5.14 A linear viscoelastic specimen is subjected to the constant stress rate load
and unload history, which is shown in Figure-Problem 5.14. Develop an expression for
the strain history
!
“(t)
for any time
!
t“4T*
. Calculate this expression for the strain using
the creep compliance
!
J(t) =(1“0.9e“t 20 )#10“2
Does the specimen recover its original length? If so, estimate the time required for
substantial recovery to occur.
Figure-Problem 5.14
SOLUTION
Let
“=#o
T*
. Since the stress history is specified and the stress rate is piece-wise constant,
the strain is expressed in terms of stress history in the form
!
!
!
“(t) =#J(t $s)ds $J(t $s)ds +J(t $s)ds
3T *
4 T *
%
T*
3T *
%
0
T*
%
&
(
)
+
!
Thus,
!
“(t) #0
as
!
t” #
. When
!
t“4T*
20 #4
or
!
t“4T*+80
, then
!
“(t)
“(4T*)
#e$4=0.0187
and the current strain is less than 2% of
!
“(4T*)
5.15 A testing program which is used to study the response of linear viscoelastic
materials is as follows:
(1) constant strain rate deformation from the undeformed state during the interval
!
0“t“T
1
,
(2) no jump in strain or stress at time
!
T
1
,
(3) constant stress rate unloading until time
!
T2
, when the specimen is completely
unloaded,
(4) the specimen is completely unloaded for times
!
T2“t
.
(a) Show this program on stress – time and strain – time plots
(b) Develop an expression for the strain for times
!
T2“t
in terms of a general creep
compliance
!
J(t)
or stress relaxation function
!
G(t)
.
SOLUTION
(a)
(b) For
0“t“T
1
, the strain increases at a constant rate. Recall
!
“(t) =#G(x)dx
0
t
$
and
˙
“ (t) =#G(t)
!
For
!
T
1“t“T2
, the stress decreases at a constant rate,
!
For
!
T2“t
,
!
“(t) =0
and
!
˙
“ (t) =0
. Then
“(t) =#(0)J(t) +J(t $s) ˙
# (s)ds
0
t
%
!
!