18_ch 10 Mechanical Design budynas_SM_ch10

# 18_ch 10 Mechanical Design budynas_SM_ch10 - 10-27 As in...

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278 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 10-25 This is the same as Prob. 10-23 since S se = S sa = 35 kpsi. Therefore, design the spring using: A313, as wound, un-peened, squared and ground, d = 0.915 in, OD = 0.971 in, N t = 15 . 59 turns. 10-26 For the Gerber fatigue-failure criterion, S su = 0 . 67 S ut , S se = S sa 1 ( S sm / S su ) 2 , S sa = r 2 S 2 su 2 S se 1 + ± 1 + ² 2 S se rS su ³ 2 The equation for S sa is the basic difference. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. d = 0.105 d = 0.112 d = 0.105 d = 0.112 S ut 278.691 276.096 N a 8.915 6.190 S su 186.723 184.984 L s 1.146 0.917 S se 38.325 38.394 L 0 3.446 3.217 S sy 125.411 124.243 ( L 0 ) cr 6.630 8.160 S sa 34.658 34.652 K B 1.111 1.095 α 23.105 23.101 τ a 23.105 23.101 β 1.732 1.523 n f 1.500 1.500 C 12.004 13.851 τ s 70.855 70.844 D 1.260 1.551 n s 1.770 1.754 ID 1.155 1.439 f n 105.433 106.922 OD 1.365 1.663 fom 0.973 1.022 There are only slight changes in the results.
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Unformatted text preview: 10-27 As in Prob. 10-26, the basic change is S sa . For Goodman, S se = S sa 1 − ( S sm / S su ) Recalculate S sa with S sa = rS se S su rS su + S se Calculations for the last 2 diameters of Ex. 10-5 are given below. d = 0.105 d = 0.112 d = 0.105 d = 0.112 S ut 278.691 276.096 N a 9.153 6.353 S su 186.723 184.984 L s 1.171 0.936 S se 49.614 49.810 L 3.471 3.236 S sy 125.411 124.243 ( L ) cr 6.572 8.090 S sa 34.386 34.380 K B 1.112 1.096 α 22.924 22.920 τ a 22.924 22.920 β 1.732 1.523 n f 1.500 1.500 C 11.899 13.732 τ s 70.301 70.289 D 1.249 1.538 n s 1.784 1.768 ID 1.144 1.426 f n 104.509 106.000 OD 1.354 1.650 fom − 0.986 − 1.034 There are only slight differences in the results....
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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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