Geology Reference
In-Depth Information
2
level) stiffness matrix of the element
e
can be
expressed as
where
λ
, , ,
and
N
is the number
of elements. The second constraint restricts the
volume of the design to an upper limit denoted
by
v
. In this statement
v
e
is the volume of the
element
e
.
=
ω
j
=
1
N
j
j
d
(
)
p
K
x
)
=
K
+
x
K K
−
(1.12)
(
e
e
e
e
e
e
in which
K
e
and
K
e
are the stiffness matrices of
the element
e
when it is made of the base mate-
rial and void (in its solid and void states) respec-
tively. Similarly for the density and the mass
matrix of the element
e
we can write
4.1. Sensitivity Analysis
Solving Problem (1.15) requires finding the sen-
sitivities of
λ
1
with respect to design variables
x
e
.
Differentiating
Kφ
j
=
λ
j
Mφ
j
we can write
ρ
x
)
=
x
ρ
(1.13)
(
∂
∂
ϕ
=
∂
∂
λ
∂
∂
ϕ
e
e
e
∂
∂
K
∂
∂
M
j
j
j
K
M
M
ϕ
+
ϕ
+
λ
ϕ
+
λ
j
j
j
j
j
x
x
x
x
x
e
e
e
e
e
(1.16)
and
Premultiplying by
ϕ
T
and rearranging the
terms we obtain
M
x
)
=
x
M
(1.14)
(
e
e
e
e
Where
ρ
is the density of the base material and
M
e
is the mass matrix of the element
e
in its
solid state.
Using these material interpolation schemes
one can change element
e
from solid to void and
backwards by changing the value of
x
e
. Thus by
choosing
x
e
-s as design variables, one can produce
different topologies without altering the finite
element mesh.
We can now formulate the optimization prob-
lem. The fundamental frequency optimization
problem can be stated as finding the best topol-
ogy of a structure to maximize its fundamental
frequency given a fixed amount of material. The
problem can thus be formulated as
=
∂
∂
λ
)
∂
∂
ϕ
∂
∂
K
∂
∂
M
(
T
j
T
T
j
ϕ
−
λ
ϕ
ϕ
M
ϕ
−
ϕ
K
−
λ
M
j
j
j
j
j
j
j
x
x
x
x
e
e
e
e
(1.17)
Using the symmetry of
K
and
M
, we can read-
ily conclude that
T
(
)
=
(
)
T
ϕ
K
−
λ
M
K
−
λ
M
ϕ
=
0
.
j
j
j
j
We also use Equation (1.6) in Equation (1.17)
to finally express the sensitivities as
∂
∂
λ
∂
∂
K
∂
∂
M
j
T
=
ϕ
−
λ
ϕ
(1.18)
j
j
j
x
x
x
e
e
e
{
}
max
λ
=
min
λ
The stiffness and mass derivatives in Equa-
tion (1.18) can be calculated using Eqs. (1.12)
and (1.14).
1
j
x x
,
,
,
x
j
1 2
, ,
,
N
=
1
2
N
d
such that
K
ϕ
=
λ
M
ϕ
,
j
=
1 2
,
,
,
N
j
j
j
d
N
∑
≤
x v
v
4.2. Solution Method
e e
e
=
1
0
≤ ≤
x
1
,
e
=
1 2
,
,
,
N
e
Having the sensitivities in Equation (1.18), the
optimization problem (1.15) can be solved using
(1.15)
Search WWH ::
Custom Search