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B_{e}= equivalent foundation diameter =\sqrt{\frac{4 B L}{\pi}}=\sqrt{\frac{4 \times 3 \times 2}{7}}=2.764 \mathrm{~m}\begin{array}{l}9 .=125 \mathrm{kr} / \mathrm{m}^{2} \\H=3 m \\S_{e}=\frac{9_{0} B_{e} I_{a} I_{F} \cdot I_{E}}{E}\left(1-\mu^{2}\right) \\E_{f}=25 \times 10^{3} \mathrm{mN} / \mathrm{m}^{2} \\E_{0}=12 \mathrm{mN} / \mathrm{m}^{2}, K=400 \mathrm{kN} / \mathrm{m}^{2} / \mathrm{m}, \mu=\overline{0} .2 \\\end{array}Calculation of InFrom graph of In vs \betaFor, \frac{H}{B_{e}}=\frac{3}{2.764}=1.08\beta=\frac{E_{0}}{K B_{e}} \Rightarrow \beta=\left(\frac{12 \times 10^{3} \mathrm{kN1/ \textrm {m } ^ { 2 }}}{400 \times 2.769} \Rightarrow \beta=10.854\right.\begin{array}{l}I_{n}=f\left(\frac{H}{B_{e}}, \beta\right)^{=1} \\I_{n}=f(1.0,10.854) \\I_{n}=0.6\end{array}\begin{array}{l}I_{F}=\frac{\pi}{4}-\frac{1}{4.6+10\left[\frac{E_{f}}{E_{\sigma}+\frac{B_{e}}{2} k}\right]\left[\frac{2 t}{B_{e}}\right]^{3}} \\t=0.25 \mathrm{~m} \\I_{F}=\frac{n}{4}-\frac{1}{4.6+10\left[\frac{25 \times 10^{3} 1}{12+\frac{2.769}{2} \times 0.4}\right]\left[\frac{2 \times 0.25}{2.769}\right]^{3}} \\I_{F}=0.777 \\\end{array}\begin{array}{l}I_{E}=1-\frac{1}{3.5 \times \exp (1.22 \times 0.2-0.4) \times\left(\frac{B_{e}}{D_{f}}+1.6\right)} \\I_{E}=1-\frac{1}{3.5 \times \exp (1.22 \times 0.2-0.4)\left(\frac{2.764}{1}+1.6\right)}=0.9235 \\S_{e}=\frac{125 \times 2.764 \times 0.6 \times 0.777 \times 0.9235 \times\left(1-0.2^{2}\right)}{12 \times 10^{3} \mathrm{kN1/ \textrm {m } ^ { 2 }}} \\S_{e}=0.0119 \mathrm{~m}=11.9 \mathrm{~mm}\end{array}elastic settlement is instantaneous. Influence factors are calculated using empirical relationship.   ...