Biomedical Engineering Reference
In-Depth Information
with
C
(
t
) to be determined. Substitution into Eq. (
5.46
) yields
dC
C
τ
C
τ
dC
dt
=
dt
e
−
t
/τ
−
e
−
t
/τ
e
−
t
/τ
1
τ
c
e
t
/τ
ε
+
=
c
ε
−→
,
(5.52)
dF
p
dt
F
p
hence
c
e
ξ/τ
ε
C
=
(
ξ
)
d
ξ
.
(5.53)
Because the strain rate
ε
=
0 for all
t
<
0, it follows that
t
e
−
t
/τ
.
c
e
ξ/τ
ε
(
ξ
)
d
ξ
F
p
=
(5.54)
0
Combining Eqs. (
5.49
) and (
5.54
), the solution
F
is given by
t
e
−
t
/τ
.
F
=
c
1
e
−
t
/τ
+
c
e
ξ/τ
ε
(
ξ
)
d
ξ
(5.55)
0
Requiring that for all
t
<
0 the force satisfies
F
=
0 leads to
F
(
t
=
0)
=
c
1
,
c
1
=
0.
(5.56)
Consequently, the solution of the first-order differential equation Eq. (
5.46
), is
given by
=
c
t
0
e
−
(
t
−
ξ
)
/τ
ε
ξ
)
d
ξ
F
(
t
)
(
.
(5.57)
Apparently, the integral equation as introduced in the previous section, Eq. (
5.26
),
can be considered as a general solution of a differential equation. In the present
case the relaxation spectrum, as defined in Eq. (
5.37
) is built up by just one single
Maxwell element and in this case:
G
(
t
)
e
−
t
/τ
.
To understand the implications of this model, consider a strain history as spec-
ified in Fig.
5.11
, addressing a spring-dashpot system in which one end point is
fixed while the other end point has a prescribed displacement history. The force
response is given in Fig.
5.12
in case
t
∗
=
5
τ
. Notice that in this figure the time
has been scaled with the relaxation time
τ
, while the force has been scaled with
c
η
r
, with
r
the strain rate, see Fig.
5.11
. Two regimes may be distinguished.
(i) For
t
<
t
∗
the strain proceeds linearly in time leading to a
constant
strain rate
r
.In
this case the force response is given by (recall that
c
η
=
τ
=
c
)
t
τ
) .
e
−
F
=
c
η
r
(1
−
(5.58)
For
t
τ
it holds that
t
τ
t
τ
e
−
≈
1
−
,
(5.59)
