Civil Engineering Reference
In-Depth Information
minimized by reducing the size of the element. However, an accurate or
fine displacement assumption can reduce the error of a large element so that
even a coarse mesh still can get accurate and convergent results.
By using displacement functions, the displacement of any point in the
element can be expressed as Equation 3.18.
n
n
n
∑
∑
∑
u
N u v
,
N v w
,
N w
,…
=
=
=
(3.18)
i
i
i
i
i
i
i
=
1
i
=
1
i
=
1
where:
n
is the number of element nodes
u v w
i
, ,
are the node displacements at node
i
N
i
is the displacement function of node
i
and describes how a known
displacement at node
i
will influence or contribute to the displace-
ment at any point within an element
i
i
From its definition, the displacement function must satisfy the following
conditions:
1.
N
i
= 1 at node
i
and
N
i
= 0 at all other nodes
2. Ensures any of the unknown displacement is continuous at element
boundaries, that is, displacements at any point on an element bound-
ary interpolated by nodal displacements of any adjacent elements
should be the same
3. Contains linear term so it is able to represent constant strain
=
∑
1
1, so it can represent rigid displacement, that is, displace-
ment at any point should be the same as that at any node when all
nodes have the same displacements
n
N
i
4.
i
In developing displacement functions for a type of element, the more com-
plicated the shape of the element, the higher the polynomial order of the
displacement function is required. An element with a higher order of dis-
placement functions will lead to a higher accuracy. Therefore, a coarser
mesh will produce relatively higher accurate results. Or, in other words,
a finer mesh is needed when a simple element with a lower-order displace-
ment function is used.
Taking a commonly used 2D rectangle element as an example, as shown
in Figure 3.2; the displacement functions of a four-node element are
1
4
1
1
4
1
(
)
(
)
(
)
(
)
N
=
+
ξ
1
+
η
,
N
=
−
ξ
1
+
η
1
2
(3.19)
1
4
1
1
4
1
(
)
(
)
(
)
(
)
N
=
−
ξ
1
−
η
,
N
=
+
ξ
1
η
−
3
4
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