`
`MECHANISM DESIGN;
`
`Analysis and Synthesis
`
`ARTHUR G. ERDMAN
`
`GEORGE N. SANDOR
`
`PRENTICE-HALL, INC., Englewood Cliffi', New Jersey 07632
`
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`Library of Com Cataloging in Publication Data
`Emum. Annual G.
`(date)
`Muhunisrn dedgn.
`Includes bibliognphim and indexes.
`l. Mnehincry—Dmign.
`I. Sandor, George N.
`II. Title.
`621.3'15
`TJZ30.E67 1984
`ISBN 0—13—572396—5 (v. 1)
`
`83—3148
`
`Editorial/production supervision
`and interior design: Karen Skrable
`Manufacturing buyer: Anthony Caruso
`Cover design: Photo Plus Art, Celine Brandes
`
`© 1984 by Arthur G. Erdmnn and George N. Sandor
`
`Volume 2 pubiishcd under the title Advanced Mechanism Design:
`Analysis and Synthesis. Voi. 2.
`
`All rights reserved. No part of this book
`may be reproduced in any form or
`by any means without permission in writing
`from the pnbiisher.
`
`Printed in the United States of America
`
`10987654321
`
`ISBN D-l3-57E'EI‘II2-5
`
`PRENTICE-HALL INTERNATIONAL, INC., London
`PRENTICE-HALL OF AUSTRALIA PTY. LIMITED, Sydney
`EDITOFIA PRENTICE-HALL DO BFIASIL, LTDA.. Rio do Janeiro
`PRENTICE—HALL CANADA INC., Toronto
`PRENTICE—HALL OF INDIA PRIVATE LIMITED, New Delhi
`PRENTICE-HALL OF JAPAN, |NC., Tokyo
`PRENTICE-HALL OF SOUTHEAST ASIA PTE. LTD.. Singapore
`WHITEHALL BOOKS LIMITED, Wellington, New Zeaiand
`
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`Contents
`
`PREFACE
`
`INTRODUCTION TO KINEMATICS AND MECHANISMS
`1.1
`Introduction
`1
`1.2
`Motion
`2
`1.3
`1.4
`1.5
`1.6
`1.7
`1.8
`
`2
`The Four-Bar Linkage
`The Science of Relative Motion
`
`5
`
`Kinematic Diagrams
`Six-Bar Chains
`10
`
`5
`
`Degrees of Freedom 16
`Analysis versus Synthesis
`Problems
`25
`
`24
`
`MECHANISM DESIGN PHILOSOPHY
`2.1
`Introduction
`49
`2.2
`2.3
`2.4
`2.5
`2.6
`
`The Seven Stages of Engineering Design
`How the Seven Stages Relate to This Text
`The Mechanism Synthesis Process
`58
`6]
`Design Categories and Mechanism Parameters
`Troubleshooting Guide: Symptoms, Causes, and Sources of Assistance
`
`51
`
`56
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`1
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`49
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`Contents
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`68
`
`175
`
`218
`
`271
`
`vi
`
`3.2
`3.3
`3.4
`3.5
`3.6
`3.7
`3.8
`
`3.9
`
`83
`
`105
`
`DISPLACEMENT AND VELOCITY ANALYSIS
`3.1
`Displacement Analysis: Useful Indices
`for Position Analysis of Linkages
`68
`Displacement Analysis
`75
`Concept of Relative Motion
`Velocity Analysis
`84
`Instant Centers
`96
`Velocity Analysis Using Instant Centers
`Mechanical Advantage
`110
`Analytical Method for Velocity and Mechanical
`Advantage Determination
`121
`Computer Program for the Kinematic Analysis
`of a Four-Bar Linkage
`126
`Appendix: Review of Complex Numbers
`Problems
`142
`Exercises
`174
`
`134
`
`‘
`
`ACCELERATION ANALYSIS
`4.1
`Introduction
`175
`4.2
`176
`Acceleration Difference
`4.3
`180
`Relative Acceleration
`4.4
`185
`Coriolis Acceleration
`4.5
`Mechanisms with Curved Slots
`Problems
`202
`
`201
`
`INTRODUCTION TO DYNAMICS OF MECHANISMS
`5.1
`Introduction 218
`5.2
`5.3
`5.4
`5.5
`5.6
`5.7
`
`223
`Inertia Forces in Linkages
`Kinetostatic Analysis by Complex Numbers
`The Superposition Method
`233
`Design Example: Analysis of a Variable-Speed Drive
`The Matrix Method 250
`Discussion of the Superposition and Matrix Approach
`to Kinetostatics
`257
`Problems
`259
`
`232
`
`240
`
`CAM DESIGN
`6.1
`Introduction 271
`6.2
`6.3
`6.4
`6.5
`
`Cam and Follower Types
`Cam Synthesis
`276
`278
`Displacement Diagrams
`Advanced Cam Profile Techniques
`
`272
`
`291
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`Contents
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`6.6
`6.7
`6.8
`6.9
`
`302
`Graphical Cam Profile Synthesis
`306
`Analytical Cam Profile Synthesis
`Cam Synthesis for Remote Follower
`319
`Cam-Modulated Linkages
`322
`Problems
`331
`
`345
`
`GEARS AND GEAR TRAINS
`7.1
`Introduction 340
`7.2
`Gear Tooth Nomenclature
`7.3
`7.4
`7.5
`7.6
`7.7
`7.8
`7.9
`
`Forming of Gear Teeth
`Gear Trains
`350
`
`346
`
`353
`Planetary Gear Trains
`356
`The Formula Method
`The Tabular Method 361
`
`364
`The Instant Center Method (or Tangential Velocity Method)
`Tooth Loads and Power Flow in Branching Planetary Gear Systems
`Problems
`379
`
`Vii
`
`340
`
`369
`
`391
`
`418
`420
`421
`423
`
`INTRODUCTION TO KINEMATIC SYNTHESIS:
`GRAPHICAL AND LINEAR ANALYTICAL METHODS
`8.1
`Introduction
`391
`8.2
`8.3
`8.4
`8.5
`8.6
`8.7
`8.8
`8.9
`
`394
`Tasks of Kinematic Synthesis
`Number Synthesis: The Associated Linkage Concept
`Tools of Dimensional Synthesis
`417
`Graphical Synthesis—Motion Generation: Two Prescribed Positions
`Graphical Synthesis—Motion Generation: Three Prescribed Positions
`Graphical Synthesis for Path Generation: Three Prescribed Positions
`Path Generation with Prescribed Timing: Three Prescribed Positions
`Graphical Synthesis for Path Generation (without Prescribed Timing):
`Four Positions
`424
`. ' '
`
`406
`
`8.10
`8.11
`8.12
`8.13
`8.14
`8.15
`8.16
`
`8.17
`8.18
`
`Function Generator: Three Precision Points
`
`427
`
`429
`The Overlay Method
`43]
`Analytical Synthesis Techniques
`Complex-Number Modeling in Kinematic Synthesis
`The Dyad or Standard Form 434
`Number of Prescribed Position versus Number of Free Choices
`
`432
`
`436
`
`Three Prescribed Positions for Motion, Path, and
`Function Generation
`439
`
`Three-Precision-Point Synthesis Program for Four-Bar Linkages
`Three-Precision-Point Synthesis: Analytical versus Graphical
`452
`
`445
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`viii
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`8.19
`8.20
`8.21
`8.22
`8.23
`8.24
`
`Extension of Three-Precision-Point Synthesis to Multiloop Mechanisms
`Circle-Point and Center-Point Circles
`456
`
`454
`
`Contents
`
`464
`Ground-Pivot Specification
`Freudenstein’s Equation for Three-Point Function Generation
`Loop-Closure-Equation Technique
`472
`475
`Order Synthesis: Four-Bar Function Generation
`Appendix: Case Study—Type Synthesis of Casement Window Mechanisms
`Problems
`499
`
`469
`
`478
`
`519
`
`527
`
`REFERENCES
`
`INDEX
`
`'
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`110
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`Displacement and Velocity Analysis
`
`Chap. 3
`
`3.7 MECHANICAL ADVANTAGE
`
`One of the major criteria of which a designer must be aware is the ability for a
`particular mechanism to transmit torque or force.
`Some mechanisms, such as a
`gear train, transmit a constant torque ratio between the input and output because
`there is a constant speed ratio between input and output (see Chap. 7).
`In a linkage,
`however, this is not the case. How might we determine a relationship between force
`out and force in? Two observations can be made without further analysis.
`
`1. As hinted in the gear train mentioned above, the torque ratio is a function of
`the speed or angular velocity ratio of the mechanism.
`2. The torque ratio is a function of geometric parameters, which, in the case of
`a linkage, will generally change during the course of the ‘mechanism motion.
`
`If we assume that a mechanism is a conservative system (i.e., energy losses
`due to friction, heat, etc., are negligible compared to the total energy transmitted
`by the system), and if we assume that there are no inertia forces, power in (Pin) is
`equal to power out (Pout). Thus the torque in times the angular velocity in is equal
`to the torque out times the angular velocity out:
`
`01'
`
`Pin = Tinwin : Toutwout = Pout
`
`Pin = Fin ' Vin = Font ' Vout = Pout
`
`(363)
`
`(3'64)
`
`where Tin and Fin are torque and force exerted on the linkage, and Tout and Font
`are those exerted by the linkage; where Vin and Vout are the velocities of the points
`through which Fm and Font, respectively, act; and where
`
`V - F = VF cos (arg F — arg V)
`
`(3.64a)
`
`Also,
`
`V - F = VIFI + VyFy
`
`(3.64b)
`
`(For a proof, see Exer. 3.4.)
`Notice that the units of torque times angular velocity and the scalar product
`of force and velocity both represent power. From Eq. (3.63),
`
`Tou
`in
`‘= ‘”
`Tin
`wout
`
`(3.65)
`
`By definition, mechanical advantage (M.A.) is the ratio of the magnitudes of
`force out over force in:
`
`
`
`MA. =
`
`(3.66)
`
`where F = lFl.
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`Sec. 3.7
`
`Mechanical Advantage
`
`111
`
`
`
`torque, and power
`Figure 3.59 Force,
`transmission ratios of a four-bar mecha-
`nism change in the course of motion.
`
`Referring to Fig. 3.59 as an illustration, and noting that
`
`Tin = zinIFiny — Zing/Fin:
`
`and
`
`Tout = zouttFouty — zoutyFoqu
`
`where the right sides represent the cross or vector product z X F in this planar,
`complex-number representation, bracketed to indicate that it is the signed (positive
`or negative) real,
`
`(see Fig. 3.59). Also,
`
`[z X F] = zF sin (arg F — arg z)
`
`[z X F] = [IFy — lyFI,
`
`(3.64c)
`
`(3.64d)
`
`where l is a vector perpendicular to the line of action of F drawn from the fulcrum
`to that line of action.
`(For a proof, see Exer. 3.5.)
`In general, therefore, the following
`relationship for mechanical advantage develops (see Fig. 3.59):
`
`
`————_—___ _
`Tout _ znutFout Sin (arg Foul. _ arg zuut) _ win
`_
`Tm
`zinFin sin (arg Fin — arg 23,)
`(”out
`
`(
`
`3.67
`
`)
`
`Thus
`
`M.A. = Fout =
`F1
`
`
`Zin sin (arg Fin — arg zin) win
`zout Sln (arg Font — arg zout) wont
`
`Vin cos (arg Fin — arg Vin)
`
`Vout cos (arg Font — arg Vout)
`
`(3.68)
`
`where the last form is derived from Eqs. (3.64) and (3.643.). Also, if Rm and Rout
`are vectors from the pivots of input and output links, respectively, to the line of
`action of forces Fin and Font, then
`
`
`
`Tout _ [Rout X Font]
`Ti _ Rin X Fin
`Note that, if Rm and Rout are perpendicular to Fin and Fout respectively,
`
`(3 67a)
`
`and
`
`Rm = zin sin (arg Fin — arg zin)ei(“3 Fin—7””
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`Displacement and Velocity Analysis
`
`Chap. 3
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`Rout: zout sin (arg Fout — arg zout)e““g Four”)
`
`which agrees with Eq. (3.67).
`Thus the mechanical advantage is a product of two factors: (1) a ratio of distances
`that depend on the placement of the input and output forces and (2) an angular
`velocity ratio. The first factor may not change in value as the mechanism moves,
`but the second one will change in most linkage mechanisms.
`Since the angular
`velocity ratio can be expressed entirely in terms of directed distances (based on the
`instant center development), the mechanical advantage can be expressed entirely in
`terms of ratios of distances.
`
`Expressions in the form of Eq. (3.68) are powerful design tools and can usually
`be verified by inspection.
`In many design situations, the mechanical advantage expres-
`sion for a mechanism will allow the optimal redesign of that device for improved
`mechanical advantage. Practical considerations such as the maximum permitted size
`of the mechanism will usually limit the amount of change allowable from an original
`design (see Chap. 8 appendix).
`For example, suppose that the four-bar linkage of Fig. 3.60 is being used as
`the driving mechanism of a manually operated pump.
`In the position shown, the
`handle is being pulled left with force Fin. Meanwhile the pressure difference across
`the piston in the cylinder is resisting the movement by a force equal and opposite
`to Font. What is the mechanical advantage of this device in the position shown?
`(The piston rod is instantaneously perpendicular to link 4.)
`If the input link is identified as link 2 and the output link as link 4, then
`according to Eq. (3.65),
`
`B _ 9.2:
`T2
`m4
`
`3.69
`
`)
`
`(
`
`Instant center (2,4) is found to lie between (1,2) and (1,4), so that
`
`Figure 3.60 Mechanical advantage of this four-bar handpump drive
`mechanism increases as the toggle position (A0, A, and B collinear) is
`approached.
`
`(3,4)
`
`B a
`
`
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`Sec. 3.7
`
`Mechanical Advantage
`
`113
`
`retically infinite: a small force at P can
`
`Figure 3.61 Toggle position of the pump-
`drive mechanism of Fig. 3.60: M.A. is theo-
`
`1
`
`1
`
`overcome a very large force at Q.
`
`5 _ g _ (1,4 — 2,4)
`2
`“’4
`(1,2 — 2,4)
`
`Therefore,
`
`and
`
`
`E = ZontFout Sin (Mg Fout _ 3T3 zuut)
`T2
`zinFin sin (arg Fin — arg zin)
`
`
`
`(3.70)
`
`
`
`MA, = 1:3“:zz‘" M W (171)
`in
`out
`(1’2 _ 2,4)
`g out
`a g
`1;
`
`
`
`l
`
`a
`
`_
`
`zou
`
`Notice the factors that make up the mechanical advantage. The mechanical advantage
`is greater if the ratio zin/zout is greater. This checks intuitively with Fig. 3.60.
`The second ratio can be checked intuitively also. Notice that as point P is moved
`to the left to the position shown in Fig. 3.61, where the four-bar linkage is in its
`“toggle” position, centers (1,2) and (2,4) coincide. According to the expression [Eq.
`(3.71)], the mechancial advantage goes to infinity in this position.
`Since links 2
`and 3 line up, (ideally) no force is required at P to resist an infinite force at Q.
`Of course, bending of link 4 would occur before an infinite force could be applied.
`The ratio of sines is a measure of the relative closeness to perpendicularity of each
`force to its arm vector. With these considerations, Eq. (3.71) is intuitively verified.
`If more mechanical advantage is required with AoP in the position shown in
`Fig. 3.60, then Eq. (3.71) dictates the following possibilities:
`
`9995-893:—
`
`Increase 2m.
`
`Decrease zout.
`
`Move Bo away from (2,4) (keeping the rest of the linkage the same).
`Move A0 toward (2,4) (keeping the rest of the linkage the same).
`Move point A until links 2 and 3 line up.
`Make Fin perpendicular to Z1".
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`114
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`Example 3.8
`
`Displacement and Velocity Analysis
`
`Chap. 3
`
`Determine the mechanical advantage of the adjustable toggle pliers in Fig. 3.62.
`Why is this device designed this way?
`
`Let us designate input link = 3, output link = 4, and ground link = 1 (see
`Solution
`Fig. 3.63). Thus Tacos = Tum; since F0“, is perpendicular to rout and Fin is perpendicular
`to I'm,
`
`T4 = [rout X Font] = routFout $111 (arg Foul: _ al'g 1'out) = routFout Sin (_900)
`
`= —l.9Fout
`
`Similarly, T3 = [rin X Fin] = (5.1)F,n and
`
`
`
`$201,4—l,41-)=—-1.6=_2_29
`(1)4
`0.7
`(3,4 — 1,3)
`
`‘
`‘
`
`from which
`
`M.A. =
`
`
`
`Font
`rin (1)3
`l5.1
`
`
`=
`— = —‘ —2.29 = .15
`Fin
`rout 0’4
`1-9 (
`)
`6
`In the position shown, the pliers have only a 6.15:1 mechanical advantage. As we
`clamp down on the pliers, however,
`instant center (2,4) approaches center (1,2).
`
`ity.
`
`Figure 3.62 See Ex. 3.8. The stop prevents
`excessive overtravel (branching) beyond the
`toggle position of the upper handle and the
`coupler link.
`
`Figure 3.63 As (3,4), (2,3), and (1,2) ap-
`proach collinearity, M.A. approaches infin-
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`Sec. 3.7
`
`Mechanical Advantage
`
`115
`
`When (1,2), (2,3), and (3,4) are close to being in a straight line, (2,4) is nearly coincident
`with (1,2), (1,3) approaches (3,4) and the mechanical advantage approaches infinity.
`The screw adjustment should be set so that the maximum mechanical advantage
`occurs at the required distance between the jaws of the pliers.
`In fact, in some brands,
`there is a stop located in the “over center” position (just past the “dead center”), as
`shown in Fig. 3.62. This gives both a very high mechanical advantage and stable “grip”
`for the linkages, since it would take an ideally infinite force at the jaws to move the
`linkage back through its toggle position.
`
`Example 3.9
`
`Determine the mechanical advantage of the slider-crank mechanism shown in Fig. 3.64.
`
`If link 2 is considered the input link and link 4 the output link, the procedure
`Solution
`described above must be modified (because link 4 is a slider so that an = 0). We
`know that (power),n = (power)out. Thus
`
`T202 = Flout ° VB
`
`(3.72)
`
`where T2 = [rin X Fin]. Notice that since the output link 4 is constrained to translate
`in the horizontal slot, any point considered as part of link 4 must have a velocity in
`the horizontal direction. Moreover, any point considered as part of link 4 has the
`same velocity, including the point of the extended plane of link 4 momentarily coincident
`with the center (2,4). This point has velocity iw2(1,2 — 2,4); therefore,
`
`
`V5 = iw2(1,2 — 2,4)
`
`(3.73)
`
`Combining (3.72) and (3.73), from (power in) = (power out), we get
`
`[fin X Fin]w2 = Tzwz = Fa ' VB
`
`
`Fa - (iw2(1,2 — 2,4)) = Fa |w2(1,2 — 2,4)|cos (arg Fa — arg VB)
`
`
`wzrinFm sin (arg Fin — arg rm) = Fa |w2(1,2 — 2,4)|cos (arg Fa - arg VB)
`
`from which, noting that
`
`'
`
`Fa cos (arg Fa — arg V15)= Font
`
`2
`and m = —1
`
`Figure 3.64 Geometric constructions to-
`ward finding the M.A. of a slider-crank
`mechanism.
`
`
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`116
`
`we have
`
`Displacement and Velocity Analysis
`
`Chap. 3
`
`
`MA = 123m = _MW =_in_
`‘“
`|(1,2 — 2,4)|
`|(1,2 — 2,4)|
`
`(3.74)
`
`which is positive (as it should be) because sin (arg Fm - erg rm) is negative.
`This expression makes sense: The longer the arm rm on the input link, provided
`that its direction remains the same, the higher the mechanical advantage. Note also
`that, as the input link rotates cw, point A moves toward the toggle position and center
`(2,4) moves toward (1,2)—increasing the mechanical advantage.
`
`Input torque Tin = [rm X Fin] cw.
`Another verification of Eq. (3.74) follows.
`Tm is resisted by the moment of a pin force FE. at joint A.
`Since link 3 is a
`“two-pin” link, E} must act along link 3.
`Its resisting momentjs [(0,2) — A') X F3]
`(cow) where ((1,2) — A’) is perpendicular to link 3. Link 3 transmits F; to slider
`4 at (3,4), where it is resisted by Fa . The vertical component Fm, of Fa is perpendicular
`to the motion of output link 4 and therefore does no work. Thus, the output force
`Font is the horizontal component of Fa, pointing to the right in Fig. 3.64. By similar
`triangles it is easy to show that
`
`where, for instance,
`
`But
`
`
`Font _ (1,2) — A'
`Fa
`(1,2 - 2,4)
`
`
`
`(1,2) — A’ = [(1,2) — A'l
`
`lin
`Fa
`— = ——
`Fin
`(1,2) - A’
`
`3.75
`
`(
`
`)
`
`3.76
`)
`
`(
`
`and by multiplication of these two ratios we find that
`
`Fout _
`Fin
`
`[in
`(1)2 _ 294)
`
`Which agrees with Eq. (3.74).
`Notice that link 3 could also be considered as the output link because point
`B is also a part of the connecting rod.
`In this case centers (1,2), (2,3), and (1,3)
`would be used to obtain an expression for mechanical advantage that would be numeri-
`cally equivalent to Eq. (3.74) (see Exer. 3.6).
`For still another verification of Eq. (3.74), as well as for the verification of
`the graphical (geometric) construction to determine the mechanical advantage, we
`resort to free-body diagrams as follows. For the free-body diagram of link 2 (see
`Fig. 3.65), summing moments about A0 yields
`
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`Sec. 3.7
`
`Mechanical Advantage
`
`117
`
` Figure 3.65
`
`rmXFm+r2X F1, =0
`
`rinF1n sin Bin + rth', sin [3,, = O
`
`rmFin sin (arg Fin — arg rm) + rzFé sin (arg F"a — arg r2) = 0
`
`where we expressed 31,. and ,BA in terms of the arguments of the vectors involved.
`Note that .81“ is cw, and therefore negative, which checks with the fact that
`the input torque is cw. Also, rm sin [31,, is the perpendicular arm of Fin about A0.
`and similarly, r2 sin [32 is that of F; about A0. Solving for F;,
`
`I
`sin gin
`_
`5m B;
`
`because
`
`< 0
`
`/‘
`
`Fa
`
`rinFin sin .3121
`——— > 0
`,
`r2 $111 82
`
`Continuing in this manner, Eq. (3.74) is verified in still another way (see Exer. 3.7).
`
`Example 3.10
`
`(a) Find the location of
`A six-link function-generator linkage is shown in Fig. 3.66.
`all the instant centers for this mechanism; (b) If the velocity of point P is known to
`
`Figure 3.66 Six-link force transmission mechanism. See Ex. 3.10.
`
`
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`Mylan Exhibit - 1033
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`
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`"8
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`Displacement and Velocity Analysis
`
`Chap. 3
`
`be 10 m/sec, find as and V3 by the instantaneous center method; (0) If force Fm is
`acting at P (see Fig. 3.66), find F0“, by instant centers.
`Solution
`
`(a) Figure 3.67 shows the location of all the instant centers for this mechanism.
`(b) Method 1: Figure 3.68 shows how the instant center graphical technique can be
`used to solve for an and VB. From this figure
`
`VA = 17.0 m/see
`
`V,
`(1)3—
`
`1'(1,3 — 2,3)
`
`V2.4 = 39 m/sec
`
`17.0
`= m = 321 rad/sec
`
`VB =(15 m/sec)i(—eilars<BoB>1)
`
`‘
`
`Method 2: Using instant center equations only (Fig. 3.68), since
`
`
`
`E = —(1’2_ 2’3) = —1.19
`“’2
`(1,3 — 2,3)
`
`(02 =i = ;—-v; = —270.3 rad/sec (cw)
`rtAoP)
`°
`
`we =
`
`
`_. VP
`(—1.l9) = 321.6 rad/sec (ccw)
`Au.”
`
`1 2 —
`004 = anM = (—270.3)(3.10)= —838.0 rad/sec (cw)
`(194 _ 2,4)
`
`then
`
`Thus
`
`v1, = iw4(BoB) = (17.2 m/sec)i(—eiarsw‘o§))
`
`(c) Method 1: Using instant centers (1,2), (1,5), and (2,5), (Fig. 3.67):
`
`
`g :92 = (1,5 — 2,5) = _0.054 — = —0.844
`T2
`(05
`(1’2 _ 25)
`0.064
`
`I
`
`and (Figs. 3.66 and 3.67):
`
`then
`
`T5 = Fout(0.048);
`
`T2 = —F1n(0.03 1)
`
`
`_
`Fin
`_
`Fm — (0.043} (0.031)(—0.844) — 0.5451:1n
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`Mylan Exhibit - 1033
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`Mylan V. Sanofi
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`Mylan Exhibit - 1033
`Mylan v. Sanofi
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`.EmEEooExii”?5E850ESE:5d95!...—
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`Mylan Exhibit - 1033
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`119
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`Mylan V. Sanofi
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`120
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`Displacement and Velocity Analysis
`
`Chap. 3
`
`(1,5)
`
`.
`
`(5'5)
`C
`
`1 —> (1,6) =0
`
`e,
`
`.‘x\\
`1
`
`P"
`P
`V"
`
`\
`
`\‘x
`
`.
`30 h‘
`k \
`1
`
`Via
`
`Figure 3.68 Velocity construction using instant centers and gage lines in six-link
`mechanism.
`
`Method 2: Using instant centers (1,2), (1,6), and (2,6), from Eq. (3.74),
`
`_
`
`
`Fm— (1,2—6,2)|
`
`T2
`
`then
`
`—F 0.031
`Fout=|——ggs7) =0.541~"1n
`
`.
`
`'-
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`Mylan Exhibit - 1033
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`Mylan V. Sanofi
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`.
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`Mylan Exhibit - 1033
`Mylan v. Sanofi
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