Production Paper 1 Solution

PRODUCTION
(PE - I : SOLUTIONS)

1) (c)

2) (c)

3) (b)

4) (a)

5) Involute

6) (c)

7) (a)

8) (b)

9) (c)

10) (c)

11) Pearlite = Cementite is finely dispersed in ferrite
Extremely heard & brittle phase
Martensite = can exist only above 7230C
Equation between 3 solid phases.
Austenite =

Eutectoid =

12) Cylinder , h = D
V cylinder = V sphere

(ts)shpere K (V / A)2
(ts) cylinder  K (V / A)2

(A)2 cylinder
 (A)2 sphere

Vcylinder  Vshpere

 D2h  4  x D3
4 3 8

D3 Ds3
4 6

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 D 3  4  2
 Ds  6 3

D   2 1/ 3
Ds  3 

(ts)sph. ( A cyl.)2
(ts) cyl.  (A spher)2

  D2  Dh
4 Ds2

  D2  D2  5 D2
4 Ds2 4
Ds2

5 D 2 5  2 2 / 3
4  Ds 4 
   x 
 3 

(ts)sp.  1.14
(ts)cyl.

13) Net force = weight of fluid displaced – weight of core

  .g.v1    gv2

= 11300 x 9.81 x

 11300 x 9.81 x  x 0.122 x 0.18 - 1600 x 9.81 x x  x 0.122 x 0.18
4 4

193.7 N

14) For maximum power
3V + I = 240
I = 240 – 3V
P = V.I = (240 – 3V)V

dP  0, for maximum power
dv

240 – 6V = 0

V  40 Volt

15) L = 900 mm, I = 150A, V = 20v, u = 300 mm/min,

n  0.8 , R = 36

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n  Hm
Hs

Hs  V .I u . n
Ab x

 150 x 20
300

60

= 480 J/mm

16) t1 = 20 mm, b1 = 100 mm
t2 = 18 mm, R = 250 mm
N = 10 RPM
 F  300MPa

P = 2T.w

LP  R.h  22.36mm

AP = Lp x b = 2236.06

F   0 x Lp x b

= 670820 N
A = 11.18

Torque / roller  F x a N .m
1000

= 7499.76 N.m

Power = 2 x T x w

 2 x 7499.76 x 2 x 10
60

= 15.7 KW

17)
  150 ,  450

V = 35 m/min
Vc = ?

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Vc  Sin(90 V  )
Sin  (

Vc V
Sin 150  Cos(30)0

Vc  28.577 m
min

18) (d)

19) It is a blanking process so clearance provided on punch
Clearance = 3% of thickness
= 0.06
Punch = dia of disc – 2 x clearance
= 19.88 mm

20) Length of spure = 20 c m
C.S area of Base of sprue = 1 cm 2 = A3 = Ac

Volume of casting = 1000 CC

Time required for filling = Volume
Ac x V

1000 1000

1 x 2gh 1x 2x981x20

= 5.05 sec

21) Allowance = Min clearance
= Minimum of hole – Maximum of pin
= (2 – 0.0015) – (1.996 + 0.0015)
= 0.001 mm

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22) (a)
23) (d)
24)

Min. clearance = Min of Hole – Max. of shaft
Max. clearance = Max. of hole – Min of shaft
Assuming Hole torerance as x
So 0.3 = x + 0.03 + 0.08

X = 0.19
Max. size of hole = 30 + 0.19

= 30.19

25)
 x  dt = F

4000 x  x 10 x 2 = F

F = 25.13

26) (a)

27) The shear angle (in degree)

V = 20m/min, α = 150 , tc = 0.4 mm t = 0.2 mm r = chip thickness

ratio  t
tc

r  0.2  0.5
0.4

tan   r cos  0.5 cos15
1 r sin 1 0.5sin15

tan  0.5547

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  29.01

28)

The chip velocity Vc  sin 
V cos(  )

Vc  20 x sin 29.01
cos(29.01  15)

Vc  9.997  10m / min

29)

tan   r cos 
1 r sin r

r  t  0.25  0.33
tc 0.75

tan   0.33 cos 00
1 0.33sin 00

tan   0.33

  18.4350

30)

Fs   . bt
sin 

FS  FC cos   FT Sin

= 950 cos 18.40 – 475 sin 18.40

= 751.04 N

s  Fs  751
bt 2.5 x 0.25

sin  sin18.4

 s  379.28  380 N
mm2

************

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