**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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