Ch16_example2.ppt

# Ch16_example2.ppt - Manufacturing Methods Chapter 16...

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Unformatted text preview: Manufacturing Methods Chapter 16 Examples Example 16-10 Operation: Drilling Geometry: 10-mm holes, 20 mm deep Tool: HSS Twist Drill Material: Free-machining steel - HB180 Find: Cutting speed and feed FIG. 16-45 From Fig. 16-45: for HSS, vs = 0.75 m/s From Section 16-7-1, Part 4: v = .7vs for ferrous materials v = .525 m/s convert to rps: v = (525 mm/s) / ( 10 mm/rev) = 16.71 revolutions/second FIG. 16-45 From Section 16-7-1, Part 4: feed for free-machining materials = 0.02 D f = 0.02 (10) = 0.2 mm/revolution = 0.1 mm/edge (two cutting edges) feed rate = f • v = 3.342 mm/s Remember to check for speed and feed reductions found in the table at the bottom of Part 4 (referred to above). Note: The first column is Hole Depth / D! Example 16-9 Operation: Boring Geometry: 130-mm hole as-cast 138-mm hole finished Tool: Disposable carbide insert Material: Cast steel (ASTM A 27-77, 70-36) Find: Cutting speed, feed, power, and cutting force From metals handbook: minimum TS = 485 MPa Convert to HB: (485 N/mm2)(3) / (9.8 m/s2) = 148.5 kg/mm2 Boring is equivalent to turning; rough turning in this case since the depth of cut = (138 - 130) / 2 = 4 mm Z f 138 mm X 130 mm feed depth of cut = 4 mm FIG. 16-45 TABLE 16-5 From Fig. 16-45: for WC, vs = 1.8 m/s fs = 0.5 mm/rev From the text in Section 16-7-1 and the caption on Fig. 16-45: increase vs by 20% for a WC insert vs = (1.2)(1.8) = 2.16 m/s From Table 16-5: Zv = 1, Zf = 1 v = (2.16)(1) = 2.16 m/s f = (0.5)(1) = 0.5 mm/rev To find the power requirements, we first need to find the material removal rate, Vt [mm3/s]. Think of the process and what is happening to get this . Use the figure to help visualize the process. Depth of cut = 4 mm (x-dir) Feed = 0.5 mm/rev (z-dir) These two form a cross-sectional area that is revolved about the z-axis. This revolution correlates to the cutting speed. Vt = (4 mm)(0.5 mm/rev)(2160 mm/s) = 4320 mm 3/s (Hint: check for correct units) Z f 138 mm X 130 mm feed depth of cut = 4 mm Find the adjusted specific cutting energy, E From Fig. 16-1: E1 = 1.4 Ws/mm3 a = 0.3 href = 1mm (remember to convert to the English units when necessary) h = undeformed chip thickness = f cos C s (assume Cs = 0 for this problem, usually assume Cs = 10 - 15 when not given) E = (1.4)[(0.5)/(1)]-0.3 = 1.72 W s/mm3 P = E Vt / = (1.72)(4320) / (0.7) = 10.64 kW ( = machine tool efficiency; assume = 0.7) Cutting force, Pc = P / v = 10,640/2.16 = 4.92 kN Hints for Milling Obtain vs & fs from Fig. 16-45 or 16-46 – Note: These values are for a single cutting edge tool. Also remember to adjust the speed and feed according to the captions and text in Section 16-7-1. Obtain Zv & Zf from Table 16-5 – Note: Zf is also given per tooth; to obtain feed per revolution, multiply by the number of cutting edges. i.e. f = fs Zf N, where N = # teeth To calculate the cutting time, use the feed rate times the distance of the cut. When finding the distance, start and finish the tool ¼ of its radius from the workpiece ...
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• Fall '06
• El-Gizawy
• Trigraph, zf

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