First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Majmaah University
College of Engineering
Civil and Environmental Engineering Department
Soil Mechanics and Foundation Engineering I (CE210)
Name:
Academic Number:
Time allowed: 2 hrs
Level (5):
Model Answer Total marks = 30 ….Counted marks = 15
Question #1
[7 marks]
1.1) Classify the following soil samples using the Triangular
Classification System.
(3 marks)
a) 30% sand, 35% clay, 35% silt, clay loam
b) A sample of soil is classified as silty loam, (shown in the
figure below as Dot) estimate the percentage of the soil
sample (silt, clay and sand). Clay 10%, silt 60%, sand 30%
1
SaMeH
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
1.2) Put a circle on the correct answer:
(4 marks)
1) Which of the following minerals cause problem if construction takes
place over it?
a.
b.
c.
d.
Illite
Gypsum
Montmorillonite
Kaolin
2) Which of following formula is correct to determine the soil porosity?
a.
𝑒=
b.
𝑒=
c.
𝑛=
d.
𝑒=
𝑛
1+𝑣
𝑛
1−𝑛
𝑊𝑠
𝑉𝑠
𝑉𝑤
𝑉𝑇
3) Grain-Size Distribution of the soil fine particles is usually conducted
using…
a.
b.
c.
d.
Hydrometer
Moisture device
Microscope
Sieve analysis
4) Saturated soil consists of (only……..):
a.
b.
c.
d.
2
SaMeH
Solid + Air + Water
Solid + Air
Solid + Water
Water + Air
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Question #2:
A soil core sample has the following specifications:
Weight of soil sample = 1.025 kg
Vol. of soil sample = 590.0 cm3
γs = 2.64 gm/cm3
Dry weight of soil = 0.898 kg




Determine:
1)
2)
3)
4)
5)
6)
7)
8)
9)
Weight of water 𝑊𝑤 (gm)
Volume of solid 𝑉𝑠 (cm3)
Volume of voids 𝑉𝑣 (cm3)
Volume of water 𝑉𝑤 (cm3)
Volume of Air 𝑉𝐴 (cm3)
Moisture content or water content
Void ratio
Porosity
Degree of saturation of a soil core sample
Help:
3
SaMeH
(6 marks)
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Solution of Question #2
1)
Weight of water 𝑊𝑤 (gm)
𝑤𝑇 = 𝑤𝑎 + 𝑤𝑤 + 𝑤𝑠
𝑤𝑇 = 𝑤𝑎 + 𝑤𝑤
𝑤𝑤 = 𝑤𝑇 − 𝑤𝑠 = 1025 – 898.0 = 127 gm
2)
Volume of solid Vs (cm3)
𝛾𝑠 =
𝑉𝑠 =
3)
𝑊𝑠
𝛾𝑠
𝑊𝑠
𝑉𝑠
898
3
=
340.152
cm
2.64
=
Volume of voids Vv (cm3)
𝑉𝑇 = 𝑉𝑣 + 𝑉𝑠
𝑉𝑣 = 𝑉𝑇 − 𝑉𝑠
𝑉𝑣 = 590 − 340.152 = 249.848
4)
Volume of water Vw (cm3)
𝛾𝑤 =
𝑉𝑤 =
5)
cm3
𝑊𝑤
𝛾𝑤
=
𝑊𝑤
𝑉𝑤
127
1
= 127 cm3
Volume of Air VA (cm3)
VA = Vv - Vw
VA = 249.848 – 127.0 = 122.848 cm3
4
SaMeH
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
6)
Moisture content or water content
𝑊𝑤
𝑤𝑐 =
7)
𝑊𝑠
127
= 898 x 100 = 14.14%
Void ratio
𝑉
249.848
e= 𝑉𝑣 = 340.152x100 = 73.45%
𝑠
8)
Porosity
𝑉
n= 𝑉𝑣 =
𝑇
9)
590.00
x100 = 42.35%
Degree of saturation of a soil core sample.
S=
5
249.848
SaMeH
𝑊𝑤
𝑉𝑣
127.00
= 249.848x100 = 50.83%
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Question #3:
(5 marks)
3.1) A soil has the following specifications:
e = 0.86, w = 28 % , and Gs = 2.72, Calculate:
(a)
Moist unit weight (gm/cm3)
Moisture content or water content
𝛾𝑚𝑜𝑖𝑠𝑡 =
(b)
(1+𝑤)𝐺𝑠
1+𝑒
(1+0.28)2.72
1+0.86
1= 1.87%
Dry Unit Weight (gm/cm3)
𝛾𝑑 =
(c)
𝛾𝑤 =
𝛾
1.87
3
=
=
1.46
gm/cm
1+𝑤 1.28
Degree of saturation %
𝑆𝑒 = 𝐺𝑠 . 𝑤
𝑆=
6
SaMeH
𝐺𝑠 .𝑤
𝑒
=
2.72 𝑥 0.28
0.86
x 100 = 88.5 %
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Question #4:
(12 marks)
Following are the results of a sieve analysis. Make the necessary
calculations to:
U.S. sieve Opening Mass of soil retained
size
(mm)
on each sieve (gm)
4
10
20
40
60
80
100
200
Pan
4.75
2.00
0.850
0.425
0.250
0.180
0.150
0.075
-
5
45
65
94
145
127
215
61
17
a) Determine the percentage finer than each sieve size and plot a
grain- size distribution curve.
b) Determine D10, D30 and D60 from the grain- size distribution
curve.
c) Calculate the uniformity coefficient, Cu
d) Calculate the coefficient of gradation, Cc
7
SaMeH
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
Solution
U.S. sieve size Opening Mass of soil retained cumulative Percentage
(mm)
on each sieve (g)
4
4.75
5
99.35
5
10
2.00
45
93.54
50
20
0.850
65
85.14
115
40
0.425
94
72.99
209
60
0.250
145
54.26
354
80
0.180
127
37.86
481
100
0.150
215
10.07
696
200
0.075
61
2.19
757
Pan
17
774
0
774
e)
D60 = 0.285
D30 = 0.175
D10 = 0.150
D25 = 0.165
D75 = 0.485
Cu =
CC =
8
SaMeH
𝐷60
𝐷10
2
𝐷30
𝐷60 ∗ 𝐷10
=
=
0.285
0.150
= 1.9
(0.175)2
0.285∗0.15
= 0.716
First midterm exam (CE 210) – Second Semester 34/35 – 16/3/2014
You are allowed to use this table in the exam.
(γw = 62.4 Ib/ft3 = 1 gm/cm3 = 1000 kg/m3 = 10 kN/m3)
Good Luck
9
SaMeH