Phát triển hệ thống lưu kho tự động

Phát triển hệ thống lưu kho tự động.

TẠP CHÍ PHÁT TRIÊN KH&CN, TẬP 13, SÓ K4 - 2010

DEVELOPMENT OF AN

'ITOMATED STORAGE/RETRIEVAL SYSTEM

Chung Tan Lam", Nguyen Tuong Long”, Phan Hoang Phung”, Le Hoai Quoc”

(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 09", 2009, Manuscript Revised December 29", 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

1 INTRODUCTION

An AS/RS is a robotic material handling

system (MHS) that can pick and deliver

material in a direct - access fashion The

selection of a material handling system for a

given manufacturing system is often an

important task of mass production in industry

One must carefully define the manufacturing

environment, including nature of the product,

manufacturing process, production volume,

operation types, duration of work time, work

station characteristics, and working conditions

in the manufacturing facility

Hence, manufacturers have to consider

several specifications: high throughput

capacity, high IN/OUT rate, hight reliability

and better control of inventory, improved

safety condition, saving investerment costs, managing professionally and efficiently This system has been used to supervise and control for automated delivery and picking [1], [2], (31

In this paper, several design hypothesis is given to propose a mathematical model and emulate to iditificate oscillation modes and kinematic displacement of system based on initial conditions of force of load As a results,

we decrease error and testing effort before manufacturing [4], [5] No existing AS/RS met all the requirements Instead of purchasing an existing AS/RS, we chose to design a system for our need of study period and present manufacturing in VietNam

This works was implemented at Robotics Division, National Laboratory of Digital Control and System Engineering (DCSELAB)

2 MODELLING OF AS/RS

An ASIRS is a robot that composed of (1)

a carriage that moves along a linear track (x-

axis), (2) one/two mast placed on the carriage,

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

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

that can extend its length in both direction is

put on the table, The motion of picking/placing

an object by the shuttle-picking device is

performed horizontally on the z-axis

In this paper, an AS/RS is considered a

none angular deflection construction in cross

section in place where having concentrated

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

as follows:

The weight of construction post is

concentrated mass in floor level (Fig 1)

Structural deformation is not depend on bar axial force Assume that the mass of each part

in AS/RS is given as mj, m;, mạ, and mụ is lifting mass

When operation, there are two main motions: translating in horizontal direction with load f,; translating in vertical direction with load f; The initial conditions of AS/RS are lifting mass, lifting speed, lifting height, moving speeds, inertia force, resistance force, which can be used to establish mathematical model of AS/RS and verify the system behavior

The assumed parameters of the AS/RS are given in Table 1

H

———

Bari

ELHL mi]

||

Kix fxs

Fig 1: Model of AS/RS

TẠP CHÍ PHÁT TRIÊN KH&CN, TẬP 13, SÓ K4 - 2010

Table 1 Parameters of the AS/RS

6EI 6EL

Case 1: Horizontal moving along steel Soy oT

rail with load f, [7]

It is assume that (1) Structural deformation with x, -— , then kị=k;

is not depend on bar axial force; (2) The mass

where E: elastic coefficient of

in each part of automated warehouse cranes is -

material given as m,, m;, m3, in there, m, is lifting

I: second moment of area mass; (3)When the system moves, there are

two main motions: travelling along steel rail

proportionality underload f, and lifting body vertical direction

underload f;

The following model for traveling can be

obtained:

my + hy (xy =x;)+ Cj#

⁄

yyy + hy (a — xị) + K2 (xạ — xã) =0

mys + ky (x3 — 9) +Cyk 3=0

where m,3 = m,+ my +m, +m,

Case 2: Vertical moving with load f, [8]

mr

k

D : stiffness of cable ,

my pty +k oxy + Cye g= fy Where my, = m+

: diameter of cable 2.2 Solution of motion equation

a, Travelling along steel rail underload f,

m3 0 0 Ất ky “ky 0

0 mạ 0 X + “ky ky +k, “ky

0 0 mạ #3 0 “ky ky

or in the matrix form M¥ + kx = F

Solution of Eq (1) can be solved by

superposition method [9] as the followings:

Eq

Eigen problem: 46 = afgo” = (kM) =0

that satisfy det (« ~Mø` ) =0

where $ :nlevel vector

o : vibration frequency (rad/s)

2

ky-m) 30 “ky 0

2 det “ky ky + ky M730 ky =0

2

L

At the position x4 = ý

Substituting constant values in Table | into

(3), we have

afb) G29 Ms} Keka 1:2 |ees‡=9

T 2

~ 3) ky = 0,

bldsr 62 tss}|Ar Kịtha kp ie = oy

0 ky ky | |#y

=> b= 41.183x10>

Ifa and b are positive, g; values is as follow

5 = {0025 0.031 -0.033}" 5 45 = {0.001 -0041 01817

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

oy =0(rad/s),

@y = 1.065 (rad/s), @, = 4.254 (rad/s)

The solutions ¢, from the equation (k- Moy? )g, = 0 are as follows:

Q)

0 (rad/s) : @, values is any

5 = 1.135 (rad/s)

1

by = {I -1252 -1338}

@)

2 =18.095(rad/S)

153.073)"

These @, need to be satisfied 9,” kg,

“by Hy = 09

2 > a = 40.025

TẠP CHÍ PHÁT TRIÊN KH&CN, TẬP 13, SÓ K4 - 2010

It can be seen that the condition

67 Mg =1 is satisfied

If resistance force is skipped, the motion

equation will be written as the followings:

H+ = 9" FO

and n individual equation can be written:

3;0)+@j x¡(0) = Ril)

t

x= 1 Jaesn @j(L—+)đ+ + g; SÌn øj/ + ị €0 jf

%

R

x (0 = i)

0; sin @;f — a;0; Cos @;f — B;@; Sin @;t

a,and f; can be specified from initial

conditions

T

x|~o =úy Mều

Tờ,

i,| <9 = 9, Mek

Geometric inversion can be defined by

principle of superposition:

UO=LO IOP Iie OM 4g x9 OF +1 4p In 0]

Displacement of point is defined by

principle of superposition [9]

a

¡(9= % 4x0 ic

If resistance forces are considered

Using integral Duhamel to find motion

equation [9]:

Boje

it £0; (t)

x¡(Ù=— Í Rị()e sin, (t-t)dr +

5; 0

Eat

e “FT (a,sinØ,t+B,cos,t)

#Œ)=# 70)

Rs() =04025/,(0)

Rạ() = 0.001f,(0)

Using integral Duhamel to find motion equation [9]

(6)

7)

where : @ =;

6;: damping ratio

St (£2 —2

Rie ý a jie:

OE OF

e**" (Seg, +5) sind, t+ (EoB.- «48, )cosat)

We find a; and 8, value based on initial condition

Displacement of point is defined by principle of superposition (Eq (8))

Influential dynamic load act (on warehouse cranes in some cases

The acting force is a constant and system has influential resistance force

It is assumed that f, = W, = 423.6 (N) From Eq (4) and Eq (5), we have

Ry (1) = 0.025 f, = 0.025* 423.6 = 10.59 (N) Ry() = 0.001 f; = 0.001* 423.6 = 0.42 (N9)

From Eq (10) x7 (0-9.42(1-6° 0t (cos1,06t+0,02sin1 ost))

7 =øs\ll~ế22 2 =1.065j1— 2 = ds

By = Wy \1- Ey" = 1.0651 — 0.027 = 1.06 (rad/s) with 04 =425:

3 fl 83° = 4.254V1-0.02" = 4.25 (rad/s) Substituting R(t), 3,3 into Eq (9) and

Substituting the values: 8 (z),Ø„,ša into from Eq (9) and Eq (11), we have a, =0,

Eq (9), and from initial condition: By =0:

4, |,-9 = 627 Mey 0p =0

derived as:

From Eq (6) and Eq (7) with a, =

x] |°

X(t) p=} 9.42(1-¢° (cost.06t+0.02sin1.06t)) (12)

x, 0.023(1-e°*" (cos4.25t+0.02sin4.25t))

With the force of load is periodic, resistance force of the system is assumed to be i = Acosat From Eq (4) and (5), we have

Ry(r) = 0.025 fy = 0.0254 cos ot

R3(r) = 0.001f, = 0.001 A coset

Solution x; (t)

Substituting R, (r),,,E) into Eq (9) and initial condition into Eq (9) and Eq (11), we have

—0.02t

x= 22.24x10Ở 4eosø(1—e (cos 1.06 + 0.02 sin 1.06r))

Solution x(t)

Substituting R;(t),G3,£5 into Eq (9) and initial condition into Eq (9) and Eq, (11), we have

-0.085t X3(1)°5.534x10™ Acoseat(1-¢ (cos4.25t-0.02sin4.25t))

The motion equation under periodic load can be derived as folows:

TẠP CHÍ PHÁT TRIÊN KH&CN, TẬP 13, SÓ K4 - 2010

xO) {°

x, (b= (1-e°°"(cos1.06t+0.02sin 1.06t)) 5.534x10 Acosot#

x,(t) (1-e*"™*(cos4.25t-0.02sin4.25t)) Itis assumed that f; = 423.6c0s 401

Equation (13) becomes

x0) 0

xp (0) }=49.42c0s401(1-0°-9" (cos! 06t+0.02sin!.06t)) (4)

-0.085t

x3()} [0.023cos40t(1-e (cos4.25t-0.02sin4.25t))

b Lifting carrier in vertical direction under and a4,64are defined from initial

The model of liffting carrier can be written when t = 0: x4 = 10 (m), iy =1(m/s)

AS my) Z4 +k oxy+Cyi ga fy (15) Substituting the values into Eq (16), Eq (17), Skipping resistance force and f, = 0, we we have g,=10 a4 = 0.083

have x,())=a,sin@,t+B,cosa,t (16) Oscillation system is of a harmonic motion

%,() =a,0, cos@,t - B,o, sinw,t (17) as

x, (t) = 0.083sin | 1.989 + 10cos 11.9891

k, — 6594

where @4 = = = 11.989 (rad/s)

my), 550

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

Lt

x, (=— | R, (te sina, (t-t)dt+

e*° (a,sinG,t+B,cosd,t)

where i = ON 1? > © = 11.9891 ~0.022 =11.987 (rad/s)

a8; can be derived from the initial condition

R,(t) 4 — ,, 6,0, —

x, (== +55] 1-e* (coso,t+ a(t) Seo? ( SH + sino, t | + 4 20)

e*"" (a, sind ,t+B ,cos@,t)

X,(= Ru@eSS" ( Sim) +0, |sind,t -

e*° ((E,0,0,+B,G, )sind, t+ (£,0,B,- 4,5, )cosd,t)

when t= 0: x= 10(m), &4 =| (m/s)

Substituting above values into Eq (20), we have By = 10

Substituting above values into Eq (21), we have 1= (E4e4Bq- a454)

_ §40484 —! _ 0.02 *11.989 4

11.987

4

Substituting a4,f 4 ,4,G4 into Eq (20), we have

R,@)

143.75

e 92 (<64.3x102sin11.987t+10cosl 1.9871)

With the force of load is a costant, the resistance force is assumed to be f = Sax = 1736.76 N Substituting R ; (t) = f; = 1736.76(N) into Eq (22):

x, ()=12.08(1-e"* (cos! 1.987t+0.02sin11.987t) +

e°* (-64,3x10“sin11.987t+10cosl 1.987t)

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

fa, = 1736.76 cos 401

Substituting R; (t)= fy = 1736.76 cos 401 into Eq (22)

1736.76 cos 40t

143.75

e°*" (-64.3x10"sin1 1.987t+10cos1 1.987t)

x, (0= (1-2? (cos1 1.987t+0.02sin!1.987t) +

Or x4 (t)=12.08cos40t (120 (0.172cos!1.987t+0.02sin1 1.9871)) (24)

2.3 Simulation Results From the motion equation, the system can

a, The carrier travelling along rail under

(10) to define displacement of point [10]

TẠP CHÍ PHÁT TRIÊN 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 f, = 423.6 (N) follows:

x) |°

x(t) }=49.42(1-¢"" (cos 1.06t+0.02sin1.06t))

x;() 0.023(1-e"* (cos4.25t+0.02sin4.25t))

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

ee

bè Sco

Tweet) ANA

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

XG()=0.023"(exp-0.085).“cosi4.25*}+0,02*sin(4 25°)

Tae L

Fig, 3 System oscillation under constant load with @ = 4.25 (rad/s)

The point’s displacement is defined by the principle of superposition

From Eq (8), we have

The point’s displacemei

u(t)

us| _

u(t)

-0.29(1-e””?! (cos1.06t+0.02sin1.06t))-

nts are given in Table 2

0.24 (1-e°°* (cos!.06t+0.02sin! 06t))+

0.23x10% (1-0 (cos4.25t+0.02sin4.25t))

25)

9.43x10% (Ie (cos4.25t+0.02sin4.25t))

-0.3(1-e°°" (cos! 06t+0.02sin1.06t))+

41.63x10" (1-e°"** (cos4.25t+0.02sin4.25t))

Table 2 The displacement of points

t(s) -ment ul (m) -ment u2 (m) -ment u3 (m)

Table 3 Point Displacement

t(s) -ment ul (m) -ment u2 (m) -ment u3 (m)

0.02 3.3624 x10" -4.3007 x10™ -3.2709 x10™

0.06 -3.455 x10 4.3995 x10" 3.4483 x10"

0.08 -8.387 x10" 0.0011 8.4055 x10"

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

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

Ay = 423.6 cos 40¢

From Eq (18), we have

x0] [0

x;( }=49/42cos40(1-e Ð*? (cos! 06t+0.02sin.068))

x3(0)] {0.023cos400(1-c°7-985* (co54.25t-0.02sin4.251))

The oscillation plot of x(t) and x;(t) are described in Fig 4 and Fig 5

“X2I)=9.42'cos40'f(.ckp(0 029) “(605(1.06%)+0.02"si 7.08")

True

TRet

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

x31)=0023'e0540"(1-er01-0.085% *eos(4.25¢}+0.02"sinié 254)

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

The point’s displacement is defined by principle of superposition

From Eq (8), we have

He 0.24cos40t (1 2 0 024 (cos1.06t+0.02sin1.061))+

0.23x107%eos40t (Le 9083! (cos4.251+0.02sin4.25t))

"a0 -0.29cos40t(t-e 0021 (cos1.06t+0.02sin1.061))- 06)

— Ìs74x10®eos4t (1 = 0 085t (cos4.25tr0.02sin4.251))

uạ(9 -0.31cos40t (1 "¬ .06tr0.02sin1.061))+

42.36x10ˆ 'cos40t (1 = 0 085t (cos4.25tr0.02sin4.251))

The point displacement are given in Table 3

b Lifting the table in vertical direction under load f,

Resistance force is skipped and f= 0 From Eq (18), the oscillation system is of harmonic motion: X,(¢) = 0.083sin 11.989¢+10cos 11.989