EXISTENCE, UNIQUENESS, AND COMPUTATION OF THE RESPONSE OF
AN ARRAY OF ELASTOPLASTIC SPRINGS TO PERTURBATIONS:
A COMPLEMENTARITY APPROACH 1
Ivan Gudoshnikov
[email protected]
Department of Mathematical Sciences, The University of Texas at Dallas
1. Introduction
In section 3 we consider a mechanical system which is a lattice of m elastoplasctic bars, but
some of the bars have rigid controlled pistons, attached to their endpoints (see figure 1). Also
there are external forces, applied to the nodes(joints) of the lattice. We assume, that we know
lengths of pistons and external forces for each moment of time t ∈ [0, T ].
Figure 1. An example of the lattice with pistons. There are 13 elastoplastic
bars and 3 rigid controlled pistons
As suggested in [1, 6a], we describe state of each elastoplastic bar by two real values: elastic
enlongation ek and plastic enlongation pk , so the sum ek + pk is total enlongation of the bar
with respect to some fixed zero length (which corresponds to the origin in space of enlongations
Rm ), see figure 2. The vectors e and p of, respectively, elastic and plastic enlongations of all bars
describe a configuration of the enitre system at any moment of time.
Figure 2. Elastic and plastic enlongations of a bar.
We are looking for the functions e(t) and p(t), such that at any moment of time the resultant
forces at all nodes are zero (it is called quasistatic evolution of the system). Besides this, the
system of bars have to be geometrically valid, i.e. the total enlongation e + p has to be such that
a) it is physically possible for the bars to stay connected, b) length of those bars, which have
an attached piston, has to be the same as the length of the piston. The condition a) may be
written as an inclusion of e + p into a secon-order hypersurface in Rm , but under assumption
of small enlongation we may linearize it (similarly to the Finite Element Method) and obtain a
linear constraint (which we call geometric constraint), of the type
e(t) + p(t) ∈ Im D,
1
The research is partially supported by National Science Foundation grant CMMI-1436856
1
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
where D is a linear operator. The condition b) produces affine constraint (which we call additional
constraint)
e(t) + p(t) ∈ U + g(t),
(1.1)
m
where U ⊂ R is a hyperplane and the vector g represents changing lengths of the pistons.
Clearly, the intersection of th geometric constraint and the additional constraint also has the
form (1.1) with some different U and g.
Physically, a resultant force at a node is a composition of a stresses of all bars, connected to
the node, external force, applied to the node, and reactions of the pistons. The stresses of bars
are generated by the elastic enlongations according to the Hooke’s law (so only this part of a
total enlongation of a bar generates stress). We collect stresses of all bars into a vector s(t), so
the Hookes law becomes
s(t) = Ae(t),
(1.2)
where A is a diagonal matrix of Hooke’s koefficients.
To have quasistatic evolution it is clearly needed to have external forces such that they can be
compensated by others mentioned forces. This implies that the external forces can be represented
in form of a vector h(t) ∈ Rm , where each component hk is a magnitude of two forces, applied to
both endpoints of a k−th bars in opposite directions (similarly to how the stresses are applied),
see figure 3 for example. The resultant force vanishes at all nodes iff
(1.3)
∃(rt ∈ U ⊥ )[h(t) + s(t) + rt = 0],
where rt is reaction of the constraint.
Figure 3. External forces, applied to the nodes, can be represented as h =
(h1 , h2 , h3 ) ∈ R3 .
+
We also assume that a possible stress value of each bar is restricted by an interval [c−
k , ck ].
The plastic component p is required to maintain such restriction. Assuming that the plastic
enlongation of a bar changes only when its stress reaches the boundary, we can derive the
following law of plasticity:
(1.4)
d
p(t) ∈ NC (s(t)),
dt
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
where NC (s(t)) is an outwards normal cone to the rectangular set C of all admissible vectors of
stress values, taken at current stress point s(t).
We also need initial conditions e0 , p0 , such that (1.1)-(1.4) are satisfied for t = 0. We call a
lattice with pistons, described by (1.1)-(1.4) Physical Elastoplastic Model.
The plasticity law (1.4) is a particular type of so-called resistance laws (see [1, Chapter 4]) and
the same resistance law also corresponds to the Coloumb’s law of friction (see [1, 4e]). This gives
us the opportunity to write another model, utilizing dry friction phenomena instead of plasticity.
The new model consists of just two point in a vector space, connected by an elastic spring. The
stress of the spring is denoted by s. The point p (we call it “plastic” component) experience dry
friction, such that the force of friction always belongs to a closed convex set C. The other point
x (the “visible” component) is locked in a translationally moving plane U + g(t) and submitted
to an external force c(t) (see figure 4). For this model we also consider the quasistatic evolution
problem. Then we show in the section 4, that the mathematical description of the Physical
Elastoplastic Model will completely fit into this new model (h corresponds to c, e + p to x and
others variables are the same up to a sign).
f
p
c(t)
s
C
−s
x
r
U + g(t)
Figure 4. Abstract elastoplastic problem.
This model may be, in some sence, easier to imagine then the Physical Elastoplastic Model.
But it also may be seen without any relation to the physical intuition, just as a purely abstract
mathematical problem. It will be formulated in terms of abstract linear spaces, so we call it
Abstract Elastoplastic Problem.
In order to solve it we will make several assumptions and reformulate the problem in terms of
a separable Hilbert space (in case of a Physical Elastoplastic Model we will still have it in terms
of enlongations space Rm ). We do this by identifying the space of forces with the configurations
space by the bijection A, so the stress s is identified with the elastic deviation e = x − p, external
force c is identified with a vector A−1 c and so on. We introduce the new variables
y := s − c − g,
z := x − c − g.
Then the Moreau Sweeping Process
−y 0 ∈ N(C−c(t)−g(t))∩U ⊥ (y),
y(0) = y0 .
allows us to find the elastic part of solution. Further, the differential inclusion
0
z ∈ (NC (y + c + g) + y 0 ) ∩ U,
z(0) = z0 .
always has a solution, which is directly connected to the plastic part. Section 5 contains the
proof of this method, which was also taken from [1] and [2] and filled with details.
Notice, that the moving set in the Moreau Sweeping Process is a cross-section of the set C,
translated on c + g, by the fixed orthogonal complement of the constraint plane (the ortogonality
3
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
here is understood in sence of a specific inner product). Recall, that the vectors c and g, may
represent the applied external forces and lengths of the pistons respectively, so their change leads
to the change of the cross-section, which sweeps the elastic part. Section 6 of the paper shows
this process in detail by manually solving a simple concrete example of an Physical Elastoplastic
Model. Also the analysis of the example highlights several general properties of such models.
2. Preliminaries
2.1. Normal cones and convex analysis.
Definition 2.1. For a Hilbert space H with an inner product < ·, · > and a convex set C ⊂ H
the support function of C is defined as
∗
δC
: H → R ∪ {+∞}
∗
δC
: x 7→ sup{< x, c >: c ∈ C}
Definition 2.2. For a Hilbert space H with an inner product < ·, · >,a closed convex set C ⊂ H
and y ∈ C the set
NC (y) = {ξ ∈ H : ∀(c ∈ C)[< ξ, c − y >6 0]}
is called an outwards normal cone to C at y.
Remark 2.1. Sometimes we will use the concept of the normal cone in more general case, when
< ·, · >: E × F → R is a bilinear form, defined on two linear spaces E, F , which not necessaraly
satisfies all inner product axioms.
Lemma 2.1. For any y, a ∈ H s.t. y + a ∈ C
NC (y + a) = NC−a (y).
Proof.
NC (y + a) = {ξ ∈ H : ∀(c ∈ C)[< ξ, c − y − a >6 0]} =
= {ξ ∈ H : ∀(c ∈ C − a)[< ξ, c − y >6 0]} = NC−a (y).
Lemma 2.2.
NC (y) = −N−C (−y)
Proof.
NC (y) = {ξ ∈ H : ∀(c ∈ C)[< ξ, c − y >6 0]} = {ξ ∈ H : ∀(c ∈ −C)[< ξ, −c − y >6 0]} =
= {ξ ∈ H : ∀(c ∈ −C)[< ξ, (−y) − c >6 0]} =
= −{ξ ∈ H : ∀(c ∈ −C)[< ξ, c − (−y) >6 0]} = −N−C (−y).
Lemma 2.3. For closed convex sets C1 , C2 ⊂ H, s.t. int C1 ∩ C2 6= ∅ and y ∈ C1 ∩ C2
NC1 (y) + NC2 (y) = NC1 ∩C2 (y).
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Proof.
Embedding NC1 (y) + NC2 (y) ⊂ NC1 ∩C2 (y):
Put ξ1 ∈ NC1 (y), ξ2 ∈ NC2 (y), i.e.
∀(c ∈ C1 )[< ξ1 , c − y >6 0]
hence
^
∀(c ∈ C2 )[< ξ2 , c − y >6 0],
h
i
^
∀(c ∈ C1 ∩ C2 ) (< ξ1 , c − y >6 0) (< ξ2 , c − y >6 0)
which implies
∀(c ∈ C1 ∩ C2 )[< ξ1 , c − y > + < ξ2 , c − y >6 0]
which is equivalent to
∀(c ∈ C1 ∩ C2 )[< ξ1 + ξ2 , c − y >6 0]
i.e. ξ1 + ξ2 ∈ NC1 ∩C2 (y). For a proof of the reversed embedding see [2, Lemma 1(b)].
Lemma 2.4. For a Hilbert space H1 =H × R with the inner product
< (x1 , ξ1 ), (x2 , ξ2 ) >1 =< x1 , x2 > +ξ1 ξ2 ,
for any ξ ∈ R we have
NC (x) × {0} = NC×R ((x, ξ)).
Proof. Let y ∈ NC (x), i.e.
∀(c ∈ C)[< y, c − y >6 0].
Observe, that for any c1 ∈ C, c2 ∈ R
< (y, 0), (c1 , c2 ) − (x, ξ) >1 =< y, c1 − x > +0 · (c2 − ξ) =< y, c1 − x >6 0,
so
(y, 0) ∈ NC×R ((x, ξ))
.
Now let
(y1 , y2 ) ∈ NC×R ((x, ξ)),
i.e. for all c1 ∈ C, c2 ∈ R
< (y1 , y2 ), (c1 , c2 ) − (x, ξ) >1 6 0
< y1 , c1 − x > +y2 (c2 − ξ) 6 0
y2 (c2 − ξ) 6 − < y1 , c1 − x > .
Two cases are possible:
(a) y2 = 0, hence < y1 , c1 − x >6 0, so we have
(y1 , y2 ) ∈ NC (x) × {0}.
(b) y2 6= 0,
< y1 , c1 − x >
,
y2
< y1 , c1 − x >
+ ξ.
c2 6 −
y2
But we can choose any c1 ∈ C and then choose
< y1 , c1 − x >
c2 > −
+ ξ,
y2
since the right-hand side is just a number. So we have a contradiction and this case is not
possible. 5
c2 − ξ 6 −
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Lemma 2.5. [2, Lemma 1(d)] If C ∈ H is nonempty, closed and convex, and ρ > 0 then the set
Cρ := {x ∈ H : dist(x, H \ C) > ρ}
is convex.
Lemma 2.6. [2, Lemma 1(c)] If C ⊂ H is nonempty, closed and convex, ρ > 0 and x ∈ Cρ then
∗
∀(y ∈ H) [ρ · kyk+ < y, x >6 δC
(y)] .
2.2. Moreau sweeping process.
Definition 2.3. ([2, Section 3.1]) Let H be a hilbert space, a map
C : [0, T ] → 2H ,
and u0 ∈ C(0). We say that a function u : [0, 1] → H is a solution of the Moreau Sweeping
Process
(
−u0 (t) ∈ NC(t) (u(t)),
u(0) = u0 ;
iff u satisfies the following 4 properties:
(1)
(2)
(3)
(4)
u(0) = u0 ,
for all t ∈ [0, T ] we have u(t) ∈ C(t),
for almost all t ∈ (0, T ) there exists u0 (t),
for almost all t ∈ (0, T ) we have −u0 ∈ NC(t) (u(t)).
Theorem 2.1. ([1, Chapter 5]),[2, Section 3.1]) If C(t) : [0, T ] → 2H is absoluely continous
in sence of Hausdorff distance and for every t ∈ [0, T ] the set C(t) is nonempty, closed and
convex, then there exists a unique soluton u of the Moreau Sweeping Process. The function u is
absolutely continous and u0 ∈ L2 ([0, T ]; H). If C(t) is Lipscitz-continous with a constant L in
sence of Hausdorff distance, then then u is also Lipschitz-continous with L.
2.3. Measurable functions and multifunctions.
Lemma 2.7. (see [6, Lemma 6.4.2]) Let X be a Hausdorff topological space and Y be a separable
metric space. Then Borel σ-algebra, generated by product topology on X × Y is equal to product
σ-algebra of Borel σ-algebras of X and Y :
B(X × Y ) = B(X) ⊗ B(Y )
Definition 2.4. (see [4, Def. III.10], cf. [5, Def. 1.§3.1])
Let (T, S ) be a measurable space (i.e. S is a σ-algebra of subsets of T), X — separable metric
space,
Φ : T → {F ⊂ X : F − complete} ⊂ 2X
then Φ is said to be S -measurable if
T0 := {t ∈ T : Φ(t) = ∅} ∈ S
and
∀(U ⊂ X : U − open) Φ−1 (U ) := {t ∈ T : Φ(t) ∩ U 6= ∅} ∈ S .
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Theorem 2.2. (see [4, Th. III.9]) Let (T, S ) be a measurable space, X — separable metric
space,
Φ : T → {F ⊂ X : F − complete} ⊂ 2X \ {∅}
Φ is S -measurable iff it admits a pointwise dense sequence of strongly measurable selections:
i
h
∃({ϕi }i∈N : (ϕi : T → X) − S -measurable)∀(t ∈ T ) Φ(t) = {ϕi (t)}i∈N
Theorem 2.3. (see [4, Prop. III.13]) If (T, S ) is a measurable space, X is a separable metric
space,
Φ : T → {F ⊂ X : F − complete} ⊂ 2X
is S -measurable, then the graph of Φ satisfies
G(Φ) := {(t, x) ∈ T × X : x ∈ Φ(t)} ∈ S ⊗ B(X).
Theorem 2.4. (Projection theorem, see [5, Prop. 1.§3.6]) Let (Ω, A , ν) be a complete-measure
space and (X, d) be a complete separable metric space(Polish space). Then for every A ∈ A ⊗
B(X)
projΩ (A) := {ω ∈ Ω : ∃(x ∈ X)[(ω, x) ∈ A]} ∈ A .
2.4. Some properties of Bochner integral. Since we are dealing with absolutely continous
functions and integration in a Hilbert space it may be useful to recall the following. For absolutely
continous vector-valued functions and the Bochner integral the statement of the Fundamental
Theorem of Lebesgue integral calculus (see [8, Th. 4.4.1] for real-valued version) is not generally
true, see [9, Example is section XIII.4] for a counter-example. But in case of reflexive separable
Banach space E an absolutely continous function f : [0, T ] → E has Bochner-integrable derivative
f 0 , defined almost everywhere and
Zb
f (b) − f (a) =
f 0 (s)ds
a
for a, b : 0 6 a 6 b 6 T , see [10, Th. 3.8.6], [11, Appendix]. Notice, that absolute continuity
implies weak absolute continuity, since a continous linear functional is Lipschitz-continous with its
norm and a composition of a Lipschitz-continous function and an absoluteny continous function
is absolutely continous.
In the opposite way, we always have a Bochner integral absolutely continous as a set function
(see [10, Th. 3.7.11]).
3. Development of the Physical Elastoplastic Model
3.1. Configuration of the elastic model. For n ∈ N we consider a set of n points(nodes)
A1 , A2 , ..An and let m pairs of them be connected by bars (where 1 6 m 6 21 n(n − 1)). We
denote a bar, connecting nodes Aik and Ajk , by Bik jk or by Bk , where k = 1..m. We make a
convention, that ik < jk for all k = 1..m and put
K := {(ik , jk ) : k = 1..m}.
Of course, any configuratiuon of the system may be described in terms of nodes coordinates,
but in this text we assume, that movements of vertices Ai are small with respect to lengths of
bars Bij . This allows us to linearize the relation (see further) between coordinates of the nodes
and enlongations of bars Bk with respect to their lengths at a “zero” configuration, which we
7
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
artificially prefer. The choice of this zero state doesn’t play a role until we specify a relation
between a configuration of the system and forces, applied to it. But, for example, the zero state
may be s.t. all of the bars Bij experience no internal stress. We describe the system in terms of
the following variables:
• ξi ∈ R3 — coordinate vector of Ai at “zero” state,
• xi ∈ R3 — small displacement vector of Ai from “zero” state, the total configuration of
the system is denoted by the vector x = (xi )ni=1 ∈ X := R3n ,
• ξˆi ∈ R — coordinate vector of Ai , so ξˆi = ξi + xi ,
• θij = θk ∈ R — length of Bij (for i = ik , j = jk ) when the system is in zero state, i.e.
when all xi = 0,
• eij = ek ∈ R — enlongation of Bij (for i = ik , j = jk ) with respect to zero state, so
length of Bij equals θij + eij when Ai , Aj are at ξˆi , ξˆj . The configuration of all bars is a
m
vector e = (ek )m
k=1 ∈ E := R .
The figure 5 shows how such system may look like.
Figure 5. Configuration of an elastic physical model.
3.2. Linearization. In order to find a relation between x and e we consider a function
ϕ : R3 × R3 → R,
which gives a length of a bar θij + eij by the coordinates of its endpoints ξˆi , ξˆj :
!1
3 2 2
X
(3.1)
ϕ(ξˆi , ξˆj ) = kξˆj − ξˆi k =
ξˆjp − ξˆip
,
p=1
0
ϕ =
∂φ ∂φ ∂φ ∂φ ∂φ ∂φ
,
,
,
,
,
∂ ξˆi1 ∂ ξˆi2 ∂ ξˆi3 ∂ ξˆj1 ∂ ξˆj2 ∂ ξˆj3
!
1
=
2
2 21
3 P
ξˆjk − ξˆik
·
k=1
· −2(ξˆj1 − ξˆi1 ), −2(ξˆj2 − ξˆi2 ), −2(ξˆj3 − ξˆi3 ), 2(ξˆj1 − ξˆi1 ), 2(ξˆj2 − ξˆi2 ), 2(ξˆj3 − ξˆi3 ) .
We evaluate the derivative at the zero configuration ξi , ξj and put it as
ϕ0 (ξi , ξj ) = (−αij1 , −αij2 , −αij3 , αij1 , αij2 , αij3 ) .
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
ξ −ξ
Observe, that αij = kξjj −ξii k is a normalized vector along the bar Bij directed from ξi to ξj . Next,
for ξi , ξj and small displacements xi ,xj we have by the definition of derivative:
eij = ϕ(ξi + xi , ξj + xj ) − ϕ(ξi , ξj ) = ϕ0 (ξi , ξj ) · (xi , xj ) + o(k(xi , xj )k) =
= −αij · xi + αij · xj + o(k(xi , xj )k) = αij · (xj − xi ) + o(k(xi , xj )k).
So for small xi , xj we can assume
(3.2)
m
e = (ek )m
k=1 = (αik jk · (xjk − xik ))k=1 =: Dx,
where D : R3n → Rm is a linear map. In the other words, for in case of small displacements
rotations of the bars are negligible and don’t count for enlongations of the bars. Figures 6 and
7 can give an image of how the linearization works.
Figure 6. Possible positions of a node Aj for xi = 0 and some fixed eij as
described by the original relation (3.1)(solid) and as the linearized relation
(3.2)(dashed).
Figure 7. The real length of Bij and its linerized appriximation. The enlongation eij = αij · (xj − xi ) is the sum of lengths of the bold segments.
3.3. External and internal forces. Let each node Ai is submitted to a given external force
yi ∈ R3 . Let the work done by force y = (yi )ni=1 ∈ Y for displacement x = (xi )ni=1 ∈ X of the
system is given by the bilinear form
·, · : X × Y → R,
(3.3)
x, y =
n
X
i=1
9
xi · yi .
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Here spaces X and Y are just two exemplars of (R3 )n , placed in seprating duality (see, for
example, [3, II.40.§6.1]) by the form (3.3), and · is the usual inner product in R3 .
Let each bar Bij generates force, applied at its endpoints Ai , Aj and directed along αij and
−αij respectively. We call this force tension or stress. Since the magnitude of stress has to be
the same for both endpoints, we can describe it as a number sij . Stress values for all bars form
the linear space S := Rm 3 (sik jk )m
k=1 = s.
For small enlongations e ∈ E the work, done by s ∈ S is given by the bilinear form
< ·, · >: E × S → R,
< e, s >= −
(3.4)
m
X
eik jk sik jk .
k=1
Here spaces E and S are two exemplars of Rm , placed in separating duality by the form (3.4).
We have the minus sign in (3.4) since for both eij and sij positive (or negative) the corresponding
displacement of endpoints of the bar has opposite to applied at the endponts stress direction, so
the work, done by stress has to be negative in such case.
In terms of the configuration spaces X and E we can write, that
D:X→E
Then substituting Dx to (3.4) we obtain a linear functional
s ◦ D : x 7→< Dx, s >,
(3.5)
The following lemma shows, how to represent this functional in the form ·, ŷ , where ŷ is
some element of Y , depending on s. In other words, we represent an operator, adjoint to D in
terms of elements of Y and form (3.3).
Lemma 3.1. Let i = 1..n be an index of a vector component of an element of Y , then
X
X
(3.6)
(D∗ s)i =
sij αij −
sji αji .
j:(i,j)∈K
j:(j,i)∈K
defines a linear operator
D∗ : S → Y
such that
∀(x ∈ X, s ∈ S)[< Dx, s >= x, D∗ s ].
(3.7)
Proof. Substituting definitions (3.3),(3.4) and (3.2) into (3.7) we obtain
(3.8)
−
m
X
(αik jk · (xjk − xik )) sik jk =
n
X
xi · (D∗ s)i
i=1
k=1
n
We fix an arbitrary i = 1..n, p = 1..3 and choose x = (xi0 p0 )3p0 =1 i0 =1 ∈ X such that the righthand side is equal to the real number (D∗ s)ip . Namely, all vectors components of x, s.t. i0 6= i
are set to 0; vector xi is set to (δpp0 )3p0 =1 .
Then in the left-hand side of (3.8) for jk = i we have
X
−
αjip sji
j:(j,i)∈K
and for ik = i we have
−
X
αijp (−1)sij
j:(i,j)∈K
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Putting all together we obtain the numerical equality
X
(D∗ s)ip =
αijp sij −
j:(i,j)∈K
X
αjip sji ,
j:(j,i)∈K
which may be written in vector form as (3.6). Lemma 3.2. D∗ = −DT .
Proof. The definition (3.3) implies that for all x ∈ X, s ∈ S
Dx · s = x, DT s ,
and by definition (3.4) we have
−(Dx · s) =< Dx, s > .
Thus
x, −DT s = − x, DT s = x, D∗ s ,
Hence for all s ∈ S
−DT s = D∗ s,
−DT = D∗ .
Lemma 3.3. Superposition of all forces vanishes iff
y + D∗ s = 0.
(3.9)
Proof. Recall that the stress of a bar Bij , applied to a point Ai has a direction αij and
magnitude sij and assume, that the superposition of tensions and external forces, applied to any
Ai vanishes:
X
X
yi +
sij αij +
sji (−αji ) = 0.
j:(i,j)∈K
j:(j,i)∈K
We substitute (3.6) and conclude, that the equilibrium of forces is equivalent to
y + D∗ s = 0.
3.4. Geometric constraint. Now we are interested in representing the model trough elements
of spaces E and S only. If the load y : [0, T ] → Y is a known function of time, (3.9) implies that
there is at least one function h(t) : [0, T ] → S, s.t. for all t ∈ [0, T ]
(3.10)
D∗ h(t) = y(t),
i.e. h is a representation of external load in terms of the space S. It means that at any moment
(3.11)
y(t) ∈ D∗ S.
This may be also interpreted as follows: external forces have to be such that they may be
compensated by stresses of the bars to maintain the quasistatic character of the evolution.
Next, notice that the equilibrium criterion (3.9) is equivalent to
D∗ (h + s) = 0,
which may be written as
∀(x ∈ X) [ x, D∗ (h + s) = 0] ,
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
which is equivalent (by (3.7)) to
∀(x ∈ X) [< Dx, h + s >= 0] .
(3.12)
So h + s has to be orthogonal to the subspace of possible configurations DX in sence of the form
(3.4), and this has the physical meaning: the system is at rest when the resultant force produces
zero work.
Moreover, if we want to consider evolution of the system in time in terms of enlongations,
we have to watch not only the forces, but also the configuration of enlongations e ∈ E itself. It
sould be consistent from the geometrical point of view, i.e. it should be realizable for at least
one x ∈ X. In our linearized framework this condition can be written as the following inclusion:
e ∈ DX,
(3.13)
which we call the geometric constraint.
Notice, that (3.12) is also equivalent to
h
i
^
∃(r ∈ F )∀(x ∈ X) (< Dx, r >= 0) (h + s + r = 0) ,
or
∃(r ∈ DX ⊥ )[h + s + r = 0],
where the symbol ⊥ is understood in sense of the work form < ·, · >, but by the definition (3.4)
the euclidean orthogonality is equivalent to ⊥.
We call such r the reaction of the constraint (3.13).
Example 3.1. Consider the system with three nodes at
ξ1 = (0, 0, 0), ξ2 = (1, 0, 0), ξ3 = (0, 1, 0),
connected by 3 bars. Then we have
α12 = (1, 0, 0), α13 = (0, 1, 0), α23
1
1
= −√ , √ , 0 .
2
2
So

 

(1, 0, 0) · (x21 − x11 , x22 − x12 , x23 − x13 )
x21 − x11
 (0, 1, 0) · (x31 − x11 , x32 − x12 , x33 − x13 )  
=
x32 − x12
Dx = 
=
1
1 √1
√
√
(−x
+
x
+
x
−
x
)
− 2 , 2 , 0 · (x31 − x21 , x32 − x22 , x33 − x23 )
31
21
32
22
2
 
x11
x12 
 
 

 x13 

−1 0 0 1
0
0
0
0 0 
x21 

0
0
0
1 0 
x
=  0 −1 0 0
 22 

0
0 0 √12 − √12 0 − √12 √12 0 
x
 23 
x31 
 
x32 
x33
Observe, that by changing x11 ,x12 and x22 the one can independently adjust all three components
of Dx to any desired value (see the figure 8), so DX = E.
We also should calculate D∗ S to know admissible values of external load y. For the first time
we use the definition (3.7) to find the matrix D∗ (but we could just take the negative transpose
of D, according to the lemma 3.2):
K = {(1, 2), (2, 3), (1, 3)},
12
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 8. By moving nodes in such directions it is possible to adjust linerized
distances between nodes independently.
(D∗ s)1 = s12 α12 + s13 α13 = (s12 , s13 , 0)
∗
i = 2 : (D s)2 = s23 α23 − s12 α12 = − s√232 − s12 , s√232 , 0
i = 3 : (D∗ s)3 = −s13 α13 − s23 α23 = s√232 , − s√232 − s13 , 0 .
i=1:
We write h instead of s, following the formula (3.10):
 
  
h12
1
y11
 0
h13
y12  
 
  
 0
0
y13  


  − h√23 − h 
 −1
12
y21  
2


  
h
 0
√23
=
y
=
D∗ h = 


22
2
  
 
y23  
0
0

  
 
h
y31  
√23
0

  
2
 
y32  − h√23 −
0

h13
2
y33
0
0
0
1
0
0
0
0
0
−1
0
0
0
0




 
− √12 
 h11
√1  h  .
13
2 

0  h23

√1 
2 
1
− √2 
0
Observe, that the matrix of D∗ is the negative transpose of D. Notice, that the image of D∗ is
described by the following equation:

y13 = 0,




y

23 = 0,


y33 = 0,
y22 = y31 ,




y21 = −y11 − y22 ,



y32 = −y12 − y22 .
It is reasonable, that the third components has to be zeroes, since all nodes lay in the xy−plane.
The last pair of equations can be transformed to
y11 + y21 + y22 = 0,
y12 + y32 + y31 = 0;
which mean a kind of symmetrical balance(see figure 9).
Example 3.2. Consider the system with three nodes at
ξ1 = (−1, 0, 0), ξ2 = (0, 0, 0), ξ3 = (1, 0, 0),
13
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 9. Admissible external forces, decomposed on directions. Same color
means same magnitude.
connected by 3 bars (figure 10).
Figure 10. A system with three aligned bars.
Observe, that
α12 = α23 = α13 = (1, 0, 0),
 

 
 
e12
α12 · (x2 − x1 )
x21 − x11
−1 0 0 1
e23  = Dx = α23 · (x3 − x2 ) = x31 − x21  =  0 0 0 −1
e13
α13 · (x3 − x1 )
x31 − x11
−1 0 0 0
Notice, that we have a relation
e12 + e23 = e13
e12 + e23 − e13 = 0
< e, (1, 1, −1) >= 0
So we have the geometric constraint
e ∈ DX = (1, 1, −1)⊥ .
14
 
x11
x12 
 
 
 x13 

0 0 0 0 0 
x21 

0 0 1 0 0 
x
22
 

0 0 1 0 0 
x
23
 
x31 
 
x32 
x33
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 11. Space E = S = R3 with involved vectors and subspaces.
Hence the reaction r and the superposition of stress and external force s + h = −r has to
belong to the one-dimensional subspace V = span{(1, 1, −1)}, see figure 11. In other words,
r12 = r23 = −r13 ,
i.e.
s12 + h12 = s23 + h23 = −s13 − h13 .
This conclusion coinsides with the observable physical requirement of having all forces compensating each other at each node.
Let’s find admissible external forces.


1
0
1
0
0
0


0
0
0

 
−1 1
0

 h12
 h23 
0
0
0
(3.14)
D∗ h = 


0
 h13
0
0


 0 −1 −1


0
0
0
0
0
0
As expected from the one-dimensional nature of the configuration, all second and third components of external forces have to be zero. So we are left with y11 , y21 , y31 , such that

 y11 = h12 + h13 ,
y21 = −h12 + h23 ,

y31 = −h23 − h13 .
15
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Fix arbitrary y11 , y31 and get from their formulae
h12 = y11 − h13 ,
h23 = −y31 − h13 .
By substituting this into the formula for y21 and cancelling h13 we finally obtain
y21 = −y11 − y31
which means
y11 + y21 + y31 = 0.
3.5. Additional constraints. The model also allows us to enforce additional artificial constraints of the form
(3.15)
e(t) ∈ U + g(t),
where U is a fixed in time linear subspace of E and g : [0, T ] → E is a given function. Since
(U + g(t)) ∩ DX may also be represented in a form (3.15) of a moving affine constraint with some
different U and g, we will consider (3.15) as a general form of constraint on e, replacing (3.13).
If we don’t have any additional constraint, we just put U = DX and g ≡ 0.
Having additional constraints on the configuration gives more possibilities for forces h + s.
Instead of (3.12) we put
∀(u ∈ U ⊂ DX)[< u, h + s >= 0],
to have zero work produced by force on allowed displacements. In terms of reaction this may be
written as
(3.16)
∃(r ∈ U ⊥ )[h + s + r = 0].
Replacement of (3.12) by (3.16) requires a proper formal justification, which we will provide
further as Lemma 3.4 for the following particular case of the constraint.
A particular example of the affine moving constraint (3.15) may be imagined as a rigid telescopic piston, attached to two nodes along with elastic bar Bij . We suppose, that we can control,
how the piston changes its length, so the length as a function of time is given by gij : [0, T ] → R.
The function gij is a part of input data in our model. The following examples illustrate, how the
telescopic pistons are interpreted as affine constraints in the space of enlongations E.
Example 3.3. Consider the system from the example 3.1 with a rigid piston attached to the
nodes A2 and A3 (figure 12).
Figure 12. Three-nodes system with moving affine constraint.
16
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Let the length of the piston be given by g23 (t). Then we have the moving affine constraint
in the form (3.15), where U = R × R × {0} and g(t) = (0, 0, g23 (t)). At a moment of time such
constraint may look like figure 13.
Figure 13. The moving constraint in the space E.
Example 3.4. Consider the system from the example 3.2 with a rigid telescopic piston, attached
to the nodes A1 and A3 , see figure 14.
Figure 14. The system with three aligned bars and the constraint.
For the original system from the example 3.2 we already had a geometric constraint of the
plane DX = (1, 1, −1)⊥ (figure 11). Similarly to the exlample 3.3, the telescopic piston gives us
the additional constraint of the form
e ∈ Ua + ga (t),
where Ua := {(e12 , e23 , 0) : e12 , e23 ∈ R},
ga (t) := (0, 0, g13 (t)),
which looks like the one on the figure 13, except the third coordinate axis corresponds now to
the bar B13 .
17
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
We have to find a linear subspace U ⊂ E and a function g : [0, T ] → E such that for all
t ∈ [0, T ]
U + g(t) = DX ∩ (Ua + ga (t)).
From the obvious geometry we have
U := span {(1, −1, 0)},
g(t) ∈ {(g12 , g23 , g13 (t)) : (g12 , g23 , g13 (t)) · (1, 1, −1) = 0}.
Without loss of generality we may choose g(t) ⊥ U , i.e.
g13 (t) g13 (t)
g(t) :=
,
, g13 (t) .
2
2
The figure 15 shows this process graphically.
Figure 15. Intersection of the constraints.
Lemma 3.4. If the system has pistons between nodes Aik , Ajk , where ik < jk , k = 1..p then the
vanishing of forces criterion is (3.16), where U is intersection of the geometric constraint and
constraints, induced by the pistons.
Proof. Each piston can generate a reaction force, applied at its endpoints Aik and Ajk as
vectors rk αik jk and −rk αik jk respectively for some scalars rk . Since the pistons are assumed to
be perfectly rigid rk have no restrictions on magnitude. Each reaction can be also written as a
vector
rk∗ := (0, . . . , 0, rk , 0 . . . , 0) ∈ F,
where rk has k-th position. Clearly rk∗ ⊥ Uk , where
Uk := Rk−1 × {0} × Rm−k ⊂ E
is a part of the constraint Uk + gk (t), generated by the k−th piston. Put
r∗ :=
p
X
k=1
18
rk∗ .
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
By applying the same argument as in the proof of the lemma 3.3 we can see, that resultant forces
at all nodes vanish iff
!
p
X
∗
∗
y+D s+
rk = y + D∗ (s + r∗ ) = 0.
k=1
Similarly to what we had before, this means that
D∗ (h + s + r∗ ) = 0,
∀(x ∈ X) [ x, D∗ (h + s + r∗ ) = 0] ,
∀(x ∈ X) [< Dx, h + s + r∗ >= 0] ,
∃(r ∈ DX ⊥ )[h + s + r + r∗ = 0],
We put
r
∗∗
∗
⊥
:= r + r ∈ DX +
p
X
Uk⊥
=
DX ∩
k=1
U := DX ∩
p
\
!⊥
Uk
,
k=1
p
\
Uk .
k=1
Thus we obtain
(3.17)
∃(r∗∗ ∈ U ⊥ )[h + s + r∗∗ = 0]
Notice, that U is the intersection of all constraint planes, which we have, so (3.17) is exactly
(3.16).
Remark 3.1. The moving constarints of the type (3.15) may be produced in a different way.
We will briefly describe it here and refer the reader to [1, 3c] for better justification. Let some of
the nodes Ai be locked in translationally moving smooth surfaces {x : ϕi (t, x) = 0}, in such way
that a node can move inside the correspinding surface without any friction, but cannot leave it.
Under the assumption of small displacements, we can replace the surfaces with their tangents,
so the constraint on nodes becomes
∇ϕi (t, ξi ) · xi + ϕi (t, ξi ) = 0.
Since the surfaces are moving translationally, the gradient ∇ϕ(t, ξi ) =: ∇ϕ(ξi ) does not depend
on time, and we can write the constraint as
xi ∈ Li + gi (t),
3
where Li is the linear subspace of R , orthogonal to ∇ϕi (ξi ) and gi : [0, T ] → R3 is such that
∇ϕi (ξi ) · (gi (t) − ξi ) = ϕi (t, ξi ),
e.g.
ϕi (t, ξi )
∇ϕi (ξi ).
k∇ϕi (ξi )k2
See figure 16 for illustration. Putting all of the inclusions together we have for some subspace
L ⊂ X, and a function ψ : [0, T ] → X
gi (t) =
x ∈ L + ψ(t).
Thus
e = Dx ∈ DL + Lψ(t).
Now put U := DL and g(t) := Lψ(t) to obtain (3.15). Notice that the one can model a large
variety of mechanisms using this method.
19
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 16. Moving constraints with points Ai and Aj at some moment of
time. The solid curves represent actual frictionless constraints, the dashed ones
are used for the linearization procedure.
3.6. Split of e and the stress formula. In the previous sections we had established the
constraints on a configuration and stress. Now we will define the dependence between them. We
say that the stress of each bar is given by the Hooke’s law, but it is generated not by the entire
enlongation, but only by its part, which we will call elastic enlongation. For each bar Bij we
replace the single enlongation value eij with two variables: elastic enlongation of a bar, denoted
again by eij and plastic enlongation pij . We will use the plastic enlongation later, while defining
the law of plasticity. Length of a bar Bij is now the sum of three terms:
θij + pij + eij ∈ R,
where θij is again a length of the bar in zero configuration. The figure 17 illustrates two components of enlongation for each bar for a system, similar to example 3.3.
y1
ξ1
x1
ξ3
α13
x3
y3
e23 p23
α12
g23 (t)
ξ2
x2
α23
y2
Figure 17. A three-node example of elastoplastic system.
The stress sij is produced by the elastic enlongation only, according to the Hooke’s law
(3.18)
sij = aij · eij
where aij ∈ R is a fixed characteristic of the bar Bij , such that aij > 0.
20
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
To describe this in terms of the configuration spaces we put
e = (eik jk )m
k=1 ∈ E,
p = (pik jk )m
k=1 ∈ S,
A : E → S,

ai1 j1
0
...
 0
a
...
i
j
2
2

A= .
..
..
.
 .
.
.
0
0
...

0
0,
..
.


,

aim jm
and we obtain
(3.19)
s = Ae.
In other words, to describe a bar as an elastic material we have to specify its relaxed length
(s.t. the bar produces no stress) and the Hooke’s coefficient. In the model they are respectively
θij + pij and aij .
Notice, that the constraint (3.15) now applied to the total enlongation and should be written
as follows:
(3.20)
e(t) + p(t) ∈ U + g(t).
3.7. Plasticity law. Now we have to explain the role of the plastic component p. We assume
that there is a known initial value of p, and we will derive the plasticity law as a differential
d
inclusion of the derivalive dt
p.
First, we explain the plasticity law for a single bar Bij . Assign to the bar two fixed values: c−
ij
−
+
− +
and c+
,
s.t.
c
6
0
6
c
.
So
we
want
to
have
s
∈
[c
,
c
]
during
all
evolution
and
change
p
ij
ij
ij
ij
ij
ij ij
accordingly to maintain this inclusion. At any moment of time there are three different possible
scenarios:
+
(1) if sij (t) ∈ (c−
ij , cij ) then there is no need to change pij , i.e. ṗij = 0.
−
(2) if sij (t) = cij < 0, then this corresponds to a maximal elastic compression of the bar.
Indeed, due to (3.18) and aij > 0 a stress is negative iff elastic enlongation is negative.
Geometrically, this may be seen on the figure 18. In this situation we say that p may
decrease with any rate, i.e. ṗij 6 0.
(3) Analogously, if sij (t) = c+
ij > 0, then this corresponds to a maximal elastic stretch of the
bar and ṗ > 0.
A possible scenario, which leads to change of pij is an increase of a total enlngation due to a
moving constraint. The elastic enlongation will increase and pij will remain constant until the
stress reaches the maximal value. Then the behavior of eij and pij will switch to the opposite.
Notice, that all three cases can be summarizes in the following rule:
(3.21)
ṗij ∈ N[c− ,c+ ] (sij ),
ij
ij
where the normal cone is understood in sence of an inner product in R (usual multiplication of
numbers). The figure 19 illustrates this.
Now we need to couple plasticity laws of all bars together. For ewach k = 1..m we rewrite
(3.21) as
(0, . . . , 0, ṗik jk , 0, . . . , 0) ∈ {0}k−1 × N[c− ,c+ ] (sij ) × {0}m−k ,
ij
ij
where the power is in the sence of Cartesian product. By summation over all k we obtain
m
X
ṗ = (ṗi1 j1 , . . . , ṗim jm ) ∈
{0}k−1 × N[c− ,c+ ] (sij ) × {0}m−k .
ij
k=1
21
ij
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 18. A negative value of sij means an elastic compression of the bar.
Figure 19. Normal cones, corresponding to all three cases.
Notice, that written component-wisely, it implies the conjunction of all (3.21). Further applying lemma 2.4 to each term in the sum (m − 1) times and choosing the components of s as ξ
from the lemma, we get the following:
m
X
ṗ ∈
NRk−1 ×[c− ,c+ ]×Rm−k (s),
ij
ij
k=1
where the normal cone is understood in sence of the usual dot product in Rm .
Now put
m
\
+
+
−
+
−
+
m−k
]
×
.
.
.
×
[c
,
c
]
=
Rk−1 × [c−
⊂ S.
,
c
C := [c−
,
c
]
×
[c
ij , cij ] × R
im jm im jm
i2 j 2 i2 j 2
i1 j 1 i1 j 1
k=1
The figure 20 shows an example of C.
Since int C 6= ∅, we apply lemma 2.3 and get
m
X
NRk−1 ×[c− ,c+ ]×Rm−k (s) = NC (s).
ij
ij
k=1
So we finally have the law of plasticity in the form
ṗ ∈ NC (s).
Recall, that the normal cone here is understood in sence of the usual dot product in Rm . We
may rewrite it, using the normal cone in sence of the work form < ·, · >. By the definition of the
normal cone and (3.4) we obtain
(3.22)
− ṗ ∈ NC (s).
22
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 20. C and s for the three-node example.
At the end we shall say, that the set C here describes the plastic properties of the material,
as long as the matrix A describes its elastic properties.
3.8. Formulation of the problem. Now we will put all parts of the model together. We
consider evolution of the system on the time-interval [0, T ] with some initial state e0 , p0 ∈ E.
The evolution is assumed to be quazistatic, i.e. at each node at each moment of time the resultant
force is zero. External load y may be given as a function y : [0, T ] → Y , satisfying (3.11), but we
find its representation h : [0, T ] → S, such that for all t ∈ [0, T ] we have D∗ h(t) = y(t). Putting
all relations together we obtain the system, which we have to satisfy for all t ∈ [0, T ]:

e(t) + p(t) ∈ U + g(t)
— moving affine constraint (3.20),



∃(rt ∈ U ⊥ )[h(t) + s(t) + rt = 0] — quasistaticity criterion (3.16),
— stress formula (3.19),
 s(t) = Ae(t)


d
− dt
p(t) ∈ NC (s(t))
— law of plasticity (3.22).
In addition, we have the initial condition
e(0) = e0
p(0) = p0 .
Recall, that U, g, h, A and e0 , p0 are given here, everything else we have to consider as unknowns.
3.9. Dependence on choice of h. Generally, the matrix D∗ is not one-to-one mapping, so
it may happen, that for the same physical external force y(t) there are multiple h(t). Here we
consider such case and show, that the choice of a different h doesn’t play any role. The proof is
split on several lemmas since each of them may be useful by itself.
Lemma 3.5. In the setting of the problem (3.20)-(3.22) let h1 and h2 be two representations of
the force y(t). Then for all t ∈ [0, T ]
h1 − h2 ∈ U ⊥ .
23
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Proof. By premise, for all t ∈ [0, T ]
D∗ h1 = D∗ h2 = y
Thus
D∗ (h1 − h2 ) = 0,
h1 − h2 ∈ Ker D∗ = DX ⊥ .
Since
U ⊥ = (DX ∩ Ua )⊥ = DX ⊥ + Ua⊥ ,
we have proven the statement of the lemma.
Lemma 3.6. For any r∗ ∈ U ⊥ let (s, p) be a solution of the problem (3.20)-(3.22) with h(t) =
h1 (t) + r∗. Then the problem (3.20)-(3.22) with h(t) = h1 (t) also has the same solution (s, p).
Proof. The proof is trivial. Indeed, we have (3.16) as
∃(rt ∈ U ⊥ )[(h(t) + r∗ ) + s(t) + rt = 0].
For each t ∈ [0, T ] we put rt∗ = rt + r∗ , so, clearly, rt∗ ∈ U ⊥ and h(t) + s(t) + rt∗ = 0. Since the
only changed terms are rt and h, and all other inclusions in the system don’t contain them, the
same (p, s) is a solution of the system with h(t) = h1 (t).
Theorem 3.1. Fix an external force y : [0, T ] → Y . The solution of the problem (3.20)(3.22) does not depend on the choice of different functions h, if for all t ∈ [0, T ] they all satisfy
D∗ h(t) = y(t).
Proof. Combine the lemmas 3.5 and 3.6.
4. Formulation of the Abstract Elastoplastic Problem
4.1. Formulation of the model. The system at any moment of time from [0, T ] is represented
by two components, both from an abstract linear space of configurations U:
• x — the visible (or exposed) component, submitted to external forces and a perfect affine
bilateral constraint, i.e.
(4.1)
x ∈ L = U + g(t),
where U is a closed linear subspace of U and g : [0, T ] → U is a given function called
guiding or drivng. In terms of the physical interpetation, the point x is locked in a planar
device. Inside the plane it can move freely without any friction, but it cannot leave the
plane.
• p — the hidden (or plastic) component. The difference
x − p =: e ∈ U
is called the elastic deviation.
Assume that we also have an abstract linear space of forces F, and the bilinear form to calculate
work, which places U and F in separating duality:
(4.2)
< ·, · >: U × F → R.
The figure 4 illustrates all involved components, forces and constraints.
Overall, there are five forces, acting in the system.
Forces, applied to p:
24
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
• s — elastic force(produced by the spring), directed towards x, given by the formula
s = A(e) = A(x − p),
(4.3)
where
A:U →F
is a given linear bijection, satisfying the following propreties:
(4.4)
∀(u ∈ U \ {0})[< u, Au >> 0].
(4.5)
∀(u1 , u2 ∈ U)[< u1 , Au2 >=< u2 , Au1 >].
• f — the force of plastic resistance, satisfying the condition
dp
(4.6)
∈ NC (−f ) := {ξ ∈ U : ∀(c ∈ C)[< ξ, c − (−f ) >6 0]},
dt
where C is given nonempty closed convex subset of F. Physically, the set C describes
admissible values of a force of friction. This is a formulation of the law of friction,
experienced by p. It is derived from the “Principle of Maximal Dissipation”, which says
that a direction of a fiction force has to be s.t. its power is minimal over all possible
directions(notice, that the power is negative due to the dissipative nature of a friction
force). For details the reader may look at [1, 4.e] and [2, 3.2]. Notice, that lemma 2.2
and replacement C := −C allow us to rewrite the relation in the form
−ṗ ∈ NC (f )
which is derived in [1, 4.e].
Forces, applied to x:
• r — reaction of the constraint L. It should satisfy the condition
(4.7)
− r ∈ NL (x) := {ξ ∈ F : ∀(l ∈ L)[< l − x, ξ >6 0]}.
Notice, that since L = U + g(t) and U is a linear space, we have
NL (x) = L⊥ = U ⊥ = {ξ ∈ F : ∀(u ∈ U)[< ξ, u >= 0]}.
• c — external force, called load, defined explicitely by a given function c : [0, T ] → F.
• (−s) — elastic force towards p.
Using quazistatic approach, i.e. assuming that the evolution is slow enough to neglect inertia,
we obtain the two following equations, which physical meaning is vanishing of the superposition
of the forces:
(4.8)
s + f = 0,
(4.9)
r + c − s = 0.
Definition 4.1. To summarize the above, we say, that in the Abstract Elastoplastic problem
we are given with:
• driving function g : [0, T ] → U and a closed linear subspace U ⊂ U;
• load function c : [0, T ] → F;
• nonempty closed convex set C ⊂ F;
• linear bijection A : U → F, s.t. (4.5) and (4.4) holds;
• initial condition for x and p.
We have to find:
• x, p : [0, T ] → U, s.t. for all t ∈ [0, T ] the value of x(t) satisfies the constraint (4.1),
• s, f, r : [0, T ] → F, satisfying formulae (4.3),(4.7) and equations (4.8),(4.9) for all t ∈
[0, T ], and (4.6) for almost all t ∈ [0, T ].
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
4.2. Connection with the physical model. In the following table we will show a possible
substitution, which turns the Physical Elastoplastic Model into the Abstract Elastoplastic Problem:
Abstract problem Physical elastoplastic model
U
E
F
S
< ·, · >
< ·, · >
x
total enlongation e + p
p
p
e
e
s
−s
A
−A
C
−C
c
h
r
r
U
U
g
g
Several comments must be made here.
(1) The relations (3.20) and (4.1) are identical.
(2) We define s in the Abstract Problem as −s from the Physical Model. This turns (3.16)
into (4.9) together with (4.7).
(3) The substitution of s and A into the formula (3.19) leads to cancellation of the minus
sign and (4.3). Clearly −A satisfies (4.5) as a diagonal matrix and (4.4) is true for the
form (3.4).
(4) The plasticity law (3.22) means
∀(c1 ∈ C)[< −ṗ, c1 − s >] 6 0
in terms of the Physical Model. Applying the change of variables, in terms of the Abstract
Problem we get for all c1 ∈ C
< −ṗ, −c1 + s >6 0
< ṗ, c1 − s >6 0
The relation (4.8) may be understood here as a definition of f , so we have
< ṗ, c1 − (−f ) >6 0,
and by the definition of a normal cone
ṗ ∈ NC (−f ),
which is precisely (4.6).
5. Solution of the Abstract Elastoplastic Problem through the Moreau
Sweeping Process.
This section contains a mathematical solution of the Abstract Elastoplastic Problem. Under
certain assumptions a trajectory of the elastic component can be found through the Moreau
Sweeping Process with a special moving set, and the trajectory of the plastic component can be
calculated as an integral of a measurable selection from a special multifunction.
26
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
5.1. Transition to a Hilbert space. At first step, we will get rid of the space F in our
calculations, using the bijection A : U → F as an identification. Let H be a space of the same
nature, as U.
We define the bilinear form
(·, ·) : H × H → R
(5.1)
(u1 , u2 ) =< u1 , Au2 > .
This bilinear form defines the inner product in H and a pre-Hilber norm
p
|u|H = (u, u).
Indeed, we can verify all axioms:
(1) ∀(u1 , u2 ∈ H) [(u1 , u2 ) =< u1 , Au2 >=< u2 , Au1 >= (u2 , u1 )] by (4.5),
(2) linearity of (·, ·) follows from linearity of A and the bilinear form < ·, · >,
(3) ∀(u ∈ H)[(u, u) =< u, Au >> 0], (u, u) = 0 ⇔ u = 0 by (4.4),
Assumption 0. H with the norm | · |H is complete and separable.
Now we redefine all values, which we have in F, as themselves, but bijectively mapped to H:
(5.2)
c(t) := A−1 c(t), C := A−1 C, s(t) := A−1 s(t), f (t) := A−1 f (t), r(t) := A−1 r(t).
Notice, that after this replacement an outwards normal cone in H, build up on the inner
product (· , ·) according to the Definition 2.2 coincides with the sets, which we used in (4.6) and
(4.7). Indeed, for arbitrary closed convex nonempty sets C1 ⊂ F, C2 ⊂ U and a ∈ F, b ∈ U we
have
∀(c ∈ C1 )[< b, c − a >6 0] ⇐⇒ ∀(c ∈ C1 )[< b, A(A−1 c − A−1 a) >6 0] ⇐⇒
⇐⇒ ∀(c ∈ A−1 C1 )[(b, c − A−1 a) 6 0] ⇐⇒ b ∈ NA−1 C1 (A−1 a),
∀(c ∈ C2 )[< c − b, a >6 0] ⇐⇒ ∀(c ∈ C2 )[< c − b, AA−1 a >6 0] ⇐⇒
⇐⇒ ∀(c ∈ C2 )[(A−1 a, c − b) 6 0] ⇐⇒ A−1 a ∈ NC2 (b).
Observe, that since U is linear subspace of U and L = U + g(t), for any x ∈ L the normal
cone NL (x) is linear subspace of F, ortogonal to U . From the preceding argument we can see,
that
V := A−1 NL (x)
is the ortogonal complement to U in the sence of (· , ·)
Now we will rewrite the problem through elements of the space H only and get rid of several
variabels:
• (4.3) becomes s = x − p
• ((4.9) and (4.7)) is equivalent to s − c ∈ V
So we obtain the new problem, which is equivalent to the original one: find functions x, p, s :
[0, T ] → H, satisfying for almost all t ∈ [0, T ]:

(5.3)
x∈U +g





(5.4)
s∈V +c
(5.5)
(5.6)

x=p+s



p0 ∈ N (s)
C
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
and initial conditions
x(0) = x0 ,
s(0) = s0 .
5.2. Assumptions. Now we introduce additional assumptions:
Assumption 1. The given functions g and c are absolutely continous.
Assumption 1a. The given functions g and c are Lipschitz-continous.
Assumptions 1 and 1a are two possible variants.
Remark 5.1. Since V the is ortogonal complement of U , we can represent
g(t) = gU (t) + gV (t),
where
c(t) = cU (t) + cV (t),
h
i
^
∀(t ∈ [0, T ]) (gU (t), cU (t) ∈ U ) (gV (t), cV (t) ∈ V )
But U and V are linear spaces, so
x ∈ U + gU + gV = U + gV ,
s ∈ V + cU + cV = V + cU .
and we can assume without loss of generality, that for any t ∈ [0, T ]
g(t) ∈ V,
(5.7)
c(t) ∈ U.
values of given g and c lay in V and U respectively. The equivalent statement for the Physical
Elastoplastic Model was presented as Theorem 3.1.
Assumption 2. The initial conditions x0 , s0 are s.t.
x0 ∈ U + g(0),
s0 ∈ (V + c(0)) ∩ C.
Assumption 3. For any t ∈ [0, T ]
(V + c(t)) ∩ int C 6= ∅
Notice, that (V + c(t)) ∩ C 6= ∅ is automatically required by (5.4) and (5.6) for the problem
to be solvable (which means that the external load can be compensated by stress within given C
and the system can develop quazistatically). The strengthened assumption with int C is called
safe load hypothesis.
Assumption 4. The set C is bounded.
The figure 21 provides a two-dimensional(for simplicity) example of the problem (5.3) - (5.6)
under the assumptions, described above. But the reader should remember, that a nontrivial
Physical Elastoplastic Model necessarily has more than two dimensions.
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 21. A two-dimentional example of the problem (5.3) - (5.6).
5.3. Change of variables. We can rearrange (5.5) to x − s = p and by adding c − g to both
parts obtain
(x − g) + (c − s) = p + c − g,
then (5.3) and (5.4) imply
x − g = proj (p + c − g, U )
c − s = proj (p + c − g, V )
and by (5.7) we can continue:
x − g = proj (p, U ) + proj (c, U ) − proj (g, U ) = proj (p, U ) + c − 0
c − s = proj (p, V ) + proj (c, V ) + proj (g, V ) = proj (p, V ) + 0 − g
So we introduce new variables:
(5.8)
y := s − c − g = −proj (p, V ),
(5.9)
z := x − c − g = proj (p, U )
hence
p = z − y,
and we can reformulate the problem again as
 0
 z − y 0 ∈ NC (y + c + g),
z ∈ U,
(5.10)

y∈V
for almost all t ∈ [0, T ].
Remark 5.2. Notice, that by (5.8) two normal cones, mentioned in (5.6) and (5.10) are the
same set:
NC (y + c + g) = NC (s).
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The response of elastoplastic springs to perturbations
Theorem 5.1. Under assumptions 0, 1 (resp. 1a), 3, 4 for
y0 ∈ (C − c(0) − g(0)) ∩ V,
z0 ∈ U
there exist unique function y : [0, T ] → H and at least one function z : [0, T ] → H, s.t. (5.10)
holds, y(0) = y0 , z(0) = z0 and y, z are absolutely continous (resp. Lipschitz-continous).
5.4. Proof of the theorem 5.1. We will split the proof on several following propositions.
Step 5.1.1. Under the conditions of theorem 5.1 the functions y,z are solutions of (5.10) if and
only if y is a solution of the Moreau Sweeping Process
−y 0 ∈ N(C−c(t)−g(t))∩V (y),
(5.11)
y(0) = y0 .
and for this y there exists z : [0, T ] → H, satisfying (5.10).
Proof. By assumption 3
int C ∩ (V + c(t)) 6= ∅,
i.e.
h
i
^
∃(x ∈ H) (x ∈ int C) (x ∈ V + c(t))
so
(x − c(t) ∈ int C − c(t))
^
(x − c(t) ∈ V )
and since g(t) ∈ V
(x − c(t) − g(t) ∈ int C − c(t) − g(t))
^
(x − c(t) − g(t) ∈ V − g(t) = V )
so
(int C − c(t) − g(t)) ∩ V = (int (C − c(t) − g(t))) ∩ V 6= ∅.
Hence by lemmas 2.3 and 2.1 we obtain
NC (y + c + g) + NV (y) = NC−c−g (y) + NV (y) = N(C−c−g)∩V (y).
Notice, that for absolutely continous function z the condition z ∈ U is equivalent to
−z 0 ∈ U = NV (y)
for almost all t ∈ [0, 1] and z(0) ∈ U . So
(5.12)
− y 0 = z 0 − y 0 − z 0 ∈ NC (y + c + g) + NV (y) = N(C−c−g)∩V (y).
The multimapping t 7→ (C − c − g) ∩ V is absolutely continous under assumptions 1, 3 and 4 (see
[1, Section 5c]) and Lipschitz-continous under assumptions 1a, 3 and 4 (see [2, Lemma 6]). Step 5.1.2. From the step 5.1.1 we obtain well-posed Moreau Sweeping Process. Existence and
uniqueness of its solution are guaranteed by theorem 2.1. Recall, that, by definition (5.8), finding
y means finding the stress s an projection of plastic component p on V . The figure 22 shows the
terms of the Moreau Sweeping Process for the configuration from the figure 21.
But, we still have to find suitable function z, such that the pair (y, z) is a solution of (5.10).
Observe, that all we need is a summable function z 0 , such that
(5.13)
z 0 ∈ (NC (y + c + g) + y 0 ) ∩ U.
The following steps show that there is at least one such function, by constructing a summable
function η = z 0 − y 0 first.
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 22. The moving set (black), the variable (y) and the normal cone (orange) for the Moreau Sweeping Process, generated by Abstract Elastoplastic
Problem.
Step 5.1.3. Put
∗
e
Φ(t)
:= {w ∈ H : δC
(w) = (w, y(t) + c(t) + g(t))}.
The set
e ∩ (U − y 0 (t))
Φ(t) := Φ(t)
is nonempty for almost all t ∈ [0, T ].
Proof. By the inclusion (5.12) for almost any t ∈ [0, T ] we may decompose
− y 0 (t) = w − u,
(5.14)
where u ∈ U and w ∈ NC (y(t) + c(t) + g(t)), so
w ∈ U − y 0 (t)
and
∀(ĉ ∈ C)[(w, ĉ − y(t) − c(t) − g(t)) 6 0]
∀(ĉ ∈ C)[(w, y(t) + c(t) + g(t)) > (w, ĉ)]
∗
(w, y(t) + c(t) + g(t)) > sup{(w, ĉ) : ĉ ∈ C} = δC
(w)
and also by (5.12) we can put
y(t) + c(t) + g(t) =: ĉ ∈ C
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The response of elastoplastic springs to perturbations
so we obtain
(w, y(t) + c(t) + g(t)) = δ ∗ (t).
Step 5.1.4. There exists strongly measurable selection from Φ
η : [0, T ] → H,
i.e. for almost all t ∈ [0, T ]
η(t) ∈ Φ(t)
and the function kη(·)k : [0, T ] → R is Lebesgue-measurable.
Proof.
∗
(1) Since C is closed, convex and nonempty, δC
is lower semicontinous(see e.g. [1, Chapter
2]). Hence (see e.g.[7, Th. 3.3.1])
∗
∀(a ∈ R) [{w ∈ H : δC
(w) > a} is open] ,
i.e.
∗
∀(a ∈ R) [{w ∈ H : δC
(w) > a} ∈ B(H)] .
∗
Therefore the function δC is B(H)-measurable ([7, 4.4]), and for the function
∗
δ̃C
: [0, T ] × H → R,
∗
δ̃C
: (t, w) 7→ δ ∗ (w),
we have
h
i
∗
∗
∀(a ∈ R) {(t, w) ∈ [0, T ] × H : δ̃C
(w) > a} = [0, T ] × {w ∈ H : δC
(w) > a} ∈ B([0, T ]) ⊗ B(H) ,
∗
so δ̃C
is B([0, T ]) ⊗ B(H)-measurable.
(2) Consider the function
v : [0, T ] × H → R,
v : (t, w) 7→ (w, y(t) + c(t) + g(t)).
As composition of absolutely continous functions and inner product is continous mapping
w.r. to the product topology on [0, T ] × H, and since (a, +∞] := {b ∈ R : b > a} is open
for any a ∈ R,
∀(a ∈ R) v −1 ((a, +∞]) ∈ B([0, T ] × H) .
Assumption 0 and lemma 2.7 imply that
B([0, T ] × H) = B([0, T ]) ⊗ B(H),
so v is B([0, T ]) ⊗ B(H)-measurable too.
(3) By [7, Th. 4.4.5]
∗
e = {(t, w) : δ̃C
G(Φ)
(t, w) = v(t, w)} ∈ B([0, T ]) ⊗ B(H) ⊂ L ([0, T ]) ⊗ B(H).
(4) Since y 0 ∈ L2 ([0, T ]; H), there exists L ([0, 1])-measurable function ψ : [0, T ] → R, s.t.
∀(t ∈ [0, T ]\Z) [y 0 (t) = ψ(t)] ,
where Z has measure zero. Since H is separable and U is closed,
h
i
∃ ({ai }i∈N ⊂ U ) {ai }i∈N = U .
Therefore each function t 7→ (ai − ψ(t)) is L ([0, T ])-measurable and
h
i
∀(t ∈ [0, T ]) {ai − ψ(t)}i∈N = U − ψ(t) ,
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
so by theorem 2.2 the multifunction t 7→ U − ψ(t) is L ([0, T ])-measurable, hence by
theorem 2.3 its graph
G(U − ψ) ∈ L ([0, T ]) ⊗ B(H).
0
(5) Denote by Z the measure-zero set of t ∈ [0, 1], where Φ(t) = ∅. Notice, that Z ×
H, Z 0 × H ∈ L ([0, T ]) ⊗ B(H) by the definition of product σ-algebra. It follows from
the definition of σ − algebra, that
e ∩ G(U − ψ)) ∪ (Z × H) ∪ (Z 0 × H) ∈ L ([0, T ]) ⊗ B(H)
F := (G(Φ)
The multifunction, defined by F as its graph is different from Φ only on the measure-zero
set Z ∪ Z 0 and is L ([0, T ])-measurable. Indeed, for any open U ∈ H we have
([0, T ] × U ) ∩ F ∈ L ([0, T ]) ⊗ B(H)
and by theorem 2.4
proj[0,1] (([0, T ] × U ) ∩ F ) ∈ L ([0, T ]).
So by theorem 2.2 there exists a strongly measurable selection η : [0, T ] → H, s.t for
almost all t ∈ [0, T ]
η(t) ∈ Φ(t).
Finally observe, that the numerical function kη(t)k is Lebesgue-measurable since H is
separable(see [10, Th. 2.8.5, th. 3.5.2, th. 3.5.3]).
Step 5.1.5. Observe, that for almost all t ∈ [0, T ]
η(t) + y 0 (t) ∈ U
(5.15)
and for all ĉ ∈ C(t)
(η(t), y(t) + c(t) + g(t)) = δ ∗ (η(t)) > (η(t), ĉ)
so
(η(t), ĉ − y(t) − c(t) − g(t)) 6 0
i.e
(η(t) + y 0 (t)) − y 0 (t) = η(t) ∈ NC (y(t) + c(t) + g(t)),
which, together with (5.15) means (cf. (5.13))
η(t) + y 0 (t) ∈ (NC (y(t) + c(t) + g(t)) + y 0 (t)) ∩ U.
Step 5.1.6. There exists ρ > 0 and an absolutely continous function h : [0, T ] → H, s.t. for all
t ∈ [0, T ]
h(t) ∈ C ∩ (V + c(t)),
(5.16)
and the closed ball
Bρ [h(t)] ⊂ C.
(5.17)
Proof. For any ρ > 0 define a set
Cρ := {w ∈ H : dist (w, H \ C) > ρ} ⊂ C.
First, we prove, that
(5.18)
∃(ρ > 0)∀(t ∈ [0, T ]) [Cρ ∩ (V + c(t)) 6= ∅] .
Assume the contrary,
∀(ρ > 0)∃(t ∈ [0, T ]) [Cρ ∩ (V + c(t)) = ∅] .
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Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Then take a sequence ρi → 0 and find the corresponding {ti } ⊂ [0, T ]. By compactness of [0, T ]
we can choose a convergent to some t0 ∈ [0, T ] subsequence: tij → t0 . Notice that ρij → 0. Now
take an arbitrary v ∈ V and observe, that
v + c(tij ) ∈
/ Cρ ,
i.e.
dist(v + c(tij ), H \ C) < ρij
Since for any A ∈ H the function dist(·, A) : H → R is continous, we have
dist(v + c(t0 ), H \ C) = 0.
This means, that
v + c(t0 ) ∈ H \ C = H \ int C,
so
(V + c(t0 )) ∩ int C = ∅,
which contradicts the assumption 3 (“safe load hypothesis”). Hence (5.18) is correct. Cρ is
closed as continous inverse image of a closed set and convex by the lemma 2.5. Moreover, the
multimapping
t 7→ Cρ ∩ (V + c(t))
(5.19)
is absolutely continous(again, by [1, Section 5c]), convex-valued and closed-valued. Take any
h0 ∈ Cρ ∩ (V + c(0)) and put the function h to be the solution of the Moreau Sweeping Process
with the moving set (5.19) and initial condition h0 . By the definition of Cρ the condition (5.17)
is satisfied for all t ∈ [0, T ]. Step 5.1.7. The function
kη(·)k : [0, T ] → R
is Lebesgue-summable.
Proof. By lemma 2.6 we obtain for all t ∈ [0, T ]
∗
ρ · kη(t)k + (η(t), h(t)) 6 δC
(η(t))
Since η(t) ∈ Φ(t),
∗
ρ·kη(t)k = δC
(η(t))−(η(t), h(t)) = (η(t), y(t)+c(t)+g(t))−(η(t), h(t)) = (η(t), y(t)+c(t)+g(t)−h(t)) =
= (η(t) + y 0 (t), y(t) + c(t) + g(t) − h(t)) − (y 0 (t), y(t) + c(t) + g(t) − h(t))
By (5.16) we have h(t) ∈ V + c(t), i.e. h(t) − c(t) ∈ V , by step 5.1.1 we have y(t) ∈ V , and recall,
that g(t) ∈ V (5.7). From the other side, we have (5.15), so since U ⊥ V we obtain
(η(t) + y 0 (t), y(t) + c(t) + g(t) − h(t)) = 0.
Hence
(5.20)
ρ · kη(t)k = −(y 0 (t), y(t) + c(t) + g(t) − h(t)) 6 ky 0 (t)k · ky(t) + c(t) + g(t) − h(t)k.
The function t 7→ ky(t) + c(t) + g(t) + h(t)k is continous on the compact interval [0, T ], so
let M be the maximum. Recall, that, by properties of Moreau Sweeping Process, y 0 (t) ∈
L2 ([0, T ], H) ⊂ L1 ([0, T ], H), so the Lebesgue integral
ZT
ZT
kη(t)kdt 6
0
M 0
M
ky (t)kdt =
ρ
ρ
0
ZT
0
34
ky 0 (t)kdt < ∞.
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Step 5.1.8. Finally we have proven, that η is strongly measurable and kη(·)k measurable and
summable. So, by [10, Th. 3.7.4] the function η is Bochner-integrable, so we put
Zt
z(t) := z0 +
(η(s) + y 0 (s))ds.
0
Observe, that step 5.1.5 shows, that such z satisfies conditions (5.10). By [10, Th. 3.7.11] the
function z is absolutely continous.
In case of Lipschitz-continous input the function y is Lipschitz-continous as a solution of
Moreau Sweeping Process,(see step 5.1.1), hence its derivative y 0 is bounded by some constant
Ly and from (5.20) we have for almost all t ∈ [0, T ]
kz 0 (t)k 6
M T Ly
.
ρ
Since z is absolutely continous as an integral, for t1 , t2 , s.t. 0 6 t1 6 t2 6 T we have
t
Zt2
Z 2
M T Ly
0
0
kz(t2 ) − z(t1 )k = · (t2 − t1 ),
z (t)dt 6 kz (t)kdt 6
ρ
t1
t1
so z is also Lipschitz-continous.
This concludes the proof of the theorem 5.1. 6. Example 1
In this section we will develop and analyze an example of Physical Elastoplastic Model. The
example is simple enough, so a physical intuition is applicable, but the example is still nontrivial
in sence that it contains all essential components of the general model.
Besides giving an illustration for the general model, the development of this example was
motivated by the following questions:
Question 6.1. Having a pair of input functions (c, g) of an Abstract Elastoplastic Problem and
a response of the system (x, p), can we change both of the response components by alternating g
only? Can we do it by alternating c only?
Question 6.2. Does Question 6.1 make sence if the Abstract Elastoplastic Problem corresponds
to some Physical Elastoplastic Model? In particular, is it possible for the plastic component of
such system to change in response to external forces only, i.e. when g ≡ 0?
It is quite easy to see, that we can, indeed, influence both components by changing g. But,
since magnitudes of external forces are limited due to the quazistatic approach (moreover, by
the safe load hypothesis), the question about c is less trivial. And, regarding a physical model,
it may be hard to figure out immidiately, how elastoplastic bars may change ther plastic length,
while external forces remain inside the elastic domain.
6.1. Example setup. Consider the mechanism, presented in example 3.4. Recall, that it consists from three aligned elastoplastic bars between the nodes at ξ1 = (−1, 0, 0), ξ2 = (0, 0, 0), ξ3 =
(1, 0, 0) at the zero configuration. Also there is a controlled rigid piston between nodes A1 and
A3 (see figure 14). Intersection of the geometric constraint and the moving constraint, induced
by the piston gives us (see figure 15) the moving affine constraint of the type (3.20):
e(t) + p(t) ∈ U + g(t),
35
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
where
(6.1)
U = span {(1, −1, 0)},
g(t) =
g13 (t) g13 (t)
,
, g13 (t) .
2
2
in the space of enlongations E = R3 .
We put the elastic boundaries for each bar as
−
+
+
−
+
c−
12 = c23 = −2, c12 = c23 = 2, c13 = −1, c13 = 1
and, for simplicity, their stiffness as
a12 = a23 = a13 = 1.
So, in terms of the Physical Elastoplastic Model, we have
C = [−2, 2] × [−2, 2] × [−1, 1],
A = I.
Applying the change of variables from the table in the section 4.2 and the transition to a Hilbert
space (5.2) we get the system of the form (5.3)-(5.6):

x ∈ U + g, (5.3)



s ∈ V + c,
(5.4)
x
=
p
+
s,
(5.5)


 0
p ∈ NC (s). (5.6)
in terms of the space H = R3 with the usual dot product. Indeed, by substituting A = −I
and the form (3.4) into (5.1) we obtain the usual dot product in a three-dimensional space,
so Assumption 0 is clearly satisfied. The sign for s and C is changed to the opposite twicely.
Notice, that Assumption 4 (boundness of C) is clearly satisfied here. The external load is now
represented as
(6.2)
c(t) = −h(t).
The terms U and g of the affine constraint for the space H are the same as (6.1), and
V = U ⊥ = (1, −1, 0)⊥ = {s ∈ H : s12 = s23 }.
Recall, that by the remark 5.1 we can take c(t) ∈ U, g(t) ∈ V without loss of generality. In
our case the vector g(t) from (6.1) is already taken orthogonal to U . Inclusion c(t) ∈ U means
that we can take a parameter function
cpar : [0, T ] → R
and put
(6.3)
c(t) = (cpar (t), −cpar (t), 0).
The figure 23 illustrates the inclusions (5.3),(5.4) and (5.6).
Observe from the figure 23, that Assumption 3 (the safe load hypothesis) means, that for all
t ∈ [0, T ] the vector c(t) belongs to the interior of C, so the smaller rectangle (V + c(t)) ∩ C
doesn’t degenerate to a segment. This means, that
(6.4)
∀(t ∈ [0, T ]) [|cpar (t)| < 2] .
6.2. Physiscal meaning of the problem. Now we have the example problem formulated in
terms of the enlongations space H. Before we will solve it, it seems useful to go back to the
physical representation to check again the physical meaning of some elements of the problem.
36
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 23. The exposed component x belongs to the subspace U , shifted on
the vector g (e.g. the black dot). The stress s has to be in the small rectangle,
which is section of the set C by the subspace V , shifted on the green vector c.
Derivative p0 must be a vector from a normal cone of the green frame box C,
taken at s.
6.2.1. Applied external forces. Since we consider the external load from the subspace U only, i.e.
it is in the form (6.3), where |cpar | < 2, we find the actual forces y ∈ Y , applied to the nodes
A1 , A2 , A3 . To do this we substitute (6.3) to (6.2) and then submit h to the operator D∗ from
(3.14):


cpar (t)
y(t) = −D∗ −cpar (t) ,
0

y11 = −cpar ,



y21 = 2cpar ,
 y31 = −cpar ,


y12 = y13 = y22 = y23 = y32 = y33 = 0.
Observe, that these forces stretch one of the bars B12 , B23 , compress another, and the whole
mechanism remains at the same place (see figure 24).
Figure 24. Possible forces, applied to the nodes.
37
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
6.2.2. Balance of forces. The inclusion (5.4), which right-hand part is shown on the figure 23 as
the small rectangle, stands for the vanishing criterion of resultant forces at all nodes. We can
verify this as follows. First, notice, that by (6.1) and (6.3) the inclusion (5.4) may be written
here as
(1, −1, 0) · s = (1, −1, 0) · (cpar , −cpar , 0),
(6.5)
s12 − s23 = 2cpar .
From the other side, the balance of forces at A2 is
s23 − s12 + y21 = 0,
2cpar = s12 − s23 ,
which coincides with (6.5). Assume, that a scalar r̂ is a reaction of the piston (which may be
any number, since we assumed that the piston is perfectly rigid). This reaction is applied along
α13 = (1, 0, 0) at A1 and along −α13 at A3 (similarly to stresses of elastoplastic bars). Then we
can write the balance of forces at A1 and A3 as
s12 + s13 + r̂ + y11 = 0,
−s23 − s13 − r̂ + y13 = 0
respectively. We add the equations together to get rid of r̂ and get
s12 − s23 − 2cpar = 0,
which again coincides with (6.5).
Remark 6.1. Notice, that when cpar = 0, i.e. the external force is zero, we have s ∈ V , which
means that s12 = s23 , i.e. the bars B12 and B23 are stressed equally.
6.2.3. The moving affine constraint. Recall, that the inclusion (5.3) was obtained as an intersection of the geometric constraint and the constraint, generated by the rigid piston. Observe,
that (5.3) together with (5.5) (which means, that the total enlongation is the sum of elastic and
plastic ones) may be written as
e + p − g = (τ, −τ, 0)
where τ ∈ R is an arbitrary number. Coordinate-wisely:

g
 e12 + p12 − 213 = τ,
g13
e23 + p23 − 2 = τ,

e13 + p13 − g13 = 0;
e12 + p12 + e23 + p23 = g13 ,
e13 + p13 = g13 .
Observe, that these are exactly properties of enlongations of bars in the mechanism on the figure
14.
6.3. Input functions and initial conditions. For simplicity, we put
ċpar ≡ 1,
cpar (0) = 0;
i.e.
cpar (t) = t,
c(t) = (t, −t, 0).
To satisfy (6.4) (i.e. Assumption 3) we put T such that
1 < T < 2.
38
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
For sake of Question 6.2 we put
g ≡ 0,
i.e.
g13 ≡ 0.
Such input functions are Lipschitz-continous, so Assumption 1a is satisfied.
We also put the initial condtions as follows:
x0 = 0,
s0 = (1, 1, 0).
Observe, that
x0 ∈ U + 0 = U + g(0),
s0 ∈ V + 0 = V + c(0),
s0 ∈ C;
so Assumption 2 is also satisfied.
Remark 6.2. Notice, that Remark 6.1 applies here, i.e. the bars B12 and B23 have equal positive
stress. But, since the visible configuration x of the system at the moment t = 0 coincides with
the zero configuration, the positive elastic enlongations have to be compensated with nonzero
plastic enlongations
p(0) = x − s = (−1, −1, 0).
(6.6)
6.4. Solution and the Moreau Sweeping Process. For now we have specified all elements
of the problem and confirmed all assumptions, so we are ready to solve the problem.
Recall, that to do this we perform the change of variables (5.8)-(5.9):
y := s − c − g = s(t) − (t, −t, 0),
z := x − c − g = x(t) − (t, −t, 0),
y0 := s0 − 0 = (1, 1, 0),
z0 := x0 − 0 = 0;
to obtain the problem (5.10):
 0
 z − y 0 ∈ NC (y + c)),
z ∈ U,

y ∈V;
for which we find the function s from the Moreau Sweeping Process (5.11):
−y 0 ∈ N(C−c)∩V (y),
y(0) = y0 .
and the function z from the differential inclusion (5.13)
0
z ∈ (NC (y + c) + y 0 ) ∩ U,
z(0) = z0 .
In particular, for t ∈ [0, 1) we have the point y resting at y0 (see figure 25).
Indeed, the plane V doesn’t change in time, and y0 remains in the interior of C until t = 1,
so y is not pulled by any edge of the section (C − c(t)) ∩ V . Hence for t ∈ [0, 1) we have
s(t) = y(t) + c(t) = (1, 1, 0) + (t, −t, 0) ∈ int C,
NC−c (y) = NC (s) = {0}.
39
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 25. Points s (left) and y together with the Moreau Sweeping Process
moving set (right) at moments t = 0, 0.5, 1.
Thus ṗ|[0,1) ≡ 0 by (5.6), so, by (6.6) and (5.5), we have for t ∈ [0, 1)
p(t) ≡ (−1, −1, 0),
x(t) = s(t) + p(t) = (t, −t, 0),
z(t) = x(t) − c(t) = 0.
We also get by continuity
y(1) = (1, 1, 0), s(1) = (2, 0, 0), p(1) = (−1, −1, 0), x(1) = (1, −1, 0), z(1) = 0.
For t ∈ [1, T ] the point y is pulled by the face of C, which is a part of the plane
y12 = 2 − t.
The normal cone to this face is
NC (s) = NC−c (y) = [0, +∞) × {0} × {0}.
The figure 26 illustrates the process at this stage. Since y ∈ V we also have
y23 = y12 = 2 − t.
And, since both U = NV (y) and NC−c (y) in (5.12) belong to the plane y13 = 0, we have
y(t) = (2 − t, 2 − t, 0).
40
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 26. Points s (left) and y (right) at moments t = 1, 1.6 in top orthographic projection.
Thus
s(t) = y(t) + c(t) = (2, 2 − 2t, 0),
y 0 = (−1, −1, 0).
Figure 27. Unique solution of the differential inclusion (5.13).
Figure 27 shows, that the differential inclusion (5.13) implies
z 0 = (1, −1, 0).
Indeed, if we put
w := (2, 0, 0) ∈ NC (y + c),
then
u := z 0 = y 0 + w = (1, −1, 0) ∈ U.
41
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Observe, that this coincides with the decomposition (5.14) in the Step 5.1.3 of the Theorem 5.1
proof. This decomposition may be illustrated here by the figure 28.
Figure 28. Decomposition −y 0 = w − u; w ∈ NC (y + c + g), u ∈ U .
Finally, we obtain for t ∈ [1, T ]
z(t) = (t − 1, 1 − t, 0),
x(t) = z(t) + c(t) = (2t − 1, 1 − 2t, 0),
p(t) = x(t) − s(t) = (2t − 3, −1, 0).
If we put all parts of the solution together we get the following piecewise-linear responce in terms
of H:
Function:
y =s−c−g
z =x−c−g
s
x
p
t ∈ [0, 1]
(1, 1, 0)
(0, 0, 0)
(1 + t, 1 − t, 0)
(t, −t, 0)
(−1, −1, 0)
t ∈ (1, T]
(2 − t, 2 − t, 0)
(t − 1, 1 − t, 0)
(2, 2 − 2t, 0)
(2t − 1, 1 − 2t, 0)
(2t − 3, −1, 0)
Remark 6.3. This result provides us the answer for the Questions 6.1 and 6.1. First, if we
put both g and c to be constant-zero, then we clearly have trivial constant responce. Second,
the calculated example of the Physical Elastoplastic Model shows, that we may force both of
the components s and p to change even if g ≡ 0 without braking any of the assumptions, but
this requires some bars to be stressed from the beginning. Third, putting c ≡ 0, we clearly may
increase the magnitude of g beyond elastic boundaries of the bars (see figure 14), forcing them
to change their elastic and plastic lengths.
6.5. Decomposition of the system. Observe from the figure 14 that the behavior of the
bar B13 is independent from the behavior of the couple of bars B12 , B23 . The system can be
considered as two independent systems as on the figure 29.
This may also be seen from figure 25 or derived analytically as follows. We utilize the description from section 6.1, without specifying the functions cpar and g13 and exact values of elastic
42
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 29. Decomposition of the system from the example 1
boundaries. Put
+
−
+
2
C1 := [c−
12 , c12 ] × [c23 , c23 ] ⊂ R ,
+
C2 := [c−
13 , c13 ] ⊂ R,
+
−
+
−
+
C = [c−
12 , c12 ] × [c23 , c23 ] × [c13 , c13 ] = (C1 × R) ∩ (R × R × C2 ),
so, similarly to the derivation of the plasticity law in section 3.7, by lemmas 2.3 and 2.4
NC−c(t)−g(t) (y) = NC1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)) ((y12 , y23 ))×{0}+{0}×{0}×NC2 −c13 (t)−g13 (t) (y13 )
Put
U1 := span {(1, −1)} ⊂ R2 ,
V1 := (1, −1)⊥ ⊂ R2
and notice, that by (6.1) we have U = U1 × {0}, so
N(C−c(t)−g(t))∩V (y) = NC−c(t)−g(t) (y) + U =
= (NC1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)) ((y12 , y23 )) + U1 ) × {0} + {0} × {0} × NC2 −c13 (t)−g13 (t) (y13 ) =
= N(C1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)))∩V1 ((y12 , y23 )) × {0} + {0} × {0} × NC2 −c13 (t)−g13 (t) (y13 ).
Notice, that the last equality relies on the safe load hypothesis, which is independent of the
third component for this system due to (6.3) and (6.4). So the Moreau Sweeping Process (5.11)
is decomposed on two independent parts:
(
−(y12 , y23 )0 ∈ N(C1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)))∩V1 ((y12 , y23 )),
(6.7)
0
−y13
∈ NC2 −c13 (t)−g13 (t) (y13 ).
In its turn, the inclusion (5.13) is
z 0 ∈ (NC−c(t)−g(t) (y) + y 0 ) ∩ U =
0
0
0
= (NC1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)) ((y12 , y23 )) × {0} + {0} × {0} × NC2 −c13 (t)−g13 (t) (y13 ) + (y12
, y23
, y13
)) ∩ (U1 × {0}),
which is equivalent to
(
0
0
(z12 , z23 )0 ∈ (NC1 −(c12 (t),c23 (t))−(g12 (t),g23 (t)) ((y12 , y23 )) + (y12
, y23
)) ∩ U1 ,
0
z13 = 0.
Observe, that the decomposition (6.7) consists from the two sweeping processes: the one is by
a (possibly changing) interval on a plane, another one is by a fixed interval on a line. Next, we
will describe another example, which has its sweeping process in R3 and cannot be decomposed
further.
43
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 30. Example 2
7. Example 2
7.1. Model description. We consider the one-dimensional mechanical model, shown on the
figure 30. It has 3 elastoplastic bars an 2 rigid controlled pistons. Similarly to section 6.5, the
bars B13 and B24 , which could be placed along with the pistons, do not contribute to the behavior
of the others bars. Their geometric constraints, composed with the additional constraints of the
pistons will be enforced below directly as an additional constraints on the bars B12 , B23 , B34 .
The lengths of the pistons are given as functions l13 (t), l24 (t). For simplicity we assume again,
that all bars have stiffness equal to 1. Let the reference length of each bar be 1.
Thus α = (1, 0, 0) and (omitting unused dimensions) we have

 
x2 − x1
−1
Dx = x3 − x2  =  0
x4 − x3
0
1
−1
0
0
1
−1
 
 x
0  1
x2 
0 
 x3  .
1
x4
Clearly the matrix D has rank 3, so DX = R3 and there is no geometric constraint. Moreover,
we can solve
Dx = 0


x2 − x1 = 0,
x3 − x2 = 0,


x4 − x3 = 0;
x1 = x2 = x3 = x4 ,
so
(7.1)
Ker D = span((1, 1, 1, 1)).
Now let x be the vector of total enlongations: x := e + p = (x12 , x23 , x34 ) The additional
constraints, produced by the pistons are
(
x12 + x23 = l13 ,
x23 + x34 = l24
(
(1, 1, 0) · x = l13 ,
(0, 1, 1) · x = l24
44
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Thus x ∈ U + g, where U ⊥ span((1, 1, 0), (0, 1, 1)). Since (1,1,0) and (0,1,1) are linearly independent, U is 1-dimensional. Notice, that (1 − 1, 1) · (1, 1, 0) = (1 − 1, 1) · (0, 1, 1) = 0, so
U = span((1, −1, 1))
.
To find g, observe, that x − g ∈ U means
(
(1, 1, 0) · (x − g) = 0,
(0, 1, 1) · (x − g) = 0.
Recall, that we also want g ⊥ U , so we have the following system:


(1, 1, 0) · (x − (x − g)) = l13 ,
(0, 1, 1) · (x − (x − g)) = l24 ,


(1, −1, 1) · g = 0;


(1, 1, 0) · g = l13 ,
(0, 1, 1) · g = l24 ,


(1, −1, 1) · g = 0;


g12 + g23 = l13 ,
g23 + g34 = l24 ,


g12 − g23 + g34 = 0;


g12 + g23 = l13 ,
g23 + g34 = l24 ,


3g23 = l13 + l24 ;

1

g23 = 3 (l13 + l24 ),
g12 = l13 − 31 (l13 + l24 ),


g34 = l24 − 31 (l13 + l24 );

  
−1
2
1
1
g = (2l13 − l24 , l13 + l24 , −l13 + 2l24 ) =  1  l13 +  1  l24  .
3
3
2
−1

(7.2)
Performing step of transition to Abstract Elastoplasic Problem, we get (meaning the notation
of the Pysical Model to the left) A = −I, c = h, C = −C, and after transition to a Hilbert space
we get −c = h, C = C and external forces, applied to the nodes are DT c = −DT (−c) = D∗ h.
This means, that, as long as the safe load hypothesis is satisfied, we can apply forces from
Im(DT ) = (Ker D)⊥ = (1, 1, 1, 1)⊥
by (7.1). This only means, that the sum of all external forces has to be zero, which is expectable
requirement. But, as mentioned before, in order to produce different behaviors it makes sence
to consider external forces from DT U only, i.e. we take them from the one-dimesional subspace.
So for a parameter function cpar : [0, T ] → R we have external forces of the type





−1 0
0 
−cpar
 1 −1 0  cpar


 −cpar  =  2cpar  .
DT · (1, −1, 1) · cpar = 
0
−2cpar 
1 −1
cpar
0
0
1
cpar
The figure 31 illustrates such forces for positive cpar .
45
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
Figure 31. Forces in example 2
The corresponding Moreau Sweeping Process (5.11) is
(
−y 0 ∈ N(C−c−g)∩(1,−1,1)⊥ (y),
y(0) = y0 .
As shown on the figure 32, by taking different elastic boundaries of the bars we can get various
shapes of the moving set, from a triangle to a hexagon.
Figure 32. The moving sets under zero external force applied (h = 0)
for C (the green boxes) equal (from left to right): [−1, 2.5] × [−2.5, 1] ×
[−1, 2.5], [−1, 2.5] × [−1, 1] × [−1, 2.5], [−1, 2.5] × [−1, 1] × [−1, 1], [−1, 1] ×
[−1, 1] × [−1, 1]. The vectors, denoted here by l13 and l24 , are directions from
(7.2) for the corresponding coefficients.
46
Ivan Gusoshnikov
The response of elastoplastic springs to perturbations
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[2] Kunze M., Monteiro Marques, M. D. P. An introduction to Moreau’s sweeping process. (English summary)
Impacts in mechanical systems (Grenoble, 1999), 1-60, Lecture notes in phys., 551, Springer, Berlin, 2000.
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