TABLE OF CONTENT

NO CONTENT 1 OBJECTIVE 2 LEARNING OUTCOME 3 THEORY 4 REST EQUIPMENTS 5 PROCEDURES 6 RESULT 7 DATA ANALYSIS 8 DISCUSSION 9 CONCLUSION10 QUESTION

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, student 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 :

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 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.

Figure 1.1 : Graph shear stress versus normal stress

4.0 TEST EQUIPMENTS

1. Shear box carriage2. Loading pad3. Perforated plate4. Porous plate 5. Retaining plate

Figure 1.2 : Shear box carriage Figure 1.3 : Loading pad, perforated

plate, porous plate and retaining plate

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

Apparatus of experiment.

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, 1.75kg, 2.5kg and 3.25kg.

6.0 RESULT

Specimen No : 1Loading : 1.75kg

Displacement Proving Ring Shear Stress (kN/m2) Strain

Dail Gauge ∆L (mm) Dail Gauge Load, P (kN)

50 0.10 1 0.0020 0.57 0.0017100 0.20 9 0.0184 5.10 0.0033150 0.30 13 0.0265 7.37 0.0050200 0.40 20 0.0408 11.33 0.0067250 0.50 24 0.0490 13.60 0.0083300 0.60 28 0.0571 15.87 0.0100350 0.70 31 0.0632 17.57 0.0117400 0.80 34 0.0694 19.27 0.0133450 0.90 37 0.0755 20.97 0.0150500 1.00 39 0.0796 22.10 0.0167550 1.10 42 0.0857 23.80 0.0183600 1.20 44 0.0898 24.93 0.0200650 1.30 46 0.0938 26.07 0.0217700 1.40 48 0.0979 27.20 0.0233750 1.50 49 0.1000 27.77 0.0250800 1.60 51 0.1040 28.90 0.0267850 1.70 52 0.1061 29.47 0.0283900 1.80 54 0.1102 30.60 0.0300950 1.90 55 0.1122 31.17 0.0317

1000 2.00 55 0.1122 31.17 0.03331050 2.10 56 0.1142 31.73 0.03501100 2.20 57 0.1163 32.30 0.03671150 2.30 58 0.1183 32.87 0.03831200 2.40 58 0.1183 32.87 0.04001250 2.50 59 0.1204 33.43 0.04171300 2.60 60 0.1224 34.00 0.04331350 2.70 60 0.1224 34.00 0.04501400 2.80 61 0.1244 34.57 0.04671450 2.90 61 0.1244 34.57 0.04831500 3.00 62 0.1265 35.13 0.05001550 3.10 62 0.1265 35.13 0.05171600 3.20 62 0.1265 35.13 0.0533

Specimen No : 2

Loading : 2.5kg

Displacement Proving Ring Shear Stress (kN/m2) Strain

Dail Gauge ∆L (mm) Dail Gauge Load, P (kN)

50 0.10 18 0.0367 10.20 0.0017100 0.20 21 0.0428 11.90 0.0033150 0.30 24 0.0490 13.60 0.0050200 0.40 31 0.0632 17.57 0.0067250 0.50 35 0.0714 19.83 0.0083300 0.60 39 0.0796 22.10 0.0100350 0.70 43 0.0877 24.37 0.0117400 0.80 46 0.0938 26.07 0.0133450 0.90 48 0.0979 27.20 0.0150500 1.00 51 0.1040 28.90 0.0167550 1.10 53 0.1081 30.03 0.0183600 1.20 55 0.1122 31.17 0.0200650 1.30 57 0.1163 32.30 0.0217700 1.40 59 0.1204 33.43 0.0233750 1.50 61 0.1244 34.57 0.0250800 1.60 63 0.1285 35.70 0.0267850 1.70 65 0.1326 36.83 0.0283900 1.80 66 0.1346 37.40 0.0300950 1.90 68 0.1387 38.53 0.0317

1000 2.00 69 0.1408 39.10 0.03331050 2.10 70 0.1428 39.67 0.03501100 2.20 71 0.1448 40.23 0.03671150 2.30 73 0.1489 41.37 0.03831200 2.40 74 0.1510 41.93 0.04001250 2.50 75 0.1530 42.50 0.04171300 2.60 75 0.1530 42.50 0.04331350 2.70 76 0.1550 43.07 0.04501400 2.80 77 0.1571 43.63 0.04671450 2.90 77 0.1571 43.63 0.04831500 3.00 78 0.1591 44.20 0.05001550 3.10 78 0.1591 44.20 0.05171600 3.20 78 0.1591 44.20 0.0533

Specimen No : 3Loading : 3.25kg

Displacement Proving Ring Shear Stress (kN/m2) Strain

Dail Gauge ∆L (mm) Dail Gauge Load, P (kN)

50 0.10 0 0.0000 0.00 0.0017100 0.20 15 0.0306 8.50 0.0033150 0.30 21 0.0428 11.90 0.0050200 0.40 28 0.0571 15.87 0.0067250 0.50 36 0.0734 20.40 0.0083300 0.60 44 0.0898 24.93 0.0100350 0.70 51 0.1040 28.90 0.0117400 0.80 57 0.1163 32.30 0.0133450 0.90 59 0.1204 33.43 0.0150500 1.00 65 0.1326 36.83 0.0167550 1.10 69 0.1408 39.10 0.0183600 1.20 72 0.1469 40.80 0.0200650 1.30 76 0.1550 43.07 0.0217700 1.40 79 0.1612 44.77 0.0233750 1.50 81 0.1652 45.90 0.0250800 1.60 84 0.1714 47.60 0.0267850 1.70 87 0.1775 49.30 0.0283900 1.80 88 0.1795 49.87 0.0300950 1.90 90 0.1836 51.00 0.0317

1000 2.00 92 0.1877 52.13 0.03331050 2.10 93 0.1897 52.70 0.03501100 2.20 94 0.1918 53.27 0.03671150 2.30 96 0.1958 54.40 0.03831200 2.40 98 0.1999 55.53 0.04001250 2.50 99 0.2020 56.10 0.04171300 2.60 99 0.2020 56.10 0.04331350 2.70 100 0.2040 56.67 0.04501400 2.80 101 0.2060 57.23 0.04671450 2.90 102 0.2081 57.80 0.04831500 3.00 103 0.2101 58.37 0.05001550 3.10 105 0.2142 59.50 0.05171600 3.20 106 0.2162 60.07 0.05331650 3.30 106 0.2162 60.07 0.05501700 3.40 107 0.2183 60.63 0.05671750 3.50 109 0.2224 61.77 0.05831800 3.60 109 0.2224 61.77 0.06001850 3.70 110 0.2244 62.33 0.06171900 3.80 111 0.2264 62.90 0.06331950 3.90 111 0.2264 62.90 0.0650

2000 4.00 112 0.2285 63.47 0.06672050 4.10 112 0.2285 63.47 0.06832100 4.20 112 0.2285 63.47 0.0700

7.0 DATA ANALYSIS

Shear Stress ( 20mm dial gauge reading )

σ = P/A = [ ( dial gauge x 0.00204) / Area ]

Strain ( 20mm dial gauge reading )

τ = ( L / L ) = [ ( Dail Gauge x 0.01) / Total Length ]

Example calculation to find shear stress and strain for specimen 1 :

L (mm) = 150 x 0.002= 0.30

Load, P (kN) = 13 x 0.00204= 0.0265

Shear Stress ( 20mm dial gauge reading ) :

= 13 x 0.00204 0.06 x 0.06

= 7.37 kN/m²

Strain ( 20mm dial gauge reading ) :

= 150 x 0.002 60

= 0.0050

Normal Stress ( kN/m²)

Specimen No : 1Loading: 1.75 kg

Normal Stress = P A

= 1.75 x 10 x 9.81 (0.06 x 0.06)1000

= 47.69 kN/m²

Specimen No : 2Loading: 2.5 kgNormal Stress = P A

= 2.5 x 10 x 9.81 (0.06 x 0.06)1000

= 68.13 kN/m²

Specimen No : 3Loading: 3.25 kgNormal Stress = P A

= 3.25 x 10 x 9.81 (0.06 x 0.06)1000

= 88.56 kN/m²

Specimen 1 Loading 1.75kg :

Specimen 2 Loading 2.5kg :

Specimen 3 Loading 3.25kg :

Shear StrengthFrom the graph, data obtained : Ø = 17 °, c = 0

1. σ = 47.69 kN/m2

τ = 0 + 47.69 tan 17 ° = 14.58 kN/m2

2. σ = 68.13 kN/m2

τ = 0 + 68.13 tan 17 ° = 20.83 kN/m2

3. σ = 88.56 kN/m2

τ = 0 + 88.56 tan 17 ° = 27.08 kN/m2

8.0 DISCUSSION

Direct shear test is simple and faster to operate. As thinner specimens are used in shear box, they facilitate drainage of pore water from a saturated sample in less time. This test is also useful to study friction between two materials one material in lower half of box and another material in the upper half of box.

For this experiment we use sand soil as the specimen. As we know, the sand soil does not have any cohesion. The friction between sand particle is due to sliding and rolling friction and interlocking action.

Significance and Applicationsa. Unlike materials like steel, most of the soils are visco-elastic, meaning the failures are

time dependantb. For most of the geotechnical designs concerning foundations, earthworks and slope

stability issues the soils are required to withstand shearing stresses along with compressive stresses

c. Shear stresses tend to displace a part of soil mass relative to rest of the soil massd. Shear strength is the capacity of the soil to resist shearing stressese. Relative sliding between soil particles is the major factor contributing to the shear

resistancef. If the normal forces increase, the number of contact points also increase thus increasing

the resistance.g. The reverse may happen if the normal loads decrease (which is the case in excavations)h. Hence the shear strength is a function of normal load, angle of friction (amount of

interlocking among the soil particles) and cohesion (intrinsic property of clays due to which they stay close to each other even at zero normal load).

The advantages of the direct shear test are:i. Cheap, fast and simple - especially for sands.

ii. Failure occurs along a single surface, which approximates observed slips or shear type failures in natural soils.

Disadvantages of the test include:i. Difficult or impossible to control drainage, especially for fine-grained soils.ii. Failure plane is forced--may not be the weakest or most critical plane in the field

iii. Non-uniform stress conditions exist in the specimen.iv. The principal stresses rotate during shear, and the rotation cannot be controlled.

Principal stresses are not directly measured.

9.0 CONCLUTION

As the conclution, the objective of this experiment is to determine the parameters that involved such as shear strength of soils, cohesion and angle of friction is achieved. Four graph has been plotted and the value of cohesion and angle of friction had been obtained. From this experiment, the value of cohesion, c is 0 kN/m² and the value of angle of friction is 17 °.

10.0 QUESTIONS

Question 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. This is also to ensure the soil does not slide away from the metal plate. When the load is applied on the soil, the perforated plate will grip the soil and push the soil.

b. What maximum value of displacement before stop the test?

The maximum value of displacement before we stop the test is when the values are constant for more than three times or we can stop the test when the incline value suddenly dropped.

Question 2

a. What is the purpose of a direct shear test? Which soil properties does it measure?

The direct shear test is one of laboratory experiment and normally used by geotechnical engineers to find and calculate the shear strength parameters of any soil that involved. The direct shear experiment measures the shear strength parameters which included the soil cohesion (c) and the angle of friction (friction angle).

b. Why do we use fixing screw in this test? What will happen if you do not removed them during test?

We used fixing screw in this direct shear test because in order to avoid shearx1 for happening before the experiment is carried out. If we don’t remove them during the test, the friction can not occur at the screw and there have will be no shear on the sample and thus the result will be not accurate.