1 / 53

Manufacturing Processes

Manufacturing Processes. Chap. 21 - Fundamentals of Machining Processes. Fundamentals of Cutting. Definition: Machining: the removal of unwanted material from a workpiece in the form or chips. High precision machining can involve tolerances of .0001” or less.

wdebra
Download Presentation

Manufacturing Processes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Manufacturing Processes Chap. 21 - Fundamentals of Machining Processes

  2. Fundamentals of Cutting • Definition: • Machining: the removal of unwanted material from a workpiece in the form or chips. • High precision machining can involve tolerances of .0001” or less. • There are 7 basic chip formation processes: • Shaping • Turning • Milling • Drilling • Sawing • Broaching • Grinding

  3. Many factors influence cutting process • Independent • Tool material, coatings, condition. • Tool shape, finish, sharpness. • Workpiece material, condition, temperature. • Cutting parameters: feed, speed, depth of cut. • Cutting fluids. • Machine tool properties: stiffness & damping. • Workholding & fixturing.

  4. Many factors influence cutting process • Dependent • Type of chip produced. • Force / Energy dissipated in cutting process. • Temperature rise in workpiece, chip, tool. • Tool Wear / Failure. • Final surface finish produced.

  5. Metal Cutting Processes • Complex process due to many variables involved: • different materials behave differently • very large strains occur • sensitivity to tool geometry, material, environment, other. • One of the most common is turning: • workpiece is rotated and cutting tool removes material as it moves to the left after setting a depth of cut. • A chip is then produced which moves up the face of the tool.

  6. Fundamentals of Cutting • Basic cutting operating parameters: SpeedFeed Depth of Cut

  7. Fundamentals of Cutting • Speed (V) • Velocity of the rotating workpiece to stationary tool. (velocity of outermost point of workpiece) • Units: sfpm or ipm • Feed (fr) • Amount of material removed per rev. • Units: in/rev*, ipm (* in turning) • Depth of Cut (d) • Distance plunged into work. Units: in.

  8. Basic Formulas • d = (D1 - D2)/2 = d.o.c. (depth of cut) • V = ( D1 Ns / 12) = surface speed; D1 in inches, Ns in rpm, V in surface ft/min. • MRR ~ 12 Vfrd = Material Removal rate (MRR in3/min); fr = feed rate (in/rev) MRR is also viewed as Volume removed per Unit time = V / CT • Cutting Time CT (in minutes) = (L + Allowance) / fr Ns • L is the length of the cut. A length allowance is added for tool entry and exit. • The allowance formulas vary by process.

  9. How to select parameters? A. Select cutting tool: most critical component. • Select its material & geometry: depends on(material to be cut) B. Determine input parameters: • Tables have been developed to associate the parameters. • They are conservative starting points for parameter selection. • RPM depends on V; • V, f and d.o.c. depend on many parameters: • (process, material, hardness, cutting tool material)

  10. How to select parameters? B. Determine input parameters: (cont.) • Table will show initial values for V, f, d.o.c.(depth of cut) • fr, d.o.c.: usually larger for roughing, smaller for finishing. • V: usually smaller for roughing, larger for finishing. • They are conservative starting points for parameter selection. • Once V is selected from table, solve for Ns via V = (D1 Ns / 12) • Now, estimate MRR via MRR ~ 12 Vfrd

  11. Understanding Chip Formation • Most cutting processes are three-dimensional. • A 2D model is a simplified version of this process (orthogonal model).

  12. Terminology used in Orthogonal Cutting • Shear Plane • Shear angle  • Shear Strain  • Rake side of Tool • Rake angle  • Plate thickness w • Relief side of Tool • Relief Angle  • Uncut Chip thickness t • Chip thickness tc • Depth of Cut to

  13. Understanding Chip Formation • Refer to figure 20.3 and 20.4 • Rake angleand Relief angle are the most critical elements of the tool geometry. • Workpiece moves at velocity V. • Uncut chip thickness is to, cut chips have thickness tc • Chip has velocity Vc. • Shearing takes place along a very narrow region called shear zone or shear plane at a shear angle .

  14. 2D Model: Orthogonal Cutting • Below shear plane, work is undeformed. • Above shear plane, chip is formed and moving up the face of the tool over a length lc • Computing the shear angle: • 1. Find Chip Thickness or Cutting Ratio: • r = to / tc = sin  / (cos  -  ) where (0 < r < 1); • to = depth of cut; tc = chip thickness • tc is always greater that to so r < 1 • (to is usually measured with a micrometer)

  15. 2D Model: Orthogonal Cutting 2. Replace rc in: • Tan  = (rc cos ) / (1 – rc sin )  and solve for shear angle . However, notice that rc = to/tc = sin  / cos ( – ) = Vc / V  and Vs / V = cos  / cos ( – )  The shear strain  = 2 cos  / ( 1 + sin )  depends exclusively on the rake angle.

  16. Shear angle influences force and power requirements, chip thickness and temperature. Therefore, relationship between shear angle, material properties and cutting process variables are important. A relationship between shear angle and workpiece props./ cutting process variables was developed.  = 45o + /2 - /2 where  = friction angle Orthogonal Cutting - Shear Angle 

  17. Coefficient of friction at the interface =>  = tan  Equation  = 45o + /2 - /2 shows: As rake angle decreases / friction at tool-chip interface increases The shear angle decreases and chip becomes thicker. More energy has to be dissipated by the chip because the shear strain required is higher. Temperature also increases. Orthogonal Cutting - Friction Angle 

  18. Recall that: r = to / tc From continuity: Vc tc = V to Since chip thickness (tc) > depth of cut (to), chip velocity Vc has to be lower than the cutting speed V. So Vc = Vr = V sin  / (cos  - ) Orthogonal Cutting - Cutting Speed V

  19. To calculate Vs (velocity at which shearing takes place): V / cos ( - ) = Vs / (cos  ) = Vc / sin  and r = to / tc = Vc / V Orthogonal Cutting - Cutting Speed V

  20. Cutting Forces and Power  

  21. Cutting Forces and Power • The figure shows force diagram in orthogonal cutting. • Cutting Force Fc acts in direction of cutting speed (supplies energy for cutting). • Thrust Force Ft acts normal to cutting velocity (normal to workpiece). • Computing the resultant force of Fc and Ft we obtain R. • R is balanced by an equal and opposite force along the shear plane.

  22. Cutting Forces and Power • R can be decomposed along the shear plane: F = R sin  and N = R cos  • Each is balanced by a normal force Fn and a shear force Fs respectively, where Fs = Fc cos - Ft sin  Fn = Fc sin  - Ft cos  • F / N Ratio is m = F/N = (Ft + Fc tan ) / (Fc - Ft tan ) where typically 0.5 <  < 2

  23. Cutting Forces and Power • Fsis used to compute the shear stress along the shear stress, defined as: ts = Fs / As where As = to w / sin  and t = the uncut chip thickness, w = workpiece width Substituting, ts = ( Fc sin  cos  - Ft sin2/ t w ) (in psi)

  24. Cutting Forces and Power • Thrust force is important: • tool holder, fixturing & machine tool must be stiff enough to withstand thrust force. Ft = R sin ( - ) or Ft = Fc tan ( - ) where  = tan-1= tan-1 ( F / N ) Note: when rake angle is 0o, F = Ft and N = Fc • Ft can be positive or negative: • If  > , Ft is positive (downward). • If  <, Ft is negative (upward). • Conclusion: it is possible to have an upward thrust force if rake angles are high and/or with low friction between tool and chip.

  25. Cutting Forces and Power • Power requirements are important for proper machine tool selection. • Cutting force data is used to: • properly design machine tools to maintain desired tolerances. • determine if the workpiece can withstand cutting forces without distortion.

  26. Cutting Forces and Power In Handout 21-11: • Fc = primary cutting force in the direction of velocity vector • (largest force; accounts for ~99% of power requirements) • Ff = feed force acting in direction of tool feed. • (usually 50% if Fc ; accounts for a small % of power requirements) • Fr = radial/thrust force acting perp. to machined surface. • (usually 50% if Ff ; accounts for very little % of power requirements)

  27. How do cutting forces vary with cutting parameters? • As V is increased, all forces drop some and then stay constant. • As d.o.c. is increased, Fc and Ff increase, Fr stays constant*. • As feed rate is increased, Fc, Ff and Fr increase*. (*) due to change in As • Conclusions: • In general, increasing speed, feed or d.o.c. will increase power reqs. • Doubling speed doubles required HP. • Doubling feed or d.o.c. doubles Fc. • In general, increasing speed does not increase Fc, but increases chip length.

  28. Cutting Forces and Power • Power = Force x Velocity P = Fc. V (ft-lb/min) • Power is mainly dissipated in the shear zone. Shearing Power = Fs. Vs • If w is the width of cut, specific energy for shearing is Us = FsVs / w d V = Fs cos  / w d cos ( - )(psi)

  29. Cutting Forces and Power • Power dissipated by friction: Pfriction = F . Vc (units: ft-lb/min) • Specific energy for friction Uf = F Vc / w d V = F rc / w d (psi) • Total specific energy Ut = Fc V / w d V = Fc / w d and Ut = Us + Uf (psi) Note: when rake angle is 0, F = Ft and N = Fc

  30. Cutting Forces and Power • Horsepower at the spindle: • HP = FcV / 33,000 • Specific Horsepower: • HPs = HP/MRR (hp/in3/min) = Fc / 396,000 frd • Motor Horsepower: • HPm = (HPs x MRR x CF) / E • (CF is a correction factor ~1.25, E is machine efficiency ~80%) • Max. d.o.c. = HPm E / 12 HPs V fr CF • Primary cutting force can be estimated by • Fc ~ (HPs x MRR x 33,000) / V

  31. Cutting Forces and Power • Horsepower conversions 1 Horsepower = 42.43 British Thermal Units per minute 33000 foot-pounds-force per minute 550 foot-pounds-force per second 10.69 kilocalories per minute 0.7457 kilowatts 745.7 watts 1.0139 horsepower (metric)

  32. Different materials will generate different types of chips depending on their properties (ductility, hardness, structure, composition). The type of chip produced significantly influences the surface finish of the workpiece and the overall cutting operation. There are two sides to a chip: a shiny burnished side, which is in contact with the cutting tool. the other side does not come in contact with the tool, it is jagged and rough. Chips produced in Metal Cutting

  33. Types include: Continuous Chips Discontinuous Built-up edge (BUE) Serrated / segmented Types of Chips produced in Metal Cutting

  34. Continuous Chips Ductile materials produce a Continuous chip (i.e. aluminum) Chip rubs against tool longer and generates added heat. Chip is long and curly. usually associated with good surface finish not always desirable: can become tangled in machine tool holder or fixture if too long. Types of Chips produced in Metal Cutting

  35. Discontinuous Chips Brittle materials produce a Discontinuous chip (i.e. gray cast iron) Chip fails not in shear but in brittle fracture. Segments are attached to each other loosely or firmly. Usually form with: materials contains inclusions/impurities very low/high cutting speeds. large docs low rake angles poor cutting fluid Types of Chips produced in Metal Cutting

  36. Build-Up Edge Chips form when there is bonding affinity between tool and work material. affinity is higher in cold work than in annealed metals. layers of workpiece material are deposited on tool. build-up becomes larger and breaks off. part is carried by tool side of chip, other part is deposited on work surface: bad for surface finish. Types of Chips produced in Metal Cutting

  37. Build-Up Edge Chips usually undesirable, a thin stable BUE protects the rake face of the tool. Can reduce by: decreasing d.o.c. increasing rake angle using sharp tools using effective cutting fluid Types of Chips produced in Metal Cutting

  38. Serrated Chips Semi-continuous chips with zones of high / low shear strain. Usually found in materials with low thermal conductivity. Have sawtooth like appearance. Types of Chips produced in Metal Cutting

  39. Surface Finish Integrity • Surface finish affects dimensional accuracy & geometric tolerances. • Surface integrity affects properties: fatigue life, corrosion resistance. • Build-up edge (BUE) has the greatest influence on surface finish. • Ceramic tools have less tendency to form BUE.

  40. Chips develop a natural curl as they leave the work surface. increases with smaller d.o.c. larger rake angle decreases tool-chip friction Chip Curl

  41. Used to avoid long continuous chips to become entangled in machine tool/fixturing. Traditional chip breaker consists of piece of metal attached to rake face of tool which bends chip & breaks it. Cutting tools / inserts now have built-in chip breaker features. They increase effective rake angle and therefore shear angle. Chip Breakers

  42. Tool Life, Wear and Failure • Tool wear is caused by: • high localized stresses • high temperatures • sliding of chip along rake face • sliding of tool along cut surface

  43. Tool Life, Wear and Failure • Tool wear: • Affects quality of machined surface & dimensional accuracy. • Is a gradual process. • Wear rate depends on material being cut, tool shape, cutting fluids, cutting parameters, among others. • Two types: • flank wear and crater wear.

  44. Tool Life, Wear and Failure • Flank Wear • Occurs on relief face of tool. (See Fig 20.15a) • Due to rubbing of tool on machined surface, causing abrasive wear. • Due to high temperatures that affect tool material.

  45. Taylor Tool Life Formula VTn = C • V: cutting speed • T: time in minutes to develop a flank wear land • n: depends on tool/work materials and cutting conditions • C: constant Notes: • When T=1, C = cutting speed. • Each combo of tool material/cutting condition has its own n & C values. • These values are determined experimentally.

  46. Predicting Tool Life & Wear • Cutting speed is the most significant process variable in tool life, along with d.o.c. and feed rate. • Introducing into the Taylor formula... VTn dx fy = C • d: depth of cut • f: feed rate (in/rev in turning) • x,y: determined experimentally for each cutting condition. • Typical values are: n=0.15, x=0.15, y=0.6.

  47. Taylor Tool Life Formula • Rearranging, T = C 1/n V -1/n d -x/n f -y/n • Plugging in the typical values, T = C7 V-7 d-1 f -4 • Notes: • If feed rate or d.o.c. are increased, must decrease cutting speed V. • A reduction in speed can result in an increase in material removed due to increased d.o.c. & feed rate.

  48. Tool Life Curves • Are plots of experimental data obtained by performing cutting tests at different conditions with different tool / work material combinations. • Cutting parameters are varied. • Generally tool life decreases with cutting speed. • Work material has a strong impact on tool life. • Harder materials will wear tool faster. • Tool life curves have a valid range and should be applied within it.

  49. Tool Life Curves • Recommended cutting speed for HSS tools yields a tool life of 60-120 min. • For carbide tools, 30-60 min. • There is an optimum cutting speed for each work / tool material combination. • Increasing cutting speed reduces tool life. Reducing speed increases tool life but decreases material removal rate. • Material removed is directly proportional to distance traveled by the tool: decreasing cutting speed helps remove more material between tool changes.

  50. Tool Life • Example: • Let n = 0.5, C = 400. Calculate % increase in tool life is cutting speed is reduced by 50%. • Solution: • Taylor Formula: VTn = C • V1 = initial speed • V2 = reduced speed • C = constant V2T2n = V1T1n

More Related