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).