1
Geotechnical Engineering–I [CE-221]
BSc Civil Engineering – 4th Semester
by
Dr. Muhammad Irfan
Assistant Professor
Civil Engg. Dept. – UET Lahore
Email: mirfan1@msn.com
Lecture Handouts: https://groups.google.com/forum/#!forum/geotech-i
Lecture # 7
14-Feb-2017
2
US STANDARD SIEVE SIZES
Sieve No. Opening (mm) Sieve No. Opening (mm)
3 inch 76.200 20 0.850
2 inch 50.800 25 0.710
1.5 inch 38.100 30 0.600
1 inch 25.400 35 0.500
3/4 inch 19.000 40 0.425
3/8 inch 9.520 50 0.355
4 4.750 60 0.250
5 4.000 70 0.212
6 3.350 80 0.180
7 2.800 100 0.150
8 2.360 120 0.125
10 2.000 140 0.106
12 1.700 170 0.090
14 1.400 200 0.075
16 1.180 270 0.053
18 1.000
3
SIEVE ANALYSIS – Procedure
• Soil used in sieve analysis is oven-dried and all lumps are
broken.
• A stack of sieves (sieve nest), with sieve opening of
decreasing size from top to bottom, is arranged.
• A pan is placed below the stack.
• The soil is then shaken through this sieve nest.
• Mass retained on each sieve is determined.
Wet Sieving Technique
Breaking lumps in clayey soils may be difficult. If soils contain silts and
clays, the wet sieving is usually used to preserve the fine content.
In this case, the soil may be mixed with water to make a slurry and then
washed through sieves.
Portions retained on each sieve are collected separately and oven-dried.
4
76.2 mm
4.75 mm
0.075 mm
Pan
100 %
Finer (Passing)
93 %
62 %
0 %
Gravel
Sand
Silt and Clay
100 – 93 = 7%
93 – 62 = 31%
62 – 0 = 62%
Gravel
Sand
Silt and Clay
SIEVE ANALYSIS – Procedure
5
CumulativePercentagepassing
througheachsieve(%)
Grain Diameter (mm)
Log scale
SIEVE ANALYSIS – Results
(Gradation Curve)
6
SIEVE ANALYSIS – Calculations
Sieve
No.
Diameter
(mm)
Wt. of soil
retained
(gm)
Cumulative
soil weight
retained on
each sieve
(gm)
Cumulative
percentage
retained (%)
Cumulative
percentage
passing (%)
(Col. 1) (Col. 2) (Col. 3) (Col. 4) (Col. 5) (Col. 6)
(Col. 4) = (Col. 3) + (Col. 4) of previous line
(Col. 5) = [(Col. 4)/Total wt.] x 100
(Col. 6) = 100 – (Col. 5)
7
SIEVE ANALYSIS – Example
8
SIEVE ANALYSIS – Example
9
0
10
20
30
40
50
60
70
80
90
100
0.010.1110
PercentFiner%
Particle Size (mm)
SIEVE ANALYSIS – Example
10
SOIL GRADATIONS
• Well graded soils
• Poorly graded soils
• Uniformly graded soils
• Gap graded soils
11
GRAIN SIZE DISTRIBUTION CURVE
Well-graded soil
Poorly-graded
(uniformly graded) soil
12
D10, D30, and D60
10
Percentpassing
30
60
Grain Diameter
D60 D30 D10
D10= Diameter corresponding to 10% passing
D30= Diameter corresponding to 30% passing
D60= Diameter corresponding to 60% passing
D50= Diameter corresponding to 50% passing
= Mean diameter of soil sample
13
COEFFICIENTS OF GRADATION
Coefficient of Uniformity
Coefficient of Curvature
10
60
D
D
Cu 
 1060
2
30
DD
D
Cc


)(6
)(4
31
sandsforC
gravelsforC
and
C
u
u
c



For a well-graded soils
14
What is the Cu for a soil with only one
grain size?
POINT TO PONDER!!!
15
D
%Finer
1
D
D
C
uniformityoftCoefficien
10
60
u 
Grain Size Distribution
Cu of Uniformly Graded Soil
16
GRAIN SIZE DISTRIBUTION CURVE
100 10 1 0.1 0.01 0.001
A
D
C
E
B
#200#43”
A: fine-grained soils
B: coarse-grained soils
Cu = 5
Cc = 0.8
Cu = 13
Cc = 0.83
17
100 10 1 0.1 0.01 0.001
A
D
C
E
B
#200#43”
C: poorly graded or
uniformly graded
Cu = 1.63
Cc = 0.96
GRAIN SIZE DISTRIBUTION CURVE
18
100 10 1 0.1 0.01 0.001
A
D
C
E
B
#200#43”
D: well-graded soils
E: gap-graded soils
Cu = 190
Cc = 1.18
Cu = 111
Cc = 0.18
GRAIN SIZE DISTRIBUTION CURVE
19
SIEVE ANALYSIS – Calculations
Sieve
No.
Diameter
(mm)
Wt. of soil
retained
(gm)
Cumulative
soil weight
retained on
each sieve
(gm)
Cumulative
percentage
retained (%)
Cumulative
percentage
passing (%)
(Col. 1) (Col. 2) (Col. 3) (Col. 4) (Col. 5) (Col. 6)
(Col. 4) = (Col. 3) + (Col. 4) of previous line
(Col. 5) = [(Col. 4)/Total wt.] x 100
(Col. 6) = 100 – (Col. 5)
20
SIEVE ANALYSIS
(Practice Problem)
21
SIEVE ANALYSIS
(Practice Problem)
22
SIEVE ANALYSIS
(Practice Problem)
0
10
20
30
40
50
60
70
80
90
100
0.010.1110
%Finer
Grain size (mm)
D10=0.09 mm
D30=0.27 mm
D60=0.58 mm
Cu = 6.44
Cc = 1.40
23
SIEVE ANALYSIS
(Practice Problem)
0
10
20
30
40
50
60
70
80
90
100
0.010.1110
%Finer
Grain size (mm)
98%
49%
Medium sand → (2.0 mm – 0.425 mm)
%age of medium sand = 98-49 = 49%
%age of gravel (75 - 4.75 mm)
%age of coarse sand (4.75 - 2.0 mm)
%age of medium sand (2.0 – 0.425 mm)
%age of fine sand (0.425 – 0.075 mm)
%age of silt and clay (passing sieve #200)
= nil
= 2
= 49
= 43
= 6
24
0
20
40
60
80
100
0.0010.010.1110
Percentpassing
Grain diameter (mm)
Grain-size distribution curve
GRAVEL SAND
SILT or CLAY
Coarse Fine Coarse Medium Fine
3/4 in. No.4 No.10 No.40 No.200 Hydrometer3 in.
D10 = 0.10 mm
D30 = 0.18 mm
D60 = 0.25 mm
Cu = 2.50
Cc = 1.30
25
Coarse-grained soils:
Gravel Sand
Fine-grained soils:
Silt Clay
0.075 mm (USCS)
0.06 mm (BS)
Sieve Analysis Hydrometer Analysis
MECHANICAL ANALYSIS OF SOIL
Mechanical analysis is the determination of the size range of particles present
in a soil, expressed as a percentage of the total dry weight.
26
HYDROMETER ANALYSIS –
Stoke’s Law
For a sphere moving in a fluid, the drag force acting on it is a
function of its diameter.
Stoke’s Law
V1 V2<
→ Silt will settle faster compared to
clay.
→ Why soil particles settle in water?
27
HYDROMETER ANALYSIS –
Conceptual Basis
→ Soil settles down
→ Gs of the solution decreases
→ Hydrometer settles in water
28
• At any time t, hydrometer measures the
specific gravity of the suspension near the
vicinity of its bulb (depth L).
• For pure water hydrometer reading will be 1.
• It indirectly gives the amount of soil that is
still in suspension.
• By knowing the amount of soil in
suspension, L, and t, we can calculate the
%age of soil by weight finer than a given
diameter.
HYDROMETER ANALYSIS
29
R
L
HYDROMETER ANALYSIS
(cm)164.029.16(cm) RL 
where, R = hydrometer reading corrected for meniscus
For ASTM 152H hydrometer
How to correlate L with particle size?
Corrected hydrometer reading;
Rc = Ractual – Zero Correction + Temp. Correction
100


s
c
W
aR
finerPercent
a = correction factor for specific gravity
Ws = Total weight of soil specimen
Zero correction = correction due to the use of deflocculating agent
30
CONCLUDED
REFERENCE MATERIAL
Principles of Geotechnical Engineering (7th Edition)
By Braja M. Das
Chapter #2

Geotechnical Engineering-I [Lec #7: Sieve Analysis-2]

  • 1.
    1 Geotechnical Engineering–I [CE-221] BScCivil Engineering – 4th Semester by Dr. Muhammad Irfan Assistant Professor Civil Engg. Dept. – UET Lahore Email: mirfan1@msn.com Lecture Handouts: https://groups.google.com/forum/#!forum/geotech-i Lecture # 7 14-Feb-2017
  • 2.
    2 US STANDARD SIEVESIZES Sieve No. Opening (mm) Sieve No. Opening (mm) 3 inch 76.200 20 0.850 2 inch 50.800 25 0.710 1.5 inch 38.100 30 0.600 1 inch 25.400 35 0.500 3/4 inch 19.000 40 0.425 3/8 inch 9.520 50 0.355 4 4.750 60 0.250 5 4.000 70 0.212 6 3.350 80 0.180 7 2.800 100 0.150 8 2.360 120 0.125 10 2.000 140 0.106 12 1.700 170 0.090 14 1.400 200 0.075 16 1.180 270 0.053 18 1.000
  • 3.
    3 SIEVE ANALYSIS –Procedure • Soil used in sieve analysis is oven-dried and all lumps are broken. • A stack of sieves (sieve nest), with sieve opening of decreasing size from top to bottom, is arranged. • A pan is placed below the stack. • The soil is then shaken through this sieve nest. • Mass retained on each sieve is determined. Wet Sieving Technique Breaking lumps in clayey soils may be difficult. If soils contain silts and clays, the wet sieving is usually used to preserve the fine content. In this case, the soil may be mixed with water to make a slurry and then washed through sieves. Portions retained on each sieve are collected separately and oven-dried.
  • 4.
    4 76.2 mm 4.75 mm 0.075mm Pan 100 % Finer (Passing) 93 % 62 % 0 % Gravel Sand Silt and Clay 100 – 93 = 7% 93 – 62 = 31% 62 – 0 = 62% Gravel Sand Silt and Clay SIEVE ANALYSIS – Procedure
  • 5.
    5 CumulativePercentagepassing througheachsieve(%) Grain Diameter (mm) Logscale SIEVE ANALYSIS – Results (Gradation Curve)
  • 6.
    6 SIEVE ANALYSIS –Calculations Sieve No. Diameter (mm) Wt. of soil retained (gm) Cumulative soil weight retained on each sieve (gm) Cumulative percentage retained (%) Cumulative percentage passing (%) (Col. 1) (Col. 2) (Col. 3) (Col. 4) (Col. 5) (Col. 6) (Col. 4) = (Col. 3) + (Col. 4) of previous line (Col. 5) = [(Col. 4)/Total wt.] x 100 (Col. 6) = 100 – (Col. 5)
  • 7.
  • 8.
  • 9.
  • 10.
    10 SOIL GRADATIONS • Wellgraded soils • Poorly graded soils • Uniformly graded soils • Gap graded soils
  • 11.
    11 GRAIN SIZE DISTRIBUTIONCURVE Well-graded soil Poorly-graded (uniformly graded) soil
  • 12.
    12 D10, D30, andD60 10 Percentpassing 30 60 Grain Diameter D60 D30 D10 D10= Diameter corresponding to 10% passing D30= Diameter corresponding to 30% passing D60= Diameter corresponding to 60% passing D50= Diameter corresponding to 50% passing = Mean diameter of soil sample
  • 13.
    13 COEFFICIENTS OF GRADATION Coefficientof Uniformity Coefficient of Curvature 10 60 D D Cu   1060 2 30 DD D Cc   )(6 )(4 31 sandsforC gravelsforC and C u u c    For a well-graded soils
  • 14.
    14 What is theCu for a soil with only one grain size? POINT TO PONDER!!!
  • 15.
  • 16.
    16 GRAIN SIZE DISTRIBUTIONCURVE 100 10 1 0.1 0.01 0.001 A D C E B #200#43” A: fine-grained soils B: coarse-grained soils Cu = 5 Cc = 0.8 Cu = 13 Cc = 0.83
  • 17.
    17 100 10 10.1 0.01 0.001 A D C E B #200#43” C: poorly graded or uniformly graded Cu = 1.63 Cc = 0.96 GRAIN SIZE DISTRIBUTION CURVE
  • 18.
    18 100 10 10.1 0.01 0.001 A D C E B #200#43” D: well-graded soils E: gap-graded soils Cu = 190 Cc = 1.18 Cu = 111 Cc = 0.18 GRAIN SIZE DISTRIBUTION CURVE
  • 19.
    19 SIEVE ANALYSIS –Calculations Sieve No. Diameter (mm) Wt. of soil retained (gm) Cumulative soil weight retained on each sieve (gm) Cumulative percentage retained (%) Cumulative percentage passing (%) (Col. 1) (Col. 2) (Col. 3) (Col. 4) (Col. 5) (Col. 6) (Col. 4) = (Col. 3) + (Col. 4) of previous line (Col. 5) = [(Col. 4)/Total wt.] x 100 (Col. 6) = 100 – (Col. 5)
  • 20.
  • 21.
  • 22.
    22 SIEVE ANALYSIS (Practice Problem) 0 10 20 30 40 50 60 70 80 90 100 0.010.1110 %Finer Grainsize (mm) D10=0.09 mm D30=0.27 mm D60=0.58 mm Cu = 6.44 Cc = 1.40
  • 23.
    23 SIEVE ANALYSIS (Practice Problem) 0 10 20 30 40 50 60 70 80 90 100 0.010.1110 %Finer Grainsize (mm) 98% 49% Medium sand → (2.0 mm – 0.425 mm) %age of medium sand = 98-49 = 49% %age of gravel (75 - 4.75 mm) %age of coarse sand (4.75 - 2.0 mm) %age of medium sand (2.0 – 0.425 mm) %age of fine sand (0.425 – 0.075 mm) %age of silt and clay (passing sieve #200) = nil = 2 = 49 = 43 = 6
  • 24.
    24 0 20 40 60 80 100 0.0010.010.1110 Percentpassing Grain diameter (mm) Grain-sizedistribution curve GRAVEL SAND SILT or CLAY Coarse Fine Coarse Medium Fine 3/4 in. No.4 No.10 No.40 No.200 Hydrometer3 in. D10 = 0.10 mm D30 = 0.18 mm D60 = 0.25 mm Cu = 2.50 Cc = 1.30
  • 25.
    25 Coarse-grained soils: Gravel Sand Fine-grainedsoils: Silt Clay 0.075 mm (USCS) 0.06 mm (BS) Sieve Analysis Hydrometer Analysis MECHANICAL ANALYSIS OF SOIL Mechanical analysis is the determination of the size range of particles present in a soil, expressed as a percentage of the total dry weight.
  • 26.
    26 HYDROMETER ANALYSIS – Stoke’sLaw For a sphere moving in a fluid, the drag force acting on it is a function of its diameter. Stoke’s Law V1 V2< → Silt will settle faster compared to clay. → Why soil particles settle in water?
  • 27.
    27 HYDROMETER ANALYSIS – ConceptualBasis → Soil settles down → Gs of the solution decreases → Hydrometer settles in water
  • 28.
    28 • At anytime t, hydrometer measures the specific gravity of the suspension near the vicinity of its bulb (depth L). • For pure water hydrometer reading will be 1. • It indirectly gives the amount of soil that is still in suspension. • By knowing the amount of soil in suspension, L, and t, we can calculate the %age of soil by weight finer than a given diameter. HYDROMETER ANALYSIS
  • 29.
    29 R L HYDROMETER ANALYSIS (cm)164.029.16(cm) RL where, R = hydrometer reading corrected for meniscus For ASTM 152H hydrometer How to correlate L with particle size? Corrected hydrometer reading; Rc = Ractual – Zero Correction + Temp. Correction 100   s c W aR finerPercent a = correction factor for specific gravity Ws = Total weight of soil specimen Zero correction = correction due to the use of deflocculating agent
  • 30.
    30 CONCLUDED REFERENCE MATERIAL Principles ofGeotechnical Engineering (7th Edition) By Braja M. Das Chapter #2