Shear Direct Wani

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DIRECT SHEAR TESTGROUP

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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


(kN/mm2)

Strain,

(kN/mm2)


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


(kN/mm2)

Strain,

(kN/mm2)


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


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


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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Shear Direct Wani