Chip Formation Mechanics -Power, Energy, Forces, TemperatureTemperaturever. 1ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20091OverviewOverview• Chip formation mechanics– Power– Energy–Forces–Torquesq– Temperature–Oblique cuttingOblique cutting ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20092Idealized Chipformation ProcessIdealized Chip-formation Processchipcutting toolshear zonecutting toolworkpieceME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20093Chip-formation Geometrypychipprimary shear zonetcBtooltoφζαtoolV (cutting velocity)ζAworkpieceME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20094Cutting Energy, Power, Forces & Torques• To get forces and torques, calculate power (energy), and p(gy),back calculateP = F x sPTP = T x ωME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20095Specific Cutting Energy(u)Specific Cutting Energy (u)chipofvolumechipformto(work)energy u =VlWorkVlEnergyuchipofvolume==VolumeVolumeME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20096Cutting Work(W)Cutting Work (W)Work(W)done by cutting force(F)=Work (W)done by cutting force (F)=cutting power (P) x cutting time (t)W = P t⎟⎞⎜⎛=VolumePowerVolumeWorkMRRtimeVolume=⎟⎠⎞⎜⎝⎛timeVolumeVolumetimeME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20097Cutting Power(P)Cutting Power (P)MRRPVWuVE===PMRRor P = u x MRRME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20098Cutting Power (P)g()• So, we can calculateP = u x MRR• Hence, the force or torque from:PFP = F x sP = T x ωME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 20099u is composed of:p•us, the shear energy per unitus, the shear energy per unit volume•ufrictional energy per unit•uf, frictional energy per unit volumechip curl energy•chip curl energy• chip acceleration kinetic energy• surface energy of new surfacesME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200910Where does cutting energy go?Where does cutting energy go?• 90% to chip•5% to tool5% to tool• 5% to workpieceME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200911ucan be obtained in two waysucan be obtained in two ways• Tabulated•Estimated•Estimated–u ~ HBME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200912Cutting Energy(u)Cutting Energy (u)As a first approximation:As a first approximation:u ≈ us+ uf07508us≈0.75 -0.8 uhenceu ≈ 1.5 usME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200913Estimation of u• Assume a simple, rigid-perfectly plastic material:plastic material:∫σσ0HBτεσdεσufmos⋅=⋅=⋅=∫γε042γ21ε3HBσo≈⇒≈≈for heavily cold-worked metals42γ21ε−≈⇒−≈ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200914Estimation ofuEstimation of uHB612στofm≈≈62HBHB3231us→≈33ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200915What number to use?What number to use?From aboveu ≈ 1.5 usus≈ (1/3 to 2/3) HBu ≈ (1/2 to 1) HBso, if no data, u ≈ HBME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200916Comparison of Tensile and Cutting (Ex. 1-1)• 304 stainless steel rod•d=05in df=048in l=6indo 0.5 in, df 0.48 in, lo 6 inWh t i th i d i•What is the energy required using tension?• What is the energy required using cutting?ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200917Tension: Ex 12Tension: Ex. 1-2σσ111⋅==⋅+∫Kudnεεσεε111+∫nudεσEnergyvolumeu=×K = 185,000 psi; n = 0.45ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200918TensionEx 13Tension -Ex. 1-3lAKEnergyn 11⋅=+ε0816.04805.0ln21=⎟⎠⎞⎜⎝⎛=ε()()()()lAnEnergyoo62500816.0000,1851245.1⋅=+=π48.01⎠⎝()()Jlbfin 449970,3625.045.1=−==πME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200919Cutting - Ex. 1-4gEnergy = (specific cutting energy) * (volume removed)()lDDuEfi⋅−×=224πuss≈ 1.5 hp min/in3x 550 x 60 x 12= 594 000 inlbf/in3()fi4= 594,000 in-lbf/in3()648.05.04000,59422⋅−×=πE= 54,650 in-lbf (6,176 J)()4ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200920ComparisonEx 15Comparison -Ex. 1-5• Tension: E = 3,940 in-lbf (449 J)• Cutting: E = 54,650 in-lbf (6,176 J)ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200921Turning vs Orthogonal CuttingTurning vs. Orthogonal CuttingTerminology used in a turning•Terminology used in a turning operation on a lathe, where f is the feed rate (in./rev or mm/rev) and d is the depth of cut. •In turning the“orthogonality”is•In turning, the orthogonality is to the left in the drawing, hence a change of coordinate system is needed. If you were doing a diametral cut-off (plunge cut) (p g )operation, no change would be needed.• Note that feed in turning is equivalent to the depth of cut in th l tti d thorthogonal cutting, and the depth of cut in turning is equivalent to the width of cut in orthogonal cutting.ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200922Relation between u and tRelation between u and t0•As the depth of cut decreases, the p,surface area to volume ratio increases, hence friction (energy) increases(gy)• Sinceuuu+≈• Hence1fsuuu+≈01tu∝ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200923Turning Power and ForceEx 21Turning Power and Force-Ex. 2-1Turning Titanium:Turning Titanium:• speed = 1 m/s = 200 sfpmf d t 0 1 / 0 0004” /•feed rate = 0.1 mm/rev = 0.0004” / rev• depth of cut = 3 mm = 0.1”What is cutting power and cutting force?ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200924Power-Ex. 2-2Power Ex. 22u≈0.06 kW/cm3/min=3.6 W/mm3/secu 0.06 kW/cm/min 3.6 W/mm/secMRR=Q=sfd=1000x01x3MRR =Q = sfd = 1,000 x 0.1 x 3 = 300 mm3/secP = u x MRR = 3.6 x 300 = 1,080 Wu 3 6 300,080= 1.45 hpME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200925ForceEx 23Force -Ex. 2-3F = P/s = 1,080 W ÷ 1 m/s =1 080 N (243 lbf)1,080 N (243 lbf)ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 200926uvs tin TurningEx 31u vs. t0in Turning –Ex. 3-1 •If()34.00/12.2 mmsWtu −≈•And1/0t–s = 1 m/s– f