The Saint-Venant Solution of a 3D Tapered Beam

Abstract

In this article, an extension of the well-known Saint-Venant solution for prismatic beams to tapered beams, i.e., straight beams with linearly varying cross-sections, is presented. The geometry of the tapered beam will be mapped to a reference constant beam, where the 3D equilibrium equations will be expressed, using a differential geometry framework. Then, the asymptotic expansion method will be used to solve these equations, leading to a series of hierarchical problems to solve. The final solution hence obtained is written as a combination of displacement and stress modes, which can be employed as a stress-recovery method for the tapered beam. It is proven in this work that the classical Saint-Venant solution is still valid for the tapered beam, but with the addition of correction modes, representing the effect of the taper.

Appendices

Appendix 1

In this appendix, we demonstrate that the matrix $\boldsymbol{A}$ defined in (71), is symmetric definite positive.

For the symmetry, only the demonstration of $A_{34} = A_{43}$ is detailed, since it can be reproduced in the same way for the other necessary relations. By replacing $\left ( \delta \boldsymbol{\sigma },\delta \boldsymbol{u} \right )$ in (63) with $\left ( \boldsymbol{\Psi }_{\mathit{II}},\ \boldsymbol{0} \right )$ and in (64) with $\left ( \boldsymbol{\Psi }_{I},\ \boldsymbol{0} \right )$, the following relations are obtained:

$$ A_{43}:= \int _{S} \psi _{\alpha }^{\mathrm{tor}} \Psi _{\mathit{II}}^{\alpha 3} dS = \int _{S} C_{\mathit{ijkl}} \Psi _{\mathit{II}}^{ij} \Psi _{I}^{kl} dS, $$

(107)

$$ A_{34}:= \int _{S} \Psi _{I}^{33} dS = \int _{S} C_{\mathit{ijkl}} \Psi _{I}^{ij} \Psi _{\mathit{II}}^{kl} dS. $$

(108)

Using the symmetry of the elasticity tensor ($C_{\mathit{ijkl}} = C_{\mathit{klij}}$), we obtain $A_{34} = A_{43}$. Following the same demonstration, it can be shown that the matrix $\boldsymbol{A}$ can be expressed as follows:

$$ \boldsymbol{A}= \int _{S} C_{\mathit{ijkl}} \left [ \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \Psi _{I}^{ij} \Psi _{I}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{II}}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{IV}}^{kl}\\ & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{II}}^{kl} & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{IV}}^{kl}\\ & & \Psi _{\mathit{III}}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{\mathit{III}}^{ij} \Psi _{\mathit{IV}}^{kl}\\ & \mathrm{sym} & & \Psi _{\mathit{IV}}^{ij} \Psi _{\mathit{IV}}^{kl} \end{array}\displaystyle \right ] dS. $$

(109)

To prove the definite positiveness of $\boldsymbol{A}$, let us consider an arbitrary vector $\boldsymbol{r}^{T} = \left \{ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} r_{1} & r_{2} & r_{3} & r_{4} \end{array} \right \} \neq \boldsymbol{0}$ and show that $\boldsymbol{r}^{T} \boldsymbol{Ar}> 0$. The scalar $\boldsymbol{r}^{T} \boldsymbol{Ar}$ is expressed as followed:

$$ \boldsymbol{r}^{T} \boldsymbol{Ar}= \boldsymbol{r}^{T} \left ( \int _{S} C_{\mathit{ijkl}} \left [ \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \Psi _{I}^{ij} \Psi _{I}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{II}}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{I}^{ij} \Psi _{\mathit{IV}}^{kl}\\ & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{II}}^{kl} & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{\mathit{II}}^{ij} \Psi _{\mathit{IV}}^{kl}\\ & & \Psi _{\mathit{III}}^{ij} \Psi _{\mathit{III}}^{kl} & \Psi _{\mathit{III}}^{ij} \Psi _{\mathit{IV}}^{kl}\\ [4pt] & \mathrm{sym} & & \Psi _{\mathit{IV}}^{ij} \Psi _{\mathit{IV}}^{kl} \end{array}\displaystyle \right ] dS \right ) \boldsymbol{r}, $$

(110)

$$ \boldsymbol{r}^{T} \boldsymbol{Ar}= \int _{S} C_{\mathit{ijkl}} \boldsymbol{r}^{T} \left \{ \textstyle\begin{array}{c} \Psi _{I}^{ij}\\ \Psi _{\mathit{II}}^{ij}\\ \Psi _{\mathit{III}}^{ij}\\ \Psi _{\mathit{IV}}^{ij} \end{array}\displaystyle \right \} \left \{ \textstyle\begin{array}{c} \Psi _{I}^{kl}\\ \Psi _{\mathit{II}}^{kl}\\ \Psi _{\mathit{III}}^{kl}\\ \Psi _{\mathit{IV}}^{kl} \end{array}\displaystyle \right \} ^{T} \boldsymbol{r} dS. $$

(111)

By noting $s^{ij}:= \boldsymbol{r}^{T} \left \{ \begin{array}{c} \Psi _{I}^{ij}\\ \Psi _{\mathit{II}}^{ij}\\ \Psi _{\mathit{III}}^{ij}\\ \Psi _{\mathit{IV}}^{ij} \end{array} \right \}$, we obtain:

$$ \boldsymbol{r}^{T} \boldsymbol{Ar}= \int _{S} C_{\mathit{ijkl}} s^{ij} s^{kl} dS > 0, $$

(112)

which concludes the demonstration.

Appendix 2

To prove the equation in (67): $T_{(0)}^{\alpha } = \int _{S} \sigma _{(0)}^{\alpha 3} dS =0$, it is necessary to prove that for all the four stress modes $\boldsymbol{\Psi }_{I,\dots ,\mathit{IV}}$ describing $\boldsymbol{\sigma }_{(0)}$ in (62), we have:

$$ \int _{S} \Psi _{I,\dots ,\mathit{IV}}^{\alpha 3} dS =0. $$

(113)

The equation above will be proven only for the first stress mode, the same demonstration can be used for the others. To this aim, let us write the variational problem leading to $\boldsymbol{\Psi }_{I}$:

$$ \int _{S} C_{\mathit{ijkl}} \Psi _{I}^{ij} \delta \sigma ^{kl} dS - \int _{S} \Psi _{I}^{i\alpha } \delta u_{i,\alpha } dS - \int _{S} \phi _{i,\alpha }^{I} \delta \sigma ^{i\alpha } dS = \int _{S} \psi _{\alpha }^{\mathrm{tor}} \delta \sigma ^{3\alpha }\, dS. $$

(114)

By choosing $\delta \boldsymbol{\sigma } =\boldsymbol{0}$ and $\left ( \delta u_{1}, \delta u_{2}, \delta u_{3} \right ) = \left ( 0,0, X^{\alpha } \right )$, it is straightforward from the above equation that $\int _{S} \Phi _{I}^{\alpha 3} \, dS =0$.