GPa-1
ln D(t) = ln 2 + t/1000
Sample was first equilibrated at 35 C for 48 hours, then stressed at
0.020 GPa until 200 hrs: so stress time is 200-48 = 152 hrs.
ln D(t) = ln 2 + 152/1000
ln D(t) = 0.6931 + 0.1520 = 0.8451
D(t) = 2.328
= 2.328 x 0.020 = 0.0466
Second case: sample was first equilibrated at 70 C for 48 hours, then
stressed at 0.020 GPa until 200 hrs:
ln D(t) = ln 2 + (7.3 x 152)/1000
ln D(t) = 0.6931 + 1.1096 = 1.8027
D(t) = 6.066
= 6.066 x 0.020 = 0.1213
Chemistry Homework Solutions
#36909
Look over a creep compliance problem I did involving time temp superposition
I have attached my solution to a creep compliance problem. I need someone to go over it and point out any mistakes I made. Thanks!
The problem involves a creep compliance function of a polymer at 35 C:
D(t) = 2 - e ^ (-t/1000) GPa ^ -1 where t is in hours. The polymer is made into a tensil specimen and heated to 70 C for 48 hrs. Determine the strain in the specimen at 200 hours after being subjected to 20 MPa at 70 C. Assume that the time temp shift factor aT is 7.3. Compare the result to the same load history at 35 C.
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