Phát triển hệ thống lưu kho tự động

TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
DEVELOPMENT OF AN AUTOMATED STORAGE/RETRIEVAL SYSTEM
Chung Tan Lam(1), Nguyen Tuong Long(2), Phan Hoang Phung(2), Le Hoai Quoc(3)

(1) National Key Lab of Digital Control and System Engineering (DCSELAB), VNU-HCM
(2) University Of Technology, VNU–HCM
(3) HoChiMinh City Department of Science and Technology
(Manuscript Received on July 09th, 2009, Manuscript Revised December 29th, 2009)

ABSTRACT: This paper shows the mathematical model of an automated storage/retrieval system

(AS/RS) based on innitial condition. We iditificate oscillation modes and kinematics displacement of
system on the basis model results. With the use of the present model, the automated warehouse cranes
system can be design more efficiently. Also, a AS/RS model with the control system are implemented to
show the effectiveness of the solution. This research is part of R/D research project of HCMC
Department of Science and Technology to meet the demand of the manufacturing of automated
warehouse in VIKYNO corporation, in particular, and in VietNam corporations, in general.
Keywords: automated storage/retrieval system , AS/RS model

safety condition, saving investerment costs,
1. INTRODUCTION

managing professionally and efficiently. This

An AS/RS is a robotic material handling

system has been used to supervise and control

system (MHS) that can pick and deliver

for automated delivery and picking [1], [2], [3].

material in a direct - access fashion. The

In this paper, several design hypothesis is

selection of a material handling system for a

given to propose a mathematical model and

given manufacturing system is often an

emulate to iditificate oscillation modes and

important task of mass production in industry.

kinematic displacement of system based on

One must carefully define the manufacturing

innitial conditions of force of load. As a results,

environment, including nature of the product,

we decrease error and testing effort before

manufacturing process, production volume,

manufacturing [4], [5]. No existing AS/RS met

operation types, duration of work time, work

all the requirements. Instead of purchasing an

station characteristics, and working conditions

existing AS/RS, we chose to design a system

in the manufacturing facility.

for our need of study period and present

Hence, manufacturers have to consider
several

specifications:

high

throughput

manufacturing in VietNam.
This works was implemented at Robotics

capacity, high IN/OUT rate, hight reliability

Division,

and better control of inventory, improved

Control and System Engineering (DCSELAB).

National Laboratory of Digital

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Science & Technology Development, Vol 13, No.K4- 2010
Structural deformation is not depend on bar

2. MODELLING OF AS/RS

An AS/RS is a robot that composed of (1)
a carriage that moves along a linear track (xaxis), (2) one/two mast placed on the carriage,

axial force. Assume that the mass of each part
in AS/RS is given as m1, m2 , m3, and mL is
lifting mass.

(3) a table that moves up and down along the

When operation, there are two main

mast (y-axis) and (4) a shuttle-picking device

motions: translating in horizontal direction with

that can extend its length in both direction is

load f1; translating in vertical direction with

put on the table. The motion of picking/placing

load f2. The innitial conditions of AS/RS are

an object by the shuttle-picking device is

lifting mass, lifting speed, lifting height,

performed horizontally on the z-axis.

moving speeds, inertia force, resistance force,

In this paper, an AS/RS is considered a
none angular deflection construction in cross
section in place where having concentrated

which can be used to establish mathematical
model of AS/RS and verify the system
behavior.
The assumed parameters of the AS/RS are

mass [4], [5], [6]. There are several assumtions

given in Table 1.

as follows:
The

weight

concentrated mass

of

construction

post

is

in floor level (Fig. 1).

x3

m3

Bar 1
Bar 2

f2

K1
x1

Fig. 1: Model of AS/RS

Trang 26

L

x2

K2
mL
m2

m1

X4
f1


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
Table 1 Parameters of the AS/RS
Parameter

Value

m1

150

[kg]

m2

30

[kg]

mL

100 – 500

[kg]

m3

20

[kg]

ξ

2

[%]

L

20

[m]

E

21x106

[N/cm2]

I

2.8x103

[cm4]

k1

352.8

[N/cm]

k2

352.8

[N/cm]

kc

6594

[N/cm]

d

20

[mm]

2.1 Mathemmatical Model

k1 =

6 EI
3
x4

,

k2 =

6 EI
( L − x4 )

3

Case 1: Horizontal moving along steel
rail with load f1 [7]

with

It is assume that (1) Structural deformation
is not depend on bar axial force; (2) The mass

x4 =

where

L

, then k1= k2

2

E:

elastic

coefficient

of

material

in each part of automated warehouse cranes is

I: second moment of area

given as m1, m2 , m3, in there, mL is lifting

k1,

mass; (3)When the system moves, there are

k2

:

stiffness

proportionality

two main motions: travelling along steel rail
underload f1 and lifting body vertical direction

Case 2: Vertical moving with load f2 [8]

underload f2.

&
m2 L &
x&
4 + k c x4 + C3 x 4 = f2

The following model for traveling can be
obtained:

mL
kc

: stiffness of cable

D

: diameter of cable

&
m13&
x&

1 + k1 ( x1 − x 2 ) + C1x 1 = f1
m23 &
x&
2 + k1 ( x2 − x 1 ) + k 2 ( x2 − x3 ) = 0

&
m3 &
x&
3 + k2 ( x3 − x 2 ) + C2 x 3 = 0

where m13 = m1+ m2 + mL + m3

where m2L = m2 +

kc =

AE
π d 2E
=
4( L − x4 )
l

2.2. Solution of motion equation
a. Travelling along steel rail underload f1

m23 = m2 + mL + m3
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Science & Technology Development, Vol 13, No.K4- 2010

If resistance force is skipped, the motion

 m13 0 0  


0 m 23 0  
0 0 m  

3 

  k1 -k1
0   x1
 

+
-k
k
+
k
-k
  1 1 2
2   x2
  0
-k 2
k 2   x 3
 

&
x&1
&

x&2
&
x&3

&+ kx = F
or in the matrix form Mx&

superposition method [9] as the followings:
Eigen problem: kφ = M φω 2 ⇒ (k − M ω 2 )φ = 0

)

that satisfy det k − M ω 2 = 0
where φ

: vibration frequency (rad/s).

x4 =

 k1

} -k1
 0


⇒ b = ±1.183x10

:

2

:

ω 2 = 1.135 (rad/s)

These



 φ 
  23 

-k1
k1 + k 2
-k 2





0  φ31
  
2
-k 2  b φ32  = ω3
k2

 φ 
  33 

−3

T

φ2 = {0.025 −0.031 −0.033}

, φ = {0.001
3

T
−0.041 0.181}

φi

:

(3)

need to be satisfied φiT kφi = ωi 2

- φ T kφ = ω 2
2
2
2

0
φ21
  
2 ⇒ a = ±0.025
-k 2  a φ22  = ω2
k2

(2)

φ1 values is any

T

If a and b are positive, φi values is as follows:

Trang 28

2

φ3 = {1 −34.9 153.073}

2

 

from the equation

2
( k − M ωi )φi = 0 are as follows:

L

-k1
k1 + k 2
-k 2

φi

2

- φ T kφ = ω 2
3
3
3

{

The solutions

ω 3 = 18.095 (rad/s)

Eq. (3), we have

b φ31 φ32 φ33

ω2 = 1.065 (rad/s), ω3 = 4.254 (rad/s).

T

Substituting constant values in Table 1 into

 k1

a {φ21 φ22 φ23} -k1
 0


(1)


0 


φ 2 = {1 −1.252 −1.338}

 k -m ω 2 -k

0
1
 1 13

2
det  -k1
k1 + k 2 -m 23ω
-k 2  = 0


2
-k 2
k 2 − m3ω 
 0


At the position

f1 


0

ω 1 = 0 (rad/s)

: n level vector

ω

 
 
=
 


ω1 = 0 (rad/s),

Solution of Eq. (1) can be solved by

(

equation can be written as:


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
It

can

be

seen

that

the

condition

T
φ M φ = I is satisfied.

T
Ri (τ ) = φi f (t )
R2 (τ ) = 0.025 f1 (t )

If resistance force is skipped, the motion
equation will be written as the followings:

R3 (τ ) = 0.001 f1 (t )

Using integral Duhamel to find motion

2
T
&
(t ) + Ω x(t ) = φ f (t )
x&

equation [9]

and n individual equation can be written:
2
&
x&
i (t ) + ωi xi (t ) = Ri (τ )
x i (t ) =

1 t

(6)

∫ R (τ ) sin ωi (t − τ ) dτ + α i sin ωi t + β i cos ωi t
ωi 0 i

R (τ )
x&i (t ) = i
sin ωi t − α iωi cos ωi t − βiωi sin ωi t
ωi

αi and βi can be specified from initial

(7)
where : ωi = ωi 1 − ξi 2

conditions

ξi : damping ratio

T

xi t =0 = φi M °u

x&i (t)=

T
x&i t =0 = φi M °u&

Geometric inversion can be defined by
principle of superposition:
u(t)=[φ ][x(t)]=[φ1 ][x1 (t)]+[φ2 ][x 2 (t)]+....+[φn ][x n (t)]

Displacement of point is defined by
principle of superposition [9]

e-ξiωit ( ( ξiωα
i i +βiωi ) sinωi t+( ξiωβ
i i - αiωi ) cosωi t )
We find αi and βi value based on initial
condition
Displacement of point is defined by
principle of superposition (Eq. (8))

n
u i (t)= ∑ φi x i (t)
i=1

Influential

Using integral Duhamel to find motion

equation [9]:
-ξ ω (t-τ)
1 t
sinωi (t-τ)dτ +
∫ R i (τ)e i i
ωi 0
-ξ ω t
e i i αi sinωi t+βi cosωi t

(

dynamic

load

act

(8)
upon

warehouse cranes in some cases

If resistance forces are considered

x i (t)=


Ri (τ)e-ξiωit  ξi2ωi2
+ωi sinωi t 2
2 2 

ωi +ξi ωi  ωi


The acting force is a constant and system
has influential resistance force
It is assumed that f1 = Wt = 423.6 (N)
From Eq. (4) and Eq. (5), we have

)

R2 (τ ) = 0.025 f1 = 0.025 * 423.6 = 10.59 (N)

(9)
R3 (τ ) = 0.001 f1 = 0.001* 423.6 = 0.42 (N)
From Eq. (10)
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Science & Technology Development, Vol 13, No.K4- 2010
2

2

Substituting R3 (τ ), ω3 , ξ3 into Eq. (9) and

2

2

from Eq. (9) and Eq. (11), we have α 3 = 0 ,

ω2 = ω2 1 − ξ 2 = 1.065 1 − 0.02 = 1.06 (rad/s)
ω3 = ω3 1 − ξ3 = 4.254 1 − 0.02 = 4.25 (rad/s)

Substituting the values: R2 (τ ), ω2 , ξ 2 into

β3 = 0 :

(

x 3 (t)=0.023 1-e

Eq. (9), and from initial condition:

0
 
T
x2 t =0 = φ2 M ° 0 = 0
0

-0.085t

( cos4.25t+0.02sin4.25t ))

with

{ω1

T
T

ω2 ω3} = {0 1.06 4.25} (rad/s)

As the results, the motion equation can be

0
 
T
x&2 t =0 = φ2 M ° 0 = 0
0

derived as:

From Eq. (6) and Eq. (7) with α 2 = 0 ,

β2 = 0 :

(

x 2 (t)=9.42 1-e

-0.02t

( cos1.06t+0.02sin1.06t ))

with ω3 = 4.25 :

 x1 (t)  0

 
-0.02t

( cos1.06t+0.02sin1.06t )
 x 2 (t)  = 9.42 1-e
 x (t)  
 3  0.023 1-e-0.085t ( cos4.25t+0.02sin4.25t )


(

)

(

)








(12)

With the force of load is periodic, resistance force of the system is assumed to be f1 = A cos ω t
From Eq. (4) and (5), we have
R2 (τ ) = 0.025 f1 = 0.025 A cos ωt
R3 (τ ) = 0.001 f1 = 0.001 A cos ωt

Solution x2 (t)
Substituting R2 (τ ), ω2 , ξ 2 into Eq. (9) and initial condition into Eq. (9) and Eq. (11), we have

x 2 (t ) = 22.24x10

−3

A cos ωt (1 − e

−0.02t

(cos1.06t + 0.02 sin1.06t ))

Solution x3(t)
Substituting R3 (τ ), ω3 , ξ3 into Eq. (9) and initial condition into Eq. (9) and Eq. (11), we have
-5
-0.085t
x 3 (t)=5.534x10 Acosωt(1-e
(cos4.25t-0.02sin4.25t))

Trang 30


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
The motion equation under periodic load can be derived as folows:

 x1 (t)  0


  22.24x10-3 Acosωt*


 


(1-e-0.02t (cos1.06t+0.02sin1.06t)) 
 x 2 (t)  = 

 

-5

 5.534x10 Acosωt*

(1-e-0.085t (cos4.25t-0.02sin4.25t)) 
 x 3 (t)  

(13)

It is assumed that f = 423.6 cos 40t
1
Equation (13) becomes

 x1 (t)  0


 

-0.02t
(cos1.06t+0.02sin1.06t)) 
 x 2 (t)  = 9.42cos40t(1-e
 x (t)  0.023cos40t(1-e-0.085t (cos4.25t-0.02sin4.25t)) 

 3  

and

b. Lifting carrier in vertical direction under
load f2

defined

from

initial

condition.
when t = 0: x4 = 10 (m), x& = 1 (m/s).
4

The model of liffting carrier can be written
as

α 4 , β 4 are

(14)

(15)

&
m2 L &
x&
4 + k c x4 + C3 x 4 = f2

Substituting the values into Eq. (16), Eq. (17),

Skipping resistance force and f2 = 0, we
have x 4 (t ) = α 4 sin ω4t + β 4 cos ω4t

(16)

x&4 (t ) = α 4ω4 cos ω4t − β 4ω4 sin ω4t (17)
where ω 4 =

x i (t)=

1
ωi

kc

=

6594

m2 L

550

Oscillation system is of a harmonic motion
as

x 4 (t ) = 0.083sin11.989t + 10cos11.989t
= 11.989 (rad/s)

If resistance forces are considered, using
integral Duhamel to find the motion equation

t

∫ R (τ)e
i

-ξ i ω i (t-τ)

sinω i (t-τ)dτ+

0

e
where

we have β = 10 , α = 0.083
4
4

-ξ i ω i t

(19)

( α i sinω i t+β i cosω i t )

2
2
ωi = ωi 1 − ξi , ωi = 11.989 1 − 0.02 = 11.987 (rad/s)

α i , β i can be derived from the initial condition.

x 4 (t)=

R 4 (τ)
ω 24 +ξ 24 ω 24

 -ξ 4 ω 4 t

ξ ω
(cosω 4 t+ 4 4 sinω 4 t  +
 1-e
ω
4


e -ξ 4 ω 4 t ( α 4 sinω 4 t+β 4 cosω 4 t )

(20)

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Science & Technology Development, Vol 13, No.K4- 2010

R 4 (τ)e -ξ 4 ω 4 t  ξ 42 ω 42
+ω 4  sinω 4 t 2
2 2 
ω 4 +ξ 4 ω 4  ω 4



x&4 (t)=

(21)

e -ξ 4 ω 4 t ( ( ξ 4 ω 4 α 4 +β 4 ω 4 ) sinω 4 t+ ( ξ 4 ω 4 β 4 - α 4 ω 4 ) cosω 4 t )
when t = 0: x4 = 10 (m), x&4 = 1 (m/s).
Substituting above values into Eq. (20), we have β 4 = 10 .

(

Substituting above values into Eq. (21), we have 1= ξ 4ω4β 4 - α 4ω4

)

ξ ω β − 1 0.02 *11.989 − 1
−4
⇒ α4 = 4 4 4
=
= −63.4x10
ω4
11.987
Substituting α 4 , β 4 , ω4 , ω4 into Eq. (20), we have
x 4 (t)=

R 4 (τ)
1-e-0.24t (cos11.987t+0.02sin11.987t +
143.75

(

e

-0.24t

)

( -64.3x10

-4

sin11.987t+10cos11.987t

(22)

)

With the force of load is a costant, the resistance force is assumed to be f2 = Smax = 1736.76 N
Substituting R i (τ) = f2 = 1736.76(N) into Eq. (22):

(
( -64.3x10

)
sin11.987t+10cos11.987t )

x 4 (t)=12.08 1-e-0.24t (cos11.987t+0.02sin11.987t +
e

-0.24t

-4

(23)

Or

The force of load is periodic, the resistance force of the system is assumed to be
f2 = 1736.76 cos 40t

Substituting R i (τ) = f2 = 1736.76 cos 40t into Eq. (22)
x 4 (t)=

1736.76 cos 40t
1-e-0.24t (cos11.987t+0.02sin11.987t +
143.75
e

-0.24t

(
( -64.3x10

-4

)
sin11.987t+10cos11.987t )

(

Or x (t)=12.08cos40t 1-e-0.24t ( 0.172cos11.987t+0.02sin11.987t )
4
2.3 Simulation Results
a. The carrier travelling along rail under
load f1

Trang 32

)

(24)

From the motion equation, the system can
be simulated to describe the oscillation and
displacement of the robot on time and use Eq.
(10) to define displacement of point [10].


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
The force of load is constant, the resistance

From Eq. (12), the system motion is as

force is assumed to be f1 = 423.6 (N)

follows:


 x1 (t)  0


 
-0.02t
( cos1.06t+0.02sin1.06t )
 x 2 (t)  = 9.42 1-e
 x (t)  
-0.085t
 3  0.023 1-e
( cos4.25t+0.02sin4.25t )


(

)

(

)








x1(t) = 0, the plot x2(t) and x3(t) is shown in Fig. 2.

Fig. 2. System oscillation under constant with ω = 1.06 (rad/s)

Fig. 3. System oscillation under constant load with ω = 4.25 (rad/s)

Trang 33


Science & Technology Development, Vol 13, No.K4- 2010
The point’s displacement is defined by the principle of superposition.
From Eq. (8), we have
 u 1 (t) 








u
(t)
 2 

=






u
(t)

 3 





(

)

 0 .2 4 1 -e -0 .0 2 t ( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t ) +

 0 .2 3 x 1 0 -4 1 -e -0 .0 8 5 t ( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )

 -0 .2 9 1 -e -0 .0 2 t ( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t ) 

-4
-0 .0 8 5 t
( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )
 9 .4 3 x 1 0 1 -e

-0 .0 2 t
( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t ) +
 -0 .3 1 -e

-4
 4 1 .6 3 x 1 0 1 -e -0 .0 8 5 t ( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )

(

)

(

)

(

)

(

)

(

)














The point’s displacements are given in Table 2.
Table 2. The displacement of points

Time

Displace

Displace

Displace

t (s)

-ment u1 (m)

-ment u2 (m)

-ment u3 (m)

0.02

x10
0.04

-5

0.08

-5

-2.602

-4

x10

-4

-4.5204
x10-5
-1.952
x10-4

4.6777

-5.956

-4.505

x10-4

x10-4

x10-4

8.3882

-0.0011

-8.112

x10
0.1

x10

2.0415
x10

0.06

-6.1618

4.8178

-4

x10-4

0.0013

-0.0017

-0.0013

Table 3. Point Displacement

Trang 34

Time

Displace

Displace

Displace

t (s)

-ment u1 (m)

-ment u2 (m)

-ment u3 (m)

0.02

3.3624 x10-5

-4.3007 x10-5

-3.2709 x10-5

0.04

-5.971 x10-6

7.6123 x10-6

5.9202 x10-6

0.06

-3.455 x10-4

4.3995 x10-4

3.4483 x10-4

0.08

-8.387 x10-4

0.0011

8.4055 x10-4

0.1

-8.622 x10-4

0.0011

8.6695 x10-4

(25)


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010

With the force of load is periodic, resistance force of the system is assumed to be

f1 = 423.6 cos 40t
From Eq. (18), we have

x1 (t)  0


 

-0.02t
(cos1.06t+0.02sin1.06t)) 
x 2 (t) = 9.42cos40t(1-e
x (t)  0.023cos40t(1-e-0.085t (cos4.25t-0.02sin4.25t)) 

 3  
The oscillation plot of x2(t) and x3(t) are described in Fig. 4 and Fig. 5.

Fig. 4. System oscillation under periodic load with ω = 1.06 (rad/s)

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Science & Technology Development, Vol 13, No.K4- 2010

Fig. 5. System oscillation under periodic load with ω = 4.25 (rad/s)

The point’s displacement is defined by principle of superposition.
From Eq. (8), we have

(

)

 u 1 (t)  
-0 .0 2 t
( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t ) +

  0 .2 4 co s4 0 t 1 -e

 
-4
-0 .0 8 5 t
( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )

  0 .2 3 x 1 0 co s4 0 t 1 -e
 u (t)  
-0 .0 2 t
( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t )  2   -0 .2 9 co s4 0 t 1 -e
=

 

  9 .7 4 x 1 0 -4 co s4 0 t 1 -e -0 .0 8 5 t ( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )

 
 u 3 (t)   -0 .3 1 co s4 0 t 1 -e -0 .0 2 t ( co s1 .0 6 t+ 0 .0 2 sin 1 .0 6 t ) +

 


  4 2 .3 6 x 1 0 -4 co s4 0 t 1 -e -0 .0 8 5 t ( co s4 .2 5 t+ 0 .0 2 sin 4 .2 5 t )

 

(
(

(

)

(

(

)

)

)

)
















(26)

The point displacement are given in Table 3.
b. Lifting the table in vertical direction under load f2

Resistance force is skipped and f2 = 0. From Eq. (18), the oscillation system is of harmonic motion:

x 4 (t ) = 0.083sin11.989t + 10 cos11.989t

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TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010

Fig. 6. Harmonic motion of system with ω = 11.989 (rad/s)

Fig. 7. System oscillation under constant load with ω = 11.978 (rad/s)

With the force of load is costant, the
resistance force is assumed to be f2 = Smax =
1736.76 N.
From

Eq.

(

-0.24t

x 4 (t)=12.08 1-e

(23),

we

With the force of load is periodic,
resistance force of the system is assumed to be

f2 = 1736.76 cos 40t

have

( 0.172cos11.987t+0.02sin11.987t ) )

Trang 37


Science & Technology Development, Vol 13, No.K4- 2010

(

From Eq. (24), we have x 4 (t)=12.08cos40t 1-e

-0.24t

( 0.172cos11.987t+0.02sin11.987t ))

Fig. 8. System oscillation under periodic load with ω = 11.987 (rad/s)

From the above plots, it can be realized

There are three computers are used to

that if we change the vibration frequency or

implement the control logic throughout the

load, the system oscillation and displacement

factory: host computer, client computer, and

will

vibration

station computer. The host computer’s function

frequency is depends on lifting body mass,

is managing the database of the system, the

lifting height, stiffness proportionality…Hence,

client computer’s function is handling in/out

if we change initial condition design, we will

operations, and the station computer’s function

iditificate oscillation modes and kinematic

is

displacement of system.

AS/RSsystem.

be

change.

Alternatively,

monitoring

and
The

controlling
control

the
system

architechture is designed to meet the demand of
3. CONTROL SYSTEM DEVELOPMENT

Trang 38

a AS/RS is shown in Fig. 9.


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
Client computer with
Shopfloor Management software

Host computer with
Warehouse Management Software

LAN Network

Station Computer n
with SCADA software

Station Computer
with SCADA software

...

AS/RS
Line 1

AS/RS

Line n

Fig. 9. System control architecture for AS/RS

Fig. 10. Warehouse Management software Interface

In other words, the control system is
composed of two control levels: management
control

and

machine

control.

The

communication between them is via LAN
network. As for management control, a server
host computer is installed with Warehouse
Management software which connect to the
warehouse database using Microsoft SQL
Server framework. The server host can perform
tasks, such as supplier management, customer
management, items management, warehouse

structure management. A barcode system is
used for the item’s identification in warehouse.
The interface of Warehouse Management

software is shown in Fig. 10.
As for the machine control, a PAC
5010KW with SCADA system is implemented
to control the motion of robot for in/out
operations as shown in Fig. 11, and the control
panel on AS/RS, Fig. 12. The design has
allocated for VIKYNO company’s warehouse
as shown in Fig. 13.

Trang 39


Science & Technology Development, Vol 13, No.K4- 2010

Fig. 11. SCADA Interface and PAC 5510KW

implemented on the AS/RS model

Fig. 13. The allocation warehouse of VIKYNO for
Fig. 12. Control panel of AS/RS model

4. CONCLUSION

AS/RS implementation.

calculate program was established to verify the
behavior of the system and the system

In this paper, mathematical model of the
AS/RS is established with several oscillation

modes and the kinematic displacement of the
system are found respectively. The generalized

Trang 40

identification process. Finally, the development
of this system have been done, but the
experimental data yet to finish at this time of
this writing. It is our works in the future.


TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010
PHÁT TRIỂN HỆ THỐNG LƯU KHO TỰ ĐỘNG
Chung Tấn Lâm(1), Nguyễn Tường Long(2), Phan Hoàng Phụng(2), Lê Hoài Quốc(3)

(1) PTN Trọng điểm Quốc gia Điều khiển số & Kỹ thuật hệ thống (DCSELAB), ĐHQG-HCM
(2) Trường Đại học Bách Khoa, ĐHQG-HCM
(3) Sở Khoa học Công nghệ Tp.HCM

TÓM TẮT: Bài báo này trình bày mô hình toán học của hệ thống lưu kho tự động (Automated

Storage/Retrieval System - AS/RS). Các chế độ giao động và chuyển vị của hệ thống được khảo sát dựa
trên mô hình cơ sở trên. Với mô hình này, việc thiết kế hệ thống robot đưa vào lấy ra sẽ hiệu quả hơn
trước khi chế tạo. Ngoài ra, một mô hình hệ thống kho hàng tự động cùng với hệ thống điều khiển đầy
đủ được thiết kế và cài đặt để thấy được sự hiểu quả của giải pháp đưa ra. Nghiên cứu này là một phần
dự án nghiên cứu chế tạo thử nghiệm của Sở khoa học Công nghệ để đáp ứng yêu cầu sản xuất kho
hàng tự động cho công ty VIKYNO nói riêng, và đáp ứng nhu cầu của các công ty Việt nam nói chung.
Từ khóa: AS/RS, Automated storage/retrieval system.

[5]. Sangdeok Park1 and Youngil Youm2,

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