Geography Reference
In-Depth Information
U
2
(
t
l
)=
Λ
(
t
l
)
=
constant
.
constant
t
r
=
λ
(
t
r
)
=
u
2
(
t
r
)
.
(1.137)
Φ
(
t
l
)
t
l
φ
(
t
r
)
End of Example.
r
, respectively, we enter Box
1.21
. Here, we compute
Φ
−
l
and
Φ
−
r
in parameterized form, namely (
Λ, Φ
)
l
and
With these parameterized curves in
M
M
3
and (
λ, φ
)
→
X
(
Λ, Φ
)
∈
R
→
x
(
λ, φ
)
∈
3
, respectively. The left and the right displacement field is used to derive the tangent vectors
R
X
1
,
X
2
}
{
of type “right”. The inner products vanish according to
our test computations in Example
1.3
. In consequence,
Ψ
l
=
Ψ
r
=
π/
2andΣ
l
=Σ
r
= 0, i.e. no
angular distortion appears.
of type “left” and
{
x
1
,
x
2
}
Box 1.21 (Angular shear or angular distortion).
Left vector field:
Right vector field:
X
(
Λ, Φ
)=
E
1
A
1
cos
Φ
cos
Λ
√
1
−E
2
sin
2
Φ
+
x
(
λ, φ
)=
e
1
r
cos
φ
cos
λ
+
(1.138)
√
1
−E
2
sin
2
Φ
+
E
3
(1
−
E
2
)
A
1
sin
Φ
√
1
−E
2
sin
2
Φ
.
+
e
2
r
cos
φ
sin
λ
+
e
3
r
sin
φ.
+
E
2
A
1
cos
Φ
cos
Λ
Left displacement field:
Right displacement field:
(1.139)
d
t
r
+
∂
∂φ
d
φ
d
X
=
∂
∂Λ
d
Λ
d
t
l
+
∂
∂Φ
d
Φ
d
t
l
.
d
x
=
∂
∂λ
d
λ
d
t
r
.
d
t
l
d
t
l
d
t
r
d
t
r
1st left parameterized curve: 1st right parameterized curve:
Λ
=1
,
(1.140)
Φ
=0
,
d
d
t
l
=
∂
∂Λ
.
λ
=1
,
φ
=0
,
d
x
d
t
r
=
∂
∂λ
.
2nd left parameterized curve:
2nd right parameterized curve:
Λ
=0
,
Φ
=1
,
d
d
t
l
=
∂
∂Φ
.
λ
=0
,
φ
=1
,
d
x
d
t
r
=
∂
∂φ
.
(1.141)
Left angular shear:
Right angular shear:
∂Φ
=0
,
0=
∂
∂λ
|
∂φ
=
=
∂
∂Λ
|
X
1
X
2
∂
X
∂
x
|
{
x
1
|
x
2
}
,
(1.142)
π
π
cos
Ψ
l
=0
⇔
Ψ
l
=
±
2
,
±
2
=
Ψ
r
⇔
cos
Ψ
r
=0
,
Σ
l
=
Ψ
l
−
Ψ
r
=0
.
Σ
r
=
Ψ
r
−
Ψ
l
=0
.
1-8 Relative Angular Shear
A third multiplicative measure of deformation: relative angular shear, Cauchy-Green defor-
mation tensor, Euler-Lagrange deformation tensor.
The third multiplicative measure of deformation is the ratio
Q
l
and
Q
r
, respectively. This ratio
is also called
relative angular shear
. In particular
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