Q5.1:
Given : Gs = 2.72
Req. :
= ?? "lb/ft3" @ w = 5%, 8%, 10%, 12% and 15%.
Soln.:
.
Note: γ
"zero air voids unit weight" which means that voids contain no air
but only water, i.e. saturaon case S =100% .
=
,
= 62.4
w%
5
8
10
12
15
/
.
149.4
139.39
133.43
127.96
120.55
Q5.5:
Given : Gs = 2.72
Vol. of Proctor mold
"ft3"
1/30
1/30
1/30
1/30
1/30
Req. :
Weight of wet soil in mold
"lb"
3.26
4.15
4.67
4.02
3.63
Moisture content
"%"
8.4
10.2
12.3
14.6
16.8
void ratio "e" and degree of saturation "S" at optimum moisture
content.
Soln.:
To determine the optimum moisture content we have to sketch the
relationship between water content "w%" and dry unit weight "
To find
find
=
for each case we have to find
due to given data
"
, then by knowing the moisture content for each case we can
=
Vol. of Proctor
mold "ft3"
1/30
1/30
1/30
1/30
1/30
.
Weight of wet soil
in mold "lb"
3.26
4.15
4.67
4.02
3.63
Moisture
content "%"
8.4
10.2
12.3
14.6
16.8
W
V
=
3.63
1/30
97.8
124.5
140.1
1206.6
108.9
90.22
112.98
124.76
105.24
93.24
1+
=
108.9
1 + 0.168
Moisture content "%"
8.4
10.2
12.3
14.6
16.8
90.22
112.98
124.76
105.24
93.24
130
125
120
115
110
105
100
95
90
85
80
,
w%
20
`
15
10
5
From curve:
. " = 125 lb/
"
=
S.e =
S=
.
1+
3
@ w% = 12.5%.
……. 125 =
.
∗
.w
∗ .
.
= 0.94 = 94% ......(2)
.
, then e = 0.358 ……(1)
0
Q5.9:
Given :
Embankment fill requires 5000 m3 of compacted soil.
Required void rao = 0.7.
Four borrow pits are available as described in the following table:
Borrow Pit
A
B
C
D
Void ratio
0.85
1.2
0.95
0.75
Cost "$/m3"
9
6
7
10
Req. : Choose the best alternative with the lowest cost .
Soln. :
Vtot = 5000 m3 , ereq = 0.7.
Vtot = Vv + Vs , 5000 = Vv+Vs , we know that Vs is constant and will not change
after compaction.
Vv = 5000 - Vs …………… (1)
e=
= 0.7 …………(2)
from (1) and (2)
0.7 =
then Vs = 2941.18m3
,
Alternative A :
e = 0.85 =
=
+
=
.
…….
= 2500
= 2941.18 + 2500 = 5441.18
= 5441.18 ∗ 9 $ = 48,971 $ …………………………………………………………(A)
Alternative B :
e = 1.2 =
=
+
=
…….
.
= 3529.42
= 2941.18 + 3529.42 = 6470.6
= 6470.6 ∗ 6 $ = 38,824 $ …………………………………………………………(B)
Alternative C :
e = 0.95 =
=
+
=
.
…….
= 2794.12
= 2941.18 + 2794.12 = 5735.3
= 5735.3 ∗ 7 $ = 40,147 $ …………………………………………………………(C)
Alternative D :
e = 0.75 =
=
+
=
.
…….
= 2205.89
= 2941.18 + 2205.89 = 5147.1
= 5147.1 ∗ 10 $ = 51,471 $ …………………………………………………………(D)
From alternative costs we can choose Alt. B with lowest cost = 38,824 $
Q5.13:
Given :
"ottawa sand" = 1667 kg/m3
mass "Oa wa sand " to fill the cone = 0.117 kg
mass ( jar + sand + cone ) before filling hole = 5.99 kg
mass ( jar + sand + cone ) ae r fil ling hol e = 2.81 kg
mass of moist soil from the hole = 3.331 kg
w% = 11.6% for moist soil from the hole.
w%
6
8
9
11
12
14
Req. :
14.8
17.45
18.52
18.9
18.6
16.9
of compaction in the field
Relative compaction in the field
Soln.:
Mass of soil to fill the hole and cone = mass ( jar + sand + cone ) before filling hole - mass ( jar + sand +
cone
) after filling hole = 5.99 – 2.81 = 3.18 kg.
Mass of soil to fill hole only = Mass of soil to fill the hole and cone - mass of soil to fill the cone =
3.18 – 0.117 = 3.063 kg.
Vol. of the hole =
Moist density of soil from the hole =
dry unit weight of soil from the hole (
.
=
.
)=
= 0.00184
=
.
=
∗ .
3
.
.
.
3
= 1812.86 kg/m .
3
= 17.78 KN/m .
(
)
=
.
.
= 15.94 KN/m ……….……………..………………………………… (1)
Relative compaction in the field (R) =
We have to find
,
,
,
.
, so we draw the relationship between w% and γ
.
w%
6
14.8
8
17.45
9
18.52
11
18.9
12
18.6
14
16.9
20
,
19
18
17
16
15
14
13
12
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
From curve
Then, R =
.
,
3
= 19KN/m
∗ 100% = 83.94%
…………….......................................................................(2)