Actuators & Mechanisms Actuator sizing

Transcription

1 Course Code: MDP 454, Course Nae:, Second Seester 2014 Actuators & Mechaniss Actuator sizing

2 Contents - Modelling of Mechanical Syste - Mechaniss and Drives

3 The study of Mechatronics systes can be divided into five areas of specialty: 1. Physical systes odeling 2. Sensors and actuators 3. Signals and systes 4. Coputers and logic systes 5. Software and data acquisition 6. Controller design - Logic controller - Microprocessor - Microcontroller - Prograable controller - PC based controller

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5 Hardware, Software and Firware Hardware is the nae given to the physical devices and circuitry of the coputer. Software refers to the progras written for the coputer. Firware is the ter given to progras stored in ROMs or in Prograable devices which peranently keep their stored inforation. In other words, firware is the cobination of persistent eory and progra code and data stored in it. Typical exaples of devices containing firware are ebedded systes.

6 Robot Platfors (1) Indoor Robots DLR Gripper NASA Mars Rover Asio Huanoid Outdoor Robots Robot Base Station KUKA Manipulator

7 Robot Platfors (2) Aibo 4 legged Robot Qurio Huanoid Robocup Tea

8 Robot Platfors (3) Robot educational kits Robot sensors

9 Stepper, AC and DC otors

10 PLC and Microcontrollers

11 PC-based Measureent and Control Pc Board CAN BUS GPIB Serial/paralell

12 Engineering Software Matlab Labview HP-VEE IDL Linux Qt

13 Modeling of Mechanical Syste Displaceent Velocity Acceleration d k v a c F = k d F = c v F = a

14 Acceleration, Velocity and Displaceent M a v M d a v d Acceleration a a v v Velocity d d Tie (Siple vibration) Frequency (real achine) Displaceent

15 FFT Transforation Displaceent d = D sin n t Displaceent D Tie 1 T Frequency T Period, T n in [sec] k Frequency, f n = n = 2 f n = 1 T n k in [Hz = 1 /sec]

16 Aplitude FFT Transforation tie = + + Freq.(Hz)

17 Actuator sizing

18 Types of Motion and Motion Conversion Linear and Angular Motion The linear otion induced in a rigid object is governed by Newton s second law of otion F = a F is the resultant of all forces acting on the object, is the ass of the object and a is the resulting linear acceleration. The constant force F produces a constant acceleration a and oves the object of ass a certain distance s according to s =1/2 at 2 s is the displaceent and t is the tie Thus, the tie required to ove ass through distance s by eans of a constant force F is given by t 2s F

19 Types of Motion and Motion Conversion For angular otion, Newton s law reads:.. T J q T is the resultant of all torques acting on a ass rotating about a fixed axis, J is the oent of inertia of the ass about its axis of.. rotation and q is the angular acceleration and the angular displaceent equation analogous to that of linear otion is q =1/2 at 2 q is the angular displaceent. Solving for t yields. t 2Jq T

20 Exaple: Consider a rotary otion axis driven by an electric servo otor. The rotary load is directly connected to the otor shaft without any gear reducer (Fig. 1). The rotary load is a solid cylindrical shape ade of steel aterial, d=75, l=50, ρ=7800kg/ 3.The desired otion of the load is a periodic otion (Fig. 2). Fig. (1) Fig.(2)

21 Exaple (Cont.): The total distance to be traveled is 1/4 of a revolution. The period of otion is t cyc =250 sec., and dwell portion of it is t dw =100 sec., and the reaining part of the cycle tie is equally divided between acceleration, constant speed and deceleration periods, t a =t r =t d =50 sec. Deterine the required otor size for this application.

22 Displaceent 1 4 revolution θ d θ r θ total θ a 0 ta ta+tr ta+tr+td tcyc tie (sec.) θ' ax

23 Actuator Sizing Algorith: 1. Define the geoetric relationship between the actuator and load. In other words, select the type of otion transission echanis between the otor and load (N=reduction ratio). 2. Define the inertia and torque\force characteristics of the load and transission echaniss, i.e. define the inertia of the tool as well as the inertia of the gear reducer echaniss (J l, T l ). 3. Define the desired cyclic otion profile in the load speed versus tie (θ' l (t)). 4. Using the reflection equations developed above, calculate the reflected load inertia and torque/force (J eff, T eff ) that will effectively act on the actuator shaft as well as the desired otion at the actuator shaft (θ' (t)). 5. Guess a actuator/otor inertia fro an available list (catalog) (or ake the first calculation with zero otor inertia assuption), and calculate the torque history, T (t), for the desired otion cycle. Then calculate the peak torque and RMS torque fro T (t).

24 Actuator Sizing Algorith Cont.: 6. Check if the actuator size eets the required perforance in ters of peak and RMS torque, and axiu speed capacity (T p, T rs, θ' ax ). If the above selected actuator/otor fro the available list does not eet the requireents (i.e. too sall or too large), repeat the previous step by selecting a different otor. It should be noted that if a stepper otor is used, the torque capacity of the stepper otor is rated only in ters of the continuous rating, not peak. Therefore, the required peak and RMS torque ust be saller than the continuous torque capacity of the step otor. 7. Most servo otor continuous torque capacity rating is given for 25 o C abient teperature is different than 25 o C, the continuous (RMS) torque capacity of the otor should be derated using the following equation for a teperature, T T (25 rs rs o C) (155 Tep 130 o C)

25 Solution: 1) Deterine the Net inertia: where: J total =the total inertia reflected on the otor axis. J eff =the load inertias reflected on the otor shaft J =the otor rotor inertia. J eff J L J eff d r ( ( ) l) r ( (75 10 ) ) 75 ( ) kg. The ratio of otor inertia to reflected load inertia should be between one-toone and up to one-to-ten. J 1 1 ~ J 1 10 eff 2

26 The one-to-one is considered the optial atch (an ideal case), where the otor drives a purely inertia load and this inertia ratio results in iniu heating of the otor. Let us assue that we will pick a otor which has a rotor inertia sae as the load so that there is an ideal load and otor inertia atch. J J J total eff kg. J J eff kg. 2) Deterine the Net Torque: T( t) T ( t) T ( t) T ( t) total R 2 where: T total (t)=the total torque. T (t)=the torque generated by the otor. T R (t)=the resistive load torques on the syste, where T R (t) represent the su of all external torques. If the load torque is in the direction of assisting the otion, it will be negative, and net result will be the addition of two torques. The T R (t) ay include friction (T f ), gravity (T g ), and process related torque and forces (i.e. an assebly application ay required the echanis to provide a desired force pressure (T R ).

27 Displaceent T R ( t) 0 Ttotal ( t) T ( t) 3) Fundaental Equations for torque calculation: The torque and otion relationship is: J total q T ( J J ) q T eff The required torque to ove the load through the desired cyclic otion given in the figure can be calculated if the value of calculated. q 1 revolution 4 θ d θ r θ total θ a 0 ta ta+tr ta+tr+td tcyc tie (sec.)

28 4) Define the desired cyclic otion profile in the for of load (otor) speed versus tie: Fro the desired otion profile specification, we can deterine the velocity and acceleration of the actuator can be deliver using the kineatic relations ta tr td 50sec. 3 where: t a =acceleration ode tie. t r =constant speed ode tie. t d =deceleration speed ode tie.

29 Velocity Velocity Displaceent 1 revolution 4 θ d θ r θ total θ a 0 ta ta+tr ta+tr+td tcyc tie (sec.) θ' ax θ' ax tie (sec.) tie (sec.)

30 Velocity θ' ax tie (sec.) 418 rad/sec Acceleration diagra rad/sec 0 ta ta+tr ta+tr+td tcyc

31 Velocity Velocity Displaceent 5) Calculate the axiu speed q required to define this profile fro the otor: ax Fro displaceent otion profile: q 1 2 total q a q r q d revolution 4 note that: q( t) q( t) dt Fro velocity otion profile: 1 1 qtotal q ax ta q ax tr q ax t 2 2 ta 2t r td qtotal q ax ( ) 2 2q total q ax ta tr td q ax 20.9rad / sec q 60 n ax rp 2 d θ a 0 ta ta+tr ta+tr+td tcyc θ' θ' ax ax tie (sec.) tie (sec.) θ r θ d θ total tie (sec.)

32 Velocity 6) Calculate the angular acceleration : Note that: dq q dt Fro velocity otion profile: d q a q ax 0 q ax 20.9 q a dt ta 0 ta d q r q ax q ax 0 q r 0 dt ( t t ) t t d qd qd dt ( t 3 d a t r r 418rad a 0 qax t ) ( t a / sec r 2 t a r ) q t ax d 3 θ' ax 418rad / sec rad/sec tie (sec.) Acceleration diagra -418 rad/sec 0 ta ta+tr ta+tr+td tcyc

33 7) Use Fundaental equation for torque calculation at each ode: ( J T T J eff ) q T ( J J ) q N a eff a. ( J J ) q 0 T r eff r d ( J J eff ) q 0 ta ta+tr ta+tr+td tcyc d N. The torque diagra profile is shown in Fig rad/sec Acceleration diagra -418 rad/sec 8) The Peak torque (axiu torque): Hence, the peak torque requireent is T ax N N. Torque diagra N. 0 ta ta+tr ta+tr+td tcyc

34 9) T rs = root ean square torque over entire cycle: T rs t cyc 0 T t ( t) cycle 2 dt N. Torque diagra Fro torque diagra: N. T rs T a 2 t a T t a r 2 t t r r T t d d 2 t t d dw T 2 H t dw 0 ta ta+tr ta+tr+td tcyc where : T H =holding torque required in dwell ode=0 T rs (1.0032) Therefore, a otor which has rotor inertia of about kg. 2, axiu speed capability of 20.9 rad/sec(199.6 rp)or better, peak and RMS torque rating in the range of N. and N. range would be sufficient for the task ( ) N.

35 Mechaniss and Drives

36 Mechaniss and Drives

37 Rotating ass driven through a gear reducer Taking gear 1 as a free body gives The equivalent oent of inertia as

38 How to calculate equivalent inertia?

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41 Conversion of Rotary to linear Motion 1. Rack and pinion drives, 2. Power (lead) screws, 3. Linkages. If the load attached to the rack has ass, then, total equivalent oent of inertia equals Conversely, if the rack is the driver, then the oent of inertia J1 attached to the pinion shaft ust be reflected back to the rack, and the equivalent linear inertia as felt by the pinion driving the rack is

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49 Thank You For Your Attention! Questions?