Machinery's Handbook, 31st Edition
Boring
1155
Example 1(b): A workpiece is AISI 4140 chromium-molybdenum steel, 220 HB; depth of cut d = 2.03mm; feed rate f = 0.20 mm/rev. Calculate the cutting force F in metric units. Power constant K p = 0.0328 kW/cm 3 /min. (from Table 1b, page 1147) Feed factor C = 1.08 (from Table 2, page 1148) Adjusted power constant, K pa = K p × C = 0.0328 × 1.08 ≅ 0.0354 kW/cm 3 /min. Calculating: F = 60,000 dfK pa = 60,000 × 2.03 × 0.20 × 0.0354 = 863.6 N Comparing the cutting forces calculations of Example 1(a) and 1(b), the cutting forces are approximately equivalent within about 2 percent. Moduli of Elasticity ( E ) of Boring Bar Materials.— Boring bar shanks are made of steels, tungsten-base metals, or cemented carbides. Most commonly used boring bar ma- terials are alloy steels. Some boring bar manufacturers use AISI 1144 free machining medium-carbon steel. Regardless of grades, all carbon and alloy steels have approxi- mately the same modulus of elasticity: A common mistake is to assume that a steel shank with a higher hardness, or one made from a higher quality of steel will deflect less. As can be seen from Equation (1), the mate- rial property that determines deflection is the modulus of elasticity. Hardness does not appear in this equation. Tungsten heavy alloys for boring bars, E = (45–48) × 106 psi (customary US units) and E = (31–33) × 10 4 N/mm 2 (metric units). Boring bars made of tungsten heavy alloys will deflect less than steel boring bars of the same diameter and overhang by 50 to 60 percent when cutting at the same depth of cut and feed rate. Cemented carbides for boring bars, E = (84–89) × 10 6 psi (customary US units) and E = (52–61) × 10 4 N/mm 2 (metric units). Boring bars made of cemented carbides provide minimum deflection because their moduli of elasticity are higher than those of steels and tungsten heavy alloys. Moment of Inertia ( I ) of a Boring Bar Cross-Sectional Area.— Moment of inertia is a property of areas. Since boring bars are available in a variety of diameters, it is important to calculate the moment of inertia of a bar cross section using appropriate formulas. Bor- ing bars are usually round with a solid or tubular cross section. Moments of inertia of solid or tubular cross-sectional areas are calculated by: (4a) (4b) where D is the diameter of the bar in inches or mm, D o is the outside diameter of the bar and D i is the inside diameter, in inches or mm. Example 2: A boring bar diameter is 1 inch. Moment of inertia I = π × (1) 4 /64 = 0.0491 in 4 . Moments of inertia of cross sections are given starting on page 241. Deflection ( y ) of the Boring Bar.— To calculate the deflection of the boring bar, it is necessary to enter one more data point in Equation (1). It is the unsupported length of the boring bar ( L ). Example , Boring Bar Deflection: The unsupported length of a 1-inch diameter boring bar is 4 inches. Using F calculated in Example 1(a) and I calculated in Example 2: Customary US units, E = 30 × 10 6 psi Metric units, E = 20.6 × 10 4 N/mm 2
(197.7)4 3 3 (30 × 10 6 ) 0.0491
FL 3 3 EI
y =
=
= 0.0028 inch
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