The locations of the first and last mass are fixed. The equilibrium location of the other masses is
the one that minimizes
E
.
(a) Show how to find the equilibrium positions of the masses 2
,...,n
−
1 using convex optimization.
Be sure to justify convexity of any functions that arise in your formulation (if it is not obvious).
The problem data are
m
i
,
k
i
,
l
i
,
g
,
x
1
,
y
1
,
x
n
, and
y
n
.
(b) Carry out your method to find the equilibrium positions for a problem with
n
= 10,
m
i
= 1,
k
i
= 10,
l
i
= 1,
x
1
=
y
1
= 0,
x
n
=
y
n
= 10, with
g
varying from
g
= 0 (no gravity) to
g
= 10
(say). Verify that the results look reasonable. Plot the equilibrium configuration for several
values of
g
.
14.3
Elastic truss design.
In this problem we consider a truss structure with
m
bars connecting a set
of nodes. Various external forces are applied at each node, which cause a (small) displacement in
the node positions.
f
∈
R
n
will denote the vector of (components of) external forces, and
d
∈
R
n
will denote the vector of corresponding node displacements. (By ‘corresponding’ we mean if
f
i
is,
say, the
z
-coordinate of the external force applied at node
k
, then
d
i
is the
z
-coordinate of the
displacement of node
k
.) The vector
f
is called a
loading
or
load
.
The structure is linearly elastic,
i.e.
, we have a linear relation
f
=
Kd
between the vector of
external forces
f
and the node displacements
d
. The matrix
K
=
K
T
≻
0 is called the
stiffness
matrix
of the truss. Roughly speaking, the ‘larger’
K
is (
i.e.
, the stiffer the truss) the smaller the
node displacement will be for a given loading.
We assume that the geometry (unloaded bar lengths and node positions) of the truss is fixed; we
are to design the cross-sectional areas of the bars. These cross-sectional areas will be the design
variables
x
i
,
i
= 1
,...,m
. The stiffness matrix
K
is a linear function of
x
:
K
(
x
) =
x
1
K
1
+
· · ·
+
x
m
K
m
,
where
K
i
=
K
T
i
followsequal
0 depend on the truss geometry. You can assume these matrices are given or
known. The total weight
W
tot
of the truss also depends on the bar cross-sectional areas:
W
tot
(
x
) =
w
1
x
1
+
· · ·
+
w
m
x
m
,
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where
w
i
>
0 are known, given constants (density of the material times the length of bar
i
). Roughly
speaking, the truss becomes stiffer, but also heavier, when we increase
x
i
; there is a tradeoff between
stiffness and weight.
Our goal is to design the stiffest truss, subject to bounds on the bar cross-sectional areas and total
truss weight:
l
≤
x
i
≤
u, i
= 1
,...,m,
W
tot
(
x
)
≤
W,
where
l
,
u
, and
W
are given. You may assume that
K
(
x
)
≻
0 for all feasible vectors
x
. To obtain
a specific optimization problem, we must say how we will measure the stiffness, and what model of
the loads we will use.
(a) There are several ways to form a scalar measure of how stiff a truss is, for a given load
f
. In
this problem we will use the
elastic stored energy
E
(
x,f
) =
1
2
f
T
K
(
x
)
−
1
f
to measure the stiffness. Maximizing stiffness corresponds to minimizing
E
(
x,f
).