Question

In a standard proctor test, 1.8 kg of moist soil was filled the mould (volume = 944 cc) after compaction. A soil sample weighing 23 g was taken from the mould and ovendried for 24h at a temperature of ${110}^{\circ }C$. Weight of the dry sample was found to be 20 g. Specific gravity of soil solids is G = 2.7. The theoretical maximum value of the dry unit weight of the soil at that water content is equal to

• A. 4.67 kN/$m^3$
• B. 11.5 kN/$m^3$
• C. 16.26 kN/$m^3$
• D. 18.85 kN/$m^3$

Answer 1

The correct option is D 18.85 kN/$m^3$

Weight of soil sample (dry weight of soil + weight of water) = 23g

weight of dry sample = 20g ; weight of water = 23 - 20

= 3g

water content,

$w=\frac{\text{Weight of water}}{\text{Weight of solids}}\times 100$

$=\frac{3}{20}\times 100$

= 15%

$\text{Specific gravity of soil solids}{G}_s=2.7$

Theoretical maximum value of dry unit weight of the soil at particular water content will be when soil is so compacted that no air void is left within the soil means soil is fully saturated at that water content. Hence degree of saturation, S = 1

${\gamma }_{max}=\frac{G_s{\gamma }_w}{1+e}$

$=\frac{G_s{\gamma }_w}{1+\frac{w{G}_s}{S}}=\frac{2.7\times 9.81}{1+\frac{0.15\times 2.7}{1}}$

$=18.852kN/m^3$