Mechanics of Orthogonal Cutting

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

1. MECHANICS OF ORTHOGONAL CUTTING BY RAMDAS BHUKYA

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 A ssumptions 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* V f t1/t2 = Vf / V c Mechanics of orthogonal cutting Vc is cutting velocity V f is chip flow velocity Vs is shear velocity WORKPIECE t1 t2 α Shear plane Friction plane Vertical plane Vs Vc V f TOOL width φ

3. WORKPIECE TOOL t1 t2 α Shear plane Friction plane Vertical plane Vs Vc V f Mechanics of orthogonal cutting φ α φ A C B t1 t2 sin Ф = t1 / AB 90- Ф sin(90- Ф + α ) = sin (90-( Ф - α )) = cos ( Ф - α ) = t 2/AB t1 = AB*sin Ф t2 = AB cos ( Ф - α ) Therefore t1/t2 = (AB*sin Ф ) / (AB* cos ( Ф - α ) ) r = sin Ф / cos ( Ф - α ) r = sin Ф / ( cos Ф * cos α + sin Ф *sin α ) r = ( sin Ф / cos Ф ) / [( cos Ф * cos α + sin Ф *sin α ) / cos Ф ] r = tanФ / ( cos α + tan Ф *sin α ) rcos α + r*tan Ф *sin α = tanФ tanФ -r*tan Ф *sin α = rcos α tanФ (1- rsin α ) = rcos α tanФ = rcos α / (1- rsin α ) From triangle ABC & ACD D 90- Ф + α Ф - α RELATION BETWEEN R, Φ AND α

4. WORKPIECE TOOL t1 t2 α Shear plane Friction plane Vertical plane Vs Vc V f φ Mechanics of orthogonal cutting Vc V f Vs φ 90- α 90-( Ф - α ) By applying SINE rule ( V f / sin Ф ) =[Vs / sin(90- α )] = [ Vc /sin(90-( Ф - α )] ( V f / sin Ф ) = (Vs / cos α ) = [ Vc / cos ( Ф - α )] V f = [ Vc *sin α / cos ( Ф - α )] V f = Vc *r Vs = [( Vc * cos α / cos ( Ф - α )] VELOCITY RELATIONSHIPS

5. Mechanics of orthogonal cutting work piece tool R2 F N Fc Ft R1 F S Fn φ α F S R F N β 90- 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 F s is Shear Force Fn Normal Force to F s R1 is also Resultant Force of Fs & Fn Weknow F = µN From diagram tan β = F/N F = tan β *N Therefore µ= tan β β is Angle of friction µ is coefficient of friction Merchant Circle

6. Mechanics of orthogonal cutting Ft Fc β - α R F S R F N β 90- β φ Ft Fc Fn α β - α R = ( Fc ^ 2) + (Fv ^ 2) Tan( β - α ) = Fc /Ft R φ Fn Β - α R = ( Fs ^ 2 ) + (Ns ^ 2) F S Tan( β - α + φ ) = F s /Ns β β - α 90- β 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) : 2 Φ + β - α Merchant Circle Energy for Cutting ( Ec ) = Fc * V C Energy for friction ( Ef ) = F * V F 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. Mechanics of orthogonal cutting R F N β 90- β φ Ft Fc α β - α A O D C E G B α α 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 sin α F S R 90- β Φ Ft Fc Fn α β - α O B A C G E 9O- Φ Φ 9O- Φ

8. Mechanics of orthogonal cutting SHEAR STRESS AND SHEAR STARIN Shear area = As = W*t1 / sin Φ Shear stress = τ = Fs / As τ = Fs sin Φ / (w*t1) Shear strain = Ƴ = Cot Φ + tan ( Φ - α ) = cos / [sin Φ * cos ( Φ - α )] Shear strain rate = Vs/ t s 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/d Φ ) {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 2 Φ )/2] - [(1-sin 2 Φ )/2] =0 2cos 2 Φ = 0 2 Φ =90 Φ =45 For minimum value 2 Φ - 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 Mechanics of orthogonal cutting α i s equal to 0 µ = Tan β = 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

10. Taylors tool life equation :- V T n = C Tool life equation (generalized)– V T n f n1 d n2 = C Tool life exponents n, n1, n2 are found by plotting experimental data on log V – log T, log T – log f and log T – log d scales. Mechanics of orthogonal cutting

11. Determination of tool life constants n, n1, n2 Long and expensive test, involves considerable amount of material, labor and machining time. Recourse is taken to experimental design techniques such as factorial design, multiple regression analysis and response surface methodology to reduce cost and no. of observations. Mechanics of orthogonal cutting

12. RAMDAS BHUKYA mighty engineer Wishing is Not Enough , We must Do…, Learning is Not Enough , We must Apply…,