Civil Engineering Reference
In-Depth Information
Thus, in matrix form
.
/
!
.
/
!
ε
x
ε
y
γ
xy
1
−
ν
0
σ
x
σ
y
τ
xy
1
E
{
ε
}=
=
(17.78)
−
ν
1
0
0
"
0
"
0
0
2(1
+
ν
)
It may be shown that
.
/
!
.
/
!
σ
x
σ
y
τ
xy
1
ν
0
ε
x
ε
y
γ
xy
E
{
σ
}=
=
ν
1
0
(17.79)
0
"
ν
2
0
"
1
−
00
2
(1
−
ν
)
which has the form of Eq. (17.55), i.e.
{
σ
}=
[
D
]
{
ε
}
δ
e
Substituting for
{
ε
}
in terms of the nodal displacements
{
}
we obtain
δ
e
{
σ
}=
[
D
][
B
]
{
}
(see Eq. (17.56))
In the case of plane strain the elasticity matrix [
D
] takes a different form to that defined
in Eq. (17.79). For this type of problem
νσ
y
E
σ
x
E
−
νσ
z
E
ε
x
=
−
σ
y
E
−
νσ
x
E
−
νσ
z
E
ε
y
=
σ
z
E
−
νσ
x
E
−
νσ
y
E
=
ε
z
=
0
τ
xy
G
=
2(1
+
ν
)
E
γ
xy
=
τ
xy
Eliminating
σ
z
and solving for
σ
x
,
σ
y
and
τ
xy
gives
ν
1
0
.
/
!
.
/
!
1
−
ν
σ
x
σ
y
τ
xy
ε
x
ε
y
γ
xy
ν
E
(1
−
ν
)
1
0
{
}=
=
σ
(17.80)
1
−
ν
0
"
+
−
0
"
(1
ν
)(1
2
ν
)
(1
−
2
ν
)
0
0
2(1
−
ν
)
which again takes the form
{
σ
}=
[
D
]
{
ε
}
Step six, inwhich the internal stresses
{
σ
}
are replaced by the statically equivalent nodal
F
e
forces
{
}
proceeds, in an identical manner to that described for the beam-element.
Thus
[
B
]
T
[
D
][
B
] d(vol)
F
e
δ
e
{
}=
{
}
vol
