Tài liệu Manipulator has 3 basic transmissions

aDANANG UNIVERSITY UNIVERSITY OF SCIENCE AND TECHNOLOGY P.F.I.E.V UNIVERSITY GRADUATIONUNDERGRADUATE THESIS SubjectTitle: DESIGN ROBOT CARTESIAN COORDINATE ROBOT SYSTEM AND MANUFACTURE A MODEL Đã định dạng: Giãn cách dòng: Ké Supervisor : MSc. Nguyen Đac Luc Student : Tran Quoc Bao Class : 10CLC1 Major : Automation Production Chapter I: Introduction I.1 Arguments for choosing the subjectObjective Technical sciences today are growing very strongly, led to great changes in production. That is the change of productive forces in all sectors by replacing human labor with machinery to ensure that labor productivity growth, yield and product quality. Hence the use of the machine, also known as handusing industrial robots in production are very popular because they meet the above requirements. As we all know robots have many advantages, especially quality Đã định dạng: Tiếng Anh (Mỹ) and accuracy, in addition to mention the high economic efficiency, can work in hazardous environments where people do not swear, the work requires careful not to be confused, gentle subtle manipulations require a level of skilled workers, and it is important not stress Robot as humans should be able to work all day. The learning, researching Robot within the framework robot design calculations will be the basis for us to calculate, design and control all types of industrial robots for production. Specifically, here we chose a subjects in calculatedion, designed of a robot that picks objects to serve arrangement, moving industrial products. I.2 Research Content  Analysis and choice select of design plans.  Set the kinetic equation for the system structure.  Design of mechanical components.  Design of controls.  Concludesion. Chapter II: Analysis and select design plans II.1 Some of the textures tTypical manipulators a) Cartesian coordinates Robot type Manipulator has 3 basic transmissions; translate according to the direction of the coordinate system (configuration TTT),. Wworkspace work with is rectangular blocks. Due to simple textures, this type of machine armmanipulator has high rigidity, precision mechanics, so it's easy to ensure often used to transport raw draftworkpices, assembly, welding plane … b) Cylindrical coordinates Robot type Workspace of Robot has hollow cylindrical shape. Normally, the first joint is rotation.. Example 3 degrees of freedom robot, has configure RTT configurationas that is shown as following picture. There are cylindrical coordinates type robots such as its robot Versatran robot of AMF (USA). c) Spherical coordinates Robot type Workspace of Robot has the globular shape. Normally, the stiffness of this robot is lower than the above two types. Example 3 degrees of freedom robot has, configuration RRR or RRT configuration. d) Coordinates biomimicry Robot type The first three movements are rotary motions;, The the first axis of rotation is perpendicular to the other two axes. The other oriented movements are rotary motion. Workspace of manipulators is almost identical a part of sphere. e) SCARA Robot type SCARA name stands for "Selective Compliant Articulated Robot Arm". This robot type is used in assembly work so SCARA is sometimes been interpreted as an acronym forcalled "Selective Compliance Assembly Robot Arm". Three first joints of this type robot of configuration Robotare RRT, the joint axes are vertical. II.2 Analysis and choice machine texturesselect structure a) Screw - Nut structure Chú thích [B1]: Hiệu chỉnh ký hiệu tiế Anh Advantages:  Speed to move quickly by attaching coaxial. Chú thích [B2]: Cần xem lại ưu điểm này!  High accuracy.  The common plan in the currentPopular in industrial CNC machines. DefectDisavantages:  Translation of each sectional screw nut depends on steps of screw thread so althought fast moving, accuracy is not high, depends very much on the fitting and assembling, construction products.  For more accurate replaceable screw actuator with ball screw. Chú thích [B3]: Chỉ cần nói: cần lắp r với độ chính xác cao This transmission decreases friction between the nut and the screw, advanced accuracy motion. b) Tooth belt – Slider structure Advantages:  Simple structure, low cost.  Work smoothly, without noise thanks to the flexibility of the belt, suitable for high-speed transmission.  Transmission between widely separated axes.  High performance transmission.  Convenient, easy to control. Defect:  There may be problems with size, great framework.  The force exerted on the drive shaft is large and must tighten the belt initial at fisrt.  The lifespan of the belt is usually lower.  Requires high precision, the construction and assembly meet more dificult.  The cost of making relatively high.  Difficulty in adjusting of the belt parallel to drag the axes the most accurate. c) Pneumatic Cylinder. Advantages:  Abundant energy source.  Used for many different purposes such as cleaning work, transmission in the machine …  Ability to transmit away by pipeline systems with small losses;  Pneumatic can drain out without causing harm to the environment.  High-speed transmission, flexible;  Easy to control with the reliability and high accuracy. Defect:  Power transmission is not large.  The accuracy is not high.  Pneumatic line is released into the environment can cause noise. d) Structure of navigation. Request:  The surface of the slider requires high corrosion resistance, 0,004÷0,008 (mm/year).  Request about the hardness of the slider relatively important.  Slider of the camera body, office machines, ironing machine should be fine processed by grinding or scraping method by means of polishing, grinding.  Flatness allows the slider 0.02 ÷ 0.03 mm / m.  Glossy of surface from level 6 to level 8.  The openings of the slider and the more fitting for it to be checked by painting and by 0.04 mm thick gauge. The guide slider: rectangular slider, swallow tail slider, roller slider cylindrical, V slider, cylinder slider. The transmission used To save in economic terms, while ensuring the requirements of the problem posed, we choose design Robot 3 degrees of freedom TTT, Robot has 3 translational links locate a point on the space in coordinates descarde, and a hand grip for gripping objects. Based on the analysis of transmission alternatives above, combined with the ability to calculate the supply and purchase of the necessary components and economic conditions, plans for tooth belt - slider transmission and pneumatic cylinders are selected to perform robot arm. Chapter III: Set the kinetic equation III.1 Attach the coordinates to the links III.2 Parameters tabulated DH Link θi αi di ai 1 0 90 d1 a1 Đã định dạng: Giãn cách dòng: Ké 2 -90 90 d2 a2 Đã định dạng: Giãn cách dòng: Ké 3 0 0 d3 a3 Đã định dạng: Giãn cách dòng: Ké Đã định dạng: Giãn cách dòng: Ké III.3 Dynamical equations 0 0 T3    1  0 0 1 1 0 0 0 0 0 a1  d3 d 2    d1  a2  a3   1  Chapter IV: Design of mechanical components IV.1 Choose axis guide We choose slided circular axis SDK Crom-plated with Ф8, hardness 58HRC. IV.2 Calculate and choose motor IV.2.1 Link 1  Total mass of belt and load m1 = 5 [kg]  Roller diameter D = 20 [mm]  Roller mass m2 = 0,1 [kg]  Friction coefficient of sliding surface µ = 0.04  Belt and roller efficiency η = 0,9  Frequency of power supply 60 Hz (Motor speed: 72 r / min). Belt speed v  .D.N 60   .2.72 60  7,5 mm/s Calculate the Required Torque TL  F .D m1.g..D   2,18.103 N.m 2. 2. Load inertia of belt and load 2   .2.103    .D  2 6 J m1  m1.    5.10 kg.m   5.   2.   2.  2 Load Inertia of Roller 1 1 2 J m 2  .m2 .D 2  .0,1.(2.103 ) 2  0, 05.106 kg.m 8 8 The load inertia is calculated as follows: J L  J m1  2.J m2  5,1.106 kg.m 2 Calculate the Acceleration Torque Ta  ( J 0  J L )   . s 180.n  f 2  905.J 0  4, 6.103 N.m Here θs = 1,8o, f = 60Hz, n = 3,6o / θs Jo : Rotor Inertia Required Torque TM  (TL  Ta )  2  1810.J 0  13,56.103 N.m Choose a Motor Select a motor that satisfies both the required torque and the permissible load inertia. Motor Rotor Inertia Permissible Load Output Torque Inertia [kg·m2] [kg·m2] 57BYGH2500B [N·m] 1.4x10-4 7x10-4 1,00 When the required torque is calculated by substituting the rotor inertia, TM is obtained as 0,267 Nm, which is below the output torque. Next, check the permissible load inertia. Since the load inertia calculated in above is also below the permissible load inertia, 57BYGH2500B can be used in this application. IV.2.2 Link 2  Total mass of belt and load m1 = 2 [kg].  Roller diameter D = 20 [mm].  Roller mass m2 = 0,1 [kg].  Friction coefficient of sliding surface µ = 0,04.  Belt and roller efficiency η = 0,9.  Frequency of power supply 60 Hz (Motor speed: 72 r / min). Belt speed v  .D.N 60   .2.72 60  7,5 mm/s Calculate the Required Torque TL  F .D m1.g..D   0.87.103 N.m 2. 2. Load inertia of belt and load Đã định dạng: Giãn cách dòng: Ké Đã định dạng: Giãn cách dòng: Ké 2   .2.103    .D  2 6 J m1  m1.   2.    2.10 kg.m  2.   2.    2 Load Inertia of Roller 1 1 2 J m 2  .m2 .D 2  .0,1.(2.103 ) 2  0, 05.106 kg.m 8 8 The load inertia is calculated as follows J L  J m1  2.J m2  2,1.106 kg.m 2 Calculate the Acceleration Torque Ta  ( J 0  J L )   . s  f 2  905.J 0  1,9.103 N.m 180.n Here θs = 7,2o, f = 60Hz, n = 3,6o / θs Jo : Rotor Inertia Required Torque TM  (TL  Ta )  2  1810.J 0  2,77.103 N.m Choose a Motor Select a motor that satisfies both the required torque and the permissible load inertia. Motor Rotor Inertia [kg·m2] Permissible Load Output Torque Inertia [kg·m2] [N·m] YH42BYGH47 1.4x10-4 7x10-4 0.55 When the required torque is calculated by substituting the rotor inertia, TM is obtained as 0,256 Nm, which is below the output torque. Next, check the permissible load inertia. Since the load inertia calculated in above is also below the permissible load inertia, YH42BYGH47 can be used in this application. IV.2.3 Link 3 Conditions for the piston can lift objects as: FP >= Mob =10N Then FP = p.S I choose pressure air compressor: p = 6bar = 6.105N/m2 Here : S  And S  FP 10   1, 67.105 m2 p 6.105  .d 2 4 d  4.S   4,6.10-3 m =4,6 mm We choose 2 axial plunger d=5(mm), Cruise 20mm IV.2.4 Hand grip  The weight will be required to pick up: 10N.  Size : long : L=15mm=0,015m. broad: a=25mm=0,025m. tall :b=10mm=0,01 m. Frictional force acting on the object is: Fms = k.FP, we choose sliding friction coefficient k = 0,3 Đã định dạng: Giãn cách dòng: Ké Đã định dạng: Giãn cách dòng: Ké Conditions for the piston do not drop object is: Fms >= Mob =10N hay k.FP>=10 Mà FP = p.S I choose pressure air compressor p = 6bar = 6.105N/m2. Here : S  And S  k.FP 0,3.10   0,5.105 m2. p 6.105  .d 2 4 d  4.S   4,5.10-3 m =2,5 mm. We choose plunger d= 4mm. Cruise 20mm Chapter V: Design of controls V.1 Construction algorithm flowchart V.2 The device and control program Solenoid valve 5/2 When electric signal doesn’t impact entrance (0), the gate (3) is blocked, gate (1) connected to the gate (2) and gate (4) connected to the gate (5). When there is the electrical signal that impacts at the door (0), the barrel valve will shift to the right, the gate (2) connected to the gate (3) and gate (1) connected to the gate (4) while the door (5) is blocked. When the impact signal at the door (0) is lost, under the action of the barrel valve springs return the original location. PIC16F887 PIC is a microcontroller RISC architecture, running a command a machine cycle (4 oscillator cycles). PIC16F887 they have 40 legs, each leg has a different function. Including some multiple-use leg (compound), each leg can act as a way to export / import (I / O) independently or as a special function used to communicate with peripheral devices.