Appendix A: Computation of the Rigid Motion
1.1 A.1 Axial Load
Equations (72) and (91) are hereby recalled as
$$\begin{aligned} &\check{\mathbf{p}}^{(DSV)}_{test} = \check{ \mathbf{p}}^{(DSV)}_{ax} = -\frac{\check{N}}{A}\mathbf{k}, \end{aligned}$$
(193)
$$\begin{aligned} &\check{\mathbf{u}}^{(DSV)}_{test} = \check{ \mathbf{u}}^{(DSV)}_{ax\, \varOmega_{0}} = - \nu \frac{\check{N}}{E A} \hat{ \boldsymbol{\rho}}, \end{aligned}$$
(194)
and are substituted into (88). The first and fourth terms on the LHS of this last equation vanish, as well as the third and fourth terms on the RHS, since the dot products involve orthogonal vectors. Moreover the third term on the LHS also vanishes since on Ω
0
r coincides with \(\hat {\boldsymbol{\rho}}\) and, recalling that the origin is placed at the centroid,
$$ \int_{\varOmega_{0}} \hat{\boldsymbol{\rho}} dA = \textbf{0} $$
(195)
so that one obtains, upon simplifying the dummy parameter \(\check{N}\), an explicit integral expression for the longitudinal component of the rigid translation
$$ u_{z0} = \frac{1}{GA} \mathbf{J}_{G}^{-1}{\mathbf{t}}\cdot \int _{\varOmega_{0}} \biggl\{ \biggl[ \boldsymbol{\psi}+(1-\bar{\nu}) \biggl( \frac{\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}}{8} \biggr) \hat{\boldsymbol{\rho}} \biggr] \biggr\} dA - \frac{\nu}{E A} \int_{\varOmega_{0}} \tilde{ \mathbf{p}}_{cant} \cdot \hat{\boldsymbol{\rho}} dA. $$
(196)
It is useful to separate in the last expression the terms depending on the short wavelength solution from the terms that only depend on the Saint-Venant long wavelength part:
$$ u_{z0}= \bar{u}_{z0}+\tilde{u}_{z0}, $$
(197)
with
$$ \bar{u}_{z0} = \frac{1}{GA} \biggl( \mathbf{J}_{G}^{-1} \int_{\varOmega_{0}} \biggl[ \boldsymbol{\psi}+\frac{1-\bar{\nu}}{8}(\hat{ \boldsymbol {\rho}}\cdot\hat{\boldsymbol{\rho}})\hat{\boldsymbol{\rho }} \biggr] dA \biggr) \cdot{\mathbf{t}} $$
(198)
and
$$ \tilde{u}_{z0} = - \frac{\nu}{E A} \int_{\varOmega_{0}} \tilde{\mathbf{p}}_{cant} \cdot \hat{\boldsymbol{\rho}} dA. $$
(199)
1.2 A.2 Bending
Equations (73) and (92), hereby recalled as
$$\begin{aligned} &\check{\mathbf{p}}^{(DSV)}_{test} = \check{ \mathbf{p}}^{(DSV)}_{bend} = - ( \check{\mathbf{g}}\cdot\hat{ \boldsymbol{\rho}} ) \mathbf{k}, \end{aligned}$$
(200)
$$\begin{aligned} &\check{\mathbf{u}}^{(DSV)}_{test} = \check{ \mathbf{u}}^{(DSV)}_{bend\, \varOmega_{0}} = \frac{\nu}{E} \biggl[ \frac{(\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }})}{2}\check{\mathbf{g}}-(\check{\mathbf{g}}\cdot\hat {\boldsymbol{ \rho}}) \hat{\boldsymbol{\rho}} \biggr], \end{aligned}$$
(201)
are substituted into (88). The LHS of (88) becomes
$$\begin{aligned} &- \hat{\mathbf{u}}_{0} \cdot \int_{\varOmega_{0}} ( \check{\mathbf{g}}\cdot\hat{\boldsymbol{\rho}} ) \mathbf{k} dA - u_{z0} \mathbf{k} \cdot \int_{\varOmega_{0}} ( \check{\mathbf{g}}\cdot\hat{\boldsymbol{\rho}} ) \mathbf{k} dA \\ &\quad - \hat{\boldsymbol{\omega}}_{0} \cdot \int _{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times ( \check{\mathbf{g}}\cdot \hat{\boldsymbol{\rho}} ) \mathbf{k} dA - \omega_{z0} \mathbf{k} \cdot \int_{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times ( \check{ \mathbf{g}}\cdot\hat{\boldsymbol{\rho}} ) \mathbf{k} dA. \end{aligned}$$
(202)
The first and second integrals on the LHS of (88) both vanish, due to (195), as the origin is located at the centroid. The fourth term on the LHS also vanishes since it is a triple product containing k twice. The only nonvanishing third term on the LHS, as a triple product, can be manipulated on account of the identity \(\hat{\boldsymbol{\omega }}_{0} \cdot \hat{\boldsymbol{\rho}}\times\mathbf{k}=\mathbf{k}\times\hat {\boldsymbol{\omega}}_{0} \cdot\hat{\boldsymbol{\rho}}\) so as to make the inertia tensor, defined by (12), appear as
$$\begin{aligned} - \hat{\boldsymbol{\omega}}_{0} \cdot \int _{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times ( \check{\mathbf{g}}\cdot \hat{\boldsymbol{\rho}} ) \mathbf{k} dA =& - \mathbf{k}\times\hat{\boldsymbol{ \omega}}_{0} \cdot \int_{\varOmega_{0}} ( \hat{\boldsymbol{ \rho}}\cdot\check{\mathbf{g}} ) \hat{\boldsymbol{\rho}} dA \\ =& - \mathbf{k}\times\hat{\boldsymbol{\omega}}_{0} \cdot \biggl( \int _{\varOmega_{0}} \hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}} dA \biggr) \check{\mathbf{g}} = - \mathbf{k}\times\hat{\boldsymbol{ \omega}}_{0} \cdot \mathbf{J}_{G} \check{\mathbf{g}} \\ =&- \mathbf{J}_{G} ( \mathbf{k}\times\hat{\boldsymbol{\omega}}_{0} ) \cdot \check{\mathbf{g}}. \end{aligned}$$
(203)
The third and fourth terms on the RHS of (88) are also zero since they contain the vanishing scalar products g
t
⋅k and \(\hat{\boldsymbol{\rho}}\cdot\mathbf{k}\). Summarizing, weighting with bending functions, (88) specializes as follows:
$$\begin{aligned} - \mathbf{J}_{G} ( \mathbf{k}\times\hat{\boldsymbol{ \omega}}_{0} ) \cdot \check{\mathbf{g}} = & \frac{1}{G} \biggl\{ \int_{\varOmega_{0}} \biggl\{ \mathbf{g}_{{\mathbf{t}}}\cdot \biggl[ \boldsymbol{\psi}+(1-\bar{\nu}) \frac{1}{8}(\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})\hat{ \boldsymbol {\rho}} \biggr] \biggr\} \hat{\boldsymbol{\rho}} dA \biggr\} \cdot \check{\mathbf{g}} \\ &{} - \frac{\nu}{E} \biggl( \int_{\varOmega_{0}} \biggl[ \frac{(\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }})}{2}\hat{\mathbf{I}}-(\hat{\boldsymbol{\rho}}\otimes\hat { \boldsymbol{\rho}}) \biggr] \tilde{\mathbf{p}}_{cant} dA \biggr) \cdot \check{\mathbf{g}}, \end{aligned}$$
(204)
and, since \(\check{\mathbf{g}}\) is arbitrary, one infers the vector identity
$$\begin{aligned} \mathbf{J}_{G} ( \mathbf{k}\times\hat{\boldsymbol{ \omega}}_{0} ) = &- \frac{1}{G} \biggl\{ \int _{\varOmega_{0}} \biggl\{ \biggl[\hat{\boldsymbol{\rho}} \otimes\boldsymbol{\psi}+(1-\bar {\nu})\frac{1}{8}(\hat{\boldsymbol{\rho}} \cdot\hat{\boldsymbol {\rho}} )\hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{ \rho}} \biggr] \biggr\} dA \biggr\} \mathbf{g}_{{\mathbf{t}}} \\ &{} + \frac{\nu}{E} \biggl( \int_{\varOmega_{0}} \biggl[ \frac{(\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }})}{2}\hat{\mathbf{I}}-(\hat{\boldsymbol{\rho}}\otimes\hat { \boldsymbol{\rho}}) \biggr] \tilde{\mathbf{p}}_{cant} dA \biggr), \end{aligned}$$
(205)
which provides
$$\begin{aligned} \hat{\boldsymbol{\omega}}_{0} =& \mathbf{k}\times \frac{1}{G} \mathbf{J}_{G}^{-1} \biggl( \int _{\varOmega_{0}} \biggl( \biggl\{ \biggl[ \hat{\boldsymbol{\rho}}\otimes \boldsymbol{\psi}+(1-\bar {\nu})\frac{1}{8}(\hat{\boldsymbol{\rho}}\cdot \hat{\boldsymbol {\rho}} )\hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}} \biggr] \biggr\} \biggr) dA \biggr) \mathbf{g}_{{\mathbf{t}}} \\ &{} - \mathbf{k}\times \frac{\nu}{E} \mathbf{J}_{G}^{-1} \biggl( \int_{\varOmega_{0}} \biggl[ \frac{(\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }})}{2}\hat{ \mathbf{I}}-(\hat{\boldsymbol{\rho}}\otimes\hat {\boldsymbol{\rho}}) \biggr] \tilde{\mathbf{p}}_{cant} dA \biggr). \end{aligned}$$
(206)
Splitting (206) into terms depending on the short wavelength solution and terms that only depend on the Saint-Venant long wavelength part, one gets
$$ \hat{\boldsymbol{\omega}}_{0} = \hat{\bar{\boldsymbol{ \omega}}}_{0} + \hat{\tilde{\boldsymbol{\omega}}}_{0}, $$
(207)
where, recalling that \(\mathbf{g}_{{\mathbf{t}}}=-\mathbf {J}_{G}^{-1}{\mathbf{t}}\), it turns out that
$$ \hat{\bar{\boldsymbol{\omega}}}_{0} = - \mathbf{k}\times \frac{1}{G} \mathbf{J}_{G}^{-1} \biggl( \int _{\varOmega_{0}} \biggl( \biggl\{ \biggl[ \hat{\boldsymbol{\rho}}\otimes \boldsymbol{\psi}+(1-\bar {\nu})\frac{1}{8}(\hat{\boldsymbol{\rho}}\cdot \hat{\boldsymbol {\rho}} )\hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}} \biggr] \biggr\} \biggr) dA \biggr) \mathbf{J}_{G}^{-1} { \mathbf{t}} $$
(208)
and
$$ \hat{\tilde{\boldsymbol{\omega}}}_{0} = - \mathbf{k}\times \frac{\nu}{E} \mathbf{J}_{G}^{-1} \biggl( \int _{\varOmega_{0}} \biggl[ \frac{(\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }})}{2}\hat{\mathbf{I}}-(\hat{ \boldsymbol{\rho}}\otimes\hat {\boldsymbol{\rho}}) \biggr] \tilde{ \mathbf{p}}_{cant} dA \biggr). $$
(209)
1.3 A.3 Torsion
Substituting (74) and (93), i.e.,
$$\begin{aligned} &\check{\mathbf{p}}^{(DSV)}_{test} = \check{ \mathbf{p}}^{(DSV)}_{tor} =-G\check{\theta}' (\mathbf {k}\times\hat{\boldsymbol{\rho}} +\boldsymbol{\nabla}\varphi_{t} ), \end{aligned}$$
(210)
$$\begin{aligned} & \check{\mathbf{u}}^{(DSV)}_{test} = \check{ \mathbf{u}}^{(DSV)}_{tor\, \varOmega_{0}} =\check{\theta}' \varphi_{t} \mathbf{k}, \end{aligned}$$
(211)
into (88), it is recognized that the sum of the first two terms on the LHS of (88) is equal to zero for translational equilibrium:
$$ \mathbf{u}_{0} \cdot \int_{\varOmega_{0}} \check{ \mathbf{p}}^{(DSV)}_{tor} dA = 0 $$
(212)
since \(\check{\mathbf{p}}^{(DSV)}_{tor}\) has zero cross sectional force resultant
$$ \int_{\varOmega_{0}} \check{\mathbf{p}}^{(DSV)}_{tor} dA = \mathbf{o}. $$
(213)
The second term on the LHS also vanishes, i.e.,
$$ \hat{\boldsymbol{\omega}}_{0} \cdot \int_{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times\check{\mathbf{p}}^{(DSV)}_{tor} dA = 0 $$
(214)
since \(\hat{\boldsymbol{\rho}}\times\check{\mathbf {p}}^{(DSV)}_{tor}\) is directed along k, because both \(\check{\mathbf{p}}^{(DSV)}_{tor}\) and \(\hat {\boldsymbol{\rho}}\) are transverse fields, and their cross product is thus orthogonal to the transverse vector \(\hat{\boldsymbol{\omega}}_{0}\). The only term remaining on the LHS can be manipulated on account of (12)3 and of the property of the triple product as follows:
$$\begin{aligned} \omega_{z0} \mathbf{k} \cdot \int_{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times\check{\mathbf{p}}^{(DSV)}_{tor} dA =& \omega_{z0} \int_{\varOmega_{0}} \mathbf{k} \times\hat{ \boldsymbol{\rho}} \cdot\check{\mathbf{p}}^{(DSV)}_{tor} dA \\ =& - G \check{\theta}' \omega_{z0} \int _{\varOmega} \mathbf{k}\times\hat{\boldsymbol{\rho}} \cdot \bigl[ ( \mathbf{k}\times\hat{\boldsymbol{\rho}} )+\varphi_t \boldsymbol{ \nabla}\bigr] dA \\ =& - G \check{\theta}' I_q \omega_{z0}. \end{aligned}$$
(215)
Turning to the RHS, the first and second terms are equal to zero since they both contain the zero term \(( \check{\mathbf{p}}^{(DSV)}_{tor} \cdot \mathbf{k} ) \). The sum of the third and fourth terms on the RHS is computed as
$$\begin{aligned} &- \frac{\nu l}{E} \int_{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \mathbf{g}_{t} - ( \mathbf{g}_t\cdot\hat{\boldsymbol{\rho}} ) \hat{\boldsymbol{\rho}} \biggr] \cdot \check{\mathbf{p}}^{(DSV)}_{tor} dA \\ &\quad = G\check{\theta}' \frac{\nu l}{E} \int _{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \mathbf{g}_{t} - (\mathbf{g}_t\cdot\hat{\boldsymbol{\rho}} ) \hat{\boldsymbol{\rho}} \biggr] \cdot (\mathbf{k}\times\hat{\boldsymbol{\rho}}+ \boldsymbol{\nabla }\varphi_{t} ) dA \\ &\quad = G\check{\theta}' \frac{\nu l}{E} \int _{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \mathbf{g}_{t} \cdot (\mathbf{k}\times\hat{\boldsymbol{\rho}}+ \boldsymbol{\nabla }\varphi_{t} ) - (\mathbf{g}_t\cdot \hat{\boldsymbol{\rho}} ) \hat{\boldsymbol{\rho}} \cdot (\boldsymbol{\nabla} \varphi_{t} ) \biggr] dA, \end{aligned}$$
(216)
where the last equality is obtained in view of the simplification \(\hat {\boldsymbol{\rho}} \cdot\mathbf{k}\times\hat{\boldsymbol{\rho}}=0\). Finally, the last integral on the RHS of (88) is computed as
$$ - \int_{\varOmega_{tot}} \tilde{\mathbf{p}}_{cant} \cdot \check{ \mathbf{u}}^{(DSV)}_{tor} dA = \int_{\varOmega_{0}} \check{\theta}' \tilde{\sigma}_{z\, cant} \varphi_{t} dA. $$
(217)
Summarizing and simplifying considering the dummy parameter \(\check{\theta}'\), we infer from (88) that
$$ \omega_{z0}=\bar{\omega}_{z0}+\tilde{\omega}_{z0}, $$
(218)
where
$$\begin{aligned} &\bar{\omega}_{z0} = - \frac{\nu l}{E I_q} \mathbf{g}_{t} \cdot \int_{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) (\mathbf{k}\times\hat{\boldsymbol{ \rho}}+\boldsymbol{\nabla }\varphi_{t} ) - \bigl[ \hat{\boldsymbol{ \rho}} \cdot (\boldsymbol{\nabla}\varphi_{t} ) \bigr] \hat{\boldsymbol{ \rho}} \biggr] dA, \end{aligned}$$
(219)
$$\begin{aligned} &\tilde{\omega}_{z0} = - \frac{1}{G I_q} \int _{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \varphi_{t} dA. \end{aligned}$$
(220)
However, (219) can be further manipulated since, on account of (46), one has
$$ \int_{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \hat{ \mathbf{I}} - ( \hat{\boldsymbol{\rho}} \otimes \hat{\boldsymbol{\rho}} ) \biggr] (\boldsymbol{\nabla}\varphi_{t} ) dA = \int_{\varOmega_{0}} \bigl[ 4 (\mathbf{p}\otimes\boldsymbol{\nabla} ) - 2 ( \hat{ \boldsymbol{\rho}} \otimes \hat{\boldsymbol{\rho}} ) \bigr] (\boldsymbol{\nabla} \varphi_{t} ) dA, $$
(221)
and, due to (222), one infers
$$\begin{aligned} \int_{\varOmega_{0}} \biggl[ \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \hat{\mathbf{I}} - \hat{\boldsymbol{\rho}} \otimes \hat{ \boldsymbol{\rho}} \biggr] (\boldsymbol{\nabla}\varphi_{t} ) dA = & - \int_{\varOmega} \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ( \mathbf{k}\times\hat{\boldsymbol{\rho}} ) dA \\ &{} - 2 \int_{\varOmega} ( \hat{\boldsymbol{\rho}} \otimes \hat{ \boldsymbol{\rho}} ) (\boldsymbol{\nabla}\varphi_{t} ) dA, \end{aligned}$$
(222)
so that (219) becomes
$$ \bar{\omega}_{z0} = 2 \frac{\nu l}{E I_q} \mathbf{g}_{t} \cdot \int_{\varOmega} ( \hat{\boldsymbol{\rho}} \otimes \hat{\boldsymbol{\rho}} ) (\boldsymbol{\nabla} \varphi_{t} ) dA. $$
(223)
1.4 A.4 Reversed Cantilever
Equations (75) and (95) are recalled as
$$\begin{aligned} & \check{\mathbf{p}}^{(DSV)}_{test} = \check{ \mathbf{p}}^{(DSV)}_{r.cant} = - \biggl[ \boldsymbol{\nabla}\otimes \boldsymbol{\psi} + \frac{1+\bar{\nu}}{4} ( \hat{\boldsymbol{\rho}}\otimes\hat{ \boldsymbol{\rho }} ) + \frac{1-3\bar{\nu}}{8} ( \hat{\boldsymbol{\rho}}\cdot\hat{ \boldsymbol{\rho }} ) \mathbf{I} \biggr] \check{\mathbf{g}}_{t}, \end{aligned}$$
(224)
$$\begin{aligned} &\check{\mathbf{u}}^{(DSV)}_{r.cant\, \varOmega_{0}} = \biggl[ \frac{1}{G} ( \check{\mathbf{g}}_t \cdot\boldsymbol{\psi} )+ \frac{1}{4E} ( \check{\mathbf{g}}_t \cdot\hat{\boldsymbol{\rho}} ) ( \hat{ \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }} ) \biggr] \mathbf{k} + \frac{1 }{E} \biggl( \frac{2}{3} \biggr) l^{3} \check{ \mathbf{g}}_t + \frac{1}{E} \frac{l^{2}}{2} (\check{ \mathbf{g}}_t \cdot\boldsymbol{\rho})\mathbf{k}. \end{aligned}$$
(225)
By substituting these equations into (88) we see that the integral appearing in the first term on the LHS of (88) is simply equal to the shear stress resultant. As a consequence, in view of (20), one has
$$ \hat{\mathbf{u}}_{0} \cdot \int_{\varOmega_{0}} \check{ \mathbf{p}}^{(DSV)}_{r.cant} dA = \hat{\mathbf{u}}_{0} \cdot \int_{\varOmega_{0}} (- \check{\boldsymbol{\tau}}_{sh} ) dA = - \hat{\mathbf{u}}_{0} \cdot \check{{\mathbf{t}}}_{sh} = \mathbf{J}_{G} \hat{\mathbf{u}}_{0} \cdot \check{ \mathbf{g}}_{t}. $$
(226)
Analogous considerations lead to the conclusion that the second term on the LHS of (88) is equal to zero, as well as the third term which vanishes since it contains a dot product between orthogonal vectors.
The fourth term on the LHS is computed below accounting of the property of the shear center \(\hat{\boldsymbol{\rho}}_{C}\) stated by (8) and the properties of the triple product:
$$\begin{aligned} \omega_{z0} \mathbf{k} \cdot \int_{\varOmega_{0}} \hat{\boldsymbol{\rho}}\times\check{\mathbf{p}}^{(DSV)}_{r.cant} dA =& \omega_{z0}\mathbf{k} \cdot \int_{\varOmega_{0}} \hat{ \boldsymbol{\rho}}\times (-\check{\boldsymbol{\tau}}_{sh} ) dA = \omega_{z0}\mathbf{k} \cdot \hat{ \boldsymbol{\rho}}_{C}\times\check{{\mathbf{t}}} \\ =& \omega_{z0} ( \hat{\boldsymbol{\rho}}_{C}\times \mathbf{J}_{G} \check{\mathbf{g}}_{t} ) \cdot \mathbf{k} = \omega_{z0} \mathbf{k}\times\hat{\boldsymbol{\rho}}_{C} \cdot ( \mathbf{J}_{G} \check{\mathbf{g}}_{t} ) \\ =& \omega_{z0} \mathbf{J}_{G} ( \mathbf{k}\times\hat{ \boldsymbol{\rho}}_{C} ) \cdot \check{\mathbf{g}}_{t}. \end{aligned}$$
(227)
At the RHS the first and second terms vanish because they contain zero dot products. The third term is computed as
$$\begin{aligned} &- \frac{\nu l}{E} \int_{\varOmega_{0}} \biggl( \frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) \bigl( \check{\mathbf{p}}^{(DSV)}_{r.cant} \cdot \mathbf{g}_{t} \bigr) dA \\ &\quad= \frac{\nu l}{E} \int_{\varOmega_{0}} \biggl[ \biggl( \frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ( \boldsymbol{\psi}\otimes\boldsymbol{\nabla} ) + \frac{1+\bar{\nu}}{4} \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ( \hat{\boldsymbol{\rho}} \otimes\hat{\boldsymbol{\rho }} ) \\ &\quad\quad{}+ \frac{1-3\bar{\nu}}{8} \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ( \hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }} ) \mathbf{I} \biggr] dA \mathbf{g}_{t} \cdot \check{\mathbf{g}}_{t}. \end{aligned}$$
(228)
The fourth term on the RHS is computed as
$$\begin{aligned} &\frac{\nu l}{E} \int_{\varOmega_{0}} ( \mathbf{g}_t\cdot\hat{\boldsymbol{\rho}} ) \bigl( \check{ \mathbf{p}}^{(DSV)}_{test} \cdot \hat{\boldsymbol{\rho}} \bigr) dA \\ &\quad= - \frac{\nu l}{E} \int_{\varOmega_{0}} \biggl( \biggl[ \boldsymbol{\psi}\otimes\boldsymbol{\nabla} + \frac{1+\bar{\nu}}{4} ( \hat{ \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho }} ) + \frac{1-3\bar{\nu}}{8} ( \hat{ \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }} ) \mathbf{I} \biggr] (\hat{ \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho }} ) \mathbf{g}_t \biggr) dA \cdot \check{\mathbf{g}}_{t} \\ &\quad= - \frac{\nu l}{E} \int_{\varOmega_{0}} \biggl( \biggl[ ( \boldsymbol{\psi}\otimes\boldsymbol{\nabla} ) (\hat{\boldsymbol{\rho}}\otimes\hat{ \boldsymbol{\rho }} ) + \frac{1+\bar{\nu}}{4} ( \hat{\boldsymbol{\rho}}\otimes\hat{ \boldsymbol{\rho }} ) (\hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho }} ) \\ &\quad\quad{}+ \frac{1-3\bar{\nu}}{8} ( \hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }} ) (\hat{ \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho }} ) \biggr] \mathbf{g}_t \biggr) dA \cdot \check{\mathbf{g}}_{t}. \end{aligned}$$
(229)
The sum of (228) and (229) turns out to be, upon rearranging terms,
$$\begin{aligned} &\frac{\nu}{E}l \biggl[ \int_{\varOmega_{0}} \biggl( \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) - ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) (\hat{\boldsymbol {\rho}} \otimes\hat{\boldsymbol{\rho}}) \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \cdot \check{\mathbf{g}}_{t} \\ &\quad{} + \frac{\nu}{E}l \biggl[ \int_{\varOmega_{0}} \biggl( \frac{(\bar{\nu}-1)}{4} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}) (\hat { \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}}) + \frac{(1-3\bar{\nu})}{16} (\hat{ \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})^{2} \hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \cdot \check{\mathbf{g}}_{t}. \end{aligned}$$
(230)
Concerning the SW dependent part of the rigid motion, the last term to be computed is the fourth on the RHS. Thus,
$$ - \int_{\varOmega_{0}} \tilde{\mathbf{p}}_{cant\, 0} \cdot \check{ \mathbf{u}}^{(DSV)}_{r.cant} dA = \int_{\varOmega_{0}} ( \tilde{\sigma}_{z\, cant} \mathbf{k} + \tilde{\boldsymbol{ \tau}}_{cant} ) \cdot \check{\mathbf{u}}^{(DSV)}_{r.cant} dA $$
(231)
and one infers
$$\begin{aligned} - \int_{\varOmega_{0}} \tilde{\mathbf{p}}_{cant} \cdot \check{\mathbf{u}}^{(DSV)}_{r.cant} dA = & \biggl( \int _{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \biggl[ \frac{1}{G} \boldsymbol{\psi}+ \frac{1}{4E} ( \hat{\boldsymbol{\rho}} \cdot\hat{\boldsymbol{\rho }} ) \hat{\boldsymbol{\rho}} \biggr] \mathbf{k} dA \biggr) \cdot \check{\mathbf{g}}_t \\ &{} + \frac{1}{E} \frac{l^{2}}{2} \check{\mathbf{g}}_t \cdot \int_{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \boldsymbol{ \rho} dA + \frac{1 }{E} \biggl( \frac{2}{3} \biggr) l^{3} \check{\mathbf{g}}_t \cdot \int _{\varOmega_{0}} \tilde{\boldsymbol{\tau}}_{cant} dA. \end{aligned}$$
(232)
It is recognized that the second and third integrals in the last expression vanish since \(\tilde{\mathbf{p}}_{cant}\), as a cross-sectional equilibrated field, has both zero axial force resultant and bending momentum resultant on Ω
0. Consequently, one has
$$ - \int_{\varOmega_{0}} \tilde{\mathbf{p}}_{cant} \cdot \check{ \mathbf{u}}^{(DSV)}_{r.cant} dA = \frac{1}{G} \biggl( \int _{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \biggl[\boldsymbol{\psi}+ \frac{1-\bar{\nu}}{8} (\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}} ) \hat{ \boldsymbol{\rho}} \biggr] dA \biggr) \cdot \check{\mathbf{g}}_t. $$
(233)
Hence, recapitulating, by weighting with the stress field associated with the reversed cantilever solution, we find the following
$$\begin{aligned} &\omega_{z0} \mathbf{J}_{G} ( \mathbf{k} \times\hat{\boldsymbol{\rho}}_{C} ) \cdot \check{\mathbf{g}}_{t} + \mathbf{J}_{G} \hat{\mathbf{u}}_{0} \cdot \check{ \mathbf{g}}_{t} \\ &\quad= \frac{\nu}{E}l \biggl[ \int_{\varOmega_{0}} \biggl( \biggl(\frac{\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho }}}{2} \biggr) ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) - ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) (\hat{\boldsymbol {\rho}} \otimes\hat{\boldsymbol{\rho}}) \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \cdot \check{\mathbf{g}}_{t} \\ &\quad\quad{}+ \frac{\nu}{E}l \biggl[ \int_{\varOmega_{0}} \biggl( \frac{(\bar{\nu}-1)}{4} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}) (\hat { \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}}) + \frac{(1-3\bar{\nu})}{16} (\hat{ \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})^{2} \hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \cdot \check{\mathbf{g}}_{t} \\ &\quad\quad{}+ \frac{1}{G} \biggl( \int_{\varOmega_{0}} \tilde{ \sigma}_{z\, cant} \biggl[\boldsymbol{\psi}+ \frac{1-\bar{\nu}}{8} (\hat { \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}} ) \hat{\boldsymbol{\rho}} \biggr] dA \biggr) \cdot \check{\mathbf{g}}_t. \end{aligned}$$
(234)
Moreover, since \(\check{\mathbf{g}}_{t}\) is arbitrary, we obtain the identity
$$\begin{aligned} \mathbf{J}_{G} \hat{\mathbf{u}}_{0} =& - \frac{\nu}{E}l \biggl[ \biggl( \int_{\varOmega_{0}} ({\boldsymbol{ \psi}}\otimes\boldsymbol{\nabla}) (\hat{\boldsymbol {\rho}}\otimes\hat{ \boldsymbol{\rho}}) - ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) \frac{(\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})}{2}\hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \\ &{}+ \frac{\nu}{E}l \biggl[ \int_{\varOmega_{0}} \biggl( \frac{(\bar{\nu}-1)}{4} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}) (\hat { \boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}}) + \frac{(1-3\bar{\nu})}{16} (\hat{ \boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})^{2} \hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{g}_{{\mathbf{t}}} \\ &{} - \omega_{z0} \mathbf{J}_{G} ( \mathbf{k}\times\hat{ \boldsymbol{\rho}}_{C} ) + \frac{1}{G} \biggl( \int _{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \biggl[\boldsymbol{\psi}+ \frac{1-\bar{\nu}}{8} (\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}} ) \hat{ \boldsymbol{\rho}} \biggr] dA \biggr). \end{aligned}$$
(235)
Introducing the usual separation of the terms depending solely on the long wavelength part of the solution, (235) provides at last
$$ \hat{\mathbf{u}}_{0} = \hat{\bar{\mathbf{u}}}_{0} + \hat{ \tilde{\mathbf{u}}}_{0} , \qquad \omega_{z0}=\bar{ \omega}_{z0}+\tilde{\omega}_{z0}, $$
(236)
where, taking also into account (20),
$$\begin{aligned} \hat{\bar{\mathbf{u}}}_{0} =& \frac{\nu}{E}l \mathbf{J}_{G}^{-1} \biggl[ \biggl( \int_{\varOmega_{0}} ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) (\hat{\boldsymbol {\rho}}\otimes \hat{\boldsymbol{\rho}}) - ({\boldsymbol{\psi}}\otimes\boldsymbol{\nabla}) \frac{(\hat {\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}})}{2}\hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{J}_{G}^{-1} {\mathbf{t}} \\ &{} - \frac{\nu}{E}l \mathbf{J}_{G}^{-1} \biggl[ \int _{\varOmega_{0}} \biggl( \frac{(\bar{\nu}-1)}{4} (\hat{\boldsymbol{\rho}}\cdot \hat{\boldsymbol{\rho}}) (\hat {\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}}) + \frac{(1-3\bar{\nu})}{16} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{ \rho}})^{2} \hat{\mathbf{I}} \biggr) dA \biggr] \mathbf{J}_{G}^{-1} {\mathbf{t}} \\ &{}- \bar{\omega}_{z0} ( \mathbf{k}\times\hat{\boldsymbol{ \rho}}_{C} ), \end{aligned}$$
(237)
$$\begin{aligned} \hat{\tilde{\mathbf{u}}}_{0} =& - \tilde{\omega}_{z0} ( \mathbf{k}\times\hat{\boldsymbol{\rho}}_{C} ) - \frac{1}{G} \mathbf{J}_{G}^{-1} \biggl( \int_{\varOmega_{0}} \tilde{\sigma}_{z\, cant} \bigl[\boldsymbol{\psi}+ (1-\bar{\nu})\mathbf{p} \bigr] dA \biggr). \end{aligned}$$
(238)
Appendix B: Further Identities
2.1 B.1 Identity 1
It is shown that the tensor
$$ \int_{\varOmega} (\mathbf{p}\otimes\boldsymbol{\nabla}) ( \boldsymbol{\nabla}\otimes\boldsymbol{\psi}) dA $$
(239)
admits the following symmetric representation:
$$ \int_{\varOmega} (\mathbf{p}\otimes\boldsymbol{\nabla}) ( \boldsymbol{\nabla}\otimes\boldsymbol{\psi}) dA = - \frac{1}{4} \int _{\varOmega} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}) \hat{\boldsymbol{\rho}}\otimes\hat{\boldsymbol{\rho}} dA - \frac{1-3\bar{\nu}}{64} \int_{\varOmega} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{ \rho}})^{2} \hat{\mathbf{I}} dA, $$
(240)
so that by virtue of the symmetry of the RHS of (240) one has
$$ \int_{\varOmega} (\mathbf{p}\otimes\boldsymbol{\nabla}) ( \boldsymbol{\nabla}\otimes\boldsymbol{\psi}) dA = \int_{\varOmega} (\boldsymbol{\psi}\otimes\boldsymbol{\nabla}) (\boldsymbol{\nabla}\otimes \mathbf{p}) dA = \int_{\varOmega} (\boldsymbol{\psi}\otimes \boldsymbol{\nabla}) (\mathbf{p}\otimes\boldsymbol{\nabla}) dA. $$
(241)
The proof of (240) is shown below with the aid of index notation. Due to the harmonicity of ψ
j
and the divergence theorem, we see that
$$ \int_{\varOmega} p_{i,k}\psi_{j,k} dA = \int _{\varOmega} (p_{i}\psi_{j,k})_{,k} dA = \int_{\partial\varOmega} (p_{i}\psi_{j,k})n_{k} ds = \int_{\partial\varOmega} -(p_{i}A_{jk})n_{k} ds, $$
(242)
where the last equality stems from the boundary property (11)2. Moreover, invoking again the divergence theorem one infers
$$ \int_{\partial\varOmega} -(p_{i}A_{jk})n_{k} ds = \int_{\varOmega} -(p_{i}A_{jk})_{,k} dA = - \int_{\varOmega} p_{i,k}A_{jk} dA - \int_{\varOmega} p_{i}A_{jk,k} dA, $$
(243)
so that we obtain
$$ \int_{\varOmega} p_{i,k}\psi_{j,k} dA = - \int_{\varOmega} p_{i,k}A_{jk} dA - \int _{\varOmega} p_{i}A_{jk,k} dA. $$
(244)
Finally, recalling (46), which in index notation reads
$$ p_{i,k}=\frac{1}{4}\rho_{i}\rho_{k}+ \frac{1}{8}\rho_{h}\rho _{h}\delta_{ik}, $$
(245)
and substituting this relation in (244), together with (6) and (48), one obtains the sought relation (240), i.e.,
$$ \int_{\varOmega} p_{i,k}\psi_{j,k} dA = - \frac{1}{4} \int_{\varOmega} \rho_{k} \rho_{k} \rho_{i}\rho_{j} dA - \frac{1-3\bar{\nu}}{64} \int_{\varOmega} (\rho_{k} \rho_{k})^{2} \delta_{ij} dA. $$
(246)
2.2 B.2 Identity 2
An alternate representation is derived for the integral
$$ \int_{\varOmega} \boldsymbol{\rho}\otimes\boldsymbol{\psi} dA $$
(247)
by expressing \(\hat{\boldsymbol{\rho}}\) as the divergence of A
p on account of (48). Accordingly, turning to index notation (in index notation (48) reads A
ik,k
=ρ
i
) one infers
$$ \int_{\varOmega} \rho_{i} \psi_{j} dA = \int_{\varOmega} A_{ik,k} \psi_{j} dA = \int _{\varOmega} (A_{ik} \psi_{j})_{,k} dA - \int_{\varOmega} A_{ik} \psi_{j,k} dA. $$
(248)
However, by the divergence theorem one has
$$ \int_{\varOmega} (A_{ik} \psi_{j})_{,k} dA = \int_{\partial\varOmega} (A_{ik} \psi_{j})n_{k} ds, $$
(249)
and, due to the boundary condition (11)2 for A
p, which in index notation reads A
ik
n
k
=−ψ
i,k
n
k
, one infers
$$ \int_{\partial\varOmega} (A_{ik} \psi_{j})n_{k} ds = - \int_{\partial\varOmega} (\psi_{i,k} \psi_{j})n_{k} ds = - \int_{\varOmega} ( \psi_{i,k} \psi_{j})_{,k} dA. $$
(250)
Moreover, since ψ is harmonic one has
$$ (\psi_{i,k} \psi_{j})_{,k}=\psi_{i,kk} \psi_{j}+\psi_{i,k} \psi _{j,k}= \psi_{i,k}\psi_{j,k} $$
(251)
and, consequently,
$$ \int_{\varOmega} \rho_{i} \psi_{j} dA = -\int_{\varOmega}\psi_{i,k}\psi_{j,k} dA - \int_{\varOmega} A_{ik} \psi_{j,k} dA. $$
(252)
It is easily recognized that the second integrand function on the RHS of (252) turns out to be equal to
$$ A_{ik}=(1+\bar{\nu})p_{i,k}-\frac{1}{2}\bar{\nu} \rho_{h}\rho _{h}\delta_{ik}. $$
(253)
Substituting (253) into (252) one finally obtains
$$ \int_{\varOmega} \rho_{i} \psi_{j} dA = -\int_{\varOmega}\psi_{i,k}\psi_{j,k} dA - (1+\bar{\nu}) \int_{\varOmega} p_{i,k} \psi_{j,k} dA + \frac{1}{2}\bar{\nu} \int_{\varOmega} \rho_{h}\rho_{h} \psi_{j,i} dA, $$
(254)
which in implicit notation reads
$$\begin{aligned} \int_{\varOmega} \boldsymbol{\rho}\otimes\boldsymbol{ \psi} dA =& -\int_{\varOmega} (\boldsymbol{\psi}\otimes \boldsymbol{\nabla}) (\boldsymbol{\nabla }\otimes\boldsymbol{\psi}) dA - (1+\bar{\nu}) \int_{\partial\varOmega} ( \mathbf{p}\otimes\boldsymbol{\nabla}) (\boldsymbol{\nabla}\otimes\boldsymbol{ \psi}) dA \\ &{}+ \frac{\bar{\nu}}{2} \int_{ \varOmega} (\boldsymbol{\rho}\cdot \boldsymbol{\rho}) (\boldsymbol{\nabla }\otimes\boldsymbol{\psi}) dA. \end{aligned}$$
(255)
2.3 B.3 Identity 3
The tensor
$$ \int_{\varOmega} (\mathbf{p}\otimes\boldsymbol{\nabla}) \boldsymbol{\nabla}\varphi_{t} dA $$
(256)
admits the following representation:
$$ \int_{\varOmega} (\mathbf{p}\otimes\boldsymbol{\nabla}) \boldsymbol{\nabla}\varphi_{t} dA = - \frac{1}{8} \int _{\varOmega} (\hat{\boldsymbol{\rho}}\cdot\hat{\boldsymbol{\rho}}) ( \mathbf{k}\times\hat{\boldsymbol{\rho}} ) dA. $$
(257)
The proof of (257) is shown below, again with the aid of index notation. Due to the harmonicity of φ
t
and the divergence theorem, we see that
$$ \int_{\varOmega} p_{i,k}\varphi_{t,k} dA = \int_{\varOmega} (p_{i}\varphi_{t,k})_{,k} dA = \int_{\partial\varOmega} p_{i}\varphi_{t,k}n_{k} ds = \int_{\partial\varOmega} -p_{i}\rho_{k}^{\bot}n_{k} ds, $$
(258)
where the last equality in (258) stems from the boundary property (10)2. Moreover, invoking again the divergence theorem one infers
$$ \int_{\partial\varOmega} -p_{i}\rho_{k}^{\bot}n_{k} ds = \int_{\varOmega} -\bigl(p_{i} \rho_{k}^{\bot}\bigr)_{,k} dA = - \int_{\varOmega} -\bigl(p_{i,k}\rho_{k}^{\bot}+p_{i} \rho_{k,k}^{\bot}\bigr)_{,k} dA. $$
(259)
However, since the divergence of \(\hat{\boldsymbol{\rho}}^{\bot}\) is zero, recalling (46) one finally obtains
$$ p_{i,k}\varphi_{t,k} dA = - \int_{\varOmega} \biggl( \frac{1}{4} \rho_{i}\rho_{k} + \frac{1}{8} \rho_{l}\rho_{l} \delta_{ik} \biggr) \rho_{k}^{\bot} dA = - \int_{\varOmega} \frac{1}{8} \rho_{l}\rho_{l} \rho_{i}^{\bot} dA, $$
(260)
which is (257) in index notation.
