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cohesive soil and non cohesive soil
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written 5.4 years ago by |
Example:- Porosity (n) = 0.7, degree of saturation = 40= 0.4
vol. of soil = 100$m^3$
To find:- Vol. of air.
Porosity $\eta = \dfrac{V_v}{V}$
$0.7\times100 = V_v$
$V_v = 70$
$a_c = ?$
$a_c+s = 1$
$a_c = 1-0.4$ $= 0.6$
Example A 588$cm^3$ vol of moist sand weights 1010gm. Its dry weight is 918gm and specific gravity of solids. G is 2.67. Asuuming density of water 1gm/$cm^3$ , the void ratio.
Answer:$V = 588 cm^3$
$W_d = 918gm$
$w = 1010gm$
$g = 2.67$
$\gamma_w = 1gm/cm^3$
$\gamma_d = \dfrac{G\gamma_w}{1+e}$
$\gamma_d = \dfrac{2.67\times 19}{1+e}$
$\gamma_d = \dfrac{\gamma_s}{V} = \dfrac{918}{588} = 1.56$
$1.56 = \dfrac{2.67\times 19}{1+e}$
$e=0.71$
SR. NO. | Relationship in mass Density | Relationship in weights |
---|---|---|
1 | $\eta = \dfrac{e}{1+e}$ | $\eta = \dfrac{e}{1+e}$ |
2 | $e = \dfrac{\eta}{1-\eta}$ | $e = \dfrac{\eta}{1-\eta}$ |
3 | $\eta a = \eta a_c$ | $\eta a = \eta a_c$ |
4 | $\rho = \dfrac{(G+S_e)\rho_w}{1+e}$ | $\gamma = \dfrac{(G+S_e)\gamma_w}{1+e}$ |
5 | $\rho_d = \dfrac{G\rho_w}{1+e}$ | $\gamma_d = \dfrac{G\gamma_w}{1+e}$ |
6 | $\rho_{sat} = \dfrac{(G+e)\rho_w}{1+e}$ | $\gamma_{sat} = \dfrac{(G+e)\gamma_w}{1+e}$ |
7 | $\rho^! = \dfrac{(G-1)\rho_w}{1+e}$ | $\gamma^! = \dfrac{(G-1)\gamma_w}{1+e}$ |
8 | $e = \dfrac{WG}{s}$ | $e = \dfrac{WG}{s}$ |
9 | $\rho_d = \dfrac{\rho}{1+w}$ | $\gamma_d = \dfrac{\gamma}{1+w}$ |
$\rho_w = 1000Kg/m^3$ = density of water
$\gamma_w = 9810N/m^2 = 9.81kN/m^3$ = unit weight of water
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