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Fig 318: The cutting insert nomenclature according to the ISO
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Mechanics of Materials Cutting
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Fig 319: The tool shank nomenclature according to the ISO
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The tensile strength of the tool shank (high carbon steel) is 500 600 N/mm2: Design stress S =
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Ultimate tensile stress 500 to 600 = = 25 to 60 N/mm2 Design Factor 10 to 20
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Referring to Figure 320, the bending moment BM = Fz L , and this is resisted by the material because of its strength Assuming the overhanging length L = 30 mm and Fz = 1160 N from experimental data, Resisting moment R = S Section modulus The bending moment of the tool shank can be expressed as Fz L = l =
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Fig 320: Shank design [10]
l = 1735 20 mm Not only must the shank be strong but it must also be rigid enough to resist deflection ( y) that is a source of chatter The stiffness of the tool shank can be calculated as Fy = 244 N (for this calculation) E = Elastic modulus = 200000 N/mm2 I = y =
l 4 204 = = 1333333mm4 12 12 224 303 Fy L3 = = 00007mm 3 E I 3 200000 1333333
As y is not greater than 002, the selection of the shank dimension is valid for this example The selected tool holder geometry would be MTJN 20 20 R 20 S 11 Calculation of the True Back Rake Angle, lt = i = tan 1 (tan l cos K sin K tan g ) = tan 1 (tan ( 5 ) cos (0 ) sin (0 ) tan ( 5 ) = 5 12 Calculation of the True Side Rake Angle, gt = a = tan 1 (cos K tan g + sin K tan l) = tan 1 (cos (0 ) tan (5 ) + sin (0 ) tan ( 5 )) = 5
Mechanics of Materials Cutting 13 Calculation of the Normal Side Rake Angle, n an = tan 1 (tan a cos i ) = tan 1 [tan ( 5 ) cos ( 5 )] = 49811 14 Calculations of the Friction Force, FF FF = =
{ (1160 cos ( 5 ) + 244 sin ( 5 ) ) sin ( 49811 )
Fp cos i + Fr sin i sin n + Fq cos n + ( Fp sin i Fr cos i )2
+ 535 cos ( 49811 ) 2 + (1160 sin ( 5 ) 244 cos ( 5 ) )
= {[(113432) ( 008683) + (53298)] 2 + ( 344172)2}1/2 = 554286 N 15 Calculation of Normal Force, FN FN = (FP cos i + Fr sin i ) cos an Fq sin an = [1160 cos ( 5 ) + 244 sin ( 5 )] cos ( 49811 ) 535 sin ( 49811 ) = 1130036 ( 46453 ) = 117649 N 16 Calculation of Coefficient of Friction, F m = N = 17 Calculation of tan t = t = = =
2 12
= 0471
554286 117649
m tan 1 m tan 1 (0471) 2523
18 Calculation of Chip ratio, rc rc =
Undeformed chip thickness, t1 Deformed chip thickness, t2 025 = 036
= 069444
Precision Engineering
19 Calculation of Shear Angle, n
r cos n Fn = tan 1 c 1 rc sin n 069444 cos ( 49811 ) = tan 1 1 069444 sin ( 49811 ) = 33124
110 Calculation of Shear Strain, e = cot Fn + tan (Fn an) = cot (32122) + tan [32122 ( 49811)] = 23168 111 Machinability Coefficient, C C = 2Fn + t g = 2(33124) + 2523 ( 5) = 9647 20 Sample Calculation 2 (Work Done Calculation) Symbols V Cutting Velocity Vs Shear Velocity Ps Shear Work Po Output Work hs Shear flow angle Vc Pi Pf hc Friction Velocity Input Work Friction Work Chip flow angle
Forces The Force Component of FP, FQ and FR is estimated from the following equations: FP FQ FR FQ FR FR = = = = = = Fz = 1160 N Fx = 535 N (for this calculation) Fy = 244 N (for this calculation) Fy sin SCEA + Fx cos SCEA (for both negative and positive SCEA) Fx sin SCEA Fy cos SCEA (for positive SCEA) Fy cos SCEA Fx sin SCEA (for negative SCEA)
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