Class-4-Velocity-Analysis1 (1)

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Transcript Class-4-Velocity-Analysis1 (1)

Mechanics of Machines Dr. Mohammad Kilani

Class 4 Velocity Analysis

DERIVATIVE OF A ROTATING VECTOR

Derivative of a Rotating Unit Vector

 A unit vector in the vector of unity magnitude and an angle

θ θ

direction, u with the x- axis. It is written as:

, is a

u θ

 cos 

 sin 

 If u

rotates, it angle

changes with time, the time derivative of u rules on the expression of u

θ θ

is found by applying the standard differentiation above

u θ

dt d

u θ

  

dt d



  sin sin 



 



cos 



sin θ j

u θ

θ cos θ i

Derivative of a Rotating Unit Vector

 Noting that sin cos         2 2     cos   sin  the time derivative of the vector u

u θ

dt d

u θ

dt d

u θ



 

dt d

 

dt d



  sin 

 cos 

  cos  

    2    2 

 sin     2 



u θ

 cos 





 sin 

  sin 

 cos 



sin θ j

u θ

θ cos θ i

Derivative of a Rotating Unit Vector

 The derivative of a unit vector whose angle with the x-axis is

, (

changes in time), is a vector whose angle with the x-axis is

θ+π/2

and whose magnitude is

d θ/dt.

 If we define a vector ω as a vector in the k direction of magnitude ω = d

θ/dt

, then



u θ

dt d

 

dt d



    2  



u θ u θ

u θ

 cos 





 sin 

j u

    2 

/dt θ + π/2 θ

u θ

Derivative of a Rotating Vector

 A vector r

= r u

pointing in the

, is a general vector of magnitude r direction. The time derivative of r

found by applying the normal differentiation rules on is the expression for r

θ r θ



u θ

r θ

dt d

r θ

dt d

r θ

dt d

r θ



u θ



u θ



u θ



r d

u θ



r d dt



    2  





u θ

 

u θ

 



r θ



/dt θ + π/2 θ

u θ

Derivative of a Rotating Vector

 Given a vector r

= ru

which rotates relative to the reference coordinates, the time derivative of

r θ

has two components; a component in the direction of u

and a component normal to u

in the direction of u

θ+π/2

 The magnitude of the component of dr

/dt in the direction of u

equal to

dr/dt;

that is the time derivative of the length of r

is  The magnitude of the component of dr

/dt in the direction of direction of u

θ+π/2

is equal to

ωr

ω x r θ

(

dr/dt

)

u θ

θ + π/2

r θ

VELOCITY ANALYSIS OF FOUR BAR MECHANISMS

Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain

 The loop closure equation of a 4-bar kinematic chain is written as 

2 

2   

3 

3   

1 

1   

4 

4  0

 

2  

dt r d dt

2  

 2

   2

  2 2    2   

 3

 3   

4 

 2 

 2  

 1

 1  0 

 4

 4

3    3

  3     2  0 

  3

 3  

1    1

  1   2  

  1

 1   

4    4

  4   2  

  4

 4   0  When all the links in the chain are of constant lengths, the equation above reduces to

2   2

   2   2  

3    3

  3 2  

1   1

   1  

2 

2 

3 

3 

1 

1 

4 

4  2  0 

4   4 

  4   2   0

Derivative of the Loop Closure Equation for a Four Bar Mechanism

 For a four bar mechanism with link 1 fixed we have d

θ 1 / dt = ω 1

= 0.

2    2

  2   2  

3    3

  3   2  

4    4

  4   2   The vector equation above contains two scalar equations and can be solved for two unknowns. Knowledge of d

θ 2 /

dt allows the calculation of

d θ 3 /

dt and d

θ 4 / dt .

2   2

2   2 sin  2 cos  2  

3   3

3   3 sin  3 cos  3  

4   4

4   4 sin  4 cos  4

Derivative of the Loop Closure Equation for a Four Bar Mechanism

2   2 

  2   2  

3    3

  3   2  

4   4 

  4   2    Eliminate d

θ 3 /

dt by carrying out a dot product with

u θ3

on both sides of the equation

2   2

2   2  4  cos   sin   4    2 2 

4   2  3 sin sin       2 4   3  

4   4     3 3 sin     2

4   4   4   cos   3   4

4 sin sin      2 4    2   3 3     3  2  Alternatively, eliminate d

θ 4 /

dt by carrying out a dot product with u

θ4

on both sides of the equation

2   2

2   2  3  cos sin     2 2   3     

3  4 2   sin sin       2 3  4   

3   3   4  4

3   3 sin      2  3  cos     3  4 

3  sin sin      2 3 2     4  4  4      2 0 

3 sin sin     2 4     4 3    2

Angular Velocity Ratio and Mechanical Advantage

  The angular velocity ratio m

is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism with link 2 as the input and link 4 as the output this is expressed as

m V

  

out in

   4 2 The efficiency of a four bar linkage is defined as the output power over the input power,  

P out P in



T out T in

 

out in

 Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have

T out T in in

 



out



m T

 1

m V

Angular Velocity Ratio and Mechanical Advantage

 The mechanical advantage is defined as the ratio between the output force to the input force

m A



F out F in



T out T in r out r in



m T r in r out

VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS

Velocity Analysis of a Slider-Crank Mechanism

 Design parameters: r

, r

θ 1

Position analysis parameters:

θ 2

θ 3

Velocity analysis parameter.

Find dr

/dt, d

θ 2

/dt , d

θ 3

/dt

r r

 2 2   2 

3 

  2    

1 2   

 4

3  3 

  3   2  

  1

 1  To eliminate

ω 3

sides by u

θ 3

dot product both

2  2 

 1 cos

2  2 

2 sin sin  cos   2    2      3 3     1  3  2 2       2

 1 3  

 1 cos  cos   1  1   3    3 

r 2 r 3 r p r 1 r 4

 To eliminate dr

/dt dot product both sides by u(

θ 1+π/2

)

2  2  3  cos  

3  2  cos cos    1   2 3      1  1   

3  3  2 cos   3   1 

Velocity Analysis of an Inverted Slider-Crank Mechanism

 Given r

, r

θ 1

θ 2

ω 2

Position analysis: Find r

θ 3

Velocity analysis: Find dr

/dt, d

θ 3

/dt 

2 

 

 1  2 2 

   2 

 3

 1

 1   2   

r r

 3 

  3 3

 3 

3  3 

  3   2   To eliminate

ω 3

sides by u

θ 3

and find dr

/dt dot product both

2  2

2  2 cos sin     2 3     2  2  

 3  3  

 3  To eliminate dr

/dt and find

ω 3

sides by u(

θ 3+π/2

)

2  2  3

3 dot product both  cos

2   2 cos    3   3    2 

3  3  2

r 2

θ 2

r 1

θ 1

r 3

θ 3

Velocity Analysis of an Inverted Slider-Crank Mechanism

 Given r

, r

θ 1

θ 2

ω 2

Position analysis: Find r

θ 3

θ 4

Velocity analysis: Find dr

/dt, d

θ 3

/dt , d

θ 4

/dt

 2  



2  

3 2 2 

   2 

 1

 3

 2 3     

 1

 1

  3

 3 with

2  2  

  4 2    2   3  

   3

 2 3   

 4



3  4 3 

  3  

3  3 

  3   2  

4  4 

  4   2   

4  4 

 3 2 

r 3 r 2 r 1 r 4

Velocity Analysis of an Inverted Slider-Crank Mechanism

2  2 

  2   Dot both 2  

 3 

 sides by 3 

  

2  2  3  

2 cos  2   Dot both

3 sides by 

 3 3

2  2   cos cos 

2  2 But  4

2  2  sin  sin     2 2   3  2 2         3   3 2     3 

2  

3  3   2  3

 3 3

3  3 

     

 3 cos   

3  2   

4 0

 3 3    2  

4  4  3 

2  2 

4  4  2 cos   

3 2 

4  4   3

 3   

2 cos  

3 2   3  

2 sin   2   3   

    2 3  2

r 2 r 3 r 1 r 4

Example

Loop 

 2 I  



 

 3 

 1

 1 

 4

 4 Loop 

 2 II  

a u

 3 

 5

 5 

 1

 1  

s u

 6  Loop Closure



 

 3  Equation (1)

 1

 1 

4 

 4 

 2

 2 Dot both sides of the equation by itself 



 

b b

 2  2

cos  4 

1 2 

2 2 

4 2  2

4  

1 2

 sin

2 2  4  

4 2

  2

4 0 cos   cos  4 4    1  2

4  2

4 cos  2 cos    2 2

4   1   cos   2 2

4 cos  4 cos   2  sin   4   2 sin  4  





2  2

 



  0 

1 , 2 

   Also from loop closure rquation (2)

a a

 

b b

  cos sin   3 3  

1 cos sin  1  1  

4 cos cos   4 4  

2 cos  cos  2 2   3 1 , 2  tan  1  

1 sin cos   1 1  

4 cos cos   4 4  

2 cos cos   2 2  

2 



C C

2 

2 ,   4 1 , 2  2 tan  1

t A



4  2

4 sin  1 cos  2  

1 2 tan    4

2 2 2  

4 2  2

2 cos  2  



 2

a r 2 r 5 b s r 4 r 1 a r 2 r 1 r 5 b r 4 s

Example

Loop 

 5 Closure 

1 

 1  Equation (2) 

s u

 6 

 2

 2 



 3 Dot both sides of the equation by itself

5 2   2

s r

1 2  cos 

 2 6  

2 2  2   

2  2

cos  cos    3  6  1   Let  1  0 ,  6   2  1    2

2 cos  cos  6     2 2    1  2  

cos  cos  2     3 3    1 

5 2   

2 

s r

1 2  cos 



 2  2

1 2 6  

2 2  2   

2 2 



4 2 2  2

2 cos  cos    3 2  1    2

4 cos  4  2 

2 2

4 cos 

cos  3  cos  2 6   2   2

cos   2   3  





2  2

 



  0 

1 , 2 

   Also from loop closure rquation

a a

 

b b

  cos sin   3 3  

1 cos sin  1  1  

4 cos  cos  4 (2) 4  

2 cos cos   2 2   3 1 , 2  tan  1  

1 sin cos   1 1  

4 cos cos   4 4  

2 cos cos   2 2  

2 





2 ,   4 1 , 2  2 tan  1   1 , 2

t A



4  2

4 sin  1 cos  2  

1 2 tan    4

2 2 2  

4 2  2

2 cos  2  



 2

a r 2 r 5 b s r 4 r 1 a r 2 r 1 r 5 b r 4 s

HW#3 6-17, 6-18 (b+c), 6-21 (b), 6-44.

METHOD OF INSTANT CENTERS

Relative Velocities Between Two Points on a Rigid Body

 Given any two points A and B that lie on a rigid body, let the line AB be a line passing through A and B, then the components of the velocity of A and the velocity of B on the line AB must be equal.

 The velocity of point B relative to A must be normal to the line AB

v A



v B



v B A

B A B

Instant Center Relative to the Ground

 If line AA’ is drawn normal to the direction of v

, then the velocity of any point on the rigid body that falls on line AC must be perpendicular to the line AA’  In a similar manner, if line BB’ is drawn perpendicular to the direction of v body that falls on this line must be

, then the velocity of any point on the rigid perpendicular to the line BB’  If point I is the intersection of lines AA’ and BB’, then the velocity of point I must be perpendicular to both AA’ and BB’. This can only happen when the velocity of point I is zero.

A B I B’ A’



A v



A v A v B

 





   

 

r A I

 

r A

 

r A

 

r B I I

  

I v A r A I

Instant Center Relative to the Ground

 Point I determined as before is called the instant center of zero velocity of the rigid body with respect to the ground. The direction of the velocity of point A is normal to the line IA. The direction of the velocity of line B is normal to the line IB.  The direction of the velocity of any other point C on the same body must be normal to the line IC.

 The magnitude of the velocity of any point on the link is proportional to its distance from the point I.

A B I C B’ A’

v A IA



v B IB



v A IC

Velocity of a Point on a Link by Instant Center Method

 Consider two points A and B on a rigid link. Let v

and v

be the velocities of points A and B at a given instant. If v

is known in magnitude and direction and v

in direction only, then the magnitude of v

may be determined by the instantaneous centre method.

 Draw AI and BI perpendiculars to the directions v

and v

respectively. Let these lines intersect at I, which is known as instantaneous centre or virtual centre of the link. The complete rigid link rotates about the centre I at the given instant. The relations shown may be used to determine the magnitude of the velocity of point B.



A v



A v A v B

 





   

 

r A I

 

r A

 

r A I

  

I v A

 

r B I r A I

Instant Center Between Two Rigid Bodies

 In the foregoing discussion, the velocities of points A and B were assumed to be reference to a coordinate system attached to the ground. The resulting center is called an instant center relative to the ground.

 The velocities of points A and B could as well be taken relative to coordinate system attached to another body. The resulting point in this case will have a zero velocity with respect to that body. In other words, point I will have the same velocity for both bodies.

A B I B’ A’

Instant Center Between Two Rigid Bodies

 A instant center between two bodies is a:  A point on both bodies  A point at which the two bodies have no relative velocity.

 A point about which one body may be considered to rotate around the other body at a given instant.

A B I B’ A’

Instant Center Between Two Rigid Bodies

 When two links are connected to one another by a revolute joint, the center of the connecting joint is an instant center for the two links.

 When two links are not connected, an instant center between the two links will also exist and can be determined if the velocities of both links is known.

A’

Instant Center Between Two Rigid Bodies

 The number of instant centers in a constrained kinematic chain is equal to the number of possible combinations of two links.

 The number of pairs of links or the number of instantaneous centers is the number of combinations of L links taken two at a time. Mathematically, number of instant centers is,





 1  2

Types of Instantaneous Centers

 The instant centers for a mechanism are of the following three types : 1. Primary (Permanent) instant centers. They can be fixed or moving 2. Secondary Instant centers (Not permanent)





 1  2

Instant Centers of a Four Bar Mechanism

 Consider a four bar mechanism ABCD as shown. The number of instant centers (N) in a four bar mechanism is given by 

 4  4  1  2  12 2  6 The instant centers I

and I

are fixed instant centers as they remain in the same place for all configurations of the mechanism. The instant centers I

and I

are permanent instant centers as they move when the mechanism moves, but the joints are of permanent nature. The instantaneous centres

I 13

and I

are neither fixed nor permanent as they vary with the configuration of the mechanism.

2 A 1 (Ground) D 4 B 3 C





 1  2

Location of Instant Centers

 When the two links are connected by a pin joint (or pivot joint), the instant center lies on the center of the pin as. Such an instant center is of permanent nature. If one of the links is fixed, the instant center will be of fixed type.

 When the two links have a pure rolling contact (without slipping), the instantaneous centre lies on their point of contact, as this point will have the same velocity on both links.

 When the two links have a sliding contact, the instant center lies at the center of curvature of the path of contact. This points lies at the common normal at the point of contact.

Kennedy (or Three Centers in Line) Theorem

 Kennedy’s theorem states that any three bodies in plane motion will have exactly three instant centers, and the three centers will lie on the same straight line.

 Note that this rule does not require that the three bodies be connected in any way. We can use this rule, in conjunction with the linear graph, to find the remaining lCs which are not obvious from inspection.

Example: Instant Centers of a Four Bar Mechanism

 Draw a circle with all links numbered around the circumference  Locate as many ICs as possible by inspection. All pin joints will be permanent ICs . Connect the links numbered on the circle to create a linear graph and record those found.

 Identify a link combination for which the IC has not been found, and draw a dotted line connecting those two link numbers. Identify two triangles on the graph which each contains the dotted line and whose two other sides are solid lines representing the ICs the already found. Use Kennedy’s theorem to locate the needed IC.

Example: Instant Centers of a Four Bar Mechanism

Example: Instant Centers of a Slider-Crank Mechanism

Example: Instant Centers of a Cam and Follower: Common Normal Method

Example: Instant Centers of a Cam and Follower: Effective Link Method

Velocity Analysis with Instant Centers

 Once the instant centers (ICs) of a linkage with respect to the ground link have been found, they can be used for a very rapid graphic analysis for that link.

 Note that some of the ICs may be very far removed from the links. For example, if links 2 and 4 are nearly parallel, their extended lines will intersect at a point far away and not be practically available for velocity analysis.

Velocity Analysis with Instant Centers

 From the definition of the instant center, both links sharing the instant center will have identical velocity at that point.

 Instant center I

involves the coupler (link 3) which is in general plane motion, and the ground link which is stationary. All points on the ground link have zero velocity in the global coordinate system, which is embedded in link 1. Therefore, I

must have zero velocity at this instant, and it can be considered to be an instantaneous "fixed pivot" about which link 3 is in pure rotation with respect to link 1.  A moment later, I

will move to a new location and link 3 will be "pivoting" about a new instant center.

Velocity Analysis with Instant Centers

 If ω

is known for the mechanism shown, the magnitude of the velocity of point A can be computed as v

= ω

2 O 2

A Its direction and sense can be determined by inspection.

 Note that point A is also instant center and it has the same velocity as part of link 2 and as part of link 3. Since link 3 is effectively pivoting about I

at this instant, the angular velocity ω

can be found by ω

= v

/AI

. Once ω

is known, the magnitude of v

can also be found from v

= ω

3 AI 13

 Once v

is known, ω

can also be found from

ω 4

= v

/BO

= v

/BO can be found from v

C 4

. Finally, v = ω

3 CI 13 C

(or the velocity of any other point on the coupler)  3 

v A AI

v B

  3