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Analytical Chem Soil HW Solutions 12

# Analytical Chem Soil HW Solutions 12 - Q1.9 Solution We...

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12 Compaction Q1.9 The following results were obtained from a standard (2.5 kg) Proctor compaction test: Mass of tin empty, g 14 14 14 14 14 Mass of tin + sample wet, g 88 68 98 94 93 Mass of tin + sample dry, g 81 62 87 82 80 Density, kg/m 3 1730 1950 2020 1930 1860 Plot a graph to determine (i) the maximum dry density, (ii) the optimum water content and (iii) the actual density at the optimum water content. If the particle relative density (grain specific gravity) G s = 2.65, calculate (iv) the specific volume and (v) the saturation ratio at the maximum dry density.
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Unformatted text preview: Q1.9 Solution We need to plot a graph of water content w against dry density ρ dry , where w = m w /m s (main text Equation 1.5) and dry = /(1+w) (main text Equation 1.27) The water content of each sample is calculated as in Example E1.1: ( ) ( ) { } ( ) { } ) ( t m s m t m s m t m w m s m t m s m w m w − + + − + + = = where (m t ) = mass of tin, empty (m t + m s + m w ) = mass of tin + wet soil sample (m t +m s ) = mass of tin + dry soil sample...
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