DIRECT SHEAR TEST

1. 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 mould. 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 pre-

determined 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. OBJECTIVE

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

3. 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

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

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. TEST EQUIPMENTS

1. Shear box carriage.

2. Loading pad.

3. Perforated plate.

4. Porous plate.

5. Retaining plate.

6. Grease.

Shear box carriage Loading page

Perforated plate, Porous plate, Grease

Retaining plat

5. PROCEDURES

1. Internal measurement is verify 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.

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

Displacement Proving ring Shear stress(KN/m2)(x 10-3)

Strain(x 10-6)Dial

gauge∆L (mm)(X10-4) Dial

gaugeLoad, P (kN)

(x 10-5)0.2 4 14 2.9 8.1 6.70.4 8 25 5.1 14.2 13.30.6 12 34 6.9 19.2 20.00.8 16 41 8.4 23.3 26.71.0 20 44 9.0 25.0 33.31.2 24 50 10.2 28.3 40.01.4 28 55 11.2 31.1 46.71.6 32 59 12.0 33.3 53.31.8 36 63 12.9 35.8 60.02.0 40 65 13.3 36.9 66.72.2 44 67 13.7 38.1 73.32.4 48 69 14.1 39.2 80.02.6 52 70 14.3 39.7 86.72.8 56 70 14.3 39.7 93.33.0 60 70 14.3 39.7 100.03.2 64 70 14.3 39.7 106.7

6.0 RESULT AND CALCULATION

Specimen No . 1

Loading : 1.75 kg 1.75 kg × 9.81 N × _1kN_ = 0.017 kN1kg 1000N

Length : 60mm = 0.06m

Area : 0.06m × 0.06m = 3.6×10-3m2

Specimen No. 2

Loading : 2.5 kg 2.5 kg × 9.81 N × _1kN_ = 0.025 kN1kg 1000N

Displacement Proving ring Shear stress(KN/m2)(x 10-3)

Strain(X 10-6)Dial

gauge∆L (mm)(X10-4) Dial

gaugeLoad, P (kN)

(x 10-5)0.2 4 20 4.1 11.4 6.70.4 8 32 6.5 18.1 13.30.6 12 28 5.7 26.0 20.00.8 16 60 12.2 33.9 26.71.0 20 71 14.5 40.3 33.31.2 24 82 16.7 46.4 40.01.4 28 92 18.8 52.2 46.71.6 32 97 19.8 55.0 53.31.8 36 100 20.4 56.7 60.02.0 40 102 20.8 57.8 66.72.2 44 102 20.8 57.8 73.32.4 48 102 20.8 57.8 80.02.6 52 102 20.8 57.8 86.7

Length : 60mm = 0.06m

Area : 0.06m × 0.06m = 3.6×10-3m2

Specimen No.3

Loading : 3.25 kg 3.25 kg × 9.81 N × _1kN_ = 0.032 kN1kg 1000N

Length : 60mm = 0.06m

Area : 0.06m × 0.06m = 3.6×10-3m2

Displacement Proving ring Shear stress(KN/m2)(x 10-3)

Strain(X 10-6)Dial

gauge∆L (mm)(X10-4) Dial

gaugeLoad, P (kN)

(x10-5)0.2 4 28 5.7 15.8 6.70.4 8 47 9.6 26.7 13.30.6 12 64 13.1 36.4 20.00.8 16 88 17.9 49.7 26.71.0 20 102 20.8 57.8 33.31.2 24 115 23.5 65.3 40.01.4 28 121 24.7 68.6 46.71.6 32 127 25.9 71.9 53.31.8 36 134 27.3 75.8 60.02.0 40 135 27.5 76.4 66.72.2 44 137 27.9 77.5 73.32.4 48 138 28.2 78.3 80.02.6 52 138 28.2 78.3 86.72.8 56 138 28.2 78.3 93.33.0 60 138 28.2 78.3 100.0

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

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

6.

3.1 cm

4.4 cm

Tan ϕ = 3.1 / 4.4

Φ = 35ᵒ

7.0 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

8.0 CONCLUSION

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.

9.0 QUESTIONS

Question 1

a. Why perforated plate in this test with teeth?

Plate with teeth is used to increase fiction and to produce a grip forces between the

plate and the sample and assists in distributing the shear stress evenly. The direction of the

plate is moving oppositely which is occur to friction.

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 (ø). 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.

10.0 REFERENCE

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