A 27.50 lb soil sample has a volume of 0.220 ft^3 a moisture content of 10.2%, and a specific gravity of solids of 2.65. Compute the unit weight, dry unit weight, degree of saturation, void ratio, and porosity.

Question

A 27.50 lb soil sample has a volume of [latex]0.220 ft^{3}[/latex] a moisture content of 10.2%, and a specific gravity of solids of 2.65. Compute the unit weight, dry unit weight, degree of saturation, void ratio, and porosity.

Accepted Answer

Solution using fundamental and derived equations

[latex]\gamma =\frac{W}{V} =\frac{27.50 lb}{0.220 ft^{3}} =125.0 lb/ft^{3} ightarrow 125 lb/ft^{3}[/latex]

[latex]\gamma _{d}=\frac{\gamma }{1+w} =\frac{125.0 lb/ft^{3}}{1+0.102} =113.4 lb/ft^{3} ightarrow 113 lb/ft^{3}[/latex]

[latex]S=\frac{w}{\frac{\gamma _{w}}{\gamma _{d}} -\frac{1}{G_{s}} } imes 100[/latex]% [latex]=\frac{0.102}{\frac{62.4 lb/ft^{3}}{113.4 lb/ft^{3}} -\frac{1}{2.65} } imes 100[/latex]% [latex]=59[/latex]%

[latex]e=\frac{G_{s}\gamma _{w}}{\gamma _{d}} -1=\frac{\left(2.65 ight) \left(62.4 lb/ft^{3} ight) }{113.4 lb/ft^{3}} -1=0.4582 ightarrow 0.458[/latex]

[latex]n=\frac{e}{1+e} imes 100[/latex]% [latex] =\frac{0.4582}{1+0.4582} imes 100[/latex]% [latex]=31.4[/latex]%

Solution using a phase diagram

Although the fundamental and derived equations were sufficient to solve this problem, and would be the easiest method, we also will illustrate a solution using a phase diagram.

Step 1: Draw and annotate a phase diagram (see Figure 4.4).

Step 2: Assume dimension.

Not applicable to this problem, because weights and volumes have been given.

Step 3: Determine dimensions in phase diagram.

[latex]W_{s}=\frac{W}{1+w} =\frac{27.50 lb}{1+0.102} =24.95 lb[/latex]

[latex]W_{w}=W - W_{s}=27.50 lb-24.95 lb=2.55 lb[/latex]

[latex]G_{s}=\frac{W_{s}}{V_{s}\gamma _{w}} [/latex]

[latex]2.65=\frac{24.95 lb}{V_{s}\left(62.4 lb/ft^{3} ight) } ightarrow V_{s}=0.1509 ft^{3}[/latex]

[latex]\gamma _{w}=\frac{W_{w}}{V_{w}}[/latex]

[latex]62.4 lb/ft^{3}=\frac{2.55 lb}{V_{w}} ightarrow V_{w}=0.0409 ft^{3}[/latex]

[latex]V_{a}=V - V_{w} - V_{s}=0.220 ft^{3} - 0.0409ft^{3} - 0.1509 ft^{3}=0.0282 ft^{3}[/latex]

Step 4: Compute parameters.

[latex]\gamma =\frac{W}{V} =\frac{27.50 lb}{0.220 ft^{3}} =125 lb/ft^{3}[/latex]

[latex]n=\frac{V_{v}}{V} imes 100[/latex]% [latex]=\frac{0.0282 ft^{3}+0.0409 ft^{3}}{0.220 ft^{3}} imes 100[/latex]% [latex]=31.4[/latex]%

[latex]e=\frac{V_{v}}{V_{s}} =\frac{0.0409 ft^{3}+0.0282 ft^{3}}{0.1509 ft^{3}} =0.458[/latex]

[latex]\gamma _{d}=\frac{W_{s}}{V} =\frac{24.95 lb}{0.220 ft^{3}} =113 lb/ft^{3}[/latex]

[latex]S=\frac{V_{w}}{V_{v}} imes 100[/latex]% [latex]=\frac{0.0409 ft^{3}}{0.0409 ft^{3}+0.0282 ft^{3}} imes 100[/latex]% [latex]=59[/latex]%

Question 4.1: A 27.50 lb soil sample has a volume of 0.220 ft^3 a moisture...

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