
182 Applied calculus of variations for engineers
where
˙q
=
⎡
⎣
˙q
1
...
˙q
n
⎤
⎦
,
we obtain
E
k
=
V
1
2
ρ ˙u
T
˙udV =
1
2
˙q
t
V
N
T
ρNdV ˙q.
Introducing the mass matrix
M =
V
N
T
ρNdV,
the final form of the kinetic energy becomes
E
k
=
1
2
˙q
T
M ˙q.
Now let’s focus on the strain energy. Note that the strain is now also ex-
pressed in terms of the basis functions. Hence
(N)=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
n
i=1
q
t
i
∂N
∂x
n
i=1
q
t
i
∂N
∂y
n
i=1
q
t
i
∂N
∂z
n
i=1
q
t
i
(
∂N
∂z
+
∂N
∂y
)
n
i=1
q
t
i
(
∂N
∂z
+
∂N
∂x
)
n
i=1
q
t
i
(
∂N
∂y
+
∂N
∂x
)
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
,
or in matrix form
(N)=Bq,
where the columns of B are
B
i
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∂N
i
∂x
∂N
i
∂y
∂N
i
∂z
∂N
i
∂z
+
∂N
i
∂y
∂N
i
∂z
+
∂N
i
∂x
∂N
i
∂y
+
∂N
i
∂x
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
.
With this, the integral becomes
V
T
(N)D(N)dV =
V
q
T
B
T
DBqdV.