Civil Engineering Reference
In-Depth Information
This demonstrates that the change in area depends only on the stretching part of metric
tensor M but not on the rotational part of metric tensor M. The ratio of the area between
the transformed object and the original object of any shape and size is always equal to det(F) =
det(U) = λμ. Indeed, by the property of rotational (orthogonal) matrix R, we have
uv
2
(
RR
⋅
u
) (
⋅
⋅ =⋅
v
)
uv
andR
(
⋅ ×⋅ =× ∀∈
u
)
(
R
v
)
uv
,
In other words, the length and the area are preserved under rotation.
4.2.8 Interpolation of metric
For complicated surfaces, the evaluation of metric is quite computationally intensive, and
very often the metric is only evaluated and available at some strategic nodal points of a
mesh. For sufficient fine meshes, at the interior of an FE, the metric can be estimated by
interpolation (Laug and Borouchaki 2003a).
4.2.8.1 Metric interpolation over a line segment
The metric tensors are given at the end points of a line segment; our task is to estimate the
metric tensor at any point on the line segment, as shown in Figure 4.14.
Metric tensor M, which is symmetric and positive definite, admits a polar decomposition
into two fundamental parts, pure stretch U and rotation R. Let M
1
and M
2
be the metrics at
the end points of line segment AB; metric M at point P = (1 − t)A + tB, t ∈ [0,1] can be defined
by the following interpolation scheme. By polar decomposition of metric M
1
, we have
λ
0
cos(
θ
) i
n(
θ
)
T
1
1
1
MFFFUR
=
,
=
,
U
=
,
R
=
1
1
1
1
11
1
1
0
−
sin(
θ
)cos()
θ
1
1
1
π
where λ
1
and μ
1
are the principal stretches, and
θ
∈ ,
0
characterises the principal direc-
1
2
tion associated with stretch λ
1
, and similarly we have for metric tensor M
2
λ
0
cos(
θ
) i
n(
θ
)
T
2
2
2
MFFFUR
=
,
=
,
U
=
,
R
=
2
2
2
2
22
2
2
0
−
sin(
θ
)cos()
θ
2
2
2
The rotation part for metric M at point P can be defined by a linear interpolation such that
θ = sθ
1
+ tθ
2
where s = 1 − t
2
M
M
2
2
M
1
2
1
1
1
Figure 4.14
Interpolation of metric on a line segment.
