1. Following are the results of a sieve analysis: US Seive 4 10 20 40 60 100 200 Pan No. Mass of soil 0 21.6 49.5 102.6 89.1 95.6 60.4 31.2 retained (gm) 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. e. Classify the soil as per ASTM, USCS and AASHTO. Solution: a. The percentage finer is determined in the table below and the grain size distribution curve (semi-log graph) is also shown. US Openin Mass of Total mass Total Mass Percentage Seive g (mm) Soil of soil of soil Finer than No Retained retained passsing (gm) (gm) (gm) I II III IV V = TotalVI = IV (V/Total)* 100 Pan 0.000 31.2 450.0 0.00 0.0 200 0.075 60.4 418.8 31.2 6.933 100 0.152 95.6 358.4 91.6 20.356 60 0.251 89.1 262.8 187.2 41.6 40 0.422 102.6 173.7 276.3 61.4 20 0.853 49.5 71.1 378.9 84.2 10 2.000 21.6 21.6 428.4 95.2 4 4.750 0.00 0.0 450.0 100.0 Total = 450.00 gm b. From the Grain-size distribution curve, For 10 % finer, D10= 0.09 mm For 30 % finer, D30= 0.2 mm For 60 % finer, D60= 0.4 mm c. So, Coefficient of Uniformity (Cu) = D60/D10 = 0.4/0.09 = 4.44 d. And, Coefficient of gradation (Cc) = (D30)2 /(D60xD10) = 0.22/(0.4x0.09) = 1.11 e. For USCS classification, Percentage passing No. 200 sieve = 6.93 % .So, the soil is coarse grained soil All of the soil (100%) passes through No 4 sieve so the soil is categorized as sand. As the percentage of fines (passing No 200 sieve = 6.93%) lies between (5-12) % it is a borderline case requiring dual symbol.
Since Cu=4.44 < 6 and Cc=1.11 > 1 the soil is poorly graded sand. For dual symbol the atterberg limits of the soil are not provided so it may be SP-SC or SP-SM. 2. The moist mass of 2.8x10-3m3 of soil is 5.53 Kg. If the moisture content is 10% and the specific gravity of soil solids is 2.72, determine the following: a. Moist density; b. Dry density; c. Void ratio; d. Porosity; e. Degree of Saturation; f. Volume occupied by water. Solution: Mass of Soil (M) = 5.53 Kg; Volume of Soil (V) = 2.8 x 10 -3 m3;
Grain-Size distribution Curve (Semi-log Plot) 100
100.00 95.20
90 84.20
80 70 61.40
60
Percentage finer than
50 41.60
40 30 20 10 0 0.010
0.100
1.000
Seive Sizes
10.000
Moisture Content (Wc) = 10%; Specific Gravity of soil solids (Gs) = 2.72 a. Moist/Bulk density of Soil (ρ) = M/V = 5.53/2.8*10-3 = 1975 Kg/m3 b. Moisture content (Wc) = Mw/Ms = (M-Ms)/Ms= (M/Ms)-1
⇒ Ms= M/(1+Wc) = 5.53/ (1+0.1) = 5.027 Kg ∴ Dry density (ρd) = Ms/V = 5.027/2.8*10-3= 1795 Kg/m3
c. Also, Dry density (ρd) = Gs.ρw/(1+e)
⇒
e= (Gs.ρw/ρd)-1= (2.72x1000/1795)-1 =0.5153 = 51.53 %
d. Porosity (η) = e/(1+e) = (0.5153/1.5153) = 0.34 = 34% e. Degree of saturation (Sr) = Wc.Gs/e = (0.1x2.72)/0.5153 = 0.5278 =52.78% f. Volume occupied by water (Vw) = (M-Ms)/ρw = (5.53-5.027)/1000=5.03x104 m3 3. For a sandy soil emax=0.75, emin=0.52 and Gs=2.7. What are the void ratio and the dry unit weight at Dr=80%. Solution: We have, Relative Density(Dr) = (emax-e)/(emax-emin)
⇒ e=emax-Dr.(emax-emin)= 0.75-0.8x(0.75-0.52) = 0.566
Dry Unit Weight (γd) = Gs.γw/(1+e) = 2.7x9.81/(1+0.566) =16.913 KN/m3 4. Classify the following soils using the Unified Soil Classification System. Give the Group Symbols and Group Names (ASTM) Sieve Analysis % Liquid Plastic Cu Cc finer Limit Limit No. 4 No. 200 1 70 30 33 12 2 48 20 41 19 3 95 70 52 24 4 100 82 30 11 5 88 78 69 31 6 71 4 NP 3.4 2.6 7 99 57 54 28 8 71 11 32 16 4.8 2.9 9 100 2 NP 7.2 2.2 10 90 8 39 31 3.9 2.1 Solution: Soil 1: For USCS classification, Percentage passing No. 200 sieve = 30 % < 50%. So, the soil is coarse grained. Percentage passing No. 4 sieve = 70 % > 50%. So, the soil is Sand. Soi l
Percentage passing No. 200 sieve = 30 % > 12%
⇒ Sands with fines.
PI= LL-PL = 33-12 = 21 From A-line Equation, PI=0.73x (LL-20) = 0.73x (33-20) = 9.49 < 21
So, it plots above A-line.
⇒ Sand with clay
Hence, USCS Classification: SC For ASTM classification, Percentage of Gravel = 30% > 15% Hence, ASTM Classification: Clayey sand with Gravel Soil 2: For USCS classification, Percentage passing No. 200 sieve = 20 % < 50%. So, the soil is coarse grained. Percentage passing No. 4 sieve = 48 % < 50%. So, the soil is Gravel. Percentage passing No. 200 sieve = 20 % > 12%
⇒ Gravel with
fines. PI= LL-PL = 41-19 = 22 From A-line Equation, PI=0.73x (LL-20) = 0.73x (41-20) = 15.33 < 22 So, it plots above A-line.
⇒ Gravel with clay
Hence, USCS Classification: GC For ASTM classification, Percentage of Sand = 48% > 15% Hence, ASTM Classification: Clayey Gravel with Sand Soil 3: For USCS classification, Percentage passing No. 200 sieve = 70 % > 50%. So, the soil is fine grained. Liquid Limit = 52 % > 50%. So, the soil is High Plasticity silt or clay. PI= LL-PL = 52-24 = 28 From A-line Equation, PI=0.73x (LL-20) = 0.73x (52-20) = 23.36 < 28 So, it plots above A-line.
⇒ Clay with High Plasticity
Hence, USCS Classification: CH For ASTM classification, Percentage passing No. 200 sieve = 70 % > 30% And, Percentage of Sand = 95% > Percentage of Gravel (5%) < 15% Hence, ASTM Classification: Sandy Fat Clay Soil 4: For USCS classification, Percentage passing No. 200 sieve = 82 % > 50%. So, the soil is fine grained. Liquid Limit = 30 % < 50%. So, the soil is Low Plasticity silt or clay. PI= LL-PL = 30-11 = 19 From A-line Equation, PI=0.73x (LL-20) = 0.73x (30-20) = 7.3 < 19 So, it plots above A-line.
⇒ Clay with Low Plasticity
Hence, USCS Classification: CL For ASTM classification, Percentage passing No. 200 sieve = 82 % > 30% And, Percentage of Sand = 100% > Percentage of Gravel (0%) < 15%
Hence, ASTM Classification: Sandy Lean Clay. Soil 5: For USCS classification, Percentage passing No. 200 sieve = 78 % > 50%. So, the soil is fine grained. Liquid Limit = 69 % > 50%. So, the soil is High Plasticity silt or clay. PI= LL-PL = 69-31 = 38 From A-line Equation, PI=0.73x (LL-20) = 0.73x (69-20) = 35.77 < 38 So, it plots above A-line.
⇒ Clay with High Plasticity
Hence, USCS Classification: CH For ASTM classification, Percentage passing No. 200 sieve = 78 % > 30% And, Percentage of Sand = 88 % > Percentage of Gravel (12%) < 15% Hence, ASTM Classification: Sandy Fat Clay Soil 6: For USCS classification, Percentage passing No. 200 sieve = 4 % < 50%. So, the soil is coarse grained. Percentage passing No. 4 sieve = 71 % > 50%. So, the soil is Sand. Percentage passing No. 200 sieve = 4 % < 5%
⇒ Clean Sand.
Cu= 2.4 < 4 and 1 ≤Cc = 2.6 ≤ 3 So, the Sand is poorly graded Hence, USCS Classification: SP For ASTM classification, Percentage of Gravel = (100 – 71) % = 29% > 15% Hence, ASTM Classification: Poorly graded Sand with Gravel Soil 7: For USCS classification, Percentage passing No. 200 sieve = 57 % > 50%. So, the soil is fine grained. Liquid Limit = 54 % > 50%. So, the soil is High Plasticity silt or clay. PI= LL-PL = 54-28 = 26 From A-line Equation, PI=0.73x (LL-20) = 0.73x (54-20) = 24.82 < 26 So, it plots above A-line.
⇒ Clay with High Plasticity
Hence, USCS Classification: CH For ASTM classification, Percentage passing No. 200 sieve = 57 % > 30% And, Percentage of Sand = 99 % > Percentage of Gravel (1 %) < 15% Hence, ASTM Classification: Sandy Fat Clay Soil 8: For USCS classification, Percentage passing No. 200 sieve = 11 % < 50%. So, the soil is Coarse grained. Percentage passing No. 4 sieve = 71 % > 50%. So, the soil is Sand. Percentage passing No. 200 sieve = 11 % lies between (5 – 12) % Dual Symbol.
⇒
Cu= 4.8 > 4 and 1 ≤ Cc = 2.9 ≤ 3 So, the Sand is poorly graded (SP). PI=LL-PL=32-16=16 From A-line Equation, PI=0.73x (LL-20) = 0.73x (32-20) = 8.76 < 16 So, it plots above A-line.
⇒ Sand contains clay (SC).
Hence, USCS Classification: SP-SC For ASTM classification, Percentage of Gravel = (100 – 71) % = 29% > 15% Hence, ASTM Classification: Poorly graded Sand with Clay and Gravel Soil 9: For USCS classification, Percentage passing No. 200 sieve = 2 % < 50%. So, the soil is Coarse grained. Percentage passing No. 4 sieve = 100 % < 50%. So, the soil is Sand. Percentage passing No. 200 sieve = 4 % < 5%
⇒ Clean Sand.
Cu= 7.2 > 6 and 1 ≤Cc = 2.2 ≤ 3 So, the Sand is poorly graded Hence, USCS Classification: SP For ASTM classification, Percentage of Gravel = (100 – 100) % = 0% < 15% Hence, ASTM Classification: Poorly graded Sand Soil 10: For USCS classification, Percentage passing No. 200 sieve = 8 % < 50%. So, the soil is Coarse grained. Percentage passing No. 4 sieve = 90 % > 50%. So, the soil is Sand. Percentage passing No. 200 sieve = 8 % lies between (5 – 12) % Dual Symbol. Cu= 3.9 < 4 and 1 ≤ Cc = 2.1 ≤ 3 So, the Sand is well graded (SW). PI=LL-PL=39-31= 8 From A-line Equation, PI=0.73x (LL-20) = 0.73x (39-20) = 13.87 >8 So, it plots below A-line.
⇒ Sand contains silt (SM).
Hence, USCS Classification: SW-SM For ASTM classification, Percentage of Gravel = (100 – 90) % = 10% < 15% Hence, ASTM Classification: Well graded Sand with Silt So il 1 2 3 4
Summary Unified Soil Classification American Society of Testing System (USCS) Materials (ASTM) Classification SC Clayey sand with Gravel GC Clayey Gravel with Sand CH Sandy Fat Clay CL Sandy Lean Clay
⇒
5 6 7 8
CH SP CH SP-SC
9 10
SP SW-SM
Sandy Fat Clay Poorly graded Sand with Gravel Sandy Fat Clay Poorly graded Sand with Clay and Gravel Poorly graded Sand Well graded Sand with Silt
5. Laboratory compaction test results on a clayey soil are listed in the table Moisture Dry Unit Weight Content (%) (KN/ m3) 6 14.80 8 17.45 9 18.52 11 18.9 12 18.5 14 16.9 Following are the results of a field unit weight determination test on the same soil with the sand cone method.
Calibrated dry density of Ottawa sand = 1570 Kg/ m3 Calibrated mass of Ottawa sand to fill the cone = 0.545 Kg Mass of Sand+ Cone +Jar (before use) = 7.59 Kg Mass of Sand+ Cone +Jar (after use) = 4.78 Kg Mass of moist soil from hole = 3.007 Kg Moisture content of Moist soil = 10.2%
Determine a. Dry unit weight in field b. Relative compaction in field Solution: Given, Mass of soil from hole (M) = 3.007 Kg Mass of Sand+ Cone +Jar (before use) (M 1) = 7.59 Kg Mass of Sand+ Cone+ Jar (after use) (M2) = 4.78 Kg Mass of Ottawa sand to fill the cone (M3) = 0.545 Kg Calibrated dry density of Ottawa sand (ρsand) = 1570 Kg/ m3 Moisture content of moist soil (Wc) = 10.2% Volume of the excavated hole (V) = (M1-M2-M3)/ ρsand = (7.59-4.78-0.545)/ 1570 = 2.265/ 1570 = 1.443*10-3 m3 Density of soil (ρfield) = M/V = 3.007/ 1.443*10-3 = 2083.85 Kg/ m3 a. Dry unit weight in field (γd,field) = ρfield*g/ (1+Wc)
Moisture Content V/s Dry Unit Weig 20
18.90
18.52
19
17.45
18 17
Dry Unit Weight (KN/ m3)
16
14.80
15 14 13 12 5
6
7
8
9
10
11
Moisture Content (%
= 2083.85x 9.81/ (1.102) = 18.55 KN/ m3 b. For max dry unit weight, From the graph of Moisture content V/s Dry unit weight, the dry unit weight corresponding to the Optimum Moisture Content is γd = 18.9 KN/m3 Relative Compaction = γd,field / γd = 18.55/ 18.9 = 0.9815 = 98.15 % 6. Calculate the seepage loss per unit length of the sheet pile (at right angles to the cross section shown) using flow net and analytical method. H1= 5m; D= 4m; H2 = 0.7m; D1 =10 m Solution: The flow-net diagram for the given problem is as shown in the attached graph sheet. From the graph, The no. of flow channels (Nf) = 5 No. of potential drop (Nd) = 10 Head Loss (h) = 5 - 0.7 = 4.3 m
Nf
∴ Seepage Loss (Q)
= k.h.
Nd
= 6.5x10-4x4.3x100x = 0.13975 cm3/
5 10
(sec.cm)
= 1.2074 m3/ (day. m) From analytical method (Method of fragments),
2 1
As shown in figure alongside both the fragments 1 and 2 are type II fragments as per Pavlovsky, For type II fragment, S = D = 4m T = D1 = 10m S/T = 4/10 = 0.4 From the graph for S/T = 0.4 (1/2φ) = 0.58
∴
Form factor (φ) = 0.86
h
Discharge (Q)
= k.
4.3 x10 2 2 * 0.86
= 6.5x 10-4x = 0.1625 cm3/ (sec.cm) = 1.404 m3/ (day. m) 7. A gravity dam 35m wide is retaining 20m of water. The dam was constructed 2m deep in a silty sand layer of 30m thickness overlying an impermeable basalt. In order to reduce seepage a cut-off wall 15m deep was constructed exactly at the downstream face. The horizontal coefficient of permeability of the silty sand layer is 2.25x10-3m/ sec while in the vertical direction is 10-3 m/sec. i) Construct the flow net in graph adopting a scale of 1:400 in vertical direction. ii) Calculate seepage. iii) Investigate safety factor at the exit due piping. iv) Determine the distribution of hydraulic pressure across the base of the dam at interval distance of 5m. v) Solve ii-iv using method of fragments. vi) If instead of downstream face the cut off wall is placed at the upstream face of the dam, determine the seepage using method of fragments. Determine also the distribution of hydraulic pressure at the base of the dam at 5m intervals. vii) What effect does the position of cut-off wall have on the seepage and hydrostatic pressure? Solution: Here, Permeability in Vertical direction, Kz = 2.25x10-3m/ sec Permeability in Horizontal direction, Kx = 10-3 m/sec Scale adopted in Vertical direction = 1:400 For Anisotropic soil, Reduction in horizontal scale = √ (K z/Kx) = √2.25 = 1.5 Horizontal Scale = 1: (400x1.5) = 1: 600 i) The flow-net diagram for given problem adopting above scale is shown in attached graph sheet. ii) From the flow-net diagram No of flow channels (Nf) =
No. of potential drops (Nd) = Head Loss (h) = 20m
Nf
∴ Seepage Loss (Q)
= k.h.
Nd
k =√ K x . K z= √2.25 x 10−3 x 10−3 =1.5 x 10−3 5 10
= 1.5x10-3x20x = 0.13975 m3/ (sec. m) = 1.2074 m3/ (day. m) iii) iv) v) vi)
m/ sec
20 m
2m
35 m
1
15 m
30 m
2