Dynamics of Machines Two Marks
Dynamics of Machines Two Marks
Dynamics of Machines Two Marks
IV SEMESTER
DYNAMICS OF MACHINES
TWO MARKS QUESTION AND ANSWERS
1. What are the conditions for a body to be in static and dynamic equilibrium?
Necessary and sufficient conditions for static and dynamic equilibrium are
1. Vector sum of all forces acting on a body is zero.
2. The vector sum of the moments of all forces acting about any arbitrary point or axis is
zero.
First condition is the sufficient condition for static equilibrium together with
second condition is necessary for dynamic equilibrium.
2. Define static force analysis.
If components of a machine accelerate, inertia is produced due to their masses.
However, the magnitudes of these forces are small compared to the externally applied
loads. Hence inertia effect due to masses are neglected. Such an analysis is known as
static force analysis.
3. Define force and applied force.
Force is a push or pull, which acts on a body changes or tends to change, the state
of rest or of uniform motion of the body. A force is completely characterized by its point
of application, its magnitude and direction.
The external force acting on a system of body from outside the system are called
applied force. The applied forces are classified as active and reactive force.
4. Rate of change of momentum of a body is directly proportional to Force acting on it.
5. Which law helps to measure a force quantitatively?
Newtons second law helps us to measure a force quantitatively.
6. Distinguish between space diagram and free body diagram.
A space diagram is a graphical description of the system. It generally shows the
shape and size of the system, the weight, the externally applied loads, the connection and
the supports of the system.
A free body diagram is a sketch of the isolated or free body which shows all the
pertinent weight forces, the externally applied loads, and the reaction from its supports
and connections acting upon it by the removed elements.
7. When will the two force member is in equilibrium?
The member under the action of two force will be in equilibrium if,
The two forces are of same magnitude,
1.
The forces act along the same line, and
2.
The forces are in opposite direction
3.
8. Give any three advantages of free body diagram.
Free body diagram assist in seeing and understanding all aspects of problem.
1.
They help in planning the approach to the problem.
2.
They make mathematical relations easier to the problem.
3.
9. When will the three force member is in equilibrium.
A body or member will be in equilibrium under the action of three forces if,
the resultant of the forces is zero, and
1.
the line of action of the forces intersect at a point.
2.
10. Differentiate between static force analysis and dynamic force analysis.
If components of a machine accelerate, inertia forces are produced due to their
masses. If the magnitude of these forces are small compared to the externally applied
loads, they can be neglected while analyzing the mechanism. Such an analysis is known
as static force analysis.
11. What do you mean by inertia? And inertia force
The property of matter offering resistance to any change of its state of rest or of
uniform motion in a straight line is known as inertia.
12. State DAlemberts principle.
DAlemberts principle states that the inertia forces and torques, and the external
forces and torques acting on a body together result in statical equilibrium.
In other words, the vector sum of all external forces acting upon a system of rigid
bodies is zero. The vector sum of all external moments and inertia torques acting upon a
system of rigid bodies is also separately zero.
13. How you will reduce a dynamic analysis probloem into an equivalent problem of
static equilibrium?
By applying DAlemberts principle to a dynamic analysis problem, we can
reduce it into an equivalent problem of static equilibrium.
14. What do you mean by Equivalent offset inertia force?
Equivalent offset inertia force is the force which can replace both inertia force and
inertia torque.
15. State the principle of super position.
The principle of super position states that for linear systems the individual
responses to several disturbances or driving functions can be superposed on each other to
obtain the total response of the system.
16. Give one example each for linear and non-linear system.
Linear system: Example: Spring system.
Non-linear system: Example: Systems with static or coulomb friction, backlash.
17. Principle of super position has a limitation that it cannot be applied for non-linear
systems.
18. When the connecting rod is large the piston executes simple harmonic motion.
19. Define Piston effort.
Piston effort is defined as the net or effective force applied on the piston, along
the line of stroke. It is also known as effective driving force (or) net load on the gudgeon
pin.
20. What do you mean by crank effort or turning moment on the crank shaft?
It is the product of the crank-pin effort (FT) and crank pin radius (r).
21. Define compound pendulum or torsional pendulum.
A rigid body suspended vertically at a point and oscillating with very small
amplitude under the action of gravitational force is known as compound pendulum or
torsional pendulum.
22. What are the requirements of an equivalent dynamical system?
The mass of the rigid body must be equal to the sum of masses of two
1.
concentrated masses. i.e. m1 + m2 = m
The centre of gravity of the two masses must coincide with the centre of gravity
2.
of the rigid body. i.e. m1l1 = m2l2
The sum of mass moment of inertia of two masses about their centre of gravity is
3.
equal to the mass moment of inertia of the rigid body. i.e. l1 . l2 = (kG)2
23. What are the forces acting on the connecting rod?
Inertia force of the reciprocating parts (FI) acting along the line of stroke.
1.
The side thrust between the cross head and the guide bars acting at P and right
2.
angles to line of stroke.
The inertia force is an imaginary force, which when acts upon a rigid body, brings
it in an equilibrium position.
Inertia force = - Accelerating force = - ma
26. Differentiate the function of flywheel and governor.
The function of flywheel is to reduce the fluctuations of speed during a cycle
1.
above and below the mean value for constant load from prime mover. The
function of governor is to control the mean speed over a period for output load
variations.
Flywheel works continuously from cycle to cycle. Governor works intermittently,
2.
i.e. only when there is change in the load.
Flywheel has no influence on mean speed of the prime mover. Governor has no
3.
influence over cyclic speed fluctuations.
27. Define Inertia Torque.
The inertia torque is an imaginary torque, which when applied upon the rigid
body, brings it in equilibrium position. It is equal to the accelerating couple in magnitude
by opposite in direction.
28. What do you understand by the fluctuations of energy and maximum fluctuation of
energy?
The variations of energy above and below the mean resisting torque line are called
fluctuations of energy.
The difference between the maximum and the minimum energies is known as
maximum fluctuation of energy.
29. Define coefficient of fluctuation of energy.
It is the rat6io of maximum fluctuation of energy to the work done per cycle.
30. What is meant by maximum fluctuation of speed?
The difference between the maximum and minimum speeds during a cycle is
called maximum fluctuation of speed.
31. Define coefficient of fluctuation of speed.
The ratio of the maximum fluctuation of speed to the mean speed is called the
coefficient of fluctuation of speed.
32. List out the few machines in which flywheel is used.
1. Punching machines, 2. Shearing machines, 3. Rivetting machines, and
4. Crushing machines.
33. What is cam dynamics?
Cam dynamics is the study of cam follower system with considering the dynamic
forces and torques developed in it.
33.a.Define unbalance and spring surge.
A disc cam produces unbalance because its mass is not symmetrical with axis of
rotation.
Spring surge means vibration of the retaining spring.
34. Write the importance of Balancing?
If the moving part of a machine are not balanced completely then the inertia
forces are set up which may cause excessive noise, vibration, wear and tear of the system.
So balancing of machine is necessary.
35. Why balancing of dynamic forces is necessary?
If dynamic force are not balanced, they will cause worse effects such as war and
tear on bearings and excessive vibrations on machines. It is very common in cam shafts,
steam turbine rotors, engine crank shafts, and centrifugal pumps, etc.,
36. Write the different types of balancing?
1. Balancing of rotating masses
Static balancing
1.
Dynamic balancing
2.
2. Balancing of reciprocating masses.
37. Write the condition for complete balancing?
The resultant centrifugal force must be zero and
1.
The resultant couple must be zero.
2.
38. Write the equation for balancing a single rotating mass by a single mass?
For balancing single rotating mass by a single rotating mass, the equation is
m1r1 = m2r2.
39. Whether grinding wheels are balanced or not? If so why?
Yes, the grinding wheels are properly balanced by inserting some low density
materials. If not the required surface finish wont be attained and the vibration will cause
much noise.
40. Why complete balancing is not possible in reciprocating engine?
Balancing of reciprocating masses is done by introducing the balancing mass
opposite to the crank. The vertical component of the dynamic force of this balancing
mass gives rise to Hammer blow. In order to reduce the Hammer blow, a part of the
reciprocating mass is balanced. Hence complete balancing is not possible in
reciprocating engines.
40.Can a single cylinder engine be fully balanced?why?
NO.A single cylinder engine cannot be fully balanced.
Because the unbalanced force due to a reciprocating masses remains constant in
direction but varies in magnitude.
41. Differentiate between the unbalanced force due to a reciprocating mass and that due to a
revolving masses.
Complete balancing of revolving mass can be possible. But fraction of
1.
reciprocating mass only balanced.
The unbalanced force due to reciprocating mass varies in magnitude but constant
2.
in direction. But in the case of revolving masses, the unbalanced force is constant
in magnitude but varies in direction.
42. What are the various cases of balancing of revolving masses?
Balancing of single rotating mass by a single mass rotating in the same plane.
1.
Balancing of single rotating mass by two masses rotating in different planes.
2.
Balancing of several rotating masses in a single plane.
3.
Balancing of several rotating masses in different planes.
4.
43. What are the effects of an unbalanced primary force along the line of stroke of two
cylinder locomotive?
Variation in tractive force along the line of stroke, and
1.
Swaying couple.
2.
44. Define tractive force.
The resultant unbalanced force due to the two cylinders along the line of stroke is
known as tractive force.
45. What is swaying couple?
The unbalanced force acting at a distance between the line of stroke of two
cylinders, constitute a couple in the horizontal direction. This couple is known as
swaying couple.
46. What is the effect of hammer blow and what is the cause of it?
The effect of hammer blow is to cause the variation in pressure between the wheel
and the rail, such that vehicle vibrates vigorously. Hammer blow is caused due to the
effect of unbalanced primary force acting perpendicular to the line of stroke.
47. What are in-line engines?
Multi-cylinder engines with the cylinder centre lines in the same plane and on the
same side of the centre line of the crank shaft, are known as in-line engine.
48. What are the condition to be satisfied for complete balance of in-line engine?
The algebraic sum of the primary and secondary forces must be zero, and
1.
The algebraic sum of the couples due to primary and secondary forces must be
2.
zero.
49. Why radial engines are preferred?
In radial engines the connecting rods are connected to a common crank and hence
the plane of rotation of the various cranks is same, therefore there are no unbalanced
primary or secondary couples. Hence radial engines are preferred.
50. What are the causes of vibration?
The causes of vibration are unbalanced forces, elastic nature of the system, self
excitations, winds and earthquakes.
51. Define Period and cycle of vibration.
Period is the time interval after which the motion is repeated itself. Cycle is
defined as the motion completed during one time period.
52. Define frequency of vibration.
It is the number of cycles described in one second. Its unit is Hz.
53. How will you classify vibration?
Free vibrations
1.
a) Longitudinal vibration,
b) Transverse vibration, and
c) Torsional vibration.
Forced vibrations, and
2.
Damped vibration.
3.
54. What is meant by free vibration and forced vibrations?
When no external force acts on the body, after giving it an initial displacement,
then the body is said to be under free or natural vibration.
When the body vibrates under the influence of external force, then the body is
said to be under forced vibrations.
55. What do you meant by damping and damped vibration?
The resistance against the vibration is called damping.
When there is a reduction in amplitude over every cycle of vibration, then the
motion is said to be damped vibration.
56. Define resonance.
When the frequency of external force is equal to the natural frequency of a
vibrating body, the amplitude of vibration becomes excessively large. This phenomenon
is known as resonance.
57. What do you mean by a degree of freedom or movability?
The number of independent coordinates required to completely define the motion
of a system is known as degree of freedom of the system.
270, -390, 190, -340, 270, -250. The scale for the turning moment is 1 mm = 500 N-m,
and for crank angle is 1 mm = 5 o. If the fluctuation of speed is not to exceed 1.5 % of
the mean, determine a suitable diameter and cross-section of the rim of the flywheel
assumed with axial dimension (i.e., width of the rim) equal to 1.5 times the radial
dimension (i.e., thickness of the rim). The hoop stress is limited to 3 Mpa and the density
of the material of the flywheel is 7500 kg/m3.
(Ans. Refer Prob. No.3.19, Page No. 3.43 Dynamics of Machines by V. JAYAKUMAR)
6. Three masses are attached to a shaft as follows: 10 kg at 90 mm radius, 15 kg at 120 mm
radius and 9 kg at 150 mm radius. The masses are to be arranged so that the shaft is in
complete balance. Determine the angular position of masses relative to 10 kg mass. All
the masses are in the same plane.
(Ans. Refer Prob. No.5.2, Page No. 5.9 Dynamics of Machines by V. JAYAKUMAR)
7. A shaft has tree eccentrics, each 75 mm diameter and 25 mm thick, machined in one
piece with the shaft. The central planes of the eccentric are 60 mm apart. The distance of
the centers from the axis of rotation are 12 mm,18 mm and 12 mm and their angular
positions are 120o apart. The density of metal is 700 kg/m 3. Find the amount of out-ofbalance force and couple at 600 rpm. If the shaft is balanced by adding two masses at a
radius 75 mm and at distance of 100 mm from the central plane of the middle eccentric,
find the amount of the masses and their angular positions.
(Ans. Refer Prob. No.5.8, Page No. 5.21 Dynamics of Machines by V. JAYAKUMAR)
8. The cranks of a three-cylinder locomotive are set at 120o. The reciprocating masses are
450 kg for the inside cylinder and 390 kg for each outside cylinder. The pitch of the
cylinder is 1.2 m and the stroke of each piston 500 mm. The planes of rotation of the
balance masses are 960 mm from the inside cylinder. If 40% of the reciprocating masses
are to be balanced, determine
1. The magnitude and the position of the balancing masses required at a radial
distance of 500 mm; and
2. The hammer blow per wheel when the axle rotates at 350 rpm.
(Ans. Refer Prob. No.6.5, Page No. 6.17 Dynamics of Machines by V. JAYAKUMAR)
9. An air compressor has four vertical cylinders 1, 2, 3 and 4 in line and the driving cranks
at 90o intervals reach their upper most positions in this order. The cranks are of 150 mm
radius, the connecting rods 500 mm long and the cylinder centre line 400 mm apart. The
mass of the reciprocating parts for each cylinder is 22.5 kg and the speed of rotation is
400 rpm. Show that there are no out-of-balance primary or secondary forces and
determine the corresponding couples, indicating the position of No.1 crank for maximum
values. The central plane of the machine may be taken as reference plane.
(Ans. Refer Prob. No.6.9, Page No. 6.31 Dynamics of Machines by V. JAYAKUMAR)
10. The firing order of a six cylinder, vertical, four-stroke, in-line engine is 1-4-2-6-3-5. The
piston stroke is 80 mm and length of each connecting rod is 180 mm. The pitch distances
between the cylinder centre lines are 80 mm, 80 mm, 120 mm, 80 mm and 80 mm
respectively. The reciprocating mass per cylinder is 1.2 kg and the engine speed is 2400
rpm. Determine the out-of-balance primary and secondary forces and couples on the
engine taking a plane mid-way between the cylinders 3 and 4 as the reference plane.
(Ans. Refer Prob. No.6.14, Page No. 6.41 Dynamics of Machines by V. JAYAKUMAR)
11. Determine the equivalent spring stiffness and the natural frequency of the following
vibrating systems when
a) the mass is suspended to a spring
b) the mass is suspended at the bottom of two springs in series
c) the mass is fixed in between two springs
d) the mass is fixed to the mid point of a spring
(Ans. Refer Prob. No. 18.1, Page No. 598 Theory of Machines by S.S. RATTAN)
12. A vibrating system consists of a mass of 50 kg, a spring of stiffness 30 kN/m and a
damper. The damping provided is only 20 % of the critical value. Determine
1. the damping factor
2. the critical damping coefficient
3. the natural frequency of damped vibrations
4. the logarithmic decrement
5. the ratio of two consecutive amplitudes.
(Ans. Refer Prob. No. 18.5, Page No. 609 Theory of Machines by S.S. RATTAN)
13. The machine mounted on springs and fitted with a dashpot has a mass of 60 kg. There
are three springs, each of stiffness 12 N/mm. The amplitude of vibrations reduces from
45 to 8 mm in two complete oscillations. Assuming that the damping force varies as the
velocity, determine
i)
the damping coefficient,
ii)
the ratio of frequencies of damped and undamped vibrations, and
iii)
the periodic time of damped vibrations.
(Ans. Refer Prob. No. 18.8, Page No. 611 Theory of Machines by S.S. RATTAN)
14. A single cylinder vertical diesel engine has a mass of 400 kg and is mounted on a steel
chassis frame. The static deflection owing to the weight of the chassis is 2.4 mm. The
reciprocating masses of the engine amounts to 18 kg and the stroke of the engine is 160
mm. A dashpot with a damping coefficient 2 N/mm/s is also used to dampen the
vibrations. In the steady-state of the vibrations, determine
1. the amplitude of the vibrations if the driving shaft rotates at 500 rpm.
2. the speed of the driving shaft when the resonance occurs.
(Ans. Refer Prob. No. 18.12, Page No. 619 Theory of Machines by S.S. RATTAN)
15. A machine supported symmetrically on four springs has a mass of 80 kg. The mass of the
reciprocating parts is 2.2 kg which move through a vertical stroke of 100 mm with simple
harmonic motion. Neglecting damping, determine the combined stiffness of the springs
so that the force transmitted to the foundation is 1/20 th of the impressed force. The
machine crank shaft rotates at 800 rpm
If under actual working conditions, the damping reduces the amplitudes of
successive vibrations by 30 %, find,
1. the force transmitted to the foundation at 800 rpm,
2. the force transmitted to the foundation at resonance, and
3. the amplitude of the vibrations at resonance.
(Ans. Refer Prob. No. 18.15, Page No. 624 Theory of Machines by S.S. RATTAN)
16. A shaft supported freely at the ends has a mass of 120 kg placed 250 mm from one end.
Determine the frequency of the natural transverse vibrations if the length of the shaft is
700 mm, E = 200 GN/m2 and shaft diameter is 40 mm.
(Ans. Refer Prob. No. 18.16, Page No. 626 Theory of Machines by S.S. RATTAN)
17. A shaft 40 mm diameter and 2.5 m long has a mass of 15 kg per meter length. It is
simply supported at the ends and carries three masses 90 kg, 140 kg and 60 kg at 0.8 m,
1.5 m and 2 m respectively from the left support. Taking E = 200 GN/m 2, find the
frequency of the transverse vibrations.
(Ans. Refer Prob. No. 18.17, Page No. 633 Theory of Machines by S.S. RATTAN)
18. The following data relate to a shaft held in long bearings.
Length of shaft
Diameter of shaft
Mass of a rotor at mid point
Eccentricity of centre of mass of rotor from centre of rotor
Modulus of elasticity of shaft material
Permissible stress in shaft material
Determine the critical speed of the shaft and the range of
unsafe to run the shaft. Assume the shaft to be mass less.
= 1.2 m
= 14,
= 16 kg,
= 0.4 mm
= 200 GN/m2
= 70 X 106 N/m2
speed over which it is
(Ans. Refer Prob. No. 18.20, Page No. 636 Theory of Machines by S.S. RATTAN)
19. The following data refer to the transmission gear of a motor ship:
Moment of inertia of flywheel
= 4800 kg m2
Moment of inertia of propeller
= 3200 kg m2
Modulus of rigidity of shaft material
= 80 X 109 N/m2
Equivalent MOI per cylinder
= 400 kg m2
Assuming the diameter of the torsionally equivalent crankshaft to be 320 mm and
treating the arrangement as a three rotor system, determine the frequency of free torsional
vibrations.
(Ans. Refer Prob. No. 18.27, Page No. 656 Theory of Machines by S.S. RATTAN)
20. A reciprocating IC engine is coupled to a centrifugal pump through a pair of gears. The
shaft from the flywheel of the engine to the gear wheel has a 48 mm diameter and is 800
mm long. The shaft from the pinion to the pump has a 32 mm diameter and is 280 mm
long. Pump speed is four times the engine speed. Moments of inertia of flywheel, gearwheel, pinion and pump impeller are 1000 kg m2, 14 kg m2, 5 kg m2 and 18 kg m2
respectively. Find the natural frequency of the torsional oscillation of the system. G = 80
G N/m2.
(Ans. Refer Prob. No. 18.28, Page No. 660 Theory of Machines by S.S. RATTAN)
21. Each arm of a Porter governor is 250 mm long. The upper and lower arms are pivoted to
links of 40 mm and 50 mm respectively from the axis of rotation. Each ball has a mass of
5 kg and the sleeve mass is 50 kg. The force of friction on the sleeve of the mechanism is
40 N. Determine the range of speed of the governor for extreme radii of rotation of 125
mm and 150 mm.
(Ans. Refer Prob. No. 16.3, Page No. 540 Theory of Machines by S.S. RATTAN)
22. The mass of each ball of a Proell governor is 7.5 kg and the load on the sleeve is 80 kg.
Each of the arms is 300 mm long. The upper arms are pivoted on the axis of rotation
whereas the lower arms are pivoted to links of 40 mm from the axis of rotation. The
extensions of the lower arms to which the balls are attached are 100 mm long and are
parallel to the governor axis at the minimum radius. Determine the equilibrium speeds
corresponding to extreme radii of 180 mm and 240 mm/
(Ans. Refer Prob. No. 16.4, Page No. 543 Theory of Machines by S.S. RATTAN)
23. In a spring loaded Hartnell type of governor, the mass of each ball is 4 kg and the lift of
the sleeve is 40 mm. The governor begins to float at 200 rpm when the radius of the ball
path is 90 mm. The mean working speed of the governor is 16 times the range of speed
when friction is neglected. The lengths of the ball and roller arms of the bell-crank lever
are 100 mm and 80 mm respectively. The pivot centre and the axis of governor are 115
mm apart. Determine the initial compression of the spring, taking into account the
obliquity of arms.
Assuming the friction at the sleeve to be equivalent to a force of 15 N, determine
the total alteration in speed before the sleeve begins to move from the mid- position
(Ans. Refer Prob. No. 16.6, Page No. 548 Theory of Machines by S.S. RATTAN)
24. The controlling force curve of a spring controlled governor is a straight line. The weight
of each governor ball is 40 N and the extreme radii of rotation are 120 mm and 180 mm.
If the values of the controlling force at the above radii be respectively 200 N and 360 N
and the friction of the mechanism is equivalent to 2 N at each ball. Find a) the extreme
equilibrium speeds of the governor, b) the equilibrium speed and the coefficient of
insensitiveness at a radius of 150 mm.
(Ans. Refer Prob. No.10.32, Page No. 10.68 Dynamics of Machines by V. JAYAKUMAR)
25. In a Porter governor, each arm is 200 mm long and is pivoted at the axis of rotation. The
mass of each ball is 5 kg and the load on the sleeve is 30 kg. The extreme radii of
rotation are 80 mm and 140 mm. Plot a graph of the controlling force vs. radius of
rotation and set off a speed scale along the ordinate corresponding to a radius of 160 mm.
(Ans. Refer Prob. No. 16.10, Page No. 563 Theory of Machines by S.S. RATTAN)