Decomposition of Rod Displacements via Bernoulli–Navier Displacements. Application: A Loading of the Rod with Shearing

Abstract

Within the framework of linear elasticity, we show that any displacement of a straight rod is the sum of a Bernoulli–Navier displacement and two terms, one for shearing and the other for warping. Then, we load a straight rod so that bending and shear contribute the same to the rotations of the cross-section.

Appendix: Extension of a Rod Displacement

Let $u$ be a displacement belonging to $W^{1,p}({\mathcal{P}}_{\delta})^{3}$, $p\in (1,\infty )$, decomposed as (2.1). The terms ${\mathcal{U}}^{\ast}$, ${\mathcal{R}}^{\ast}$ and $\overline{U}^{\ast}$ of this decomposition satisfy (2.3).

Set

$$ \begin{aligned}[b] {\mathcal{U}}^{\ast \ast}(x_{1})&\doteq \left \{ \begin{aligned} &{\mathcal{U}}^{\ast}(x_{1})\quad &&\text{if}\;\; x_{1}\in [0,L], \\ &2{\mathcal{U}}^{\ast}(L)-{\mathcal{U}}^{\ast}(2L-x_{1})\quad &&\text{if}\;\; x_{1} \in [L,2L], \end{aligned} \right . \\ {\mathcal{R}}^{\ast \ast}(x_{1})&\doteq \left \{ \begin{aligned} &{\mathcal{R}}^{\ast}(x_{1})\quad &&\text{if}\;\; x_{1}\in [0,L], \\ &{\mathcal{R}}^{\ast}(2L-x_{1})\quad &&\text{if}\;\; x_{1}\in [L,2L], \end{aligned} \right . \\ \overline{u}^{\ast \ast}(x)&\doteq \left \{ \begin{aligned} &\overline{u}^{\ast}(x)\quad &&\text{for a.e. }\;\; x\in (0,L)\times \omega _{\delta}, \\ &\overline{u}^{\ast}(2L-x_{1},x_{2},x_{3})\quad &&\text{for a.e. }\;\; x \in (L,2L)\times \omega _{\delta}. \end{aligned} \right . \end{aligned} $$

(7.1)

Now, we define the extension of $u$ denoted $u^{\ast \ast}$ by

$$ u^{\ast \ast}(x)={\mathcal{U}}^{\ast \ast}(x_{1})+{\mathcal{R}}^{\ast \ast}(x_{1}) \land \big(x_{2}{\mathbf{e}}_{2}+x_{3}{\mathbf{e}}_{3}\big)+\overline{u}^{\ast \ast}(x)\quad \text{for a.e. }\; x\in (0,2L)\times \omega _{\delta}. $$

So, we have

$$ u^{\ast \ast}\in W^{1,p}((0,2L)\times \omega _{\delta})^{3},\quad { \mathcal{U}}^{\ast \ast},\;\; {\mathcal{R}}^{\ast \ast}\in W^{1,p}(0,2L),\quad \overline{u}^{\ast \ast}\in W^{1,p}((0,2L)\times \omega _{\delta})^{3}. $$

Moreover, using the estimates (2.3) we easily check that

$$ \begin{aligned}[b] &\|\overline{u}^{\ast \ast}\|_{L^{p}((0,2L)\times \omega _{\delta})} \le C\delta \|e(u)\|_{L^{p}({\mathcal{P}}_{\delta })},\quad \|\nabla \overline{u}^{\ast \ast}\|_{L^{p}((0,2L)\times \omega _{\delta})}\le C \|e(u)\|_{L^{p}({\mathcal{P}}_{\delta })}, \\ &\delta \Bigl\| {\frac{d{\mathcal{R}}^{\ast \ast}}{dx_{1}}}\Big\| _{L^{p}(0,2L)}+ \Bigl\| {\frac{d{\mathcal{U}}^{\ast \ast}}{dx_{1}}}-{\mathcal{R}}^{\ast \ast}\land { \mathbf{e}}_{1}\Big\| _{L^{p}(0,2L)}\le {\frac{C}{\delta ^{2/p}}} \|e(u)\|_{L^{p}({ \mathcal{P}}_{\delta }}. \end{aligned} $$

(7.2)

The constants do not depend on $\delta$ and $L$.