Rough and finish turning of a C45 shaft on a geared-head lathe
A 320 mm long bar of C45 (black, Ø62 mm) must be turned to Ø56 h6 over its whole length on a geared-head centre lathe. The headstock gives 63, 90, 125, 150, 200, 250, 315 and 400 rpm; the feed gearbox gives 0.10, 0.15, 0.20 and 0.25 mm/rev. The toolholder accepts a maximum depth of cut of 2 mm. The process sheet fixes roughing at 30 m/min and 0.25 mm/rev, and finishing at 45 m/min and 0.10 mm/rev, with 0.5 mm of stock (on diameter) left for the finishing pass. The lathe motor is 3.7 kW with 85% overall efficiency; for C45 take the specific cutting energy as 2.5 J/mm³. The operator proposes to skip the roughing pass, take the whole 3 mm of radial stock in a single cut at 63 rpm, and to size the shaft at the same setting so that a separate finishing set-up can be avoided.
1. Select the spindle speed actually used for roughing and for finishing, state the true cutting speed at each setting, and calculate the machining time for one shaft.
Read the worked answer
Roughing. N = 1000V/(πD) = 1000 × 30/(π × 62) = 154 rpm. The next lower available step is 150 rpm (rounding up to 200 would exceed the specified surface speed).
- True V = π × 62 × 150/1000 = 29.2 m/min
- Time per pass = L/(f × N) = 320/(0.25 × 150) = 8.53 min
Stock split. Total radial stock = (62 − 56)/2 = 3 mm; 0.25 mm radial (0.5 mm on diameter) is reserved for finishing, so roughing removes 2.75 mm. With a 2 mm holder limit this is two passes: 2.00 + 0.75 mm. Roughing time = 2 × 8.53 = 17.07 min.
Finishing. N = 1000 × 45/(π × 56) = 256 rpm → 250 rpm; true V = π × 56 × 250/1000 = 44.0 m/min; time = 320/(0.10 × 250) = 12.8 min.
Total cutting time ≈ 29.9 min (add handling, gauging and a spring pass if the h6 band demands it).
2. Check the power demand of the proposed 3 mm radial cut at 63 rpm against the 3.7 kW motor, and then give your verdict on the proposal.
Read the worked answer
Power check. V = π × 62 × 63/1000 = 12.3 m/min.
- MRR = 1000 × V × f × d = 1000 × 12.3 × 0.25 × 3 = 9 225 mm³/min = 154 mm³/s
- Power at the tool = 2.5 × 154 = 385 W
- Power at the motor terminals = 385/0.85 = 453 W ≈ 0.45 kW
That is only about 12% of the 3.7 kW rating, so power is not the limiting factor and the cut would not trip the motor.
Verdict: still reject it, for four reasons.
- The toolholder is rated 2 mm; 3 mm overloads the tool and the compound slide.
- 12.3 m/min is far below the 30 m/min specified. At that speed a carbide insert rubs, builds up an edge, tears the surface and actually raises the specific cutting force.
- A 320 mm bar at Ø62 has L/D ≈ 5.2, so it is flexible; radial force scales with depth of cut, and a 3 mm cut deflects the work to give a barrel-shaped, size-drifted shaft.
- Ø56 h6 (tolerance 19 µm) and a fine finish cannot come from a roughing feed of 0.25 mm/rev - the 0.25 mm radial finishing allowance exists precisely to be cut by a sharp tool at 0.10 mm/rev.
3. The shop considers running the roughing pass at 200 rpm instead of 150 rpm on a 40-piece batch. Quantify the saving and judge whether the process sheet should be changed.
Read the worked answer
At 200 rpm: V = π × 62 × 200/1000 = 38.96 m/min (about 33% above the sheet value).
- Time per roughing pass = 320/(0.25 × 200) = 6.40 min → two passes = 12.80 min
- Saving = 17.07 − 12.80 = 4.27 min per shaft, i.e. 171 min (about 2.9 h) over 40 pieces
- As a share of the 29.9 min floor-to-floor time the saving is only about 14%
Tool-life reasoning. Using Taylor with n = 0.25: T₂/T₁ = (29.2/38.96)^(1/0.25) = 0.75⁴ ≈ 0.32, so tool life falls to roughly one third unless the insert grade is rated for the higher speed. At 200 rpm the 40-piece batch needs about 8.5 h of continuous cutting, so a tool change (plus the size drift while the edge wears) can easily cost more than 2.9 h.
Judgement: run a 10-piece trial at 200 rpm, measure flank wear (reject the edge at 0.3 mm) and the diameter trend. Change the sheet only if one edge survives the batch; otherwise keep 150 rpm and attack the 12.8 min finishing pass instead.