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TEST TITLE : DIRECT SHEAR TEST

1.0

OBJECTIVE To determine the parameter of shear strength of soil, cohesion, c and angle of friction, ø.

2.0

LEARNING OUTCOME At the end of this experiment, students are able to:

3.0

•

Determine the shear strength parameter of the soil

•

Handle shear strength test, direct shear test

THEORY The general relationship between maximum shearing resistance, Շf and normal stress, σn for soils can be represented by the equation and known as Coulomb’s Law:

τ f = c + σ tan φ 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 remains constant. If there is no normal stress the friction

disappears. This relationship is shown in Figure 1. 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, Ø and its intercept on the vertical (shear stress) axis being the apparent cohesion, denoted by c.

Figure 1: Shear stress versus normal stress

4.0

TEST EQUIPMENTS 1. Shear box carriage 2. Loading pad 3. Perforated plate 4. Porous plate 5. Retaining plate

Figure 3: Loading pad, Perforated plate, Porous plate, Retaining plate

Figure 2: Shearbox carriage 5.0

PROCEDURES 1.

The internal measurement is verified by using the vernier calipers. The length of the sides, L and the overall depth, B.

2.

The base plate is fixed inside the shear box. Then porous plate is put on the base plate. Next, perforated grid plate is fitted over porous so that the grid plates should be at right angles to the direction shear.

3.

Two halves of the shear box is fixed by means of fixing screws

4.

For cohesive soils, the soil sample is transferred from square specimen cutter to the shearbox by pressing down on the top grid plate. For sandy soil, soil is compacted in layers to the required density in shear box

5.

The shear box assembly is mounted on the loading frame

6.

The dial is set of the proving ring to zero

7.

The loading yoke is placed on the loading pad and the hanger is lifted carefully onto the top of the loading yoke.

8.

The correct loading is then applied to the hanger pad.

9.

The screws clamping the upper half to the lower half is carefully removed.

10.

The test is conducted by applying horizontal shear load to failure. Rate strain should be 0.2mm/min

11.

record readings of horizontal and force dial gauges at regular intervals.

12.

Finally the test is conducted on three identical soil samples under different vertical compressive strsses, 1.75 kg, 2.5 kg and 3.25 kg

6.0

RESULTS

Specimen Number

=

1

Loading

=

1.75 kg

Displacement ∆ L Dail gauge (mm) 0 0.0 50 0.1 100 0.2 150 0.3 200 0.4 250 0.5 300 0.6 350 0.7 400 0.8 450 0.9 500 1.0 550 1.1 600 1.2 650 1.3 700 1.4 750 1.5 800 1.6 850 1.7 900 1.8 950 1.9 1000 2.0 1050 2.1 1100 2.2 1150 2.3 1200 2.4 1250 2.5 1300 2.6 1350 2.7

Proving Ring Dail Load,P gauge (kN) 12 0.105 20 0.175 25 0.219 30 0.263 35 0.306 43 0.376 47 0.411 52 0.455 58 0.508 60 0.525 64 0.560 65 0.569 67 0.586 69 0.604 71 0.621 72 0.630 75 0.656 76 0.665 77 0.674 78 0.683 79 0.691 79 0.691 81 0.709 82 0.718 83 0.726 84 0.735 84 0.735 84 0.735

Shear Stress,σ ( kN/mm²) x 10ˉ5 2.92 4.86 6.08 7.31 8.50 10.44 11.42 12.64 14.11 14.58 15.56 15.81 16.28 16.78 17.25 17.50 18.22 18.47 18.72 18.97 19.19 19.19 19.69 19.94 20.17 20.42 20.42 20.42

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00

Specimen Number

=

2

Loading

=

2.5 kg

Displacement Dail gauge 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

∆ L (mm) 0.0 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

Proving Ring Load,P Dail (kN) gauge x 10ˉ3 1 8.75 1 8.75 5 43.75 15 131.25 22 192.50 25 218.75 30 262.50 35 306.25 40 350.00 45 393.75 50 437.50 55 481.25 60 525.00 63 551.25 66 577.50 70 612.50 72 630.00 74 647.50 76 665.00 78 682.50 80 700.00 82 717.50 83 726.25 85 743.75 88 770.00 89 778.75 90 787.50 90 787.50 91 796.25 91 796.25 91 796.25

Shear Stress,σ ( kN/mm²) x 10ˉ5 0.24

0.24 1.22 3.65 5.35 6.08 7.29 8.51 9.72

10.94 12.15 13.37 14.58 15.31 16.04 17.01 17.50 17.99 18.47 18.96 19.44 19.93 20.17 20.66 21.39 21.63 21.88 21.88 22.12 22.12 22.12

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00 46.67 48.33 50.00

Specimen Number

=

3

Loading

=

3.25 kg

Displacement Dail gauge 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 7.0

∆ L (mm) 0.0 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

DATA ANALYSIS

Proving Ring Load,P Dail (kN) gauge x 10ˉ3 35.00 4 35.00 4 43.75 5 43.75 5 87.50 10 280.00 32 393.75 45 455.00 52 516.25 59 568.75 65 612.50 70 656.25 75 682.50 78 726.25 83 770.00 88 805.00 92 831.25 95 866.25 99 892.50 102 918.75 105 945.00 108 962.50 110 971.25 111 1006.25 115 1032.50 118 1041.25 119 1058.75 121 1067.50 122 1067.50 122 1067.50 122

Shear Stress,σ ( kN/mm²) x 10ˉ5 0.97

0.97 1.22 1.22 2.43 7.78 10.94 12.64 14.34

15.80 17.01 18.23 18.96 20.17 21.39 22.36 23.09 24.06 24.79 25.52 26.25 26.74 26.98 27.95 28.68 28.92 29.41 29.65 29.65 29.65

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00 46.67 48.33

1. Shear stress ( 20mm dial gauge reading ) σ = P/A = [ ( dial gauge x 0.00875) / Area ] 2. Strain ( 20mm dial gauge reading ) τ = ( ∆ L / L ) = [ ( Dail Gauge x 0.002) / Total Length ] The example calculation to find shear stress and strain a) Specimen 1 Dial gauge reading

=

4

Area

=

60 x 60 = 3600mm2

=

(4 x 0.00875) / 3600

=

0.97 x 10ˉ5kN/mm2

Dial gauge

=

50

Length

=

60mm

=

( 50 x 0.002 ) / 60

=

1.67 x 10ˉ3

Shear stress, σ

Strain, τ

b) Specimen 2

Dial gauge reading

=

5

Area

=

60 x 60 = 3600mm2

=

(5 x 0.00875) / 3600

=

1.22 x 10ˉ5kN/mm2

Dial gauge

=

100

Length

=

60mm

=

( 100 x 0.002 ) / 60

=

3.33 x 10ˉ3

Dial gauge reading

=

5

Area

=

60 x 60 = 3600mm2

=

(5 x 0.00875) / 3600

=

1.22 x 10ˉ5kN/mm2

=

150

Shear stress, σ

Strain, τ

3) Specimen 3

Shear stress, σ

Dial gauge

Length

=

60mm

=

( 150 x 0.002 ) / 60

=

5.0 x 10ˉ3

Strain, τ

Normal stress Load = 1.75 kg P = 1.75 x 10 x 9.81 / 1000 = 0.17 kN Normal stress =

P = 0.17 A

= 4.72 x 10ˉ5 kN/m2

3600

Load = 2.5 kg P = 2.5 x 10 x 9.81 / 1000 = 0.25 kN Normal stress =

P = 0.25 A

= 6.94 x 10ˉ5 kN/m2

3600

Load = 3.25 kg P = 3.25 x 10 x 9.81 / 1000 = 0.32 kN Normal stress =

P = 0.32 A

Specimen Number

=

1

3600

= 8.89 x 10ˉ5 kN/m2

Loading

=

1.75 kg

Shear Stress VS Strain 25

20

Shear Stress

15

10

5

0 0

5

10

15

20

25 Strain

Specimen Number

=

2

30

35

40

45

50

Loading

=

2.5 kg

Shear Stress VS Strain 25

20

Shear Stress

15

10

5

0 0

10

20

30 Strain

Specimen Number

=

3

40

50

60

Loading

=

3.25 kg

Shear Stress VS Strain 35

30

Shear Stress

25

20

15

10

5

0 0

10

20

30 Strain

Shear Stress VS Normal Stress

40

50

60

Shear Stress VS Normal Stress 35

30

Shear Stress

25

20

15

10

5

0 0

1

2

3

4

5 Norm al Stress

Φ = 28˚ c =0

Shear strength

6

7

8

9

10

τ = c + σ tan φ

Data obtained: Φ = 28˚ , c = 0

σ = 4.72 x 10ˉ5 kN/m2 τ = 0 + 4.72 x 10ˉ5tan 28˚ = 2.51 x 10ˉ5 kN/m2 σ = 6.94 x 10ˉ5 kN/m2 τ = 0 + 6.94 x 10ˉ5 tan 28˚ = 3.69 x 10ˉ5 kN/m2 σ = 8.89 x 10ˉ5 kN/m2 τ = 0 + 8.89 x 10ˉ5 tan 28˚ = 4.73 x 10ˉ5 kN/m2

10.0

DISCUSSION

From the test of direct shear, graphs of shear stress versus strain and graph of shear stress and normal stress are plotted. The shear strength parameters are determined from the graphs plotted. The values are angle of friction and the cohesion of soil. From the graph of shear stress versus strain, the shear stress for load 1.75 kg is 2.51 x 10ˉ5 kN/m2, the shear stress for load 2.5 kg is 3.69 x 10ˉ5 kN/m2 and the shear stress for load 3.25 kg is 4.73 x 10ˉ5 kN/m2. With the determined value from graph of shear stress versus strain, graph of shear stress against normal stress is plotted. The cohesion of soil and the angle of friction of soil are determined. The cohesion of soil is the intercept of y- axis and the angle of friction is the angle of the linear line produced (line’s slope). From the graph, the cohesion of soil is 0.0 kN/m2 as the sample of soil used is sand. As we know that sand is a type of coarse grained soil and it is assume cohesion less. Form the graph, the angle of friction is 28°. The direct shear test has advantages and disadvantages. It is simple and fast especially for sands. The failure that occurs is along a single surface, which approximates observed slips or shear type failure in natural soils.

10.0

CONCLUSION

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 0.0 kN/m2 as the soil used for the experiment is coarse- grained soil which is sand and the value of friction of angle is 28°. 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 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.

10.0

QUESTIONS

Question 1 a. Why perforated plate in this test with teeth? The perforated plate in this test with teeth because we want to produce a grip forces between the plate and the soil and eventually assists in distributing the shear stress evenly. b. What maximum value of displacement before stop the test? The maximum value of displacement before stop the test is when the values are constant for more than three times and also when the incline value suddenly dropped.

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 (friction angle). The results of the test are plotted on a graph with the peak stress on the xaxis 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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1.0

OBJECTIVE To determine the parameter of shear strength of soil, cohesion, c and angle of friction, ø.

2.0

LEARNING OUTCOME At the end of this experiment, students are able to:

3.0

•

Determine the shear strength parameter of the soil

•

Handle shear strength test, direct shear test

THEORY The general relationship between maximum shearing resistance, Շf and normal stress, σn for soils can be represented by the equation and known as Coulomb’s Law:

τ f = c + σ tan φ 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 remains constant. If there is no normal stress the friction

disappears. This relationship is shown in Figure 1. 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, Ø and its intercept on the vertical (shear stress) axis being the apparent cohesion, denoted by c.

Figure 1: Shear stress versus normal stress

4.0

TEST EQUIPMENTS 1. Shear box carriage 2. Loading pad 3. Perforated plate 4. Porous plate 5. Retaining plate

Figure 3: Loading pad, Perforated plate, Porous plate, Retaining plate

Figure 2: Shearbox carriage 5.0

PROCEDURES 1.

The internal measurement is verified by using the vernier calipers. The length of the sides, L and the overall depth, B.

2.

The base plate is fixed inside the shear box. Then porous plate is put on the base plate. Next, perforated grid plate is fitted over porous so that the grid plates should be at right angles to the direction shear.

3.

Two halves of the shear box is fixed by means of fixing screws

4.

For cohesive soils, the soil sample is transferred from square specimen cutter to the shearbox by pressing down on the top grid plate. For sandy soil, soil is compacted in layers to the required density in shear box

5.

The shear box assembly is mounted on the loading frame

6.

The dial is set of the proving ring to zero

7.

The loading yoke is placed on the loading pad and the hanger is lifted carefully onto the top of the loading yoke.

8.

The correct loading is then applied to the hanger pad.

9.

The screws clamping the upper half to the lower half is carefully removed.

10.

The test is conducted by applying horizontal shear load to failure. Rate strain should be 0.2mm/min

11.

record readings of horizontal and force dial gauges at regular intervals.

12.

Finally the test is conducted on three identical soil samples under different vertical compressive strsses, 1.75 kg, 2.5 kg and 3.25 kg

6.0

RESULTS

Specimen Number

=

1

Loading

=

1.75 kg

Displacement ∆ L Dail gauge (mm) 0 0.0 50 0.1 100 0.2 150 0.3 200 0.4 250 0.5 300 0.6 350 0.7 400 0.8 450 0.9 500 1.0 550 1.1 600 1.2 650 1.3 700 1.4 750 1.5 800 1.6 850 1.7 900 1.8 950 1.9 1000 2.0 1050 2.1 1100 2.2 1150 2.3 1200 2.4 1250 2.5 1300 2.6 1350 2.7

Proving Ring Dail Load,P gauge (kN) 12 0.105 20 0.175 25 0.219 30 0.263 35 0.306 43 0.376 47 0.411 52 0.455 58 0.508 60 0.525 64 0.560 65 0.569 67 0.586 69 0.604 71 0.621 72 0.630 75 0.656 76 0.665 77 0.674 78 0.683 79 0.691 79 0.691 81 0.709 82 0.718 83 0.726 84 0.735 84 0.735 84 0.735

Shear Stress,σ ( kN/mm²) x 10ˉ5 2.92 4.86 6.08 7.31 8.50 10.44 11.42 12.64 14.11 14.58 15.56 15.81 16.28 16.78 17.25 17.50 18.22 18.47 18.72 18.97 19.19 19.19 19.69 19.94 20.17 20.42 20.42 20.42

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00

Specimen Number

=

2

Loading

=

2.5 kg

Displacement Dail gauge 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

∆ L (mm) 0.0 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

Proving Ring Load,P Dail (kN) gauge x 10ˉ3 1 8.75 1 8.75 5 43.75 15 131.25 22 192.50 25 218.75 30 262.50 35 306.25 40 350.00 45 393.75 50 437.50 55 481.25 60 525.00 63 551.25 66 577.50 70 612.50 72 630.00 74 647.50 76 665.00 78 682.50 80 700.00 82 717.50 83 726.25 85 743.75 88 770.00 89 778.75 90 787.50 90 787.50 91 796.25 91 796.25 91 796.25

Shear Stress,σ ( kN/mm²) x 10ˉ5 0.24

0.24 1.22 3.65 5.35 6.08 7.29 8.51 9.72

10.94 12.15 13.37 14.58 15.31 16.04 17.01 17.50 17.99 18.47 18.96 19.44 19.93 20.17 20.66 21.39 21.63 21.88 21.88 22.12 22.12 22.12

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00 46.67 48.33 50.00

Specimen Number

=

3

Loading

=

3.25 kg

Displacement Dail gauge 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 7.0

∆ L (mm) 0.0 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

DATA ANALYSIS

Proving Ring Load,P Dail (kN) gauge x 10ˉ3 35.00 4 35.00 4 43.75 5 43.75 5 87.50 10 280.00 32 393.75 45 455.00 52 516.25 59 568.75 65 612.50 70 656.25 75 682.50 78 726.25 83 770.00 88 805.00 92 831.25 95 866.25 99 892.50 102 918.75 105 945.00 108 962.50 110 971.25 111 1006.25 115 1032.50 118 1041.25 119 1058.75 121 1067.50 122 1067.50 122 1067.50 122

Shear Stress,σ ( kN/mm²) x 10ˉ5 0.97

0.97 1.22 1.22 2.43 7.78 10.94 12.64 14.34

15.80 17.01 18.23 18.96 20.17 21.39 22.36 23.09 24.06 24.79 25.52 26.25 26.74 26.98 27.95 28.68 28.92 29.41 29.65 29.65 29.65

Strain, τ (kN/mm2) x 10ˉ3 0 1.67 3.33 5.00 6.67 8.33 10.00 11.67 13.33 15.00 16.67 18.33 20.00 21.67 23.33 25.00 26.67 28.33 30.00 31.67 33.33 35.00 36.67 38.33 40.00 41.67 43.33 45.00 46.67 48.33

1. Shear stress ( 20mm dial gauge reading ) σ = P/A = [ ( dial gauge x 0.00875) / Area ] 2. Strain ( 20mm dial gauge reading ) τ = ( ∆ L / L ) = [ ( Dail Gauge x 0.002) / Total Length ] The example calculation to find shear stress and strain a) Specimen 1 Dial gauge reading

=

4

Area

=

60 x 60 = 3600mm2

=

(4 x 0.00875) / 3600

=

0.97 x 10ˉ5kN/mm2

Dial gauge

=

50

Length

=

60mm

=

( 50 x 0.002 ) / 60

=

1.67 x 10ˉ3

Shear stress, σ

Strain, τ

b) Specimen 2

Dial gauge reading

=

5

Area

=

60 x 60 = 3600mm2

=

(5 x 0.00875) / 3600

=

1.22 x 10ˉ5kN/mm2

Dial gauge

=

100

Length

=

60mm

=

( 100 x 0.002 ) / 60

=

3.33 x 10ˉ3

Dial gauge reading

=

5

Area

=

60 x 60 = 3600mm2

=

(5 x 0.00875) / 3600

=

1.22 x 10ˉ5kN/mm2

=

150

Shear stress, σ

Strain, τ

3) Specimen 3

Shear stress, σ

Dial gauge

Length

=

60mm

=

( 150 x 0.002 ) / 60

=

5.0 x 10ˉ3

Strain, τ

Normal stress Load = 1.75 kg P = 1.75 x 10 x 9.81 / 1000 = 0.17 kN Normal stress =

P = 0.17 A

= 4.72 x 10ˉ5 kN/m2

3600

Load = 2.5 kg P = 2.5 x 10 x 9.81 / 1000 = 0.25 kN Normal stress =

P = 0.25 A

= 6.94 x 10ˉ5 kN/m2

3600

Load = 3.25 kg P = 3.25 x 10 x 9.81 / 1000 = 0.32 kN Normal stress =

P = 0.32 A

Specimen Number

=

1

3600

= 8.89 x 10ˉ5 kN/m2

Loading

=

1.75 kg

Shear Stress VS Strain 25

20

Shear Stress

15

10

5

0 0

5

10

15

20

25 Strain

Specimen Number

=

2

30

35

40

45

50

Loading

=

2.5 kg

Shear Stress VS Strain 25

20

Shear Stress

15

10

5

0 0

10

20

30 Strain

Specimen Number

=

3

40

50

60

Loading

=

3.25 kg

Shear Stress VS Strain 35

30

Shear Stress

25

20

15

10

5

0 0

10

20

30 Strain

Shear Stress VS Normal Stress

40

50

60

Shear Stress VS Normal Stress 35

30

Shear Stress

25

20

15

10

5

0 0

1

2

3

4

5 Norm al Stress

Φ = 28˚ c =0

Shear strength

6

7

8

9

10

τ = c + σ tan φ

Data obtained: Φ = 28˚ , c = 0

σ = 4.72 x 10ˉ5 kN/m2 τ = 0 + 4.72 x 10ˉ5tan 28˚ = 2.51 x 10ˉ5 kN/m2 σ = 6.94 x 10ˉ5 kN/m2 τ = 0 + 6.94 x 10ˉ5 tan 28˚ = 3.69 x 10ˉ5 kN/m2 σ = 8.89 x 10ˉ5 kN/m2 τ = 0 + 8.89 x 10ˉ5 tan 28˚ = 4.73 x 10ˉ5 kN/m2

10.0

DISCUSSION

From the test of direct shear, graphs of shear stress versus strain and graph of shear stress and normal stress are plotted. The shear strength parameters are determined from the graphs plotted. The values are angle of friction and the cohesion of soil. From the graph of shear stress versus strain, the shear stress for load 1.75 kg is 2.51 x 10ˉ5 kN/m2, the shear stress for load 2.5 kg is 3.69 x 10ˉ5 kN/m2 and the shear stress for load 3.25 kg is 4.73 x 10ˉ5 kN/m2. With the determined value from graph of shear stress versus strain, graph of shear stress against normal stress is plotted. The cohesion of soil and the angle of friction of soil are determined. The cohesion of soil is the intercept of y- axis and the angle of friction is the angle of the linear line produced (line’s slope). From the graph, the cohesion of soil is 0.0 kN/m2 as the sample of soil used is sand. As we know that sand is a type of coarse grained soil and it is assume cohesion less. Form the graph, the angle of friction is 28°. The direct shear test has advantages and disadvantages. It is simple and fast especially for sands. The failure that occurs is along a single surface, which approximates observed slips or shear type failure in natural soils.

10.0

CONCLUSION

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 0.0 kN/m2 as the soil used for the experiment is coarse- grained soil which is sand and the value of friction of angle is 28°. 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 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.

10.0

QUESTIONS

Question 1 a. Why perforated plate in this test with teeth? The perforated plate in this test with teeth because we want to produce a grip forces between the plate and the soil and eventually assists in distributing the shear stress evenly. b. What maximum value of displacement before stop the test? The maximum value of displacement before stop the test is when the values are constant for more than three times and also when the incline value suddenly dropped.

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 (friction angle). The results of the test are plotted on a graph with the peak stress on the xaxis 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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