optimal design of engineering structures
TRANSCRIPT
Truss Topology Design
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Minimax MSE estimator(Eldar, Ben-Tal, Nemirovski, IEEE Tran. Signal Proc. 2004)
Assumption: x is norm bounded ‖x‖T ≤ L.
The minimax MSE linear estimator minimizes the
worst case MSE across all bounded signals x, i.e.,
x̂mmx = Gy, where G is the solution to the following
optimization problem:
minG
max‖x‖T ≤L
xT (I − GH)T (I − GH)x + Tr(GCGT )︸ ︷︷ ︸
MSE
(6)
NOTICE for fixed G, the optimal value of the inner
max problem is
L2λmax(T−1/2(I −GH)T (I −GH)T−1/2) + Tr(GCGT )
so problem (6) reduces to:
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The simplest TTD problem is
Compliance = 12fT x → min
s.t.
[m∑
i=1
tibibTi
]
︸ ︷︷ ︸
A(t)�0
x = f
m∑
i=1
ti ≤ w
t ≥ 0
• Data:
— bi ∈ Rn, n – # of nodal degrees of freedom
(for a 10 × 10 × 10 ground structure, n ≈ 3, 000)
— m – # of tentative bars (for 10 × 10 × 10
ground structure, m ≈ 500, 000)
• Design variables: t ∈ Rm, x ∈ Rn
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Multi-Load TTD
(P) minxj ,t
maxj=1,...,k
{fTj xj}
s.t.
m∑
i=1
tiAixj = fj
m∑
i=1
ti = v
ti = 0 , i = 1, . . . , m
xj ∈ IRn, j = 1, . . . , k
minxj ,...,xk
λ∈Rk
k∑
j=1
fTj xj + v · max
i=1,...,m
K∑
j=1
xTj Aixj
λj
s.t.k∑
j=1
λj = 1, λ ≥ 0 .
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CONVEX PROGRAM
min
x1,...,xk
λ∈IRk
τ∈IR
{∑
fTj xj + vτ
}
∑
λj = 1, λ ≥ 0
dual var.t∗j
⇒∑
j
(
xTj Ajxj
λj
)
≤ τ
︸ ︷︷ ︸
a conic quadratic constraint
∀ i
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04-115
Example: Assume we are designing a planar truss { acantilever; the 9� 9 nodal structure and the only loadof interest f� are as shown on the picture:
9� 9 ground structure and the load of interest
The optimal single-load design yields a nice truss asfollows:
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Optimal cantilever (single-load design)the compliance is 1.000
04-116
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Optimal cantilever (single-load design)the compliance is 1.000
� It turns out that when the \load of interest" f � isreplaced with the small (k f k= 0:005 k f� k) \badlyplaced" occasional load f , the compliance jumps from1.000 to 8.4 (!)
� In order to improve the stability of the design, let usreplace the single load of interest f� by the ellipsoidcontaining this load and all loads f of magnitude k f knot exceeding 10% of the magnitude of f �:
F = ff = u1f� + u2f2 + f3u3 + ::: + u20f20 j uTu � 1g;
where f2; :::; f20 are of the norm 0:1 k f� k and f �; f2; :::; f20is an orthogonal basis in V = R20.
04-117
� Passing from the single-load to the robust design, wemodify the result as follows:
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Optimal cantilever (single-load design)
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\Robust" cantilever
Compliances DesignSingle-load Robust
Compliance w.r.t. f � 1.000 1.0024
max compliance w.r.t. 32000 1.03loads f : k f k� 0:1 k f � k
