Direct Shear Test

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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:

• 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, Շf and

normal stress, σn for soils can be represented by the equation and known as

Coulomb’s Law:

φ σ τf =c+ tan 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

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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.

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

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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.

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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

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6.0 RESULTS

Specimen Number = 1

Loading = 1.75 kg

Displacement Proving Ring

Shear Stress,σ ( kN/mm²) (kN/mmStrain, τ 2₎ Dail gauge ∆ L (mm) Dail gauge Load,P (kN) x 10ˉ5 _{x 10ˉ}3 0 0.0 12 0.105 2.92 0 50 0.1 20 0.175 4.86 1.67 100 0.2 25 0.219 6.08 3.33 150 0.3 30 0.263 7.31 5.00 200 0.4 35 0.306 8.50 6.67 250 0.5 43 0.376 10.44 8.33 300 0.6 47 0.411 11.42 10.00 350 0.7 52 0.455 12.64 11.67 400 0.8 58 0.508 14.11 13.33 450 0.9 60 0.525 14.58 15.00 500 1.0 64 0.560 15.56 16.67 550 1.1 65 0.569 15.81 18.33 600 1.2 67 0.586 16.28 20.00 650 1.3 69 0.604 16.78 21.67 700 1.4 71 0.621 17.25 23.33 750 1.5 72 0.630 17.50 25.00 800 1.6 75 0.656 18.22 26.67 850 1.7 76 0.665 18.47 28.33 900 1.8 77 0.674 18.72 30.00 950 1.9 78 0.683 18.97 31.67 1000 2.0 79 0.691 19.19 33.33 1050 2.1 79 0.691 19.19 35.00 1100 2.2 81 0.709 19.69 36.67 1150 2.3 82 0.718 19.94 38.33 1200 2.4 83 0.726 20.17 40.00 1250 2.5 84 0.735 20.42 41.67 1300 2.6 84 0.735 20.42 43.33 1350 2.7 84 0.735 20.42 45.00

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Loading = 2.5 kg

Shear Stress,σ ( kN/mm²) (kN/mmStrain, τ 2₎ Dail gauge ∆ L (mm) Dail gauge Load,P (kN) x 10ˉ5 _{x 10ˉ}3 x 10ˉ3 0 0.0 1 8.75 0.24 0 50 0.1 1 8.75 0.24 1.67 100 0.2 5 43.75 1.22 3.33 150 0.3 15 131.25 3.65 5.00 200 0.4 22 192.50 5.35 6.67 250 0.5 25 218.75 6.08 8.33 300 0.6 30 262.50 7.29 10.00 350 0.7 35 306.25 8.51 11.67 400 0.8 40 350.00 9.72 13.33 450 0.9 45 393.75 10.94 15.00 500 1.0 50 437.50 12.15 16.67 550 1.1 55 481.25 13.37 18.33 600 1.2 60 525.00 14.58 20.00 650 1.3 63 551.25 15.31 21.67 700 1.4 66 577.50 16.04 23.33 750 1.5 70 612.50 17.01 25.00 800 1.6 72 630.00 17.50 26.67 850 1.7 74 647.50 17.99 28.33 900 1.8 76 665.00 18.47 30.00 950 1.9 78 682.50 18.96 31.67 1000 2.0 80 700.00 19.44 33.33 1050 2.1 82 717.50 19.93 35.00 1100 2.2 83 726.25 20.17 36.67 1150 2.3 85 743.75 20.66 38.33 1200 2.4 88 770.00 21.39 40.00 1250 2.5 89 778.75 21.63 41.67 1300 2.6 90 787.50 21.88 43.33 1350 2.7 90 787.50 21.88 45.00 1400 2.8 91 796.25 22.12 46.67 1450 2.9 91 796.25 22.12 48.33 1500 3 91 796.25 22.12 50.00

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Loading = 3.25 kg

Shear Stress,σ ( kN/mm²) (kN/mmStrain, τ 2₎ Dail gauge ∆ L (mm) Dail gauge Load,P (kN) x 10ˉ5 _{x 10ˉ}3 x 10ˉ3 0 0.0 4 35.00 0.97 0 50 0.1 4 35.00 0.97 1.67 100 0.2 5 43.75 1.22 3.33 150 0.3 5 43.75 1.22 5.00 200 0.4 10 87.50 2.43 6.67 250 0.5 32 280.00 7.78 8.33 300 0.6 45 393.75 10.94 10.00 350 0.7 52 455.00 12.64 11.67 400 0.8 59 516.25 14.34 13.33 450 0.9 65 568.75 15.80 15.00 500 1.0 70 612.50 17.01 16.67 550 1.1 75 656.25 18.23 18.33 600 1.2 78 682.50 18.96 20.00 650 1.3 83 726.25 20.17 21.67 700 1.4 88 770.00 21.39 23.33 750 1.5 92 805.00 22.36 25.00 800 1.6 95 831.25 23.09 26.67 850 1.7 99 866.25 24.06 28.33 900 1.8 102 892.50 24.79 30.00 950 1.9 105 918.75 25.52 31.67 1000 2.0 108 945.00 26.25 33.33 1050 2.1 110 962.50 26.74 35.00 1100 2.2 111 971.25 26.98 36.67 1150 2.3 115 1006.25 27.95 38.33 1200 2.4 118 1032.50 28.68 40.00 1250 2.5 119 1041.25 28.92 41.67 1300 2.6 121 1058.75 29.41 43.33 1350 2.7 122 1067.50 29.65 45.00 1400 2.8 122 1067.50 29.65 46.67 1450 2.9 122 1067.50 29.65 48.33 7.0 DATA ANALYSIS

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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 Shear stress, σ = (4 x 0.00875) / 3600 = 0.97 x 10ˉ5_kN/mm2 Dial gauge = 50 Length = 60mm Strain, τ = ( 50 x 0.002 ) / 60 = 1.67 x 10ˉ3 b) Specimen 2

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Dial gauge reading = 5 Area = 60 x 60 = 3600mm2 Shear stress, σ = (5 x 0.00875) / 3600 = 1.22 x 10ˉ5_kN/mm2 Dial gauge = 100 Length = 60mm Strain, τ = ( 100 x 0.002 ) / 60 = 3.33 x 10ˉ3 3) Specimen 3

Dial gauge reading = 5

Area = 60 x 60 = 3600mm2

Shear stress,

σ = (5 x 0.00875) / 3600 = 1.22 x 10ˉ5_kN/mm2

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Length = 60mm Strain, τ = ( 150 x 0.002 ) / 60 = 5.0 x 10ˉ3 Normal stress Load = 1.75 kg P = 1.75 x 10 x 9.81 / 1000 = 0.17 kN Normal stress = P = 0.17 = 4.72 x 10ˉ5_kN/m2 _{A 3600} Load = 2.5 kg P = 2.5 x 10 x 9.81 / 1000 = 0.25 kN Normal stress = P = 0.25 = 6.94 x 10ˉ5_kN/m2 _{A 3600} Load = 3.25 kg P = 3.25 x 10 x 9.81 / 1000 = 0.32 kN Normal stress = P = 0.32 = 8.89 x 10ˉ5_kN/m2 _{A 3600} Specimen Number = 1

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Loading = 1.75 kg

Shear Stress VS Strain

0 5 10 15 20 25 0 5 10 15 20 25 30 35 40 45 50 Strain S h ea r S tr es s

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Loading = 2.5 kg

0 5 10 15 20 25 0 10 20 30 40 50 60 Strain S h ea r S tr es s

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Loading = 3.25 kg

Shear Stress VS Normal Stress

0 5 10 15 20 25 30 35 0 10 20 30 40 50 60 Strain S h ea r S tr es s

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Φ = 28˚ c = 0

Shear strength

Shear Stress VS Normal Stress

0 5 10 15 20 25 30 35 0 1 2 3 4 5 6 7 8 9 10 Norm al Stress S h ea r S tr es s

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φ σ

τ =c+ tan _{Data obtained: Φ = 28˚ , c = 0}

σ = 4.72 x 10ˉ5_kN/m2 τ = 0 + 4.72 x 10ˉ5_{tan 28˚ = 2.51 x 10ˉ}5_kN/m2 σ = 6.94 x 10ˉ5_kN/m2 τ = 0 + 6.94 x 10ˉ5_{tan 28˚ = 3.69 x 10ˉ}5_kN/m2 σ = 8.89 x 10ˉ5_kN/m2 τ = 0 + 8.89 x 10ˉ5_{tan 28˚ = 4.73 x 10ˉ}5_kN/m2 10.0 DISCUSSION

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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.

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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.

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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 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.

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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.