Faculty: Civil and Environmental Engineering Department: Geotechnical and Transportation Engineering Title: Consolidation Test
Objective:
To determine the consolidation characteristics characteristics low permeability soils. Learning Outcome:
At the end of this experiment, students are able to: -
Conduct one dimensional consolidation test Identify the factors causes soil consolidation Determine the consolidation parameters ( C V, mV, CC, PC)
Theory:
When a fully saturated soil is compressed with stress, the volume will decrease because of the soil grains are compacted and reduce the void volume. The void volume occupied by water will then lost and cause escape of water. For the clay case, the movement of water is very slow because of its low permeability. Time needed for the test might be long or it required r equired excess water to break the permeable boundary. Settlement is the result of soil volume decreasing while the consolidation means the rate of volume decreasing with time. The test is use to estimate the amount of settlement and time of consolidation. From this test some consolidation parameters: -
Coefficient of Consolidation (CV), Coefficient of volume compressibility (mV), Compression Index (CC), Pre-consolidation pressure (PC)
Two methods for determining the coefficient of consolidation: i) ii)
Casagrande or log (time) or 50% consolidation Taylor or √time or 90% consolidation
Coefficient of consolidation can be determined by this equation, cv
T v H t
Where, CV = coefficient of consolidation (m2/year)
2
TV = Time factor H = Maximum length of drainage path (m) t = Time to achieve 50% or 90% consolidation (year or minute)
Square Root Time (minute) 0
5
√t90
10
15
20
25
30
35
40
0
5
)
m 10 m( t n
e 15 m el tt
e 20 S
x 25
1.15x
1
30
2
Figure 3.1: Settlement versus log Time Time (minute) 0.1
1
t 50
10
100
1000
10000
0% consolidation line
0
δ
20
δ
40
m(
60
n
80
m
e
t
m
)
A 50% consolidation line
B = 4A
el
tt 100 e
S 120
100% consolidation line
140 160
Figure 3.2: Settlement versus square root time
Test Equipments:
1. Consolidation Apparatus -
Consolidation Ring
-
Corrosion – Resistant Porous Plate
-
Consolidation Cell
-
Dial Gauge
-
Loading Device
2. Balance readable to 0.1g 3. Vernier Caliper 4. Stop – Clock readable to 1s Procedures:
1. The internal diameter and height of ring is measures with vernier calliper. 2. Mass of ring is weighed to nearest 0.01g (mR). 3. Two specimens of peat soil is compacted and filled into the ring. 4. Initial moisture content is determined from trimming soil. 5. The weight of ring and specimen (m 1) is weighed and recorded. 6. The mass of bulk specimen (m) is weighed to the nearest 0.01g using this equation. m = m1 - mR 7. Consolidation ring with soil specimen is placed centrally on the porous disc. 8. Ring retainer and cell body is fitted and then place the upper porous disc centrally on the top of the specimen. 9. The consolidation cell is then placed centrally in position on the platform of the machine base. 10. End of beam is lifted to allow the loading yoke to be raised to the vertical position and adjust the loading stem by screwing downward until the end engages closely in the recess on the top of the loading cap. 11. The compression dial gauge is attached to the arm on the support post. 12. Weight of 2.0kg is added carefully to the load hanger for one specimen and 5kg for another specimen.
13. Water in room temperature is added to the cell and make sure that the specimen and upper porous disc are completely submerged. 14. Beam support is let downs and starts the stop watch simultaneously. 15. Compression gauge readings is observed and recorded in selected time. 16. Graph of reading of the compression against time to a logarithmic scale and against square root time is plotted.
Result: 1) 2kg Load
CONSOLIDATION TEST – CALCULATION SHEET
Date started: 5 October 2011 Soil Type: Peat soil
Sample No.: 1 Cell No: 2kg load
BEFORE TEST Moisture content from trimming: 23.5(%) Weight of ring : 322.5 (g) Weight of sample + ring: 416.3 (g) Weight of sample : 93.8 (g) Weight of dry sample: 82.5 (g) Weight of initial moisture: 11.3 (g) Initial moisture content: 13.7 (%) Initial void ratio, = 1.909
S.G. (Assumed) : 2.7 Diameter of ring : 74.9 (mm) Area of ring: 4406 (mm2) Thickness of ring: 20.18 (mm) Volume of ring : 88914.927 (mm3) Density, : 1.05 (Mg/m3) Dry density: 0.928 (Mg/m3)
SETTLEMENT READINGS
hr
min
sec
Time(min)
√ time
Clock time
Gauge reading
0 10 20 30 40
0 0.17 0.33 0.50 0.67
0 0.41 0.57 0.71 0.82
3.00pm
0 390 401 409 415
Cumulative compression, H (mm) 0 0.790 0.802 0.818 0.83
50 1 2 4 8 15 30 1
0.83 1 2 4 8 15 30 60
0.91 1.00 1.41 2.00 2.83 3.87 5.48 7.75
419 424 439 455 472 487 505 522
3.01pm 3.02pm 3.04pm 3.08pm 3.15pm 3.30pm 4.00pm
0.838 0.848 0.878 0.910 0.944 0.974 1.010 1.044
2) 5kg Load CONSOLIDATION TEST – CALCULATION SHEET
Date started: 5 October 2011 Soil Type: Peat soil
Sample No.: 2 Cell No: 5kg load
BEFORE TEST Moisture content from trimming: 23.5(%) Weight of ring : 312.5 (g) Weight of sample + ring: 392.0 (g) Weight of sample : 79.5 (g) Weight of dry sample: 69.6 (g) Weight of initial moisture: 9.9 (g) Initial moisture content: 14.22 (%) Initial void ratio, = 2.45
S.G. (Assumed) : 2.7 Diameter of ring : 74.9 (mm) Area of ring: 4406 (mm2) Thickness of ring: 19.52 (mm) Volume of ring : 88914.927 (mm3) Density, : 0.890 (Mg/m3) Dry density : 0.783 (Mg/m3)
SETTLEMENT READINGS
hr
min
sec
Time(min)
√ time
Clock time
Gauge reading
0 10 20 30
0 0.17 0.33 0.50
0 0.41 0.57 0.71
3.00pm
0 690 720 750
Cumulative compression, H (mm) 0 1.38 1.44 1.50
40 50 1 2 4 8 15 30 1
0.67 0.83 1 2 4 8 15 30 60
0.82 0.91 1.00 1.41 2.00 2.83 3.87 5.48 7.75
3.01pm 3.02pm 3.04pm 3.08pm 3.15pm 3.30pm 4.00pm
Analysis Data:
For 2kg load, Weight of sample = [Weight of sample + ring] - Weight of ring = 416.3 g – 322.5 g = 93.8 g
Initial moisture content = =
Weight of initial moisture Weight of dry sample 11.3
82.5 = 13.70 % Volume
2
r
(
h
74.9 2
) 2 (19.52)
88914.927mm3
765 778 788 825 864 963 994 1024 1058
1.53 1.56 1.58 1.65 1.73 1.93 1.99 2.05 2.12
Density = =
Weight of sample Volume of ring 93.8g
8.89 10 -5 m
= 1.055 Mg m -3
Initial void ratio = =
S.G Dry density 2.7
0.928 = 1.909
-1
-1
There are two methods for determining the coefficient of consolidation: (i) (ii)
Casagrande or log (time) or 50% consolidation Taylor or √ time or 90% consolidation
The coefficient of consolidation can be determined by this equation,
cv
Where,
T v H
2
t
cv = coefficient of consolidation (m2/year)
Tv = Time factor H = Maximum length of drainage path (m) t = Time to achieve 50% or 90% consolidation (year or minute)
Discussion:
Soil specimens prepared should be compacted well and same in the ring, so that it will not give a result of consolidation in big differ since the type of soil used is same. During the weighing of sample, the scale must not be surrounded with disturbance (included wind from fan and window) because the sensitivity of scale used is 0.01g. The loading stem should be ensured truly horizontal with bubble pond system before released and subjected to the soil specimen. This is to prevent external load applied on the sample. High consolidation rate is due to the high volume of void whether is occupied by water or air. When the soil is applied to stress such as structure in the real practice, the soil particles are
compacted and void reduced. If the saturated soil if applied with load, the water will resist the load initially but after been applied to load water is squeeze from sample. It is the same concept as the Spring Analogy where the spring represents the soil and assisted with water to resist stress. After a moment of the soil is free of stress, it will rebound back and contains void and higher in volume compared to compacted volume. This is cause by the air and water occupied the void volume again.
Question: QUESTIONS 1
1)
From your experimental data, determine the coefficient of consolidation, cv (m2/year) using Casagrande Method. Please comment your answer.
For t50; cv
2)
T v H
2
(0.848)(10 x10
t
3
m)
2
(0.90 min)
0.94 x10
4
m
2
/ min
A Clay sample collected from 5 metres deep in Batu Pahat has a unit weight of 18 kN/m3. The following data were recorded during an oedometer test. Effective Stress 25 50 100 200 400 800 200 (kN/m2) Void ratio (e)
0.85
0.82
0.71
0.57
0.43
0.3
0.4
50 0.5
(i) Plot the graph of void ratio against effective stress on semi-log graph and determine the compression index (Cc), Preconsolidation pressure (Pc) and coefficient of volume compressibility (mv). Compression index (Cc) = 0.3124 The compression index (Cc) is the slope of the graph Cc = gradient of the graph
e1
log
e2
P 2 P 1
1.20 0.40 1100 log 11 0.400
Preconsolidation pressure (Pc) = 73kN/m2 From graph, we obtained: Pre-consolidation pressure, Pc=73kN/m2 Coefficient of volume compressibility (mv) = 0.256
Coefficient of volume compressibility, mv=
e
'
eavg
e1
1 eavg
es
2
mv
1 '
Slope of the graph
e
0.76 0.42 2 0.59
e
1 '
1 eavg
1 (0.40) 1 0.59
0.256
(ii) Define whether the soil is normally consolidated or over consolidated. 3 d = 5m, =18 kN m P o
d
5m 18kNm
90kNm
3
2
Over consolidation, OCR
P C P O 73
90 0.811
1
Since, OCR<1, the soil is under consolidated. It means that the stress had been applied to the sample of soil previously is less than the stress applied during that test.
QUESTIONS 2
1) From the experimental data , determine the coefficient of consolidation, cv (m2/year) using Taylor Method. Please comment your answer. H 0
H ring 20.18mm
H 2.40mm
H fin al H 0
H ( av )
C v
20.18
2.40
17.78mm
H 0
H
H fina l
2 20.18 17.78 2
18.98mm 0.01898m
0.848( H av ) 2
4t 90 0.848(0.01898) 2
1.30 4
2
2.301 10
120.91m / year
2
m / min
2) Clay samples collected from 10 metres deep in Parit Raja has a unit weight of 20 kN/m3. The following data were recorded during an oedometer test. Effective Stress 50 100 200 400 800 1600 400 100 (kN/m2) Void ratio (e)
0.95
0.92
0.81
0.67
0.53
0.4
0.5
0.6
(i) Plot the graph of void ratio against effective stress on semi-log graph and determine the compression index (Cc), Preconsolidation pressure (Pc) and coefficient of volume compressibility (mv). The compression index (Cc) is the slope of the graph Cc = gradient of the graph Compression index (Cc) = 0.256 The compression index (Cc) is the slope of the graph Cc = gradient of the graph
e1
log
e2
P 2 P 1
1.1 0.60 1800 log 20 0.256
Preconsolidation pressure (Pc) = 136 kN/m 2 From graph, we obtained: Pre-consolidation pressure, Pc=136 kN/m 2
Coefficient of volume compressibility (mv) = 0.156
Conclusion:
In the real practice of construction, consolidation test is important to determine the rate of settlement of the soil on site. Settlement is one of the big enemy in construction and harmful to safety. If stress is applied to the soft soil with big bulk volume, the soil particles will packed tightly. If the soil is saturated with water, the water will squeeze out from the void and the magnitude of consolidation can be identified through out many tests.
Reference:
http://en.wikipedia.org/wiki/Consolidation_%28soil%29
