24_ch 13 Mechanical Design budynas_SM_ch13

# 24_ch 13 Mechanical Design budynas_SM_ch13 - W = − 637 i...

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356 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Solving gives F x C = 1987 + 1188 1610 = 1565 lbf F y C = 746 499 + 425 = 672 lbf F z D =− 532 + 686 = 154 lbf Ans . 13-41 V W = π d W n W 60 = π (0 . 100)(600) 60 = π m/s W Wt = H V W = 2000 π = 637 N L = p x N W = 25(1) = 25 mm λ = tan 1 L π d W = tan 1 25 π (100) = 4 . 550° lead angle W = W Wt cos φ n sin λ + f cos λ V S = V W cos λ = π cos 4 . 550° = 3 . 152 m/s In ft/min: V S = 3 . 28(3 . 152) = 10 . 33 ft/s = 620 ft/min Use f = 0 . 043 from curve A of Fig. 13-42. Then from the ±rst of Eq. (13-43) W = 637 cos 14 . 5°(sin 4 . 55°) + 0 . 043 cos 4 . 55° = 5323 N W y = W sin φ n = 5323 sin 14 . = 1333 N W z = 5323[cos 14 . 5°(cos 4 . 55°) 0 . 043 sin 4 . 55°] = 5119 N The force acting against the worm is
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Unformatted text preview: W = − 637 i + 1333 j + 5119 k N Thus A is the thrust bearing. Ans. R AG = − . 05 j − . 10 k , R AB = − . 20 k ± M A = R AG × W + R AB × F B + T = R AG × W = − 122 . 6 i + 63 . 7 j − 31 . 85 k R AB × F B = . 2 F y B i − . 2 F x B j Substituting and solving gives T = 31 . 85 N · m Ans . F x B = 318 . 5 N, F y B = 613 N So F B = 318 . 5 i + 613 j N Ans . B G A x y z Worm shaft diagram 100 100 W r W t W a 50...
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