1. ME 482 - Manufacturing Systems Machining Operations by Ed Red Machining Operations by Ed Red

2. ME 482 - Manufacturing Systems Objectives Introduce machining operations terminology Introduce machining efficiency measures Reconsider cutting parameters as they apply to efficiency Review a machining efficiency example Consider modern machine operations (papers) Objectives Introduce machining operations terminology Introduce machining efficiency measures Reconsider cutting parameters as they apply to efficiency Review a machining efficiency example Consider modern machine operations (papers)

3. ME 482 - Manufacturing Systems Machining terms Chatter – interrupted cutting usually at some frequency Down milling – cutting speed in same direction as part feed Up milling – cutting speed in opposite direction as part feed Peripheral milling – tool parallel to work Face milling – tool perpendicular to work Ideal roughness – geometrically determined roughness Machinability – machining success determined by tool life, surface finish Optimal machining – parameter choices that increase machining throughput or reduce operational costs Machining terms Chatter – interrupted cutting usually at some frequency Down milling – cutting speed in same direction as part feed Up milling – cutting speed in opposite direction as part feed Peripheral milling – tool parallel to work Face milling – tool perpendicular to work Ideal roughness – geometrically determined roughness Machinability – machining success determined by tool life, surface finish Optimal machining – parameter choices that increase machining throughput or reduce operational costs

4. ME 482 - Manufacturing Systems Machining operations on lathe (other than normal turning) Machining operations on lathe (other than normal turning) Chamfer Taper Contour Form Facing Cutoff Threading Boring Drilling Knurling

5. ME 482 - Manufacturing Systems Two types of milling operations Peripheral Face

6. ME 482 - Manufacturing Systems Face milling operations FacingPartial facing End milling Profiling Pocketing Surface contouring

7. ME 482 - Manufacturing Systems Face milling movements Peripheral milling cutting positions Face milling cutting positions Full face cut Offset face cut

8. ME 482 - Manufacturing Systems Milling cutter time analysis Spindle rpm related to cutter diameter and speed: N (rpm) = v/(  D) Feedrate in in/min: f r = N n t f where f = feed per tooth n t = number of teeth MRR is MRR =w d f r Spindle rpm related to cutter diameter and speed: N (rpm) = v/(  D) Feedrate in in/min: f r = N n t f where f = feed per tooth n t = number of teeth MRR is MRR =w d f r

9. ME 482 - Manufacturing Systems Milling time analysis Slab milling: Approach distance, A : A = d (D-d) Time to mill workpiece, T m : T m = (L + A)/f r Face milling: Allow for over-travel O where A = O: Full face A = O = D/2 Partial face A = O = w (D – w) Machining time: T m = (L + 2A)/f r Slab milling: Approach distance, A : A = d (D-d) Time to mill workpiece, T m : T m = (L + A)/f r Face milling: Allow for over-travel O where A = O: Full face A = O = D/2 Partial face A = O = w (D – w) Machining time: T m = (L + 2A)/f r

10. ME 482 - Manufacturing Systems Milling time analysis - example Problem statement: A face milling operation is performed to finish the top surface of a steel rectangular workpiece 12 in. long by 2 in. wide. The milling cutter has 4 teeth (cemented carbide inserts) and is 3 in. in diameter. Cutting conditions are 500 fpm, f = 0.01 in./tooth, and d = 0.150 in. Determine the time to make one pass across the surface and the metal removal rate during the cut. Problem statement: A face milling operation is performed to finish the top surface of a steel rectangular workpiece 12 in. long by 2 in. wide. The milling cutter has 4 teeth (cemented carbide inserts) and is 3 in. in diameter. Cutting conditions are 500 fpm, f = 0.01 in./tooth, and d = 0.150 in. Determine the time to make one pass across the surface and the metal removal rate during the cut.

11. ME 482 - Manufacturing Systems Milling time analysis - example Solution?Numbers? Full face A = O = D/2 Machining timeT m = (L + 2A)/f r Metal removal rateMRR = w d f r Feedrate in in/minf r = N n t f N (rpm) = v/(  D) Full face A = O = D/2 Machining timeT m = (L + 2A)/f r Metal removal rateMRR = w d f r Feedrate in in/minf r = N n t f N (rpm) = v/(  D)

12. ME 482 - Manufacturing Systems Tolerance by process

13. ME 482 - Manufacturing Systems Surface finish by process

14. ME 482 - Manufacturing Systems Surface finish by geometry Ideal roughness, R i = f 2 /(32 NR) where NR = tool nose radius Ideal roughness, R i = f 2 /(32 NR) where NR = tool nose radius Actual roughness, R a = r ai R i (about 2 x R i ) because of edge effects, chip interactions, surface tearing, etc. Actual roughness, R a = r ai R i (about 2 x R i ) because of edge effects, chip interactions, surface tearing, etc.

15. ME 482 - Manufacturing Systems Machinability is a measure of machining success or ease of machining. Suitable criteria: tool life or tool speed level of forces surface finish ease of chip disposal Machinability is a measure of machining success or ease of machining. Suitable criteria: tool life or tool speed level of forces surface finish ease of chip disposal Machinability What is a free machining steel? http://www.sandmeyersteel.com/303.html#1

16. ME 482 - Manufacturing Systems Problem statement: A series of tool life tests is conducted on two work materials under identical cutting conditions, varying only speed in the test procedure. The first material, defined as the base material, yields the Taylor tool life equation v T 0.28 = 1050 and the other material (test material) yields the Taylor equation v T 0.27 = 1320 Determine the machinability rating of the test material using the cutting speed that provides a 60 min. tool life as the basis of comparison. This speed is denoted by v 60. Problem statement: A series of tool life tests is conducted on two work materials under identical cutting conditions, varying only speed in the test procedure. The first material, defined as the base material, yields the Taylor tool life equation v T 0.28 = 1050 and the other material (test material) yields the Taylor equation v T 0.27 = 1320 Determine the machinability rating of the test material using the cutting speed that provides a 60 min. tool life as the basis of comparison. This speed is denoted by v 60. Machinability - example

17. ME 482 - Manufacturing Systems Solution: The base material has a machinability rating = 1.0. Its v 60 value can be determined from the Taylor tool life equation as follows: v 60 = 1050/60 0.28 = 334 ft/min The cutting speed at a 60 min. tool life for the test material is determined similarly: v 60 = 1320/60 0.27 = 437 ft/min Accordingly, the machinability rating can be calculated as MR (for the test material) = 437/374 = 1.31 (or 131%) Solution: The base material has a machinability rating = 1.0. Its v 60 value can be determined from the Taylor tool life equation as follows: v 60 = 1050/60 0.28 = 334 ft/min The cutting speed at a 60 min. tool life for the test material is determined similarly: v 60 = 1320/60 0.27 = 437 ft/min Accordingly, the machinability rating can be calculated as MR (for the test material) = 437/374 = 1.31 (or 131%) Machinability - example

18. ME 482 - Manufacturing Systems Optimized machining Cutting speed can be chosen to maximize the production rate or minimize the cost per part (or unit) produced. This is referred to as optimized machining because more than one production variable contributes to the production rate and costs. Variables: T h - part handling timeC o (C g ) – operator (grinder’s) cost rate/min T m – machining timeC h – cost of part handling time T t – tool change timeC m – cost of machining time n p – number of parts cut by C tc – cost of tool change time tool during tool life T c – cycle time per part C t – cost per cutting edge T – tool life C tp = C t /n p - tool cost per part Cutting speed can be chosen to maximize the production rate or minimize the cost per part (or unit) produced. This is referred to as optimized machining because more than one production variable contributes to the production rate and costs. Variables: T h - part handling timeC o (C g ) – operator (grinder’s) cost rate/min T m – machining timeC h – cost of part handling time T t – tool change timeC m – cost of machining time n p – number of parts cut by C tc – cost of tool change time tool during tool life T c – cycle time per part C t – cost per cutting edge T – tool life C tp = C t /n p - tool cost per part

19. ME 482 - Manufacturing Systems Maximum production rate - turning Total time per part produced (cycle time): T c = T h + T m + T t /n p where T t /n p is the tool change time per part. Consider a turning operation. The machining time is given by T m =  D L/(v f) The number of parts cut per tool is given by n p = T/T m = f C (1/n) /(  D L v (1/n -1) ) Total time per part produced (cycle time): T c = T h + T m + T t /n p where T t /n p is the tool change time per part. Consider a turning operation. The machining time is given by T m =  D L/(v f) The number of parts cut per tool is given by n p = T/T m = f C (1/n) /(  D L v (1/n -1) )

20. ME 482 - Manufacturing Systems Maximum production rate - turning Substituting, we get the total cutting time T c = T h +  D L/(v f) + T t [  D L v (1/n -1) /( f C (1/n) )] Minimizing cycle time (dT c /dv = 0 ) gives optimum (max) cutting speed and tool life: v max = C/[(1 - n) T t /n] n T max = (1 - n) T t /n Substituting, we get the total cutting time T c = T h +  D L/(v f) + T t [  D L v (1/n -1) /( f C (1/n) )] Minimizing cycle time (dT c /dv = 0 ) gives optimum (max) cutting speed and tool life: v max = C/[(1 - n) T t /n] n T max = (1 - n) T t /n

21. ME 482 - Manufacturing Systems Minimum cost per unit - turning Cost of part handling time: C h = C o T h Cost of machining time: C m = C o T m Cost of tool change time: C tc = C o T t /n p Cost of part handling time: C h = C o T h Cost of machining time: C m = C o T m Cost of tool change time: C tc = C o T t /n p

22. ME 482 - Manufacturing Systems Minimum cost per unit - turning Tool cost per part: C tp = C t /n p Tooling cost per edge: Disposable inserts C t = P t /n e n e = num of edges/insert P t = original cost of tool Single point grindable C t = P t /n g + T g C g “includes purchase price” n g = Num tool lives/tool T g = time to grind tool Tool cost per part: C tp = C t /n p Tooling cost per edge: Disposable inserts C t = P t /n e n e = num of edges/insert P t = original cost of tool Single point grindable C t = P t /n g + T g C g “includes purchase price” n g = Num tool lives/tool T g = time to grind tool

23. ME 482 - Manufacturing Systems Minimum cost per unit - turning Total cost per part: C c = C o T h + C o T m + C o T t /n p + C t /n p Substituting for T m and n p : C c = C o T h + C o  DL/fv + (C o T t + C t )  DLv (1/n -1) /( f C (1/n) ) Minimizing cost per part (dC c /dv = 0) gives cutting speed and tool life to minimize machining costs per part: v min = C{n C o /[(1 – n)(C t + C o T t )]} n T min = (1 – n) (C t + C o T t )/(n C o ) Total cost per part: C c = C o T h + C o T m + C o T t /n p + C t /n p Substituting for T m and n p : C c = C o T h + C o  DL/fv + (C o T t + C t )  DLv (1/n -1) /( f C (1/n) ) Minimizing cost per part (dC c /dv = 0) gives cutting speed and tool life to minimize machining costs per part: v min = C{n C o /[(1 – n)(C t + C o T t )]} n T min = (1 – n) (C t + C o T t )/(n C o )

24. ME 482 - Manufacturing Systems Minimum cost per unit - example Problem statement: Suppose a turning operation is to be performed with HSS tooling on mild steel (n = 0.125, C = 200 from text table). The workpart has length = 20.0 in. and diameter = 4.0 in· Feed = 0.010 in./rev. Handling time per piece = 5.0 min and tool change time = 2.0 min. Cost of machine and operator = $30.00/hr, and tooling cost = $3.00 per cutting edge. Find (a) cutting speed for maximum production rate and (b) cutting speed for minimum cost Problem statement: Suppose a turning operation is to be performed with HSS tooling on mild steel (n = 0.125, C = 200 from text table). The workpart has length = 20.0 in. and diameter = 4.0 in· Feed = 0.010 in./rev. Handling time per piece = 5.0 min and tool change time = 2.0 min. Cost of machine and operator = $30.00/hr, and tooling cost = $3.00 per cutting edge. Find (a) cutting speed for maximum production rate and (b) cutting speed for minimum cost

25. ME 482 - Manufacturing Systems Minimum cost per unit - example Solution: Cutting speed for maximum production rate is v max = C/[(1 - n) T t /n] n = 200/[(.875) 2/0.125] 0.125 = 144 ft/min Converting C o from $30/hr to $0.5/min, the cutting speed for minimum cost is given by v min = C{n C o /[(1 – n)(C t + C o T t )]} n = = 200{(0.125)(0.5)/[(0.875)(3.00 + (0.5)(2))]} 0.125 = 121 ft/min Solution: Cutting speed for maximum production rate is v max = C/[(1 - n) T t /n] n = 200/[(.875) 2/0.125] 0.125 = 144 ft/min Converting C o from $30/hr to $0.5/min, the cutting speed for minimum cost is given by v min = C{n C o /[(1 – n)(C t + C o T t )]} n = = 200{(0.125)(0.5)/[(0.875)(3.00 + (0.5)(2))]} 0.125 = 121 ft/min

26. ME 482 - Manufacturing Systems Machining operations What did we learn?