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**Important Relationships**

Followings are the important relationships between the various quantities defined in the previous section.

\[n={e \over {1 + e}}\] (2.11)

\[{V_s}={V \over {1 + e}}\] (2.12)

\[{V_v}={e \over {1 + e}}V\] (2.13)

\[e={{{G_s}w} \over S}\] (2.14)

\[{\gamma _{bulk}}={\gamma _t}={{{G_s} + Se} \over {1 + e}}{\gamma _w}\] (2.15)

For completely saturated soil, S =1, Thus,

\[{\gamma _{sat}}={{{G_s} + e} \over {1 + e}}{\gamma _w}\] (2.16)

For completely dry soil, S =0, Thus,

\[{\gamma _d}={{{G_s}} \over {1 + e}}{\gamma _w}\] (2.17)

\[\gamma '={{{G_s} - 1} \over {1 + e}}{\gamma _w}\] (2.18)

\[{\gamma _d}={{{\gamma _{bulk}}} \over {1 + w}}\] (2.19)

**Problem 1: In a partially saturated soil, moisture or water content is 20% and \[{\gamma _{bulk}}\] = 18 kN/m ^{3}. Determine the degree of saturation and void ratio. G_{s }= 2.65. Take the unit weight of the water as 10 kN/m^{3}.**

**Solution: **

\[{\gamma _d}={{{\gamma _{bulk}}} \over {1 + w}}={{18} \over {1 + 0.2}} = 15\;kN/{m^3}\]

\[{\gamma _d}={{{G_s}} \over {1 + e}}{\gamma _w}\]

Thus, \[e={{{G_s}{\gamma _w}} \over {{\gamma _d}}}-1={{2.65 \times 10} \over {15}}-1=0.767\]

Thus, the void ratio is 0.767.

\[e={{{G_s}w} \over S}\]

Thus, \[S={{{G_s}w} \over e}={{2.65 \times 0.2} \over {0.77}}=0.691\]

Thus, the degree of saturation is 69.1%.

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