Tool-life economics on a CNC turning cell
A CNC turning cell finish-machines 60 mm diameter, 300 mm long mild-steel shafts at 90 m/min and 0.25 mm/rev feed. Trials on the coated-carbide insert gave the Taylor relation VT^0.25 = 300, with V in m/min and T in min. Machine-plus-operator cost is Rs 2.0/min, each indexable edge costs Rs 40, and an edge change takes 1.5 min. The plant engineer proposes raising the speed to 110 m/min to cut cycle time; the shop supervisor objects that tool changes will eat the gain. A second proposal is to switch to a tougher, cheaper edge costing Rs 18 but with the same Taylor constants; the vendor claims this will lower cost per piece at the existing 90 m/min. The cell runs 1,800 h/year and annual demand is 24,000 shafts. The current tool-change policy replaces an edge only when flank wear reaches the rejection limit, so no premature changes occur.
1. Compute tool life and cost per piece at 90 m/min and at 110 m/min. Does the supervisor's objection hold?
Read the worked answer
At V = 90 m/min: T = (300/90)^4 = 123.5 min. Spindle speed N = 1000V/(πD) = 1000×90/(π×60) = 477.5 rpm. Machining time tm = 300/(0.25×477.5) = 2.51 min. Tool changes per piece = 2.51/123.5 = 0.0203. Cost = 2.0×2.51 + 0.0203×(2.0×1.5 + 40) = 5.03 + 0.87 = Rs 5.90.
At V = 110 m/min: T = (300/110)^4 = 55.3 min. N = 583.5 rpm. tm = 300/(0.25×583.5) = 2.06 min. Changes per piece = 2.06/55.3 = 0.0372. Cost = 2.0×2.06 + 0.0372×43 = 4.11 + 1.60 = Rs 5.71.
The supervisor is only partly right. Tool cost per piece rises from Rs 0.87 to Rs 1.60, but cycle time falls enough that total cost drops by about Rs 0.19/pc, or 3.2%. The objection would be valid only if the speed were pushed far above the optimum.
2. Find the cutting speed for minimum cost per piece and the resulting cost. Compare it with 110 m/min.
Read the worked answer
For minimum cost per piece under Taylor tool life, Vopt = C[Cm/((1/n − 1)(Cm tc + Ct))]^n, where Cm is machine cost per minute, tc is tool-change time and Ct is tool cost.
Here n = 0.25, C = 300, Cm = 2.0, tc = 1.5, Ct = 40. Then 1/n − 1 = 3 and Cm tc + Ct = 3 + 40 = 43. So Vopt = 300[2/(3×43)]^0.25 = 300(2/129)^0.25 = 105.6 m/min.
At 105.6 m/min: T = (300/105.6)^4 = 65.1 min; N = 560.2 rpm; tm = 2.14 min; cost = 2.0×2.14 + (2.14/65.1)×43 = 4.28 + 1.42 = Rs 5.70.
This is essentially the same as the 110 m/min result. The 110 m/min proposal is therefore near-optimal and acceptable; the flat cost curve means the exact speed is not critical. Pushing much beyond 110 m/min would raise tool cost faster than it saves cycle time.
3. Evaluate the Rs 18 edge at 90 m/min and at its own optimum. Should the plant switch?
Read the worked answer
At 90 m/min with Ct = 18: T remains 123.5 min and tm = 2.51 min. Tool cost per piece = (2.51/123.5)(3 + 18) = 0.0203×21 = Rs 0.43. Total cost = 2.0×2.51 + 0.43 = Rs 5.45, versus Rs 5.90 for the Rs 40 edge. That is a saving of Rs 0.45/pc, or about Rs 10,800/year at 24,000 shafts.
At its own optimum: Vopt = 300[2/(3×(3 + 18))]^0.25 = 300(2/63)^0.25 = 126.6 m/min. T = (300/126.6)^4 = 31.6 min; tm = 2.51×90/126.6 = 1.79 min; cost = 2.0×1.79 + (1.79/31.6)×21 = 3.57 + 1.19 = Rs 4.76.
On the stated Taylor constants, the cheaper edge is economically better, especially at its optimum. The judgement is to trial it first: a cheaper grade may not maintain the same Taylor constants, edge strength or surface finish at 126.6 m/min. Switch only if shop trials confirm the wear behaviour.