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The assumed displacement now appears as
u
x
=
N
u
Gv
x
=
Nv
x
(6.141)
Similarly, it is found that
u
y
=
N
u
Gv
y
=
Nv
y
.
The interpolation (shape) function matrix
N
is found to be
] (6.142)
For an isoparametric element, the assumed displacements are also used to describe the
geometry of the slave element. Then the coordinate transformation from the master element
to the slave element is
N
u
G
=
N
=
[
N
1
N
2
N
3
]
=
[
η
1
−
ξ
−
ηξ
x
=
N
1
x
1
+
N
2
x
2
+
N
3
x
3
(6.143)
y
=
N
1
y
1
+
N
2
y
2
+
N
3
y
3
where
x
i
,y
i
,i
1
,
2
,
3 are the coordinates of the corners of the triangle forming the iso-
parametric element.
The element stiffness matrix is obtained by inserting the strains and stresses, derived
from the assumed displacements, in the principle of virtual work. Use
k
i
=
of Eq. (6.138). The
quantities
D
u
and
N
are obtained from the strain expressions
u
x
u
y
D
u
N
v
x
v
y
x
y
γ
xy
∂
x
0
(ξ
,
η)
0
=
=
0
∂
y
=
0
N
(ξ
,
η)
(6.144)
∂
y
∂
x
D
u
u
=
B
u
v
The calculation of
B
u
involves the derivatives of
N
i
(ξ
,
η)
,
with respect to
x
and
y
.
Since the
relationship
are complicated and difficult to use in this calculation,
the procedure based on Eq. (6.136), using
J
−
1
,
the inverse of the Jacobian, will be employed.
For the three-node triangular element,
ξ
=
ξ(
x, y
)
,
η
=
η(
x, y
)
=
i
∂
N
i
x
i
i
∂
N
i
−
+
−
+
x
2
x
3
y
2
y
3
y
i
∂ξ
∂ξ
J
=
i
∂
N
i
x
i
i
∂
N
i
y
i
x
1
−
x
2
y
1
−
y
2
∂η
∂η
The inverse of
J
can be written as
y
1
−
y
2
y
2
−
y
3
β
3
β
1
1
2
A
1
2
A
J
−
1
=
=
x
2
−
x
1
x
3
−
x
2
γ
3
γ
1
1
2
where
A
=
|
J
|
. Then
1
2
A
(β
∂
=
∂
ξ
+
β
∂
η
)
x
3
1
1
2
A
(
−
γ
∂
=
∂
ξ
−
γ
∂
η
)
y
3
1
Thus,
B
u
is evaluated as
N
1
N
2
N
3
000
000
N
1
β
3
∂
ξ
+
β
1
∂
η
0
1
2
A
B
u
=
D
u
N
=
0
−
γ
3
∂
ξ
−
γ
1
∂
η
N
2
N
3
−
γ
3
∂
ξ
−
γ
1
∂
η
β
3
∂
ξ
+
β
1
∂η
β
1
−
β
3
−
β
1
β
3
0
0
0
1
2
A
=
0
0
0
−
γ
1
γ
3
+
γ
1
−
γ
3
(6.145)
−
γ
γ
+
γ
−
γ
β
−
β
−
β
β
1
3
1
3
1
3
1
3
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