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0% found this document useful (0 votes)
253 views34 pagesKOM Unit 4
kom subject chapter this related to mechanical engineer
Uploaded by
jagtapatharva37Copyright
© © All Rights Reserved
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/ 34
SYNTHESIs o,
MECHANIs)4¢
Print
Steps In synthesis : Type synthesis, number synthesis and dimensional synthesis, Tasks of Kineratio
‘synthesis : Path, function and motion generation (Body guidance). Precision Positions, Chebychey
‘Spacing, Mechanical and structural eros.
Graphical Synthesis: Inversion and relative pole method for three position synthesis of Four-Bar and
Single Slider Crank Mechanisms.
Analytical Synthesis: Three postion synthesis of Fou-Bar mechanism using Freudenstein's equation,
Blotch synthesis,
Classification of Synthesis Problems (Tasks of Kinemati Synthesis).
43° Dimensional Syutbesie
44
‘Analytical Method for Dimensional Synthesis of Four Bar Chaln Mechanism (Freudenstein's Equation) ~..
Synthesis of Function Generation nn
45
‘46 Three Postion Motion Synthesis of Four Bar Chal Mechanism (Body Guidance):..
4.7 Bloc’s Synthesis wv kinematics of Machinery (SPPU)
4d Introduction
earlier unlts we discussed about analysis of
In anisms, It was seen that analysis of mechantsmee
mips at determining the performance or moti;
atpracteristles ofa given mechanisms,
ce synthesis of mechanisms may be treated as a
everse problem of analysis of mechanisms, and alms at
Jesigning a mechanism which can satisfy the preseribca
trotion characteristics.
Kinematle Analysis \s the process of determination of
~ {elucty and acceleration of the various links of an
existing mechanisms.
= Kinematle Synthesls deals with the determination of
the lengths and orientation of the various lengths of the
links so that a mechanism could be evolved to satisfy
certain conditions.
= Thereore, ‘The synthests of mechanisms the design of
the mechanism to produce a desired output motion fora
given input motion”.
= In thi chapter, some techniques for the design of four
bar mechanism and slider crank mechanism are
discussed.
~ In synthesis of mechanism, the problem divides itself
into the following three steps.
4.1.1 Type Synthesis
Explain the term Type synthesis
ISPPU ~ Dec. 17; Dec 19
Type synthesis refers to selection of ‘mechanism, ie.
‘DPE of mechanism to be used for a required output. It
may be gear drive, or belt and chain drive or cam and
follower mechanism,
During selection of type of mechanism one may have to
ign aspects like manufacturing process,
safety
consider desi
material selection,
Bar ration etc,
41.2 Number Synthesis
space consideration,
Slain the term Number synthesis
ISPPU ~ Dec. 17, Dec 19}
Number
lnk and maaithesls refers to determination of number of
M60 Indugar Pe Of joint’s required for specified motion. It
Ration "tS the degree of freedom required for specified
42
Synthests of Mechanisms
4.1.3, Dimensional Synthesis
Cia
Explain the tarm Dimensional Synthesis
Dimensional synthesis refers to find out the
dimensions of mechanism required. This includes length of
links, the distance between pvc: points on the links, angle
between arms cf bell-crank levers, cam profile and
diameter of roller of follower, gear ratios.
4.2 Classification of Synthesis
Problems (Tasks of Kinematic
Synthesis)
Synthesis problems are classified into three types which
areas follows:
4.2.1 Function Generation
5
. Explain the term Function generation.
SEES eA
~ Itis frequently required in synthesis of mechanism that
an output link must rotate, oscillate, and reciprocate
according to a prescribed function of time or function of
{input motion, This is called function generation.
A simple example is that of a synthesizing of four-bar
mechanism to generate the function y = f (x). Where
xwould represent the motion or angle of input link and
mechanism is to be designed such that the motion or
angle of output link would be the approximate function
y=f().
Fig. 4.2.1: Four bar chain mechanism to generate the
function y = f(x)
(Co,
PWIGht Reg No, 1-98408/2021) NY Niners oy Machinery (AN)
2 Path Generation
Q. Exnain the torm Path peneration
~ ti peek generation, a point on coupler link oF Moating
Unk (link which fs not eonnected ditwetly to fied Link) Is
to trace a deseribe path with reference to fved frame of
reference,
‘The path ts generally an are of a circle, ellipse or a
straight line, An example of path generation fsa four bar
‘mechanism with coupler link, where a point on coupler
Tink ts required to follow a path y' © £(Q), as Input link
routes
/ SBN hy pat
Fig. 4.2.2: Four bar chain mechanism with coupler link
to follow a pathy = f(x)
4.2.3, Body Guldance (Motion
Generation)
Core
Q. Explain the term Body guidance (Mation generation).
SEC ESS aA A
In body guidance or motion generation the entire body
is guided through a pre selected sequence of motion.
Such a body is guided usually as a part of coupler link.
For example, in construction industry heavy parts such
as buckets and blades of bull dozer must be moved
through a series of prescribed positions.
Ke must be noted that in path generation one is
concerned with the path of a trace point on a coupler
link in body guidance or motion generation the entire
‘motion ofthe coupler link is of concern.
at
Graph
elatively fast In prowtuclng results and.
unifetstand than analytteal methous,
anehods have advantage he chy
P
aslo
4.3.1 Dimensional Synthesis by
Inversion Method
In dimensional synthesis by Inversion method th
concept of inversion Is used to determine th
dimensions of mechanism requlred.
Dimensional synthesis of four bar chaln and slider eran
chain mechanism by inversion method is as follows,
3.1.1 Two-position Synthesis of
Four Bar Chain Mechanism by
Inversion Method
4.
In this method the dimensions of links are determined
with two specified position of input link. Fig. 43.1 shows
2 four bar mechanism with two positions in which link
is fixed, Let A, By C, Dy be the first position of
mechanism and A, By Cz D, Is the second position of
‘mechanism which is shown by dotted line.
cy
4.3 Dimensional Synthesis
= Dimensional synthesis deals with the determination of
@imensions of links angle between links etc. of the
mechanism to satisfy the motion characteristics.
‘= Both graphical and analytical methods are availabe for
@imensional synthesis The choice of method depends
Fig. 4.3.1 : Two-position of four bar chain mechanism
— When input link 2 is rotates through O12 angle then, the
‘output link 4 has to move through @i2 angle. Let lengt
oflink 1 and link 21s known and angle 0, Gp, 2 and ¢5
known itis required to obtain the length of ink 3204
link 4.
upka the type of problem tobe solved.
(Copyright Reg. No. L-98408/2021)
I emattex of Machinery (SPPU)
van for synthesising the
TM ensiont of the links of four bar mechanism by two
vation method Ts as follows ¢
ejay Ay Dy equal to Fength of lnk 1 as shown In
Vig 432.
praw Ay Bi, equal to length of input link 2 at angle 0
easured from Ay Dy. Draw Ay By equal to length of
Input link 2 at angle Oj, measured from A, By,
raw a line from D; at angle 6, the magnitude ts not
known.
Draw an arc of radlus equal to D; A, with D, asa
center and angle equal to $42. We get point A;.
raw another arc of radlus equal to D, B, with D, as
center and angle equal to Oy. We get point B.
2
66 Taking centers By and B,, draw an arcs having equal
radius such that they will Intersect at line drawn
through polnt Dy at angle 6. We get potnt C,.
7, Join B, C; by dark line and B,C; by dotted line.
8, The linkage Ar B; Cy Dy Is shown the first position of
mechanism and the linkage A;B‘C,D, shows the
second position of mechanism.
= Thus B, C; or B, C; will give the length of tink 3 and D; Cy
‘will give the length of link 4,
®
ae
Fig. 4.3.2: Two-position synthesis for four bar chain
mechanism by Inversion method
43.1.2. Three-position Synthesis for
Four Bar Chain Mechanism by
Inversion Method
~ In this method the dimensions of links are determined
with three specified position of input link Fig. 433
poet @ four bar mechanism with three positions.
"tA, By Cy D, be the first position and Ay Bz C2 Dy Ar By
GD, be the second and third position of mechanism
7 Pectvely. Let 8, 6,2 and Oy, be the angle of input ink
ta 912 and ¢yybe the corresponding angle of output
(ConsrightReg No. L-98408/2021)
a
Synthesis of Mechanisms
{tls required to determine the length of link 3, link 4 and
angle $ of output link 4.
f Dat re,
Fig. 4.3.3: Three-position of four bar chain
mechanism
The graphical procedure for synthesising the
dimensions of the links of four bar mechanism by three
Position method is as follows :
1. Draw AyD; equal to length of link 1 as shown in
Fig. 434.
2 Draw A; B, equal to length of link 2 at angle @
measured from Ay Dj. Draw Ay By Ay Bs at angle O22
and 63 from A; By for second and third position of
mechanism respectively.
3. Draw an arc of radius D, By with D; 2s a center and
‘angle equal to $,2 we get point B.
4. Draw another arc of radius D, By with D, asa center
and angle equal to 63. We get point B
5. Join B,B, and 8,8,
6 Draw perpendicular bisectors of lines B,B; and B/B,,
‘These bisectors will intersect at point C..
7. Join B, Cy and D, C, which give the length of link 3
and link 4 respectively and also we get angle ¢ of
output link 4,
Fig. 43.4: Three-position synthesis for four bar chain
mechanism by inversion method
Ve RE Kinematics of Machinery (SPPU)
‘Ex. 43.1: Synthesize a four bar mechanism by the mothod
‘ol laversion, Assume the following data :
Q Length between fx points 10 emit) 2-positions of the
‘4gpUt link from the initial position 80° and 60",
the
‘The graphical
dimensions of the links of four bar mechanism Is.as follows:
1) Draw A; Dy equal to length of 10 cm as shown in
Fig P43.
2) Draw A, B, at any arbitrary length and angle @ measured
procedure for synthesising
from A; Dj. Draw A; Bs, A; By at angle 30° and
ABs.
3) Join B; D; and B,D,
4) Draw an arc of radius D, B: with D, as center and angle
‘equal to 20°, we get point B.
from
5) Draw an are of radius D, By with D, as center and angle
equal to 40°, we get point B,.
6) Join 8,8, and 8'8;,
7) Draw perpendicular bisectors of lines 8,8, and BB,
‘These bisections will intersect at point C.
8) Join B, C; and D, C, which gives the length of link 3 and
link 4 respectively.
Fig. P.4.3.1.
From Fig, P.43.1
AyDy wAns,
A:B, = h=34em wwns.
BC. = h=10.5em Ans.
C.D, = 4=Sem wns,
45
Synthess
3 of Mechaniin,
ESS eee
Baa + Bou a ob mocha bya
‘ot inversion
‘Aseumo the following data,
1) Longth of fixed lnk Is BO mm ad INU tnk fongth
25 mm,
2) Inial poston of input link 90° and 2-postions of he np
link from the intial position 30° and Go",
'8) 2epostions of the ouput lnk from the inal postion age
and 40°. And determine the length of couplor link, outpy,
link and initia! postion of output tink.
Ex. 49.3 ¢ Synthesis a four bar mochanism by the methed gl
Inversion
‘Assume the following data,
1) Length of fixed Ink is 100 mm and Input tink length is
30mm. 4
li) Initial position of Input link 30° and 2-positions of the Input
link from the Initia position 30° and 60°. 3
ii) 2:positions of the output lik trom the Initial positor
and 40".
‘And determine the length of coupler link, output link and
initial position of output tink. EER
20
a ASA 1 Syed « fox ber mecharion ETRE
Bisse tins ctor nettle Gato :
Positions 1 2 3 |
8 30 90 180 4
o 40 115 175
‘Consider L, = Grounded Link, L, = Input Link, L, = Coupler
Unk, Ly = Output Link, @ = Input link angle, ¢ = output link
angle. ff
It the grounded link of length 100 mm is horizontal and Input
Jink Is of 20 mm length, synthesize the mechanism using
Precision positions o! the input link and precision positions of
‘the output link. Ground the pivot of Input link on left hand
side and ground the pivot of output link on right hand side,
Input and output links are rotating in opposite directions. Use
‘the method of Inversion. Draw the mechanism in its first
precision position, |
‘Comment on the mechanism obtained.
a
SEPEREOS
Soln. :
AyD; = 100mm,
A,B; = 20mm,
m= 78°,
on = = 135°
(Copyright Reg. No. L-98408/2021) Ay aneates of Machinery (SHPO)
Fig. P. 4.
From Fig. P.4.3.4
BC = Couplerlinklength=100mm Ans,
CD = Output link length = 20mm Ans.
AsS+1 = p+quitis "crossed double crank
mechanism”
ee ————————
Ex 43:8 : Design a four bar mechanism to co-ordinate three
positions of the Input and output links es foliows =
SECS es
Ex, 4.3.6 : Synthesis the four bar mechanism ABCD, in
which the fangth of the fixed link AD is 400 mm and the crank
AB is of 120 mm long. The Initial position makes an angle
60° with link AD, the angle between the fixed link AD and 2°
position is 120° and the angle between the third position and
link AD Is 180°. The angla between the first and second and
second and third position of the output link are the 60° and
30° respectively. Draw the mechanism in the frst position by
lnversion meth
Ex. 4,3.7 : Synthesize a four bar mechanism in
method for three precision position using the following
Qy2= 30", wiz = 40°, 81g = 50°, Yyy = 100°.
ey
Take length of fixed link as 10 units and length of crank as,
2 units. The crank has turned through 10° measured
‘anticlockwise {rom horizontal In its first position.
a
Ex. 4.9.8 : Design a four link mechanism to coordinate three,
Positons of Input and oulput links for the following angular
displacements by Inversion method. a
D2 = 35", oo = 50" 0,3 Org = 80% oe
SEES:
(Copyright Reg, No. L-98408/2021)
46
4.3.1.3
Synthesis of Mechanisms
Four Position Synthesis for
Four Bar Chain Mechanism by
Inversion Method
~ Inthis method the dimenslons of links are determined
‘with four specified position of input link. Fig 435
shows four position of input link in which the first
Position and second position of input
symmetrical about the frame center line so as to cause
two of the equivalence points 8, and 2 of pasivion one
and position two will enincides.
The effect of this is only two equivalent points 2 and 8,
15 required to produce for synthesising the problem.
‘Thus we are converting the problem of four-position
synthesis into three-position synthesis problem Sach
method Is called as point position reduction
Fig. 4.3.5 : Four-position of input link of four bar
chain mechanism
Let 8,82, 0:3 and 8,, be the angle of input link for four
positions and 0, O12 Gz» and oy, be the corresponding
angle of output link It is required to find the length of
link3 and link 4 and angle 6 of the link 4.
Ifany two links are made symmetrical about the frame
center line then, the procedure is same as three-position
synthesising of links of four bar mechanism which is as
follows:
1. Draw A; Dy equal to length of link 1 as shown in
Fig. 43.6.
Draw A; By, A; B: Ar By and A; B, at an angle @, 6,2,
6, and 6,, respectively and A, B, and A; By should be
symmetric about frame center line.
Draw an arc of radius D; By and D; as a center and
angle equal to $1». We get point By,
Draw another arc of radius D,B, with D, as a center
and angle equal to ¢,4. We get point By.
Join B,B, and BB.
Draw perpendicular bisectors of lines ByB, and B, By,
6
‘These bisectors will intersect at Cy.
¥ Kinematics of Machinery (SPPU)
7. Join By G, and D, G, which gives the length of link 3
and length of link 4 respectively and also we get
angle 6 of output link 4.
Fig. 4.3.6 : Four-pesition synthesis for four bar chain
‘mechanism by inversion method
3.1.4 Two-position Synthesis for
Slider Crank Mechanism by
Inversion Method
= In this method the length of connecting rod is
determined with two specified position of crank
Fig. 43.7 shows two position of slider crank mechanism
having eccentricity e’.
= When crank moves through angle @., the slider will
move linearly through distance S,3.
Fig, 4.3.7 : Two-position of slider crank mechanism
= While synthesising the slider crank mechanism by two
position method it is required to obtain the length of
connecting rod so as to get displacement of slider
through S,: and crank will move through angle 8,
= The graphical procedure to synthesizing the slider crank
mechanism by two position method Is zs follows
1. Draw two parallel lines 1 and 2 ata distance'e' apart,
as shown in Fig. 43.8.
(Copyright Reg, No.1-98408/2021)
Synthesis of Mechanisms
“Take an arbitrary point A; online 1.
‘Take an arbitrary point Con line 2,
Mark point Cy on line I2 such that Cy C= Si
Join A; C; and rotate polnt Cy about Ay through an
angle, to obtain C}
Join C,C3 and draw perpendicular bisector,
ince A; B, Is length of crank which is known
therefore, mark B, on perpendicular bisector,
Join A: By and By Cx.
. The linkage Ay By C; is the required slider crank
mechanism, Thus length By Cy will give the length of
connecting rod.
+7
me
Fig. 4.3.8 : Two:position synthesis for slider crank
mechanism by inversion method
4.3.1.5 Three-position Synthesis for
Slider Crank Mechanism by
Inversion Method.
Crs
@. Explain we pater |opiaie ot
‘mechanism by using inversion method. |
In this method the length of connecting rod is
determined with three specified position of crank.
Fig. 43:9 shows three position of slider crank
‘mechanism having eccentricity‘.
= Let Ay By C; be the first position and A, B; C, and
‘Ay By C, be the second and third position of slider crank
‘mechanism respectively. And @,, and @ be the angle of
crank for second and third position and Siz, Sis be the
linear displacement of slider for second and third
Position
fan memates of Machinery (SPPU)
‘ +8 Synthesis of Mechanisms
"mnte synthesising the slider crank mechanism by three
postion methods, I Is requlred to obtaln the length of
fonnecting Fod so as to get displacement of slider
through Siz and Sis during second and third position and
‘ankwill move through angle 8:2 and 6; respectively
Fig. 43.10 : Two-position synth
‘mechanism by inversion method
~ The graphical procedure to synthesising the slider crank
‘mechanism by three position method is as follows :
41. Draw two parallel lines 1 and 2 ata distance’e' apart
as shown in Fig. 4.3.10,
Take an arbitrary point A, on line 1.
Take an arbitrary point C, on line 2
Mark Cz and C3 om line 2 such that C; Cz = Sand Cy
GsSp,
Toin A, Cand A, C3.
Rotate point Cz about Ay through an angle @2 to
btain C; rotate point Cy about Ay through an angle
Suto obtain c;,
a7
Join CC; and C,C; and draw perpendicular bisectors
fF CiC; and CC’. These bisectors will intersect at
Point B,.
§ Hla By and B,C,
The linkage A, B, C, is the required slider crank
mechanism. Thus length A, By will give the length of
ae and B, C, will give the length of connecting
Note
nih
lenge Poewon ‘synthesis of slider crank mechanism
Bt eo Sank must be known for determining the
jg amecting rod. But in case of tree-postion
OF slider crank mechanism the length of crank
99 tod both will be determine by graphical.
Ex 43.9 : Design a sider crark macharism to coordinate
three positions of the Input and outs links for the following
ata by inversion method. }
6290, S.2=40mm; 85-60, :
Si. 96mm, Ecoenttcy « 20 mma
Fig. P.43.9
Follow the procedure as explained in 4.3.15
From Fig.P.439
AB\=7Smm_
B,C, = 90mm
43.2
Ans,
ensional Synthesis by
Relative Pole Method
= The pole isa centre of rotation of moving link relative to
a fixed link of the mechanism. To locate the pole point,
let us considera four bar mechanism A,B,C\D; in which
link AD; is fixed and crank A,B, is at initial rotation
angle 0.
Ifthe crank A,B; Is rotated by angle 6,3 in anticlockwise
direction to take new position AiB,, the qutput link C.D,
rotates by angle ¢u to reorient the coupler link B,C, to
the new position B,C as shown in Fig. 4.3.11.
‘The normal’ drawn at the mid-points of B,B, and C,C,
intersect at point R. This point is centre of rotation of
coupler link relative to the fixed link AyD, and is called
pole. WF Kinematics of Machinery (SPPU)
49
Synthesis of Mechanisms
~ Arelative pole is a centre of rotation of a link relative to
‘other moving links such as crank and output link. In a
four bar mechanism, if the crank and the output links
rotate by angles 612 and 62 respectively from their
initial positions @ and @, then the relative pole coupler
link relative to crank and output link can be found by
the following procedure (Fig, 4.3.12).
1. Join nodes A; and D, ofthe fixed link AyD.
2. Rotate the link A.D, about point A; through an angle
€12/2 In the direction opposite to that of the link
AB:
larly, rotate the link A,D; about point D, through
an angle $12/2 in the direction opposite to that of
the link C,D.
‘The point of intersection of above two positions of A,D,
is known as relative pole of coupler link and is
represented by R
Fig. 4.3.12 : Relative pole of coupler link
(Copyright Reg. No. L-98408/2021)
~~ Dimensional synthesis of four bar chain and slider crank
chain mechanism by relative pole method Is as follows,
4.3.2.1 Two Position Synthesis of
Four Bar Chain Mechanism by
Relative Pole Method
= Let us assume that two positions of input link crank
separated by angle @j2 are given. The corresponding
Positions of the output link are separated by angle 612
(Fig. 43.13()), Both links are required to move in the
anticlockwise direction. The procedure of synthesis is a5
follows:
~ Let length of ink 1 and tink 2 eran is known and angle
8, O12, @ and Ox is known, itis required to obtain the
length of ink 3 coupler and output ink 4.
1, Draw 4D; equal to length of link 1 as shown in
Fig. 4.3.13(b).
2 Draw a line A,X, passing through point A and
{inclined at an angle 62/2 from link A:D; in the
direction opposite to the direction of rotat
link A,B.
3, Similarly, draw another line D.X: passing through
Point D; inclined at angle $,2/2 from link A;D; in the
direction opposite to the direction of rotation of link
CD,
4. The intersection of lines A,X, and D;X, at point R is
known as relative pole of coupler link 3.
5. Draw A,B, equal to length of input link 2 at angle @
‘measured from A,D;,
6. Join the points B, and R by a straight line.
7. Draw aline RY from point R such that angle A,RD Is
‘equal to angle ByRY, ie. the angle subtended by the
fixed link A,D, fs equal to the angle subtended by the
coupler link B,C,
8. Draw a line from D, at angle which intersect the
line RY at point C,
9. Join ByCy. Thus B,C, will give length of coupler link 3
and D.C; give length of output link 4.
&
*
(a) Two positions of four bar chain mechanism
ran W Kinematics of Machinery (SPPU)
—
{b) Two position synthesis of four bar chain
mechanism by relative pole method
Fig. 4.3.13
4.3.2.2 Three Position Synthesis of
Four Bar Chain Mechanism by
Relative Pole Method
~ In three position synthesis, two pairs of the crank and
output link rotations are required to be coordinated,
let the two pairs of coordinates indicating three
Positions of crank and output link is shown in
Fig. 4.3.14,
&
oe
{ Cy
i
Position synthesis of four bar chain
‘Mechanism by relative pole method
4:10
Synthesis of Mechanisms
‘The procedure of synthesis Is as follows:
~ Let 6, 8,2 and 0, be the angle of input link 2 and 0, x2
and 0,3 be the corresponding angle of output link 4. It is
Fequired to determine the length of link 3 and link 4
and angle 6 of output link 4.
1. Draw A,D, equal to length of link 1.
2. Locate the relative poles R, and R: as described
previously for two pairs of coordinates (8,2, 6:3) and
(81,00).
3. Draw A,B, equal to length of input link 2 at angle @
measured from A,D;.
4. As the crank and output link subtend equal angles at
the relative pole, so draw a line R,T such that angle
D,R,T is equal to angle ARB,
5. Join Rand B; bya straight line.
6. Draw a line R,S such that angle A:R:B; is equal to
angle D,R,S.
7. Locate the point C; at the intersection of lines R,T
and RS.
8, Join links BC, and C;D, which give length of link 3
and link 4 respectively and also we get angle 6 of
output link 4,
4.3.2.3 Two Position Synthesis of
Slider Crank Mechanism by
Relative Pole Method
Consider a slider-crank mechanism in which when
crank takes turn by angle @ in anticlockwise direction from
the first position, the slider changes its position by a
distance S. Let the slider axis is offset by distance e from the
axis of fixed link.
position synthesis of slider crank mechanism
by TWoP Fig. 4.3.15
Fig. 4.3.14
18408/20211
«
“ork Reg. No. Lo YW Kinematics of Machinery (SPFU)
‘The detailed procedure of synthesis is as follows :
1. Choose a point A; on the fixed link (frame) and draw a
line A.N along the slider movement, Draw another line
ALY perpendicular to AX.
Draw a line parallel to A,X at distance equal to the
required offset distance e.
Select a line segment A,E of length S,2/2 on the line AX
such that the distance A,E is measured in a direction
opposite to the motion of the slider.
. Atpoints A; and E, draw perpendicular lines PA and PE.
Draw a line PR, at an angle @:: with line PA in the
direction opposite to the rotation of crank.
‘The intersection of line PR, with PE is relative pole Ri.
Draw line A,B, and mark its first position at angle 8.
Join B with R, by a straight line.
Construct angle B,R,T equals to angle A,R,E. The line
R,T intersects the offset line at point C,.
10.Join B,
jC; which give length of connecting rod and
linkage A,B,C, is the required slider crank mechanism.
4.3.2.4 Three Position Synthesis of
Slider Crank Mechanism by
Relative Pole Method
Q. Expizin three position synthesis of single sider
mechanism by using relative pole method.
sa
In the three position synthesis, the two pairs of
movement of crank and slider are coordinated. Let (8:2, S::)
and (@:s, Sis) be coordinates of the crank and the slider. The
crank is assumed to rotate in anticlockwise direction. The
stepwise procedure of synthesis is as under.
Choose point A; on the fixed link and draw a line AX
along the movement of slider.
L
Locate the relative poles R; and Re for two palrs of
coordinates (812S:2) (Su) respectively
(as described in two position synthesis method).
and
Drawa line A,B, and mark its first position at angle 6.
Join B, with Ry and Rz by straight lines.
R,Q making an angle B,R,Q equals to angle
411
Snthesis, Mec
Similarly, draw another line RT which
B,R;T equals to angle A,R,F. maken
Make the position of slider C, at the inters
ect
twolines. ma
Join ByC, which give length of connect
1B rod.
A,B,C; is the required slider crank mechanism
Fig. 4.3.16(a) : Three position of slider crank
mechanism
Fig. 4.3.16(b) : Three position synthesis of slider-crank
mechanism by relative pole method
Ex. 43.10 : Design a slider crank mechanism to coo
three positions of crank and slider for the following data by
relative pole method. 0;2=40°, Sjz = 180mm 61s
Sra = 300 mm. Take eccentricity of slider as 20
Saenrs
Mark
G12 = 40°, 81s = 120",
Siz= 180 mm, S,= 300 mm,
e=20mm
From Fig. P.4.3.10,
A\By=211.9 mm ;B, C= 253.6 mm
{Copyright Reg. No. L-98408/2021)
he re
rnematles of Machinery (SPPU)
412
Synthests of Mechanisms
Fig. P. 4.3.10
4.4 Analytical Method for
Dimensional Synthesis of Four
Bar Chain Mechanism
(Freudenstein’s Equation)
Tr «
Q. Derive the Frudenstein's equation of four bar
mechanism.
Q. Discuss “analytical © synthesis _ using - kinematic
Coefficient in four bar mechanism, EGQuUEITENEE!
~ Consider a four bar chain mechanism ABCD having
Jength I, l,l and ly, The link AD is fixed and the link AB,
‘Cand DC make angles 8, Band ¢ respectively long the
link AD or the x-axis as shown in Fig. 44.1.
~ Considering the links as vectors and writing the vector
‘isplacement relationship along x-axis and y-axis.
ABez9me
3,C,22880m
Riso
‘Alongx-axs,
088+ cos Pl, cose-k = 0 -0
‘Along y-axis,
hsinO+hsinB-Lsing = 0 m0)
Rewriting Equations (7) and (i)
hesB = h+lcoso-kos® (il)
hsinB = Lsing-hsind Ov)
Squaring and adding Equations (iif) and (iv),
Fcos'B +f sin'B= [1+ 4,cos 9-4 cos®}*+ [l, sin &-I,sin]?
E [eos*B + sintB] =[F + Fcos*9 + Ecos*@+ 2 hycose
~2h hoes 0-24 kos eas orf sito + f sv 2 44 sin 8 sin g)
E (cos*B + sin’) = + & (cos + sin*9)+ & (cos*6 + sino)
+ 2hl.cos @-2 hh cos 8-2 hl, (cos @ cos 6 + sin 8 sin 6)
Ref + R48 +2 hl cos 9-2 hk cos @-2 hl, cos (8-9)
Dividing by 2 byl, on both sides and rearranging
fetGeh hk
Ti 7709S $7, 005 B= cos (8-4) ov)
‘The above equation may be written as,
Kycos-Kyc0s8+K; = cos (8-4) (442)
where, (44.2)
h on(4.4,3)
F Fig. 4.4.1 : Four bar mechanism
© :
*urlght Reg, No, L-98408/202i) Vet
bh WW Kinematics of Machinery (SPPU)
fet
R+h
2h
The Equation (4.4.1) is known as Freudenstein's
equation.
~ Consider three position of four bar chain mechanism as
shown in Fig, 4.4.2. Let @;, 8; and @; be the angle of input
link and @;. @; and @; be the corresponding angle of
output link. It is required to determine the dimensions
bylesls and of the four bar mechanism.
k= wol44.4)
Fig. 4.4.2 : Three position of four bar chain
— For the three different position of the four bar chain
mechanism the Equation (4.4.1) may be written as,
K, 005 6;-Kz cos ®,+Ks = cos (0,-#,) (445)
K; cos ;~ Kz cos 8; + Kz = cos (8-2) (4.4.6)
K, cos 6: Ks cos @;+K; = cos (85-43) (447)
— From above equations the values of Ky, Kz and Ky are
obtained and hence the length of links can be
determined from Equations (4.4.2), (4.4.3) and (4.4.4).
.5 Synthesis of a Function
Generation
Q. Explain the foll terms related to Synthesis of
mechanisms :
1) Precision postions fi) Structural error
SES RAE
= In synthesising the mechanism it is required that output,
link should rotate, oscillate or reciprocate according to a
specified function of time or the motion of input link, for
example a synthesising of four bar mechanism to
generate the function y = f (x), where x would represent
the motion or angle of input link and mechanism is
designed such that the motion or angle of output link
would be the approximate function y =f (x).
— It is usually impossible to accurately produce the
function y = f (x) at more than a few positions, The
positions at which the generated function produced by
(Copyright Reg. No. L-98408/2021)
Synthesis of Mechanisms
thesized linkage and required function agreed arg
known as prectston positions or precision points,
Fig, 4.5.1: Function Generation
4.5.1 Types of Errors in Synthesis of
Mechanisms
— There are three types of errors present in the design of
linkages for function generation. These errors are as
follows.
1. Structural error
2. Mechanical error and
3.Graphical error.
— Structural error is the difference between the
generated function and the desired function for a certain
value of input variable. Therefore, the precision points
are spaced in such a way as to minimize the structural
error of the linkage.
~ Mechanical errors are caused because of mechanical
faults such as improper machining, casting error of
components of the linkage, clearance in the components
because of rubbing, overloading of linkages,
‘manufacturing tolerances on the link dimensions etc.
— Graphical Error is caused because of incorrectness in
drawing of perpendicular or parallel lines. It may occur
because of wrong graphical construction and wrong
choice of scale. Also, there may be human errors in
drawing work.
4.5.2 Chebyshev Spacing Method
Q. Write short note on : Chebychev spacing,
TEE
— Since, the structural error is the difference between the
generated function and the desired function fora certain
value of input variable, therefore precision points are
spaced in such a way that to minimise the structural
error of the linkage.
anaes
¥ jnomates of Machinery (SPPU) Z
Kl If Synthests of Mechanisms
—febyshev spacing method Is used to locating the
: mon position In four bar mechanism,
i very good ttalfor the spacing of Precision poston is
called a Chebyshev spacing,
According to Freudenstein and Sandor the chebyshey
- 1gfor n positions n the range x35 xs given by,
spacln ‘
1 ‘n (2)-1)
4p Bd teen) oS
wo(4.5.0)
where, % = Precislon positions
212,300
x, = Starting position
xj = Finishing position
n = Number of precision position
for example, if we want to synthesis the linkage to
generate the function
y = x0 (9
over the range 1x3 using three precision position,
emeh, 423, j=1,2and3
‘Then from Equation (4.5.1) the three value of x are:
1 (2x1 =1
% =5G+D4 (3=1) cos GN]
= 1136
1 1 n(2x2—
2 = 76+ Fees eos 2 292=H]
= 2.000
1 1 n(2x3-1
% =7G+ 403-1) one 2S=H]
= 2.866
The corresponding values of y are,
Ya = x8 = (1.134)°* = 1.078
a = Es (2) = 1.515
Yo = x¥= (2,866) = 1.880
~ Gebyshev spacing of the precision posions is also
tained by graphical method, The procedure Is as
follows:
1. Draw a circle of diameter equal to the range Ox
2, S2XO8) as shown in Fig 45.2.
* Daw a polygon having number of sides equal to
f Ice the number of precision positions required Le.
for three Precision position draw hexagon inside the
cleee,
pa Perpendicular from each comer which
me the diagonal of circle at precision
los xy, and xy as shown in Fig, 45.2.
Fig. 4.5.2: Graphical method for Chebyshev spacing
For range 15x 3, the x,= 1,%=3 and
x= 3-1=2and we get
x= 1.134,
x;=2.00and
xy 2.866
= In case of angular function of input and output link, the
angles of rotation @ and 6 are the linear analogy of x and
y respectively as shown in Fig. 4.53.
— Let AO, 40, Ax and Ay are the desired ranges of the
respective variables 6, 6,x and y respectively.
Therefore,
48 = 6-6,
Fig. 4.5.3 : Angle relationship for function generation
Since there is linear relationship between the angular
and linear changes, therefore we can write,
O-% _ a8
(xrx) ~ Ax
2.88, = a &-x)
&
PYtIBht Rep No \rosanOTOTS
¥ aS aa
If, is datum. Then @, = 0
ag
8 = Ay OX)
or & = x Gx) wu(4.5.2)
where,j=1,2,..0
Similarly the relationship between y and ¢ may be
written as,
a6
&
or 6 (453)
‘The Equations (4.5.2) and (4.5.3) are considered as
angle relationship for function generation.
Steps to be followed while solving the problems on
Chebyshev spacing method :
Step 1 :Calculate values of x, x, x, in the given range
X%< x < x by using Freudenstein and sandor
equation.
(0e33-Ftor-aycos| "99
Step 2: Calculate values of ys, Yo, Yo, .— on substituting
values of x in the given function.
Write the precision points ic. (xy yi), (xp ya),
(ys)
Step 4: Calculate the desired ranges ie,
8 =6,-8,; 49 = 6-4,
AX =x-X; Ay =yr-y.
Step 5 : Calculate angle relationship for function generation
Le. 81, 82, Os, om and Os, Oa Psem- by using
A
= 0,65 CH= XI50* Ot gy O-¥D
Ex. 4.5.1 1 Determine the chebyshev spacing for function
y=x'" for the range 0 < x < 3 where three precision points
ate required. For these position points, determine @,; 8, and
da O
1A8 = 40° and Ag = 90°,
ISPPU - Dec 07, May 12, Dec. 19 10 Marks
Step3:
3oln. :
jiven
The given function fs, y= x45
The range 1s.0 $ x $3,
lumber of precision points is, n= 3,
Starting positionis,x, = 0.
Finishing position is, x,
1S
ynthests of Mechanisms
Calculate valies of xy Xa Xs
VANRC XS XS Xp,
The chebyshev spacing for n number of precision pointy
Inthe range xy p +4, therefore above mechanism is class It
four bar linkage and opposite to shortest link is fixed, so
mechanism obtained is double rocker mechanism,
Please download e-book for detail
Solution: for.Ex. 4.5.19
‘Ex, 4.5.19: A four revolute mechanism Is to be synthesized
by using three precision points, to generate the function y =
2x 4 3, for the range 0 < x < 8, Assuming 50° starting
position and 130° finishing position for input link as well as
for the output link, find out values of x, y, 8 (input angles) and
4 (output angles) corresponding to the three precision points
with Chebyshev spacing. (Use graphical method for
‘obtaining the precision points and round off the precision
points to nearest single decimal)
Uf the grounded link is horizontal and its length is 100 mm,
synthesize the mechanism using initial, final and the
Precision point positions of the input link and the precision
Point positions of the output link. Assume crank length as
‘50 mm, ground pivot of input link on left hand side and
‘ground pivot of output link on right hand side. Use the
‘method of inversion. :
Draw the mechanism in its first precision position and
‘comment on the mechanism obtained.
ISPPU - May 09, 10 Marks}
Ex. 45.20 : Design a four bar mechanism to generate the
function y = sin x, x varies from 0 to 90°. Angle of input link
varies from 30° to 160° and angle of output link varies from
160° to 120°. Assume length for fixed link as a one unit. Use
three precision positions from Chebyshev spacing.
: [SPPU - Dec. 17; 12 Marks]
Soln, :
Give
‘The given function is y = sinx
The range is, o*sx<90°
Starting angle for inputlinkis, @, = 30°
Finishing angle of outputlinkis, = 150°
Starting angle for outputlink is, = 60°
Finishing angle for output links, = 120°
‘Number of precision points is, n=3
Calculate values of x1, X2, Xs, »m in the given
Fange x,< x <%
The chebyshev spacing for n number of precision points
{in the range x,< xx; is given by,
n(2) 4]
» = dora tocnden[
Step1:
2n
forj=
Synthesis of Mechanisms
aan m(2x1=1)
we Foren Form of EE]
X= AS-45° cos § = 6.029"
forj=2,
=]
xa = 5 (00" +07) 4 (00° 0") eos| 2 2X2= 1]
Lx = 45° 45 cos Es 45°
for) =3,
n(2¥3=1
xy = $(90" + 0°)-4 (90°07) cos| eeeaa
ay = 45-459 cos 52 203,79"
Calculate values of ys, yx Ys — om
substituting values of x In the given function.
On substituting above values of x in the given function,
y=5 sin x, we get the corresponding value of y are,
=sinx,= sin (0°) =0
inx,= sin (6.029°) = 0.1050
sinx, = sin (45°) = 0.7071
Ys = sinxs= sin (63.979) = 0.9944
Yr = sinxy= sin (90°) =1
Step 3: Write the precision points.
The three precision points are as follows :
(6.025%, 0.1050), (45°, 0.7071), (83.97°, 0.9944),
(90,1)
Step
Ans.
Step4: Calculate the desired ranges
‘The desired ranges are,
AB = 6-8, = 150° 30°= 120°
AG = O14, = 120% 60° = 60°
Ax = xx, =90°-0" = 90°
ay = Yr-¥s =1-0=1
Calculate angle relationship for function
generation
Angle relationship for function generation is given by,
0, = 0,442 x9
Step 5:
and = ef ory
forj=1
0, +A 2Gu-x)
0 4 120° 5.
= 30° +Gge (6.029
Ans,
WF 4:25 Synthesis of Mechanisn
W Kinematics of M
Ao
and * hy’ Oryd
= 60°+ 5° (0.1050) = 66.3° Ans,
forj=2
AO
O: = B+ Ay (ar%)
= 307+ ae (45°- 0°) = 90° Ans.
Ao
and 2 = thy Ory)
= 60°+ 2 (0.7071-0°)=102.42" Ans.
Ag
8 = 8,450 Om)
(83.97°-0°)= 141,96" ...Ans.
= 30°#5
4o
a = Ot Ry Or ¥5)
60° + (0.99440) = 119.66°
and
swAns,
‘The values of input and output angles are as follows,
jachinery (SPPU)
‘The Freudenstein’s equation for the first position g
8,038" and >,
input and output link (ie. at 63°) om
be written as,
K; cos (66.3°) ~ K, (38.038°) + Ks = cos (38.038 66.35
0.40 K,~ 0.787 Kz + Ks = 0.88 ~{
Similarly, Freudensteins equation for the secong
position of input and output link. (ie. at 8; = 90°, ang
z= 102.42°) can be written as,
K, cos (102,42°) — K; cos (90°) + Ks = cos (90°- 102.42)
— 0.21 K, + K3= 0.97¢
i)
Similarly, Freudenstein’s equation for the third position
of input and output link. (le. at 6; = 141.96° ang
5 = 119.66°) can be written as,
K, cos (119.66*) ~ K, cos (141.96°) + Ky = cos (141.96°- 119.660)
0.495 K, + 0.787 Kz + Ks= 0.925 i)
By solving Equations (1), (ii) and (Iii) simultaneously by
elimination method, we can find out values of Ky, K; and Ky
Substituting Equation (i) from Equation (i), we get
0.40 K,- 0.787 Kz+ Ks = 0.88
- 20K, +K = 0976
+
0.61 K,- 0.789 Kz .096 nfl)
Substituting Equation (ii) from Equation (iil), we get
6, = 38.038", $1= 66.3°
Cae n= 10242" = 0495 Ky + 0.787 Kz +Ky = 0.925
0) = 141.96, oo= 119.66 -021K, +K = 0.976
Step 6: Calculate values of Ky, K; and Ks. * = =
‘Weikaow that - 0.285 K; + 0.787K; = -0.051 @)
The Freudenstein's equation for four bar chain Adding Equation (iv) and Equation (v), we get
mechanism is given by + 0.61 K,-0.787 K, = - 0.096
Ky cos-Kycos®+Ks = cos (8-9) *
= 285K, +0.787K, = ~0.051
where,
0325K, = -0.147
os Ky = -0.452 ole)
Using value of K; in Equation (v), we get
~ (0.285) (~ 0.452) + 0.787 K, =- 0.051
0.12882 + 0.787 Kz = - 0.051
0.787 K = -0.17982
K, = -0228 (vi)
Using take of K, in Equation (ii), we get
~ (0.21) (-0.452)+K, = 0.976
0.09492+K, = 0.976
K = 0881 itd
Fig. P. 4.5.20
(Copyright Reg. No. L-98408/2021) eran ¥F Kinematics of Machinery (SPPU)
4.26 Synthesls of Mechanism
a
Step7: Calculate the length of links of four bar AK = /-% = 4-2 =D
mechanism,
‘Assuming the et of fixed link as th oe eee
xed link as the unit i.e. y=
ara elanithe n= Ay ‘Angle relationship for functlon generation Is given by,
From Equation (vi), Ky *E
h
ri 1
slo = Tpggg 72212 unktawAns. | Forj=1 0, = +A Gx -
From Equation (vii), K, = 0 = 30+ 40/2 (2-2)
0; = 30
~4.305 unltotns. | popje2 9, = 30+ 42 (%-m) =20+90(3-2) 250
From Equation (viii), 40
5 = 30475 (x -%) = 30+ 40/2 (4-2)
feftek Forj=3 0; = 3047) (x)-%) = 30+ 40/2 (4-2)
ok = 30440
(P+ 2217HE Ea: 2 a
+ 0881 = C2212) (- 4385) Now = Ru-v +%
= 2.78 unit Ans.
ms. | rorj=d 1 = Geng -y) +40
Ex. 4.5.21 : Synthesize a four-bar mechanism to generate a :
function y = 2 logy (x) and x varies from 2 to 4 with an
interval of 1. Assume 0 to vary from 30° to 70° and 6 from
40° to 100°. Starting position of input and output link is 30°
‘and 40°. If length of fixed link is 1 unit determine other link
engths and draw mechanism in its first pos!
Soin. :
Given function is,
¥=Zlogi0%
‘The range is,2< x < 4
x= 4
4. = 40",
= 100°
For three precision position
‘Since x varies from 2 to 4 with an interval of 1,
Xp = 3, x= 4
Corresponding values of y are ys, Y2,¥s,on substituting x In
the given function, y = 2 login
= 60/0.602 (0.602- 0.602) + 40
o: = 40
42 = 60/0.602 (ys-y,) +40
60/0.602 (0.954 ~ 0.602) + 40
92 = 75.08
62 = 60/0.602 (ys- y,) +40
0 /0.602 (1.204 - 0.602) + 40
100
Forj=2
Forj=3
o
‘The values of input and output angles are,
0, = 40,
75.08,
100
‘The Freudenstein equation is,
K, cos $-K cos + Ky = cos (0-9)
When, 0, =30and = 40
Ky c08 y= Ke 605 0, + Ky= cos (8 ~ 0)
«i Ky €08 40 ~ K; cos 30+ Ky= cos (30-40)
when x yi = 2logio2 = 0.602 ©. 0.766K, ~ 0.066 K, + Ky = 0.9848 (1)
When = 3; ya = 2logio3 = 0.954 When 0;= S0and 75.08
oe ys = 2logyo4 = 1.204 Kycosds~Ky cos; #Ky = cos (0;~ 43)
The hres eect poled, +. Ky c08 75.00 Ky 608 50+ Ky= cos (50 - 75.08)
175), (41.204). 0.257K, 0.643 Kz+Ky = 0.906 (2)
(2,0.602), (3, 0.954), (3.66.1. wpgi'ereipounayayen00
The desired ranges are, Kycoss~Kyc0s®)# Ky = €08(0s~9s)
0 = 0-8, = 70-30 = 40 +. Kycos 100K; cos 70+ Ky = cos (70-100)
Ag = 1-6, = 100-40 = 60
[Canevinke Daw Na 1-98408/2021) WF Kinematics of Machinery (SPPU)
= 0.1736 Ky - 0.342 Ky + Kj= 0.866 3)
Solve simultaneously,
= 0.262,
3. K K,=0241, — K,= 0.994
K
* b
h
4.149 unit
we eaeter— B49"
0.994 = —>y58167x4149
2x3B167 x 4149
0.994x2x3,8167x4.149 = 32.781-5
B= 32781-31481
B 213 = 114unt
k= Lunit = 3.8167 unit,
ky = 144unit, 4 = 4.149 unit.
Mechanism in first position is shown in Fig. P. 4.5.21.
‘outpist link ‘c’, Angles 6 and for three
pene positions are given in table below :
Ee 4 2 3 !
e 20° | 35° 50°
a5e | 45° 60"
link lengtha =1.0
EYSPPU - Dec. 06, Dec. 12, 10 Marks}
427
Synthesis of Mechanin,
ic
Fig. P. 4.5.22
Soin. : The Freudenstein’s equation for four bar mechanism
is
Rashes Rieus@ eh
and K,=
= cos (8-9)
2 kya (bree dy/2ac
~@)
0)
Where a, b, c,d, @ and 6 are as shown in Fig, P. 45.22
Now, the Freudenstein's equation for the first position of
the input and output link.
(ie. when 6, = 20° and 6 = 35°) may be written as:
K, cos 35° K; cos 20° + Ks = cos (20°-35°)
ie, 0.819 K,- 0.940 Kp + Ks= 0.966 0
Similarly, the Freudenstein’s equation for the second
position (ie. when 6; = 35°, ¢2 = 45°) may be written as
K, cos 45°- Kz cos 35°+ Ky = cos (35° 45°)
Le, 0.707 K:- 0.819 K; +Ky= 0.985 -@
And the Freudenstein's equation for the third position
(ie. when @; = 50° and 65 = 60°) may be written as,
Ky cos 60° K; cos 50° + Ky = cos (50° 60°)
05 K,- 0.643 Kz + Ky = 0.985
Then, by solving simultaneously Equations (i, (i) and
(iii, by elimination method, we can find the value of Ky, K
and Ks,
Subtracting Equation (Ii) from Equation (2), we can
eliminate K, and get,
(0.819 K,~ 0.940 Kz + Ks)
(0.707 K,~ 0.819 Ke + Ks)
Pa cord
le.
= 0.966
0.985
0.112 Ky 0.121 Ky a)
= 222 Hae 9 oy
Substituting value of K; in Equation (iil), we get
[ 112 Ky + 0.019
5 K,- 0.643 > —
le Ky = 0.095K, +1086 (4)
Substituting values of K; and Ky in Equation (i), we get
(Copyright Reg. No. 1-98408/2021) Kinematics of Machinery (SPPU)
os
Kk; 0.94 [peeeoe | (0.095 K; + 1.086) = 0.968
0,819 K- 870 Ky~ 0.148 + 0.095 K, + 1.086 =
0.044K, = 0.028
Ky = 0.636 n(vii)
_ Substituting values of K; from Equation (vi) in Equation
(i), we get
F K; = 0.095 (0.636) + 1.086=1.146 (viii)
_ Also, substituting values of K; from Equation (vii) in
iquation (v), we get
112
966
n(x)
ollows :
a = 1unit
From Equation (b),
4
ne vi m4
abated
dy = SEE
and hence,
a
Ky = &
sd = ak, =(1) (0.636) =0.636 units ..Ans.
4
Ree
a
= = 0.853 units Ans.
_sibeeed
953)? + (0.636)
2 (1) (0.853)
0.177
0.421 unit
" the grounded ink f= 80 mm, using Frudenstoin's
“tation, find out lengths of other links to satisfy the given
Restional conditions. Also draw the synthesized mechanism
's frst position and comment on the mechanism obtained.
Soln. :
The Frudenstein's equation given by
K, cos -KzcosO+K3 = cos (0-9)
For first position
6, = 40, 6-50
K, cos 50 ~ Ky cos 40+ Ky
~ K; [0.6428] - Ke [0.7661] + Ky
For second position
= 55, $2560
K, cos 60 ~K, cos 55 + Ky
< Ky [0.5] ~ Ke [0.5736] + Ks
For third position 03=70 ; 3=75
K, cos 75 ~ Kz cos 70 + Ky
K; [0.2588] - K; [0.3420] + Ks
Solving Equations (1), (2) and (3) we get,
K, = 0.2
Ky = 1.016
K,
cos (40 - 50)
0.9848
cos (55 - 60)
0.9962
cos (75-70)
0.9962
= 0.208
Synthesis of Mechani
itl
m2
(3
(Copyright Rea NoL-96408/2021) 4.29 Synthesis of Mechanien
& Kinematics of Machinery (SPPU)
x oz08 = 2 h Ane
ee no : Now, Ke = o2062 =
I, = 144.23 mm wAns.
Peb-B+k sh = 145mm “aie
Ky = yy Rag-f+e
Ce Nowks = at
302+ 150? - Fh + 144.237 oe
= ; 10" + 1517-15 + 145°
- 2x 150 x 144,23 i
eo cans, | Ot = Ox Ts
pate GS ABSA e829 agit
Ex. 4.5.24 : Design a four-bar mechanism with input link
Jy coupler link J, and output link 4, Angles @ and 4 for
three successive positions are given in table below :
ageidownloddie-book tordeta
NEE (Pines eee eee)
Position > 1 2 3
8 40° se | 70° Ex, 4.8.25 : Synthesize a four bar mechanism with input
$ eee spe are link ‘a, coupler link‘, output link '’ and grounded link
‘Angles 0 and ¢ for three successive positions are given in
It the grounded link =.30 mm, using Frudenstein’s | the table below:
equation find out lengths of other links to satisfy the given ae |
Positional conditions. Also draw the synthesized ait(eameal asso
mechanism in its first position.
¢ | ase | 45 | 60°
If the length of grounded link is 40 mm, using
Seln Freudenstein's equation find out other link lengths to satsty
Weikmaw that: the given positional conditions. Draw the synthesized
K, cos Ke cos 8 + Ky = cos (8-9) “mechanism in its second positor
When, 0= 40° and @ = 50° !
K, cos 50K; cos 40°+ Ky = cos (40°-50°) wl) | Ex. 4.8.28
‘When, @ = 55° and = 60° ‘a’, coupler link ‘b', output link ‘cand grounded link ‘¢’. Ang
@ and @ for three successive positions are given in the table
Ky cos 60°-K; cos 55° + Ks = cos (55° 60") Gi) | below
When, @ = 70° and 6 = 75° il 2 3
K, cos 75° Kz cos 70°+ Ks = cos (70° 75°) (ii) o 55° 0° | 125
g 2 4c ¢ 55° sor | 125°
If the length of grounded link is 100 mm, using
Froudenstein's equation find out other link lengths to satisly
the given positional conditions. Comment on tha link lengths
and also in the mechanism obtained.
Ex, 4.5.27 : Fig, P, 4.5.27 shows a schematic of a fout
bar mechanism with Input tink ‘a’ and output tink‘
angles 0 and ¢ for three successive positions are given in
$ the table bolow :
Fig, P.4.5.24 1 2 2
0 [ses =25
Solving Equations (0, (i) and (ii) we get, ¢ [ite [ao = 50
K,= 0.198, Ke = 0.2062, W the length of grounded tink Is 40 mm, ol
K,= 1.0184 Freudensteln’s equation find out other link lengths !2
h 40 satisfy the given positional conditions. a
but, Ky = 7 0198 = 2 Draw the synthesized mechanism in its second ;
SNR
(Copyright Reg. No.L-9840872023) ee reno, matics of MACnInery (94FU)
Fig. P. 4.5.27
(4528 + Synthesize a four bar mechanism for three
ccossive positions given in the table below :
| 1 2 3
t
e | 4 60 82
L3 | 98 120 195
era’ is input link, ‘b’ as coupler link, 'c’ as. output
‘g as grounded link, ‘é" as input angle, 'g' as output
nde. If the fength of grounded link is 50 mm, using
) satisty the given positional conditions. Draw the
cherism in its second position and comment on the
henism obtained. SAIS
4,529 : Design a four bar mechanism with input link.
, coupler link , output link c. Angle 8 and 9 for three
ecessive positions are given in the table below : Use
Freudenstein method :
Position | 1] 2 | 3
6 |-30| 50 | 70
| 40 | 75 | 100
W/the mechanism in second position.
SOLE ERMC
4520 : Synthesis a four bar mechanism using
Psudenstein's equation to satisty In one of iis positions for
P folowing specifications assuming fixed link length as 1
et EERE
y= 2rad/sec’
q = rad/sec
y= 7rad/sect
4-30
Synthesis of Mechanisms
4.6 Three Position Motion
Synthesis of Four Bar Chain
Mechanism (Body G
Whiversity Question]
Q. Explain the term Body guidance related to Synthesis
of mechanisms,
The graphical method for synthesis the four bar cl
mechanism considering body guidance or motion
generation for three specified positions of coupler link is
as follows.
Consider the three positions of coupler link which
moves from B,C, to B,C; to BsCs as shown in
Fig. 4.6.1(a).
Draw four lines connecting B, to By, By to Bs, C; to Cz and
Cy to Cs
Locate the centre of a circle which passes through the
three points B,, Bz, B; by drawing perpendicular
bisectors of B,B2 and B,Bs. Let these two bisectors
Intersects at point A, which is first pivot point of four
bar chain mechanism as shown in Fig. 4.6.1(b)-
Similarly, locate the centre of a circle which passes
through the three points C, C, C; by drawing
perpendicular bisectors of C,C: and C;C3. Let these two
bisectors intersects at point D; which is second pivot
point of four bar chain mechanism as shown in
Fig. 4.6.1(b).
The complete linkage, A:B,C,D; is shown in Fig. 4.6.1(c)
Which guides the coupler link B,C; through three
specified positions. WF Kinematics of Machinery (SPPU)
Fig. 4.6.1 : Three position synthesis for four bar chain
mechanism (Body Guidance)
Ex. 4.6.1 : Fig. P. 4.6.1 shows three positions of link AB its
length is 80 mm. Itis to be moved through the successive
positions A,B, A,B, and A,B, with co-ordinates as shown.
Position A,B, is horizontal and point A, is 80 mm vertically
above point B,. Graphically synthesize a four bar mechanism
with link AB as its coupler. Find-out the co-ordinates of the
‘ground pivots in the same reference frame. Also find lengths
pt and output links. Draw mechanism in its second
BSPPU - May 08, 6 Marks]
of
Fig. P. 4.6.1
Soln. : The graphical procedure for synthesising the four
bar chain mechanism is as follows;
1, Draw three given positions of link AB i.e. A;By, AzBy and
AsB; in xy plane with suitable scale,
2. Join points AyA, and AzAs and draw two perpendicular
bisectors which intersect at point C;.
(Copyright Reg. No. 1-98408/2021)
431
Synthesis of Mechanisn,
0, (110,200
A830,
Second pollo ot rar
hain mechan
Fig. P. 4.6.1(a)
3. Similarly join points BB, and B,B, and draw two |
perpendicular bisectors which intersect at point D;. |
4, Point C, and Dj is two pivot points of mechanism and |
C\A,BD, is the first position of four bar chain |
mechanism. Similarly CyAzB,D; and C,AsBsD; is second |
and third position of four bar chain mechanism
respectively.
5, From Fig. P. 4.6.1(a)
= The coordinates of ground pivots C; and D: is,
C, (60, 110), Ds (110, 200)
— The length of input link C,Az = 100 mm and oxtput
link D,B2 = 140 mm.
Ex. 4.6.2 : Fig. P.4.6.2 shows three positions of link AB. tis
length is 100 mm. It is to be moved through the successive
positions A;B,, AB. and A,B, with co-ordinates as shown.
Graphically synthesize a four bar mechanism with link AB 2s
its coupler. Find out the co-ordinates of the ground pivets in
‘the same reference frame. Also find lengths of input and
‘tout inks, Drew the mechanism in ts second postion,
ISPPU - May 08. 6 Marks
B:
By
A 8, (140.120)
(203920) (20.420)
s
(160,90)
ai A 14020)
Fig. P. 4.6.2
Soin. ; From Fig. P. 4.6.2(a).
Correct condition of ground pivot (83, 74) and (130, 301).
Length of input link = 78 mm. ee 32
¢ ofoutput link = 181 mm. Bear Synthesis of Mechanisms
‘ nay for the the Russian Scientist developed a method
Synthesis of four bar chain mechanism using
complex numbers,
Im this method the, the a
Velocities and an,
input and out
ingular positions i.e. angular
gular accelerations are prescribed for
put links and relative links dimensions are
determined,
~ Fig. 4.7.1 shows four bar chain mechanism ABCD.
8 Now, for the loop ABCD, the loop closure equation in the
complex form is,
orzo
>>>
Ry+Ry+Ry+R,=0 oD)
Where, R = nef,
(379) a
R, = nee
R= ne,
eo x R = rye”
Fig. P. 4.6.2(0) smeenetene ene = 0 i)
4.7 Bloch’s Synthesis ee
Let Gp = O(= « (Angular velocity)
2
£9 612.0 (angular acceleration)
a?
‘Taking the first and second derivatives with respect to
time of equation (il) we get (for i= 2, 3 and 4)
iS 18
tyqaz e? + i303 €°3 + yg = 0
8
iwgize? +iaayg e+ ionyge* = 0
oo
dr. i@gRz+i@gR3tiagRy = 0
2 > -
a @gRzragR3+OyRg = 0 fil)
Fig. 4.7.1
Second derivative a é ; 104 a
: ts
fag (w5+ fg) €2 + 173 (034 iar) © +1 Y4 (04 IO "
' 24104) Ra = 0
a at) win elaborates) (okt Be
2
> tia) Re ©
Or (wb fxg) Rat (05¢ toa) Rls a ine i
nag? te thee equations (0, (i) and (Iv) and four uno ons
‘hy MY then be carried out by determinants. It we ©
* simultaneous equations as
a
Ri
j i > 4
Ry 3 R3 a
es
2 Ra =
(o +a R, (024103) Rs (ost!)
Coy
Reg. No, L-98408/2021) oe de ee
oC)
the
ywns may be chosen arbitrarily and
a and (iv) back into vector notations, we
0
a - Res! ~@)
+R +R3 Synthesis of Mecthanin,
Hs 33
YF Kinematics of Machinery (SPPU) fs :
3 ah
Equation (v) can be written in matrix form as Re = 003 («9+ 12) ~ 2 (3+ 10)
Zz —< 2-3
1 1 1 Ra and Ry = - Ra Rar Re
2 3 me BR {=| ° Putting the values given in above equations
3
aptiag ostias wytiog J] _, Be = 10(87+i 100) - 8 [107 + i(-150)]
= 640+ 1000i-800+1200i
ot) = -160+2200i
.
Solving equation (vi) for R 2 we get. A, = 160? + 2200? = 2205.81 and
8 coto = 2200. g-- 415°
- 160
0 03 m4
5 4 = 2205.81 2-4,15° =» Ans,
0 artis writ
Ps 5 a >
Ra = Determinant D Ry = 20 [102 +i (- 150)] - 10 [207 + (0)]
‘The value of determinant D is given as, = 2000 - 3000i-4000
7 1 1 = -2000-3000i
D = e m4 or = ¥2000? + 30007 = 3605.55
2 e
oftiag ostiag of ing and cot = Son; 0= 33.69"
Zz °
We can obtain similar expressions for R's and Rg. Since = 3605.55 33.69
D is common to all so the value of D can be assumed to be 3, = epz0% 1 (-20 162+ 1 C2001
Es
equal to unit. Its effect will be to change the valves of Ro,
> >
Ra and Ry by the same factor. We can be get
27>
dimensionless vector links once the values of Ro, R3 and
3
Rgare calculated,
‘Thus, we can write,
2 2 2
Ra = 04 (05+ a5) - 3 (04+ ig)
> 2 a
R3 = @2 (@4+ i or) ~ @g(w3+ ig) soli)
> 2 Zs
Ra = 05 (03412) - 0 (w5+ is)
2 > oo
Ry = -Ry- Ry Ry
ee Ry a FS
Ex, 4.7.1: Synthesize a four bar chain mechanism that wil in
one of its positions satisty the following values for the
‘angular velocities and accelerations.
(02 = 20 rad/sec. sap = 0 rad / sec? ;
3 = B radisec, ; ag = 100 rad / sec? ;
(4 = 10 rad/sec. jag =~ 150 rad / sec;
Soin.
According to Bloch’s synthesis method,
~
Ro
2 2
Ms (3+ i 3) ~ 3 (w+ i erg)
2:3 eeal 2
Ra = 02 (04+ 104) - (03+ ig)
Copyright Reg, No. L-98408/2021)
(viii)
= 3200 - 1280-20001
= 1920-20001
'1920? + 20007 = 272.43,
and ~43.83°
= 2772.73 £-43.83°
2 oo >
Now Ri =-Rz- Ry Ry
~ © 160 + 2200 i) - (- 2000 - 3000 i)
= (1920 - 2000)
160 + 2000 - 1920-2200 i + 30001 +2000
= 240+2800i
or 1240? + 2800? = 2810.26
and cot
Theory
Q.1 Explain the following terms :
(0. Type synthesis
(i) Number synthesis
(il) Dimensional synthesis (Section 4.1)
© ine
aties of Machinery (SPPU)
t
Geian th follwing Terms related wo ayageae
problem
(y Function generation
{i Path generation
(i) Body guidance (motion generation)
ection 4.2)
explain two-position synthesis for four bar
mechanism. (Section 4.3.1.1)
giplain three-postion synthesis for four bar
mechanism. (Section 4.3.1.2)
Byplain fourposition synthesis for four bar
mechanism. (Section 4.3.1.2)
Bxplein two-position synthesis for slider crank
mechanism. (Section 4.3.1.4)
Explain three-position synthesis for slider crank
mechanism. (Section 4.3.1.5)
Explain the following terms :
()) Precision positions
(i) Structural error. (Section 4.8)
Explain chebyshev spacing method for locating
precision position in four bar mechanism.
(Section 4.5.1)
Explain the graphical procedure to synthesis the
‘our bar mechanism with coupler link. (Section 4.6)
umericais
M
‘A four ber mechanism is used to generate the
lunetion y = 4/x for the range 1 < x $3. Find the
a1
13
Synthesis of Mechanisms
three precision postions from chebyshev spacing, if
‘Re Inia values of the crank angle and folower
angle is 30° and 200° respectively.
Take A0 = Ad = 90°,
of 6 and 9,
Find the corresponding values
5°, 0, = 113.979,
$1= 183.62°, @ = 132.38°, 6 = 112.01°]
Determine the chebyshev spacing for function
¥ = log4o x for the range 1