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Fundamentals of Metal cutting and Machining Processes THEORY OF METAL MACHINING Akhtar Husain Ref: Kalpakjian & Groover.

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Presentation on theme: "Fundamentals of Metal cutting and Machining Processes THEORY OF METAL MACHINING Akhtar Husain Ref: Kalpakjian & Groover."— Presentation transcript:

1 Fundamentals of Metal cutting and Machining Processes THEORY OF METAL MACHINING Akhtar Husain Ref: Kalpakjian & Groover

2 THEORY OF METAL MACHINING
Overview of Machining Technology Theory of Chip Formation in Metal Machining Force Relationships and the Merchant Equation Power and Energy Relationships in Machining Cutting Temperature

3 Material Removal Processes
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

4 Machining 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).

5 Why Machining is Important
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

6 Disadvantages with Machining
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

7 Machining in Manufacturing Sequence
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

8 Speed and Feed 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

9 Machining Operations Most Important Machining Operations:
Turning Drilling Milling Other Machining Operations: Shaping and Planing Broaching Sawing

10 Turning Single point cutting tool removes material from a rotating workpiece to form a cylindrical shape Three most common machining processes: (a) turning,

11 Drilling Used to create a round hole, usually by means of a rotating tool (drill bit) with two cutting edges

12 Milling 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.

13 Cutting Tool Classification
Single-Point Tools One dominant cutting edge Point is usually rounded to form a nose radius Turning uses single point tools 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

14 Cutting Tools (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.

15 Cutting Conditions in Machining
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

16 Cutting Conditions for Turning
Speed, feed, and depth of cut in turning.

17 Roughing vs. Finishing 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

18 Machine Tools 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

19 Chip Thickness Ratio 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

20 Determining Shear Plane Angle
tc to

21 Determining Shear Plane Angle
Based on the geometric parameters of the orthogonal model, the shear plane angle  can be determined as: where r = chip ratio, and  = rake angle r = t1/t2

22 Shear Strain in Chip Formation
Figure 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.

23 Shear Strain Shear strain in machining can be computed from the following equation, based on the preceding parallel plate model:  = tan( - ) + cot  where  = shear strain,  = shear plane angle, and  = rake angle of cutting tool

24 Forces Acting on Chip 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

25 Resultant Forces 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

26 Cutting Force and Thrust Force
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 Figure Forces in metal cutting: (b) forces acting on the tool that can be measured ©2007 John Wiley & Sons, Inc. M P Groover, Fundamentals of Modern Manufacturing 3/e

27 Coefficient of Friction
Coefficient of friction between tool and chip: Friction angle related to coefficient of friction as follows:

28 Fs F R Fn F N

29 F R F N

30 F R Ft F Fc N

31 F F = Fc sin a + Ft cos a R Ft F Fc N N = Fc cos a - Ft sin a

32 Fs F R Fn F N

33 Fs F Fn R

34 Fs F Fc Fn R Ft

35 Fs = Fc cos f - Ft sin f Fs F Fc Fn R Ft Fn = Fc sin f + Ft cos f

36 Forces in Metal Cutting
Thus equations can be derived to relate the forces that cannot be measured to the forces that can be measured: F = Fc sin + Ft cos N = Fc cos ‑ Ft sin Fs = Fc cos ‑ Ft sin Fn = Fc sin + Ft cos Based on these calculated force, shear stress and coefficient of friction can be determined

37 Significance of Cutting forces
In the set of following force balance equations:- F = Fc sin a + Ft cos a F = friction force; N = normal to chip force N = Fc cos a - Ft sin a Fc = cutting force; Ft = thrust force Fs = Fc cos f - Ft sin f Fs = shear force; Fn = normal to shear plane force Fn = Fc sin f + Ft cos f Forces are presented as function of Fc and Ft because these can be measured. Friction angle = b tan b = m = F/N Shear plane stress: t = Fs/As where As = to w/sin f

38 Shear Stress Shear stress acting along the shear plane:
where As = area of the shear plane Shear stress = shear strength of work material during cutting

39 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 ( b - a)/[cos ( f + b - a)] Ft = Fs sin ( b - a)/[cos ( f + b - a)] * The other forces can be determined from the equations on the previous slide.

40 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 = 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 f = r cos a /[1 – r sin a] tan f = cos 10 /[1 – sin 10] => f = 25.4° Now calculate shear strain from g = tan(f - a) + cot f g = tan( ) + cot 25.4 = in/in answer!

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 = 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 f - Ft sin f Fs = 350 cos sin 25.4 = 194 lb Determine shear plane area from As = to w/sin f As = (0.02) (0.125)/sin 25.4= in2 The shear stress is t = 194/ = 33,276 lb/in2 answer!

42 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 = 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!

43 Formula Sheet Forces Fc and Ft in terms of Fs:
Shear Plane Angle Ф = tan-1 [(r cos α )/(1 – r sin α)] Shear Strain  = tan( - ) + cot  Forces in Cutting: F = Fc sin + Ft cos N = Fc cos ‑ Ft sin Fs = Fc cos ‑ Ft sin Fn = Fc sin + Ft cos Forces Fc and Ft in terms of Fs: Fc = Fs cos ( b - a)/[cos ( f + b - a)] Ft = Fs sin ( b - a)/[cos ( f + b - a)] Merchant Relation: f = 45 + a/2-b/2 Shear Stress: t = Fs/As where As = to w/sin f 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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