{"id":277,"date":"2022-01-13T23:17:39","date_gmt":"2022-01-13T23:17:39","guid":{"rendered":"https:\/\/books.gw-project.org\/land-subsidence-and-its-mitigation\/chapter\/exercise-1-solution\/"},"modified":"2022-01-17T19:52:01","modified_gmt":"2022-01-17T19:52:01","slug":"exercise-1-solution","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/land-subsidence-and-its-mitigation\/chapter\/exercise-1-solution\/","title":{"raw":"Exercise 1 Solution","rendered":"Exercise 1 Solution"},"content":{"raw":"
\r\n

Let\u2019s denote with \u03b3<\/em>w<\/em><\/sub> = 1000 kg\/m3<\/sup> the water specific weight and with \u03b3<\/em>grain<\/em><\/sub> = 2700 kg\/m3<\/sup> the specific weight of the solid. The specific weight of the saturated clay \u03b3<\/em>s<\/em><\/sub> can be computed as follows:<\/p>\r\n

\u03b3<\/em>s<\/em><\/sub> = \u03d5<\/em> \u03b3<\/em>w<\/em><\/sub> + (1 \u2212 \u03d5<\/em>)\u03b3<\/em>grain<\/em><\/sub><\/span><\/p>\r\n

\u03b3<\/em>s<\/em><\/sub> = 0.4 (1000) + (1 \u2212 0.4)2700 = 2020 kg\/m3<\/sup><\/span><\/p>\r\n

With a 6 m deep lake, pressure p<\/em> at 15 m below the lake bottom amounts to<\/p>\r\n

p<\/em> = (6 + 15) m 1000 kg\/m3<\/sup> = 21,000 kg\/m2<\/sup><\/span><\/p>\r\n

Pressure is often reported in bars. There are 9.8067 \u00d7 10-<\/sup>5<\/sup> bars per 1 kg\/m2<\/sup><\/span>, so<\/p>\r\n

p<\/em> = 21,000 kg\/m2<\/sup> = 2.06 bars.<\/span><\/p>\r\nThe <\/span>geostatic stress is:\r\n

\u03c3<\/em>c<\/em><\/sub> = 6 m 1000 kg\/m3<\/sup> + 15 m 2020 kg\/m3<\/sup> = 6000 kg\/m2<\/sup> + 30,300 kg\/m2<\/sup> = 36,300 kg\/m2<\/sup><\/span><\/p>\r\n

\u03c3<\/em>c<\/em><\/sub> = 3.56 bar<\/p>\r\n

Rearranging Terzaghi\u2019s principle (Equation 3), the effective vertical stress \u03c3<\/em>z<\/em><\/sub> is equal to:<\/p>\r\n

\u03c3<\/em>z<\/em><\/sub> = \u03c3<\/em>c<\/sub> \u2212 p<\/em> = 3.56 bar \u2212 2.06 bar = 1.50 bar<\/span><\/p>\r\n

If the water level drops to 4 m, both the pressure and the total vertical stress reduce the same amount ([latex]2\\ \\textup{m}\\ 1000\\frac{\\textup{kg}}{\\textup{m}^{3}}\\frac{9.8067\\times 10^{-5}\\textup{bars}}{1\\ \\textup{kg}\/\\textup{m}^{2}}[\/latex] = 0.2 bar). Hence, \u03c3<\/em>z<\/em><\/sub> does not vary.<\/p>\r\n

Return to Exercise 1<\/span><\/a><\/p>\r\n

<\/p>\r\n\r\n<\/div>","rendered":"

\n

Let\u2019s denote with \u03b3<\/em>w<\/em><\/sub> = 1000 kg\/m3<\/sup> the water specific weight and with \u03b3<\/em>grain<\/em><\/sub> = 2700 kg\/m3<\/sup> the specific weight of the solid. The specific weight of the saturated clay \u03b3<\/em>s<\/em><\/sub> can be computed as follows:<\/p>\n

\u03b3<\/em>s<\/em><\/sub> = \u03d5<\/em> \u03b3<\/em>w<\/em><\/sub> + (1 \u2212 \u03d5<\/em>)\u03b3<\/em>grain<\/em><\/sub><\/span><\/p>\n

\u03b3<\/em>s<\/em><\/sub> = 0.4 (1000) + (1 \u2212 0.4)2700 = 2020 kg\/m3<\/sup><\/span><\/p>\n

With a 6 m deep lake, pressure p<\/em> at 15 m below the lake bottom amounts to<\/p>\n

p<\/em> = (6 + 15) m 1000 kg\/m3<\/sup> = 21,000 kg\/m2<\/sup><\/span><\/p>\n

Pressure is often reported in bars. There are 9.8067 \u00d7 10–<\/sup>5<\/sup> bars per 1 kg\/m2<\/sup><\/span>, so<\/p>\n

p<\/em> = 21,000 kg\/m2<\/sup> = 2.06 bars.<\/span><\/p>\n

The <\/span>geostatic stress is:<\/p>\n

\u03c3<\/em>c<\/em><\/sub> = 6 m 1000 kg\/m3<\/sup> + 15 m 2020 kg\/m3<\/sup> = 6000 kg\/m2<\/sup> + 30,300 kg\/m2<\/sup> = 36,300 kg\/m2<\/sup><\/span><\/p>\n

\u03c3<\/em>c<\/em><\/sub> = 3.56 bar<\/p>\n

Rearranging Terzaghi\u2019s principle (Equation 3), the effective vertical stress \u03c3<\/em>z<\/em><\/sub> is equal to:<\/p>\n

\u03c3<\/em>z<\/em><\/sub> = \u03c3<\/em>c<\/sub> \u2212 p<\/em> = 3.56 bar \u2212 2.06 bar = 1.50 bar<\/span><\/p>\n

If the water level drops to 4 m, both the pressure and the total vertical stress reduce the same amount (\"2\ \textup{m}\ 1000\frac{\textup{kg}}{\textup{m}^{3}}\frac{9.8067\times 10^{-5}\textup{bars}}{1\ \textup{kg}/\textup{m}^{2}}\" = 0.2 bar). Hence, \u03c3<\/em>z<\/em><\/sub> does not vary.<\/p>\n

Return to Exercise 1<\/span><\/a><\/p>\n

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