On Cesàro-Volterra Method in Orthotropic Saint-Venant Beam

Abstract

The linearly elastic and orthotropic Saint-Venant beam model, with a spatially constant Poisson tensor and fiberwise homogeneous elastic moduli, is investigated by a coordinate-free approach. A careful reasoning reveals that the elastic strain, fulfilling the whole set of differential conditions of integrability and a differential condition imposed by equilibrium, is defined on the whole ambient space in which the beam is immersed. At this stage the shape of the beam cross-section is inessential and Cesàro-Volterra formula provides the general integral of the differential conditions of kinematic compatibility. The cross-section geometrical shape comes into play only when differential and boundary equilibrium conditions are imposed to evaluate the warping displacement field. The treatment of an orthotropic Saint-Venant beam is applied to investigate about the locations of the shear and twist centres. It is shown that the position of the shear centre can be expressed in terms of the sole cross-section twist warping. The advantage with respect to treatments in the literature is that the solution of a single Neumann-like problem is required.

Appendix

The issue of kinematic compatibility between strain and displacement fields was first investigated by Kirchhoff [17], Saint-Venant [34], Beltrami [3]. A coordinate-free formulation was provided by Gurtin [15]. For the readers convenience, the results referred to in the paper are recalled hereafter.

Lemma 1

Curl (∇u)^T=(∇curl u)^T, with $\mathbf{u}:\textsc{Dom}\subseteq \mathcal{S}\mapsto V$.

Proof

It is enough to perform the computation:

The result follows. □

Proposition 11

(Necessary Condition of Integrability)

Let $\mathbf{u}:\textsc{Dom}\subseteq \mathcal{S}\mapsto V$ be a 3-D vector field.

$$\mathbf{D}:=\mathrm{sym}\,\nabla \mathbf{u}\quad \implies\quad \mathrm{Curl}\,(\mathrm{Curl}\,\mathbf{D})^T=\mathbf{O}. $$

Proof

By Lemma 1 and because Curl ∇u=O, a direct computation gives:

$$\mathrm{Curl}\,\mathbf{D}=\mathrm{Curl}\,(\mathrm{sym}\,\nabla \mathbf{u}) =\mathrm{Curl}\,\biggl(\frac{1}{2}\nabla \mathbf{u}+\frac{1}{2}\nabla \mathbf{u}^T\biggr) =\frac{1}{2}\mathrm{Curl}\,(\nabla \mathbf{u})^T =\frac{1}{2}(\nabla \mathrm{curl}\,\mathbf{u})^T , $$

so that $\mathrm{Curl}\,(\mathrm{Curl}\,\mathbf{D})^{T} =\mathrm{Curl}\,\nabla(\frac{1}{2}\mathrm{curl}\,\mathbf{u})=\mathbf{O}$. □

Lemma 2

Let A:Dom↦L(V;V) be a tensor field fulfilling the differential equation Curl A(x)=O ∀x∈Dom, with $\textsc{Dom}\subseteq \mathcal{S}$ a simply connected 3-D domain. Then, there exists a vector field u:Dom↦V such that ∇u=A.

Proof

Curl A=O, so that o=(Curl A)^T h=curl (A ^T h), ∀h∈V. Since the domain $\textsc{Dom}\subseteq \mathcal{S}$ is simply connected, then there exists a scalar field $f:\textsc{Dom}\mapsto \mathcal{R}$ such that ∇f=A ^T h. Setting f:=u⋅h, with u:Dom↦V a vector field, we get the formula: ∇f=(∇u)^T h. Accordingly, the relation ∇u=A follows. □

Proposition 12

(Integrability Criterion)

Let $\textsc{Dom}\subseteq \mathcal{S}$ be a simply connected 3-D domain. The differential condition:

$$\mathrm{Curl}\,(\mathrm{Curl}\,\mathbf{D})^T(\mathbf{x})=\mathbf{O}, \quad\forall \mathbf{x}\in \textsc{Dom}, $$

ensures the existence of a vector field u:Dom↦V such that sym ∇u=D.

Proof

Resorting to Lemma 2, we infer that there exists a vector field ω:Dom↦V such that: (Curl D)^T=∇ω. Then, being I ₁(Curl D)^T=0, we have that I ₁(∇ω)=div ω=0. Let us consider the skew-symmetric tensor field W:Dom↦L(V;V) related to the axial vector field ω:Dom↦V, i.e., W(x)h=ω(x)×h, for any x∈Dom and h∈V. By resorting to the formula curl (ω×h)=(∇ω)h−(div ω)h and recalling that div ω=0, a direct computation gives:

Accordingly the formula holds: Curl D+Curl W=Curl (D+W)=O, so that by Lemma 2 we can write ∇u=D+W, where u:Dom↦V is a vector field. The result follows. □