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 agoRelated Solutions
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