Linear Programming: Chapter 15
Structural Optimization
Robert J. Vanderbei
November 4, 2007
Operations Research and Financial Engineering
Princeton University
Princeton, NJ 08544
http://www.princeton.edu/∼rvdb
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Linear Programming Structural Optimization, Lecture Notes - Mathematics - Prof. J Vanderbei.pdf, Prof. J Vanderbei, Mathematics, Linear Programming, Structural Optimization, Minimum Weight Structural Design, Redundant Equations, Skew Symmetric Matrices, Conservation Laws, Torque Balance, Trusses, AMPL Model
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Linear Programming: Chapter 15
Structural Optimization
Robert J. Vanderbei
November 4, 2007
Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 http://www.princeton.edu/∼rvdb
Structural Optimization
Forces: xij = tension in member {i, j}.
- xij = xji.
- Compression = -Tension.
Force Balance: Look at joint 2:
x 12
[
]
- x 23
[
]
- x 24
[
]
[
b^12 b^22
]
3
1 2
4
5
b 1 b 2
b 5
u 21
u 24 u 23
Notations:
pi = position vector for joint i
uij =
pj − pi ‖pj − pi‖
( Note uji = −uij )
Constraints: ∑
j: {i,j}∈A
uij xij = −bi i = 1,... , m.
Minimum Weight Structural Design
minimize
{i,j}∈A
lij |xij |
subject to
j: {i,j}∈A
uij xij = −bi i = 1, 2 ,... , m.
Not quite an LP. Use a common trick:
xij = x+ ij − x− ij , x+ ij , x− ij ≥ 0 |xij | = x+ ij + x− ij
Reformulated as an LP:
minimize
{i,j}∈A
(lij x+ ij + lij x− ij )
subject to
j: {i,j}∈A
(uij x+ ij − uij x− ij ) = −bi i = 1, 2 ,... , m
x+ ij , x− ij ≥ 0 {i, j} ∈ A.
Redundant Equations
Recall network flows: number of redundant equations = number of connected components.
Row combinations: yiT uij + yjT uji
Sum of “x”-component rows:
[ 1 0
]
[
u(1) ij u(2) ij
]
[
]
[
u(1) ji u(2) ji
]
Sum of “y”-component rows, “z”-component rows, etc. is similar.
Skew Symmetric Matrices
Definition. RT^ = −R
For d = 1: no nonzero ones.
For d = 2: (^) [ 0 − 1 1 0
]
For d = 3: (^)
Structure is stable if the redundancies just identified represent the only redundan- cies.
Conservation Laws
Suppose a combination of rows of A vanishes. Then the same combination of elements of b must vanish.
Force Balance: (^) ∑
i
b(1) i = 0 and
i
b(2) i = 0
What is meaning of the other redundancies? ∑
i
(Rpi)T^ bi = 0
Answer...
Trusses
Definition.
- Stable
- Has md − d(d + 1)/ 2 members (d is dimension).
Anchors
No force balance equation at anchored joints.
Earth provides counterbalancing force.
If enough (d(d + 1)/ 2 ) independent constraints are dropped (due to anchoring), then no force balance or torque balance limitations remain.
AMPL Model
param m default 26; # must be even param n default 39;
set X := {0..n}; set Y := {0..m};
set NODES := X cross Y; # A lattice of Nodes
set ANCHORS within NODES := { x in X, y in Y : x == 0 && y >= floor(m/3) && y <= m-floor(m/3) };
param xload {(x,y) in NODES: (x,y) not in ANCHORS} default 0; param yload {(x,y) in NODES: (x,y) not in ANCHORS} default 0;
param gcd {x in -n..n, y in -n..n} := (if x < 0 then gcd[-x,y] else (if x == 0 then y else (if y < x then gcd[y,x] else (gcd[y mod x, x]) )));
subject to Ybalance {(xi,yi) in NODES: (xi,yi) not in ANCHORS}: sum { (xi,yi,xj,yj) in ARCS } ((yj-yi)/length[xi,yi,xj,yj]) * (comp[xi,yi,xj,yj]-tens[xi,yi,xj,yj])
sum { (xk,yk,xi,yi) in ARCS } ((yi-yk)/length[xk,yk,xi,yi]) * (tens[xk,yk,xi,yi]-comp[xk,yk,xi,yi]) = yload[xi,yi]; ;
let yload[n,m/2] := -1;
solve;
printf: "%d \n", card({(xi,yi,xj,yj) in ARCS: comp[xi,yi,xj,yj]+tens[xi,yi,xj,yj] > 1.0e-4})
structure.out;
printf {(xi,yi,xj,yj) in ARCS: comp[xi,yi,xj,yj] + tens[xi,yi,xj,yj] > 1.0e-4}: "%3d %3d %3d %3d %10.4f \n", xi, yi, xj, yj, tens[xi,yi,xj,yj] - comp[xi,yi,xj,yj]
structure.out;
The Michel Bracket
Constraints: 2, Variables: 31, Time: 193 secs
Click here for parametric self-dual simplex method animation tool. Click here for affine-scaling method animation tool.