SOIL MECHANICS
AND FOUNDATION ENGINEERING-I (CE-210)
•
BOOKS
1.
SOIL MECHANICS by T WILLIAM LAMBE & ROBERT V. WHITMAN, Wiley ASTERN
2.
FOUNDATION ANALYSIS & DESIGN by JOSEPH E. BOWLES, Mc Graw Hills
3.
GEOTECHNICAL ENGINEERING by SHASHI K GULHATI & MANOJ DATTA, TATA Mc
Graw Hill
4.
SOIL MECHANICS AND FOUNDATION ENGINEERING by B C PUNMIA, A K JAIN & A
K JAIN, Laxmi Publications Pvt. Ltd.
THREE PHASE SYSTEM
PRELIMINARY RELATIONSHIPS
𝑊𝑤
𝑊𝑠
•
Water Content
•
Unit Weights
•
Specific Gravity
•
Void Ratio
𝑒=
•
Porosity
𝑛=
•
Degree of Saturation
𝑆=
•
Percentage air voids
•
Air Content
𝑤=
𝑉𝑣
𝑉𝑠
𝑉𝑣
𝑉
𝑉𝑤
𝑉𝑣
UNIT WEIGHTS RELATIONSHIPS
•
Unit Weight of Water
𝛾𝑤 =
𝑊𝑤
𝑉𝑤
•
Unit Weight of Soil Solids
𝛾𝑠 =
𝑊𝑠
𝑉𝑠
•
Specific Gravity of Soil Solids
𝐺
•
Total/ Bulk Unit Weight of Soil
𝛾𝑡 =
•
Dry Unit Weight of Soil
𝛾𝑑 =
•
Submerged Unit Weight of Soil
𝛾’ = 𝛾𝑡 - 𝛾𝑤
𝛾
= 𝛾𝑠
𝑤
𝑊
𝑉
𝑊𝑠
𝑉
INTER-RELATIONSHIPS
•
𝛾=
𝑆𝑒+𝐺
1+𝑒
•
𝛾=
(1+𝑤)
𝐺
(1+𝑒)
•
𝛾𝑑 =
•
𝛾𝑑 = (1+𝑤)
•
𝛾′
𝛾𝑤
1
𝐺
(1+𝑒)
𝛾𝑤
𝛾𝑤
𝛾
=
𝐺 −1 + 𝑆−1 𝑒
𝛾𝑤
(1+𝑒)
(if S=100%, 𝛾′
=
𝐺 −1
(1+𝑒)
𝛾𝑤 )
PROBLEMS
1. A soil sample has a diameter of 38mm and a height of 76mm. Its wet weight is 1.15N.
Upon drying its weight reduced to 0.5N. G is 2.7. In the wet state what was the Degree of
Saturation and the water content of the soil sample. Also determine the void ratio, the
porosity, total unit weight and the dry unit weight.
2. A soil deposit has a void ratio of 10.0 and the specific gravity of soil solids is 2.0.
I.
If S=100% determine its g’
II.
If S=90% determine its g’
3. If the soil has void ratio of 0.43 and a G of 2.7. Determine total unit weights of the soil
when S=100%, 90%, 75% and 50%.
4. A soil sample has a porosity of 40%. The specific gravity of solids is 2.70. Calculate void
ratio, dry and total unit weight if the soil is 50% saturated and unit weight if the soil is
completely saturated.
5. The unit weight of a soil equals 16.4 kN/m3. The specific gravity of solids is 2.70.
Determine the void ratio under the assumption that the soil is perfectly dry. What would be
the void ratio, if the sample has water content as 8%.
PROBLEMS
6. A natural soil deposit has a bulk unit weight of 18.44 kN/m3 and water content of 5%.
Calculate the amount of water required to be added to 1 m 3 of soil to raise the water
content to 15%. Assume the void ratio remains constant. What will then be the degree of
saturation? Assume G=2.67.
7. Calculate the unit weights and specific gravities of solids of (a) a soil composed of quartz
and (b) a soil composed of 60% quartz, 25% mica and 15% iron oxide. Assume that both
soils are saturated and have void ratio of 0.63. Take G for quartz = 2.66 for mica = 3.0
and for iron oxide = 3.8.
8. A soil has a bulk unit weight of 20.11 kN/m3 and water content of 15%. Calculate the water
content if the soil partially dries to a unit weight of 19.42 kN/m3 and the void ratio remains
unchanged.
SOLUTION OF PROBLEM 1
•
Ww = 1.15-0.5 = 0.65 N
•
w=
•
V = × 𝑑 2 = 86.2 x 10 3 mm3
•
Vs = 𝐺 𝛾𝑠 = 2.7×10−5 = 18.5 x 10 3 mm3 (gw=10-5 N/mm3)
•
Vv = V-Vs= 67.7 x 10 3 mm3
•
Vw =
𝑊𝑤
𝑊𝑠
=
0.65
0.5
= 130%
𝜋
4
𝑊
0.5
𝑤
𝑊𝑤
𝛾𝑤
0.65
= 10−5 = 65 x 10 3 mm3
65×103
67.7×103
•
S=
•
Comment: Degree of saturation is less than 100% even though the water content is very
high at 130%.
= 96%
SOLUTION CONTD.
•
Ww = 1.15-0.5 = 0.65 N
𝑉𝑣
𝑉𝑠
67.7×103
18.5×103
•
e=
•
𝑛 = 1+𝑒 = 4.66 = 0.79
•
Comment: Void ratio is more than 1.0 as 3.66 but porosity can never be more than 1.0
and is only 0.79.
=
𝑒
= 3.66
3.66
SOLUTION CONTD.
𝑊
𝑉
1.15×10−3
•
gt = g =
•
gd =
𝑊𝑠
𝑉
•
OR
•
𝑡
gd = (1+𝑤)
= (1+1.3) = 5.8 𝑘𝑁/𝑚 3
•
Comment: The dry unit weight is less than the total unit weight. The dry unit weight is
even less than the unit weight of water because in this soil sample the void space is much
more than the space occupied by the solids-its void ratio is 3.66.
𝛾
=
= 86.2×10−6 = 13.3 𝑘𝑁/𝑚3
0.5×10−3
86.2×10−6
= 5.8 𝑘𝑁/𝑚3
13.3
HINTS TO PROBLEMS 2, 3 & 5
PROBLEM 2
e = 10% = 0.1; G=2.0;
(i) S=100% = 1
(ii) S = 90% = 0.9
𝛾𝑏 =
𝐺 − 1 + (𝑆 − 1)
𝛾𝑤
1+𝑒
PROBLEM 3
e = 0.43, G = 2.7
at various S
𝛾𝑡 =
𝑆𝑒 + 𝐺
𝛾
1+𝑒 𝑤
PROBLEM 5
𝛾𝑑 =
1
𝐺𝛾
(1 + 𝑒) 𝑠 𝑤
SOLUTION OF PROBLEM 4
𝑛
1−𝑛
0.4
1−0.4
•
𝑒=
•
𝛾𝑑 =
•
w=
•
𝛾 = 𝛾𝑑 1 + 𝑤 = 15.89 × 1.124 = 17.85𝑘𝑁/𝑚 3
•
w=
•
𝛾 = 𝛾𝑑 1 + 𝑤 = 15.89 × 1.247 = 19.81𝑘𝑁/𝑚 3
=
𝐺 𝛾𝑤
1+𝑒
𝑒𝑆
𝐺
𝑒𝑆
𝐺
=
=
=
= 0.667 𝑂𝑅 66.7%
2.7×9.81
1+0.667
0.667×0.5
2.70
0.667×1
2.70
= 15.89 𝑘𝑁/𝑚3 𝛾𝑤 = 9.81 𝑘𝑁/𝑚 3
= 0.124
= 0.247
SOLUTION OF PROBLEM 6
𝛾
1+𝑤
18.44
1+0.05
•
𝛾𝑑 =
•
Earlier : w=5%
=
= 17.56 𝑘𝑁/𝑚3
𝑊𝑠 = 𝛾𝑑 𝑉 = 17.56 × 1 = 17.56 𝑘𝑁
𝑊𝑤 = 𝑊𝑠 × 𝑤 = 17.56 × 0.05 = 0.878 𝑘𝑁
𝑊𝑤 0.878
𝑉𝑤 =
=
= 0.0895 𝑚3
𝛾𝑤
9.81
•
Later : w=15%
𝑊𝑤 = 𝑊𝑠 × 𝑤 = 17.56 × 0.15 = 2.634 𝑘𝑁
𝑊𝑤 2.634
𝑉𝑤 =
=
= 0.2685 𝑚3
𝛾𝑤
9.81
Hence additional water required to raise the water content from 5% to 15% = 0.2685 – 0.0895
= 0.179 m3 = 179 litres
Void ratio, 𝑒 =
𝐺𝛾𝑤
𝛾𝑑
−1=
2.67×9.81
17.56
− 1 = 0.49
After the water has been added, e remains the same
S=
𝑤𝐺
𝑒
=
0.15×2.67
0.49
= 0.817 𝑜𝑟 81.7%
SOLUTION OF PROBLEM 7
a) For the soil composed of pure quartz, G = 2.66
𝛾𝑠𝑎𝑡 =
𝐺+𝑒
2.66 + 0.63
𝛾𝑤 =
× 9.81 = 19.8 𝑘𝑁/𝑚 3
1+𝑒
1 + 0.63
b) For the soil composed of pure quartz,
G = (2.66 x 0.6) + (3.0 x 0.25) + (3.8 x 0.15) = 2.92
𝛾𝑠𝑎𝑡 =
𝐺+𝑒
2.92 + 0.63
𝛾𝑤 =
× 9.81 = 21.36 𝑘𝑁/𝑚3
1+𝑒
1 + 0.63
SOLUTION OF PROBLEM 8
•
Before drying, g = 20.11 kN/m3
𝛾𝑑 =
•
20.11
= 17.49 𝑘𝑁/𝑚 3
1 + 0.15
Since after drying e does not change, V and gd will remain unaltered,
1+𝑤 =
Hence
𝛾
19.42
=
= 1.11
𝛾𝑑 17.49
w = 1.11-1 = 0.11 or 11%