Solve Problem 24.21 except that in parts (a) and (b) determine the cutting speeds and tool lives for maximum production rate. Use of a spreadsheet calculator is recommended.

Problem 24.21

Three tool materials are to be compared for the same finish turning operation on a batch of 150 steel parts: high-speed steel, cemented carbide, and ceramic. For the high-speed steel tool, the Taylor equation parameters are: n = 0.130 and C = 80 (m/ min). The price of the HSS tool is $20 and it is estimated that it can be ground and reground 15 times at a cost of $2 per grind. Tool change time is 3 min. Both carbide and ceramic tools are in insert form and can be held in the same mechanical tool holder. The Taylor equation parameters for the cemented carbide are: n = 0.30 and C = 650 (m/min); and for the ceramic: n = 0.6 and C = 3,500 (m/min). The cost per insert for the carbide is $8 and for the ceramic is $10. There are six cutting edges per insert in both cases. Tool change time is 1.0 min for both tools. The time to change a part is 2.5 min. The feed is 0.30 mm/rev, and depth of cut is 3.5 mm. The cost of machine time is $40/hr. The part is 73.0 mm in diameter and 250 mm in length. Setup time for the batch is 2.0 hr. For the three tooling cases, compare:

(a) cutting speeds for minimum cost,

(b) tool lives,

(c) cycle time,

(d) cost per production unit,

(e) total time to complete the batch and production rate.

(f) What is the proportion of time spent actually cutting metal for each tooling? Use of a spreadsheet calculator is recommended.