MAE455 hw 2 solution - MAE 4550 Intro to Composites Fall...

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MAE 4550 – Intro to Composites Fall 2009 HW 2 Steps for finding ρ and x 0 : 1) Arrange strengths (x) from least to greatest (MPa) 2) Calculate F i for each piece of data where F i = (i-0.5)/n 3) Plot x vs. Fi 4) Find x 0 from plot where F=0.632 5) Plot ln(x/x 0 ) vs. ln(-ln(1-Fi)) 6) Find ρ from slope of plot Carbon 10mm x (MPa) Fi ln(x/xo) ln(-ln(1-Fi) 2982 0.02 -0.27985 -3.90194 2992 0.06 -0.27651 -2.78263 3130 0.1 -0.23142 -2.25037 3160 0.14 -0.22188 -1.89165 3167 0.18 -0.21966 -1.61721 3233 0.22 -0.19904 -1.39247 3375 0.26 -0.15605 -1.2003 3427 0.3 -0.14076 -1.03093 3437 0.34 -0.13785 -0.87824 3457 0.38 -0.13205 -0.73807 3547 0.42 -0.10635 -0.60747 3560 0.46 -0.10269 -0.48421 3641 0.5 -0.08019 -0.36651 3698 0.54 -0.06466 -0.25292 3699 0.58 -0.06439 -0.14214 3870 0.62 -0.01919 -0.03295 3933 0.66 -0.00305 0.075858 3978 0.7 0.00833 0.185627 3989 0.74 0.011092 0.297935 3994 0.78 0.012344 0.41484 4033 0.82 0.022062 0.539296 4401 0.86 0.109383 0.676058 4447 0.9 0.119781 0.834032 4777 0.94 0.191364 1.034398 4789 0.98 0.193873 1.364055
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Plot of x vs. Fi y = 0.00055663x - 1.56432256 0 0.2 0.4 0.6 0.8 1 1.2 0 1000 2000 3000 4000 5000 6000 0.632=.00055663x-1.56432256 x 0 =3945 MPa Plot of ln(x/x 0 ) vs. ln(-ln(1-Fi)) y = 8.6767x + 0.0475 -5 -4 -3 -2 -1 0 1 2 3 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 ρ = 8.6767
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Carbon 100mm X (MPa) Fi ln(x/xo) ln(-ln(1-Fi)) 2101 0.02 -0.51714 -3.90194 2103 0.06 -0.51619 -2.78263 2231 0.1 -0.4571 -2.25037 2265 0.14 -0.44198 -1.89165 2346 0.18 -0.40684 -1.61721 2813 0.22 -0.2253 -1.39247 2982 0.26 -0.16696 -1.2003
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This note was uploaded on 02/01/2012 for the course M&AE 455 at Cornell University (Engineering School).

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MAE455 hw 2 solution - MAE 4550 Intro to Composites Fall...

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