Volumetric Relationships :
1 Answer

The following five volumetric relationship is only widely used in soil Engineering.

(1) Void Ratio (e) =Volume of voids/ Volume of solids.

$\boxed{e = \dfrac{V_v}{V_s}}$

The void ratio is expressed in the decimal. Ex:- 0.4,0.5

(2) $Porosity(\mu)$ = Volume of voids/Total Volume

$\boxed{\mu = \dfrac{V_v}{V}}$

Porosity is expressed in the %.

However in equations,it is used as ratio, for example, a porosity of 50% will be used as 0.5 in equation.

The porosity of soil cannot exceeds 100% as it would mean Vv(Volume of Voids) is greater than V, which is absurd, it will have much smaller value.

An inter-relationship can be found between the void ratio and porosity as under.

$\dfrac{1}{\eta} = \dfrac{V}{V_v} = \dfrac{V_{v} + V_s}{V_v}$


$\dfrac{1}{\eta} = 1 + \dfrac{1}{e} = \dfrac{1+e}{e}$ ...(a)


$\eta = \dfrac{e}{1+e}$

also from Eqn (a),

$\dfrac{1}{e} = \dfrac{1}{\eta} - 1 = \dfrac{1-\eta}{\eta}$


$\boxed{e = \dfrac{\eta}{1-\eta}}$

(3) Degree of saturation:- (S)

$S_r$ or S = $\dfrac{Volume of water}{Volume of voids} = S = \dfrac{V_w}{V_v}$

The degree of saturation is generally expressed in %.

It is equal to zero when is absolutely dry and equal to 100% when the soil is fully saturated.

(4) Percentage Air Voids$(\eta_a)$

It is the ratio of volume of air to the total volume.

Thus, $\eta_a = \dfrac{V_a}{V}$

As the name indicates, it is represented as a%.

(5) Air content :-

It is the ratio of volume of air to the volume of voids.

Thus, $a_c$ = $\dfrac{V_a}{V_v}$

Air content is usually expressed in %

Both air content, the % air voids are zero when the soil is saturated $(V_a = 0) As inter relationship can be found between the void ratio and the porosity as under. From Eqn of Porosity, we get $\eta = \dfrac{V_v}{V}$ But, $\dfrac{1}{\eta} = \dfrac{V}{V_v}$ But total (V) = Volume of voids + Volume of Solids $\dfrac{1}{\eta} = \dfrac{V_v + V_s}{V_v}$ or $\dfrac{1}{\eta} = \dfrac{V_v}{V_v} + \dfrac{V_s}{V_v}$ $\dfrac{1}{\eta} = 1 + \dfrac{V_s}{V_v}$ $e = \dfrac{V_v}{V_s}$ $\dfrac{1}{\eta} = 1 + \dfrac{1}{\eta}$ $\dfrac{1}{\eta} = \dfrac{e+1}{e}$ $\eta = \dfrac{e}{e+1}$ or $\eta = \dfrac{e}{1+e}$

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