Theory of metal cutting

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this is 2nd presentation of manufacturing processes in this presentation we discuss in detail about the theory of metal cutting, machiening processes,cutters etc

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  • r = t1/t2
  • Theory of metal cutting

    1. 1. 1. Overview of Machining Technology 2. Theory of Chip Formation in Metal Machining 3. Force Relationships and the Merchant Equation 4. Power and Energy Relationships in Machining 5. Cutting Temperature
    2. 2. A family of shaping operations, the common feature of which is removal of material from a starting workpart so the remaining part has the desired geometry  Machining – material removal by a sharp cutting tool, e.g., turning, milling, drilling  Abrasive processes – material removal by hard, abrasive particles, e.g., grinding  Nontraditional processes - various energy forms other than sharp cutting tool to remove material
    3. 3. Cutting action involves shear deformation of work material to form a chip  As chip is removed, new surface is exposed (a) A cross sectional view of the machining process, (b) tool with‑ negative rake angle; compare with positive rake angle in (a). Machining
    4. 4.  Variety of work materials can be machined ◦ Most frequently used to cut metals  Variety of part shapes and special geometric features possible, such as: ◦ Screw threads ◦ Accurate round holes ◦ Very straight edges and surfaces  Good dimensional accuracy and surface finish
    5. 5.  Wasteful of material ◦ Chips generated in machining are wasted material, at least in the unit operation  Time consuming ◦ A machining operation generally takes more time to shape a given part than alternative shaping processes, such as casting, powder metallurgy, or forming
    6. 6.  Generally performed after other manufacturing processes, such as casting, forging, and bar drawing ◦ Other processes create the general shape of the starting workpart ◦ Machining provides the final shape, dimensions, finish, and special geometric details that other processes cannot create
    7. 7.  Speed is the relative movement between tool and w/p, which produces a cut  Feed is the relative movement between tool and w/p, which spreads the cut
    8. 8.  Most Important Machining Operations: ◦ Turning ◦ Drilling ◦ Milling  Other Machining Operations: ◦ Shaping and Planing ◦ Broaching ◦ Sawing
    9. 9. Single point cutting tool removes material from a rotating workpiece to form a cylindrical shape Three most common machining processes: (a) turning, Turning
    10. 10. Used to create a round hole, usually by means of a rotating tool (drill bit) with two cutting edges Drilling
    11. 11. Rotating multiple-cutting-edge tool is moved across work to cut a plane or straight surface  Two forms: peripheral (side) milling and face (end) milling (c) peripheral milling, and (d) face milling. Milling
    12. 12. 1. Single-Point Tools ◦ One dominant cutting edge ◦ Point is usually rounded to form a nose radius ◦ Turning uses single point tools 2. Multiple Cutting Edge Tools ◦ More than one cutting edge ◦ Motion relative to work achieved by rotating ◦ Drilling and milling use rotating multiple cutting edge tools
    13. 13. (a) A single point tool showing rake face, flank, and tool point; and (b)‑ a helical milling cutter, representative of tools with multiple cutting edges. Cutting Tools
    14. 14.  Three dimensions of a machining process: ◦ Cutting speed v – primary motion ◦ Feed f – secondary motion ◦ Depth of cut d – penetration of tool below original work surface  For certain operations, material removal rate can be computed as RMR = v f d where v = cutting speed; f = feed; d = depth of cut
    15. 15. Cutting Conditions for Turning Speed, feed, and depth of cut in turning.
    16. 16. In production, several roughing cuts are usually taken on the part, followed by one or two finishing cuts  Roughing - removes large amounts of material from starting workpart ◦ Creates shape close to desired geometry, but leaves some material for finish cutting ◦ High feeds and depths, low speeds  Finishing - completes part geometry ◦ Final dimensions, tolerances, and finish ◦ Low feeds and depths, high cutting speeds
    17. 17. A power‑driven machine that performs a machining operation, including grinding  Functions in machining: ◦ Holds workpart ◦ Positions tool relative to work ◦ Provides power at speed, feed, and depth that have been set  The term is also applied to machines that perform metal forming operations
    18. 18. where r = chip thickness ratio; to = thickness of the chip prior to chip formation; and tc = chip thickness after separation  Chip thickness after cut is always greater than before, so chip ratio always less than 1.0 c o t t r =
    19. 19. to tc
    20. 20.  Based on the geometric parameters of the orthogonal model, the shear plane angle φ can be determined as: where r = chip ratio, and α = rake angle α α φ sin cos tan r r − = 1
    21. 21. Figure 21.7 Shear strain during chip formation: (a) chip formation depicted as a series of parallel plates sliding relative to each other, (b) one of the plates isolated to show shear strain, and (c) shear strain triangle used to derive strain equation. Shear Strain in Chip Formation
    22. 22. Shear strain in machining can be computed from the following equation, based on the preceding parallel plate model: γ = tan(φ - α) + cot φ where γ = shear strain, φ = shear plane angle, and α = rake angle of cutting tool
    23. 23.  Friction force F and Normal force to friction N  Shear force Fs and Normal force to shear Fn Forces in metal cutting: (a) forces acting on the chip in orthogonal cutting Forces Acting on Chip
    24. 24.  Vector addition of F and N = resultant R  Vector addition of Fs and Fn = resultant R'  Forces acting on the chip must be in balance: ◦ R' must be equal in magnitude to R ◦ R’ must be opposite in direction to R ◦ R’ must be collinear with R
    25. 25.  F, N, Fs , and Fn cannot be directly measured  Forces acting on the tool that can be measured: ◦ Cutting force Fc and Thrust force Ft ©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e Figure 21.10 Forces in metal cutting: (b) forces acting on the tool that can be measured Cutting Force and Thrust Force
    26. 26. Coefficient of friction between tool and chip: Friction angle related to coefficient of friction as follows: N F =µ βµ tan=
    27. 27. F F N Fn Fs R
    28. 28. F F N R
    29. 29. F F N R Ft Fc
    30. 30. F F N R Ft Fc F = Fc sin α + Ft cos α N = Fc cos α - Ft sin α
    31. 31. F F N Fn Fs R
    32. 32. F Fn Fs R
    33. 33. F Fn Fs R Fc Ft
    34. 34. F Fn Fs R Fc Ft Fs = Fc cos φ - Ft sin φ Fn = Fc sin φ + Ft cos φ
    35. 35.  Thus equations can be derived to relate the forces that cannot be measured to the forces that can be measured: F = Fc sinα + Ft cosα N = Fc cosα ‑ Ft sinα Fs = Fc cosφ ‑ Ft sinφ Fn = Fc sinφ + Ft cosφ  Based on these calculated force, shear stress and coefficient of friction can be determined
    36. 36. Significance of Cutting forces In the set of following force balance equations:- F = Fc sin α + Ft cos α F = friction force; N = normal to chip force N = Fc cos α - Ft sin α Fc = cutting force; Ft = thrust force Fs = Fc cos φ - Ft sin φ Fs = shear force; Fn = normal to shear plane force Fn = Fc sin φ + Ft cos φ Friction angle = β tan β = µ = F/N Shear plane stress: τ = Fs/As where As = to w/sin φ Forces are presented as function ofForces are presented as function of FFcc and Fand Ftt because these can bebecause these can be measured.measured. Forces are presented as function ofForces are presented as function of FFcc and Fand Ftt because these can bebecause these can be measured.measured.
    37. 37. Shear stress acting along the shear plane: φsin wt A o s = where As = area of the shear plane Shear stress = shear strength of work material during cutting s s A F S =
    38. 38. Cutting forces given shear strength Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*: Fs = S As Fc = Fs cos ( β − α)/[cos ( φ + β − α)] Ft = Fs sin ( β − α)/[cos ( φ + β − α)] * The other forces can be determined from the equations on the previous slide.
    39. 39. Machining example In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 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 φ = r cos α /[1 – r sin α] tan φ = 0.444 cos 10 /[1 – 0.444 sin 10] => φ = 25.4° Now calculate shear strain from γ = tan(φ - α) + cot φ γ = tan(25.4 - 10) + cot 25.4 = 2.386 in/in answer!
    40. 40. Machining example (cont.) In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 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 Fs = Fc cos φ - Ft sin φ Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb Determine shear plane area from As = to w/sin φ As = (0.02) (0.125)/sin 25.4= 0.00583 in2 The shear stress is
    41. 41. Machining example (cont.) In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 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 hpc = Fc v/33,000 hpc = (350) (200)/33,000 = 2.12 hp answer!
    42. 42.  Shear Plane Angle Ф = tan-1 [(r cos α )/(1 – r sin α)]  Shear Strain γ = tan(φ - α) + cot φ  Forces in Cutting: F = Fc sinα + Ft cosα N = Fc cosα ‑ Ft sinα Fs = Fc cosφ ‑ Ft sinφ Fn = Fc sinφ + Ft cosφ  Forces Fc and Ft in terms of Fs: Fc = Fs cos ( β − α)/[cos ( φ + β − α)] Ft = Fssin ( β − α)/[cos ( φ + β − α)]  Merchant Relation: φ = 45 + α/2−β/2  Shear Stress: τ = Fs/As where As = to w/sin φ  Cutting Power: P = V Fc / 33,000 hp (V in ft /s and Fc in lb) P = V Fc / 1000 kW (V in m /s and Fc in N) PG = Pc / E

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