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Date Submitted: 09/22/2012 01:39 AM
LOSITAÑO, MARVIN JOHN CANTES BU BSCE 4A
ASSIGNMENT #1 (CE411c)
3.1) For a given soil, show that
a. γsat = γd + nγw
b. γsat = n γw
where wsat = moisture content at saturated state
c. γd =
sol’n:
a) γsat = γd + nγw
since:
γsat = γw γd = n=
so:
γw = + γw
=
γw = γw
b) γsat = n γw
where wsat = moisture content at saturated state
since:
γsat = γw Gsw = eS (but since s = 1) w =
γw = n γw
= γw
= γw
γw = γw
c) γd =
from :
γd = Gsw = eS
=
=
=
3.3) The moist mass of 2.8 x 10-3 m3 of soil is 5.53 kg. If the moisture content is 10 % and the specific gravity of soil solids is 2. 72, determine the following:
a. Moist density
b. Dry density
c. Void ratio
d. Porosity
e. Degree of Saturation
f. Volume occupied by water
sol’n:
a) ρb = m/v
= 5.53 kg/2.8 x 10-3 m3
= 1975 kg/m3
For b and c:
ρb = Gsρw
1975 kg/m3 = (2.72)(1000kg/m3)
1975 + 1975e = 2992
e = 0.515
ρdry =
=
ρdry = 1795.38
d) n = x 100%
= x 100%
n = 34%
e) Gsw = eS
S =
= x 100%
S = 52.8%
f) e =
Vs =
V = Vv + Vs
V = eVs + Vs = Vs(1 + e) =
Ms = ρsVs Vs = 1.848 x 10-3m3
= GsρwaterVs
= (2.72)(1000kg/m3)( 1.848 x 10-3m3)
= 5.027 kg
MT = Ms + Mw + Ma ρwater = Mw/Vw
Mw = 5.53 kg – 5.027 kg Vw = 0.503 kg/(1000...