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 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. 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:
 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
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
Perforated plate, Porous plate,
Retaining plat
Loading page
Grease
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
6.0 RESULT AND CALCULATION
Specimen No . 1
1.75 kg × 9.81 N × _1kN_ = 0.017 kN
1kg
1000N
Length : 60mm = 0.06m
Loading : 1.75 kg
Area
: 0.06m × 0.06m = 3.6×10-3m2
Displacement
Dial
∆L (mm)(X10-4)
gauge
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
Dial
gauge
14
25
34
41
44
50
55
59
63
65
67
69
70
70
70
70
Proving ring
Load, P (kN)
(x 10-5)
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
Shear stress
(KN/m2)
(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
Strain
(x 10-6)
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
Specimen No. 2
Loading : 2.5 kg
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)
gauge
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
Dial
gauge
20
32
28
60
71
82
92
97
100
102
102
102
102
Proving ring
Load, P (kN)
(x 10-5)
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
Shear stress
(KN/m2)
(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
Strain
(X 10-6)
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
Specimen No.3
3.25 kg × 9.81 N × _1kN_ = 0.032 kN
1kg
1000N
Length : 60mm = 0.06m
Loading : 3.25 kg
Area
: 0.06m × 0.06m = 3.6×10-3m2
Displacement
Dial
∆L (mm)(X10-4)
gauge
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
Dial
gauge
28
47
64
88
102
115
121
127
134
135
137
138
138
138
138
Proving ring
Load, P (kN)
(x10-5)
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
(KN/m2)
(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
Strain
(X 10-6)
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
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 yintercept 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.