Transcript Cont….

INTRODUCTION TO
DYNAMICS
Prepare By:130360119070 – Pansuriya Mayank
Sanjaybhai Raguru College of engineering
Mechanical Engineering Department
Inertia Forces
• The inertia force is an imaginary force.
• Which acts upon a rigid body, brings it in an equilibrium
position.
• It is numerically equal to the accelerating force in
magnitude and Opposite in direction.
• Mathematically,
• Inertia force = – Accelerating force = – m.a
• Where,
m = Mass of the body, and
a = Linear acceleration of the centre of gravity of the
body.
Cont….
• Similarly, 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 but
opposite in direction.
Resultant Effect of a System of Forces
Acting on a Rigid Body
• Consider a rigid body acted upon
by a system of forces.
• These forces may be reduced to a
single resultant force F whose line of
action is at a distance h from the
centre of gravity G.
• Now let us assume two equal and
opposite forces (of magnitude F)
acting through G, and parallel to the
resultant force, without influencing
the effect of the resultant force F, as
shown in Fig.
Cont…
• A little consideration will show that the body is now
subjected to a couple (equal to F × h) and a force, equal
and parallel to the resultant force F passing through G.
• The force F through G causes linear acceleration of the c.g.
and the moment of the couple (F × h) causes angular
acceleration of the body about an axis passing through G
and perpendicular to the point in which the couple acts.
• Let,
α = Angular acceleration of the rigid body due to couple,
h = Perpendicular distance between the force and centre
of gravity of the body,
m = Mass of the body,
Cont……
k = Least radius of gyration about an axis through G, and
I = Moment of inertia of the body about an axis passing
through its centre of gravity and perpendicular to the
point in which the couple acts
= m.k2
Force, F = Mass × Acceleration = m.a
...(i)
F.h = m.k2.α = I.α ...( I = m.k2)
...(ii)
and
• From equations (i) and (ii), find the values of a and α, if
the values of F, m, k, and h are known.
D-Alembert’s Principle
• Thus, D-Alembert’s principle states that “the resultant force
or resultant torque acting on a body together with the
reversed effective force or reverse inertia torque (or inertia
force), will keep the body in equilibrium position.”
• According to Newton’s second law of motion,
F = m.a
...(i)
Where,
F = Resultant force acting on the body,
m = Mass of the body, and
a = Linear acceleration of the centre of mass of the
body.
• The equation (i) may also be written as:
F – m.a = 0
...(ii)
Cont……
• If the quantity –m.a be treated as a force, equal, opposite
and with the same line of action as the resultant force F.
• And include this force with the system of forces of which
F is the resultant, then the complete system of forces will
be in equilibrium.
• This principle is known as D-Alembert’s principle.
• The equal and opposite force –m.a is known as reversed
effective force or the inertia force (briefly written as F1).
• The equation (ii) may be written as
F + F1 = 0
...(iii)
Simple Harmonic Motion
• Consider a particle, moving round the circumference of a
circle of radius r, with a uniform angular velocity ω rad/s,
as shown in Fig.
Cont…
• Let P be any position of the particle after t seconds and θ be
the angle turned by the particle in t seconds.
• If N is the projection of P on the diameter X X ′, then
• Displacement of P is given by,
x = r. cos θ
... (i)
• The velocity of point P is given by
V = ω.r
• The centripetal acceleration of the P is given by
a = ω2.r.
• The component of the velocity and acceleration parallel to OY
- axis, i.e.
VN = V sin θ = ω.r sin θ
• And,
aN = a cos θ = ω2.r cos θ
Cont…
• Substituting the value of r cos θ = x from equation (i), we
get,
• Periodic time of motion is given by,
• Frequency of simple harmonic motion,
Cont…
• Therefore, when the particle moves with angular simple
harmonic motion, then the periodic time,
Simple Pendulum
Let,
L = Length of the string,
m = Mass of the bob in kg,
W = Weight of the bob in
newtons,
= m.g, and
θ = Angle through which the
string is displaced.
Cont…
• The couple tending to restore the bob to the equilibrium
position or restoring torque,
T = m.g sinθ × L
• Since angle θ is very small, therefore sinθ = θ radians.
∴ T = m.g.L.θ
• We know that the mass moment of inertia of the bob about an
axis through the point of suspension,
I = mass × (length)2 = m.L2
• Angular acceleration of the string,
Cont….
• The periodic time,
... (i)
• And frequency of oscillation,
... (ii)
• The periodic time and the frequency of oscillation of a
simple pendulum depends only upon its length and
acceleration due to gravity.
• The mass of the bob has no effect on it.
Cont….
Notes :
1. The motion of the bob from one extremity to the other
(i.e. from B to C or C to B) is known as beat or swing.
Thus one beat = ½ oscillation.
∴ Periodic time for one beat =
2. A pendulum, which executes one beat per second (i.e.
one complete oscillation in two seconds) is known as a
second’s pendulum.
Compound Pendulum
• When a rigid body is suspended vertically,
and it oscillates with a small amplitude under
the action of the force of gravity, the body is
known as compound pendulum, as shown in
Fig.
• Let,
m = Mass of the pendulum in kg,
W = Weight of the
newtons = m.g,
pendulum
in
kG = Radius of gyration about an axis
through the centre of gravity G,
h = Distance of point of suspension
O from the centre of gravity G of
the body.
Cont…
• If the pendulum is given a small angular displacement θ,
then the couple tending to restore the pendulum to the
equilibrium position OA,
T = mg sinθ × h = mgh sinθ
• Since θ is very small, therefore substituting sinθ = θ
radians, we get
T = m.g.h.θ
• Now, the mass moment of inertia about the axis of
suspension O,
• Angular acceleration of the pendulum,
Cont…..
• The angular acceleration is directly proportional to angular
displacement, therefore the pendulum executes simple
harmonic motion.
• We know that the periodic time,
• Frequency of oscillation,
Cont….
Notes :
1. Comparing this equation with equation simple pendulum
• we see that the equivalent length of a simple pendulum,
which gives the same frequency as compound pendulum,
is,
Cont….
Notes :
2. The periodic time of a compound pendulum is minimum
when the distance between the point of suspension and the
centre of gravity is equal to the radius of gyration of the
body about its centre of gravity.
∴ Minimum periodic time of a compound pendulum, (h= KG)
Bifilar Suspension
• The moment of inertia of a body
may be determined experimentally
by an apparatus called bifilar
suspension.
• The body whose moment of inertia
is to be determined (say AB) is
suspended by two long parallel
flexible strings as shown in Fig.
• When the body is twisted through a
small angle θ about a vertical axis
through the centre of gravity G, it
will vibrate with simple harmonic
motion in a horizontal plane.
Cont….
• Let,
m = Mass of the body,
W = Weight of the body in newtons = m.g,
kG = Radius of gyration about an axis through the centre of
gravity,
I = Mass moment of inertia of the body about a vertical axis,
G = m.kG2,
l = Length of each string,
x = Distance of A from G (i.e. AG),
y = Distance of B from G (i.e. BG),
θ = Small angular displacement of the body from the equilibrium
position in the horizontal plane,
φA and φB = Corresponding angular displacements of the strings,
α = Angular acceleration towards the equilibrium position.
Cont….
• When the body is stationary, the tension in the strings are
given by taking moments about B and A respectively,
• When the body is displaced from its equilibrium position in a
horizontal plane through a small angle θ, then the angular
displacements of the strings are given by ФA and ФB,
Cont…
• Component of tension TA in the horizontal plane, acting
normal to A′B′ at A′ as shown in Fig.
• Component of tension TB in the horizontal plane, acting
normal to A′B′ at B′ as shown in Fig.
Cont….
• These components of tensions TA and TB are equal and
opposite in direction, which gives rise to a couple.
• The couple or torque applied to each string to restore the body
to its initial equilibrium position, i.e. restoring torque
... (i)
• And accelerating (or disturbing) torque
... (ii)
Cont……
• Equating equations (i) and (ii),
• We know that periodic time,
Trifilar Suspension
(Torsional Pendulum)
• It is also used to find the moment
of inertia of a body experimentally.
• The body (say a disc or flywheel)
whose moment of inertia is to be
determined is suspended by three
long flexible wires A, B and C, as
shown in Fig.
• When the body is twisted about its
axis through a small angle θ and
then released, it will oscillate with
simple harmonic motion.
Cont…
• Let,
m = Mass of the body,
W = Weight of the body in Newton = m.g,
kG = Radius of gyration about an axis through the centre of
gravity,
I = Mass moment of inertia of the disc about an axis
through O and perpendicular to it = m.k2,
l = Length of each wire,
r = Distance of each wire from the axis of the disc,
θ = Small angular displacement of the disc,
φ = Corresponding angular displacements of the wire, and
α = Angular acceleration towards the equilibrium position.
Cont…
• Then, for small displacements,
• Since the three wires are attached symmetrically with respect to
the axis, therefore the tension in each wire will be one-third of
the weight of the body.
∴ Tension in each wire = m.g/3
• Component of the tension in each wire perpendicular to r,
• Torque applied to each wire to restore the body to its initial
equilibrium position i.e. restoring torque,
Cont….
• Total restoring torque applied to three wires,
... (i)
• We know that disturbing torque
= I.α = m.kG2 .α
• Equating equations (i) and (ii),
... (ii)
Cont….
• We know that periodic time,
Equivalent Dynamical System
• In order to determine the motion of a rigid body, under the action of
external forces.
• It is usually convenient to replace the rigid body by two masses
placed at a fixed distance apart, in such a way that,
Cont..
1. The sum of their masses is equal to the total mass of the
body;
2. The centre of gravity of the two masses coincides with
that of the rigid body; and
3. The sum of mass moment of inertia of the two masses
about their centre of gravity must be equal to the mass
moment of inertia of the body.
• When these three conditions are satisfied, then it is said to
be an equivalent dynamical system.
• Consider a rigid body, having its centre of gravity at G, as
shown in Fig.
Cont…..
• Let,
m = Mass of the body,
kG = Radius of gyration about its centre of gravity G,
m1 and m2 = Two masses which form a dynamical equivalent
system,
l1 = Distance of mass m1 from G,
l2 = Distance of mass m2 from G,
And L = Total distance between the masses m1 and m2.
• Thus, for the two masses to be dynamically equivalent,
m 1 + m2 = m
m1.l1 = m2.l2
And
m1.(l1)2 + m2.(l2)2 = m.(kG)2
...(i)
...(ii)
...(iii)
Cont…..
• From equations (i) and (ii),
...(iv)
...(v)
• Substituting the value of m1 and m2 in equation (iii), we have
∴
l1.l2 = (kG)2
...(vi)
• These equations gives the essential condition of placing the two
masses, so that the system becomes dynamical equivalent.
Cont….
• The distance of one of the masses (i.e. either l1 or l2) is
arbitrary chosen and the other distance is obtained from
equation (vi).
 NOTE:• When the radius of gyration kG is not known, then the position
of the second mass may be obtained by considering the body as
a compound pendulum.
• The length of the simple pendulum which gives the same
frequency as the rigid body (i.e. compound pendulum) is
Cont….
• We also know that
• This means that the second mass is situated at the centre of
oscillation of the body, which is at a distance of,
l2 = (kG)2/l1.
Determination of Equivalent Dynamical
System of Two Masses by Graphical Method
• Consider a body of mass m, acting at G as shown in Fig.
Cont…
• This mass m, may be replaced by two masses m1 and m2 so that
the system becomes dynamical equivalent.
• The position of mass m1 may be fixed arbitrarily at A.
• Now draw perpendicular CG at G, equal in length of the radius
of gyration of the body, kG.
• Then join AC and draw CB perpendicular to AC intersecting
AG produced in B.
• The point B now fixes the position of the second mass m2.
• A little consideration will show that the triangles ACG and
BCG are similar.
• Therefore,
Correction Couple
• For equivalent dynamical system of two mass we consider
three conditions.
• But sometimes the situation arise in practice such that the
locations for both masses m1 and m2 have to be selected
arbitrarily.
• For example, in case of slider crank mechanism, while analysis
the inertia effect of connecting rod, it is convenient to assume
the masses m1 and m2 be placed at gudgeon pin and the crank
pin respectively.
• A little consideration will show that when the two masses are
placed arbitrarily, then only the first two conditions will only
satisfied without the satisfying the third condition of equivalent
dynamical system.
Cont..
• In order to satisfy the third condition of dynamical equivalence
also, it is necessary to apply some couple on the arbitrarily
chosen two mass system, called correction couple which can be
determined as follows:
• Consider two masses m1 and m2 which are placed at points A
and B respectively to satisfy the three conditions of dynamic
equivalence of connecting rod as shown in Fig.(b).
• In case the mass m2 is required to be placed at point D instead
of B as shown in Fig.(c), then the third condition will not be
satisfied.
• In order to make the system dynamically equivalent, we have to
apply some correction couple so as the system becomes
dynamically equivalent.
Cont..
• Let,
I = Mass moment of inertia of a dynamically equivalent
system;
kG = Radius of gyration about its centre of gravity G,
I1 = New mass moment of inertia of non dynamically
equivalent system;
k1 = New radius of gyration of non dynamically equivalent
system;
α = Angular acceleration of the body;
l1 = Distance of mass m1 placed at A from G,
l2 = Distance of mass m2 placed at B from G,
And l3 = Distance of mass m3 placed at D from G,
Cont…
• The torque required to accelerate the dynamically equivalent
system,
T = I.α
∴ T = m (kG)2 α
(i)
• Similarly, the torque required to accelerate the non dynamically
equivalent system,
T1 = I1 .α
∴
T1 = m (k1)2 α
(ii)
• The correction couple is the difference between the torques
required to accelerate the non dynamically equivalent system
and the torque required to accelerate the dynamically equivalent
system,
Tc = T1 – T
Cont…
• Substituting the value of T1 and T in above equation we get,
Tc = m (k1)2 α – m (kG)2 α
∴
Tc = m (k12 – kG2) α
(iii)
• This couple must be applied, when the masses are placed
arbitrarily to make the system dynamical equivalent.
• It should be noted that the direction of the correction couple
may be same or reverse of the angular acceleration of
connecting rod.
Cont…
• On substituting value of (kG)2 = l1.l2, and (k1)2 = l1.l3, in the
equation (iii)
Tc = m (k12 – kG2) α
∴ Correction couple, Tc = m (l1.l3 – l1.l2) α
= m.l1 (l3 – l2) α
• But,
l3 – l2 = l – L
∴ Tc = m.l1 (l – L) α
• where l = Distance between the two arbitrarily masses, and
• L = Distance between the two masses for a true dynamically
equivalent system.
• It is the equivalent length of a simple pendulum when a body is
suspended from an axis which passes through the position of mass
m, and perpendicular to the plane of rotation of the two mass system.
Force Analysis in IC Engine Mechanism
• The study of forces in IC engine mechanism are classified
into two types as follows :
1. Steady force analysis,
2. Dynamic force analysis.
•
In static force analysis, we do not consider the effect of
inertia forces arising due to the mass of the connecting
rod.
•
In dynamic force analysis, we also consider the effect of
inertia forces caused due to the mass of connecting rod.
•
The force analysis can be done both by analytical and
graphical methods.
Static Force Analysis of IC Engine
Mechanism
• Consider an IC engine mechanism shown in Fig. in which
the crank OC rotates at an angular speed of ω in clockwise
direction.
• At an instant it is inclined at angle θ from I.D.C.
• Let,
r = Radius of crank,
l = Length of connecting rod,
n = Obliquity ratio = l/r
mR = Mass of the reciprocating parts, e.g. piston, crosshead
pin or gudgeon pin etc., in kg,
WR = Weight of the reciprocating parts in newtons = mR.g
m = Mass of the connecting rod,
θ = Angle made by crank with I.D.C.,
Φ = Angle made by connecting rod with line of
reciprocation (i.e. line OP) when line is inclined at
angle θ
1. Piston effort.
• It is the net force acting on the piston or crosshead pin, along the
line of stroke. It is denoted by FP in Fig.
• The acceleration of the reciprocating parts,
• Accelerating force or inertia force of the reciprocating parts,
• Therefore,
• Piston effort, FP = Net load on the piston ± Inertia force
= FL ± FI ...(Neglecting frictional resistance)
= FL ± FI − RF ...(Considering frictional resistance)
where, RF = Frictional resistance.
• The –ve sign is used when the piston is accelerated, and +ve sign is used
when the piston is retarded.
• In a reciprocating steam engine, net load on the piston,
FL = p1A1 – p2 A2 = p1 A1 – p2 (A1 – a)
• Where,
p1, A1 = Pressure and cross-sectional area on the back end side
of the piston,
p2, A2 = Pressure and cross-sectional area on the crank end side
of the piston,
a = Cross-sectional area of the piston rod.
• Notes :
1. If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the
piston, then
2. In case of a vertical engine,
2. Force acting along the connecting rod.
• It is denoted by FQ in Fig. From the geometry of the figure, we find that
3. Thrust on the sides of the cylinder walls or normal
reaction on the guide bars.
• It is denoted by FN in Fig. From the figure, we find that
4.Crank-pin effort and thrust on crank shaft
bearings.
• The force acting on the connecting rod FQ may be resolved into two
components, one perpendicular to the crank and the other along
the crank.
• The component of FQ perpendicular to the crank is known as crankpin effort and it is denoted by FT in Fig.
• The component of FQ along the crank produces a thrust on the
crank shaft bearings and it is denoted by FB in Fig.
• Resolving FQ perpendicular to the crank,
• and resolving FQ along the crank,
5. Crank effort or turning moment or torque on the crank
shaft.
• The product of the crankpin effort (FT) and the crank pin radius (r) is known as
crank effort or turning moment or torque on the crank shaft.
• Mathematically, Crank effort,