hmwk_3_Problem_solutions

# hmwk_3_Problem_solutions - Problems 21.1 In an orthogonal...

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Problems 21.1 In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation. Solution : (a) r = t o / t c = 0.30/0.65 = 0.4615 φ = tan -1 (0.4615 cos 15/(1 - 0.4615 sin 15)) = tan -1 (0.5062) = 26.85 ° (b) Shear strain γ = cot 26.85 + tan (26.85 - 15) = 1.975 + 0.210 = 2.185 21.2 In Problem 21.1, suppose the rake angle were changed to α = 0°. Assuming that the friction angle remains the same, determine (a) the shear plane angle, (b) the chip thickness, and (c) the shear strain for the operation. Solution : From Problem 21.1, α = 15° and φ = 26.85°. Using the Merchant Equation, Eq. (21.16): φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 15 – 2(26.85) = 51.3° Now, with α = 0 and β remaining the same at 51.3°, φ = 45 + 0/2 – 51.3/2 = 19.35 ° (b) Chip thickness at α = 0: t c = t o /tan φ = 0.30/tan 19.35 = 0.854 mm (c) Shear strain γ = cot 19.35 + tan (19.35 - 0) = 2.848 + 0.351 = 3.199 21.5 The cutting force and thrust force in an orthogonal cutting operation are 1470 N and 1589 N, respectively. The rake angle = 5°, the width of the cut = 5.0 mm, the chip thickness before the cut = 0.6, and the chip thickness ratio = 0.38. Determine (a) the shear strength of the work material and (b) the coefficient of friction in the operation. Solution : (a) φ = tan -1 (0.38 cos 5/(1 - 0.38 sin 5)) = tan -1 (0.3916) = 21.38° F s = 1470 cos 21.38 – 1589 sin 21.38 = 789.3 N A s = (0.6)(5.0)/sin 21.38 = 3.0/.3646 = 8.23 mm 2 S = 789.3/8.23 = 95.9 N/mm 2 = 95.9 MPa (b) φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 5 – 2(21.38) = 52.24° μ = tan 52.24 = 1.291 21.9 In an orthogonal cutting operation, the rake angle = -5°, chip thickness before the cut = 0.2 mm and width of cut = 4.0 mm. The chip ratio = 0.4. Determine (a) the chip thickness after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain. Solution : (a) r = t o / t c , t c = t o / r = 0.2/.4 = 0.5 mm (b) φ = tan -1 (0.4 cos(–5)/(1 - 0.4 sin(–5))) = tan -1 (0.3851) = 21.1 ° (c) β = 2(45) + α - 2(φ) = 90 + (-5) - 2(21.8) = 42.9 ° (d) μ = tan 42.9 = 0.93 (e) γ = cot 31.8 + tan(31.8 - 15) = 2.597 + 0.489 = 3.09 21.12 A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 lb/in 2 and a shear strength of 45,000 lb/in 2 . The diameter is reduced using a turning operation at a cutting speed of 400 ft/min. The feed is 0.011 in/rev and the depth of cut is 0.120 in. The rake angle on the tool in the direction of chip flow is 13°. The cutting conditions result in a chip ratio of 0.52. Using the orthogonal model as an approximation of turning, determine

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(a) the shear plane angle, (b) shear force, (c) cutting force and feed force, and (d) coefficient of friction between the tool and chip. Solution
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## This note was uploaded on 12/07/2009 for the course E ISE 515 taught by Professor Olaharryson during the Fall '09 term at N.C. State.

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hmwk_3_Problem_solutions - Problems 21.1 In an orthogonal...

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