INTRODUCTION A direct shear test is a laboratory or field test used by geotechnical engineers to measure the shear strength properties of soil or rock material or of discontinuities in soil or rock masses. The test is performed on three or four specimens from a relatively undisturbed soil sample. The test is carried out on either undisturbed samples or remoulded samples. To facilitate the remoulding 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. From the results, the shear strength parameter can be determined.
Direct shear test is quick and Inexpensive. Shortcoming is that it fails the soil on a designated plane which may not be the weakest one. It used to determine the shear strength of both cohesive as well as non-cohesive soils according to ASTM D 3080.
THEORY Shear strength is composed of: 1 Internal friction which is the resistance due to friction between individual particles at their contact points and interlocking of particles. This interlocking strength is indicated through parameter φ. 2 Cohesion which resistances due to inter-particle force which tend hold the particles together in a soil mass. The indicative parameter is called Cohesion intercept (c). The strength of a soil depends of its resistance to shearing stresses. It is made up of basically the components : 1. Frictional – due to friction between individual particles. 2. Cohesive - due to adhesion between the soil particles The two components are combined in Coulomb’s shear strength equation,
τ
f
=c+
σ
n
tan
∅
Where : τ
f
= shearing resistance of soil at failure
c = apparent cohesion of soil σ
n
= total normal stress on failure plane
ø = angle of shearing resistance of soil (angle of internal friction) This equation can also be written in terms of effective stresses. Shear strength of a soil is the maximum resistance to shearing stress at failure on the failure plane.
Apparatus 1 Soil sample 2 Shear box equipment 3 Weight set
Procedure 1 Pushing a cutting ring of size 100 mm in diameter 200 mm of the undisturbed specimen. The square specimen of size 60 mm x 60 mm is then cut from the circular specimen obtained. 2 Carefully insert the test specimen. Place the loading block in place. Position the horizontal and vertical displacement indicator. Apply appropriate normal load. 3 By a gap about 1.0 mm, separate the upper and lower halves of the shear box frames. 4 Remember to remove the locking screw. 5 The rate of the load should be on the order of 0.6 mm/s. 6 Stop the test apparatus when the shearing force has reached it failure. 7 Take reading of the load, shear displacement and vertical displacement dials. 8 Back of the applied normal pressure, dismantle and clean the shear box. 9 Repeat the test using differential load.
RESULT. Data and Observation Table 1: Proving ring calibration factor WF No.
: 14120
Ring Number
: 15615
Capacity
: 2 kN
1 Division
: 0.002 mm
Load (kN)
Compression (Division)
0
0
0.2
146
0.4
291
0.6
439
0.8
587
1.0
736
1.2
883
1.4
1033
1.6
1183
1.8
1335
2.0
1487
4kg Shear Displa y Dial
Shear Disp
Norm al Disp dial
Norm al Disp
Providi ng Ring dial
Shear Force, Pᵇ
28 75 123 172
0.28 0.75 1.23 1.72
-6 -9 -8 -5.2
-0.06 -0.09 -0.08 -0.052
31 41 51 59
0.212 0.281 0.349 0.411
Crosssection al Area, A 0.0036 0.0036 0.0036 0.0036
222
2.22
-0.5
-0.005 60
0.414
0.0036
274
2.74
6
0.06
75
0.514
0.0036
322
3.22
14
0.14
79
0.541
0.0036
375
3.75
23.5
0.235
80
0.548
0.0036
423
4.23
31
0.31
81
0.555
0.0036
474
4.74
38
0.38
80
0.548
0.0036
92 142
0.92 1.42
-11 -11.5
-0.11 46 -0.115 58
0.315 0.397
0.0036 0.0036
193
1.93
-10
-0.1
0.473
0.0036
Shear Stress, ( ᵹ)
Strain ,ε
58.980 78.006 97.032 112.25 3 132.15 5 142.69 4 150.30 4 152.20 7 154.11 0 152.20 7
0.005 0.013 0.021 0.029
87.519 110.35 0 131.27 9
0.015 0.024
0.037 0.046 0.054 0.063 0.071 0.079
6kg
69
0.027
198
1.98
-9.5
-0.095 70
0.489
0.0036
201
2.01
-9.3
-0.093 72
0.499
0.0036
215
2.15
-8.5
-0.085 73
0.500
0.0036
220
2.20
-7.6
-0.076 75
0.511
0.0036
234
2.34
-7.2
-0.072 77
0.518
0.0036
238
2.38
-7.0
-0.070 76
0.520
0.0036
240
2.40
-6.8
-0.068 77
0.530
0.0036
241
2.41
-6.5
-0.065 77
0.527
0.0036
291
2.91
-1.5
-0.015 85
0.582
0.0036
340
3.4
4.3
0.043
87
0.596
0.0036
390
3.9
11
0.11
89
0.610
0.0036
440
4.4
16.5
0.165
89
0.610
0.0036
491
4.91
22
0.22
89
0.610
0.0036
41
0.41
-8
-0.08
40
0.274
91
0.91
-6
-0.06
61
0.418
141
1.41
0
0
78.5
0.538
162
1.62
-11
-0.11
91
0.623
176
1.76
-11
-0.11
91.5
0.633
0.003 6 0.003 6 0.003 6 0.003 6 0.003 6
134.39 0 137.44 0 139.12 2 142.55 1 143.78 9 144.34 2 145.10 0 146.49 9 161.72 0 165.52 5 169.33 0 169.33 0 169.33 0
0.029
76.104
0.007
116.05 8 149.35 3 173.13 5 175.14 4
0.015
0.030 0.032 0.035 0.037 0.038 0.039 0.040 0.049 0.057 0.065 0.073 0.082
8kg
0.024 0.032 0.034
186
1.86
-12
-0.12
92
0.640
198
1.98
-13
-0.13
93
0.654
200
2.00
20
0.20
97
0.668
220
2.20
23
0.23
98
0.670
240
2.4
25
0.25
99
0.678
290
2.9
41
0.41
100
0.685
339
3.39
56
0.56
100
0.685
389
3.89
68
0.68
92
0.630
0.003 6 0.003 6 0.003 6 0.003 6 0.003 6 0.003 6 0.003 6 0.003 6
176.22 2 178.40 2 182.78 9 184.23 4 188.35 6 190.25 9 190.25 9 190.25 9
0.037 0.038 0.038 0.039 0.040 0.048 0.057 0.065
ANALYSIS OF RESULTS
Given data: 1. Specimen size (length x width) = 60 x 60mm 2. Test load = 4kg 3. Calibration factor = 0.0068 Cross-sect. area of specimen, A
Shear force, Ph
Normal stress Area
Shear displacement
Normal displacement
Shear stress,
=
60 x 60
=
3600mm²
=
3.6 x 10-3m²
=
Proving ring dial x Calibration factor
=
31 x 0.0068
=
0.212kN
=
Load x acceleration by gravity,g /
=
4 x 9.81 / (3600 x 10-3)
=
10.9 kPa
=
Horizontal displacement
=
Shear displacement dial x 0.01mm
=
28 x 0.01mm
=
0.28mm
=
Vertical displacement
=
Normal disp. dials x 0.001mm
=
-6 x 0.001mm
=
-0.06 mm
=
Ph / A
=
0.212 kN/ (3.6 x 10-6 ) mm²
Strain,
=
58.980 kPa
=
Shear displacement / length
=
0.28mm / 60mm
=
0.005
0.0011❑
Plot shearing stress at failure versus normal stress and show the angle of internal friction, Ø and intercept,c. From equation of the graph we get, y = 5.5x + 42 The gradient of the best line is 5.5. Therefore, tan Ø = 5.5 Ø=
tan −1
5.5
Ø = 79.70 ° Ƭf = c + σn tan Ø Where c is the interception of the above graph which is 42kPa.
DISCUSSION. -
-
-
From the experiment, the value of Ø that we get from the graph of Ƭ vs σn is 79.70 while the c value is 42. The Ø value is get by considering the gradient of the graph and using tangent find the angle of friction (Ƭ vs σ n) and the c value is get at the interception of the graph with the y-axis. Since the value can be find directly from the graph itself, this method is named as DIRECT SHEAR TEST. From the graph plotted, the standard pattern for the graph shear stress versus normal stress should be linear that starts from origin, however, in this experiment, it is not. This is may be due to error or mistakes that been done doing the experiment. The direct shear test is a laboratory test used to find the shear strength parameters of an undisturbed soil sample. The shear strength, τf of a soil is a point where the soil starts to fail or displace by the acting of external forces. The shear strength of a soil can be expressed as following: τf = c + σf tan ø where c = shear strength parameters as cohesion intercept and ø = angle of internal friction.
-
There are some errors occurs during the experiment. These mistakes can be minimized by some precautions. Before shearing a sample, check that the various alignment pins or screws are properly set. The large pair of screws used to align the shear box halves must be removed before shearing the sample or you will also shear the pins. The other set is used to create a thin gap between the top and bottom halves. If this gap is too large, your sample will run out; if too small, the top of the box will drag on the bottom. During shear, there are two readouts to watch: horizontal displacement and vertical displacement. One person should be assigned to just read one dial gauge in order to avoid mistakes. Therefore, there will be two person that will read each horizontal and vertical dial gauge respectively. When announced, the two person should quickly record the reading
Question :To conduct this experiment, we must prepare soil sample, shear box equipment which is a square specimen of size 60mm x 60mm and weight set. Firstly, we must put the upper and lower halves of the shear box
frame together and lock it. Then, we put 3 layers of soil sample in the shear box which each of layer must be smooth surface with the help of square holder. After that, we put the shear box in the shear box apparatus to collect the reading data with different load apply which are 4kg, 6kg and 8kg. Before we switch on the button, we must unlock the screws to ensure the shearing force is going smoothly. We take shear, normal and proving ring dial as a reading data for every minute until the reading is constant or 3 drop readings. 1 What is the purpose of a direct shear test ? Which soil properties does it measure ? The purpose of a direct shear test is to determine the shear parameters of a soil using shear box apparatus. The soil properties that measure are cohesion, angle of internal friction and shear modulus. 2 Can you predict what will happen if you don’t remove the locking screws during the test ? The shearing process will be not going smoothly. 3 What is the purpose of the porous stone in the direct shear test ? The specimen is located between 2 porous stones that serves as drains during the first and second steps of the test. The surfaces of contact between sample and porous stones are grooved, to prevent slippage between sample and stones during shear. 4 List 4 possible errors that would cause inaccurate determinations of strength and stress deformation characteristics. a Parallax error which when we take reading data. b The screws are lock when the shear box apparatus is in the process c Permeability of porous inserts too low d Leakage of pore water out of specimen
CONCLUSION We can conclude that our experiment meets the objective, which is to find the angle of internal friction and the cohesion value for kaolin soil type. From this test, the shear strength parameters are cohesion intercept, c and the angle of shearing resistance, ø can be determined using tangent formula. Therefore, this experiment is called DIRECT SHEAR TEST which mean we can directly find the cohesion and angle of friction value from the graph itself. The shear strength, τf of a soil sample is also called as the shear failure; where failure occurs when a critical combination of shear stress and effective normal stress develop.
