Introduction and soil properties

This document provides an introduction to a site and its soil properties. It includes information about the types of soil present, their characteristics, and recommendations for leveling the top soil. The document also discusses laboratory tests conducted on the soil.

Contents
1 Introduction and soil properties....................................................................................................2
a) Laboratory Test:.........................................................................................................................3
2 Foundation pressure under the initial design................................................................................6
3 Proposed improve foundation design............................................................................................8
4 Calculation of the settlements.....................................................................................................10
5 Calculation of the loads on the culvert........................................................................................14
1

1 Introduction and soil properties
The site is having an area of 7918.18 m2. The length and width of the plot is 150.64 m and 52.87 m
respectively. Over the site 20 bore holes were dug up. The depth of each bore hole is approximately
25 m. The soil profile is consisting of mainly 4 types of soil.
1. first layer is consisting of sandy silt with low plasticity. The moist field density of the soil is 1698
kg/m3. The specific gravity of soil is 2.66. Due to the presence of coarse-grained soil, the water
content of the sample retrieved from the site is very less (2.6%).
2. Second layer of soil is consisting of poorly graded sand with clay. The moist field density of the soil
is 1654 kg/m3. The specific gravity and water content of the sample is 2.65 and 9.3% respectively.
3. Third layer of soil is consisting of fat clay with sand. The moist field density of the soil is 1960
kg/m3. The water content of the sample is 28.3%. Large water content might be due to the presence
of clay which holds water.
4. The final layer consists of clayey sand with gravel. The moist field density of the soil is 2038 kg/m3
and the water content is 22.3%.
The cross section shows the undulation in ground at the top. It is recommended to do some levelling
at the top soil throughout.
2

a) Laboratory Test:
Sieve Analysis:
0.01 0.1 1 10
0
10
20
30
40
50
60
70
80
90
100
Sieve Analysis
diameter
Percent finer
3

Standard Compaction:
The compaction of the soil is done with the help of standard proctor.
The maximum dry density was come out to be 1780.8 kg/m3 and optimal moisture content was
19.75 %.
4

CBR Test Results:
5

2 Foundation pressure under the initial design
Column
Load
(kN) x (m) y (m) Load*x Load*y
6

C1 2100 2 2 4200 4200
C2 1900 2 6 3800 11400
C3 1900 6 2 11400 3800
C4 2600 8 8 20800 20800
C5 1500 18 8 27000 12000
C6 1900 22 8 41800 15200
C7 2500 22 14 55000 35000
C8 1500 16 10 24000 15000
C9 1900 16 14 30400 26600
C10 2600 20 18 52000 46800
C11 2100 12 24 25200 50400
C12 2500 8 26 20000 65000
C13 2800 2 24 5600 67200
Wall-1 950 2.5 20 2375 19000
Wall-2 3000 4.5 20 13500 60000
Wall-3 1800 7 20 12600 36000
Wall-4 1200 9 20 10800 24000
Wall-5 400 10.5 20 4200 8000
Wall-6 350 11.5 20 4025 7000
Net 35500 368700 527400
Centroids of given footing
X’ = ΣLoad*x/ΣLoad = 368700/35500 = 10.4
Y’ = ΣLoad*y/ΣLoad = 527400/35500 = 14.9
Resultant forces:
In X Direction = 368700 kN
In Y Direction = 527400 kN
Moments:
In X Direction (Mx) = (X – X’)*Load
In Y Direction (My) = (Y – Y’)*Load
Column
Load
(kN) X-X' Y-Y' Mx My
C1 2100 -8.4 -12.9 -17610.4 -26998.3
C2 1900 -8.4 -8.9 -15933.2 -16827.0
C3 1900 -4.4 -12.9 -8333.2 -24427.0
C4 2600 -2.4 -6.9 -6203.4 -17826.5
C5 1500 7.6 -6.9 11421.1 -10284.5
7

C6 1900 11.6 -6.9 22066.8 -13027.0
C7 2500 11.6 -0.9 29035.2 -2140.8
C8 1500 5.6 -4.9 8421.1 -7284.5
C9 1900 5.6 -0.9 10666.8 -1627.0
C10 2600 9.6 3.1 24996.6 8173.5
C11 2100 1.6 9.1 3389.6 19201.7
C12 2500 -2.4 11.1 -5964.8 27859.2
C13 2800 -8.4 9.1 -23480.6 25602.3
Wall-1 950 -7.9 5.1 -7491.6 4886.5
Wall-2 3000 -5.9 5.1 -17657.7 15431.0
Wall-3 1800 -3.4 5.1 -6094.6 9258.6
Wall-4 1200 -1.4 5.1 -1663.1 6172.4
Wall-5 400 0.1 5.1 45.6 2057.5
Wall-6 350 1.1 5.1 389.9 1800.3
Net 35500 0 0
Pressure Distribution based on resultant forces:
P = ΣLoad/Area
= 35500/(22*26) = 62.06 kN/m2
3 Proposed improve foundation design
Divide the whole footing in three parts:
Part – 1: Consisting of column C11, C12 and C13 and wall loads
Column
Load
(kN) x (m) y (m) Load*x Load*y
C11 2100 12 24 25200 50400
C12 2500 8 26 20000 65000
C13 2800 2 24 5600 67200
Wall-1 950 2.5 20 2375 19000
Wall-2 3000 4.5 20 13500 60000
Wall-3 1800 7 20 12600 36000
Wall-4 1200 9 20 10800 24000
Wall-5 400 10.5 20 4200 8000
Wall-6 350 11.5 20 4025 7000
Net 15100 98300 336600
Centroids of given spread footing
X’ = ΣLoad*x/ΣLoad = 98300/15100 = 6.5
Y’ = ΣLoad*y/ΣLoad = 336600/15100 = 22.3
Bearing pressure = 100 kpa
8

Area of footing = Net Load / Bearing Pressure
= 15100 / 100 = 151 m2
Assume square dimension = SQRT(151) = 12.3m
Assume 13 m
Net bearing pressure = load / area = 15100 / (13 * 13) = 89.35 kN/m2 < 100 kN/m2 (Hence Safe)
Part – 2: Consisting of column C1, C2, C3 and C4
Column
Load
(kN) x (m) y (m) Load*x Load*y
C1 2100 2 2 4200 4200
C2 1900 2 6 3800 11400
C3 1900 6 2 11400 3800
C4 2600 8 8 20800 20800
Net 8500 40200 40200
Centroids of given spread footing
X’ = ΣLoad*x/ΣLoad = 40200/8500 = 4.7
Y’ = ΣLoad*y/ΣLoad = 40200/8500 = 4.7
Bearing pressure = 100 kpa
Area of footing = Net Load / Bearing Pressure
= 8500 / 100 = 85 m2
Assume square dimension = SQRT(85) = 9.3m
Assume 10 m
Net bearing pressure = load / area = 8500 / (10 * 10) = 85 kN/m2 < 100 kN/m2 (Hence Safe)
Part – 3: Consisting of column C5, C6, C7, C8, C9 and C10
Column
Load
(kN) x (m) y (m) Load*x Load*y
C5 1500 18 8 27000 12000
C6 1900 22 8 41800 15200
C7 2500 22 14 55000 35000
C8 1500 16 10 24000 15000
C9 1900 16 14 30400 26600
C10 2600 20 18 52000 46800
Net 11900 230200 150600
Centroids of given spread footing
9

X’ = ΣLoad*x/ΣLoad = 11900/230200 = 19.3
Y’ = ΣLoad*y/ΣLoad = 11900/150600 = 12.7
Bearing pressure = 100 kpa
Area of footing = Net Load / Bearing Pressure
= 11900 / 100 = 119 m2
Assume square dimension = SQRT(119) = 10.9 m
Assume 11 m
Net bearing pressure = load / area = 11900 / (11 * 11) = 98.34 kN/m2 < 100 kN/m2 (Hence Safe)
4 Calculation of the settlements
Settlement
Mass of solids in specimen Ms = 121.0 g 121 g
Inside diameter of the ring Di = 6.3 cm 6.3 cm
Specific gravity of solids Gs = 2.65
Hight of specimen H0 = 25.4 mm 25.4 mm
Area of specimen A = cm2
31.1724
5 cm2
Height of solids Hs = Ms A×Gs×ρw =
mm
14.6476
7 mm
Pressur
e (kPa)
Deformati
on
(0.001mm)
Heigh
t H
(mm)
Hv=H-
Hs
e=Hv/
Hs
0 0 25.4
10.7523
3
0.73406
4
6 644
24.75
6
10.1083
3
0.69009
8
12.5 660 24.74
10.0923
3
0.68900
6
25 714
24.68
6
10.0383
3
0.68531
9
50 797 24.60 9.95533 0.67965
10

3 1 3
100 948
24.45
2
9.80433
1
0.66934
4
200 1164
24.23
6
9.58833
1
0.65459
8
400 1674
23.72
6
9.07833
1 0.61978
400 1674
23.72
6
9.07833
1 0.61978
200 1700 23.7
9.05233
1
0.61800
5
100 1592
23.80
8
9.16033
1
0.62537
8
50 1480 23.92
9.27233
1
0.63302
4
50 1480 23.92
9.27233
1
0.63302
4
100 1517
23.88
3
9.23533
1
0.63049
8
200 1543
23.85
7
9.20933
1
0.62872
3
400 1688
23.71
2
9.06433
1
0.61882
4
800 2343
23.05
7
8.40933
1
0.57410
7
1600 3224
22.17
6
7.52833
1
0.51396
1
3200 3962
21.43
8
6.79033
1
0.46357
8
1 10 100 1000 10000
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
Pressure
e
Cc = ∆e / log σ2 – log σ1 = (0.513961 – 0.574107) / log 1600 – log 800 = 0.434
11

For 1st Case:
q0 = 89.35 kN/m2
B = 13m; L = 13m
e0 = 0.734; Cc = 0.434; Hc = 30 m
z B+z L+z ∆σ Sc
1.5 14.5 14.5
71.8192
6
0.00192
4
3 16 16
58.9843
8
0.00165
3
4.5 17.5 17.5
49.3061
2
0.00143
3
6 19 19
41.8282
5
0.00125
2
7.5 20.5 20.5
35.9309
9
0.00110
2
9 22 22
31.1983
5
0.00097
7
10.5 23.5 23.5
27.3426
9
0.00087
1
12 25 25 24.16 0.00078
13.5 26.5 26.5
21.5023
1
0.00070
3
15 28 28 19.2602 0.00063
12

7
16.5 29.5 29.5
17.3513
4
0.00057
9
18 31 31 15.7128
0.00052
8
19.5 32.5 32.5
14.2958
6
0.00048
4
21 34 34
13.0622
8
0.00044
5
22.5 35.5 35.5
11.9817
5 0.00041
24 37 37
11.0299
5 0.00038
25.5 38.5 38.5
10.1872
2
0.00035
2
Total settlement = 14.51 mm
For 2nd case:
q0 = 85 kN/m2
B = 10m; L = 10m
e0 = 0.734 ; Cc = 0.434; Hc = 30 m
z B+z L+z ∆σ Sc
1.5 11.5 11.5
64.2722
1
0.00183
6
3 13 13
50.2958
6
0.00151
6
4.5 14.5 14.5
40.4280
6
0.00126
9
6 16 16
33.2031
3
0.00107
5
7.5 17.5 17.5 27.7551
0.00092
1
9 19 19
23.5457
1
0.00079
7
10.5 20.5 20.5
20.2260
6
0.00069
6
12 22 22
17.5619
8
0.00061
2
13.5 23.5 23.5
15.3915
8
0.00054
3
15 25 25 13.6
0.00048
4
16.5 26.5 26.5 12.1039 0.00043
13

5 4
18 28 28
10.8418
4
0.00039
1
19.5 29.5 29.5
9.76730
8
0.00035
5
21 31 31
8.84495
3
0.00032
3
22.5 32.5 32.5
8.04733
7
0.00029
5
24 34 34
7.35294
1
0.00027
1
25.5 35.5 35.5
6.74469
4
0.00024
9
Total settlement = 12.07 mm
For 3rd case:
q0 = 98.35 kN/m2
B = 11 m; L = 11 m
e0 = 0.734; Cc = 0.434; Hc = 30 m
z B+z L+z ∆σ Sc
1.5 12.5 12.5 76.16 0.00187
3 14 14
60.7142
9
0.00156
8
4.5 15.5 15.5
49.5317
4 0.00133
6 17 17
41.1764
7 0.00114
7.5 18.5 18.5
34.7699
1
0.00098
7
9 20 20 29.75
0.00086
2
10.5 21.5 21.5
25.7436
5
0.00075
8
12 23 23
22.4952
7
0.00067
2
13.5 24.5 24.5
19.8250
7
0.00059
9
15 26 26
17.6035
5
0.00053
7
16.5 27.5 27.5
15.7355
4
0.00048
4
18 29 29
14.1498
2
0.00043
8
19.5 30.5 30.5
12.7922
6
0.00039
9
14

21 32 32
11.6210
9
0.00036
4
22.5 33.5 33.5 10.6037
0.00033
4
24 35 35
9.71428
6
0.00030
7
25.5 36.5 36.5
8.93225
7
0.00028
3
Total settlement = 12.93 mm
5 Calculation of the loads on the culvert
Dead Loads on culvert:
Assume γ = 20kN/m3
BC = 3.6 m
BD = 4.0 m
H = 1.9 m
15

Angle of internal friction (Φ) = 30°
a. Vertical earth pressure due to fill (WFV) (AS 1597.2:2013, § 3.3.2)
Embankment Installation:
WFV = FE γ H
FE = 1+0.2 (H/BC)
= 1+0.2*(1.9/3.6) = 1.105
WFV = 1.105 * 20 * 1.9 = 41.99 kPa
Trench Installation:
WFV = FT γ H
FT = CDBD2/ H*BC
CD = 1-e-0.3848*H/BD / 0.3848 (for granular soil without cohesion)
= 1- e-(0.3848*1.9/4) / 0.3848 = 1-0.833 / 0.3848
= 0.434
FT = 0.434 * 4 * 4 / 1.9 * 3.6 = 1.015 ≤ FE (1.11)
WFV = 1.015 * 20 * 1.9 = 38.57 kPa
b. Horizontal earth Pressure due to fill (AS 1597.2:2013, § 3.3.3)
WFH = ko γ H
= 0.5 * 20 * 1.9 = 19 kPa
c. Horizontal pressure due to compaction (WAH) (AS 1597.2:2013, § 3.3.4)
ht
WA
H
WF
H
WFV
(trench)
Tot
al
0 0 19 38.57
57.
6
0.
1 3 19 38.57
60.
6
0.
2 6 19 38.57
63.
6
0.
3 9 19 38.57
66.
6
0.
4 12 19 38.57
69.
6
0.
5 15 19 38.57
72.
6
16

0.
6 14 19 38.57
71.
6
0.
7 13 19 38.57
70.
6
0.
8 12 19 38.57
69.
6
0.
9 11 19 38.57
68.
6
1 10 19 38.57
67.
6
1.
1 9 19 38.57
66.
6
1.
2 8 19 38.57
65.
6
1.
3 7 19 38.57
64.
6
1.
4 6 19 38.57
63.
6
1.
5 5 19 38.57
62.
6
1.
6 4 19 38.57
61.
6
1.
7 3 19 38.57
60.
6
1.
8 2 19 38.57
59.
6
1.
9 1 19 38.57
58.
6
2 0 19 38.57
57.
6
ht
WA
H
WF
H
WFV
(embankment)
Tot
al
0 0 19 41.99 61
0.
1 3 19 41.99 64
0.
2 6 19 41.99 67
0.
3 9 19 41.99 70
0.
4 12 19 41.99 73
0.
5 15 19 41.99 76
0.
6 14 19 41.99 75
0.
7 13 19 41.99 74
0.
8 12 19 41.99 73
17

0.
9 11 19 41.99 72
1 10 19 41.99 71
1.
1 9 19 41.99 70
1.
2 8 19 41.99 69
1.
3 7 19 41.99 68
1.
4 6 19 41.99 67
1.
5 5 19 41.99 66
1.
6 4 19 41.99 65
1.
7 3 19 41.99 64
1.
8 2 19 41.99 63
1.
9 1 19 41.99 62
2 0 19 41.99 61
Traffic load
Ultimate design load relevant to strength limit state
Consider W 80 and A 160 loads
Wheel contact length, a = 400 mm; b = 500 mm
Distribution of road wheel load through fill
A = L1L2 = (b+1.15H) (a+1.15H)
= (0.5 + 1.15 * 2) (0.4 + 1.15 * 2) = 2.8 * 2.7 = 7.56 m2
Vertical loads due to road traffic loadings (WLV) (AS 1597.2:2013, § 3.3.5.5.4)
ΣP = 80 kN (AS 5100.2:2017, § 7.2.2)
18

A = 7.56 m2 (calculated above)
DLA = 0.4 (AS 1597.2:2013, § 3.3.5.2)
WLV = (1+0.4) * 80 / 7.56 = 14.81 kPa
Horizontal loads due to road traffic loadings (WLH) (AS 1597.2:2013, § 3.3.5.5.5)
WLV = 14.81 kPa
ko = 0.5
WLV = 14.81 * 0.5 = 7.41 kPa
19