Velocity Analysis | Mechanics Of Machines

Ch # 6

Velocity Analysis

Theory of Machines

Kinematics Roadmap Dynamics

Acceleration Velocity Position

Synthesis  Graphical Method  Algebraic Method  Analytical Method Theory of Machines

6.1 Velocity • The rate of change of position with respect to time • Position (R) is a vector quantity and so is velocity • Velocity can be angular or linear • Angular velocity will be denoted by  and linear as V

d  dt dR V dt

Theory of Machines

Pure Rotation • Link PA in pure rotation • RPA can be represented as a complex number in polar form RPA = pej • Where p is the scalar length of the vector • This can be easily differentiated to obtain

dRPA d  pje j  pje j dt dt  pj cos  j sin    p  sin   j cos 

VPA  VPA

The velocity VPA is referred to as an absolute velocity since it is referred to A, which is the origin of the Global coordinate axes in that system. In such case it can be referred to as VP, with the absence of second subscript implying reference to the global coordinate system

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Rotation + Translation • If the pivot A is not stationary • It has a known linear velocity VA as part of the translating carriage, link 3 • If  is unchanged the velocity of point P versus A will be the same as before, but VPA can no longer be considered an absolute velocity • If is now velocity difference and must carry the second subscript as VPA

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Velocity Difference • The absolute velocity VP must now be found from the velocity difference equation VPA = VP – VA • Rearranging: VP = VA + VPA

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Relative velocity • Two independent bodies P and A, are shown. • if their independent velocities VP and VA are known, their relative velocity VPA can be found from VPA = VP – VA

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6.2 Graphical velocity analysis • CASE I: two points in the same body: velocity difference • CASE II: two points in different bodies: relative velocity

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Graphical velocity analysis • For solving any velocity analysis problem graphically, two equations are needed

VP  VA  VPA V  v  r

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Example 6-1 • Given 2, 3, 4, 2, find 3, 4, VA, VB, VC by graphical method

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Graphical Velocity Analysis (3 & 4) • Given linkage configuration, 2, find 3 and 4 • It is known that VA and direction of VB and VBA are perpendicular to their corresponding links

VBA VB

VBA Direction VBA

VA VB

VB Direction

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Graphical Velocity Analysis (3 & 4)

• After finding VB and VBA, find 3 & 4 vB 4  BO4 vBA 3  BA

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Graphical Velocity Analysis (VC) • After finding 3 and 4, find VC • VC=VA+VCA

VC

Double Scale VCA

VA VCA VC

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6.3 Instant Centers • A point common to two bodies in plane motion, which has the same instantaneous velocity in each body. • The velocity center ij is the common point on i (or its extension) and on j (or its extension), which has the same velocity. • Both links sharing the instant center will have identical velocity at that point • Instant centers are sometimes called centros or poles • It takes two bodies or links to create an instant center (IC), one can easily predict the quantity of instant centers to expect from any collection of links • For a combination of n things taken r at a time is

nn  1n  2 ......n  r  1 C r! for our case r  2 and its reduces to : n(n  1) C 2

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Instant Centers • Kennedy’s rule: any three bodies (links) will have exactly three instant centers and they will lie on the same straight line • The pins are instant centers • I13 is from links 1,2,3 and 1,3,4 • I24 is from links 1,2,4 and 2,3,4

I24

I13

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Instant centres in pin jointed linkage

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Instant Centres (Linear Graph) • A linear graph is useful for keeping track of which IC’s have been found • Can be created by drawing a circle on which we mark off as many points as there are links in assembly • Lines are then drawn between the dots representing the links pair each time an instant circle is found • The resulting linear graph is a set of lines connecting the dots • The circle is not included which was used to place the dots

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Rectilinear slider’s instant centre at infinity

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Instant Centres in the slider-crank linkage

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Velocity Analysis using instant centres

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Velocity Analysis using instant centres • The magnitude of VA can be computed as 2 and length of link 2 are known • Note that point A is also instant centre I2,3 • It has the same velocity as part of the link 2 and as part of link 3 • Since link 3 is effectively pivoting about I1,3 at this instant, the angular velocity 3 can be found by 3=vA/AI1,3 • Once 3 is known, the magnitude of VB can also be found; vB = (BI1,3)3 • Once VB is known, 4 can also be found as 4=vB/BO4 • Finally the magnitude of VC (or any other point on the coupler) can be found vC = (CI1,3)w3 Theory of Machines

Velocity Analysis using instant centres • Since the location of I1,3, which is instantaneous “fixed” pivot for link 3, all of that link’s absolute velocity vectors for this instant will be perpendicular to their radii from I1,3 to the point in question • VB and VC can be seen to be perpendicular to their radii from I1,3 • VB is also perpendicular to the radius from O4 because B is also pivoting about that point as part of link 4 • Also the magnitude of velocities of VB and VC can be found by drawing arcs from point B and C to line extended from I1,3 I1,2 and then extending their tips to line I1,3 VA Theory of Machines

Analytical Solutions for velocity analysis: The fourbar Pin-Jointed Linkage

Theory of Machines

Analytical Solutions for velocity analysis: The fourbar Pin-Jointed Linkage • The vector loop equation is R2+R3-R4-R1 = 0 • As before, substitute the complex number notation for the vectors, denoting their scalar length as a, b, c, d aej2+bej3-cej4-dej1=0 To get an expression for velocity, differentiate this equation w.r.t. time j  d 3 j  d 2 j d 4 jae

2

but d d 2 d   2 ; 3  3 ; 4   4 dt dt dt and : ja2 e j 2  jb3e j3  jc4 e j 4  0

dt

 jbe

3

dt

 jce

4

dt

0

Theory of Machines

Analytical Solutions for velocity analysis: The fourbar Pin-Jointed Linkage

• 1 term has dropped out because that angle is a constant, and thus its derivative is zero • The equation is infact the relative velocity or velocity difference equation VA + VBA – VB = 0

V A  ja 2 e

j 2

VBA  jb3e VB  jc 4 e

j 3

j 4 Theory of Machines

Analytical Solutions for velocity analysis: The fourbar Pin-Jointed Linkage

ja2 e j 2  jb3e j3  jc4 e j 4  0 Expand the complex part

ja2 (cos 2  j sin  2 )  jb3 (cos3  j sin 3 )  jc4 (cos 4  j sin  4 )  0

Multiply by j

a2 ( j cos 2  j 2 sin  2 )  b3 ( j cos 3  j 2 sin  3 )  c4 ( j cos 4  j 2 sin  4 )  0

a2 ( j cos 2  sin  2 )  b3 ( j cos3  sin 3 )  c4 ( j cos 4  sin  4 )  0 Separate the real and imaginary part

 a2 sin  2  b3 sin 3  c4 sin  4  0 a2 cos 2  b3 cos3  c4 cos 4  0 Solve for 3 and 4

3 

a2 sin( 4   2 ) b sin( 3   4 )

a2 sin( 2   3 ) 4  c sin( 4   3 )

Theory of Machines

Analytical Solutions for velocity analysis: The fourbar Pin-Jointed Linkage • Once 3 and 4 have been sorted, linear velocities

VA  ja2 (cos 2  j sin  2 )  a2 ( sin  2  j cos 2 ) VBA  jb3 (cos3  j sin 3 )  a3 ( sin 3  j cos3 ) VB  ja4 (cos 4  j sin  4 )  a4 ( sin  4  j cos 4 ) • There are also two solutions to this velocity problem, corresponding to open and cross branches of the linkage. They are found by the substitution of the open or crossed branch values of 3 and 4 obtained in chapter 4

Theory of Machines

The fourbar Slider-Crank • R2 – R3 – R4 – R1 = 0 • aej2-bej3-cej4-dej1=0 • a, b, c, 1 and 4 are constant but the length of link d varies with time .

ja2 e j 2  jb3e j3  d  0

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The fourbar inverted Slider-Crank ja2 e

j 2

 jb3e

j 3



 b e j3  jc4 e j 4  0

3   4   Differentiate it w.r.t time

3   4

Theory of Machines

Velocity analysis of the geared fivebar linkage      R2  R3  R4  R5  R1  0 ae j 2  be j3  ce j 4  de j5  fe j1  0 a2 je j 2  b3 je j3  c4 je j 4  d5 je j5  f1 je j1  0

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Problem 6-16

a) Graphical method b) Instant center method c) Analytical method

Theory of Machines

Problem 6-8(a)

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Assignment session + Quiz Attempt all related problems

For Assignment session: 6-4 to 6-12, 6-16 to 6-18

Theory of Machines