hw6_sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II SOLUTIONS FOR HOMEWORK NO. 6 Problem 1 (40 points) Part A: Results are shown in Figure 1. Matlab scripts for Part A and B are attached. Graphically estimated fitting constants are: A =9 . 5737 × 10 12 [ m/cycle ( MP a m ) m ], m =3 . 17144 Part B: Least squares fitted constants are: A =8 . 8497 × 10 12 [ m/cycle ( MP a m ) m ], m = 3 . 21473 Part C: With the data for Part B, the“Paris-law” is given as following: da (∆ K ) 3 . 21473 = A (∆ K I ) m =8 . 8497 × 10 12 (1) dN “Paris-law” can be rewritten as : da K I =∆ a 0 ( ) m (2) dN K I 0 where a 0 ( da dN ) 0 is the corresponding reference crack growth rate, and K I 0 is a reference crack driving force. K I 0 and a 0 are the values of any point on the power law growth rate curve. If we choose K I 0 =6 . 0 MP a m , the corresponding a 0 = A × K m = I 0 2 . 8085 × 10 9 m/cycle ,and m does not change. 1
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da/dN [m/cycle] F igure 1: da/dN vs . DK I 10± 10± 10± 10± DK [MP a *m 0.5 ]
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% Problem 1 da_dN=1e-6*[4.26,9.12,17.5,35.1]'; % unit (mm/cycle) deltaKI=[6.84,8.76,10.35,13.3]';
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.002 taught by Professor Davidparks during the Spring '04 term at MIT.

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hw6_sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT...

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