# hw_09_sol - MechEng 382 Winter 2009 Homework#9 Solutions...

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MechEng 382 Winter 2009 Homework #9 Solutions 1. [ Dowling , Prob. 11.1] Estimate the constants C and m for the straight line portion of the data of Fig. 11.3. Solution Using a pair of points that span the straight-line region in Fig. 3 as indicated, ± Δ K, ∂a ∂N ² 1 (20MPa m , 2 . 5 × 10 - 5 mm / cyc ) , ± Δ K, ∂a ∂N ² 2 (30MPa m , 9 . 5 × 10 - 5 mm / cyc ); these two equations give the requisite information to approximate the parameters C and m : ∂a/∂N | 2 ∂a/∂N | 1 = C Δ K m 2 C Δ K m 1 = ± Δ K 2 Δ K 1 ² m (11.10) = m = log ∂a/∂N | 2 - log ∂a/∂N | 1 log Δ K 2 - log Δ K 1 = log 9 . 5 × 10 - 5 - log 2 . 5 × 10 - 5 log 30 - log 20 = 3 . 29 ; = C = ∂a/∂N | 1 Δ K m 1 = 2 . 5 × 10 - 5 20 3 . 29 = 1 . 3 × 10 - 9 mm / cyc / (MPa m) m . 1

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MechEng 382 Winter 2009 Homework #9 Solutions Figure 1: Data for Prob. 1 [ Dowling , Fig. 11.3]. 2
MechEng 382 Winter 2009 Homework #9 Solutions 2. [ Dowling , Prob. 11.4] Representative data points are given in Table P11.4 (Fig. 3 ) from the results of tests at R = 0 . 1 on 2124-T851 aluminum. Figure 2: Data for Prob. 2 , after [ Dowling , Table P11.4]. (a) Plot these points on log-log coordinates, and obtain approximate values of constants C and m for Eq. 11.10. Solution The tabulated values are plotted in Fig. 3 . Using the endpoints, which represent the line well, ± Δ K, ∂a ∂N ² 1 (2 . 99MPa m , 1 . 26 × 10 - 6 mm / cyc ) , ± Δ K, ∂a ∂N ² 2 (15 . 9MPa m , 1 . 77 × 10 - 4 mm / cyc ); C and m follow as above: ∂a/∂N | 2 ∂a/∂N | 1 = C Δ K m 2 C Δ K m 1 = ± Δ K 2 Δ K 1 ² m (11.10) = m = log ∂a/∂N | 2 - log ∂a/∂N | 1 log Δ K 2 - log Δ K 1 = log 1 . 77 × 10 - 4 - log 1 . 26 × 10 - 6 log 15 . 9 - log 2 . 99 = 2 . 96 ; = C = ∂a/∂N | 1 Δ K m 1 = 1 . 26 × 10 - 6 2 . 99 2 . 96 = 4 . 93 × 10 - 8 mm / cyc / (MPa m) m . Remark The approximations for C and m will vary considerably. (b) Use a log-log least-squares ﬁt to obtain reﬁned values of C and m . Solution The ﬁt is plotted with the data in Fig. 3 . The Paris law can be manipulated: da dN = C Δ K m ⇐⇒ log ± da dN ² | {z } y = m |{z} m log Δ K | {z } x +log C | {z } b . 3

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MechEng 382 Winter 2009 Homework #9 Solutions A least-squares linear regression of this modiﬁed equation to the data gives slope m = 2 . 90 and intercept b = - 7 . 29, so m = m = 2 . 90 = 2 . 90 C = 10 b = 10 - 7 . 29 = 5 . 18 × 10 - 8 mm / cycle / ( MPa m ) m 10 0 10 1 10 2 10 -6 10 -5 10 -4 10 -3 Δ K (MPa m) da dN (mm/cycle) Data Fit Figure 3: Curve-ﬁtted fatigue crack growth, Prob. 2 . 4
MechEng 382 Winter 2009 Homework #9 Solutions 3. [ Dowling , Prob. 11.37] Consider the center-cracked plate of AISI 4340 steel of Ex. 11.4, with the same initial crack size of a i = 1mm , but let the cyclic forces be half as large—that is, P min = 40 and P max = 120kN . (a)

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## This note was uploaded on 09/28/2010 for the course MECHENG 382 taught by Professor Thouless during the Winter '08 term at University of Michigan.

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hw_09_sol - MechEng 382 Winter 2009 Homework#9 Solutions...

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