4_ch 10 Mechanical Design budynas_SM_ch10

4_ch 10 Mechanical Design budynas_SM_ch10 - d = . 006 in,...

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264 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (10-19): 3 N a 15, N a = 7 . 5 O . K . y 1 = F 1 k = 75 2643(10 3 ) = 28 . 4mm ( y ) for yield = 81 . 12(1 . 2) 2643(10 3 ) = 36 . 8mm y s = 81 . 12 2643(10 3 ) = 30 . 69 mm ξ = ( y ) for yield y 1 1 = 36 . 8 28 . 4 1 = 0 . 296 Eq. (10-20): ξ 0 . 15, ξ = 0 . 296 O . K . Table 10-6: S sy = 0 . 45 S ut O . K . As-wound τ s = K B ± 8 F s D π d 3 ² = 1 . 135 ³ 8(81 . 12)(20) π (2) 3 ´³ 10 3 (10 3 ) 3 (10 6 ) ´ = 586 MPa Eq. (10-21): S sy τ s = 703 . 4 586 = 1 . 2 O . K . (Basis for Prob. 10-5 solution) Table 10-1: L s = N t d = 8 . 5(2) = 17 mm L 0 = F s k + L s = 81 . 12 2 . 643 + 17 = 47 . 7mm 2 . 63 D α = 2 . 63(20) 0 . 5 = 105 . 2mm ( L 0 ) cr L 0 = 105 . 2 47 . 7 = 2 . 21 which is less than 5. Operate over a rod? Not O.K. Plain and ground ends have a poor eccentric footprint. Ans. 10-7 Given: A228 (music wire), SQ&GRD ends,
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Unformatted text preview: d = . 006 in, OD = . 036 in, L = . 63 in, N t = 40 turns. Table 10-4: A = 201 kpsi in m , m = . 145 D = OD d = . 036 . 006 = . 030 in C = D / d = . 030 / . 006 = 5 K B = 4(5) + 2 4(5) 3 = 1 . 294 Table 10-1: N a = N t 2 = 40 2 = 38 turns S ut = 201 (0 . 006) . 145 = 422 . 1 kpsi S sy = . 45(422 . 1) = 189 . 9 kpsi k = Gd 4 8 D 3 N a = 12(10 6 )(0 . 006) 4 8(0 . 030) 3 (38) = 1 . 895 lbf/in...
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