Title Direct shear test Introduction Direct shear test is commonly used to determine the shear strength properties of cohesion less soils (i.e.: sands and gravels) because of simplicity in strength testing. Also it is a advantageous for shearing cohesive soil into large strain of deformation and the residual strength can be determined. The shear strength properties are needed for engineering analysis such as determining the stability of slope, and finding th bearing capacity of foundations. Objective To determine the internal friction angle of a fine, dry sand. Standard reference: British standard 1377-part 7: shear strength tests (total stress) Apparatus
1. 500g of fine, dry sand 2. Wood tamper 3. Direct shear machine 4. Shear box: a) Loading cap b) Top half c) Separating screws d) Locking pins
Test procedures 1. The inner dimensions of the shear box and the inner area (A) was measured and calculated. 2. The top and the bottom halves of shear box was fixed together by locking pins. 3. 127g of sand was weight out to the nearest 0.1g 4. The soil was placed in three layers into the shear box so that the surface of the specimen coincide with the level mark inside the box (the height of specimen, h=14.115mm). 25 times tamping was applied on each layer using a wood tamper. 5. Any spilled or removed material was collected, the total of the unused soil was weight to the nearest 0.1g, and the initial mass of the specimen (Mo) was determined. 6. The porous plate was placed on the specimen. The plate was gently bed down to form a level surface. 7. The shear box was placed in the direct shear machine. The shear box was clamp in, and advances the screw manually so that all moving part was seared snugly against one another. 8. A normal force (N) was applied by putting a 2kg load on the dead weight system. 9. Position and zero the deformation indicators (horizontal and vertical dial gauges) 10. The locking pins from the shear box was removed and turn the separating screws one-quarter of a turn to separate the top and bottom halves of the shear box. 11. The shearing of the specimen was begun at a deformation rate
(∆ H / ∆t )
of
approximately 0.5 mm/min. 12. Reading of the force measuring device (F) was recorded, the horizontal displacement gauge ( ∆ H ), the vertical deformation gauge ( ∆ v ), and elapse time, at a regular intervals of
0.1mm horizontal displacement (i.e take the readings for every
∆ H=0.1mm ¿ . please
take note that 1 Div = 0.01 mm for dial gauge, while 1 Div = 0.0074 kN for force measurement proving ring. 13. The test after the shear force (F) readings drops significantly was stopped or remains constant for a continuous 3 readings. 14. The machine was reverse to release shear force. Take the normal load off. 15. The shear box was removed carefully from machine and empties the shear box.
16. The test by changing the normal load in step (8) to 4kg was repeated.
Results: Height before fill in sand = 37.34mm Height after fill in sand = 7.845mm Thickness of porous plate = 6.83mm Sample height, h=¿ 22.665mm mo=¿
Mass of soil,
129.8g
Dimension of porous plate = 59.88 mm× 59.82 mm
I) Calculation a) Bulk density,p ρ=
mo A ×h
−3 2 A ¿ 0.05988 ×0.05982=3.58 ×10 m
ρ=
129.8 ×10−3 ( 3.58 ×10−3 ) × 0.022665 3
−3
ρ=1.599 ×10 kg m
b) Normal stress, σ n =k
( NA )
σn
; k is the factor of the dead weight system, k =1
σ n =100 kN /m2 100=
N −3 3.58 ×10
2
−1
N=3.58 ×10 kN /m
c) Shear stress, t τ=
F A
For shear force, τ=
F=0. 320 kN ,
0.320 3.58 ×10−3
τ =89.39 kN / m
2
Test Data 1 Horizontal Displacement, ∆H DIV mm 0 0
Vertical Displacement, ∆V DIV mm 0 0
Shear Force, F kN 0
20
0.2
-1
-0.01
0.108
40
0.4
-2
-0.02
0.156
60
0.6
-2.2
-0.022
0.196
80
0.8
-1.9
-0.019
0.231
100
1
0
0
0.252
120
1.2
2.5
0.025
0.267
140
1.4
5
0.05
0.281
160
1.6
7.5
0.075
0.292
180
1.8
10
0.1
0.301
200
2
12.5
0.125
0.306
220
2.2
15
0.15
0.308
240
2.4
17.5
0.175
0.31
260
2.6
19.5
0.195
0.31
shear stress 0 30.15057 196 43.55082 616 54.71770 466 64.48872 335 70.35133 456 74.53891 4 78.44732 148 81.51821 307 84.03076 073 85.42662 054 85.98496 447 86.54330 839 86.54330 839
280
2.8
21.5
0.215
0.309
300
3
23.5
0.235
0.309
320
3.2
26
0.26
0.31
86.26413 643 86.26413 643 86.54330 839
100 90 80 70 60 shear stress,t (kN/m²
50 40 30 20 10 0 0
0.5
1
1.5
2
2.5
3
Horizontal displacement, ΔH (mm)
Shear stress at failure, τf = 85.946 kPa
3.5
0.3 0.25 0.2 0.15 vertical displacement ,Δv (mm 0.1 0.05 0 -0.05
0 0.5 1 1.5 2 2.5 3 3.5
Horizontal displacement, ΔH (mm)
Graph of Normal Stress (�n) vs. Shear Stress (�) 200 150 Shear Stress, � (kN/m2) 100 50 0 0
50
100
150
200
250
Normal Stress, �n (kPa
σ n' =σ n−u ; givenu=0 , pore water pressure remains zero throughout the test. '
σ n =σ n−0 σ n' =σ n τ =c+ σ n tanφ '
c=c
' For sand, c=c =0
86.543=0+100 tanφ '
φ=φ =40.874
φ=φ '
Height before fill in sand = 37.34mm Height after fill in sand = 6.066mm Thickness of porous plate = 6.83mm Sample height, h=¿ 24.444mm mo=¿
Mass of soil,
137.4g
Dimension of porous plate = 59.88 mm× 59.82 mm
a) Bulk density,p ρ=
mo A ×h
−3 2 A ¿ 0.05988 ×0.05982=3.58 ×10 m
ρ=
137.4 × 10−3 ( 3.58 ×10−3 ) × 0.02444
ρ=1.569 ×10 3 kg m−3
b) Normal stress, σ n =k
( NA )
σn
; k is the factor of the dead weight system, k =1
σ n =200 k N /m2 200=
N 3.58 ×10−3
2
−1
N=7.16 ×10 kN /m
c) Shear stress, t τ=
F A
For shear force, τ=
F=0.6396 kN ,
0. 6396 3.58 ×10−3
τ =178.66 kN /m
Horizontal Displacement,∆H DIV mm 0 0
2
Vertical Displacement, ∆V DIV mm 0 0
Shear Force, F kN 0
20
0.2
-2.5
-0.025
0.198
40
0.4
-4.5
-0.045
0.312
60
0.6
-5
-0.05
0.386
80 100
0.8 1
-4.5 -3
-0.045 -0.03
0.442 0.481
120
1.2
-1.5
-0.015
0.532
140
1.4
0
0
0.539
160
1.6
2.5
0.025
0.556
180
1.8
4.5
0.045
0.566
200
2
7
0.07
0.573
220
2.2
8.5
0.085
0.577
240 260
2.4 2.6
11 13
0.11 0.13
0.581 0.582
shear stress 0 55.2760485 9 87.1016523 2 107.760377 5 123.394007 5 134.281714 148.519484 1 150.473687 8 155.219611 2 158.011330 8 159.965534 5 161.082222 4 162.198910 2 162.478082
280
2.8
14.5
0.145
0.579
300
3
16
0.16
0.578
320
3.2
17
0.17
0.578
180 160 140 120 100 shear stress,t (kN/m²
80 60 40 20 0 0
0.5
1
1.5
2
2.5
3
3.5
Horizontal displacement, ΔH (mm)
Shear stress at failure, τf = 161.974 kPa
2 161.640566 3 161.361394 4 161.361394 4
0.2 0.15 0.1 vertical displacement ,Δv (mm
0.05 0 0 0.5 1 1.5 2 2.5 3 3.5 -0.05 -0.1
Horizontal displacement, ΔH (mm)
Graph of Normal Stress (�n) vs. Shear Stress (�) 200 178.61 150 Shear Stress, � (kN/m2) 100 50 0 0 0
50
100
150
200
250
Normal Stress, 𝞼n (kPa
σ n' =σ n−u ; givenu=0 , pore water pressure remains zero throughout the test. σ n' =σ n−0 For sand,
'
c=c =0
162.478=0+ 200tanφ
φ=φ' =39.09
'
σ n =σ n τ =c+ σ n tanφ c=c' φ=φ '
Discussion : In this experiment, two sets of data is obtained by using two different normal stress which are 32.0 kPa and 64.0 kPa and φ value is calculated to be 40.87 and 39.09 respectively. The typical φ value can be obtained using the graph below.
As both of the results obtained are around 40, we can categories the soil we use as sand. In the graph of shear stress (�) vs normal stress (�n), the larger φ value implies a denser soil. In the experiment, as a higher force is applied, the particles of the soil will pack tightly, forcing air out and increasing its shear strength. Since the particles are closer to each other, it means that the soils is denser relating a higher density will yield a higher shear strength From the shear stress (�) vs horizontal displacement (H) graphs plotted above showing that the maximum of shear stresses of these samples are 85.946kN/m^2 and 161.974kN/m2 under different normal stresses of 32.0kPa and 64.0kPa respectively. Also, from the graph, it is seen that when a higher normal stress acts upon the soil, the higher the maximum shear stress is recorded. The drop in vertical displacement might be due to the soil forcing air out, condensing it. As the horizontal displacement continue, the soil will be unable to condense anymore, pushing
the soil upward . Conclusion From our experiment and the shear stress graph above, the maximum value of the shear 2 soil stresses which are 96.944 kN /m and 169.167kN/m2. And angle of the internal friction,
∅' values for 2 different max shear stresses are 25.86º and 40.22° respectively calculated by the formula stated above.
