Geoscience Reference
In-Depth Information
A = 50
cm
2
free
water
b
a
measured
a = 4.35 cm
b = 3.00 cm
c = 7.35 cm
d = 6.00 cm
dry
sand
saturated
sand
c
1
c
d
water
weight 970 gr
volume 350 cc
weight 675 gr
volume 300 cc
weight 420 gr
Buret
weight 125
gr
Figure 2.6 Simple classification test (laboratory)
The unit weight of dry sand is
d
= (675 - 125) / 350 = 1.57 gr/cc = 15.7 kN/m
3
Volume of grains and replaced water provide porosity
dA = bA + ncA
6 = 3 + 7.35
n
n
= 0.408
Weight of the saturated sand gives the unit weight of solids
s
and the wet unit
soil weight
s
= 2.505 gr/cc = 25.05 kN/m
3
bA
w
+ cA((1 - n)
s
+ n
w
) = 970-125 gr
w
= 1.89 gr/cc = 18.9 kN/m
3
= (1 - n)
s
+ n
m =
160x10
-6
m. Then, the so-called specific surface
11
(grain surface/grain volume)
becomes
S
= 6/
D
50
= 3.75x10
4
m
-1
. For the volume of 350 cc, the total granular
volume is
V
s
= (1 -
n
) 350 = 207.2 cc = 207.2x10
-6
m
3
and the total inner surface
becomes
SV
s
= 3.75x10
4
x207.2x10
-6
= 7.77 m
2
.
The number of equivalent particles in 350 cc becomes 350x10
-6
/(
Consider the sand grains as uniform spheres with diameter
D
50
= 160
D
50
3
) =
27.2x10
6
. So, 30 cm
3
of sand with
D
50
of 160
m contains 27.2 million particles
with a surface of 7.77 square metre! Soil consists of many particles, many
intergranular contacts, which determine the shear resistance, and a relatively large
internal surface, which determines the permeability.
11
The volume of a sphere is
V =
D
3
/6; the surface is
A =
D
2
. Hence,
S = A/V
= 6/
D
.
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