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where
B
is obtained by applying the derivatives in
D
to the trial functions
N
1
a
∂
ξ
0
(
1
−
ξ)(
1
−
η) ξ(
1
−
η) ξη η(
1
−
ξ)
0
0
0
0
1
b
0
∂
η
0
0
0
0
(
1
−
ξ)(
1
−
η) ξ(
1
−
η) ξη η(
1
−
ξ)
1
1
b
∂
η
a
∂
ξ
D
N
1
−
η
a
1
−
η
a
a
−
a
−
0
0
0
0
1
−
ξ
b
−
b
b
1
−
ξ
b
0
0
0
0
−
=
(6.26)
−
η
a
−
η
a
a
−
a
1
−
ξ
b
−
b
b
1
−
ξ
b
1
1
−
−
B
Material Law
Equation (6.25) is the strain-displacement relationship needed in integral
I
of Eq. (6.21).
Also, in order to express the stress in terms of displacements in integral
I,
a stress-strain
relationship, i.e., the material law, is necessary. Assume a plane stress condition for this
case of an in-plane problem. Then [Chapter 1, Eq. (1.39a)] the material law is
1
ν
0
σ
x
x
E
=
ν
10
00
1
−
ν
2
σ
(6.27)
y
y
1
−
ν
2
τ
γ
xy
xy
σ
=
E
Formation of the Element Stiffness Matrix
The material law, the kinematics, and the trial solution can be introduced into Eq. (6.21)
to express the principle of virtual work in terms of the nodal displacements. For a single
element the volume integral I becomes
V
δ(
T
σ
dV
T
E
T
ED
u
u
dV
V
δ
=
V
δ
=
)
dV
D
u
u
u
T
u
D
T
ED
u
u
dV
u
T
k
D
u
dV
=
V
δ
=
V
δ
(6.28)
D
u
u
and the operator notation
k
D
of Chapter 2 have been introduced. The
subscript index to the left of the operator matrix
D
designates that the operator is applied
to the preceding quantity which, in this case, is
where
=
u
T
δ
.
The element stiffness matrix can be
obtained from this expression.
Consider next the details of the development of the element stiffness matrix
k
i
For a
single element, it follows from Eqs. (6.28) that the contribution of the volume integral I of
Eq. (6.21) can be expressed as
.
V
δ
V
δ
T
σ
dV
u
T
k
D
u
dV
=
(6.29)
Next, the integral containing
k
D
needs to be expressed in discrete form. For this two-
dimensional problem, let
t
be the constant thickness of the element and
dV
=
tdxdy
=
tabd
ξ
d
η.
The contribution to the volume integral for a single element is obtained by
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