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Density of water, ${\rho}_{w}=1g/c{m}^{3}$
$Bulkdensityofsoil,{\rho}_{b}=\frac{massofsoil}{volumeofsoil}\phantom{\rule{0ex}{0ex}}Here,\phantom{\rule{0ex}{0ex}}volumeofsoil=volumeofmould=1000c{m}^{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Drydensityofsoil,{\rho}_{d}=\frac{{\rho}_{b}}{1+w}=\frac{G{\rho}_{w}}{1+{\displaystyle \frac{wG}{S}}}=\frac{(1-{n}_{a})G{\rho}_{w}}{1+wG}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}ForZeroairvoidline,Airvoid\left({n}_{a}\right)=0\phantom{\rule{0ex}{0ex}}Therefore,Drydensityforzeroairvoidline,{(\rho}_{}$_{d}
Mass of wet soil + Mould | Mass of mould |
Mass of wet soil (g) |
Volume of mould (cm^{3}) |
Bulk Density, ${\rho}_{b}$ (g/cm^{3}) |
Water content, w (%) |
Dry density, ${\rho}_{d}$ (g/cm^{3}) |
(ρ_{d})_{ZAV} (g/cm^{3}) |
(kg) | (kg) | ||||||
3.69 | 2 | 1690 | 1000 | 1.69 | 6.2 | 1.04 | 2.33 |
3.75 | 2 | 1750 | 1000 | 1.75 | 8 | 1.62 | 2.23 |
3.82 | 2 | 1820 | 1000 | 1.82 | 9.8 | 1.65 | 2.15 |
3.92 | 2 | 1920 | 1000 | 1.92 | 11.5 | 1.72 | 2.07 |
3.98 | 2 | 1980 | 1000 | 1.98 | 12.4 | 1.76 | 2.03 |
3.93 | 2 | 1930 | 1000 | 1.93 | 13.3 | 1.703 | 1.99 |
(a)
(b)
Optimum moisture content = 12.4 %
(c)
Maximum dry density =1.76 g/cm^{3 }=1.76 x 10^{-3} kg/cm^{3}
(d)
Zero air void line has been plotted in the above figure.
(e)
Let the degree of saturation, water content and the void ratio at maximum dry density be S, w and e respectively.
Now,
${\rho}_{d}=\frac{G{\rho}_{w}}{1+{\displaystyle \frac{wG}{S}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1.76=\frac{2.72x1}{1+{\displaystyle \frac{12.4x2.72}{S}}}\phantom{\rule{0ex}{0ex}}Solvingtheaboveequation\phantom{\rule{0ex}{0ex}}S=61.83\%$
Therefore, the required degree of saturation at maximum dry density is 61.83%.