Comportamento macroscopico e microscopico di un

sottile nastro elastico.

Giuseppe Tomassetti

Arezzo, 25 gennaio 2019

Collaborazione con Riccardo Barsotti e Roberto Paroni

1

Controlled shape change in thin elastic strips

From: S. Armon et al., Science, 2011.

Thin elastic strips are interesting because

they exhibit radical shape changes upon

small changes of texture or environmental conditions.

2

Elastic ribbons

A ribbon is body having the shape of a rectangular parallelepiped with

thickness ε << width h << length `

The body prefers to accommodate constraints, loads, or geometrical incompatibilitiesby undergoing bending rather than stretching.

For this reason, most effects can be captured by modeling elastic ribbons as unstretchableplates.

3

Elastic ribbons

A ribbon is body having the shape of a rectangular parallelepiped with

thickness ε << width h << length `

The body prefers to accommodate constraints, loads, or geometrical incompatibilitiesby undergoing bending rather than stretching.

For this reason, most effects can be captured by modeling elastic ribbons as unstretchableplates.

3

Elastic ribbons

A ribbon is body having the shape of a rectangular parallelepiped with

thickness ε << width h << length `

The body prefers to accommodate constraints, loads, or geometrical incompatibilitiesby undergoing bending rather than stretching.

For this reason, most effects can be captured by modeling elastic ribbons as unstretchableplates.

3

Application example: a deployable structure

4

Michael Sadowsky

5

Sadowsky’s quest for the Mobius band

In a 1930 paper∗ Sadowsky addresses the question whether a Mobius band could be actuallyrealized from a sheet of paper.

• he models paper as an elastic plate which can bend but cannot stretch;

• by assembling pieces of flat and cylindrical surfaces, he constructs a constrained Mobiusband;

From: Schwarz, G. E. (Am. Math. Mon. 1990) The Dark Side of the Moebius Strip.

• Sadowsky poses the problem of what would be the shape of an unconstrained Mobiusband.

∗ See also: D.F. Hinz and E. Fried (2014) Translation of Michael Safowsky’s Paper “An Elementary Prooffor the Existence of a Developable Mobius Band and the Attribution of the Geometric Problem to a VariationalProblem.

6

What is the shape of an unconstrained Mobius band?

Sadowsky assumes that in the absence of loads or constraints the shape of the band can befound by solving

min∫ `

0

[∫ +h/2

−h/2(tr K)2dx2

]dx1,

where ` is the length, h is the width, and K is the second fundamental form.

Considering this formulation too hard to attack, he turns his attention to an infinitesimally-narrow band, characterized by:

h/` << 1.

He argues that for such a band the equilibria can be found by solving:

min∫ `

0WS(τ, κ)dx1,

where τ is the torsion and κ is the principal curvature of the midline, and where

WS(τ, κ) =(τ2 + κ2)2

κ2 if κ 6= 0,

and WS(τ, κ) = 0 otherwise.

7

What is the shape of an unconstrained Mobius band?

Sadowsky assumes that in the absence of loads or constraints the shape of the band can befound by solving

min∫ `

0

[∫ +h/2

−h/2(tr K)2dx2

]dx1,

where ` is the length, h is the width, and K is the second fundamental form.

Considering this formulation too hard to attack, he turns his attention to an infinitesimally-narrow band, characterized by:

h/` << 1.

He argues that for such a band the equilibria can be found by solving:

min∫ `

0WS(τ, κ)dx1,

where τ is the torsion and κ is the principal curvature of the midline, and where

WS(τ, κ) =(τ2 + κ2)2

κ2 if κ 6= 0,

and WS(τ, κ) = 0 otherwise.

7

What is the shape of an unconstrained Mobius band?

Sadowsky assumes that in the absence of loads or constraints the shape of the band can befound by solving

min∫ `

0

[∫ +h/2

−h/2(tr K)2dx2

]dx1,

where ` is the length, h is the width, and K is the second fundamental form.

Considering this formulation too hard to attack, he turns his attention to an infinitesimally-narrow band, characterized by:

h/` << 1.

He argues that for such a band the equilibria can be found by solving:

min∫ `

0WS(τ, κ)dx1,

where τ is the torsion and κ is the principal curvature of the midline, and where

WS(τ, κ) =(τ2 + κ2)2

κ2 if κ 6= 0,

and WS(τ, κ) = 0 otherwise.

7

What is the shape of an unconstrained Mobius band?

Sadowsky assumes that in the absence of loads or constraints the shape of the band can befound by solving

min∫ `

0

[∫ +h/2

−h/2(tr K)2dx2

]dx1,

where ` is the length, h is the width, and K is the second fundamental form.

Considering this formulation too hard to attack, he turns his attention to an infinitesimally-narrow band, characterized by:

h/` << 1.

He argues that for such a band the equilibria can be found by solving:

min∫ `

0WS(τ, κ)dx1,

where τ is the torsion and κ is the principal curvature of the midline, and where

WS(τ, κ) =(τ2 + κ2)2

κ2 if κ 6= 0,

and WS(τ, κ) = 0 otherwise.

7

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

Isometric deformations of the band

The band initially occupies a rectangular domain of length ` and width h, in the (x1, x2)plane:

ω = (0, `)× (−h/2, h/2).

The typical configuration of the strip is specified by the deformation χ : ω → R3.

In-plane inextensibility yields∂αχ · ∂βχ = δαβ.

For n = ∂1χ ∧ ∂2χ and F = ∇χ, the second fundamental form is

K = −F>∇n.

Theorema Egregium implies thatdet K = 0.

8

The building blocks of Sadowsky’s construction

The constraint det K = 0 implies that K is proportional to

K(θ) = I− b(θ)⊗ b(θ).

whereb(θ) = cos θe1 + sin θe2,

for some θ.

From: D. Hinz and E. Fried (J. Elasticity 2015).

In Sadowsky’s construction

θ = ±π

3.

9

The material frame

By the inextensibility constraint, the spatial curve defined by

r(x1) := χ(x1, x2)

is the arc-length parametrization of the midline in the deformed configuration.

In-plane inextensibility also implies that the material frame

d1(x1) = r′(x1), d2(x) = ∂2χ(x1, 0), d3(x) = n(x1, 0)

constitute an orthonormal triad.

The material frame uniquely identifies the bending-twist vector k by the property

d′i = k ∧ di.

10

The Frenet frame is material

The components of k are:

k1 = d′2 · d3 twist;

k2 = d′3 · d1 curvature around d2;

k3 = d′1 · d2 curvature around d3.

Unstretchability impliesk3 = 0,

Thus, if k2 6= 0, then d3 coincides with the Frenet normal and

τ2 = k21, κ2 = k2

2.

Sadowsky refers to this property by writing that the Frenet frame of a band is material.

Moreover, he uses this property as a kinematic characterization of a band.

11

The Frenet frame is material

The components of k are:

k1 = d′2 · d3 twist;

k2 = d′3 · d1 curvature around d2;

k3 = d′1 · d2 curvature around d3.

Unstretchability impliesk3 = 0,

Thus, if k2 6= 0, then d3 coincides with the Frenet normal and

τ2 = k21, κ2 = k2

2.

Sadowsky refers to this property by writing that the Frenet frame of a band is material.

Moreover, he uses this property as a kinematic characterization of a band.

11

The Frenet frame is material

The components of k are:

k1 = d′2 · d3 twist;

k2 = d′3 · d1 curvature around d2;

k3 = d′1 · d2 curvature around d3.

Unstretchability impliesk3 = 0,

Thus, if k2 6= 0, then d3 coincides with the Frenet normal and

τ2 = k21, κ2 = k2

2.

Sadowsky refers to this property by writing that the Frenet frame of a band is material.

Moreover, he uses this property as a kinematic characterization of a band.

11

The Frenet frame is material

The components of k are:

k1 = d′2 · d3 twist;

k2 = d′3 · d1 curvature around d2;

k3 = d′1 · d2 curvature around d3.

Unstretchability impliesk3 = 0,

Thus, if k2 6= 0, then d3 coincides with the Frenet normal and

τ2 = k21, κ2 = k2

2.

Sadowsky refers to this property by writing that the Frenet frame of a band is material.

Moreover, he uses this property as a kinematic characterization of a band.

11

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Sadowsky’s energy

The restriction K◦ to the midline of second fundamental form K is determined by k1, k2,except for γ := K◦22:

K◦ =(−k2 k1k1 γ

).

In particular,(tr K◦)2 = (−k2 + γ)2.

If k2 6= 0 then

det K◦ = 0 ⇔ −γ =k2

1k2

,

and

(tr K◦)2 = k22 +

k41

k22+ 2k2

1 =(k2

1 + k22)

2

k22

=(τ2 + κ2)2

κ2 .

If k2 vanishes also k1 does.

Thus, Sadowsky strips cannot twist.

12

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

Equilibria

We incorporate the constraint det K◦ = 0 in the (renormalized) strain energy by letting

W(K◦) :={

(tr K◦)2 if det K◦ = 0,+∞ otherwise.

Neglecting self-contact, we formulate an equilibrium problem as follows:

• find a curve r(x1), an orthonormal frame di(x), and an internal parameter γ(x1) whichsolve

min∫ `

0W([−k2 k1k1 γ

])where k1 = d′2 · d3 and k2 = d′3 · d1,

under the constraints d1 = r′ and d′1 · d2 = 0,

and

r(0) = r(`), d1(0) = d1(`),∫ `

0k1dx1 = π.

Observe that W is nonconvex.

13

An example of microstructure

Nonconvexity of the integrand can imply non-existence of a minimizer.

As an example, consider the problem

min|y′ |=1

∫ 1

0y2(x)dx. (1)

We incorporate the constraint |y′| = 1 in the integrand by defining the non-convex function

w(y′) =

{0 if |y′| = 1;+∞ otherwise.

We can then replace (1) with

min∫ 1

0[y2 + w(y′)]dx

14

There exists a

minimizing sequence (yn)

such that |y′n| = 1 andyn → y = 0.

We construct yn by alternating on a scale 1/n the slopes +1 and −1 in such a waythat yn is as close as possible to 0 on average.

the limit y does not comply with the constraint |y′| = 1.

15

There exists a

minimizing sequence (yn)

such that |y′n| = 1 andyn → y = 0.

We construct yn by alternating on a scale 1/n the slopes +1 and −1 in such a waythat yn is as close as possible to 0 on average.

the limit y does not comply with the constraint |y′| = 1.

15

Convexification

We replace w by its the largest convex minorant:

w∗∗(y′) =

{0 if |y′| ≤ 1+∞ otherwise

The limit y = 0 solves

min∫ 1

0[y2 + w∗∗(y′)]dx

We interpret

y2 + w∗∗(y′) as the macroscopic energy density

because it captures the macroscopic behavior of minimizing sequences.

16

The macroscopic energy of a ribbon

Instead of solving

min∫ `

0W([−k2 k1k1 γ

]),

we solve

min∫ `

0W∗∗

([−k2 k1k1 γ

]).

After the relaxation the kinematical constraint between γ, k1, and k2 is lost.

Thus we can minimize first with respect to γ to obtain the effective macroscopic energy

min∫ `

0W(k1, k2) where W(k1, k2) = min

γW∗∗

([−k2 k1k1 γ

]).

This procedure is supported by rigorous results obtained by Γ-convergence.∗

∗L. Freddi, P. Hornung, M.G. Mora, R. Paroni (J. Elasticity, 2016) A corrected Sadowsky functional forinextensible elastic ribbons.

17

The relaxed version of Sadowsky’s energy

The microscopic bending energy is

(tr K)2 = |K|2 + 2 det K.

The macroscopic bending energy turns out to be:

W∗∗(K) = (tr K)2 + 4(det K)− = |K|2 + 2|det K|.

In particular,

W(k1, k2) = minγ

W∗∗([−k2 k1k1 γ

])=

(k2

2 + k21)

2

k22

= WS(k1, k) if |k2| > |k1|,

4k21 otherwise.

18

Identikit of the relaxed Sadowsky energy

Since W∗∗(K) is isotropic, it depends only on the eigenvalues of K, thus there exists w∗∗ :R2 → R such that:

w∗∗(κ1, κ2) = W∗∗(K) whenever K = κ1b1 ⊗ b1 + κ2b2 ⊗ b2,

with bα · bβ = δαβ.

Graph of w∗∗.

19

Identikit of the relaxed Sadowsky energy

Since W∗∗(K) is isotropic, it depends only on the eigenvalues of K, thus there exists w∗∗ :R2 → R such that:

w∗∗(κ1, κ2) = W∗∗(K) whenever K = κ1b1 ⊗ b1 + κ2b2 ⊗ b2,

with bα · bβ = δαβ.

Graph of w∗∗.

19

Explicit form of the convex envelope: laminations

Theorem 1. ∗ For Q : R2×2sym → R a quadratic form, let

W(K) =

{Q(K) if det K = 0,+∞ otherwise.

Then for each K there exist A, B satisfying det A = det B = 0 and t ∈ [0, 1] such that

W∗∗(K) = (1− t)Q(A) + tQ(B), with K = (1− t)A + tB.

Explicit formulas for A, B and t are available.

∗R. Paroni and G. Tomassetti (J. Elasticity, 2018) Macroscopic and microscopic behavior of narrow elasticribbons.

20

Twisted states: macroscopic solution

We conside the following twisted state:

r(x1) = x1e1,d2(x1) = + cos(Θx1/`)e2 + sin(Θx1/`)e3,d3(x1) = − sin(Θx1/`)e2 + cos(Θx1/`)e3.

In this state the ribbon is uniformly twisted under a controlled terminal rotation Θ.

The twist isk1 = Θ/`,

while k2 = 0 and k3 = 0.

21

Twisted states: microscopic solution

On the midline, the second fundamental form is

K◦(γ) =(

0 k1k1 γ

).

SinceW∗∗(K) = |K|2 + 2|det K|,

the macroscopic energy density is

minγ

W∗∗ (K◦(γ)) = W∗∗([

0 k1k1 0

]).

It turns out that the lamination is(0 k1k1 0

)=

12(A + B) with A = k1

(−1 11 −1

), B = k1

(1 11 1

).

and that

W(k1, 0) =12(Q(A) + Q(B)).

22

Interpretation

For a twisted state the second fundamental form associated to the macroscopic defor-mation has non-vanishing determinant, hence is microscopically inaccessible.

At the microscopic level, this deformation can be generated by alternating, on a finescale, two deformations with null Gaussian curvature.

The developability constraint survives only as a constraint that the ribbon cannot bendabout the normal d3

23

Interpretation

For a twisted state the second fundamental form associated to the macroscopic defor-mation has non-vanishing determinant, hence is microscopically inaccessible.

At the microscopic level, this deformation can be generated by alternating, on a finescale, two deformations with null Gaussian curvature.

The developability constraint survives only as a constraint that the ribbon cannot bendabout the normal d3

23

Interpretation

For a twisted state the second fundamental form associated to the macroscopic defor-mation has non-vanishing determinant, hence is microscopically inaccessible.

At the microscopic level, this deformation can be generated by alternating, on a finescale, two deformations with null Gaussian curvature.

The developability constraint survives only as a constraint that the ribbon cannot bendabout the normal d3

23

A closer look at the microscopic states

A microscopic state is must be a multiple of

K(θ) = I− b(θ)⊗ b(θ), with b(θ) = cos θe1 + sin θe2.

Twisted states are a fine mixture of

A = k1K(π/2), B = k1K(−π/2).

flat state

→ state A

state B

The actual construction requires some work because A and B are kinematically incompatible.

24

A closer look at the microscopic states

A microscopic state is must be a multiple of

K(θ) = I− b(θ)⊗ b(θ), with b(θ) = cos θe1 + sin θe2.

Twisted states are a fine mixture of

A = k1K(π/2), B = k1K(−π/2).

flat state

→ state A

state B

The actual construction requires some work because A and B are kinematically incompatible.

24

Assembling the microscopic states

25

General case

One period of the function fn.

Let θA and θB such thatA = K(θA), B = K(θB).

We letθ(n)(x1) = fn(nx1)θA + (1− fn(nx2))θB,

andK(n)(x1) = K(θ(n)(x1)).

26

Thanks for your attention

References

G.T., V. Varano (Meccanica, 2017) Capturing the helical to spiral transitions in thin ribbonsof nematic elastomers.

R. Paroni, G.T. (J. Elasticity, 2018) Macroscopic and Microscopic Behavior of Narrow ElasticRibbons.

27