Mechanism Synthesis & Analysis unit 2…..

ME 262 –Unit 2
Mechanism Synthesis and Analysis
Dr. F. W. ADAM
MECHANICAL ENGINEERING DEPARTMENT
KNUST

Mechanism Synthesis and Analysis
Mechanism: is a mechanical device that
has the purpose of transferring motion
from a source to an output.
• Analysis – determination of position,
velocity, acceleration, etc. for a given
mechanism
• Synthesis – design of mechanism to do a
specific task.

Logical synthesis of mechanisms
• Type, number and dimensional synthesis
 Type synthesis: Kind of mechanism
selected
 Number synthesis: Number of joints or
links
 Dimensional synthesis: The determination
of the proportions (lengths) of the links
necessary to accomplish the desired
motions

Terminology and Definitions
• Kinematics: the study of the motion of bodies without reference to
mass or force
• Links: one of the rigid bodies or members joined together to form a
kinematic chain considered as rigid bodies
• Kinematic pair/ mechanical joint : a connection between two bodies
that imposes constraints on their relative movement.
• Ground/frame: a fixed or stationary link in a mechanism.
• Lower pair: the two lower pair have surface contact with one on
another.
• Higher pair: the two lower pair have point or line contact with one
another.
• Degree of freedom (DOF)/Mobility: of a mechanical system is the
number of independent parameters that define its configuration.
• Frame: the fixed link in a mechanism.

• Closed-Loop Kinematic Chains: a kinematic
chain is an assembly of links and joints. Each
link in a closed-loop kinematic chain is
connected to two or more other links.
• Open-Loop Kinematic Chains: at least one link is
connected to only on other link.

• Manipulators: manipulators designed to
simulate human arm and hand motion are an
example of open-loop kinematic chains.
• Robots: programmable manipulators

• Linkage/mechanism/machine/engine: an
assemblage of rigid bodies connected by
kinematic pairs.
• Planar motion: motion of all linkages is 2D
(projected onto a common plane).
• Spacial motion: at least the motion of one link
is 3D.

Inversion: if two
otherwise identical
linkages have different
fixed links, then each
is an inversion of the
other.

Link Types
A link is a rigid connection between two or more elements
of different kinematic pairs.

Mechanisms for specific applications
• Drafting Instrument
• Pantograph Linkages
• Slider-Crank
Mechanism
• Rotating
Combustion(Wankel)
Engine

Fluid Links
Gear trains
Lamination-Type Impulse Drive
Swash Plate

Power Screws
Differential Screws
Ball Screws

Universal Joints Special-Use Clutches
Sprag-Type Reverse-Locking Clutches
One-Way Clutches
Universal Joints
Automotive Steering Linkage
Computer Controlled Industrial Robots

Classification of mechanisms
No Description
1 Screw
2 Roller
3 Wheel
4 Cam
5 Linkage
6 Ratchet

Kinematic Pair(Joints)
• Lower pair(six types)
• Higher pair(cam and its follower have line or
point contact between the elements)
• Wrapping(chains and belts)

Degree of freedom
For example the bead shown in A has only
one coordinate needed to specify its
location.
The bar shown in B needs two
coordinates(complete, independent and
with admissible variation) to locate the
end point. (x, y) or (r, θ)
A
B
C
The bar shown in D needs three
coordinates(complete, independent and
with admissible variation) to locate the
end point. (x, y, θ)

Degree of freedom(dof)/Mobility
• Any unconstrained rigid body has six dof
• A body moving in a plane has three dof
• A cluster of n bodies not connected have dof=3n
• A fixed link(frame) has no degree of freedom
• Each lower pair reduces total dof by 2
• Each higher pair reduces total dof by 1
• Grubbler’s Criterion: the total dof of a planar
mechanism is
dof=3(n-1)-2j-h
Where n=number of links
j=number of lower pairs
h=number of higher pairs

Classification of Closed Planar Four-Bar
Linkages-The Grashof Criterion
• Closed planar linkages of four pin-connected
rigid links are usually identified as four-bar
linkages.
• If one of the links can perform a full rotation
relative to the other three links, the linkage is
called Grashof mechanism

Four-Bar Linkages
Grashof condition
A simple relationship which
predicts the behavior of a
four-link mechanism inversion
based only on link lengths.
S=length of shortest link
L=length of longest link
P=length of one link
Q=length of other link
If S+L≤P+Q then at least one
link will crank, and the
mechanism is called a Grashof
mechanism.

Criteria of motion for each class of
Four-Bar Linkages
Type of Mechanism Shortest link Relationship between link
lengths
Grashof Any S+L≤P+Q
crank rocker Driver crank S+L< P+Q
drag link Fixed link S+L< P+Q
double rocker coupler S+L< P+Q
crossover
position(change point)
S+L= P+Q
Non-Grashof S+L< P+Q
double rocker of the
second kind(triple rocker)
S+L>P+Q

Straight line mechanisms
watt’s linkage as used in his steam engine

Slider-crank mechanism
2
2
2
2
sin
1
cos
sin
1
cos
1
sin
cos
cos
cos
sin
sin



































l
r
l
r
x
l
r
l
r
x
l
r
h

EXAMPLE OF MECHANISM
Ken Youssefi MECHANICAL ENGINEERING DEPT. 30
Can crusher
Simple press
Rear-window wiper

EXAMPLE OF MECHANISMS
31
Moves packages from an assembly
bench to a conveyor
Lift platform
Microwave carrier to assist
people on wheelchair

32
Lift platform
Front loader
Device to close the
top flap of boxes

33
Stair climbing mechanism
A box that
turns itself off
Airplane landing
gear mechanism

34
Conceptual design for an
exercise machine
Rowing type exercise machine

36
Six-bar linkage prosthetic
knee mechanism
Extension position
Flexed position

Dimensional Synthesis
• The determination of the proportions (length)
of the links necessary to accomplish the
desired motions.
• Function generation, path generation and
Motion Generation (body guidance).

Function generation, path generation
and Motion Generation (body
guidance).
• Function Generation: the correlation of an input
motion with an output motion in a mechanism
• Path Generation: Control of a point in the plane
such that it follows some prescribed path
• Motion Generation: Control of a line in the plane
such that is follows some prescribed set of
sequential positions

39
Mechanism Categories
Function Generation Mechanisms
A function generator is a linkage in which the relative
motion between links connected to the ground is of interest.
A four-bar hand actuated wheelchair
brake mechanism

40
A four-bar drive linkage for a lawn sprinkler

41
A four-bar function generation mechanism to
operate an artificial hand used for gripping.
A four-bar function generation
mechanism to lower an attic stairway.

42
Motion Generation Mechanisms
In motion generation, the entire motion of the coupler
link is of interest (rigid body guidance).
New Rollerblade brake system

43
Four-bar automobile hood linkage design

44
Rotating a monitor into a storage position
Moving a storage bin from an
accessible position to a stored position

45
Lifting a boat out of water
Moving a trash pan from the floor up over
a trash bin and into a dump position

46
Path Generation Mechanisms
In path generation, we are concerned only with the path of a tracer
point and not with the motion (rotation) of the coupler link.
Crane – straight line motion

47
Path Generation Mechanisms
A four-bar path generation mechanism as
part of an arm-actuated propulsion system
for a wheelchair
A four-bar path generation mechanism to guide a
thread in an automatic sewing machine

Slider-Crank Mechanism
• The slider-crank mechanism below has stroke
𝐵1𝐵2 = 2𝑟2
• The extreme positions 𝐵1 𝑎𝑛𝑑 𝐵2 are found
by constructing circle arcs through 𝑂2 of
length 𝑟3 − 𝑟2 and 𝑟3 + 𝑟2 respectively.
Centered slider-crank mechanism General/ offset slider-crank mechanism

Rocker Output- Two Position with
Angular Displacement
Example 1
Design a four bar Grashof crank-rocker
mechanism to give 45° of rocker motion with
equal time forward and back, from a constant
speed motor input.

Example 1
• Draw the output link O4B in both extreme
positions, B1 and B2 in any convenient location,
such that the desired angle of motion θ4 is
subtended.
• Draw the chord B1B2 and extend it in either
direction.
• Select a convenient point O2 on the line B1B2
extended.
• Bisect line segment B1B2, draw a circle of that
radius about O2.
• Label the two intersection of the circle and B1B2
extended, A1 and A2.
• Measure the length of the coupler as A1 to B1 or
A2 to B2.
• Measure ground length 1, crank length 2, and
rocker length 4.
• Find the Grashof condition. If non-Grashof, redo
steps 3 to 8 with O2 further from O4.
• Make model of the linkage and check its
function and transmission angles.

Example 2
• Rocker Output- Two Position with Complex
Displacement (Motion)
– Design a fourbar linkage to move link CD from
C1D1 to C2D2.

Example 2
• Draw the link CD in its two desired
positions, C1D1 and C2D2 in plane as
shown.
• Draw construction line from point C1 to
C2 and from point D1 to D2.
• Bisect line C1C2 and line D1D2 and
extend their perpendicular bisectors to
intersect at O4. Their intersection is the
rotopole.
• Select a convenient radius and draw an
arc about the rotopole to intersect both
lines O4C1 and O4C2. Label the
intersection B1 and B2.
• Do steps 2 to 8 of example 1 to
complete the linkage.
• Make a model of the linkage and
articulate it to check its function and its
transmission angles.

Coupler Output- Two Position with
Complex Displacement (Motion)
Example 3: Design a fourbar linkage
to move link CD from C1D1 to C2D2
(with moving pivots at C and D).

Example 3
• Draw the link CD in its two desired positions, C1D1
and C2D2 in plane as shown.
• Draw construction line from point C1 to C2 and from
point D1 to D2.
• Bisect line C1C2 and line D1D2 and extend their
perpendicular bisectors in convenient directions. The
rotopole will not be used in this solution.
• Select any convenient point on each bisector as the
fixed pivots O2 and O4, respectively.
• Connect O2 with C1 and call it link 2. Connect O4 with
D1 and call it link 4.
• Line C1D1 is link 3. Line O2O4 is link 1.
• Check the Grashof condition, and repeat steps 4 to 7
if unsatisfied. Note that any Grashof condition is
potentially acceptable in this case.
• Construct a model and check its function to be sure
it can get from the initial to final position without
encountering ant limit (toggle) positions.
• Check the transmission angles.

Review Example 3
Design a dyad to control and limits the
extremes of motion of the linkages in the
previous example to its two design positions

Review Example 3
• Select a convenient point on link 2 of the
linkage designed in Example 3. Note that
it need not be on the line O2C1. Label
this point B1.
• Draw an arc about center O2 through B1
to intersect the corresponding line O2B2
in the second position of link 2. Label
this point B2. The chord B1B2 provides
us with the same problem in Example 1.
• Do steps 2-9 of Example 1 to complete
the linkage, except add links 5 and 6 and
center O6 rather than links 2 and 3 and
center O2 . Link 6 will be the driver
crank. The fourbar subchain of the links
O6, A1, B1, O2 must be a Grashof crank-
rocker.

Three-Position Synthesis
• Example 4 – Coupler Output – 3 position with
Complex Displacement
• Design a fourbar linkage to move the link CD
shown from position C1D1 to C2D2 and then to
position C3D3. Moving pivots are C and D. Find
the fixed pivot locations.

Example 4
• Draw link CD in its three position
• C1D1, C2D2 , C3D3 in the plane as shown.
• Draw construction lines from point C1 to C2 and
from C2 to C3.
• Bisect line C1C2 and line C2C3 and extend their
perpendicular bisector until they intersect. Label
their intersection O2.
• Repeat steps 2 and 3 for lines D1D2 and D2D3. Label
the intersection O4.
• Connect O2 with C1 and call link 2. Connect O4 with
D1 and call link 4.
• Line C1D1 is link 3. Line O2O4 is link 1.
• Check the Grashof condition. Note that any
Grashof condition is potentially acceptable in this
case.
• 8.Construct a model and check its function to be
sure it can get from initial to final position without
encountering any limits positions.

Three-Position Synthesis – Example 5
• Coupler Output – 3 position
with Complex Displacement
(Alternate Attachment
Points for Moving Pivots)
• Design a fourbar linkage to
move the link CD shown
from position C1D1 to C2D2
and then to position C3D3.
Use different moving pivot
than CD. Find the fixed
pivot locations.

Example 5
• Draw link CD in its three position C1D1, C2D2 , C3D3 in the
plane as shown.
• Define new attachment points E1 and F1 that have a fixed
relationship between C1D1 and E1F1 within the link. Now
use E1F1 to define the three position of the link.
• Draw construction lines from point E1 to E2 and from E2 to
E3.
• Bisect line E1E2 and line E2E3 and extend their
perpendicular bisector until they intersect. Label their
intersection O2.
• Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the
intersection O4.
• Connect O2 with E1 and call link 2. Connect O4 with F1 and
call link 4.
• Line E1F1 is link 3. Line O2O4 is link 1
• Check the Grashof condition. Note that any Grashof
condition is potentially acceptable in this case.
• Construct a model and check its function to be sure it can
get from initial to final position without encountering any
limits positions. If not, change locations of point E and F
and repeat steps 3 to 9.

Example 6
• Example 6 – Three –Position Synthesis with
Specified Fixed Pivots - Inverting the 3-position
problem
• Invert a fourbar linkage which move the link CD
shown from position C1D1 to C2D2 and then to
position C3D3. Use specified fixed pivots O2 and O4.

Example 6
• Draw link CD in its three position C1D1, C2D2 , C3D3 in the plane as
shown.
• Draw the ground link O2O4 in its desired position in the plane with
respect to the first coupler position C1D1
• Draw construction arc from point C2 to O2 and from D2 to O2
whose radii define the side of triangle C2O2D2 This define the
relationship of the fixed pivot O2 to the coupler line CD in the
second coupler position.
• Draw construction arc from point C2 to O4 and from D2 to O4
whose radii define the side of triangle C2O4D2 This define the
relationship of the fixed pivot O4 to the coupler line CD in the
second coupler position.
• Now transfer this relationship back to the first coupler position
C1D1 so that the ground plane position O2’O4’ bears the same
relationship to C1D1 as O2O4 bore to the second coupler position
C2D2. In effect, you are sliding C2 along the dotted line C2C1 and D2
along the dotted D2D1. By doing this, we have pretended that
ground plane moved from O2O4 to O2’O4’ instead of the coupler
moving from C1D1 to C2D2. We have inverted the problem.
• Repeat the process for the third coupler position as shown in the
figure and transfer the third relative ground link position to the
first, or reference, position.
• The three inverted position of the ground plane that correspond
to the three desired coupler positions are labeled O2O4,O2’O4’ ,
and O2’’O4’’ and have also been renamed
• E1F1, E2F2 and E3F3 as shown in the figure.

Example 7
• Example 7 – Finding the
Moving Pivots for Three
Positions and Specified
Fixed Pivots
• – Design a fourbar linkage
to move the link CD shown
from position C1D1 to C2D2
and then to position C3D3.
Use specified fixed pivots
O2 and O4. Find the
required moving pivot
location on the coupler by
inversion.

Example 7
• Start with inverted three positions plane as shown in the figures. Lines E1F1, E2F2 and E3F3 define
the three positions of the inverted link to be moved.
• Draw construction lines from point E1 to E2 and from point E2 to E3.
• Bisect line E1E2 and line E2E3 extend the perpendicular bisector until they intersect. Label the
intersection G.
• Repeat steps 2 and 3 for lines F1F2 and line F2F3. Label the intersection H.
• Connect G with E1 and call it link 2. Connect H with F1 and call it link 4.
• In this inverted linkage, line E1F1 is the coupler, link 3. Line GH is the “ground” link1.
• We must now reinvert the linkage to return to the original arrangement. Line E1F1 is really the
ground O2O4 and GH is really the coupler. The figure shows the reinversion of the linkage in which
points G and H are now the moving pivots on the coupler and E1F1 has resumed its real identity as
ground link O2O4.
• The figure reintroduces the original line C1D1 in its correct relationship to line O2O4 at the initial
position as shown in the original example 3. This form the required coupler plane and defines a
minimal shape of link 3.
•

Example 7
• The angular motions required to reach the second and third position of
line CD shown in the figure are the same as those defined in figure b for
the linkage inversion. The angle F1HF2 in the figure b is the same as the
angle H1O4H2 in the figure and F2HF3 is the same as angle H2O4H3. The
angular excursions of link 2 retain the same between figure b and e as
well. The angular motions of links 2 and 4 are the same for both inversion
as the link excursions are relative to one another.
• Check the Grashof condition. Note that any Grashof condition is
potentially acceptable in this case provided that the linkage has mobility
among all three position. This solution is a non-Grashof linkage.
• 11. Construct a model and check its function to be sure it can get from
initial to final position without encountering any limit (toggle) positions.
• In this case link 3 and 4 reach a toggle position between points H1 and H2.
This means that this linkage cannot be driven from link 2 as it will hang up
at that toggle position. It must driven from link 4.

Quick – Return Mechanism
• Fourbar Quick-Return
• Time ratio (TR) defines the degree of quick-return of the linkage.
𝑇𝑅 =
𝑓𝑜𝑤𝑎𝑟𝑑 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒/𝑎𝑛𝑔𝑙𝑒
𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑠𝑡𝑟𝑜𝑘𝑒 𝑡𝑖𝑚𝑒/𝑎𝑛𝑔𝑙𝑒
𝑇𝑅 =
𝛼
𝛽
𝛼 + 𝛽 = 360°
𝛿 = 180 − 𝛼 = 180 − 𝛽
• If 𝑇𝑅 >1 the mechanism is a quick return mechanism
• Works well for time ratios down to about 1.5

Synthesis of Crank-and-Rocker
Mechanism with unit time ratio
180
min
max 
 


Example 1
• Fourbar Crank-Rocker Quick-
Return Linkage for Specified
Time Ratio
• Redesign Example 1 (two
position) to provide a time
1:1.25 with 45° output rocker
motion.

Example 1
• Draw the output link O4B in both
extreme position, in any convenient
location, such that the desired angle
of motion, θ4 , is subtended.
• Calculate α, β, δ using the equations.
In this example α=160°, β=200°,
δ=20°.
• Draw a construction line through
point B1 at any convenient angle.
• Draw a construction line through
point B2 at an angle δ from the first
line.
• Label the intersection of the two
construction line O2.
• The line O2O4 define ground.

Example 1
• Calculate the length of crank and coupler by
measuring O2B1 and O2B2 and solve
simultaneously;
• coupler + crank=O2B1 ; coupler - crank=O2B2
or you can construct the crank length by
swinging an arc centered at O2 from B1 to cut
line O2B2 extended.
• Label that intersection B1’. The line B2B1’ is
twice the crank length. Bisect this line
segment to measure crank length O2A1.
• Calculate the Grashof condition.
• If non-Grashof, repeat steps 3 to 8 with O2
further O4.
• Make a model of the linkage and articulate it
to check its function.
• Check the transmission angle.

Example 2
• Sixbar Drag Link Quick-Return Linkage for
Specified Time Ratio.
• Provide a time ratio of 1:1.4 with 90 degree
rocker motion.

Example 2
• Calculate α, β, δ. In this example α=150°, β=210°.
• Draw a line of centers XX at any convenient location.
• Choose a crank pivot location O2 on line XX and draw
an axis YY perpendicular to XX.
• Draw a line of convenient radius O2A about center O2
• Lay out angle α with vertex at O2, symmetrical about
quadrant one
• Label points A1 and A2 at the intersection of the lines
subtending angle α and the circle of radius O2A.

Example 2
• Set the compass to a convenient radius AC long enough to cut XX in two places
on either side of O2 when swung from A1 and A2. Label the intersection C1 and
C2.
• The line O2A1 is the driver crank, link 2, and line A1C1 is the coupler, link 3.
• The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate
the fixed pivot O4.
• The line O2O4 now defines the ground link. Link O4C1 is the driven crank, link 4.
• Calculate the Grashof condition. If Non-Grashof, repeat steps 7-11 with a shorter
radius in step 7.
• invert the method of example 1 (two positions) to create the output dyad using
XX as the chord and O4C1 as the driving crank. The point B1 and B2 will lie on line
XX and be spaced apart a distance 2O4C1. The pivot O6 will lie on the
perpendicular bisector of B1B2, at a distance from line XX which subtends the
specified output rocker angle.
• Check the transmission angle.

Synthesis of Rocker-and-Rocker
Mechanism
Three-Position Synthesis
• Given ψ12, ψ23, ψ13, ∅12, ∅23, ∅13, 𝐴, 𝜃2, 𝑟1, 𝑟2 𝑎𝑛𝑑 (𝜃4, 𝑟4, 𝑟3? )
• Draw 𝑂2𝐴 in its three specified positions and locate 𝑂4
• Draw a ray from 𝑂4 𝑡𝑜 𝐴2 and rotate it through ∅12 to locate 𝐴2
′
• Draw a ray from 𝑂4 𝑡𝑜 𝐴3 and rotate it through ∅13 to locate 𝐴3
′
• Draw mid-normals to the lines 𝐴1
′
𝐴2
′
and 𝐴2
′
𝐴3
′
these intersect at 𝐵1

Coupler Curves
For the mechanisms considered, the
displacement of the links joined with the fixed
link was the input or output of the simple
mechanisms. In great number of applications
the output from a simple mechanism is the path
traced by one of the points on the coupler link.
These paths are generally called “coupler point
curves” or “coupler paths”.

Review of Vector Analysis
i. Cartesian/rectangular coordinate system
i
k
j

Example
The motion of a particle is described in
Cartesian coordinates as
𝑥 𝑡 = 2𝑡2
+ 4𝑡
𝑦 𝑡 = 0.1𝑡3
+ cos 𝑡 𝑧 𝑡 = 3𝑡
Find the velocity and acceleration of the
particle.

ii. Cylindrical coordinate system
𝒓 𝑡 = 𝑅𝒆𝒓 + 𝑧𝒌 = 𝑅𝑐𝑜𝑠 𝜃 𝒊 + 𝑅𝑠𝑖𝑛 𝜃𝒋 + 𝑧𝒌
𝒗 𝑡 = ሶ
𝒓 𝑡 = ሶ
𝑅𝒆𝒓 + 𝑅 ሶ
𝜃𝒆𝜽 + ሶ
𝑧𝒌 𝒂 𝑡 = ሷ
𝑅 − 𝑅 ሶ
𝜃2
𝒆𝒓 + 𝑅 ሷ
𝜃 + 2 ሶ
𝑅 ሶ
𝜃 𝒆𝜽 + ሷ
𝑧𝒌
𝒆𝒓 = 𝑐𝑜𝑠𝜃𝒊 + 𝑠𝑖𝑛𝜃𝒋 𝑎𝑛𝑑𝒆𝜽 = −𝑠𝑖𝑛 𝜃𝒊 + 𝑐𝑜𝑠𝜃 𝒋

Example
At the instant shown the length of
the boom AB is being decreased at
the constant rate of 0.2 m/s and the
boom is being lowered at the
constant rate of 0.08 rad/s. Using
cylindrical coordinate system,
Determine (a) the velocity of point B,
(b) the acceleration of point B.

Solution
𝑅 = 6𝑚; ሶ
𝜃 = −0.08 𝑟𝑎𝑑/𝑠;
ሶ
𝑅 = −0.2𝑚/𝑠;
ሶ
𝑧 = ሷ
𝑧 = 0; ሷ
𝑅 = 0; ሷ
𝜃 = 0
∴ 𝒗 𝑡 = (−0.2𝒆𝒓 − 0.48𝒆𝜽) m/s
𝑎𝑛𝑑 𝒂 𝑡 = −𝑅 ሶ
𝜃2
𝒆𝒓 + 2 ሶ
𝑅 ሶ
𝜃𝒆𝜽
𝒂 𝑡 = −38.4𝒆𝒓 + 32𝒆𝜽 𝑚𝑚/𝑠2

iii. Spherical coordinate system
𝒆𝑹 × 𝒆𝝓 = 𝒆𝜽
𝒓 𝑡 = 𝑅𝒆𝒓
𝑥 = 𝑅𝑠𝑖𝑛𝜙𝑐𝑜𝑠 𝜃 𝑦 = 𝑅𝑠𝑖𝑛𝜙𝑠𝑖𝑛 𝜃 𝑧 = 𝑅𝑐𝑜𝑠𝜙
𝒆𝑹 = 𝑠𝑖𝑛𝜙𝑐𝑜𝑠 𝜃 𝒊 + 𝑠𝑖𝑛𝜙𝑠𝑖𝑛 𝜃𝒋 + 𝑐𝑜𝑠𝜙𝒌
𝒆𝜽 = −𝑠𝑖𝑛 𝜃 𝒊 + 𝑐𝑜𝑠 𝜃𝒋
𝒆𝝓 = 𝑐𝑜𝑠𝜙 𝑐𝑜𝑠 𝜃 𝒊 + 𝑠𝑖𝑛 𝜃𝒋 − 𝑠𝑖𝑛𝜙𝒌
𝑑𝒆𝑹
𝑑𝑡
= ሶ
𝜃𝑠𝑖𝑛𝜙𝒆𝜽 + ሶ
𝝓𝒆𝝓
𝑑𝒆𝜽
𝑑𝑡
= − ሶ
𝜃𝑠𝑖𝑛𝜙𝒆𝑹 − ሶ
𝜃𝒄𝒐𝒔𝜙𝒆𝝓
𝑑𝒆𝝓
𝑑𝑡
= ሶ
𝜃𝑐𝑜𝑠𝜙𝒆𝜽 − ሶ
𝜙𝒆𝑹
𝒗 𝑡 = ሶ
𝑅𝒆𝑹 + 𝑅 ሶ
𝜃𝑠𝑖𝑛𝜙𝒆𝜽 + 𝑅 ሶ
𝜙𝒆𝝓
𝒂 𝑡 = ( ሷ
𝑅 − 𝑅 ሶ
𝜃2𝑠𝑖𝑛2𝜙 − 𝑅 ሶ
𝜙2)𝒆𝑹+ 𝑅 ሷ
𝜃𝑠𝑖𝑛𝜙 + 2 ሶ
𝑅 ሶ
𝜃𝑠𝑖𝑛𝜙 + 2𝑅 ሶ
𝜃 ሶ
𝜙𝑐𝑜𝑠𝜙 𝒆𝜽 +
(𝑅 ሷ
𝜙 + 2 ሶ
𝑅 ሶ
𝜙 − 𝑅 ሶ
𝜃2𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜙)𝒆𝝓

Example
At a given instant, the satellite
dish has an angular motion 𝜔1
= 6 rad/s and ሶ
𝜔1 = 3 𝑟𝑎𝑑/𝑠2
about the z axis. At this same
instant 𝜃 = 25𝑜
, the angular
motion about the x axis is 𝜔2 =
2 rad/s, and ሶ
𝜔2 = 1.5 𝑟𝑎𝑑/𝑠2.
Using spherical coordinate
system, determine the velocity
and acceleration of the signal
horn A at this instant.

Solution
𝑅 = 1.4 𝑚; 𝜃 = 90𝑜
; 𝜙 = 65𝑜
ሶ
𝑅 = ሷ
𝑅 = 0;
ሶ
𝜙 = 𝜔2 = 2 𝑟𝑎𝑑/𝑠; ሷ
𝜙 = ሶ
𝜔2 = 1.5 𝑟𝑎𝑑/𝑠2
;
ሶ
𝜃 = 𝜔1 = 6 𝑟𝑎𝑑/𝑠; ሷ
𝜃 = ሶ
𝜔1 = 3 𝑟𝑎𝑑/𝑠2
∴ 𝒗 𝑡 = (2.8 𝒆𝝓 + 8.16 𝒆𝜽) m/s
𝑎𝑛𝑑 𝒂 𝑡 = −𝑅 ሶ
𝜃2
𝒆𝒓 + 2 ሶ
𝑅 ሶ
𝜃𝒆𝜽
𝒂 𝑡 = −47 𝒆𝒓 + 34.25 𝒆𝜽 + 21.4𝒆𝝓 𝑚/𝑠2

Transport Theorem
• Consider a vector A observed
from a moving coordinate
system xyz, which is rotating
with angular velocity Ω
90
Figure a
Figure b

General Motion
91
A
B
A
B r
r
r /


Position Vector
Velocity
     
A
B
A
B r
dt
d
r
dt
d
r
dt
d
/


  B/A
/ r
x



 xyz
A
B
A
B v
v
v
Acceleration
       
B/A
/ r
x




dt
d
v
dt
d
v
dt
d
v
dt
d
xyz
A
B
A
B
     
B/A
B/A
B/A
B/A r
x
x
v
x
2
r
x
a 







 xyz
xyz
A
B a
a 
    B/A
2
B/A
B/A
B/A r
v
x
2
r
x
a 






 xyz
xyz
A
B a
a 
For 2D motions, the above equation reduces to

Example 1
Crank AB of the engine system shown has a constant
clockwise angular velocity of 2000 rpm. For the crank
position shown, determine the angular velocity as well
as the angular acceleration of the connecting rod BD
and the linear velocity and acceleration of point D.

solution
Since the crank rotates about A with constant 𝜔𝐴𝐵 = 2000 𝑟𝑝𝑚 =
209.4 𝑟𝑎𝑑/𝑠, The velocity of B can be calculated from
𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴 + Ω𝐴𝐵 x 𝑟𝐵/𝐴
𝑣𝐵/𝐴 = 𝑣𝐴 = 0
𝑣𝐵 = Ω𝐴𝐵 x 𝑟𝐵/𝐴 = 209.4 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
𝑣𝐵 = − 4.04 𝒊 − 4.81 𝒋 𝑚𝑠−1
𝑣𝐷 = 𝑣𝐵 + 𝑣𝐷/𝐵 + Ω𝐷𝐵 x 𝑟𝐷/𝐵
𝑣𝐷/𝐵 = 0
𝑣𝐷𝒊 = −4.04 𝒊 + 4.81 𝒋 + Ω𝐷𝐵 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
Expanding and comparing coefficients
𝑣𝐷 = −4.04 − 0.019Ω𝐷𝐵
0 = 4.81 + 0.023 Ω𝐷𝐵
Ω𝐷𝐵 = −6.28 𝑟𝑎𝑑/𝑠; 𝑣𝐷 = −209.13 𝑚𝑠−1

Solution
Since 𝛼𝐴𝐵 = 0. The acceleration of B can be calculated from
𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴 + 2Ω𝐴𝐵 x 𝑣𝐵/𝐴 + 𝛼𝐴𝐵 x 𝑟𝐴𝐵 − Ω𝐴𝐵
2
𝑟𝐵/𝐴
𝑣𝐵/𝐴 = 𝑎𝐴 = 𝑎𝐵/𝐴 = 𝛼𝐴𝐵 = 0
𝑎𝐵 = −Ω𝐴𝐵
2
𝑟𝐵/𝐴 = −209.42 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
𝑎𝐵 = − 1007.63 𝒊 + 845.56 𝒋 𝑚𝑠−2
𝑎𝐷 = 𝑎𝐵 + 𝑎𝐷/𝐵 + 2Ω𝐷𝐵 x 𝑣𝐷/𝐵 + 𝛼𝐷𝐵 x 𝑟𝐷/𝐵 − Ω𝐷𝐵
2
𝑟𝐷/𝐵
𝑣𝐷/𝐵 = 𝑎𝐷/𝐵 = 0
𝑎𝐷
= − 1007.63 𝒊 + 845.56 𝒋 + 𝛼𝐷𝐵 𝒌 x 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
− 6.282 0.03 cos 40 𝒊 + 0.03 sin 40𝒋
Expanding and comparing coefficients
𝑎𝐷𝒊
= −1007.63 − 0.019𝛼𝐷𝐵 − 0.91 𝒊
+ −845.56 +0.0236𝛼𝐷𝐵 −0.75 𝒋
0 = −846.33 + 0.023 𝛼𝐷𝐵
𝑎 = −1037.84 − 0.019𝛼

Example 2
Determine the velocity and acceleration
of the collar C at this instant
Solution
𝑣𝐷 = 𝑣𝐵 + 𝑣𝐷/𝐵 + Ω𝐷𝐵 x 𝑟𝐷/𝐵
𝑣𝐷 = 3𝒋 + 5 𝒌 x 0.2 𝒋
= −𝒊 + 3 𝒋 𝑚/𝑠
𝑎𝐷
= 𝑎𝐵 + 𝑎𝐷/𝐵 + 2Ω𝐷𝐵 x 𝑣𝐷/𝐵
+ 𝛼𝐷𝐵 x 𝑟𝐷/𝐵 − Ω𝐷𝐵
2
𝑟𝐷/𝐵
𝑎𝐷
= 2𝒋 + 10 𝒌 x 3𝒋 − 6 𝒌 x 0.2𝒋
− 25(0.2 𝒋)
−2

Relative Velocity Method
The crank of a slider crank mechanism rotates clockwise at a
constant speed of 300 rpm. The crank is 150 mm and the
connecting rod is 600 mm long. Determine:
1. Linear velocity and acceleration of the midpoint of the connecting
rod, and
2. Angular velocity and angular acceleration of the connecting rod, at a
crank angle of 45° from inner dead centre position.

solution
Velocity analysis
𝑣𝐵 = 𝜔𝑟𝐵 = 0.15 × 10𝜋 = 4.713 𝑚/𝑠
From the velocity diagram
𝑣𝐴𝐵 = 3.4 𝑚/𝑠
𝑣𝐴 = 4 𝑚/𝑠
𝑣𝐷 = 4.1 𝑚/𝑠
Acceleration analysis
𝑎𝐵
𝑟 =
𝑣𝐵
2
𝑂𝐵
=
4.7132
0.15
= 148.1 𝑚/𝑠2
𝑎𝐴𝐵
𝑟 =
𝑣𝐴𝐵
2
𝐴𝐵
=
3.42
0.6
= 19.3 𝑚/𝑠2
From the acceleration diagram
𝛼𝐴𝐵 =
𝑎𝐴𝐵
𝑡
𝐴𝐵
=
1032
0.6
= 171.67 𝑟𝑎𝑑/𝑠2

Example 2
In the mechanism shown, the slide QD is driven with a uniform
angular velocity of 10 rad/s and the block C has a mass of 1.5 kg.
Draw the velocity and acceleration diagrams for the mechanism in
the position shown. OQ = 1 m; OC = 2 m.

solution
Velocity analysis
𝑣𝑐 = 28.8 𝑚/𝑠
From the velocity diagram
𝑣𝑐𝑐′ = 7.5 𝑚/𝑠
Acceleration analysis
𝑎𝑐
𝑟
=
𝑣𝑐
2
𝑂𝐶
= 414 𝑚/𝑠2
𝑎𝐶′𝑄
𝑟
= 288 𝑚/𝑠2
𝑎𝐶′𝐶
𝐶
= 2𝑣𝑐𝑐′𝜔𝐶′𝑄
𝑎𝐶′𝐶
𝐶
= 150 m/𝑠2
From the acceleration diagram
𝑎𝐶 = 417 𝑚/𝑠2

Assignment 1
Determine the angular velocity and
angular acceleration of the rod DE
at the instant shown. The collar at
C is pin connected to AB and slides
over rod DE.

Assignment 2
The rocker of a crank-rocker linkage is to have a length
of 500 mm and swing through a total angle of 45° with
a time ratio of 1.25. Determine a suitable set of
dimensions for 𝑟1, 𝑟2 𝑎𝑛𝑑 𝑟3.

Assignment 3
The figure shows two positions of a folding seat used in the
aisles of buses to accommodate extra seated passengers. Design
a four-bar linkage to support the seat so that it will lock in the
open position and fold to a stable closing position along the side
of the aisle.
3. Determine the link lengths of a slider-crank linkage to have
a stroke of 600 mm and a time ratio of 1.2.

Assignment 4
Design a fourbar linkage
to move the object below
from position 1 to 2 using
points A and B for
attachment. Add a driver
dyad to limit its motion to
the range of positions
shown making it a sixbar.
All fixed pivots should be
on the base.

Assignment 5
Owusu House of Flapjacks wants to automate
their flapjack production. They need a
mechanism which will automatically flip the
flapjacks “on the fly” as they travel through the
griddle on a continuously moving conveyor. This
mechanism must track the constant velocity of
the conveyor, pick up a pancake, flip it over and
place it back onto the conveyor.

Assignment 6
𝑂2𝐴 = 8 𝑐𝑚
𝐴𝐵 = 32 𝑐𝑚
𝑂4𝐵 = 16 𝑐𝑚
𝑂2𝑂4 = 16 𝑐𝑚
𝑂4𝐶 = 12 𝑐𝑚
𝐴𝐷 = 16 𝑐𝑚
Draw the coupler curve of point D at the middle of the
coupler AB for a complete revolution of link 2.

Assignment 7
If link CD has an angular velocity of = 6 rad/s, determine
the velocity and acceleration of point E on link BC and the
angular velocity and angular acceleration of link AB at the
instant shown.

Assignment 8
At the instant shown rod AB has an angular velocity 𝜔𝐴𝐵 = 4
rad/s and an angular acceleration 𝜔𝐶𝐷 = 2 𝑟𝑎𝑑/𝑠2. Determine
the angular velocity and angular acceleration of rod CD at this
instant. The collar at C is pin connected to CD and slides freely
along AB.

Assignment 9
The"quick-return" mechanism
consists of a crank AB, slider
block B, and slotted link CD. If
the crank has the angular
motion shown, determine the
angular motion of the slotted
link at this instant.

Assignment 10
Ball C moves with a speed of 3 m/s,
which is increasing at a constant
rate of 1.5 m/𝑠2
, both measured
relative to the circular plate and
directed as shown. At the same
instant the plate rotates with the
angular velocity and angular
acceleration shown. Determine the
velocity and acceleration of the
ball at this instant.

Mechanism Synthesis & Analysis unit 2…..

More Related Content

What's hot (20)

Similar to Mechanism Synthesis & Analysis unit 2….. (20)

Recently uploaded (20)

Mechanism Synthesis & Analysis unit 2…..