Environmental Engineering Reference
In-Depth Information
L
∫
T
uu u
The projected equations defi ne the weak formulation of the problem. Note that
demanding zero virtual work for any virtual displacements is equivalent to per-
forming the projection with respect to a properly defi ned function basis. Proper
in this sense means that boundary conditions must be taken into account. For all
constrained DOF corresponding to kinematic conditions, we must set
d u
p
=0. Note
that static or load conditions will naturally appear in the weak formulation when
integration by parts is carried out. Considering the bending term as an example, a
double integration results in
d
Fy
()d
=∀
0
d
0
L
L
L
⎡
⎤
′′
′
∫
T
T
T
∫
T
d
u Ku
(
′′
) d
y
=
d
u Ku
(
′′
)
−
d
u Ku
′
(
′′
)
+
d
u
′′
(
Ku
′′
) d
y
⎢
⎥
22
22
22
22
⎣
⎦
0
The underlined terms correspond to the boundary terms and represent the virtual
work done by the reacting force (
k
22
u
0
0
) at the support points
of the beam. If the displacement or rotation is specifi ed then the term equals zero
because
d u
or
d u
″
)
′
and moment (
k
22
u
″
is zero. On the contrary, if the load is specifi ed then either the
force or the moment is set to its given value. All elastic terms in eqn (7) are inte-
grated in the same way.
The most popular method for solving dynamic equations is the Finite Element
Method [12-14]. It consists of projecting the equations using a basis of fi nite
dimension. To this end fi rst the beam is divided into elements. Then for each ele-
ment the same local approximations are defi ned for
u
and
d
u
. In doing so, specifi c
polynomial shape functions and discrete DOF
û
are chosen (Fig. 3).
The choice of the shape functions depends on the order of the problem. The beam
equations are second order for the tension and torsion so we can choose linear shape
functions (
g
n
,
n
= 1,2), and fourth order for the bending so we use cubic functions
(
′
). The discrete DOF's usually correspond to the nodal values of
u
but can also include the values of its space derivative as in the case of bending.
Taking as nodes the two ends of each element
e
:
a
n
n
b
,
=
, ,
a
=
,1
e
T
e
u
() (
yuvw
=
q
)
=
N
()
y
u
ˆ
h
y
e
ˆ
ˆ
T
u
ˆ
=
(
uuvww
ˆˆˆˆ ˆ
,
′
,
,
,
′
,
q
,
u u
ˆ ˆ ˆ
,
′
,
v
,
w
ˆ
,
w
ˆ
′
,
q
) ,
e
e
e
e
e
e
e
e
+
1
e
+
1
e
+
1
e
+
1
e
+
1
e
+
1
⎡
0
1
0
1
⎤
bb
00 0 0
bb
0 0 0 0
1
1
2
2
⎢
⎥
00
g
00000
g
000
,
⎢
⎥
e
1
2
N
=
⎢
⎥
0
1
0
1
000
bb
0000
bb
0
⎢
⎥
1
1
2
2
( 8)
⎢
00000
g
00000
g
⎥
⎣
⎦
1
2
yy
L
−
e
0
2
3
1
2
3
x
=
,
b
= −
1
3
x
+
2,
x
b
=
L
( 2
x
−
x
+
x
,
1
1
e
e
0
2
3
1
2
3
Ly y
=−
,
b
=−
3
x
2
x
,
b
=−+
L
(
x
x
),
e
e
+
1
e
2
2
e
g
=−
1,
x
g
=
x
1
2
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