Tools › Mechanical Design › Machining and Manufacturing › Machining Parameters › Example
Turning a 2-in 4140 bar at 500 rpm: cutting speed against the TC 9-524 and NAVEDTRA tables, removal rate, unit-power horsepower, Taylor tool life and the minimum-cost and maximum-production speeds
| Tool | Machining Parameters |
| Source | U.S. Department of the Army (1996). Fundamentals of Machine Tools. Training Circular TC 9-524, Washington, DC. Chapter 7 (lathe speeds and feeds, RPM = CS × 4 / D) and Appendix A Tables 7-2, 7-3; Naval Education and Training Program Management Support Activity (1990). Machinery Repairman 3 & 2. NAVEDTRA 12204, Pensacola, FL. Table 8-1; Taylor, F. W. (1907). "On the Art of Cutting Metals." Transactions of the ASME, Vol. 28, paragraph 700. |
| Level | Undergraduate |
Problem
A 2-in-diameter 4140 steel bar is turned at 500 rpm with a carbide tool at 0.010 in/rev and 0.100 in depth over 8 in (0.25 in approach), illustrative unit power 1.4 hp·min/in³, drive efficiency 0.85, Taylor n = 0.25 and C = 400 ft/min, machine rate 60 per hour, 2-min tool change, 5 per edge, 1.5 min handling, 1/32-in nose radius. Find the cutting speed and compare it with the manuals' recommended range, the removal rate and cutting time, the power, force and torque, the tool life and pieces per edge, the cost and time per piece, the two economic speeds and the ideal roughness.
Inputs to Enter
- Turning, speed given as 500 rpm; D 2 in, L 8 in, depth 0.100 in, feed 0.010 in/rev, extra travel 0.25 in, 1 pass, nose radius 0.03125 in
- Alloy steel (4130, 4140), carbide tool; unit power 1.4 hp·min/in³, drive efficiency 0.85; Taylor n 0.25, C 400 ft/min
- Machine rate 60 per hour, tool change 2.0 min, tool cost 5.0 per edge, handling 1.5 min; rake 10° (chip ratio and friction left blank)
Answers
- v = π D N = π × 2 × 500 / 12 = 261.8 ft/min (the manuals' shorthand 4 CS / D would call for 523.6 rpm at this speed); recommended range for 4140 with carbide: NAVEDTRA Table 8-1 70 ft/min (HSS) × 2 to 3 = 140 to 210 ft/min, so the entered speed is above the range
- Feed rate 5 in/min; MRR = π (D − d) d f N = π × 1.9 × 0.1 × 0.010 × 500 = 2.985 in³/min (handbook form π D d f N = 3.142); t_m = (8 + 0.25) / (0.010 × 500) = 1.65 min
- P_c = u × MRR = 1.4 × 2.985 = 4.178 hp, motor power 4.916 hp; F_c = P_c / v = 526.7 lb; T = P_c / ω = F_c D / 2 = 526.7 lb-in
- Taylor life T = (400 / 261.8)^4 = 5.45 min, 3.30 pieces per edge; cost per piece 5.269 (machining 1.65 + tools 2.12 + handling 1.5), time per piece 3.756 min
- Minimum-cost speed v = C [n/(1 − n) · C_o/(C_o t_c + C_t)]^n = 186.9 ft/min (357 rpm, life 21 min, cost 4.582); maximum-production speed v = C [n/((1 − n) t_c)]^n = 255.6 ft/min (488 rpm, life 6 min, time 3.754 min)
- Ideal roughness R_a = f² / (32 r) = 0.010² / (32 × 0.03125) = 100 µin (R_t 400 µin)
Notes
The tool's default inputs are this cut; press Submit. The case is the tool's own default (the brief's 4140 bar with an illustrative unit power and Taylor pair, stated as such on the page), checked in test_machining_parameters.php (218 checks) against a numpy / scipy re-derivation (References/Machining-Parameters/checks/machining_check.py: the economic speeds equal the scipy minima of the cost and time functions to 1.6e-8 at 20 parameter sets); the same test reproduces every worked example of TC 9-524 chapters 4, 7 and 8 and NAVEDTRA 12204 chapters 5, 8 and 11 and Taylor's 1907 ratios 0.92 and 0.84.