The moisture content of an undisturbed sample of clay belonging to a volcanic region is 265% under 100% saturation. The specific gravity of the solids is 2.5. The dry unit weight is 21 Ib/ft^3. Determine (i) the saturated unit weight, (ii) the submerged unit weight, and (iii) void ratio.

Question

The moisture content of an undisturbed sample of clay belonging to a volcanic region is 265% under 100% saturation. The specific gravity of the solids is 2.5. The dry unit weight is 21 [latex]lb / ft ^{3}[/latex]. Determine (i) the saturated unit weight, (ii) the submerged unit weight, and (iii) void ratio.

Accepted Answer

(i) Saturated unit weight, [latex]\gamma_{ ext {sat }}=\gamma_{t}[/latex]

[latex]W=W_{w}+W_{s}=w W_{s}+W_{s}=W_{s}(1+w)[/latex]

From Fig. Ex. 3.3, [latex]\gamma_{t}=\frac{W}{V}=\frac{W}{1}=W[/latex]. Hence

[latex]\gamma_{t}=21(1+2.65)=21 imes 3.65=76.65 lb / ft ^{3}[/latex]

(ii) Submerged unit weight, [latex]\gamma_{b}[/latex]

[latex]\gamma_{b}=\gamma_{ sat }-\gamma_{w}=76.65-62.4=14.25 lb / ft ^{3}[/latex]

(iii) Void ratio, e

[latex]V_{s}=\frac{\gamma_{d}}{G_{s} \gamma_{w}}=\frac{21}{2.5 imes 62.4}=0.135 ft ^{3}[/latex]

Since 5 = 100%

[latex]V_{v}=V_{w}=\frac{w imes W_{s}}{\gamma_{w}}=2.65 imes \frac{21}{62.4}=0.89 ft ^{3}[/latex]

[latex]e=\frac{V_{v}}{V_{s}}=\frac{0.89}{0.135}=6.59[/latex]

Question 3.3: The moisture content of an undisturbed sample of clay belong...

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