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In-Depth Information
Then
G
T
0
G
T
1
0
k
i
B
u
(
B
u
(ξ )
=
x
)
EI
B
u
(
x
)
dx
G
=
EI
B
u
(ξ )
d
ξ
G
G
from Eq
.(
4
.
45
)
G
from Eq
.(
4
.
47a
)
B
u
(
x
)
from Eq
.(
4
.
54
)
B
u
(ξ )
from Eq
.(
4
.
69
)
(4.70)
00 0
0
000 0
000 0
004 6
00612
G
T
EI
1
0
1
00 0
0
EI
G
T
=
d
ξ
G
=
G
00 4
12
ξ
4
3
0012
ξ
36
ξ
2
Pre-and post multiplication by
G
of Eq. (4.47a) leads to the stiffness matrix
12
−
6
−
12
−
6
G
T
1
0
2
2
EI
−
6
4
6
2
k
i
B
u
(ξ )
=
EI
B
u
(ξ )
d
ξ
G
=
(4.71)
−
12
6
12
6
3
2
2
−
6
2
6
4
which, of course, is the same as
k
i
of Eq. (4.65).
Note that for the case of a simple beam element for which the polynomials of Eq.
(4.41) or (4.47) are employed as trial series, the resulting stiffness matrix of Eq. (4.71) is
the correct stiffness matrix [of Eq. (4.12)], rather than an approximate stiffness matrix. If
fewer terms in the polynomial are retained, a different stiffness matrix results.
Loading Vector
To
evaluate the loading vector
p
i
0
use Eq. (4.58), which corresponds to
p
i
0
=
V
a
M
a
V
b
M
b
T
. We choose to utilize the normalized coordinate at
ξ
=
x
/
.
b
G
T
0
G
T
1
0
p
i
0
N
T
N
u
(
N
u
(ξ )
=
(
x
)
p
z
(
x
)
dx
=
x
)
p
z
(
x
)
dx
=
p
z
(ξ )
d
ξ
a
G
from Eq
.(
4
.
45
)
N
u
and
G
N
u
from Eq
.(
4
.
54
)
from Eq
.(
4
.
47a
)
1
−
3
ξ
2
+
2
ξ
3
1
1
(
−
ξ
+
2
ξ
2
−
ξ
3
)
N
T
=
(ξ )
p
z
(ξ )
ξ
=
p
z
(ξ )
ξ
d
d
(4.72)
ξ
2
−
ξ
3
3
2
0
0
(ξ
2
−
ξ
3
)
If the applied distributed load
p
z
is constant of magnitude
p
0
,
then the integral of Eq. (4.72)
results in
2
−
/
1
/
12
p
i
0
=
p
0
(4.73)
1
/
2
/
12
If
p
z
varies linearly from
ξ
=
0to
ξ
=
1
,
where its magnitude is
p
0
,
then
p
z
=
p
0
ξ
and Eq.
(4.72) provides
9
−
p
0
60
21
3
2
p
i
0
=
(4.74)
Table 4.2 lists the vector
p
i
0
for a variety of loading conditions.
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