Direct Shear Test 1k72j

TABLE OF CONTENTS NO. 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

CONTENT Introduction Objective Theory Background Equipment Procedures Results Sample Calculation Discussion Conclusion Reference

PAGES 2 2 2-3 3-4 4-5 5-7 8-9 9-10 10-11 11

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1.0 Introduction The test is carried out on either undisturbed samples or remolded samples. To facilitate the remolding purpose, a soil sample may be compacted at optimum moisture content in a compaction mold. Then specimen for the direct shear test could be obtained using the correct cutter provided. Alternatively, sand sample can be placed in a dry state at a required density, in the assembled shear box. A normal load is applied to the specimen and the specimen is sheared across the predetermined horizontal plane between the two halves of the shear box. Measurements of shear load, shear displacement and normal displacement are recorded. The test is repeated foe two or more identical specimens under different normal loads. From the results, the shear strength parameters can be determined.

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

3.0 Theory Background 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:

 f  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

τf

= shearing resistance of soil at failure

σf

= total normal stress on failure plane

The friction components increase with increasing normal stress but the cohesion components remains constant. If there is no normal stress the friction disappears. This 2

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, Ø and its intercept on the vertical (shear stress) axis being the apparent cohesion, denoted by c.

Graph of Shear Stress vs Normal Stress

4.0 Equipment 1. Shear box carriage. 2. Loading pad. 3. Perforated plate. 4. Porous plate. 5. Retaining plate. 6. Grease.

Shear box carriage

Loading page

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Perforated plate, Porous plate,

Grease

Retaining plat

5.0 Procedures 1. Internal measurement is by using vernier calipers. The length of the sides, L and the overall depth, B. 2. Base plate is fixed inside the shear box. Then porous plate is put on the base plate. 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 transfer 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. 6. The dial is set of the proving ring to zero 7. The loading yoke is placed on the loading pad and carefully lift the hanger onto the top of the loading yoke. 8. The correct loading is applied to the hanger pad. 9. Carefully the screws clamping the upper half is removed to the lower half. 10. The test is conducted by applying horizontal shear load to failure. Rate strain should be 0.2mm/min 11. Readings of horizontal is recorded and force dial gauges at regular intervals. 12. Conduct test on three identical soil samples under different vertical compressive stresses, 1.75kg, 2.5kg and 3.2kg

6.0 Results Specimen No . 1 Loading : 1.75 kg

1.75 kg × 9.81 N × _1kN_ = 0.017 kN 1kg

1000N

Length : 60mm = 0.06m Area

: 0.06m × 0.06m = 3.6×10-3m2

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Displacement Dial ∆L (mm)(X10-4)

Proving ring Dial Load, P (kN) -5

gauge

gauge

(x 10 )

0.2 4 0.4 8 0.6 12 0.8 16 1.0 20 1.2 24 1.4 28 1.6 32 1.8 36 2.0 40 2.2 44 2.4 48 2.6 52 2.8 56 3.0 60 3.2 64 Specimen No. 2

14 25 34 41 44 50 55 59 63 65 67 69 70 70 70 70

2.9 5.1 6.9 8.4 9.0 10.2 11.2 12.0 12.9 13.3 13.7 14.1 14.3 14.3 14.3 14.3

Loading : 2.5 kg

Shear stress 2

(KN/m )

Strain (x 10-6)

(x 10-3) 8.1 14.2 19.2 23.3 25.0 28.3 31.1 33.3 35.8 36.9 38.1 39.2 39.7 39.7 39.7 39.7

6.7 13.3 20.0 26.7 33.3 40.0 46.7 53.3 60.0 66.7 73.3 80.0 86.7 93.3 100.0 106.7

Shear stress

Strain

(KN/m2)

(X 10-6)

(x 10-3) 11.4 18.1 26.0 33.9 40.3 46.4 52.2 55.0 56.7 57.8 57.8 57.8 57.8

6.7 13.3 20.0 26.7 33.3 40.0 46.7 53.3 60.0 66.7 73.3 80.0 86.7

2.5 kg × 9.81 N × _1kN_ = 0.025 kN 1kg

1000N

Length : 60mm = 0.06m Area

: 0.06m × 0.06m = 3.6×10-3m2

Displacement Dial ∆L (mm)(X10-4)

Proving ring Dial Load, P (kN) -5

gauge

gauge

(x 10 )

0.2 4 0.4 8 0.6 12 0.8 16 1.0 20 1.2 24 1.4 28 1.6 32 1.8 36 2.0 40 2.2 44 2.4 48 2.6 52 Specimen No.3

20 32 28 60 71 82 92 97 100 102 102 102 102

4.1 6.5 5.7 12.2 14.5 16.7 18.8 19.8 20.4 20.8 20.8 20.8 20.8

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Loading : 3.25 kg

3.25 kg × 9.81 N × _1kN_ = 0.032 kN 1kg

1000N

Length : 60mm = 0.06m Area

: 0.06m × 0.06m = 3.6×10-3m2

Displacement Dial ∆L (mm)(X10-4) gauge 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

4 8 12 16 20 24 28 32 36 40 44 48 52 56 60

Proving ring Dial Load, P (kN) -5

gauge

(x10 )

28 47 64 88 102 115 121 127 134 135 137 138 138 138 138

5.7 9.6 13.1 17.9 20.8 23.5 24.7 25.9 27.3 27.5 27.9 28.2 28.2 28.2 28.2

Shear stress

Strain

(KN/m2)

(X 10-6)

(x 10-3) 15.8 26.7 36.4 49.7 57.8 65.3 68.6 71.9 75.8 76.4 77.5 78.3 78.3 78.3 78.3

6.7 13.3 20.0 26.7 33.3 40.0 46.7 53.3 60.0 66.7 73.3 80.0 86.7 93.3 100.0

7.0 Sample Calculation

1. Displacement = dial gauge x 0.002 = 0.2 x 0.002 = 4 x 10-4 mm

2. Proving ring = dial gauge x 0.00204/1000 = 14 x 0.00204/1000 = 2.9×10-5 kN

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3. Shear stress (0.2 mm dial gauge) = Dail gauge x 0.00204/1000 Area = 14(0.00204)/1000 kN 0.06 m x 0.06 m = 8.1×10-3 kN/m2

4. Strain (0.2 mm dial gauge) = displacement / total length = 4 x 10-4 mm / 60 mm = 6.7×10-6

5. Normal Stress, ( kN/ mm2 ) a) For 1.75kg load. = Load , P Area, A = _0.017 kN___ 0.06m × 0.06m = 4.7 kN / m2 b) For 2.5 kg load = 0.025 kN_____ 0.06m × 0.06m = 6.9 kN / m2

c) For 3.25 kg load = 0.032 kN______ 0.06m × 0.06m = 8.9kN / m2

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6.

3.1 cm

Tan ϕ = 3.1 / 4.4 Φ = 35ᵒ

4.4 cm

8.0 Discussion The direct shear test is suited to the relatively rapid determination of the the parameter of the shear strength of soil, to find the value of cohesion and also to find the angle of friction. At the end of result we had plot the graph, which is the graph of shear stress versus strain. The graph will gained us to value of friction angle. (Refer to the graph). At these 3 samples which are 1.75kg, 2.5 kg and 3.25kg there are no error data obtained. The value obtained from the dial gauge showed increases directly. This is because the dial gauge reading has increased the time by the time. The cohesion of soil and the angle of friction of soil are determined. 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 type of coarse grained soil and it is assume cohesion less. Form the graph, the angle of friction is 35°.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

9.0 Conclusion

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Direct shear test is useful when cohesion less soils are to be tested. In this test the failure plane is forced to occur at a predetermined location where both normal and shear stresses are acting; the sample is placed in a closed shear box, fixed at the base with the top free to translate under a horizontal force. The two portions of the box are spaced by using spacing screws to reduce the friction. The space should be at least as large as the largest sand particle. The box is then placed in the direct shear apparatus, and increasing horizontal load is applied with constant corresponding vertical load, and the horizontal deformation shall be recorded by using the dial gage. For each test shear stress-strain diagram is drawn in order to find out the ultimate stress, then the shear failure envelope is drawn by relating each ultimate shear stress to the normal stress corresponding to it in at least three tests. 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 REFERENCE



Braja M. Das, Principles of Geotechnical Engneering. Seventh Edition. SI Edition. Cengage Learning. 10

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