computer aided analysis and design of machine elements, computer aided analysis and design of electromagnetic devices, computer aided analysis and design of machine elements pdf
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1002 1002 Theory of Machines Theory of Machines
25
F F F F Fea ea ea ea eatur tur tur tur tures (Main) es (Main) es (Main) es (Main) es (Main) Computer Computer
Computer
Computer Computer
1. Introduction.
2. Computer Aided Analysis for
Four Bar Mechanism
Aided Aided
Aided Aided Aided
(Freudenstein’s Equation).
3. Programme for Four Bar
Mechanism.
Analysis and Analysis and Analysis and
Analysis and Analysis and
4. Computer Aided Analysis for
Slider Crank Mechanism.
6. Coupler Curves.
7. Synthesis of Mechanisms.
Synthesis of Synthesis of
Synthesis of
Synthesis of Synthesis of
8. Classifications of Synthesis
Problem.
9. Precision Points for Function
Mechanisms Mechanisms
Mechanisms
Mechanisms Mechanisms
Generation.
10. Angle Relationship for
function Generation.
25.1. Introduction
11. Graphical Synthesis of Four
Bar Mechanism.
We have already discussed in chapters 7 and 8, the
12. Graphical Synthesis of Slider
graphical methods to determine velocity and acceleration
Crank Mechanism.
analysis of a mechanism. It may be noted that graphical
13. Computer Aided (Analytical)
method is only suitable for determining the velocity and
Synthesis of Four Bar
acceleration of the links in a mechanism for a single position
Mechanism.
of the crank. In order to determine the velocity and
15. Least square Technique.
acceleration of the links in a mechanism for different
16. Programme using Least
positions of the crank, we have to draw the velocity and
Square Technique.
acceleration diagrams for each position of the crank which
17. Computer Aided Synthesis of
is inconvenient. In this chapter, we shall discuss the analytical
Four Bar Mechanism With
Coupler Point. expressions for the displacement, velocity and acceleration
18. Synthesis of Four Bar in terms of general parameters of a mechanism and
Mechanism for Body
calculations may be performed either by a desk calculator
Guidance
or digital computer.
19. Analytical Synthesis for
slider Crank Mechanism
1002
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1003
25.2. Computer Aided Analysis for Four Bar Mechanism (Freudenstein’s
Equation)
Consider a four bar mechanism ABCD, as shown in Fig. 25.1 (a), in which AB = a, BC = b,
CD = c, and DA = d. The link AD is fixed and lies along X-axis. Let the links AB (input link), BC
(coupler) and DC (output link) make angles θβ,andφ respectively along the X-axis or fixed link
AD.
(a) Four bar mechanism. (b) Components along X-axis and Y-axis.
Fig. 25.1
The relation between the angles and link lengths may be developed by considering the
links as vectors. The expressions for displacement, velocity and acceleration analysis are derived
as discussed below :
1. Displacement analysis
For equilibrium of the mechanism, the sum of the components along X-axis and along
Y-axis must be equal to zero. First of all, taking the sum of the components along X-axis as shown
in Fig. 25.1 (b), we have
ab cosθ+ cosβ−c cosφ−d = 0 . . . (i)
or bc cosβ= cosφ+d −a cosθ
Squaring both sides
22 2
cos ( cos cos )
bc β= φ+d −a θ
22 2 22
cd cos φ+ + 2cd cosφ+a cos θ
=
−φ 2ac cos cosθ− 2a d cosθ . . . (ii)
Now taking the sum of the components along Y-axis, we have
ab sinθ+ sinβ−csinφ = 0 . . . (iii)
or bc sinβ= sinφ−a sinθ
Squaring both sides,
22 2
sin ( sin sin )
bc β= φ−a θ
22 22
sin sin 2 sin sin . . . (iv)
=φca+θ−acφθ
Adding equations (ii) and (iv),
2 22 2 22 2 2 22
(cos sin ) (cos sin ) 2 cos (cos sin )
bc β+ β = φ+ φ +d +cd φ+a θ+ θ
−φ 2( accoscosθ+sinφsinθ)−2a dcosθ
1004 Theory of Machines
222 2
or
bc=+d+ 2cd cosφ+−a 2ac(cosφcosθ+ sinφsinθ)− 2ad cosθ
22 2 2
or
2ac(cosφcosθ+φ sin sinθ)=− a b+c++ d 2c d cosφ− 2a d cosθ
22 2 2
ab−+c+d d d
cosφcosθ+ sinφsinθ= +φ cos−θ cos
. . . (v)
2ac a c
222 2
−++
dd abcd
Let . . . (vi)
==kk;;and =k
12 3
ac 2ac
Equation (v) may be written as
cos cos sin sin cos cos
φθ+ φ θ=kk φ− θ+k . . . (vii)
12 3
or cos (φ – θ) or cos(θ−φ) =kk cosφ− cosθ+k
12 3
The equation (vii) is known as Freudenstein’s equation.
Since it is very difficult to determine the value of φ for the given value of , from
θ
equation (vii), therefore it is necessary to simplify this equation.
From trigonometrical ratios, we know that
2
1t−φ an ( /2)
2tan(φ/2)
cos
and φ=
sinφ=
2
2
1t+φ an ( /2)
1tan ( /2)
+φ
Substituting these values of sin φ and cosφ in equation (vii),
2
1−φ tan ( / 2) 2 tan(φ / 2)
cos sin
×θ+ × θ
22
1t+φ an ( /2) 1t+φ an ( /2)
2
1tan ( /2)
−φ
=×kk− cosθ+k
123
2
1tan ( /2)
+φ
2
cosθ− 1 tan (φ / 2)+ 2sinθ tan (φ / 2)
22 2
=kk 1− tan (φ− / 2) cosθ 1+ tan (φ+ / 2)k 1+ tan (φ / 2)
12 3
2
cosθ− cosθ tan (φ / 2)+ 2sinθ tan (φ / 2)
22 2
=− kk tan (φ / 2)−k cosθ−k cosθ tan (φ / 2)+k+k tan (φ/ 2)
11 2 2 3 3
Rearranging this equation,
22 2 2
−θ cos tan (φ / 2)+kk tan (φ+ / 2) cosθ tan (φ / 2)−k tan (φ / 2)+ 2sinθ tan (φ / 2)
12 3
=− cosθ+kk − cosθ+k
12 3
2
−φ tan ( / 2) cosθ−kk− cosθ+k + 2sinθtan (φ/ 2)−k−k+ cosθ(1+k )= 0
12 3 1 3 2
2
(1)cos tan /2(2sin)tan/2 (1)cos0
−kk θ+ −k φ+− θ φ+k+k− +k θ=
231 13 2
(By changing the sign)
2
or tan ( / 2) tan( / 2) 0 . . . (viii)
AB φ+ φ +C=
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1005
Ak =− (1 ) cosθ+k+k
where . . . (ix)
231,
B =−2sinθ,and
Ck=+k− (1+k ) cosθ
13 2
Inner view of an aircraft engine.
Note : This picture is given as additional information and is not a direct example of the current chapter.
The equation (viii) is a quadratic equation in tan(φ / 2). Its two roots are
2
−±BB − 4AC
tan (φ= / 2)
2A
2
−±BB − 4AC
−1
2tan
φ=
or . . . (x)
2A
From this equation (x), we can find the position of output link CD (i.e. angle φ ) if the
length of the links (i.e. a, b, c and d) and position of the input link AB (i.e. angle θ ) is known.
If the relation between the position of input link AB (i.e. angle ) and the position of
θ
coupler link BC (i.e. angle β ) is required, then eliminate angle φ from the equations (i) and (iii).
The equation (i) may be written as
ca cosφ= cosθ+b cosβ−d ... (xi)
Squaring both sides,
22 22 22
ca cos φ= cos θ+b cos β+ 2ab cosθ cosβ
2
+ 2 cos 2 cos . . . (xii)
da−θ d −bdβ
Now equation (iii) may be written as
ca sinφ= sinθ+bsinβ . . . (xiii)
1006 Theory of Machines
Squaring both sides,
2 2 22 22
... (xiv)
ca sin φ= sin θ+b sin β+ 2ab sinθsinβ
Adding equations (xii) and (xiv),
22 2 22 2 22 2
ca (cos φ+ sin θ) = (cos θ+ sin θ)+b (cos β+ sin β)
2
2 (cos cos sin sin ) 2 cos 2 cos
+θ ab β+θβ+ d− a dθ− b dβ
22 2
or
ca=+b+ 2ab(cosθcosβ+ sinθsinβ)
2
+−da 2d cosθ− 2bd cosβ
22 2 2
or
2ab(cosθcosβ+θ sin sinβ)=c−a−b−+ d 2ad cosθ+ 2bd cosβ
22 2 2
ca−−b−d d d
cosθβ cos+ sinθsinβ= + cosθ+ cosβ
. . . (xv)
2
ab b a
22 2 2
c−− abd−
dd
Let and = k . . . (xvi)
==kk;;
14 5
2ab
ab
Equation (xvi) may be written as
∴
cosθβ cos+ sinθsinβ=kk cosβ+ cosθ+k . . . (xvii)
14 5
From trigonometrical ratios, we know that
2
2tan( /2) 1t−β an ( /2)
β
cos
sinβ= , and β=
2 2
1t+β an ( /2) 1t+β an ( /2)
Substituting these values of sin β and cosβ in equation (xvii),
2
1−β tan ( / 2) 2 tan(β / 2)
cosθ+ sinθ
22
1t+β an ( /2) 1t+an (β/2)
2
1t−β an ( /2)
=+ kk cosθ+k
145
2
1t+β an ( /2)
2
cos 1 tan ( / 2) 2sin tan( / 2)
θ− β + θ β
22 2
=−kk 1 tan (β / 2)+ cosθ 1+ tan (β / 2) ++ k 1tan (β/2)
14 5
2
cosθ− cosθ tan (β/ 2)+ 2sinθ tan(β/ 2)
22
=
kk− tan (β / 2)+θ k cos+θ k cos tan (β / 2)
11 4 4
2
tan ( / 2)
++ kk β
55
22 2 2
cos tan ( / 2) tan ( / 2) cos tan ( / 2) tan ( / 2)
−θ β +kk β− θ β−k β
14 5
+θ 2sin tan(β / 2)−kk− cosθ−k+ cosθ= 0
14 5
2
− tan (β/ 2)(kk +1)cosθ+−k + 2sinθtan(β/ 2)−(kk −1)cosθ++k = 0
451 451
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1007
2
or
(kk +θ 1) cos+−k tan (β / 2)+ (−2sinθ) tan(β/ 2)+(kk −θ 1) cos++k = 0
451 451
(By changing the sign)
2
or . . . (xviii)
DE tan (β+ / 2) tan(β / 2)+F= 0
Dk=+(1)cosθ+k−k,
451
where E =−2sinθ, and ... (xix)
F= (k−θ 1)cos+k+k
451
The equation (xviii) is a quadratic equation in tan(β/ 2) . Its two roots are
2
−±EE − 4DF
tan(β= / 2)
2D
2
4
−±EE −DF
−1
or . . . (xx)
β= 2tan
2 D
From this equation (xx), we can find the position of coupler link BC (i.e. angle β ).
Note: The angle may be obtained directly from equation (i) or (iii) after determining the angle φ.
α
2. Velocity analysis
Let ω = Angular velocity of the link AB=θ d / dt ,
1
BC=β d / dt
ω = Angular velocity of the link , and
2
ω = Angular velocity of the link CD=φ d / dt .
3
Differentiating equation (i) with respect to time,
dd θβ dφ
−θabc sin× −β sin× + sinφ× = 0
dt dt dt
or −ωab sinθ− ω sinβ+ωc sinφ= 0 ... (xxi)
12 3
Again, differentiating equation (iii) with respect to time,
dd θβ dφ
ab cosθ× + cosβ× −c cosφ× = 0
dt dt dt
or ab ωθ cos+ω cosβ−cω cosφ= 0 ... (xxii)
12 3
cosβ sinβ
Multiplying the equation (xxi) by and equation (xxii) by ,
−ab ω sinθβ cos−ω sinββ cos+cω sinφβ cos = 0 ... (xxiii)
12 3
and ab ω cosθsinβ+ωβ cos sinβ−cωφ cos sinβ= 0 ... (xxiv)
12 3
Adding equations (xxiii) and (xxiv),
ac ωβ sin(−θ)+ω sin (φ−β)= 0
13
−ω a sin(β−θ)
1
... (xxv)
∴ ω=
3
csin(φ−β)
1008 Theory of Machines
Again, multiplying the equation (xxi) by cosφ and equation (xxii) by sinφ ,
−ωab sinθcosφ− ω sinβcosφ+cω sinφcosφ= 0 ... (xxvi)
12 3
and ab ω cosθsinφ+ ω cosβsinφ−cω cosφsinφ = 0 ... (xxvii)
123
Adding equations (xxvi) and (xxvii),
ab ωφ−θ+ωφ−β=
sin( ) sin( ) 0
12
−ω a sin(φ−θ)
1
... (xxviii)
∴
ω=
2
bsin(φ−β)
From equations (xxv) and (xxviii), we can find ω and ω , if ab ,,c,θφ , ,β and ω are
3 2 1
known.
3. Acceleration analysis
Let α = Angular acceleration of the link AB=ω d dt ,
/
1 1
α = Angular acceleration of the link BC=ω d / dt , and
2 2
α = Angular acceleration of the link CD=ω d / dt .
3 3
Differentiating equation (xxi) with respect to time,
dd θβdd ωω
12
−ωab cosθ× + sinθ× − ω cosβ× + sinβ×
12
dt dt dt dt
dφ dω
3
+ω c cosφ× + sinφ× = 0
3
dt dt
dd
vdu
∵ ()uuvv =× +×
. . .
dx dx dx
22
or
−ωaa cosθ− sinθα −bb ω cosβ− sinβα
112 2
2
... (xxix)
+ωcc φ+ φα =
cos sin 0
33
Again, differentiating equation (xxii) with respect to time,
dd θβdd ωω
12
ab ω×−sinθ× + cosθ× + ω ×−sinβ× + cosβ×
12
dt dt dt dt
dφ dω
3
−ω c ×−sinφ× + cosφ× = 0
3
dt dt
22
or
−ωaa sinθ+ cosθα −bω sinβ+b cosβα
112 2
2
sin cos 0 ... (xxx)
+ωcc φ− φα =
33
cosφ
Multiplying equation (xxix) by , and equation (xxx) by sin φ ,
22
−aa ω cosθφ cos−α sinθφ cos−bω cosβφ cos
11 2
22
... (xxxi)
−αbc sinβcosφ+ ω cos φ+cα sinφcosφ= 0
23 3
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1009
22
and
−aa ω sinθφ sin+ α cosθφ sin −bω sinβφ sin
112
22
... (xxxii)
+αbc cosβsinφ+ω sin φ−cα cosφsinφ= 0
23 3
Adding equations (xxxi) and (xxxii),
2
−ωaa (cosφcosθ+ sinφsinθ)+ α (sinφcosθ− cosφsinθ)
11
2
−ωbb (cosφcosβ+ sinφsinβ)+ α (sinφcosβ− cosφsinβ)
22
22 2
+ω c (cos φ+ sin φ)= 0
3
22 2
−aω cos(φ−θ)+aα sin (φ−θ)−bbc ω cos(φ−β)+ α sin (φ−β) + ω = 0
11 2 2 3
22 2
−αaa sin (φ−θ)+ ω cos(φ−θ)+ωb cos(φ−β)−cω
11 2 3
... (xxxiii)
∴
α=
2
bsin (φ−β)
cosβ sin β
Again multiplying equation (xxix) by and equation (xxx) by ,
222
−aa ω cosθβ cos−α sinθβ cos−bω cosβ−bα sinβcosβ
11 2 2
2
... (xxxiv)
+ωcc cosφcosβ+ α sinφcosβ= 0
33
222
and
−ωaa sinθsinβ+ α cosθsinβ−bω sin β+bα cosβsinβ
11 2 2
2
... (xxxv)
+cc ω sinφβ sin−α cosφβ sin = 0
33
Adding equations (xxxiv) and (xxxv),
2222
−ωaa (cosβcosθ+ sinβsinθ)+ α (sinβcosθ− cosβsinθ)−bω (cos β+ sin β)
112
2
+ωcc (cosφcosβ+sinφsinβ)+ α (sinφcosβ− cosφsinβ)= 0
33
222
−ωaa cos (β−θ)+ α sin (β−θ)−bω +cω cos(φ−β)+cα sin (φ−β)= 0
11 23 3
222
−αaa sin (β−θ)+ ω cos (β−θ)+bω −cω cos(φ−β)
11 23
... (xxxvi)
∴ a =
3
sin ( )
c φ−β
From equations (xxxiii) and (xxxvi), the angular acceleration of the links BC and CD (i.e.
α and α ) may be determined.
2 3
25.3. Programme for Four Bar Mechanism
The following is a programme in Fortran for determining the velocity and acceleration of
the links in a four bar mechanism for different position of the crank.
C PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A FOUR-BAR
C MECHANISM
DIMENSION PH (2), PHI (2), PP (2), BET (2), BT (2), VELC (2), VELB (2), ACCC (2),
ACCB (2), C1 (2), C2 (2), C3 (2), C4 (2), B1 (2), B2 (2), B3 (2), B4 (2)
READ (, ) A, B, C, D, VELA, ACCA, THETA
PI = 4.0 ATAN (1.0)
THET = 0
IHT = 180/THETA
DTHET = PI/IHT
1010 Theory of Machines
DO 10 J = 1, 2 IHT
THET = (J – 1) DTHET
AK = (A A – B B + C C + D D) 0.5)
TH = THET 180/PI
AA = AK – A (D – C) COS (THET) – (C D)
BB = – 2.0 A C SIN (THET)
CC = AK – A (D + C) COS (THET) + (C D)
AB = BB 2 – 4 AA CC
IF (AB . LT . 0) GO TO 10
PHH = SQRT (AB)
PH (1) = – BB + PHH
PH (2) = – BB – PHH
DO 9 I = 1, 2
PHI (I) = ATAN (PH (I) 0.5/AA) 2
PP (I) = PHI (I) 180/PI
BET (I) = ASIN ((C SIN (PHI (I)) – A SIN (THET)) / B)
BT (I) = BET (I) 180/PI
VELC (I) = A VELA SIN (BET (I) – THET) / (C SIN (BET (I) – PHI (I)))
VELB (I) = (A VELA SIN (PHI (I) – THET) ) / (B SIN (BET (I) – PHI (I))))
C1 (I) = A ACCA SIN (BET (I) – THET)
C2 (I) = A VELA 2 COS (BET (I) – THET) + B VELB (I) 2
C3 (I) = C VELC (I) 2 COS (PHI (I) – BET (I) )
C4 (I) = C SIN (BET (I) – PHI (I))
ACCC (I) = (C1 (I) – C2 (I) + C3 (I) ) / C4 (I)
B1 (I) = A ACCA SIN (PHI (I) – THET )
B2 (I) = A VELA 2 COS (PHI (I) – THET )
B3 (I) = B VELB (I) 2 COS (PHI (I) – BET (I) ) – C VELC (I) 2
B4 (I) = B (SIN (BET (I) – PHI (I))))
9 ACCB (I) = (B1 (I) – B2 (I) – B3 (I)) / B4 (I)
IF (J . NE . 1) GO TO 8
WRITE (, 7)
7 FORMAT (4X,’ THET’, 4X,’ PHI’, 4X,’ BETA’, 4X,’ VELC’, 4X,’ VELB’, 4X,’ ACCC’, 4X,’
ACCB’)
8 WRITE (, 6) TH, PP (1), BT (1), VELC (1), VELB (1), ACCC (1), ACCB (1)
6 FORMAT (8F8 . 2)
WRITE (, 5) PP (2), BT (2), VELC (2), VELB (2), ACCC (2), ACCB (2)
5 FORMAT (8X, 8F8 . 2)
10 CONTINUE
STOP
END
The various input variables are
A, B, C, D = Lengths of the links AB, BC, CD, and DA respectively in mm,
THETA = Interval of the input angle in degrees,
VELA = Angular Velocity of the input link AB in rad/s, and
2
ACCA = Angular acceleration of the input link in rad/s .
The output variables are :
THET = Angular displacement of the input link AB in degrees,
PHI = Angular displacement of the output link DC in degrees,
BETA = Angular displacement of the coupler link BC in degrees,
VELC = Angular velocity of the output link DC in rad/s,
VELB = Angular velocity of the coupler link BC in rad/s,
2
ACCC = Angular acceleration of the output link DC in rad/s ,
2
ACCB = Angular acceleration of the coupler link BC in rad/s .
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1011
Example 25.1. ABCD is a four bar mechanism, with link AD fixed. The lengths of the links
are
AB = 300 mm; BC = 360 mm; CD = 360 mm and AD = 600 mm.
2
The crank AB has an angular velocity of 10 rad/s and an angular retardation of 30 rad/s ,
both anticlockwise. Find the angular displacements, velocities and accelerations of the links BC
and CD, for an interval of 30° of the crank AB.
Solution.
Given input :
A = 300, B = 360, C = 360, D = 600, VA = 10, ACCA = –30, THETA = 30
OUTPUT :
THET PHI BETA VELC VELB ACCC ACCB
.00 – 114.62 – 65.38 – 10.00 – 10.00 – 61.67 121.67
114.62 65.38 – 10.00 – 10.00 121.67 – 61.67
30.00 – 144.88 – 82.70 – 8.69 – .84 101.52 181.43
97.30 35.12 – .84 – 8.69 181.43 101.52
60.00 – 166.19 – 73.81 – 6.02 6.02 38.02 77.45
106.19 13.81 6.02 – 6.02 77.45 38.02
90.00 174.73 – 47.86 – 8.26 12.26 – 180.18 216.18
132.14 – 5.27 12.26 – 8.26 216.18 – 180.18
270.00 – 132.14 5.27 12.26 – 8.26 – 289.73 229.73
– 174.73 47.86 – 8.26 12.26 229.73 – 289.73
300.00 – 106.19 – 13.81 6.02 – 6.02 – 113.57 – 1.90
166.19 73.81 – 6.02 6.02 – 1.90 – 113.57
330.00 – 97.30 – 35.12 – .84 – 8.69 – 170.39 – 49.36
144.88 82.70 – 8.69 – .84 – 49.36 – 176.39
25.4. Computer Aided Analysis For Slider Crank Mechanism
A slider crank mechanism is shown in Fig. 25.2 (a). The slider is attached to the connecting
rod BC of length b. Let the crank AB of radius a rotates in anticlockwise direction with uniform
(a) (b)
Fig. 25.2 Slider crank mechanism.
2
angular velocity ω rad/s and an angular acceleration α rad/s . Let the crank makes an angle
θ
1 1
with the X-axis and the slider reciprocates along a path parallel to the X-axis, i.e. at an eccentricity
CD = e, as shown in Fig. 25.2 (a).
1012 Theory of Machines
The expressions for displacement, velocity and acceleration analysis are derived as
discussed below :
1. Displacement analysis
For equilibrium of the mechanism, the sum of the components along X-axis and along Y-
axis must be equal to zero. First of all, taking the sum of the components along X-axis, as shown in
Fig. 25.2 (b), we have
ab cosθ+ cos(−β)−x = 0 ... ( β in clockwise direction from X-axis is taken – ve)
or bx cosβ= −a cosθ ... (i)
Squaring both sides,
22 2 22
... (ii)
bx cos β= +a cos θ− 2xa cosθ
Now taking the sum of components along Y-axis, we have
be sin(−β)+ +a sinθ = 0
or −βbe sin+=a sinθ
be sinβ= −a sinθ ... (iii)
∴
Squaring both sides,
22 2 22
sin sin 2 sin ... (iv)
be β= +a θ−ea θ
Adding equations (ii) and (iv),
22 2 2 2 22 2
(cos sin ) (cos sin ) 2 cos 2 sin
b β+ β = x +e +a θ+ θ − xa θ− ea θ
22 2 2
bx=+e+a−2cxaosθ−2easinθ
2222
or
xa +−(2 cosθ)x+a −b +e −2easinθ=0
2
or ... (v)
xk++x k= 0
12
22 2
where ka =−2cosθ , and 2sin ... (vi)
ka= −+− b e ea θ
1
2
The equation (v) is a quadratic equation in x. Its two roots are
2
−±kk − 4k
11 2
... (vii)
x =
2
From this expression, the output displacement x may be determined if the values of a, b, e
and are known. The position of the connecting rod BC (i.e. angle β) is given by
θ
sinθ−
ae
sin (−β) =
b
ea−θ sin
or
sinβ=
b
ea−θ sin
−1
... (viii)
∴
β= sin
b
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1013
Note : When the slider lies on the X-axis, i.e. the line of stroke of the slider passes through the axis of
rotation of the crank, then eccentricity, e = 0. In such a case, equations (vi) and (viii) may be written as
22
ka =−2cosθ , and
ka=−b
1
2
−θ asin
−1
β= sin
and
b
2. Velocity analysis
Let ω = Angular velocity of the crank AB=θ d /, dt
1
ω = Angular velocity of the connecting rod BC=β d / dt , and
2
dx/. dt
v = Linear velocity of the slider =
S
Differentiating equation (i) with respect to time,
ddβθ x d
ba ×−sinβ× = − ×−sinθ×
dt dt dt
dx
or ... (ix)
−ωab sinθ− ω sinβ− = 0
12
dt
Again, differentiating equation (iii) with respect to time,
dd βθ
ba cosβ× = − cosθ×
dt dt
or ... (x)
ab ωθ cos+ω cosβ= 0
12
cosβ
Multiplying equation (ix) by and equation (x) by sinβ ,
dx
... (xi)
−ωab sinθcosβ−ω sinβcosβ− × cosβ= 0
12
dt
and ab ωθ cos sinβ+ω cosβsinβ= 0 ... (xii)
12
Adding equations (xi) and (xii),
dx
aωβ (sin cosθ− cosβsinθ)− × cosβ= 0
1
dt
dx
aωβ sin(−θ)= × cosβ
1
dt
dx aωβ sin(−θ)
1
=
... (xiii)
∴
dt cosβ
From this equation, the linear velocity of the slider (v ) may be determined.
S
The angular velocity of the connecting rod BC (i.e. ω ) may be determined from equa-
2
tion (x) and it is given by
−ω a cosθ
1
ω=
2
... (xiv)
cos
b β
1014 Theory of Machines
3. Acceleration analysis
Let α = Angular acceleration of the crank AB=ω d / dt ,
1 1
α = Angular acceleration of the connecting rod = dd ω/,t and
2 2
22
dx /dt
a = Linear acceleration of the slider =
S
Differentiating equation (ix) with respect to time,
2
dd θβdd ωωdx
12
−ωab cosθ× + sinθ× − ω cosβ× + sinβ× − = 0
12
2
dt dt dt dt
dt
2
dx
22
−αab sinθ+ω cosθ− α sinβ+ω cosβ− = 0
... (xv)
11 2 2
2
dt
The chain-belt at the bottom of a bulldozer provides powerful grip, spreads weight
and force on the ground, and allows to exert high force on the objects to be moved.
Note : This picture is given as additional information and is not a direct example of the current chapter.
Differentiating equation (x) with respect to time,
dd θβdd ωω
12
ab ω×−sinθ× + cosθ× + ω ×−sinβ× + cosβ× = 0
12
dt dt dt dt
22
ab α cosθ−ω sinθ + α cosβ−ω sinβ = 0
... (xvi)
11 2 2
cos β
Multiplying equation (xv) by and equation (xvi) by sin β ,
222
−αab sinθcosβ+ω cosθcosβ−α sinβcosβ+ω cosβ
11 2 2
2
dx
−× cosβ= 0
... (xvii)
2
dt
222
ab α cosθsinβ−ω sinθsinβ + α cosβsinβ−ω sin β = 0
and ... (xviii)
11 2 2
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1015
Adding equations (xvii) and (xviii),
2
aα (sinβcosθ− cosβsinθ)−ω (cosβcosθ+βθ sin sin )
11
2
dx
22 2
−ω b (cos β+ sin β)− ×cosβ= 0
2
2
dt
2
dx
22
sin ( ) cos ( ) cos 0
aa α β−θ − ω β−θ −bω − × β =
11 2
2
dt
2 22
dxaa α sin (β−θ)− ω cos (β−θ)−bω
11 2
... (xix)
∴
=
2
cosβ
dt
From this equation, the linear acceleration of the slider (a ) may be determined.
S
The angular acceleration of the connecting rod BC (i.e. α ) may be determined from
2
equation (xvi) and it is given by,
22
ab (αθ cos−ω sinθ)−ω sinβ
11 2
α=
... (xx)
2
b cosβ
25.5. Programme for a Slider Crank Mechanism
The following is a programme in Fortran to find the velocity and acceleration in a slider
crank mechanism.
c PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A SLIDER
c CRANK MECHANISM
READ (, ) A, B, E, VA, ACC, THA
PI = 4 ATAN (1.)
TH = 0
IH = 180/THA
DTH = PI / IH
DO 10 I = 1, 2 I H
TH = (I – 1) DTH
BET = ASIN (E – A SIN (TH) ) / B)
VS = – A VA SIN (TH – BET) / (COS (BET) 1000)
VB = – A VA COS (TH) / B COS (BET)
AC1 = A ACC SIN (BET – TH) – B VB 2
AC2 = A VA 2 COS (BET – TH)
ACS = (AC1 – AC2) / (COS (BET) 1000)
AC3 = A ACC COS (TH) – A VA 2 SIN (TH)
AC4 = B VB 2 SIN (BET)
ACB = – (AC3 – AC4) / (B COS (BET) )
I F (i . EQ . 1) WRITE (, 9)
9 FORMAT (3X,’ TH’, 5X,’ BET’, 4X,’ VS,’ 4X,’ VB,’ 4X,’ ACS’, 4X,’ ACB’)
10 WRITE (, 8) TH 180 / P I , BET 180 / P I, VS, VB, ACS, ACB
8 FORMAT (6 F 8 . 2)
STOP
END
1016 Theory of Machines
The input variables are :
A, B, E = Length of crank AB (a), connecting rod BC (b) and offset (e) in mm,
VA = Angular velocity of crank AB (input link) in rad/s,
2
ACC = Angular acceleration of the crank AB (input link) in rad/s , and
THA = Interval of the input angle in degrees.
The output variables are :
THA = Angular displacement of the crank or input link AB in degrees,
BET = Angular displacement of the connecting rod BC in degrees,
VS = Linear velocity of the slider in m/s,
VB = Angular velocity of the crank or input link AB in rad/s,
2
ACS = Linear acceleration of the slider in m/s , and
2
ACB = Angular acceleration of the crank or input link AB in rad/s .
Example 25.2. In a slider crank mechanism, the crank AB = 200 mm and the connecting
rod BC = 750 mm. The line of stroke of the slider is offset by a perpendicular distance of 50 mm.
2
If the crank rotates at an angular speed of 20 rad/s and angular acceleration of 10 rad/s , find at
an interval of 30° of the crank, 1. the linear velocity and acceleration of the slider, and 2. the
angular velocity and acceleration of the connecting rod.
Solution.
Given input :
A = 200, B = 750, E = 50, VA = 20, ACC = 10, THA = 30
OUTPUT :
T H B E T V S V B ACS AC B
.00 3.82 .27 – 5.32 – 101.15 – .78
30.00 – 3.82 – 2.23 – 4.61 – 83.69 49.72
60.00 – 9.46 – 3.80 – 2.63 – 35.62 91.14
90.00 – 11.54 – 4.00 .00 14.33 108.87
120.00 – 9.46 – 3.13 2.63 44.71 93.85
150.00 – 3.82 – 1.77 4.61 55.11 54.35
180.00 3.82 – .27 5.32 58.58 4.56
210.00 11.54 1.29 4.53 62.42 – 47.90
240.00 17.31 2.84 2.55 57.93 – 93.34
270.00 19.47 4.00 .00 30.28 – 113.14
300.00 17.31 4.09 – 2.55 – 21.45 – 96.14
330.00 11.54 2.71 – 4.53 – 75.44 – 52.61
25.6. Coupler Curves
It is often desired to have a mechanism to guide a point along a specified path. The path
generated by a point on the coupler link is known as a coupler curve and the generating point is
called a coupler point (also known as tracer point). The straight line mechanisms as discussed in
chapter 9 (Art. 9.3) are the examples of the use of coupler curves. In this article, we shall discuss
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1017
the method of determining the co-ordinates of the cou-
pler point in case of a four bar mechanism and a slider
crank mechanism.
1. Four bar mechanism
Consider a four bar mechanism ABCD with an
offset coupler point E on the coupler link BC, as shown
in Fig. 25.3. Let the point E makes an angle with BC
α
in the anticlockwise direction and its co-ordinates are E
(x , y ).
E E
γ
First of all, let us find the value of BD, and β .
From right angled triangle BB D,
1
Fig. 25.3. Four bar mechainsm
BB BB asin θ
11
tan
γ= = =
with a coupler point.
B D AD−− AB d a cosθ
11
sin
a θ
−
1
or
γ= tan
da−θ cos
22 22 2
and ( ) () ( ) () ( )
BD=+ BB B D=+ BB AD− AB
11 1 1
22
(sin ) ( cos )
=θad+−a θ
22 2 2 2
=θad sin++a cosθ− 2ad cosθ
22 2 2
(sin cos ) 2 cos
=θad + θ+−adθ
22
=+ ad−2c adosθ
Now in triangle DBC,
222
()()( BD+− BC CD)
cos(γ+β) =
... (cosine law of triangle)
2 BC × BD
22 2
fb+−c
=
2bf
22 2
fb+−c
−1
or γ+β= cos
2bf
22 2
fb+−c
−1
... (i)
∴ β= cos−γ
bf
2
Let us now find the co-ordinates x and y . From Fig. 25.3, we find that
E E
x== AEAB+BEA= B+BE ... ( BE = BE )
∵ 12 1
E2 1 12 1 1
=θae cos+ cos(α+β) ... (ii)
and y== EEEE+EEB= B+EE ... ( EE =BB )
∵
21 1
E2 21 1 1 1
=θae sin+ sin (α+β) ... (iii)
From the above equations, the co-ordinates of the point E may be determined if a, e, , α
θ
and β are known.
1018 Theory of Machines
2. Slider crank mechanism
Consider a slider crank mechanism with an offset coupler point E, as shown in Fig. 25.4.
Let the point E makes an angle α with BC in the anticlockwise direction and its co-ordinates are
E (x , y ).
E E
β
First of all, let us find the angle . From right angled triangle BC C,
1
BC BB−θ B C a sin− e
11 11 1
sinβ= = =
BC BC b
ae sinθ−
−1 1
... (iv)
∴ β=
sin
b
Now xA==E AB+BE=AB+BB
E1 1 11 1 2
=θae cos+ cos(α−β) ... (v)
and y== EEEB+BEB= B+BE
E1 12 2 1 2
=θae sin+ sin (α−β) ... (vi)
Fig. 25.4 Slider crank mechanism with
From the above equations, the co-ordinates of the
coupler point.
β
point E may be determined, if a, b, e, e , θ , α and
1
are known.
Note : When the slider lies on the X-axis, i.e. the line of stroke of the slider passes through the axis of
rotation of the crank, then eccentricity e = 0. In such a case equation (iv) may be written as
1
asin θ
−1
β= sin
b
25.7. Synthesis of Mechanisms
In the previous articles, we have discussed the computer-aided analysis of mechanisms, i.e.
the determination of displacement, velocity
and acceleration for the given proportions
of the mechanism. The synthesis is the
opposite of analysis. The synthesis of
mechanism is the design or creation of a
mechanism to produce a desired output
motion for a given input motion. In other
words, the synthesis of mechanism deals
with the determination of proportions of a
mechanism for the given input and output
motion. We have already discussed the
application of synthesis in designing a cam
(Chapter 20) to give follower a known
motion from the displacement diagram and
in the determination of number of teeth on
the members in a gear train (Chapter 13)
to produce a desired velocity ratio.
Roller conveyor.
In the application of synthesis, to
Note : This picture is given as additional information and is
the design of a mechanism, the problem
not a direct example of the current chapter.
divides itself into the following three parts:
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1019
1. Type synthesis, i.e. the type of mechanism to be used,
2. Number synthesis, i.e. the number of links and the number of joints needed to produce
the required motion, and
3. Dimensional synthesis, i.e. the proportions or lengths of the links necessary to satisfy
the required motion characteristics.
In designing a mechanism, one factor that must be kept in mind is that of the accuracy
required of the mechanism. Sometimes, it is possible to design a mechanism that will theoretically
generate a given motion. The difference between the desired motion and the actual motion produced
is known as structural error. In addition to this, there are errors due to manufacture. The error
resulting from tolerances in the length of links and bearing clearances is known as mechanical
error.
25.8. Classifications of Synthesis Problem
The problems in synthesis can be placed in one of the following three categories :
1. Function generation ; 2. Path generation ; and 3. Body guidance.
These are discussed as follows :
1. Function generation. The major classification of the synthesis problems that arises in
the design of links in a mechanism is a function generation. In designing a mechanism, the frequent
requirement is that the output link should either rotate, oscillate or reciprocate according to a
specified function of time or function of the motion of input link. This is known as function genera-
tion. A simple example is that of designing a four bar mechanism to generate the function y = f (x).
In this case, x represents the motion of the input link and the mechanism is to be designed so that
the motion of the output link approximates the function y.
Note : The common mechanism used for function generation is that of a cam and a follower in which the
angular displacement of the follower is specified as a function of the angle of rotation of the cam. The
synthesis problem is to find the shape of the cam surface for the given follower displacements.
2. Path generation. In a path generation, the mechanism is required to guide a point (called
a tracer point or coupler point) along a path having a prescribed shape. The common requirements
are that a portion of the path be a circular arc, elliptical or a straight line.
3. Body guidance. In body guidance, both the position of a point within a moving body
and the angular displacement of the body are specified. The problem may be a simple translation
or a combination of translation and rotation.
25.9. Precision Points for Function Generation
In designing a mechanism to generate a particular function, it is usually impossible to
accurately produce the function at more than a few points. The points at which the generated and
desired functions agree are known as precision points or accuracy points and must be located so
as to minimise the error generated between these points.
The best spacing of the precision points, for the first trial, is called Chebychev spacing.
According to Freudenstein and Sandor, the Chebychev spacing for n points in the range xx ≤≤x
SF
(i.e. when x varies between x and x ) is given by
S F
11 π−(2 j1)
xx=+()x−(x−x)cos
j SF F S ... (i)
22 2n
11 π−(2 j1)
=+() xx−×∆×xcos
SF
22 2
n
1020 Theory of Machines
where x = Precision points
j
∆ x = Range in xx=−x , and
FS
j = 1, 2, ... n
The subscripts and indicate start and finish positions respectively.
S F
The precision or accuracy points may be easily obtained by using the graphical method as
discussed below.
1. Draw a circle of diameter equal to the range ∆=xx− x .
FS
2. Inscribe a regular polygon having the number of sides equal to twice the number of
precision points required, i.e. for three precision points, draw a regular hexagon inside the circle,
as shown in Fig. 25.5.
3. Draw perpendiculars from each corner which intersect the diagonal of a circle at preci-
sion points x , x , x .
1 2 3
Now for the range 13 ≤≤xx, =1;x =3, and
SF
∆=xx− x= 31 − = 2
∴
FS
or radius of circle, rx =∆ /2= 2/2= 1
∆ x 2
xx=+r=x+ =12 + =
∴
2S S
22
∆ x
xx=−r cos30°=−x cos30°
12 2
2
2
=− 2 cos30°= 1.134
2
Fig. 25.5. Graphical method for
∆ x
determining three precision
xx=+r cos30°=x+ cos30°
and
32 2
2
points.
2
=+ 2 cos30°= 2.866
2
25.10. Angle Relationships for Function Generation
(a) Four bar mechanism. (b) Linear relationship between x and θ.
Fig. 25.6
Chapter 25 : Computer Aided Analysis and Synthesis of Mechanisms 1021
Consider a four bar mechanism, as shown in Fig. 25.6 (a) arranged to generate a function
y = f (x) over a limited range. Let the range in x is (x – x ) and the corresponding range in is
θ
F S
() θ−θ () yy − φ() φ−φ
. Similarly, let the range in y is ( and the corresponding range in is .
FS FS FS
The linear relationship between x and is shown in Fig. 25.6 (b). From the figure, we find
θ
that
θ−θ
FS
θ=θ + () xx − ... (i)
SS
xx −
FS
Similarly, the linear relationship between y and φ may be written as
φ−φ
FS
φ=φ + () yy −
SS ... (ii)
yy −
FS
An automatic filling and sealing machine.
Note : This picture is given as additional information and is not a direct example of the current chapter.
For n points in the range, the equation (i) and (ii) may be written as
θ−θ ∆θ
FS
θ=θ + () − =θ + () −
xx xx
jjSSSS j
− ∆
xx x
FS
φ−φ ∆φ
FS
φ=φ + () yy − =φ + () yy −
jjSSSS j
and
yy − ∆ y
FS
where j = 1, 2, ... n,
∆=xx− x; ∆θ = θ − θ ,
FS FS
∆=yy− y ; and ∆φ = φ −φ
FS FS
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