3_ch 10 Mechanical Design budynas_SM_ch10

3_ch 10 Mechanical Design budynas_SM_ch10 - . (e) Table...

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Chapter 10 263 10-5 Static service spring with: HD steel wire, d = 2mm, OD = 22 mm, N t = 8 . 5 turns plain and ground ends. Preliminaries Table 10-5: A = 1783 MPa · mm m , m = 0 . 190 Eq. (10-14): S ut = 1783 (2) 0 . 190 = 1563 MPa Table 10-6: S sy = 0 . 45(1563) = 703 . 4MPa Then, D = OD d = 22 2 = 20 mm C = 20 / 2 = 10 K B = 4 C + 2 4 C 3 = 4(10) + 2 4(10) 3 = 1 . 135 N a = 8 . 5 1 = 7 . 5 turns L s = 2(8 . 5) = 17 mm Eq. (10-21): Use n s = 1 . 2 for solid-safe property. F s = π d 3 S sy / n s 8 K B D = π (2) 3 (703 . 4 / 1 . 2) 8(1 . 135)(20) ± (10 3 ) 3 (10 6 ) 10 3 ² = 81 . 12 N k = d 4 G 8 D 3 N a = (2) 4 (79 . 3) 8(20) 3 (7 . 5) ± (10 3 ) 4 (10 9 ) (10 3 ) 3 ² = 0 . 002 643(10 6 ) = 2643 N/m y s = F s k = 81 . 12 2643(10 3 ) = 30 . 69 mm (a) L 0 = y + L s = 30 . 69 + 17 = 47 . 7mm Ans . (b) Table 10-1: p = L 0 N t = 47 . 7 8 . 5 = 5 . 61 mm Ans . (c) F s = 81 . 12 N (from above) Ans . (d) k = 2643 N/m (from above) Ans
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Unformatted text preview: . (e) Table 10-2 and Eq. (10-13): ( L ) cr = 2 . 63 D = 2 . 63(20) . 5 = 105 . 2 mm ( L ) cr / L = 105 . 2 / 47 . 7 = 2 . 21 This is less than 5. Operate over a rod? Plain and ground ends have a poor eccentric footprint. Ans. 10-6 Referring to Prob. 10-5 solution: C = 10, N a = 7 . 5, k = 2643 N/m, d = 2 mm, D = 20 mm, F s = 81 . 12 N and N t = 8 . 5 turns . Eq. (10-18): 4 C 12, C = 10 O . K ....
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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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