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Mechanics of Orthogonal Cutting

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Ramdas Bhukya

it is power point presentation on orthogonal cutting, in which i explained as easy as the student can understand... and related to GATE

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Mechanics of Orthogonal Cutting
t1 is un-cut chip thickness t2 is cut chip thickness r is chip thickness ratio r = t1/t2 < 1 ( t1 < t2) k = 1/r = chip reduction coefficient α is rake angle φ is shear angle Assumptions 1. No contact at the flank. 2. Width of chip remains constant. 3. Uniform cutting velocity. 4. A continues chip is produced. 5. Volumetric changes of material during machining is zero. That is Volume before cutting = volume after cutting t1 *b*l1 = t2*b*l2 t1/t2 = l2/l1 = r Also we can say that volumetric flow rate is also equal t1*b*Vc = t2*b*Vf t1/t2 = Vf/ Vc Vc is cutting velocity Vf is chip flow velocity Vs is shear velocity WORKPIECE t1 α Shear plane Friction planeVertical plane Vs Vc Vf TOOL width φ
WORKPIECE TOOL t1 α Shear plane Friction plane Vertical plane Vs Vc Vf φ α φ AC B t1 sinФ = t1 / AB 90-Ф sin(90- Ф+ α ) = sin (90-(Ф-α)) = cos(Ф-α) = t2/AB t1 = AB*sinФ t2 = AB cos (Ф-α) Therefore t1/t2 = (AB*sinФ) / (AB*cos(Ф-α) ) r = sinФ / cos (Ф-α) r = sinФ / ( cosФ*cosα+ sinФ*sinα ) r = ( sinФ /cos Ф ) / [( cosФ*cosα+ sinФ*sinα ) / cosФ] r = tanФ/ (cosα + tanФ*sinα ) rcosα + r*tanФ*sinα = tanФ tanФ -r*tanФ*sinα = rcosα tanФ (1- rsinα) = rcos α tanФ = rcosα / (1- rsinα) From triangle ABC & ACD D 90-Ф+α Ф-α RELATION BETWEEN R, Φ AND α
WORKPIECE TOOL t1 α Shear plane Friction plane Vertical plane Vs Vc Vf φ Vc Vf Vs φ90-α 90-(Ф-α) By applying SINE rule (Vf / sinФ) =[Vs / sin(90-α)] = [Vc/sin(90-(Ф-α)] (Vf / sinФ) = (Vs / cosα) = [Vc/cos(Ф-α)] Vf = [Vc*sin α /cos(Ф-α)] Vf = Vc*r Vs = [(Vc*cos α /cos(Ф-α)] VELOCITY RELATIONSHIPS
work piece tool R2 F N Fc Ft R1 FS Fn φ α FS R F N β φ Ft Fc Fn α β-α Fc is Cutting Force Ft is Thrust Force R1 is Resultant Force of Fc & Ft F is Friction Force N is Normal Force of F R2 is Resultant Force of F& N Fs is Shear Force Fn Normal Force to Fs R1 isalso Resultant Force of Fs & Fn Weknow F = µN From diagram tanβ = F/N F = tanβ*N Therefore µ= tanβ β is Angle of friction µ is coefficient of friction
Ft Fc β-α R FS R F N β φ Ft Fc Fn α β -α R = (Fc ^2) + (Fv^2) Tan(β-α) = Fc /Ft R φ Fn R = (Fs^2) + (Ns^2) FS Tan(β-α+φ) = Fs/Ns β β-α α R = (F^2) + (N^2) R F N Theories of Angles Lee & Shaffer theory : Φ+β-α = 45 Stabler theory : Φ+β-(α/2)= 45 Merchant Constant (Cm) : 2Φ+β-α Energy for Cutting (Ec) = Fc * VC Energy for friction (Ef) = F * VF Energy for shearing (Es) = Fs * Vs Percentage of energy loss in friction = (Ec/Ef)*100 Percentage of energy loss in shearing = (Ec/Ef)*100
R F N β φ Ft Fc α β -α A O D C E G B α 9O-α 9O-α Relationship of Fs & Fn with Fc & Ft Fn = AE = AD+DE = DE+CB = Fc sin φ + Ft cos φ Fs = OA = OB-AB = OB-BC = Fc cos φ - Ft sin φ Relationship of F & N with Fc & Ft F= OA = CB = CG+GB = ED+GB = Fc sin α + Ft cos α N= AB = OD –CD = OD- GE = Fc COS α - Ft sinα FS R Φ Ft Fc Fn α β -α O B A C G E 9O-Φ 9O-Φ
Shear area = As = W*t1 / sin Φ Shear stress = τ = Fs / As τ = Fs sinΦ / (w*t1) Shear strain = Ƴ = Cot Φ + tan (Φ-α) = cos / [sin Φ*cos (Φ-α)] Shear strain rate = Vs/ts WORKPIECE t1 Fs φ W Vs The minimum value of shear strain when rake angle is zero Shear strain = Ƴ = Cot Φ + tan Φ (d/dΦ) {cot Φ + tan Φ } = 0 -cosec^2 Φ + sec^2 Φ =0 -(1/sin^2 Φ)+ (1/cos^2 Φ) =0 cos^2 Φ-sin^2 Φ=0 [(1-cos 2Φ)/2] - [(1-sin 2Φ)/2] =0 2cos 2 Φ = 0 2Φ =90 Φ =45 For minimum value 2Φ - α =90 For orthogonal cutting Depth of cut = t1 = feed*θ ( θ is side cutting edge angle ) Width of cut = t1/ sin θ
Fc Ft N F µ = Tanβ = F/N = Ft/Fc But when Ft > Fc, β > 45 µ > 1 In this case use formulae for finding µ The classical friction theory µ = [ln ( 1/r)] / [( π/2) – α] Actuvally the value of µ is always comes less than one Ft < Fc, β < 45 µ < 1
• Taylorstool life equation:- •VTn=C • Tool life equation (generalized)– •VTnfn1dn2=C • Tool life exponents n,n1, n2 are found by plotting experimental data on log V – log T,log T– log fand log T– log d scales.
Determination of toollife constants n,n1, n2 • Longand expensive test, involves considerableamount of material, labor and machiningtime. • Recourseis taken to experimental design techniques suchasfactorial design, multiple regression analysis and response surface methodology to reduce cost and no. of observations.
RAMDAS BHUKYA mighty engineer

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Mechanics of Orthogonal Cutting

  • 2. t1 is un-cut chip thickness t2 is cut chip thickness r is chip thickness ratio r = t1/t2 < 1 ( t1 < t2) k = 1/r = chip reduction coefficient α is rake angle φ is shear angle Assumptions 1. No contact at the flank. 2. Width of chip remains constant. 3. Uniform cutting velocity. 4. A continues chip is produced. 5. Volumetric changes of material during machining is zero. That is Volume before cutting = volume after cutting t1 *b*l1 = t2*b*l2 t1/t2 = l2/l1 = r Also we can say that volumetric flow rate is also equal t1*b*Vc = t2*b*Vf t1/t2 = Vf/ Vc Vc is cutting velocity Vf is chip flow velocity Vs is shear velocity WORKPIECE t1 α Shear plane Friction planeVertical plane Vs Vc Vf TOOL width φ
  • 3. WORKPIECE TOOL t1 α Shear plane Friction plane Vertical plane Vs Vc Vf φ α φ AC B t1 sinФ = t1 / AB 90-Ф sin(90- Ф+ α ) = sin (90-(Ф-α)) = cos(Ф-α) = t2/AB t1 = AB*sinФ t2 = AB cos (Ф-α) Therefore t1/t2 = (AB*sinФ) / (AB*cos(Ф-α) ) r = sinФ / cos (Ф-α) r = sinФ / ( cosФ*cosα+ sinФ*sinα ) r = ( sinФ /cos Ф ) / [( cosФ*cosα+ sinФ*sinα ) / cosФ] r = tanФ/ (cosα + tanФ*sinα ) rcosα + r*tanФ*sinα = tanФ tanФ -r*tanФ*sinα = rcosα tanФ (1- rsinα) = rcos α tanФ = rcosα / (1- rsinα) From triangle ABC & ACD D 90-Ф+α Ф-α RELATION BETWEEN R, Φ AND α
  • 4. WORKPIECE TOOL t1 α Shear plane Friction plane Vertical plane Vs Vc Vf φ Vc Vf Vs φ90-α 90-(Ф-α) By applying SINE rule (Vf / sinФ) =[Vs / sin(90-α)] = [Vc/sin(90-(Ф-α)] (Vf / sinФ) = (Vs / cosα) = [Vc/cos(Ф-α)] Vf = [Vc*sin α /cos(Ф-α)] Vf = Vc*r Vs = [(Vc*cos α /cos(Ф-α)] VELOCITY RELATIONSHIPS
  • 5. work piece tool R2 F N Fc Ft R1 FS Fn φ α FS R F N β φ Ft Fc Fn α β-α Fc is Cutting Force Ft is Thrust Force R1 is Resultant Force of Fc & Ft F is Friction Force N is Normal Force of F R2 is Resultant Force of F& N Fs is Shear Force Fn Normal Force to Fs R1 isalso Resultant Force of Fs & Fn Weknow F = µN From diagram tanβ = F/N F = tanβ*N Therefore µ= tanβ β is Angle of friction µ is coefficient of friction
  • 6. Ft Fc β-α R FS R F N β φ Ft Fc Fn α β -α R = (Fc ^2) + (Fv^2) Tan(β-α) = Fc /Ft R φ Fn R = (Fs^2) + (Ns^2) FS Tan(β-α+φ) = Fs/Ns β β-α α R = (F^2) + (N^2) R F N Theories of Angles Lee & Shaffer theory : Φ+β-α = 45 Stabler theory : Φ+β-(α/2)= 45 Merchant Constant (Cm) : 2Φ+β-α Energy for Cutting (Ec) = Fc * VC Energy for friction (Ef) = F * VF Energy for shearing (Es) = Fs * Vs Percentage of energy loss in friction = (Ec/Ef)*100 Percentage of energy loss in shearing = (Ec/Ef)*100
  • 7. R F N β φ Ft Fc α β -α A O D C E G B α 9O-α 9O-α Relationship of Fs & Fn with Fc & Ft Fn = AE = AD+DE = DE+CB = Fc sin φ + Ft cos φ Fs = OA = OB-AB = OB-BC = Fc cos φ - Ft sin φ Relationship of F & N with Fc & Ft F= OA = CB = CG+GB = ED+GB = Fc sin α + Ft cos α N= AB = OD –CD = OD- GE = Fc COS α - Ft sinα FS R Φ Ft Fc Fn α β -α O B A C G E 9O-Φ 9O-Φ
  • 8. Shear area = As = W*t1 / sin Φ Shear stress = Ï„ = Fs / As Ï„ = Fs sinΦ / (w*t1) Shear strain = Ƴ = Cot Φ + tan (Φ-α) = cos / [sin Φ*cos (Φ-α)] Shear strain rate = Vs/ts WORKPIECE t1 Fs φ W Vs The minimum value of shear strain when rake angle is zero Shear strain = Ƴ = Cot Φ + tan Φ (d/dΦ) {cot Φ + tan Φ } = 0 -cosec^2 Φ + sec^2 Φ =0 -(1/sin^2 Φ)+ (1/cos^2 Φ) =0 cos^2 Φ-sin^2 Φ=0 [(1-cos 2Φ)/2] - [(1-sin 2Φ)/2] =0 2cos 2 Φ = 0 2Φ =90 Φ =45 For minimum value 2Φ - α =90 For orthogonal cutting Depth of cut = t1 = feed*θ ( θ is side cutting edge angle ) Width of cut = t1/ sin θ
  • 9. Fc Ft N F µ = Tanβ = F/N = Ft/Fc But when Ft > Fc, β > 45 µ > 1 In this case use formulae for finding µ The classical friction theory µ = [ln ( 1/r)] / [( Ï€/2) – α] Actuvally the value of µ is always comes less than one Ft < Fc, β < 45 µ < 1
  • 10. • Taylorstool life equation:- •VTn=C • Tool life equation (generalized)– •VTnfn1dn2=C • Tool life exponents n,n1, n2 are found by plotting experimental data on log V – log T,log T– log fand log T– log d scales.
  • 11. Determination of toollife constants n,n1, n2 • Longand expensive test, involves considerableamount of material, labor and machiningtime. • Recourseis taken to experimental design techniques suchasfactorial design, multiple regression analysis and response surface methodology to reduce cost and no. of observations.