MATERIAL REMOVAL PROCESSES Theory of Metal Machining 1
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MATERIAL REMOVAL PROCESSES Theory of Metal Machining 1. Overview 2. Theory of Chip Formation 3. Force Relationships 4. Power and Energy Relationship 5. Cutting Temperature 1 Introduction • Everyday Experience: Scraping the ice from your windshield – Edge angle of the ice scraper – Force required depending on the characteristics of ice • Incentives: Making a ceramic vase out of clay – Shaping – Removal of excess materials - ‘machining’ • Powder Metal or Cast – Exact dimension – Tolerance & Surface Finish 2 Classification Turning and Related Operations Drilling and Related Conventional Operations Machining Milling Other Machining Operations Material Removal Grinding Processes Abrasive Processes Processes Other Abrasive Processes Mechanical Energy Processes Electrochemical Nontraditional Processes machining Thermal Energy Processes Chemical Machining 3 Material Removal Processes A family of shaping operations through which undesired excess material is removed from a starting workpart so the remaining part become closer to the desired shape • Categories: – 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 • A shearing process in which excess materials is removed by cutting tools. – A variety of work materials – ‘Repeatable’ regular geometries – Close tolerance (<0.025μm) – Smooth surface finish (0.4μm) – Waste, Expensive: Cost and Time – Other processes such as casting, forging, and bar drawing create the general shape – Machining provides the final shape, dimensions, finish, and special geometric details 5 1. Overview Work • Types New Surface Speed motion Feed Motion – Turning - Lathe (tool) (work) Speed motion – Drilling – Drill press Cutting tool (Tool) – Milling – Milling Machine Feed Motion Drill bit • Peripheral (tool) • Face • Cutting Tool Work Speed motion Speed motion Milling Cutter Milling Cutter New Surface New Surface Feed Motion (work) Feed Motion (work) Work material Work Peripheral (End) Milling Face (Slab) Milling 6 Cutting condition • Relative motion between tool and work • Cutting conditions – Cutting speed, v (m/s) – Surface speed – Feed f (m): the lateral distance traveled by the tool during one revolution. – Depth of cut d (m) • Material Removal Rate: = dfvMRR – Roughing - removes large amounts of material, at high feeds and depths, low speeds – Finishing - Achieves final dimensions, tolerances, and finish, Low feeds and depths, high cutting speeds 7 Machine Tools • A power-driven machine that performs a machining operation – Holds workpart – Positions tool relative to work – Provides power and controls speed, feed, and depth. – Pumps a Cutting fluid v d f 8 2. Theory of Chip Formation • Orthogonal Cutting Model Rake angle: α α Shear angle: φ Shear Chip thickness ratio: Plane Tool t l sinφ r=o = s tc tc ls cosφ()− α to φ By rearranging l Work s r cosα tanφ = 1− r sinα r is always less than 1.0. 9 Shear Strain in chip α B A D C sinsin cos cos sin φ Using (α ± β ) = α β ± α β αcos() βcos± = α cos βm sin α sin β B A φ ACDC+ AD φ−α γ= = =tanφ() − αcot + φ BD BD α sinφ()− α cosφ cosα D = + = cosφ()− α sinφsinφ cos() φ− α C As φ (from 10° to 35°) increase, γ (from 5 to 2) decreases. 10 Velocity α Vc Vs α 90-φ+α φ-α oolT 90-α φ Vc V V φ Vs V VV Work = s = c ⎛π ⎞ ⎛π ⎞ sinφ sin⎜ −φ + α⎟ sin⎜ −α ⎟ V cosα V ⎝ 2 ⎠ ⎝ 2 ⎠ γ& = s = Δy cosφ()− α Δy V VVs c = = where Δy is the finite thickness of φcos() α− cos α sin φ shear plane, typically 0.03mm.the Strain rate is around 10Shear 3-105sec-1 11 Actual Chip Formation (a) Discontinuous chip •Brittle materials at low cutting speed •High tool-chip friction and large feed and Secondary depth Shear Zone (b) Continuous chip •Ductile materials with high speeds and small feed and depth of cut (c) Continuous chip with built-up edge •Ductile material at low to medium speeds (d) Serrated chip Tool •Difficult-to-machine metals at high cutting speeds Effective φ Primary Shear Zone Work12 The ‘Real’ Cutting Force Cutting Forces are measured with Dynamometer. Normal Ft Shear Fr •Area: A=bh σ Fn where b=chip width F t h=chip thickness •Temperature (500-1000oC) R • Pressure (1000-3000 MPa) Sticking Zone Sliding Zone 13 3. Force Relationships α Tool Tool F φ R Fs β Fc R’ N Work F Work Fn R” t F: Friction Force Fs: Shear Force Fc: Cutting Force N: Normal Force Fn: Normal Force Ft: Thrust Force 14 Force Diagram F F φ s c FFF= c sinα + t cosα β−α NFF=c cosα − t sin α FFFs= ccosφ − t sin φ Fn FFF=sinφ + cos φ Ft n c t F F R = s β sinφ φ cos()+ β − α τt w α = s o sinφ φ cos()+ β − α α F cosβ()− α N F = s csinφ φ cos()+ β − α Fs sinβ()− α Ft = sinφ φ cos()+ β − α 15 Cutting Force ⎡ cosβ( − α) ⎤ 2 • Cutting Force:Fc = bhτ s ⎢ K⎥ bh= c Kc [] N/ mm sin⎣φ φ cos()+ β − ⎦ α • Thrust Force: ⎡ cosβ()− α ⎤ Ft = bhτ s ⎢sin ⎥K= t bh ⎣sinφ φ cos()+ β − ⎦ α τ sbh =FRcosφ +() β − αFF =cosφ −sin φ = FRc = cos(β −α) s c t sinφ FR=t sinβ() − α =FRn φsin +() β − αFF =c sin φt cos + φ Kc and Kt must be calibrated α experiments.through machining t b c oolT d φ h to f Fc R F Work 16 t iewop VT Chatter Analysis • Mechanical vibration k c ibration:Freem V x&&&+ c+ x= 0 kx ibration:Forced V m x&&&+ c x + kx =o sinω F t Assumex t= ( X )() wt +φ sin F kx c& x Or using complex harmonic functions jα j ω t m x&&&+ c x + kx =o F e e Φ w F Assumex t= Xe (jω()) t+ φ () x() t 2 2 jφ j ω t jα j ω t k m()−ω j c + Xe ω e=() F = o t F e e 1 X 1 1 Magnitude ratio:Φ()w = = F k 2 2 2 o 1()−r +()ζ 2 r r φ 1 − 2ζr Phase: φ = tan −1 +α 0 1− r 2 where r = ω ζc= 2 km -90 ωn k 17 and ω = n m -180 r The Merchant Equation F • Shear stress: τ = s As t w • Shear Plane Area: A = o s sinφ FFcosφ − sinφ • Shear stress: τ = c t to wsinφ • Shear plane angle will Merchant’s Assumption: form to minimize energy • After differentiating τ w.r.t φ, Merchant’s Equation: α β =φ 45 + − 2 2 18 Implication of Merchant’s Eq. α β =φ 45 + − 2 2 • An increase in rake angle causes the shear plane angle to increase. • A decrease in friction angle cause the shear plane angle to increase. • The analysis from orthogonal cutting can be used in a typical turning if the feed is small relative to depth of cut. Effect of shear plane angle φ : (a) higher with a resulting φ Tool lower shear plane area; Tool (a) smaller φ with a resulting larger shear plane area. 19 Turning vs. Orthogonal Feed f Uncut Chip thickness to Depth d Width of cut w Cutting speed v Cutting speed v Cutting force Fc Cutting force Fc Feed force Ff Thrust force Ft α d t c Tool t φ f oF c Ft Work 20 4. Power & Energy Relation PFV= • Power (energy per unit time) c c FV HP = c 33c , 000 P hp = 33c , 000P in ft-lb/min P hp • Horse power P = c or hp = c g E g E ficiency E=90%with mechanical ef P • Unit Power P = c u MRR PcF c v Fc • Specific energyUP=u = = = MRR vto wto w 21 Size Effect & Energy Distribution 0.01 0.04(in) 1.6 100 1.4 Tool 1.2 Work 1.0 0.8 50 0.6 Correction Factor Chip 0.4 Proportion of Energy 0 0.125 0.25 0.5 1.0 1 2 3 Chip thickness before cut (mm) Cutting Speed (m/s) 22 Specific Energy for various work materials (to=0.25mm) Materials Brinell Specific Energy (U) Hardness N·m/mm3 In-lb/in3 Hp/in3/min Carbon Steel 150-200 1.6 240,000 0.6 200-250 2.2 320,000 0.8 251-300 2.8 400,000 1.0 Alloy Steels 200-250 2.2 320,000 0.8 251-300 2.8 400,000 1.0 301-350 3.6 520,000 1.3 351-400 4.2 640,000 1.6 Cast iron 125-175 1.1 160,000 0.4 175-250 1.6 240,000 0.6 Stainless steel 150-200 2.8 400,000 1.0 Aluminum 50-100 0.7 100,000 0.25 Aluminum Alloys 100-150 0.8 120,000 0.3 Magnesium Alloys 50-100 0.4 60,000 0.15 23 Problem 21.30 lathe performs A a turning operation on a work piecdiameter. shear strength of the of 6in The e work=40,000lb/in2. The rake angle of the tool =10o. The machine settings are: rotational speed=500rev/min, feed=0.0075in/rev. and depth=0.075in. The chip thickness after the cut requi0.015in. Determine: (a) the horsepower is red (b) for the unit horsepower this material, with the correction factor (1 for (c) the unit horsepower tothe orthogonal model.=0.01in.) Use , Fo get HPT(a) c and v are needed 0t .