1
2.21
某办公楼工程地质勘探中取原状土做试验?/p>
用天平称
50cm
3
湿土质量?/p>
95.15g
,烘干后质量?/p>
75.05g
,土粒比重为
2.67
。计算此土样的天然密度、干
密度、饱和密度、天然含水率、孔隙比、孔隙率以及饱和度?/p>
【解?/p>
m
= 95.15g
?/p>
m
s
= 75.05g
?/p>
m
w
= 95.15 - 75.05 = 20.1g
?/p>
V
= 50.0
cm
3
?/p>
d
s
= 2.67
?/p>
V
s
= 75.05/(2.67
?/p>
1.0) = 28.1 cm
3
?/p>
g
= 10 m/s
2
,则
V
w
= 20.1 cm
3
V
v
= 50.0 - 28.1 = 21.9 cm
3
V
a
= 50.0
?/p>
28.1
?/p>
20.1 = 1.8 cm
3
于是?/p>
?/p>
=
m
/
V
= 95.15 / 50 = 1.903g/ cm
3
?/p>
d
=
m
s
/
V
= 75.05 / 50 = 1.501g/ cm
3
?/p>
s
a
t
= (
m
s
+
?/p>
w
?/p>
V
v
)/
V
= (75.05 + 1.0
?/p>
21.9) / 50 = 1.939g/ cm
3
w
=
m
w
/
m
s
= 20.1 / 75.05 = 0.268 = 26.8%
e
=
V
v
/
V
s
= 21.9 / 28.1 = 0.779
n
=
V
v
/
V
= 21.9 / 50 = 0.438 = 43.8%
S
r
=
V
w
/
V
v
= 20.1 / 21.9 = 0.918
2.22
一厂房地基表层为杂填土,厚
1.2m
,第二层为粘性土,厚
5m
,地下水
位深
1.8m
。在粘性土中部取土样做试验,测得天然密?/p>
?/p>
=
1.84g/
cm
3
,土粒比
重为
2.75
。计算此土样的天然含水率
w
、干密度
?/p>
d
、孔隙比
e
和孔隙率
n
?/p>
【解】依题意知,
S
r
= 1.0
?/p>
?/p>
s
a
t
=
?/p>
= 1.84g/ cm
3
?/p>
?
,得
n
=
e
/(1 +
e
) = 1.083 /(1 + 1.083) = 0.520
g/cm
3
?/p>
2.23
某宾馆地基土的试验中,已测得土样的干密度
?/p>
d
= 1.54g/ cm
3
,含?/p>
?/p>
w
=
19.3%
,土粒比重为
2.71
。计算土的孔隙比
e
、孔隙率
n
和饱和度
S
r
。又
测得该土样的液限与塑限含水率分别?/p>
w
L
= 28.3%
?/p>
w
p
= 16.7%
。计算塑性指?/p>
I
p
和液性指?/p>
I
L
,并描述土的物理状态,为该土定名?/p>
【解】(
1
?/p>
?/p>
=
?/p>
d
(1 +
w
) = 1.54
?/p>
(1 + 0.193) = 1.84g/ cm
3
n
=
e
/(1 +
e
) = 0.757 /(1 + 0.757) = 0.431
?/p>
2
?/p>
I
p
=
w
L
-
w
p
= 28.3
?/p>
16.7 = 11.6
I
L
= (
w
L
-
w
) /
I
p
= (28.3
?/p>
19.3)/11.6 = 0.776