Computational modelling of the mechanics of tubular structures composed of helical rods

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Universit`

a di Pisa

Dipartimento di Ingegneria dell’Informazione

Corso di Laurea Magistrale in Bionics Engineering

Computational modelling of the mechanics of

tubular structures composed of helical rods

Supervisori

Prof. Antonio De Simone

Dr.

Alessandro Lucantonio

Candidato

Jacopo Quaglierini

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Acknowledgements

Desidero ringraziare i miei relatori, il professor De Simone e il dottor Lucantonio, per la disponibilit`a, la gentilezza, e l’interesse sia umano che professionale dimostrati in questi mesi, prima in aula e poi nello svolgimento di questo lavoro.

Ringrazio i miei genitori e tutta la mia famiglia, che mi hanno sempre sostenuto nelle mie scelte e creduto in me e nelle mie capacit`a.

Ringrazio i miei amici di sempre Alessandro, Daniele, Gabriele, Mirko e Claudia, che mi hanno sempre accettato per come sono.

Ringrazio i miei amici e colleghi dell’Universit`a di Pisa e del corso di Bionics Engineering, che mi hanno accompagnato in questo percorso di studi e di crescita personale e accademica. Un grazie in particolare a Federico e Matteo, con cui ho condiviso questa sfida negli ultimi mesi, e con cui conto di condividere quella prossima del dottorato di ricerca.

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Abstract

This thesis studies the mechanical properties of tubular structures com-posed of assemblies of helical rods, which are found in nature and in several engineering applications.

In particular, we investigate the response under compression of as-semblies of pin-jointed helical springs via numerical simulations, and we compare the behavior of the assemblies as the number of springs composing the assembly increases.

The assemblies are modelled as 3D Cosserat rods in the large deforma-tion regime. We parametrize the rotadeforma-tions of directors using quater-nions. The equilibrium equations in weak form are derived from the Principle of Virtual Work and they are solved numerically through a custom-made implementation in the software COMSOL Multiphysics R

(Weak Form PDE mode).

The numerical results allow to compute the load-displacement curve and show that:

1. the mutual support between helices of an assembly stabilizes its behavior with respect to the single helix case: the buckling ob-served in the compression of a single helix is suppressed in the ensemble response;

2. the collective behavior of the assemblies can be described in terms of a bulk behavior, where each rod deforms as a perfect circular helix, and a boundary behavior, where rods deviate from perfect helices in a way that is affected by the specific boundary condi-tions applied.

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A.1 Left-handed Helices . . . 57 B Dirichlet boundary conditions 67 B.1 Right-handed helices . . . 67 B.2 Left-handed helices . . . 71 C Calculation of the location of pin-joints 75 C.1 Intersection between helices of the same chirality . . . 75 C.2 Intersection between right-handed and left-handed helices . . . 77

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Chapter 1 Introduction and motivations

The mechanical behavior of tubular structures made of helical rods is of great interest in several fields, ranging from engineering to biology. Indeed, from a technological view point these structures are interesting for their shape-shifting capabilities, which make them adaptable. This means that they can change con-formation — and hence properties — according to changing functional needs. Exploiting this characteristic, they have been used for applications in a broad range of domains, e.g. deployable antennas in aerospace engineering (Olson et al. (2013)), the sheaths of McKibben artificial muscles in soft robotics and bio-robotics (Tondu (2012); Hassan et al. (2019); Boxerbaum et al. (2012), see Figure 1.2), and tubular vascular stents in biomedical engineering (Kleinstreuer et al. (2008); Douglas (2012); Zunino et al. (2009)).

Moreover, there are many examples of biological structures that can be modeled as assemblies of rods, such as the bundles of microtubules and motors in all eukaryotic flagella and cilia (9+2 structure, see Alberts et al. (6)), the tail sheaths structure of bacteriophage viruses (Falk et James (2006); Kostyuchenko et al. (2005)), or the pellicle of euglenids, a family of unicellular algae (Noselli et al. (2019), see also Figure 1.1).

In all above examples the cited system consists of a tubular structure, whose enve-lope is a network of 1D structures, namely threads, fibers, or rods. Understanding the behavior of those systems requires the underlying knowledge of how single 1D rods spanning the whole structure respond to external loads, and how they interact with each other, that is the ensemble response.

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1. INTRODUCTION AND MOTIVATIONS

Figure 1.1 a) Neck protrusion with shape changes in the microtubules meshwork of Lacrymaria Olor and b) metaboly of Euglena gracilis based on a sliding helices mechanism. Courtesy of Professor A. De Simone, from a pre-print of Cicconofri et al. (2019).

Figure 1.2 A robot with a continously deformable envelope, capable of peristaltic motion (Boxerbaum et al. (2012)).

In the perspective of developing novel applications based on such structures, this thesis focuses on the study and understanding of the mechanical response of a single helix and of tubular assemblies for increasing helices number. In particular, the structure behavior under compression will be investigated, highlighting the differences for different number of helices forming the assembly. In order to do so, numerical simulation techniques, based on a Cosserat 3D rod model, will be adopted. For this purpose, the software package COMSOL Multiphysics R _v5.4

will be used in equation mode, implementing custom weak form formulations of the Principle of Virtual Work to model the nonlinear response of the interacting 1D helical rods in the large deformation regime, under prescribed loads and dis-placements. Assemblies made of 1, 2, 4, and 8 pin-jointed helices will be studied and numerically simulated.

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Chapter 2 Mechanics of elastic rods in the

large deformation regime: a 3D

Cosserat rod model and its

numerical implementation

2.1 Kinematics

Let us consider a rod, of length L, described in 3D space. We assume that that the rod is inextensible and unshearable. Since the rod is inextensible, its axis arc-length parameter s in the deformed configuration coincides with the arc-arc-length parameter of the undeformed one, sR. We introduce the concept of directors, to

describe the orientation of the rod cross sections with respect to its axis. d₃(s) is defined as the tangent vector and, together with d₁(s) and d₂(s), it forms an orthonormal reference frame, with d₁(s) and d₂(s) in particular defining the plane of the rod cross section. Also after deformations, they keep their orientation and they remain unitary (i.e. they remain an orthonormal basis) due to unsherability and inextensibility, as well as the rigidity of the cross section.

Each point of the cross section is identified by s, ξ1, ξ2, these last two scalar

coordinates describing the point position in the cross section’s plane:

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2. MECHANICS OF ELASTIC RODS IN THE LARGE DEFORMATION REGIME e₂ e₁ e₃ d₁ d₃ d₂ r(s, ξ1, ξ2 ) r₀(s) ξ1d₁(s) + ξ2d₂(s) s

Figure 2.1 Generic rod of arc-length s, with global (black) and local (red) refer-ence frames.

So the position of a point of the rod with respect to the global reference frame is given by the sum of r₀(s), that is the absolute position of its cross section’s corresponding axis point, and a vector ξ1d1(s)+ξ2d2(s), that describes the position

of that point in the rod’s cross section, with respect to its local reference frame defined by d1, d2, d3 (see Figure 2.1).

Directors dk with k = 1, 2, 3 form the local orthonormal frame and, as they are an

orthonormal basis, they are not independent. The tangent vector is: d₃(s) = dr0(s)

ds = r

0

0(s) (2.2)

where we are adopting the notation (·)0 in place of d(·) ds .

The tangent vector d₃(s) remains unitary due to inextensibility; while searching the whole three vectors, we have the advantage of actually not needing 9 parame-ters but fewer, thanks to orthonormality: we just need 3 angles (Euler’s angles) to obtain the local frame from the global one, making the composition of 3 rotations:

{d₁(s), d₂(s), d₃(s)} = R(θ(s), φ(s), ψ(s)){e₁, e₂, e₃} (2.3) The directors can then be arranged as columns of the rotation matrix:

R = {d₁(s), d₂(s), d₃(s)} =      d11 d21 d31 d12 d22 d32 d13 d23 d33      (2.4)

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2.1 Kinematics

Since R(s) is a rotation for ∀s ∈ [0, L], we have that:

K(s) := R0(s)RT(s) (2.5) is a skew-symmetric matrix. Indeed, by differentiating:

R(s)RT(s) = I (2.6) we obtain:

R0(s)RT(s) + R(s)R0T(s) = 0 (2.7) and hence:

R(s)R0T(s) = (R0(s)RT(s))T = −R0(s)RT(s) . (2.8) Curvatures are obtained from the rate at which directors change orientation along the axis line r₀(s). Indeed, by differentiating the law of transformation from global to local reference frame d_i(s) = R(s)e_i, we obtain:

d0_i(s) = R0(s)e_i = R0(s)RT(s)R(s)e_i = R0(s)RT(s)d_i(s), i = 1, 2, 3 (2.9) which we can rewrite as the Darboux equations:

d0_i(s) = K(s)d_i(s), i = 1, 2, 3 . (2.10) Setting: K(s) =      0 −Ω κ2 Ω 0 −κ1 −κ2 κ1 0      , κ(s) = κ1(s)d1(s) + κ2(s)d2(s) + Ω(s)d3(s) (2.11)

where K(s) is expressed in the local reference frame and κ(s) is defined as the curvature vector, we can rewrite Equation 2.10 as:

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2. MECHANICS OF ELASTIC RODS IN THE LARGE DEFORMATION REGIME e₂ e₁ e₃ d₁ t(s)=d₃(s) d₂ s q(s)

Figure 2.2 Geometry of a planar curve.

or, more explicitly:              d0₁(s) = Ω(s)d₂(s) − κ2(s)d3(s) d0₂(s) = −Ω(s)d₁(s) + κ1(s)d3(s) d0₃(s) = κ2(s)d1(s) − κ1(s)d2(s) (2.13)

κ1(s), κ2(s), Ω(s) are defined as the two flexural curvatures and the torsional one,

respectively. They can be found as:              κ1(s) = d02(s) · d3(s) = −d2(s) · d 0 3(s) κ2(s) = d03(s) · d1(s) = −d 0 1(s) · d3(s) Ω(s) = d0₁(s) · d₂(s) = −d₁(s) · d0₂(s) (2.14)

In the special case of a planar curve in the plane d2− d3 (see Figure 2.2), we have:

             d₃(s) = cos(θ)e₃+ sin(θ)e₁ d₂(s) = − sin(θ)e₃+ cos(θ)e₁ d₁(s) = −e₂ (2.15)

Substituting Equation 2.15 in Equations 2.14: κ1(s) = d02(s) · d3(s) = [− cos(θ)θ

0

e₃− sin(θ)θ0e₁] · [cos(θ)e₃+ sin(θ)e₁] = = − cos2(θ)θ0− sin2_(θ)θ0

= −θ0

(2.16)

Sign apart, this is the familiar result found in 2D models. Moreover:

κ2(s) = Ω(s) = 0 (2.17)

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2.2 Parametrization of rotations via quaternions

Quaternions represent a way of parametrizing rotations avoiding singularities (e.g. gimbal lock), employing four redundant parameters. Considering the equation:

{d₁(s), d₂(s), d₃(s)} = R(s){e₁, e₂, e₃} (2.18)

the rotation R can be expressed, according to Rodrigues’s formula, in terms of an axis a (with kak = 1) and an angle α:

R = cos(α)P + sin(α)A + a⊗ a , Ab = a ∧ b ∀ b (2.19) where A is the skew-symmetric matrix representing the vector product with a. P is the projection matrix on the plane orthogonal to a:

P = I − a ⊗ a . (2.20) a ⊗ a is the projection matrix along a, defined via the dyadic product as:

(a⊗ b)c = (b · c)a ∀ a, b, c . (2.21) Knowing a, A can be easily found as:

     Ae₁ = a ∧ e₁ = −a2e3+ a3e2 Ae₂ = a ∧ e₂ = a1e3− a3e1 (2.22)

where Ae₁, Ae₂ are the first and second column of A, respectively. Since A is skew-symmetric we obtain:

A =      0 −a3 a2 a3 0 −a1 −a2 a1 0      (2.23)

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2. MECHANICS OF ELASTIC RODS IN THE LARGE DEFORMATION REGIME

Let us replace a and α, using other four parameters q1, q2, q3, q4, defined as:

                   q1 = a1sin α₂ q2 = a2sin α₂ q3 = a3sin α₂ q4 = cos α₂ (2.24)

We can thus define the vector part of the quaternions as: q = a sinα

2

. (2.25)

Since a is a unit vector, we can write that: q· q = (a · a) sin2α 2 = sin2 α 2 . (2.26) Then: q· q + q2 4 = 1 (2.27)

that implies that the norm of the vector (q1,q2,q3,q4) is unitary, ensuring the

de-pendency of one component on the others.

Finally, knowing that:      cos(α) = cos2α 2 − sin2α 2 = q2₄− q · q sin(α) = 2 cosα 2 sinα 2 = 2q4sin α 2 (2.28)

we can rewrite Rodrigues’s formula as:

R = (q₄2− q · q)I + 2q · q + 2q4Q , Qb = q ∧ b (2.29)

where Q is the skew-symmetric matrix representing the vector product with q.

Thus, we can express the directors, which correspond to the columns of the rotation matrix, in terms of q1, q2, q3, and q4.

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2.3 Equilibrium equations of rods via the Principle of Virtual Work

2.3 Equilibrium equations of rods via the

Prin-ciple of Virtual Work

Once the rod’s curvatures are defined, we can apply the Principle of Virtual Work. Let us write the elastic energy of the rod:

Eel = Z L 0 EI1 2 (κ1− κs1) 2₊EI2 2 (κ2− κs2) 2₊GJ 2 (Ω − Ωs) 2 ds (2.30) where κs1, κs2, Ωs are the intrinsic curvatures that the rod exhibits in the absence

of loads. Then, we apply the virtual variation operator δ: δEel = Z L 0 h EI1(κ1− κs1) ˜κ1+ EI2(κ2− κs2) ˜κ2+ GJ (Ω − Ωs) ˜Ω i ds (2.31)

where ˜κ1, ˜κ2, ˜Ω are the virtual perturbations of the curvatures. The external

virtual work is:

δW = Z L

0

f · ˜r₀+ m · ˜θds . (2.32) where vectors f , m identify the distributed forces and distributed torques, respec-tively, and ˜r₀, ˜θ are the virtual variations of r₀(s) and θ(s), defined as:

θ(s) = Z s 0 κ(σ) dσ (2.33) r₀(s) = r₀(0) + Z s 0 d₃(σ) dσ (2.34) The Principle of Virtual Work states that:

δEel = δW ∀ ˜θ(s), ˜φ(s), ˜ψ(s) (2.35)

where ˜θ(s), ˜φ(s), ˜ψ(s) are virtual variations of Euler’s angles.

In principle, we would have to express ˜κ1, ˜κ2, ˜Ω, ˜r0 in terms of ˜θ(s), ˜φ(s), ˜ψ(s).

However, since we will work with a parametrization of rotations based on quater-nions, we will recast them directly as a function of ˜q1, ˜q2, ˜q3, ˜q4, so that:

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The additional constraint of Equation 2.27 can be imposed, instead of explicitly writing an additional equation, using a Lagrange multiplier:

E = Eelastic+

Z L

0

η(q · q + q₄2− 1) ds . (2.37) Taking the virtual energy variation:

δEel+ Z L 0 η(q·q+q42−1)(2q·˜q+2q4q˜4)+˜η(q·q+q42−1) ds = δW ∀ ˜q, ˜q4, ˜η (2.38) where: ˜ q = (˜q1, ˜q2, ˜q3) . (2.39)

2.4 Numerical implementation in COMSOL

Mul-tiphysics

The governing equations of the rod model (Equation 2.38) along with the boundary conditions were implemented and solved in the finite elements software COMSOL Multiphysics R _{v5.4. In particular, the Weak Form PDE mode was used, which}

allows for a custom definition of the equations of the model.

Regarding the mesh, upon checking that the numerical solution was insensitive to further refinement, each rod of the assembly was divided into about 48 elements. Quadratic shape functions were employed to discretize the quaternions fields. In all the boundary value problems numerically studied, a continuation approach was followed so that the position of one boundary point was varied gradually. Indeed, at each boundary displacement step the system of non-linear algebraic equations, resulting from the finite elements procedure, was solved by a quasi-Newton iterative algorithm, where the initial guess was provided by the numerical solution at the previous displacement step.

A direct solver (MUMPS) was chosen for the solution of the linear system obtained through the quasi-Newton algorithm at each iteration.

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2.5 Boundary conditions

To set the problems discussed in Section 4, we imposed boundary conditions of the following type:

• Fixed rotations at s = 0, L. In order to impose these conditions, since rotations are expressed as a function of quaternions, Dirichlet boundary conditions on the value of the quadruplet q1, q2, q3, q4 were set at both ends

of the rod, specifying quaternions values in those points (see section 4). • Fixed Position r₀(s) = r1(s), r2(s), r3(s) at s = 0, L. The condition at

s = 0 is implicit in the choice of integration constants to reconstruct r₀(s) from the tangent vector (see Equation 2.34). The condition at s = L is enforced by imposing in the software weak constraints for r₀(s) coordinates using Lagrange multipliers, which corresponds to adding to Equation 2.38:

λi(ri(L) − ri) ˜ri(L) + ˜λi(ri(L) − ri), i = 1, 2, 3 , (2.40)

where λi and ri are the Lagrange multiplier and the imposed value for i-th

coordinate of r₀(L), respectively. In particular, ˜r0(L) is expressed in terms

of (˜q, ˜q4) by applying the virtual operator δ to Equation 2.34. Since the

resulting equation must hold for every λi, the constraint equation ri(L) = ri

is enforced for every i.

• No relative translation at crossing points between helices of the assembly, so that those points act like rotational joints connecting the helices. These constraints are of the same nature as the position constraints at s = L and therefore we apply here the same procedure.

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Chapter 3 Geometry of circular helices and

of helical rods in natural

configuration

A cylindrical (or general) helix is a 3D curve, characterized by the property that its tangent vector at any point makes a constant angle with a fixed line called axis of the helix. This is possible if and only if its curvature to torsion ratio is constant.

Figure 3.1 Circular helix of radius 1 and pitch 6, where arrows indicate the direction of the tangent vector. A circular helix is a cylindrical helix

charac-terized by constant curvature and constant torsion (see Figure 3.1).

The radius of a helix, given one of its points, is the distance from that point and the axis of the helix and it corresponds to the inverse of its curvature, as well.

The pitch of a helix is the height of one complete helix turn, defined as the projec-tion, along its axis, of the difference be-tween two points that differ by one turn. Helices can be right-handed or left-handed, that is they have two possible chiralities.

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3. GEOMETRY OF CIRCULAR HELICES AND OF HELICAL RODS IN NATURAL CONFIGURATION

proceeds along the helix in a counter-clockwise fashion, the helix is right-handed, while it is left-handed if it proceeds clockwise.

The chirality of the helix is a property of the curve, that is it does not depend on the adopted perspective: a right-handed helix and a left-handed one cannot be superimposed through simple roto-translations.

In the Cartesian space, circular helices can be parametrized, assuming that their axis is parallel to one of the principal axes, in the following way:

r₀(s) =a cos s_c, a sin s_c, b_cs

(3.1) which represents, in particular, a right-handed helix of radius a and pitch 2πb, with c = √a2_{+ b}2 _{as a normalization parameter. Its curvature and torsion are}

equal to a/c2 _{and b/c}2_{, respectively.}

3.1 Naturally free helices

Circular helices can be obtained as natural configurations of rods in the absence of external loads: we call these helices “naturally free helices”. Setting κs1 = κ∗1,

κs2 = κ∗2, Ωs = Ω∗ in Equation 2.30, we observe that the configuration with

κ1 = κ∗1, κ2 = κ∗2, Ω = Ω

∗ _{minimizes the rod’s elastic energy, and hence that}

it is a natural equilibrium configuration. In particular, circular helices arise by setting κ∗₂ = 0. Indeed, using Equations 2.14 to express curvatures as functions of directors:              −d0₃(s) · d₂(s) = κ∗₁ d0₃(s) · d₁(s) = 0 d0₁(s) · d₂(s) = Ω∗ (3.2)

and Frenet’s equations:              t0(s) = Knn(s) n0(s) = −Knt(s) + τ b(s) b0(s) = −τ n(s) (3.3)

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3.1 Naturally free helices

it can be shown that configurations satisfying Equations 3.2 are circular helices. Theorem 3.1.1. Assume that Equations 3.2 hold. Then r₀(s) =

Z s

0

d₃(σ) dσ is a circular helix of curvature Kn = kκ∗1k and torsion τ = Ω

∗_{; for κ}∗ 1 > 0

we have {d₁(s), d₂(s), d₃(s)} = {b(s), −n(s), t(s)}, while for κ∗₁ < 0 we have {d₁(s), d₂(s), d₃(s)} = {−b(s), n(s), t(s)}.

Proof. We rewrite Equation 3.2 replacing d₃(s) with t(s):

−t0(s) · d₂(s) = κ∗₁ (3.4) t0(s) · d₁(s) = 0 (3.5) d0₁(s) · d₂(s) = Ω∗ (3.6) d₁(s) · d₂(s) ∧ t(s) = 1 (3.7) where Equation 3.7 comes from orthonormality. Considering t0(s) · t(s) = 0 and Equation 3.7, we infer that t0(s) is in the plane defined by d₁(s) and d₂(s), which means that t0(s) = αd₁(s) + βd₂(s). From Equation 3.5 we have that:

α = 0 ⇒ t0(s) = βd₂(s) . Considering Frenet equations, we have:

t0(s) = Knn(s) = βd2(s) .

Therefore, β = kKnk and d2(s) = ±n(s), since both d2(s) and n(s) have unit

norm.

1. case d₂(s) = +n(s): From Equation 3.4 we have:

−t0(s) · d₂(s) = −Knn(s) · n(s) = −Kn= κ∗1 ⇒ Kn= −κ∗1

that is admissible only if κ∗₁ < 0 (no solution otherwise). Moreover: d₁(s) = d₂(s) ∧ d₃(s) = n(s) ∧ t(s) = −b(s)

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d0₁(s) · d₂(s) = (−b0(s)) · n(s) = τ n(s) · n(s) ⇒ τ = Ω∗ In conclusion:

{d₁(s), d₂(s), d₃(s)} = {−b(s), n(s), t(s)}, Kn = −κ∗1 > 0, τ = Ω ∗_.

2. Case d₂(s) = −n(s): From Equation 3.4 we have:

−t0(s) · d₂(s) = Knn(s) · n(s) = Kn= κ∗1 ⇒ Kn= κ∗1

that is admissible only if κ∗₁ > 0 (no solution otherwise). Considering d₁(s):

d₁(s) = d₂(s) ∧ d₃(s) = −n(s) ∧ t(s) = b(s) d0₁(s) · d₂(s) = b0(s) · (−n(s)) = (−τ n(s)) · (−n(s)) ⇒ τ = Ω∗ In conclusion: {d₁(s), d₂(s), d₃(s)} = {b(s), −n(s), t(s)}, Kn = κ∗1 > 0, τ = Ω ∗ .

In this thesis, we adopted {d₁(s), d₂(s), d₃(s)} = {−b(s), n(s), t(s)}. Let us find the explicit expressions of directors and curvatures of a helical rod considering, without loss of generality, the right-handed helix of Equation 3.1 as its axis:

d₃(s) = t(s) = r0₀(s) = −a csin s c, a ccos s c, b c (3.8) According to Frenet’s equations:

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3.2 Geometry of helical assemblies and location of pin-joints

Differentiating Equation 3.8, t0(s) can be written as: t0(s) = kn(s)n(s) = _ca2 − cos s c, − sin s c, 0 (3.10) We choose d₂(s) = +n(s), so that: d₂(s) =− cos s c, − sin s c, 0 (3.11) and d₁(s) is therefore found taking the vector product:

d₁(s) = d₂(s) ∧ d₃(s) =₋b csin s c, b ccos s c, − a c (3.12) The directors are then:

               d₁(s) = −b csin s c, b ccos s c, − a c d₂(s) = − cos s c, − sin s c, 0 d₃(s) = −a csin s c, a ccos s c, b c (3.13)

and the curvatures can be obtained as:              κs1 = d02(s) · d3(s) = − a c2 κs2 = d03(s) · d1(s) = 0 Ωs = d01(s) · d2(s) = b c2 (3.14)

3.2 Geometry of helical assemblies and location

of pin-joints

In this thesis, we studied helical assemblies formed by a number of helices n = 1, 2, 4, 8, arranged as in Figure 3.2: for n ≥ 2, pairs of right-handed and left-handed helices have their bases positioned on the x-y plane where an axis intersects the circle of radius a. Points of intersection between two different helices are modelled as rotational joints. The 8 helices forming the largest assembly can be

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Figure 3.2 Helices assembly for n = 4. parametrized in the Cartesian space as:

r(s) =a cos s_c, a sin s_c, b_cs

(3.15) r(s) =

a cos s_c, −a sin s_c, b_cs (3.16) r(s) =−a sin s c, a cos s c, b cs (3.17) r(s) = a sin s_c, a cos s_c, b_cs (3.18) r(s) =−a cos s c, −a sin s c, b cs (3.19) r(s) = −a cos s c, a sin s c, b cs (3.20) r(s) =a sin s_c, −a cos s_c, b_cs

(3.21) r(s) = −a sin s c, −a cos s c, b cs (3.22) Let us determine a θ0 that allows us to rewrite the right-handed helices, described

by Equations (3.15, 3.17, 3.19, 3.21), in the form:

r(s) =a cos s_c + θ0, a sin s_c+ θ0, b_cs

(3.23) For Equation 3.15 the solution is trivially θ0 = 0. In the other three cases we have:

• Equation 3.17:      − sin s c = cos s_c +π₂ cos s_c = sin s_c +π₂ (3.24)

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3.2 Geometry of helical assemblies and location of pin-joints that leads to θ0 = π 2. • Equation 3.19:      − cos s c = cos s c+ π − sin s c = sin s c+ π (3.25) that leads to θ0 = π. • Equation 3.21:      sin s_c = cos s_c −π 2 − cos s c = sin s_c −π 2 (3.26) that leads to −π 2.

Similarly, let us determine a θ0 that allows us to rewrite left-handed helices,

de-scribed by Equations (3.16, 3.18, 3.20, 3.22), in the form: r(s) =a cos s_c + θ0, −a sin s_c+ θ0, b_cs

(3.27) For Equation 3.16 the solution is trivially θ0 = 0. In the other three cases we have:

• Equation 3.18:      sin s c = cos s c − π 2 cos s c = − sin s c − π 2 (3.28) that leads to θ = −π 2. • Equation 3.20:      − cos s c = cos s_c + π sin s_c = − sin s_c + π (3.29) that leads to θ = π. • Equation 3.22:      − sin s c = cos s c+ π 2 − cos s c = − sin s c+ π 2 (3.30) that leads to θ = π 2.

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Summing up, for right-handed helices:

r_right(s) =a cos s_c + θ0, a sin s_c+ θ0, b_cs

θ0 ∈ h 0,π 2, π, − π 2 i (3.31) while for left-handed helices:

r_left(s) =a cos s_c + θ0, −a sin s_c+ θ0, b_cs

θ0 ∈ h 0, −π 2, π, π 2 i (3.32)

Let us see where the helices intersect, that is where the pin joints are located along the helices’ axis lines. All the helices lie on a cylinder of radius a and axis z. Since a cylinder is a developable surface, we can imagine to unravel the cylinder around which the helices revolve into a flat rectangular surface. In this representation the vertical direction corresponds to the z coordinate of our Cartesian parametrization and the horizontal direction is θ, that represents the azimuth of the point r₀(s) in cylindrical coordinates. Since the Cartesian coordinates x and y of a point of the helix are 2π periodic in θ, in order to indentify the intersection between helices it is convenient to represent z as a function of the azimuth module 2π (see Figure 3.3). For the right-handed helix described in Equation 3.1:

q z

2p 2pb

4pb

Figure 3.3 Plane rapresentation of a right-handed helix.

z(s) = bs

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where we replaced s/c with the azimuth coordinate θ.

Right-handed helices with basis in different points of the circumference are rotated versions of Equation 3.1 by an angular shift θ0, which has the same meaning as in

Equations (3.31, 3.32): zright(θ, θ0) = b(θ − θ0) . (3.34) q z 2p 2pb 4pb q 0

Figure 3.4 Plane rapresentation of two right-handed helices, one shifted by θ0.

Left-handed helices would be naturally parametrized by zleft= −bθ. On the other

hand, since we are interested only in angles module 2π and we want to represent intersection points with right-handed helices with θ ∈ [0 2π], we can reparametrize them as:

zleft(θ) = b(2π − θ) = −bθ + 2πb (3.35)

For shifted left-handed helices the same reasoning applies:

zleft(θ, θ0) = b(2π − (θ − θ0)) = −b(θ − θ0) + 2πb (3.36)

Since the helix pitch is 2πb, we can study the intersection between helices in the interval z ∈ [0 2πb], as they will periodically repeat along z.

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q z 2p 2pb 4pb q 0sx

Figure 3.5 Plane rapresentation of two left-handed helices, one delayed.

q z 2p 2pb 4pb q 0dx q0sx

Figure 3.6 Plane rapresentation of four helices, two right-handed (one shifted) and two left-handed (one shifted).

Thus, we can find the cross-points by checking the intersections between the lines describing the generic right-handed helix (Equation 3.34) and the generic left-handed one (Equation 3.36). It can be shown that helices with the same chirality, arranged along a circumference, do not cross (see Appendix C.1). Notice that, points having the same z and with θ either 0 or 2π are also intersection points, which are not represented by line crossings as in Figure 3.6 and have to be treated separately. Here we report the cross-points calculated for each of the 8 modelled

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helices (see Appendix C.2 for complete calculations). First right-handed helix, θ0 = 0:

• crosses the first left-handed helix in s

c = kπ, k = 0, 1, 2 . . . • crosses the second left-handed helix in s

c = π

4 + kπ, k = 0, 1, 2 . . . • crosses the third left-handed helix in s

c = π

2 + kπ, k = 0, 1, 2 . . . • crosses the fourth left-handed helix in s

c = 3π

4 + kπ, k = 0, 1, 2 . . . Second right-handed helix, θ0 = +

π 2:

• crosses the first left-handed helix in s c =

3π

4 kπ, k = 0, 1, 2 . . . • crosses the second left-handed helix in s

c = kπ, k = 0, 1, 2 . . . • crosses the third left-handed helix in s

c = π

2 + kπ, k = 0, 1, 2 . . . Third right-handed helix, θ0 = π:

π

2 + kπ, k = 0, 1, 2 . . . • crosses the second left-handed helix in s

c = 3π

c = kπ, k = 0, 1, 2 . . . • crosses the fourth left-handed helix in s

c = π

4 + kπ, k = 0, 1, 2 . . . Fourth right-handed helix, θ0 = −

π 2:

π

4 + kπ, k = 0, 1, 2 . . . • crosses the second left-handed helix in s

c = π

c = 3π

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3.3 Description of helical assemblies based on

quaternions

We recall the Rodrigues’s formula expressed in terms of quaternions:

R = (q2₄ − q · q)I + 2q · q + 2q4(s)Q, Qb = q ∧ b (3.37)

Since the directors correspond to the columns of R, they can be expressed as functions of quaternions q1, q2, q3, q4: d₁ =      1 − 2(q2 2 + q32) 2(q1q2+ q3q4) 2(q1q3− q2q4)      d₂ =      2(q1q2− q3q4) 1 − 2(q2₁+ q2₃) 2(q2q3+ q1q4)      d₃ =      2(q1q3 + q2q4) 2(q2q3− q1q4) 1 − 2(q2 1 + q22)     

Let us now find the description, in terms of quaternions, of generic right-handed and left-handed helices defined in Equations (3.31, 3.32), respectively.

3.3.1 Generic right-handed helix

                                                       d11= 1 − 2(q22+ q32) = −bcsin s c+ θ0 d12= 2(q1q2+ q3q4) = +b_ccos s_c+ θ0 d13= 2(q1q3− q2q4) = −a_c d21= 2(q1q2− q3q4) = − cos s_c+ θ0 d22= 1 − 2(q12+ q32) = − sin sc+ θ0 d23= 2(q2q3+ q1q4) = 0 d31= 2(q1q3+ q2q4) = −a_csin s_c + θ0 d32= 2(q2q3− q1q4) = +a_ccos s_c+ θ0 d33= 1 − 2(q12+ q22) = +bc d₁(s) =      −b csin s c+ θ0 +b_ccos s_c+ θ0 −a c      d₂(s) =      − cos s c+ θ0 − sin s c+ θ0 0      d₃(s) =      −a csin s c + θ0 +a_ccos s_c + θ0 +b_c     

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3.3 Description of helical assemblies based on quaternions

Summing the equations of d12 and d21 we obtain:

4q1q2 = cos s_c + θ0

b c− 1

. (3.38)

Summing the equations of d23 and d32:

4q2q3 = a c cos s c + θ0

and re-arranging the terms we get: cos s

c+ θ0 =

4c

a q2q3. (3.39) From the equation of d23 we have:

q2q3 = −q1q4. (3.40)

Inserting Equation 3.39 in Equation 3.38: 4q1q2 = 4c aq2q3 b − c c

since c 6= 0, it results that:

q1q2 = q2q3 b − c a (3.41) and if q2 6= 0, we get: q1 = q3 b − c a . (3.42)

Inserting Equation 3.40 in Equation 3.41: q1q2 = −q1q4 b − c a (3.43) and if q1 6= 0, we get: q2 = −q4 b − c a . (3.44)

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The cases where either q1 or q2 (or both) are equal to zero have to be checked

separately. Considering the equation of d13, the case q1 = q2 = 0 leads to:

−a

c = 0 ⇒ a = 0

that represents a limit case, i.e. a straight rod, not of our interest. Let us check the case q1 = 0, q2 6= 0. From the equation of d23 we have:

q2q3 = 0 ⇒ q3 = 0

which verifies Equation 3.42. From the equation of d13 we get:

2q2q4 =

a c and re-arranging the terms:

q2q4 =

a

2c. (3.45)

From the equation of d33:

1 − 2q2₂ = b c bringing all constant terms on the right-side:

q2₂ = 1 − b c 2 = c − b 2c and taking the square root we obtain:

q2 = ±

r c − b

2c . (3.46)

Given q2, we obtain q4 from Equation 3.45:

q4 = a 2cq2 = a 2c ± √ 2c √ c − b ! = √ c2_{− b}2 2c ± √ 2c √ c − b ! = = p(c − b)(c + b) 2c ± √ 2c √ c − b ! = ± r c + b 2c (3.47)

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We can therefore verify that the relationship between q2 and q4 given by Equation

3.44 is satisfied: ± r c − b 2c = − b − c a ! ± r c + b 2c ! = ∓ b − c p(c − b)(c + b) ! √ c + b √ 2c ! = = ∓ − c − b p(c − b)(c + b) !√ c + b √ 2c ! = ± c − b p(c − b)2c = ± r c − b 2c Let us check the remaining case q1 6= 0, q2 = 0. From the equation of d23we have:

q1q4 = 0 ⇒ q4 = 0

which verifies Equation 3.44. From the equation of d13 we get:

2q1q3 = −

a c and re-arranging the terms:

q1q3 = −

a

2c. (3.48)

From the equation of d33:

1 − 2q2₁ = b

c (3.49)

bringing all constant terms on the right-side: q₁2 = 1 − b c 2 = c − b 2c and taking the square root we obtain:

q1 = ±

r c − b

2c . (3.50)

Given q1, we obtain q3 From Equation 3.48:

q3 = − a 2cq1 = −a 2c ± r 2c c − b ! = = −p(c − b)(c + b) 2c ± √ 2c √ c − b ! = ∓ r c + b 2c (3.51)

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where we have used that c =√a2_{+ b}2_{, c − b 6= 0.}

We can therefore verify that the relationship between q1 and q3 given by Equation

3.42 is satisfied: ± r c − b 2c = b − c a ! ∓ r c + b 2c ! = ∓ b − c p(c − b)(c + b) ! √ c + b √ 2c ! = = ∓ − c − b p(c − b)(c + b) ! √ c + b √ 2c !! = ± c − b p(c − b)2c = ± r c − b 2c Thus, it is true that:

       q1 = q3 b − c a q2 = −q4 b − c a ∀s ∈ [0, L]

By squaring and re-arranging Equation 3.42 we have that: q₃2 = a

2

(b − c)2q 2

1. (3.52)

Substituting Equation 3.52 in the equation of d22:

1 − 2 q₁2+ a 2 (b − c)2q 2 1 = − sin s c+ θ0

and re-arranging the equation’s terms leaving only q₁2 on the left-side we get: q₁2 = (b − c)

2

2 (b − c)2_{+ a}2 1 + sin

s

c+ θ0 . (3.53)

Considering that c2 _{= a}2_{+ b}2_{, (b − c)}2 _{= (c − b)}2_{, the found expression can be}

simplified as: q₁2 = 1 + sin s c+ θ0 (c − b) 4c

and by taking the square root we obtain:

q1 = ± v u u t 1 + sin s_c + θ0 (c − b) 4c . (3.54)

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Substituting Equation 3.54 in Equation 3.42 we get the expression of q3.

In order to find the expression of q4, let us then consider the equation of d12 and

replace q1 e q2 using the relationships of Equations (3.42, 3.44):

2 − q3q4 b − c a 2 + q3q4 = b ccos s c+ θ0 .

Re-arranging the equation’s term leving only q4 on the left-side we get:

q4 = a2_b 2cq3 a2 − (b − c)2 cos s c+ θ0 .

Considering that c2 = a2+ b2, b 6= 0, the found expression can be simplified as: q4 =

c + b 4cq3

cos s_c + θ0 . (3.55)

Replacing q3 exploiting again Equation 3.42:

q4 = − (c + b)(c − b) 4caq1 cos s_c+ θ0 = − a 4cq1 cos s_c+ θ0 (3.56) where we have used that (c + b)(c − b) = c2_{− b}2 _{= a}2_.

Using Equation 3.44, q2 can be found as:

q2 = −q4 b − c a = a(b − c) 4acq1 cos s_c+ θ0 = − c − b 4cq1 cos s_c+ θ0 (3.57)

However, due to sqare root operation q1 sign is lost, leading to singularities in its

first derivative. Thus, to obtain a continous first derivative, the sign of q1 has to

be reversed whenever it reaches a zero.

In the points where q1 becomes zero, we have singularities for q2 and q4 too, given

thier definitions. Using logic operators, we can impose to q2 and q4 the expressions

calculated in Equations (3.46, 3.47) whenever q1 reaches a zero.

This operation depends on the particular expression describing q1, that differs for

different θ0, and thus it has to be done separately for each of the four right-handed

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Let us consider the right-handed helix with θ = 0:                        q1 = ± r 1 + sin(s_c)(c − b) 4c q2 = − c − b 4cq1 cos(s_c) q3 = q1 a b − c q4 = − a 4cq1 cos(s_c) (3.58)

q1 zeroes are located at those points such that:

1 + sin(s_c) = 0 ⇒ sin(s_c) = −1 that is true for:

s c =

3π

2 + 2kπ k = 0, 1, 2 . . . Considering the interval s_c ∈ [0, 4π]:

1. From 0 to 3π

2 we have e.g. sign -2. From 3π 2 to 7π 2 we have sign + 3. From 7π 2 to 4π we have sign

-Thus we can take the 4π module of s/c to impose the proper sign, based on the interval in which s/c is, obtaining the correct curve. Regarding the sign of q2,q4

in the zeroes of q1, let us consider the Equation 3.39:

4q2q3 = a ccos( s c + θ0) θ0=0 = a ccos( s c)

which implies that:

q2q3 → 0− when

s c →

3π

2 + 2kπ .

From Equation 3.42 we have that q1 and q3 are always opposite in sign, so q1 and

q2 have the same sign whenever q1 approaches zero with s/c increasing. Thus,

if the first half-wave is negative, q2 in s/c = 3π/2 is negative. From Equation

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(corresponding to minimum and maximum peaks of q2 and q4) the sign has to

be reversed half-wave by half-wave. Analogous consideration can be made for the other right-handed helices of the assembly, defined by θ0 = +π/2, π, −π/2.

Indeed, in all four cases the sign of q1 can be imposed according to the value of s/c

module 4π, and the considerations on the sign of q2,q4 hold in all four cases (for

an explicit derivation, see Appendix A). We report the expressions of quaternions for the other three right-handed helices and where singularities show up.

• For θ0 = + π 2:                        q1 = ± r 1 + cos(s_c)(c − b) 4c q2 = c − b 4cq1 sin(s_c) q3 = q1 a b − c q4 = a 4cq1 sin(s_c) q1 = 0 in s c = π + 2kπ k = 0, 1, 2 . . . (3.59) • For θ0 = π:                        q1 = ± r 1 − sin(s_c)(c − b) 4c q2 = c − b 4cq1 cos(s_c) q3 = q1 a b − c q4 = a 4cq1 cos(s_c) q1 = 0 in s c = π 2 + 2kπ k = 0, 1, 2 . . . (3.60) • For θ0 = − π 2:                        q1 = ± r 1 − cos(s_c)(c − b) 4c q2 = − c − b 4cq1 sin(s_c) q3 = q1 a b − c q4 = − a 4cq1 sin(s_c) q1 = 0 in s c = 0 + 2kπ k = 0, 1, 2 . . . (3.61)

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3.3.2 Generic left-handed helix

                                                       d11= 1 − 2(q22+ q32) = +bcsin s c+ θ0 d12= 2(q1q2+ q3q4) = +b_ccos s_c+ θ0 d13= 2(q1q3− q2q4) = +a_c d21= 2(q1q2− q3q4) = − cos s_c+ θ0 d22= 1 − 2(q12+ q32) = + sin sc+ θ0 d23= 2(q2q3+ q1q4) = 0 d31= 2(q1q3+ q2q4) = −a_csin s_c + θ0 d32= 2(q2q3− q1q4) = −a_ccos s_c + θ0 d33= 1 − 2(q12+ q22) = +bc d₁(s) =      +b_csin s_c+ θ0 +b_ccos s_c+ θ0 +a_c      d₂(s) =      − cos s c + θ0 + sin s_c+ θ0 0      d₃(s) =      −a csin s c+ θ0 −a ccos s c + θ0 +b_c     

Following analogous steps (Appendix A.1), for left-handed helices we have that:        q1 = −q3 b − c a q2 = q4 b − c a ∀s ∈ [0, L]                        q1 = ± s 1−sin s c+θ0 (c−b) 4c q2 = − c − b 4cq1 cos s c+ θ0 q3 = q1 a c − b q4 = _4cqa 1 cos s c + θ0 (3.62)

As in the right-handed case, the square root operation leads to singularities in the first derivative of q1, so that its sign has to be reversed at each of its zeroes. As

in the right-handed case, q2 and q4 have singularities in those points, so that it

is necessary to impose the expression calculated in Equations (3.46, 3.47). The considerations on the sign of q2 and q4in left-handed cases are reversed with respect

to right-handed ones. In the following we report the expressions of quaternions for all the four left-handed helices (θ0 = 0, −π/2, π, +π/2) and where singularities

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3.3 Description of helical assemblies based on quaternions • For θ0 = 0:                        q1 = ± r 1 − sin(s_c)(c − b) 4c q2 = − c − b 4cq1 cos(s_c) q3 = q1 a c − b q4 = a 4cq1 cos(s_c) q1 = 0 for s c = π 2+2kπ k = 0, 1, 2 . . . (3.63) • For θ0 = − π 2:                        q1 = ± r 1 + cos(s c)(c − b) 4c q2 = − c − b 4cq1 sin(s_c) q3 = q1 a c − b q4 = a 4cq1 sin(s_c) q1 = 0 for s c = π+2kπ k = 0, 1, 2 . . . (3.64) • For θ0 = π:                        q1 = ± r 1 + sin(s_c)(c − b) 4c q2 = c − b 4cq1 cos(s c) q3 = q1 a c − b q4 = − a 4cq1 cos(s_c) q1 = 0 for s c = 3π 2 +2kπ k = 0, 1, 2 . . . (3.65) • For θ0 = + π 2:                        q1 = ± r 1 − cos(s_c)(c − b) 4c q2 = c − b 4cq1 sin(s_c) q3 = q1 a c − b q4 = − a 4cq1 sin(s_c) q1 = 0 for s c = 0+2kπ k = 0, 1, 2 . . . (3.66)

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Chapter 4 Examples of structural behavior:

compression of cylindrical

assemblies of helical springs

In this section we consider the behavior under compression of helical springs clamped at both ends. The starting condition in absence of external loads corre-sponds to naturally free helices. Compression is applied by imposing a shortening, i.e. a relative displacement of helices ends along their axis. In this frame, we define the compression ratio as:

ε = ∆h h0

= h − h0 h0

(4.1) where h is the current height of the assembly, that is the coordinate of helices tips along e₃, and h0 is the starting height. In the reported Figures, the triplet X-Y-Z

corresponds to the global reference frame {e₁, e₂, e₃}.

We will compare the behavior of a single helix with the one of an assembly of n = 2, 4, 8 pin-jointed helices, described in section 3.2.

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4. EXAMPLES OF STRUCTURAL BEHAVIOR: COMPRESSION OF CYLINDRICAL ASSEMBLIES OF HELICAL SPRINGS

4.1 Initial values and Dirichlet boundary

condi-tions

The initial condition for the quaternions field on each rod of the assembly corre-sponds to a naturally free helix of radius a = 0.06 m, pitch b = 0.025 m, with chirality and base position according to the specific helix in exam.

Rods length L is set to 7πc, which means that the helix takes 3 and half turns. Their bending stiffnessess along d₁(s), d₂(s) are B1 = 1 Nm2, B2 = 1 Nm2,

respectively, whereas their torsional stiffness along d₃(s) is B3 = 2 Nm2.

Coordinates in the x-y plane of both ends of each helix are fixed to their initial values, whereas the z coordinate of helices tips is gradually decreased via a weak constraint, from the value of the initial naturally free helices (0.54978 m) to half of it (0.27489 m).

Relative displacements between helices at their crossing points (calculated in sub-section 3.2) are prevented via weak constraints.

Dirichlet boundary conditions on quaternions field are imposed at both ends of each helix, enforcing that the orientation of the directors at those points remains equal to the one of the corresponding initial naturally free helix. To calculate them, we simply have to substitute s = 0, 7πc in Equations (3.58-3.61, 3.63-3.66) obtaining quaternions value at the base and at the tip, respectively.

The values assigned to each helix are reported in following (for explicit calculation, see Appendix B).

• First right Helix:

q1base = q1tip = q2tip = ±

r

(c − b)

4c (4.2)

q2base = −q1base (4.3)

q3base = q4base = q3tip = ∓

r

(c + b)

4c (4.4)

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4.1 Initial values and Dirichlet boundary conditions

• Second Right Helix:

q1base = q2tip = ±

r

(c − b)

2c (4.6)

q2base = q4base = q1tip = q3tip = 0 (4.7)

q3base = ∓

r

(c + b)

2c (4.8)

q4tip = −q3base (4.9)

• Third Right Helix:

q1base = q2base = q1tip = ±

r

(c − b)

4c (4.10)

q3base = q3tip = q4tip = ∓

r

(c + b)

4c (4.11)

• Fourth Right Helix:

q2base = q1tip = ± r (c − b) 2c (4.15) q4base = ± r (c + b) 2c (4.16) q3tip = −q4base (4.17)

• First Left Helix:

r

(c − b)

4c (4.18)

r

(c + b)

4c (4.20)

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• Second Left Helix:

q1base = q2tip = ±

r

(c − b)

2c (4.22)

q3base = ±

r

(c + b)

2c (4.24)

• Third Left Helix:

r

(c − b)

4c (4.26)

r (c + b) 4c (4.27) q4base = −q3base (4.28) q2tip = −q1base (4.29) (4.30)

• Fourth Left Helix:

q2base = q1tip = ± r (c − b) 2c (4.32) q4base = ∓ r (c + b) 2c (4.33) q3tip = −q4base (4.34)

4.2 Results

We report the results on the response under compression of a single helix and of the assemblies of helices described in section 3.2. We compare the curvatures profiles and the normalized load-displacement curve between the different cases.

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4.2 Results

4.2.1 The case n = 1: behavior of a single helix

Figure 4.1 Y-X view of the single helix assembly at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.2 Z-Y view of the single helix (n = 1) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.3 Z-X view of the single helix (n = 1) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

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Figure 4.4 Curvatures of the single helix (n = 1) as a function of the normalized arc-length for compression ratio ε = 0.5.

As shown in Figures (4.1-4.3), the single right-handed helix exhibits buckling as the structure is compressed with both ends clamped. The deviation from a perfect helix is particularly marked at compression ratio equal to 0.5. Coherently with such deviation, κ2 is different from zero and κ1 and Ω are far from being constant

(see Figure 4.4).

4.2.2 The case n ≥ 2: behavior of helical assemblies

The assembly formed by two helices, a right-handed and a left-handed one, starts to exhibit a more regular behavior. The buckling in the z-y plane is absent: the two helices, having different chiralities, would buckle in the same plane but in opposite directions. This results in their reciprocal counter-balancing, as they prevent each other from buckling in their preferred direction.

The collective behavior of the two helices can be rationalized as follows: each helix prevents the other one from following what can be interpreted as the first mode of bifurcation for a single helix, described in section 4.2.1. Another higher mode of bifurcation is entered, as shown in Figures 4.5-4.7.

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4.2 Results

Figure 4.7 Z-X view of the double-helix assembly (n = 2) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.5 Y-X view of the double-helix assembly (n = 2) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.6 Z-Y view of the double-helix assembly (n = 2) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

It can be appreciated that the helix pitch does not remain constant across the entire helix, which corresponds to an oscillatory bifurcation mode in the z-x plane.

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Figure 4.8 Curvatures of the double-helix assembly as a function of the nor-malized arc-length on the left-handed (top) and right-handed (bottom) helices for compression ratio ε = 0.5. The star markers indicate the position of pin-joints.

Another interesting result emerging from the collective behavior of helical assem-blies is that the value of κ2and the oscillations of κ1and Ω along the arc-length are

depressed. This phenomenon can be greatly appreciated looking at the curvatures profile of Figure 4.8, where two zones start to be distinguishable: a boundary zone near the edges s = 0, L, where curvatures oscillates the most, and a bulk zone in the interior of the rods, where they tend to flatten.

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4.2 Results

Similarly to the passage from 1 to 2 helices, in the quadri-helix assembly the two pairs of opposite chirality helices counterbalance each other, further reducing the overall instability seen in the case n = 2.

Figure 4.9 Y-X view of the quadri-helix (n = 4) assembly at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.10 Z-Y view of the quadri-helix (n = 4) assembly at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.11 Z-X view of the quadri-helix (n = 4) assembly at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

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Figure 4.12 Curvatures of the quadri-helix (n = 4) assembly as a function of the normalized arc-length on one left-handed helix (top) and one right-handed helix (bottom) for compression ratio ε = 0.5. The star markers indicate the position of pin-joints.

The rods exhibit a less variable pitch (see Figure 4.9-4.11) and a wider bulk zone in their interior, where the curvatures profiles become flatter (see Figure 4.12). If one further increases n, the trends highlighted so far are confirmed, that is for n = 8 the bulk zone is even more prominent, although its additional growth is reduced (see Figures 4.13-4.15).

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4.2 Results

Figure 4.13 Y-X view of the octa-helix assembly (n = 8) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.14 Z-Y view of the octa-helix assembly (n = 8) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

Figure 4.15 Z-X view of the octa-helix assembly (n = 8) at compression ratio ε equal to 0 (left) and to 0.5 (right). The color bar refers to κ1.

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Figure 4.16 Curvatures of the octa-helix assembly (n = 8) as a function of the normalized arc-length on one left-handed helix (top) and one right-handed helix (bottom) for compression ratio ε = 0.5. The star markers indicate the position of pin-joints.

In the bulk zone κ2 and Ω appear almost flat while only κ1 retains some

oscilla-tions (see Figure 4.16). Indeed it can be noted that, among the three principal curvatures, κ1 is the one displaying the slowest settlement to a constant value as

n increases.

We explicitly remark that the bulk behavior, where κ2 = 0, κ1 = κ1, Ω = Ω, is

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4.2 Results

4.2.3 Load-displacement curves

We report the curves of reaction forces versus compression ratio ε at helices ends, for compressions up to half of their initial height. Figures 4.17-4.19 report on the vertical axis the absolute value of the components of the average reaction force at helices ends, normalized by the Euler’s critical load for a double-clamped rod, and on the horizontal axis the compression ratio ε. The behavior of the assemblies is compared as the number of helices n increases. In Figure 4.17 the plotted expressions are: 1 nPcrit n X i=1 Fxi , 1 nPcrit n X i=1 Fyi , 1 nPcrit n X i=1 Fzi , Pcrit = B1 2π L 2

where Fxi, Fyi, Fzi are the reaction force components on the i-th helix and Pcrit is

the Euler’s critical load.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3

Figure 4.17 Comparison for different n of the curves of the average absolute value of tip forces along e₁, normalized by Euler’s critical load for clamped rods.

Assemblies with n ≥ 2 show smaller and smaller deviations, as n increases, from a linear relationship for the component along e₃ and null along the other axes. The cases n = 4 and n = 8 already show a relationship which is practically perfectly

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linear for the component along e₃ and with null components along e₁, e₂.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 -5 -4 -3 -2 -1 0 1 2 3 4 5 10 -4

Figure 4.18 Comparison for different n of the curves of the average absolute value of tip forces along e₂, normalized by Euler’s critical load for clamped rods.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Figure 4.19 Comparison for different n of the curves of the average absolute value of tip forces along e₃, normalized by Euler’s critical load for clamped rods.

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Chapter 5 Conclusions and outlook

We have set up a computational model for the behavior of assemblies of helical rods in the large deformation regime. This model was then used to investigate the concrete example of compression of helical assemblies.

Inspired by the experimental results of Haringx (1950) on the behavior of a sin-gle helix, we examined the behavior of assemblies made of n = 1, 2, 4, 8 helices, generalizing the study to the case of the collective behavior of an assembly of springs.

The numerical results allow to compute the load-displacement curve and show that:

1. the mutual support between helices of an assembly stabilizes its behavior with respect to the single helix case: the buckling observed in the compres-sion of a single helix is suppressed in the ensemble response;

2. the collective behavior of the assemblies can be described in terms of a bulk behavior, where each rod deforms as a perfect circular helix, and a boundary behavior, where rods deviate from perfect helices in a way that is affected by the specific boundary conditions applied.

In the future we plan to extend our work to treat more general boundary condi-tions:

• tips of the helices unconstrained in the plane perpendicular to the helices axis, as in the first case discussed in Haringx (1950), where experimental data are available;

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5. CONCLUSIONS AND OUTLOOK

• Free rotations and translation in the radial direction at bases of the helices, in order to model the helical antennas in Olson et al. (2013).

Moreover, we plan to validate experimentally our findings using small prototypes of the studied assemblies, formed by naturally free helices pin-jointed together at their crossing points. These helices could be fabricated using modern additive manufacturing technologies, such as 3D printing. Assemblies formed by a number of helices n > 8 will be studied as well, with the objective of generalizing our simulations findings. Moreover, assemblies of different height-diameter ratio will be used, as in Haringx (1950).

Finally, we will explore the applicability of our findings to more complex structures. In particular, we will investigate the existence of a distinctive and regular bulk behavior, described by perfect helices.

We will study whether this feature can be used to devise simplified models for structures like the ones reported in the Introduction: biological structures, such as the flagella of eukaryotic cells or the microtubule meshwork of Lacrymaria Olor, and artificial ones such as McKibben-like structures.

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Bibliography

Alberts, B., Johnson, A., Lewis, J., Morgan, D., Raff, M., Roberts, K. et Walter, P. (6). ed. garland science; 2014. Molecular Biology of the Cell. Boxerbaum, A. S., Shaw, K. M., Chiel, H. J. et Quinn, R. D. (2012).

Contin-uous wave peristaltic motion in a robot. The international journal of Robotics Research, 31(3):302–318.

Cicconofri, G., Arroyo, M., Noselli, G. et DeSimone, A. (2019). Mor-phable structures from unicellular organisms with active, shape-shifting en-velopes: variations on a theme by gauss. Pre-print.

Douglas, G. R. (2012). Design of stent expansion mechanisms. Th`ese de doc-torat, University of British Columbia.

Falk, W. et James, R. D. (2006). Elasticity theory for self-assembled protein lattices with application to the martensitic phase transition in bacteriophage t4 tail sheath. Physical Review E, 73(1):011917.

Haringx, J. (1950). Instability of springs. Phillips Technical Review, 100(8). Hassan, T., Cianchetti, M., Moatamedi, M., Mazzolai, B., Laschi, C. et

Dario, P. (2019). Finite-element modeling and design of a pneumatic braided muscle actuator with multifunctional capabilities. IEEE/ASME Transactions on Mechatronics, 24(1):109–119.

Kleinstreuer, C., Li, Z., Basciano, C. A., Seelecke, S. et Farber, M. A. (2008). Computational mechanics of nitinol stent grafts. Journal of biomechan-ics, 41(11):2370–2378.

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BIBLIOGRAPHY

Kostyuchenko, V. A., Chipman, P. R., Leiman, P. G., Arisaka, F., Mesyanzhinov, V. V. et Rossmann, M. G. (2005). The tail structure of bacteriophage t4 and its mechanism of contraction. Nature structural & molec-ular biology, 12(9):810.

Noselli, G., Beran, A., Arroyo, M. et DeSimone, A. (2019). Swimming euglena respond to confinement with a behavioural change enabling effective crawling. Nature Physics, page 1.

Olson, G., Pellegrino, S., Banik, J. et Costantine, J. (2013). Deployable helical antennas for cubesats. In 54th AIAA/ASME/ASCE/AHS/ASC Struc-tures, Structural Dynamics, and Materials Conference, page 1671.

Tondu, B. (2012). Modelling of the mckibben artificial muscle: A review. Journal of Intelligent Material Systems and Structures, 23(3):225–253.

Zunino, P., D’Angelo, C., Petrini, L., Vergara, C., Capelli, C. et Migli-avacca, F. (2009). Numerical simulation of drug eluting coronary stents: chanics, fluid dynamics and drug release. Computer Methods in Applied Me-chanics and Engineering, 198(45-46):3633–3644.

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Appendix A

Calculation of quaternions fields

describing the helices of the

assemblies

Second right helix Let us consider the right-handed helix with θ = +π 2:                        q1 = ± r 1 + cos(s_c)(c − b) 4c q2 = c − b 4cq1 sin(s_c) q3 = q1 a b − c q4 = a 4cq1 sin(s_c) (A.1)

1 + cos(s_c) = 0 ⇒ cos(s_c) = −1 that is true for:

s

c = π + 2kπ k = 0, 1, 2 . . . Considering the interval s_c ∈ [0, 4π]:

1. From 0 to π we have e.g. sign + 2. From π to 3π we have sign

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-A. CALCULATION OF QUATERNIONS FIELDS DESCRIBING THE HELICES OF THE ASSEMBLIES

3. From 3π to 4π we have sign +

Then we can take the 4π module of s/c and impose the proper sign, based on the interval in which s/c is, to obtain the correct curve. Regarding the sign of q2, q4

in the zeroes of q1, let us consider Equation 3.39:

4q2q3 = a ccos( s c+ θ0) θ0=+π₂ = −a c sin( s c)

which implies that:

q2q3 → 0− when

s

c → π + 2kπ

From Equation 3.42 q1 and q3 are always opposite in sign, so q1 and q2 have the

same sign whenever q1 approaches zero with s/c increasing. Thus, if the first

half-wave is negative, q2 in s/c = pi is negative. From Equation 3.44 q4 has always

the same sign of q2 and also for these values (corresponding to minimum and

maximum peaks of q2 and q4) the sign has to be reversed half-wave by half-wave.

Third right helix Let us consider the right-handed helix with θ = π:                        q1 = ± r 1 − sin(s_c)(c − b) 4c q2 = c − b 4cq1 cos(s_c) q3 = q1 a b − c q4 = a 4cq1 cos(s_c) (A.2)

1 − sin(s_c) = 0 ⇒ sin(s_c) = 1 that is true for:

s c =

π

2 + 2kπ k = 0, 1, 2 . . . Considering the interval s_c ∈ [0, 4π]:

1. From 0 to π

2 we have e.g. sign + 2. From π

2 to 5π

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-3. From 5π

2 to 4π we have sign +

Then we can take the 4π module of s/c and impose the proper sign, based on the interval in which s/c is, to obtain the correct curve. Regarding the sign of q2, q4

in the zeroes of q1, let us consider Equation 3.39:

4q2q3 = a ccos( s c+ θ0) θ0=π = −a ccos( s c)

which implies that:

q2q3 → 0− when

s c →

π

2 + 2kπ

From Equation 3.42 q1 and q3 are always opposite in sign, so q1 and q2 have the

same sign whenever q1 approaches zero with s/c increasing. Thus, if the first

half-wave is negative, q2 in s/c = π/2 is negative. From Equation 3.44 q4 has always

the same sign of q2 and also for these values (corresponding to minimum and

maximum peaks of q2 and q4) the sign has to be reversed half-wave by half-wave.

Fourth right helix Let us consider the right-handed helix with θ = −π 2:                        q1 = ± r 1 − cos(s_c)(c − b) 4c q2 = − c − b 4cq1 sin(s_c) q3 = q1 a b − c q4 = − a 4cq1 sin(s_c) (A.3)

1 − cos(s_c) = 0 ⇒ cos(s_c) = 1 that is true for:

s

c = 0 + 2kπ k = 0, 1, 2 . . . Considering the interval s_c ∈ [0, 4π]:

1. From 0 to 2π we have e.g. sign + 2. From 2π to 4π we have sign

Computational modelling of the mechanics of tubular structures composed of helical rods

Universit`

a di Pisa

Dipartimento di Ingegneria dell’Informazione

Corso di Laurea Magistrale in Bionics Engineering

Computational modelling of the mechanics of

tubular structures composed of helical rods

Supervisori

Prof. Antonio De Simone

Dr.

Alessandro Lucantonio

Candidato

Jacopo Quaglierini

Acknowledgements

Abstract

Contents

Chapter 1

Introduction and motivations

Chapter 2

Mechanics of elastic rods in the

large deformation regime: a 3D

Cosserat rod model and its

numerical implementation

2.1

Kinematics

2.2

Parametrization of rotations via quaternions

2.3

Equilibrium equations of rods via the

Prin-ciple of Virtual Work

2.4

Numerical implementation in COMSOL

Mul-tiphysics

2.5

Boundary conditions

Chapter 3

Geometry of circular helices and

of helical rods in natural

configuration

3.1

Naturally free helices

3.2

Geometry of helical assemblies and location

of pin-joints

3.3

Description of helical assemblies based on

quaternions

3.3.1

Generic right-handed helix

3.3.2

Generic left-handed helix

Chapter 4

Examples of structural behavior:

compression of cylindrical

assemblies of helical springs

4.1

Initial values and Dirichlet boundary

condi-tions

4.2

Results

4.2.1

The case n = 1: behavior of a single helix

4.2.2

The case n ≥ 2: behavior of helical assemblies

4.2.3

Load-displacement curves

Chapter 5

Conclusions and outlook

Bibliography

Appendix A

Calculation of quaternions fields

describing the helices of the

assemblies