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The Boussinesq Problem

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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 Boussinesq Problem (Joseph Valentin B., 1842-1929) consists in finding the elastic state in a linearly elastic isotropic half-space, subject to a concentrated load applied in a point of its boundary plane and perpendicular to it. This problem has wide geotechnical applications.

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Notes

  1. 1.

    The information items needed for this calculation are:

    $$\begin{aligned} {\varvec{S}}\varvec{g}^1=\sigma {\varvec{r}},\quad {\varvec{S}}\varvec{g}^2={\varvec{S}}\varvec{g}^3={\varvec{0}},\quad {\varvec{S}}{\varvec{h}}=(\sin \vartheta )\sigma {\varvec{r}};\quad {\varvec{r}},_{\vartheta \vartheta }=-{\varvec{r}},\quad {\varvec{r}},_{\varphi \varphi }=-(\sin \vartheta ){\varvec{h}}. \end{aligned}$$

    With this, one finds:

    $$\begin{aligned} {\varvec{S}},_\rho \varvec{g}^1+{\varvec{S}},_\vartheta \varvec{g}^2+{\varvec{S}},_\varphi \varvec{g}^3&=({\varvec{S}}\varvec{g}^1),_\rho +\;({\varvec{S}}\varvec{g}^2),_\vartheta -\rho ^{-1}{\varvec{S}}{\varvec{r}},_{\vartheta \vartheta } +\;({\varvec{S}}\varvec{g}^3),_\varphi -(\rho \sin ^2\vartheta )^{-1}{\varvec{S}}{\varvec{r}},_{\varphi \varphi }\nonumber \\&=(\sigma {\varvec{r}}),_\rho +\;\rho ^{-1}\sigma {\varvec{r}}\;+ (\rho \sin ^2\vartheta )^{-1}(\sin \vartheta ){\varvec{S}}{\varvec{h}},\nonumber \end{aligned}$$

    whence (5.4) easily follows.

  2. 2.

    On differentiating (5.3) with respect to \(\rho \), we quickly find that

    $$\begin{aligned} \int _{-\pi /2}^{+\pi /2}\big ( 2\widehat{\sigma }(\rho ,\vartheta )+\rho \widehat{\sigma },_\rho (\rho ,\vartheta ) \big )\cos \vartheta \,d\vartheta =0. \end{aligned}$$

    We are then driven to choose a mapping \(\widehat{\sigma }\) that satisfies the partial differential equation (5.5).

  3. 3.

    Point \(\vartheta =0\) is the only one in the interval \((-\pi /2,+\pi /2)\) where Eq. (5.7) is singular. The other fundamental solution of this equation being singular at that point is:

    Here is a method to construct this solution. It is not difficult to show that (5.7) is equivalent to

    $$\begin{aligned}(\sin \vartheta \,W(\vartheta ))^\prime =0, \end{aligned}$$

    where

    is the wronskian of \(\widehat{\tau }\) and . Hence, modulo a constant,

    $$\begin{aligned} W(\vartheta )=\frac{1}{\sin \vartheta }\,, \end{aligned}$$

    and the combination of the last two relations yields the following first order ODE for :

    which can be re-written in the form

    The last bit of information needed is:

    $$\begin{aligned} \int \frac{1}{\sin \vartheta \cos ^2\vartheta }=\frac{1}{\cos \vartheta }+\log \tan \frac{\vartheta }{2}.\end{aligned}$$
  4. 4.

    In an attempt to satisfy (2.68) with the field (5.9), it is found that

    $$\begin{aligned} \varDelta \widetilde{\varvec{S}}+\frac{1}{1+\nu }\nabla \nabla (\mathrm tr \,\widetilde{\varvec{S}})\ne {\varvec{0}}. \end{aligned}$$
  5. 5.

    To obtain the last two relations, it is useful to recall that

    $$\begin{aligned} \nabla {\varvec{u}}={\varvec{u}},_z\otimes \ {{\varvec{e}}}_1+{\varvec{u}},_r\otimes \ {\varvec{h}}+r^{-1}{\varvec{u}},_\varphi \otimes \ {\varvec{h}}^\prime , \end{aligned}$$

    and that the physical components of \({\varvec{u}}\) are:

    $$\begin{aligned} u_z:={\varvec{u}}\cdot {{\varvec{e}}}_1,\quad u_r:={\varvec{u}}\cdot {\varvec{h}},\quad u_\varphi :={\varvec{u}}\cdot {\varvec{h}}^\prime . \end{aligned}$$
  6. 6.

    For an alternative way to deduce this condition, one writes (5.13) and (5.14) in the form:

    $$\begin{aligned}(r\sigma _4),_r=-r\sigma _1,z,\quad (r\sigma _4),z=\sigma _3-(r\sigma _2),_r; \end{aligned}$$

    differentiates the first equation with respect to \(z\), the second with respect to \(r\): and finishes by eliminating \((r\sigma _4),_{zr}\).

  7. 7.

    To take the last step in the calculation, use has been made of the following alternative version of (5.13):

    $$\begin{aligned} r^{-1}\sigma _4,_r+\;r^{-2}\sigma _4=-r^{-1}\sigma _1,z. \end{aligned}$$
  8. 8.

    This condition is arrived at by adding (5.34) and (5.35)\(_{1}\) and by taking into account of (5.30).

  9. 9.

    \(R\) is star-shaped if there is a point \(p_0\in R\) such that the line segment from \(p_0\) to any point \(p\in \partial R\) intersects \(\partial R\) only at \(p\) itself.

  10. 10.

    Revert to the footnote in Sect. 5.2.2.

  11. 11.

    On taking both (5.51) and (5.52) into account, (5.40) becomes:

    $$\begin{aligned} f=-2\pi \int _{0}^{+\pi /2}\!\!\!(\cos \vartheta \,\widetilde{\tau }_1(\vartheta )+|\sin \vartheta |\widetilde{\tau }_4(\vartheta ))|\sin \vartheta |d\vartheta =-\pi \alpha _0. \end{aligned}$$
  12. 12.

    In [7], Flamant himself recognizes his debts to Boussinesq.

  13. 13.

    This result follows from the fact that

    $$\begin{aligned}\int \frac{1}{x_2^2+x_3^2}\,dx_3=|x_2|^{-1}\arctan \frac{x_3}{|x_2|}.\end{aligned}$$

References

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  8. Gurtin ME (1972) The linear theory of elasticity. In: Flügge S (ed) Handbuch der Physik, vol VIa/2. Springer, Berlin

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Authors and Affiliations

  1. Department of Civil Engineering and Information Engineering, University of Rome Tor Vergata, via del Politecnico 1, 00173, Rome, Italy

    P. Podio-Guidugli

  2. Institute of Continuum Mechanics and Material Mechanics, Technical University of Hamburg, Eißendorfer Straße 42, 21073, Hamburg, Germany

    A. Favata

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  1. P. Podio-Guidugli
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Correspondence to P. Podio-Guidugli .

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Podio-Guidugli, P., Favata, A. (2014). The Boussinesq Problem. In: Elasticity for Geotechnicians. Solid Mechanics and Its Applications, vol 204. Springer, Cham. https://doi.org/10.1007/978-3-319-01258-2_5

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  • DOI: https://doi.org/10.1007/978-3-319-01258-2_5

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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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