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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 stressfree 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 onehalf 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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Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw5sol.doc
and that onehalf 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 onehalf 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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 Spring '05
 Anderson
 Materials Science And Engineering

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