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Geometric and Analytic Tools

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Elasticity for Geotechnicians

Part of the book series: Solid Mechanics and Its Applications ((SMIA,volume 204))

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Abstract

The problems in classical elasticity we tackle have intrinsic symmetries that are best exploited with the use of ad hoc coordinate systems, because the associated vector and tensor bases allow for convenient representations of the fields of interest and their transformations under the action of differential operators. In this chapter we collect a modicum of basic material from differential geometry and analysis.

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Notes

  1. 1.

    It follows from (3.1) and (3.2) that, for each index \(i\), vectors \(g_i\) and \(g^i\) have inverse physical dimensions:

    $$\begin{aligned} \mathrm{dim }(g_i) =\big (\mathrm{dim }(g^i)\big )^{-1}. \end{aligned}$$

    Thus, in particular, the identity tensor is dimensionless, as required implicitly by (3.3)\(_1\); moreover, in view of (3.4),

    $$\begin{aligned} \mathrm{dim }(v_i) =\big (\mathrm{dim }(v^i)\big )^{-1}. \end{aligned}$$
  2. 2.

    On differentiating the first of (3.5), we find that

    $$\begin{aligned} 2\rho \nabla \rho =2{{\varvec{x}}}\quad \Rightarrow \quad \nabla \rho =\rho ^{-1}{{\varvec{x}}},\;\mathrm{with }\;\,\rho =|{{\varvec{x}}}|; \end{aligned}$$

    on differentiating the second, that

    $$\begin{aligned} \frac{1}{\cos ^2\vartheta }\nabla \vartheta =\frac{({{\varvec{x}}}\cdot {{\varvec{e}}}_1){{\varvec{e}}}_2-({{\varvec{x}}}\cdot {{\varvec{e}}}_2){{\varvec{e}}}_1}{({{\varvec{x}}}\cdot {{\varvec{e}}}_1)^2}\,,\;\mathrm{con }\;\,\cos \vartheta =\rho ^{-1}({{\varvec{x}}}\cdot {{\varvec{e}}}_1),\;\,\sin \vartheta =\rho ^{-1}({{\varvec{x}}}\cdot {{\varvec{e}}}_2), \end{aligned}$$

    whence (3.6)\(_2\).

  3. 3.

    To derive (3.7), recall that:

    $$\begin{aligned} g^1:=\nabla z,\quad z={{\varvec{x}}}\cdot {{\varvec{e}}}_1;\qquad g^2:=\nabla r,\quad r^2=({{\varvec{x}}}\cdot {{\varvec{e}}}_2)^2+({{\varvec{x}}}\cdot {{\varvec{e}}}_3)^2;\qquad g^3:=\nabla \varphi ,\quad \tan \varphi =\frac{{{\varvec{x}}}\cdot {{\varvec{e}}}_3}{{{\varvec{x}}}\cdot {{\varvec{e}}}_2}\,. \end{aligned}$$
  4. 4.

    To derive (3.12), recall that:

    $$\begin{aligned} \begin{aligned}&g^1=\nabla \rho ,\quad \nabla ({{\varvec{x}}}\cdot {{\varvec{x}}})=\nabla (\rho ^2),\quad \rho =|{{\varvec{x}}}|;\\&g^2=\nabla \vartheta ,\quad {{\varvec{e}}}_1=\nabla ({{\varvec{x}}}\cdot {{\varvec{e}}}_1)=\nabla (\rho \cos \vartheta )=(\cos \vartheta )\nabla \rho +\rho \nabla (\cos \vartheta )=\cos \vartheta \,{{\varvec{r}}}-\rho \sin \vartheta \nabla \vartheta ;\\&g^3=\nabla \varphi ,\quad \frac{\cos \varphi {{\varvec{e}}}_3-\sin \varphi {{\varvec{e}}}_2}{\rho \sin \vartheta \cos ^2\varphi }=\frac{({{\varvec{x}}}\cdot {{\varvec{e}}}_2){{\varvec{e}}}_3-({{\varvec{x}}}\cdot {{\varvec{e}}}_3){{\varvec{e}}}_2}{({{\varvec{x}}}\cdot {{\varvec{e}}}_2)^2}=\nabla \big (\frac{{{\varvec{x}}}\cdot {{\varvec{e}}}_3}{{{\varvec{x}}}\cdot {{\varvec{e}}}_2}\big )=\nabla (\tan \varphi )=\frac{1}{\cos ^2\varphi }\nabla \varphi \,. \end{aligned} \end{aligned}$$

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Correspondence to P. Podio-Guidugli .

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© 2014 Springer International Publishing Switzerland

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Podio-Guidugli, P., Favata, A. (2014). Geometric and Analytic Tools. In: Elasticity for Geotechnicians. Solid Mechanics and Its Applications, vol 204. Springer, Cham. https://doi.org/10.1007/978-3-319-01258-2_3

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  • DOI: https://doi.org/10.1007/978-3-319-01258-2_3

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  • Publisher Name: Springer, Cham

  • Print ISBN: 978-3-319-01257-5

  • Online ISBN: 978-3-319-01258-2

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