CIVENG 3A03 Assignment #1
Solutions
1. A 0.5 m3 sample of moist soil weights 9.81 kN. The water content is 10%. Assume s =
2.70Mg/m3.
(1) With this information, fill in all blanks in the phase diagram of Fig. 1.
Figure 1
(2) Calculate the void ratio, the porosity, the degree of saturation, the dry unit weight and
the saturated unit weight.
2. Verify the relation  =
 s +  w Sr e
1+ e
(Hint: You may assume Vs=1)
, fill in all the blanks in the phase diagram of Fig. 1.
3. What is the water content of a fully saturated soil with a dry density of 1.70 Mg/m3?
Assume Gs = 2.71.
4. Problem 1.2 in the text.
5. Problem 1.4 in the text.
(You are required to fill in all the blanks of the phase diagrams for all problems).
Solutions:
Q1:
Wt = 9.81kN , Vt = 0.5m3 , water content w = 10%, s = 2.70Mg/m3
Wt = Ws + Ww = Ws + Ws  w  Ws =
Wt
= 8.92kN
1+ w
Weight of water Ww = Wt  Ws = 0.89 kN
W
W
8.92kN
= 0.337 m3
Volume of solids: Vs = s = s =
 s  s w 2.70  9.81kN / m3
W
0.89
= 0.091 m3
Volume of water: Vw = w =
 w 9.81
(1 Mg = 9.81 kN)
Volume of voids: Vv = Vt  Vs = 0.5  0.337 = 0.163 m3
Volume of air: Va = 0.072 m3
0.072m3
0.163 m3
0.091m3
0.89 kN
0.5 m
9.81 kN
0.337m
8.92 kN
e=
Vv 0.163
=
= 0.48;
Vs 0.337
Sr =
n=
Vw 0.091
=
= 0.558 = 55.8%
Vv 0.163
Vv 0.163
=
= 0.33
Vt
0.5
d =
Ws 8.92
=
= 17.84 kN / m3
Vt
0.5
 sat =
Ws + Vv w 8.92 + 0.163  9.81
=
= 21.04 kN / m3
Vt
0.5
1+e
eSr
Q2: Assuem Vs = 1
Wt  s +  w S r e
=
Vt
1+ e
Q3: Assume Vs =1
According the defination of void ratio: Vv = eVs = e
Weight of soil particles: Ws = (Gs w ) 1 =  d (1 + e)
e=
Gs w
1 =
w (eSr)
2.71 9.81
 1 = 0.594
1.70  9.81
For fully saturated soil: Sr = 100%, Vw = Vv
Ws + Vv w Gs w + e w 2.71 + 0.594
=
=
 9.81 = 20.33 kN / m3
Vt
1+ e
1 + 0.594
w e
1+e
 sat =
(1+e) d = (1)(s)
Q4:
 = 1.91 9.81 = 18.74kN / m3 , w = 9.5%, Gs = 2.70   s = Gs w = 26.49kN / m3
e=
 s (1 + w)
26.49  1.095
1 =
 1 = 0.55
18.74
wGs 2.70  0.095
=
= 46.8%
0.55
e
G +e
2.70 + 0.55
w =
= s
 9.81 = 20.57 kN / m3
1+ e
1 + 0.55
Sr =
 sat
Q5. M=168 g, Ms=130.5 g, Mw = M - Ms=37.5 g , Gs=2.73,
W=1.65 N, Ws=1.28 N, Ww = W - Ws=0.67 N
1
V = D 2 H =  38 2  76 = 8.62  10 4 mm 3 = 8.62  10 5 m 3
4
4
M
Ms
130.5
Vs = s =
=
= 47.8 cm 3 = 4.78  10 5 m 3
where  w = 1g / cm 3
 s G s  w 2.73  1
Vv=V-Vs=3.84*10-5 m3, Vw =
w=
M w 37.5
=
= 37.5 cm 3 = 3.75  10 5 m 3
w
1
M w 37.5
=
= 28.7%
M s 130.5
Vw 3.75  10 5
=
= 97 .7. %
Vv 3.84  10 5
V v 3.84  10 5
e=
=
= 0.80
Vs 4.78  10 5
Sr =
Ws 1.28  10 3 kN
d =
=
= 14.8 k N / m 3
5
3
V
8.62  10 m
W + Vv  w
= 19.2 k N / m 3
 sta = s
V
 ' =  sta   w = 9.41 k N / m 3
e
e
0.80
water content at saturation: w = w =
=
= 29.3%
s
G s 2.73