Chapter 9. Introduction to Geotechnical Earthquake Design
1. For each of the following multiple-choice questions, select all of the correct answers.
The correct answers are marked by underline.
(1) The movement of a fault is horizontal (or lateral) and each block moves to the left of
the opposite block. Which of the following faults fit(s) this type of movement?
A. Normal fault.
B. Reverse fault (thrust fault).
C. Right lateral fault.
D. Left lateral fault.
E. Left-lateral normal fault.
(2) Which of the following movements describes the “left-lateral normal fault”?
A. The movement is vertical, the hanging wall moves up relative to the foot wall.
B. The movement is horizontal (or lateral), each block moves to the right of the
opposite block.
C. The movement is horizontal (or lateral), each block moves to the left of the
opposite block.
D. The movement is combined by the normal fault movement and left-lateral
fault movement.
E. The movement is combined by the reverse fault movement and left-lateral
fault movement.
(3) An epicenter is:
A. The location where a fault rupture first occurs.
B. The point vertically above the focus and on the ground surface.
C. The location where the seismometer is installed.
D. The position after the fault rupture.
E. The location where the body wave originates.
(4) A primary wave is a:
A. Shear wave.
B. Body wave.
C. Surface wave.
D. Compressional wave.
E. Longitudinal wave.
(5) A surface wave is a(n):
A. Interaction between P-waves and S-waves.
B. Interaction between SV waves and SH waves.
C. Body wave.
D. Love wave and Rayleigh wave.
E. Primary wave.
(6) The following conditions contributes to the occurrence of liquefaction:
A. Sandy or silty soils.
B. Saturation.
C. Loose or medium compaction.
D. Seismic stress.
E. High plasticity index of soil.
2. A soil-retaining wall is shown in Figure 9.26. Under the design earthquake, the
horizontal acceleration coefficient is 0.25, and the vertical acceleration coefficient is
0.1. The external friction angle between the granular backfill and the wall is twothirds of the internal friction angle.
(1) Calculate the total static earth force behind the retaining wall and determine the
location and direction of the resultant force.
(2) Calculate the total dynamic earth force under the design earthquake and determine
its point of application.
α = 15°
Sand
H=6m
θ = 10°
γ = 18.5 kN/m3
f =35°
c=0
Figure 9.26 Schematic for Problems 2 and 4
Solution:
Active earth force acts on the retaining wall.
(1) Static earth force:
The Coulomb active earth force per unit length of a wall is:
1
Pa = g H 2 K a
2
where Ka is the Coulomb active earth pressure coefficient:
Ka =
cos2 (f ¢ - q )
é
sin (f ¢ + d ¢) sin (f ¢ - a ) ù
ú
cos q cos (d ¢ + q ) ê1+
cos (q + d ¢) cos (q - a ) úû
êë
2
2
where:  = 35,  =10,  =
2
´ 35 = 23.3 ,  = 15.
3
Calculate Ka = 0.400
1
Pa = g H 2K a = 0.5´18.5´62 ´ 0.400 =133.2 kN/m
2
Pa acts on the 1/3 of H from the bottom of the wall, the direction is shown in the figure
below.
α=15°
Pa
H=6m
θ=10°
23.3°
H/3
(2) Dynamic earth force:
æ kh ö
æ 0.25 ö
= tan -1 ç
= 15.5
÷
è 1- 0.1 ÷ø
è 1- kv ø
b = tan -1 ç
The dynamic active earth pressure coefficient, KAE, is:
K AE =
cos 2 (f - q - b )
é
sin (d + f ) sin (f - a - b ) ù
cos b cos q cos (d + q + b ) ê1+
ú
cos (d + q + b ) cos (a - q ) úû
êë
cos 2 ( 35 - 10 - 15.5 )
2
2
=
é
sin ( 23.3 + 35 ) sin ( 35 - 15 - 15.5 ) ù
cos15.5 ´ cos 2 10 ´ cos ( 23.3 + 10 + 15.5 ) ê1+
ú
cos ( 23.3 + 10 + 15.5 ) cos (15 - 10 ) úû
êë
= 0.908
The total dynamic active earth force per unit length of a wall is:
1
PAE = g H 2 1- kv K AE = 0.5´18.5´62 ´ 1-0.1 ´0.908 = 272.1 kN/m
2
(
)
(
)
To calculate the point of application, z0, from the bottom of the wall:
2
PAE = PA + DPAE
The active earth force under static condition, PA = 133.2 kN/m
So: DPAE = PAE - PA = 272.1-133.2= 138.9 kN/m
So: z0 =
PA
(
)
H
6
+ DPAE 0.6H 133.2´ +138.9´ 0.6´ 6
3
3
=
= 2.82 m
PAE
272.1
3. A cantilever retaining wall is shown in Figure 9.27.
The backfill is granular soil. The wall geometry and the characteristics of the backfill
and foundation soils are shown in the figure. Under the design earthquake, the
horizontal acceleration coefficient is 0.2, and the vertical acceleration coefficient is
0.05. The external friction angle between the granular backfill and the wall is twothirds of the internal friction angle.
(1) Calculate the total static earth force behind the retaining wall and determine the
location and direction of the resultant force.
(2) Calculate the total dynamic earth force under the design earthquake and determine
its point of application.
1.0 m
Granular backfill
γsat = 19 kN/m3
ϕ = 30°
H = 10 m
1.5m
1.0m 1.0m
1.0m
6.0m
Figure 9.27 Schematic for Problems 3 and 5
Solution:
In the design, the lateral earth pressure (static or dynamic) on a cantilever retaining wall
is calculated on the vertical profile AB, as shown in the figure below.
1.0 m
A
Granular backfill
γsat = 19 kN/m3
ϕ = 30°
H = 10 m
1.5m
1.0m 1.0m
1.0m
6.0m
B
Active earth force acts on the retaining wall.
(1)
Static earth force:
The Coulomb active earth force per unit length of a wall is:
1
Pa = g H 2 K a
2
where Ka is the Coulomb active earth pressure coefficient:
Ka =
cos2 (f ¢ - q )
é
sin (f ¢ + d ¢) sin (f ¢ - a ) ù
ú
cos2 q cos (d ¢ + q ) ê1+
cos (q + d ¢) cos (q - a ) úû
êë
2
where:  = 30,  =0,  =  = 30 (because profile AB is the interface between soil
and soil, the friction on AB is soil’s internal friction angle), and  = 0.
Calculate Ka = 0.297
1
Pa = g H 2K a = 0.5´19´102 ´0.297=282.2 kN/m
2
Pa acts on the 1/3 of H from the bottom of the wall, the direction horizontal.
(2) Dynamic earth force:
The dynamic active earth pressure coefficient, KAE:
cos2 (f - q - b )
K AE =
2
é
sin (d + f ) sin (f - a - b ) ù
2
ú
cos b cos q cos (d + q + b ) ê1+
cos (d + q + b ) cos (a - q ) úû
êë
where:  = 30,  =0,  =  = 30, and  = 0.
Calculate KAE = 0.484
The total dynamic active earth force per unit length of a wall is:
1
PAE = g H 2 1- kv K AE = 0.5´19´102 ´ 1-0.05 ´0.484 = 436.8 kN/m
2
(
)
(
)
To calculate the point of application, z0, from the bottom of the wall:
PAE = PA + DPAE
The active earth force under static condition, PA = 282.2 kN/m
So: DPAE = PAE - PA = 436.8-282.2= 154.6 kN/m
So: z0 =
PA
(
)
H
10
+ DPAE 0.6H 282.2´ +154.6 ´ 0.6´10
3
3
=
= 4.28 m
PAE
436.8
4. The problem statement is the same as in Problem 2. Assume the lateral soil pressure
is passive.
(1) Calculate the total static earth force behind the retaining wall and determine the
location and direction of the resultant force.
(2) Calculate the total dynamic earth force under the design earthquake and determine
its point of application.
Solution:
Passive earth force acts on the retaining wall.
(1)
Static earth force:
The Coulomb passive earth force per unit length of a wall is:
1
Pp = g H 2K p
2
where Kp is the Coulomb passive earth pressure coefficient:
cos2 f ¢ + q
Kp =
é
sin f ¢ + d ¢ sin f ¢ + a
2
cos q cos q - d ¢ ê1ê
cos q - d ¢ cos q - a
ë
(
(
)
(
(
)
) (
) (
)
)
ù
ú
ú
û
2
where:  = 35,  = 10,   = 35 2/3 = 23.3,  = 15. Plug in the values of the four
angles in Kp and find Kp = 16.34
1
Pp = g H 2K p = 0.5´18.5´62 ´16.34 = 5441.2 kN/m
2
The point of application of Pp is at H/3 from the bottom of the wall, the direction is
shown in the graph below.
α=15°
H=6m
θ=10°
23.3°
H/3
Pp
(2) Dynamic earth force:
æ kh ö
æ 0.25 ö
= tan -1 ç
= 15.5
÷
è 1- 0.1 ÷ø
è 1- kv ø
b = tan -1 ç
The dynamic active earth pressure coefficient, KAE, is:
K PE =
cos2 (f + q - b )
é
sin (d + f ) sin (f + a - b ) ù
ú
cos b cos2 q cos (d - q + b ) ê1cos (d - q + b ) cos (a - q ) úû
êë
where:  = 35,  = 10,   = 35 2/3 = 23.3,  = 15, and  = 15.5. Plug in the
values of the five angles in KPE and find KPE = 14.00
The total dynamic active earth force per unit length of a wall is:
2
(
)
(
)
1
PPE = g H 2 1- kv K PE = 0.5´18.5´62 ´ 1-0.1 ´14.00 = 4195.8 kN/m
2
To calculate the point of application, z0, from the bottom of the wall:
PPE = PP + DPPE
From (1), PP = 5441.2 kN/m
So: DPPE = PPE - PP = 4195.8  5441.2 = 1245.4 kN/m
So: z0 =
PP
(
)
(
)
H
6
+ DPPE 0.6H 5441.2´ + -1254.4 ´ 0.6 ´ 6
3
3
=
= 1.52 m
PPE
4195.8
5. The problem statement is the same as in Problem 3. Assume the lateral soil pressure
is passive.
(1) Calculate the total static earth force behind the retaining wall and determine the
location and direction of the resultant force.
(2) Calculate the total dynamic earth force under the design earthquake and determine
its point of application.
Solution:
Passive earth force acts on the retaining wall.
(1)
Static earth force:
The Coulomb passive earth force per unit length of a wall is:
1
Pp = g H 2K p
2
where Kp is the Coulomb passive earth pressure coefficient:
cos2 f ¢ + q
Kp =
é
sin f ¢ + d ¢ sin f ¢ + a
2
cos q cos q - d ¢ ê1ê
cos q - d ¢ cos q - a
ë
(
(
)
)
(
(
) (
) (
)
)
ù
ú
ú
û
2
where:  = 30,  = 0,   =  = 30,  = 0. Plug in the values of the four angles in Kp
and find Kp = 10.10
1
Pp = g H 2K p = 0.5´19´102 ´10.10 = 9595.0 kN/m
2
Pp acts on the 1/3 of H from the bottom of the wall, the direction horizontal.
(2) Dynamic earth force:
The dynamic active earth pressure coefficient, KAE, is:
cos2 (f + q - b )
K PE =
é
sin (d + f ) sin (f + a - b ) ù
ú
cos b cos2 q cos (d - q + b ) ê1cos (d - q + b ) cos (a - q ) úû
êë
where:  = 30,  = 0,   =  = 30,  = 0, and  = 11.9. Plug in the values of the
five angles in KPE and find KPE = 7.80
2
The total dynamic active earth force per unit length of a wall is:
(
)
(
)
1
PPE = g H 2 1- kv K PE = 0.5´19´102 ´ 1-0.05 ´7.80= 7039.5 kN/m
2
To calculate the point of application, z0, from the bottom of the wall:
PPE = PP + DPPE
From (1), PP = 9595.0 kN/m
So: DPPE = PPE - PP = 7039.5  9595.0 = 2555.5 kN/m
So: z0 =
PP
(
)
(
)
H
10
+ DPPE 0.6H 9595.0´ + -2555.5 ´ 0.6´10
3
3
=
= 2.36 m
PPE
7039.5
6. An earth embankment has a slope angle of 38 degrees and a height of 10 m. The soil
has unit weight of 18.8 kN/m3, cohesion of 43 kN/m2, and internal friction angle of 30
degrees. The slope is subjected to a horizontal acceleration of 0.25 g and vertical
acceleration of 0.1 g. What is the factor of safety of the slope on a potential plane
failure surface inclined at 25 degrees with respect to the horizontal?
Solution:
The earth embankment with plane failure surface is shown in the figure below.
B
142°
kv W
H = 10m
khW
W
13°
38° Θ=25°
A
C
25°
γ = 18.8 kN/m3
c = 43 kN/m2
ϕ = 30°
D
First calculate the area of the triangle of ABC. Using trigonometry:
Since:
AB
BC
AC
=
=
0
0
sin25 sin13 sin1420
Find: BC = 6.76 m, AC = 18.49 m
1
Area ABC = ´ 6.76´10 = 33.80 m2
2
W = 33.80  18.8 = 635.44 kN/m
Use Equation (9.15):
FS =
=
S ( Resisting forces) c × LAC + éë(W - kvW ) cosq - khW sinq ùû tan f
=
S ( Driving forces)
(W - kvW ) sinq + khW cosq
43 ´18.49 + éë( 635.44 - 0.1´ 635.44 ) ´ cos 25 - 0.25 ´ 635.44 ´ sin 25 ùû ´ tan 30
( 635.44 - 0.1´ 635.44 ) ´ sin 25 + 0.25 ´ 635.44 ´ cos 25
= 2.73 > 1.5
7. As shown in Figure 9.28, a natural slope has a slope angle of 45 degrees and a height
of 12 meters. The soil is saturated clay, its unit weight is 18 kN/m3, and cohesion is
50 kN/m2. The slope is subjected to a horizontal acceleration of 0.3 g and vertical
acceleration of 0.1 g. A potential sliding surface is shown in the figure. Determine the
factor of safety for the seismic slope stability along the circular failure surface.
Center (3m, 18m)
R = 21 m
H = 12 m
45°
Toe (0, 0)
Figure 9.28 Schematic for Problem 7
Solution:
For a circular (rotational) failure surface in a saturated clay slope with  = 0, as shown in
Figure 9.12(b), the factor of safety for an assumed failure plane is:
The area of the sliding section can be found to be 208.5 m2. The weight of the sliding
section per unit length of the slope is:
W =  A = 18  208.5 = 3753 kN/m
Assume:  = 0. Therefore, the factor of the safety is:
æ a ×p ö
cç R ×
R
S ( Resisting moment )
180 ÷ø
è
FS =
=
S ( Driving moment ) (W - kvW ) L1 + khWL2
where: L1 = 7.6 m, L1 = 13.7 m, and  = 104
æ a ×p ö
cç R ×
R
S Resisting moment
180 ÷ø
è
FS =
=
S Driving moment
W - kvW L1 + khWL2
(
(
)
) (
)
æ
104 ´ p ö
50 ´ ç 21´
´ 21
180 ÷ø
è
=
( 3753 - 0.1´ 3753) ´ 7.6 + 0.3 ´ 3753 ´ 13.7
= 0.97 < 1.0
Failure will occur.
8. The problem statement is the same as Problem 7. Using the ordinary method of slices,
determine the factor of safety for the seismic slope stability along the specified
circular failure surface.
Solution: Not provided.
9. As shown in Figure 9.29, a 15 m high embankment with a slope angle of  = 40 is to
be constructed in an earthquake-active region. The embankment soil is sandy silt with
strength parameters of  = 25 and c = 15 kPa, and the unit weight is 17 kN/m3. The
foundation soil is sandy clay with strength parameters of  = 20 and c = 20 kPa,
and the unit weight is 18 kN/m3. Assume a toe failure with circular failure surface as
shown in the figure. Given  = 30o and R = 20 m, the center O can be determined.
Assume the horizontal coefficient of acceleration kh = 0.2. Determine the factor of
safety of the slope on the circular failure plane.
Center O
R
H = 15 m
Toe circle
β=40° α=30°
Figure 9.29 Schematic for Problem 9
Solution: Not provided.
10. A natural slope is shown in Figure 9.30. The slope angle is 40 degrees. The slope is
subjected to a horizontal acceleration of 0.2 g. Using the ordinary method of slices,
determine the factor of safety for the seismic slope stability. Several trial failure
surfaces may be needed.
Sand
γ1 = 18.0 kN/m3
c1 = 0
ϕ1 = 38°
H = 30 m
H = 20 m
Sandy clay
γ2 = 19.5 kN/m3
c2 = 60 kN/m2
ϕ2 = 15°
Figure 9.30 Schematic for Problem 10
Solution: not provided.
11. A natural slope is to be evaluated for permanent displacement under a design
earthquake. The slope is 33 m tall and is mainly comprised of sandy soil. The slope
angle is 38. Its unit weight is 18.2 kN/m3, and its friction angle is 40. The peak
ground acceleration was 0.5 g, and the peak ground velocity of 0.5 m/sec. Evaluate
the permanent displacement of the slope using the Newmark sliding block method.
Solution:
Maximum velocity of ground motion: vmax = 0.5 m/sec
Peak horizontal ground acceleration: amax = 0.5g
The yield acceleration coefficient can be approximated using:
ky = tan (f - b )
where: internal friction angle  = 40, and the slope angle  = 38.
So: ky = tan (f - b ) = tan ( 40 - 38 ) = 0.035
So ay = 0.035g
The upper-bound permanent displacement of the slope is:
umax =
2
vmax
amax
2a y a y
(0.5 m/sec)
2
=
2´ 0.035´ 9.81 m/sec
2
´
0.5g
= 5.20 m
0.035g
12. The problem statement is the same as in Problem 11. The average shear wave
velocity in the soil slope was measured to be 90 m/sec. The earthquake magnitude in
the Richter scale is 7.5. Evaluate the earthquake-induced permanent displacement of
the slope using the Makdisi-Seed method.
Solution:
The earthquake magnitude in the Richter scale is 7.5
The maximum velocity of ground motion vmax = 0.5 m/sec
The peak horizontal ground acceleration amax = 0.5g
The yield acceleration ay = 0.035g
The average shear wave velocity v s = 90 m/sec
Height of slope H = 33 m
The first natural frequency:
v
90
w 1 = 2.404 s = 2.404 ´ = 6.556 rad/sec
H
33
The natural period of the embankment:
2p
2p
T0 =
=
= 0.958 sec
w 1 6.556
ay
amax
=
0.035g
= 0.07
0.5g
From Figure 9.19, using M = 7.5, find:
U
amaxT0
» 0.8sec
So: U  0.8 sec  0.5  9.81 m/sec2  0.958 sec = 3.76 m
13. What are the conditions necessary for liquefaction to occur?
Answer:
The conditions required for liquefaction to occur include:
(1) the soil deposit is sandy or silty soil;
(2) the soil must be saturated or nearly saturated (usually below groundwater table);
(3) the soil is loose or medium compact;
(4) the soil is subjected to seismic stress (such as from earthquake).
14. Is it true that cohesive soils absolutely do not liquefy? How to qualitatively evaluate
the liquefaction potential of cohesive soils?
Answer:
No, cohesive soils can liquefy.
To qualitatively evaluate cohesive soils, the so-called Chinese criteria defined by Seed
and Idriss (1982) can be used  liquefaction can only occur in cohesive soils if all three
of the following conditions are met:
(1) the clay content (particles smaller than 5 ) < 15% by weight;
(2) the liquid limit < 35%;
(3) the natural moisture content > 0.9 times the liquid limit.
15. A multi-story building is to be built on a level ground at a seismically active site. The
standard penetration test is performed during the subsoil exploration phase. A safety
hammer is used, and the outside diameter of the modified California sampler used is 3
inches. The peak ground horizontal acceleration that was recorded in the seismic
history in this area is 0.2 g, and the design earthquake magnitude is 6.5. The mat
foundation is designed to be 8 m below ground where the fine content is 15%. At 8 m
depth, the soil sample is retrieved by the sampler. The blow counts at the first,
second, and third six-inch penetrations are 8, 9, and 8, respectively. The boring log
indicates the following subsurface profile. Assess the liquefaction hazard at the
footing of the building at this site.
0
CS, γ = 19.0 kN/m3
GWT
5m
SM, γsat = 19.8 kN/m3
15 m
SP, γsat = 19.5 kN/m3
Figure 9.31 Subsoil Profile of the site for liquefaction evaluation, Problem 15
Solution:
(1) Calculate CSR:
CSR =
where:
t av
a
s
= 0.65 max × vo × rd
¢
¢
s vo
g s vo
Total overburden stress: s v0 = 19.0 ´ 5 +19.8 ´ 3 = 154.4 kN/m2
Effective overburden stress: s v0
¢ = 19.0 ´ 5 + (19.8 - 9.81) ´ 3 = 125.0 kN/m2
Peak horizontal acceleration: amax = 0.2 g
Since z = 8 m  9.15 m,
Stress reduction resistance: rd = 1.0 - 0.00765 ´ 8 = 0.939
So: CSR = 0.65
amax s v0
0.2g 154.4
rd = 0.65 ´
´
´ 0.939 = 0.151
g s v0
g
125.0
¢
(2) Calculate CRR using the SPT method:
First calculate the corrected SPT blow count:
(N1)60 = NmC N CE CB CR CS
Nm is the measured SPT blow count, which is the total blow count of the 2nd and 3rd sixinch penetrations. So Nm = 9+8=17.
The correction factor based on effective stress (note: Pa = 1atm = 101 kN/m2):
2.2
2.2
CN =
=
= 0.902
s v0
125.0
¢
1.2 +
1.2 +
Pa
101
From Table 9.5, given the safety hammer, the bore hole diameter of 3 inch (or 76.2 mm),
and the rod length of 8 m, choose the correction factor for SPT hammer energy ratio CE =
1.0, the correction factor for borehole diameter CB = 1.0, the correction factor for SPT
rod length CR = 0.95, and the correction factor for samplers with liner CS = 1.1 (assume a
liner is not used).
So:
(N1)60 = 17  0.902  1.0  1.0  0.95  1.1 = 16
Then calculate the equivalent clean-sand SPT blow count (N1 ) 60:
(N1)60CS = a + b (N1 )60
Since fines content (FC) = 15%,
a = exp éë1.76 - (190 / FC 2 ) ùû = 2.498
b = éë 0.99 + ( FC1.5 /1000 ) ùû = 0.99 + (151.5 /1000 ) = 1.048
( N1 )60CS = a + b ( N1 )60 = 2.498 +1.048 ´16 = 19.27
From Figure 9.22, find CRR7.5 = 0.21.
(3) Calculate MSF:
From Figure 9.24, at M=6.5, the MSF is chosen as the middle value of the hatched area:
MSF = 1.5
(4) Factor of safety against liquefaction:
æ CRR7.5 ö
æ 0.21 ö
FS = ç
MSF = ç
´1.5 = 2.08 > 1.3
÷
è CSR ø
è 0.151 ÷ø
Conclusion: The SPT analysis concludes that the site will not liquefy under the design
earthquake.
16. A hospital is built in an earthquake-prone region. To mitigate the liquefaction hazard,
dynamic compaction is performed to improve the soil. Then the cone penetration test
is conducted to evaluate the mitigation effort. The peak ground horizontal
acceleration is 0.35 g and the design earthquake magnitude is 8.0. The design depth
of the hospital’s foundation is at 6 m; at this depth, the cone penetration resistance at
the tip is 11 MPa and the sleeve resistance is measured to be 200 kPa. Assume the
groundwater table is at the ground surface, and saturated unit weight is 19.6 kN/m3.
Calculate the factor of safety against liquefaction.
Solution:
(1)
CSR = 0.6
amax s v0
r
g s v0
¢ d
s v0 = 20´6 = 120 kN/m2
(
)
s v0
¢ = 20- 9.81 ´6 = 61.1 kN/m2
rd = 1.0-0.00765´6 = 0.954
amax s v0
0.35g 120
rd = 0.65´
´
´ 0.954 = 0.407
g s v0
g
61.1
¢
(2) CRR using CTP method:
q
qc1N = CQ c
pa
CSR = 0.6
qc = 11´103 kN/m2
æ p ö
CQ = ç a ÷
è s v¢0 ø
n
pa = 101 kN/m2
s v0
¢ = 61.6 kN/m2
(
) (
)
2
2
Ic = é 3.47- logQ + 1.22+ log F ù
êë
úû
q - s v0 æ pa ö
Q= c
pa çè s v0
¢ ÷ø
0.5
n
æ fs ö
F =ç
÷ ´100%
è qc - s v0 ø
f s = 200 kPa
To find n:
(1) Assume n = 1.0
n
q - s v0 æ pa ö 11´103 -120 æ 101 ö
Q= c
=
çè 61.1 ÷ø
pa çè s v0
101
¢ ÷ø
1.0
= 178
æ fs ö
200
F =ç
´100% =
´100% = 1.838%
÷
11000-120
è qc - s v0 ø
(
) (
)
2
2
Ic = é 3.47 - logQ + 1.22+ log F ù
úû
ëê
(
0.5
) (
)
0.5
2
2
= é 3.47 - log178 + 1.22+ log1.838 ù
êë
úû
= 1.92 < 2.6
Since Ic < 2.6, the soil is most likely granular, use n = 0.5, recalculate Ic
(2) Use n = 0.5:
n
q - s v0 æ pa ö 11´103 -120 æ 101 ö
Q= c
=
çè 61.1 ÷ø
pa çè s v0
101
¢ ÷ø
0.5
= 138
æ fs ö
200
F =ç
´100% =
´100% = 1.838%
÷
11000-120
è qc - s v0 ø
(
) (
)
2
2
Ic = é 3.47 - logQ + 1.22+ log F ù
êë
úû
(
0.5
) (
)
2
2
= é 3.47 - log138 + 1.22+ log1.838 ù
úû
ëê
= 1.992 < 2.6
n
æ p ö æ 101 ö
CQ = ç a ÷ = ç
¢ ø è 61.1 ÷ø
è s v0
qc1N = CQ
So:
0.5
0.5
= 1.285
qc
11000
= 1.285´
= 140.0
pa
101
(q )
c1N cs
= K c qc1N
and:
K c = -0.403Ic 4 +5.581Ic 3 - 21.63Ic 2 + 33.75Ic -17.88
= -6.345+ 44.114 - 85.829+66.230-17.88
= 1.29
(q )
c1N cs
= K cqc1N = 1.29´140.0 = 180.6
From Figure 9.25, find CRR7.5 = 
æ CRR7.5 ö
So: FS = ç
÷ MSF = ¥
è CSR ø
Conclusion: No liquefaction.