4020-F2011-HW12Solutions - 4l 3 E " t 3 = 1 32 # 1 4...

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1 MSE 4020 HW12-SOLUTION 11/16/2011 Problem 1 (a) ! max = P " l 3 3E " I (b) I = b ! h 3 12 = 1 12 ! 0.01 ! 0.01 3 = 8.333 ! 10 " 10 m 4 From the figure in the handout, we can estimate, E = 2.2 GPa ! " max = P # l 3 3E # I = 0.98 # 0.5 3 3 # 2.2 # 10 9 # 8.333 # 10 $ 10 = 0.02227 m = 2.227 cm (c) 5 years in log units in seconds is logt 2 = 8.2 ! max ' = 0.06 = P " l 3 3E " I # E = 0.98 " 0.5 3 0.06 " 3 " 8.333 " 10 $ 10 = 0.8 " 10 10 dyn cm 2 From the figure in the handout, we can estimate, at 100°C, logt 1 = 4.2 Thus, shift factor can be obtained as, log(a(T)) = logt 2 ! logt 1 = 8.2 ! 4.2 = 4.0 From given data listed in the table below: x y (which is the shift factor) 23°C 4.8 60°C 2.8 Assuming y is linearly dependent on x, then, when y = 4.0, x = 38°C
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2 Problem 2 At T = 40°C, log(a(40 ° C)) = 7.65 t 2 = 10 years logt 2 = 4.94 ! logt 1 = logt 2 " log(a(40 ° C)) = 4.94 " 7.65 = " 2.7075 which is corresponding time at master curve (115°C) in log units. Thus, the relaxation modulus at 115°C is log(E r ) = 8.75 ! E r = 8156.2 psi From the equation in Problem 1, ! max = P " l 3 3E " I = P " l 3 3E " b " t 3 12 = P b "
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Unformatted text preview: 4l 3 E " t 3 = 1 32 # 1 4 " 4 1 2 $ % & ' ( ) 3 8156.2 " t 3 = 1 32 # t = 0.07886 in 3 Problem 3 From model shown above, the strain can be written in term of time as, ! = " 1 E g + 1 E s # 1 E s exp # E s $ t % & ' ( ) * % & ' ' ( ) * * where E g ,E s and ! are to be fitted by Matlab script attached. E g = 4.16666 ! 10 9 E s = 3.5861 ! 10 9 " = 2.68 ! 10 9 4 Problem 4 (a) From the sketch above, ! = k 2 + " 2 # 2 $ % $ = ! k 2 + " 2 # 2 # = 2 &' f % $ = ! k 2 + 4 " 2 '& 2 ' f 2 = 10 3 10 7 ( ) 2 + 4 ' 5 ' 10 3 ( ) 2 '& 2 ' 10 2 = 1.0 ' 10 ( 4 cm (b) tan ! = tan45 = 1 = "#$ E % $ = E " = 10 7 5 # 10 3 = 2000 rad s (c) Again, & = k 2 + " 2 $ 2 ' % ' = & k 2 + " 2 $ 2 = 10 3 10 7 ( ) 2 + 5 # 10 3 ( ) 2 # 2000 2 = 7.07107 # 10 ( 5 cm...
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This note was uploaded on 12/05/2011 for the course MSE 4020 at Cornell University (Engineering School).

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4020-F2011-HW12Solutions - 4l 3 E " t 3 = 1 32 # 1 4...

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