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ENB272 - Shear Strength of Soils Report.docx
ENB272 - Shear Strength of Soils Report.docx

ENB272 - Shear Strength of Soils Report.docx

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Queensland University of Technology

ENB272

Chaminda

Fall

Description

Shear Strength of Soils
Practical
ENB272 – Geotechnical Engineering 1
QUT Engineering Student 1)
Table 1
Description Value
Internal diameter of the shear box 63.0mm
Height of the shear box 36.15mm
Thickness of the top porous stone 6.5mm
Unfilled height of the shear box after placing the 10.98mm
top porous stone
Mass of the sand + pan (before use) 174.80g
Mass of the sand + pan (after use) 71.98g
Vertical normal load 400N
2)
The dry density may be calculated using the equation;
Dry Density (P ) = Dry Mass (M) / Volume (V)
d
3
Where mass is in grams, and volume is in g/cm . The dry mass (M) is calculated below, subtracting
the ‘Mass of the sand + pan (after use)’ from the ‘mass of the sand + pan (before use)’;
Dry Mass (M) = 174.80 – 71.98
= 102.82g
The volume (V) is calculated below;
2
Volume (V) = [(πD )/4](h)
2
= [(π63 )/4](h)
h = (height of the shear box) – (thickness of the top porous
stone) – (unfilled height of the shear box after placing the top
porous stone)
h = 36.15 – 6.5 – 10.98
h = 18.67mm
2
Volume (V) = [(π63 )/4](18.67)
= 58198.9mm 3
3
= 58.1989cm
Therefore the Dry Density (P d can be calculated to be;
Dry Density (P d = 102.82/58.1989
= 1.7667g/cm 3 3)
The vertical load may be determined using the equation below;
Normal Stress (σ ) n Vertical Load (N) / Area (mm ) 2
The Vertical Load is known to be 400N (refer to Table 1). The area of the shear box is determined
using the equation below;
2
Area (A) = [(πD )/4]
= [(π63 )/4]
2
= 3117.25mm
The Normal Stress (σ )ncan then be calculated, as illustrated below;
Normal Stress (σ )n= 400/3117.25
= 0.1283MPa
= 128.3kPa
4)
Refer to Appendix (Table 3.0).
5)
Refer to Appendix (Table 3.0).
6)
Refer to Appendix (Table 4.0). 7)
Figure 1.0: Shear Stress vs. Shear Displacement
8)
As illustrated in Figure 1.0 the maximum shear stress was calculated to be 93.01kPa, which had a
corresponding shear displacement of 1.022mm. 9)
Normal Stress (σ ) = Vertical Load (N) / Area (mm )
n
Normal Stress (σ n = 400/3117.25
= 0.1283MPa
= 128.3kPa
Maximum Shear Stress vs. Normal Stress
100
90
80
(kPa)
τ 60
50
40
30
Shear Stress,
20
10
0
128.3
Normal Stress, n (kPa)
Figure 2.0: Maximum Shear Stress vs. Normal Stress
10)
Maximum Shear Stress vs. Normal Stress
100
90
80
(kPa)
τ 60
50
40
30
Shear Stress,
20
10
0
128.3
Normal Stress, n (kPa)
Figure 3.0: Maximum Shear Stress vs. Normal Stress with Mohr-Coulomb failure envelope In order to determine the slope on the Mohr-Coulomb failure envelope, the following equation must
be used;
T = c + σ tan(Φ)
n
Where; T is maximum shear stress in kilopascals, c is cohesion (which in this case σsnis normal
stress in kilopascals, aΦd is the angle of internal friction in degrees. This equation can be
manipulated to solve for the angle of internal friction, as illustrated below;
T = c + n tan(Φ)
-1
Φ = tan (T/σ )n
Φ = tan (93.01/128.3)
o
Φ = 35.94
11)
o An application of the internal friction angle of the soil is that the internal friction angle may
be used to determine the soil type. Refer to Table2.0 below.
From this data, it may be suggested that the soil analysed is a ‘Sand and Gravel Mixture’.
o The internal friction angle may be used to determine the angle in which a retaining wall is
most likely to fail
o The internal friction angle provides an insight into the potential angle of failure, therefore
allowing for necessary foundations to be implemented to increase the shear strength of the
soil
o The internal friction angle allows for the determination of slope stability, which is critical
throughout a range of civil engineering applications.
Table 2.0: Soil Type Classification Chart
SOIL TYPE ANGLE
(DEGREES)
Sand and gravel 33 – 36
mixture
Well-graded 32 – 35
sand
Fine to medium 29 – 32
sand
Silty sand 27 – 32
Silt (non-plastic) 26 – 30 12)
Horizontal Displacement vs. Shear
Displacement
0.3
0.25
0.2
0.15
0.1
0.05
Horizontal Displacement (mm)
0
-0.05
Shear Displacement (mm)
Figure 4.0: Horizontal Displacement vs. Shear Displacement
13)
Refer to Figure 4.0.
14)
The angle of dilation may be determined by the following equation within the initial vertical
displacement increases, before reaching the critical state;
Angle of Dilation (Ψ) = tan [(vert. diss 2) – (vert. diss 1)] / [(hor. diss 2) – (hor.
diss 1)]
Therefore;
Angle of Dilation (Ψ) = tan [(0.0789 – 0.0448)/(0.7991-0.5823)]
o
= 8.94 15)
Through analysing Figure 1.0, it is clear that this sand specimen may be classified as dense. This is
characterised by the initially large increase in shear stress (compared to shear displacement),
resulting in a peak of 93.01MPa. Peaks in shear stress are typical only in dense substrates. This is
followed by the shear stress plateauing between 80MPa and 90MPa in the critical state, as the shear
displacement increases. The dense characteristic of the sand is also illustrated in Figure 4.0, where
the initially the horizontal displacement increases rapidly, before plateauing at approximately
0.25mm. If the substrate was not dense, this horizontal displacement would have increased
signific

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