Geotechnical Engineering Calculations and Rules of Thumb Copyright © 2016 Elsevier Inc. All rights reserved.
2. Hydrometer 3. Water content
4. Atterberg limit tests (liquid limit and plastic limit) 5. Permeability test
6. UU tests (undrained unconfined tests) 7. Density of soil
8. Consolidation test 9. Triaxial tests 10. Direct shear test
4.1 Sieve analysis
Sieve analysis is conducted to classify soil into sands, silts, and clays. Sieves are used to separate soil particles and group them based on their size. This test is used for the purpose of classification of soil.
Standard sieve sizes are shown in Tables 4.1 and 4.2.
Hypothetical sieve analysis test based on selected group of sieves is given in Figure 4.1 as an example.
If we know the percentage of soil passed from a given sieve, we can find the per- centage of soil retained in that sieve.
• Sieve no. 4 (size 4.75 mm): All soil went past Sieve no. 4.
Percent retained at this sieve is 0%.
Percent passed = 100%.
• Sieve no. 16 (1.18 mm): 20% of soil was retained in Sieve no. 16.
Percent retained at Sieve no. 16 = 20%
Percent passed = 100 – 20 = 80%.
• Sieve no. 50 (0.30 mm): 25% of soil was retained in Sieve no. 50.
Total retained, so far = 20 + 25 = 45%
Percent passed = 100 – 45 = 55%
• Sieve no. 80 (0.18 mm): 20% retained in Sieve no. 80.
Total retained so far = 20 + 25 + 20 = 65%
Percent passed = 100 – 65 = 35%.
• Sieve no. 200 (0.075 mm): 30% retained in Sieve no. 200.
Total retained so far = 20 + 25 + 20 + 30 = 95%
Percent passed through Sieve no. 200 = 100 – 95 = 5%.
Now it is possible to draw a graph indicating percent passing at each sieve (Figure 4.2).
4.1.1 D60
D60 is defined as the size of the sieve that allows 60% of the soil to pass. This value is used for soil classification purposes and it frequently appears in geotechnical engi- neering correlations.
Table 4.2
British sieve number and mesh size
British sieve number Size (mm)
8 2.057
16 1.003
30 0.500
36 0.422
52 0.295
60 0.251
85 0.178
100 0.152
200 0.076
300 0.053
Notes: gravel, sizes greater than #4 sieve is considered to be gravel; sands, #4–#200; silts and clays, smaller than #200.
Table 4.1
US sieve number and mesh size
Sieve number Mesh size (mm)
4 4.75
6 3.35
8 2.36
10 2.00
12 1.68
16 1.18
20 0.85
30 0.60
40 0.425
50 0.30
60 0.25
80 0.18
100 0.15
200 0.075
270 0.053
Notes: gravel, sizes greater than #4 sieve is considered to be gravel; sands, #4–#200; silts and clays, smaller than #200.
If you look at the 0.3 mm sieve, 55% of the soil would pass though this sieve. If we make the sieve bigger, more soil would pass. On the other hand, if we make the sieve smaller, then less than 55% of the soil would pass.
To find the D60 value, draw a line at 60% passing point. Then drop it down to obtain the D60 value (Figure 4.3).
In this case, D60 is closer to 0.5 mm.
4.1.2 Find D30
As earlier, draw a line at 30% passing line. In this case, D30 happened to be at approxi- mately 0.1 mm (Tables 4.3 and 4.4).
Figure 4.1 Sieve analysis.
Figure 4.2 Percent passing versus particle size.
4.2 Hydrometer
Hydrometer tests are conducted to classify particles smaller than #200 sieve.
(0.075 mm).
When soil particles are mixed with water, larger particles would settle fast. On the other hand, smaller particles tend to float or settle at a low velocity (Figure 4.4).
Since D2 is greater than D1, the velocity of sand particles (V2) will be greater than the velocity of silt particles (V1). Similarly, V3 will be greater than V2.
Figure 4.3 Finding D60.
Table 4.3
Size ranges for soils and gravels
Soil Size in (in.) Size in (mm) Comments
Boulders 6 or larger 150 or larger
Cobbles 3–6 75–150
Gravel 0.187–3 4.76–75 Greater than #4 sieve size
Sand 0.003–0.187 0.074–4.76 Sieve #200 to sieve #4
Silt 0.00024–0.003 0.006–0.074 Smaller than sieve #200 Clay 0.00004–0.00008 0.001–0.002 Smaller than sieve #200 Colloids Less than 0.00004 Less than 0.001
Table 4.4
Specific gravity (G
s)
Soil Specific gravity
Gravel 2.65–2.68
Sand 2.65–2.68
Silt (inorganic) 2.62–2.68
Organic clay 2.58–2.65
Inorganic clay 2.68–2.75
Velocity of settling particles is given by the “Stokes Law”.
= × −η × V 980 (G G ) D
30
w 2
G, specific gravity of soil; Gw, specific gravity of water; D, particle diameter (mm);
V, velocity of particles (mm/s); η, absolute viscosity of water (in Poises).
“V” and “D” are unknown quantities in the given equation.
If the settling velocity “V” can be found by an experiment then the particle diameter
“D” can be computed.
4.2.1 Hydrometer test procedure
The reader is referred to the ASTM D422 for a full explanation of the hydrometer test procedure. Overview of the test is provided here (Figure 4.5).
V=980×(G−Gw)×D230η Figure 4.4 Settling soil particles.
Figure 4.5 Hydrometer.
Hydrometer reading is dependant upon the density of the liquid. When soil is mixed with water very fine particles will suspend in water while the heavy particles will settle at the bottom. If more soil were to suspend in water hydrometer reading,
“L” will be smaller. The reading “L” is an indication of the amount of fine particles suspended in water.
4.2.1.1 Procedure
Mix 50 g of oven dried soil and 1000 mL of distilled water.
• Mix soil and water thoroughly.
• Insert the hydrometer and obtain the hydrometer readings, “L”.
• The hydrometer reading provides grams of soil in suspension per liter of solution.
• Obtain readings for different time periods. Typically 2, 5, 15, 30, 60, 250, and 1440 min.
• Compute “D” for every reading.
• Plot a graph between grams of soil per liter of water versus “D”.
• Soil particles will increase the density of water/soil mixture. Hence, initially “L” would be a lower value. When time passes, the larger diameter particles would settle. The density of the soil/water mixture would go down and the hydrometer would sink. (“L” value would increase).
• Lets say at time T1, the length was measured to be L1. Use the Stokes equation to find D1.
= = −η V L
T
G G D
980( )
1 1 30
1
1 w 1
2
• Since L1 and T1 are known, D1 can be computed.
• L1 is the hydrometer reading and T1 is the time passed.
• G1 is the specific gravity of soil and has to be measured separately.
• According to the Stokes equation, all particles larger than D1 have settled below the hydrom- eter level.
• Obtain the hydrometer reading (grams of soil in suspension). The “D” value computed from the Stokes equation gives the largest particles that possibly could be in suspension. In other words, all the particles in the solution are smaller than the “D” value obtained using the Stokes equation.
• Hence, the hydrometer reading is similar to the percent weight passing reading given by a sieve.
• Use the weights passing readings to obtain the percent passing readings by dividing the weight per each size by the total sample.
• Fill the table below.
Sieve analysis
Sieve size number Percent passing
4 ——————–
10 ——————–
40 ——————–
200 ——————–
V1=L1T1=980(G1−Gw)D1230η
Hydrometer analysis
0.074 mm ——————–
0.005 mm ——————–
0.001 mm ——————–
Combine the hydrometer readings with sieve analysis readings and obtain one graph (Figure 4.6).
4.3 Liquid limit, plastic limit, and shrinkage limit (Atterberg limit)
4.3.1 Liquid limit
Liquid limit is the water content where the soil starts to behave as a liquid.
Liquid limit is measured by placing a clay sample in a standard cup and making a separation (groove) using a spatula. The cup is dropped till the separation vanishes.
The water content of the soil is obtained from this sample. The test is performed again by increasing the water content. Soil with low water content would yield more blows and soil with high water content would yield less blows.
A graph is drawn between number of blows and the water content (Figure 4.7).
Liquid limit of a clay (LL) is defined as the water content that corresponds to 25 blows.
What’s the significance of liquid limit?
Liquid limits for two soils are shown in Figure 4.8. Soil 1 would reach a liquid-like state at water content of LL1. On the other hand, soil 2 would attain this state at water content LL2.
In Figure 4.8, LL1 is higher than LL2. In other words, soil 2 loses its shear strength and becomes liquid-like at a low water content than soil 1.
Figure 4.6 Hydrometer readings incorporated to sieve analysis curve.
4.3.2 Plastic limit
Plastic limit is measured by rolling a clay sample to a 3 mm diameter cylindrical shape. During continuous rolling at this size, the clay sample tends to lose moisture and cracks start to appear. The water content where cracks start to appear is defined to be the plastic limit (Figure 4.9).
4.3.3 Practical considerations of liquid limit and plastic limit The water content where a soil converts to a liquid-like state is known as the liquid limit. Consider two slopes as shown in Figure 4.10. Assuming all other factors to be equal, which slope would fail first?
Figure 4.8 Liquid limits for two soils.
Figure 4.7 Graph for liquid limit test.
Figure 4.9 Plastic limit test.
During a rain event, soil 1 would reach the liquid limit prior to soil 2. Hence, soil 1 would fail before soil 2.
During earthquakes, water tends to rise. If soils with low liquid limit were to be present, those soils would lose their strength and fail.
The plastic limit indicates the limit of plasticity. When the water content goes be- low the plastic limit of a soil, then cracks would start to appear in that soil. Soils lose their cohesion below the plastic limit.
4.3.4 Shrinkage limit
When the moisture content of a cohesive soil reduces due to evaporation, the soil would shrink or a reduction of volume would occur(Figure 4.11). The volume re- duction would come to a stop below the shrinkage limit. In other words, below shrinkage limit, reduction in moisture content will not cause reduction in volume (Figure 4.12).
The liquidity index is calculated as: LI = (NM−PL)/PI
Where, NM, the soil’s natural moisture content in percent; PL, plastic limit;
PI, plasticity index.
Figure 4.11 Volume versus water content.
4.4 Permeability test
Transport of water through soil media depends on the pressure head, velocity head, and the potential head due to elevation. In most cases, the most important parameter is the potential head due to elevation.
In Figure 4.13, water travels from “A” to “B” due to high potential head. The veloc- ity of traveling water is given by the Darcy equation.
= ×
v k i (Darcy’s equation)
v, velocity; k, coefficient of permeability (cm/s or in./s); i, hydraulic gradient = h/L;
L, length of soil.
= = ×Q A v Volume of water flow
A, area; v, velocity.
v=k×i (Darcy's equation)
Volume of water flow=Q=A×v
Figure 4.12 Liquid limit and plastic limit.
Figure 4.13 Water flowing through soil.
Design Example 4.1
Find the volume of water flowing in the pipe shown in Figure 4.14. Soil permeability is 10−5 cm/s.
Area of the pipe is 5 cm2. Length of soil plug is 50 cm.
Solution
Apply the Darcy equation.
v k i= × (Darcy’s equation)
v, velocity; k, coefficient of permeability (cm/s or in./s); i, hydraulic gradient = h/L; L, length of soil.
v k h L v
( / )
105 20/50 4 10 cm/s6
= ×
= − × = × −
Volume of water flow = A × v = 5 × 4 × 10−6 cm3/s = 2 × 10−5 cm3/s
4.4.1 Seepage rate
Water movement in soil occurs through the voids inside the soil fabric. More voids mean more water can flow through (Figure 4.15).
We could see that the velocity of water seepage through a soil mass is dependant upon the void ratio or porosity of the soil mass (Figure 4.16).
= = ×Q v A Water volume travel throughsoil
A, Area; v, Velocity.
v=k× i (Darcy's equation)
v=k×(h/L)v=10−5×20/50=4×10− 6 cm/s
W a t e r v o l u m e t r a v - el through soil=Q=v×A
Figure 4.15 Seepage path. Figure 4.16 Seepage velocity.
Total cross sectional area, A; void area, Av; porosity (n) is defined as Vv/V.
= n V Vv/
Vv,volume of voids = L × Av; Av, area of voids; V, total volume = L × A; L, length; A, total cross sectional area.
Hence,
= × × =
= ×= ×
n L A L A A A A n A
Q v A
( v)/( ) v/
v
Velocity of water traveling through voids (vs) is known as the seepage velocity.
Q = vs × Av
Q = v × A = vs × Av = vs × (n × A) v × A = vs × (n × A)
= vs v n/
4.5 Unconfined–undrained compressive strength tests (UU tests)
Unconfined compressive strength test is designed to measure the shear strength of clay soils. This is the easiest and most common test done to measure the shear strength.
Since the test is done with the sample in an unconfined state and the load is ap- plied fast so that there is no possibility of draining, the test is known as unconfined–
undrained test (UU test) (Figures 4.17, 4.18, and 4.19).
n=Vv/V
n=(L×Av)/(L×A)=Av/AAv=n×AQ=v×A
vs=v/n
Figure 4.17 UU test apparatus.
A soil sample is placed in a compression machine and compressed till failure.
Stresses are recorded during the test and plotted.
4.6 Tensile failure
When a material is subjected to a tensile stress, it would undergo tensile failure.
Figure 4.20 shows material failure under tension. The tensile failure of soil is not as common as shear failure and tensile tests are rarely conducted. On the other hand, Figure 4.18 UU apparatus and stress–strain curve. Stress at failure, q.
Figure 4.19 Mohr’s circle for the UU test. Stress at failure, q; C, cohesion = q/2.
Figure 4.20 Tensile strength test.
tensile failure is common in tunnels. Rocks in tunnel roofs are subjected to tensile forces and proper supports need to be provided (Figure 4.21).
Figure 4.21 Rock under tension.
