Chapter 2 Planar Kinematic Analysis.pdf

Microsoft PowerPoint Chapter 2 PLANAR KINEMATIC ANALYSIS 15/9/2021 1 15 Sep 21 Mechanisms and Machine Components Design CHAPTER 2 KINEMATIC ANALYSIS OF PLANAR MECHANISMS 1 DEPARTMENT OF EDUCATION AND[.]

Trang 1

Mechanisms and Machine Components

Design

CHAPTER 2: KINEMATIC ANALYSIS OF

PLANAR MECHANISMS

DEPARTMENT OF EDUCATION AND TRAINING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY

AND EDUCATION

TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com

Theoretical contents

I Position Analysis

II Velocity Analysis

III Acceleration Analysis

2

15-Sep-21

Trang 2

II Links rotate around point O ( Hình 1)

• Velocity vector : 𝑉 = 𝜔 𝑙

Value : 𝑉 = 𝜔 𝑙

Direction: ┴ OA and aligned with 𝜔

• Acceleration : 𝑎 = 𝑎 + 𝑎

Normal Acc : 𝑎 = 𝜔 𝑙 = : Hướng từ A → O

Tangent Acc : 𝑎 = 𝜀 𝑙 : ┴ OA and aligned with 𝜀̅

15-Sep-21

I Link translating motion (tịnh tiến):

• Velocity of all point are the same

• Tangent with trajectory

• Acceleration Vectors are collinear

Kinematic theory

Hình 1

ω A

A

O

A O

V 

A

𝑎 𝑎

𝜀



B

4

III Planar kinematics (song phẳng): ( Hình 2)

• Velocity : 𝑉 = 𝑉 + 𝑉

• Acceleration : 𝑎 = 𝑎 + 𝑎 + 𝑎

Fig 2

IV Coincident A ( A1& A2≡ A ) ( Hình 3)

 Khâu 1 quay quanh trục cố định hoặc chuyển

động song phẳng với vận tốc 𝜔

• Velocity : 𝑉 = 𝑉 + 𝑉

• Acceleration : 𝑎 = 𝑎 + 𝑎 + 𝑎

where,

Relative Acc 𝑎 : // sliding direction Link 1&2

Coriolix Acc 𝑎 = 𝜔 ˄𝑉 : Phương chiều của

𝑉 đã quay 90° theo chiều 𝜔

A

B

B A

V

A

V 

ωA

B

V 

𝑎

𝑎 𝑎 Kinematic theory

1

A

Trang 3

TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com 15-Sep-21 15-Sep-21

1.1 Position analysis of mechanism

15-Sep-21

- As shown in the given mechanism, determine the tracking of points and links

belong to mechanism

Ex: determine the tracking point C if link AB rotates 1 revolution

I Position Analysis

e

A

B

C

2

3

B8

B1

B2

B3

B4

B5

B6

B7

C1

C3

e

A

B

C

2

3

B8

B1

B2

B3

B4

B5

B6

B7

C8

C1

C7

C2

C4 C5

15-Sep-21

6

15-Sep-21

1.2 Example

I Position Analysis

Trang 4

EX2: Determine the trajectory of point B,C and swing angle γ α of link C if

link AB rotates 1 revolution

AB

BC CD AD

0,5m

0,8m





  

B1

B2

B3

B4

B5

B6

B7

D

C

3 1

2

4

B

C1

C3

1.2 Example

I Position Analysis

8

B 1

B 2

B 3

B 4

D

C

3 1

2

4

B

γ α

C 1

C 8

C 2

C 3

C 4

C 5

C 6

C 7

1.2 Example

I Position Analysis

Trang 5

VD: Given mechanism ABCD

𝑙 = 𝑙 = 0,5𝑚; ω1= 10 rad/s

a) Draw diagram

b) Determine 𝑉 , ω2, ω3

Solve:

𝑉 = ω1.𝑙 =5m/s

2.2 Example

𝑉

𝑉 𝑉

A

B

C

D

ω1

300

1

2

3

4

ω2

ω3

A

D

ω1

300

1

2

3

4

B and C belong to link 2

𝑉 = 𝑉 +𝑉 (1)

C and D belong to link 3

𝑉 = 𝑉 +𝑉 (2)

P

c

┴ CD

300 b

15-Sep-21

10

15-Sep-21

V  B

V  CB  BC

V  CD

2.2 Example

* Draw diagram tutorial

• Set 1 point P ( Coordinate)

• S𝑒𝑡 𝑠𝑐𝑎𝑙𝑒 𝑉 µ =

Velocity diagram

• From (1) draw Pb =

µ and draw bx┴ BC

• From (2) draw Pd =

µ =0=>d≡P draw Py ┴CD Then, bx& Py cut at point c

Trang 6

Velocity diagram

2.2 Example

0 2

2

3 cos 30 10 ( / )

2 0.5

10 3( / )

CB

m

V

rad s





  



3

10( / )

10(rad/ s)

CD CD C

CD AD

CD

m V





• Từ họa đồ ta có:

Pc = 2 Pb ⟺ V c =2V b =10m/s

𝑉

P

c

300

b

𝑉

12

1

10

AB BC CD

rad

s





a) Draw velocity diagram

b) Determine 𝑉 , ω2, ω3

4

3

2

ω1

A

D

B

1

2

3

300

Exercise 2:

1

2

, ?

AB

c

rad

V





C B CB

V   V   V 

Exercise 1:

2.2 Example

Trang 7

A

D

ω1

300

1

2

3

4

3



1 10

0,5

AB BC

rad const

s

m

a) Draw Acceleration Diagram

b) a c,  2, 3 ?

t

n

2

B BA t BA 1 AB

BA

ω =const=>ε =0 a =ε l =0;













 



B & C belong to Link 2

n t

C B CB CB

C D CD CD

a =a +a +a     (2)

2

3

ω =10 3 (rad / s)

ω =10 (rad / s)

n 2

2

B BA 1 AB

n 2

2

CB 2 CB

m

s m

a =ω l =150

s







2 D

n 2

2

CD 3 CD

m

s

m

s





𝑎 𝑎

𝑎

𝑎 𝑎

π

300

600

14

Draw Acceleration Diagram

• π : gốc (cực)

b

a

  



BC



CB

CD t t

t

(1)   a = a +a +a   

(2)   a = a +a +a   

CB t

a

CD

a  a / cos 60  437.2m / s

CB

a  a tan 60  410m / s

a  a  a

𝑎

𝑎 𝑎

𝑎

CD

 III Acceleration Analysis

Trang 8

// AC

┴BC

P

ω1

A

B

C

1

2

3

300

4

ω2

C 2

rad 0,6m; 10 s

a)v , ?

b)a , ?

  

 

 





B 1 AB m

V     6 s

Solution

* C3& C4≡ C (Link 3 sliding in Link 4 )

C3 C4 C3C4

V   V   V  (2)

c b

m

C B CB

V   V   V (1) 

* B & C belong to Link 2

1 Velocity

𝑉 𝑉

𝑉

b

c

// AC

π

300

ω1

A

B

C

1

2

3

300

4

16

n t

C B CB CB

a   a   a   a 

* C3& C4≡ C (Link 3 is sliding in Link 4 )

(khâu 3 trượt trên khâu 4 cố định)

t

C C3 C4 C3C4

a   a   a   a 

t CB a

* B & C belong to Link 2

1

n 2

B BA 1 AB

const

 

n 2

CB 2 BC t

CB











2 Acceleration



𝑎

𝑎 𝑎

𝑎

b

c

c B

3

a a tan 30 60 m / s

2

III Acceleration Analysis EX2:

Trang 9

ω1

A

C

1

4 B

2

vB

1

C 2

rad

 

 



3

C B CB

C C3 C4 C3C4

C

CB 2 BC B

2

m

v 50.0,4 20 s

rad

  





Solve:

p

C

//AC b

3.2 Example

TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com 15-Sep-21

18

15-Sep-21

Exercise 3:

t

BA 1 AB

n BA

B BA t

BA

n 2

2

BA 1 AB

a

m

s







* B and C belong link 2

n t

C B CB CB

n 2

2

CB 2 BC

m

s

3 4

C 3 C4 C3C4

2 C

t

3 2 BC

2

m

s

0



  





b C

┴BC

3.2 Example

Trang 10

TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com 15-Sep-21

Tiêu đề	Kinematic Analysis Of Planar Mechanisms
Tác giả	TS. Phan Cụng Bỡnh
Trường học	Hochiminh City University Of Technology And Education
Chuyên ngành	Mechanisms And Machine Components Design
Thể loại	thesis
Năm xuất bản	2021
Thành phố	Hochiminh City

Định dạng
Số trang	10
Dung lượng	1,99 MB