eng mektan ii 1.handout
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Mekanika Tanah merupakan salah satu mata kuliah di teknik sipil. Pdf merupakan buatan dosen oleh sipil UGM.TRANSCRIPT
Soil Mechanics II
(Teuku Faisal Fathani)
1. Application of soil mechanics
2. Stress Distribution
3. Consolidation
4. Settlement
5. Shear Strength Parameter of Soil
6. Slope Stability
References:
- Principles of Geotechnical Engineering (Braja M.
Das, 2002)
- Soil Mechanics (R. F. Craig, 1987)
- Mekanika Tanah I (Hary Christady H., 2002)
Shear Strength Parameters
Shear strength parameter:
Internal resistant force per unit area
Failure of shear at a slip surface due to applied force to the soil.
Shear resistant:
1. Cohesion (c): depend on the type of soil and its density,
independent from normal stress ( ) at the shear surface.
2. Friction inter material ( tan ): depend on the normal
stress ( ) at the shear surface and internal friction angle ( )
3. Combination of c and
MOHR-COULOMB Failure Criteria
Mohr (1900): Failure of a material due to the combination of critical
condition between normal stress ( ) and shear stress ( )
Coulomb (1776) f ( ) :
N
F
A
= shear strength (kN/m2)
c = cohesion (kN/m2)
= internal friction angle ( 0)
= normal stress at the failure
surface (kN/m2)
MOHR-COULOMB Failure Criteria
Mohr
Mohr-Coulomb
c
A
B
C
y
x
f
A Failure does not occur
B Failure occurs
C Failure never happen
In effective stress condition (Terzaghi, 1925):
= shear strength (kN/m2)
c’ = effective cohesion (kN/m2)
’ = effective internal friction angle ( 0)
’ = effective normal stress on the failure plane (kN/m2)
u = pore water pressure (kN/m2)
Saturated soil
Mohr’s Circle and failure envelope
’
Failure
envelope
c f
3’ 1’ f’
2
1’
1’
3’ 3’ f’
f
1’ = effective major principle stress
3’ = effective minor principle stress
= theoretical angle between the failure plane and
major principal plane
Relationship between effective principle
stress at failure and shear strength
parameter c - :
Laboratory Test for Determination of
Shear Strength Parameter
1. Direct shear test
2. Triaxial test UU, CU CD
3. Unconfined compression test
4. Vane shear test
Direct Shear Test
N
T
Shear box
Porous stone
Sample
Loading
plate
h
L
Method of test :
Stress-controlled Test
Peak shear strength
Strain-controlled Test
Peak shear strength and Residual shear strength
Direct test is appropriate to be used for sandy soil.
Shearing the sample up to failure or the maximum strain
reaches max = 20%
Direst shear tests are repeated on similar specimens at
various normal stresses (min. 3 times).
1
2
3
c
1 2 3
Pure sand c = 0 ; maka
Dry condition :
Stress-strain characteristics of sand:
L
= constant
dense
loose
L H
+
-
Exp
an
sio
n
Co
mp
ressio
n
Dense sand
Loose sand
Peak shear strength
Residual shear strength
Shear displacement
Factors affected shear strength of sand:
- Particle size
- Water inter the particle
- Roughness of the surface of particle
- Grain size distribution
- Shape of particle
- Pore number (e) or relative density (Dr)
- Main principal stress
- Stress history
Example
Direct shear test on a clean compacted sand. Shear box with
dimension of 250 x 250 mm2. The results as follow:
Calculate shear strength parameter of the sand in dense condition
and loose condition
m dense sand peak stress
t loose sand residual stress
Normal force (kN) 5,00 10,00 11,25
Peak shear force (kN) 4,90 9,80 11,00
Residual shear force (kN) 3,04 6,23 6,86
Normal stress (kN/m2) 80 160 180
Peak shear stress (kN/m2) 78,4 156,8 176
Residual shear stress (kN/m2) 48,6 99,7 109,8
(kN/m2)
(kN/m2)
Peak stress
Residual
stress
From the above figure
m dense sand = 450
t loose sand = 320
Triaxsial Test
• The most reliable method to determine the shear strength
parameter of soil
• Commonly use for site investigation for civil construction and for research
h
h h
v
v
Applied stresses:
1 = major principal stress
3 = minor principal stress
2 = 3 = confining stress
= 1 - 3 = deviator stress
z
y
x 3
1
1
2
Correction of the sample area at a
certain strain (A)
Axial Load and Confining Pressure
3 3
3
3
3 3
3 + = 1
3 + = 1
L
Loading on vertical direction:
1. Apply the dead load gradually until failure. Axial deformation
is measured by dial gage.
2. Apply the axial deformation with a constant increment (strain-
controlled). Axial load is measured by proving ring.
Types of Standard Triaxial Test
Test condition 3 (confining
pressure)
Deviator stress,
Output
U.U.
(Unconsolidated-
Undrained)
Drainage
connection close
Drainage
connection
close
Total stress
C.U.
(Consolidated-
Undrained)
3 consolidation
Drainage
connection open
Drainage
connection
close
u
Total stress
Effective
stress
C.D.
(Consolidated-
Drained)
3 consolidation
Drainage
connection open
Drainage
connection
open
Effective
stress
C.D. Test
3 3
3
3
uc=0 3 3
ud=0
3 + = 1
3 + = 1
B = pore pressure parameter by
Skempton (1954)
B = 1 for saturated soft soil
Confining
pressure Deviator
stress
t Vc
+
-
Because of the pore water pressure (u) during the
implementation of deviator stress is totally dissipated
Total stress = Effective stress
The same test is conducted at the same soil sample with
different 3 (confining pressure).
After getting 1 and 3 Mohr’s circle + failure envelope can be drawn
Failure envelope of Triaxsial CD
For sand dan clay NC
Failure envelope
3= 3’
2
1
1
3 3
f
f
2
1= 1’
A
B
f
f
C.U. Test
3 3
3
3
uc=0 3 3
ud 0
3 + = 1
3 + = 1
A = pore water pressure parameter
by Skempton (1954)
Confining
pressure Deviator
stress
t Vc
+
-
Pore water pressure during the application of deviator stress = ud
Total stress Effective stress
Pore water pressure at failure can be measured = udf
Major principal stress at failure (total) :
Minor principal stress at failure (total) :
Pore water pressure at failure :
Major principal stress at failure (effective) :
Minor principal stress at failure (effective) :
In order to determine the shear strength parameter triaxial test
on soil samples should be done with different confining pressure 3
(%)
31
32
33
c
3 1 1
f1
1 3 3
3 3
3 + = 1
3 + = 1
Failure envelope of Triaxsial CU
For sand dan clay NC
Failure envelope
of effective stress
3
1
1
3 3
f
f
1
A
B
udf
C
D
cu
1’ 3’
udf
Failure envelope
of total stress
A parameter for clay:
• Lempung NC : 0,5 – 1
• Lempung OC : -0,5 – 0
A value depends on OCR
c’ = c = maximum confining pressure when the soil sample consolidates
A = pore water pressure parameter by Skempton
(1954) at failure:
U.U. Test
3 3
3
3
uc 0 3 3 ud 0
3 + = 1
3 + = 1
Confining
pressure Deviator
stress
Total pore water pressure =
and
Hence
Triaxial UU test on saturated clay :
f will be the same for different confining pressure 3
Failure envelope
1
1
3 3
f
f
cu
3 1 1 3 3 1
The application of Triaxial UU, CU and CD:
UU :
• Foundation on soft soils
• Embankment on soft soils
• Dam on soft soils
The loading apply so fast, hence the consolidation and drainage
did not occur yet at the soil layers end of construction
CU :
• Slope stability, where the soil has been consolidated and stable
• Rapid draw-down at of reservoir
• Embankment construction (several phase / stage)
CD :
• Embankment long time construction
• Earth dam affected by steady
• Clay excavation
• Practically, it is difficult to implement this test for clay,
because the time to get u = 0 will be too long,
p = need to be small it takes a long time and very easy to seep out.
Example 1
Triaxial CD has been done for normally consolidated clay.
The result of test as follow:
3 = 276 kN/m2
f= 276 kN/m2
Calculate :
a. Internal friction angle,
b. The angle between failure surface and major principal plane,
c. Normal stress ’ and shear stress f at failure surface
For Normally Consolidated Clay, the failure envelope equation
Principal stress:
Failure envelope
3’=276 kN/m2
2
’1
’1
’3 ’3
f
f
B
1’=552 kN/m2 A O
Example 2
Consolidated-Drained Triaxial test for example no. 1 :
a. Calculate effective normal stress ( ’) acting on the plane with
maximum shear stress ( ).
Maximum shear stress occurs at the plane with = 450
b. Why does the failure occur at the plane of = 54,730, not at the
plane having maximum shear stress?
Shear stress that causes the failure at = 450:
Shear stress acting on that plane :
Example 3
The result of triaxial test (UU) as follow:
Calculate the shear strength parameter by using Mohr’s circle.
No of test Confining pressure
(kg/cm2)
Deviator stress (kg/
cm2)
1 1 0,57
2 1,4 0,71
3 1,8 0,76
4 2,2 0,84
(kg/cm2)
Failure envelope
c
33 13 14 34 32 12
1 2 3
(kg/cm2)
1
No test 3 (kg/cm2) (kg/cm2) 1= 3+
(kg/cm2)
( 1+ 3)/2
(kg/cm2)
1 1 0,57 1,57 1,285
2 1,4 0,71 2,11 1,753
3 1,8 0,76 2,56 2,182
4 2,2 0,84 3,04 2,620
From the figure :
Example 4
Triaxial test (CU) for normally consolidated clay, with the result:
3 = 260 kN/m2
Deviator stress : f= 200 kN/m2
Pore water pressure : udf = 120 kN/m2
Calculate :
a. Internal friction angle at Consolidated-Undrained (CU)
condition
b. Internal friction angle at Consolidated-Drained (CD) condition
Unconfined Compression
Test
(Uji Tekan Bebas)
Unconfined
Compression Test
as a special type of
Triaxial UU for
saturated clay
ASTM D2166
AASHTO T208
1
1
3=0 cu
3=0 1=qu
qu = unconfined compression
strength
Consistency qu (kN/m2)
Very soft 0 – 25
Soft 25 – 50
Medium 50 – 100
Stiff 100 – 200
Very stiff 200 – 400
Hard >400
Theoretical failure envelope
of total stress
cu
3 1 1 3 0 1=qu
The real failure envelope
of total stress
1 2
3
Result of Unconfined Compression Test (Mohr’s circle-1)
vs. Triaxial UU (Mohr’s circle- 2 and 3) for saturated clay
Vane shear Test
To determine shear strength at undrained condition cu ( = 0)
for saturated clay.
Method:
• Vane shear equipment is driven to
the borehole with minimum depth of
3 x borehole diameter
• Rotate with the velocity of 6o – 12o per minute.
• Every 15-30 seconds, record T value
Bjerrum (1974) : cu form vane shear test is too high, because of
the shear zone expanded.
Vane shear result in the field need to be corrected:
= correction factor = 1,7 – 0,54 log (PI) See HCH MT1 :
5.2.4 page 305
PI = Plasticity index = LL - PL
cu vs o’
Clay NC :
(Skempton, 1957)
o’ = effective overburden pressure
IP = plasticity Index (%)
Clay OC :
(Ladd et al., 1977)
Sensitivity and Thixotropy of clay
1
Axial strain
undisturbed
remolded
qu
qu
Sensitivity Classification
4 – 8 Sensitive clay
8 – 16 Extra sensitive
clay
> 16 Quick clay
From unconfined
compression test
Example :
Result of Unconfined Compression Test:
Calculate the shear strength parameters.
Solution: Draw the relation between Strain ( ) vs Stress ( )
Strain (%) Stress (kg/cm2)
1 0,268
5 0,450
7 0,46
8 0,47
9 0,46
12 0,45
Ko (Lateral Earth Pressure Coefficient)
(Lateral earth pressure coefficient)
Depend on and stress history
Ko for sand
Ko value for sand OC > sand NC
Sand OC
h = 0,4 – 0,5
h = 0,6 (dense sand)
(Jaky, 1944)
(Schmidt,1967 & Alphan,1967)
Slope stability analysis
1. Transitional movement
2. Rotational movement (circular slip
surface)
3. Method of slice
4. Slope stability software for 2D analysis
5. 3D landslide movement simulation
Slope Stability Analysis
• Slope stability concept: limit plastic equilibrium
• Purpose to determine the safety factor (FS) of the potential slip
surface
• Assumption:
Sliding occurs at a certain slip surface 2 dimensional
problem
Sliding material massif
Isotropic shear strength
FS is analyzed based on average shear strength at the slip
surface
Difference between landslide and slope failure Landslides Slope Failures
Geology Occur in places with particular
geology or geological formation
Slightly related to geology
Soils Are mainly active on cohesive
soil such as slip surface
Frequently occur even in sandy
soils
Topography Occur on gentle slopes of 5° to
20°
Frequently occur on the slopes
steeper than 30°
Situation of
activities
Continuous, or repetitive
occurrences
Occur suddenly
Moving velocity Low at 0.001 to 10 mm/day High speed > 100 mm/day
Masses Have little disturbed masses Have greatly disturbed mass
Provoking causes Greatly affected by groundwater Affected by rainfall intensity
Scale Have a large scale between 1
and 100 ha
Have a small scale. Average
volume is about 440 m3
Symptom Have cracks, depressions,
upheavals, groundwater fluctuation, before occurrence
Have few symptoms and
suddenly slip down
Gradient 10° to 25° 35° to 60°
Tsaoling Landslide
Induced by 1999 Chi-chi
Earthquake, Taiwan
Volume: 1.4 x 108 m3
Affected area: 698 ha
Total length: 4 km
Source area:
Length: 1.5 km
Width: 2 km
Depth: < 200 m
Destruction of 5 houses,
resulting in 29 deaths.
Causes of Landslide
• Rainfall or storm rainfall the rising of
groundwater level
• Construction works Earthwork, Cutting,
Filling, Tunnel construction,
• Reservoir induced landslide the rising and
drawdown of reservoir level
• Earthquake horizontal acceleration gx, gy
Shear Strength Parameters
Shear strength parameter:
Internal resistant force per unit area
Failure of shear at a slip surface due to applied force to the soil.
Shear resistant:
1. Cohesion (c): depend on the type of soil and its density,
independent from normal stress ( ) at the shear surface.
2. Friction inter material ( tan ): depend on the normal
stress ( ) at the shear surface and internal friction angle ( )
3. Combination of c and
MOHR-COULOMB Failure Criteria
Mohr (1900): Failure of a material due to the combination of critical
condition between normal stress ( ) and shear stress ( )
Coulomb (1776) f ( ) :
N
F
A
= shear strength (kN/m2)
c = cohesion (kN/m2)
= internal friction angle ( 0)
= normal stress at the failure
surface (kN/m2)
MOHR-COULOMB Failure Criteria
Mohr
Mohr-Coulomb
c
A
B
C
y
x
f
A Failure does not occur
B Failure occurs
C Failure never happen
In effective stress condition (Terzaghi, 1925):
c’ = effective cohesion (kN/m2)
’ = effective internal friction angle ( 0)
’ = effective normal stress (kN/m2)
u = pore water pressure (kN/m2)
= average shear stress
d = average shear stress at the critical slip surface due to the weight of
sliding material
Mohr-Coulomb
c- = shear strength parameter at the critical slip surface
SLOPE STABILITY
Safety factor for cohesion component
Safety factor for friction component
In general FS 1,2
FS = Fc = F
Analysis on a translational slip surface
A. Infinite slope
A.1. Without seepage
W
P
Na
Ta
Tr
Nr
E
E
A
B
P
Q
S
T H
b
Bedrock
If F = 1 (critical), so H = Hc:
Granular soil (c = 0)
Cohesive soil ( = 0)
F = 1 (critical), ( = 0): Stability number
W
P
Na
Ta
Tr
Nr
E
E
A
B
P
Q
S
T H
b
Bedrock
Example 1
H
Bedrock
= 1,86 t/m3
c =1,8 t/m2
=20o
a) H = 8 m ; = 22o calculate FS & Hc
b) H = 8 m ; = 25o , Calculate FS and Hc
Analysis on a translational slip surface
B. Finite slope
B.1. Culman’s method
W
P
Na
Nr
Ta
Tr
H
A
B C
W
P
Na
Nr
Ta
Tr
H
A
B C
Shear resistant ( d) at AB:
Critical condition F=1
= d
Critical condition
F=1 cd = c ; d =
Example 1
H = ?
Previous
landfill
New landfill
timb = 1,96 t/m3
c =2,5 t/m2
=17o
= 48,5o
= 40o Calculate the maximum height of
new landfill, if the safety factor is
decided to be F=2
B. Finite slope
B.2. Analysis on a circular slip surface
O
(a) Toe circle
O
(b) Slope circle
bedrock
Slope stability analysis on cohesive soil
OWithout
groundwater
y
W R
A
B C
C
= 0
W = weight of sliding material (kN)
LAC = length of circular plane (m)
c = cohesion (kN/m2)
R = radius of slip surface circle (m)
y = distance between W to point O (m)
OWith the
existence of
groundwater
W
R
A
B C
C
= 0
W’ = effective soil weigth (kN)
LAC = length of circular slip surface (m)
c = cohesion (kN/m2)
R = radius of slip surface circle (m)
y = distance between W’ to point O (m)
W’
U
Slope stability analysis on cohesive soil
Slope stability analysis on cohesive
soil, using Taylor Diagram (1948)
W1 = area (EFCB) x x 1
W2 = area (EFDA) x x 1
O
y1
W1
R
A
B C
Cd
= 0
D F
E
W2
y2
Nd
= angle from the figure in radian
O
y1
W1
R
A
B C
Cd
= 0
D F
E
W2
y2
Nd
Critical condition (F=1); H = Hc ; cd = cu
Taylor Method (1948)
Stability number:
Nd value is decided by using STABILITY DIAGRAM : =0 Taylor (1948)
Example:
Excavation of 10 m depth in saturated cohesive soil. Unit volume of
saturated clay is 18,5 kN/m3 and cohesion is 40 kN/m2. Bedrock located
at 12 m below the surface. If the internal friction angle of the soil is =
0, calculate the inclination of the slope if the safety factor is decided to
be F=1,5.
Depth factor : D = 12/10 = 1,2
Slope stability analysis for the soil > 0,
using Taylor Diagram (1948)
If the soil has the friction component ( ) normal stress distribution (N)
affect the distribution of shear stress
Normal stress resultant and friction component have the inclination of
based on normal line direction
O
R
A
B C
D F
E
Contoh:
An earthfill H =12,2 m, = 30o. Bedrock at infinite depth. C = 38,3 kN/m2,
= 10o and total unit weigth = 15,7 kN/m3. Calculate safety factor of
cohesion (Fc), to internal friction (F ) and overall safety factor (F).
a. Assuming all works = 10o ; = 30o
From figure cd/ H = 0,075 cd = 14,4 kN/m2
Fc = c/cd = 38,3 / 14,4 = 2,67
b. Assuming all c works c = 38,3 kN/m2 ; = 30o ; cd/ H = 0,2
From figure < 0 F =
(Resistant moment due to cohesion > driving moment)
c. F to shear strength by trial and error
All works F = 1 Fc = 2,67
Fc = c/cd = 2 cd = 38,3/2 = 19,2 kN/m2 cd/ H = 0,1 From figure d = 7o F = tan 10o / tan7o = 1,44
Fc = c/cd = 1,8 cd = 38,3/1,8 = 21,3 kN/m2 cd/ H = 0,11 from figure d = 5o F = tan 10o / tan5o = 2,02
Stability Analysis of Landslide Slope
The design of a slope should ideally be based on an
allowable deformation
The difficulty with deformation analysis stress-strain
relationship, peak and residual strengths, anisotropic, pore
pressure distribution, the non-homogeneity, and the effect due to initial stress.
Finite element method reflecting all of the factors.
As an alternative, a limit equilibrium analysis stability of a
slope, in terms of a safety factor F.
Limit equilibrium method analysis of natural & artificial
slopes (cut and fill)
No Method Equilibrium equation
Force Moment
Horizontal Vertical
1 Fellenius (1927) – –
2 Bishop’ Simplified (1955) –
3 Janbu’s Simplified (1954) –
4 Corps of Engineering (1982) –
5 Lowe and Karfiath (1960) –
6 Spencer (1967)
7 Sarma (1973)
8 Morgenstern and Price (1975)
Slope stability analysis based on the limit equilibrium
and slice method
Remarks :
” ” : The equilibrium of horizontal forces, vertical forces or moments are taken into account for analysis.
Circular slip surface
Bishop Method
Fellenius Method
These methods are currently being widely used in
the field of landslide analysis.
O x
R
n n+1
W
Xn En
Xn+1
En+1
h
b
P C
B
ls
A
D
l
W
Xn-Xn+1
En-En+1
S
P
P’
ul
tan =1/F.tan ’
Bishop method
The Bishop method is a method for analyzing the equilibrium of a sliding
block, which slumps in a single movement about a given point.
The equilibrium equation for moments about the center of rotational
movement is expressed as :
The Mohr-Coulomb failure criterion is :
Bishop method
In solving stability problems determine statically indeterminate elements,
obtaining equilibrium among the slice in horizontal and vertical directions.
In the simplified Bishop method, horizontal forces are ignored, and only the
vertical forces in each slice are taken into account:
Since both sides of expression contain F, the safety factor has to
be obtained by a series of calculations.
O x
R
n n+1
W
Xn En
Xn+1
En+1
h
b
P C
B
ls
A
D
l
Fellenius
method Internal forces applied to the wall
of each slice are ignored:
The moments of the entire
sliding block are in equilibrium:
Bishop method
Fellenius method
F : the safety factor,
c’ : cohesion of the slip surface (kN/m2),
W : weight of each slice (kN), W = A
: unit volume weight (kN/m3),
A : area of each slice (m2),
u : water pressure on the slip surface of each slice (kN/m2),
l : length of slip surface of each slice (m),
’ : friction angle of slip surface,
: angle between the center bottom of each slice and the vertical line of the
center of sliding surface circle.
Janbu method : an analytical method for analyzing the stability of a
landslide block sliding surface has a complex shape
Janbu Method
where fo is a modifying coefficient representing the influence of the
shearing force acting on the walls of each slice.
The coefficient can be decided from analysis of soil and other
conditions, covering more than 40 different cross sections.