MACHINING ECONOMETRICS Machinery's Handbook, 31st Edition
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Table 2b. Variation of ECT , Chip Thickness and f z / f z 0 with ar / D
Face Milling
End Milling (straight)
ECT 0 corrected for f z / f z 0
ECT 0 corrected for f z / f z 0
ar / D
f z / f z 0 1.398 1.107 1.036 1.000 0.958 0.952 0.932 0.897 0.892 0.891 0.890 0.890 0.890
f z / f z 0 1.000 0.884 0.880 0.892 0.934 0.945 1.000 1.422 1.840 2.574 3.694 6.377 20.135
hm
ECT 0.411 0.370 0.332 0.303 0.252 0.243 0.206 0.089 0.051 0.026 0.012 0.004 0.002
hm
ECT 0.103 0.093 0.083 0.076 0.063 0.061 0.051 0.022 0.013 0.006 0.003 0.001 0.001
1.0000 0.9000 0.8080 0.7360 0.6137 0.5900 0.5000 0.2170 0.1250 0.0625 0.0300 0.0100
0.108 0.137 0.146 0.151 0.158 0.159 0.162 0.169 0.170 0.170 0.170 0.170 0.170
0.575 0.410 0.344 0.303 0.242 0.231 0.192 0.080 0.046 0.023 0.011 0.004 0.002
0.108 0.122 0.123 0.121 0.116 0.115 0.108 0.076 0.059 0.042 0.029 0.017 0.005
0.103 0.082 0.073 0.067 0.059 0.057 0.051 0.032 0.024 0.017 0.011 0.007
0.0010 0.005 In face milling, the approximate values of aa / CEL = 0.95 for lead angle LA = 0 ° (90 ° in the metric system); for other values of LA , aa / CEL = 0.95 3 sin( LA ), and 0.95 3 cos( LA ) in the metric system. Example, Face Milling: For a cutter with D = 250 mm and ar = 125 mm, calculate ECT F for f z = 0.1, z = 12, and LA = 30 degrees. First calculate ar / D = 0.5, and then use Table 2b and find ECT 0 F = 0.2. Calculate ECT F with above formula: ECT F = 0.2 3 (12⁄8) 3 (0.1⁄0.17) 3 0.95 3 sin 30 = 0.084 mm. End milling: ECT E = ECT 0 E 3 ( z ⁄2) 3 ( f z ⁄0.17) 3 ( aa / CEL ), or if ECT E is known calculate f z from: f z = 0.17 3 ( ECT E / ECT 0 E ) 3 (2/ z )) 3 ( CEL / aa ) The approximate values of aa / CEL = 0.95 for lead angle LA = 0 ° (90 ° in the metric system). Example, High-Speed End Milling: For a cutter with D = 25 mm and ar = 3.125 mm, calculate ECT E for f z = 0.1 and z = 6. First calculate ar / D = 0.125, and then use Table 2b and find ECT 0 E = 0.0249. Calculate ECT E with above formula: ECT E = 0.0249 3 (6⁄2) 3 (0.1⁄0.17) 3 0.95 3 1 = 0.042 mm. Example, High-Speed End Milling: For a cutter with D = 25 mm and ar = 0.75 mm, cal culate ECT E for f z = 0.17 and z = 2 and 6. First calculate ar / D = 0.03, and then use Table 2b and find f z / f z 0 = 3.694 Then, f z = 3.694 3 0.17 = 0.58 mm/tooth and ECT E = 0.0119 3 0.95 = 0.0113 mm and 0.0357 3 0.95 = 0.0339 mm for 2 and 6 teeth respectively. These cutters are marked HS2 and HS6 in Fig. 26a, Fig. 26d, and Fig. 26e. Example, High-Speed End Milling: For a cutter with D = 25 mm and ar = 0.25 mm, calculate ECT E for f z = 0.17 and z = 2 and 6. First calculate ar / D = 0.01, and then use Table 2b and find ECT 0 E = 0.0069 and 0.0207 for 2 and 6 teeth respectively. When obtaining such small values of ECT , there is a great danger to be far on the left side of the H -curve, at least when there are only 2 teeth. Doubling the feed would be the solution if cutter design and material permit. Example, Full Slot Milling: For a cutter with D = 25 mm and ar = 25 mm, calculate ECT E for f z = 0.17 and z = 2 and 6. First calculate ar / D =1, and then use Table 2b and find ECT E = 0.108 3 0.95 = 0.103 and 3 3 0.108 3 0.95 = 0.308 for 2 and 6 teeth, respectively. These cutters are marked SL2 and SL6 in Fig. 26a, Fig. 26d, and Fig. 26e.
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