Phát triển hệ thống lưu kho tự động

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.

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, 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 Trang 25 Science & Technology Development, Vol 13, No.K4- 2010 Structural deformation is not depend on bar 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 as follows: The weight concentrated mass of construction post is in floor level (Fig 1) x3 m3 Bar Bar 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 Parameters of the AS/RS Parameter Value m1 150 [kg] m2 30 [kg] mL 100 – 500 [kg] m3 20 [kg] ξ [%] 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 = EI x4 , k2 = EI ( L − x4 ) 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 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& + k c x4 + C3 x = f2 The following model for traveling can be obtained: mL kc : stiffness of cable D : diameter of cable & m13& x& + k1 ( x1 − x ) + C1x = f1 m23 & x& + k1 ( x2 − x ) + k ( x2 − x3 ) = & m3 & x& + k2 ( x3 − x ) + C2 x = 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 Trang 27 Science & Technology Development, Vol 13, No.K4- 2010 If resistance force is skipped, the motion  m13 0     0 m 23   0 m    3    k1 -k1   x1    + -k k + k -k   1 2   x2   -k k   x   & 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 ⇒ (k − M ω )φ = ) that satisfy det k − M ω = where φ : vibration frequency (rad/s) x4 =  k1 } -k1   ⇒ b = ±1.183x10 : : ω = 1.135 (rad/s) These   φ    23  -k1 k1 + k -k    φ31    -k  b φ32  = ω3 k2  φ    33  −3 T φ2 = {0.025 −0.031 −0.033} , φ = {0.001 T −0.041 0.181} φi : (3) need to be satisfied φiT kφi = ωi - φ T kφ = ω 2 2 φ21    ⇒ a = ±0.025 -k  a φ22  = ω2 k2 (2) φ1 values is any T If a and b are positive, φi values is as follows: Trang 28 φ3 = {1 −34.9 153.073}   from the equation ( k − M ωi )φi = are as follows: L -k1 k1 + k -k φi - φ T kφ = ω 3 { The solutions ω = 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 into  k1  a {φ21 φ22 φ23} -k1   (1)    φ = {1 −1.252 −1.338}  k -m ω -k   13  det  -k1 k1 + k -m 23ω -k  =   -k k − m3ω    At the position f1   ω = (rad/s) : n level vector ω     =    ω1 = (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 T & (t ) + Ω x(t ) = φ f (t ) x& equation [9] and n individual equation can be written: & x& i (t ) + ωi xi (t ) = Ri (τ ) x i (t ) = t (6) ∫ R (τ ) sin ωi (t − τ ) dτ + α i sin ωi t + β i cos ωi t ωi 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 − ξi 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 (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) t sinωi (t-τ)dτ + ∫ R i (τ)e i i ωi -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  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) Trang 29 Science & Technology Development, Vol 13, No.K4- 2010 2 Substituting R3 (τ ), ω3 , ξ3 into Eq (9) and 2 from Eq (9) and Eq (11), we have α = , ω2 = ω2 − ξ = 1.065 − 0.02 = 1.06 (rad/s) ω3 = ω3 − ξ3 = 4.254 − 0.02 = 4.25 (rad/s) Substituting the values: R2 (τ ), ω2 , ξ into β3 = : ( x (t)=0.023 1-e Eq (9), and from initial condition: 0   T x2 t =0 = φ2 M ° 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 derived as: From Eq (6) and Eq (7) with α = , β2 = : ( x (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 (t)  = 9.42 1-e  x (t)     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 into Eq (9) and initial condition into Eq (9) and Eq (11), we have x (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 (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 (t)  =      -5   5.534x10 Acosωt*  (1-e-0.085t (cos4.25t-0.02sin4.25t))   x (t)   (13) It is assumed that f = 423.6 cos 40t Equation (13) becomes  x1 (t)  0      -0.02t (cos1.06t+0.02sin1.06t))   x (t)  = 9.42cos40t(1-e  x (t)  0.023cos40t(1-e-0.085t (cos4.25t-0.02sin4.25t))      and b Lifting carrier in vertical direction under load f2 defined from initial condition when t = 0: x4 = 10 (m), x& = (m/s) The model of liffting carrier can be written as α , β are (14) (15) & m2 L & x& + k c x4 + C3 x = f2 Substituting the values into Eq (16), Eq (17), Skipping resistance force and f2 = 0, we have x (t ) = α sin ω4t + β cos ω4t (16) x&4 (t ) = α 4ω4 cos ω4t − β 4ω4 sin ω4t (17) where ω = x i (t)= ωi kc = 6594 m2 L 550 Oscillation system is of a harmonic motion as x (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+ e where we have β = 10 , α = 0.083 4 -ξ i ω i t (19) ( α i sinω i t+β i cosω i t ) 2 ωi = ωi − ξi , ωi = 11.989 − 0.02 = 11.987 (rad/s) α i , β i can be derived from the initial condition x (t)= R (τ) ω 24 +ξ 24 ω 24  -ξ ω t  ξ ω (cosω t+ 4 sinω t  +  1-e ω   e -ξ ω t ( α sinω t+β cosω t ) (20) Trang 31 Science & Technology Development, Vol 13, No.K4- 2010  R (τ)e -ξ ω t  ξ 42 ω 42 +ω  sinω t 2  ω +ξ ω  ω  x&4 (t)= (21) e -ξ ω t ( ( ξ ω α +β ω ) sinω t+ ( ξ ω β - α ω ) cosω t ) when t = 0: x4 = 10 (m), x&4 = (m/s) Substituting above values into Eq (20), we have β = 10 ( Substituting above values into Eq (21), we have 1= ξ 4ω4β - α 4ω4 ) ξ ω β − 0.02 *11.989 − −4 ⇒ α4 = 4 = = −63.4x10 ω4 11.987 Substituting α , β , ω4 , ω4 into Eq (20), we have x (t)= R (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 (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 (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 ) 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 (t)  = 9.42 1-e  x (t)   -0.085t   0.023 1-e ( cos4.25t+0.02sin4.25t )  ( ) ( )       x1(t) = 0, the plot x2(t) and x3(t) is shown in Fig Fig System oscillation under constant with ω = 1.06 (rad/s) Fig 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 (t)          u (t)    =       u (t)       ( )  -e -0 t ( co s1 t+ 0 sin t ) +   x -4 -e -0 t ( co s4 t+ 0 sin t )   -0 -e -0 t ( co s1 t+ 0 sin t )   -4 -0 t ( co s4 t+ 0 sin t )  x 1 -e  -0 t ( co s1 t+ 0 sin t ) +  -0 -e  -4  x 1 -e -0 t ( co s4 t+ 0 sin t ) ( ) ( ) ( ) ( ) ( )             The point’s displacements are given in Table Table 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 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 (t) = 9.42cos40t(1-e x (t)  0.023cos40t(1-e-0.085t (cos4.25t-0.02sin4.25t))      The oscillation plot of x2(t) and x3(t) are described in Fig and Fig Fig System oscillation under periodic load with ω = 1.06 (rad/s) Trang 35 Science & Technology Development, Vol 13, No.K4- 2010 Fig 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 (t)   -0 t ( co s1 t+ 0 sin t ) +    co s4 t -e    -4 -0 t ( co s4 t+ 0 sin t )    x co s4 t -e  u (t)   -0 t ( co s1 t+ 0 sin t )    -0 co s4 t -e =       x -4 co s4 t -e -0 t ( co s4 t+ 0 sin t )     u (t)   -0 co s4 t -e -0 t ( co s1 t+ 0 sin t ) +       x -4 co s4 t -e -0 t ( co s4 t+ 0 sin t )    ( ( ( ) ( ( ) ) ) )              (26) The point displacement are given in Table b Lifting the table in vertical direction under load f2 Resistance force is skipped and f2 = From Eq (18), the oscillation system is of harmonic motion: x (t ) = 0.083sin11.989t + 10 cos11.989t Trang 36 TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 13, SỐ K4 - 2010 Fig Harmonic motion of system with ω = 11.989 (rad/s) Fig 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 (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 (t)=12.08cos40t 1-e -0.24t ( 0.172cos11.987t+0.02sin11.987t )) Fig 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 CONTROL SYSTEM DEVELOPMENT Trang 38 a AS/RS is shown in Fig 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 AS/RS Line n Fig 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 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 TĨM TẮT: Bài báo trình bày mơ hình tốn học hệ thống lưu kho tự động (Automated Storage/Retrieval System - AS/RS) Các chế độ giao động chuyển vị hệ thống khảo sát dựa mô hình sở Với mơ hình này, việc thiết kế hệ thống robot đưa vào lấy hiệu trước chế tạo Ngồi ra, mơ hình hệ thống kho hàng tự động với hệ thống điều khiển đầy đủ thiết kế cài đặt để thấy hiểu giải pháp đưa Nghiên cứu phần dự án nghiên cứu chế tạo thử nghiệm Sở khoa học Công nghệ để đáp ứng yêu cầu sản xuất kho hàng tự động cho cơng ty VIKYNO nói riêng, đáp ứng nhu cầu công ty Việt nam nói chung Từ khóa: AS/RS, Automated storage/retrieval system [5] Sangdeok Park1 and Youngil Youm2, REFERENCE Motion Analysis of a Translating Flexible [1] Ashayeri, J and Gelders, L F, Warehouse design optimization, European Journal of Operational Research, 21: 285-294, 1985 [2] Jingxiang Gu, The forward reserve warehouse sizing problem, PhD Industrial and and Thesis, System dimensioning School of Engineering, Georgia Institute of Technology, 2005 [3] David E Mulcahy, Materials Handling Handbook, Published by McGraw-Hill Professional, 1999 [4] Kim H-S (Korea Maritime Univ.), Ishikawa H (Kongo Corp.), Kawaji S (Kumamoto Univ.), Control High for Stacker Rack Crane Warehouse Systems, Vol 2000; No Pt.1, 2000 Beam Carrying a Moving Mass, International Journal Korean Society of Precision Engineering, Vol 2, No 4, November, 2001 [6] GS TSKH Võ Như Cầu, Tính tốn kết cấu theo phương pháp động lực học, Nhà xuất xây dựng Hà Nội, 2006 [7] Ing.J.Verschoof, Crane Design, Practice, and Maintenance, Professional Engineering Publishing Limited London and Bury StEdmunds, UK, 2002 [8] Huỳnh Văn Hoàng, Trần Thị Hồng, Nguyễn Hồng Ngân, Nguyễn Danh Sơn, Lê Hồng Sơn, Nguyễn Xuân Hiệp, Kỹ thuật nâng chuyển, Nhà xuất Đại Học Quốc Gia TPHCM, 2004 Trang 41 Science & Technology Development, Vol 13, No.K4- 2010 [9] Nguyễn Tuấn Kiệt, Động lực học kết cấu khí, Nhà xuất Đại học quốc gia Tp Hồ Chí Minh, 2002 [10] Nguyễn Hồi Sơn, Vũ Như Phan Thiện, Đỗ Thanh Việt, Phương pháp phần tử Trang 42 hữu hạn với Matlab, Nhà xuất Đại học quốc gia Tp Hồ Chí Minh, 2001  ... độ giao động chuyển vị hệ thống khảo sát dựa mô hình sở Với mơ hình này, việc thiết kế hệ thống robot đưa vào lấy hiệu trước chế tạo Ngồi ra, mơ hình hệ thống kho hàng tự động với hệ thống điều... số & Kỹ thuật hệ thống (DCSELAB), ĐHQG-HCM (2) Trường Đại học Bách Khoa, ĐHQG-HCM (3) Sở Khoa học Công nghệ Tp.HCM TĨM TẮT: Bài báo trình bày mơ hình tốn học hệ thống lưu kho tự động (Automated... 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),