Bouyant/Submerged Soil (Effective Unit Weight)
Problem 1. In an undisturbed one cubic foot volume of soil obtained from a test pit is found to have a wet weight. of 103.2 lbs. The dry weight of the sample 84.5 lbs. What would be the effective unit weight of such a soil if were submerged below water table. The specific gravity of the soil is determined to be 2.65 and void ratio
Given data-
The volume of soil sample (V) = 1 ft^{3}
Wet weight (W_{w}) = 103.2 lbs
Dry weight (W_{d}) = 84.5 lbs
Specific gravity (G) = 2.65
Solution-
$\mathbf{Dry}\mathbf{}\mathbf{Unit}\mathbf{}\mathbf{Weight}\mathbf{-}\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{d}}=\frac{{\mathrm{W}}_{\mathrm{d}}}{\mathrm{V}}=\frac{84.5}{1}\phantom{\rule{0ex}{0ex}}\overline{){\mathrm{\gamma}}_{\mathrm{d}}=84.5\mathrm{lbs}/{\mathrm{ft}}^{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{d}}=\frac{{\mathrm{G\gamma}}_{\mathrm{w}}}{1+\mathrm{e}}\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{w}}=\mathrm{Unit}\mathrm{weight}\mathrm{of}\mathrm{water}=62.4\mathrm{lbs}/{\mathrm{ft}}^{3}\phantom{\rule{0ex}{0ex}}84.5=\frac{2.65\times 62.4}{1+\mathrm{e}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{e}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9569}}\phantom{\rule{0ex}{0ex}}\mathbf{Hence}\mathbf{,}\mathbf{}\mathbf{The}\mathbf{}\mathbf{void}\mathbf{}\mathbf{ratio}\mathbf{}\mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{9569}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathbf{Effective}\mathbf{}\mathbf{Unit}\mathbf{}\mathbf{Weight}\mathbf{-}\phantom{\rule{0ex}{0ex}}\mathrm{\gamma}\text{'}={\mathrm{\gamma}}_{\mathrm{saturated}}-{\mathrm{\gamma}}_{\mathrm{w}}\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{saturated}}=\frac{\left(\mathrm{G}+\mathrm{e}\right){\mathrm{\gamma}}_{\mathrm{w}}}{\left(1+\mathrm{e}\right)}\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{saturated}}=\frac{\left(2.65+0.9569\right)62.4}{\left(1+0.9569\right)}\phantom{\rule{0ex}{0ex}}{\mathrm{\gamma}}_{\mathrm{saturated}}=115.01\mathrm{lbs}/{\mathrm{ft}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{\gamma}\text{'}={\mathrm{\gamma}}_{\mathrm{saturated}}-{\mathrm{\gamma}}_{\mathrm{w}}\phantom{\rule{0ex}{0ex}}\mathrm{\gamma}\text{'}=\left(115.01-62.4\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{\gamma}\mathbf{\text{'}}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{61}\mathbf{}\mathbf{lbs}\mathbf{/}{\mathbf{ft}}^{\mathbf{3}}}\phantom{\rule{0ex}{0ex}}\mathbf{Hence}\mathbf{,}\mathbf{}\mathbf{Effective}\mathbf{}\mathbf{Unit}\mathbf{}\mathbf{Weight}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{61}\mathbf{}\mathbf{lbs}\mathbf{/}{\mathbf{ft}}^{\mathbf{3}}\phantom{\rule{0ex}{0ex}}$