Question

In: Civil Engineering

The subsoil at the site consists of a 10 m thick homogeneous layer of dense sand...

The subsoil at the site consists of a 10 m thick
homogeneous layer of dense sand having the following properties:

Dry unit weight = 16 kN/m3,
Gs = 2.68,
angle of friction = 35 degrees

The natural ground water table lies at 2 m below the ground surface.

Determine the change in shear strength
at 10 m depth (kPa).

Solutions

Expert Solution

d = 16 KN/m^3

G. = 2.68

=350

d. =10 m

The effective vertical stress at the plane of interest is :

= 2d + 8b

d = 16 KN/m^3. And G = 2.68

d = 16 = [G/1+e]w = [2.68/1+e]9.81

16e = 10.2908

e = 0.64

b = [(G-1)/(1+e)]w = [(2.68-1)/(1+0.64)]9.81 = 9.75 KN/m^3

= 2*16 + 8*9.75 = 110 KN/m^2

Hence the shearing strength of the sand is

S = tan =110*tan35 = 77 KN/m2

so the shear strength at 10 m depth is 77 KN/m2

(change in shear strength cannot be found as no other condition is given)

Ally Wells answered 3 months ago
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