shear direct wani
TRANSCRIPT
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Figure 1 : Direct Shear Machine
1.0OBJECTIVE
To determine the parameter of shear strength of soil, cohesion, c and angle of friction,
2.0LEARNING 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
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3.0 THEORY
The general relationship between maximum shearing resistance, f and normal stress, n for
soils can be represented by the equation and known as Coulombs Law:
tan cf
where:
c = cohesion, which is due to internal forces holding soil particles together in a solid mass
= 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
remains constant. If there is no normal stress the friction disappears. This relationship shown in the
graph below. This graph generally approximates to a straight line, its inclination to the horizontal axis
being equal to the angle of shearing resistance of the soil, and its intercept on the vertical (shear
stress) axis being the apparent cohesion, denoted by c.
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4.0 TEST EQUIPMENTS AND MATERIALS
Figure 3 : Shear box carriage
Figure 4: Loading pad
Figure 5 : Perforated plate, Porous plate, Retaining plate and vernier caliper
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5.0PROCEDURES
1. Verify internal measurement using vernier calipers. The length of the sides, L and the overalldepth, B.
2. Fix base plate inside the shear box. Then put porous plate on the base plate. Fit perforated gridplate over porous so that the grid plates should be at right angles to the direction shear.
3. Fix two halves of the shear box by means of fixing screws4. For cohesive soils, transfer the soil sample from square specimen cutter to the shearbox by
pressing down on the top grid plate. For sandy soil, compact soil in layers to the required density
in shear box
5. Mount the shear box assembly on the loading frame.
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6. Set the dial of the proving ring to zero7. Place the loading yoke on the loading pad and carefully lift the hanger onto the top of the
loading yoke.
8. Apply the correct loading to the hanger pad.
9. Carefully remove the screws clamping the upper half to the lower half10.Conduct the test by applying horizontal shear load to failure. Rate strain should be 0.2mm/min11.Record readings of horizontal and force dial gauges at regular intervals.
12.Conduct test on three identical soil samples under different vertical compressive strsses, 1.75kg,2.5kg and 3.2kg
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6.0RESULT
Result Sample 1
Loading : 1.75 kg
Displacement Proving Ring Shear Stress,
(kN/mm2)
Strain,
(kN/mm2)
Dail Gauge L (mm) Dail Gauge Load, P (kN)
50 0.1 101 0.8838 24.55 0.0017
100 0.2 209 1.8288 50.80 0.0033
150 0.3 308 2.6950 74.86 0.0050
200 0.4 377 3.2988 91.63 0.0067
250 0.5 423 3.7013 102.81 0.0083
300 0.6 448 3.9200 108.89 0.0100
350 0.7 467 4.0863 113.51 0.0117
400 0.8 472 4.1300 114.72 0.0133
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Result Sample 2
Loading : 2.5 kg
Displacement Proving Ring Shear Stress,
(kN/mm2)
Strain,
(kN/mm2)
Dail Gauge L (mm) Dail Gauge Load, P (kN)
50 0.1 77 0.6738 18.72 0.0017
100 0.2 97 0.8488 23.58 0.0033
150 0.3 140 1.2250 34.03 0.0050
200 0.4 197 1.7238 47.88 0.0067
250 0.5 261 2.2838 63.44 0.0083
300 0.6 323 2.8263 78.51 0.0100
350 0.7 366 3.2025 88.96 0.0117
400 0.8 408 3.5700 99.17 0.0133
450 0.9 440 3.8500 106.94 0.0150
500 1.0 457 3.9988 111.08 0.0167
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Result Sample 3
Loading : 3.25 kg
Displacement Proving Ring Shear Stress,
(kN/mm2)
Strain,
(kN/mm2)
Dail Gauge L (mm) Dail Gauge Load, P (kN)
50 0.1 54 0.4725 13.13 0.0017
100 0.2 84 0.7350 20.42 0.0033
150 0.3 111 0.9713 26.98 0.0050
200 0.4 151 1.3213 36.70 0.0067
250 0.5 188 1.6450 45.69 0.0083
300 0.6 222 1.9425 53.96 0.0100
350 0.7 245 2.1438 59.55 0.0117
400 0.8 270 2.3625 65.63 0.0133
450 0.9 288 2.5200 70.00 0.0150
500 1.0 300 2.6250 72.92 0.0167
550 1.1 307 2.6863 74.62 0.0183
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Data Analysis and Calculation
1. Shear stress ( 50mm dial gauge reading )
2. Strain ( 50mm dial gauge reading )
The example calculation to find shear stress and strain
a) Specimen 1
Proving Ring
Dial gauge reading = 50
Area = 0.06 x 0.06 = 0.0036 m2
Shear stress,
Load = {101 X 0.00875}/(0.06 X 0.06)
= 24.55 kN/ m2
= P/A = [(Dial gauge x 0.00875) / Area]
= [ ( Dail Gauge x 0.002) / Total Length ]
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Displacement
Dial gauge = 50
Length = 60mm
Strain,
= (50 x 0.002) / 60
= 0.0017
b) Specimen 2
Dial gauge reading = 77
Area = 0.06 x 0.06 = 0.0036 m2
Shear stress,
Load = {77 X 0.00875}/(0.06 X 0.06)
= 18.72kN/m2
Displacement
Dial gauge = 50
Length = 60mm
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Strain,
= (50 x 0.002) / 60
= 0.0017
c) Specimen 3
Dial gauge reading = 54
Area = 0.06 x 0.06 = 0.0036 m2
Shear stress
Load = {54 X 0.00875}/(0.06 X 0.06)
= 13.13 kN/ m2
Displacement
Dial gauge = 50
Length = 60mm
Strain,
= (50 x 0.002) / 60
= 0.0017
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f = c + tan 20
Shear strength
From graph,
C = 0.33
= 20
1. = 4.78 kN/m2
f = 0.005+ 4.78 tan 27 = 2.44 kN/m2
2. = 6.81 kN/m2
f = 0.005 + 6.81 tan 27 = 3.47 kN/m2
3. = 8.72 kN/m2
f = 0.005 + 8.72 tan 27 = 4.45kN/m2
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The graph of Shear stress () versus Strain () for specimen 1
The graph of Shear stress () versus Strain () for specimen 2
0
20
40
60
80
100
120
140
1 2 3 4 5 6 7 8 9
shearstress(kn/mm2)
strain,(kn/mm2)
Shear Stress,
(kN/mm2)
Strain, (kN/mm2)
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0 50 100 150
shearstress(kn/mm2)
strain(kn/mm2)
Strain, (kN/mm2)
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7.0DISCUSSION
The experiment of the Direct Shear Stress is to determine the parameter of the shear strength of
soil, to find the value of cohesion and also to find the angle of friction. By plotting the graph, which is the
graph of shear stress versus strain, we can find for these values. (Refer to the graph)
Erratum often occurs while taking the readings of the data. As it happens, because the dial gauges
have some technical problem So, the readings that we took may have differed from what we weresupposed to have taken. Another reason is, it may also happen when the equipments were not fixed in
properly enough hence the data obtained as the result may slightly run from the one that we were
supposed to obtain. The equipment we used were not cleaned properly before it was used to undergo
the experiment. Old particles of sand and soil from previous experiments were still stuck onto the
bottom of some of the plates hence the result of the experiment may be affected a little by this incident
8.0 CONCLUSIOIN
As a conclusion, we can know that the objective of the experiment is to determine the
parameter of shear strength of soil, cohesion and angle of friction was achieved. From the
experiment that we have done, the value of cohesion, c is 0.005kN/m 2 and the value of friction
of angle is 27.
The direct shear test can be used to measure the effective stress parameters of any type
of soil as long as the pore pressure induced by the normal force and the shear force can
dissipate with time. For the experiment we use the clean sands as a sample, so there is no
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problem as the pore pressure dissipates readily. However, in the case of highly plastic clays, it is
merely necessary to have a suitable strain rate so that the pore pressure can dissipate with time.
Direct shear tests can be performed under several conditions. The sample is normally
saturated before the test is run. The test can be run at the in-situ moisture content. Before we
find the value of cohesion and friction angle, we must plot the graph from the data that we get
from the experiment. The results of the tests on each specimen are plotted on a graph with the
peak (or residual) stress on the x-axis and the confining stress on the y-axis. The y-intercept of
the curve which fits the test results is the cohesion, and the slope of the line or curve is the
friction angle.
9.0 QUESTIONS 1
a.Why perforated plate in this test with teeth?The perforated plate in this test with teeth because by the teeth, the experiment can be
produce a grip forces between the involved plate and the sand and can assists in distributing the
shear stress.
Perforated
Perforated
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b. What maximum value of displacement before stop the test?
The maximum value of displacement before stop the test is when the values from dial gauge are
constant at least three times continuously or no more increase data and also when the incline
value suddenly dropped so we stop the test.
Question 2
a. What is the purpose of a direct shear test? Which soil properties does it measure?
A direct shear test is a laboratory test used by geotechnical engineers to find the shear strength
parameters of soil. The direct shear test measures the shear strength parameters which are the soil
cohesion (c) and the angle of friction (). The results of the test are plotted on a graph with the peak
stress on the x- axis and the confining stress on the y- axis. The y- intercept of the curve which fits
the test results is the cohesion and the slope of the line or curve is the friction angle.
b) Why do we use fixing screw in this test? What happen if you do not removed them during test?
We use fixing screw in this direct shear test because in order to avoid shear for happening
before the experiment is carried out. If we do not remove them during the test, they will be no
friction and the there will be no shear on the sample and thus the result will be not accurate.
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REFERENCE
BOOK NAME : GEOTECHNICAL ENGINEERING, SOIL MECHANICS
AUTHOR : JOHN N. CERNICA
PUBLISHER : JOHN WILEY & SONS, INC
BOOK NAME : FUNDAMENTALS OF GEOTECHNICAL ENGINEERING, FOURTH EDITION
AUTHOR : BRAJA M.DAS
PUBLISHER : GLOBAL ENGINEERING: CHRISTOPHER M.SHORTT
BOOK NAME : MEKANIK TANAH EDISI KE EMPART
AUTHOR : R.F. CRAIG
TRANSLATOR : AMINATON MARTO
FATIMAH MOHD NOOR
FAUZIAH KASIM
PUBLISHER : UTM
BOOK NAME : BASIC SOIL MECHANICS
AUTHOR : ROY WHITLOW
PUBLISHER : PRENTICE HALL
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