Various issues raised by the mechanics of an octopus’s arm

Abstract

Various issues raised by the mechanics of an octopus’s arm are presented from the point of view of continuum mechanics. First, we study the conditions for a mixture of a number of fiber bundles to span the space of stress tensors. The geometry of a fiber bundle, as reflected in the direction and density of the fibers, is described by a vector field, or, more precisely, by a differential two-form. For the stress analysis, we propose a three-dimensional continuum model of the arm that is based on constitutive data available for muscle fibers. An analysis of small deformations superimposed on an activated configuration of the muscle bundles reveals the stiffening mechanism which enables the arm to support various external loadings, even if the geometry and the number of the fiber groups does not allow the tensions in them to span the space of stress tensors.

Keywords

Octopus continuum mechanics stresses muscular hydrostat incompressibility activation

1. Introduction

The octopus’s arm belongs to a family of organs termed muscular hydrostats which also contains the human tongue and elephant’s trunk. Muscular hydrostats are characterized by the absence of rigid skeletons. They consist of densely packed three-dimensional array of muscle fibers, and are usually considered incompressible [1]. Incompressibility, together with the particular arrangement of the muscle fibers, enables the octopus to control the arm, compensating for the lack of rigid links and the inability of the muscles to support compression. The efficiency of such a structure and the prospect of mimicking it in a continuous robotic manipulator has motivated the interest in the mechanical aspects of the arm. (See, for example, [2 –5].)

The octopus’s arm contains five main groups of muscle fibers: longitudinal fibers, two groups of mutually orthogonal transverse fibers and two groups of right-handed and left-handed helicoidal fibers. (See Figure 1. For a detailed study see [6].) Thus, using incompressibility, the octopus can extend the arm by contracting the transverse muscles. Alternatively, the arm may be extended by using helicoidal muscles. Helicoidal fibers exist in three different layers: internal, median and external, where the last two are positioned close to the circumference of the arm. The mean pitch angle for the external and median oblique muscle fibers is 62° whereas internal oblique muscle fibers have a lower mean pitch angle of 49°.

Figure 1.

The muscular structure of an octopus’s arm.

One may easily compute the pitch angle for which the extension of the arm will not induce any strain in the helicoidal fibers. Let F(t), t ∈ [0, 1], be the parameterized, transversely isotropic deformation gradient

[F] (t) = [\begin{matrix} \frac{1}{\sqrt{1 + λ t}} & 0 & 0 \\ 0 & \frac{1}{\sqrt{1 + λ t}} & 0 \\ 0 & 0 & 1 + λ t \end{matrix}] .

(1)

Letting u be a unit vector tangent to the helicoidal fibers, one requires that

\frac{d}{dt} {[F (t) (u) \cdot F (t) (u)] |}_{t = 0} = 0,

(2)

which gives a pitch angle of 54.73°, the magic angle. Thus, activating the internal helicoidal muscles will contract the arm and activating the external and median oblique fibers will extend the arm.

In this work we consider various aspects of the mechanics of an octopus’s arm from the point of view of continuum mechanics. Specifically, we make an attempt to capture the way the morphology of the arm enables the arm to support external loadings. Conceiving the arm as a mixture of groups of muscle fibers in a number of directions, we first wish to study the conditions under which tensions in the fibers span the space of stress tensors at a point. This geometric problem is considered in some detail in Section 2. We observe that, with the incompressibility constraint, the arm functions similarly to a tensegrity structure.

In Section 3 we propose a convenient geometric description of a bundle of (muscle) fibers and the kinematics of the bundle. Then, we examine the various stress tensors induced by a mixture of intertwining fiber bundles as a model of a muscular hydrostat.

The constitutive theory for muscle fibers (e.g. [7, 8]) is rather complicated. Thus, after a short review, we propose in Section 4 a simplified loading scenario in which the arm of the octopus is first brought from a reference configuration to an inactive intermediate configuration under no external loading and negligible activation. In the second stage, the arm is activated isometrically, and, finally, the arm is loaded under fixed activations of the muscles and it is assumed that the resulting superimposed deformations are small. Linearization of the expressions for the stress shows how the arm stiffens due to the prestress generated by the activation. Moreover, this geometric stiffening enables the arm to support various external loadings, even if the geometry and the number of fiber groups do not allow the tensions in them to span the space of stress tensors.

Finally, we consider in Section 5 the stress analysis problem for a mixture of activated muscle bundles. The expression for the divergence of the stress is specialized to the situation under consideration. The stress analysis problem for the isometric activation stage is indeterminate. Thus, the octopus has the freedom to activate the muscle fibers based on additional considerations such as the stability and rigidity of the arm when acted upon by additional external loading. For given activations of the muscle groups and a given external load, the stress analysis problem is determinate, as expected.

2. Bases for the space of symmetric tensors induced by mixtures of fibers

It is assumed in this work that the stress tensor at a point in the muscular hydrostat is the sum of uniaxial stress tensors produced by tensions in the five groups of muscle fibers with the addition of hydrostatic pressure as implied by the incompressibility constraint. A natural question one may pose in this context is what is the subspace of symmetric 3 × 3 tensors that may be spanned this way and how does this subspace depend on the geometry of the fiber bundles. In particular, if the arm of the octopus is to support various loads, it seems natural to find out the conditions for the geometry of the fibers so that the whole space of stresses may be spanned. This section deals with this problem and related issues. We begin with a study of the implications of the structure for a general mixture of six fiber bundles on the possible states of stress. Then, we move on in our analysis and consider five fiber bundles with additional hydrostatic pressure.

2.1. Bases for the space of symmetric tensors

In the following analysis we consider the possibility of spanning the space of all stress tensors using six fiber bundles. Given the vectors w_n, n = 1,…,6, in $ℝ^{3}$ , we study geometric conditions so that {w₁ ⊗ w₁, …, w₆ ⊗ w₆} span the six-dimensional space of symmetric 3 × 3 tensors. Hence, the vectors w_n should satisfy

\sum_{n = 1}^{6} α^{n} w_{n} \otimes w_{n} = 0, only if all α^{n} = 0 .

(3)

Evidently, if there is no triplet of vectors that are linearly independent, the family of six vectors will not span the space of tensors. Thus, we may choose three linearly independent base vectors which we enumerate as w_i, i = 1, 2, 3, spanning $ℝ^{3}$ , where we use wⁱ to denote the corresponding contravariant (dual, reciprocal) base vectors. Applying (3) to wⁱ ⊗ w^j, i, j = 1, 2, 3, using double contraction, we obtain

\sum_{n = 1}^{6} α^{n} {(w_{n})}^{i} {(w_{n})}^{j} = 0,

(4)

where (w_n)ⁱ = w_n · wⁱ is the ith contravariant component of w_n. As a result of the symmetry of {w_n ⊗ w_n}, the last nine equations reduce to the six independent equations

\underset{A}{\underset{}{\underset{︸}{[\begin{matrix} 1 & 0 & 0 & {(w_{4})}^{1} {(w_{4})}^{1} & {(w_{5})}^{1} {(w_{5})}^{1} & {(w_{6})}^{1} {(w_{6})}^{1} \\ 0 & 1 & 0 & {(w_{4})}^{2} {(w_{4})}^{2} & {(w_{5})}^{2} {(w_{5})}^{2} & {(w_{6})}^{2} {(w_{6})}^{2} \\ 0 & 0 & 1 & {(w_{4})}^{3} {(w_{4})}^{3} & {(w_{5})}^{3} {(w_{5})}^{3} & {(w_{6})}^{3} {(w_{6})}^{3} \\ 0 & 0 & 0 & {(w_{4})}^{1} {(w_{4})}^{2} & {(w_{5})}^{1} {(w_{5})}^{2} & {(w_{6})}^{1} {(w_{6})}^{2} \\ 0 & 0 & 0 & {(w_{4})}^{1} {(w_{4})}^{3} & {(w_{5})}^{1} {(w_{5})}^{3} & {(w_{6})}^{1} {(w_{6})}^{3} \\ 0 & 0 & 0 & {(w_{4})}^{2} {(w_{4})}^{3} & {(w_{5})}^{2} {(w_{5})}^{3} & {(w_{6})}^{2} {(w_{6})}^{3} \end{matrix}]}}} {\begin{matrix} α^{1} \\ α^{2} \\ α^{3} \\ α^{4} \\ α^{5} \\ α^{6} \end{matrix}} = {\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}},

(5)

where the matrix of coefficients is denoted by A. Thus, one may write the necessary and sufficient condition as

\det A = | \begin{matrix} {(w_{p})}^{i} {(w_{p})}^{j} & {(w_{q})}^{i} {(w_{q})}^{j} & {(w_{r})}^{i} {(w_{r})}^{j} \\ {(w_{p})}^{i} {(w_{p})}^{k} & {(w_{q})}^{i} {(w_{q})}^{k} & {(w_{r})}^{i} {(w_{r})}^{k} \\ {(w_{p})}^{j} {(w_{p})}^{k} & {(w_{q})}^{j} {(w_{q})}^{k} & {(w_{r})}^{j} {(w_{r})}^{k} \end{matrix} | \neq 0,

(6)

where i, j, k and p, q, r are permutations of 1, 2, 3 and 4, 5, 6, respectively. The condition for the determinant in (6) may be rewritten as

\begin{array}{r} [{(w_{p})}^{i} {(w_{p})}^{j} {(w_{q})}^{k} {(w_{r})}^{k} (w_{q} \times w_{r}) + {(w_{p})}^{k} {(w_{q})}^{i} {(w_{q})}^{j} {(w_{r})}^{k} (w_{r} \times w_{p}) \\ + {(w_{p})}^{k} {(w_{q})}^{k} {(w_{r})}^{i} {(w_{r})}^{j} (w_{p} \times w_{q})] \cdot w_{k} \neq 0, \end{array}

(7)

by expanding the determinant by the first row.

Let C denote the set of all sextuplets of vectors in $ℝ^{3}$ , and let D denote the set of sextuplets that satisfy condition (6). It is easy to show that the following subsets contain D, so that the respective conditions defining them are necessary. (Throughout this section, by a vector, we refer always to a vector in a sextuplet under consideration.)

N₁ There are no pairs of collinear vectors.

N₂ No collection of four vectors is situated in a single plane.

N₃ There are no two distinct planes containing all six vectors where each plane contains three distinct vectors.

Since ${\bar{N}}_{2} \cup {\bar{N}}_{3}$ is the collection of sextuplets of vectors for which two planes (or less) contain all of the vectors, N₂ ∩ N₃ is the collection of sextuplets for which no two planes (or less) contain all of the vectors. Note also that the subset of N₁, which contains elements for which there is more than one pair of collinear vectors, is covered by N₂ ∩ N₃. We conclude that N₁ ∩ N₂ ∩ N₃ is the collection of sextuplets of noncollinear vectors for which no two planes (or less) contain all of the vectors.

We now let C₁ ⊂ C be the subset containing elements for which at least one collection of three linearly dependent vectors exist. Using (7), it is easy to show that C₁ ∩ N₁ ∩ N₂ ∩ N₃ ⊂ D, so that it determines a sufficient condition. Next, we consider $C_{2} : = {\bar{C}}_{1} \subset N_{1} \cap N_{2} \cap N_{3}$ , containing sextuplets of vectors such that every triplet of vectors is linearly independent. By expanding the determinant by the first column, (6) may be rewritten as

[{(w_{p})}^{j} {(w_{p})}^{k} {(w_{q})}^{i} {(w_{r})}^{i} w_{i} + {(w_{p})}^{i} {(w_{p})}^{k} {(w_{q})}^{j} {(w_{r})}^{j} w_{j} + {(w_{p})}^{i} {(w_{p})}^{j} {(w_{q})}^{k} {(w_{r})}^{k} w_{k}] \cdot (w_{q} \times w_{r}) \neq 0,

(8)

where none of the coefficients vanish since all triplets chosen from the collection {w_n} are linearly independent. Hence, we may divide both sides by (w_p)ⁱ(w_p)^j(w_p)^k. Interchanging the indices p and r for clarity, it follows that

\underset{v}{\underset{}{\underset{︸}{[\frac{{(w_{p})}^{i} {(w_{q})}^{i}}{{(w_{r})}^{i}} w_{i} + \frac{{(w_{p})}^{j} {(w_{q})}^{j}}{{(w_{r})}^{j}} w_{j} + \frac{{(w_{p})}^{k} {(w_{q})}^{k}}{{(w_{r})}^{k}} w_{k}]}}} \cdot (w_{p} \times w_{q}) \neq 0 .

(9)

The last inequality represents the following sufficient condition for the vector v given by the square brackets above:

v \neq 0 is not situated in the w_{p} w_{q} - plane .

(10)

The form of condition (9) implies that singular solutions do exist. As an example, let w _i = [1, 0, 0], w _j = [0, 1, 0], w _k = [0, 0, 1], w _p = [1, 1, 1] and w _q = [3, 2, 1]. Then, some possible singular solutions are w _r ∈ {[9, 3, 1], [7, 2.8, 1], [5, 2.5, 1], [−1, −2, 1],…}.

We now partition C ₁ into the following subsets containing the elements satisfying the respective conditions.

C₁₁: Only one plane contains three vectors. We observe that C₁₁ ⊂ N₁ ∩ N₂ ∩ N₃ and hence C₁₁ is a sufficient condition.

C₁₂: There are exactly two planes each containing three vectors. We further partition C₁₂ into the following subsets.

C₁₂₁: Each of the two planes contains three distinct vectors. We note that $C_{121} = {\bar{N}}_{3}$ .

C₁₂₂: Each of the two planes contains three vectors and one of the six vectors is shared by both planes. We have, C₁₂₂ ⊂ N₁ ∩ N₂ ∩ N₃; hence, C₁₂₂ determines a sufficient condition.

C₁₂₃: Each of the two planes contains more than one common vector. Clearly, C₁₂₃ ∩ N₁ = ∅.

C₁₃: There are three planes each of which contains three vectors. Note that for the elements of C₁₃, there are three planes each of which contains three vectors. Clearly, C₁₃ ⊂ N₁ ∩ N₂ ∩ N₃ so it determines a sufficient condition.

C_1n: There are more than three planes each containing three vectors. It is evident that C_1n ∩ N₁ ∩ N₂ ∩ N₃ = ∅.

Next, we consider the subset C₂ containing sextuplets for which there are no planes containing three vectors. Clearly, one may consider the partition C₂ = C₂₁ ∪ C₂₂, where C₂₁ contains the elements in D, C₂₁ = C₂ ∩ D, whereas $C_{22} = C_{2} \cap \bar{D}$ contains sextuplets that are linearly dependent. That is, according to (10), we have the following subsets.

C₂₁: A collection of sextuplets such that v ≠ 0 is not situated in the w_pw_q -plane, a subset of D.

C₂₂: A collection of sextuplets such that v ≠ 0 is situated in the w_pw_q-plane, a subset of $\bar{D}$ .

Note that, using set subtraction (\), we may finally write

D = N_{1} \cap N_{2} \cap N_{3} ∖ C_{22} = C_{11} \cup C_{122} \cup C_{13} \cup C_{21},

(11)

which describes a necessary and sufficient condition.

Remark 2.1. Due to the inability of fibers to produce or sustain compression, a mixture of six fiber bundles (assuming a linearly independent basis ${{\hat{w}}_{n} \otimes {\hat{w}}_{n}}$ ) can produce only stress tensors $σ = \sum σ^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n},$ for which σⁿ > 0. It is clear that for such stress tensors and any unit vector $\hat{ν}$ ,

σ_{ν ν} = σ : (\hat{ν} \otimes \hat{ν}) = \sum σ^{n} {({\hat{w}}_{n} \cdot \hat{ν})}^{2} > 0,

(12)

so that σ is positive definite and each internal plane is under tension. Evidently, linear independence of the base tensors implies that there is no activated self-equilibrated state of stress.

2.2. Bases that contain I for the space of symmetric tensors

We recall that the arm of an octopus is composed of five families of muscle fibers and is assumed to be incompressible. Here, we study the implications of such a structure on the possible states of stress using a procedure analogous to that presented in the preceding section. Thus, taking into account the possible hydrostatic pressure and assuming that five unit vectors, ${{\hat{w}}_{n}}$ , representing the directions of the fibers are given, we consider the representation of the stress tensor in the form

σ = \sum_{n = 1}^{5} σ^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n} - p I .

(13)

For the problem at hand, we will use notation which is analogous to that used in Section 2.1 modified by the superscript p. (For example, the set of all quintuplets of vectors will be denoted by C^p.) Hence, the condition that {w₁ ⊗ w₁, …, w₅ ⊗ w₅, I} span the space of tensors is

\sum_{n = 1}^{5} α^{n} w_{n} \otimes w_{n} - p I = 0, only if all α^{n} = 0 and p = 0 .

(14)

By the same algebraic computations as used in Section 2.1, one obtains the necessary and sufficient condition

\det A^{p} = | \begin{matrix} g^{i j} & g^{i k} & g^{j k} \\ {(w_{p})}^{i} {(w_{p})}^{j} & {(w_{p})}^{i} {(w_{p})}^{k} & {(w_{p})}^{j} {(w_{p})}^{k} \\ {(w_{q})}^{i} {(w_{q})}^{j} & {(w_{q})}^{i} {(w_{q})}^{k} & {(w_{q})}^{j} {(w_{q})}^{k} \end{matrix} | \neq 0,

(15)

where i, j, k are the permutations of 1, 2, 3, p, q are the permutations of 4, 5 and g^ij are the components of the contravariant metric tensor for the basis {w_i, w_j, w_k}. The expansion of the determinant by the second row, using the triple scalar product, gives

[{(w_{p})}^{i} {(w_{p})}^{j} {(w_{q})}^{k} (w_{k} \times w^{k}) + {(w_{p})}^{k} {(w_{p})}^{i} {(w_{q})}^{j} (w_{j} \times w^{j}) + {(w_{p})}^{j} {(w_{p})}^{k} {(w_{q})}^{i} (w_{i} \times w^{i})] \cdot w_{q} \neq 0 .

(16)

Using equation (16) one may define the following subsets, the elements of which satisfy the respective necessary conditions.

$N_{1}^{p}$ : There are no pairs of collinear vectors.

$N_{2}^{p}$ : No collection of four vectors is situated in a single plane.

$N_{3}^{p}$ : There are no triplets of mutually perpendicular vectors.

$N_{4}^{p}$ : No collection of five vectors is situated in two perpendicular planes.

It is easy to show that $C_{1}^{p} \cap N_{1}^{p} \cap N_{2}^{p} \cap N_{4}^{p} \subset D^{p}$ , for which there are two non-perpendicular planes (one containing three vectors and the other containing two vectors), determines a sufficient condition. Assume now that the four necessary conditions above hold, and that the last sufficient condition is not satisfied. That is, we consider $C_{2}^{p} \cap N_{3}^{p} \subset N_{1}^{p} \cap N_{2}^{p} \cap N_{4}^{p}$ for which each triplet of vectors is linearly independent (the five vectors cannot be situated in two planes) and, in addition, no triplet is orthogonal. Denoting the vector in the square brackets of (16) by u, it may be expressed as

u = [(w_{p} \times w_{q}) \cdot w_{i}] {(w_{p})}^{i} g_{j k} w_{i} + [(w_{p} \times w_{q}) \cdot w_{j}] {(w_{p})}^{j} g_{i k} w_{j} + [(w_{p} \times w_{q}) \cdot w_{k}] {(w_{p})}^{k} g_{i j} w_{k} .

(17)

Hence, (16) gives us the following sufficient condition:

u \neq 0 is not perpendicular to w_{q} .

(18)

As an example for which (18) is satisfied, consider a quintuplet that contains a linearly independent triplet for which one of the vectors is perpendicular to the plane spanned by the remaining two, which are not orthogonal. It is still assumed that all the triplets are linearly independent and are not mutually orthogonal.

As a counterexample, one may consider the case where there is a quadruplet of vectors such that, for each partition into two pairs of vectors, the two planes spanned by the two pairs are perpendicular. Again, it is still assumed that all the triplets are linearly independent and are not mutually orthogonal.

We now partition $C_{1}^{p} \cap N_{1}^{p} \cap N_{2}^{p}$ into the following disjoint subsets defined by the corresponding conditions.

$C_{11}^{p}$ : The two planes, one containing three vectors and the other containing two vectors, are not perpendicular. That is, we consider $C_{1}^{p} \cap N_{1}^{p} \cap N_{2}^{p} \cap N_{4}^{p}$ , so that it determines a sufficient condition.

$C_{12}^{p}$ : The two planes are perpendicular. This condition determines a singular subset $C_{1}^{p} \cap N_{1}^{p} \cap N_{2}^{p} \cap {\bar{N}}_{4}^{p}$ . We partition $C_{12}^{p}$ into the following subsets:¹

$C_{121}^{p}$ : One plane contains three vectors. We further consider the following.

$C_{1211}^{p}$ : All of the vectors in the first plane are not perpendicular to any vector in the second plane.

$C_{1212}^{p}$ : One of the two vectors in the first plane is perpendicular to the three vectors in the second plane.

$C_{1213}^{p}$ : One of the three vectors in the first plane is perpendicular to the two vectors in the second plane.

$C_{122}^{p}$ : Two planes contain three vectors. For elements of $C_{122}^{p}$ , each of the two planes contains three vectors and one (of the five vectors) is situated in both planes. Note that ${\bar{N}}_{3}^{p}$ intersects this subset.

$C_{12 n}^{p}$ : More than two planes contain three vectors. Evidently, $C_{12 n}^{p} = \emptyset$ .

We next consider the subset $C_{2}^{p}$ which we partition in to the following subsets containing a collection of quintuplets which satisfy the respective conditions.

$C_{21}^{p}$ : There is a triplet of mutually perpendicular vectors so that $C_{21}^{p} = C_{2}^{p} \cap {\bar{N}}_{3}^{p}$ . This is a singular subset.

$C_{22}^{p}$ : There are no triplets of mutually perpendicular vectors so that $C_{22}^{p} = C_{2}^{p} \cap N_{3}^{p}$ . Clearly, one may partition $C_{22}^{p}$ as $C_{22}^{p} = C_{221}^{p} \cup C_{222}^{p}$ , where $C_{221}^{p} = C_{22}^{p} \cap D^{p}$ and $C_{222}^{p} = C_{22}^{p} \cap {\bar{D}}^{p}$ .

$C_{221}^{p}$ : The vector u ≠ 0 is not perpendicular to w_q.

$C_{222}^{p}$ : The vector u ≠ 0 is perpendicular to w_q.

We finally conclude that the collection of quintuplets {w_n} of vectors, such that {w₁ ⊗ w₁, …, w₅ ⊗ w₅, I} span the set of all symmetric tensors, is

D^{p} = N_{1}^{p} \cap N_{2}^{p} ∖ C_{12}^{p} \cap C_{21}^{p} \cap C_{222}^{p} = C_{11}^{p} \cup C_{221}^{p} .

(19)

2.3. Active stresses equilibrating hydrostatic pressure

In this section we examine the conditions for the tensions in the activated muscle bundles to equilibrate the pressure in an unloaded octopus’s arm. See Section 4.3 for the particular relevance of this situation in our loading scenario. In such a case,

σ = \sum_{n = 1}^{5} σ^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n} - p I = 0, for some σ^{n} \neq 0 and p \neq 0 .

(20)

Thus, it is necessary that {w_n ⊗ w_n, I} are linearly dependent so that the quintuplets of vectors belong to the set ${\bar{D}}^{p}$ defined in the foregoing subsection.

Firstly, assuming linear independence of ${{\hat{w}}_{n} \otimes {\hat{w}}_{n}}$ , n = 1, …, 6, we want to find the conditions for the active stress tensor to produce hydrostatic pressure, i.e.

\sum_{n = 1}^{6} σ^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n} = p I .

(21)

Since the linear independence conditions are indifferent to the magnitude of the vectors w_n, we consider the unit vectors ${{\hat{w}}_{n}}$ . We denote the corresponding fiber basis tensors by ${D_{n}} = {{\hat{w}}_{n} \otimes {\hat{w}}_{n}}$ . The corresponding contravariant basis is {D^m} satisfying $D_{n} : D^{m} = δ_{n}^{m}$ , and the metric tensor is D_nm = D_n : D_m for the six-dimensional space. (See also [9] for an analogous analysis.) Note that D_nn = 1 since $D_{n m} = ({\hat{w}}_{n} \otimes {\hat{w}}_{n}) : ({\hat{w}}_{m} \otimes {\hat{w}}_{m}) = {({\hat{w}}_{n} \cdot {\hat{w}}_{m})}^{2}$ . The contravariant metric tensor D^nm = Dⁿ : D^m is the inverse of D_nm. By the action of (21) on D^m, we obtain

σ^{m} = p \sum_{l} D^{m l} .

(22)

Clearly, D^ml depends only on the geometry of the fibers. The condition that the fibers cannot support compression implies that σ^m > 0 and p > 0 for all m, and, hence, the fibers should satisfy

\sum_{l} D^{m l} > 0 .

(23)

Next, we consider the conditions for the prestress in the incompressible structure consisting of five fiber bundles to balance the pressure p for an unloaded arm. Thus, equation (20) holds and a necessary condition for nonvanishing values of p is that a sufficient condition for ${{\hat{w}}_{n} \otimes {\hat{w}}_{n}, I}$ to span the space of tensors is violated. Assuming that the subspace of stress tensors spanned by ${{\hat{w}}_{n} \otimes {\hat{w}}_{n}}$ , n = 1, …, 5, contains I, one can compute the metric tensor D_ml, m, l = 1, …, 5, for this subspace. In analogy with (23), the condition for tensions in the fibers will be

\sum_{l} D^{m l} > 0, m, l = 1, \dots, 5 .

(24)

Remark 2.2. Using (24) and the subsets defined in Subsection 2.2, it is easy to show that hydrostatic pressure in a muscular hydrostat can be equilibrated by appropriately arranged mixtures of either three, four or five muscle groups.

2.4. The corresponding dual bases for the space of symmetric tensors

Consider a given fiber basis, {D_n} = {w_n ⊗ w_n}, n = 1, …, 6, so that one may write $σ = \sum_{n = 1}^{6} {\tilde{σ}}^{n} D_{n}$ . Using the same notation as in (15), we consider now a natural basis ${E_{n^{'}}}$ , n′ = 1,…,6, for the space of symmetric tensors, the elements of which are

\begin{matrix} E_{1} = w_{1} \otimes w_{1}, E_{2} = w_{2} \otimes w_{2}, E_{3} = w_{3} \otimes w_{3}, E_{4} = w_{1} \otimes w_{2} + w_{2} \otimes w_{1}, \\ E_{5} = w_{1} \otimes w_{3} + w_{3} \otimes w_{1}, E_{6} = w_{2} \otimes w_{3} + w_{3} \otimes w_{2} . \end{matrix}

(25)

It is easy to show that corresponding contravariant basis is given by

\begin{matrix} E^{1} = w^{1} \otimes w^{1}, E^{2} = w^{2} \otimes w^{2}, E_{3} = w^{3} \otimes w^{3}, E^{4} = \frac{1}{2} [w^{1} \otimes w^{2} + w^{2} \otimes w^{1}], \\ E^{5} = \frac{1}{2} [w^{1} \otimes w^{3} + w^{3} \otimes w^{1}], E^{6} = \frac{1}{2} [w^{2} \otimes w^{3} + w^{3} \otimes w^{2}] . \end{matrix}

(26)

The transformations between the covariant bases may be determined by the 6 × 6 matrix [B] such that $D_{n} = B_{\cdot n}^{n^{'}} E_{n^{'}}, E_{n^{'}} = B_{\cdot n^{'}}^{n} D_{n}$ , where $B_{\cdot n}^{n^{'}}$ is the inverse of $B_{\cdot n^{'}}^{n}$ . One may also represent the transformations between the contravariant bases using the matrix [C], where $D^{n} = C_{n^{'}}^{\cdot n} E^{n^{'}}$ and $E^{n^{'}} = C_{n}^{\cdot n^{'}} D^{n}$ . It is well known that $C_{n^{'}}^{\cdot n} = B_{\cdot n^{'}}^{n}$ , that is, [C] = [B]^−T and it is observed here that [B]^T = [A]. Using the transformation for the contravariant bases with the results in (26), one obtains the duals to be

\begin{array}{l} D^{1} & = w^{1} \otimes w^{1} - a_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) \\ - a_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) - a_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \\ D^{2} & = w^{2} \otimes w^{2} - b_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) \\ - b_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) - b_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \\ D^{3} & = w^{3} \otimes w^{3} - c_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) \\ - c_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) - c_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \\ D^{4} & = d_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) + d_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) \\ + d_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \\ D^{5} & = e_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) + e_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) \\ + e_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \\ D^{6} & = f_{3} \frac{1}{2} (w^{1} \otimes w^{2} + w^{2} \otimes w^{1}) + f_{2} \frac{1}{2} (w^{1} \otimes w^{3} + w^{3} \otimes w^{1}) \\ + f_{1} \frac{1}{2} (w^{2} \otimes w^{3} + w^{3} \otimes w^{2}), \end{array}

(27)

where

{\begin{matrix} \begin{array}{l} a_{i} = & \frac{1}{| A |} [{(w_{4})}^{1} {(w_{4})}^{1} {(w_{5})}^{i} {(w_{6})}^{i} (w_{5} \times w_{6}) + {(w_{4})}^{i} {(w_{5})}^{1} {(w_{5})}^{1} {(w_{6})}^{i} (w_{6} \times w_{4}) \\ + {(w_{4})}^{i} {(w_{5})}^{i} {(w_{6})}^{1} {(w_{6})}^{1} (w_{4} \times w_{5})] \cdot w_{i}, \\ b_{i} = & \frac{1}{| A |} [{(w_{4})}^{2} {(w_{4})}^{2} {(w_{5})}^{i} {(w_{6})}^{i} (w_{5} \times w_{6}) + {(w_{4})}^{i} {(w_{5})}^{2} {(w_{5})}^{2} {(w_{6})}^{i} (w_{6} \times w_{4}) \\ + {(w_{4})}^{i} {(w_{5})}^{i} {(w_{6})}^{2} {(w_{6})}^{2} (w_{4} \times w_{5})] \cdot w_{i}, \\ c_{i} = & \frac{1}{| A |} [{(w_{4})}^{3} {(w_{4})}^{3} {(w_{5})}^{i} {(w_{6})}^{i} (w_{5} \times w_{6}) + {(w_{4})}^{i} {(w_{5})}^{3} {(w_{5})}^{3} {(w_{6})}^{i} (w_{6} \times w_{4}) \\ + {(w_{4})}^{i} {(w_{5})}^{i} {(w_{6})}^{3} {(w_{6})}^{3} (w_{4} \times w_{5})] \cdot w_{i}, \\ d_{i} = & \frac{1}{| A |} {(w_{5})}^{i} {(w_{6})}^{i} (w_{5} \times w_{6}) \cdot w_{i}, e_{i} = \frac{1}{| A |} {(w_{4})}^{i} {(w_{6})}^{i} (w_{6} \times w_{4}) \cdot w_{i}, \\ f_{i} = & \frac{1}{| A |} {(w_{4})}^{i} {(w_{5})}^{i} (w_{4} \times w_{5}) \cdot w_{i} . \end{array} \end{matrix}

(28)

Note that the determinant |A| has been computed in Subsection 2.1.

Remark 2.3. One may give a mechanical interpretation for the contravariant base tensors Dⁿ, n = 1, …, 6. While $D_{n} = {\hat{w}}_{n} \otimes {\hat{w}}_{n}$ represents a unit uniaxial stress in the n th bundle, we may interpret D^m as a variation of the strain tensor. Thus, the condition $D_{n} : D^{m} = δ_{n}^{m}$ implies that D^m is the variation of the strain tensor for which the tension in the mth fibers performs a unit amount of power while the tensions in the other fibers do not perform any power.

3. Geometry, tension in fibers and stresses

In this section, we propose the mathematical object for the representation of the geometry, kinematics and stresses for bundles of fibers. For the objects analyzed, we study the consequences of the incompressibility of the material.

3.1. Kinematics of fiber bundles

We identify a material body with a fixed reference configuration and, assuming that a reference frame has been chosen, a generic material point X is identified with the radius vector X pointing at it. Consider now a bundle of fibers embedded in the body. For a point X in the body, we let W be a vector in the direction of the fibers, where we assume that the fibers have some specific orientation, or sense. In the case of muscle fibers for example, this may be taken as the sense in which an activation signal is propagating. The magnitude of W is defined to be the density of the fibers, that is, the number of fibers per unit area normal to W. We will refer to W as the fiber density vector. Viewing X as an independent variable, one obtains a vector field W(X) where we abuse the notation by using the same symbol for both the variable W and the field. Let d A be an infinitesimal element of area at X with a unit normal $\hat{N}$ . Then, the number of fibers crossing d A is $W \cdot \hat{N} d A$ .

Consider now a deformation κ of the body in space and the deformation gradient F at the point X. Under the action of F, d A transforms to an infinitesimal deformed area d a with a unit normal $\hat{n}$ , originating at x = κ(X). We let w denote the fiber density vector for the image of the fiber bundle under κ.

Assuming that the fibers are material quantities carried with the material points at each configuration of the body, it follows that the amount of fibers crossing the area element dA in the reference configuration is equal to the amount crossing its image da. Thus,

W \cdot \hat{N} d A = w \cdot \hat{n} d a .

(29)

Since (29) holds for all $\hat{N} d A$ , using Nanson’s formula one has

w = \frac{1}{J} F (W),

(30)

where J ≡ det F. It is observed that this transformation implies that the fiber density is not a genuine vector. In fact, similarly to flux fields, it is a differential two-form in the three-dimensional body manifold as we observe in the following subsection.

In case the deformation is isochoric, J = 1, and the expression above simplifies to

w = F (W) .

(31)

Hence, for isochoric deformations the fiber density vector is strictly transformed according to the same rules as the vectors.

Let ρ₀ = |W| and ρ = |w| denote the density of the fibers per unit area at the reference and current configurations, respectively, and let λ be the extension of the fibers due to deformation, that is, the ratio λ = |F(W)|/|W|. Then, according to (30),

ρ = \frac{λ ρ_{0}}{J} .

(32)

For the incompressible case,

λ = \frac{ρ}{ρ_{0}},

(33)

so that the stretch of the fibers is equal to the ratio between current and reference densities.

Finally, the total amount of fibers that intersect a surface S is

Φ = \int_{S} w \cdot \hat{n} d a .

(34)

The assumption that fibers are neither created nor destroyed in R implies that, for the boundary S = ∂R of any region R inside the fiber bundle,

Φ = 0 = \int_{\partial R} w \cdot \hat{n} d a .

(35)

Using Stokes’ theorem,

\int_{R} div w d v = 0,

(36)

and since the last equation holds for any region R,

div w = 0 .

(37)

3.2. Differential geometry of fibrous bodies

The structure of a fibrous body may be described in a very general framework in which no particular reference configuration is used and in which the body manifold has no specific Riemannian metric. In this work, we use the word fibers in the mechanical sense, as in muscle fibers, rather than in the differential geometric sense.

Let $ℬ$ be the n-dimensional manifold representing the body. The geometry of the fiber structure is specified by an (n − 1)-differential form ω, that is, a smooth section of $^{n - 1} T^{*} ℬ$ : the vector bundle of completely antisymmetric tensors of order (n − 1). We will refer to ω as the structure form. For a collection of n − 1 vectors, $v_{1}, \dots, v_{n - 1} \in T_{X} ℬ$ , the value ω(v₁, …, v_n₋₁) is interpreted as the amount of fibers intersecting the infinitesimal parallelepiped determined by the vectors with the orientation determined by the order of the vectors. Thus, the integral of ω over an oriented (n − 1)-dimensional submanifold $S$ ,

Φ = \int_{S} ω,

(38)

represents the total amount of fibers that penetrate $S$ in a particular sense, which depends on the orientation chosen for it.

Consider a n-dimensional submanifold with boundary $ℛ$ of $ℬ$ . Assuming that there are neither sources nor sinks of fibers insider $ℛ$ , it is concluded that

\int_{\partial ℛ} ω = 0 .

(39)

Using Stokes’ theorem, the last equation implies that

\int_{ℛ} d ω = 0,

(40)

where dω denotes the exterior derivative of the structure form. Since equation (40) holds for an arbitrary region $ℛ$ , we conclude that

d ω = 0 .

(41)

In fact, dω represents the distribution of sources of the fibers. Alternatively, one may think of dω as the distribution of defects in the structure of the fibers.

For a point $X \in ℬ$ , the direction of the fiber is determined by the one-dimensional vector subspace V_X of the tangent space, $T_{X} ℬ$ at X, defined by

V_{X} = {v \in T_{X} ℬ | v ⌟ ω = 0} .

(42)

Here, v ⌟ ω denotes the contraction (or inner product) of the form ω with the vector v, that is, the (n − 2)-form such that,

v ⌟ ω (v_{1}, \dots, v_{n - 2}) = ω (v, v_{1}, \dots, v_{n - 2}) .

(43)

It follows that in the case where V_X is a subspace of the vector space spanned by a collection of vectors, v₁,…,v_n₋₁, the amount of fibers crossing the parallelepiped generated by the vectors v₁, …, v_n₋₁ is identically zero. For the Euclidean three-dimensional case, in the analogous situation, the area element determined by the two vectors v₁ and v₂ contains the fiber vector w and has no fibers cross it.

To describe the correspondence between the differential geometric framework and that described in Section 3.1, we restrict ourselves to the three-dimensional case. Given a volume element, a nowhere vanishing three-form θ, the structure form ω may be represented by a unique vector field w satisfying

ω = w ⌟ θ

(44)

or, equivalently,

ω (X) (v, u) = θ (X) (w, v, u)

(45)

for any pair of vectors v and u in $T_{X} ℬ$ . Assume now that a configuration of the body in a Euclidean physical space is given. One may set

θ (w, u, v) = w \cdot (u \times v) = \det (w, u, v) .

(46)

Thus, the relation between the structure form and the fiber density vector field is

ω (u, v) = w \cdot (u \times v) = w \cdot \hat{n} d a,

(47)

where da is the infinitesimal area of the parallelogram determined by u and v and $\hat{n}$ is a unit vector in the direction of the cross product. It is observed that while ω is independent of the configuration, the vector field w depends on the configuration. In particular, both fiber density vectors w and W defined in Section 3.1 are representations of a two-form ω defined on the body manifold.

3.3. Tension in fiber bundles and stresses

According to Cauchy’s theorem, the infinitesimal surface force df on an infinitesimal area d a with a unit normal $\hat{n}$ can be expressed by the action of the Cauchy stress tensor σ as

d f = σ (\hat{n} d a) .

(48)

For a fiber bundle, it is assumed that the fibers can support only tension; hence, d f and the direction of the fibers $\hat{w}$ are parallel independently of $\hat{n}$ . We assume that there is a positive scalar function T that specifies the tension per fiber at each point in the bundle. This is the continuum mechanical counterpart of the microscopic picture of a tension in a myofibril. We recall that the number of fibers that cross the area element d a which have a normal $\hat{n}$ is $w \cdot \hat{n} d a$ and that the force acting on d a is $d f = (T \hat{w}) w \cdot \hat{n} d a$ . Thus, the total force exerted on a bundle of fibers across a surface S is

f = \int_{S} (T \hat{w}) w \cdot \hat{n} d a = \int_{S} T \hat{w} \otimes w (\hat{n}) d a .

(49)

It is clear from (48) and (49) that the Cauchy stress tensor is

σ = T \hat{w} \otimes w = T ρ \hat{w} \otimes \hat{w},

(50)

representing a uniaxial stress in the direction of the fibers. Substituting (50) to (48), using Nanson’s formula and (30), we find that

\begin{matrix} d f & = \frac{T}{ρ} w \otimes W (\hat{N} d A) . \end{matrix}

(51)

Equation (51) describes the action of the first Piola–Kirchhoff stress tensor P satisfying $d f = P (\hat{N} d A)$ , so that

P = T \hat{w} \otimes W = T ρ_{0} \hat{w} \otimes \hat{W} .

(52)

It is reasonable to assume that the values of T and ρ₀ are those obtained in experiments, since the tension T is known at the current state whereas the density ρ₀ is measured at the reference state.

Let df_R = F⁻¹(df) be the ‘pull back’ of the force df onto the reference configuration of the body. Using equations (30) and (51), the second Piola–Kirchhoff stress tensor S satisfies

d f_{R} = S (\hat{N} d A), S = \frac{T}{λ} \hat{W} \otimes W = \frac{T ρ_{0}}{λ} \hat{W} \otimes \hat{W} .

(53)

It is easy to verify that the expressions for the Piola–Kirchhoff stresses are in accordance with the usual formulas given in terms of the Cauchy stress.

Finally, for incompressible deformations, the expressions for the stresses simplify to

σ = \frac{T}{ρ} w \otimes w, P = \frac{T}{ρ} w \otimes W, S = \frac{T}{ρ} W \otimes W .

(54)

We view the arm of the octopus as a body consisting of N intertwining fiber bundles and model it as a mixture with non-interacting constituents, where N fiber density fields are present at each point. (At this point we do not yet impose the incompressibility constraint.) Thus, the Cauchy stress may be written as

σ= \sum_{n = 1}^{N} \frac{T^{n}}{ρ^{n}} w_{n} \otimes w_{n} .

(55)

By the linearity of the first and second Piola transformations in σ , one obtains

P = \sum_{n = 1}^{N} \frac{T^{n}}{ρ^{n}} w_{n} \otimes w_{n}, S = \sum_{n = 1}^{N} \frac{T^{n}}{ρ^{n}} w_{n} \otimes w_{n} .

(56)

Remark 3.1. Consider the stress at a given point in a continuous body. The Cauchy stress tensor σ may be represented using the unit eigenvectors, ${\hat{w}}_{i}$ , as

σ = \sum_{i = 1}^{3} σ_{i} {\hat{w}}_{i} \otimes {\hat{w}}_{i},

(57)

which is identical to the stress in a mixture of three bundles of fibers directed along the principle axes. Therefore, according to (30),

S = \sum_{i = 1}^{3} J σ_{i} F^{- 1} ({\hat{w}}_{i}) \otimes F^{- 1} ({\hat{w}}_{i}) = \sum_{i = 1}^{3} J \frac{σ_{i}}{λ_{i}^{2}} {\hat{W}}_{i} \otimes {\hat{W}}_{i} .

(58)

Thus, the second Piola–Kirchhoff stress tensor may be interpreted as resulting from a family of three fiber bundles that are mutually orthogonal at the current configuration, but are no longer orthonormal at the reference configuration.

4. Constitutive theory and its linearization

In this section we adapt a standard constitutive theory for active muscular fiber bundles to our model of the octopus’s arm and linearize it for small deformations superimposed on a finite deformation. Muscular hydrostats such as the octopus’s arm, the human tongue, and the elephant’s trunk use incompressibility to overcome their lack of rigid links and the disability of the fibers to extend actively. For example, active contraction of the transverse fibers will produce passive elongation of the arm, and the capability to support compression along its axis. Here, we study the way the geometry of the activated fibers together with the incompressibility result in the stiffening of the arm and enable it to support various external loads. Unlike some other works formulated in terms of a strain energy function (e.g. [10 –13]) we propose a model which is based on the experimentally available data: the dependence of the tension in a myofibril on the extension and activation.

4.1. The constitutive model for muscular fiber bundles

The tension developed by a muscle fiber during its activation depends, in general, on the history of its stretch, stretch rate and level of activation. In our discussion, studying only quasi-static deformations, the dependence on the stretch rate is omitted. We first consider a constitutive model for the activation of a single muscle fiber which assumes additive decomposition [14 –16]. In such a constitutive model, the total force T in the fiber is assumed to be the sum of an active contribution, T_act, and a passive component, T_pas, which does not depend on the activation, in the form

T = T_{act} + T_{pas} = {\hat{T}}_{a} (λ, α) + {\hat{T}}_{p} (λ) .

(59)

Here, λ is the stretch of the fiber, the ratio l/l_r between the current length and the unloaded and inactive length, and α is the activation level. The activation parameter takes values in the interval [0, 1], α = 0 for the passive state for which T_act vanishes whereas α = 1 describes the tetanized state of the muscle producing the maximal force for the current length. The experimental data concerning the dependence of ${\hat{T}}_{a}$ on the history of the stretch and the level of activation is very limited. Nevertheless, such data is available for the isometric force–stretch relation. For the isometric force–stretch data, the variable λ in ${\hat{T}}_{a} (λ, α)$ indicates the length to which the fiber is stretched while passive prior to activation [7, 17, 18]. (We use the same notation, ${\hat{T}}_{a} (λ, α)$ , although a specific form of the history, as described in the last sentence, is considered.) The continuous curve in Figure 2 describes ${\hat{T}}_{a}$ for a single fiber and some fixed level of activation as a function of the fixed length to which the fiber is stretched prior to the activation. The dashed curve represents ${\hat{T}}_{p}$ for a single inactive fiber under variable stretch.

Figure 2.

A typical force-length relation for a single muscle fiber for some fixed activation.

The data provided by equation (59) is insufficient to describe how the force is changed if an additional stretch is imposed on the fiber for non-vanishing activation. Additional data of this type may be found in works which focus on ‘force produced after stretch’ as in [19, 20]. Such experiments measure the force in the form of

T = {\hat{T}}_{l} (λ, λ_{0}, α_{0}),

(60)

where λ is the additional stretch and λ₀ is the initial stretch prior to the activation α₀. Roughly, the force tends to follow approximately a straight line with a positive slope. Our ‘three-stage model’ described in Section 4.3 uses such a loading scenario.

4.2. Linearization of the constitutive model for small superimposed deformations

As a background and for the introduction of the notation, we include a short review of the relevant kinematics and analysis of stress for the general case of infinitesimal deformations superimposed upon a finite deformation.

We consider a configuration κ₀ relative to the reference configuration and we refer to it as the intermediate configuration. The notation X₀ = κ₀(X) will be used for position of the body point X in the intermediate configuration. The body is deformed further to the current state by the deformation χ relative to the intermediate configuration so that x = χ(X₀) is the current location of the point $X = κ_{0}^{- 1} (X_{0})$ . Thus, the total deformation from the reference configuration to the current configuration is κ = χ ∘ κ₀ and, accordingly, F₀, F and F_tot ≡ F ∘ F₀ are the corresponding deformation gradients.

Let σ ₀ denote the stress tensor at the intermediate configuration which we view as given relative to the geometry of the intermediate configuration. From this intermediate state, we superimpose a deformation F on the body. One can represent the stress tensor at the current configuration, using either the Cauchy stress tensor σ or the Piola–Kirchhoff tensors P and S, using the corresponding constitutive relations $σ = \hat{σ} (F), P = \hat{P} (F)$ and $S = \hat{S} (F)$ . We now restrict our attention to the case of small deformations, F = I + δF, where the superimposed displacement gradient δF is assumed to be infinitesimal. Linearizing the constitute relation for the second Piola–Kirchhoff stress tensor in a small neighborhood of the identity I, one has

S = \hat{S} (I + δ F) ≅ \hat{S} (I) + D \hat{S} (δ F) = σ_{0} + D \hat{S} (δ F) .

(61)

Here, $D \hat{S} (δ F)$ is the directional derivative of the constitutive relation at the identity.

The linearized relation between the Cauchy and the second Piola–Kirchhoff stresses is

σ ≅ (1 - tr (δ F)) σ_{0} + δ F \circ σ_{0} + σ_{0} \circ δ F^{T} + D \hat{S} (δ F) .

(62)

If the body is incompressible tr(δF) = 0, and one adds a hydrostatic pressure term −pI which cannot be determined by the constitutive relation.² Hence,

σ ≅ σ_{0} + δ F \circ σ_{0} + σ_{0} \circ δ F^{T} + D \hat{S} (δ F) - p I .

(63)

4.3. The three-stage model

Consider now a three-dimensional continuous body, possibly incompressible, which consists of a mixture of N bundles of muscle fibers. In accordance with the foregoing scheme of notation, variables corresponding to the intermediate configuration are indicated by a zero subscript. The fiber density vector at the reference configuration will be denoted by W and its magnitude by ρ_R = |W|. For the intermediate configuration κ₀, we let w_n, $ρ_{0}^{n}$ and $λ_{0}^{n}$ denote the nth fiber density vector, its magnitude and the stretch, respectively. Correspondingly, v_n, ρⁿ and λⁿ are their analogs under the current configuration χ.

In order to describe the state of stress in a muscular hydrostat, we use the constitutive relations as in Section 4.1 according to our earlier assumptions concerning a general mixture of fiber bundles as in Section 3.3. Focusing on the role of the fibers alone, and in order to illustrate some of the principles which make it possible for such organs to function, we neglect here the intramuscular resistance of the connective tissues. Since experimental data is available mostly for the isometric activation and for the extension under constant activation, we conceive the following three-stage model loading scenario.

First, we consider κ₀ as a deformation of the organ, in the absence of activation and with negligible forces in the fibers, from some stress-free reference state to an intermediate configuration. Despite of the fact that due to the passive deformation of the fibers stresses may exist, our model neglects this passive part.

Next, in this passively deformed intermediate unloaded configuration κ₀, the muscle fibers are activated isometrically. We denote the active isometric tension in the fibers as $T_{0} = {\hat{T}}_{a} (λ_{0}, α_{0})$ , where α₀ represents the level of activation in this activated intermediate configuration. One may express the Cauchy stress tensor at a point X₀ by

σ_{0} = \sum_{n} T_{0}^{n} ρ_{0}^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n} - p_{0} I,

(64)

where p₀ is the hydrostatic pressure in the intermediate configuration.

Finally, from this activated intermediate state, with fixed activation levels $α_{0}^{n}$ in the muscles, the body is infinitesimally deformed under external loading to the current state by the deformation χ relative to the intermediate configuration. The tension in a muscle fiber is given by (60) as $T = {\hat{T}}_{l} (λ, λ_{0}, α_{0})$ for additional stretch λ from the isometric condition under the external loading. For each n = 1,…,N, one has for the initial condition of the loading $T_{0}^{n} = {\hat{T}}_{l}^{n} (λ^{n} = 1, λ_{0}^{n}, α_{0}^{n}) = {\hat{T}}_{a}^{n} (λ_{0}^{n}, α_{0}^{n})$ . The Cauchy stress tensor at a point x in the current configuration χ is

σ = \sum_{n} T^{n} ρ^{n} {\hat{v}}_{n} \otimes {\hat{v}}_{n} - p I,

(65)

where p is the hydrostatic pressure in the current configuration.

4.4. Linearization of the constitutive relation for the case of muscle fibers

Recalling that the second Piola–Kirchhoff stress obtained in (56) is

S = \sum_{n} \frac{T^{n}}{λ^{n}} {\hat{w}}_{n} \otimes w_{n},

(66)

the process of linearization is further simplified and one arrives at

D \hat{S} (δ F) = \sum_{n} [({\frac{\partial {\hat{T}}_{l}^{n}}{\partial λ^{n}} |}_{λ_{0}^{n}} - T_{0}^{n}) (δ F ({\hat{w}}_{n}) \cdot {\hat{w}}_{n}) w_{n} \otimes {\hat{w}}_{n}] .

(67)

Using the linearized transformation (63), we obtain

σ = \sum_{n} [T_{0}^{n} w_{n} \otimes {\hat{w}}_{n} + ({\frac{\partial {\hat{T}}_{l}^{n}}{\partial λ^{n}} |}_{λ_{0}^{n}} - T_{0}^{n}) (δ F ({\hat{w}}_{n}) \cdot {\hat{w}}_{n}) w_{n} \otimes {\hat{w}}_{n} + T_{0}^{n} (δ F (w_{n}) \otimes {\hat{w}}_{n} + {\hat{w}}_{n} \otimes δ F (w_{n}))] - p I .

(68)

Introducing the infinitesimal strain tensor $e = \frac{1}{2} (δ F + δ F^{T})$ and the infinitesimal rotation tensor $ω = \frac{1}{2} (δ F - δ F^{T})$ , one may rewrite (68) as

σ = \sum_{n} [σ_{0}^{n} + e \circ σ_{0}^{n} + σ_{0}^{n} \circ e + ω \circ σ_{0}^{n} - σ_{0}^{n} \circ ω + B^{n} (e)] - p I,

(69)

where

σ_{0}^{n} = T_{0}^{n} ρ_{0}^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n},

(70)

and Bⁿ is the fourth rank tensor,

B^{n} = ({\frac{\partial {\hat{T}}_{l}^{n}}{\partial λ^{n}} |}_{λ_{0}^{n}} - T_{0}^{n}) ρ_{0}^{n} {\hat{w}}_{n} \otimes {\hat{w}}_{n} \otimes {\hat{w}}_{n} \otimes {\hat{w}}_{n} .

(71)

Here, Bⁿ(e) = Bⁿ : e and clearly $D \hat{S} = \sum_{n} {\hat{B}}^{n}$ . For a muscular hydrostat, we have

σ = \sum_{n} [σ_{0}^{n} + e \circ σ_{0}^{n} + σ_{0}^{n} \circ e + ω \circ σ_{0}^{n} - σ_{0}^{n} \circ ω + B^{n} (e)] - p I .

(72)

The fact that the body is prestressed due to activation causes geometric contributions to the stiffness in addition to the elastic terms. In order to distinguish them from the elastic stiffness, a material property, these stiffness terms are usually referred to as the geometric stiffness.

In the special case where $ω \circ \sum_{n} σ_{0}^{n} = \sum_{n} σ_{0}^{n} \circ ω$ , it is observed that (72) reduces to

σ = \sum_{n} [T_{0}^{n} w_{n} \otimes {\hat{w}}_{n} + ({\frac{\partial {\hat{T}}_{l}^{n}}{\partial λ^{n}} |}_{λ_{0}^{n}} - T_{0}^{n}) (e ({\hat{w}}_{n}) \cdot {\hat{w}}_{n}) w_{n} \otimes {\hat{w}}_{n} + T_{0}^{n} (e (w_{n}) \otimes {\hat{w}}_{n} + {\hat{w}}_{n} \otimes e (w_{n}))] - p I .

(73)

As a very simple example, let F₀ = I and let the fiber density of three activated mutually orthogonal bundles, w₁, w₂ and w₃, be $ρ_{0}^{1}$ , $ρ_{0}^{2}$ and $ρ_{0}^{3}$ , respectively. In the intermediate configuration, being self-equilibrated, $\sum_{n} σ_{0}^{n} = p_{0} I$ , n ∈ {1, 2, 3}. Here, 1 and 2 indicate the two bundles of vertical and horizontal transverse fibers and 3 indicates the longitudinal fibers. From (73), one has

\begin{array}{l} σ_{11} & = [T_{a}^{1} + (T_{a}^{1} + {\frac{\partial {\hat{T}}_{l}^{1}}{\partial λ^{1}} |}_{λ_{0}^{1}}) e_{11}] ρ_{0}^{1} - p, \\ σ_{22} & = [T_{a}^{2} + (T_{a}^{2} + {\frac{\partial {\hat{T}}_{l}^{2}}{\partial λ^{2}} |}_{λ_{0}^{2}}) e_{22}] ρ_{0}^{2} - p, \\ σ_{33} & = [T_{a}^{3} + (T_{a}^{3} + {\frac{\partial {\hat{T}}_{l}^{3}}{\partial λ^{3}} |}_{λ_{0}^{3}}) e_{33}] ρ_{0}^{3} - p, \\ σ_{12} & = σ_{21} = (T_{a}^{1} ρ_{0}^{1} + T_{a}^{2} ρ_{0}^{2}) e_{12}, \\ σ_{13} & = σ_{31} = (T_{a}^{1} ρ_{0}^{1} + T_{a}^{3} ρ_{0}^{3}) e_{13}, \\ σ_{23} & = σ_{32} = (T_{a}^{3} ρ_{0}^{3} + T_{a}^{2} ρ_{0}^{2}) e_{23} . \end{array}

(74)

It is observed that the geometric stiffness together with incompressibility, enables the muscle fibers to resist both tension and compression, increasing the stiffness along the directions of the fibers. The fibers can also support shear due to the change in the direction of the deformed fibers. In spite of the simple geometry and a small amount of muscle groups which cannot span the space of symmetric stress tensors alone in the intermediate configuration, the prestress in the three mutually orthogonal fiber groups allows the muscular hydrostat to support any stress field undergoing some deformation.

5. The stress analysis problem

In this section, we summarize the equations for the analysis of stress pertaining to our model of the mechanics of the octopus’s arm. The resulting equations specialize the general equations of stress analysis, and in particular, we consider the three-stage scenario and the stress tensor is given in terms of the values of the tension in the incompressible mixture of five fiber bundles.

5.1. The equilibrium conditions

The Cauchy stress at the current configuration according to (50) and (65) may be represented as

σ = \sum_{n} T^{n} {\hat{v}}_{n} \otimes v_{n} - p I,

(75)

where we should use the constitutive relation $T^{n} = {\hat{T}}_{l}^{n} (λ, λ_{0}^{n}, α_{0}^{n})$ . Therefore, the divergence of the Cauchy stress in (75) satisfies

div σ = \sum_{n} [(v_{n} \cdot grad T^{n}) {\hat{v}}_{n} + T^{n} (v_{n} \cdot grad {\hat{v}}_{n} + {\hat{v}}_{n} div v_{n})] - grad p = - b .

(76)

Here, the operators div and grad denote the divergence and gradient computed relative to the current configuration. It is observed that the second term in the square brackets is related to the curvature of the fibers. The absence of sources and sinks, as in (37), implies that the last term in the square brackets vanishes identically for all n. The first term in (76) contains the directional derivative, v_n · grad Tⁿ, of the tension in the direction of the fibers. According to [21, 22] it is quite reasonable to assume that the activation level is constant along a fiber. Thus, assuming that the current extension λⁿ and activation $α_{0}^{n}$ do not change along a fiber, the tension Tⁿ is uniform along the fiber as well. It follows that the directional derivative vanishes.

We conclude that (76) assumes the form

div σ = \sum_{n} T^{n} v_{n} \cdot grad {\hat{v}}_{n} - grad p = - b,

(77)

and it is expected that curvature of the fibers will be accompanied by pressure variation.

The boundary condition for the stress, $σ (\hat{n}) = t$ , must hold on the boundary of the arm in its current configuration, where $\hat{n}$ is a unit normal to the boundary and t is the boundary loading. Substituting (75) to the boundary condition gives,

\sum_{n} T^{n} (\hat{n} \cdot v_{n}) {\hat{v}}_{n} - p \hat{n} = t,

(78)

where we keep in mind that the conditions Tⁿ > 0 and p > 0.

The corresponding conditions for the intermediate configurations are obtained by replacing the various variables with their analogs, e.g. w_n, $T_{0}^{n}$ , $λ_{0}^{n}$ and p₀ use the differential operators Grad,Div, and require that the loads b and t vanish.

5.2. The stress analysis problems for the three-stage model

The isometric activation stage. Following the unloaded activation stage, the arm is at some given intermediate state given by κ₀, and the Cauchy stress is presented as in (55). The configuration κ₀ determines the displacement vector d and the deformation gradient tensor F₀, so that the variables $λ_{0}^{n}, ρ_{0}^{n}$ and $\hat{w}$ may be specified. We assume that the octopus determines the hydrostatic pressure p₀ by suitable activation of the muscle groups, and thus we consider this field as known.

The unknown fields are therefore the five values of the tension $T_{a}^{n}$ in the fibers, and the corresponding five levels of activations $α_{0}^{n}$ , a total of 10 unknown scalar fields. Note that we have a total of eight independent equations: three equations of equilibrium as in (76), and five constitutive relations (64), for the various fiber bundles.

Thus, two additional conditions are required to solve the problem. These conditions may be thought of as the degrees of freedom the octopus has in activating the muscles. For example, while it might be possible to produce the pressure in the intermediate configuration by activating only three muscle groups, the activation of additional groups may result in additional stability and rigidity of the arms for applied loadings.

The final loading stage. We now consider the third and final stage: the response to an external loading under fixed activation of the fibers. The final linearized Cauchy stress, following from the superimposed infinitesimal deformation δF, is presented by (68). In accordance with our scenario, we consider the loading, b and t, and the fixed activation levels, $α_{0}^{n}$ , as given. Here, we have eighteen unknown fields in total: three components for displacement field d, nine components of the infinitesimal deformation field δF, five final values for the tensions in the fibers Tⁿ and the final pressure p.

We have the following independent equations: three equilibrium equations, five constitutive relations (65) for passive deformation of the fibers under constant activation, the incompressibility condition e_ii = 1 and nine relations for the infinitesimal strain in terms of the displacements.

Footnotes

Acknowledgements

We would like to express our gratitude to W Herzog, M Epstein, SS Blemker and L Dorfmann for their help with the constitutive model of the arm and to M Epstein for his advice with the bases for the space of tensors.

This article is dedicated by the authors to Marcelo Epstein, on the occasion of his 70th birthday. Reuven Segev wishes to express his deep gratitude for the inspiration, mentoring, and friendship over a period of more than 35 years.

Declaration of Conflicting Interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

This work was supported by the Perlstone Center for Aeronautical Engineering Studies and the H Greenhill Chair for Theoretical and Applied Mechanics at Ben-Gurion University.

Notes

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