1. ME 482 - Manufacturing Systems Metal Machining Metal Machining

2. ME 482 - Manufacturing Systems Objectives Introduce cutting terminology and principles Review modern machining technologies and new methods (papers) Introduce cutting parameters Develop cutting models Analyze a cutting example Objectives Introduce cutting terminology and principles Review modern machining technologies and new methods (papers) Introduce cutting parameters Develop cutting models Analyze a cutting example

3. ME 482 - Manufacturing Systems Machining types Turning Drilling Milling Shaping Planing Broaching Machining types Turning Drilling Milling Shaping Planing Broaching

4. ME 482 - Manufacturing Systems Machining tools Single point Multiple point Machining tools Single point Multiple point

5. ME 482 - Manufacturing Systems Machining tool materials Most modern cutting tool materials are a matrix of materials designed to be very hard. These materials will be covered in the next chapter. Machining tool materials Most modern cutting tool materials are a matrix of materials designed to be very hard. These materials will be covered in the next chapter.

6. ME 482 - Manufacturing Systems Machining surface finish

7. ME 482 - Manufacturing Systems Machining terminology Speed – surface cutting speed (v) Feed – advance of tool through the part (f) Depth of cut – depth of tool into part (d) Rake face – tool’s leading edge Rake angle – slant angle of tool’s leading edge (  ) Flank – following edge of cutting tool Relief angle – angle of tool’s following edge above part surface Machining terminology Speed – surface cutting speed (v) Feed – advance of tool through the part (f) Depth of cut – depth of tool into part (d) Rake face – tool’s leading edge Rake angle – slant angle of tool’s leading edge (  ) Flank – following edge of cutting tool Relief angle – angle of tool’s following edge above part surface

8. ME 482 - Manufacturing Systems Machining terminology (cont.) Chip thickness – thickness of machined chip (t c ) Depth of cut = t o Shear plane length – measured along shear plane chip (l s ) Chip width (not shown) – width of machined chip (w ) Shear angle – angle of shearing surface measured from tool direction (  ) Machining terminology (cont.) Chip thickness – thickness of machined chip (t c ) Depth of cut = t o Shear plane length – measured along shear plane chip (l s ) Chip width (not shown) – width of machined chip (w ) Shear angle – angle of shearing surface measured from tool direction (  ) lsls Orthogonal model

9. ME 482 - Manufacturing Systems Cutting conditions Note: - Primary cutting due to speed - Lateral motion of tool is feed - Tool penetration is depth of cut The three together form the material removal rate (MRR): MRR = v f d with units of (in/min)(in/rev)(in) = in 3 /min/rev (or vol/min-rev) Types of cuts: Roughing: feeds of 0.015 – 0.05 in/revdepths of 0.1 – 0.75 in Finishing: feeds of 0.005 – 0.015 in/revdepths of 0.03 – 0.075 in Cutting conditions Note: - Primary cutting due to speed - Lateral motion of tool is feed - Tool penetration is depth of cut The three together form the material removal rate (MRR): MRR = v f d with units of (in/min)(in/rev)(in) = in 3 /min/rev (or vol/min-rev) Types of cuts: Roughing: feeds of 0.015 – 0.05 in/revdepths of 0.1 – 0.75 in Finishing: feeds of 0.005 – 0.015 in/revdepths of 0.03 – 0.075 in

10. ME 482 - Manufacturing Systems Cutting geometry Chip thickness ratio = r = t o / t c From the shear plane geometry: r = l s sin  /[l s cos(  -  )] which can be arranged to get tan  = r cos  /[1 – r sin  ] Chip thickness ratio = r = t o / t c From the shear plane geometry: r = l s sin  /[l s cos(  -  )] which can be arranged to get tan  = r cos  /[1 – r sin  ] Obviously, the assumed failure mode is shearing of the work along the shear plane.

11. ME 482 - Manufacturing Systems Cutting geometry Note from the triangles in (c) that the shear strain (  ) can be estimated as  = AC/BD = (DC + AD)/BD = tan(  -  ) + cot  Note from the triangles in (c) that the shear strain (  ) can be estimated as  = AC/BD = (DC + AD)/BD = tan(  -  ) + cot  Thus, if know r and  can determine , and given  and , can determine .

12. ME 482 - Manufacturing Systems Cutting forces Since R = R’ = R’’, we can get the force balance equations: F = F c sin  + F t cos  F = friction force; N = normal to chip force N = F c cos  - F t sin  F c = cutting force; F t = thrust force F s = F c cos  - F t sin  F s = shear force; F n = normal to shear plane force F n = F c sin  + F t cos  Since R = R’ = R’’, we can get the force balance equations: F = F c sin  + F t cos  F = friction force; N = normal to chip force N = F c cos  - F t sin  F c = cutting force; F t = thrust force F s = F c cos  - F t sin  F s = shear force; F n = normal to shear plane force F n = F c sin  + F t cos  Friction angle =  tan  =  = F/N Shear plane stress:  = F s /A s where A s = t o w/sin  Friction angle =  tan  =  = F/N Shear plane stress:  = F s /A s where A s = t o w/sin  Forces are presented as function of F c and F t because these can be measured.

13. ME 482 - Manufacturing Systems Cutting forces given shear strength Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*: F s = S A s F c = F s cos (  cos (  F t = F s sin (  cos (  * The other forces can be determined from the equations on the previous slide. Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*: F s = S A s F c = F s cos (  cos (  F t = F s sin (  cos (  * The other forces can be determined from the equations on the previous slide.

14. ME 482 - Manufacturing Systems Merchant equations Combining the equations from the previous slides:  = (F c cos  - F t sin  t o w/sin  Merchant eqn The most likely shear angle will minimize the energy. Applying d  /d  = 0 gives:  = 45° +  Merchant reln What does the Merchant relation indicate? Combining the equations from the previous slides:  = (F c cos  - F t sin  t o w/sin  Merchant eqn The most likely shear angle will minimize the energy. Applying d  /d  = 0 gives:  = 45° +  Merchant reln What does the Merchant relation indicate? - increase in friction angle decreases shear angle - increase in rake angle increases shear angle - increase in friction angle decreases shear angle - increase in rake angle increases shear angle If we increase the shear angle, we decrease the tool force and power requirements! The Merchant reln is a function of  and . Where did these variables come from? Answer - Although the Merchant eqn is not shown as a direct function of  and , these enter from the equations for F c and F t from the previous slide!

15. ME 482 - Manufacturing Systems Cutting models The orthogonal model for turning approximates the complex shearing process: Cutting models The orthogonal model for turning approximates the complex shearing process: t o = feed (f) w = depth of cut (d) t o = feed (f) w = depth of cut (d)

16. ME 482 - Manufacturing Systems Cutting power Power is force times speed: P = F c v(ft-lb/min) The cutting horsepower is hp c = F c v/33,000(hp) The unit horsepower is hp u = hp c /MRRunits? Due to efficiency losses (E about 90%), the gross hp is hp g = hp c /E Power is force times speed: P = F c v(ft-lb/min) The cutting horsepower is hp c = F c v/33,000(hp) The unit horsepower is hp u = hp c /MRRunits? Due to efficiency losses (E about 90%), the gross hp is hp g = hp c /E

17. ME 482 - Manufacturing Systems Cutting energy Specific energy is U = F c v/(v t o w) = F c /(t o w) (in-lb/in 3 ) The table shown contains power and specific energy ratings for several work materials at a chip thickness of 0.01 in. For other chip thicknesses, apply the figure to get a correction factor multiply U by correction factor for thickness different than 0.01”). Specific energy is U = F c v/(v t o w) = F c /(t o w) (in-lb/in 3 ) The table shown contains power and specific energy ratings for several work materials at a chip thickness of 0.01 in. For other chip thicknesses, apply the figure to get a correction factor multiply U by correction factor for thickness different than 0.01”).

18. ME 482 - Manufacturing Systems Machining example In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear strain): Determine r = 0.02/0.045 = 0.444 Determine shear plane angle from tan  = r cos  /[1 – r sin  ] tan  = 0.444 cos  /[1 – 0.444 sin  ] =>  = 25.4 ° Now calculate shear strain from  = tan(  -  ) + cot   = tan(25.4 - 10) + cot 25.4 = 2.386 in/inanswer! In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear strain): Determine r = 0.02/0.045 = 0.444 Determine shear plane angle from tan  = r cos  /[1 – r sin  ] tan  = 0.444 cos  /[1 – 0.444 sin  ] =>  = 25.4 ° Now calculate shear strain from  = tan(  -  ) + cot   = tan(25.4 - 10) + cot 25.4 = 2.386 in/inanswer!

19. ME 482 - Manufacturing Systems Machining example (cont.) In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear stress): Determine shear force from F s = F c cos  - F t sin  F s = 350 cos 25.4 - 285 sin 25.4 = 194 lb Determine shear plane area from A s = t o w/sin  A s = (0.02) (0.125)/sin  = 0.00583 in 2 The shear stress is  = 194/0.00583 = 33,276 lb/in 2 answer! In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear stress): Determine shear force from F s = F c cos  - F t sin  F s = 350 cos 25.4 - 285 sin 25.4 = 194 lb Determine shear plane area from A s = t o w/sin  A s = (0.02) (0.125)/sin  = 0.00583 in 2 The shear stress is  = 194/0.00583 = 33,276 lb/in 2 answer!

20. ME 482 - Manufacturing Systems Machining example (cont.) In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (cutting horsepower): Determine cutting hp from hp c = F c v/33,000 hp c = (350) (200)/33,000 = 2.12 hpanswer! In orthogonal machining the tool has rake angle 10 °, chip thickness before cut is t o = 0.02 in, and chip thickness after cut is t c = 0.045 in. The cutting and thrust forces are measured at F c = 350 lb and F t = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (cutting horsepower): Determine cutting hp from hp c = F c v/33,000 hp c = (350) (200)/33,000 = 2.12 hpanswer!

21. ME 482 - Manufacturing Systems Cutting temperatures In machining 98% of the cutting energy is converted into heat. This energy flows into the work part, chip, and tool. Cook determined an experimental equation for predicting the temperature rise at the tool-chip interface during machining:  T = 0.4 U (v t o /K) 0.333 /(  c ) where  T = mean temperature rise (°F) U = specific energy (in-lb/in 3 ) v = cutting speed (in/s) t o = chip thickness before cut (in)  c  = volumetric specific heat of the work material (in-lb/(in 3 -°F)) K = thermal diffusivity of the work material (in 2 /s) Note - To get total temperature at tool-chip interface, must add in ambient temperature! In machining 98% of the cutting energy is converted into heat. This energy flows into the work part, chip, and tool. Cook determined an experimental equation for predicting the temperature rise at the tool-chip interface during machining:  T = 0.4 U (v t o /K) 0.333 /(  c ) where  T = mean temperature rise (°F) U = specific energy (in-lb/in 3 ) v = cutting speed (in/s) t o = chip thickness before cut (in)  c  = volumetric specific heat of the work material (in-lb/(in 3 -°F)) K = thermal diffusivity of the work material (in 2 /s) Note - To get total temperature at tool-chip interface, must add in ambient temperature! Example in text calculates  T = 936° total tool temperature, given v = 200 ft/min,  c = 120 in- lb/(in 3 - °F) and K = 0.125 in 2 /s

22. ME 482 - Manufacturing Systems Cutters Toroid

23. ME 482 - Manufacturing Systems Cutters

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