Direct Shear Test

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:

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.

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 1: Shear stress versus normal stress

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 Proving RingShear Stress,kN/mm²)

Strain, τ (kN/mm2)

Dail gaugeL

(mm)Dail

gaugeLoad,P

(kN) x 10ˉ5 x 10ˉ3

0 0.0 12 0.105 2.92 050 0.1 20 0.175 4.86 1.67100 0.2 25 0.219 6.08 3.33150 0.3 30 0.263 7.31 5.00200 0.4 35 0.306 8.50 6.67250 0.5 43 0.376 10.44 8.33300 0.6 47 0.411 11.42 10.00350 0.7 52 0.455 12.64 11.67400 0.8 58 0.508 14.11 13.33450 0.9 60 0.525 14.58 15.00500 1.0 64 0.560 15.56 16.67550 1.1 65 0.569 15.81 18.33600 1.2 67 0.586 16.28 20.00650 1.3 69 0.604 16.78 21.67700 1.4 71 0.621 17.25 23.33750 1.5 72 0.630 17.50 25.00800 1.6 75 0.656 18.22 26.67850 1.7 76 0.665 18.47 28.33900 1.8 77 0.674 18.72 30.00950 1.9 78 0.683 18.97 31.671000 2.0 79 0.691 19.19 33.331050 2.1 79 0.691 19.19 35.001100 2.2 81 0.709 19.69 36.671150 2.3 82 0.718 19.94 38.331200 2.4 83 0.726 20.17 40.001250 2.5 84 0.735 20.42 41.671300 2.6 84 0.735 20.42 43.331350 2.7 84 0.735 20.42 45.00

Specimen Number = 2

Loading = 2.5 kg


Strain, τ (kN/mm2)

Dail gaugeL

(mm)Dail

gauge

Load,P (kN)

x 10ˉ5 x 10ˉ3x 10ˉ3

0 0.0 1 8.75 0.24 050 0.1 1 8.75 0.24 1.67100 0.2 5 43.75 1.22 3.33150 0.3 15 131.25 3.65 5.00200 0.4 22 192.50 5.35 6.67250 0.5 25 218.75 6.08 8.33300 0.6 30 262.50 7.29 10.00350 0.7 35 306.25 8.51 11.67400 0.8 40 350.00 9.72 13.33450 0.9 45 393.75 10.94 15.00500 1.0 50 437.50 12.15 16.67550 1.1 55 481.25 13.37 18.33600 1.2 60 525.00 14.58 20.00650 1.3 63 551.25 15.31 21.67700 1.4 66 577.50 16.04 23.33750 1.5 70 612.50 17.01 25.00800 1.6 72 630.00 17.50 26.67850 1.7 74 647.50 17.99 28.33900 1.8 76 665.00 18.47 30.00950 1.9 78 682.50 18.96 31.671000 2.0 80 700.00 19.44 33.331050 2.1 82 717.50 19.93 35.001100 2.2 83 726.25 20.17 36.671150 2.3 85 743.75 20.66 38.331200 2.4 88 770.00 21.39 40.001250 2.5 89 778.75 21.63 41.671300 2.6 90 787.50 21.88 43.331350 2.7 90 787.50 21.88 45.001400 2.8 91 796.25 22.12 46.671450 2.9 91 796.25 22.12 48.331500 3 91 796.25 22.12 50.00

Specimen Number = 3

Loading = 3.25 kg


Strain, τ (kN/mm2)

Dail gaugeL

(mm)Dail

gauge

Load,P (kN)

x 10ˉ5 x 10ˉ3x 10ˉ3

0 0.0 4 35.00 0.97 050 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

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 10ˉ5kN/mm2

Dial gauge = 50

Length = 60mm

Strain,

τ = ( 50 x 0.002 ) / 60

= 1.67 x 10ˉ3

b) Specimen 2


Area = 60 x 60 = 3600mm2

Shear stress,

σ = (5 x 0.00875) / 3600

= 1.22 x 10ˉ5kN/mm2

Dial gauge = 100

Length = 60mm

Strain,

τ = ( 100 x 0.002 ) / 60

= 3.33 x 10ˉ3

3) Specimen 3


Area = 60 x 60 = 3600mm2

Shear stress,

σ = (5 x 0.00875) / 3600

= 1.22 x 10ˉ5kN/mm2

Dial gauge = 150

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


A 3600

Load = 3.25 kg

P = 3.25 x 10 x 9.81 / 1000 = 0.32 kN


A 3600

Specimen Number = 1

Loading = 1.75 kg

Specimen Number = 2

Loading = 2.5 kg

Specimen Number = 3

Loading = 3.25 kg

Shear Stress VS Normal Stress

Φ = 28˚c = 0

Shear strength

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 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.

Direct Shear Test

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