11_ch 10 Mechanical Design budynas_SM_ch10

11_ch 10 Mechanical Design budynas_SM_ch10 -...

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Chapter 10 271 The spring is solid safe. With n s = 1 . 2, y ± s = ( S sy / n )( π d 3 ) 8 K B kD = (737 . 8 / 1 . 2)( π )(3 . 4) 3 8(1 . 095)(3 . 75)(47 . 4) = 48 . 76 mm L ± 0 = L s + y ± s = 17 . 85 + 48 . 76 = 66 . 61 mm Wind the spring to a free length of 66.61 mm. Ans. 10-16 Given: B159 (phosphor bronze), SQ&GRD ends, d = 3.7 mm, OD = 25.4 mm, L 0 = 95 . 3mm, N t = 13 turns. Table 10-4: A = 932 MPa · mm m , m = 0 . 064 Table 10-5: G = 41 . 4GPa D = OD d = 25 . 4 3 . 7 = 21 . 7mm C = D / d = 21 . 7 / 3 . 7 = 5 . 865 K B = 4(5 . 865) + 2 4(5 . 865) 3 = 1 . 244 N a = N t 2 = 13 2 = 11 turns S ut = 932 (3 . 7) 0 . 064 = 857 . 1MPa Table 10-6: S sy = 0 . 35(857 . 1) = 300 MPa k = d 4 G 8 D 3 N a = (3 . 7) 4 (41 . 4) 8(21 . 7) 3 (11) ± (10 3 ) 4 (10 9 ) (10 3 ) 3 ² = 0 . 008 629(10 6 ) = 8629 N/m or 8 . 629 N/mm L s = dN t = 3 . 7(13) = 48 . 1mm F s = ky s y s = L 0 L s = 95 . 3 48 . 1 = 47 . 2mm τ s = K B ± 8( ky s ) D π d 3 ²
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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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