Kinematics of Machines
TE2030E
Class code: TE2030E - 157243
Instructor: Minh Bau Luong, Ph.D
bau.luongminh@hust.edu.vn
Chapter 7
Accelerations in Mechanisms
7.1 Introduction
7.2 Linear Acceleration
7.3 Angular Acceleration
7.4 Equivalent Linkages
7.5 Coriolis Acceleration
7.6 Euler-Savary Equation
Chapter 7. ACCELERATIONS IN MECHANISMS
7.1 Introduction
Acceleration analysis is important due to its effect on inertia forces, F = m.A,
influencing stresses in the parts of a machine, bearing loads, vibration, and noise .
VB/O1
Recall:
AB/O1
B
AnB/O1
AtB/O1
If ω = const, 𝛼 = 0 or At = 0
Coriolis acceleration, AC = 2Vω
Coriolis motion refers to simultaneous
translational and rotational motion of an object.
O1
Acceleration components
Relative acceleration
Acceleration components
Relative acceleration – General case
Coriolis Acceleration
Simultaneous translational
and rotational motion
Coriolis Acceleration
When a sliding joint is present on a rotating link, an additional component of
acceleration will be present, called the Coriolis component
As the linkage moves, the slider is both slipping and rotating ➞ slider has a two-degree-offreedom system
VP_Slip is the relative linear slip velocity
The Coriolis component of acceleration with a positive (CCW) 𝛼 and a negative (CW) ω
Acceleration analysis for a four-bar link
Aʹ
Acceleration analysis for a four-bar link
Position vector loop for a four-bar linkage showing acceleration vectors
Acceleration components
Position vector loop for a four-bar crank-slider linkage showing acceleration vectors
7.2 Linear Acceleration - A slider-crank mechanism
Find the acceleration of point C
First, VC is found by:
and
Then, from
= constant
AC is found by:
From O2”:
AtC: direction of the path of motion of point C
AnB: parallel to O2B
Double prime is used for points on the acceleration polygon.
AnC/B: parallel to BC, direction from C to B
AC/B
From the terminus of AnC/B, drawing a line perpendicular to
BC. Its point of intersection with the horizontal line from
O2” determines the magnitudes of AtC/B and AtC.
B”C” is acceleration image of the link BC.
For any mechanism there is an image in the acceleration polygon for each link,
which is similar to image for each link in the velocity polygon.
Considering point D on the link 3 of the previous mechanism.
If AD is determined by the same way as
determination of AC,
then we obtain the triangle B”C”D” that
is similar to the triangle BCD.
That means:
Triangle B”C”D” is acceleration image of the link BCD.
Note: If B, C, and D are in a clockwise order on the
link, then B", C", and D" must be in a clockwise order.
AC/B is represented by B”C”:
𝐴𝐶/𝐵 = 𝐴𝑛𝐶Τ𝐵 +→ 𝐴𝑡𝐶Τ𝐵
𝐴𝐶/𝐵 =
(𝐴𝑛𝐶Τ𝐵 )2 + (𝐴𝑡𝐶Τ𝐵 )2 =
(𝐵𝐶 𝜔2 )2 + (𝐵𝐶 𝛼)2
= 𝐵𝐶 𝜔4 + 𝛼 2
7.3 Angular Acceleration
From
𝐴𝑡
Angular acceleration can be calculated: 𝛼 =
𝑅
Example in the previous slider- crank
mechanism:
= constant
𝐴𝑡 C/B is directed upward, then
point C is accelerating upward in a
direction tangent to its path of
motion relative to B,
and hence 3 is counterclockwise.
AC/B
Example: The angular velocity and acceleration of link 2 are given. The
acceleration of points C, D, and E are to be determined along with
the angular accelerations of links 3 and 4 .
Trial-and-error method
Example: Suppose VE and AE are known and AB is desired.
First, we find AC as follows:
𝐴𝐶 =
𝐴𝐵 +→ 𝐴𝐶Τ𝐵
𝐴𝐶 = 𝐴𝐷 +→ 𝐴𝐶Τ𝐷
Since there are more than two
unknowns in the vector equation,
this equation we cannot be
solved alone. However, the
acceleration image for link 3 can
be found by trial and error
method.
Homework: exercise 7.1, 7.4, 7.9 in the textbook
7.4 Equivalent Linkages
For a direct-contact mechanism, the problem can be simplified by replacing the
mechanism by an equivalent four-bar linkage.
An equivalent four-bar linkage has links whose angular velocities and accelerations
are identical at the instant to the corresponding links of the original mechanism.
N-N: common normal at the contact point.
C2 and C4: centers of curvature of bodies 2
and 4, respectively, for their points of
contact P2 and P4.
If the pin connection for links 2' and 3' is
located at point C2 and the pin connection
for links 3' and 4’ is located at C4, then 1,
2', 3', and 4' constitute an equivalent fourbar linkage. That means:
If
2’ = 2
2’ = 2
then
4’ = 4
4’ = 4
This is because C2 and C4 belong both the original linkage and the equivalent four-bar linkage,
C2C4 can be seen as a rigid link. Therefore, the propertiers of motion of C2 and C4 as belonging
to original linkage is similar to that as belonging to quivalent four-bar linkage at the instant.
Any direct-contact mechanism has an infinite number of equivalent four-bar linkages.
Example: O2C2C4O4 is an equivalent four-bar linkage.
Another equivalent four-bar linkage can be determined as follows:
• Point E is any point on link 4 or an extension of it.
• Point D is the center of curvature of this path of E on 2.
• If link 4 is a point follower (shown shaded) and the path is used for the outline
of body 2, then the new body 2 and the point follower have angular velocities
and accelerations identical to those of the original mechanism.
• And the equivalent mechanism is now O 2DEO4.
Some other equivalent four-bar linkages
Some other equivalent four-bar linkages
Any link having an infinite length can be represented by a slider having rectilinear motion
Some other equivalent four-bar linkages
Homework: exercise 7.6 in the textbook
7.5 Accelerations for members in rolling contact
Considering disk 2 which rolls on stationary body 1.
Suppose 2 and 2 are known.
To be found: acceleration of P2, a fixed point on body 2
P2 is an instant center, VP2 = 0.
Where:
𝑑𝑉𝐶𝑡 𝑅2 𝑑ω
𝑡
𝐴𝐶 =
=
𝑑𝑡
𝑑𝑡
→
From the pole O”, drawing the vectors on
the right hand of the vector equations.
Then, the vector AP2 is obtained which is
directed along the normal at the point of
contact.
- In case the surface of body 1
is concave.
Similar vector equation can be
observed, except that:
- In case the surface of body 1 is flat, then R1 will be infinite:
- In case the outlines of either or both bodies are not circular, but have
changing curvature along their length, then Rl and R2 are taken as the radii
of curvature at the point of contact.
7.6 Coriolis acceleration
A point in one body moves along a path on a second body, if the second body is
rotating, then the acceleration of the point in the first body relative to a coincident point
in the second body will have a Coriolis component.
Let’s consider the figure on the left:
P3: on slider 3, moves along OF, OF in body 2.
P2: fixed point on OF, coincident with P3 at
the instant.
In a time interval dt ( d): P3 → P3’
Considering P3 → P3’ as the sum of motion
P2 → P2’, P2’ → B, B → P3’.
but
then
On the other hand:
Because 2 is constant and OP3 is increasing at a constant rate, the velocity
of P3 perpendicular to line OF is increasing at a constant rate. Thus the
acceleration of P3 perpendicular to OF is constant.
This Eq can be withdrawn
or
A is called the Coriolis component of acceleration for point P 3.
The Coriolis acceleration has the direction of VP3/P2 after VP3/P2 is rotated 90° in
the direction of the angular velocity of the path.
Example of Coriolis component of acceleration:
P2 is fixed point in 2
P3 is in 3 which is moving relative to 2.
Let’s consider the figure on the left:
P3: on slider 3, moves along the path of body 2
P2: fixed point on body 2, coincident with P3
at the instant:
the instant.
or
in which:
2VP3/P2.2: Coriolis component, part of the
acceleration of P3 relative to P2.
AtP3/P2 is either known at the beginning or can be found by drawing acceleration polygon.
Example of Coriolis component of acceleration:
A quick-return mechanism is shown, link 2 is the driver, and the angular velocity and
acceleration of link 4 are to be found.
Let P2 and P4 be fixed points on links 2 and 4 which are coincident at the instant.
We can find: VP2 = 0.151 m/s and (from the velocity polygon) 4 = 0.144 rad/s ccw
Acceleration of P4:
or for more convenient:
where:
can be found by drawing acceleration polygon
Acceleration polygon
Example of Coriolis component of acceleration:
A cam with oscillating follower is shown below in which the angular velocity and
acceleration of the cam are indicated, the angular acceleration of link 4 needs to be
found.
Let P2 and P4 be points on links 2 and 4
which are coincident at the instant.
P4 is the axis of the roller, and the path
which it describes on 2 is shown.
From the velocity polygon:
Acceleration of P4:
Then by the same way as previous
example, angular acceleration of link 4 is
determined:
Acceleration polygon
Homework: exercise 7.11, 7.12 in the textbook
7.7 Hartmann’s construction and Euler- Savary equation
From these examples above, we note that to find the angular acceleration of driven
link by acceleration polygon it is normally to know the path of its point that
describes on the driver link or the radius of this path.
For example:
One component needs to calculated:
Radius of curvature for the path
of P3 describing on body 2.
Center of curvature for the path of P3 describing on body 2.
Hartmann's construction is a graphical method for finding the center of curvature for
the path of any point of one body describing on another .
Euler- Savary equation is used to calculate radius of that curvature.
Hartmann’s construction
The path which a point in one body describes on another depends only on the relative
motion of the bodies. Hence when using the Hartmann’s construction, we can hold one
of the bodies fixed and consider the motion of the other body relative to it.
t and n axes: common tangent and common normal of
the centrodes. (Direction of line connecting centers of
2 curvatures that contact at contact point is common
n
P3
normal direction-đường thẳng đi qua tâm tức thời 23
và nối tâm C2C3. Line perpendicular to common
VP3
normal direction is common tangent)
P3: an arbitrary point on body 3 (body 3 moves in
V23
23
t relation to body 2)
23: instant center of 2 bodies
VP3: velocity of point P3 of the moving plane
V23: transfer velocity of the instant center (not the
V’23
same as velocity of the instant center).
As body 3 moves on 2, a new instant center 23 exists
C
for each point of contact. The velocity with which 23
propagates along the outline of body 2 is V23.
V’23: component of V23 parallel to VP3.
Then, the center C of path curvature for the path of P3 is the intersection point of the
line through the tips of vectors VP3 and V’23 and P3-23.
Example: Body 3 rolls on body 2, assume that body 2 is fixed. C2 and C3 are centers of
cicular oulines of the bodies. P3 is any point on 3.
The center C of the curvature of the path of P can be determined as follows:
n
t
t
n
Determining instant center 23
Drawing VC3 at any convenient scale,
VC3 = [23-C3]. = e.
Drawing VP3, VP3 = [23-P3]. =f. (or
determining VP3 by rotation of radius
method as we have known VC3 or via
angle )
V23 goes through 23 and parallel to VC3.
Its magnitude is determined by drawing
a line from C2 to the tip of VC3 (coi C3 và
23 quay quanh tâm C2).
V’23 is determined by projecting V23 on
the direction of VP3.
A line from the terminus of VP3 and V’23
intersects line P3 -23 at C.
C is the center of curvature for the path
which P3 describes on body 2, then CP3
is the radius of curvature of the path.
Example: A mechanism is shown below, need to find the center C of curvature of the
path which P3 describes on body 2.
C2 and C3 are the centers of curvature for the
outlines of bodies 2 and 3.
C can be determines in the same manner as
explained in the previous slide.
Note: in this case, 2 is not given then VC3 is
assumed in either direction at any convenient
length; and angle  = 900 then V’23  V23.
Example: A mechanism is shown below, need to find the center C of curvature of the
path which P4 describes on body 2.
C can be determines in the same manner as
explained in the previous slide.
Note: C4 is the center of curvature for link 4
and lies at infinity.
We hold link 2 fixed and consider the motion
of link 4 relative to 2.
Line from 24 to the tip of VC4 can be drawn
at any convenient angle to line 24-C4.
Euler- Savary equation
This equation gives the location for the center C of curvature of the path of the point
considered in the previous examples.
Let C2 - 23 = d; 23 - C3= e; 23-P3= f; C -23 = g,  is the angle which P3-23 makes
with T-T, then
and
Each of the lengths is to be considered positive if it extends in the same direction
that we would move if we went from body 2 to body 3 at the point of contact.
Homework: exercise 7.8 in the textbook