Question

A sand layer found at sea floor under 20m water depth is characterized with relative density = 40% maximum void ratio = 1.0, minimum void ratio = 0.5, and specific gravity of soil solids = 2.67. Assume the specific gravity of sea water to be 1.03 and the unit weight of fresh water to be 9.81 kN/$m^3$

What would be the effective stress (rounded off to the nearest integer value of kPa) at 30m depth into the sand layer?

• A. 77 kPa
• B. 273 kPa
• C. 268 kPa
• D. 281 kPa

Answer 1

The correct option is C 268 kPa

Relative density,

$D_r=40$


Maximum void ratio,

$e_{max}=1$

Minimum void ratio

$e_{min}=0.5$

Specific gravity of soil solids,

$G_s=2.67$

Specific gravity of sea water = 1.03

Unit weight of sea water

= specific gravity $\times$ ${\gamma }_{freshwater}$

${\gamma }_{sea}=1.03\times 9.81$

= 10.1 kN/$m^3$

Relative density = $\frac{e_{max}-e_{natural}}{e_{max}-e_{min}}$

$\Rightarrow 0.4=\frac{1-e}{1-0.5}$

$\Rightarrow e=0.8$

submerged unit weight,

${\gamma }_{sub}=\left(\frac{G_s-1}{1+e}\right){\gamma }_w$

In this formula, $G_s$ is with respect to fresh water but sand is submerged in sea water, so value of specific gravity of solids w.r.t sea water

$G_s{}^{\prime }=\frac{G_s}{Specificgravityofseawater}$

$=\frac{G_s}{1.03}=\frac{2.67}{1.03}=2.592$

$\therefore$ Submerged unit weight in sea water

${\gamma }_{sub}=\left(\frac{G_s{}^{\prime }-1}{1+e}\right){\gamma }_{sea}$

$=\left(\frac{2.592-1}{1+0.8}\right)\times 10.1$

= 8.934 kN/$m^3$

Effective stress at 30 m depth into sand layer

$\overline{\sigma }={\gamma }_{sub}\times H$

= 8.934 $\times$ 30

= 268.02

= 268 kPa

Alternative :


Weight of sea water,

$w_w=e\times unitwt.ofseawater$

= e $\times$ 1.03 ${\gamma }_w$

Submerged unit wt. in sea water

${\gamma }_{sub}^{\prime }={\gamma }_{sat}-{\gamma }_{seawater}$

$=\frac{G_s{\gamma }_w+1.03e{\gamma }_w}{1+e}-1.03{\gamma }_w$

$=\frac{2.67\times 9.81+1.03\times 0.8\times 9.81}{1+0.8}-1.03\times 9.81$

= 8.938 kN/$m^3$

Effective stress at 30 m depth into sand layer

$\overline{\sigma }={\gamma }_{sub}^{\prime }\times H$

= 8.938 $\times$ 30

= 268.14

= 268 kPa