15/9/2021

1

DEPARTMENT OF EDUCATION AND TRAINING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY
AND EDUCATION

Mechanisms and Machine Components
Design

CHAPTER 2: KINEMATIC ANALYSIS OF
PLANAR MECHANISMS

15-Sep-21

2

Theoretical contents

I. Position Analysis

II. Velocity Analysis

III. Acceleration Analysis

TS. Phan Cơng Bình 15-Sep-21

1

15/9/2021

3 A

Kinematic theory V AO

I. Link translating motion (tịnh tiến):

• Velocity of all point are the same ωA
• Tangent with trajectory
• Acceleration Vectors are collinear O

II. Links rotate around point O ( Hình 1) A
• Velocity vector : 𝑉 = 𝜔 . 𝑙
Value : 𝑉 = 𝜔 . 𝑙 𝑎
𝑎

Direction: ┴ OA and aligned with 𝜔

• Acceleration : 𝑎 =𝑎 +𝑎 O 𝜀 Hình 1

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

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

TS. Phan Cơng Bình 15-Sep-21

4 B

Kinematic theory 

VB
III. Planar kinematics (song phẳng): ( Hình 2) 
V BA

• Velocity : 𝑉 =𝑉 +𝑉 ωA

• Acceleration : 𝑎 = 𝑎 + 𝑎 + 𝑎 
IV. Coincident A (A1 & A2 ≡ A) ( Hình 3)
A VA
 Khâu 1 quay quanh trục cố định hoặc chuyển
động song phẳng với vận tốc 𝜔 B

• Velocity : 𝑉 = 𝑉 + 𝑉 𝑎𝑎

• Acceleration : 𝑎 = 𝑎 + 𝑎 + 𝑎 𝜀 𝑎 𝑎
where, 𝑎 Fig. 2

Relative Acc 𝑎 : // sliding direction Link 1&2 A1 ≡ A2 ≡ A 1

Coriolix Acc 𝑎 = 𝜔 ˄𝑉 : Phương chiều của

𝑉 đã quay 90° theo chiều 𝜔

 Khâu 1 tịnh tiến hoặc cố định 𝑎 =0 ; 𝑎 =0 2Hình 3

TS. Phan Cơng Bình 15-Sep-21

2

15/9/2021


5

I. Position Analysis

1.1 Position analysis of mechanism

- 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

B2

B3 B B1

2

B4 A B8 3C
B5 B6 C1
e

C3
B7

TS. Phan Cơng Bình 15-Sep-21

6 I. Position Analysis

1.2 Example Solve

Ex1 :

B2

B3 B B1

2

B4 A B8 3C

e

B5 C4 C3C5 C6 C2 C7 C1 C8
B6
B7
TS. Phan Cơng Bình
15-Sep-21

3

7 I. Position Analysis 15/9/2021
4
1.2 Example

EX2: Determine the trajectory of point B,C and swing angle γα of link C if
link AB rotates 1 revolution

AB  0,5m

BC  CD  AD  0,8m C3


B2 C1

B3 B1 2 C

1B
3

B4 A 4 B8 D

B5 B7

TS. Phan Cơng Bình B6



8 I. Position Analysis

1.2 Example Solution
EX2:
C7 C3 C2 C8
B3
B2 C4 C1
B4
C6 C5 B 2 C 1

1B 3
γα

A 4 B8 D


B5 B7

B6

TS. Phan Cơng Bình

15/9/2021

9

II. Velocity Analysis

2.2 Example B 2 C

VD: Given mechanism ABCD

𝑙 = 𝑙 = 0,5𝑚; ω1= 10 rad/s 1 3

a) Draw diagram ω1 300

b) Determine 𝑉 , ω2, ω3 A D

4
Solve:

𝑉 =ω1. 𝑙 =5m/s 𝑉B 2 C
B and C belong to link 2
𝑉 =𝑉 +𝑉 (1) ω2
C and D belong to link 3

𝑉 =𝑉 +𝑉 (2) 1 𝑉𝑉 3

TS. Phan Công Bình ω1 ω3

300

A 4 D

15-Sep-21

10

II. Velocity Analysis

2.2 Example 
* Draw diagram tutorial
• Set 1 point P ( Coordinate) b VB P
• S𝑒𝑡 𝑠𝑐𝑎𝑙𝑒 𝑉 µ =

• From (1) draw Pb= µ and draw bx ┴ BC V  BC

CB


300 VCD┴ CD

• From (2) draw Pd = µ =0=>d≡P c
draw Py ┴CD. Then, bx & Py cut at point c

Velocity diagram


TS. Phan Cơng Bình 15-Sep-21

5

15/9/2021

11 II. Velocity Analysis

2.2 Example

• Từ họa đồ ta có: b𝑉
P
Pc = 2Pb ⟺Vc=2Vb=10m/s
300
VCB  2CB  VC cos 300  10 32 (m / s) 𝑉
CB  0.5m 𝑉

 2  VCB  10 3(rad / s) c
 CB
Velocity diagram
VCD  3CD  VC  10(m / s)
CD  2. AD  1(m)

 3  VCD  10(rad/ s)
 CD

TS. Phan Cơng Bình 15-Sep-21

12


II. Velocity Analysis

2.2 Example B 2C

Exercise 1: ω1 3
2 AB   BC  CD  98"
A 4
1  10 rad s

a) Draw velocity diagram

b) Determine 𝑉 , ω2, ω3

D

B Exercise 2:

2  AB  0, 6m 1  10 rad s
1
2 ,Vc ?

3  

300 C VC  VB  VCB

A Hint:  
VC3  VC 4  VC3C 4
4


TS. Phan Cơng Bình 15-Sep-21

6

15/9/2021

13

III. Acceleration Analysis

EX 1: B 2 2 C

1  10 rad s  const 𝑎 𝑎
 AB   BC  0,5m 1𝑎 𝑎
3

𝑎 3

ω1

a) Draw Acceleration Diagram

b) ac , 2,3  ? 300
A D

4

ω2 =10 3 (rad / s) ω1=const=>ε1 =0  a BA t =ε1.lAB =0;
ω3 =10 (rad / s)
n

  aBA
a B =a BA  t n 2 m
=a BA =ω1 .lAB =50 2
aBA  0 s

B & C belong to Link 2 C & D belong to Link 3

  n t   n t
aC =aB +aCB +aCB (1) aC =a D +aCD +aCD (2)

a B  a n =ω12 .l AB =5 0 m 2 aD  0 m 2 s
BA
s aCD n =ω32. CD =100 m 2

a n = ω 2 .l = 1 50 m
 C B 2 C B 2  s
 s
TS. Phan Cơng Bình

14

III. Acceleration Analysis

Draw Acceleration Diagram  BC π  CD
𝑎
• π: gốc (cực) 𝑎 a 𝑎 300
aB nCB
nBA 600
• a  
ab 𝑎 𝑎


  n t 𝑎
(1)  aC = a B +aCB +aCB

  n t
(2)  aC = aD +aCD +aCD

a  aCB n  aCD n cos 300  236.6 m / s2

aCB t  a tan 600  410 m / s2 tCB tCD
aCD t  a / cos 600  437.2 m / s2
t
ac  a CD t 2  a CD n 2 aCB  2. CB  2
t
TS. Phan Cơng Bình aCD  3.CD  3



7

15/9/2021

15

III. Acceleration Analysis

EX2: AB  0,6m;1  10 rad s

𝑉 B
a)vC ,2  ?

 ω2

b)aC,2  ?

Solution 𝑉 2

1 Velocity 1 300
ω1
A C3
VB  1.AB  6 m s ┴BC
𝑉4
* B & C belong to Link 2 P
 
VC  VB  VCB (1)

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

 c

VC3  VC4  VC3C4 (2)

b  c  vc  vb  6m s
vCB  2.BC  0  2  0

TS. Phan Cơng Bình

16

III. Acceleration Analysis


EX2: B

2 Acceleration 𝑎

1  const 1𝑎 2 C3
 aB  a BA n  12. AB  60
ω1 300 4𝑎

* B & C belong to Link 2 A
n
  n t aCB  2. BC  0 2

aC  aB  aCB  aCB t
aCB  CB

* C3 & C4 ≡ C (Link 3 is sliding in Link 4) π c // AC

(khâu 3 trượt trên khâu 4 cố định)

   t t
aC  aC3  aC4  aC3C4 300 aCB

𝑎 ┴BC
ac  aB tan 300  60 32 m / s2 b
t
aB 120 a CB
aCB t  0   2.CB  2  
cos30 3  CB

TS. Phan Công Bình


8

15/9/2021

17

III. Acceleration Analysis

3.2 Example AB  0, 4m, BC  0,8m

vB 1  50 rad s  const

ω1 B 

A 4 a) vC , 2  ?

1 2

C 
b) aC,2  ?
Solve:
v3  50.0, 4  20 m s b
 
vC  vB  vCB C
   p //AC
vC  vC3  vC4  vC3C4
C  p  vC  0
vCB  2. BC  vB  2


 2  2,5rad s

TS. Phan Cơng Bình 15-Sep-21

18

III. Acceleration Analysis

3.2 Example

Exercise 3:

1  const  1  0

a BA t  1. AB  0 Cb bπ
C
 n
a B  a BA a BA
t
aBA  0

n 2 m
 a BA   1. AB  10 2
s

┴BC C3, C4  C
  
* B and C belong link 2 aC  a3  aC4  aC3C4

  n t   15m

aC  a B  aCB  aCB aC

n 2 m  2
aCB  2. BC  5 2 s
s
a3t  2. BC  0

 2  0

TS. Phan Cơng Bình 15-Sep-21

9

15/9/2021

19

TS. Phan Cơng Bình 15-Sep-21

10