Mechanism and Machine Theory Vol. 19, No. 3, pp. 349-356, 1984 0094--114X/84 $3.00+.00 Printed in Great Britain. © 1984 Pergamon Press Ltd.

VELOCITY AND ACCELERATION ANALYSIS OF COMPLEX MECHANISMS BY GRAPHICAL ITERATION

N. RAKESH, T. S. MRUTHYUNJAYA and G. D. GIRISH Department of Mechanical Engineering, Indian Institute of Science, Bangalore 560012, India

(Received 1 December 1982)

Abstract--A simple graphical method is presented for velocity and acceleration analysis of complex mechanisms possessing low or high degree of complexity. The method is iterative in character and generally yields the solution within a few iterations. Several examples have been worked out to illustrate the method.

I. INTRODUCTION

In view of the fact that they are generally simpler and enable a better visualization of the working of the mechanism, graphical methods for kinematic analysis of plane linkage mechanisms have continued to serve a useful purpose from the pedagogical standpoint. The method most commonly employed involves drawing the velocity and acceleration diagrams based on a sequential application of the relative velocity and relative acceleration equations. A mechanism which cannot be solved by this method directly is referred to as a complex mechanism[l]. Confining our attention to graphical methods applicable for such mechanisms Rosenauer and Willis[2] presented a method which makes use of centros and auxiliary points for velocity analysis and the plan of relative normal accelerations for acceleration analysis. Hall and Ault[3] utilized the concept of auxiliary points for both velocity and acceleration analysis. Goodman[4] showed that the principle of inversion can be used to advantage in velocity and acceleration analysis of complex mechanisms. However, the method[4] fails when no inversion of a given complex mechanism turns out to be simpler. Modrey's[5] method resolves the analysis of complex mechanisms into a superposition of solutions of two or more simpler mechanisms. Other methods include the three-line construction [6], Carter's [7] method and the method of normal accelerations [8]. These are suitable essentially for low complexity mechanisms. The pa- per presents a method, believed to be new, for the velocity and acceleration analysis of low and high complexity mechanisms. The method is conceptually simple but iterative in character and in most cases the solution is obtained within a few iterations. Several examples have been worked out to illustrate the method.

2. BASIS OF THE METHOD

Let us consider some points, say, A, B, C, D and E belonging to the links of the complex mechanism such that there exists a closed path ABCDEA in the mechanism any two adjacent points in this path being joined by a rigid link (Fig. IA). Then we can write

down the following vector equations relating their velocities:

Vs = VA + Van (1)

Vc = Vs + Vcs = VA + VBA + Vca (2)

Vo = Vc + VDc = VA + VB,~ + Vcn + Voc (3)

Ve -- VD + V ~ = VA + VaA + VcB + V~c + V ~ (4)

Va = VE + VAE = VA + VsA + VcB + VDc + VED + V,~E. (5)

Hence from eqn (5) we get,

VBA + VCB + VDc + VED + VAt = 0. (6)

Similarly we can derive the following vector equation relating the relative accelerations of the points

aBA + acB + ape + aE~ + aAE = 0. (7)

Equation (6) implies that the relative velocity vectors Vs~, Vcs . . . . V~E form a dosed polygon, i.e. abcde must be a closed polygon in the velocity diagram where a, b, c, etc. represent the tips of velocity vectors VA, Vs, Vc., etc. respectively. In addition we know that the relative velocity vector for any two points on a rigid link is oriented at right angles to the line joining the two points in the configuration diagram. These facts can be utilized to determine the exact locations of a, b, c,. etc. in the velocity diagram by an interative procedure as shown in Fig. I(B). In this diagram, the directions along which points a, b, c, etc. lie are marked as a-line, b-line, c-line, etc. re- spectively. We start with an arbitrary point al on the a-line. We then locate b] as the intersection with the b-line of the line drawn from al at right angles to AB. In a similar manner we locate c~, dl, el and then a2. If a2 coincides with a~ points, a~, bl, el, dl and el represent the exact locations of a, b, c, d and e respectively. If a 2 does not coincide with a] we start the next iteration using a2 as the starting point and locating b2, c2 , . . . , etc. We continue this procedure till two consecutive locations of the starting point coincide. If the procedure does not result in con-

349

350

B

(A)

N. RAKESH et al.

X•-Iine ~ , ,~n e

(B)

Fig. I(A). Portion of a complex mechanism, (B) determination of velocities for the portion in (A) by graphical iteration.

vergence in the chosen sequence a - , b - , c - * d - - , e - , a , we follow the reverse sequence, i.e. a ~ e ~ d - * c - - * b - - * a and repeat the procedure.

A similar procedure can be employed to determine the accelerations of the points by making use of eqn (7) which implies that the relative acceleration vectors aaA, aca . . . . , a~e form a closed polygon and the fact that the relative acceleration vector for any two points on a rigid link is oriented to the line joining the two points in the configuration diagram at an angle which depends on the angular velocity of the link.

In the particular case of all the points considered belonging to the same link, the polygons of relative velocity vectors and relative acceleration vectors re- ferred to above reduce to the velocity and acceler- ation images respectively of the link.

3. APPLICATION OF THE METHOD TO SOME TYPICAL COMPLEX MECHANISMS

In this section the method of graphical iteration is illustrated by applying it to the kinematic analysis of several typical complex mechanisms.

3.1 Illustrative example 1 (adopted from[I] In the mechanism of Fig. 2(A) link 6 has an

instantaneous velocity of 45 in./sec and an acceler- ation of 400in./sec 2 in the directions indicated. Determine the velocities and accelerations of the remaining links. The proportions of the mechanism are: OzA = 4.75 in. 04B = 6.5 in. A B = 12 in. A C = 13.25in. B C = 2 i n . and CD = 14.5in.

Solution. The given velocity of point D is repre- sented in direction and magnitude by od (Fig. 2B). We now have to determine the triangle abc in the velocity diagram corresponding to the triangle A B C in Fig. 2(A). The a-, b- and c-lines are drawn as shown in Fig. 2(B). Any point on c-line is chosen as c~ the first trial location for c. In Fig. 2(B) d itself is chosen as c~. From c~ a line is drawn perpendicular to CA to intersect the a-line at a~ and from a~ a line is drawn perpendicular to A B to intersect the b-line

at bl. The line drawn from bl perpendicular to B C intersects the c-line at c2. Continuing this procedure the triangle abc is exactly determined in the fourth iteration. The velocity diagram is thus completed. The velocity of A from Fig. 2(B) is 68.8 in./sec. This agrees with the answer found in[l].

Figure 2((2) shows the construction for acceler- ation diagram. Here o ' d ' represents to scale the given acceleration of point D. All the normal acceleration components are calculated from the known velocities. From d' we lay out to scale the normal acceleration component a~D and draw a line at fight angles to this to represent the c'-line. The a ' - and b'-lines are similarly drawn as shown in Fig. 2(C). An arbitrary point on the a'-line is chosen as a;. In Fig. 2(C) the tip of normal acceleration component aM n is itself chosen as a;. From a; the normal acceleration com- ponent a~A is laid out and a line is drawn at right angles to this to intersect the b'-line at b;. The component a~a is laid out from b~ and a line drawn at fight angles to this intersects the c'-line at c;. Then from c~ the component a~c is laid out and the line drawn at right angles to this intersects the a'-line at at. Repeating this procedure we find that the acceler- ation diagram is completed in the fourth iteration. The tangential acceleration of A is found from the diagram as 389 in./sec 2. From this the angular accel- eration of link 2 is determined as 81.9 rad/sec 2, ccw. This agrees with the value found in[1] employing Goodman's method.

In this example the points considered for velocity and acceleration determination are three in number and they all belong to the same rink. Hence the triangles abc and a'b'c" reduce to the velocity and acceleration images respectively. For solving this category of problems the three-line construction[6] also makes use of the a-, b- and c-lines shown in Fig. 2(B) and the a'- , b '- and c'-lines shown in Fig. 2(C). The procedure of determining the exact location of abc and a'b'c" is however different.

In the following illustrative examples we confine

Analysis of complex mechanisms by graphical iteration

\ I3~S~ 13in

~ x V6 m,

__~o ~_ -,,,;,,-

oy~...- ~o, ~ne(J. O~A)

ao "~ ~ 'cLl ine

351

t y ° ~ b'(b', i

( C ) ~ a ' ( a~ )

, a ' - l ine

Fig. 2. Illustrative example-l: (A) configuration diagram, (13) velocity diagram and (C) acceleration diagram.

352 N. RAKFSH et al.

ourselves to the velocity analysis of the respective mechanisms since the acceleration analysis follows similar lines and does not pose any difficulty.

3.2 Illustrative example 2 The mechanism shown in Fig. 3(A) is driven by

link 6. We are required to find the velocities of the remaining links.

Solution. From the given angular velocity 0)6 of link 6 the velocity of point D is calculated and represented to scale by od in Fig. 3(B). The a-, b- and c-lines are drawn as shown in Fig. 3(B). Since the points consid- ered viz. A, B and C belong to the same link and all lie on a line a direct application of the method fails to give results. However, the method can be success- fully applied if we choose an additional point on the link. The additional point can be any point belonging to the link but not collinear with A and B. This should be distinguished from an auxiliary point em- ployed in the auxiliary point method[3]. The choice of an auxiliary point is subject to greater restrictions as it has to be an intersection of two auxiliary rays along which the velocity and acceleration com- ponents are known or can be determined easily.

After choosing the additional point X as shown in Fig. 3(A) we select any point on the c-line (Fig. 3B), say, d itself as cl. From cl a line is drawn perpendic- ular to CA (and CB) intersecting the a-line at a~ and the b-line at b~. The point xt is obtained as the intersection of the line from al perpendicular to A X and the line from bt perpendicular to BX. Then c2 is located as the intersection with the c-line of the line drawn from x~ perpendicular to XC. Continuing this procedure, we find that the velocity diagram is com- pleted in the fourth iteration.

3.3 Illustrative example 3 The six-link mechanism of Fig. 4(A) incorporating

sliding pairs is driven by link 2 (i.e. 02C) which rotates at constant angular velocity 0) 2 . It is required to find the velocities of the other links.

Solution. Let S be a point belonging to link 4 but coincident with C in the phase considered. The velocity of point C is represented to scale by oc in Fig. 4(B). From c a line parallel to the direction of motion of the sliding block 3 is drawn to represent the s-line. The a- and b-lines are drawn as shown in Fig. 4(B). As shown in the figure, the triangle abs in the velocity

D

J /V 2 / \ , V "q,

• I

I

(

I

~ c l )

~,A) a ~ 1 04B)

Fig. 3. Illustrative example-2: (A) configuration diagram, (B) velocity diagram.

Analysis of complex mechanisms by graphical iteration 353

0 6 oO I

5 B ,4

1

s - l i n e

b ( b T ~ b - line (.L 0 s B )

Fig. 4. Illustrative example-3: (A) configuration diagram, (B) velocity diagram.

diagram is exactly determined in the seventh iteration choosing c itself as the starting point s t and following the s - , b - - , a - , s sequence. The acceleration analysis of this mechanism can also be carried out by graph- ical iteration taking into account the presence of Coriolis component.

3.4 Illustrative example 4 (adopted f rom [1]) In the mechanism of Fig. 5(A) o~ 6 = 0.4 rad/sec and

0~-~ 1.5 rad/sec 2. Determine co 2 and ~'2- The propor- tions of the mechanism are: O206--- 15in., 02A = 02B = 3.5 in., O6E = 5.5 in., BD = ED = 9 in., E C = 4.5 in., D C = 10.5 in., angle AO2B = 135 °.

Solution. The same general procedure as employed in the previous examples is used here. As shown in Fig. 5(B) the polygon acdb in the velocity diagram can be determined exactly in the sixth iteration choosing the origin 0 itself as the starting point bj and following the b - - , d - , c - , a - - , b sequence. The velocity values given by Fig. 5(B) agree with those found in[l]. It may be noted that this mechanism is one of high degree of complexity[l]. As pointed out by Hirschhorn[1] it cannot be solved by the auxiliary

points method[3] directly but can be solved by the Goodman's[4] method considering an inversion of the mechanism. The solution of the mechanism for velocities and accelerations by the present method of graphical iteration, on the other hand, is straight- forward.

3.5 Illustrative example 5 (adopted f rom the exercises on p. 143 of[ l])

In the mechanism of Fig. 6(A), co 2 = 8 rad/sec, constant. Determine co6, (o7, ~6 and %. The propor- tions of the mechanism are: 02A = 1.5in., 03B = 2.5 in., A C = 6 in., BD = 8 in., CD = 6 in., C E = 4 in., pitch radius of gear 2 = 2 in., pitch radius of gear 3 --- 3 in., distance of 07 from 0203 = 11 in., angle A0203 -~ 120 °, angle 0 2 0 3 B = 150 °, angle DCE = 90 °, and CA makes 90 ° with 0203.

Solution. This mechanism is another example of one with a high degree of complexity[l]. Good- man's[4] method is not applicable here as no in- version of the mechanism is kinematically simpler. The solution of the mechanism by the present method is particularly simple as shown in the case of veloci-

354 N. ~ et al.

\ , ' % . °2 I/

(A) / c - l i n e (.L.EC)

~..~,c(c6)

[ /"4 /a s - ~ /" ~...~1 ~, \~,\ b-tine(J.O2B)

d-line(& ED)

a(a~>... ( B ) el_ line (J.O2A)~.

Fig. 5. Illustrative example-4: (A) configuration diagram, (B) velocity diagram.

ties in Fig. 5(B) where the triangle cde in the velocity diagram is obtained in the fourth iteration itself choosing an arbitrary point on the c-line as cl and following the c-- .e- .d- .c sequence. For verification the velocity diagram is also drawn by the auxiliary points method [3] in Fig. 6(C). The auxiliary point X of link 6 is chosen at the intersection of BD (line I) and O~E (line III) and auxiliary point Y of link 6 is chosen at the intersection of AC (line H) and 07E. Point x in the velocity diagram is found by simulta- neous solution of the equations

(V,~) I = (V~,)' = (VB)' (8)

( v x ) " = Or~) t' = C¢o,)" (9)

Here (Vx) ~ indicates the component of the velocity of point X along direction L Point y in the velocity diagram is similarly located on the basis of the equations

( v ~ ) " = (VE)" = (Vo,)" (10)

( V r ) m = ((Vc) m = (V~) m (11)

With x and y determined the points c, d and e are located by using the principle of velocity image. It

may be seen that the velocity diagram obtained by the auxiliary points method in Fig. 6C is identical to the one found by graphical iteration in Fig. 6(B).

3.6 Illustrative example 6 Figure 7(A) shows an eight-link mechanism in

which link 2 is the driving link. A velocity analysis of the mechanism is to be carried out.

Solution. Velocity of F is represented to scale by of in Fig. 7(B). The problem is to locate the points a, b, c, d and e in the velocity diagram. We draw the a-, b-, d- and e-lines as shown in Fig. 7(B). The c-line is not known. We can, however, use the fact that c is common to the triangles cab and ced in the velocity diagram. We start with any point el on the e-line. In the construction shown in Fig. 7(B)fitself is chosen as el. Point d~ is located as the intersection with the d-line of the line drawn from e~ perpendicular to ED. We simultaneously choose an arbitrary point on the a-line as a~ and locate b~ as the interaction with the b-line of the line drawn from a~ perpendicular to AB. Now ci is located as the intersection of the line from d~ perpendicular to DC and the line from bl perpen- dicular to BC. This procedure is continued from c~ following the c--*e~d~c sequence for the triangle ced and the c ~ a ~ b ~ c sequence for the triangle cab.

/ c-

IlneU

AC

)

Fig.

6.

Illu

stra

tive

exam

ple-

5 (A

) co

nfig

urat

ion

diag

ram

, (B

) ve

loci

ty d

iagr

am

by g

raph

ical

ve

loci

ty

diag

ram

by

aux

iliar

y po

ints

m

etho

d.

itera

tion,

(C

)

I e-lin

e (I

EF

)

\d-li

ne(lO

,D) _

/+

)

- .‘.

._,.

&

h_lin

sf

1 n-

RI,

Fig.

7.

Illu

stra

tive

exam

ple-

6 (A

) co

nfig

urat

ion

diag

ram

, (B

) ve

loci

ty

diag

ram

.

356 N. RAtC.~SH et al.

B

7

G

A( D

1

Fig. 8. An eight-link mechanism with only five-bar loops.

As shown in Fig. 7(B) these triangles are determined exactly and the velocity diagram is completed in the sixth iteration.

The foregoing examples illustrate the analysis of some typical complex mechanisms by the graphical iteration method. The method can similarly be ap- plied to the analysis of more complex mechanisms. For example, Fig. 8 shows an eight-link mechanism with only five-bar loops which as per Modrey[5] is of the highest degree of complexity. If link 2 is the input link we can draw the a-, c-, f - and g-lines in the velocity diagram and the problem reduces to a type considered under illustrative Example 6. If link 5 is the input link we can draw the b-, c-, d-, e- and g-lines and the f-line is not known. We can solve the mechanism by considering f as the common point between the polygons, for example, bcdefand bcdefg and following the sequence e -~d--*c ~ b --*f~e in one case and e - - , d ~ c ~ b ~ g - ~ f o e in the other case. Alternatively we can use the polygons gbf (sequence g ~ b ~ f - ~ g ) and bcdefg (sequence g ~ b ~ c ~ d ~ e -*f ~g).

4. CONCLUSIONS

The paper presents a graphical iterative method for kinematic analysis of complex mechanisms. Several examples have been worked out to illustrate the application of the method. The iterative nature of the method may seem to be a disadvantage. It is, how- ever, to be recognized that the element of trial and

error cannot be entirely eliminated while carrying out kinematic analysis of complex mechanisms by graph- ical methods. This is because of the fact that regard- less of the method used to find the velocities and accelerations the position determination in such mechanisms has to be carried out only by graphical trial and error. It is, however, surprising to find that with the exception of Hain [9] other texts do not even mention this fact. The advantages of the method presented in the paper are that it is conceptually simple and is applicable for mechanisms with any degree of complexity. As the illustrative examples show the method generally yields the solution within a few iterations.

REFERENCES

1. J. Hirschhorn, Kinematics and Dynamics of Plane Mech- andros. McGraw-Hill, New York (1962).

2. N. Rosenauer and A. H. Willis, Kinematics of Mech- an/sins. Associated General Publications, Sydney (1953).

3. A. S. Hall, Jr. and E. S. Ault, Machine Design 15, 120 (Nov. 1943).

4. T. P. Goodman, Trans. ASME, 80, 1676 (1958). 5. J. Modrey, J. AppL Mech., Trans. ASME, 7,6, 184 (1959). 6. J. S. Beggs, Mechan£sm, p. 46. McGraw-Hill, New York

(1955). 7. W. J. Carter, J. Applied Mechanics, Trans. ASME, 17,

142 (1950). 8. J. Hirschhorn Product Engng 32(19), 26 (1961). 9. K. Hain, Applied Kinematics, 2nd Edn, p. 54. McGraw-

Hill, New York (1967).

GESCHWIINDIGKEITS- UND BESCHLEUNIGUNGSANALYSE EBENER EDMPLEXER EDPPELGETRIEBE MITTELS

GRAFISCHER ITERATION

N. Rakesh, T. S. Mruthyunjaya und G. D. Girish

Kurzfassun~ - Es wird eine zeichnerische Methode zur Ermittlung der Geschwindigkeiten

und Beschleunigungen in ebenen komplexen Koppelge~rieben mit miederer eder h~herer Kom-

plexit~t - wie Koppelge~riebe ohne Gelenkviereck - vorgestellt. Die Methode ist itera~iv,

und im allgsmeinen erEib~ sich die LSsung dutch wenige Iteratloneno Sie wird mit der Me-

thode der Hilfsp~tmkte und andere~J in der Literatur vorgestellten Methoden verglichen.

Verschiedene ausgefihhrte Beispiele ±llustrieren die Handhabung dieser Methode.