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FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF GEOTECHNICAL AND TRANSPORTATION ENGINEERING LAB GEOTECHNIC
FULL REPORT Subject Code
BFC 31901 U2 DIRECT SHEAR TEST & UNCONFINED Code & Experiment Title COMPRESSION TEST Course Code 3 BFF th Date 18 SEPTEMBER 2012 Section / Group SECTION 11 / GROUP 2 Name Mohd Safwan bin Mohd Yusoff AF 100112 of Group 1. Eric Cheah Keng Yang AF 100226 2. Fong Kar Guan AF 100240 3. Iryanie binti Rosli AF 100151 4. Lee Chee Aun AF 100246 5. Maryam binti Ramli AF 100203 6. Wan Mohd Ikhwan Shafiq AF 100076 EN. MUSTAFFA BIN ANJANG AHMAD Lecturer/Instructor/Tutor EN. BASIL DAVID DANIEL Received Date 24th SEPTEMBER 2012
Comment by examiner
Received
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TEST 1: DIRECT SHEAR TEST 1.0 OBJECTIVE To determine the parameter of shear strength of soil, cohesion, c and angle of friction, ø for sand. 2.0 LEARNING OUTCOME At the end of this experiment, students are able to: • Determine the shear strength parameter of the soil • Handle shear strength test, direct shear test 3.0 THEORY The general relationship between maximum shearing resistance, τ and normal stress, σ for soils can be represented by the equation and known as Coulomb’s Law: τ = c + σ tan ø where: c = cohesion, which is due to internal forces holding soil particles together in a solid mass ø = friction, which is due to the interlocking of the particles and the friction between them when subjected to normal stress. The friction components increase with increasing normal stress but the cohesion components remain constant. If there is no normal stress the friction disappears. This relationship shown in the graph below that is equal to the angle of shearing resistance of the soil, ø and its intercept on the vertical (shear stress) axis being the apparent cohesion, denoted by c.
4.0 EQUIPMENTS & MATERIALS 4.1. Shear box carriage 4.2. Loading pad 4.3. Perforated plate 4.4. Porous plate 4.5. Retaining plate 4.6. Weights
4.7. Sand sample 5.0 PROCEDURES 5.1. The internal measurements of the shear box were verified by using vernier calipers. The length of the sides, L and the overall depth, B. 5.2. The base plate was fixed inside the shear box. Then porous plate was put on the base plate. Next, perforated grid plate was fitted over porous so that the grid plates were at right angle to the direction of shear.
Figure 5.1: Shear box carriage
Figure 5.2: Arranging the plates in shear box 5.3. Two halves of the shear box was fixed by means of fixing screws. 5.4. For cohesive soils, the soil sample was transferred from square specimen cutter to the shear box by pressing down on the top grid plate. For sandy soil, soil was compacted in layers to the required density in shear box.
Figure 5.3: Preparation of sand specimen using square cutter
Figure 5.4: Sand specimen prepared before being loaded into shear box
Figure 5.5: Placing the square specimen into shear box 5.5. The shear box assembly was mounted on the loading frame. 5.6. The dial of the proving ring was set to zero. 5.7. The loading yoke was placed on the loading pad and hanger was lifted carefully onto the top of the loading yoke. 5.8. The correct loading was then applied to the hanger pad.
5.9. The screws clamping the upper half to the lower half were carefully removed. 5.10. The test was conducted by applying horizontal shear load to failure. Rate of strain were set as 1.6 mm/min. 5.11. Readings of horizontal and force dial gauges were recorded at regular intervals. 5.12. Finally the test was conducted on the three identical soil samples under different vertical compressive stress, 1.75kg, 2.5kg and 3.25kg.
6.0 RESULTS From the experiment, the dimensions of the test samples were measured as follows (dimensions are same for all three samples as the method of preparing them is fixed): Length, L: 60.00 mm Width, W: 60.00 mm Thickness, B: 24.20 mm a) To find ∆ L (mm): Example:
Dial Gauge x 0.002 mm 50 x 0.002 = 0.1 mm
b) To find Load, P (kN): Example:
Dial Gauge x 0.00204 kN 8 x 0.00204 = 0.0163 kN 2
c) To find Shear Stress:
Area (m τ = Load, P (kN) ¿ ¿
τ=
Example:
0.0163 kN 0.06 m x 0.06 m
= 4.5278 kNm-2 d) To find Strain:
∆L (mm) ε = Total Length (mm) 0.1 mm ε = 60 mm
Example: Specimen No
:1
Loading
: 1.75 kg
= 0.0017
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600
ΔL (mm)
Proving Ring Dial Gauge
Load, P (kN)
Shear Stress, τ (kN/m2)
0.1 0 0.0000 0.0000 0.2 8 0.0163 4.5278 0.3 14 0.0286 7.9444 0.4 18 0.0367 10.1944 0.5 23 0.0469 13.0278 0.6 27 0.0551 15.3056 0.7 32 0.0653 18.1389 0.8 36 0.0734 20.3889 0.9 40 0.0816 22.6667 1.0 44 0.0898 24.9444 1.1 47 0.0959 26.6389 1.2 49 0.1000 27.7778 1.3 51 0.1040 28.8889 1.4 54 0.1102 30.6111 1.5 55 0.1122 31.1667 1.6 57 0.1163 32.3056 1.7 59 0.1204 33.4444 1.8 60 0.1224 34.0000 1.9 61 0.1224 34.5556 2.0 62 0.1265 35.1389 2.1 63 0.1285 35.6944 2.2 64 0.1306 36.2778 2.3 65 0.1326 36.8333 2.4 66 0.1346 37.3889 2.5 66 0.1346 37.3889 2.6 66 0.1346 37.3889 2.7 67 0.1367 37.9722 2.8 67 0.1367 37.9722 2.9 68 0.1387 38.5278 3.0 68 0.1387 38.5278 3.1 68 0.1387 38.5278 3.2 68 0.1387 38.5278 Table 6.1: Data tabulation for Specimen 1
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417 0.0433 0.0450 0.0467 0.0483 0.0500 0.0517 0.0533
Specimen No
:2
Loading
: 2.50 kg
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550
ΔL (mm)
Proving Ring Dial Gauge
Load, P (kN)
Shear Stress, τ (kN/m2)
0.1 13 0.0265 7.3611 0.2 25 0.0510 14.1667 0.3 40 0.0816 22.6667 0.4 50 0.1020 28.3333 0.5 58 0.1183 32.8611 0.6 66 0.1346 37.3889 0.7 72 0.1469 40.8056 0.8 77 0.1571 43.6389 0.9 82 0.1673 46.4722 1.0 85 0.1734 48.1667 1.1 87 0.1775 49.3056 1.2 89 0.1816 50.4444 1.3 91 0.1856 51.5556 1.4 92 0.1877 52.1389 1.5 93 0.1897 52.6944 1.6 95 0.1938 53.8333 1.7 95 0.1938 53.8333 1.8 96 0.1958 54.3889 1.9 97 0.1979 54.9722 2.0 97 0.1979 54.9722 2.1 97 0.1979 54.9722 2.2 97 0.1979 54.9722 2.3 98 0.1999 55.5278 2.4 98 0.1999 55.5278 2.5 99 0.2020 56.1111 2.6 99 0.2020 56.1111 2.7 100 0.2040 56.1111 2.8 101 0.2060 57.2222 2.9 101 0.2060 57.2222 3.0 101 0.2060 57.2222 3.1 101 0.2060 57.2222 Table 6.2: Data tabulation for Specimen 2
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417 0.0433 0.0450 0.0467 0.0483 0.0500 0.0517
Specimen No
:3
Loading
: 3.25 kg
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250
ΔL (mm)
Proving Ring Dial Gauge
Shear Stress,
Load, P (kN)
0.1 0 0.0000 0.2 40 0.0816 0.3 55 0.1122 0.4 64 0.1306 0.5 71 0.1448 0.6 80 0.1632 0.7 87 0.1775 0.8 92 0.1877 0.9 97 0.1979 1.0 101 0.2060 1.1 105 0.2142 1.2 107 0.2183 1.3 110 0.2244 1.4 112 0.2285 1.5 113 0.2305 1.6 114 0.2326 1.7 115 0.2346 1.8 116 0.2366 1.9 117 0.2387 2.0 118 0.2407 2.1 119 0.2428 2.2 118 0.2407 2.3 118 0.2407 2.4 118 0.2407 2.5 118 0.2407 Table 6.3: Data tabulation for Specimen 3
τ (kN/m2) 0.0000 22.6667 31.1667 36.2778 40.2222 45.3333 49.3056 52.1389 54.9722 57.2222 59.5000 60.6389 62.3333 63.4722 64.0278 64.6111 65.1667 65.7222 66.3056 66.8611 67.4444 66.8611 66.8611 66.8611 66.8611
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417
7.0 DATA ANALYSIS 7.1. Graph of Shear Stress, τ (kN/m2) versus Strain, ε
Shear Stress, τ vs Strain, ε 80 70 60 50
Shear Stress, τ (kN/m2)
Specimen 1 Specimen 2 Specimen 3
40 30 20 10 0 0
0.01
0.02
0.03
0.04
0.05
0.06
Strain, ε
Figure 7.1: Graph of Shear stress, τ against Strain, ε for all specimens 7.2. Graph of Shear Stress, τ (kN/m2) versus Normal Stress, σ (kN/m2) Normal stress, σ is the element of pressure created when a force acts directly perpendicular to a surface area. Therefore, its equation can be described as σ=
Force, F Area, A , where in this experiment, the force acting perpendicular equals to
the applied loading on the hanger. 1.75 kg x 9.81 ms -2 For specimen 1, σ = 0.06 m x 0.06 m = 4.769 kNm-2 2.50 kg x 9.81 ms -2 For specimen 2, σ = 0.06 m x 0.06 m = 6.813 kNm-2 3.25 kg x 9.81 ms -2 For specimen 3, σ = 0.06 m x 0.06 m = 8.856 kNm-2
When plotting shear stress, τ against normal stress, σ the result will be a straight vertical line for all specimens since the loading does not gradually increase with the same specimen. Therefore, we use the shear stress value achieved by each specimen when the displacement dial gauge reads 150, to be plotted with the specimen’s normal stress calculated with their respective weights.
Shear Stress, τ vs Normal Stress, σ 35 30 25 20
Shear Stress, τ (kN/m2) 15 10 5 0 4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
Normal Stress, σ (kN/m2)
Figure 7.2: Graph of Shear stress, τ against Normal stress, σ for all specimens From Figure 7.2, it is found that the angle of the gradient of the graph is 36° whereas the gradient line touches the y-axis at a value of 0 kN/m 2. Therefore experimentally, cohesion, c = 0 kN/m2 and the angle of friction, ø = 36°. In theory, the cohesion, c value should be approximately zero because the sample used in this test is none other than loose and coarse sand grains, not soil. As such, cohesion factor rarely exists when sand particles are involved unless under special conditions. There are random factors which will affect the results such as human’s limitations, not uniformly compacted sand and nature of the sand itself and so on. Thus, the results obtained may not be as ideal as one imagined. Nevertheless, it is still proven through this experiment that sand has little or no cohesion at all (c = 0 kN/m2). To calculate the shear strength of the soil/sand, use Coulomb’s Law:
f c tan Specimen 1: Shear strength, τf1 = 0 + 4.769 tan 36° = 3.465 kN/m2 Specimen 2: Shear strength, τf2 = 0 + 6.813 tan 36° = 4.950 kN/m2 Specimen 3: Shear strength, τf3 = 0 + 8.856 tan 36° = 6.434 kN/m2
8.0 DISCUSSION 8.1. In loose sand, the resisting shear stress increases with shear displacement until a failure shear stress value is reached. After that, the shear resistance remains approximately constant with any further increase in the shear displacement. This is clearly visible in Figure 7.1 where all three specimens reached a maximum value of shear stress 38.5278 kN/m2, 57.2222 kN/m2 and 66.8611 kN/m2 respectively. 8.2. In Figure 7.2, the line of best fit in the graph intersects the y-axis at a value of 0 kN/m2. The angle of this line measured to the horizontal is 36°, simply by using a protractor. We may say the cohesion property of sand is 0 kN/m 2 in this experiment because in truth, cohesion rarely exists in a coarse-grained soil such as sand. Hence, we can conclude that the sand has less or none cohesion properties. 8.3. From the trends of these 3 specimens, it can be deducted that a higher applied load will enable the sand to reach its maximum shear stress faster. Moreover, a higher load also inevitably means a higher maximum shear stress. Thus, the larger the load applied on the sand under a constant strain environment, the greater the maximum shear stress achieved. 8.4. Results may vary for each sample depending on the accuracy and precision of the data observed. Since obtaining data in this experiment requires manual observation, some inaccuracies may be contributed by human’s incapability to immediately read the gauge when required to.
8.5. Some precautions that can be taken to minimise the errors that might be created are as follows: 8.5.1. Before starting the test, the upper half of the box should be brought in proper with the proving ring. 8.5.2. The arrangement of the plates should be such that porous plate’s grids are perpendicular to perforated grid plates. 8.5.3. Another important factor is to ensure the horizontal position of the hanger at the bottom of the shear box. This is done by using a leveller where levelling bubbles are provided to enable the object to achieve truly horizontal position. Once this is done, the leveller should be removed so that it would not disrupt the test. 8.5.4. Before subjecting the specimen to shear, the fixing screws should be taken out. In other words, the fixing screws have to be released before turning on the shear box at the same time. 8.5.5. Similarly, ensure the dial gauges are adjusted to zero value before beginning the test. 8.5.6. When conducting the test, no external force or vibrations should be transmitted onto the shear box. 8.6. The benefit of conducting this type of test includes the ease of sample preparation and its swift and inexpensive procedures. The shear strength parameters of both fine and coarse grained soils can be obtained either in undisturbed or remolded state. 8.7. However, to every pro there exists con. For instance, drainage cannot be controlled and pore water pressure cannot be measured. Moreover, the failure plane is always forced horizontal which is not the weakest plane in the case of in situ conditions.
9.0 CONCLUSION The aim of this direct shear test is to get the ultimate shear resistance, peak shear resistance cohesion, angle of shearing resistance and stress-strain characteristics of the soils. Shear parameters are used in the design of earthen dams and embankments. These are usually used in calculating the bearing capacity of soil-foundation systems and also help in estimating the earth pressures behind the retaining walls. The values of these parameters are used in checking the stability to natural slopes, cuts and fills. As acquired, the shear strength of soil, cohesion and the friction angle of sand have all been computed from the results of this experiment. The maximum shear strength of sand acquired is 6.434 kN/m2, being the highest among all three samples. The other two values
of shear strength consist of 3.465 kN/m 2 and 4.950 kN/m2 for specimen 1 and 2 respectively. Cohesion of sand on the other hand is 0 kN/m 2 by referring to Figure 7.2 and its angle of friction amounts to 36°. The direct shear test is rather simple to perform but it has some inherent shortcomings. The reliability of the results may be questioned. This is due to the fact that in this test the soil is not allowed fail along the weakest plane but is forced to fail along the plane of split of the shear box. Furthermore, the shear stress distribution over the shear surface of the specimen is not uniform. In spite of these weaknesses, the direct shear test is the easiest and most cost-saving for a dry or saturated sandy soil.
10.0 QUESTIONS & ANSWERS 10.1. Why perforated plates in this test with teeth? The perforated plates have teeth because they are required to produce a firm gripping force between the plate and the sand sample and they are also helpful in distributing the shear stress evenly among the surface of the specimen. 10.2. What maximum value of displacement before stop the test? The maximum value of displacement is when the proving ring readings become constant for three or more times. Besides that, the maximum value can also be identified when the value suddenly drops or in other words, the proving ring dial gauge turns the opposite way/reduces. 10.3.
What is the purpose of a direct shear test? Which soil properties does it
measure? The purpose of a direct shear test is to obtain the shear strength parameters of both fine and coarse grained soils either in undisturbed or remolded state. Often it is used for most of the geotechnical designs concerning foundations, earthworks and slope stability issues. In general, this test measures the angle of friction (ø), cohesion (c), and shear strength (τ) parameters of the material. 10.4.
Why do we use fixing screw in this test? What happen if you do not removed
them during test? Fixing screw is used so that shear does not occur before the experiment is conducted. If there is no fixing screw to counter the applied loading, there will be a hard time controlling the experiment and eventually, things might go wrongly. The fixing screw is needed so that when the test starts, there will be shear on the sample
and the results obtain will be accurate given that no other malevolent factors take place.
11.0 REFERENCES …Braja M. Das. (2005). Fundamentals of Geotechnical Engineering (2nd ed.). United States of America, US: Thomson Canada Limited. …Braja M. Das. (2010). Principles of Geotechnical Engineering (7th ed.). United States of America, US: Cengage Laerning. …http://virtual-labs.ac.in/labs/CEVL/CEVL02-SM/Data %20files/09%20Direct_Shear_Test.pdf …http://www.idoub.com/doc/39758363/Direct-Shear-Test …http://www.uta.edu/ce/geotech/lab/Main/Soil%20Lab/Direct%20Shear%20test/DS.pdf
TEST 2: UNCONFINED COMPRESSION TEST 1.0 OBJECTIVE To determine the shear strength of the cohesive soil sample. We will measure this with the unconfined compression test, which is an unconsolidated undrained (UU or Q-type) test where the lateral confining pressure is equal to zero (atmospheric pressure).
2.0 LEARNING OUTCOME At the end of this experiment, students are able to: • Describe the deflection of the jet generates forces on the vane. • Identify the relationship between force and rate of momentum flow in the jet • Measure the force generated by a jet of water striking a plate.
3.0 INTRODUCTION The unconfined compression test is by far the most popular method of soil shear testing because it is one of the fastest and cheapest methods of measuring shear strength. The method is used primarily for saturated, cohesive soils recovered from thin-walled sampling tubes. The unconfined compression test is inappropriate for dry sands or crumbly clays because the materials would fall apart without some land of lateral confinement. To perform an unconfined compression test, the sample is extruded from the sampling tube. A cylindrical sample of soil is trimmed such that the ends are reasonably smooth and the length-to-diameter ratio is on the order of two. The soil sample is placed in a loading frame on a metal plate; by turning a crank, the operator raises the level of the bottom plate. The top of the soil sample is restrained by the top plate, which is attached to a calibrated proving ring. As the bottom plate is raised, an axial load is applied to the sample. The operator turns the crank at a specified rate so that there is constant strain rate. The load is gradually increased to shear the sample, and readings are taken periodically of the force applied to the sample and the resulting deformation. The loading is continued until the soil develops an obvious shearing plane or the deformations become excessive. The measured data are used to determine the strength of the soil specimen and the stress strain characteristics. Finally, the sample is oven dried to determine its water content. The maximum load per unit area is defined as the unconfined compressive strength, qu. In the unconfined compression test, we assume that no pore water is lost from the sample during set-up or during the shearing process. A saturated sample will thus remain
saturated during the test with no change in the sample volume, water content, or void ratio. More significantly, the sample is held together by an effective confining stress that results from negative pore water pressures (generated by menisci forming between particles on the sample surface). Pore pressures are not measured in an unconfined compression test; consequently, the effective stress is unknown. Hence, the undrained shear strength measured in an unconfined test is expressed in of the total stress.
4.0 THEORY The unconfined compressive strength, qu is defined as the maximum unit axial compressive stress at failure or at 20% strain, whichever occurs first. The unconfined compression test is very popular and used worldwide. It is simple test where atmospheric pressure surrounds the soil sample. The test is also called an unconsolidated-undrained (U or UU) test. The unconfined compression test is a form of triaxial test in which the major principal stress (σ1) is equal to the applied axial stress and the minor principal stresses (σ3) is equal to zero. At failure, the relationship between the two principal stresses is given by:
Where,
As σ3 = 0 for an unconfined compression test,
For clayed soil, Ø = 0°, σ1 = 2c
The vertical stress s1 at failure is known as the unconfined compressive strength (qu) Hence, qu = 2c qu is obtained by dividing the normal load at failure by the corrected area as given by:
Where,
The axial load may be applied to the specimen either by the controlled strain procedure, in which the stress is applied to produce a pre-determined rate of strain, or by the controlled stress procedure, in which the stress is applied in pre-determined increments of load. IS: 2720 (Part 10) -1973 recommend use of controlled strain test.
5.0 EQUIPMENTS & MATERIALS 5.1. Compression device of any suitable type (loading frame of capacity 2t, with constant rate of movement)
5.2. Sample extractor
5.3. Split mould 3.5 cm diameter and 7 cm long
5.4. Frictionless end plates of 7.5 cm diameter (Perspex plate with silicon grease coating) 5.5. Oven 5.6. Balance Sensitive to weigh 0.01 g
5.7. Containers for moisture content determination 5.8. Proving ring of 0.01 kg sensitivity for soft soils and 0.05 kg for stiff soils. 5.9. Dial gauge (sensitivity 0.01 mm) 5.10. Vernier calipers 5.11. Soil sample = Wet clayed soil
5.12.
Wooden hammer and wooden compactor
6.0 PREPARATION OF SAMPLE 6.1. The split mould is oiled lightly from inside.
6.2. Remoulded soil sample is prepared by compacting the soil at desired water content and dry density in the split mould.
6.3. Split mould is opened carefully and sample is taken out. 6.4. Place this soil sample in an air-tight container for 24 hrs. 6.5. Minimum three soil samples should be prepared for test.
7.0 PROCEDURES 7.1. The initial length and diameter of the soil specimen were measured. 7.2. The specimen on the base plate of the load frame was placed (sandwiched between the end plates). 7.3. Hardened steel ball was placed on the bearing plate. The centreline of specimen was 7.4. 7.5. 7.6. 7.7.
adjusted such that the proving ring and the steel ball are in the same line. Dial gauge was fixed to measure vertical compression of the specimen. Gear position on the load frame was adjusted to give suitable vertical displacement. The reading of proving ring and dial gauge were set to zero. The load was start being applied and the readings of the proving ring dial and strain dial were recorded for every 5 mm compression.
7.8. The loading was continued till failure occurs or 20% vertical deformation is reached. 7.9. The failure pattern was sketched; measure the angle between the cracks and the horizontal if possible.
8.0 EXPERIMENTAL DATA
9.0 EXPERIMENTAL RESULTS Sample 1 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850
Strain dial reading (ΔL) (mm) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Axial Strain (ε) ΔL/ L0 0 0.0014 0.0028 0.0042 0.0056 0.0070 0.0085 0.0099 0.0113 0.0127 0.0141 0.0155 0.0169 0.0183 0.0198 0.0212 0.0226 0.0240 0.0254 0.0268 0.0282 0.0296 0.0310 0.0325 0.0339 0.0353 0.0367 0.0381 0.0395 0.0409 0.0423 0.0437 0.0452 0.0466 0.0480 0.0494 0.0508 0.0522
1-ε
Corrected area Ac =( A0
1 0.9986 0.9972 0.9958 0.9944 0.9930 0.9915 0.9901 0.9887 0.9873 0.9859 0.9845 0.9831 0.9817 0.9802 0.9788 0.9774 0.9760 0.9746 0.9732 0.9718 0.9704 0.9690 0.9675 0.9661 0.9647 0.9633 0.9619 0.9605 0.9591 0.9577 0.9563 0.9548 0.9534 0.9520 0.9506 0.9492 0.9478
Proving ring readings (div.)
Axial load P (kg)
Compressive stress A σ = P/ c
0 0 1 2 3 4 5 7 8 11 15 22 26 28 30 32 44 46 48 49 50 52 54 56 58 59 60 61 62 64 65 66 68 69 70 71 72 72
0 0 0.002 0.004 0.006 0.008 0.010 0.014 0.016 0.022 0.030 0.044 0.052 0.056 0.060 0.064 0.088 0.092 0.096 0.098 0.100 0.104 0.108 0.112 0.116 0.118 0.120 0.122 0.124 0.128 0.130 0.132 0.136 0.138 0.140 0.142 0.144 0.144
0 0 0.0002 0.0004 0.0006 0.0008 0.0010 0.0014 0.0016 0.0023 0.0031 0.0045 0.0053 0.0057 0.0061 0.0065 0.0089 0.0093 0.0097 0.0099 0.0101 0.0105 0.0109 0.0112 0.0116 0.0118 0.0120 0.0122 0.0124 0.0127 0.0129 0.0131 0.0135 0.0137 0.0138 0.0140 0.0142 0.0142
/1-
ε) (cm²) 9.6376 9.6511 9.6647 9.6782 9.6919 9.7055 9.7202 9.7340 9.7477 9.7616 9.7754 9.7893 9.8033 9.8173 9.8323 9.8463 9.8604 9.8746 9.8888 9.9030 9.9173 9.9316 9.9459 9.9613 9.9758 9.9903 10.0048 10.0193 10.0339 10.0486 10.0633 10.0780 10.0938 10.1087 10.1235 10.1384 10.1534 10.1684
(kg/cm2)
1900 1950 2000 2050 2100 2150
3.8 3.9 4.0 4.1 4.2 4.3
0.0536 0.0550 0.0564 0.0579 0.0593 0.0607
Strain dial reading (ΔL) (mm)
Axial Strain (ε) ΔL/ L0
0.9464 0.9450 0.9436 0.9421 0.9407 0.9393
10.1834 10.1985 10.2136 10.2299 10.2451 10.2604
73 74 74 75 75 75
0.146 0.148 0.148 0.150 0.150 0.150
0.0143 0.0145 0.0145 0.0147 0.0146 0.0146
1-ε
Corrected area Ac =(
Proving ring reading s (div.)
Axial load P (kg)
Compressive stress A σ = P/ c
0 2 6 9 13 15 29 32 34 40 44 46 50 51 54 56 59 61 63 65 67 69 72 73 74 74 75 75 75
0 0.004 0.012 0.018 0.026 0.030 0.058 0.064 0.068 0.080 0.088 0.092 0.100 0.102 0.108 0.112 0.118 0.122 0.126 0.130 0.134 0.138 0.144 0.146 0.148 0.148 0.150 0.150 0.150
Sample 2 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
0 0.0014 0.0028 0.0042 0.0056 0.0070 0.0085 0.0099 0.0113 0.0127 0.0141 0.0155 0.0169 0.0183 0.0197 0.0211 0.0226 0.0240 0.0254 0.0268 0.0282 0.0296 0.0310 0.0324 0.0338 0.0352 0.0367 0.0381 0.0395
A0 1 0.9986 0.9972 0.9958 0.9944 0.9930 0.9915 0.9901 0.9887 0.9873 0.9859 0.9845 0.9831 0.9817 0.9803 0.9789 0.9774 0.9760 0.9746 0.9732 0.9718 0.9704 0.9690 0.9676 0.9662 0.9648 0.9633 0.9619 0.9605
/1-ε)
(cm²) 9.6652 9.6788 9.6923 9.7060 9.7196 9.7333 9.7481 9.7618 9.7757 9.7895 9.8034 9.8174 9.8313 9.8454 9.8594 9.8735 9.8887 9.9029 9.9171 9.9314 9.9457 9.9600 9.9744 9.9888 10.0033 10.0178 10.0334 10.0480 10.0627
(kg/cm2) 0 0.0004 0.0012 0.0019 0.0027 0.0031 0.0059 0.0066 0.0070 0.0082 0.0090 0.0094 0.0102 0.0104 0.0110 0.0113 0.0119 0.0123 0.0127 0.0131 0.0135 0.0139 0.0144 0.0146 0.0148 0.0148 0.0150 0.0150 0.0149
Sample 3 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850
Strain dial reading (ΔL) (mm) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Axial Strain (ε) ΔL/ L0 0 0.0014 0.0028 0.0042 0.0057 0.0071 0.0085 0.0099 0.0113 0.0127 0.0142 0.0156 0.0170 0.0184 0.0198 0.0212 0.0226 0.0241 0.0255 0.0269 0.0283 0.0297 0.0311 0.0326 0.0340 0.0354 0.0368 0.0382 0.0396 0.0410 0.0425 0.0439 0.0453 0.0467 0.0481 0.0495 0.0510 0.0524
1-ε
Corrected area Ac =( A0
1 0.9986 0.9972 0.9958 0.9943 0.9929 0.9915 0.9901 0.9887 0.9873 0.9858 0.9844 0.9830 0.9816 0.9802 0.9788 0.9774 0.9759 0.9745 0.9731 0.9717 0.9703 0.9689 0.9674 0.9660 0.9646 0.9632 0.9618 0.9604 0.9590 0.9575 0.9561 0.9547 0.9533 0.9519 0.9505 0.9490 0.9476
/1-ε)
(cm²) 9.6486 9.6621 9.6757 9.6893 9.7039 9.7176 9.7313 9.7451 9.7589 9.7727 9.7876 9.8015 9.8155 9.8295 9.8435 9.8576 9.8717 9.8869 9.9011 9.9153 9.9296 9.9439 9.9583 9.9737 9.9882 10.0027 10.0172 10.0318 10.0464 10.0611 10.0769 10.0916 10.1064 10.1213 10.1361 10.1511 10.1671 10.1821
Proving ring reading s (div.)
Axial load P (kg)
0 0 2 3 4 5 9 16 18 20 22 24 26 28 30 39 46 50 52 53 56 58 59 60 61 62 64 66 68 69 70 71 72 73 74 75 75 75
0 0 0.004 0.006 0.008 0.010 0.018 0.032 0.036 0.040 0.044 0.048 0.052 0.056 0.060 0.078 0.092 0.100 0.104 0.106 0.112 0.116 0.118 0.120 0.122 0.124 0.128 0.132 0.136 0.138 0.140 0.142 0.144 0.146 0.148 0.150 0.150 0.150
Compressive stress A σ = P/ c (kg/cm2) 0 0 0.0004 0.0006 0.0008 0.0010 0.0018 0.0033 0.0037 0.0041 0.0045 0.0049 0.0053 0.0057 0.0061 0.0079 0.0093 0.0101 0.0105 0.0107 0.0113 0.0117 0.0118 0.0120 0.0122 0.0124 0.0128 0.0132 0.0135 0.0137 0.0139 0.0141 0.0142 0.0144 0.0146 0.0148 0.0148 0.0147
9.1 EXPERIMENTAL DATA Sample 1
Sample 2
Sample 3
35.03
35.08
35.05
70.86
70.94
70.65
963.76
966.52
964.86
Initial diameter of specimen (mm) Initial length of specimen (mm) Initial c/s area of specimen (mm²)
Figure 9.1: Failure of soil sample under test
10.0 DATA ANALYSIS Curtailment, ∆L = Strain dail (DIV) x 0.002 = 50 x 0.002 = 0.1 mm ε
=
∆L L0
= 0.10 70.86 = 0.0014 1 – ε = 1 – 0.0014 = 0.9986
where
L0 = length of specimen ∆L = curtailment
AC
=
A0
where
A0 = area of specimen
1–ε = 963.76 0.9986 = 965.1112 mm2 = 9.651112 cm2 Load, P = proving ring reading X 0.002 = 10 X 0.002 = 0.02 kg σ = P Ac =
0.02 9.651112
= 2.07 x 10-3 kg/cm2 (all the calculations for sample 2 and 3 are same as above)
Graph of Compressive Stress vs Axial Strain 0.02 0.01 0.01 0.01 Compressive stress, σ (kg/cm2)
Sample 1 Sample 2 Sample 3
0.01 0.01 0 0 0 0
0.02 0.04 0.06 0.08 Axial Strain, ε
Figure 10.1: Graph of compressive stress, σ against axial strain, ε
Sample 1:
Sample 2:
Sample 3:
c = 0.0146
c = 0.0149
c = 0.0147
qu = 0.0292
qu = 0.0298
qu = 0.0294
The graph may not look smooth because we use the compression device manually. By doing it in manual mode, our data will contain some reading error and recording error because we cannot achieve constant speed when rotating the wheel. However, the line of the graph is still within its range of value, so it is acceptable.
11.0 DISCUSSION In the unconfined test, no radial stress is applied to the sample (σ3 = 0). The plunger load, P is increased rapidly until the soil sample fail, that is cannot any additional load. The loading is applied quickly so that the pore water cannot drain from the soil. The effect stress path is unknown since pore water pressure changes are not normally measured. This test is considered as undrained shear test assuming that there is no moisture loss from the specimen during the test. The specimen must not certain any fissures, silt seams, varves, or other defects, this mean that the specimen must be intact, homogenous clay. Rarely are over-consolidated clays intact, and often even normally consolidated clays have some fissures. 11.1.
What are the differences between unconfined compression test and confined
compression test? The differences between unconfined compression test and confined compression test are sample used for unconfined compression test is not covered by any mould or casing but sample used for confined compression test is enclosed between rigid end-caps inside a thin rubber membrane to seal it from cell water, rubber O-ring are fitted over the membrane at the cap to provide a seal. 11.2. What are the advantages of doing unconfined compression test? The advantage of doing unconfined compression test is without any calculation the confining pressure σ3 is equal to 0. From the test result we get the maximum unconfined compression strength (qu) is determined, using qu = σ1and σ3 is equal to 0 plot in graph where normal stress versus shear stress to determine the undrained strength Cu where Cu = qu/2 of the unconfined compression strength we obtained.
11.3. What are the limitations of unconfined compression test? The limitations of the unconfined compression test is applicable to the fully saturated non-fissured clays, and only the undrained strength Cu can be measured. 11.4. Give 4 common laboratory errors for unconfined compression test? The common laboratory errors for unconfined compression test are: Getting wrong reading from dial gauge during the test was running. The soil sample prepared is too wet. Insensitivity of measurements at low strains due to high early soil stiffness. The application of the load to the soil sample was not equal, either too fast or too slow.
12.0 CONCLUSION The unconfined compressive strength is considered to be equal to the load at which the failure of the soil occurs. The consistency of soil samples will have different compressive strength due to different porosity, moisture content and existing of microorganism in the soil.
13.0 REFERENCES …http://civilengineeringlaboratory.blogspot.com/2012/02/unconfined-compressiontest.html …http://spin.mohawkcollege.ca/courses/smeatonk/CV504%20PDFs/Lab %20Manual/Unconfined%20Compression%20Procedure.pdf …http://www.cyut.edu.tw/~jrlai/CE7334/Unconfined.pdf …http://www.uic.edu/classes/cemm/cemmlab/Experiment%2013-Unconfined %20Compression.pdf …http://www.uta.edu/ce/geotech/lab/Main/Soil%20Lab/09_UCS/UCS.pdf
FULL REPORT Subject Code
BFC 31901 U2 DIRECT SHEAR TEST & UNCONFINED Code & Experiment Title COMPRESSION TEST Course Code 3 BFF th Date 18 SEPTEMBER 2012 Section / Group SECTION 11 / GROUP 2 Name Mohd Safwan bin Mohd Yusoff AF 100112 of Group 1. Eric Cheah Keng Yang AF 100226 2. Fong Kar Guan AF 100240 3. Iryanie binti Rosli AF 100151 4. Lee Chee Aun AF 100246 5. Maryam binti Ramli AF 100203 6. Wan Mohd Ikhwan Shafiq AF 100076 EN. MUSTAFFA BIN ANJANG AHMAD Lecturer/Instructor/Tutor EN. BASIL DAVID DANIEL Received Date 24th SEPTEMBER 2012
Comment by examiner
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TEST 1: DIRECT SHEAR TEST 1.0 OBJECTIVE To determine the parameter of shear strength of soil, cohesion, c and angle of friction, ø for sand. 2.0 LEARNING OUTCOME At the end of this experiment, students are able to: • Determine the shear strength parameter of the soil • Handle shear strength test, direct shear test 3.0 THEORY The general relationship between maximum shearing resistance, τ and normal stress, σ for soils can be represented by the equation and known as Coulomb’s Law: τ = c + σ tan ø where: c = cohesion, which is due to internal forces holding soil particles together in a solid mass ø = friction, which is due to the interlocking of the particles and the friction between them when subjected to normal stress. The friction components increase with increasing normal stress but the cohesion components remain constant. If there is no normal stress the friction disappears. This relationship shown in the graph below that is equal to the angle of shearing resistance of the soil, ø and its intercept on the vertical (shear stress) axis being the apparent cohesion, denoted by c.
4.0 EQUIPMENTS & MATERIALS 4.1. Shear box carriage 4.2. Loading pad 4.3. Perforated plate 4.4. Porous plate 4.5. Retaining plate 4.6. Weights
4.7. Sand sample 5.0 PROCEDURES 5.1. The internal measurements of the shear box were verified by using vernier calipers. The length of the sides, L and the overall depth, B. 5.2. The base plate was fixed inside the shear box. Then porous plate was put on the base plate. Next, perforated grid plate was fitted over porous so that the grid plates were at right angle to the direction of shear.
Figure 5.1: Shear box carriage
Figure 5.2: Arranging the plates in shear box 5.3. Two halves of the shear box was fixed by means of fixing screws. 5.4. For cohesive soils, the soil sample was transferred from square specimen cutter to the shear box by pressing down on the top grid plate. For sandy soil, soil was compacted in layers to the required density in shear box.
Figure 5.3: Preparation of sand specimen using square cutter
Figure 5.4: Sand specimen prepared before being loaded into shear box
Figure 5.5: Placing the square specimen into shear box 5.5. The shear box assembly was mounted on the loading frame. 5.6. The dial of the proving ring was set to zero. 5.7. The loading yoke was placed on the loading pad and hanger was lifted carefully onto the top of the loading yoke. 5.8. The correct loading was then applied to the hanger pad.
5.9. The screws clamping the upper half to the lower half were carefully removed. 5.10. The test was conducted by applying horizontal shear load to failure. Rate of strain were set as 1.6 mm/min. 5.11. Readings of horizontal and force dial gauges were recorded at regular intervals. 5.12. Finally the test was conducted on the three identical soil samples under different vertical compressive stress, 1.75kg, 2.5kg and 3.25kg.
6.0 RESULTS From the experiment, the dimensions of the test samples were measured as follows (dimensions are same for all three samples as the method of preparing them is fixed): Length, L: 60.00 mm Width, W: 60.00 mm Thickness, B: 24.20 mm a) To find ∆ L (mm): Example:
Dial Gauge x 0.002 mm 50 x 0.002 = 0.1 mm
b) To find Load, P (kN): Example:
Dial Gauge x 0.00204 kN 8 x 0.00204 = 0.0163 kN 2
c) To find Shear Stress:
Area (m τ = Load, P (kN) ¿ ¿
τ=
Example:
0.0163 kN 0.06 m x 0.06 m
= 4.5278 kNm-2 d) To find Strain:
∆L (mm) ε = Total Length (mm) 0.1 mm ε = 60 mm
Example: Specimen No
:1
Loading
: 1.75 kg
= 0.0017
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600
ΔL (mm)
Proving Ring Dial Gauge
Load, P (kN)
Shear Stress, τ (kN/m2)
0.1 0 0.0000 0.0000 0.2 8 0.0163 4.5278 0.3 14 0.0286 7.9444 0.4 18 0.0367 10.1944 0.5 23 0.0469 13.0278 0.6 27 0.0551 15.3056 0.7 32 0.0653 18.1389 0.8 36 0.0734 20.3889 0.9 40 0.0816 22.6667 1.0 44 0.0898 24.9444 1.1 47 0.0959 26.6389 1.2 49 0.1000 27.7778 1.3 51 0.1040 28.8889 1.4 54 0.1102 30.6111 1.5 55 0.1122 31.1667 1.6 57 0.1163 32.3056 1.7 59 0.1204 33.4444 1.8 60 0.1224 34.0000 1.9 61 0.1224 34.5556 2.0 62 0.1265 35.1389 2.1 63 0.1285 35.6944 2.2 64 0.1306 36.2778 2.3 65 0.1326 36.8333 2.4 66 0.1346 37.3889 2.5 66 0.1346 37.3889 2.6 66 0.1346 37.3889 2.7 67 0.1367 37.9722 2.8 67 0.1367 37.9722 2.9 68 0.1387 38.5278 3.0 68 0.1387 38.5278 3.1 68 0.1387 38.5278 3.2 68 0.1387 38.5278 Table 6.1: Data tabulation for Specimen 1
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417 0.0433 0.0450 0.0467 0.0483 0.0500 0.0517 0.0533
Specimen No
:2
Loading
: 2.50 kg
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550
ΔL (mm)
Proving Ring Dial Gauge
Load, P (kN)
Shear Stress, τ (kN/m2)
0.1 13 0.0265 7.3611 0.2 25 0.0510 14.1667 0.3 40 0.0816 22.6667 0.4 50 0.1020 28.3333 0.5 58 0.1183 32.8611 0.6 66 0.1346 37.3889 0.7 72 0.1469 40.8056 0.8 77 0.1571 43.6389 0.9 82 0.1673 46.4722 1.0 85 0.1734 48.1667 1.1 87 0.1775 49.3056 1.2 89 0.1816 50.4444 1.3 91 0.1856 51.5556 1.4 92 0.1877 52.1389 1.5 93 0.1897 52.6944 1.6 95 0.1938 53.8333 1.7 95 0.1938 53.8333 1.8 96 0.1958 54.3889 1.9 97 0.1979 54.9722 2.0 97 0.1979 54.9722 2.1 97 0.1979 54.9722 2.2 97 0.1979 54.9722 2.3 98 0.1999 55.5278 2.4 98 0.1999 55.5278 2.5 99 0.2020 56.1111 2.6 99 0.2020 56.1111 2.7 100 0.2040 56.1111 2.8 101 0.2060 57.2222 2.9 101 0.2060 57.2222 3.0 101 0.2060 57.2222 3.1 101 0.2060 57.2222 Table 6.2: Data tabulation for Specimen 2
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417 0.0433 0.0450 0.0467 0.0483 0.0500 0.0517
Specimen No
:3
Loading
: 3.25 kg
Displacement Dial Gauge 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250
ΔL (mm)
Proving Ring Dial Gauge
Shear Stress,
Load, P (kN)
0.1 0 0.0000 0.2 40 0.0816 0.3 55 0.1122 0.4 64 0.1306 0.5 71 0.1448 0.6 80 0.1632 0.7 87 0.1775 0.8 92 0.1877 0.9 97 0.1979 1.0 101 0.2060 1.1 105 0.2142 1.2 107 0.2183 1.3 110 0.2244 1.4 112 0.2285 1.5 113 0.2305 1.6 114 0.2326 1.7 115 0.2346 1.8 116 0.2366 1.9 117 0.2387 2.0 118 0.2407 2.1 119 0.2428 2.2 118 0.2407 2.3 118 0.2407 2.4 118 0.2407 2.5 118 0.2407 Table 6.3: Data tabulation for Specimen 3
τ (kN/m2) 0.0000 22.6667 31.1667 36.2778 40.2222 45.3333 49.3056 52.1389 54.9722 57.2222 59.5000 60.6389 62.3333 63.4722 64.0278 64.6111 65.1667 65.7222 66.3056 66.8611 67.4444 66.8611 66.8611 66.8611 66.8611
Strain, ε 0.0017 0.0033 0.0050 0.0067 0.0083 0.0100 0.0117 0.0133 0.0150 0.0167 0.0183 0.0200 0.0217 0.0233 0.0250 0.0267 0.0283 0.0300 0.0317 0.0333 0.0350 0.0367 0.0383 0.0400 0.0417
7.0 DATA ANALYSIS 7.1. Graph of Shear Stress, τ (kN/m2) versus Strain, ε
Shear Stress, τ vs Strain, ε 80 70 60 50
Shear Stress, τ (kN/m2)
Specimen 1 Specimen 2 Specimen 3
40 30 20 10 0 0
0.01
0.02
0.03
0.04
0.05
0.06
Strain, ε
Figure 7.1: Graph of Shear stress, τ against Strain, ε for all specimens 7.2. Graph of Shear Stress, τ (kN/m2) versus Normal Stress, σ (kN/m2) Normal stress, σ is the element of pressure created when a force acts directly perpendicular to a surface area. Therefore, its equation can be described as σ=
Force, F Area, A , where in this experiment, the force acting perpendicular equals to
the applied loading on the hanger. 1.75 kg x 9.81 ms -2 For specimen 1, σ = 0.06 m x 0.06 m = 4.769 kNm-2 2.50 kg x 9.81 ms -2 For specimen 2, σ = 0.06 m x 0.06 m = 6.813 kNm-2 3.25 kg x 9.81 ms -2 For specimen 3, σ = 0.06 m x 0.06 m = 8.856 kNm-2
When plotting shear stress, τ against normal stress, σ the result will be a straight vertical line for all specimens since the loading does not gradually increase with the same specimen. Therefore, we use the shear stress value achieved by each specimen when the displacement dial gauge reads 150, to be plotted with the specimen’s normal stress calculated with their respective weights.
Shear Stress, τ vs Normal Stress, σ 35 30 25 20
Shear Stress, τ (kN/m2) 15 10 5 0 4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
Normal Stress, σ (kN/m2)
Figure 7.2: Graph of Shear stress, τ against Normal stress, σ for all specimens From Figure 7.2, it is found that the angle of the gradient of the graph is 36° whereas the gradient line touches the y-axis at a value of 0 kN/m 2. Therefore experimentally, cohesion, c = 0 kN/m2 and the angle of friction, ø = 36°. In theory, the cohesion, c value should be approximately zero because the sample used in this test is none other than loose and coarse sand grains, not soil. As such, cohesion factor rarely exists when sand particles are involved unless under special conditions. There are random factors which will affect the results such as human’s limitations, not uniformly compacted sand and nature of the sand itself and so on. Thus, the results obtained may not be as ideal as one imagined. Nevertheless, it is still proven through this experiment that sand has little or no cohesion at all (c = 0 kN/m2). To calculate the shear strength of the soil/sand, use Coulomb’s Law:
f c tan Specimen 1: Shear strength, τf1 = 0 + 4.769 tan 36° = 3.465 kN/m2 Specimen 2: Shear strength, τf2 = 0 + 6.813 tan 36° = 4.950 kN/m2 Specimen 3: Shear strength, τf3 = 0 + 8.856 tan 36° = 6.434 kN/m2
8.0 DISCUSSION 8.1. In loose sand, the resisting shear stress increases with shear displacement until a failure shear stress value is reached. After that, the shear resistance remains approximately constant with any further increase in the shear displacement. This is clearly visible in Figure 7.1 where all three specimens reached a maximum value of shear stress 38.5278 kN/m2, 57.2222 kN/m2 and 66.8611 kN/m2 respectively. 8.2. In Figure 7.2, the line of best fit in the graph intersects the y-axis at a value of 0 kN/m2. The angle of this line measured to the horizontal is 36°, simply by using a protractor. We may say the cohesion property of sand is 0 kN/m 2 in this experiment because in truth, cohesion rarely exists in a coarse-grained soil such as sand. Hence, we can conclude that the sand has less or none cohesion properties. 8.3. From the trends of these 3 specimens, it can be deducted that a higher applied load will enable the sand to reach its maximum shear stress faster. Moreover, a higher load also inevitably means a higher maximum shear stress. Thus, the larger the load applied on the sand under a constant strain environment, the greater the maximum shear stress achieved. 8.4. Results may vary for each sample depending on the accuracy and precision of the data observed. Since obtaining data in this experiment requires manual observation, some inaccuracies may be contributed by human’s incapability to immediately read the gauge when required to.
8.5. Some precautions that can be taken to minimise the errors that might be created are as follows: 8.5.1. Before starting the test, the upper half of the box should be brought in proper with the proving ring. 8.5.2. The arrangement of the plates should be such that porous plate’s grids are perpendicular to perforated grid plates. 8.5.3. Another important factor is to ensure the horizontal position of the hanger at the bottom of the shear box. This is done by using a leveller where levelling bubbles are provided to enable the object to achieve truly horizontal position. Once this is done, the leveller should be removed so that it would not disrupt the test. 8.5.4. Before subjecting the specimen to shear, the fixing screws should be taken out. In other words, the fixing screws have to be released before turning on the shear box at the same time. 8.5.5. Similarly, ensure the dial gauges are adjusted to zero value before beginning the test. 8.5.6. When conducting the test, no external force or vibrations should be transmitted onto the shear box. 8.6. The benefit of conducting this type of test includes the ease of sample preparation and its swift and inexpensive procedures. The shear strength parameters of both fine and coarse grained soils can be obtained either in undisturbed or remolded state. 8.7. However, to every pro there exists con. For instance, drainage cannot be controlled and pore water pressure cannot be measured. Moreover, the failure plane is always forced horizontal which is not the weakest plane in the case of in situ conditions.
9.0 CONCLUSION The aim of this direct shear test is to get the ultimate shear resistance, peak shear resistance cohesion, angle of shearing resistance and stress-strain characteristics of the soils. Shear parameters are used in the design of earthen dams and embankments. These are usually used in calculating the bearing capacity of soil-foundation systems and also help in estimating the earth pressures behind the retaining walls. The values of these parameters are used in checking the stability to natural slopes, cuts and fills. As acquired, the shear strength of soil, cohesion and the friction angle of sand have all been computed from the results of this experiment. The maximum shear strength of sand acquired is 6.434 kN/m2, being the highest among all three samples. The other two values
of shear strength consist of 3.465 kN/m 2 and 4.950 kN/m2 for specimen 1 and 2 respectively. Cohesion of sand on the other hand is 0 kN/m 2 by referring to Figure 7.2 and its angle of friction amounts to 36°. The direct shear test is rather simple to perform but it has some inherent shortcomings. The reliability of the results may be questioned. This is due to the fact that in this test the soil is not allowed fail along the weakest plane but is forced to fail along the plane of split of the shear box. Furthermore, the shear stress distribution over the shear surface of the specimen is not uniform. In spite of these weaknesses, the direct shear test is the easiest and most cost-saving for a dry or saturated sandy soil.
10.0 QUESTIONS & ANSWERS 10.1. Why perforated plates in this test with teeth? The perforated plates have teeth because they are required to produce a firm gripping force between the plate and the sand sample and they are also helpful in distributing the shear stress evenly among the surface of the specimen. 10.2. What maximum value of displacement before stop the test? The maximum value of displacement is when the proving ring readings become constant for three or more times. Besides that, the maximum value can also be identified when the value suddenly drops or in other words, the proving ring dial gauge turns the opposite way/reduces. 10.3.
What is the purpose of a direct shear test? Which soil properties does it
measure? The purpose of a direct shear test is to obtain the shear strength parameters of both fine and coarse grained soils either in undisturbed or remolded state. Often it is used for most of the geotechnical designs concerning foundations, earthworks and slope stability issues. In general, this test measures the angle of friction (ø), cohesion (c), and shear strength (τ) parameters of the material. 10.4.
Why do we use fixing screw in this test? What happen if you do not removed
them during test? Fixing screw is used so that shear does not occur before the experiment is conducted. If there is no fixing screw to counter the applied loading, there will be a hard time controlling the experiment and eventually, things might go wrongly. The fixing screw is needed so that when the test starts, there will be shear on the sample
and the results obtain will be accurate given that no other malevolent factors take place.
11.0 REFERENCES …Braja M. Das. (2005). Fundamentals of Geotechnical Engineering (2nd ed.). United States of America, US: Thomson Canada Limited. …Braja M. Das. (2010). Principles of Geotechnical Engineering (7th ed.). United States of America, US: Cengage Laerning. …http://virtual-labs.ac.in/labs/CEVL/CEVL02-SM/Data %20files/09%20Direct_Shear_Test.pdf …http://www.idoub.com/doc/39758363/Direct-Shear-Test …http://www.uta.edu/ce/geotech/lab/Main/Soil%20Lab/Direct%20Shear%20test/DS.pdf
TEST 2: UNCONFINED COMPRESSION TEST 1.0 OBJECTIVE To determine the shear strength of the cohesive soil sample. We will measure this with the unconfined compression test, which is an unconsolidated undrained (UU or Q-type) test where the lateral confining pressure is equal to zero (atmospheric pressure).
2.0 LEARNING OUTCOME At the end of this experiment, students are able to: • Describe the deflection of the jet generates forces on the vane. • Identify the relationship between force and rate of momentum flow in the jet • Measure the force generated by a jet of water striking a plate.
3.0 INTRODUCTION The unconfined compression test is by far the most popular method of soil shear testing because it is one of the fastest and cheapest methods of measuring shear strength. The method is used primarily for saturated, cohesive soils recovered from thin-walled sampling tubes. The unconfined compression test is inappropriate for dry sands or crumbly clays because the materials would fall apart without some land of lateral confinement. To perform an unconfined compression test, the sample is extruded from the sampling tube. A cylindrical sample of soil is trimmed such that the ends are reasonably smooth and the length-to-diameter ratio is on the order of two. The soil sample is placed in a loading frame on a metal plate; by turning a crank, the operator raises the level of the bottom plate. The top of the soil sample is restrained by the top plate, which is attached to a calibrated proving ring. As the bottom plate is raised, an axial load is applied to the sample. The operator turns the crank at a specified rate so that there is constant strain rate. The load is gradually increased to shear the sample, and readings are taken periodically of the force applied to the sample and the resulting deformation. The loading is continued until the soil develops an obvious shearing plane or the deformations become excessive. The measured data are used to determine the strength of the soil specimen and the stress strain characteristics. Finally, the sample is oven dried to determine its water content. The maximum load per unit area is defined as the unconfined compressive strength, qu. In the unconfined compression test, we assume that no pore water is lost from the sample during set-up or during the shearing process. A saturated sample will thus remain
saturated during the test with no change in the sample volume, water content, or void ratio. More significantly, the sample is held together by an effective confining stress that results from negative pore water pressures (generated by menisci forming between particles on the sample surface). Pore pressures are not measured in an unconfined compression test; consequently, the effective stress is unknown. Hence, the undrained shear strength measured in an unconfined test is expressed in of the total stress.
4.0 THEORY The unconfined compressive strength, qu is defined as the maximum unit axial compressive stress at failure or at 20% strain, whichever occurs first. The unconfined compression test is very popular and used worldwide. It is simple test where atmospheric pressure surrounds the soil sample. The test is also called an unconsolidated-undrained (U or UU) test. The unconfined compression test is a form of triaxial test in which the major principal stress (σ1) is equal to the applied axial stress and the minor principal stresses (σ3) is equal to zero. At failure, the relationship between the two principal stresses is given by:
Where,
As σ3 = 0 for an unconfined compression test,
For clayed soil, Ø = 0°, σ1 = 2c
The vertical stress s1 at failure is known as the unconfined compressive strength (qu) Hence, qu = 2c qu is obtained by dividing the normal load at failure by the corrected area as given by:
Where,
The axial load may be applied to the specimen either by the controlled strain procedure, in which the stress is applied to produce a pre-determined rate of strain, or by the controlled stress procedure, in which the stress is applied in pre-determined increments of load. IS: 2720 (Part 10) -1973 recommend use of controlled strain test.
5.0 EQUIPMENTS & MATERIALS 5.1. Compression device of any suitable type (loading frame of capacity 2t, with constant rate of movement)
5.2. Sample extractor
5.3. Split mould 3.5 cm diameter and 7 cm long
5.4. Frictionless end plates of 7.5 cm diameter (Perspex plate with silicon grease coating) 5.5. Oven 5.6. Balance Sensitive to weigh 0.01 g
5.7. Containers for moisture content determination 5.8. Proving ring of 0.01 kg sensitivity for soft soils and 0.05 kg for stiff soils. 5.9. Dial gauge (sensitivity 0.01 mm) 5.10. Vernier calipers 5.11. Soil sample = Wet clayed soil
5.12.
Wooden hammer and wooden compactor
6.0 PREPARATION OF SAMPLE 6.1. The split mould is oiled lightly from inside.
6.2. Remoulded soil sample is prepared by compacting the soil at desired water content and dry density in the split mould.
6.3. Split mould is opened carefully and sample is taken out. 6.4. Place this soil sample in an air-tight container for 24 hrs. 6.5. Minimum three soil samples should be prepared for test.
7.0 PROCEDURES 7.1. The initial length and diameter of the soil specimen were measured. 7.2. The specimen on the base plate of the load frame was placed (sandwiched between the end plates). 7.3. Hardened steel ball was placed on the bearing plate. The centreline of specimen was 7.4. 7.5. 7.6. 7.7.
adjusted such that the proving ring and the steel ball are in the same line. Dial gauge was fixed to measure vertical compression of the specimen. Gear position on the load frame was adjusted to give suitable vertical displacement. The reading of proving ring and dial gauge were set to zero. The load was start being applied and the readings of the proving ring dial and strain dial were recorded for every 5 mm compression.
7.8. The loading was continued till failure occurs or 20% vertical deformation is reached. 7.9. The failure pattern was sketched; measure the angle between the cracks and the horizontal if possible.
8.0 EXPERIMENTAL DATA
9.0 EXPERIMENTAL RESULTS Sample 1 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850
Strain dial reading (ΔL) (mm) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Axial Strain (ε) ΔL/ L0 0 0.0014 0.0028 0.0042 0.0056 0.0070 0.0085 0.0099 0.0113 0.0127 0.0141 0.0155 0.0169 0.0183 0.0198 0.0212 0.0226 0.0240 0.0254 0.0268 0.0282 0.0296 0.0310 0.0325 0.0339 0.0353 0.0367 0.0381 0.0395 0.0409 0.0423 0.0437 0.0452 0.0466 0.0480 0.0494 0.0508 0.0522
1-ε
Corrected area Ac =( A0
1 0.9986 0.9972 0.9958 0.9944 0.9930 0.9915 0.9901 0.9887 0.9873 0.9859 0.9845 0.9831 0.9817 0.9802 0.9788 0.9774 0.9760 0.9746 0.9732 0.9718 0.9704 0.9690 0.9675 0.9661 0.9647 0.9633 0.9619 0.9605 0.9591 0.9577 0.9563 0.9548 0.9534 0.9520 0.9506 0.9492 0.9478
Proving ring readings (div.)
Axial load P (kg)
Compressive stress A σ = P/ c
0 0 1 2 3 4 5 7 8 11 15 22 26 28 30 32 44 46 48 49 50 52 54 56 58 59 60 61 62 64 65 66 68 69 70 71 72 72
0 0 0.002 0.004 0.006 0.008 0.010 0.014 0.016 0.022 0.030 0.044 0.052 0.056 0.060 0.064 0.088 0.092 0.096 0.098 0.100 0.104 0.108 0.112 0.116 0.118 0.120 0.122 0.124 0.128 0.130 0.132 0.136 0.138 0.140 0.142 0.144 0.144
0 0 0.0002 0.0004 0.0006 0.0008 0.0010 0.0014 0.0016 0.0023 0.0031 0.0045 0.0053 0.0057 0.0061 0.0065 0.0089 0.0093 0.0097 0.0099 0.0101 0.0105 0.0109 0.0112 0.0116 0.0118 0.0120 0.0122 0.0124 0.0127 0.0129 0.0131 0.0135 0.0137 0.0138 0.0140 0.0142 0.0142
/1-
ε) (cm²) 9.6376 9.6511 9.6647 9.6782 9.6919 9.7055 9.7202 9.7340 9.7477 9.7616 9.7754 9.7893 9.8033 9.8173 9.8323 9.8463 9.8604 9.8746 9.8888 9.9030 9.9173 9.9316 9.9459 9.9613 9.9758 9.9903 10.0048 10.0193 10.0339 10.0486 10.0633 10.0780 10.0938 10.1087 10.1235 10.1384 10.1534 10.1684
(kg/cm2)
1900 1950 2000 2050 2100 2150
3.8 3.9 4.0 4.1 4.2 4.3
0.0536 0.0550 0.0564 0.0579 0.0593 0.0607
Strain dial reading (ΔL) (mm)
Axial Strain (ε) ΔL/ L0
0.9464 0.9450 0.9436 0.9421 0.9407 0.9393
10.1834 10.1985 10.2136 10.2299 10.2451 10.2604
73 74 74 75 75 75
0.146 0.148 0.148 0.150 0.150 0.150
0.0143 0.0145 0.0145 0.0147 0.0146 0.0146
1-ε
Corrected area Ac =(
Proving ring reading s (div.)
Axial load P (kg)
Compressive stress A σ = P/ c
0 2 6 9 13 15 29 32 34 40 44 46 50 51 54 56 59 61 63 65 67 69 72 73 74 74 75 75 75
0 0.004 0.012 0.018 0.026 0.030 0.058 0.064 0.068 0.080 0.088 0.092 0.100 0.102 0.108 0.112 0.118 0.122 0.126 0.130 0.134 0.138 0.144 0.146 0.148 0.148 0.150 0.150 0.150
Sample 2 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
0 0.0014 0.0028 0.0042 0.0056 0.0070 0.0085 0.0099 0.0113 0.0127 0.0141 0.0155 0.0169 0.0183 0.0197 0.0211 0.0226 0.0240 0.0254 0.0268 0.0282 0.0296 0.0310 0.0324 0.0338 0.0352 0.0367 0.0381 0.0395
A0 1 0.9986 0.9972 0.9958 0.9944 0.9930 0.9915 0.9901 0.9887 0.9873 0.9859 0.9845 0.9831 0.9817 0.9803 0.9789 0.9774 0.9760 0.9746 0.9732 0.9718 0.9704 0.9690 0.9676 0.9662 0.9648 0.9633 0.9619 0.9605
/1-ε)
(cm²) 9.6652 9.6788 9.6923 9.7060 9.7196 9.7333 9.7481 9.7618 9.7757 9.7895 9.8034 9.8174 9.8313 9.8454 9.8594 9.8735 9.8887 9.9029 9.9171 9.9314 9.9457 9.9600 9.9744 9.9888 10.0033 10.0178 10.0334 10.0480 10.0627
(kg/cm2) 0 0.0004 0.0012 0.0019 0.0027 0.0031 0.0059 0.0066 0.0070 0.0082 0.0090 0.0094 0.0102 0.0104 0.0110 0.0113 0.0119 0.0123 0.0127 0.0131 0.0135 0.0139 0.0144 0.0146 0.0148 0.0148 0.0150 0.0150 0.0149
Sample 3 Elapsed Time ‘t’ (minutes)
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850
Strain dial reading (ΔL) (mm) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Axial Strain (ε) ΔL/ L0 0 0.0014 0.0028 0.0042 0.0057 0.0071 0.0085 0.0099 0.0113 0.0127 0.0142 0.0156 0.0170 0.0184 0.0198 0.0212 0.0226 0.0241 0.0255 0.0269 0.0283 0.0297 0.0311 0.0326 0.0340 0.0354 0.0368 0.0382 0.0396 0.0410 0.0425 0.0439 0.0453 0.0467 0.0481 0.0495 0.0510 0.0524
1-ε
Corrected area Ac =( A0
1 0.9986 0.9972 0.9958 0.9943 0.9929 0.9915 0.9901 0.9887 0.9873 0.9858 0.9844 0.9830 0.9816 0.9802 0.9788 0.9774 0.9759 0.9745 0.9731 0.9717 0.9703 0.9689 0.9674 0.9660 0.9646 0.9632 0.9618 0.9604 0.9590 0.9575 0.9561 0.9547 0.9533 0.9519 0.9505 0.9490 0.9476
/1-ε)
(cm²) 9.6486 9.6621 9.6757 9.6893 9.7039 9.7176 9.7313 9.7451 9.7589 9.7727 9.7876 9.8015 9.8155 9.8295 9.8435 9.8576 9.8717 9.8869 9.9011 9.9153 9.9296 9.9439 9.9583 9.9737 9.9882 10.0027 10.0172 10.0318 10.0464 10.0611 10.0769 10.0916 10.1064 10.1213 10.1361 10.1511 10.1671 10.1821
Proving ring reading s (div.)
Axial load P (kg)
0 0 2 3 4 5 9 16 18 20 22 24 26 28 30 39 46 50 52 53 56 58 59 60 61 62 64 66 68 69 70 71 72 73 74 75 75 75
0 0 0.004 0.006 0.008 0.010 0.018 0.032 0.036 0.040 0.044 0.048 0.052 0.056 0.060 0.078 0.092 0.100 0.104 0.106 0.112 0.116 0.118 0.120 0.122 0.124 0.128 0.132 0.136 0.138 0.140 0.142 0.144 0.146 0.148 0.150 0.150 0.150
Compressive stress A σ = P/ c (kg/cm2) 0 0 0.0004 0.0006 0.0008 0.0010 0.0018 0.0033 0.0037 0.0041 0.0045 0.0049 0.0053 0.0057 0.0061 0.0079 0.0093 0.0101 0.0105 0.0107 0.0113 0.0117 0.0118 0.0120 0.0122 0.0124 0.0128 0.0132 0.0135 0.0137 0.0139 0.0141 0.0142 0.0144 0.0146 0.0148 0.0148 0.0147
9.1 EXPERIMENTAL DATA Sample 1
Sample 2
Sample 3
35.03
35.08
35.05
70.86
70.94
70.65
963.76
966.52
964.86
Initial diameter of specimen (mm) Initial length of specimen (mm) Initial c/s area of specimen (mm²)
Figure 9.1: Failure of soil sample under test
10.0 DATA ANALYSIS Curtailment, ∆L = Strain dail (DIV) x 0.002 = 50 x 0.002 = 0.1 mm ε
=
∆L L0
= 0.10 70.86 = 0.0014 1 – ε = 1 – 0.0014 = 0.9986
where
L0 = length of specimen ∆L = curtailment
AC
=
A0
where
A0 = area of specimen
1–ε = 963.76 0.9986 = 965.1112 mm2 = 9.651112 cm2 Load, P = proving ring reading X 0.002 = 10 X 0.002 = 0.02 kg σ = P Ac =
0.02 9.651112
= 2.07 x 10-3 kg/cm2 (all the calculations for sample 2 and 3 are same as above)
Graph of Compressive Stress vs Axial Strain 0.02 0.01 0.01 0.01 Compressive stress, σ (kg/cm2)
Sample 1 Sample 2 Sample 3
0.01 0.01 0 0 0 0
0.02 0.04 0.06 0.08 Axial Strain, ε
Figure 10.1: Graph of compressive stress, σ against axial strain, ε
Sample 1:
Sample 2:
Sample 3:
c = 0.0146
c = 0.0149
c = 0.0147
qu = 0.0292
qu = 0.0298
qu = 0.0294
The graph may not look smooth because we use the compression device manually. By doing it in manual mode, our data will contain some reading error and recording error because we cannot achieve constant speed when rotating the wheel. However, the line of the graph is still within its range of value, so it is acceptable.
11.0 DISCUSSION In the unconfined test, no radial stress is applied to the sample (σ3 = 0). The plunger load, P is increased rapidly until the soil sample fail, that is cannot any additional load. The loading is applied quickly so that the pore water cannot drain from the soil. The effect stress path is unknown since pore water pressure changes are not normally measured. This test is considered as undrained shear test assuming that there is no moisture loss from the specimen during the test. The specimen must not certain any fissures, silt seams, varves, or other defects, this mean that the specimen must be intact, homogenous clay. Rarely are over-consolidated clays intact, and often even normally consolidated clays have some fissures. 11.1.
What are the differences between unconfined compression test and confined
compression test? The differences between unconfined compression test and confined compression test are sample used for unconfined compression test is not covered by any mould or casing but sample used for confined compression test is enclosed between rigid end-caps inside a thin rubber membrane to seal it from cell water, rubber O-ring are fitted over the membrane at the cap to provide a seal. 11.2. What are the advantages of doing unconfined compression test? The advantage of doing unconfined compression test is without any calculation the confining pressure σ3 is equal to 0. From the test result we get the maximum unconfined compression strength (qu) is determined, using qu = σ1and σ3 is equal to 0 plot in graph where normal stress versus shear stress to determine the undrained strength Cu where Cu = qu/2 of the unconfined compression strength we obtained.
11.3. What are the limitations of unconfined compression test? The limitations of the unconfined compression test is applicable to the fully saturated non-fissured clays, and only the undrained strength Cu can be measured. 11.4. Give 4 common laboratory errors for unconfined compression test? The common laboratory errors for unconfined compression test are: Getting wrong reading from dial gauge during the test was running. The soil sample prepared is too wet. Insensitivity of measurements at low strains due to high early soil stiffness. The application of the load to the soil sample was not equal, either too fast or too slow.
12.0 CONCLUSION The unconfined compressive strength is considered to be equal to the load at which the failure of the soil occurs. The consistency of soil samples will have different compressive strength due to different porosity, moisture content and existing of microorganism in the soil.
13.0 REFERENCES …http://civilengineeringlaboratory.blogspot.com/2012/02/unconfined-compressiontest.html …http://spin.mohawkcollege.ca/courses/smeatonk/CV504%20PDFs/Lab %20Manual/Unconfined%20Compression%20Procedure.pdf …http://www.cyut.edu.tw/~jrlai/CE7334/Unconfined.pdf …http://www.uic.edu/classes/cemm/cemmlab/Experiment%2013-Unconfined %20Compression.pdf …http://www.uta.edu/ce/geotech/lab/Main/Soil%20Lab/09_UCS/UCS.pdf
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