HW-10-17-08s-Ch19

HW-10-17-08s-Ch19 - IE 215 Solutions for Problems due Oct...

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IE 215 Solutions for Problems due Oct 17, 2008 (Ch 19) 19.15 A cylindrical workpart with D = 2.5 in and h = 2.5 in is upset forged in an open die to a height = 1.5 in. Coefficient of friction at the die-work interface = 0.10. The work material has a flow curve defined by: K = 40,000 lb/in 2 and n = 0.15. Determine the instantaneous force in the operation (a) just as the yield point is reached (yield at strain = 0.002), (b) at height h = 2.3 in, (c) h = 1.9 in, and (d) h = 1.5 in. Solution: (a) V = π D 2 L /4 = π (2.5) 2 (2.5)/4 = 12.273 in 3 Given ε = 0.002, Y f = 40,000(0.002) 0.15 = 15,748 lb/in 2 and h = 2.5 - 2.5(0.002) = 2.495 A = V/h = 12.273/2.495 = 4.92 in 2 K f = 1 + 0.4(0.1)(2.5)/2.495 = 1.04 F = 1.04(15,748)(4.92) = 80,579 lb (b) Given h = 2.3, = ln(2.5/2.3) = ln 1.087 = 0.0834 Y f = 40,000(0.0834) 0.15 = 27,556 lb/in 2 V = 12.273 in 3 from part (a) above. At h = 2.3, A = V/h = 12.273/2.3 = 5.34 in 2 Corresponding D = 2.61 (from A = π D 2 /4) K f = 1 + 0.4(0.1)(2.61)/2.3 = 1.045 F = 1.045(27,556)(4.34) = 153,822 lb (c) Given h = 1.9, = ln(2.5/1.9) = ln 1.316 = 0.274 Y f = 40,000(0.274) 0.15 = 32,948 lb/in 2 V = 12.273 in 3 from part (a) above. At h = 1.9, A = V/h = 12.273/1.9 = 6.46 in 2 Corresponding D = 2.87 (from A = π D 2 /4) K f = 1 + 0.4(0.1)(2.87)/1.9 = 1.060 F = 1.060(32,948)(6.46) = 225,695 lb (d) Given h = 1.5, = ln(2.5/1.5) = ln 1.667 = 0.511 Y f = 40,000(0.511) 0.15 = 36,166 lb/in 2 V = 12.273 in 3 from part (a) above. At h = 1.5, A = V/h = 12.273/1.5 = 8.182 in 2 Corresponding D = 3.23 (from A = π D 2 /4) K f = 1 + 0.4(0.1)(3.23)/1.5 = 1.086 F = 1.086(36,166)(8.182) = 321,379 lb 19.17 A cold heading operation is performed to produce the head on a steel nail. The strength coefficient for
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HW-10-17-08s-Ch19 - IE 215 Solutions for Problems due Oct...

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