Optimization of End Mill and Operation Parameters for 3D Slot Milling

1 Optimization of End Mill and Operation Parameters for 3D Slot Milling Winter 2007, ME 555 Design Optimization Project Professor: Dr. Panos Papalambros April 19 th 2007 Hsin-Yu Kuo Jonathan Loss Joseph John Piazza

2 Abstract: Current industry practice in developing new milling tools has been largely based on experience coupled with iterative machining trials to develop a tool/process combination that achieves the manufacturing requirements, hopefully at a low cost. Several analytical and mechanistic models mentioned in the literature show some promise in reducing the need to perform machining trials to find a tool/process that not only meets manufacturing requirements, but also meet the increasing demand for low manufacturing costs. However, these models have not addressed the issue of tool design and have mainly concentrated on process parameter modification and control. Thus, optimization of end mill geometry and operation parameters to get better efficiency in a simple three-dimension slot milling will be done in this project. RESULTS OF ANALYSIS Subsystem 1 Cost ($ per Part) When applying the design rules in Section 1.7.1 and running a simulation, one can see that the objective function obtained is about $8 per part. Subsystem 2 Efficiency (Parts/Day) When applying the design rules in Section 2.7.2 and running a simulation, the objective function obtained, after rounding down, is 125 parts/day and the Eigen-values of the hessian are all positive, confirming a minimum. Subsystem 3 Quality ($ per Part) The price per part obtained following the design rules in Section 3.7.1 is always $ 79.72 per part. Overall System Profit ($ per Day) If overall system optimization was not done, and using the definition for profit (pg 84), the total profit would be approximately $6,931. This number was generated by using the solution point of the cost sub-system. If you used the efficiency or quality sub-system then the profit would be less. Using the overall system optimization, the profit is $8,172. This is an improvement of about 18%, which is a minimum profit improvement. The improvement over the efficiency and quality subsystem solution points is 63% and 141% respectively.

3 Table of Contents: 1. Subsystem 1 Cost (Cindy Huo) 1.1 Design Problem Statement..7 1.2 Nomenclature.. 8 1.3 Mathematical Model.. 10 1.3.1 Objective Function...10 1.3.2 Constraints...10 1.3.3 Design Variables and Parameters 13 1.3.4 Intermediate Model......13 1.3.5 Final Model Summary.15 1.4 Model Analysis..17 1.4.1 Feasibility Analysis..17 1.4.2 Monotonicity Analysis.....18 1.5 Numerical Result 21 1.5.1 SQP method Result..21 1.5.2 Scaling Model and Result...22 1.5.3 Discussion 23 1.6 Parametric Study.. 25 1.6.1 Objective Function vs. Tool Cost.. 25 1.6.2 Objective Function vs. Removed Material Volume.. 26 1.7 Discussion of Results.. 30 1.7.1 Design Rules for Minimizing Cost 30 1.7.2 Further Discussion 30 2. Subsystem 2 Efficiency (Joe Piazza) 2.1 Design Problem Statement.. 32 2.2 Nomenclature. 33 2.3 Mathematical Model.. 34 2.3.1 Objective Function... 34 2.3.2 Constraints 34 2.3.3 Initial Design Variables and Parameters. 37 2.3.4 Intermediate Model.. 38

4 2.3.5 Simplified Model. 39 2.4 Model Analysis.. 41 2.4.1 Feasibility Analysis. 41 2.4.2 Monotonicity Analysis. 42 2.4.3 Scaling and Constraints... 44 2.5 Optimizations Study.. 45 2.5.1 Analytical Results with Assumptions.. 45 2.5.2 Numerical Verification of Analytical Results.. 48 2.5.3 Numerical Results of Full Model.. 50 2.5.4 Numerical Results of Full Model w/ decreased Allowable Stress.. 52 2.6 Parametric Study. 54 2.6.1 Case I. Objective Fn variation vs. Tool Life and Cutter Stress... 54 2.6.2 Case II. Objective Fn variation vs. Cutter Stress and Volume... 55 2.6.3 Case III. Objective Fn variation vs. Tool Life and Volume.. 57 2.6.4 Case IV. Objective Fn Variation vs Increasing tool life in Case II.. 58 2.7 Discussion of Results... 61 2.7.1 Design Implications 61 2.7.2 Design Rules for Maximizing Efficiency.. 61 2.7.3 Further Discussion. 62 3. Subsystem 3 Quality (Jonathan Loss) 3.1 Design Problem Statement.. 63 3.2 Nomenclature.. 64 3.3 Mathematical Model... 66 3.3.1 Objective function. 66 3.3.2 Constraints. 66 3.3.3 Design Variables and Parameters.. 71 3.3.4 Final Model 72 3.4 Model Analysis.... 74 3.4.1 Feasibility Analysis 74 3.4.2 Monotonicity Analysis... 76 3.5 Numerical Results.....81

5 3.5.1 Matlab SQP Method. 82 3.5.2 Analysis of the result... 85 3.6 Parametric Study 87 3.6.1 Influence of δ max on the optimum.87 3.6.2 Influence of R max on the optimum 88 3.6.3 Influence of pph on the optimum..89 3.7 Discussion of Results..90 3.7.1 Design rules...90 3.7.2 Further Discussion.90 4. Overall System - Profit 4.1 Design Problem Statement...92 4.2 Nomenclature..93 4.3 Mathematical Model...95 4.3.1 Objective Function....95 4.3.2 Constraints 95 4.3.3 Initial Design Variables and Parameters..101 4.3.4 Intermediate Model..101 4.4 Model Analysis..104 4.4.1 Feasibility Analysis.. 104 4.4.2 Monotonicity Analysis. 104 4.4.3 Scaling and Constraints....104 4.5 Optimizations Study....105 4.5.1 Numerical Results of Full Model..105 4.6 Discussion of Results...107 4.6.1 Integral System Discussion 107 4.6.2 Next Generation Optimization...109 5. Appendix 5.1 Subsystem 1. 111 5.1.1 MatLab Programs...111 5.2 Subsystem 2..115 5.2.1 MatLab Programs.. 115

6 5.3 Subsystem 3 122 5.3.1 MatLab Programs..122 5.4 Overall System 128 5.4.1 MatLab Programs..128 5.5 Acknowledgments...133 5.6 References...133

7 Subsystem 1 Cost ($/part) Cindy Huo 1.1 Problem Statement In order to optimize the profit of product, minimizing the cost of production is one important issue. Thus, the objective of the first subsystem is to get minimum manufacturing cost per product. There are two main factor which affect costs a lot, the manufacturing time and the cost of tool. With less manufacturing time, the cost spent on machining will be less; with the longer tool life, the cost of tool will be less. However, in order to shorten the machining time, the cutting parameters need to be more aggressive, which will lead to decreasing tool life. The tradeoff of shorten machining time and longer tool life motivates the cost subsystem optimization study. A lot of study in machining economic has been done. It s also an important topic in all manufacturing book. However, most study assume cutting speed to be the main affect of tool life and machining time, thus makes cutting speed to be the only variable of optimization case. There are few studies in optimization milling process since it s one of the most complicated machining processes. Moreover, limited researches considered tool geometry as an important factor in tool life. By analyzed the simplest case of 3D slot milling, more detailed machining economic optimization work will be done in this subsystem.

8 1.2 Nomenclature C m : machining cost per piece ($/part) C l : cost of loading, unloading and material handling ($/part) C t : tooling cost, includes tool changing and regrinding of the cutter ($/part) C e : cost of machine power ($/W-hr) T m : machining time per piece (sec) T l : time involved in loading and unloading the part (sec) T c : time required to change the tool (sec) T g : grinding time of the tool (sec) L m : labor cost of production operator ($/hr) B m : burden cost ($/hr) M m : machine cost, required power cost to machine parts ($/hr) L g : labor cost of tool-grinder operator ($/hr) B g : burden rate of tool grinder ($/hr) D c : depreciation of the tool in dollars per grind ($/tool) MRR: material removal rate (mm 3 / min) u t : specific energy (total energy per unit volume) (kw/mm 3 ) e: machine efficiency N p : number of pieces per tool makes T: tool life (sec) F f : friction force on the cutting edge F n : normal force on the cutting edge F c : cutting force on the cutting edge F t : thrust force on the cutting edge F l : lateral force on the cutting edge F eq : resultant cutting force (N) σ: stress on the tooth root (N/m 2 ) σ t : tensile stress of WC tool (N/m 2 ) W: width of slot (mm) L: length of slot (mm)

9 H: depth of slot (mm) R i : inner ball end mill radius (mm) R o : outer ball end mill radius (mm) D r : distance moved per revolution (mm) f: feed rate of cutter (mm/sec) N: spindle speed (rpm) d: depth of cut (mm) α: rake angle (radians) Φ: helix angle (radians)

10 1.3 Mathematical Models 1.3.1 Objective function: The object of this subsystem is to minimize the manufacturing cost. The manufacturing cost could be divided into three portions, machining cost, loading cost and tool cost. Thus the objective function could be derived as: min Cost = Cm + Cl + Ct (1.1) 1.3.2 Constraints: Machining cost could be obtained by total machining time multiply the machining cost per unit time. The machining cost includes labor cost, burden cost and machine cost. The equality constraint is: C m Lm + Bm + Mm = Tm 3600 (1.2) The machine cost is determined by the power needed during the machining process, which is the material removal rate multiplies specific energy, then times cost of machine power: M m MRR ut 3600 = C η 1000 e (1.3) The material removal rate of milling process is to multiply the feed, spindle speed, depth and the ball diameter. MRR = 2R o f d (1.4) The loading cost part could be obtained by total loading time multiplies loading cost per unit time, which includes labor and burden cost. Lm + Bm Cl = Tl 3600 (1.5) Tooling cost is the most complicated part, including tool changing and regrinding. Each part also divides into labor and burden part. 1 Ct = Tc ( Lm + Bm) + Tg ( Lg + Bg) + D c N (1.6) p

11 Machining time for one part could be calculated by the total material removal volume divided by material removal rate. Here the material removal volume is the volume of slot. T m W H L = (1.7) 2 R f d o Number of pieces per tool could make is obtained by the tool life divided by machining time. N p T = (1.8) T m The tool life equation could be obtained from literature for cutting titanium with tungsten carbide cutter, after converge the unit. T = 5.0217 10 D d ( R N) (1.9) 29 1.6 1.7 4.95 r o If the tool changed too frequently, it will waste a lot of time and also cost a lot. So the tool life will be constrained to be larger than one hour. T 3600 (1.10) Distance moved per revolution can be defined as a function of spindle speed and feed rate. D r f 60 = (1.11) N In order not to break the tool with too large moved distance per revolution, the distance need to be constrained by the tool geometry. Ro Ri Dr (1.12) 50 There is a functional relationship between outer and inner radius determined experimentally. Ro + 3.5 Ri = (1.13) 2 The spindle speed is limited to the machine tool capability but must be within operational limits of carbide cutters given by recommended SFM values (25-300 SFM for cutting Ti). 12 SFM N = Ro 2π 12 25.4 (1.14)

12 Therefore, there s constraint about relationship between N and outer radius R o. 14561 N Ro (1.15) N R o 174726 (1.16) The cutting force could be calculated using the constructed cutting force model in literature, which built based on oblique cutting force model. 2 2 ( α ) F = F (sin Φ ) + (cos Φ ) sin + F cosφ cosα (1.17) c f n F = F cosφcosα F sinα (1.18) t f n ( α ) F = F cos Φsin Φ sin + sin Φcos Φ + F sin Φ cosα (1.19) l f n ( ) F = d/ sin Φ D exp 18.315 0.26267ln( D ) 0.10327ln( N R 2π + 1.8828ln(1 sin α) + 0.0047625ln( D ) ln( N R 2 π) (1.20) n r r o r o ( ) F = d / sin Φ D exp 15.031 0.6292ln( D ) + 0.32256ln( N R 2π + 0.34417ln(1 sin α) + 0.299585ln( D ) ln( N R 2 π) (1.21) f r r o r o 2 2 2 F eq F c F t F l = + + (1.22) With the use of beam theory, it is assumed that the cross section of tooth is a rectangular shape. The stress at the tooth root could be obtained by following equation. 6 ( ) 10 Feq Ro Ri σ = 1 d 2 t 24 sin Φ (1.23) With the use of beam theory, we can find the bending stress of the cutter at the fixed point on the shank. σ bend F + F 2H R 10 = (1.24) 2 2 6 c t 0 4 π R0 /4 There is a functional relationship between outer radius and the thickness of the tooth determined experimentally. t = 0.1974R o + 1.5592 (1.25) The stress of the tooth needs to be smaller than the tensile stress σ t (0.3448 GPa) of the tool material (WC). σ σ t (1.26) The bending stress needs to be smaller than the tensile stress σ t (0.3448 GPa) of the tool

13 material (WC) σ bend σ (1.27) t The spindle speed is limited to the machine tool capability. N 20,000 (1.28) Depth of cut need to be smaller or equal to the depth of slot. d H (1.29) Outer Radius need to be smaller or equal to the width of slot and greater than or equal to a 10 mm diameter cutter. R 5 (1.30) o 2R o W (1.31) Rake angle is not good for cutting when exceeding 15 degree from literature α 15* π /180 (1.32) Helix angle is constrained by the manufacturing process of tool. Φ 75* π /180 (1.33) Φ 30* π /180 (1.34) All parameters or variables need to be greater than zero. ALLvar iables 0 (1.35) 1.3.3 Design Variables and Parameters: The parameters are: C e, T l, T c, T g, L m, B m L g, B g, D c, u t, e, σ t, W, L, H - geometry parameters : L = 200 mm, H = 40mm, W = 100 mm - time parameters : T c =150 sec, T l =180 sec, T g =1800sec - machine tool parameters: u t = 0.066*10-3 kw/mm 3 *sec, η = 70% - Cost parameters : C e =0.08971 $/kw-hr, L m =20 $/hr (average worker), B m =0.06528*4 $/hr (assume 4 power 1kW lights are used), L g =25 $/hr (experienced worker), B g =0.06528*2 $/hr(assume 2 power 1kW lights are used), D c =10 $/tool - Tensile strength for material (WC): σ t = 344.8*10 6 Pa Intermediate variables: C m, C l, T m, C t, M m, MRR, N p, T, F eq, σ max, D r, R i

14 Variables: f, N, d, α, Φ, R o DOF = 6 1.3.4 Intermediate Model: min Cost = Cm + Cl + Ct Where W L H 2Ro f d ut Ce 3600 Cm = Lm + Bm +, which is derived from 2Ro f d 3600 η equation (1.2), (1.3), (1.4) and (1.7). ( ) C = T L + B, which is from equation (1.5) l l m m Subject to W L H Lm + Bm Lg + Bg Ct = T 27 0.7 0.6 3.35 3.95 c Tg D c 1.435 10 d f N R + + o 3600, which is 3600 derived from equation (1.6), (1.7), (1.8), (1.9), (1.11) 6 1.6 4.95 3.35 1.7 29 g 1 : 2.52 10 f R N d 5.0217 10 0, derived from (1.9), (1.10), (1.11) g 2 : ( ) o 6000 f N R o 3.5 0, derived from (1.11), (1.12), (1.13) g 3 :14561 N R o 0, from (1.14), (1.15) g 4 : N R o 174726 0, from (1.14), (1.16) g 5 : d H 0, from (1.29) g 6 :5 R o 0, from (1.30) g 7 : 2R W 0, from (1.31) o g 8 : α 15* π /180 0, from (1.32) g 9 : Φ 75* π /180 0, from (1.33) g 10 :30* π /180 Φ 0, from (1.34) F g 11 : eq Ro 3.5 10 2 1 d 2 t 24 sin Φ 6 σ 0 t g 13 : F + F 8H 10 2 2 6 c t 3 π R0 σ 0 t

15 where 2 2 2 F eq = F c + F t + F l ( ) ( ) F = d / sin Φ D exp 15.031 0.6292ln( D ) + 0.32256ln( N R 2π + 0.34417ln(1 sin α) + 0.299585ln( D ) ln( N R 2 π) f r r o r o F = d / sin Φ D exp 18.315 0.26267ln( D ) 0.10327ln( N R 2π + 1.8828ln(1 sin α) + 0.0047625ln( D ) ln( N R 2 π) n r r o r o 2 2 F = F (sin Φ ) + (cos Φ ) sinα + F cos Φcosα c f ( ) F = F cos Φcosα F sinα t f n Fl = Ff ( cosφsin Φ sinα + sin Φcos Φ ) + Fnsin Φcosα t = 0.1974R o + 1.5592, from (1.17)-(1.28) g 12 : ALLvar 0, from (1.35) iables n 1.3.5 Final Model Summary (substitute all parameters): ( ) 1 C = 2251.1 R d f + 6.766 m C = 1.0131 l o C = 1.305 10 d f R N t 20 0.7 0.6 3.95 3.35 o Objective function: ( ) 1 20 0.7 0.6 3.35 3.95 Sub to: g 1 : min Cost = 2251.1 R f d + 1.305 10 d f N R + 7.779 6 1.6 4.95 3.35 1.7 29 2.52 10 f Ro N d 5.0217 10 0 6000 f N R o 3.5 0 g 2 : ( ) g 3 :14561 N R o 0 g 4 : N R o 174726 0 g 5 : d 40 0 g 6 :5 R o 0 o o g 7 : 2Ro 100 0 g 8 : α 15* π /180 0 g 9 : Φ 75* π /180 0 g 10 :30* π /180 Φ 0 g 11 : R 3.5 F + F + F 10 2 1 d 2 ( 0.1947Ro + 1.5592) 24 sin Φ 2 2 2 o 6 c t l 6 344.8 10 0

16 g 12 : F + F 320 10 2 2 6 c t 3 π R0 where 6 344.8 10 0 2 2 ( α ) F = F (sin Φ ) + (cos Φ ) sin + F cosφ cosα c f n F = F cosφcosα F sinα t f n ( α ) F = F cos Φsin Φ sin + sin Φcos Φ + F sin Φ cosα l f n where d 60 f 60 f N Ro 2π 60 f N Ro 2π Fn = / sin Φ exp 18.315 0.26267ln( ) 0.10327ln( ) 1.8828ln(1 sin α) 0.0047625ln( ) ln( ) 1000 1000N + 1000N 1000 1000N 1000 d 60 f 60 f N Ro 2π 60 f N Ro 2π Ff = / sin Φ exp 15.031 0.6292ln( ) 0.32256ln( ) 0.34417ln(1 sin α) 0.299585ln( ) ln( ) 1000 1000N + + + 1000N 60 1000N 60 g 12 : N 0 g 13: d 0 g 14: R o 0 g 15: α 0 g 16: Φ 0 g 17: f 0

17 1.4 Model Analysis 1.4.1 Feasibility Analysis: Select a set of variables: R o = 10mm, f = 2 mm/sec, N = 2000 rpm, d = 5 mm, α = 5*pi/180 radians, Φ = 45*pi/180 radians 6 1.6 4.95 3.35 1.7 29 29 g 1 : 2.52 10 f R N d 5.0217 10 = 5.02 10 0 feasible g 2 : ( ) o 6000 f N R o 3.5 = 9000 0 feasible g 3 :14561 N R o = 45439 0 feasible g 4 : N R o 174726 = 114726 0 feasible g 5 : d 40 = 35 0 feasible g 6 :5 R o = 5 0 feasible g 7 : 2Ro 100 = 80 0 feasible g 8 : α 15* π /180 = 0.1745 0 feasible g 9 : Φ 75* π /180 = 0.5236 0 feasible g 10 :30* π /180 Φ= 0.2618 0 feasible g 11 : R 3.5 F + F + F 10 2 1 d 2 ( 0.1947Ro + 1.5592) 24 sin Φ 2 2 2 o 6 c t l = 6 6 344.8 10-204.0885 10 0 feasible g12: F + F 320 R 10 2 2 6 c t o 4 π R0 = feasible 6 6 344.8 10-333.4537 10 0 where 2 2 ( α) F = F (sin Φ ) + (cos Φ ) sin + F cosφ cosα = 207.577 c f n F = F cosφcosα F sinα = 25.68 t f n ( α ) F = F cos Φsin Φ sin + sin Φcos Φ + F sin Φ cosα = 207.577 l f n

18 where d 60 f 60 f N Ro 2π 60 f N Ro 2π Fn = / sin Φ exp 18.315 0.26267ln( ) 0.10327ln( ) + 1.8828ln(1 sin α) 0.0047625ln( ) ln( ) 1000 1000N 1000N 1000 1000N 1000 = 294.68N d 60 f 60 f N Ro 2π 60 f N Ro 2π Ff = / sin Φ exp 15.031 0.6292ln( ) 0.32256ln( ) 0.34417ln(1 sin α) 0.299585ln( ) ln( ) 1000 1000N + + + 1000N 60 1000N 60 = 1.34 10 4 The chosen variables are feasible g 12, g 13, g 14, g 15, g 16, g 17 because they are all positive. Objective function of machining cost of one part is ( ) 1 20 0.7 0.6 3.35 3.95 Cost = 2251.1 R f d + 1.305 10 d f N R + 7.779 = 8.1315 o o 1.4.2 Monotonicity Analysis: 1. Monotonicity Table: R o f d N α Φ cost - - - + g 1 + + + + g 2 - + - g 3 - - g 4 + + g 5 g 6 g 7 g 8 g 9 g 10 + - + + + - g 11?????? g 12?????? g 13 g 14 g 15 - - -

19 g 16 g 17 g 18 - Table 1.1: Full Monotonicity Table for Cost Subsystem - - 2. Monotonicity Analysis: (a) By the bound of variable R o, g 6 dominate g 15. (b) From MP1 with respect to R o, one of g 1, g 4, g 7, g 11 and g 12 is active. (c) From MP1 with respect to f, one of g 1, g 2, g 11 and g 12 is active. (d) From MP1 with respect to d, one of g 1, g 5, g 11 and g 12 is active. (e) From MP1 with respect to N, one of g 2, g 3, g 11, g 12 and g 13 is active. (f) From MP2 with respect to α, either the case which g 8 and g 11 are active, or g 8 and g 12 are active, or g 11 and g 12 are active, or g 11 and g 17 are active, or g 12 and g 17 are active. Or none of the constraints are active, which makes the rake angle maintains starting value. (g) By the bound of variable Φ, g 10 dominate g 17. (h) From MP2 with respect to Φ, either the case which g 9 and g 11 are active, or g 9 and g 12 are active, or g 11 and g 12 are active, or g 10 and g 11 are active, or g 10 and g 12 are active. Or none of the constraints are active, which makes the helix angle maintains starting value. 3. Modified Monotonicity Table: From analysis (a) and (g), g 15 and g 17 could be cancelled. R o f d N α Φ cost - - - + g 1 + + + + g 2 - + - g 3 - - g 4 + + g 5 g 6 g 7 - + +

20 g 8 g 9 g 10 + + - g 11?????? g 12?????? g 13 g 14 g 16 g 18 - - - - Table 1.2: Modified Monotonicity Table for Cost Subsystem According to the modified monotonicity table, the model is well bounded. However, no further analysis could be done.

21 1.5 Numerical Result 1.5.1 Matlab SQP method: 1. Different starting point N(rpm) F(mm) d(mm) Ro(mm) α (radians) Φ (radians) Cost cons bound starting 2000 2 5 10 0.087 0.785 result 1999.9 6.3077 40 40.35 0.262 0.78487 8.1315 3,6 starting 6000 2 5 10 0.087 0.785 result 5999.9 5.6234 40 19.185 0.262 0.785 8.5586 5 3,6 starting 1000 2 5 15 0.087 0.785 result 1000.3 4.716 40 50 0.262 0.78445 8.0429 5 3,4, 6 starting 1000 2 5 25 0.087 0.785 result 1000.3 4.7157 40 50 0.262 0.7848 8.0429 5 3,4, 6 starting 1000 2 10 20 0.174 1.047 result 1000.5 4.7167 40 50 0.262 1.0473 8.0429 5 3,4,6 Table 1.3: Result Table for Setting Different Starting Point Discussion: With different start point, it will find the nearest local minimum result, which might not be the global optimization result. The result varies a lot with different starting spindle speed N. The objective function changes a little with varying starting point, thus changing the tolerance of objective function might lead to better optimization result. However, it is shown that the upper bounds of d and α are always active. And most of the time, the force constraint is also active. 2. Different tolerance of function Fn. tol. N(rpm) F(mm) D(mm) Ro(mm) α (radians) Φ (radians) Cost cons bound starting 1000 2 5 15 0.087 0.785 1e^-6 1000.3 4.716 40 50 0.262 0.78445 8.0429 5 3,4, 6 1e^-10 1204.3 5.7746 40 50 0.262 0.55576 8.0270 5 3,4,6 starting 2000 2 5 15 0.087 0.785 1e^-6 1999.8 6.3075 40 40.35 0.262 0.78492 8.1315 3,6 1e^-10 1204.3 5.7746 40 50 0.262 0.64904 8.0270 5 3,4,6

22 Table 1.4: Result Table for Setting Different Function Tolerance Discussion: By setting smaller function tolerance, we get similar result from different starting point. From the analysis of activity of constraints, it is shown that the upper bound of d, Ro and α and force constraint are active with smaller tolerance. 1.5.2 Scaling Variables: Neither SQP and DIRECT method show good convergence on the problem. The huge different scale of variables, such as N is in the range of thousands, feed is between 1-5 and radius is something around 40-50, might be the main reason due to the bad result. Also, the different scale in constraints and cost functions also has problem of sensitivity. Thus, scaling method will be used in all the further analysis. Scale variables to value about 0 to 1. Which make Ro R o =, 50 d d =, 40 f f =, 5 N N =, substitute into models. 1000 Modified objective function and some constraints of the model: ( ) 1 0.7 0.6 3.35 3.95 o o min Cost = 0.2251 R f d + 0.02614 d f N R + 7.779 g 1 : f R N d 9.9471 0 1.6 4.95 3.35 1.7 o g 2 : f N ( R o ) 6 10 0.7 0 g 3 : 0.2912 N R 0 g 4 : N R 3.49452 0 o g 5 : d 1 0 g 6 : 0.1 R 0 o g 7 : R o 1 0 o

23 g 11 : 2 2 2 50R o 3.5 Fc + Ft + Fl 2 344.8 0 1 40d 2 ( 9.735Ro + 1.5592) 24 sin Φ g 12 : where F F R 2 2 3 c + t 2.56 10 o π R 4 0 2 2 ( α ) 344.8 0 F = F (sin Φ ) + (cos Φ ) sin + F cosφ cosα c f n F = F cosφcosα F sinα t f n ( α ) F = F cos Φsin Φ sin + sin Φcos Φ + F sin Φ cosα l f n where 5 3 3 1.2 10 / sin f exp 18.315 0.26267ln( f ) 0.10327ln(100 ) 1.8828ln(1 sin ) 0.0047625ln( f Fn = d Φ π N R o + α ) ln(100 π N R o ) N 10000N 10000N 5 3 3 1.2 10 / sin f exp 15.031 0.6292ln( f ) 0.32256ln(100 ) 0.34417ln(1 sin ) 0.299585ln( f Ff = d Φ + π N R o + α + ) ln(100 π N R o ) N 10000N 10000N Starting point: N = 2, f = 0.4, R = 0.2, d = 0.125, α = 0.087, Φ = 0.785 o Optimization result: N = 1.0159, f = 1.5747, R = 1, d = 1, α = 0.262, Φ = 0.785, which means N = 1015.9 rpm, f = 7.8735 mm/sec, d = 40 mm, Ro = 50 mm, α = 0.262 radians, Φ =0.785 radians, and Cost = 7.9581 output = iterations: 16; funccount: 145; stepsize: 1 lower upper ineqlin ineqnonlin 3,4,6 2 The numerical result shows good global convergence. As shown in the activity analysis of constraints, d, Ro and α reaches the upper bound. g 2 is also active. Because neither g 11 nor g 12 is active, thus Φ stays in the starting value and didn t move. o 1.5.6 Discussion Compare the activity of constraints to the result in monotonicity analysis. By MP1 with respect to Ro, the upper bound, which is g 7, is active. By MP1 with respect to d, the upper bound, which

24 is g 5, is active. The active inequality constraint g 2 shows the constrained relationship between Ro, f and N. Thus the activity of g 2 is reasonable by MP1 with respect to N and f. Rake angle goes to the upper bound shown as g 2. However, there are only two constraints related to rake and helix angle, which are g 11 and g 12. As shown in the listed momotonicity table, neither constraint is active. The algorithm somehow drove the rake angle reaches the upper bound, and stopped there because the value of rake angle won t affect the objective function. Helix angle have similar situation that it won t move at all and just stay at the starting point at the optimized result. R o f d N α Φ cost - - - + g 1 + + + + g 2* - + - g 3 - - g 4 + + g 5* g 6 g 7* g 8* g 9 g 10 + - + + + - g 11?????? g 12?????? g 13 g 14 g 16 g 18 - - - - Table 1.5: Monotonicity Table Analysis vs. Optimization Result (* means active)

25 1.6 Parametric Study 1.6.1 Objective Function vs. Tool Cost: 1. Changing Tool Depreciation Cost: Tungsten carbides tool are very expensive. So increasing tool depreciation cost to make tool life a more important issue in our case might change the optimization result. Tool depreciation cost D c = $30/tool. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.2251 R f d + 0.0484 d f N R + 7.779 o - optimization result : N = 897 rpm, f = 6.952 mm/sec, d = 40 mm, Ro = 50mm, α = 0.041684 radians, Φ =0.78508 radians, and Cost = 7.9819 Tool depreciation cost D c = $50/tool. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.2251 R f d + 0.0708 d f N R + 7.779 o - optimization result : N = 830.7 rpm, f = 6.438 mm/sec, d = 40 mm, Ro = 50mm, α = 0.001 radians, Φ =0.785 radians, and Cost = 7.9981 Tool depreciation cost D c = $100/tool. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.2251 R f d + 0.2924 d f N R + 2.7034 o - optimization result : N = 623.74 rpm, f = 4.834 mm/sec, d = 40 mm, Ro = 50mm, α = 0.24956 radians, Φ =0.78506 radians, and Cost = 8.0708 o o o 8.08 8.06 y = 4E-06x 2 + 0.0008x + 7.9513 cost funciton 8.04 8.02 8 7.98 7.96 7.94 0 20 40 60 80 100 120 depreciation cost Figure 1.1: Parametric Study of Objective Function vs. Tool life and Tool Depreciation Cost Discussion: Changing tool depreciation cost increase the total cost, as shown in Fig.1.1. At the same time, spindle speed and feed rate decrease. This is reasonable because they affect tool life a lot, and tool life is supposed to increase to lower price in this case. The actual tool cost might be dependent to tool size. Thus, reset the tool depreciation cost

26 to a function of radius R o. Assume D = 5+ 0.5R, objective function: c ( ) 1 0.7 0.6 3.35 3.95 0.7 0.6 3.35 4.95 min Cost = 0.2251 R f d + 0.0206 d f N R + 0.0279 d f N R + 7.779 o o o - optimization result : N = 896.7 rpm, f = 6.949 mm/sec, d = 40 mm, Ro = 50mm, α = 0.2336 radians, Φ =0.7851 radians, and Cost = 7.9820 Discussion: Even though increasing the affect of radius on the tool cost, the result still shows that the radius will exceed the upper bound. Same situation happens as the previous result: spindle speed and feed decreases in order to get longer tool life. o 2. Changing grinding time of tool: If increase the grinding time of tool, it will lead to similar optimization result as increasing the depreciation cost of tool. However, the grinding time of tool might be a function of tool geometry. Thus, reset grinding time of 1800 seconds as a parameter to a function of radius and helix angle. AssumeT = 1800 + 6 R / sin Φ. And new Objective function: g o - optimization result : N = 998.2 rpm, f = 7.7365 mm/sec, d = 40 mm, Ro = 50mm, α = 0.1943 radians, Φ =1.309 radians, and Cost = 7.9613 Discussion: In this case, Φ will appear in the objective function. As the result shows, spindle speed and feed are not increasing but decreasing to get optimized result. However, Φ reaches the upper bound in order to reduce grinding time. 1.6.2 Objective Function vs. Removed Material Volume: 1. Changing Slot Depth: The depth of cut d always exceeds upper bound, which is the depth of slot in our case. However, it is difficult to see what will happen if several paths cutting one slot is needed. Thus, by changing the slot depth H, but keeping the original upper bound of d, some parametric study has also been done. Depth of slot H= 60. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.3377 R f d + 0.0392 d f N R + 11.163 o - optimization result : N = 1016 rpm, f = 7.874 mm/sec, d = 40 mm, Ro = 50mm, α = 0.26164 radians, Φ =0.78513 radians, and Cost = 11.4317 o

27 Depth of slot H = 80. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.4502 R f d + 0.0523 d f N R + 14.546 o - optimization result : N = 1015.8 rpm, f = 7.8725 mm/sec, d = 40 mm, Ro = 50mm, α = 0.26166 radians, Φ =0.78517 radians, and Cost = 14.9043 Depth of slot H = 100. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.5628 R f d + 0.0654 d f N R + 16.915 o - optimization result : N = 1015.8 rpm, f = 7.8725 mm/sec, d = 40 mm, Ro = 50mm, α = 0.253.08 radians, Φ =0.78517 radians, and Cost = 17.3630 Depth of slot H = 100. Upper bound of d=h. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.5628 R f d + 0.0654 d f N R + 16.915 o - optimization result : N = 741.53 rpm, f = 5.747 mm/sec, d = 100 mm, Ro = 50mm, α = 0 radians, Φ =0.7852 radians, and Cost = 17.1604 o o o cost function ($) 20 18 16 14 12 10 8 6 4 2 0 30 40 50 60 70 80 90 100 110 dpeth of slot (mm) Figure 1.2: Parametric Study of Objective Function vs. Tool life and Depth of Slot Discussion: Cost increases with increasing depth, which is reasonable because it will need more paths to cut the slot, as shown in Fig. 1.2. However, we can see that the variables didn t change a lot during the parametric study. Thus the optimized variables are almost fixed to certain values in this case. If changing the upper bound of cut of depth d to be the same value of depth of slot, the optimized d is still the same as the depth of slot. Thus we can know that in order to get the minimum cost of process, the number of paths needed to cut slot is really important issue. At the same time, by varying the spindle speed and feed rate, good tool life could still be obtained. Also, by varying the angles, the optimization result will still be feasible for the stress constraints.

28 2. Changing Slot Width: The radius of ball-end mills R o always exceeds upper bound, which is half of the width of slot in our case. By changing the slot depth, but keeping the original upper bound of R o, some parametric study has also been done. Width of slot W = 150. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.3377 R f d + 0.0392 d f N R + 11.163 o - optimization result : N = 1016 rpm, f = 7.874 mm/sec, d = 40 mm, Ro = 75 mm, α = 0.26164 radians, Φ =0.78513 radians, and Cost = 11.4317 Width of slot W = 200. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.4502 R f d + 0.0523 d f N R + 14.546 o - optimization result : N = 1015.8 rpm, f = 7.8725 mm/sec, d = 40 mm, Ro = 100 mm, α = 0.26166 radians, Φ =0.78517 radians, and Cost = 14.9043 Width of slot W = 150. Upper bound of 2R o =W. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.3377 R f d + 0.0392 d f N R + 11.163 o - optimization result : N = 589.39 rpm, f = 7.0235 mm/sec, d = 40 mm, Ro = 75 mm, α = 0.10509 radians, Φ =0.78516 radians, and Cost = 11.3638 Width of slot W = 200. Upper bound of 2R o =W. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 0.4502 R f d + 0.0523 d f N R + 14.546 o - optimization result : N = 401.15 rpm, f = 6.452 mm/sec, d = 40 mm, Ro = 100 mm, α = 0.2551 radians, Φ =0.78593 radians, and Cost = 14.7646 o o o o cost function ($) 16 14 12 10 8 6 4 2 0 50 70 90 110 130 150 170 190 210 230 250 width of slot (mm) Figure 1.3: Parametric Study of Objective Function vs. Tool life and Width of Slot

29 Discussion: The result of changing width of slot is similar to the result as changing depth of slot, as shown in Fig. 1.3. Cost increases with increasing material removal volume, which needs more paths to finish one part. However, while setting the upper bound of the ball radius to be the related to the width of slot, we can see the spindle speed decreases a lot more comparing to the case of changing depth. The reason is the feed speed is constrained by the relationship between spindle speed and radius. So when radius increases to make the slot finished in one path, the spindle speed need to decrease a lot in order to make the feed speed not to reduce too much. 3. Changing Total Volume of Removed Material: In order to see more affect on the constraint, and also see the trade offs between radius and feed speed. W, H and L are all changed, with variable s upper bound limited by the parameters. Width of slot W = 200, depth of slot H = 100, length of slot L=400. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 2.251 R f d + 0.2614 d f N R + 68.68 o - optimization result : N = 292.87 rpm, f = 4.7103 mm/sec, d = 100 mm, Ro = 100 mm, α = 0.058602 radians, Φ =0.76271 radians, and Cost = 69.2789 Width of slot W = 200, depth of slot H = 100, length of slot L=800. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 4.4888 R f d + 0.522 d f N R + 136.343 o - optimization result : N = 292.78 rpm, f = 4.70895 mm/sec, d = 100 mm, Ro = 100 mm, α = 0.071 radians, Φ =0.52905 radians, and Cost = 137.5376 Width of slot W = 400, depth of slot H = 200, length of slot L=1600. Objective function: ( ) 1 0.7 0.6 3.35 3.95 min Cost = 22.5111 R f d + 2.61 d f N R + 677.683 o - optimization result : N = 91.743rpm, f = 3 mm/sec, d = 200 mm, Ro = 200 mm, α = 0 radians, Φ =0.65116 radians, and Cost = 680.0303 Discussion: The result shows that the number of paths per part is the most important issue in this case. That means depth of cut and ball-end mills radius will reach upper bound in whatever cases to get the optimization result. Also, the cost increases a lot according to the removed material volume. o o o

30 1.7 Discussion of Results 1.7.1 Design Rules: From the numerical optimization result, following design rules could be obtianed 1) Tool life should be limited to about 2 hours in order to reduce cost but still reduce producing time, which affects machining cost a lot. 2) Outer cutter radius and depth of cut should be as large as possible, 50 mm and 40 mm respectively in the case. 3) Via simulations feed should be between 7-8 mm/sec, Tool RPM should be between 1000-1100 rpm, 4) Rake angle and helix angle doesn t reach optimized value because of the inactive of stress constraints. But they might be adjusted when reaching the optimization result. Suggested rake angle should be as large as possible, and helix angle should be 0.5236-0.785 radians (30 degrees-45degrees). When applying the design rules and running a simulation, one can see that the objective function obtained is about $8 per part. 1.7.2 Further Discussion: In the real world, the ball-end mills radius is constrained by manufacturing issue. Also, it will af be affected by the work situation, such as the width of slot or the surface finish needed to be obtained during the milling process. At the same time, depth of cut will also be limited to the length of the cutting edge due to manufacturing process. There are a lot more constrained about tool life could be done to improve the model. Due to the difficulty of modeling, those constraints are not listed and considered in this case. First of all, the stress distribution all over the cutter is not modeled and being constrained. In real case, finite element stress distribution simulation will be done to check the maximum stress all over the surface of tool. In this case, only stress at the root of the tool and at the fixed point at the shank has been estimated. Second, milling process is a discontinuous cutting procedure where fatigue life is very important issue. The cycle loading will result in tool failure and reduce tool life. Moreover, temperature issue will be an interesting topic to discuss about to improve the

31 optimization case. The strength of Tungsten Carbides tools is very sensitive to temperature changes, and the temperature is dependent to the cutting condition such as the spindle speed, feed and depth of cut. However, in this case we still get reasonable result to minimize cost of producing part. By increasing feed speed, cut of depth and radius of the ball, lower cost could be obtained. At the same time, spindle speed and the rake and helix angles need to be adjusted to makes the optimization value be feasible for all the constraints.

32 Subsystem 2 Efficiency (parts/day) Joe Piazza 2.1 Problem Statement The purpose of this individual subsystem is strictly to maximize the efficiency of a milling process. As one might expect, this subsystem breaks down to only a few variable that compete with one another. Such tradeoffs can be described as speed (RPM), feed (velocity of workpeice), depth of cut, and downtime/tool life (e.g. you can feed faster if you have a smaller depth of cut, but the slower you feed the more tool life you obtain). The motivation of this system is strictly productivity, the more parts you can make in a given time frame, the more profit you can achieve.

33 2.2 Nomenclature {Name: Description (units)} E = Efficiency (Parts/Day) α = rake angle (degrees) Φ = helix angle (degrees) N = speed of cutter (rpm) d = depth of cut (mm) R o = Outside radius of cutter (mm) R i = Inside radius of cutter (mm) f = feed of cut or velocity of workpeice (mm/sec) H = Total depth of slot (mm) L = Length of slot (mm) W = total width of slot (mm) T = Tool life equation Life (sec) T c = Time to change one tool (sec) T p = Total production time (sec/part) T l = Time to Load One Part (sec/part) T m = Machining Time per part (sec/part) N p = Pieces per Tool (part/tool) H d = Hours in production day (hours/day) D r = Distance moved per revolution (mm/rev) F c = Cutting force (N) F t = Tool thrust force (N) F l = Lateral force (N) F f = Frictional force on cutting edge (N) F n = Normal force on cutting edge (N) σ = stress cutting flute, beam (N/mm^2) σ t = tensile stress of WC tool (N/mm^2)

34 2.3 Mathematical Model: 2.3.1: Objective function The objective of this subsystem is to maximize the efficiency of one production day. This can be written as directly proportional to the time it takes to produce one part. 2.3.2: Constraints Hindered by mostly physical constraints, all parameters and variables are greater or equal to zero. Total time to manufacture one part can be added up by all different items, including part loading time, machining time, and tool changing time. Because tool changing is not necessary for each part, thus the tool changing time is the average tool changing time. Total time for one part must be completed in a timely fashion therefore must be less than 14 minutes per cycle. Machining time for one part could be calculated by the total material removal volume divided by material removal rate. Here the material removal volume is the volume of slot. Machining time must be completed in a timely fashion therefore must be less than 10 minutes per part. Number of pieces per tool could make is obtained by the tool life divided by machining time. The tool life equation could be obtained from literature for cutting titanium with tungsten

35 carbide cutter and after converting into the correct units (sec): The tool life of the cutter must be greater than 1 hour so that production may run smoothly: Distance moved per revolution The tool with also break if try to take off more material than what is possible with the geometry of the cutter. The maximum distance to move per revolution can be calculated by looking at feed and speed recommendations for carbide end mills. There is function relationship between the Outer and Inner Radii of a end mill determined experimentally which is: Therefore The spindle speed is limited to the machine tool capability but must be within operational limits of carbide cutters given by recommended SFM values (25-300 SFM for cutting Ti). Therefore: The cutting force along x, y, z axis could be calculated using the constructed cutting force

36 model in previous research. The equations were obtained from an oblique cutting model that was formulated based on experiments. And, A possible failure mode of the end mill can be described by the flute breaking off from the main inner shank (tooth root). Therefore a simple beam formula, rectangular, is assumed to be accurate where t is the approximate average thickness of the flute. The maximum stress couldn t exceed the tensile stress of material. Rake angle is not good for cutting when exceeding 15 degree from literature. Helix angle is not good for cutting when exceeding 75 degrees and smaller than 30

37 degrees from literature. Depth of cut need to be smaller or equal to the depth of slot. Outer Radius need to be smaller or equal to the width of slot and greater than or equal to a 10 mm diameter cutter. Given the rpm ranges above, feed must be within a feasible limit. Another possible mode of failure is the end mill breaking off at the tool holder. Assuming that the tool holder is rigid than this failure can be evaluated by cantilever beam equations. This stress must also be lower than the tensile stress of tungsten carbide. 2.3.3: Initial Design Variables and Parameters Variables: E, T p, T m, N p, T, D r, R o, R i, σ, F f, F n, F c, F t, F l, N, d, f, α, Φ Parameters and Values : L (200 mm), H (40 mm), W (100 mm), T l (180 sec), T c (150 sec), σ t (344.8 N/mm^2), H d (7 hrs/day) Degrees of Freedom = # Variables - # equality constraints = 19-14 = 5

38 2.3.4: Intermediate Model (For parametric study, L, W, H, and T, Sigma) Sub to: g1: (Eq. 2.28) g2: (Eq. 2.29) g3: (Eq. 2.27) g4: (Eq. 2.25) g5: (Eq. 2.26) g6: (Eq. 2.24) g7: (Eq. 2.14 w/ 2.13) g8: (Eq. 2.15 w/ 2.13) g9: (Eq. 2.11 w/ 2.10, 2.12a, 2.12b) g10: (Eq. 2.9 w/ 2.8, 2.10) g11: (Eq. 2.6 w/ 2.5) g12: (Eq. 2.4 w/ 2.3, 2.5, 2.7, 2.8, 2.10) g13: (Eq. 2.23 w/ 2.21, 2.22, 2.16, 2.17, 2.18, 2.19, 2.20) g14: (Eq. 2.30) g15: (Eq. 2.32 w/ 2.31) g16 g21: (Eq. 2.2) Where:

39 And, And, 2.3.5: Simplified and Initial Summary of Model (All Parameters substituted) Sub to: g1: (Eq. 2.28) g2: (Eq. 2.29) g3: (Eq. 2.27) g4: (Eq. 2.25) g5: (Eq. 2.26) g6: (Eq. 2.24) g7: (Eq. 2.14 w/ 2.13) g8: (Eq. 2.15 w/ 2.13) g9: (Eq. 2.11 w/ 2.10, 2.12a, 2.12b) g10: (Eq. 2.9 w/ 2.8, 2.10) g11: (Eq. 2.6 w/ 2.5) g12: (Eq. 2.4 w/ 2.3, 2.5, 2.7, 2.8, 2.10) g13:

40 (Eq. 2.23 w/ 2.21, 2.22, 2.16, 2.17, 2.18, 2.19, 2.20) g14: (Eq. 2.30) g15: (Eq. 2.32 w/ 2.31) g16 g21: (Eq. 2.2) Where: And, And,

41 2.4 Model Analysis: 2.4.1: Feasibility Analysis To determine weather a solution is possible given the constraints, values for the variable were selected based on current industrial specifications. These value were applied to the constraints g1-g20 and then the objective function was calculated. The results can be seen below: R o = 25 mm, d = 20 mm, f = 10 mm/sec, N = 5,000 rpm, α = 0.1745 (10 o ), and = 0.7854 (45 o ) g1: g2: g3: g4: g5: g6: g7: g8: g9: g10: g11: g12: g13: g14: g15: g16: g17: g18: g19: g20: g21: Objective Function: E = 96.9 parts/day

42 2.4.2: Monotonicity Analysis Subsystem 2 - Monotonicity Table Ro d f N α Φ 1/E - - - + g1 + g2 - g3 + g4 + g5 - g6 + g7 - - g8 + + g9 - + - g10 + + + + g11 - - - g12 + - - + g13?????? g14 + g15?????? g16 - g17 - g18 - g19 - g20 - g21 - Table 2.1: Full Monotonicity Table for Subsystem 2 Observations: 1) If constraint g13 or g15 is not active than both the helix angle and rake angle do not matter in the design of the end mill (in the sense of this model) because they do not appear in the objective function. If constraint g13 or g15 is active than either g6 or g19

43 (alpha) and g4 or g5 (phi) are also active and the helix and rake angle will either be at its maximum or minimum based on boundedness. 2) Constraint g13 or g15 is a particularly difficult equation in which to perform monotonicity analysis which is why there are question marks. Constraint g13 or g15 also contains all design variables. 3) From feasibility analysis above it seems that one of g7 or g9 dominates the other. This can also be seen from Table 2.1 above. 4) This subsystem is obviously a well bounded problem. All variables have the possibility to be bounded from both above and below. 5) g21 will never be active because g5 obviously dominates g21 therefore it is grayed out. 6) g16 will also never be active because g2 obviously dominated g16 therefore it is grayed out. 7) By MP1 with respect to R o, either g1, g8, g10, g12 must be active (possibly g13 or g15 which is unknown) 8) By MP1 with respect to d, either g3, or g10 must be active (possibly g13 or g15 which is unknown) 9) By MP1 with respect to f, either g9, g10, or g14 must be active (possibly g13 or g15 which is unknown) 10) By MP1 with respect to N, either g7, g9, g18 must be active (possibly g13 or g15 which is unknown) 11) By MP1 and the above analysis g2, g11, g17, and g18 will never be active therefore are grayed out All constraints will be coded into optimization program as a safety precaution. An analysis of results will be presented after simulation is complete and #11 above is expected to be true.

44 2.4.3: Scaling of Variables and Constraints Because of the large values associated with this problem, both optimizer algorithms used in this analysis, Direct and SQP, had a great deal of trouble converging on any solution points. In particular, constraint g10 had to be scaled by a factor of 1x10-21 to get the constraint within the desired 0 to 1 region when running the Direct Method. In fact, all constraints had to be scaled by a factor of at least 1x10-5 to achieve the desired region. In addition to scaling the constraints to be within 0-1, the design variables were also are scaled accordingly to improve algorithm speed and accuracy. For example, all variables are individually scaled such that the lower and upper bounds are all between 0 and 1. This was accomplished by simply using the upper bound of each variable as the scaling factor. By doing these two things, it allowed the algorithm to run much faster and converge on an acceptable solution point.

45 2.5 Optimization Study: Coding for this study can be found in Appendix 5.3 (Subsys2_fmincon.m, Subsys2_fname.m, Subsys2_cname.m, Forceseq.m, and Forces1and2.m) 2.5.1: Analytical Results with Assumptions In this section, the optimization problem will be solved using hand calculations and the monotonicity table above. Since constraints g13 and g15 are particularly complex, they are assumed to not to be active. Therefore variables alpha (rake angle) and phi (helix angle) are not relevant to the solution. The non-active constraints are removed and the resulting table can be seen below. Subsystem 2 - Monotonicity Table Ro d f N 1/E - - - + g1 + g3 + g7 - - g8 + + g9 - + - g10 + + + + g12 + - - + g14 + g19 - Table 2.2: Subsystem 2 Monotonicity Table for Analytical Model Observations: 1) By Mp1 with respect to N, either g7, g9 or g19 is active a. Tool rpm being 0 does not make any physical sense because the cutter will not cut, therefore g19 cannot be active

46 b. Assume g7 is active therefore and substitute this into g9. Therefore g9 reduces to. Even if Ro is at its largest value (50 mm) then would have to be for this to be feasible. Knowing that f is a monotonic decreasing function and must be bounded from above, we know that f will be greater than 2.25 and therefore g7 is not active. c. The only logical choice left is that g9 is active and. The model can now can be reduced too: g1: g3: g8: g10: g12: g14: And the simplified Montonicity Table for the above equations appears as: Subsystem 2 - Simplified Monotonicity Table Ro d f 1/E - - - g1 + g3 + g8? (-) < feasible - g10 + + + g12 + - - g14 + Table 2.3: 1 st Simplification of Subsystem 2 Monotonicity Table for Analytical Model 2) It can be seen, in Table 2.3, that g8 is now a regional monotonicity problem with respect to Ro.

47 a. Examining the broken down regions it can be seen that if than g8 will never be less than or equal too zero and therefore infeasible. b. Therefore, to be feasible, g8 must be monotonic decreasing, and therefore not active with respect to Ro, and since, g8 becomes monotonic decreasing and not active with respect to. c. In addition, since, g14 cannot be active either. d. In conclusion both g8 and g14 are inactive and are grayed out. 3) By Observation #2 above and MP1 with respect to, g10 must be active and. Substituting into the remaining equations, it can be seen that: g1: g3: g12: And the further simplified Monotonicity Table appears as: Subsystem 2 - Simplified Monotonicity Table Ro d 1/E - - g1 + g3 + g12 - - Table 2.4: 2 nd Simplification of Subsystem 2 Monotonicity Table for Analytical Model 4) We can verify results by testing activity a. If g1 is active than g12 reduces to and this is obviously monotonic decreasing, therefore g3 would be active

48 b. If g3 is active than g12 reduced to and this is again monotonic decreasing, therefore g1 would be active c. If g1 and g3 are both not active then the problem is not well bounded. d. We can now see that. e. Now substituting into the above values for that. we see f. The final results for the analytical model are: And the Objective Function 2.5.2: Numerical Verification of Analytical Results with Assumptions We can verify these results by performing an optimization in Matlab by using the scaled version of the problem (discussed in the last section). Starting point does not matter in this particular problem; multiple starting points were tested with the same result implying a global minimum and a pure convex feasible space. Using a simple starting point of:

49 MatLab computes the optimal point as: And the Objective Function The active constraints in this simulation are equivalent to g1, g3, g9 and g10. These active constraints can tell us a great deal about the physical meaning of the important aspect of the problem. First off, it confirms that are not important to the solution of the problem, which was expected. Also, activity in g1 and g3 make sense when it comes to efficiency because the larger are, the less passes you have to take with the cutter to cut the entire slot and less passes means faster. Activity in g9 and g10 tell us something completely different, mainly that we are being bounded by Tool Life and the manufacturer recommended chip removal thickness per flute. A simpler description is as follows, we are limited by how long we want the tool to last and how fast we can push the cutter into the workpiece. These parameters will be included in the parametric study for discussion of effect on the objective function.

50 This is precisely what was obtained in the analytical solution with a small amount of error that can be attributed to round off in hand calculations. This can be verified as a minimum by evaluating the hessian. 0.0045 0.0018 0.0191 0.0122 Hessian = 0.0018 1.0013 0.0109 0.0052 0.0191 0.0109 0.1444 0.0458 0.0122 0.0052 0.0458 0.0675 Eigen-values (Hessian) = 0.0012, 0.0463, 0.1686, 1.0015 Since all e-values are larger than zero, the Hessian is Positive Definite and the solution point is therefore a minimum. 2.5.3: Numerical Results of Full Model Unfortunately, due to the complexity of g13 and g15, analytical results for the full model are not available. Therefore, only numerical results will be presented. In this model, all constraints will be added into the MatLab code. The hope is that the stress constraints will be more restrictive than either or both of the tool life and chip removal per tool constraints (discussed in previous section). Again, the starting point used in this problem is shown below.

51 And MatLab computed the optimal point as: And the Objective Function The active constraints in this simulation are equivalent to g1, g3, g9 and g10. Again, are not important to the solution of the problem due to the lack of activity of constraints g13 or g15. Also, starting point again does not matter in this particular problem; multiple starting points were tested with the same result implying a global minimum and a pure convex feasible space. There as a few conclusions that can be made by this result. 1) Adding in the force constraints did not change the objective function output. 2) The results are the same as in the above section, Numerical Verification of Analytical Results with Assumptions. 3) Tool Life and manufacturer recommended chip removal thickness per flute dominate both stress constraints. a. This result makes sense because manufacturer recommended operating conditions are based on experiments and are set such that the tool does not reach the tensile strength of the material due to fatigue concerns.

52 2.5.4: Numerical Results of Full Model with decreased Allowable Stress One might argue that allowing the stress to have an upper limit of the tensile stress would be a bad idea. The reasons behind this are things like fatigue concerns, powder metal variation, and the lack of ductility in tungsten carbide. Therefore, we will do an analysis of a reduced allowable stress state that is modeled off fatigue like. This model being, use the max allowable stress to be 1/3 rd the tensile strength. The resulting change can be seen below. The simulation was then re-run using the standard starting point of: The optimal point found by the Matlab algorithm was, And the Objective Function

53 The active constraints in this simulation were g1, g3, g6, g10 and g13. Also, due to the addition of the activity of different constraint, the objective function has dropped from 126.33 to 124.66 parts/day. Even though this is a decrease, it is a small decrease, amounting only to a drop of 1.32% in the objective function. As you can see above, this result was quite interesting. The most interesting part was that when g13 is active, there should be 6 constraints active, one for each variable, but there are only 5 active. The variable that does not have a constraint bounding it is, the helix angle of the cutter. As you can see about, the value of did not move from its starting point. This was tested for multiple starting points and the same result was found for each point. A few explanations can be offered for this result. The most likely explanation is that the direction of decent for is so small that the computer algorithm rounded it off to zero. This can be confirmed by the fmincon output. To try and make it converge, the scaling on g13 and g15 was removed to make the functions more sensitive to changes in. The value of did decrease a little bit but then stalled once again. This confirms that the constraint g5 would be active if the computer would continue the algorithm. This means the optimal would be. Subsys2_glcsolve.m in Appendix 5.3 was used to find the scaling values for this problem.

54 2.6 Parametric Study: In this section, the reader should expect to see different sets of parameter values. 3-D graphs will be used to illustrate the difference in the objective function. Three parameters were chosen to be varied in this study for specific reasons. The first parameter is desired tool life and was chosen because it shows up in the active constraint g10. It is hoped that when tool life is varied than maybe other constraints will dominate it allowing us to see a limit for tool life. The second parameter is allowable cutter stress and although it does not show up in any active constraints at the moment, it is thought that as it decreases either g13 or g15 will become active and therefore changing the solution. Finally, the last parameter is volume of material removed and was chosen for it influence in multiple constraints and the objective function. This translates into 3 different plots comparing the effect of one parameter on another. 2.6.1: Case I. Objective function variation vs Desired Tool Life and Allowable Cutter Stress Below you can see the 3-D plot with iso-value lines of the objective function verses the desired tool life and allowable cutter stress (Appendix 5.3, para1.m and graphs.m). Figure 2.1: Parametric Study of Objective Function vs. Tool life and Cutter Stress

55 Below are a few conclusions that can be made about Figure 2.1: 1) The objective function can change dramatically based on the range of parameter values. 2) As seen by the iso-value lines, the shape of the graph below and allowable cutter stress of 75 N/mm 2 can be explained by different constraints being active before transitioning into its plateau region. 3) The objective function value is relatively constant between the ranges of about 75-345 N/mm 2 regardless of how long you want the cutter to last. You can see that the gradient of the objective function is quite small in this plateau region. 4) It is best to keep the cutter stress low to avoid premature failures. One third the tensile strength still has a very good objective function value. The best part is that this will not hurt your efficiency. 2.6.2: Case II. Objective Fn Variation vs. Allowable Cutter Stress and Volume of Material Below you can see the 3-D plot with iso-value lines of the objective function verses the allowable cutter stress and volume of material removed (Appendix 5.3, para2.m and graphs.m). Figure 2.2: Parametric Study of Objective Function vs. Cutter Stress and Volume

56 Again, a few conclusions about Figure 2.2 are listed below: 1) The shape of this graph verifies that the objective function changes based on the parameter value. Also, it is similar to that of the above graph in Case I. 2) The odd spikes on the sloping region can be explained by activating different constraints at specific values of parameters. 3) As can be seen by the iso-value lines, at a given volume, increasing the allowable cutter stress does not do much good for the objective function. Again, this is similar to the result in Case I. 4) This graph has less slightly sloped plateau region, unlike the flat region in Case I. 5) The shape of the graph below and allowable cutter stress of 125 N/mm 2 can be explained by different constraints being active before transitioning into its plateau region. 6) For a given workpiece volume, it would be best to allow only a stress of 150 N/mm 2 for extended cutter life and good objective function value. This is because, there is little to no gain in the objective function for increasing the allowed cutter stress.

57 2.6.3: Case III. Objective function variation vs. Desired Tool Life and Volume of Material Below you can see the 3-D plot with iso-value lines of the objective function verses the desired tool life and volume of material removed (Appendix 5.3, para3.m and graphs.m). Figure 2.3: Parametric Study of Objective Function vs. Volume and Tool life As before, a few conclusions about Figure 2.3 are listed below. 1) This graph is straight forward, an increase in objective function value based on the decrease in volume removed. 2) Desired tool life has very little effect on the objective function at a given volume, as verified by the iso-value lines. 3) As one might suspect, the best objective function value is when you have very little material to remove from a workpeice.

58 2.6.4: Case IV. Objective function variation vs. Increasing Tool Life in Case II The purpose of this graph is to illustrate the effect of desired tool life changes on Case II. The reason this particular arrangement was selected is because in both Case I and III desired tool life had very little effect on the change in objective function at a given value for the other parameter in the graph. It is thought this would be a good way to graphically show all three variables in one graph. The way this graph is generated is desired tool life is varied from 1000 to 14000 seconds in increments of 1000 seconds. This generates 14 surfaces of objective function values verses allowable cutter stress and workpiece volume. These 14 surfaces were then put in one graph to show the change that desired tool life has on the objective function iso-value lines, seen below (Appendix 5.3, para4.m and graphs.m). Figure 2.4: Parametric Study of Objective Function vs. Volume and Cutter Stress with Varying Tool Life

59 The conclusion of Figure 2.4 on the previous page can be seen below. 1) The general shape of the surface is the same as the surface in Case II. It is shown at a different angle to show the effect of change in desired tool life on the iso-value lines of the objective function. 2) Changing the desired tool life does not change the general shape of the surface; it merely translates the surface downwards. 3) As can be seen by the text box in the above graph, the three arrows shows what happens to the objective function iso-value lines and surfaces when the required tool life is increased. a. This shows that when you increase required tool life and you want to obtain the same objective function value then you must decrease the amount of volume removed from the workpiece. b. Or at a given workpiece volume and increasing required tool life; you must increase the allowable cutter stress to obtain the same objective function value. 4) In text box 1 in Figure 2.4 refers to the high slope region of the surface. In this region constraints g1,g2, g6, g8, and g13 were active. This means that the objective function has some of the works results when constrain g8 is active. 5) In text box 2 in Figure 2.4 refers to the middle region of the surface. In this region constraints g1, g2, g5, g6, g13, and g14 were active. The jump in the object function between the high slope and middle region can be characterized by g8 becoming deactive. This region provides and interesting result, when g14 is active, corresponding to bending stress, the helix angle does reach its lower bounds (g6), this verifies that the smallest helix angle is the best for the cutter. 6) In text box 3 in Figure 2.4 refers the plateau region of the surface. In this region constraints g1, g2, g6, g10, and g13 were active. This has the same reasoning as in the section titled Numerical Results of Full Model with decreased Allowable Stress. The region has a slight gradient in the direction of workpiece volume but is relatively constant along the allowable cutter stress region.

60 7) Again, for a given workpiece volume, it would be best to allow only a stress of 150 N/mm 2 for extended cutter life and good objective function value. This is because there is little to no gain in the objective function for increasing the allowed cutter stress. This result is regardless of the desired tool life.

61 2.7 Discussion of Results: 2.7.1: Design Implications Based on engineering knowledge and the parametric study s some design implications can now be identified. First implication being that the desired tool life of your cutter has an effect on how many parts you can produce in a given day. This leads to choosing a tool life that makes sense, like the tool life being approximately 2 hours so that the cutter can be changed just before or after the operators break time. This choice also comes with knowledge of operator behavior. The second implication comes from Case I, II, and IV in the parametric study. Examining Figures 2.1, 2.2, and 2.4 one can see that not much advantage is gained in the objective function after a certain allowable cutter stress is reached. Knowing that machining has many variables not taken into consideration in this model, such as workpiece material and cutter defects, we know the lower the stress the better. The third implication comes from the simulations in Section 2.4 (Model Analysis). This gives us values of the variables that will obtain the highest objective function value. All these are summarized below in the Design Rules section. 2.7.2: Design Rules for Maximizing Efficiency 1) Tool life should be approximately 2 hours. 2) Allowable cutter stress should be about equal to 150 N/mm 2 3) Outer cutter radius and depth of cut should be as large as possible, 50 mm and 40 mm respectively 4) Via simulations feed should be between 9-10 mm/sec, Tool RPM should be between 1381-1469 rpm, rake angle should be 0.2618 radians (15 degrees), and helix angle should be 0.5236 radians (30 degrees) When applying the design rules and running a simulation, one can see that the objective function obtained, after rounding, is 125 parts/day and the Eigen-values of the hessian are all positive, confirming a minimum.

62 2.7.3: Further Discussion The above results for design implications can be considered as very reasonable. If you think about this problem, if you want to maximize efficiency, then you would want too make the smallest amount of passes possible and to go as fast as possible. This is accomplished by making the cutter as big as possible and taking a depth of cut as deep as possible. Both of which are verified multiple times in the simulations. The other 2 active constrain in the final solution g9 and g10 have practical, sensible reasoning behind why they are active. As discussed before Tool Life and manufacturer recommended chip removal thickness per flute dominate both stress constraints. This result makes sense because manufacturer recommended operating conditions are based on experiments and are set such that the tool does not reach the tensile strength of the material due to fatigue concerns. To make this problem more interesting, you could include material and workpiece defect equations and what implications that would have on the stress in the cutter. You could also have a more complicated model for the stress on the flute and the fatigue of the tool. Most of these things require a great deal of modeling and would severely increase the complexity of the overall model and the formulation of the objective function.

63 Subsystem 3 Quality ($/part) Jonathan Loss 3.1 Problem Statement This subsystem will be focused on finding the optimal selling price of a machined part, as we consider that the price at which a part will be sold is a direct function of its quality, which itself is a decreasing function of the amount of errors (deflection from planned cut). A few tradeoffs can already be anticipated, such that higher feed, while improving the efficiency, will decrease the quality if the depth of cut and rotation speed remain constant. In fact all six variables of the problem will have an influence on the part quality, in two main ways. The first way is known as chatter, and consists in vibrations of the mill greatly affecting the quality of the cut. These appear when the forced oscillations of the system, due to the periodic nature of the process (teeth attacking periodically the material of the piece), have frequencies approaching the natural frequencies of the system. The quality of the part machined in these conditions is so low that it cannot even be considered acceptable, thus only conditions when chatter does not arise can be studied. This is done using stability lobe diagrams, and choosing values of the variables such that we are always in the stable part of these diagrams. The second way in which part quality is influenced, and which will actually be used to measure part quality, is through the measurement of the deflection from the planned cut, also known as Surface Location Error (SLE). The cutting forces at work on the mill make it bend and deflect from the original cut in the direction normal to the feed direction, and the analysis of this deflection along with the movement of the mill gives us the error. This model greatly simplifies the real problem, but captures the essential tradeoffs; the main part of the error indeed happens in the direction normal to the cut, and chatter has to be avoided at all costs.

64 3.2 Nomenclature W : Width of slot [mm] H : depth of slot [mm] d : depth of cut [mm] d min : minimum depth of cut [mm] M : length of mill [mm] df : depth of final cut [mm] N H : closest integer by inferior value to H/d [no unit] R i : inner ball end mill radius [mm] R o : outer ball end mill radius [mm] t : teeth thickness [mm] Θi : Position of the i-th tooth [radians] f : feed [mm/s] D r : distance moved per revolution (mm) N: spindle rotation speed [rpm] T : period for one revolution [s] α: rake angle [radians] Φ: helix angle [radians] δ : deflection from planned cut [mm] δ max : deflection from planned cut [mm] pph : parts per hour [no units] Err : surface defined by the deflection integrated over one fourth of a period [mm^2] Errmax : maximum admissible error [mm 2 ] Qual : quality of a given cut (of depth d or df) [no units] Qual tot : total quality [no units] Price : price at which the piece is sold [$] Dprice : price variability of the piece according to quality [$] Minprice : price of the worst possible piece [$] F f : friction force on the cutting edge [N] F n : normal force on the cutting edge [N] F c : cutting force on the cutting edge [N]

65 F t : thrust force on the cutting edge [N] F l : lateral force on the cutting edge [N] F yi : force in the direction normal to the cut generated by the i th tooth [N] F ytot : total force in the direction normal to the cut [N] E : Young Modulus of the carbide mill [Pa] I : area moment of inertia of the carbide mill [mm 4 ] σ : stress on the tooth root (N/m 2 ) σ t : tensile stress of WC tool (N/m 2 ) P max : motor maximum power [W] e = efficiency ( power used for cutting / power delivered by motor ) Figure (3.1) : Mill Figure (3.2) : rake definition

66 3.3 Mathematical Model 3.3.1 Objective function: The objective function describes the price of the piece through its quality, itself a function of error, error being the difference between the planned cut and the actual cut. The chatter issue will be taken care of as a constraint, since chatter is unacceptable due to the extremely poor quality it yields. Price = Dprice*Qual tot + Minprice (3.1) 3.3.2 Constraints: The quality a given cut (depth d or df) is a function of the error. Err Qual = 1 Errmax δmax With Err max = 2 π 0.1 (3.2) The error is a function of the feed rate (f) and of the deflection (δ) L 1 Err = δ( x) * dx L 0 Hence changing the variable x to t using x = f * t : L T /4*( ) L/ f f* T/4 T /4 1 1 1 Err = t * f * dt t * f * dt t * f * dt L δ( ) = L δ( ) δ( ) f * T /4 0 0 0 N *2 π Finally changing t to θ using θ = * t : 60 π /2 2 Err = * δ( θ ) * d θ π (3.3) 0 The cutting force can be calculated based on a cutting force model found in the literature,

67 (built using an oblique cutting force model). 2 2 ( α ) F = F (sin Φ ) + (cos Φ ) sin + F cos Φ cosα (3.4) c f n F = F cosφcosα F sinα (3.5) t f n ( α ) F = F cosφsin Φ sin + sin Φcos Φ + F sin Φ cosα (3.6) l f n ( ) F = d/ sin Φ D exp 18.315 0.26267ln( D ) 0.10327 ln( N R 2π + 1.8828ln(1 sin α) + 0.0047625ln( D ) ln( N R 2 π) (3.7) n r r o r o ( ) F = d/ sin Φ D exp 15.031 0.6292ln( D ) + 0.32256ln( N R 2π + 0.34417ln(1 sin α) + 0.299585ln( D ) ln( N R 2 π) (3. 8) f r r o r o F = G( Θ )*( F sin( Θ ) + F cos( Θ )) with ( ) 1 yi i t i c i (3.9) 2 2 2 2 2 F eq F c F t F l F n F f = + + = + (3.10) G Θ = if Θ [ 0, π ] i i, 0 otherwise F ytot F ytot is a function of F yi. 4 = F i= 1 yi Knowing that only two teeth are cutting at the same time, one at θ and one at θ + π/2, we obtain the following equation (only valid for θ between 0 and π/2): F = ( F + F)cos( θ ) + ( F F )sin( θ ) (3.11) ytot c t t c Θi is a function of the spindle speed. Θ i = ( N*2 π / 60)* t+ ( i* π / 2) (3.12) The area moment of inertia of the carbide mill is a function of R o. I R o 4 = π *(2 ) /64 (3.13) The period T is a function of N. T = 60 / N (3.14)

68 Distance moved per revolution can be defined as a function of the period and feed rate. D = f * T (3.15) r The deflection δ is a function of F ytot, I and d. 3 Fytot *( M 0.5 d) δ= (3.16) 3* E* I The total quality Qual tot is a function of Qual(d) and Qual(df). Qual tot Qual( d)* NH * d + Qual( df )* df = (3.17) H N and df are functions of d. NH = floor( H / d) (3.18) df = H N * d (3.19) H There is a functional relationship between outer and inner radius determined experimentally. Ro + 3.5 Ri = (3.20) 2 With the use of beam theory, assuming the cross section of tooth is a rectangular shape, the stress at the tooth root can be obtained by following equation. 6 ( ) 10 Feq Ro Ri σ = 1 d 2 t 24 sin Φ (3.22) There is a functional relationship between R o and the thickness of the tooth determined experimentally. t = 0.1974R o + 1.5592 (3.23)

69 Practical inequalities The rotation speed has an upper bound, due to mechanical limitations of the motor/system powering the ball-end mill. N 20, 000 (3.24) For similar reasons the the feed has a lower bound. f f min (3.25) Same for maximum feed. f f max (3.26) Same for rake. α 15* π /180 (3.27) Same for helix Φ 75* π /180 (3.28) Φ 30* π /180 (3.29) Depth of cut need to be smaller or equal to the depth of slot. d H (3.30) The depth of cut has a lower bound: d d min (3.30 bis) Outer radius needs to be smaller or equal to an upper bound R max (2* R max < W). Ro R (3.31) max

70 Outer Radius need to be greater or equal to a 10 mm diameter cutter. R 5 (3.32) o The error must be inferior to a maximum value. δ Err Err max = 2 max π (3.33) Indeed, with Err max calculated as the error obtained for a maximum deflection δ max (of 10 microns for example) over a sinusoidal path we obtain: δmax Err max = 2 π The spindle speed must be within operational limits of carbide cutters given by recommended SFM values (25-300 SFM for cutting Ti). 12 SFM N = Ro 2π 12 25.4 Therefore, there s a constraint relating N to the outer radius R o. 14561 N Ro (3.34) N R o 174726 (3.35) All physical variables must be positive. ALLvar iables 0 (3.36) Physical In order not to break the tool with too large moved distance per revolution, the distance need to be constrained by the tool geometry, with the following relationship derived experimentally. Ro Ri f * T (3.37) 50

71 The variables and parameters must always be such that the system is within the stability lobes, otherwise due to chatter the quality is too poor to be even considered. A rough approximation of the region contained within the stability lobes can be: 12 d * N 0.8 (3.39) 5000 which implies N 340 (3.40) The feed rate must not exceed a certain value given by the maximum rate of material removal by the ball-end mill. This reflects a physical issue: the ball end mill can only move forward as much as it can remove the necessary amount of material. f T R R R (3.41) 2 2 * *2 o π *( o i ) The stress of the tooth needs to be smaller than the tensile stress σ t (0.3448 GPa) of the tool material (WC). σ σ t (3.42) In order to create a competition between variables, and simulate thus the tradeoff that occurs between productivity and quality, we add a constraint of a minimum productivity (this constraint should not be of any use anymore when we study the overall system) : W* H* L f * d *2Ro pph (3.43) 3600 With the use of beam theory, we can find the bending stress of the cutter at the fixed point on the shank. σ bend F + F 4H 10 = (3.44) 2 2 6 c t 3 π R0 3.3.3 Design Variables and Parameters:

72 The parameters are: W, H, M, Minprice, Dprice, E, Pmax, e, σ t, Errmax, Rmax, - geometry parameters : M =2*H = 80 mm, H = 40mm, W = 100 mm, Rmax = 15 mm - Cost parameters : Minprice=10 $, Dprice=90$ - Material properties: σ t = 344.8*10 6 Pa, E=600 GPa - Maximum deviation from cut: δ max = 10 microns - Productivity parameters: pph = 13 - Milling machine parameters: f min = 0.2 mm/s, f max = 1270 mm/s Intermediate variables: df, N H, T, t, F ytot, F eq, σ, D r, R i, Θi, δ, Err, Qual, I, Err max Variables: f, N, d, α, Φ, R o DOF = 6 3.3.4 Final Model: Qual( d)* NH * d + Qual( df )* df Min NegPrice = Price = Minprice + Dprice* H Derived from equations (3.1) and (3.17) Where N = floor( H / d) and df = H N * d from (3.18) and (3.19) H H 3 2 Ft *( M 0.5 d) And where Qual( d) = 1 3* E* I δmax Derived from equations (3.2), (3.3), (3.11), (3.13) and (3.16) 0.1 2 2 ( α ) F = F (sin Φ ) + (cos Φ ) sin + F cos Φ cosα ; F = F cos Φcosα F sinα c f n ( cos sin sinα sin cos ) sin cosα ( ) ( ) F = F Φ Φ + Φ Φ + F Φ l f n t f n F = d/ sin Φ D exp 18.315 0.26267ln( D ) 0.10327 ln( N R 2π + 1.8828ln(1 sin α) + 0.0047625ln( D ) ln( N R 2 π) n r r o r o F = d/ sin Φ D exp 15.031 0.6292ln( D ) + 0.32256ln( N R 2π + 0.34417ln(1 sin α) + 0.299585ln( D ) ln( N R 2 π) f r r o r o D = f *60/ N r

73 g1: N 20,000 0 from (3.24) g2: fmin f 0 from (3.25) g3: f fmax 0 from (3.26) g4: α 15* π /180 0 from (3.27) g5: Φ 75* π /180 0 from (3.28) g6: 30* π /180 Φ 0 from (3.29) g7: d H 0 from (3.30) g8: R Rmax 0 from (3.31) o g9: 5 R o 0 from (3.32) 3 4 Ft *( M 0.5 d) g10: Errmax 0 3 π * E* I g11: 14561 N* R o 0 from (3.34) from (3.33) g12: N* Ro 174726 0 from (3.35) g13: ( ) 6000 f N R o 3.5 0 from (3.37), (3.14) and (3.15) 2 Ro + 3.5 120* f * Ro N* π *( Ro ) 0 from (3.41) and (3.20) 2 g14: ( ) g15: ( Ro + ) Φ ( 0.1974R + 1.5592) 3F 3.5 sin *10 eq d o 2 6 2 σ 0 from (3.42), (3.20), (3.22) and (3.23) t g16: F + F 4H 10 2 2 6 c t 3 π R0 σ 0 from (3.44) and (3.42) t g17: dmin d 0 from (3.30 bis) g18: W* H* L pph ( f * d *2 Ro ) 0 from (3.43) 3600 g19: 12 d * N + 0.8 0 from (3.39) 5000 g20: 340 N 0 from (3.40)

74 3.4 Model Analysis 3.4.1 Feasibility Analysis: Select a set of variables: R o = 10mm, f = 2 mm/sec, N = 2000 rpm, d = 3 mm, α = 5*pi/180 radians, Φ = 45*pi/180 radians g1: N- 20,000 = -18,000 0 feasible g2: 0.2 - f = -1.8 0 feasible g3: f -1270 = -1268 0 feasible g4: α 15* π /180 = 0.1745 0 feasible g5: Φ 75* π /180 = 0.5236 0 feasible g6: 30* π /180 Φ= 0.2618 0 feasible g7: d 40 = 35 0 feasible g8: Ro 15 = 5 0 feasible g9: 5 R o = 5 0 feasible g10: 3 4 Ft *( M 0.5 d) 0.02 = 0.00584 0 3 π * E* I π feasible g11: 14561 N R o = 45439 0 feasible g12: N R o 174726 = 114726 0 feasible g13: ( ) 6000 f N R o 3.5 = 9000 0 feasible 2 Ro + 3.5 120* f * Ro N* π *( Ro ) = 47,965 0 2 g14: ( ) g15: ( Ro + ) Φ ( 0.1974R + 1.5592) 3F 3.5 sin *10 eq d o 2 6 2 = 6 6 344.8*10 317.4*10 0

75 g16: F + F 4H 10 2 2 6 c t 3 π R0 6 6 344.8*10 = 289.5*10 0 feasible g17: 0.1 d = 2.9 0 feasible g18: g19: W* H* L pph ( f * d *2 Ro ) = 48 0 feasible 3600 12 d N + 0.8 = 1 0 feasible 5000 g20: 340 N = 1660 0 feasible All the variables are positive and NegPrice = -31.55 (corresponding to a price of $ 31.55). Therefore the model is feasible.

76 3.4.2 Model Analysis Monotonicity Table: The monotonicity of all functions using the cutting force model are impossible to determine analytically, however empirical study of the functions behavior as well as educated guesses allow us to determine the monotonicity of several variables. These will be indicated by a star next to them. R o f d N α Φ NegPrice --* +* +* --* +* --* g 1 + g 2 -- g 3 + g 4 + g 5 + g 6 g 7 + -- g 8 + g 9 -- g 10 --* +* +*? +* --* g 11 -- -- g 12 + + g 13 -- + -- g 14?(+) + -- g 15? +*???? g 16? +*???? g 17 -- g 18 -- -- -- g 19 + -- g 20 --

77 The monotonicity of the objective function with respect to its variables can be more or less justified by the following plots, made in the neighborhood of the optimum. Figure (3.3) : Negprice(N,f) Figure (3.4) : Negprice(d,R)

78 Figure (3.5) : Negprice(α,Φ) Monotonicity Analysis: The complicated expression of the cutting force makes monotonicity analysis complicated. However, we have been able to find some simple properties of the cutting force model, such that the error decreases with increasing outer radius, but increases with increasing feed and depth. These results are intuitive, as indeed one expects the cutter to deviate more with a bigger force exerted on it. And where the cutting force is an increasing function of the feed and depth of cut, the inclination of the tool to bend is a decreasing function of its outer radius (simple beam theory). Finally, empirical studies have proven that the cutting force is a increasing function of α (rake angle), and a decreasing function of Φ (helix angle). Therefore: By MP1 applied to f, we know that either g2 or g18 is active. The fact that g2 is a simple lower bound makes obvious that g18 is the constraint that has to be active. Nevertheless, even using all this information we still cannot deduce anything concerning possible activity of constraints for all the variables except f, because of the constraints g10,

79 g15 and g16. Thus we decide to make the assumption that these constraints are not active (this will be justified a posteriori). We obtain the following simplified monotonicity table: Simplified monotonicity table R o f d N α Φ NegPrice --* +* +* --* +* --* g 1 + g 2 -- g 3 + g 4 + g 5 + g 6 g 7 + -- g 8 + g 9 -- g 11 -- -- g 12 + + g 13 -- + -- g 14? (+) + -- g 17 -- g 18 -- -- -- g 19 + -- g 20 -- By MP1 applied to R o, at least one among the constraints (g8, g12, g14) is active. By MP1 applied to d, at least one among the constraints (g17, g18) is active.

80 By MP1 applied to Φ, g5 is active: Φ = 75 degrees. By MP1 applied to α, the zero lower bound is active: α = 0 degrees. By MP1 applied to N, g12 is active:. 174726 N = R o Finally, we have chosen R max such that it is the limiting element in the choice of R o, giving us the activity of g12: R o = R max = 15 mm. This choice will be explained numerical results section.

81 3.5 Numerical Results Preamble: We chose to directly scale the constraints, since this can only improve results. Moreover, when the DIRECT algorithm was tried, it could never find feasible points, which was not a problem for the SQP algorithm, given any starting point. Therefore the scaling was done manually, in the following manner: For each constraint we considered simple upper and lower bounds of the function at the left side of the inequality. Then we scaled by dividing the constraint by: upperbound + lowerbound (the arithmetic average of the two). 2 Thus, kn being the scaling corresponding to constraint gn we obtained: k1 = 10, 000 k2 / 2 = fmax k3 / 2 = fmax 15 π /180 k4 = 2 (75 + 30) π /180 k5 = 2 (75 + 30) π /180 k6 = 2 k7 H = 2 R k8 = + R 2 max min R k9 = + R 2 max min Err k10 = 2 max 174726 + 14561 k11 = 2

82 174726 + 14561 k12 = 2 6000* f k13 = 2 max 120* f * R k14 = 2 max max σ k15 = t 2 k16 σ = t 2 W* H* L k18 = pph 2*3600 k19 H = 2 k20 = 10,000 This scaling method was justified a posteriori by looking at the constraints: all those that are inactive are between 0.002 and 10. In addition to this, all the variables were scaled between zero (their lower bound) and 1 (their upper bound). The relationship between the non scaled variable (x) and the scaled one (y) is the following: x lb y = with ub = upper bound and lb = lower bound ub lb 3.5.1 Matlab SQP method: Initially, the model didn t include constraints g2, g3, g8, g17 and g18. This proved to be a big problem, because in this case in order to achieve a perfect quality all that was needed was to have the feed f come close to zero, and keep roughly all the other variables where they were (at the starting point). This is easy to understand: a very small f implies very small cutting forces, which itself ensures low deflection and good quality, whatever the rest of the variables. But in terms of having a meaningful solution, this is useless since no productivity is achieved with no feed. This will automatically be taken care of in the overall system as a competing

83 subsystem will try to maximize efficiency (that is productivity), and thus increase feed. Therefore a logical solution was to simulate the competing subsystem by requiring a minimum productivity. This is what constraint g18 adds through the parameter pph (parts per hour). But then a similar problem arised for R o : A large diameter of the cutting tool (around the width of the slot) provides a great rigidity, which in turn decreases the deviation to something close to zero. Hence for the same reason, the diameter must be bounded in a region such that trade-offs between the other variables will be necessary. For this reason the constraint g8 was added, and it is indeed (almost) always active at the optimum. Without g8, the cutter took its diameter equal to its width, and the result was such a high rigidity that for whatever starting point, the other variables only changed slightly to get to the optimum. g2, g3 and g17 were added so that if the productivity constraint was not sufficient, there would be others to play a similar role. They also reflect physical issues: there is a minimum cutting depth below which milling doesn t really make sense anymore; the feed also has upper and lower bounds defined by the milling machine (usual values were taken in this case). Different starting points N (rpm) F (mm/s) d (mm) R o α Φ Price (mm) (deg) (deg) ($) Active constraints starting 1 2,000 2 3 7 5 45 result 1 11,648 3.55 27.16 15 0 75 79.72 g4 g5 g8 g12 g18 g19 starting 2 5,000 2 2 5 10 30 result 2 11,648 3.55 27.16 15 0 75 79.72 g4 g5 g8 g12 g18 g19 starting 3 10,000 1 10 5 10 30 result 3 11,648 3.55 27.16 15 0 75 79.72 g4 g5 g8 g12 g18 g19 starting 4 1,000 5 1 3 15 75 result 4 11,648 3.55 27.16 15 0 75 79.72 g4 g5 g8 g12 g18 g19 starting 5 17,000 5 1 3 15 75 result 5 11,648 3.55 27.16 15 0 75 79.72 g4 g5 g8 g12 g18 Table 3.1: Result Table for Different Starting Points

84 Discussion: We see that with different starting points we obtain the same solution. The exit flags differ, in some cases the first order optimality conditions are satisfied, in some cases the gradient becomes too small and the optimization is stopped. It is interesting to notice that initially, for sets of parameters that were not well picked, several optima were obtained for these different starting points, all of them minima at a hundred dollars. This boggled me for quite a while, until I realized that many variables had a very important influence on the deviation from the planned cut. Thus they were as many different ways to improve the quality, making so many path leading to a cost of roughly a hundred dollars. So either there were as many optimas (which I doubt) or they all were so close to the optima that for the algorithm it is as if they were AT the optima: the rate of change, gradient is below their detection threshold. Indeed fmincon returned exitflags saying that the gradient magnitude was below its threshold. A good picture of a similar problem would be the following: Figure (3.6) : example function

85 They are many different starting points from which a solution close to the minimum can be reached, as the area where the value of the function is close to the minimum is wide. Therefore each starting point would give a different solution, all of them almost optimal. This explains why the activity of the constraints didn t correspond exactly to what the monotonicity analysis yielded. The problem was solved by picking good parameters, such as Rmax = 15 mm (instead of higher values which offered such high rigidity the other variables had little incidence on the quality of the cut. Moreover ball-end mills rarely have diameters bigger than 15mm). Err Err Changing Qual = 1 to Qual = 1 Errmax Errmax 0.1 also improved a great deal the optimization, by un-flattening the optimal area, and enabling the algorithm to continue descent. 3.5.2 Analysis of the result All the constraints that were predicted to be active by the monotonicity analysis turned out to be also active in the numerical results: g18 is active as predicted, the lower bound of d is not reached. g5 is active, Φ reaches its upper bound of 75 degrees. g12 is active. The lower bound of α is reached. The radius R o reaches its upper bound, just as intuition and monotonicity analysis predicted. Moreover there is an additional active constraint, which is g19, the stability lobe constraint. This was not predicted by the monotonicity analysis, as both NegPrice and g19 are increasing functions of the depth d and decreasing functions of N. The explanation for this behavior is the following: 1 st step: The activity of g12 gives us the spindle speed N, since we already know the radius reaches its upper bound.

86 2 nd step: We now have only 2 variables for the optimization: the depth d and the feed f. g18 is active as predicted, so there is tradeoff between d and f, both trying to be as low as possible. 3 rd step: It turns out that decreasing f provides better quality than decreasing d, so f will be minimized, forcing thus d to increase, until it hits an upper bound. This bound is provided by g19, the stability lobes constraint, which thus explain its activity. The optimization is clearly robust as all of the six different starting points converged to the same optimum. However its has to be noted that the DIRECT algorithm was never able to find a feasible space, and thus never gave any results (except, surprisingly, when there was a mistake in the code). After having a look at the constraints while the DIRECT algorithm ran, it turned out that the offending constraint was g10, the limiting of the error to acceptable values (below Err max ). Removing this constraint enabled DIRECT to run and find a feasible space, but the value of the error remained huge in this feasible space, much bigger that Err max. I cannot explain this behavior besides the fact that DIRECT appears unable to find a feasible space, since the optimization doesn t seem to be a problem for the SQP algorithm. On the other hand the Simulated Annealing algorithm, while having problem finding the optimum, was able to confirm that the solution found by fmincon was indeed an optimum.

87 3.6 Parametric Study Preamble: The parameters chosen for the study are pph, R max and δ max. Indeed as explained above R max has a crucial importance in defining how wide is the region where the objective function is very close to the optimum, and pph creates a tradeoff with a productivity constraint. δ max corresponds to a kind of scaling of the objective function, a well as a tightening (or loosening) of the feasible space. The set of parameters considered as the point around which this study is conducted is the default set of parameters: - geometry parameters : M =2*H = 80 mm, H = 40mm, W = 100 mm, Rmax = 15 mm - Cost parameters : Minprice=10 $, Dprice=90$ - Material properties: σ t = 344.8*10 6 Pa, E=600 GPa - Maximum deviation from cut: δ max = 10 microns - Productivity parameters: pph = 13 - Milling machine parameters: f min = 0.2 mm/s, f max = 1270 mm/s 3.6.1 Influence of δ max on the optimum The parameter δ max basically defines Err max and thus influences both the constraint g10 and the objective function. Increasing δ max corresponds to making the error acceptability more lenient, accepting a lesser quality for the same price, thus increasing the price for a given set of other parameters.

88 Figure (5) : Optimum(δ max ) 3.6.2 Influence of R max on the optimum The parameter R max has a very important role, as it defines the radius at the optimum. Thus the quality will increase for a bigger R max as it increases the rigidity of the tool and thus reduces the deflection from the planned cut. Figure (6) : Optimum(R max ) We notice that the increase in quality seems to be much more linear than for δ max.

89 3.6.3 Influence of pph on the optimum The parameter pph measures the productivity of the process, and appears in the constraint g18. In a way it plays the role of the competing subsystem efficiency, by enforcing a minimum productivity, by default 13 pieces produced per hour. To meet this requirement, variables f, R o and d have to be higher than what they would be without. Thus an increase in pph results in a decrease in quality, which corresponds to intuition: A higher productivity means a lower quality as f, d and R o have to collectively increase and since R o already reaches its upper bound, f and d are the only ones increasing. Moreover the price is a decreasing function of f and d, so this will result in a price decrease. Figure (6) : Optimum(pph)

90 3.7 Discussion of Results 3.7.1 Design Rules: From the numerical optimization results, given the default parameters, the following design rules have been obtained: 5) The radius of the cutting tool should be as big as possible, in our case with our default parameters this means 15mm. 6) The depth of cut should be either 27.16 mm. 7) The spindle speed should be 11,648 rpm. 8) Rake angle should be zero, helix angle should be 75 degrees in all situations. 9) The feed rate should be 3.55 mm/s. The price per part obtained following these design rules is $ 79.72. 3.7.2 Further Discussion: The relationship between price per part and error was decided arbitrarily, it was initially linear in the first model, and it was found that putting it to the power 0.1 was more realistic and helped the algorithm. It respects the fact that the biggest price changes occur when the quality is getting closer to the best quality possible, but it is still an arbitrary decision: this relationship can be approximated in many other different ways which might be more realistic. However, for our purposes, it already yields a good understanding of the optimization problem. Indeed we have seen that our model yields good robust results, consistent with the analysis of the model, both by intuition and by monotonicity. In order to improve upon these results, a better cutting force model should be used, as it is the backbone of this optimization. The stability lobes should be properly calculated, since they also

91 change with respect to Ro and f, instead of using a very rough approximation as we do here. This would however result in a much more complicated model, as the FRF (Frequency Response Function) would have to be determined. Finally using a professional optimization software might become indispensable, given the complexity of this new model.

92 Overall System Profit ($/day) 4.1 Description: The objective of this project is to get the maximum profit from the machined product, which mainly depends on the manufacturing process. The product is a slot of dimensions W L with a depth H, which is milled by a Tungsten Carbide end mill of radius R, possibly with several paths, in titanium alloys. The profit will be affected by three major factors: manufacturing cost, machining efficiency and quality of the machined piece. Assuming that all final products are sold, manufacturing cost can be seen as the prime cost of the product, while machining quality will affect the selling price of the product. Finally, the efficiency will define the quantity of the products produced over a day. As a result, the profit could be calculated by multiplying the quantity of products with the difference between selling price and prime price. The objective function of the whole system will be: max Profit = Efficiency * (Quality Cost) The objective of three subsystems are clear, minimizing the prime cost, maximizing the machining efficiency and maximizing the quality of product. However, the optimization results of each subsystem won t lead to maximum profit of the whole system. For example, the prime cost might depend extensively on tool life since Tungsten Carbides tools have high cost and the tool life decrease dramatically if the cutting parameters such as feed and speed increase. On the other hand, the increasing cutting parameters definitely lead to increasing efficiency. The tradeoffs between subsystems will bring on a different optimization result of the whole system.

93 4.2 Nomenclature {Name: Description (units)} C m : machining cost per piece ($/part) C l : cost of loading, unloading and material handling ($/part) C t : tooling cost, includes tool changing and regrinding of the cutter ($/part) C e : cost of machine power ($/W-hr) T g : grinding time of the tool (sec) L m : labor cost of production operator ($/hr) B m : burden cost ($/hr) M m : machine cost, required power cost to machine parts ($/hr) L g : labor cost of tool-grinder operator ($/hr) B g : burden rate of tool grinder ($/hr) D c : depreciation of the tool in dollars per grind ($/tool) MRR: material removal rate (mm 3 / min) u t : specific energy (total energy per unit volume) (kw/mm 3 ) e: machine efficiency E = Efficiency (Parts/Day) α = rake angle (degrees) Φ = helix angle (degrees) N = speed of cutter (rpm) d = depth of cut (mm) R o = Outside radius of cutter (mm) R i = Inside radius of cutter (mm) f = feed of cut or velocity of workpeice (mm/sec) H = Total depth of slot (mm) L = Length of slot (mm) W = total width of slot (mm) T = Tool life equation Life (sec) T c = Time to change one tool (sec) T p = Total production time (sec/part) T l = Time to Load One Part (sec/part) T m = Machining Time per part (sec/part)

94 N p = Pieces per Tool (part/tool) H d = Hours in production day (hours/day) D r = Distance moved per revolution (mm/rev) F c = Cutting force (N) F t = Tool thrust force (N) F l = Lateral force (N) F f = Frictional force on cutting edge (N) F n = Normal force on cutting edge (N) σ = stress cutting flute, beam (N/mm^2) σ t = tensile stress of WC tool (N/mm^2) Θi : Position of the i-th tooth [radians] δ max : deflection from planned cut [mm] δ : deflection from planned cut [mm] Err : surface defined by the deflection integrated over one fourth of a period [mm^2] Errmax : maximum admissible error [mm 2 ] Qual : quality of a given cut (of depth d or df) [no units] Qual tot : total quality [no units] Price : price at which the piece is sold [$] Dprice : price variability of the piece according to quality [$] Minprice : price of the worst possible piece [$] P max : motor maximum power [W] d min : minimum depth of cut [mm] N H : closest integer by inferior value to H/d [no unit] df : depth of final cut [mm] M : length of mill [mm]

95 4.3 Mathematical Model: 4.3.1: Objective function The objective of this subsystem is to maximize the profit of one production day. This can be written as directly proportional to the cost of machining the part, efficiency of producing the part, and quality of the part (cost of selling the part). 4.3.2: Constraints Hindered by mostly physical constraints, all parameters and variables are greater or equal to zero. Total time to manufacture one part can be added up by all different items, including part loading time, machining time, and tool changing time. Because tool changing is not necessary for each part, thus the tool changing time is the average tool changing time. Total time for one part must be completed in a timely fashion therefore must be less than 14 minutes per cycle. Machining time for one part could be calculated by the total material removal volume divided by material removal rate. Here the material removal volume is the volume of slot. Machining time must be completed in a timely fashion therefore must be less than 10 minutes per part. Number of pieces per tool could make is obtained by the tool life divided by machining time.

96 The tool life equation could be obtained from literature for cutting titanium with tungsten carbide cutter and after converting into the correct units (sec): The tool life of the cutter must be greater than 1 hour so that production may run smoothly: Distance moved per revolution The tool with also break if try to take off more material than what is possible with the geometry of the cutter. The maximum distance to move per revolution can be calculated by looking at feed and speed recommendations for carbide end mills. There is a functional relationship between the Outer and Inner Radius of a end mill determined experimentally by: Therefore The spindle speed is limited to the machine tool capability but must be within operational limits of carbide cutters given by recommended SFM values (25-300 SFM for cutting Ti). Therefore:

97 The cutting force along x, y, z axis could be calculated using the constructed cutting force model in previous research. The equations were obtained from an oblique cutting model that was formulated based on experiments. And, A possible failure mode of the end mill can be described by the flute breaking off from the main inner shank (tooth root). Therefore a simple beam formula, rectangular, is assumed to be accurate where t is the approximate average thickness of the flute. The maximum stress couldn t exceed the tensile stress of material. Rake angle is not good for cutting when exceeding 15 degree from literature. Helix angle is not good for cutting when exceeding 75 degrees and smaller than 30 degrees from literature.

98 Depth of cut need to be smaller or equal to the depth of slot. Outer Radius need to be smaller or equal to the width of slot and greater than or equal to a 10 mm diameter cutter. Given the rpm ranges above, feed must be within a feasible limit. Another possible mode of failure is the end mill breaking off at the tool holder. Assuming that the tool holder is rigid than this failure can be evaluated by cantilever beam equations. This stress must also be lower than the tensile stress of tungsten carbide. Machining cost could be obtained by total machining time multiply the machining cost per unit time. The machining cost includes labor cost, burden cost and machine cost. The equality constraint is: C m Lm + Bm + Mm = Tm 3600 The machine cost is determined by the power needed during the machining process, which is the material removal rate multiplies specific energy, then times cost of machine

99 power: M m MRR ut 3600 = C η 1000 e The material removal rate of milling process is to multiply the feed, spindle speed, depth and the ball diameter. MRR = 2R o f d The loading cost part could be obtained by total loading time multiplies loading cost per unit time, which includes labor and burden cost. Lm + Bm Cl = Tl 3600 Tooling cost is the most complicated part, including tool changing and regrinding. Each part also divides into labor and burden part. 1 C = T ( L + B ) + T ( L + B ) + D N t c m m g g g c p The variables and parameters must always be such that the system is within the stability lobes, otherwise due to chatter the quality is too poor to be even considered. A rough approximation of the region contained within the stability lobes can be: 12 d * N 0.8 5000 The error must be inferior to a maximum value. δmax Err Err max = 2 π Indeed, with Err max calculated as the error obtained for a maximum deflection δ max (of 10 microns for example) over a sinusoidal path we obtain: δmax Err max = 2 π The deflection δ is a function of F ytot, I and d.

100 δ= F *( M 0.5 d) ytot 3* E* I 3 N and df are functions of d. NH = floor( H / d) and df = H N * d The area moment of inertia of the carbide mill is a function of R o. 4 *(2 ) /64 I = π R o Θi is a function of the spindle speed. Θ i = ( N*2 π / 60)* t+ ( i* π / 2) The error must be inferior to a maximum value. δmax Err Err max = 2 π The quality a given cut (depth d or df) is a function of the error. Qual = 1 ( Err / Err ) max H With δmax Err max = 2 π The error is a function of the feed rate (f) and of the deflection (δ) L 1 Err = δ( x) * dx L Hence changing the variable x to t using x = f * t : 0 L T /4*( ) L/ f f* T/4 T /4 1 1 1 Err = t * f * dt t * f * dt t * f * dt L δ( ) = L δ( ) δ( ) f * T /4 0 0 0 N *2 π Finally changing t to θ using θ = * t : 60 π /2 2 Err = * δ( θ ) * d θ π 0

101 4.3.3: Initial Design Variables and Parameters Variables: E, T p, T m, N p, T, D r, R o, R i, σ, F f, F n, F c, F t, F l, N, d, f, α, Φ Parameters and Values : L (200 mm), H (40 mm), W (100 mm), T l (180 sec), T c (150 sec), σ t (344.8 N/mm^2), H d (7 hrs/day), T g =1800sec, u t = 0.066*10-3 kw/mm 3 *sec, η = 70%, C e =0.08971 $/kw-hr, L m =20 $/hr (average worker), B m =0.06528*4 $/hr (assume 4 power 1kW lights are used), L g =25 $/hr (experienced worker), B g =0.06528*2 $/hr(assume 2 power 1kW lights are used), D c =10 $/tool, σt= 344.8*106 Pa, E=600 GPa, δ max = 10 microns, Minprice=$10, Dprice=$90 and the exponent of (Err/Err max ) was changed from 0.1 to 0.2. Degrees of Freedom = 6 4.3.4: Intermediate Model (For parametric study, L, W, H, and T, Sigma)

102 Sub to: g1: (Eq. 4.28) g2: (Eq. 4.29) g3: (Eq. 4.27) g4: (Eq. 4.25) g5: (Eq. 4.26) g6: (Eq. 4.24) g7: (Eq. 4.14 w/ 4.13) g8: (Eq. 4.15 w/ 4.13) g9: (Eq 4.11 w/4.10, 4.12a, 4.12b) g10: (Eq. 4.9 w/ 4.8, 4.10) g11: (Eq. 4.6 w/ 4.5)