Biomedical Engineering Reference
In-Depth Information
Z
ffi
dt ¼ q
Z
ffi
dt ¼ 2q
Z
t
2
t
2
t
2
ffi
dt
dqu
T
u
d u
T
u
þ
u
T
d u
d u
T
u
ð
2
:
54
Þ
t
1
t
1
t
1
And knowing that u
T
u is a scalar and u
= ou
=
ot,
Z
ffi
dt ¼
Z
dt
t
2
t
2
odu
T
ot
ou
ot
d u
T
u
ð
2
:
55
Þ
t
1
t
1
Then integrating by parts, with respect to time,
t
2
Z
dt ¼
Z
dt
þ
du
T
o
u
ot
t
2
t
2
o
du
T
ot
du
T
o
2
u
ot
2
o
u
ot
ð
2
:
56
Þ
t
1
t
1
t
1
Notice that u satisfies, by imposition, the conditions at the initial time, t
1
, and
final time, t
2
, leading to a null du at t
1
and t
2
. Therefore the last term in Eq. (
2.56
)
vanishes. Considering the last development and switching the integration order
again, Eq. (
2.53
) becomes,
dt ¼
Z
Z
t
2
Z
t
2
q
Z
ffi
dX
ffi
dt
1
2
dqu
T
u
du
T
u
ð
2
:
57
Þ
X
X
t
1
t
1
2
u
ot
2
the acceleration. The second term on Eq. (
2.52
) can also be
developed. The integrand function in the second integral term can be written as
follows,
being u
= o
ffi
¼ de
T
r + e
T
dr
d e
T
r
ð
2
:
58
Þ
as the two terms in Eq. (
2.58
) are in fact scalars, the transpose does not affect the
result, as so,
ffi
T
¼ dr
T
e
e
T
dr ¼ e
T
dr
ð
2
:
59
Þ
Using the constitutive equation r ¼ ce and the symmetric property of the
material matrix, c
T
¼ c, it is possible to write,
dr
T
e ¼ d c
ð
T
e ¼ de
T
c
T
e ¼ de
T
ce ¼ de
T
r
ð
2
:
60
Þ
