cohesive soil and non cohesive soil

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written 4.8 years ago by

shubhamshukla • 40

Example:- Porosity (n) = 0.7, degree of saturation = 40= 0.4

vol. of soil = 100$m^3$

To find:- Vol. of air.

Porosity $\eta = \dfrac{V_v}{V}$

$0.7\times100 = V_v$

$V_v = 70$

$a_c = ?$

$a_c+s = 1$

$a_c = 1-0.4$ $= 0.6$

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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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