A Frame-Independent Solution to Saint-Venant’s Flexure Problem

Abstract

The paper illustrates a solution approach for the Saint-Venant flexure problem which preserves a pure objective tensor form, thus yielding, for sections of arbitrary geometry, representations of stress and displacement fields that exploit exclusively frame-independent quantities.

The implications of the availability of an objective solution to the shear warpage problem are discussed and supplemented by several analytical and numerical solutions.

The derivation of tensor expressions for the shear center and the shear flexibility tensor is also illustrated. Furthermore, a Cesaro-like integration procedure is provided whereby the derivation of a frame-independent representation of the displacements field for the shear loading case is systematically carried out via the use of Gibbs’ algebra.

The objective framework presented in this paper is further exploited in a companion article (Serpieri, in J. Elast. (2013)) to prove the coincidence of energetic and kinematic definitions of the shear flexibility tensor and of the shear principal axes.

Appendices

Appendix A: Tensor Identities

We recall hereafter the definitions and tensor identities used throughout the paper.

1.1 A.1 Vector Product Between Vectors and Tensors

The operation of vector product between a vector a and a tensor T is defined as the unique tensor $\mathbf{a\times T}$ that satisfies the property

$$ \mathbf{(a}\times\mathbf{\mathbf{T)b}}=\mathbf{a}\times\mathbf {\mathbf {Tb}} $$

(86)

for any vector b. On account of this definition one has componentwise

$$ ({\mathbf{a}}\times {\mathbf{T}})_{ij} = \varepsilon_{ikh}a _{k}T_{hj}. $$

(87)

Setting T=I in (86) one obtains a relation between a skew-symmetric tensor W _a =a×I and its axial vector a. In indicial notation the relation between W _a and a is

$$ (W_{\mathbf{{\mathbf{a}}}})_{ij}=-\varepsilon_{ijk}\,a _{k}, \qquad a_{i}=-\frac{1}{2} \varepsilon_{ijk}(W_{{\mathbf{a}}})_{jk}. $$

(88)

In view of the properties recalled above, one also infers

$$ \mathbf{a\times T=(a}\times\mathbf{\mathbf{I)T=}W}_{\mathbf {a}}\mathbf{T}, \quad{\mathbf{a}}\times {\mathbf{T}}{\mathbf{b}}=({\mathbf{a}}\times {\mathbf{I}}){\mathbf{T}}{\mathbf{b}}={\mathbf{W}}_{{\mathbf{a}}}{\mathbf{T}}{\mathbf{b}}, $$

(89)

and

$$ \mathbf{a\times(b}\otimes\mathbf{\mathbf{c)=(}a\times b)}\otimes \mathbf{\mathbf{c}}. $$

(90)

Moreover, using the identity of double vector product,

$$ \mathbf{a\times(b}\times\mathbf{\mathbf{c)=}(a\cdot c\mathbf {)\mathbf {b}-}(a}\cdot\mathbf{\mathbf{b)c}}, $$

(91)

one proves the relation

$$ \mathbf{a\times(b}\times\mathbf{\mathbf{T)}}=\mathbf{(\mathbf{b \,\otimes\,}a\mathbf{)\mathbf{\mathbf{T}}-}(a}\cdot\mathbf{\mathbf{b)T}}, $$

(92)

from which the following identities are inferred as special cases:

$$\begin{aligned} &\mathbf{a\times(b}\times\mathbf{\mathbf{I)}}=\mathbf {W}_{\mathbf {a}}\mathbf{W}_{\mathbf{b}}=\mathbf{(\mathbf{b\,\otimes\,}a\mathbf {)-}(a}\cdot \mathbf{\mathbf{b)I}}, \end{aligned}$$

(93)

$$\begin{aligned} &({\mathbf{a}}\times {\mathbf{b}}) \times {\mathbf{I}}={\mathbf{b}}\otimes{\mathbf{a}}- {\mathbf{a}}\otimes {\mathbf{b}}. \end{aligned}$$

(94)

1.2 A.2 Gibbs’ Notation

The symbol ∇ was introduced by Gibbs [14] to denote a vector whose indicial expression is

$$[ \boldsymbol{\nabla}]^{t}=\left [ \begin{array}{c@{\quad}c@{\quad}c} \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac {\partial}{\partial z} \end{array} \right ] $$

in a Cartesian reference frame xyz. With the aid of vector ∇ it is possible to represent a large family of differential operators by means of binary operations in which ∇ is one of the operands. Actually, denoting by φ, v and S a scalar, a vector and a tensor field, respectively, one has

$$\begin{aligned} \begin{array}{ll@{\qquad}l} &\mathrm{grad\,}\varphi=\varphi \boldsymbol{\nabla},& \mathrm{grad\,}\mathbf {v}=\mathbf{v}\otimes \boldsymbol{\nabla}, \\ &\mathrm{div\,}\mathbf{v}=\mathbf{v}\cdot \boldsymbol{\nabla}, & \mathrm{div\,}\mathbf{S}=\mathbf{S\boldsymbol{\nabla}}, \\ &\mathrm{curl\,}\mathbf{v}=\boldsymbol{\nabla}\times {\mathbf{v}}={\mathbf{W}}_{\boldsymbol{\nabla}} {\mathbf{v}}, &\mathrm{curl\,}\mathbf{S}=\boldsymbol{\nabla}\times {\mathbf{S}}^{t}={\mathbf{W}}_{\boldsymbol{\nabla}} {\mathbf{S}}^{t}. \end{array} \end{aligned}$$

Similarly, the Laplacian Δc of a field c, either of scalar, vector or tensor type is expressed in Gibbs notation as Δc=(∇⋅∇)c while the Hessian of a scalar field φ is denoted as φ ∇⊗∇.

The main advantage of this notation consists in the faculty of inferring the differential identities commonly employed in elastostatics relying on the usual algebraic operations, i.e., treating ∇ as a common vector.

1.2.1 A.2.1 Product Rule

Let a, b and c be fields either of scalar, vectorial or tensor nature which are combined by means of an algebraic bilinear operation f,

$$ {\mathbf{c}}=f({\mathbf{a}}, \, {\mathbf{b}}). $$

(95)

Denoting by g, h, i, l and m bilinear operations such that the following relation is satisfied,

$$ g\bigl(f({\mathbf{a}}, \, {\mathbf{b}}), \, {\mathbf{d}}\bigr) = h\bigl({\mathbf{a}}, \, i({\mathbf{b}}, \, {\mathbf{d}})\bigr) = l\bigl({\mathbf{b}}, \, m({\mathbf{a}}, \, {\mathbf{d}})\bigr), $$

(96)

the following product derivation rule holds:

$$ g\bigl(f({\mathbf{a}}, \, {\mathbf{b}}), \, \boldsymbol{\nabla}\bigr) = h\bigl({\mathbf{a}}, \, i({\mathbf{b}}, \, \boldsymbol{\nabla})\bigr) + l\bigl({\mathbf{b}}, \, m({\mathbf{a}}, \, \boldsymbol{\nabla})\bigr). $$

(97)

To show an application of (97), consider, for instance, the gradient of the vector field φ v,

$$ (\varphi {\mathbf{v}}) \otimes \boldsymbol{\nabla}= \varphi( {\mathbf{v}}\otimes \boldsymbol{\nabla}) + {\mathbf{v}}\otimes( \varphi \boldsymbol{\nabla}). $$

(98)

This is the well known identity that in classical notation is written as

$$ \mathrm{grad\,} ( \varphi {\mathbf{v}}) = \varphi\, \mathrm{grad}\,{\mathbf{v}}+ {\mathbf{v}}\otimes( \mathrm{grad\,} \varphi). $$

(99)

The differential identities exploited in this paper, and in particular those used in the derivation of the displacement field, can be all inferred from (97); in particular, denoting by φ, a and A a scalar, a vector and a tensor field, respectively, one has

$$\begin{aligned} \begin{aligned} (\varphi{\mathbf{a}}\otimes \boldsymbol{\nabla})& = {\mathbf{a}} \otimes(\varphi \boldsymbol{\nabla}) + \varphi({\mathbf{a}}\otimes \boldsymbol{\nabla}), \\ ({\mathbf{a}}\times {\mathbf{b}})\otimes \boldsymbol{\nabla}& = {\mathbf{a}}\times({\mathbf{b}}\otimes \boldsymbol{\nabla})- {\mathbf{b}}\times({\mathbf{a}}\otimes \boldsymbol{\nabla}), \\ ({\mathbf{a}}\cdot {\mathbf{b}}) \boldsymbol{\nabla}& = (\boldsymbol{\nabla}\otimes{\mathbf{a}} ) {\mathbf{b}}+ (\boldsymbol{\nabla}\otimes {\mathbf{b}}) {\mathbf{a}}, \\ \boldsymbol{\nabla}\times( \varphi{\mathbf{a}} )& = (\boldsymbol{\nabla}\varphi) \times {\mathbf{a}}+ \varphi(\boldsymbol{\nabla}\times{\mathbf{a}} ), \\ \boldsymbol{\nabla}\times( {\mathbf{a}}\times {\mathbf{b}})& = (\boldsymbol{\nabla}\cdot {\mathbf{b}}){\mathbf{a}}- ({\mathbf{b}}\otimes \boldsymbol{\nabla}) {\mathbf{a}} + ({\mathbf{a}}\otimes \boldsymbol{\nabla}) {\mathbf{b}}- (\boldsymbol{\nabla}\cdot{ \mathbf{a}} ){\mathbf{b}}, \\ \boldsymbol{\nabla}\times( {\mathbf{a}}\otimes {\mathbf{b}})& = (\boldsymbol{\nabla}\times {\mathbf{b}}) \otimes{\mathbf{a}}- {\mathbf{b}}\times(\boldsymbol{\nabla}\otimes{\mathbf{a}} ), \\ \boldsymbol{\nabla}\times( \varphi {\mathbf{A}})^{t}& = (\boldsymbol{\nabla}\varphi) \times {\mathbf{A}}^{t} + \varphi\bigl( \boldsymbol{\nabla}\times {\mathbf{A}}^{t} \bigr). \end{aligned} \end{aligned}$$

(100)

Moreover, the following second order differential identities hold true:

$$\begin{aligned} \boldsymbol{\nabla}\cdot( \boldsymbol{\nabla}\times{\mathbf{a}} ) =&{\mathbf{a}}\cdot( \boldsymbol{\nabla}\times \boldsymbol{\nabla})=0, \\ \boldsymbol{\nabla}\times( {\mathbf{a}}\otimes \boldsymbol{\nabla}) =& ( \boldsymbol{\nabla}\times {\mathbf{a}} ) \otimes \boldsymbol{\nabla}, \\ \boldsymbol{\nabla}\times( \boldsymbol{\nabla}\otimes{\mathbf{a}} ) =& ( \boldsymbol{\nabla}\times \boldsymbol{\nabla}) \otimes{\mathbf{a}}={\mathbf{0}}. \end{aligned}$$

(101)

Appendix B: Derivation of the Displacement Field Associated with the Flexure Problem for a Free Beam

This appendix provides the details of the integration procedure, based upon the differential identity (67), which allows one to derive the displacement field ${\mathbf{u}}^{(\mathit {DSV})}_{cant, \, free}$ starting from the infinitesimal strain field E _cant, free associated, via the inverse of the linear elastic law, with the stress tensor field of formula (65). To achieve a simpler notation, in this appendix the scripts $( \cdot)^{(\mathit{DSV})}_{cant, \, free}$ are henceforth omitted so that the stress and infinitesimal strain tensors are denoted by the symbols T and E in place of T _cant, free and E _cant, free .

Firstly, it is useful to express in an alternative from the shear stress field (40) by introducing the vector

$$ {\mathbf{p}}=\dfrac{1}{8}(\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}})\hat{{\boldsymbol{\rho}}} $$

(102)

whose gradient is the symmetric tensor

$$ {\mathbf{p}}\otimes \boldsymbol{\nabla}=\boldsymbol{\nabla}\otimes {\mathbf{p}}=\dfrac{1}{8} \bigl[2(\hat{{\boldsymbol{\rho}}}\otimes \hat{{\boldsymbol{\rho}}})+(\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}) \hat{{\mathbf{I}}}\bigr]. $$

(103)

The vector p allows one to express A ^p as

$$ {\mathbf{A}}^p= (1+\bar{\nu}) (\boldsymbol{\nabla}\otimes {\mathbf{p}})-\bar{\nu}\frac {\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} \hat{{\mathbf{I}}}, $$

(104)

so that the tangential stress field (40) becomes

$$ {{ {\boldsymbol{\tau}}_{sh}^{(\mathit{DSV})} }= \biggl[ (\boldsymbol{\nabla}\otimes{{\boldsymbol{\psi}}} ) + (1+\bar{\nu}) ( \boldsymbol{\nabla}\otimes {\mathbf{p}})-\bar{\nu}\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2}\hat{{\mathbf{I}}} \biggr]{{\mathbf{g}}}_{t} }. $$

(105)

The previous relation, upon introducing the vector ξ defined as

$$ {\boldsymbol{\xi}}={\boldsymbol{\psi}}+ (1+\bar{\nu}){\mathbf{p}}, $$

(106)

becomes

$$ {\boldsymbol{\tau}}_{sh}^{(\mathit{DSV})} = \biggl( \boldsymbol{\nabla}\otimes {\boldsymbol{\xi}}- \bar{\nu }\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} \hat{{\mathbf{I}}} \biggr) {\mathbf{g}}_{t} = ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_{t} )\boldsymbol{\nabla}- \bar{\nu}\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t}. $$

(107)

The stress tensor field is

$$ {\mathbf{T}}= (l-z) {({\mathbf{g}}_{{\mathbf{t}}}\mathbf{\cdot {\mathbf{r}})}({\mathbf{k}}\otimes {\mathbf{k}})+\bigl( {\boldsymbol{\tau}}_{sh}^{(\mathit{DSV})} \otimes {\mathbf{k}}\bigr)+\bigl({\mathbf{k}}\otimes {\boldsymbol{\tau}}_{sh}^{(\mathit{DSV})} \bigr)}. $$

(108)

Applying the linear isotropic law

$$ \mathbf{E} =\frac{1+\nu}{E} {\mathbf{T}}-\frac{\nu}{E}(\mathrm{tr\,}{\mathbf{T}}){\mathbf{I}}$$

(109)

to (108), we obtain the following representation for the infinitesimal strain field:

$$\begin{aligned} \mathbf{E} =&\frac{1+\nu}{E} \bigl[ (l-z){({\mathbf{g}}_{{\mathbf{t}}} \mathbf{\cdot {\mathbf{r}})}({\mathbf{k}}\otimes {\mathbf{k}})+\bigl( {\boldsymbol{\tau}}_{sh}^{(\mathit {DSV})} \otimes {\mathbf{k}}\bigr)+ \bigl({\mathbf{k}}\otimes {\boldsymbol{\tau}}_{sh}^{(\mathit{DSV})} \bigr)} \bigr] \\ &{}-\frac{\nu}{E} \bigl[ (l-z)\mathbf{({\mathbf{g}}_{{\mathbf{t}}} \mathbf{\cdot {\mathbf{r}})}I} \bigr]. \end{aligned}$$

(110)

Substituting (107) into the previous expression and taking into account the relation $(1+\nu)\bar{\nu}=\nu$ one finally obtains

$$\begin{aligned} \mathbf{E} = & \displaystyle \frac{1+\nu}{E} \bigl\{ (l-z)\mathbf{({\mathbf{g}}_{{\mathbf{t}}}\mathbf{\cdot {\mathbf{r}})}(k\otimes k)} + \bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_{t} )\boldsymbol{\nabla}\bigr] \otimes {\mathbf{k}}+ {\mathbf{k}}\otimes\bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_{t} )\boldsymbol{\nabla}\bigr] \bigr\} \\ &{}-\frac{\nu}{E} \biggl[ (l-z)\mathbf{({\mathbf{g}}_{{\mathbf{t}}} \mathbf {\cdot {\mathbf{r}})}I} + \frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t} \otimes {\mathbf{k}}+ {\mathbf{k}}\otimes \frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t} \biggr]. \end{aligned}$$

(111)

The explicit computation of the curl of E, i.e., ∇×E ^t, required by the previously outlined integration procedure, needs the separate computation of the curl of the addends contained in the RHS of the previous expression. Thus, we have

$$\begin{aligned} \begin{aligned} \mathbf{\boldsymbol{\nabla}\times} \bigl[ (l-z)\mathbf{( {\mathbf{g}}_{{\mathbf{t}}}\mathbf{\cdot r)}(k\otimes k)} \bigr] &= (l-z)\mathbf{( {\mathbf{g}}_{{\mathbf{t}}}\mathbf{\cdot r)}\boldsymbol{\nabla}\times(k\otimes k)} \\ &= \bigl[ (l\mathbf{-}z)\mathbf{{\mathbf{g}}_{{\mathbf{t}}}-({\mathbf{g}}_t\mathbf{ \cdot r)}k} \bigr] \mathbf{\times(k\otimes k)} \\ &= (\mathbf{{\mathbf{g}}_t\times k)\otimes}(l-z){\mathbf{k}}, \boldsymbol{\nabla}\times\bigl\{ \bigl[ \mathbf{k\otimes({\boldsymbol{\xi}}}\cdot {\mathbf{g}}_t)\mathbf{ \boldsymbol{\nabla}} \bigr] \bigr\} \\ &= ( \mathbf{\boldsymbol{\nabla}\times k} ) \otimes\bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\bigr] \\ &= - ( \mathbf{k\times \boldsymbol{\nabla}}) \otimes\bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t) \boldsymbol{\nabla}\bigr] \\ &= - {\mathbf{k}}\times\bigl\{ \boldsymbol{\nabla}\otimes\bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\bigr] \bigr\} \\ &= -{\mathbf{k}}\times\bigl[( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{\boldsymbol{\nabla}\otimes \boldsymbol{\nabla}} \bigr], \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned}[t] \boldsymbol{\nabla}\times\bigl\{ \bigl[ ( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\bigr] \otimes {\mathbf{k}}\bigr\} &= \bigl\{ \bigl[ \mathbf{ \boldsymbol{\nabla}\times({\boldsymbol{\xi}}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\bigr] \bigr\} \otimes {\mathbf{k}}= \textbf{0}, \\ \mathbf{\boldsymbol{\nabla}\times} \bigl[ (l-z)\mathbf{({\mathbf{g}}_t \mathbf{\cdot r)}I} \bigr] &= (l-z)\mathbf{({\mathbf{g}}_t\mathbf{ \cdot r)}\boldsymbol{\nabla}\times I} \\ &= \bigl[ (l-z)\mathbf{{\mathbf{g}}_t-({\mathbf{g}}_t\mathbf{\cdot r)k}} \bigr] \mathbf{\times I} \\ &= (l-z){\mathbf{g}}_{t} \times {\mathbf{I}}-({\mathbf{g}}_t \cdot {\mathbf{r}}){\mathbf{k}}\times {\mathbf{I}}, \\ \boldsymbol{\nabla}\times\biggl( {\mathbf{k}}\otimes\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t} \biggr) &= - {\mathbf{k}}\times\biggl[ \boldsymbol{\nabla}\biggl(\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} \biggr) \otimes {\mathbf{g}}_{t} \biggr] = - {\mathbf{k}}\times \hat{{\boldsymbol{\rho}}}\otimes {\mathbf{g}}_{t}, \\ \boldsymbol{\nabla}\times\biggl[ \biggl( \frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} \biggr) {\mathbf{g}}_{t} \otimes {\mathbf{k}}\biggr] &= \biggl[ \boldsymbol{\nabla}\biggl( \frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} \biggr) \biggr] \times {\mathbf{g}}_{t} \otimes {\mathbf{k}}= \hat{{\boldsymbol{\rho}}}\times {\mathbf{g}}_{t} \otimes {\mathbf{k}}. \end{aligned} \end{aligned}$$

(112)

The identities above can be proved by invoking the differential identities reported in Appendix A.2.1 which, in turn, stem from the properties of the vector product (Appendix A.1).

Using the identities (112), the curl of E becomes

$$\begin{aligned} \boldsymbol{\nabla}\times {\mathbf{E}}^{t} =& \displaystyle\frac{1+\nu}{E} \bigl\{ ( \mathbf{{\mathbf{g}}_t\times k)\otimes}(l-z){\mathbf{k}}-{\mathbf{k}}\times\bigl[( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{\boldsymbol{\nabla}\otimes \boldsymbol{\nabla}} \bigr] \bigr\} \\ &{} +\frac{\nu}{E} \{ - \hat{{\boldsymbol{\rho}}}\times {\mathbf{g}}_{t} \otimes {\mathbf{k}}+ {\mathbf{k}}\times \hat{{\boldsymbol{\rho}}}\otimes {\mathbf{g}}_{t} \} \\ &{} +\frac{\nu}{E} \bigl[ (z-l){\mathbf{g}}_{t} \times {\mathbf{I}}+( {\mathbf{g}}_t \cdot {\mathbf{r}}){\mathbf{k}}\times {\mathbf{I}}\bigr]. \end{aligned}$$

(113)

Because of identity (67), we now dispose of the expression of the gradient of ω; hence, to obtain ω, it is necessary to integrate the expression above. To this end (113) is further developed as

$$\begin{aligned} {\boldsymbol{\omega}}\otimes \boldsymbol{\nabla} =& \displaystyle\frac{1+\nu}{E} \biggl\{ \biggl( lz- \frac{z^{2}}{2} \biggr) ( \mathbf{{\mathbf{g}}_t\times k)} -{\mathbf{k}}\times \bigl[( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{\boldsymbol{\nabla}} \bigr] \biggr\} \otimes \boldsymbol{\nabla} \\ &{} + \frac{\nu}{E} \bigl[ {\mathbf{g}}_{t} \times \hat{{\boldsymbol{\rho}}}\otimes {\mathbf{k}}+ {\mathbf{k}}\times \hat{{\boldsymbol{\rho}}}\otimes {\mathbf{g}}_{t} + (z-l){\mathbf{g}}_{t} \times {\mathbf{I}}+ ( {\mathbf{g}}_t \cdot {\mathbf{r}}){\mathbf{k}}\times {\mathbf{I}}\bigr] \\ =& \frac{1+\nu}{E} \biggl\{ \biggl( lz-\frac{z^{2}}{2} \biggr) ( \mathbf{ {\mathbf{g}}_t\times k)} -{\mathbf{k}}\times\bigl[( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{\boldsymbol{\nabla}} \bigr] \biggr\} \otimes \boldsymbol{\nabla} \\ &{} + \frac{\nu}{E} \biggl[ (z-l) {\mathbf{g}}_{t} \times {\mathbf{r}}+ ( {\mathbf{g}}_{t} \cdot {\mathbf{r}}) {\mathbf{k}}\times \hat{{\boldsymbol{\rho}}}+ \frac{z^{2}}{2} ({\mathbf{k}}\times {\mathbf{g}}_{t}) \biggr] \otimes \boldsymbol{\nabla}. \end{aligned}$$

(114)

In particular, the last equality in (114) relies on the identity

$$ {\mathbf{g}}_{t} \times \hat{{\boldsymbol{\rho}}}\otimes {\mathbf{k}}+ (z-l){\mathbf{g}}_{t} \times {\mathbf{I}}= \bigl[ (z-l) {\mathbf{g}}_{t} \times {\mathbf{r}}\bigr] \otimes \boldsymbol{\nabla}+ \biggl[ \frac{z^{2}}{2} ({\mathbf{k}}\times {\mathbf{g}}_{t}) \biggr] \otimes \boldsymbol{\nabla}, $$

(115)

which is inferred from the relation $\hat{{\boldsymbol{\rho}}}={\mathbf{r}}- z{\mathbf{k}}$.

The rightmost expression in (114) allows one to directly identify the expression of the axial vector ω as

$$\begin{aligned} \mathbf{{\boldsymbol{\omega}}} =&\frac{1+\nu}{E} \biggl[ \biggl( lz-\frac{z^{2}}{2} \biggr) ({\mathbf{g}}_t\mathbf{\times k)}- \mathbf{\mathbf{k\times}({\boldsymbol{\xi}}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\biggr] \\ &{}+ \frac{\nu}{E} \biggl[ (z-l) {\mathbf{g}}_{t} \times {\mathbf{r}}+ ( {\mathbf{g}}_{t} \cdot {\mathbf{r}}) {\mathbf{k}}\times {\mathbf{r}}+ \frac{z^{2}}{2} ({\mathbf{k}}\times {\mathbf{g}}_{t} ) \biggr] +{\boldsymbol{\omega}}_{0}, \end{aligned}$$

(116)

where ω ₀ is an arbitrary constant vector field. Accordingly, the skew-symmetric component of the displacement gradient W, whose axial vector is ω, turns out to be

$$\begin{aligned} {\mathbf{W}} =& {\boldsymbol{\omega}}\times {\mathbf{I}}= \frac{1+\nu}{E} \biggl\{ \biggl( lz - \frac{z^{2}}{2} \biggr) ({\mathbf{g}}_t\times {\mathbf{k}})\times {\mathbf{I}}- \bigl[ {\mathbf{k}}\times({\boldsymbol{\xi}}\cdot {\mathbf{g}}_{{\mathbf{t}}})\boldsymbol{\nabla}\bigr] \times {\mathbf{I}}\biggr\} \\ &{}+ \frac{\nu}{E} \biggl\{ {\mathbf{k}}\times\biggl[ ({\mathbf{g}}_t\cdot {\mathbf{r}}) {\mathbf{r}}+ \frac{z^{2}}{2}{\mathbf{g}}_t \biggr] \times {\mathbf{I}}+\bigl[( {\mathbf{g}}_t\times(z-l){\mathbf{r}}\bigr]\times {\mathbf{I}}\biggr\} +{\boldsymbol{\omega}}_{0} \times {\mathbf{I}}. \end{aligned}$$

(117)

The last addend ω ₀×I is the displacement gradient of an arbitrary rigid displacement field up to which the displacement field is defined. For the sake of brevity this term will be omitted in the following developments although it will be reported in the final expression of the displacement field.

Recalling identity (94) in Appendix A, one infers

$$\begin{aligned} {\mathbf{W}} =&\frac{1+\nu}{E} \biggl\{ \biggl( lz-\frac {z^{2}}{2} \biggr) \bigl[ ( \mathbf{k\otimes\mathbf{{\mathbf{g}}}_{{\mathbf{t}}})-}(\mathbf{ {\mathbf{g}}}_{{\mathbf{t}}} \mathbf{\,\otimes\, k)} \bigr] \\ &{} + \bigl[ \mathbf{\mathbf{k\,\otimes\,}({\boldsymbol{\xi}}}\cdot {\mathbf{g}}_{{\mathbf{t}}})\boldsymbol{\nabla}\bigr] -\mathbf{ \bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{\boldsymbol{\nabla}\mathbf {\otimes\, k}}\bigr] } \biggr\} \\ &{} +\frac{\nu}{E} \biggl\{ \biggl[ \mathbf{\mathbf{({\mathbf{g}}}}_{{\mathbf{t}}} \mathbf{\mathbf{\mathbf{\cdot r)r}}}+ \frac{z^{2}}{2}\mathbf{ \mathbf{\mathbf{{\mathbf{g}}_t}}}\biggr] \mathbf{\mathbf{\mathbf{ \,\otimes\, k}}}-\, \mathbf{k}\otimes\,\biggl[ ({\mathbf{g}}_{\mathbf{t}} \cdot\mathbf{r})\mathbf{r} + \frac{z^{2}}{2}\mathbf{\mathbf{\mathbf{{\mathbf{g}}_{{\mathbf{t}}}}}} \biggr] \\ &{} + \bigl[(z-l)\mathbf{r\otimes g}_{{\mathbf{t}}}\bigr]-\bigl[ {\mathbf{g}}_t\otimes(z-l)\mathbf{r\bigr]}\biggr\} . \end{aligned}$$

(118)

By definition, the sum of the expression above with that of E reported in (111) provides the displacement gradient $\mathrm{grad\,}\mathbf{u}={\mathbf{u}}\otimes \boldsymbol{\nabla}$. Comparing expressions (111) and (118) one recognizes the presence of two terms premultiplied either by the coefficient $\frac{1+\nu}{E}$ or by $\frac{\nu}{E}$. For the sake of readability, we introduce the representation

$$ \mathrm{grad\,}\mathbf{u} = \frac{1+\nu}{E} \mathrm{grad\,} \mathbf{u}_{1+\nu} + \frac{\nu}{E} \mathrm{grad\,} \mathbf{u}_{\nu} $$

(119)

in the computation of the displacement gradient. In particular, the integral of $\mathrm{grad\,}\mathbf{u}_{1+\nu}$ can be achieved more easily upon developing the integrand as follows:

$$\begin{aligned} \mathrm{grad\,}\mathbf{u}_{1+\nu} =& (l\mathbf{-}z) \mathbf{({\mathbf{g}}_t\mathbf{\cdot r)}(k\otimes k)} + \bigl[ ({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\otimes {\mathbf{k}}\bigr] + \bigl[ \mathbf{k\otimes({\boldsymbol{\xi}}}\cdot {\mathbf{g}}_{{\mathbf{t}}})\boldsymbol{\nabla}\bigr] \\ &{} + \biggl( lz-\frac{z^{2}}{2} \biggr) \bigl[ (\mathbf{k \otimes {\mathbf{g}}_t)-}({\mathbf{g}}_t\,\otimes\, \mathbf{k}) \bigr] \\ &{} + \bigl[ \mathbf{k}\,\otimes\,({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\boldsymbol{\nabla}\bigr] - \bigl[( {\boldsymbol{\xi}}\cdot {\mathbf{g}}_{{\mathbf{t}}})\boldsymbol{\nabla}\,\otimes\, \mathbf{k} \bigr] \\ = & \biggl\{ \biggl( lz-\frac{z^{2}}{2} \biggr) ( {\mathbf{g}}_t \cdot {\mathbf{r}}) {\mathbf{k}}+ 2({\boldsymbol{\xi}}\cdot {\mathbf{g}}_t)\mathbf{k+} \biggl( \frac{z^{3}}{6}-l\frac {z^{2}}{2} \biggr) {\mathbf{g}}_{{\mathbf{t}}} \biggr\} \otimes \boldsymbol{\nabla}. \end{aligned}$$

(120)

Following an analogous strategy for $\mathrm{grad\,}\mathbf{u}_{\nu}$ one has

$$\begin{aligned} \mathrm{grad\,}\mathbf{u}_{\nu} =& (z-l)\mathbf{({\mathbf{g}}_{{\mathbf{t}}} \mathbf{\cdot {\mathbf{r}})}I} - \frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t} \otimes {\mathbf{k}}- {\mathbf{k}}\otimes\frac{\hat{{\boldsymbol{\rho}}}\cdot \hat{{\boldsymbol{\rho}}}}{2} {\mathbf{g}}_{t} \\ &{}+ \biggl[ \mathbf{\mathbf{({\mathbf{g}}}}_{{\mathbf{t}}}\mathbf{\mathbf {\mathbf{ \cdot r)r}}}+ \frac{z^{2}}{2}\mathbf{\mathbf{\mathbf{ {\mathbf{g}}_t}}} \biggr] \mathbf{\mathbf{\mathbf{\otimes k}}} - \mathbf{ \mathbf{\mathbf{k\otimes}}} \biggl[ \mathbf{\mathbf{({\mathbf{g}}}}_{{\mathbf{t}} } \mathbf{\mathbf{\mathbf{\cdot r)r}}}+ \frac{z^{2}}{2}\mathbf {\mathbf{\mathbf{ {\mathbf{g}}_{{\mathbf{t}}}}}} \biggr] \\ &{} + \bigl[(z-l)\mathbf{r\otimes g}_{{\mathbf{t}}}\bigr]-\bigl[ {\mathbf{g}}_t\otimes(z-l)\mathbf{r\bigr]} \\ =& \biggl\{ (z-l)\mathbf{({\mathbf{g}}_t\mathbf{\cdot r)}r}-\biggl[ (z-l) \displaystyle\frac{(\mathbf{r\cdot r})}{2}-\frac {z^{3}}{3} \biggr] {\mathbf{g}}_{{\mathbf{t}}} - \displaystyle\frac{(\mathbf{r\cdot r})}{2}\mathbf{\mathbf{({\mathbf{g}}}}_{{\mathbf{t}}} \cdot {\mathbf{r}}) {\mathbf{k}}\biggr\} \otimes \boldsymbol{\nabla}. \end{aligned}$$

(121)

Finally, adding all displacement terms and taking into account (106), the sought displacement field is given by

$$\begin{aligned} {\mathbf{u}} =& \frac{1+\nu}{E} \biggl\{ \biggl( \frac{z^{3}}{6}-l \frac{z^{2}}{2} \biggr) {\mathbf{g}}_t + \biggl( lz-\frac{z^{2}}{2} \biggr) \mathbf{({\mathbf{g}}_t\mathbf{\cdot r)}}{\mathbf{k}}+ 2 {\mathbf{g}}_t \cdot\bigl({\boldsymbol{\psi}}+ (1+\bar{\nu}){\mathbf{p}}\bigr) {\mathbf{k}}\biggr\} \\ &{} +\frac{\nu}{E} \biggl\{ (z-l)\mathbf{({\mathbf{g}}_t\mathbf{\cdot r)}r-} \biggl[ (z-l)\frac{(\mathbf{r\cdot r})}{2}-\frac{z^{3}}{3} \biggr] {\mathbf{g}}_{{\mathbf{t}}} - \frac{(\mathbf{r\cdot r})}{2} ( {\mathbf{g}}_{t} \cdot {\mathbf{r}}) {\mathbf{k}}\biggr\} \\ &{} + {\boldsymbol{\omega}}_{0}\times {\mathbf{r}}+{\mathbf{u}}_{0}. \end{aligned}$$

(122)

A significantly simpler expression is obtained by substituting (102) and (8) in (122) and is reported in (68).