10_ch 10 Mechanical Design budynas_SM_ch10

# 10_ch 10 Mechanical Design budynas_SM_ch10 - Table 10-6 S...

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270 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design y s = L 0 L s = 19 . 1 10 . 4 = 8 . 7mm τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 287 ± 8(9 . 165)(8 . 7)(5 . 1) π (1) 3 ² = 1333 MPa (1) τ s > S sy , that is, 1333 > 995 MPa; the spring is not solid safe. Solve Eq. (1) for y s giving y ± s = ( S sy / n )( π d 3 ) 8 K B kD = (995 / 1 . 2)( π )(1) 3 8(1 . 287)(9 . 165)(5 . 1) = 5 . 43 mm L ± 0 = L s + y ± s = 10 . 4 + 5 . 43 = 15 . 83 mm Wind the spring to a free length of 15.83 mm. Ans. 10-15 d = 3.4 mm, OD = 50.8 mm, L 0 = 74.6 mm, N t = 5 . 25 . Table 10-4: A = 1855 MPa · mm m , m = 0 . 187 Table 10-5: G = 77 . 2GPa D = OD d = 50 . 8 3 . 4 = 47 . 4mm C = D / d = 47 . 4 / 3 . 4 = 13 . 94 (large) N a = N t 2 = 5 . 25 2 = 3 . 25 turns K B = 4(13 . 94) + 2 4(13 . 94) 3 = 1 . 095 S ut = 1855 (3 . 4) 0 . 187 = 1476 MPa
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Unformatted text preview: Table 10-6: S sy = . 50(1476) = 737 . 8 MPa k = d 4 G 8 D 3 N a = (3 . 4) 4 (77 . 2) 8(47 . 4) 3 (3 . 25) ± (10 − 3 ) 4 (10 9 ) (10 − 3 ) 3 ² = . 003 75(10 6 ) = 3750 N/m or 3 . 750 N/mm L s = dN t = 3 . 4(5 . 25) = 17 . 85 F s = ky s y s = L − L s = 74 . 6 − 17 . 85 = 56 . 75 mm τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 095 ± 8(3 . 750)(56 . 75)(47 . 4) π (3 . 4) 3 ² = 720 . 2 MPa (1) τ s < S sy , that is, 720 . 2 < 737 . 8 MPa...
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