9_ch 10 Mechanical Design budynas_SM_ch10

9_ch 10 Mechanical Design budynas_SM_ch10 - mm m , m = ....

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Chapter 10 269 Table 10-6: S sy = 0 . 35(2361 . 5) = 826 . 5MPa k = d 4 G 8 D 3 N a = (0 . 2) 4 (69 . 0) 8(0 . 71) 3 (38) ± (10 3 ) 4 (10 9 ) (10 3 ) 3 ² = 1 . 0147(10 3 )(10 6 ) = 1014 . 7 N/m or 1 . 0147 N/mm L s = dN t = 0 . 2(40) = 8mm F s = ky s y s = L 0 L s = 15 . 9 8 = 7 . 9 τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 446 ± 8(1 . 0147)(7 . 9)(0 . 71) π (0 . 2) 3 ²± 10 3 (10 3 )(10 3 ) (10 3 ) 3 ² = 2620(1) = 2620 MPa (1) τ s > S sy , that is,2620 > 826 . 5 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 = (826 . 5 / 1 . 2)( π )(0 . 2) 3 8(1 . 446)(1 . 0147)(0 . 71) = 2 . 08 mm L ± 0 = L s + y ± s = 8 . 0 + 2 . 08 = 10 . 08 mm Wind the spring to a free length of 10.08 mm. This only addresses the solid-safe criteria. There are additional problems. Ans. 10-14 Given: A228 (music wire), SQ&GRD ends, d = 1mm, OD = 6 . 10 mm, L 0 = 19 . 1mm , N t = 10 . 4 turns. Table 10-4: A = 2211 MPa
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Unformatted text preview: mm m , m = . 145 Table 10-5: G = 81 . 7 GPa D = OD d = 6 . 10 1 = 5 . 1 mm C = D / d = 5 . 1 / 1 = 5 . 1 N a = N t 2 = 10 . 4 2 = 8 . 4 turns K B = 4(5 . 1) + 2 4(5 . 1) 3 = 1 . 287 S ut = 2211 (1) . 145 = 2211 MPa Table 10-6: S sy = . 45(2211) = 995 MPa k = d 4 G 8 D 3 N a = (1) 4 (81 . 7) 8(5 . 1) 3 (8 . 4) (10 3 ) 4 (10 9 ) (10 3 ) 3 = . 009 165(10 6 ) = 9165 N/m or 9 . 165 N/mm L s = dN t = 1(10 . 4) = 10 . 4 mm F s = ky s...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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