Detailed Solutions

ESE-2019 Mechancial Engineering

Mains Test Series Test No : 15

Section A
Q.1 (a) Solution:
We need to find resultant of given 3 forces and its intersection with the AB.
   
Resultant force, R = F1 + F2 + F3

= 2 (90 cos 30iˆ − 90 sin 30 ˆj ) + ( −200 ˆj )

= 90 3iˆ − 290 ˆj

Where î and ĵ are unit vector in x and y direction.

y

R = (90 3 )2 + ( −290 )2 = 329.24 kN

Let’s take C to be the point of intersection of resultant force and line AB which is at ‘d’
distance in x directions from A, then moment of these three forces will be zero at C.
So, Mc = –90 cos30° × (3.6 × 3) + 90sin30° × d – 90cos30° × (3.6 × 2) + 90sin30° ×
 Test No : 15 MECHANICAL ENGINEERING | 11

⎛ 6.3 ⎞
⎜d − × 3.6 ⎟ + 200 × (d – 2.4)
⎝ 3 × 3.6 ⎠

⇒ 0 = −45 3 × 3.6 × 3 + 45d − 45 3 × 3.6 × 2 + 45 × ( d − 2.1 ) + 200 ( d − 2.4 )

⇒ d = 6.8188 m

So, R will pass through C which is at 6.8188 m distance from A.

Q.1 (b) Solution:

Due to wind there will be moment at bottom of tree.

dy

Consider a rectangular area of thickness dy at height y from bottom.

dM = dF × y
h
⎛b ⎞
M = ∫ P ⎜⎝ h (h − y ) dy ⎟⎠ × y
0

h h
Pb ⎛ hy 2 y 3 ⎞
=
Pb
h ∫
hy −(y 2
dy =) ⎜ − ⎟
0
h ⎝ 2 3 ⎠0

Pb 3 ⎛ 1 1 ⎞ 2
M = × h ⎜ − ⎟ = Pbh
h ⎝2 3⎠ 6
2
2 × 103 × 4 × ( 10 )
= = 133.33 kNm
6
M σ
Using, =
I r

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
12 | ESE 2019 : MAINS TEST SERIES

Mr 4M 4 × 133.33 × 103
⇒ σ = = = 3 = 21.2 MPa
⎛ πr 4 ⎞ πr 3 π× ( 0.2 )
⎜ ⎟
⎝ 4 ⎠
So, maximum stress at bottom will be 21.2 MPa.

Q.1 (c) Solution:

We note that BD + DE = 0.24 sinβ
0.15sin30° + 0.12 = 0.24sinβ
β = 54.34°
I13

c
I23
B 30°
m b
15
0.
A 30°
D a
0.

I12
24
m

0.12 m
β I34
E C

a = 0.24sin54.34°
a = 0.195 m
b = 0.24cos54.34°
b = 0.140 m
c = btan30° = 0.140 tan30° = 0.080 m
c 0.080
I13I23 = = = 0.16 m
sin 30° sin 30°
I13I34 = a + c = 0.195 + 0.16 = 0.355 m
Now that the instant centre for each bar has been found
V B = AB × ωAB = 0.15 × 3 = 0.45 m/s
VB 0.45
ωBC = = = 2.8125 rad/s (anticlockwise)
I 13 I 23 0.16
Vc = I13I34 × ωBC = 0.355 × 2.8125 = 0.998 m/s  1 m/s
Radial component of acceleration = BC × ωBC 2 = 0.24 × 2.81252 = 1.898 m/s2

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 13
Q.1 (d) Solution:
200 mm 600 mm

μRN
P
x mm
RN
+
100 mm O
Brake drum

Take moment about O:

Ft × x + P × 800 = RN × 200
μRN × x + P × 800 = RN × 200 ...(1)
D
Torque, T = Ft ×
2
D
T = μRN ×
2
0.3
360 = 0.3 × RN ×
2
R N = 8000 N
From Eq. (1),
μRN × x + P × 800 = RN × 200
0.3 × 8000 × 50 + P × 800 = 8000 × 200
P = 1850 N
2. For self braking P should be zero.
So, from equation (1)
μRN × x + P × 800 = RN × 200
0.3 × 8000 × x = 8000 × 200
x = 666.66 mm

Q.1 (e) Solution:

Let, d = Inner diameter of clutch
D = Outer diameter of clutch
d
= x
D
We know that, for uniform wear theory

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
14 | ESE 2019 : MAINS TEST SERIES

πμPa d ( D2 − d 2 )
T =
8
πμPa ( 2
T = D d − d3 )
8
π × μ × Pa ⎛ 2
3 D d d
3 ⎞
T = ×D ⎜ − ⎟
8 ⎝ D3 D3 ⎠
(multiplying and dividing by D3)

d3 ⎞
{ }
π× μ × Pa ⎛
3 d d
T = ×D ⎜ − ⎟ Let, =x
8 ⎝ D D3 ⎠ D

π × μ × Pa × D3 (
T = x − x3 )
8

π× μ× Pa × D3 (
x (1 − x ) )
2
T =
8
For maximum torque capacity,
d (
x − x3 ) = 0 {only x is variable here}
dx
1 – 3x 2 = 0 Torque

1
x = = 0.577
3
d
So, = 0.577
D o 0.577 1 (d/D)

Q.2 (a) Solution:

As per given information
ZB = 27, ZC = 30, ZD = 24, ZE = 21
Tabular method:

action arm A Gear D (24) Gear B/C (27/30) Gear E (21)

−24 +24 30
×
A arm fixed D + 1 rev 0 +1 27 27 21
−24 +24 30
x × x
D + x rev. 0 +x 27 27 21
24 24 30
y− x y+ × x
+ y rev.of arm y y+x 27 27 21

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 15
As per given condition gear E is fixed,
24 30
y+ × x = 0
27 21
24 30
y = − × x
27 21
Deriving shaft attached with arm A and driven shaft D,
NA y
ND = y+x

24 30
NA − × x
27 21
ND = ⎡ ⎛ 24 30 ⎞ ⎤ = 4.706
⎢1 − ⎜ 27 × 21 ⎟ ⎥ x
⎣ ⎝ ⎠⎦
Assuming that there are no frictional losses and that the members are revolving at uniform
speeds.
TA + TD + TE = 0 ...(1)
ωATA + ωDTD + ωETE = 0
But E is fixed, so that ωE = 0
ωATA + ωDTD = 0
A ω
TD = −TA × ω = – 80 × 4.706 = – 376.480 Nm
D

Holding torque, TE = ?
From equation (1)
TA + TD + TE = 0
80 – 376.480 + TE = 0
TE = 296.480 Nm

Q.2 (b) Solution:

For equilibrium moment about C will be zero.
Mc = 0
wo

wox
l
A C B
x l -x
RC

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
16 | ESE 2019 : MAINS TEST SERIES

1 ⎛ wo x ⎞ x ⎛ wo x ⎞ ( (l − x ) 1 ⎛ w x⎞ 2
⇒ ×⎜ ⎟x× =⎜ ⎟ l − x )× + × (l − x ) ⎜ w0 − 0 ⎟ × (l − x )
2 ⎝ l ⎠ 3 ⎝ l ⎠ 2 2 ⎝ l ⎠ 3
2 3
x3 x (l − x ) (l − x )
⇒ = +
6l 2l 3l
⇒ x3 = 3x(l – x)2 + 2(l – x)3
⇒ x3 = 3x(l2 + x2 – 2lx) + 2(l 3 – x3 – 3l 2x + 3lx2)
⇒ x3 = 3xl 2 + 3x3 – 6lx2 + 2l 3 – 2x3 – 6l2x + 6lx2
⇒ 0 = 2l 3 – 3l 2x
2l
⇒ x =
3
As ΣF y = 0
1
Rc = wo l
2
1 wo x w x 2 w 4l 2
(SF)c = × ×x = o = o
2 l 2l 2l × 9
2 wol
SF =
9
5 wo l
18
SFD Parabolic
variation

A B
C
Parabolic
variation
2 wol
9
BMD

C
A B

Cubic variation

Cubic variation
4 wo l 2
81

1 ⎛ wo x ⎞ x 3 2
Mc = ⎜ ⎟ x × = wo ⎛⎜ 2 l ⎞⎟ = 4 wol
2⎝ l ⎠ 3 6l ⎝ 3 ⎠ 81
From BMD it is clear maximum moment is at C.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 17

⎛ 4w l2 ⎞ ⎛ a ⎞
⎜ o ⎟
⎜ ⎟
Mc ⎛ a ⎞ ⎝ 81 ⎠ ⎝ 2 ⎠ 8 wo l 2 wo l 2
Maximum stress, σc = ⎜ ⎟ = = =
I ⎝2⎠ a4 27 a 3 ( 1.5 a )3
12
Q.2 (c) Solution:

P × 60 7.5 × 106 × 60
Torque, T = =
2 πN 2 π× 360
T = 198943.679 Nmm
For pulley 1:

φ 500 mm

φ 250 mm

P4 P1 P2 P3

(P1 – P2) × 125 = 198943.679

(P1 – P2) = 1591.55 N ...(1)
P1
Also, P2 = 2.5 ...(2)

From equation (1) and (2)

P 1 = 2652.58 N
and P 2 = 1061.03 N
Weight of pulley is given by;
W1 = m1g = 10 × (9.81) = 98.1 N
Total downward force acting on pulley 1.
= P1 + P2 + W1 = 2652.58 + 1061.03 + 98.1 = 3811.71 N
For pulley 2:
(P3 – P4) × 250 = 198943.679

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
18 | ESE 2019 : MAINS TEST SERIES

P3
and P4 = 2.5
By solving:
P 3 = 1326.29 N
P 4 = 530.52 N
Weight of pulley 2: W2 = m2g = 30 × 9.81 = 294.3 N
Total downward force on pulley 2 = P3 + P4 + W2
= 1326.29 + 530.52 + 294.3 = 2151.11 N
3811.71 2151.11

A B
250 mm 500 mm 250 mm
Mb

537777.5
952927.5
T

Bending moment diagram,

(B.M.)A = 3811.71 × 250 = 952927.5 N-mm
(B.M.)B = 2151.11 × 250 = 537777.5 N-mm
Torque is constant, T = 198943.68 Nmm
As bending moment is maximum at A, so, critical point is A.
16
τmax = 3
( Mb )2 + (T )2
πd
16
( 952927.5 )2 + ( 198943.68 )2 = 195
πd 3 3
d = 42.41 mm

Q.3 (a) Solution:

As per given information,
W = 440 N
k = 35430 N/m
c = 1240 Ns/m

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 19
Fluctuating pressure, P = 4276sin30t
Fluctuating force, F = (P × area)
F = 4276sin30t ×(0.0516)
F = 220.64 sin30t N
Maximum force, Fo = 220.64 N
k 35430
Natural frequency, ωn = =
m 440 / 9.81 = 28.1056 rad/s

c 1240
ξ = =
2mωn 2 × 440 × 28.1056
9.81
ξ = 0.4918 (hence this is case of underdamped system)
The steady state amplitude
Fo / k
x = A= 2
⎛ 2
⎛ ⎞ ⎞ 2
⎜ 1 − ⎜ ω ⎟ ⎟ + ⎛ 2ξ ω ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎝ ωn ⎠ ⎠ ⎝ ωn ⎠

220.64/35430
=
2 2 2
⎛ ⎛ 30 ⎞ ⎞ ⎛ 30 ⎞
⎜⎜ 1 − ⎜ ⎟ ⎟⎟ + ⎜ 2 × 0.4918 × ⎟
⎝ ⎝ 28.1056 ⎠ ⎠ ⎝ 28.1056 ⎠

A = X = 5.879 × 10–3 m
⎛ 2 ξω / ω n ⎞
The phase angle, φ = tan − 1 ⎜
(
⎜ 1 − (ω / ω n )
⎝
2
) ⎟
⎟
⎠

−1 ⎛ 2 × 0.4918 × 30/28.1056 ⎞
φ = tan ⎜ 2 ⎟ (IInd quadrant)
⎜ ⎛ 30 ⎞ ⎟
⎜ 1−⎜ ⎟ ⎟
⎝ ⎝ 28.1056 ⎠ ⎠
φ = (– 82.439)
π
φ = (180° − 82.439° ) ×
180°
φ = 1.7027 radian
Then the steady state displacement as a function of time
x = Xsin(ωt – φ)
x = 5.879 × 10–3 sin(30t – 1.7027) meter
The force FT transmitted to the base is the sum of the spring and damper forces.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
20 | ESE 2019 : MAINS TEST SERIES

FT = ( kx + cx )
FT = kX sin(ωt – φ) + cωX cos(ωt – φ)

2 2
The maximum value of FT = ( kX )2 + (cωX )2 = X k + (cω)

−3 2 2
= 5.879 × 10 35430 + (1240 × 30 )
FT = 302.0184 N

Q.3 (b) Solution:

As per given data

O
r′
r
P
5.5 kg M
θ
10 cm
C 20 cm
e γ
θ
a β
28 cm

I Q N
c b 4 cm

Mg ± f
2

M = 90 kg; r = 20 cm = r′, m = 5.5 kg;

⇒ h = 282 − 20 2 = 19.59 cm

a = 28 2 − 16 2 = 22.978 cm

e = (a + 10) cm
e = 22.978 + 10 = 32.978 cm
f = 0
tan β
k =
tan θ
16
tanβ = ⇒ β = 34.85°
22.978
20
tanθ = ⇒ θ = 45.59°
19.59

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 21
16
tanγ =
32.978
γ = 25.88°
β – γ = 8.96°
16 19.59
k = × = 0.682
22.978 20
Length, MN = ((10 + 22.978)2 + 162)1/2 = 36.65 cm
Given condition, (r = r′)
895 ⎛ a ⎞ ⎡ 2 mg + ( Mg ± f ) (1 + k ) ⎤
1. N2 = h ⎜ e ⎟⎢ ⎥ ...(1)
⎝ ⎠⎣ 2mg ⎦
895 ⎛ 0.22978 ⎞ ⎡ 2 × 5.5 × 9.81 + ( 90 × 9.81 )( 1 + 0.682 ) ⎤
N2 = ⎜ ⎟
0.1959 ⎝ 0.32978 ⎠ ⎢⎣ 2 × 5.5 × 9.81 ⎥⎦

N 2 = 46991.1812
N = 216.7744 rpm

O
r′
r
P
5.5 kg
M
θ1
10 cm
C 20 cm
e γ1
θ1
β1 β1
a′ 28 cm

I Q N
c′ b′ 4 cm

Mg ± f
2

20
sinγ1 =
36.65
γ 1 = 33.072°
m = 5.5 kg
M = 90 kg
β 1 = 33.072° + 8.96° = 42.032°
e = 30.711 cm

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
22 | ESE 2019 : MAINS TEST SERIES

a
cosβ 1 =
28
a′
cos 42.032° =
28
a′ = 20.797 cm

b′ = 282 − 20.797 2
b′ = 18.747 cm
r′ = 24 cm
r = b′ + 4 cm = 22.747 cm
22.747
sinθ1 =
28
⇒ θ1 = 54.33°
c′
tanθ1 =
a′
c′
tan54.33° =
20.797
c′ = 28.974 cm
Taking moment about I,
2 mg (c′ + r − r ′ ) + ( Mg ± f ) ( b′ + c ′ )
ω2 =
2 mr ′e
2 × 5.5 × 9.81 (0.28974 + 0.22747 − 0.24 ) + (90 × 9.81)(0.18747 + 0.28974 )
ω2 =
2 × 5.5 × 0.24 × 0.30711
ω2 = 556.56

556.56 × 60 2
⇒ N = = 225.282 rpm
( 2 π )2

Q.3 (c) Solution:

Given: d = 60 cm, r = 30 cm = 0.3 m, t = 2 cm
∈1 = 255 × 10–6
∈2 = 60 × 10–6
(a) γmax = ∈1 – ∈2 (In-plane)

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 23
= 195 × 10–6
From Hooke’s law for shearing stress and strain, we have
τmax = Gγmax
= 77.2 × 109 × 195 × 10–6
= 15.054 MPa
Note that this is maximum in plane shear stress as we calculated maximum in plane
strain.
Pr
As we know τmax(in plane) =
4t
P × 30
⇒ 15.054 =
4×2
⇒ P = 4.0144 MPa
γ –6
(10 )
2
D

γmax
(In-plane)
2

B A ∈(10–6)

∈2 = 60
E

∈1 = 255

(b) For a thin walled cylinder we know

τ

τmax D
τmax (in plane)
B C A
o σ2 σ1
σ1 −σ2 σ2
=
E 2 2

Pr
σ1 = = 60.216 MPa
t
Pr
σ2 = = 30.108 MPa
2t

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
24 | ESE 2019 : MAINS TEST SERIES

Pr
τmax = = 30.108 MPa (Max. absolute shear stress)
2t
Maximum shearing stress corresponding to a rotation of 45° about longitudinal axis.
Alternative Solution;
Given: Diameter of cylinder, d = 60 cm = 0.6 m
Wall thickness, t = 2 cm = 0.02 m
∈transverse = ∈1 = 255 × 10–6
∈longitudinal = ∈2 = 60 × 10–6
Shear modulus, G = 77.2 GPa
We have to first find the gauge pressure. Let’s take gauge pressure to be P and tank to
be thin cylinder vessel, then we have.
Pd
σhoop = σ1 =
2t
Pd
σlongitudinal = σ2 =
4t
Let’s take ν to be the Poisson’s ratio, then,
σ1 νσ 2 Pd νPd
∈1 = − = −
E E 2tE 4tE
Pd
= (2 − ν ) ...(1)
4tE
σ2 νσ1 Pd νPd
∈2 = − = −
E E 4tE 2tE
Pd
= (1 − 2ν ) ...(2)
4tE
Dividing eq. (1) and eq. (2)
∈1 2−ν
∈2 =
1 − 2ν
255 2−ν
⇒ =
60 1 − 2ν
⇒ 17 – 34ν = 8 – 4ν
⇒ 9 = 30ν
ν = 0.3
Given, G = 77.2 GPa

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 25
E
= 77.2 GPa
2 (1 + ν )
E = 77.2 × 2 × (1.3) = 200.72 GPa
Using eq. (1)
Pd
∈1 = (2 − ν )
4tE
P × 0.6
255 × 10–6 = ( 2 − 0.3 )
9
4 × 0.02 × 200.72 × 10
⇒ P = 4.0144 MPa
(ii) Now in second part we have to find principal stresses and maximum shearing stess.
We know in the given plane as there is no shearing force σ1 and σ2 are itself principal
stresses.

Pd 4.0144 × 10 6 × 0.6
σ1 = = = 60.216 MPa
2t 2 × 0.02
Pd
σ2 = = 30.108 MPa
4t
Also as we neglect radial stress due to pressure, we get
σ3 = 0
σ1 − σ 2
Maximum shear stress, (in plane 12) = = 15.054 MPa
2
σ1 − σ 3
Maximum shear stress(abs.) = = 30.108 MPa
2
Q.4 (a) Solution:
60 kN

50
1 2
θ
75 75
6 3
75
5 4

Primary force at each bolt:

60
P1′ = P2′ = P3′ = P4′ = P5′ = P6′ = = 10 kN
6

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
26 | ESE 2019 : MAINS TEST SERIES

As radius from centre of gravity is higher for 2 and 4 rivet and they are near to load P. So
2 and 4 rivets subjected to maximum force.

r2 = 752 + 50 2
r2 = 90.138 mm
75
tanθ =
50
θ = 56.31°
Secondary force due to moment at 2 (or 4);

P2′′ 2 2 2 2 2 2
P×e =
r2
(
r1 + r2 + r3 + r4 + r5 + r6 )
P2′′ (
60 × 200 = 90.1382 × 4 + 502 × 2 )
90.138
P2′′ = 28.844 kN
Resultant force on rivet 2:

P2′′sinθ
θ
θ

P2′′
P2′ + P2′′cosθ

2 2
P2 = ( P2' + P2'' cos θ) + ( P2'' sin θ)
P2 = (10 + 28.844 cos 56.31° )2 + ( 28.844 × sin 56.31° )2

P 2 = 35.383 kN
Diameter of rivets:
π 2
P4 = ×d τ
4

4 × 35.383 × 103
d = = 17.33 mm  18 mm
π× 150

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 27
Q.4 (b) Solution:
Different methods can be used for this problem, the given solution is using singularity
or Macualay method.
FBD of the beam,
P
a b
A B
RA RB
l

By taking moments at A and B, we get

Pa Pb
RB = and RA =
l l
Using singularity method.

d2y Pbx
EI = − P (x − a ) ...(1)
dx 2 l
On integrating,
2
dy Pbx 2 P ( x − a )
EI = − + c1 ...(2)
dx 2l 2
On integrating again,
3
Pbx 3 P ( x − a )
EIy = − + c1 x + c 2 ...(3)
6l 6
Using boundary condition, x = 0, y = 0
c2 = 0
Using another boundary condition x = l, y = 0

Pbl 2 Pb 3
So, 0 = − + c1l (as, l – a = b)
6 6
− Pb ( 2 Pab
⇒ c1 = l − b2 ) = − (l + b )
6l 6l
Now we have slope and deflection equation so we can find slope and deflection at
different location,
2
dy Pbx 2 P ( x − a ) Pab
1. EI = − − (l + b )
dx 2l 2 6l
At A, x = 0

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
28 | ESE 2019 : MAINS TEST SERIES

⎛ dy ⎞ − Pab
So, θA = ⎜ dx ⎟ = (l + b )
⎝ ⎠A 6EIl
At B, x = l
⎛ dy ⎞ 1 ⎡ Pbl Pb 2 Pab ⎤
So, θB = ⎜ dx ⎟ = ⎢ − − ( l + b )⎥
⎝ ⎠B EI ⎣ 2 2 6l ⎦
Pb ⎛ a ⎞
= ⎜ l − b − (l + b ) ⎟
2 EI ⎝ 3l ⎠
Pb ⎛ a ⎞
= ⎜ a − (l + b ) ⎟
2 EI ⎝ 3l ⎠

Pab ⎛ 3l − l − b ⎞ Pab Pab

= ⎜ ⎟ = 6EI l ( 2 l − b ) = 6EI l ( l + a )
2EI ⎝ 3l ⎠

2. As given a > b, we can observe that maximum deflection will be in the region
dy
x∈(o, a). For maximum deflection, = 0.
dx
So, using slope equation we get

Pbx 2 Pab (l + b )
0 = −
2l 6l
a (l + b )
⇒ x = (You can observe this x is less than a)
3
Putting this value of x in elastic curve or deflection curve, we get

Pb ⎛ a (l + b ) ⎞ a ( l + b ) Pab a (l + b )
EIy max = ⎜ ⎟ − (l + b )
6l ⎝ 3 ⎠ 3 6l 3

Pab ( a (l + b ) ⎛ 1 ⎞
EIy max = l + b) ⎜ − 1⎟
6l 3 ⎝3 ⎠

− Pab ( l + b ) a ( l + b )
⇒ ymax =
9lEI 3
By comparing we can observe that,
k = –9
Where minus sign shows the deflection in downward direction.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 29
Q.4 (c) Solution:
mw = 0.7 kg; r = 0.04 m, mf = 0.3 kg
N = 3000 rpm
2 πN 2 π × 3000
ω = = = 314.16 rad/s
60 60
5
Vehicle speed = 60 km/h = 60 × = 16.67 m/s
18
Radius of circular path, R =100 m
V 16.67
ωp = = = 0.1667 rad/s
R 100
Assuming wheel as a disc,
1 1
Iω = Mr 2 = × 0.7 × 0.04 2 = 560 × 10–6 kgm2
2 2
ω

θ o
ωcosθ

θ
0.02
2
mV
R

mV 2
Iωωpcosθ = 0.02 sin θ× mg + × 0.02 cos θ
R
⎛ mV 2 ⎞
cos θ ⎜ I ωωp − × 0.02 ⎟ = 0.02 × mgsinθ
⎝ R ⎠

⎛ mV 2 ⎞
⎜ I ωω p − × 0.02 ⎟
tanθ = ⎜⎜ R ⎟ =0
0.02 mg ⎟
⎝ ⎠

−6 0.3 × 16.67 2 × 0.02

560 × 10 × 0.1667 × 314.16 −
100
tanθ =
0.02 × 0.3 × 9.81
tanθ = 0.2149
θ = 12.133°

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
30 | ESE 2019 : MAINS TEST SERIES

Section B
Q.5 (a) Solution:
Given, Initial diameter, d1 = 200 mm
Initial height, h1 = 120 mm
Final height, h2 = 90 mm
Coefficient of friciton, μ = 0.25
⎛ 2μr2 ⎞
Average pressure, pavg. = σ f ⎜ 1 + 3h ⎟
⎝ 2 ⎠
{where, r2 is final radius and h2 is final height}
⎛h ⎞ ⎛ 120 ⎞
True strain, ∈1 = ln ⎜ o ⎟ = ln ⎜ ⎟ = 0.28768
⎝ h1 ⎠ ⎝ 90 ⎠

Flow stress, σf = k ∈n1 = 1000 ∈0.17

1 = 1000(0.28768)0.17

= 809.123 MPa
By volume conservation,

π 2 π
d1 h1 = d22 h2
4 4

(200)2 × 120 = d22 × ( 90 )

Final diameter, d2 = 230.94 mm

⎛ 2μr2 ⎞
Average pressure, pavg. = σ f ⎜ 1 + 3h ⎟
⎝ 2 ⎠

⎛ 2 × 0.25 × 230.94 ⎞
= 809.123 ⎜ 1 + ⎟
⎝ 2 × 3 × 90 ⎠
pavg. = 982.140 MPa
Upsetting force, F = (pavg.) × (πr22)
2
⎛ 230.94 ⎞
= (982.140 ) × π× ⎜ ⎟
⎝ 2 ⎠
F = 41139745.75 N
Upsetting force at the end of the stroke, F = 41.1397 MN
Hence, upsetting force required at the end of the stroke is 41.1397 MN

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 31
Q.5 (b) Solution:
1. Accumulator (ACC) register:
• It is an 8-bit special purpose register that is a part of the ALU. It is also identified
as register A.
• In arithmetic and logical operations the accumulator may store the operand,
execute an instruction with the help of other registers, and memory and finally
store the result of the operation. In the former case it acts as a source, and in the
latter a destination.
2. General purpose registers:
• The 8085 microprocessor contains six -bit general purpose registers. These are
identified as B, C, D, E, H and L.
• These registers are used in microprocessor for temporary storage of operands or
intermediate data in calculations.
• These registers can be used either simply for storage of 8-bit data or in pairs for
storage of 16 bit data. When used in pairs, only selected combination can be used
for pairing, i.e., B-C, D-E and H-L. When registers are used in pairs the high
order byte resides in the first register and low order byte in the second register.
3. Stack pointer(SP):
• It is a 16 bit special function register.
• The stack is a sequence of memory locations set aside by a programmer to store/
retrieve the contents of accumulator, flags, program counter and general purpose
registers during the execution of a program. Any portion of the memory can be
used as a stack.
• In this register, data is stored temporarily on first come and last go basis.
4. Program counter (PC):
• It is a 16-bit special purpose register and is used to hold the memory address of
the next instruction to be executed.
• The contents of the PC are automatically updated by the microprocessor during
the execution of an instruction so that at the end of execution it points to the
address of the next instruction in the memory.
• The microprocessor uses the PC for sequencing the execution of instructions.
5. Instruction register:
• During the execution of a program, microprocessor addresses some memory,
which supplies an 8-bit data of instruction code to the data bus which gets stored
in the register called the instruction register.
• The instruction register holds the op-code (operation code or instruction code) of
the instruction which is being decoded and executed.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
32 | ESE 2019 : MAINS TEST SERIES

Q.5 (c) Solution:

If, C = `250, Ch = 10% of C

2 DC0 2 × 15000 × 2500

EOQ = = = 1732.051 units
Ch 250 × 0.1

This calculated EOQ do not fall under offered lot size of 1-1499, so we have to go to next
unit price.
If C = `235, Ch = 0.1 × 235 = `23.5

2 DC0 2 × 15000 × 2500

EOQ = = = 1786.474 units
Ch 23.5

TIC(EOQ) = 2 DC 0C h = 2 × 15000 × 2500 × 23.5 = `41982.14

Total cost, (TC) = TIC + D × C = 41982.14 + 15000 × 235

= `3566982.14
Q D
Total cost, C(Q = 2500) = D × C + × Ch + × C0
2 Q

⎛ 2500 ⎞ ⎛ 15000 ⎞
= (15000 × 225) + ⎜⎝ × 225 × 0.1⎟ + ⎜
⎠ ⎝ 2500
× 2500⎟
⎠
2
= 3375000 + 28125 + 15000
TC = `3418125
Q D
Total cost, C(Q = 5000) = D × C + × Ch + × C0
2 Q
15000 5000
= (15000 × 220) + × 2500 + × 22
5000 2
= 3300000 + 7500 + 55000 = `3362500
As evident from total inventory cost which is minimum for Q = 5000, best lot size for
ordering is 5000 and above.

Q.5 (d) Solution:

Cast Iron: Cast iron are defined as the iron-carbon alloys that contain carbon between
2.1% and 6.67%. However commercial cast irons normally contain less than 4.5% of
carbon. They can be easily melted because of the high percent of carbon and can be
castable to required shape because of their fluidity.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 33
Gray cast iron microstructure consists of graphite flakes dispersed throughout the metal
matrix. The tips of the graphite flakes are sharp and pointed, and may act as the sites for
crack initiation, when an external tensile stress is applied.
Gray cast iron Malleable cast iron

(i) C - 2.5 – 4% C - 2.3 – 2.7%

Composition and Si - 1.3% Si - 1–1.7%
heat treatment Mn - 0.7 – 0.8% Mn - <0.55%
No special heat treatment required. Requires initial rapid cooling for formation of white
cast iron, later, white cast iron is held at 800-900°C
for more graphite formation, which leads to formation
of malleable cast iron.

(ii)

Graphite flakes Graphite clusters

(iii) → Weak in tension, but excellent → Increased ductility

Mechanical compressive strength. → Higher corrosion resistance
characteristics → Good machinability as graphite flakes → Excellent impact strength
offer lubrication and chip breaking. → Higher tensile strength
→ Very good damping characteristics as graphite
flakes absorb energy.
→ Good resistance to adhesive wear.
(iv) As damper machine base materials, engine General engineering equipments and others
Application blocks of automobiles including. connecting rods, transmission
gears and other heavy duty services.

Q.5 (e) Solution:

Given: P = 4 + 0.8L – 0.1L2
dP
For maximum power, = 0
dL
0.8 – 0.2L = 0
L = 4 mm
Maximum power, P = 4 + 0.8 × 4 – 0.1 × 42 = 5.6 kW
Efficiency, η = 75%
1
Area of bead, A = × BC × AD
2
6tan 30°
B D C

30° 30° 6 mm

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
34 | ESE 2019 : MAINS TEST SERIES

1
= × ( 2 × 6 tan 30° ) × 6 = 20.7846 mm2
2
Volume of weld bead = (A × l)
= (20.7846 × 800) = 16627.68 mm3 = 16.62768 cm3
Weight of weld bead = V × ρ
= (16.62768 × 7880 × 10–3) gram
= 131.026 gram
Heat required for melting = 131.026 × 1440 = 188677.44 Joule
= 188.67744 kJ
188.67744
Time required for welding = = 44.9232 s
0.75 × 5.6
800
Speed of welding = = 17.808 mm/s
44.9232
Result: Speed of welding will be 17.808 mm/s

Q.6 (a) Solution:

Hydraulic actuators: An actuator wherein hydraulic energy is used to impart motion is
called hydraulic actuators. All systems involving high loads are operated by hydraulic
actuators in which oil pressure is applied on mechanical actuator to give an output in
terms of rotary or linear motion. Basic example of hydraulic actuator is steering gear of
ship in which hydraulic pressure is used to move the rudder actuator. It is based on
principle of Pascal’s law.
Advantages:
1. Easy to produce transmit, store, regulate and control, maintain and transform the
hydraulic power.
2. Possible to generate high gain in force and power amplification.
3. Hydraulic systems are uniform and smooth, generate stepless motion and variable
speed and force to a greater accuracy.
4. It is used where very high speed and large forces are required. It has higher load
capacity.
5. Weight to power ratio of an hydraulic system is comparatively less than that for
an electro mechanical system.
Disadvantages:
1. The manufacturing cost of the system is high since the hydraulic elements have
to be machined to a high degree of precision.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 35
2. Hydraulic elements have to be specially treated to protect them against rust,
dirt, corrosion etc.
3. Petroleum based hydraulic oil may pose fire hazards thus limiting the upper
level of working temperature.
4. Hydraulic power is not readily available compared to electric power.

Central valve
Pressure P A
Motor Pump
regulator

Filter T B

Tank
Oil Cylinder

Block diagram of hydraulic system

Pneumatic actuators: It uses pressurized air to transmit and control power. In this type,
compressed air at high pressure is used which converts this energy into either linear or
rotary motion. Pneumatic actuators enable large forces to be produced from relatively
small pressure changes. The most common example is “Main engine pneumatic
Actuator” used for changing of roller position over cam shaft for reversing.
Advantages:
1. Pneumatic systems are fire and explosion proof whereas hydraulic system are
not, unless non flammable liquid is used.
2. In this no return pipes are used when air is used.
3. Ecological purity.
4. Low cost
5. High speed operation
6. Pneumatic system are insensitive to temperature changes in contrast to hydraulic
system in which fluid friction due to viscosity depends greatly on temperature.
7. Ease in reversal of movements.
Disadvantages:
1. The normal operating pressure of pneumatic system is lower than that of hydraulic
system.
2. Accuracy of pneumatic actuators is poor at low velocities.
3. Output power is less compared to hydraulic system.
4. Difficulties in performance at slow speed.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
36 | ESE 2019 : MAINS TEST SERIES

5. In pneumatic systems, external leakage is permissible to a certain extent, but

internal leakage must be avoided because the effective pressure difference is
rather small. In hydraulic system internal leakage is permissible to a certain extent,
but external leakage must be avoided.
ON/OFF Control
Motor

Central valve
P A
Air in

Filter Compressor Reservoir

R B
Air

Air treatment unit Exhaust (Air)

Cylinder

Block diagram of pneumatic system

Q.6 (b) Solution:

(i)

MRP Input MRP Output

Engineering Bill of MRP Processing Planned order schedule
changes materials
file
Inventory order schedule

Sales forecast Master

customer MRP
production Planning reports
orders SYSTEM
schedule

Performance reports
Service parts Inventory
requirements
record
inventory file Exception reports
status file

Primary outputs of MRP:

1. Planned order schedule.
2. Order release note.
3. Rescheduling notices.
4. Cancellation notices.
5. Reports on inventory status.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 37
Secondary outputs of MRP:
1. Exception report
2. Performance report
3. Planning report
(ii)
The functions of inventory control are:
1. to minimize the capital investment in inventory by eliminating excessive stock,
2. to ensure availability of needed inventory for uninterrupted production and for
meeting customer demand.
3. to provide a scientific basis for planning of inventory needs,
4. to minimize the risk of loss due to obsolescence, deterioration, etc. and
5. to tide over the demand fluctuations by maintaining reasonable safety stock.
In ABC analysis, the inventories of an organization are assumed to be of non equal
value. Thus, the inventory is grouped into three categories (A, B and C) in order of their
estimated importance.
‘A’ items are very important for an organization. These items are of high annual usage
value and have a significant impact on overall inventory cost. Because of the high value
of these ‘A’ items, frequent value analysis is required. These items are held under very
tight control and accurate records. In addition to that, an organization needs to choose
an appropriate order pattern (e.g. just-in-time) to avoid excess capacity. ‘A’-items have
tight inventory control, more secured storage areas and better sales forecasts. Re-orders
should be frequent and avoiding stock-outs on A-items is a priority.

Q.6 (c) Solution:

(i) Let, the true stress and true strain relationship is,
σT = k(εTn)
where, σT is true stress, ∈T true strain, k and n are constants.
We know that, σT = σ(1 + ε)
(σT)1 = 195 (1 + 0.176) = 229.32 MPa
(σT)2 = 259 (1 + 0.289) = 333.851 MPa
True strain, εT = ln(1 + ε)
(εT)1 = ln(1 + 0.176) = 0.16212
(εT)2 = ln(1 + 0.289) = 0.25387
Now, σT = kεTn
lnσT = lnk + nlnεT

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
38 | ESE 2019 : MAINS TEST SERIES

Putting values of (σT) and (εT) from above conditions,

ln(229.32) = lnk + nln(0.16212)
5.43512 = lnk + n(–1.81942)
lnk – (1.81942)n = 5.43512 ...(1)
By second given condition,
ln(333.851) = lnk + nln(0.25387)
5.81069 = lnk –(1.37093)n
lnk – (1.37093)n = 5.81069 ...(2)
On solving equation (1) and (2),
lnk = 6.95875, n = 0.83741
k = 1052.28 MPa
σT = 1052.28(εT0.83964) MPa
For engineering strain of 0.245,
True strain, εT = ln(1 + ε) = ln(1.245) = 0.2191
True stress, σT = 1052.28 × (0.2191)0.83964
σT = 294.12 MPa
σT 294.12
Engineering stress, σ = ( =
1 + ε ) 1 + 0.245
Engineering stress, σ = 236.241 MPa

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 39
Q.6 (c) Solution:

(ii)
L+δ
1535°C
δ-ferrite δ
(BCC) Liquid
γ+δ
5
L+γ L + Fe3 C
1410°C

Austenite γ 4 1150°C
(FCC)
Temperature

γ + Fe3 C
910°C
α+γ
ferrite (BCC) α
725°C
Steels 3 Cast-irons
2
Hyper α + Fe3C
1 eutectoid
steels Hypo
Hypo eutectic Hyper eutectic cast iron
cast-iron
eutectoid steels
(Fe3C)
100%Fe 0.02% 0.8% 2.1% 4.3% 6.67%C

Percentage of Carbon

Q.7 (a) Solution:

Step I: Converting the problem of maximization into equivalent minimization problem
by subtracting each element of the matrix from the highest element of the matrix.
9 3 4 2 10
12 10 8 11 9
11 2 9 0 8 (Here, Highest element = 14)
8 0 10 3 7
7 5 6 2 9

Step II: Develop opportunity cost matrix

(a) Subtract the smallest element of each row from every element of the corresponding
row.
(b) Subtract the smallest element of each column from every element of the
corresponding column .
7 1 2 0 8 3 1 2 0 7
4 2 0 3 1 0 2 0 3 0
11 2 9 0 8 7 2 9 0 7
8 0 10 3 7 4 0 10 3 6
5 3 4 0 7 1 3 4 0 6

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
40 | ESE 2019 : MAINS TEST SERIES

Step III: Make allocations in the opportunity cost matrix (i.e. put square on the zeros
and cross all zeros (if any) of the corresponding row/column).
3 1 2 0 7
0 2 0 3 0
7 2 9 0 7
4 0 10 3 6
1 3 4 0 6

Since total number of allocations is less than the size of the matrix. So, current solution
is not optimal. We have to perform optimally.
Step IV: Draw the minimum number of lines to cover all the zeros. Select the smallest
element that do not have line through them. Subtract it from all the elements that do
not have line through them. Add it to elements at the intersection of two lines and
leave the remaining element of the matrix unchanged. Then again make allocations.
3 1 2 0 7 2 0 1 0 6
0 2 0 3 0 0 2 0 4 0
7 2 9 0 7 6 1 8 0 6
4 0 10 3 6 4 0 10 4 6
1 3 4 0 6 0 2 3 0 5

Since total number of allocations is less than the size of the matrix, so current solution
is not optimal. Again we have to perform optimally.
Step V: Repeat step IV
2 0 1 0 6 2 0 0 0 5
0 2 0 4 0 1 3 0 5 0
6 1 8 0 6 6 1 7 0 5
4 0 10 4 6 4 0 9 4 5
0 2 3 0 5 0 2 2 0 4

Since total number of allocations is equal to the size of the matrix. So, current solution
is optimal.
Optimal Assignment:

Machine Job
1 3
2 5
3 4
4 2
5 1
Maximum profit = 10 + 5 + 14 + 14 + 7 = Rs. 50

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 41
Q.7 (b) Solution:
(i)
Load cells are elastic devices that can be used for measurement of force through indirect
methods i.e., through use of secondary transducers.
Load cells utilize an elastic member as the primary transducer and strain gauges as
secondary transducer. When the combination of the strain gauge-elastic member is used
for weighing, it is called a “load cell”.
While designing load cells using strain gauges the following factors should be
considered:
i. Stiffness of the elastic element.
ii. Optimum positioning of gauges on the element.
iii. Provision for compensation of the temperature.
When large loads are to be measured, the direct tensile -compressive member may be
used, whereas, in case of small loads, strain amplification provided by bending may be
used with advantage.
Strain-gauge load cells: These cells convert weight or force into electrical outputs which
are provided by the strain gauges; these outputs can be connected to various measuring
instruments for indicating, recording and controlling the weight or force.
Usually the strain gauges are directly applied to the force developing device and the
device is calibrated against strain-gauge output.
• These are excellent force measuring devices, particularly for transient and non-steady
forces.
• These are used in conjuction with CRO (for display purposes) for measurement of
rapidly changing loads.
Construction and working of the load cell:
It consists of a steel cylinder, on which four identical strain gauges are mounted. The
gauges Rg1 and Rg4 are along the direction of applied load and the gauges Rg2 and Rg3
are attached circumferentially to gauges Rg1 and Rg4. All the four gauges are connected
electrically to the four limbs of a Wheatstone bridge circuit.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
42 | ESE 2019 : MAINS TEST SERIES

Load (force) B

Rg1 Rg3

A V0 C
Steel cylinder

Rg1 Strain gauge Rg2 Rg4

Rg2 Rg3
Rg4
D

key
V
Wheatstone bridge circuit
Load cell

When there is no load on the cell, all the four gauges have the same resistance
(i.e. R g1 = R g2 = R g3 = R g4). Obviously the terminals B and D are at the same
potential, the bridge is balanced and the output voltage is zero.

V
i.e., VAB = VAD =
2
and, VAB – VAD = Vo = 0
On the application of a compressive load to the unit, the vertical gauges (Rg1 and Rg4)
undergo compression (i.e., negative strain) and, therefore, there is decrease in resistance.
The circumferential gauges Rg2 and Rg3, simultaneously, undergo tension (i.e., positive
strain) leading to increase in resistance. The two strains are not equal; these are related
to each other by a factor, μ, the Poisson’s ratio.
Voltage due to applied load will be given as:

⎛ dR V ⎞
Vo = (1 + μ ) ⎜ × ⎟
⎝ R 2⎠
Obviously, this voltage is a measure of the applied load.
The use of four identical strain gauges in each arm of the bridge provides full temperature
compensation and also increases the sensitivity of the bridge 2(1 + μ)times.
Uses: The strain gauge load cells find extensive use in the following:
i. Road vehicle weighing devices.
ii. Draw bar and tool force dynamometers.
iii. Crane load monitoring etc.

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 43
(ii)
We know that, output voltage is:

Vi
Vo = (1 + μ ) (G.F )×∈×
2
G.F = Gauge factor
μ = Poission’s ratio
∈ = Strain
Vi = Supply voltage

σ 1 × 106
We know that ∈ = = = 5 × 10–6
E 200 × 109

So, output voltage,

−6 6
Vo = ( 1 + 0.3 ) × 2 × 5 × 10 × = 3.9 × 10–5
2
Vo = 39 μV
Sensitivity of load cell wheat stone bridge,

Vo 39 × 10 −6
S = = = 13 × 10–9 V/N
P 3 × 10 3

Q.7 (c) Solution:

(i)
1. Size of abrasive: By increasing size of abrasive MRR increases because impact
will be on more area, so MRR increases. Because standoff distance is large so
MRR will never decrease.
MRR

Size of abrasive

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
44 | ESE 2019 : MAINS TEST SERIES

2. Percentage of grains: As % of grains increases MRR increases because of increased

concentration of abrasives. But after a limit grain will start colliding to each other
this will decrease the MRR.
MRR

% Grains
3. Standoff distance: Initially by increasing the standoff distance there will be more
and more momentum due to increased acceleration time. These abrasives are
thrown into atmosphere, so there will be drag between abrasives and atmospheric
air, so in the intermediate portion of curve there will be a balance between
acceleration and drag. So the curve will be flat. But after certain standoff distance
drag will overcome the acceleration and hence MRR decreases.
MRR

Standoff distance

(ii)
The flat, a glass or fused quartz disk with parallel flat surfaces, is placed on the surface
of workpiece. When a monochromatic light beam is aimed at the surface at an angle, the
optical flat splits it into two beams, appearing as light and dark bands to the naked eye.
The number of fringes that appear is related to the distance between the surface of the
part and the bottom surface of the optical flat.
A truly flat workpiece surface (that is when the angle between two surface is zero) will
not split the light beam and no fringes will appear.
L2
Optical
M flat
J
L1 G
F L
C
A
H

E
B
D K

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 45
L1 is a ray of monochromatic light, a portion of this ray will be reflected from B and a
portion will be transmitted.
The transmitted ray will reflect back from the surface and come out of optical flat from
F(as shown). For ray BC and EF there will be path difference equal to (BD + DE). If this
λ
path difference is equal to there will be a dark band if this path difference is λ, a white
2
fringe will appear.
So on the other end of optical flat the total path difference is (HK + KL).
Suppose there are ‘n’ fringes seen in optical flat (we count number of white fringes),
then this path difference is equal to nλ.
HK  KL 2HK = nλ

nλ
HK =
2
Since angle is very small so HK can be approximated to a vertical line. If l is the length
of surface(DK).
HK
= tanθ  θ
l
Using this equation inclination angle of surface can be calculated.

Q.8 (a) Solution:

Ist case:
Given: Wα = 0.78, Wβ = 0.22, C0 = 30% (wt% of Q)
We know that,
Cβ − C0
Wα = C − C
β α
Cβ − 0.30
0.78 = C − C ...(i)
β α
C0 − Cα
Wβ = C − C
β α
0.30 − Cα
0.22 = C − C ...(ii)
β α
Case IInd case:
Given: Wα = 0.36, Wβ = 0.64, C0 = 65% (wt% of Q)

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
46 | ESE 2019 : MAINS TEST SERIES

We know that,
Cβ − C0
Wα = C − C
β α
Cβ − 0.65
0.36 = C − C ...(iii)
β α
C0 − Cα
Wβ = C − C
β α
0.65 − Cα
0.64 = C − C ...(iv)
β α
Equation (i) divided by equation (iii)

0.78 Cβ − 0.30
= C − 0.65
0.36 β

2.1667 (Cβ – 0.65) = Cβ – 0.30

1.1667 (Cβ) = 1.408355 – 0.30
C β = 0.94999
 0.95
Equation (ii) divided by equation (iv)

0.22 0.30 − Cα
= 0.65 − C
0.64 α

0.34375 (0.65 – Cα) = 0.30 – Cα

0.65625 (Cα) = 0.30 – (0.34375 × 0.65)
0.0765625
Cα = = 0.1167
0.65625
P-rich α -phase composition : 88.33 wt% of P and 11.67 wt% of Q
Q-rich β -phase composition : 5 wt% of P and 95 wt% of Q

Q.8 (b) Solution:

Volume of casting, Vc = 25 × 12.5 × 5 = 1562.5 cm3
Surface area of casting, Ac = 2(25 × 12.5 + 12.5 × 5 + 5 × 25) = 1000 cm2

⎛V ⎞ 1562.5
For casting, ⎜ ⎟ = = 1.5625
⎝ A ⎠C 1000

For side riser, Diameter of side riser = Height of side riser

d = h
© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 47

π 2 ⎛1⎞ 3
V
⎛ ⎞ d h ⎜ ⎟d d
⎜ ⎟ = 4 = ⎝4⎠ =
⎝ A ⎠r 2π 2 3 2 6
d + πdh d
4 2
(i) Using Caine’s method,

⎛V ⎞
⎜ ⎟
⎝ A ⎠r
Freezing ratio, x =
⎛V ⎞
⎜ ⎟
⎝ A ⎠c
a
x = +c ...(1)
y−b

π 2
d h
Vr πd 3
where, y = = 4 =
Vc 1562.5 6250

⎛V ⎞ ⎛d⎞
⎜ ⎟ ⎜ ⎟
⎝ A ⎠r d
Now, x = = ⎝6⎠ =
⎛V ⎞ 1.5625 9.375
⎜ ⎟
⎝ A ⎠c
Now, putting value of x, y in equation (1),

⎛ a ⎞
x = ⎜ y −b ⎟+c
⎝ ⎠

d 0.1
= +1
9.375 3
πd
− 0.03
6250

⎡ 0.1 × 6250 + πd 3 − 0.03 × 6250 ⎤

d ⎣ ⎦
= 3
9.375 πd − 0.03 × 6250

πd 4 – 187.5d = 9.375(πd3 + 437.5)

πd 4 – 9.375 πd 3 – 187.5d – 4101.5625 = 0
On solving, d = 10.889 cm

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
48 | ESE 2019 : MAINS TEST SERIES

Hence dimensions of optimum cylindrical side riser,

Diameter of side rise = Height of side riser = 10.889 cm
(ii) Using modified Caine’s method:
L+W 25 + 12.5
Shape factor, S.F. = = = 7.5
T 5
From given table,
⎛ Vr ⎞
⎜ V ⎟ = 0.5
⎝ c⎠
(From table Vr/Vc value corresponding to shape factor)
π 2
d h
4 = 0.5
1562.5
πd 3 = 0.5 × 1562.5 × 4
d3 = 994.7184
Diameter = Height = d = 9.98 cm

Q.8 (c) Solution:

L3
z1 y1 z2 y2

zo yo

{1} {2}
x1
L2
{0}
xo d3

L1
{3}
x3

y3 L4
z3

{4} x4

y4
z4

© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in
 Test No : 15 MECHANICAL ENGINEERING | 49

Link i ai αi di θi qi cθi sθi cα i sα i

1 L1 0 L2 θ1 = 30° θ1 0.866 0.5 1 0
2 L3 0 0 θ2 = 45° θ2 0.707 0.707 1 0
3 0 d
180° 3 0 d3 1 0 −1 0
4 0 0 L4 θ4 = 30° θ4 0.866 0.5 1 0
We know that,

⎡C θi −SθiC α i Sθi Sα i aiC θi ⎤

⎢ Sθ C θiC αi −C θi Sαi ai Sθi ⎥⎥
⎢ i
i – 1T = ⎢ 0 Sαi C αi di ⎥
i
⎢ ⎥
⎣ 0 0 0 1 ⎦

⎡0.866 −0.5 0 50 × 0.866 ⎤

⎢ 0.5 0.866 0 50 × 0.5 ⎥
⎢ ⎥
0T (θ ) = ⎢ 0 0 1 5 ⎥
1 1
⎢ ⎥
⎣ 0 0 0 1 ⎦

⎡0.707 −0.707 0 50 × 0.707 ⎤

⎢0.707 0.707 0 50 × 0.707 ⎥⎥
⎢
1T (θ ) = ⎢ 0 0 1 0 ⎥
2 2
⎢ ⎥
⎣ 0 0 0 1 ⎦

⎡1 0 0 0 ⎤
⎢0 −1 0 0 ⎥
⎢ ⎥
2T (d ) = ⎢ 0 0 − 1 50 ⎥
3 3
⎢ ⎥
⎣0 0 0 1 ⎦

⎡0.866 − 0.5 0 0 ⎤
⎢ 0.5 0.866 0 0 ⎥
⎢ ⎥
3T (θ ) = ⎢ 0 0 1 30 ⎥
4 4
⎢ ⎥
⎣ 0 0 0 1⎦

Overall transformation matrix is 0T4 = 0T1 × 1T2 × 2T3 × 3T4



© Copyright: www.madeeasy
www.madeeasy.in
.madeeasy.in