1. If a bar of SAE 1040 is to be turned with a high speed steel tool with a feed of 0.015” per revolution, and a depth of 0.050”. Previous experiments have revealed that the following cutting velocities yielded the following tool lives,
2. Two tools are being compared for their costs. The table below summarizes the details of each tool. Find the economic tool life and cutting speed for each tool, and determine the least expensive tool.
ans. As temperatures rise both the tool and work change. Heat causes expansion, therefore the dimensions change, and accuracy decreases. Heat also causes decreased strength of the material. This causes faster wear in the tool, but also makes the work easier to cut.
4. We are going to estimate the effects of feedrate on tool life. Some simple calculations yield the Taylor tool life coefficients of n = 0.4 and a C = 400. Find the change in tool life (in %) when velocity drops by a) 20% and b) 40%. [based on Kalpakijian]
ans. Two failures typically occur; wear and fracture. If a tool is worn, and the material and geometry permit, we can recondition a tool - grinding is common. If a tool is fractured or can’t be reconditioned, it can be discarded. In some cases tools contain parts that can be reclaimed, or materials that can be recycled.
8. Consider that at a certain velocity we will get the lowest cost per piece. As the cutting velocity rises the cost per piece rises (but we will improve the production rate) what cost components rise or drop?
15. We have been asked to calculate the cutting speeds that gives the maximum possible production rate and lowest cost for an existing job. The current tool will last for 4 hours if we cut at 300 fpm and 2 hours at 345 fpm. The following things are known about the job.