12_ch 10 Mechanical Design budynas_SM_ch10

# 12_ch 10 Mechanical Design budynas_SM_ch10 - S sy / n )( π...

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272 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 10-17 Given: A232 (Cr-V steel), SQ&GRD ends, d = 4 . 3mm, OD = 76 . 2mm, L 0 = 228 . 6mm, N t = 8 turns. Table 10-4: A = 2005 MPa · mm m , m = 0 . 168 Table 10-5: G = 77 . 2GPa D = OD d = 76 . 2 4 . 3 = 71 . 9mm C = D / d = 71 . 9 / 4 . 3 = 16 . 72 (large) K B = 4(16 . 72) + 2 4(16 . 72) 3 = 1 . 078 N a = N t 2 = 8 2 = 6 turns S ut = 2005 (4 . 3) 0 . 168 = 1569 MPa Table 10-6: S sy = 0 . 50(1569) = 784 . 5MPa k = d 4 G 8 D 3 N a = (4 . 3) 4 (77 . 2) 8(71 . 9) 3 (6) ± (10 3 ) 4 (10 9 ) (10 3 ) 3 ² = 0 . 001 479(10 6 ) = 1479 N/m or 1 . 479 N/mm L s = dN t = 4 . 3(8) = 34 . 4mm F s = ky s y s = L 0 L s = 228 . 6 34 . 4 = 194 . 2mm τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 078 ± 8(1 . 479)(194 . 2)(71 . 9) π (4 . 3) 3 ² = 713 . 0MPa (1) τ s < S sy , that is, 713 . 0 < 784 . 5 ; the spring is solid safe. With n s = 1 . 2 Eq. (1) becomes y ± s = (
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Unformatted text preview: S sy / n )( π d 3 ) 8 K B kD = (784 . 5 / 1 . 2)( π )(4 . 3) 3 8(1 . 078)(1 . 479)(71 . 9) = 178 . 1 mm L ± = L s + y ± s = 34 . 4 + 178 . 1 = 212 . 5 mm Wind the spring to a free length of L ± = 212 . 5 mm. Ans. 10-18 For the wire diameter analyzed, G = 11 . 75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts ( a ) and ( b ). For N a , k = 20 / 2 = 10 lbf/in....
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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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