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Please welcome for any correction or misprint in the entire manuscript and your
valuable suggestions kindly mail us [email protected].
1. What is CORIOLIS component of acceleration? Derive an expression for evaluating
it and explain how the direction is fixed.
2. Deduce an expression for the couple that is called into play in the case of a wheel
rotating with uniform angular velocity in order to maintain a given rate of
precession.
3. Two wheels W1 and W2 of radii r1 and r2 respectively are lying in a plane. Their
centers are connected by a rod R of length (r1 + r2) and the wheels engage without
slipping. If the centre of W1 is fixed then W2 and R can rotate independently about
this centre, and W2 will roll on W1. Derive a relationship between the angular
velocities of W1, W2 and R.
4. A wheel 3 (Fig. 2) is planned to link 2 at A and rotates clock wise with angular
velocity ω3 and anti-clockwise angular acceleration α3, while link 2 is rotating
counter clockwise with positive values of angular velocity and acceleration. In Fig.
2, B is a point on link 2 while P is a point on link 3 (coincident at the particular
instant of time). Find the acceleration of P.
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2 Problems of Practices on Kinematic Analysis of Plane Mechanism
5. In a mechanism, a slider is guided along a slot of a rotating slotted lever. Explain
the various acceleration terms involved. Derive an expression for Coriolis
component of acceleration.
6. Bar AB is connected by pin C to slider D that slides along the fixed vertical rod EF
as shown in Fig. 2. Find the velocity and acceleration of the slider D if the bar AB
rotates at a constant angular velocity of 10 rad/s in counter clockwise direction.
7. A rigid rod (Fig. 2) slides over a fixed circle of radius a. The distance of rod's tip P,
measured from the contact point Q is denoted by s, and the rod makes an angle φ
with the fixed x-axis. Find the acceleration components (ax, ay) of point P with
respect to the fixed frame (X. Y), also find the components (aξ, aη) of the absolute
acceleration in the direction of moving axes (ξ, η).
8. What is the coriolis acceleration? Find the magnitude and direction of this
acceleration in the following cases.
9. Develop an equation for the relationship between the angular velocities of the
input and output cranks of a four-bar linkage.
10. In a slider crank mechanism the crank length is 10 cm and the connecting rod is 40
cm long. The crank rotates in anticlockwise direction at a constant speed of 600
rpm. Find the acceleration of the slider and the angular acceleration of the
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3 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
connecting rod at an instant when the crank is past the inner dead centre by 45º.
11. With usual notations show that the velocity of slider of slider-crank mechanism is
expressed approximately by
nrVp
2sinsin2
where n = length of the connecting rod/crank radius.
12. A revolving mass of a scooter engine crank is 2 kg at a radius of 10 cm. Determine
the magnitude of balance mass at a radius of 100 mm in two planes 10 cm apart.
One of the planes is at a distance of 5 cm from the plane of the crank. These planes
are
(i) on same side of crank
(ii) on opposite sides of crank.
13. A single cylinder horizontal reciprocating engine mechanism has a crank of 8 cm
length and connecting rod 36 cm length. The engine speed is 2000 rpm clockwise.
Determine the velocity and acceleration of piston when the crank is 315° from
inner dead centre. Also determine the angular acceleration of connecting rod and
total acceleration of its mid-point. Use relative velocity and acceleration method
only.
14. The crank length of a petrol engine is 50 mm and the connecting rod is 175 mm
long and the crank rotates at a uniform speed of 400 rpm. Calculate the velocity
and acceleration of the piston at different positions of the piston along its stroke
and plot the two curves.
Also find the crank position at which the piston's acceleration becomes zero.
15. A quick return mechanism of a shaper is shown in Fig. 2. The crank O1A rotates at
40 r.p.m. in the counter-clockwise direction. Determine the linear velocity of the
cutting tool when the crank O1A is at 45° with the horizontal. All dimensions are
given in the figure.
16. The dimensions of different links of the crank and slotted lever quick-return
mechanism shown in Figure are given below:
O1O2 = 700 mm, O1B = 250 mm
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4 Problems of Practices on Kinematic Analysis of Plane Mechanism
O2D = 1250 mm, DR = 350 mm.
The crank O1B rotates at 40 r.p.m. in the counter clockwise direction and at the
present instant of consideration makes an angle of 45° with the vertical.
Determine:
(i) Velocity of the ram R which moves in a horizontal direction.
(ii) Angular velocity of link O2D.
17. The driver link AB of a four bar mechanism is rotated at 5.0 rad/s in counter
clockwise direction as shown in the above figure. Locate and indicate the
Instantaneous Centre (I.C.) of the coupler BC with respect to fixed link AD at an
instant when BAD = 180°. Find angular velocity of the coupler using I.C. method
only.
18. Determine the maximum and minimum transmission angle of the mechanism
shown which is driven by member O1A.
19. The crank of a crank and slotted lever quick return mechanism is driven at 120
rpm clockwise. The vertical distance between the centres of rotation of the crank
and slotted lever is 50 cm. What should be the length of the crank if the quick
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5 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
return ratio is 1:2? Determine the angular velocity of the slotted lever, when the
tool-post attains maximum velocity during cutting stroke.
The objective of kinematic analysis is to determine the kinematic quantities such as
displacements, velocities, and accelerations of the elements in a mechanism when the
input motion is given. Conversely, the objective may be to determine the input motion
required to produce a specified motion of another element. In short, kinematic analysis
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6 Problems of Practices on Kinematic Analysis of Plane Mechanism
establishes the relationship between the motion of the various components or links of a
mechanism. In this chapter, we shall consider only the mechanisms with lower pairs.
The kinematic analysis of a higher-pair mechanism can also be carried out. This may be
done by converting the mechanism into an equivalent mechanism consisting of only the
lower pairs. Both the graphical and analytical methods can be used for kinematic
analysis. However, the graphical methods, providing better insight and visualization,
still occupy a prominent place in planar kinematics. Unlike in an analytical method, the
accuracy of the solution provided by a graphical method is often limited.
2.1 DISPLACEMENT ANALYSIS
When the kinematic dimensions and/the configuration(s) of the input link(s) of a
mechanism are prescribed, the configurations (linear and angular) of all the other links
are determined by displacement analysis.
Graphical Method
In a graphical method of displacement analysis, the mechanism is drawn to a
convenient scale and the desired unknown quantities are determined through suitable
geometrical constructions and calculations. No generalized approach can be discussed so
far as the graphical methods are concerned; the solution technique will vary from
problem to problem. We shall demonstrate the basic features of the methods through a
few examples. Some of these features which play a major role in displacement analysis
are as follows:
(i) The configuration of a rigid body in plane motion is completely defined by the
locations of any two points on it.
(ii) Two intersecting circles have two points of intersection and one has to be
careful, when necessary, to choose the correct point for the purpose in hand.
(iii) The use of a tracing paper, as an overlay, is very convenient and very often
provides an unabiguous and quick solution.
(iv) The graphical method of displacement analysis fails if no closed loop with four
links exists in the mechanism.
Application 2.1: As shown in Diagram 2.1(a) a six-link, Whitworth quick-return
mechanism used in slotting machines. Link 2 (= O2A) rotates with a constant angular
speed in the counter-clockwise (CCW) direction. The sliding link 6 represents the cutting
tool whose cutting motion is to the left and the return (idle) motion is to the right.
Determine the quick-return ratio which is defined as the ratio of the time intervals
taken by the tool to complete its cutting and idle motions, respectively.
Solution: Referring to Diagram 2.1b, the circles kA and kC are the paths of the points A
and C, respectively. The centres of these circles are at O2 and O4 and their radii are O2A
and O4C, respectively. The tool occupies its extreme right and extreme left positions (DR
and DL) when links 4 and 5 become collinear as shown in the diagram, i.e., OuCRDR and
OuCLDL. The distance DRDL is the stroke of the tool (= 2O4C). Since the points A, O4, and
C always remain collinear (on link 4), the locations of A (on kA) corresponding to the
extreme positions are now easily obtained as AR and AL, respectively. Thus, the rotations
of link 2 corresponding to the forward and return motions of the tool are obtained as θf
and θr, respectively.
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From measurement, θf = 239°. Since link 2 rotates at a constant angular speed, the
quick-return ratio (qrr) is obtained as
Analytical Method
Let us consider a 4R linkage (Diagram 2.2) of given link lengths, viz., li, i = 1, 2, 3, and 4.
The configuration of the input link (2) is also prescribed by the angle θ2, and we have to
obtain the configurations of the other two links, namely, the coupler and the follower,
expressed by the angles θ3 and θ4. From Diagram 2.2, all links are denoted as vectors,
viz., l1, l2, l3, and l4. All angles are measured CCW from the x-axis which is along the
fixed vector l1 rendering θ1 = 0.
Diagram 2.2 Four bar mechanism
Considering the closed loop O2O4BAO2, we can write
Using complex exponential notation with θ1 = 0, then above equation is (2.1) can be
written as
Equating the real and imaginary parts of this equation separately to zero, we get
Thus, the two unknowns, namely, θ3 and θ4, can be solved from the two equations (2.2a)
and (2.2b) as now explained.
Rearranging (2.2a) and (2.2b), we get
Squaring both sides of these two equations and adding, we obtain
or
B A
(a)
6 5
4
3 2
C
D
A (b)
Diagram 2.1
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8 Problems of Practices on Kinematic Analysis of Plane Mechanism
where
It may be noted that with the prescribed data (i.e., link lengths and θ2), the coefficients
a, b, and c of (2.3) are known.
To solve for θ4, from (2.3),
Substitute these values in Eq. (2.3), then we get
which gives
Thus, for a given position of the input link, two different values of θ4 are obtained as
follows:
These two values correspond to the two different ways in which the 4R mechanism can
be formed for any given value of θ2, as explained in Diagram 2.3 where the same
problem has been solved by a graphical method.
Diagram 2.3 Four bar mechanism
To solve for the coupler orientation θ3, we can eliminate θ4 from (2.2a) and (2.2b), we get
where
It is quite obvious that we can get c' from the expression for c by interchanging l3 and l4
but attention may be drawn to the roots of θ3 and θ4 which pertain to the same
configuration.
Since the 4R linkages are very useful in practice, it is instructive to go into further
details of the output-input (i.e., plots of θ4 vs. θ2) relationship of these linkages. The
typical nature of this relationship for various types of 4R linkages is indicated in
Diagram 2.4a-d. For crank-rocker and double-crank mechanisms, the output-input
B A
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9 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
characteristics are as shown in Diagram 2.4a and 2.4b, respectively. The same
characteristics for double-rocker mechanisms are like those shown in Diagram 2.4c or
Diagram 2.4d, depending on whether the linkage is Grashof type or non-Grashof type.
Diagram 2.4
The following distinctive features of various plots shown in Diagram 2.4 should be
noted:
(i) For all Grashof-type linkages, there are two disconnected branches, each
corresponding to one mode of assembly. The assembly modes are mirror image of
each other, with the mirror placed along the fixed link. The linkage shown in
Diagram 2.3 is a crank rocker, therefore the dashed configuration can never be
obtained from the configuration O2ABO4. The dashed configuration will be
obtained if initially the mechanism is assembled as the mirror image of O2ABO4
and link 2 then driven so as to coincide with O2A.
(ii) For a non-Grashof linkage, the plot of θ4 vs. θ2 is a single closed loop, implying
that, once assembled, it can take up the mirror image configuration without being
dismantled.
(iii) Except at the ends of the swing, all rocking links pass through the same position
twice the different orientations of the other link connected to the frame.
Of all the possible variations of a 4R mechanism, crank rocker is most commonly used in
practice. In general, for a uniform angular speed of the crank, the rocker takes a
different time interval during its forward and return motions. It is useful (at this stage)
to note the relationship between the link lengths that ensures equal time for the
forward and return strokes of the rocker.
Diagram 2.5
If the rotation of the crank has to be exactly equal to corresponding to both forward
and return strokes of the rocker, then the link lengths should be such that the outer and
inner dead-centre configurations are like those shown in Diagram 2.5, where the points
A1, A2, B1, B2, and O2 are all collinear. From the triangles O2O4B1 and O2O4B2, we can
write
These two equations, after simplification, result in
(a)
(b)
(c)
(d)
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10 Problems of Practices on Kinematic Analysis of Plane Mechanism
where, l1, l2, l3, and l4 are the lengths of frame, crank, coupler, and follower, respectively.
Thus, (2.6) constitutes the condition required for equal time interval during the forward
and return motions of the follower (link 4) when the crank (link 2) rotates with a
constant angular speed. Furthermore, we should note, from Diagram. 2.5, that for such
a 4R mechanism without quick return,
or
where is the swing angle of the rocker. Now onwards, such a 4R linkage will be
designated as one without the quick-return effect.
2.2 ALGEBRAIC POSITION ANALYSIS
In this section we present the classical approach used in the position analysis of the
slider-crank mechanism. Diagram 2.6 shows the offset version that has been chosen for
analysis. By making the offset equal to zero, the same equations can be used for the
centered or symmetrical version. The two problems that occur in the position analysis of
the slider-crank mechanism are:
Problem 1: Given the input angle θ2, find the connecting rod angle θ3 and the position xB.
Problem 2: Given the position xB, find the input angle θ2 and the connecting rod angle θ3.
Diagram 2.6
Now, starting with problem 1, we define the position of point A with the equation set
Next, we note that
From the geometry of Diagram 2.6, we see that
Then using the trigonometric identity we have,
from Eq. (2.9),
We arbitrarily select the positive sign, which we see from Diagram 2.6 corresponds
solution with the piston to the right of the crank pin.
Now using Eq. (2.11) and Eq. (2.10), we get
4
2 A
B
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Thus, with the angle θ2 given, the unknowns θ3 and xB can be obtained by solving Eqs.
(2.10) and (2.12).
Problem 2 requires that, given xB, we solve Eq. (2.12) for the angle θ2. So, now we apply
well-known Newton-Raphson method. This method can be explained by reference to
Diagram 2.7. This figure is a graph of some function f(x) versus x. Let xn be a first
approximation or a rough estimate of the root where f(x) = 0 which we wish to find. A
tangent line drawn to the curve at x = xn intersects the x axis at xn + 1, which is a better
approximation to the root. The slope of the tangent line is the derivative of the function
at x = xn and is
Diagram 2.7
Solving for xn + 1,
Now solving the single slider mechanism with the help of Newton-Raphson method.
Suppose in Eq. (2.22), replace the angle θ2 with and let r2, r3, e, and xB be given
constants. Then
and
Now squaring and adding the above equations, then we get
Solving for θ2 gives
The solution of problem 1 for the centered version is, of course, obtained directly from
Eq. (1.12) by making e = 0.
The Crank-and-Rocker Mechanism
The four-bar linkage shown in Diagram 2.8 is called the crank-and-rocker mechanism.
Thus link 2, which is the crank, can rotate in a full circle; but the rocker, link 4, can only
oscillate. We shall generally follow the accepted practice of designating the frame or
fixed link as link 1. Link 3 in Diagram 2.8 is called the coupler or connecting rod. With
the four-bar linkage the position problem generally consists of finding the positions of
the coupler and output link or rocker when the dimensions of all the members are given
together with the crank position.
Tangent line
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12 Problems of Practices on Kinematic Analysis of Plane Mechanism
Diagram 2.8
To obtain the analytical solution we designate s as the distance AO4 in Diagram 2.8. The
cosine law can then be written twice for each of the two triangles O4O2A and ABO4. In
terms of the angles and link lengths shown in the figure we then have
There will generally be two values of λ corresponding to each value of θ2. If θ2 is in the
range 0 ≤ θ2 ≤ , the unknown directions are taken as
θ3 = φ β; θ4 = λ β
However, if θ2 is in the range ≤ θ2 ≤ 2, then
θ3 = φ + β; θ4 = λ + β
Finally, the transmission angle is given by the
2.3 CONCEPT OF TRANSMISSION ANGLE
However, even at the stage of kinematic design, we should ensure that the output
member receives, along its direction of movement, a large component of the force (or
torque) from the member driving it. Assuming all binary links as two-force members
(i.e., neglecting gravity, inertia, and frictional effects), we can express the free-running
quality of simple mechanisms (like 4R or slider-crank mechanisms) through an index
termed as the transmission angle.
Diagram 2.9
For a 4R linkage, the transmission angle (μ) is defined as the acute angle between the
coupler (AB) and the follower (O4B), as shown in Diagram 2.9. If ABO4 is acute, then μ
= ABO4. On the other hand, if ABO4 is obtuse, then μ = ‒ ABO4. As explained in
this diagram, if μ = /2, then the entire coupler force is utilized to drive the follower.
For good transmission quality, the minimum value of μ (μmin) > 30°. For a crank-rocker
Component producing
(driving) torque
Coupler force
B
A
2
4
3
B
A
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13 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
mechanism, the minimum value of μ, occurs when the crank becomes collinear with the
frame, i.e.,
or
It is not at all difficult to prove the last assertion. If the swing angle of the rocker is
increased, maintaining the same quick-return ratio, then the maximum possible value of
μmin decreases. If the forward and return strokes of the rocker take equal time, then
(μmin)max is restricted to . Therefore, such a crank rocker will have a poor
transmission quality if > 120°.
Application 2.2: A crank-rocker 4R linkage without the quick-return effect has to have
a swing angle and a minimum transmission angle μmin. Determine the link-length
ratios l2/ l1, l3/ l1, and l4/ l1.
Solution: Using Eqs. (2.6) and (2.7), for such a linkage, we have
Considering the configuration θ2 = 0 in Diagram 2.9, when ABO4 = μmin, we can write
Equations (a), (b), (c) are rewritten, respectively, as
Using (d) in (f), we obtain
Using (e) in (g), we obtain
Using (d), (g), and (h), it is easy to show that
It is obvious from (i) that l3/l1 ≤ 1 and then, from (h), we get
For a slider-crank mechanism, the transmission angle is defined as the acute angle
between the connecting rod and the normal to the slider movement as indicated in
Diagram 2.10. The minimum transmission angle in this case is given by
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14 Problems of Practices on Kinematic Analysis of Plane Mechanism
Diagram 2.10
2.4 GENERAL PLANE MOTION
Before going into the details of various methods of velocity and acceleration analysis of
plane mechanisms (i.e., a series of interconnected rigid bodies), let us briefly
recapitulate the fundamentals of the velocity and acceleration of a particle and a rigid
body in plane motion.
Motion of a Link
Let us suppose a rigid link OA, has uniform angular velocity ω rad/s in the counter-
clockwise direction with radius r, rotate about a fixed point O as shown in Diagram
2.11(a). After a small time interval δt link OA turns through with a small angle δθ and
point A reach at point A' as shown in Diagram 2.11(b).
Diagram 2.11
Now, velocity of A with respect to O is
The direction of VAO is along the displacement of A is perpendicular to OA as shown in
Diagram 2.11(c). The fact that the direction of velocity vector is perpendicular to the link
also emerges from the fact that A can neither approach nor recede from O and thus, the
only possible motion of A relative to O is in a direction perpendicular to OA.
Consider a point B on the link OA. It is also moves with link OA and reaches at point B'
in time interval δt. Again it is seen that the direction of VBO is also perpendicular to OA.
Velocity of B = ω. OB perpendicular to OB.
This means that all the particles (or points) present on OA and moves with link OA,
then direction of velocity of the particles (or points) is perpendicular to link OA.
At other instants, when the link OA assumes other positions, the velocity vectors will
have their directions changed accordingly. Also, the magnitude of the instantaneous
linear velocity of a point on a rotating body is proportional to its distance from the axis
of rotation.
Hence velocity of any point on a link with respect to another point on the same link is
always perpendicular to the line joining these points on the configuration (or space)
diagram.
O
B
A
O
B'
A' B
A
A
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15 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
Now
Rubbing Velocity
The rubbing velocity at a pin joint is defined as the algebraic sum between the angular
velocities of the two links which are connected by pin joints, multiplied by the radius of
the pin.
Consider two links A and B which are connected by a pin joint at O as shown in
Diagram 2.12.
Diagram 2.12
Consider ω1 = Angular velocity of link A; ω2 = Angular velocity of link B; r = Radius of
the pin at the joint.
According to definition,
Rubbing velocity at the pin joint O
= (ω1 ‒ ω2) r, if the links move in the same direction
= (ω1 + ω2) r, if the links move in the opposite direction
At gudgeon pin of the connecting rod one member is turning and the other is sliding, the
rubbing velocity is ω. r because the angular velocity of the sliding member is zero.
Plane Motion of a Particle
The path of a particle moving in a plane is shown in Diagram 2.13a. The positions and
velocities of the particle at the instants t and (t + δt) have been indicated. Also, the
radius and centre of curvature corresponding to these instants have been shown. As
depicted in Diagram 2.13b, the change in velocity δV can be resolved into two mutually
perpendicular components δVn and δVt. When δt → 0, the direction of the tangential
component δVt coincides with that of V, and the normal component δVn will be directed
towards the centre of curvature.
Diagram 2.13
Obviously, the magnitude of the velocity of the particle at the time t can be expressed as
Path of particle
Reference line
(a)
O
P' (t + δt ) V + δV
V
P(t) δs
δθ
θ
(b)
δθ
V + δV
δ
V
V
VA
A
AB
B
VB
VA
VBA (⊥r to AB in
sense of ω) VB
B A
O
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16 Problems of Practices on Kinematic Analysis of Plane Mechanism
where θ is the inclination [with respect to a reference line (see Diagram 2.13a)] and ω is
the magnitude of the angular velocity of the radius of curvature. The velocity vector V is
always tangential to tin path and its direction is obtained by rotating ρ through 90° in
the sense of ω. In vector notation,
V = ω × ρ
The normal and tangential components of acceleration can be derived as follows. The
normal component of acceleration is
From Diagram 2.13b, the magnitude of δVn is
δVn = V δθ
So,
Using (2.13) in this relation, we obtain
The normal component of acceleration is always directed towards the centre of
curvature. In vector notation,
an = ω × (ω × ρ)
The magnitude of the tangential component of acceleration is
Using (2.13), we find the final expression for at becomes
where α is the magnitude of the angular acceleration of the radius vector.
Plane Motion of a Rigid Body
The position of a rigid body in plane motion is completely defined by specifying either
the positions of any two points on the body or the position of a point and the orientation
of a line fixed on the rigid body.
Similarly, the velocity (of all points) of a rigid body in plane motion is completely defined
by specifying the velocity of any point on the body along with the angular velocity ω of
the body. If the rigid body is in pure translation (without rotation), then the motion of
all points on it is identical. The difference in velocity of two points A and B (Diagram
2.13c) is entirely due to ω and this difference is expressed as (Diagram 2.13d)
VBA = VB VA = ω × AB
It is easy to identify that the difference in the motions of B and A, due to ω, is a circular
motion of B about A since AB remains constant, i.e., ρ in (2.13)-(2.15) is constant and
replaced by AB.
For the velocity analysis of a mechanism, the following form of the foregoing equation is
used:
VB = VA + VBA = VA + ω × AB [2.16]
It should be noted that for this equation to be valid, A and B must be on the same rigid
body and ω must be the angular velocity of the body.
The acceleration of a rigid body is defined by specifying either the accelerations of any
two points on it or the acceleration of any point on it and its angular acceleration
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17 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
(assuming that the angular velocity of the body is already known). The acceleration of a
point B on a rigid body can be expressed in terms of the acceleration of a point A and the
angular velocity and acceleration of the body. The difference in acceleration is given by
From (2.14) and (2.15), with ρ = AB = constant,
written in vector notation as
The normal component is always along BA and the tangential component is
perpendicular to AB in the sense of . The commonly-used form of the acceleration
relation is
Motion Difference between Two Instantaneously Coincident Points
In a mechanism, a link is quite often guided along a prescribed path in another moving
link. For the velocity and acceleration analyses of such a mechanism, the differences in
the velocities and accelerations of two instantaneously coincident points belonging to the
two links have to be determined. In this section, we shall derive the expressions for
these quantities.
Diagram 2.14
Diagram 2.14a shows a rotating rigid link (labeled 2) on which link 3 is moving along a
straight line. The configurations at the instants t and (t + δt) are, respectively, shown by
the symbols without and with a prime. Further, P2 and P3 represent two points on links
2 and 3, respectively, coincident at the instant t. The displacement of P3 can be written
as
where represents the displacement of P2 and
represents the
displacement of P3 with respect to link 2. Dividing both sides of the foregoing equation
by δt and taking the limit δt → 0, we get
where Vp3/2 is the velocity of P3 as seen by an observer attached to link 2. The direction
of Vp3/2 is tangential to the path of P3 in link 2.
From (2.18), it may appear that the absolute acceleration of P3 can also be obtained from
an equation similar to it, i.e., ap3 = ap2 + ap3/2, where ap3/2 is the acceleration of P3 as seen
Link 2
Link 3
A
C
2
3
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18 Problems of Practices on Kinematic Analysis of Plane Mechanism
by an observer on link 2. However, this is incorrect since an extra term has to be added
to the right-hand side of this equation, as now explained.
From Diagram 2.14b, we see that the block has moved through an additional transverse
distance because of the rotation of link 2 and the radial motion of link 3 with respect
to link 2. When δt→ 0,
From this equation, we observe that the additional displacement term is proportional to
the square of the time elapsed. Therefore, this displacement must be due to an
additional acceleration of P3 in the transverse direction. If the magnitude of this
additional acceleration is ac, then
or
In vector notation, (2.19) is written as
This extra transverse component of acceleration is known as the Coriolis component.
The final expression for ap3 will then be
It should be noted that the direction ac is obtained by rotating Vp3/2 through 90° in the
sense of ω2. For a straight-line path of P3 on link 2, the direction of ap3/2 is along the
straight line.
When link 3 moves along a curvilinear path on the rotating link 2 (see Diagram 2.14c),
equation (2.20) can be written in terms of the components of ap3/2 as
It may be noted that the magnitude of is equal to
, where ρ is the radius of
curvature of the path of P3 on link 2. The direction of Vp3/2 is obviously tangential to this
path. For a straight-line path of P3 on link 2, ρ becomes infinite and = 0.
It may be pointed out that for (2.18) and (2.21) to be valid, P2 and P3 need not belong to
two adjacent links. We can use these relations for two coincident points, Pi and Pj,
belonging to the i-th and j-th links, respectively. The important point is that the path of
Pj in the i-th link should be known (generally, the path of Pi in the j-th. link is not the
same as this one) and we can write
or
with
To identify the path of Pj in the i-th link, it is advisable to consider a kinematic
inversion of the mechanism keeping the i-th link fixed.
Brief Solution of Coriolis Component of Acceleration
Generally we know that the acceleration of fixed point with respect to another fixed
point on moving link has two components. These are normal (radial or centripetal) and
other is tangential component.
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19 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
Now suppose a case in which acceleration of moving point relative to fixed point on the
moving link. This will have three acceleration components, these are
(a) Normal, (radial, centripetal) component an;
(b) Tangential component at; and
(c) Coriolis component ac, ac.
Now, consider a link OA rotates with an angular velocity ω about a fixed point O. Point
B is slider which moves radially along OA as shown in Diagram 2.15.
Consider at time interval t at any given instant,
ω = Angular velocity of the link;
= Angular acceleration of the link
V = Linear velocity of the slider on the link
a = Linear acceleration of the slider on the link
r = Radial distance of point B on the slider.
After a short interval of time dt, the link has been taken angular displacement dθ, and
dr is the radial displacement of the slider in the outward direction.
Diagram 2.15
Now acceleration in vertical (y) direction (which is parallel to OA).
From diagram initial velocity in y-direction = V
Final velocity in y-direction = (V + a.dt) cos dθ ‒ (ω + .dt)(r + dr) sin dθ.
If angle is very small, then cos dθ ≈1, and sin dθ ≈ dθ, then acceleration will be
After neglecting the small terms, then we get
Now acceleration in horizontal (x) direction (which is ⊥r to OA).
From diagram initial velocity in x-direction = rω
Final velocity in x-direction = (V + a.dt) sin dθ ‒ (ω + .dt)(r + dr) cos dθ.
If angle is very small, then cos dθ ≈1, and sin dθ ≈ dθ, then acceleration will be
where
A
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20 Problems of Practices on Kinematic Analysis of Plane Mechanism
After neglecting the small terms, then we get
ax = 2Vω + r [b]
First term in Eq. (b) is termed as Coriolis component of acceleration
ac = 2Vω
Radial component of acceleration of point B with respect to O, is
Tangential component of acceleration of B with respect to O, is
Radial component of the slider B is
Tangential component of the slider is
Direction between VOB, ω, and ac:
Diagram 2.16
2.5 INSTANTANEOUS CENTRE OF VELOCITY
In Diagram 2.17, a rigid body 2 is shown to be in plane motion with respect to the fixed
link 1. The velocities of two points A and B of the rigid link 2 are shown by VA and VB,
respectively. Two lines drawn through A and B in directions perpendicular to VA and VB
meet at P. Let PA = r1 and PB = r2. The velocity of the point B in the direction of AB is
VB cos , and that of the point A in the same direction is VA cos θ. As the length of AB is
fixed, the component of VBA in the direction of AB is zero. Thus, VB cos = VA cos θ.
Diagram 2.17
From the triangle PAB, we have
Thus, the velocities of the points A and B are proportional and perpendicular to PA and
PB, respectively. So, instantaneously, the rigid body can be thought of as being
1
2 B
A
C VC
VB
VA
P
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21 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
momentarily in pure rotation about the point P. The velocity of any point C on the body
at this instant is given by VC = PC.VB /r2 in a direction perpendicular to PC. This point P
is called the instantaneous centre of velocity, and its instantaneous velocity is zero. So,
alternatively, the instantaneous centre of velocity can be defined as a point which has no
velocity with respect to the fixed link.
If both links 1 and 2 are in motion, in a similar manner, we can define a relative
instantaneous centre P12 (sometimes called centro) to be a point on 2 having zero relative
velocity (i.e., the same absolute velocity) with respect to a coincident point on 1.
Consequently, the relative motion of 2 with respect to 1 appears to be pure rotation (for
the instant) about P12. It is evident that P12 and P21 are identical. Thus, if a mechanism
has N links, the number of relative instantaneous centres is N(N ‒ l)/2. The absolute
instantaneous centre of velocity of a link is the relative instantaneous centre of velocity
with respect to the fixed link. Note carefully that the instantaneous centres can lie
outside the physical boundary of the links and anywhere in the plane of motion.
However, they are considered to be integral points of the concerned two links (imagined
to be extended).
How to Find Out the Location of Instantaneous Centres in a Mechanism:
In certain cases, the relative instantaneous centres are easily identified by inspection of
the geometry of motion. Examples of such situations are as follows:
1. If two links have a hinged joint, the location of the hinge is the relative
instantaneous centre because one link is in pure rotation with respect to the other
about that hinge. Such a instantaneous centre is of permanent nature, but if one
of the links is fixed, the instantaneous centre will be of fixed type. (Diagram
2.18a).
2. If the relative motion between two links is pure sliding, the relative instantaneous
centre lies at infinity on a line perpendicular to the direction of sliding. (Diagram
2.18b)
3. If one link is rolling (without slipping) over another, the point of contact is the
relative instantaneous centre. (Diagram 2.18c)
4. If a link is sliding over a curved element, the centre of curvature is the relative
instantaneous centre. (Remember the curved slider and the equivalent hinge)
(Diagram 2.18d)
5. If the relative motion between two links is both rolling and sliding, the relative
instantaneous centre lies on the common normal to the surfaces of these links
passing through the point of contact. (Diagram 2.18e)
The determination of its exact location requires some more information.
Diagram 2.18
Fixed link 1
(b)
B A
VA
VB
Link 2
(Slider)
P12 at ∞
Link 1
(Slider)
Fixed link 2
(c)
P12 VB
VA
B A
Link 1
(Slider)
Fixed link 2
(e)
B A
VB
VA
P12
Link 1
(f)
B A
VB
VA
Link 2
(Slider)
P12
(a) P12
Link 2
Link 1
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22 Problems of Practices on Kinematic Analysis of Plane Mechanism
The instantaneous centre of acceleration is defined as a point on a link having zero
relative acceleration with respect to a coincident point on the other link, and in general
is different from the instantaneous centre of velocity.
2.6 ARDNHOLD-KENNEDY THEOREM OF THREE CENTRES
The Aronhold-Kennedy theorem states that if three bodies are in relative motion with
respect to one another, the three relative instantaneous centres of velocity are collinear.
Diagram 2.19
Proof: Diagram 2.19 shows three links 1, 2, and 3 in relative motion with respect to one
another. Since we are interested only in the relative motion, without any loss of
generalities we can assume one of the three links, say, link 1, is fixed. P12(O2) and
P13(O3) are the points about which links 2 and 3 are rotating. If P23 is not on the line
joining P12P13, let it be somewhere else as shown in the diagram. Considering P23 as a
point on link 2, its velocity must be in a direction perpendicular to P12P23. If P23 is taken
as a point on link 3, its velocity must be in a direction perpendicular to P13P23. By
definition, P23 must have the same velocity whether it is considered to be on link 2 or 3.
This cannot be so unless P23 is on the line P12P13, otherwise the directions will be
different as has been shown in the diagram.
This theorem will be used very often for determining the relative instantaneous centres
of a mechanism.
Types of Instantaneous Centres
The instantaneous centres for a mechanism are of the following three types:
1. Fixed instantaneous centres, does not change at their position in space with time.
2. Permanent instantaneous centres, for two links at joint which in space when
mechanism moves, and
3. Neither fixed nor permanent instantaneous centres, also known as movable
instantaneous centres and finding by using Kannedy’s theorem.
The first two types i.e. fixed and permanent instantaneous centres are together known
as primary instantaneous centres and the third type is known as secondary
instantaneous centres.
Consider a four bar mechanism O2ABO4 as shown in Diagram 2.20. The number of
instantaneous centres (N) in a four bar mechanism is given by
The instantaneous centres P12 and P14 are called the fixed instantaneous centres as they
remain in the same place for all configurations of the mechanism. The instantaneous
centres P23 and P34 are the permanent instantaneous centres as they move when the
mechanism moves, but the joints are of permanent nature. The instantaneous centres
2 1
3 2
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23 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
P13 and P24 are neither fixed nor permanent instantaneous centres as they vary with the
configuration of the mechanism.
Diagram 2.20
Angular Velocity Ratio Theorem
It’s state that velocity of the link in a given mechanism equals the product of the angular
velocity and radius of I-centre at the instant.
When the angular velocity of a link is known and it is required to find the angular
velocity of another link, locate their common I-centre. The velocity of this I-centre
relative to a fixed third link is the same whether the I-centre is considered on the first or
the second link. First consider the I-centre to be on the first link and obtain the velocity
of the I-centre. Then consider the I-centre to be on the second link and find its angular
velocity.
Proof: Consider an example of four bar link as shown in Diagram (2.20).
Now according to theorem
VBA = ω2 AB
V2 = ω2 (P12 P23) (a)
Let ω3 be the angular velocity of link 3, then
V2 = ω3 (P13 P23) (b)
From Eqs. (a) and (b), then we have
In general
If the velocity of any link x to be determine, when angular velocity of link y is known,
then
2.7 VELOCITY AND ACCELERATION IMAGES
The concepts of velocity and acceleration images are used extensively in the kinematic
analysis of mechanisms having ternary, quaternary, and higher-order links. If the
velocities and accelerations of any two points on a link are known, then, with the help of
images, the velocity and acceleration of any other point on the link can be easily
determined. An example to illustrate this concept follows. The method used will be
better appreciated when the kinematic analysis of complex mechanisms is considered
later in this chapter.
2
A
B
4
1
3
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24 Problems of Practices on Kinematic Analysis of Plane Mechanism
Diagram 2.21
A rigid link BCDE having four hinges is shown in Diagram 2.21a. Let the angular
velocity and acceleration of this link be ω and (CCW). The absolute velocity vectors of
the points E, B, C, and D are shown in Diagram 2.21b as VE, VB, VC, and VD,
respectively. The velocity difference vectors are
eb = VBE; bc = VCB; ec = VCE; bd = VDB
and their magnitudes are, respectively,
EB; BC; EC; BD
So,
Hence, the velocity diagram bcde is a scale drawing of the link BCDE. The diagram bcde
(formed by the tip of the absolute velocity vectors) it called the velocity image of the link
BCDE. The velocity image is rotated through 90° in the direction ω, as all the velocity
difference vectors are perpendicular to the corresponding lines. The scale of the image is
determined by ω, and therefore the scale will be different for each link of a mechanism.
The letters identifying the end points of the image are in the same sequence as that in
the link diagram BCDE. The absolute velocity of any point X on the link is obtained by
joining the image of X(x) with the pole of the velocity diagram o.
The absolute acceleration vectors of the points E, B, C, D are shown in Diagram 2.21c as
aE, aB, aC, aD respectively. The acceleration difference vectors are
Thus,
Similarly,
So, the diagram ebcd (formed by the tip of the absolute acceleration vectors) is a scale
drawing of the link EBCD and is called the acceleration image. The scale will be
different for each link of a mechanism. The letters identifying the end points of the
image are in the same sequence as that in the link diagram BCDE. The absolute
acceleration of any point X on the link can be obtained by joining the image of X(x) with
(c) Acceleration diagram
(a) Space diagram
E
X B
C
D
(b) Velocity diagram
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the pole of the acceleration diagram o. The orientation of the acceleration image from
the link diagram is (180°‒ 0) in the counter-clockwise direction, where θ = tan-1
[bb1/(b1e)] = tan-1(/ω2), with a positive in the CCW sense (as explained in Diagram
2.21c). It should be noted that once the absolute velocities and accelerations of any two
points (e.g., E and C) of a rigid link are known, those of any other point on the link (such
as X, B, and D) can be determined just by drawing the respective images.
2.8 VELOCITY AND ACCELERATION ANALYSIS (GRAPHICAL)
The general principal for carrying out the kinematic analysis of most problems is to
construct the velocity and acceleration diagrams starting from input link. In these
diagrams, the fixed link is representing by a point. Such a point is termed as the pole of
velocity diagram or the acceleration diagram.
Slider-Crank Mechanism
Consider a slider-crank mechanism OAB as shown in Diagram 2.22(a). Crank OA
rotates clockwise with angular velocity ω rad/sec. It is required to draw the velocity and
acceleration diagrams.
Diagram 2.22
Velocity Diagram
Velocity of point A with respect to fixed point O is VAO = ω.OA in the clockwise sense.
First of all, velocity polygon is drawn as follows:
1. Take any point o and from it draw oa = VAO = ω.OA with some suitable scale as
shown in Diagram 2.22 (b). Now, VBO and VBA were known in direction only. VAO
was known in direction and magnitude both.
2. The velocity of point B with respect to point A. (on connecting rod) is
perpendicular to line AB. So from point ‘a’ draw a line perpendicular to AB
representing VAB.
3. The velocity of slider B relative to O (VBO) is parallel to the line OB. From o draw
a line parallel to OB to intersect VAB point b.
Measure VBA = ab and VBO = ob from the velocity diagram.
Angular velocity of coupler AB is
Acceleration Diagram
We know that radial/normal acceleration of A with respect to O is given as
(a) Single slider
mechanism
B
O
A
(b) Velocity diagram
⊥
⊥
(c) Acceleration diagram
⊥
⊥
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26 Problems of Practices on Kinematic Analysis of Plane Mechanism
The tangential acceleration of point A with respect to O is zero because crank OA rotates
with constant angular velocity.
Thus = 0. The acceleration diagram is drawn as:
1. Draw vector oa' = parallel to AO with some suitable scale as shown in
Diagram 2.22 (c).
2. From point a' draw vector a'x = which is the radial component of acceleration
of B with respect to A. It is parallel to BA. VBA can be taken from velocity
diagram.
3. From x draw vector xb' perpendicular to a'x or AB. The vector xb' represents the
tangential component of acceleration of B with respect to A i.e. . It is known in
direction only and contains b' .
4. The acceleration of point B with respect to O is parallel to the line of motion of
slider B i.e. along OB. It is not radial acceleration as the slider has reciprocating
motion. So from point o draw a line parallel to OB representing aB to intersect
vector xb' at b'.
5. Join a' to b'. Vector a'b' represents acceleration of B with respect to A i.e. aBA.
aB, , aBA., can be found by measurement with the scale.
S.No. Vector Magnitude Direction Sense
1 OA → O
2 AB → A
3 ⊥ AB
4 to line of motion B
Four-bar Chain Mechanism
Consider a four-bar chain ABCD is shown in Diagram 2.23(a). AB is the crank and
rotates at ω rad/sec in the clockwise sense. It is required to draw the velocity and
acceleration diagrams.
Diagram 2.23
Velocity diagram
The velocity of point B with respect to point A (VBA) can be written as
VBA = ω.AB in the clockwise sense
1. Take points a, d as fixed centres as shown in Diagram 2.23(b).
2. The velocity of point B with respect to A i.e. VBA is perpendicular to AB. VBA is
given. Draw vector ab = VBA perpendicular to AB with some suitable scale.
3. The velocity of point C with respect to point B i.e. VCB is perpendicular to line BC.
Only the direction of VCB is known. So from point b draw a vector bc
perpendicular to BC to represent VCB.
(a) Four-bar mechanism
D
B
A ω
C
(b) Velocity diagram
⊥
⊥
⊥
⊥
(c) Acceleration diagram
⊥
⊥
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4. The velocity of point C with respect to D i.e. VCD is known in direction only. From
a (or d) draw a vector ac perpendicular to CD to represent VCD.
5. The vectors ac and be intersect at c. The vector ac = VCD.
By measurement of vectors ac and bc the velocities VCD and VCB can be found. The
angular velocity of link BC can be found as
ωCB = VCB/BC = rad/sec
Similarly,
ωCD = VCD/CD = rad/sec
Acceleration diagram
The angular acceleration of link AB is not given, so it may be assumed that = 0 or
tangential component of acceleration of B with respect to A, will be zero. Radial
acceleration = ω2. BA is known in direction (parallel to BA) also. Diagram 2.23(c).
1. Draw vector a'b' = = ω2. AB parallel to BA. a' and d' are fixed points and
=
0.
2. From b' draw vector b'x = =
. BC = /BC. VCB is taken from velocity
diagram, b'x is parallel to CB. From x draw vector xc' perpendicular to b'x. Its
magnitude is unknown. It contains point c'.
3. From point a' draw vector a'y parallel to CD as it represents radial acceleration
of point C with respect to D i.e. . The magnitude of a'y =
= /CD and
tangential component of acceleration is perpendicular to a'y.
Draw vector yc' representing , from point y.
and intersect at point c'.
4. Join c' to y and b' to c'. Also join a' to c'.
Here from diagram the values of various components of acceleration can be measured.
Angular accelerations of links BC and CD can be determined as:
Four-bar Chain Mechanism (with off-set)
Consider a four-bar chain ABCD with off-set BFC as shown in Diagram 2.24(a). AB is
the crank and rotates with ω rad/sec in the clockwise sense and angular acceleration
rad/sec in the clockwise sense. It is required to draw the velocity and acceleration
diagrams.
Velocity Diagram:
To draw the velocity diagram same as pervious here we discuss only the construction of
velocity due to off-set BFC.
Intermediate point: The velocity of an intermediate point on any of the links can be
found easily by dividing the corresponding velocity vector in the same ratio as the point
divides the link. For point E on the link BC,
be/bc = BE/BC
ae represents the absolute velocity of E.
Offset Point: Write the vector equation for point F,
The vectors VBA and VCD are already there on the velocity diagram.
VFB is ⊥r to BF, draw a line ⊥r to BF through b;
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28 Problems of Practices on Kinematic Analysis of Plane Mechanism
VFC is ⊥r to CF, draw a line ⊥r to CF through c;
The intersection of the two lines locates the point f.
af or df indicates the velocity of F with respect to A (or D) or the absolute velocity of F.
Diagram 2.24
Acceleration diagram
S.No. Vector Magnitude Direction Sense
1 AB → A
2 ⊥ AB or ab → b
3 BC → B
4 ⊥ BC
5 DC → D
6 ⊥ DC
Construct the acceleration diagram as follows:
1. Select the pole point a' or d'.
2. Take the 1st vector from the above table, i.e. take a'x to a convenient scale in the
proper direction and sense.
3. Add the second vector to the first and then the third vector to the second.
4. For the addition of the fourth vector, draw a line perpendicular to BC through
the head y of the third vector. The magnitude of the fourth vector is unknown
and c' can lie on either side of y.
5. Take the fifth vector from d'.
6. For the addition of the sixth vector to the fifth, draw a line perpendicular to DC
through head z of the fifth vector.
The intersection of this line with the line drawn in step (4) locates the point c'.
Total acceleration of B = a'b'
Total acceleration of C with respect to B = b'c'
Total acceleration of C = d'c'.
Intermediate Point: The acceleration of intermediate points on the links can be obtained
by dividing the acceleration vectors in the same ratio as the points divide the links. For
point E on the link BC (Diagram 2.24d),
(a) Four-bar mechanism
E
F
D
B
A
C
(b) Velocity diagram
⊥
⊥
⊥
(c) Acceleration diagram
(d) Acceleration diagram of off-set
⊥
⊥
⊥
⊥
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BE/BC = b' e'/b' c'
a'e' gives the total acceleration of point E.
Offset Points: The acceleration of an offset point on a link, such as F on BC (Diagram
2.24d), can be determined by applying any of the following methods:
BA already exists on the acceleration diagram. Now,
to BF, direction towards B
⊥
Shaper (Crank and Slotted Lever) Mechanism
A shaper mechanism is shown in Diagram 2.25(a). Crank O2A rotates in anticlockwise
direction with angular velocity ω2. It is required to draw the velocity diagram of the
configuration. The velocity of point A which is located at one end of the crank O2A can be
written as VAO2 = ω2.O2A. It is perpendicular to O2A. The point B is located on link 4
(O1C) such that it coincides with point A. The velocity of point B (VBO1) is perpendicular
to O1B. The points O1 and O2 are fixed, so they may be treated as one point in the
velocity diagram, as shown in [Diagram 2.25(b)].
Diagram 2.25
Velocity diagram:
1. First of all, take point o2. From o2 draw vector o2a to some suitable scale such
that VA = VAO2 = ω2.O2A. It is perpendicular to O2A.
2. The velocity of point B with respect to point A is along line O1B. Thus, VBA is
parallel to O1B. From a draw a vector ab representing VBA.
3. The velocity of point B with respect to O1 i.e. VBO1 is perpendicular to O1B. From
o1 draw a vector o1b (i.e. VBO1) to intersect ab at point b. Extend o1b to o1c such
that
Now, point c is located.
4. The velocity of point D with respect to C i.e. VDC is perpendicular to DC. From c
draw a vector cd perpendicular to DC, it represents VDC.
(a) Shaper mechanism
Point A on crank
O2A and B on link
O1C
2
6 D
5
C
3
B
A
4
(c) Acceleration diagram
⊥
⊥
⊥
(b) Velocity diagram
⊥
⊥
⊥
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30 Problems of Practices on Kinematic Analysis of Plane Mechanism
5. The velocity of point D with respect to point O1 i.e. VDO1 is along the line of stroke
of the slider link 6. So from o1 draw a vector o1d to intersect cd at point d. Vector
o1d represents the velocity of point D on the slider. It can be written as
Angular Velocity of O1C (link 4) can be determined as
Acceleration diagram
S.No. Vector Magnitude Direction Sense
1 O2A → O2
2
O1B → O1
3 ⊥ O1B
4 O1C
5 ⊥ O1C
where ω1 is the angular velocity of O1C.
2.9 GOODMAN'S INDIRECT METHOD
Goodman's indirect approach to the acceleration analysis of a complex mechanism is
based on the following two properties of a constrained mechanism:
(i) The angular velocities and accelerations of the links are linear functions of the
respective input quantities.
(ii) The relative angular velocities and accelerations between different links of a
linkage remain unaffected by a kinematic inversion.
Basic Relations
Let i denote the input link and l denote any other link. The angular velocity of l at any
instant can be expressed in terms of that of i. Thus,
Cl is a geometrical property depending only on the configuration of the mechanism
(except at dead-centre locations, when two links are collinear) and is independent of
velocities and accelerations. Equation (2.25) is the mathematical statement of the
obvious fact that the velocity polygons of a mechanism at any instant with different
input velocities are scale drawings of (i.e., they are similar to) one another. The angular
acceleration of the link l is
is also a geometrical property. The first term on the right-hand side of (2.26)
represents the acceleration of the link l, the acceleration of the input link being zero and
the actual velocity being ωi. Thus, (2,26) can be written as
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31 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
where 0l denotes the angular acceleration of the link l obtained from an auxiliary
acceleration diagram drawn with true velocities but with zero input acceleration.
Diagram 2.26
For a sliding input link, we have
where Vi and ai denote the velocity and acceleration of the input link.
Diagram 2.26a shows two consecutive links of some mechanism moving with angular
velocities and accelerations as indicated. The corresponding velocity diagram is shown
in Diagram 2.26b. The auxiliary acceleration diagram with i = 0 is shown in Diagram
2.26c and the true acceleration diagram with αi is shown in Diagram 2.26d. Note that
the normal components of accelerations (⊥r to the corresponding velocities) remain the
same in Diagrams 2.26c and 2.26d.
Let us suppose the motion of the point C. Without losing any generality, for simplicity of
analysis, the point O is taken as fixed. ωr and ωs are the angular velocities and r and s
are the angular accelerations of the two links. Since the normal components remain the
same in Diagrams 2.26c and 2.26d, the difference between and
is entirely due to
the difference between the tangential components ( l to the corresponding velocities). So,
Using Eq. (2.27), then, we have
or
Similarly, for a sliding input link, we have
⊥
⊥
⊥
⊥
⊥
⊥
O
C
B
⊥
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32 Problems of Practices on Kinematic Analysis of Plane Mechanism
In sliders, the total acceleration is in the direction of the velocity (i.e., the tangential
component is the total acceleration). So, for a rotating input link, we can write
and for the sliding input link, we have
The absolute velocities and accelerations mean those with respect to the fixed link f (i.e.,
the frame) Thus, (2.27) and (2.28) can be written as
Applications
The form of equations (2.27a) and (2.28a) is applicable to any inversion of the
mechanism where the use of f is no longer restricted to the frame. The subscript i
denotes any alternative input link (not necessarily the actual input link) with assumed
zero acceleration, on the basis of which the auxiliary acceleration diagram should be
drawn.
The indirect approach can be applied to a mechanism with a low degree of complexity in
the following manner:
(i) Choose an alternative input link to transform the mechanism to a simple one.
The auxiliary analysis is carried out with zero acceleration of this alternative
input link.
(ii) Find the actual values by using (2.27) to (2.32).
For mechanisms with a high degree of complexity, a direct kinematic inversion is made
to transform the mechanism to a simple one. The auxiliary velocities and accelerations
are obtained first. Thereafter, using (2.27a) and (2.28a), the actual values can be
determined.
2.10 VELOCITY AND ACCELERATION ANALYSIS (ANALYTICAL)
The analytical method of velocity and acceleration analysis starts from the loop-closure
equations which were discussed during displacement analysis. These equations are valid
at all times, and therefore successive differentiations of these equations with respect to
time establish the relationships between the velocity and acceleration quantities of
various links of a mechanism. The most important point to note is that, once the
configuration of the mechanism is known (i.e., the displacement analysis is complete),
the velocity and acceleration equations are linear in the unknown quantities and,
therefore, are very easy to solve. Consequently, when the velocity and acceleration
analysis has to be carried out for a large number of configurations, the analytical
method turns out to be more advantageous than the graphical method.
We shall now derive in detail the angular velocity and acceleration of the coupler and
the follower of a 4R linkage when the configuration and the crank motions are
prescribed. Referring back to Diagram 2.2, assume that the configuration of the
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33 Problems of Practices on Theory of Mechanism and Machines By Brij Bhooshan
mechanism has already been determined, i.e., l1, l2, l3, l4 and θ2 are prescribed and θ3
and θ4 have been solved. Now we determine the angular velocity and acceleration of the
coupler and the follower if those of the crank are given.
Towards this end, differentiate (2.2a) and (2.2b) with respect to time and obtain
We should note that (2.33a) and (2.33b) are two simultaneous linear equations in the
two unknowns, viz., and , which can be easily solved to yield
Differentiating (2.33a) and (2.33b) once more with respect to time, we get
Once the velocity analysis is complete, (2.34a) and (2.34b) again provide two linear
equations in which are obtained as
The same methodology can be extended to a mechanism having prismatic pairs. The
only difference will be that all vectors appearing in the loop-closure equation will not be
constant.
2.11 KLEIN'S CONSTRUCTION
A graphical method to find the velocity and acceleration of mechanism was given by
Professor Klein.
Single Slider Crank-Mechanism
A slider crank-mechanism OCP is shown in Diagram 2.27 for which velocity and
acceleration diagrams are shown in diagram.
Diagram 2.27
O
L
K
X
Q
M C
N P
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34 Problems of Practices on Kinematic Analysis of Plane Mechanism
Klein's construction is shown in Diagram 2.27. The construction of Diagram 2.27 is as
follows:
1. The line PC to cut a line through O perpendicular to the line of action at M.
2. Draw a circle with diameter CP at centre X.
3. Draw again circle with radius CM at centre C.
4. KL, the chord common to these two circles, with cut CP at Q and QP at N.
5. Quadrilateral OCQN is acceleration diagram.
6. OCN is velocity diagram.
Four Bar Mechanism
A four-bar mechanism as shown in Diagram 2.28.
Diagram 2.28
Klein's construction for velocity and acceleration diagram is made as under:
1. Draw a circle with centre B with radius BM.
2. Draw a second circle on the link BC as diameter.
3. The chord common to these two circles intersect BC at X and is produce to Q.
4. Two further circle draw, one with centre C and radius AM, and other on CD as
diameter T.
5. The cord of common to these two circles cut CD at S along AM mark of AT equal to
CS.
6. Through T draw TQ at right angles to M.
With this construction
Q
X B
D
S
C
M
T
A b
t
c, m, q s
o
θ