Theory of Machine
Exam Duration: 45 Mins Total Questions : 30
Which of the following is a turning pair?
- (a)
Piston and cylinder of a reciprocating steam engine
- (b)
Shaft with collars are both ends fitted in a circular hole
- (c)
Ball and socket joint
- (d)
Lead screw of lathe with nut
The motion transmitted between the teeth of gears in mesh is
- (a)
rolling
- (b)
sliding
- (c)
may be rolling or sliding depending upon the shape of teeth
- (d)
partly sliding and partly rolling
In a four bar linkage, if the lengths of shortest, longest and the other two links denoted by s, I, p and q, then it would result in Grashof's linkage provided that
- (a)
l + p < s + q
- (b)
l + s < p + q
- (c)
l + p > s +q
- (d)
l + p = s + q
If n links are connected at same joint, the joint is equivalent to
- (a)
(n - 2) binary joint
- (b)
(n - 1) binary joint
- (c)
(2n - 1) binary joint
- (d)
None of these
A ball and a socket joint form a
- (a)
sliding pair
- (b)
turning pair
- (c)
rolling pair
- (d)
spherical pair
In the shown mechanism, it is given that
AB = 30 cm, BD = 28 cm
BC = 60 cm, DC = 40 cm and speed of crank AB is 500 rpm then angular velocity of link BC is
- (a)
23 rad/s
- (b)
14.7 rad/s
- (c)
19.6 rad/s
- (d)
None of these
Draw instantaneous centres
\({ \omega }_{ 2 }=\frac { 2\pi N }{ 60 } \)
\({ \omega }_{ 2 }=\frac { 2\times 31.4\times 500 }{ 60 } \)
= 52.35 rad/s
\(\Rightarrow \frac { { \omega }_{ 3 } }{ { \omega }_{ 2 } } =\frac { { I }_{ 12 }{ I }_{ 23 } }{ { I }_{ 13 }{ I }_{ 23 } } \)
\(\Rightarrow \frac { { \omega }_{ 3 } }{ 52.35 } =\frac { 30 }{ 80 } \)
\(\Rightarrow { \omega }_{ 3 }=19.6\ rad/s\)
If mechanism forms a structure then the number of degree of freedom (n) is
- (a)
0
- (b)
2
- (c)
1
- (d)
-1
The component of the acceleration, parallel to the velocity of the particle, at the given instant is called
- (a)
coriolis component
- (b)
tangential component
- (c)
radial component
- (d)
None of these
In a quick return motion mechanism, length of crank radius is 3 cm and length of fixed end is 5 cm, then the ratio of time of cutting stroke to return stroke is
- (a)
1.2
- (b)
2.38
- (c)
3.7
- (d)
4.2
\(cos\ \theta =\frac { 3 }{ 5 } \)
\(\Rightarrow \) \(\theta\) = 53.13°
Angle in return stroke
= 2 \(\times\) 53.13
= 106.26°
Angle in cutting stroke = 360 - 106.26 = 253.7°
\(\frac { Time\ of\ cutting\ stroke }{ Time\ of\ return\ stroke } =\frac { 253.7° }{ 106.26° } =2.38\)
The natural angular frequency of the system will be
- (a)
\(\sqrt { \frac { 3({ k }_{ 1 }+{ k }_{ 2 }) }{ m } } \)
- (b)
\(\sqrt { \frac { ({ k }_{ 1 }+{ k }_{ 2 }) }{ 3m } } \)
- (c)
\(\sqrt { \left( \frac { { k }_{ 1 }+{ k }_{ 2 } }{ 3m } \right) } \)
- (d)
\(\sqrt { \left( \frac { 3m }{ { k }_{ 1 }+{ k }_{ 2 } } \right) } \)
From figure, x = e
Total force acting on the bar in upward direction
= (k1 + k2)x
Then equivalent torque
= (k1 + k2) l x
\(\tau\) = (k1 + k2) l l\(\theta\)
=(k1+k2) l2\(\theta\)
The initial displacement of given bar is in downward direction
\(\tau\) = - (k1 + k2) l2\(\theta\)
1\(\alpha\)= - (k1 + k2)l2\(\theta\)
\(\frac { m{ c }^{ 2 } }{ 3 } \times \alpha \) =-(k1+k2) l2\(\theta\)
(By theorem of parallel axis
I0=IG+\(\frac { m{ l }^{ 2 } }{ 4 } =\frac { m{ l }^{ 2 } }{ 12 } +\frac { m{ l }^{ 2 } }{ 4 } =\frac { m{ l }^{ 2 } }{ 3 } \))
\(\alpha =-3\left( \frac { { k }_{ 1 }+{ k }_{ 2 } }{ m } \right) \theta \)
\(\alpha =-{ \omega }^{ 2 }\theta \)
\(\therefore \omega =\sqrt { \frac { 3({ k }_{ 1 }+{ k }_{ 2 }) }{ m } } \)
If the mass of the rod is m, then the natural angular frequency of the system will be
- (a)
\(\sqrt { \frac { k }{ m } } \)
- (b)
2\(\sqrt { \frac { k }{ m } } \)
- (c)
\(\sqrt { \left( \frac { { k }_{ 1 }+k_{ 2 } }{ 3m } \right) } \)
- (d)
\(\sqrt { \frac { 3m }{ { k }_{ 1 }+k_{ 2 } } } \)
From figure,
y = l/2\(\theta\)
x = I\(\theta\)
Net upward torque
= 2kx x Z+ 4ky + l/2
\(\tau\) = 2k Ie Z+4kxl/2\(\theta\) 1/2
\(\tau\) = 3kz2\(\theta\)
To give initially motion, the bar is in downward direction
\(I\alpha =-(3k{ l }^{ 2 }\theta )\)
\(\frac { m{ l }^{ 2 } }{ 3 } \alpha =-3k{ l }^{ 2 }\theta \)
\(\alpha =-\frac { 9k }{ m } \theta \)
\(\omega =\sqrt { \frac { 9k }{ m } } =3\sqrt { \frac { k }{ m } } \)
What will be the amplitude of vibratory motion in given figure?
- (a)
13.33 cm
- (b)
14.33 cm
- (c)
16.33 cm
- (d)
12.33 cm
F0 = F cos rot
F = 200N
\(\omega \) = 50
m = 1 kg
s = 4 X 103 N/m
We know that
x=\(\frac { F }{ \sqrt { ({ c\omega ) }^{ 2 }+(s-m{ \omega }^{ 2 }{ ) }^{ 2 } } } \)
=\(\frac { 200 }{ 4\times { 10 }^{ 3 }-2500 } \)
=\(\frac{200}{1500}\)
=\(\frac{2}{15}\)m
= \(\frac{200}{15}\)cm
=13.33 cm
\(m\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +c\frac { dx }{ dt } \)sx = 0 shows a vibration system, the angular frequency of vibration system is
- (a)
\(\sqrt { s/m } \)
- (b)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
- (c)
\(\frac{s}{m}\)
- (d)
\({ \frac { s }{ m } -\left( \frac { c }{ 2m } \right) }^{ 2 }\)
\(m\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +c\frac { dx }{ dt } \)sx = 0
We know that
\({ \omega }_{ n }=\sqrt { s/m } \)
a=\(\frac{c}{2m}\)
\({ \omega }_{ d }=\sqrt { { \omega }_{ n }^{ 2 }-{ a }^{ 2 } } \)
\(\sqrt { \frac { s }{ m } -{ \left( \frac { c }{ 2m } \right) }^{ 2 } } \)
\(3\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +6\frac { dx }{ dt } \)15x = 20 cos 4t shows a vibration system. Maximum amplitude of the system will be
- (a)
50 cm
- (b)
51 cm
- (c)
49 cm
- (d)
52 cm
Compare from the equation
\(m\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +c\frac { dx }{ dt } \)sx = 0
m = 3 kg, c = 6 N-s/m, s = 15 N/m
F = 20,\(\omega \)=4
We know that
Amplitude =xmax=\(\frac { F }{ \sqrt { (c{ \omega }^{ 2 })+(s-m{ \omega }^{ 2 }{ ) }^{ 2 } } } \)
=\(\frac { 20 }{ \sqrt { { (24) }^{ 2 }+(15-{ 48) }^{ 2 } } } \)
xmax=\(\frac { 20 }{ \sqrt { { (24) }^{ 2 }+(-33{ ) }^{ 2 } } } \)
xmax=0.49 m =49 cm
\(5\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +20\frac { dx }{ dt } \)+3125x = 0 shows a vibration system. The damping factor of this system is
- (a)
0.06
- (b)
0.07
- (c)
0.08
- (d)
0.09
\(m\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +c\frac { dx }{ dt } \)sx = 0
m = 5 kg, c = 20 N-s/m,
s = 3125 N/m
Damping factor =\(\frac { c }{ { c }_{ c } } =\frac { c }{ 2m{ \omega }_{ n } } \)
=\(\frac { c }{ 2\times 5\sqrt { \frac { 3125 }{ 5 } } } ({ \omega }_{ n }=\sqrt { s/m } )\)
=\(\frac { 20 }{ 10\times 25 } =\frac { 20 }{ 250 } \)=0.8
The maximum unbalance force on a system which is supported by five springs is 432 N. The transmitted force to base through springs is 39.27 N. If the supported mass on spring is 120 kg, then stiffness of the each spring is (circular frequency of the system is 157.1 rad/s)
- (a)
49360 N-m
- (b)
50368 N-m
- (c)
46568 N-m
- (d)
45468 N-m
Unbalanced force = 432 N
Transmitted force = 39.27 N
Transmissibility =\(\frac { { F }_{ T } }{ F } =\frac { 39.27 }{ 432 } \)
\(\varepsilon \)=.090
Damping system is absent, so c = 0, m= 157.1 rad/s
\(\varepsilon =\frac { { \omega }_{ n }^{ 2 } }{ { \omega }_{ n }^{ 2 }-{ \omega }^{ 2 } } \)
\({ \omega }_{ n }^{ 2 }-{ \omega }^{ 2 }\)
=\({ \omega }_{ n }^{ 2 }\left( 1+\frac { 1 }{ \varepsilon } \right) ={ \omega }^{ 2 }\)
\({ \omega }_{ n }^{ 2 }=\frac { { \omega }^{ 2 } }{ 1+\frac { 1 }{ \varepsilon } } \)
\({ \omega }_{ n }=\frac { \omega }{ \sqrt { 1+\frac { 1 }{ \omega } } } =\frac { 157.1 }{ \sqrt { 12 } } \) =45.35 rad/s
\({ \omega }_{ n }=\sqrt { \frac { { S }_{ eq } }{ m } } \)
\(\quad \quad { S }_{ eq }=m{ \omega }_{ n }^{ 2 }\) =120x(45.35)2
Seq=246804.1 N/n
Stiffness of each spring
=\(\frac{246804.1}{5}\)
= 49360N/m
When a body is subjected to transverse vibration, the stress induced in a body will be
- (a)
compressive stress
- (b)
shear
- (c)
tensile stress
- (d)
Both shear and tensile stresses
Magnification factor at resonance in forced vibration system is given by
- (a)
\(\frac { s }{ c{ \omega }_{ n } } \)
- (b)
\(\frac { c{ \omega }_{ n } }{ s } \)
- (c)
\(\sqrt { \frac { s }{ c{ \omega }_{ n } } } \)
- (d)
\(\sqrt { \frac { c{ \omega }_{ n } }{ s } } \)
Magnification factor
D=\(\frac { { x }_{ max } }{ { x }_{ 0 } } \)
=\(\frac { 1 }{ \sqrt { { \left( \frac { c\omega }{ s } \right) }^{ 2 }+{ \left( 1-\frac { { \omega }^{ 2 } }{ { \omega }_{ n } } \right) }^{ 2 } } } \)
At resonance
\(\omega ={ \omega }_{ n }\)
D=\(\frac { s }{ c{ \omega }_{ n } } \)
If I, and 12 are the distances of two masses from the centre of gravity of the body and kG is the radius of gyration of the body, then the essential condition of placing the two masses, so that the system becomes dynamically equivalent is
- (a)
l1l2=\({ k }_{ G }^{ 2 }\)
- (b)
l1l2=\(\sqrt{k_{G}}\)
- (c)
l1=kG
- (d)
l2=kG
The partial balancing means
- (a)
balancing partially the revolving masses
- (b)
balancing partially the reciprocating masses
- (c)
Both (a) and (b)
- (d)
None of the above
Secondary forces in reciprocating mass on engine are
- (a)
of same frequency as of primary forces
- (b)
of half frequency as of primary forces
- (c)
twice the frequency as of primary forces
- (d)
four times the frequency as of primary forces
The velocity ratio of two pulleys connected by an open belt or crossed belt is
- (a)
directly proportional to their diameters
- (b)
inversely proportional to their diameters
- (c)
inversely proportional to square of their diameter
- (d)
None of the above
Due to slip of belt, velocity ratio of belt drive
- (a)
increases
- (b)
decreases
- (c)
may increase or decreases
- (d)
not affected
The connecting rod of an engine has length 300 mm between its centres. If mass of rod is 70 kg and mass moment of inertia is 7000 kq-rnrn". Centre of gravity is at 100 mm from big end.
Radius of gyration of connecting rod is
- (a)
15 mm
- (b)
16 mm
- (c)
10 mm
- (d)
8 mm
I=m\({ k }_{ G }^{ 2 }\)
7000=70x\({ k }_{ G }^{ 2 }\)
\(\Rightarrow\)kG=10 mm
The connecting rod of an engine has length 300 mm between its centres. If mass of rod is 70 kg and mass moment of inertia is 7000 kq-rnrn". Centre of gravity is at 100 mm from big end.
For dynamical equivalent system, if one mass is placed at small end then other mass will be placed at
- (a)
1.0 mm
- (b)
1.2 mm
- (c)
0.5 mm
- (d)
0.7 mm
We know that for dynamic equivalent system
\({ k }_{ G }^{ 2 }\)=l1l2
l2=\(\frac { { k }_{ G }^{ 2 } }{ { l }_{ 1 } } =\frac { { (10) }^{ 2 } }{ (300-100) } \)
=0.5 mm
Other mass is placed at 0.5 mm distance from centre towards big end.
The connecting rod of an engine has length 300 mm between its centres. If mass of rod is 70 kg and mass moment of inertia is 7000 kq-rnrn". Centre of gravity is at 100 mm from big end.
How much mass would be placed at small end for dynamic equivalent system?
- (a)
0.048 kg
- (b)
0.037 kg
- (c)
0.050 kg
- (d)
0.017 kg
m1=\(\frac { m{ l }_{ 2 } }{ { l }_{ 1 }+{ l }_{ 2 } } \)
=\(\frac { 15\times 0.5 }{ 200+0.5 } \)
=0.037 kg
In the given figure, rope makes 2.5 turns round this drum having diameter 300 mm. The weight w is 9 kN.
Power required to raise this weight is
- (a)
2.0 kN
- (b)
3.2 kN
- (c)
2.8 kW
- (d)
Data is insufficient
Power required to raise this weight is
P = (T2 - T1) v
Here,
v=\(\frac { \pi dN }{ 60 } \)
=\(\frac { 3.14\times 0.3\times 20 }{ 60 } \)
= 0.314 rn/s
\(\therefore\)P = (9000 -176.47) x 0.314
P = 2.7 kW
A riveting machine is driven by a constant torque 2kW motor. Mass of fly wheel and all moving parts is 100 kg and radius is 0.5 m. Speed of fly wheel is 300 rpm before riveting. Speed of fly wheel immediately after riveting if one riveting operation takes 1 s and absorb 10000 J of energy
- (a)
197rpm
- (b)
176rpm
- (c)
200 rpm
- (d)
210 rpm
Energy supplied by motor per second = 2000 J
Energy absorbed by machine per second = 10000 J
Maximum fluctuation of energy = 10000 - 2000
= 8000 J
But, we know that
\(\triangle E=\frac { 1 }{ 2 } I({ \omega }_{ 1 }^{ 2 }-{ \omega }_{ 2 }^{ 2 })\)
\({ \omega }_{ 1 }=\frac { 2\pi N }{ 60 } =\frac { 2\pi \times 300 }{ 60 } \)=31.42 rad/s, \(\omega_{2}\)=?
\(\Rightarrow 7000=\frac { 1 }{ 2 } m{ k }^{ 2 }({ \omega }_{ 1 }^{ 2 }-{ \omega }_{ 2 }^{ 2 })\)
\(\Rightarrow 7000=\frac { 1 }{ 2 } \times 100\times { (0.5) }^{ 2 }[(31.42{ ) }^{ 2 }-{ \omega }_{ 2 }^{ 2 })\)
\(\Rightarrow { \omega }_{ 2 }\)=20.66 rad/s
\(\Rightarrow { N }_{ 2 }=\frac { 20.66\times 60 }{ 2\pi } \)
=197.3 rpm
A flat belt is 100 mm wide and 10 mm thick with density 1000 kg/m3. The maximum stress in belt is not to exceed 1 MPa. The speed of belt at which it transmits the maximum power is
- (a)
20 m/s
- (b)
22.25 m/s
- (c)
16 m/s
- (d)
18.75 m/s
Maximum tension in belt
=\(\sigma \) max b t
= 1 x 106 x 100 x 10x 10-6
= 1000 N
Mass of belt per unit length = Volume x Density
= btl\(\rho \)
\(\frac { 100 }{ 1000 } \times \frac { 10 }{ 100 } \times 1\times 1000\)
For maximum power,
v=\(\sqrt { \frac { T }{ 3m } } \)
=\(\sqrt { \frac { 1000 }{ 3\times 1 } } \)=18.25 m/s
Consider a crankslider mechanism in which Mass of reciprocating part is 500 kg. Crank moves with 20 rad/s and its length is 0.2 m. Inertia force on reciprocating part when crank has travelled 600 from IDC is (obliquity ratio n = 4)
- (a)
10 kN
- (b)
17 kN
- (c)
12 kN
- (d)
15 kN
F1=\({ m }_{ R }{ \omega }^{ 2 }r\left( cos\theta +\frac { cos2\theta }{ n } \right) \)
F1 = 500 x 20 x 20 x 0.2\(\left[ cos60+\frac { cos120 }{ 4 } \right] \)
=40000 x\(\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 8 } \right] \)
= 15000 N
F1 = 15 kN
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