Question A dense silt clay has the following properties: Void ratio e=0.30 Effective diameter D10=10 𝜇m Capillary constant C=0.20cm2 Free ground water level is 8.0 m below the ground surface. a) Find the capillary rise in the silt. b) Find the vertical effective stress in kPa at 5 m depth. Assume 𝛾s=26.5 kN/m3 and that the soil above the capillary rise and ground surface is partially saturated at 50%. c) Find the vertical effective stress at depth 10 m depth. Assume 𝛾s=26.5 kN/m3 and that the soil above the capillary action rise and ground surface is partially saturated at 50%.

Capillary rise can be calculated using allen hazen , please refer image for solution.  llsing Alun Hazen equation\begin{array}{rlr}h_{i} & =\frac{c}{e \cdot D_{10}} \\h_{c} & =\frac{0.2 \mathrm{~cm}^{2}}{0.3 \times 10 \times 10^{-4} \mathrm{~cm}} \\h_{c} & =666.66 \mathrm{~cm}\end{array}Given. r_{s}=26.5 \mathrm{kN} / \mathrm{m}^{3}\begin{array}{l}\gamma_{b}=21.53 \mathrm{kN} / \mathrm{m}^{3} \\r_{\text {sat }}=\left(\frac{u+e}{1+e}\right) \gamma_{w} \\=\frac{(2.65+0.3) \times 10}{1+0.3} \\\end{array}\begin{array}{l} \gamma_{\text {sat }}=22.70 \mathrm{kN} / \mathrm{m}^{3} \\ \therefore \text { upto } 5 \mathrm{~m} \text { depth }\end{array}eind capillary ueis\begin{array}{l} \text { rsat }=22.70 \mathrm{kN} / \mathrm{m}^{3} \\\therefore \text { uplo } 5 \mathrm{~m} \text { depth there is not any uwit } \\\text { end capilcary reis } \\\therefore \text { Effechve shess }(\bar{\sigma})=\text { Total stress - porewaler } \\\text { pressiere } \\=\sigma-4\end{array}\begin{aligned}& =\sigma-4 \\& =\sigma-0 \\\sigma & =H \times \gamma b=5 \times 21.53 \\& =107.65 \mathrm{kN} / \mathrm{m}^{2} \\& =107.65 \mathrm{kPa} .\end{aligned}(b)(c) Here, r_{\text {sat }}=22.70 \mathrm{kN} / \mathrm{m}^{3}r_{b}=21.53 \mathrm{kN} / \mathrm{m}^{3}\therefore Effechve stress at 10 \mathrm{~m} depth.\begin{aligned}\sigma & \bar{\sigma}=\sigma-u \\& =7.333 \times \gamma_{b}+0.667 \times \gamma_{\text {sat }}+2 \times r_{\text {sat }} \\& =218.42 \mathrm{kN} / \mathrm{m}^{2} \\u & =2 \times \gamma_{w} \\& =2 \times 10 \\& =20 \mathrm{kN}^{2}\left(\mathrm{~m}^{2}\right.\end{aligned}Effective stress, (\bar{\sigma})\begin{aligned}\sigma & =\sigma-4 \\& =218.42-20 \\& =198.42 \mathrm{kN} / \mathrm{m}^{2} \\\sigma & =198.42 \mathrm{kPa}\end{aligned}(c) ...