Question

Problem 1 Consider the drilling of Feature 3 (Thru Hole) on the flashlight head (Figure 1) Assume the feed of the drilling opcration (i.e., the distance the drill penetrates per unit revolution) is,f= 0.01 inrev. " a) Calculate the material removal rate (in in3/s or mm3/s). b) Use the Taylor Tool Life equation to estimate the tool life (in minutes) for this operation. Use Table 1 to pick appropriate values of C and n (the flashlight material is aluminum and the tools are highspeed steel). How many such holes would you be able to drill before the tool needs to be replaced? Nonsteel Cutting Steel Cutting Tool Material m/min(/min)in min 60 200 1500 Plain carhon sool steel High-speed steel Cemented carbide 20 70 0.125 0.25 Coated carbide 0.6 Compiled from 14].1),and other sources The parameter values are approximated for tarning at feed-0.25mm/rev (0010 intev) and dep 25mm (0 1000 inNonsteel cuning refers to easy to-machine metals such as aluminum beass, and cast ron Steel cutting refers to the machining of mild (unhardened) sieel. It should be noted that sgncn variations in these salues can he expected in practice Table 1. Representative values of n and C in the Taylor tool life equation for selected tool materials (from M.P. Groover, Fundamentals of Modern Manuftuing ed) Estimate the power in hp or W) required for this operation. The specific energy requirements in c) Estimate the rise in temperature at the tool-chip interface during the drilling (in F or °C). How d) If you were fabricating the flashlight from Stainless Steel instead of Aluminum, what would cutting operations (Table 21.2 in Kalpakjian, 7th ed.) can be applied to drilling processes. close is it to the melting point of the workpiece material? How close is the adiabatic temperature rise to the melting point? be (1) the tool life, (2) the power requirement, and (3) the temperature rise? .000- 0.500 0377 (V 1.500 Recommended tool: V drill bit. Recommended speed: 1200 RPM Figure I. Thru hole step for Problem I

Answer 1

Giver ten feed rate (f) = 0.01 in/rev^n cutting velocity (V) = 3 in/s From Figure 1: depth of hole (D) = 0.5 inch (a) material removal rate (mrr) = fx dx V mrr = 0.01 times 0.5 times 3 = 0.015 in^3/s (b) Tayler Tool life equation, V.T^n = C (1) From table 1: For high speed steel C = 120in/min and n = 0.125 V = 3 in/s = 3 times 60 = 180 in/min put these values in equation (1) 180 times 0.125 = 4726.8/39.39 => 70.125 = 0.26 T = (0.67)^1/0.125 => T = 0.041 minute