4.3 Consider the creep function
!
J(t) =t
600 +1
600 (6 “2e“t 4 )
.
(a) Sketch a graph of
!
J(t)
(b) Show that the corresponding relaxation function has the form
!
G(t) =C1e“t#1+C2e“t#2
.
It is not necessary to find the constants
C1,C2,“1,“2
SOLUTION
!
$
‘
$
‘
This is the Laplace transform of
!
$
‘
$
‘
!
G(t) =50e“t 8 +100e“t 2
4.4 Consider a relaxation function of the form
!
G(t) =Go1+9e“t 10
( )
.
(a) Plot
!
G(t) Go
vs.
!
t
for
!
0“t“100
.
(b) Plot
!
G(t) Go
vs.
!
log10 t
for
!
0.001 “t“100
.
(c) Describe how the log scale changes the shape of the plot .
(d) Determine the value of
!
log10 t
at the inflection point of the plot .
(e) Determine the value of
!
G(t) Go
at the inflection point and compare with the
value of
!
G(t) Go
when
!
t 10 =1
.
Note: According to the definition of logarithm,
!
t=10log10 t
. One way to plot
!
G(t) Go
vs.
!
log10 t
is to rewrite the expression as
!
G(t) =Go1+9e“10log10 t10
#
$
% &
‘
(
.
Now let
!
x=log10 t
, so that
!
G=Go1+9e“10x10
#
$
% &
‘
(
,
and plot this vs.
!
x
.
SOLUTION
(b)
(c) The log scale stretches out the plot for times less than 1 and compresses the plot for
!
The time corresponding to the inflection point is the characteristic relaxation time.
4.5 Consider the stress relaxation function
!
G(t) =Co200e“2t +100e“t+10
( )
.
(a) In a step strain test, suggest a time
!
t
when the stress can be considered close
to its fully relaxed value.
(b) Show that the corresponding creep function also is the sum of a constant and
two exponentials, i.e. it has the form
!
J(t) =J1e“d1t+J2e“d2t+J3
,
in which
!
J1,J2,J3,d1,d2
are constants.
(c) Find
!
J3
.
(d) Find
!
d1,d2
.
(e) In a step stress test, suggest a value of time
!
t
when the creep strain can be
considered close to its fully developed value.
SOLUTION
!
!
This can be written as
!
J (a) =a+1
( )
a+2
( )
10Coa 31a2+43a +2
[ ]
=a+1
( )
a+2
( )
310Coa a +d1
( )
a+d2
( )
=J1
a+d1
+J2
a+d2
+J3
a
!
!
(e) Let the smallest of
!
d1t, d2t
be about 6.
!
0.048t “6
. Then
!
t“125