Budynas ch14

# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 14 14-1 d = N P = 22 6 = 3 . 667 in Table 14-2: Y = 0 . 331 V = π dn 12 = π (3 . 667)(1200) 12 = 1152 ft/min Eq. (14-4 b ): K v = 1200 + 1152 1200 = 1 . 96 W t = T d / 2 = 63 025 H nd / 2 = 63 025(15) 1200(3 . 667 / 2) = 429 . 7 lbf Eq. (14-7): σ = K v W t P FY = 1 . 96(429 . 7)(6) 2(0 . 331) = 7633 psi = 7 . 63 kpsi Ans. 14-2 d = 16 12 = 1 . 333 in, Y = 0 . 296 V = π (1 . 333)(700) 12 = 244 . 3 ft/min Eq. (14-4 b ): K v = 1200 + 244 . 3 1200 = 1 . 204 W t = 63 025 H nd / 2 = 63 025(1 . 5) 700(1 . 333 / 2) = 202 . 6 lbf Eq. (14-7): σ = K v W t P FY = 1 . 204(202 . 6)(12) 0 . 75(0 . 296) = 13 185 psi = 13 . 2 kpsi Ans. 14-3 d = mN = 1 . 25(18) = 22 . 5 mm, Y = 0 . 309 V = π (22 . 5)(10 3 )(1800) 60 = 2 . 121 m/s Eq. (14-6 b ): K v = 6 . 1 + 2 . 121 6 . 1 = 1 . 348 W t = 60 H π dn = 60(0 . 5)(10 3 ) π (22 . 5)(10 3 )(1800) = 235 . 8 N Eq. (14-8): σ = K v W t FmY = 1 . 348(235 . 8) 12(1 . 25)(0 . 309) = 68 . 6 MPa Ans.

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350 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 14-4 d = 5(15) = 75 mm, Y = 0 . 290 V = π (75)(10 3 )(200) 60 = 0 . 7854 m/s Assume steel and apply Eq. (14-6 b ): K v = 6 . 1 + 0 . 7854 6 . 1 = 1 . 129 W t = 60 H π dn = 60(5)(10 3 ) π (75)(10 3 )(200) = 6366 N Eq. (14-8): σ = K v W t FmY = 1 . 129(6366) 60(5)(0 . 290) = 82 . 6 MPa Ans. 14-5 d = 1(16) = 16 mm, Y = 0 . 296 V = π (16)(10 3 )(400) 60 = 0 . 335 m/s Assume steel and apply Eq. (14-6 b ): K v = 6 . 1 + 0 . 335 6 . 1 = 1 . 055 W t = 60 H π dn = 60(0 . 15)(10 3 ) π (16)(10 3 )(400) = 447 . 6 N Eq. (14-8): F = K v W t σ mY = 1 . 055(447 . 6) 150(1)(0 . 296) = 10 . 6 mm From Table A-17, use F = 11 mm Ans. 14-6 d = 1 . 5(17) = 25 . 5 mm, Y = 0 . 303 V = π (25 . 5)(10 3 )(400) 60 = 0 . 534 m/s Eq. (14-6 b ): K v = 6 . 1 + 0 . 534 6 . 1 = 1 . 088 W t = 60 H π dn = 60(0 . 25)(10 3 ) π (25 . 5)(10 3 )(400) = 468 N Eq. (14-8): F = K v W t σ mY = 1 . 088(468) 75(1 . 5)(0 . 303) = 14 . 9 mm Use F = 15 mm Ans.
Chapter 14 351 14-7 d = 24 5 = 4 . 8 in, Y = 0 . 337 V = π (4 . 8)(50) 12 = 62 . 83 ft/min Eq. (14-4 b ): K v = 1200 + 62 . 83 1200 = 1 . 052 W t = 63 025 H nd / 2 = 63 025(6) 50(4 . 8 / 2) = 3151 lbf Eq. (14-7): F = K v W t P σ Y = 1 . 052(3151)(5) 20(10 3 )(0 . 337) = 2 . 46 in Use F = 2 . 5 in Ans. 14-8 d = 16 5 = 3 . 2 in, Y = 0 . 296 V = π (3 . 2)(600) 12 = 502 . 7 ft/min Eq. (14-4 b ): K v = 1200 + 502 . 7 1200 = 1 . 419 W t = 63 025(15) 600(3 . 2 / 2) = 984 . 8 lbf Eq. (14-7): F = K v W t P σ Y = 1 . 419(984 . 8)(5) 10(10 3 )(0 . 296) = 2 . 38 in Use F = 2 . 5 in Ans. 14-9 Try P = 8 which gives d = 18 / 8 = 2 . 25 in and Y = 0 . 309 . V = π (2 . 25)(600) 12 = 353 . 4 ft/min Eq. (14-4 b ): K v = 1200 + 353 . 4 1200 = 1 . 295 W t = 63 025(2 . 5) 600(2 . 25 / 2) = 233 . 4 lbf Eq. (14-7): F = K v W t P σ Y = 1 . 295(233 . 4)(8) 10(10 3 )(0 . 309) = 0 . 783 in

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Using coarse integer pitches from Table 13-2, the following table is formed. P d V K v W t F 2 9.000 1413.717 2.178 58.356 0.082 3 6.000 942.478 1.785 87.535 0.152 4 4.500 706.858 1.589 116.713 0.240 6 3.000 471.239 1.393 175.069 0.473 8 2.250 353.429 1.295 233.426 0.782 10 1.800 282.743 1.236 291.782 1.167 12 1.500 235.619 1.196 350.139 1.627 16 1.125 176.715 1.147 466.852 2.773 Other considerations may dictate the selection. Good candidates are P = 8 ( F = 7 / 8 in) and P = 10 ( F = 1 . 25 in) . Ans. 14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0 . 309 . V = π (36)(10 3 )(900) 60 = 1 . 696 m/s Eq. (14-6 b ): K v = 6 . 1 + 1 . 696 6 . 1 = 1 . 278 W t = 60(1 . 5)(10 3 ) π (36)(10 3 )(900) = 884 N Eq. (14-8): F = 1 . 278(884) 75(2)(0 . 309) = 24 . 4 mm Using the preferred module sizes from Table 13-2: m d V K v W t F 1.00 18.0 0.848 1.139 1768.388 86.917 1.25 22.5 1.060 1.174 1414.711 57.324 1.50 27.0 1.272 1.209 1178.926 40.987 2.00 36.0 1.696 1.278 884.194 24.382 3.00 54.0 2.545 1.417 589.463 12.015 4.00 72.0 3.393 1.556 442.097 7.422 5.00 90.0 4.241 1.695 353.678 5.174 6.00 108.0 5.089 1.834 294.731 3.888 8.00 144.0 6.786 2.112 221.049 2.519 10.00 180.0 8.482 2.391 176.839 1.824 12.00 216.0 10.179 2.669 147.366 1.414 16.00 288.0 13.572 3.225 110.524 0.961 20.00 360.0 16.965 3.781 88.419 0.721 25.00 450.0 21.206 4.476 70.736 0.547 32.00 576.0 27.143 5.450 55.262 0.406 40.00 720.0 33.929 6.562 44.210 0.313 50.00 900.0 42.412 7.953 35.368 0.243 Other design considerations may dictate the size selection. For the present design, m = 2 mm ( F = 25 mm) is a good selection. Ans. 352 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Chapter 14 353 14-11 d P = 22 6 = 3 . 667 in, d G = 60 6 = 10 in V = π (3 . 667)(1200) 12 = 1152 ft/min Eq. (14-4 b ): K v = 1200 + 1152 1200 = 1 . 96 W t = 63 025(15) 1200(3 . 667 / 2) = 429 . 7 lbf Table 14-8: C p = 2100 psi [Note: using Eq. (14-13) can result in wide variation in C p due to wide variation in cast iron properties] Eq. (14-12): r 1 = 3 . 667 sin 20° 2 = 0 . 627 in, r 2 = 10 sin 20° 2 = 1 . 710 in Eq. (14-14): σ C = − C p K v W t

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