Example 12.2 (Page 296) - Soil Mechanics : Principles and Practice by G.E. Barnes
From Textbook
A slope is to be cut into a soft clay with undrained shear strength of 30 kN/m2 and unit weight of 18 kN/m3• The slope is 8.0 m high and its inclination is 2: 1 (horizontal:vertical). Determine the factor of safety for the trial circle shown on Figure 12.20.
Equation 12.9 is used with values of b, α and mid-slice height h determined for each slice. The weight of each slice is obtained from W = Y*b*h.
Slip Circle tangential to the ROCK Layer at depth of 1.6 m below the toe.
initial analysis indicated following
Textbook Print errors
column \[REGION[169a817e30754497a2e04b6778308f13]]\
column \[REGION[da5d69869be34193b8981d4a3b922879]]\
From Textbook
The accuracy of the factor of safety F is affected by the number of slices adopted.
A computer analysis (SLOPE ©) of the same slope using 25 slices gave a value of F = 1.486.
This circle does not give the lowest factor of safety so it is not the critical slip circle. A computer run (SLOPE©)
obtained the F values for circles with their centres on a grid pattern and all tangential to the depth of 1.6 m below
the toe. Contours of factor of safety are plotted on this grid and the circle with the lowest value of F lies at the centre
of the contour plot, as shown on Figure 12.20. This gives the minimum factor of safety as 1.430.
METHOD of Analysis :
Note : Textbook method using eqn 12.9 is approximate.
Instead, an accurate method of finding Strip Areas (Program 21) and Full Curve Length (Program 22) is used
Slope Geometry
Define Accuracy for Findroot
Define Slope Toe Cordinates
Note: Pl do not change cordinates of A, as all cordinates read from ACAD are based on this defined value of A
Now define initial Center cordinates,
Pl. do not change this, as ACAD values shown in this worksheet have been derived from relevant ACAD drawing.
Radius of Circle from Center to touch the rock layer
Calculation of other important cordinates
Slope and Intercept Line AB
Intercept of Line AB using c = y - mx
Slope of AB
1. To find the two intersection points AA and C of the Slip Circle width slope profile
From ACAD
X = 0.35 Y = 1.00
From ACAD
X = 20.72 Y = 9.00
2. To find Top Cords of Strips intersecting the Horizontal Sections of the Slope Profile.
3. To find Bottom Cords of Any Strips intersecting the Slip Circle
Top Cords on Horizontal Section - Using Program 6
From ACAD
X = 1.00 Y = 1.00
X = 0.35 Y = 1.00
X = 17.00 Y = 9.00
X = 19.00 Y = 9.00
X = 20.72 Y = 9.00
4. To find Top Cords of Strips intersecting the Sloping Section AB of the Slope Profile.
Top Cords BC_AB, use Program 9
Bottom Cords BC_AB, use Program 4
From ACAD
From ACAD
X = 1.00 Y = 1.00
X = 1.00 Y = 0.69
X = 2.00 Y = 1.50
X = 3.00 Y = -0.04
X = 5.00 Y = 3.00
X = 5.00 Y = -0.46
X = 7.00 Y = -0.60
5 To find Bottom Cords of Strips intersecting the Horizontal Sections of the Slope Profile.
Bottom Cords BC_HOZ, use Program 11
From ACAD
X = 1.00 Y = 0.69
X = 0.35 Y = 1.00
X = 17.00 Y = 3.36
X = 19.00 Y = 5.68
X = 20.72 Y = 9.00
6 To find Top Mid Cords of Strips intersecting the Horizontal Sections and the Slope Profile.
Top Mid Cords of Strips on Horizontal lines of Slope Profile
Top Mid Cords of Strips on Slope AB
X = 1.50 Y = 1.25
X = 0.68 Y = 1.00
X = 17.50 Y = 9.00
X = 3.50 Y = 2.25
X = 19.50 Y = 9.00
X = 20.36 Y = 9.00
X = 6.50 Y = 3.75
X = 14.50 Y = 7.75
X = 16.50 Y = 8.75
7 To find Bottom Mid Cords of Strips intersecting the Horizontal Sections and the Slope Profile.
7.1 Bottom Mid Cords on Horizontal Lines
7.2 Bottom Mid Cords on Slope AB
X = 0.68 Y = 0.84
X = 1.5000 Y = 0.48
X = 17.50 Y = 3.86
X = 18.50 Y = 5.00
X = 3.5000 Y = -0.17
X = 19.50 Y = 6.46
X = 5.5000 Y = -0.52
X = 20.35 Y = 8.11
X = 6.5000 Y = -0.59
X = 12.50 Y = 0.48
X = 14.50 Y = 1.47
X = 16.50 Y = 2.91
Next step is to find strip angles from bottom mid cordinates
Note : After Adjust
- ve ----> Resisting
+ ve ----> Failing
After Adjustment
After Adjustment
Before Adjustment
Angle = -25.66°
Angle = -22.13°
Angle = 58.89°
Angle = 66.2°
Angle = 40.59°
Next step is to find Strip Areas from Top and Bottom Cordinates
From ACAD
From ACAD
Area = 0.10
Area = 0.76
Area = 5.12
Area = 3.96
Area = 3.13
Area = 0.59
Area = 4.33
Area = 6.26
Area = 6.26
Area = 5.82
Next step is to find Strip Weights from above Areas
Note : After Adjust
- ve ----> Resisting
+ ve ----> Failing
Use the length of Circlular Arc (AA_C) for better accuracy
Distance between extremes AA and CC of arc
Angle subtended at center by AA and CC and Length of Arc
ACAD List command
From ACAD
Angle = 97.05°
From ACAD
length 24.73
FS - Accurate method using R and θ
Textbook result for width = 2m
Most accurate value
Total Number of SLICES - Using PROGRAM 22
From Textbook
The accuracy of the factor of safety F is affected by the number of slices adopted.
A computer analysis (SLOPE ©) of the same slope using 25 slices gave a value of F = 1.486.
Analysis of multiple trial slip circles to find the Min FS using array of different centers
Define array of centers
For multiple cases
For multiple cases
Min FS
Center at Min FS
Radius at Min FS
Slices at Min FS
Redefine
center point, X= 7.0000 Y= 14.0000
radius 14.6000
Lowest FS
FS Center Rad Slices
Tallies with Textbook Result