Biomedical Engineering Reference
In-Depth Information
2
4
3
5
þ
ð
þ t
x
ð
t
t
0
dð
e
t=t
F
1
k
e
t=t
F
d
t
e
t=t
F
d
t
C
e
t=t
F
x
ð
t
Þ¼
h
ð
t
Þ
t
Þ
:
t
F
0
The integrals in this equation may be evaluated with ease using the definition of
the unit step function and the integral formula (A.222), while the initial condition
x
(0
+
)
0, thus the creep function
c
(
t
) for the stan-
dard linear solid (1.10) is again recovered. The relaxation function for the standard
linear solid is found by the same methods with some interchange in the roles of
F
(
t
)
and
x
(
t
). In this case the selection of
p
(
t
) and q(
t
) in (A.227) are that
p
(
t
)
¼
x
o
¼ t
x
/
k
t
F
requires that
C
¼
¼
(1/
t
x
)
and q(
t
)
¼
(
k
/
t
x
)(
x
þ t
F
(d
x
/d
t
)), then the solution of (1.8) for
F
(
t
) is given by
(A.228) as
2
4
3
5
þ
ð
þ t
F
ð
t
t
Þ¼
e
t=t
x
k
t
x
d
x
d
t
Þ
e
t=t
x
d
t
e
t=t
x
d
t
C
e
t=tx
F
ð
t
x
ð
t
:
0
0
To obtain the relaxation function one sets
x
(
t
)
¼
h
(
t
), thus
2
4
3
5
:
ð
þ t
F
ð
t
t
0
ðdð
e
t=t
x
k
t
x
e
t=t
x
d
t
e
t=t
x
d
t
F
ð
t
Þ¼
h
ð
t
Þ
t
Þ
0
The integrals in this equation may again be evaluated using the definition of the
unit step function and the integral formula (A.222). Thus setting
F
(
t
)
¼
r
(
t
) and
using the fact that the initial condition
F
(0
+
)
¼
F
o
¼
k
t
F
/
t
x
requires that
C
¼
0,
the relaxation function for the standard linear solid (1.10) is obtained again.