VIETNAM NATIONAL UNIVERSITY

HO CHI MINH UNIVERSITY OF TECHNOLOGY

FACULTY OF MECHANICAL ENGINEERING

PROJECT C-4 REPORT

 HO CHI MINH UNIVERSITY OF TECHNOLOGY

Preface
First of all, we want to express such a deep gratitude to our
lecturer, who has taken charge of Kinematics and Dynamics of
machine dedicatedly this semester. All the members have gained
a lot of knowledge that is beneficial to our schooling as well as
had a great chance to discover such an interesting topic of this
project mutually.
While making a report, it is difficult to avoid errors so we are
looking forward to your comments. Therefore, we can learn from
experience for better upcoming reports.
Thank you so much for taking time to read our
report!

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Table of content

FUNCTIONS OF STUDENT
- Fill the suitable value from the attached excel file
-Design the mechanism (Define all kinematic lengths of link)
-Analyzing kinematics of the mechanism
-Analyzing kinetics of the mechanism

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- Consumed power at the given position of the shaper

A. Dữ liệu đã cho:
1. Vị trí của tay quay:
φ1=120
2. Hệ số hồi nhanh:
Q=1.5
3. Số vòng quay mỗi phút của khâu dẫn động:
n1=550 vòng/phút
4. Chiều dài hành trình:
S=800 mm
5. Chiều cao của máy bào hhh được chọn để đạt được góc
áp lực nhỏ nhất tác động lên bàn trượt.
(Điểm III là trung điểm của GJ.)
6. Khoảng cách khung dọc: lAC= h/2
7.Chiều dài thanh nối DE: lDE= 1.5 lAB
8.Trọng tâm của các khâu:
o Trọng tâm của khâu 1 là S1 tại điểm A với khối lượng
m1=20 kg
o Trọng tâm của khâu 3 là S3 tại trung điểm với khối
lượng m3=40 kg
o Trọng tâm của khâu 5 tại khớp E với khối lượng
m5=200 kg
9.Lực cắt tác động lên dao trong hành trình cắt được
thể hiện theo biểu đồ ở trên của hình. Trong hành trình

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hồi, lực cắt bằng không. Độ hở trước và sau của phôi là

2% chiều dài hành trình.

B. Nhiệm vụ của sinh viên:

1. Điền các giá trị phù hợp từ file Excel đính kèm.
2. Thiết kế cơ cấu (xác định tất cả các chiều dài động học của
các khâu).
3. Phân tích động học của cơ cấu.
4. Phân tích động lực học của cơ cấu.
5. Tính công suất tiêu thụ tại vị trí đã cho của máy bào.

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1. Fill the suitable value from the attached Excel File

1.Position of crank 1= 1200
2.Quick return coefficient: Q=1,5
3.Number of revolution per minute of drive link n1= 550(rpm)
4.Length of Stroke S=800 (mm)
5. Height of shaper h is selected how to get the smallest angle of
pressure that applie on the ram. (I is midpoint of GJ)
6. Vertical frame distance lAC= h/2
7. Length of rod DE lDE= 1.5 lAB
8. Center of gravity of link 1 is S1 that is at A with a mass of m1=
20 kg
Center of gravity of link 3 is S3 that is at midpoint with a mass
of m3= 40 kg
Center of gravity of link 5 is joint E with a mass of m5= 200 kg
9. Cutting force that applies on cutter in the cutting stroke as in the
diagram on the top of the above figure. In the return stroke, cutting
force is null, Front and rear clearance of the workpiece is 2% of
the length of stroke.

2. Design the mechanism (Define all kinematic lengths of link)

We will set :
AC Link 0
Line through Link 1

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S1(A)
AB Link 2
CD Link 3
Slider B Link 4
DE Link 6
Length Stroke Link 5

0
180 +θ
-)From Quick return coefficient Q,we have:Q= 1800 −θ =1 ,5=¿ θ=¿36.

-) I is the middle point of GJ so

h=CI=
CG +CJ
=
l3 +l 3 cos ( θ2 ) with l = S
(S=800mm=0,8m).
()
3
2 2 θ
2sin
2

Since l3=lCD =1 , 294(m)

S S
+ cot
θ
()
()
θ 2
=>h= 2 sin 2
2

=1,263(m)
2

h
-)According to given data, l0=l AC = 2 =0,632 ( m )

-) B ^A C=180 0−φ=1800−120 0=60 0

θ
=> C B^ A=180 − 2 −70 =92
0 0 0

AB AC
= ≤¿
=> sin
()
˙ θ
2
^ A ) AB=l AB=l 2=0 ,195(m)
sin ( C B

Since, lDE =l6=1 , 5.l AB=1, 5.0,195=0,2925(m)

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 HO CHI MINH UNIVERSITY OF TECHNOLOGY

AB BC
=

()
˙ θ
sin
2
sin ( 60 ° ) =>BC= 0,546 (m/s)

3.Analyzing kinematics of the mechanism

To analyze kinematics of the mechanism,we need to decompose
system into Coulisse System and Planar Mechanism.

Planar Mechanism

Coullise System

3.1 Determine Velocity

a) Coullise System

AC Link 0
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Line through Link 1

S1(A)
AB Link 2
CD Link 3
Slider B Link 4
DE Link 6
Length Stroke Link 5
We have:B2 is connected to B4 by rotate joint and B3 is connected
to B4 by prismatic joint with B2 on link 2; B4 on link 4 and B3 on
link 3.
 ⃗
v B2=⃗
v B4 =⃗v A +⃗v B 2 A =>⃗
v B2=⃗
v B4 =⃗v B A
2

55 m
¿> v B =v B =ω 1 l AB= π .0,195=11,231 ( )
2 4
3 s
 ⃗
v B3=⃗
v B 4+ ⃗v B B 4 3

 ⃗
v B3=⃗v C +⃗v C B =¿ ⃗
v B3=⃗v C B 3 3

With vA=vC=0 because these joint are fixed according to diagram

and:
 ⃗
v B4 have some features : Align with ω1 ; perpendicular ¿ AB ;
m
v B =11,231 ( )
4
s
 ⃗v B B have some features :∥¿ BC
4 3

 v B3 ( ⃗v C B ) have some features : Align withω 3 ; perpendicular ¿

⃗ 3

BC ; v B =¿ω .l ¿3 3 BC

And now we determine the velocity scale through vB 4 which is

represented by a segment of 50mm.It means that
vB
μv =
p b4
4
=
11.231
50
=0,2246
m/s
mm ( ).Select an arbitrary point p as the origin
for the velocity diagram ⃗
v B3 .From point p,draw p b4 with a length of
50(mm) perpendicular to direction AB,and then extend the

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direction BC from point b4.Next,from point p,draw p b3

perpendicular to BC and intersecting BC.
So we can get velocity from the length and velocity scale( μv ¿

v B B =51 , 95.0,2246=11 ,668

4 3 ( ms )
m
v B =14 ,1.0,2246=3 , 167( )
3
s

v S =v A=0
1

vB
So ω 3= l 3
=3 ,167 /¿0,546=5,8(rad/s) and ω 3 align withv B (clockwise )
3
BC

b)Planar Mechanism

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We can calculate directly v Dthrough ω 3: v D =ω3 . l CD=7 ,505 (m/s)

Beside that, v S =0 ,5. v D=0 ,5.7,505=3,753

3 ( ms )
To determine v E ,we have: ⃗
v E =⃗
v D +⃗
v DE

 ⃗
v E have some features :∥¿ DE
 ⃗
v D have some features: pe rpendicular ¿CD , align with ω3 ;
vD=7,505 (m/s)
 ⃗
v DE have some features : perpendicular ¿ DE ; ω 6
At first,we need to determine angle CDE to use velocity scale
to calculate ⃗
vE

G^ ( EH
DE )
D E=arcsin with EH parallel to IG and EH perpendicular

to DG.It means that IG=h-CG=h-l ⋅ cos ( 2 )

θ
3

Thus,EH=IG=0,0315(m)=> G ^D E=arcsin ( DE )=arcsin ( 0,2925 )=6 , 192=>

EH 0,0315

^ E=G D
CD ^ E+C D
^G

<=> ^ E=6 , 192+ arcsin

CD ( 11,, 231
294 )
=78 , 24 ⁡

Now we determine the velocity scale through v E which is

represented by a segment of 20mm.It means that
( )
v E 7 , 505 m/s
μv ' = = =0,37525 .Select an arbitrary point p as the origin
pe 20 mm mm
for the velocity diagram v D .From point p,draw pd with a length of
20mm p e rpendicular ¿ directionC D,and then extend the direction which
perpendicular to DE from point d.Next,from point p,draw pe
parallel to DE and intersecting pde.

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v DE=4 , 06.0,37525=1,524 ( ms )
m
v E =19 ,58.0,37525=7,347( )
s

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 HO CHI MINH UNIVERSITY OF TECHNOLOGY

3.2 Determine Acceleration

a) Coullise System

a⃗ B = ⃗aB + ⃗arB B +a⃗ kB

3 4 4 3 4
B3

<=>a⃗ nB +a⃗ tB =a⃗ nB + ⃗arB B +a⃗ kB B

3 3 4 4 3 4 3

a⃗ nB 3
+a⃗ tB 3
= a⃗ nB 4
+ a⃗ rB 4
B3 + a⃗ kB 4
B3

=2.5,8.11,6
r
=BC. ω 23 =BC.ε 3 =AB. ω1
2
? =2. ω 3 . v⃗ B B
4 3

68
( )
2
=0,546.5 , 82 =0,195.
55 π
=18,367(m/ s2 ¿ 3

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 HO CHI MINH UNIVERSITY OF TECHNOLOGY

=646,870(m/ s2 ¿ =135,(m/ s2 ¿
From C to B Perpendicular From B to A Lin Rotate ⃗v B B 90
4 3

to CD e on following to ω 3
CD

We need to determine steps to draw acceleration diagram:

 Select an arbitrary point p’.
 From p’,draw consecutive vector a⃗ nB , ⃗a kB B ∧direction a⃗ rB B
4 4 3 4 3

 From p’,draw consecutive vector a⃗ nB and direction a⃗ tB

3 3

 Intersection of two directon of two vector a⃗ rB B ∧a⃗ tB is b3.

4 3 3

 '
a⃗ B is p b3
3

Through acceleration diagram,we have:

a B =758 , 05 (m/ s2 ¿
3

a B =18 , 37 (m/ s2 ¿
n
3

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t
t aB rad t
a B =758,05(m/ s2 ¿=> ε 3= =1388 , 37 ( 2 )
3
and align with a⃗ B
3
BC s
3

(Clockwise)
a B B =206 , 18(m/ s2 ¿
r
4 3

a D /a B =CD / BC =>a D =1796,551(m/ s2 ¿

3

a A=aC =0 because joint A and joint C are fixed

b) Planar System

It is required to determine a⃗ E in Planar System.

From above,we have: a D =1796,551(m/ s2 ¿ and
a⃗ E =⃗a D + a⃗ DE

a⃗ E =⃗a D + a⃗ nED+a⃗ tED

a⃗ E =a⃗ D + n
a⃗ ED + t
a⃗ ED
=1796,551(m/ s ¿
2

From D to C(or
D to B)
Given:
- Length of links:l AB=195 mm ; l AC =632 mm ; lCD =1294 mm ; lDE =292 ,5 mm

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 HO CHI MINH UNIVERSITY OF TECHNOLOGY

2. Reaction force of pairs RA, RB, RC, RD, RE, and RF

Consider link 5:

⇔ Σ ⃗F ❑=⃗
Σ⃗
F ❑= ⃗
R 45+ ⃗
R O 5 +⃗
P=0
R 45+ ⃗
R 45+ ⃗
R 05+ ⃗
P =0 (1)
x y

(1) on Ox:
⃗
R x45❑−⃗
P=0
==> ⃗R 45❑ =⃗
x
P=2000 ( N)❑
(1) on Oy:
⃗
R 45❑ −⃗
y
R05=0
==> ⃗ y
R 45 ⃗
❑ = R05

Consider link 4:

Σ⃗
F ❑= ⃗ R 54=⃗
R 34+ ⃗ Rτ34❑+ ⃗
Rn34 + ⃗x
R54 ⃗y
❑ + R54 ❑❑=0 (2)
We have: R54x ¿ R45x =2000 (N)

(2) on Ox:

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τ n x
−R34 . cos(1)+ R34 . cos(1)−R 54❑=0
 −R34τ + R34n −¿❑ 2000=0 ¿
==> 0+ R n34−¿❑ 2000=0 ¿
n 2000
==> R34= 1 (N)

(2) on Oy:

⇔ 0+ R n34−R54y ❑=0
τ n y
−R34 + R34 −R54❑=0

⇔ Rn34=R 54y ❑
y 2000
==> R54= 1
y y 2000
==> R54=R 45=R05= 1
=2000 (N)

Σ M ¿ E =⃗ R54∗0+⃗
Rτ34❑∗l ED + ⃗ Rn34∗0
 Rτ34 .l ED=0
==> Rτ34=0
2000
==> Rn34=R 34= 1 (N)

∑ M ❑=R12∗0+ R32∗x=0=¿> x=0

C

Consider link 3:

τ 20 00
We have: R34=R 43=R 43= 1
(N )

Σ⃗
F ❑= ⃗
R 43+ ⃗ R 03=⃗
R 23+ ⃗ Rτ43❑+⃗ R23 + ⃗
Rn43❑+ ⃗ R x03❑+ ⃗
R03y ❑❑=0 (3)

Σ M ¿ C =R03∗0+ R43∗cos ( 18 ) .l CD + R32∗l CB =0

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⇔ 0+
2000
∗cos ( 18 )∗1,294 ❑+ R32 . 0,546=0
1

(3) on Ox:

⇔ −2000+ 4507,938 .cos (18)+ R03x =0❑

τ x ❑
−R 43+ R23 cos (18)+ R 03=0

 R03x =−2287,3038(N )

(3) on Oy:

⇔ 2000+ 4507,938∗sin(18)+ R03y =0❑

n y ❑
R43 + R23 sin(18)+ R03=0

==> R03y =−3393,0295 (N )

==> R03= √¿ ¿

We have: R23=R 32=4507,938(N )

Consider link 2:
R32=R21=4507,938(N )

We have: R32=R12=4507,938(N )

Consider link 1:
Σ⃗
F ❑= ⃗
R 12+ ⃗

⇔ ⃗R01=− ⃗R 12
R01=0

==> R01=R12=4507,938(N )

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R A =RO 1=4507,938(N )
R B=R21=R12=4507,938(N )
RC =R O 3=4086,9692(N )
2000
R D=R 43= (N )
1
2000
R E=R 34= =2000(N )
1
R F=R O 5=2000(N )

3. Determine balance moment

Force analysis:
Σ M / A=M cb + R21 . cos 30.l AB=0
==> M cb =−R21∗cos ( 30 )∗l AB=−4507,938∗cos ( 30 )∗0 ,19 5=−761,278( N . m)
==> M cb is in the opposite directional rotation with ω 1

Virtual work
M cb∗ω1 + P∗V G=0
M cb∗55 π
==> 3
+ 2000∗20 , 85=0

==> M cb =−724 , 01(N . m)

==> M cb is in the opposite directional rotation with ω 1

Relative error between the two calculations:

Δ 761,278−724 , 01
Δ %= = .100 %=5 %
M c b (761,278+724.01)/2
tb

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