3.10: Example 3.1
- Page ID
- 123402
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)An H = 4 m tall water tank with radius rexternal = 5 m is going to be founded on a ring foundation, with the dimensions depicted in the following plan view. Assuming the tank is filled with water, determine the vertical stress due to the weight of the tank below the centre of the foundation, at a depth z = 5 m.
Answer
First we have to determine the maximum load, Qext on the foundation, assuming that the tank is filled with water:
\[A_{\text {tank }}=\underline{n r_{\text {externa }} I^2}=78.54 \mathrm{~m}^2 \nonumber\]
\[Q_{\text {ext }}=A \times H \times y_{\mathrm{w}}=78.54 \mathrm{~m}^2 \times 4 \mathrm{~m} \times 10 \mathrm{kN} / \mathrm{m}^3=3141 \mathrm{kN}\]
where 10 kN/m3 is the unit weight of water.
The area of the ring footing on which the load is applied is:
\[A_{\text {footing }}=n r_{\text {external }} 2^2-n r_{\text {internal }} 2=78.54 \mathrm{~m}^2-50.26 \mathrm{~m}^2=28.28 \mathrm{~m}^2 \nonumber\]
thus the pressure applied on the ring foundation of the tank will be:
\[q_{\text {ext }}= \dfrac{Q_{\text {ext }} }{A_{\text {footing }}}=111 \mathrm{kPa}\]
Employing the principle of superposition discussed in Chapter 3.5, this ring pressure is equivalent, in terms of stresses applied to the soil, to a circular pressure qext,1 = 111 kPa on a radius rexternal = 5 m, plus a circular pressure qext,2 = -111 kPa on a radius rinternal = 4 m (see also Figure 3.14). Negative pressure values correspond to tension. The additional vertical stress in the soil below the axis of symmetry of a circular pressure is provided by Equation \ref{3.13}. At depth z = 5 m, the positive pressure of radius rexternal = 5 m will result in a compressive stress:
\[ \Delta \sigma_{z, \text { ext }}=q_{\text {ext }, 1}\left[1-\left(\frac{1}{1+\left(\frac{r_{\text {external }}}{z}\right)^2}\right)^{\frac{3}{2}}\right]=111\left[1-\left(\frac{1}{1+\left(\frac{5}{5}\right)^2}\right)^{\frac{3}{2}}\right]= 71.75 \mathrm{kPa} \]
while the negative pressure of radius rinternal = 4 m in a tensile stress:
\[ \Delta \sigma_{z, \text { int }}=q_{\text {ext }, 2}\left[1-\left(\frac{1}{1+\left(\frac{r_{\text {internal }}}{z}\right)^2}\right)^{\frac{3}{2}}\right]=-111\left[1-\left(\frac{1}{1+\left(\frac{4}{5}\right)^2}\right)^{\frac{3}{2}}\right]= -58.15 \mathrm{kPa}\]
The total stress increment due to the loading applied on the ring foundation is found from Equation \ref{3.21} as:
\[\Delta \sigma_z=\Delta \sigma_{z, e x t}+\Delta \sigma_{z, \text { int }}=13.6 \mathrm{kPa}\]