

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: 41 :

Production Technology 7.3 Taper Measurement

07. Sol: (i) Ans: (b)

Ans: 19.2

01.

h  h1 sin = 2 w

Sol: 2.5

h2  h1 = 100sin30 = 50

O2

h2 = h1 + 50 = 75

50

/2

(ii) Ans: (d)

15

h  25 sin(30)  100.005

08.

h1

h2



h = 75.0025 mm



h2 = 75.0025 + 0.005 = 75.0075 mm

Ans: (a)

Sol: L = 250 mm,

d 2  d1 2h1  h 2   d 2  d1 

sin  / 2  

30  15 252.5  30  15

d = 20 mm



90 250



  = 21.2 deg 09.

 = 19.2

Sol: tan  / 2 

Sol:  = 2732

5 8.66

  = 60

o

125 h

= 27.533 h sin = 25

15 = 1/6 105  15

02. Ans: 60

Ans: 11.556 mm

 32  = 27 o     60 

O1

sin  / 2 

h = 100 – (d/2) = 100 – 10 = 90 mm sin  

30



105=5

O2 O1

h = 11.556 mm 36.345=31.34 36.34 40 50 ACE Engineering Publications

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05.

Ans: 112.41 mm

Sol:

ME – GATE_ Vol – I_Solutions Ans: 78.074 mm

Sol:

70 mm 30 mm h2

h1 C

h1+r1=O2A+r2+h2

O2

B A

D

12.5mm

O1 d1 = 100 m

A

12.5mm

C

M

 



Diameter = O1C+O1A+O2D 

d1  2

O1O 2 2  O 2 A 2

O1O 2  r1  r2 =75

d2 2

X

= 70 + 50  30  25 = 65 D  50  75 2  65 2  25

  37.5  2 

= 112.4165 mm

75   = 37.5 –  = 37.5 – 2910 2 = 820

le OBC

Ans: 43.33 mm

sin 37.5 =

Sol:

 OB 

O2

d = 25

42 O1

O

 +  = 4550 + 2910 = 75

O2A = h1 + r1 – r2 – h2

04.



A

35

BC OB

12.5 BC   20.533 sin 37.5 sin 37.5

le OAB cos 820 =

OA OB

 OA = OB cos 820 = 20.316 mm X = M – (OA + R) O1A =

25  17  18.33 2

2

D = r + O1A + r

= 110.89 – (20.316 + 12.5) = 78.074 mm

= 25 + 18.33 = 43.33 mm

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Production Technology



Ans: 1.1

Sol: d2 – d1 = 10 ;

 3   Tan 1  = 6 2  28.54 

h2 – h1 = 12.138

d 2  d1  sin     2  2  (h 2  h 1 )  d 2  d1 

 Taper angle    6 0 2 Included angle = 120

 = 88.9 Error = 90 – 88.9 = 1.1 09. 07.

Ans: 38.94

Ans: (c)

Sol: tan  

Sol:

10   = tan-1(1/3)    18.434 30 10mm

10mm

h  D2 

D

2

2



30mm 10mm

D

  2 Sin    2 hD Sin

 2

D 2h  D

Z=10 Z=0

10 – (10/3)

 D  2 2h  D

Distance at Z = 0,

If D = 0, h = 0

10   D 0  210  10 tan 30  210   3 

D = 1, h = 1 1 1    Sin     2  2 1  1 3

= 6.67  2 = 13.33 mm

   2   19.47   = 38.94   08.

Z=40

1sec



Ans: (d)

1

With probe diameter compensation

3  Sol: Tan     2  28.54

Dactual  13.334  2  r sec  = 13.334 + 2 ×(1 sec 18.435) = 15.442 mm.



2

3

15.54 + 8 + 5 = 28.54

ACE Engineering Publications

3

10.

Ans: (d)

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1  71  0.11667 2.04103

7.4 Screw Thread Measurements 01.

2 = 91 = 0.15 2.618103

Ans: (d)

P = 0.004

Sol: Major diameter = s + (R2 R1)

De = 30.6651

= 35.5 + (11.8708 9.3768)

 = 60 (metric thread)

= 37.994 mm 02.

Virtual correction VC = (0.004  cos30) + (0.0131 

Ans: (a)

3.5(0.11667 + 0.15) )

Sol: Minor diameter

VC = 0.01569

= 30.5 + (15.3768  13.5218)

VED = De + VC

= 32.355 mm

= 30.6651 + 0.01569 = 30.6807

Correct answer is (a) 03.

Ans: (a)

p  Sol: best wire diameter, d =   sec  2 2  3.5   60  =  sec  = 2  2   2 

05.

Ans: (a)

06.

Ans: (d)

R 2  R1  Sol: Sin     2  M 2  M1  R 2  R 1 

=

M = 30.5 + (12.2428  13.3768) = 29.366 mm p   De = M  d  tan  2 2  3.5    tan 30  = M  2  2  

07.

VC  P cos



2 M = 14.701 + (1.155+ tan30) =16.433 2

2

08.

 0.0131P 1   2 

Ans: 16.433 mm

p   Sol: De = M   d  tan  2 2 

Ans: (a)

Sol: VED = De  VC

Ans: (d)

Sol: Lead = pitch  no of starts

Pitch =

P = pitch error

1.4434  0.8660 22.06  20.32  1.4434  0.8660

 = 59.5566 = 593323

= 29.366 – 3.010366 = 26.355 mm 04.

ME – GATE_ Vol – I_Solutions

lead 3 = =1.5 no of starts 2

1, 2  flank angle errors in deg ACE Engineering Publications

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: 45 : 09.

Production Technology

Ans: (d)

7.5 Surface Finish Measurement

Sol: Rollers will not used to measure pitch

diameter.

01.

p  Best size diameter d =   sec  2 2

(i)

 2   60  =   sec  2  2  = 1.1547 = 1.155 10.

Common data Q 11 & 12

(ii)

Ans: (c)

Sol: CLA(Ra) = (h1+h2+h3+……+h10)/n

 2   60  =   sec  =1.155 mm 2  2 

Sol: Peaks 35 40 35 42 35

Valley 25 22 18 25 23 Rz =

peaks  valleys no of peaks (35  40  35  42  35)  (25  22  18  25  23) =15 5

(iv) Ans: (b) Sol:

RMS =

h 12  h 22  h 32  ........  h 2n = 33 n

(or) RMS = 1.1  R a=1.1  30=33

Ans: (a)

p   Sol: Deff = M –  d  tan  2 2  = 16.455 – 1.155.tan30 = 14.7226 mm Ans: 1.732 mm

Sol: The best wire size = (p/2) sec(/2)

= (3/2) sec(60/2)

(v)

Ans: (c)

Sol: If Ra value from 18.75 to 37.5 international

grade of roughness is given by N11. 02.

Ans: (c)

Sol: Ra =

= 1.732 mm = ACE Engineering Publications

300 = 30 10

(iii) Ans: (b)



Ans: (a)

p  Sol: Best size diameter, d =   sec  2 2

13.

= 4 2  18 = 24

=

= 0.2 cos30 = 0.346

12.

Sol: Rt = max. peak – min.valley

Ans: (d)

 Sol: V.C =  P.cos    0.0131 P(1+ 2) 2

11.

Ans: (c)

A  w

1 1000  HM VM

480  480 1 1000   = 0.8 0.8 100 15000

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: 46 :

03.

Ans: (d)

Sol: Rt =

04.

ME – GATE_ Vol – I_Solutions

Chapter‐ 8

0.05 =50  m tan 45

Advanced Machining Methods Numerical Control (NC)

Ans: (c)

Sol:

01.

Ans: (a)

Sol: Pitch of lead screw = 5mm

1 rev = 5mm

40

50

Am = 0.105

1mm = 1/5 rev 200mm = 1/5  200 = 40rev

10

= 40 360 = 14400 deg.

2.5

Am act = 0.105  0.01 2.5 = 0.08 K= 

A m act

02.

Ans: (b)

Sol: Pitch of lead screw = 5mm,

3

(10  2.5)  0.04

BLU = 0.005mm

0.082 1  = 0.8 3 2.5  10  0.04 1000

 Distance travelled /pulse





Length of travel = 9mm No.of pulses = L/BLU = 9 / 0.005

05.

Ans: (c)

06.

Ans: (c)

= 1800 pulse. 03.

Ans: (b)

Sol: For 1 rev of motor 360 are required 07.

 360 pulses are required

Ans: (a)

When motor is rotated by 1 rev 08.

Ans: 2

Sol: R a 

 lead screw will rotate by 1 rev

h 16  4  16  0 64    2 m n 32 32

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 ACE Engineering Publications

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Production Technology

 3.6 mm of linear movement of axis

 Pulses per min = feed / BLU

360 pulses = 3.6mm

= 50/0.01 = 5000

1 pulse = 3.6/360 = 0.01mm = 10 microns

07.

Ans: (a)

Sol: 04.

Ans: (b)

CNC drill table X axis

pulses

Sol: 10V = 100 rpm

= 100  5 = 500 mm/min That is for 500mm/min = 10V

Pulse generator

Stepper motor

Driver

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)

Sol: A, Stepper motor  200 steps / rev

 200 pulses /rev

Hence 200 pulse = 1 revolution of motor = 1 revolution of lead screw = 4mm That is 1 pulse = 4/200 = 1/50 = 0.02mm,

Pitch = 4 mm, no. of starts = 1,

hence BLU does not depends on the

Gear ratio = N0/Ni= 1/4 = U

frequency of pulse generator. But if the

F = 10000 pulses per min 200 pulses  1 rev of motor

 1/4 rev of lead screw = 1/4  4  1 mm linear distance. = 1mm linear distance 1 pulse = 1/200 = 0.005mm = 5 microns = 1 BLU

speed of the table means it will get doubled. 08. Ans: 20 Sol: p = 5 mm

1000 pulses  1 rev of motor

 1 rev of lead screw Velocity of table = 6 m/min = 6000 mm/min

Feed = BLU  pulse /min = 0.005  10000 = 50mm/min For changing BLU = 10 microns = 0.01mm

 Gear ratio has to be reduced to 1/2 Feed = BLU  pulse /min ACE Engineering Publications

= 100 mm/sec 1000 pulses  1 rev of lead screw  5 mm 1 pulse 

5  0.005 mm 1000

BLU = 0.005 mm

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Table speed = BLU × Rate of Pulses Rate of pulses =

13.

ME – GATE_ Vol – I_Solutions Ans: (d)

Sol: Appropriate answer but the correct answer

100 0.005

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.

centre

55

Ans: 60

Sol: In the combined movement, the tool is

(50,55)

moving for 50mm with a speed of 50

100mm/min. whereas in the same time tool

70

is traveling x-axis by only 30mm. Hence,

10. 11.

Ans: (b)

For 50mm 100mm/min For 30mm 

Ans: (a)

100  30  60mm / min 50

Sol: G02 – circular interpolation clockwise

G03 – circular interpolation counter clockwise

15.   Ans: (a) Sol: Because diameter of milling cutter is 16mm,

12.

Ans: (c)

the radius is 8mm. the dotted line indicates

Sol: Because the tool has to travel from P1 to P2

in clock wise. Y

mm all around the rectangular slot

P2 = (10, 15)



cutter center position, which is shifted by 8 (–8,58)

 Center (15, 15)



P1 = (15, 10) X

(–8,–8)

S

P

(0,50)

(100,50)

(0,0)

(100,0)

R

Q

(108,58)

(108,–8)

If the given shape is rectangular hole, then the answer is (8,8), (92,8), (92,42), (8,42), (8,8)

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: 49 : 16. Sol:

Ans: (a) Y

Q (4,5)

P

33.7

(1,3)

36.9

Production Technology 18.

Ans: (b)

19.

Ans: (a)

Sol: Given coordinates (0,0) to (100, 100) R

Mean, L = 100, depth, d = 2 m

D

3.2

Diameter, D = 10

C O

APC =

X

PQ  2 2  32 = 3.6055 = PC

=

PD = PC  cos 3.2 = 3.6

210  2  = 4

Time/slot =

x co-ordinate of point C = 1 + 3.6 = 4.6

17.

d D  d 

104 104 = fN 50

DC = 3.6 sin 3.2 = 0.2

= 2.08 min

y co-ordinate of point C = 3.0 – 0.2 = 2.8

= 124.8 sec  120

Ans: (a)

20.

Sol: “P” after translation = (1+2, 3+3, –5 –4)

= (3, 6, –9) Rotation about z- axis means

 x  cos   sin   y   sin  cos     z  0 0    0 1  0 0  1 1 0  0 0  0 0

0 0 1 0

0 0 0  1

0 0 1 0

0 0 0  1

3 6     9   1

0  6  0  0    6  3  0  0  0   3      0  0  9  0    9       0  0  0  1  1 

x   y   z    1 

Ans: 54.166 mm/sec, 10 micron

Sol: f = 500 pulse/rev

p = 5 mm,

N = 650 rpm

(i) v = Np =

650  5 60

v = 54.166 mm/sec Now, 1 min = 650 rev 1 sec =

 f  500 

650 rev 60

650 60

f = 5416.66 pulse/sec And, v = B.L.U. f = 54.166  BLU  5416.66 B.L.U. = 0.01 mm B.L.U. = 10 microns

Final point = [–6, 3, –9]

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: 50 : 21. Sol:

Ans: 287

 = 0.9

Chapter‐ 9 NTM, Jigs and Fixtures

0.9 = 1 pulse 360 

360 pulse  400 pulses 0.9

 1 revolution = 4 mm pitch = 400 pulses   2.87 mm = 287 pulses 22.

ME – GATE_ Vol – I_Solutions

01. Ans (c) 03.

melting and vaporization associated with cavitation and also erosion & cavitation or

Sol: Pulse rate = N  pulse/rev

Feed rate = 15 rpm  4 mm/rev

Ans: (c)

Sol: In EDM the mechanism of MR is due to

Ans: 100 pulse, 60 mm/min

400  15   100 pulse / sec 60

02. Ans: (d)

spark erosion and cavitation 04.

Ans: (d)

Sol: The high thermal conductivity of the tool

= 60 mm/min

material

will

have

high

electrical

conductivity hence the heat generated with in the tool is low and what ever heat generated it will be distributed easily therefore tool melting rate reduces and tool wear reduces. Where as due to specific heat of work material, the rise in temp of W.P is faster and more amount of MR is possible. 05.

Ans: (b)

Sol: Given w = 1 + (2  0.5) = 2

t =5, f = 20 mm/rev MRR = wtf = 2.5.20 = 200 mm/min 06.

Ans: (a)

Sol: As the thermal conductivity of tool material

is high the heat dissipation from the tool is taking place and if the specific heat is high, it needs large amount of heat for raising the temps of tool material up to MP. ACE Engineering Publications

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: 51 :

Production Technology 10.

Ans: (a)

Sol: In ECM

07. (i) Ans: (a) , (ii) Ans: (c)

MRR  gram atomic weight of material

Sol: D = 12mm, t = 50mm, R = 40 ,

C = 20 F, Vs = 220V, Vd = 110V

 Current density

 Vs  Cycle time = R.C ln    tc  Vs  Vd 



1 dis tan ce between tool and work

 Thermal conduction of electrolyte.

 220  = 40 2010–6 × ln    110 

11.

6

 554  10 sec  0.55 milli sec

Ans: (b)

Sol: I = 5000 A

A = 63, Z = 1, F = 96500

Average power input = W

 E   0.5  CVd 2  =   tc  t c   

MRR 

AI 5000  63  ZF 1  96500

 3.264 g / sec .

= 218 W = 0.218 kW 08.

12. Ans: (a)

Ans: (b)

Sol: For Rough machining i.e. stock removal the

electrolyte should have high electrical conductivity, called passivity electrolyte, where

as

for

finish

machining

the

electrolyte should have low electrical conductivity

called

non–passivity

Sol: A = 55.85, Z = 2, F = 96540

Specific resistance = 2Ω-cm Voltage = 12V Inter electrode gap = 0.2 mm Resistance R

electrolyte will be used.

 09. Ans: (b)

I

Sol: In ECM

MRR  gram atomic weight of material MRR  Current density MRR 

1 dis tan ce between tool and work

Sp. Resis tance  Inter electrode gap Suface area

2  10  0.2  0.01 20  20

V 12   1200A R 0.01

MRR 

AI 55.85  1200  ZF 2  96540

 0.3471 g / sec

MRR  Thermal conduction of electrolyte. ACE Engineering Publications

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: 52 : 13.

Ans: 51.542

18.

1  0.009 50  0.009 L  0.02  Sol: R  Area Area Area I

V 12  1.5  Area   23.333  Area 50  0.009 R

L = 3 + 6 = 9 m = 0.009 MRR 

ME – GATE_ Vol – I_Solutions Ans: (d)

Sol: Relative motion between tool and work

piece is not necessary. 19.

Ans: (c)

Sol:

AI 55.85  23.333  Area  ZF 7860  10 6  2  96500

A

B

= 0.98189  Area MRR  0.8590 mm / sec Area

If D = Dmin = 59.9 X1 = distance between center of shaft and

 0.8590  60 mm / min

59.9 corner of V – block  2  34.583 sin 60

= 51.542 mm/min 14.

Ans: 680

15.

Ans: (c)

Sol: EDM, ECM and AJM

60.1 X 2  2  34.698 sin 60

are used for

Error in depth = 2(X2 – X1) = 0.223 mm

producing straight holes only but in LBM by maneuvering or bending laser gun

20.

slightly it is possible perform the Zig – Zag

Sol: Resolving the force “F” into Horizontal

F sin   100

hole. 16.

Ans: (b) (Both are Correct)

Sol: In EBM Vacuum is provided to avoid the

dispersion of electrons after the magnetic lense, but this vacuum is giving an addition function of providing efficient shield to the weld bead. 17.

(1) 100  tan   (2) 200

1   tan 1    26.565 2 100  223.6 kg Sin 

Taking the moments about vertical axis

Sol: Out of all the NTM’s ECM will give large

MRR and EBM will give very small MRR. ACE Engineering Publications

F cos   100  100  200 …… (2)

F

Ans: (d)

………. (1)

xF cos   100  30  100  30  100  20  x = 10 mm

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Production Technology

Positional error = 30.025 – 30 = 0.025 mm

21. (i) Ans: (d), (ii) Ans: 10.6 mm Sol: P

(b)

Fixed V – block and movable rectangular block

O2

O1 A

X2

X1 Q

O1

30 30.025

O2 4

3 A

O1 O 2  4 2  3 2 = 5 O1O 2  5  x 2  x 2

(c)

x = 3.5 Block of uniform thickness is preferable because of balanced condition.

x1 

30  34.64 Sin 60

x2 

30.025  34.66 Sin 60

Positional error = x2 – x1 = 0.0298 mm The positional error is mainly depends on the fixed element. So when fixed V – block and marble V – block is used, the positional error is remains same as (b).

22.

Out of the 3 cases, case (a) is giving lower

Sol:

(a)

Clamping

Fixed rectangular block and movable V –

positional error, hence preferable.

clamp.

O1 O2 · ·

Clamping

30 30.025

ACE Engineering Publications

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