hw5sol

# hw5sol - Materials Science and Engineering 765 mechanical...

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1/7 Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw5sol.doc Materials Science and Engineering 765 mechanical behavior of materials Spring 2005 Homework 5 (correction 5.25.05) with solutions Due: Tuesday, May 31, by 5pm. Show all work necessary to arrive at your answers. 1. Time Dependent Response of a Polymer Recall from lecture that the time dependent response of a polymer under a constant applied stress can be modeled as: γ = σ b 3 4 kT 1 e t / τ ( ) n b 3 V , where = 1 2 ν e f o / kT This model assumes that the temperature is small enough so that the number of available side groups to switch from a backward to a forward position depletes with time,. The stress-free activation energy associated with a switch is f o . Consider three cases a , b , and c as discussed below. In all cases, assume that the applied stress is such that b 3 /4 kT = 10 -2 , the attempt frequency = 10 16 /s, and the number of sites per unit volume is n / V = 1/ b 3 . Put the resulting curves of vs t from these three cases onto one set of axes for easy comparison. a. Case a : Assume that all sites have an activation energy f 0 = 38 kT . Produce a plot of vs t , where ranges from 0 to 10 -2 and t ranges from 0 to 10s. Label the curve " a ". The MathCad program below plots the function: a = 0.01 1 e t / 38 kT ( ) (1) , where 38 kT = 1 2 10 16 /s e 38 b. Case b : Assume that one-half of the sites have f 0 = 37 kT and the remaining one- half have f 0 = 39 kT . Label the vs t result as curve " b ". The MathCad program below plots the function: b = 0.01 1 e t / 37 kT ( ) 1 2 + 1 e t / 39 kT ( ) 1 2 , where 37 kT = 1 2 10 16 e 37 ; 39 kT = 1 2 10 16 e 39 c. Case c : Assume that the number of sites with activation energy in the range f 0 to f 0 + d f 0 is n ( f 0 ) d f 0 where n ( f 0 ) = 2 0.83 38 kT 2 f 0 e 0.83 38 f 0 kT 2 Convince yourself that the total number of sites is n f 0 ( ) d f 0 0 = 1

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2/7 Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw5sol.doc and that one-half of the sites have an activation energy less than 38 kT and one- half have an energy greater than 38 kT . Label γ vs t result as curve " c ". The MathCad program below shows that one-half the sites have an activation energy less than 38 kT and one half the sites have an activation energy greater than 38 kT . The program also computes the function c = 0.01 1 e t / τ f n ( f ) df 0 100 kT , where f = 1 2 10 16 /s e f / kT d. Compare the results for curves a , b , and c . Note that all cases have a different distribution of f 0 , yet the mean value,
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## This note was uploaded on 07/17/2008 for the course MSE 765 taught by Professor Anderson during the Spring '05 term at Ohio State.

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hw5sol - Materials Science and Engineering 765 mechanical...

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