Precision Machine Design

Transcription

1 Precision Machine Design Topic 5 Applications of precision machine design theory Purpose: This lecture provides examples of how the use of error theory can help to characterize a design and provide insight into how a concept should evolve. Outline: Modeling philosophies Error motions caused by bearing deflections Errors caused by rigidly coupling a linear actuator to a linear motion carriage Selection of machine concept for a precision surface grinder Using error budget spreadsheets to guide the conceptual design process "The man who doesn't make up his mind to cultivate the habit of thinking misses the greatest pleasure in life. He not only misses the greatest pleasure, but he cannot make the most of himself. All progress, all success, springs from thinking" Thomas Edison 5-1

2 Modeling philosophies Practice by always trying to analyze objects you encounter in everyday life. Try to envision what would happen to a machine if each of its parts (individually) were made of rubber. Start simple and develop a model. Next, envision a very complex model and then simplify it and see if you arrive at the same simple model. Utilize superposition to model complex systems: M 5-2

3 Accuracy of models used also depends on the analysis methods Bending deformations: Learn to use singularity functions. Shear deformations: Learn to use energy methods. Boundary conditions: Play with foam models and explore upper and lower bounds. Number of intermediate coordinate frames used: Machine tools generally only need a coordinate frame applied at bearings, joints, or sudden changes in the geometry. 5-3

4 Error motions caused by bearing deflections The goal is to find the error components for a HTM lumped parameter representation of a kinematically supported carriage. Kinematic systems are considered here, but analysis is extendible to any number of contacts. Systems supported by a kinematic arrangement of bearings (i.e., not over constrained) are easier to model, manufacture, and maintain. Example: A carriage for fine positioning of an instrument head over a range of about 0.5 m: Cross section through vee rollers Cross section through outrigger roller Alumina rail, straight to 5-10 µm, costs about $500 in production quantities (1/2 the cost of stainless steel). There are many common types of kinematic bearing designs. 5-4

5 A ceramic (alumina) rail with five ABEC 7 bearings as bearings, preloaded from underneath by a friction drive roller: 5-5

6 Platten 0 Θ b5 = 1 0 Y F b5 F b4 X F b3 Bearing way F b1 1 Θ b4 = Θ b1 = 2 Θ F b3 = Θ b2 = 2 b2 Z degree dovetail slide with five bearing points. 2 2 Platten Y F b4 X F b5 F b3 Bearing way F b Θ F 1 b4 = Θ b1 = 2 b2 Θ b5 = Θ b3 = Θ b2 = 2 1 Z Platten 45 degree male Vee way. Note that with a single female Vee way, platten Abbe errors will be lower. Y F b4 0 1 Θ b3 = Θ b2 = X Vee Flat 0 1 Θ b4 = Θ b1 = F b1 Vee and Flat bearing ways. F b2 F b3 F b5 0 Θ b5 = 1 Z 0 5-6

7 The known parameters of the problem are: Pbi Bearing point coordinate vectors: P bi = X bi Y bi Z bi i = 1,5 Θfbi Direction cosine vectors of bearings: Θ bi = α bi β bi γ bi i = 1,5 Kbi Bearing stiffnesses. µvi Dynamic coefficients of friction. Ffj Generic applied force vectors. Pfj Generic forces' coordinate vectors. Γ Generic applied torques about all three axes. 5-7

8 There are 16 unknowns: Fbi Five bearing reaction forces. vx Steady state x direction carriage velocity. δbi Gap changes the five bearing points. εx Rotation about the X axis (roll). εy Rotation about the Y axis (yaw). εz Rotation about the Z axis (Pitch). δy Y direction errors. δz Z direction errors. 5-8

9 To solve for the unknowns: Force & moment balance (6 equations). ΣM X = 0 = Γ X + ΣM Y =0= Γ Y - ΣM Z =0= Γ Z + 5 i = 1 5 i = 1 ΣF X = 0 = - ΣF Y = 0 = ΣF Z = 0 = 5 i = 1 5 i = 1 5 i = 1 5 i = 1 N j = 1 µ vi v x + F fxj N j = 1 F bi β bi + F fyj N j = 1 F bi γ bi + F fzj F bi -Z bi β bi + Y bi γ bi 5 i = 1 N j = 1 µ vi v X Z bi + F bi Z bi α bi - X bi γ bi 5 i = 1 + -Z fj F fyj + Y fj F fzj N j = 1 + Z fj F fxj -X fj F fzj N j = 1 µ vi v X Y bi + F bi -Y bi α bi + X bi β bi + -Y fj F fxj +ΣX fj F fyj 5-9

10 Bearing force/deflection (5 equations). δ bi = F bi / K bi New bearing coordinates in the carriage's coordinate frame equal the old ones minus the bearing deflections: [P bi ] new = [P bi ] - [Θ fbi ]δ bi New bearing coordinates in carriage system are transferred to the bearing way system by: [P bi ] = 1 -ε Z ε Y 0 ε Z 1 -ε X δ Y -ε Y ε X 1 δ Z [P bi ] new Five equations for Y and Z position of the bearing points in the bearing way's coordinate system: Y bearing way i = - (Z bearing way i - Z bi )γ bi / β bi + Y bi The final set of 20 equations in 20 unknowns (ex to dz and x1n to z5n) is readily solved using a spreadsheet. 5-10

11 KINSLIDE.XLS To analyze of errors in a kinematically supported carriage (slide) Written by Alex Slocum. Last modified 5/26/95 by Alex Slocum Only change cells with boldface numbers. Assumes steady state motion (static friction effects are assumed to be zero) Desired steady state velocity 0.1 Contact point 1 coordinates, direction cosines, stiffness, friction coef. xb alph yb beta zb gama Kb1 100,000 md1: Dynamic mu Contact point 2 coordinates, direction cosines, stiffness, friction coef. xb alph yb beta zb gama Kb2 100,000 md2: Dynamic mu Contact point 3 coordinates, direction cosines, stiffness, friction coef. xb alph yb beta zb gama Kb3 100,000 md3: Dynamic mu Contact point 4 coordinates, direction cosines, stiffness, friction coef. xb alph yb beta zb gama Kb4 100,000 md4: Dynamic mu Contact point 5 coordinates, direction cosines, stiffness, friction coef. xb alph yb beta zb gama Kb5 100,000 md5: Dynamic mu Applied forces and locations (Default Fx1=vx*(md1+md2+md3+md4+md5)) 5-11

12 Actuator Gravity loads Cutting forces Ffx Ffx2 0 Ffx3 0 Ffy1 0 Ffy2 0 Ffy3 0 Ffz1 0 Ffz2-100 Ffz3 0 xf1 3 xf2 0 xf3 0 yf1 3 yf2 0 yf3 0 zf1 3 zf2 0 zf3 0 Applied torques Gamx 0 Gamy 0 Gamz 0 Results: Steady state velocity and bearing reaction forces vx 1.00E-01 Bearing deflections f1b 6.82E+01 d1b 6.82E-04 f2b -7.33E+01 d2b -7.33E-04 f3b -2.56E+00 d3b -2.56E-05 f4b -2.56E+00 d4b -2.56E-05 f5b 7.24E+00 d5b 7.24E-05 Error motions (microunits) HTM ex E E E+00 ey E E E-04 ez E E E E+00 dy dz

13 Errors caused by rigidly coupling a linear actuator to a linear motion carriage Moderately expensive if hand finishing is required. Provides no isolation from error motions. Provides very high axial stiffness. Y Bearing shape y Leadscrew shape y I2 y I1 X 5-13

14 A linear bearing's shape can be represented by: y 1 = y I1 + F 1 C δf1 + M 1 C δm1 α 1 = α I1 + F 1 C αf1 + M 1 C αm1 An actuator's shape can be represented by: y 2 = y I2 + F 2 C δf2 + M 2 C δm2 α 2 = α I2 + F 2 C αf2 + M 2 C αm2 After the bearing carriage and leadscrew nut are forced together and clamped, the following equilibrium conditions must exist: F 1 + F 2 = 0 M 1 + M 2 = 0 y 1 = y 2 α 1 = α

15 For purposes of simplifying terms in the analysis, assume the following notation: y = y I1 - y I2 α = α I1 - α I2 C δf = C δf1 + C δf2 C αf = C αf1 + C αf2 C δm = C δm1 + C δm2 C αm = C αm1 + C αm2 The following expressions for the equilibrium forces and moments are found: M 1 = y C αf - α C δf C αm C δf - C δm C αf F 1 = y C αm - α C δm C αf C δm - C δf C αm The initial shape of the linear bearing and leadscrew can be represented as a function of position along the bearing. 5-15

16 Example: Consider a typical machine with the following properties: Leadscrew length (m) 0.50 Leadscrew diameter (m) 0.05 Leadscrew modulus of elasticity (N/m2) 2.04E+11 Leadscrew moment of inertia (m4) 3.07E-07 Carriage force deflection compliance (m/n) 1.14E-09 Carriage moment slope compliance (rad/nm) 3.54E-08 Typical carriage deflections caused by the weight of a leadscrew and carriage deflections per micron of lateral misalignment: 1 Deflection (microns, microradians).1 Misalign lateral deflection Misalign angular deflection Weight lateral deflection Weight angular deflection Position along leadscrew (m) Some type of coupling is usually required to isolate the carriage from lateral motion error sources in the actuator. The region of minimum error in a machine is called its sweet spot. 5-16

17 Paddle coupling to isolate actuator error motions: Yoke Primary carriage Bed Leadscrew Secondary carriage Yoke Paddle The actuator is directly connected to a secondary carriage which has a "U" shaped yoke that straddles the carriage's paddle. For high stiffness and damping (machine tools): Single entry fluid film thrust bearings, which cannot resist lateral or angular motions, center the paddle. For instruments: Wire or coupling or wobble pin: The secondary carriage can also act as the anchor point for the moving end of a cable carrier. Disadvantages: The need for an extra carriage and the bearing (coupling) between the paddle and the yoke. If the actuator is a leadscrew, one still has to reckon with backlash and friction. 5-17

18 Since a carriage pitches about its center of stiffness, to minimize radial displacement effects on the ballscrew: The ballscrew nut should be located at the system's center of stiffness: Bearing blocks Bearing rails Center of stiffness As the carriage pitches, if the ballscrew nut is at the center of stiffness, it will not be subject to a radial displacement. Radial error motion of the screw will not cause pitch of the screw. In most machines, the effect is so small, that this is not a significant problem. 5-18

19 Selection of machine concept for a precision surface grinder Input will be required from all departments: Business Sales Engineering Service Manufacturing. It is wise in the design discussion process to identify the major structural accuracy issues: A design where the structural loop stiffness (tool-to-workpiece stiffness) is constant independent of position. The bearings rails are fixed, and the bearing carriages move with the spindle axis. A design where the structural loop stiffness (tool-to-workpiece stiffness) is maximized at any one position. The bearings carriages are fixed, and the bearing rails move with the spindle axis. Structural loop Constant stiffness design Maximum stiffness design 5-19

20 When designing machines, it is very instructional to Draw the machine with the axes in different positions. Take a highlighter marker and highlight the structural loop in each configuration. Look for changes in the shape and length of the loop. This can help indicate where attention should be paid. Structural loop Constant stiffness design Maximum stiffness design 5-20

21 Consider the major selection criteria for a precision surface grinder: Major criteria Manufacture Norm. Accuracy Cost ability Modularity Evolvability Ergonomics Priority priority Accuracy Cost Manufacturabi lity Modularity Evolvability Ergonomics Sum

22 The normalized priority represents the %/100 of the decision: Accuracy = 47% of the decision. For a world class machine, accuracy is very important. Cost = 16% of the decision. Cost is not that big of a factor when the customer has to make critical parts and there are few machines available to meet spec. Manufacturability = 9% of the decision. It should be manufacturable, because this also often affects the accuracy. Modularity = 9% of the decision. It would be nice, but far from critical, if the machine was made from modular components. Evolvability = 9% of the decision. It would be nice, but far from critical, if the machine could easily evolve into other types of grinders (e.g., lower cost, production, or profile). Ergonomics = 9% of the decision. It should be usable and serviceable, because this also often affects the accuracy. 5-22

23 All the accuracy parameters have equal importance. Accuracy Thermal Norm. Norm. Profile Surface finish stability Straightness Priority priority weight Profile Surface finish Thermal stability Straightness Sum Each of the accuracy represents about 12% of the design decision (Accuracy is 47% of the total decision). 5-23

24 The biggest issue is to reduce labor, even at the expense of paying more for purchased components. Purchase orders can be canceled, but employees are difficult to lay off. This can have negative effects: Suppliers can increase delivery times, which will increase delivery time for the machine. Cost Norm. Norm. Sale price Maintain Labor Components Priority priority weight Sale price Maintain Labor to produce Components Sum

25 Assembly requires lots of system's integration, and it is where most errors are made, so it is the most important factor. Manufacturability Norm. Norm. Bearings Castings Assembly Priority priority weight Bearings Castings Assembly Sum All the major components should be modular (shelf storable). Modularity Norm. Norm. X axis Y axis Z axis Spindle Priority priority weight X axis Yaxis Zaxis Spindle Sum

26 The most important evolvability issue is can the design be scaled so costs can be reduced when designing a machine series? Evolvability Norm. Norm. Profile Production Different Sizes Priority priority weight Profile Production Different sizes Sum The person who spends the most time using the machine is the most important. Ergonomics Norm. Norm. Machinist Serviceman Seal maintenance Priority priority weight Machinist Serviceman Seal maintenance Sum

27 In reality, many designs would be considered, including those of the competition. Here are the "final" results which show the two "best" designs: Property Constant K Maximum K Accuracy Profile (avg. X,Y,Z) Table: X 5 5 Vertical feed: Y 5 3 Horizontal feed: Z 5 3 Surface finish (avg. X,Y,Z) Table: X 7 7 Vertical feed: Y 5 5 Horizontal feed: Z 7 3 Thermal stability (avg. X,Y,Z) Table: X 5 5 Vertical feed: Y 5 5 Horizontal feed: Z 5 5 Straightness (avg. X,Y,Z) Table: X 5 5 Vertical feed: Y 5 3 Horizontal feed: Z 7 3 Cost Sale price 5 6 Maintain 5 5 Labor to produce 2 5 Components to buy 5 3 Manufacturability Bearings (avg. X,Y,Z) Table: X 1 1 Vertical feed: Y 7 7 Horizontal feed: Z 3 7 Castings (avg. X,Y,Z) Table: X 10 5 Vertical feed: Y 5 6 Horizontal feed: Z 5 5 Assembly (avg. X,Y,Z) Table: X 1 1 Vertical feed: Y 5 7 Horizontal feed: Z

28 Modularity Table: X 5 5 Vertical feed: Y 5 5 Horizontal feed: Z 5 5 Spindle 5 5 Evolvability Profile 7 7 Production 7 7 Different sizes 5 5 Ergonomics Machinist 5 5 Serviceman 5 5 Seal maintenance 5 2 Constant K Maximum K Desirability A sensitivity analysis can be done with the spreadsheet to see how variations in decisions affect the outcome. If small variations can make the decision change, then both concepts need to be evaluated in greater detail. If several designs are very closely rated, more details should be investigated. A preliminary error budget should be made to check the accuracy in different configurations. 5-28

29 Using error budget spreadsheets to guide the conceptual design process When the design selection process narrows the choice down to a couple concepts, how do you make the final decision? Consider the issue of a constant stiffness design verses a maximum stiffness design. Maximum stiffness design: When the grinding wheel is close to the inside edge of the table, a very tight structural loop is obtained. Maximum stiffness can improve surface finish. Maximum stiffness can make it possible to grind hard materials with minimal subsurface damage. Constant stiffness design: As the wheel moves across the table: There is a greater probability (depending on the design of the structure) that the error will be constant. Thus there will be less taper in the part's width. 5-29

30 To verify these thoughts: Geometries and stiffnesses for various elements have to be entered into an error budget spreadsheet. Different designs in different configurations must be considered. Forces input are those expected during grinding, and from gravity acting on center of mass. Error budget spreadsheets are run for each case, and results are copies into a new file for tabulation and comparison. Results 1 : Constant K design Maximum K design Wheel deflection when retracted 1.4 µm 0.6 µm Wheel deflection when extended 1.5 µm 0.9 µm Change in deflection 0.1 µm 0.3 µm 1 The large size of the spreadsheets and the proprietary nature of the design this example comes from make it impractical to show the actual sections of the spreadsheets here. 5-30