Soil Mechanics

Practice Problems PE Exam
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Part 3 : Soil Mechanics

136 Problems
284
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1) A continuous foundation is given in the following diagram. Determine the

distance of . Given equation: = + + = / .
= / , = / , = / , = , Friction angle=20
A) 10.35m B) 13.5m C) 13.0m D) 12.5m
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The Answer is A
Step 1:From this problem, the equation of bearing capacity is given as,
1
q = c N + qN + ΥBN
2
Vertical effective stress at the level of the foundation can be given,
q = γD = (16 kN/m )D = 16D
Substitute all known values into the equation of bearing capacity, we can get the
equation shown below
N , N = 5Nɣ =6.5
1
q = c N + qN + ΥBN
2
= (20kN/m )(15) + (16D )(6.5)
+ (0.5)(18kN/m − 9.8kN/m )(6m)(5) = 1500kN/m
Solve this equation,

D = 10.35m~10.35m
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2. A continuous foundation is given in the following diagram. Determine the

value of B. Given equation: q = c N + qN + ΥBN = 2300kN/m . γ =
16kN/m ,γ = 18kN/m , c = 20kN/m , Friction angle=30
= .
A) 4.6m B) 3.85m C) 5.2m D) 6.2m
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The Answer is B
1
q = c N + qN + ΥBN
2
q = γD = (16 kN/m )(4m) = 64kN/m
N =30, ɣ =22, =18
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1
q = c N + qN + ΥBN
2
= (20kN/m )(30) + (64kN/m )(18)
+ (0.5)(18kN/m − 9.8kN/m )B(22) = 2100kN/m

B = 3.85m~3.85m
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3. A continuous square foundation is given in the following diagram. Determine

the area of this square foundation. Given equation: q = 1.3c N + qN +
0.5ΥBN = 5500kN/m . γ = 16kN/m ,γ = 18kN/m , c = 15kN/m ,
D = 4m.Friction angle=37.5
A) 40m2 B) 48m2 C) 27m2 D) 36m2
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The Answer is C
q = 1.3c N + qN + 0.4ΥBN
q = γD = (16 kN/m )(4m) = 64kN/m
N =58, Nɣ =70, N =45
q = 1.3c N + qN + 0.4ΥBN
= (1.3)(15kN/m )(58) + (64kN/m )(45)
+ (0.5)(18kN/m − 9.8kN/m )B(70) = 5500kN/m
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B = 5.2m
The area of this foundation can be calculated by,
A = B = (5.2m) = 27m ~27m2
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4. A continuous foundation is given in the following diagram. Determine the

value of γ . Given equation: q = c N + qN + ΥBN = 350kN/m .
γ = 16kN/m B = 6m, c = 20kN/m ,. D = 4m.
Friction angle=10.0
A) 19kN/m3 B) 18kN/m3 C) 16kN/m3 D) 16.5kN/m3
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The Answer is D
1
= + +
2
= = (16 / )(4 ) = 64 /
1
q = c N + qN + ΥBN
2
= (20kN/m )(8.5) + (64kN/m )(2.5)
+ (0.5)(γ − 9.8kN/m )(6m)(1) = 350kN/m

γ = 16.5kN/m~16.5kN/m3
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5)A continuous foundation is given in the following diagram. Determine the value
of . Given equation: = + + = / . =
/ = , = / , = .
Friction angle=20.0
A) 19kN/m3 B) 18kN/m3 C) 10.6kN/m3 D) 17kN/m3
295
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The Answer is C
1
q = c N + qN + ΥBN
2
q = γD = γ(4m) = 4γ
1
q = c N + qN + ΥBN
2
20kN
=( )(15) + (4γ)(6.5)
m
+ (0.5)(18kN/m − 9.8kN/m )(6m)(5.0) = 700kN/m

γ = 10.6kN/m~10.6kN/m3
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6)A circular foundation is given in the following diagram. It is known that this
foundation subjects three loads as dead and live load, , self-weight load, ,
and soil load, . Assume that the size of this foundation is given as 4m.
= , = , = . Determine the gross allowable
bearing capacity of this foundation.
A) 68kN/m2 B) 72kN/m2 C) 54kN/m2 D) 63kN/m2
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The Answer is A
Step 1:From this problem, the equation of gross allowable bearing capacity of this
foundation is given as,
W +W +W
q =
A
Step 2: The area of this foundation can be calculated based on the information of
size, B, as,
πB π(4m)
A= = = 12.6m
4 4
Step 3: Solve the gross allowable bearing capacity of this foundation,

W +W +W 300kN + 100kN + 450kN
q = = = 67.5kN/m
A 12.6m
~68kN/m2
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7)A square foundation is given in the following diagram. It is known that this
and soil load, . Assume that the maximum gross allowable bearing capacity of
this foundation is 70kN/m2. = , = , = .
Determine the size of this foundation.
A) 4.0m B) 3.5m C) 3.0m D) 4.5m
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The Answer is B
W +W +W
q =
A
size, B, as,
A=B

W +W +W 300kN + 100kN + 450kN
q = = = 70kN/m
A B
Solve this equation we got,
B = 3.5m~3.5m
300
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8)A triangle foundation is given in the following diagram. The length of each side
of this triangle is the same. It is known that this foundation subjects three loads as
dead and live load, , self-weight load, , and soil load, . Assume that
the maximum gross allowable bearing capacity of this foundation is 70kN/m2.
= , = , = . Determine the size of this
foundation.
A) 4.0m B) 3.5m C) 5.3m D) 4.5m
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The Answer is C
W +W +W
q =
A
size, B, as,
√3
A= B
4

W +W +W 300kN + 100kN + 450kN
q = = = 70kN/m
A √3
B
4
B = 5.3m~5.3m
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this foundation is 70kN/m2. = , = . Determine the
maximum self-weight of this foundation if size of it is 4m.
A) 370kN B) 520kN C) 450kN D) 320kN
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The Answer is C
+ +
=
size, B, as,
= =( ) =

+ + + +
= = = /

= ~370kN
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10)A circular foundation is given in the following diagram. It is known that this
this foundation is 70kN/m2. = , = . Determine the
maximum self-weight of this foundation if size of it is 4m.
A) 150kN B) 130kN C) 170kN D) 110kN
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The Answer is B
+W +
=
size, B, as,
( )
= = = .

+ + + +
= = = /
.
= ~130kN
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this foundation is 70kN/m2. Dead load is 200kN, = . = .
Determine the maximum live load this foundation can carry if size of it is 4m.
A) 250kN B) 230kN C) 270kN D) 210kN
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The Answer is C
+ +
=
size, B, as,
= =( ) =

+ + + +
= = = /

=
Since we know the value of dead load, we can calculate the maximum live load
as,
= − = − = ~270kN
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12)A continuous foundation is given in the following. It is known that the

ultimate load, = / . If the unit weight of soil is assumed to be
3
16kN/m , = , factor of safety is 3. Determine the net allowable bearing
capacity of this foundation.
A) 454kN/m2 B) 479kN/m2
C) 433kN/m2 D) 542kN/m2
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The Answer is B
Step 1:Vertical effective stress at the level of the foundation can be calculated by,
= =( / )( )= /
Step 2: Net ultimate load can be calculated by subtracting vertical effective stress
from the ultimate load as,
( ) = − = / − / = /
Step 3: Net allowable bearing load can be calculated by,

/
~479kN/m2
( )
( ) = = = /
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ultimate load, = / and net allowable bearing load is 450kN//m2. If
the unit weight of soil is assumed to be 16kN/m3, factor of safety is 3. Determine
the value of .
A) 7.6m B) 8.3m C) 9.4m D) 10.2m
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The Answer is C
= =( / ) =
( ) = − = / −

( ) / −
( ) = = = /
Solve the equations above we got,

= . ~9.4m
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the unit weight of soil is assumed to be 16kN/m3, = . Determine the factor
of safety.
A) 2 B) 3 C) 2.5 D) 3.5
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The Answer is B
= =( / )( )= /
( ) = − = / − / = /

( ) /
( ) = = = /

= 3. ~3
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the factor of safety is 3.1, = . Determine the unit weight of this soil.
A) 17.5kN/m3 B) 16.4kN/m3
C) 15.2kN/m3 D) 18.6kN/m3
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The Answer is A
= = ( )=
( ) = − = / −

( ) / −
( ) = = = /
.

= . / ~17.5kN/m3
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16.A standard penetration test (SPT) was conducted on a ground to determine the
bearing capacity of soil. It was measured that the number of blows needed for the
sampler to penetrate each 6-inch depth is 12, 14, 16, and 18 for the 1st, 2nd, 3rd,
and 4th 6-inch depth increment. What is the SPT resistance (N-value)?
A) 12 B) 14 C) 16 D) 30
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The Answer is D
Step 1:
In a SPT, the sum of the number of blows required for the second and third 6-inch
penetration is termed the "standard penetration resistance" or the "N-value".
Step 2:
From the given information, it is known that the number of blows for the 2nd and
3rd penetration of 6-inch depth is 14 and 16, respectively. Therefore, the N-value
is
+ =
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17) As shown in the figure below, a clay layer exists between two sandy soil
layers. The underground water level is 1 m below the ground. The second sandy
soil layer contains confined water. At a point A, which is 7 m below the ground
(in the second sandy soil layer), the head of water pressure is 1 m above the
ground. It is known that the unit weight of soil above the water level is 16.5
kN/m3, the saturated unit weight of the first layer of sandy soil below the water
level is 19.2 kN/m3, the saturated unit weight of the second layer of sandy soil is
20.2 kN/m3, the saturated unit weight of the clay soil is 18.4 kN/m3, Determine
the pore water pressure at point B, which is in the middle of the clay layer.
H0=1m
1
H1=2 m
W. L. Sandy soil 1
H2=1m 1sat
H3=2 m 2 B Clay
H5=1 m
H4=2 m 3sat
Sandy soil 2
A
A) 0 kPa B)23 kPa C) 35 kPa D) 51 kPa
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The Answer is D
Since the second sandy soil layer contains confined water, and it can be seen that
the head of water pressure at point A exceeds the groundwater level, the clay
layer is an aquiclude layer, which is impermeable. Therefore, the pore water
pressure at point B is zero.
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18) As shown in the figure below, a clay layer exists between two sandy soil
layers. The underground water level is 1 m below the ground. The second sandy
soil layer contains confined water. At a point A, which is 7 m below the ground
(in the second sandy soil layer), the head of water pressure is 1 m above the
ground. It is known that the unit weight of soil above the water level is 16.5
kN/m3, the saturated unit weight of the first layer of sandy soil below the water
level is 19.2 kN/m3, the saturated unit weight of the second layer of sandy soil is
20.2 kN/m3, the saturated unit weight of the clay soil is 18.4 kN/m3, Determine
the effective vertical soil pressure at point A.
H0=1m
1
H1=2 m
W. L. Sandy soil 1
H2=1m 1sat
H3=2 m 2 Clay
H4=2 m 3sat
Sandy soil 2
A
A) 23 kPa B) 35 kPa C) 51 kPa D) 88 kPa
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The Answer is C
Effective stress  is equal to the total stress  minus pore water pressure
. =  − .
Step 1:
The total stress  is calculated by = + + + =
( . / )( )+( . / )( )+( . / )( )+
( . / )( )= .
Step 2:
The pore water pressure is calculated by = =( . / )( +
+ + + )= .
Step 3:
The effective stress  is
=− = . − . =
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19) Determine the active lateral earth pressure on the frictionless wall, as shown
in the figure below, at the base of the wall. The soil behind of the wall is clean
sand with a saturated unit weight of 18.0 kN/m3 and angle of internal friction of
32. Groundwater level is at the surface of the ground.
Groundwater Level
sat =18 kN/m3

H=4 m
=32
A) 5.2 kPa B) 10.1 kPa C)16.3 kPa D) 18.5 kPa
323
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The Answer is B
Step 1:
Based on the Rankine theory, the Rankine active earth pressure coefficient
°
= °− = °− = . .
Step 2:
The vertical effective stress at the base of the wall, , is
= − =( − ) = − . ( )= .
Step 3:
The active lateral earth pressure at the base of the wall, , is
= = . × . = .
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20) Determine the lateral force (per unit length of wall) on the frictionless
retaining wall, as shown in the figure below. The soil behind of the wall is clean
sand with a saturated unit weight of 20.0 kN/m3 and angle of internal friction of
36. Groundwater level is at the surface of the ground.
Groundwater Level
sat =20 kN/m3

H=6 m
=36
A) . / B) . /
C) . / D) . /
325
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The Answer is C
Step 1:
°
= °− = °− = . .
Step 2:
At the surface of the ground, the vertical effective stress and the pore water
pressure are both zero, = = .
At the base of the wall,
= − =( − ) = − . ( )= .
= = . ( )= .
Step 3:
The active lateral earth pressure at the base of the wall, , is
= = . × . = .
Step 4:
Based on the distributions of lateral earth pressure and pore water pressure,
15.9 kPa 58.8 kPa

Lateral earth pressure Hydrostatic pressure
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the lateral force (per unit length of wall) on the wall, , is

1
= + = + = ( + )
= ( . + . )( )= . /
where is the lateral force due to soil solids and is the lateral force due to the
pore water.
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21) A 10-inch diameter steel pile is driven 60 ft into insensitive clay, which has an
undrained shear strength = / . The groundwater table is at the
ground surface. Assume the entire pile length is effective, the base resistance
= and the adhesive stress = . , determine the allowable bearing
capacity of the pile for a factor of safety = .
A) 20 kips B) 33 kips C) 60 kips D) 72 kips
328
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The Answer is B
Step 1:
The ultimate bearing capacity of the pile due to friction, , is the product of the
adhesive stress, , and the surface area of the shaft.
= . = . =
= = = ( )≅
= ≅ .
Step 2:
The ultimate bearing capacity of the pile due to end bearing, , is the product of
the base resistance, , and the base area.
= = =
( )
= = ≅ =
≅ .
Step 3:
The allowable bearing capacity is
+ . + .
= = ≅ .
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22) For an over consolidated clay of 3 m thickness, the following data are given:
average effective pressure = ; initial void ratio = . ; average
increase of pressure in the clay layer ∆ = ; compression index =
. ; recompression index = . ; past maximum consolidation stress
= . Determine the primary consolidation settlement of the clay.
A) 0.020 m B)0.040 m C) 0.060 m D) 0.080 m
330
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The Answer is A
Step 1:
Since the clay is over consolidated, the final
effective pressure needs to be compared with the
past maximum consolidation stress to determine
which compression index should be used. Given the
initial average effective pressure = and
pressure increase ∆ = , the final effective
pressure is = +∆ = + = < = .
Therefore, the clay is in recompression under the increased pressure.
Recompression index = . should be used.
Step 2:
The primary consolidation settlement of the clay is
∆ ( . )
= = = =( )
+ + + .
≅ .
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23) As shown in the figure below, a clay layer of thickness 10 m is above a gravel
layer. The groundwater level is at the surface of the clay layer. The clay is over
consolidated and the past maximum consolidation stress is = . If a 3 m
thick fill layer is placed above the clay layer, determine the settlement of the fill
layer due to the primary consolidation of the clay layer.
Ground Surface
H1=3 m W.L. Fill, =20 kN/m3
H2=10 m Clay, sat=18 kN/m3

e0=1.2, CR=0.16, Cc=0.34
H3=2 m Gravel, sat=24 kN/m3
Rock
A) 0.4 m B) 1.2 m C) 2.2 m D) 3.4 m
332
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The Answer is A
Step 1:
Before the fill layer is placed, the vertical stress in the clay is due to the self
weight of the soil, which increases linearly with depth. Since the clay is saturated,
the effective vertical stress at center of the clay layer is
= ( − )= ( ) − . =
Step 2:
After the fill layer is placed, the effective vertical stress at center of the clay layer
is
= + = +( ) =
Step 3:
Since the clay is over consolidated, before the effective vertical stress reaches the
past maximum consolidation stress, = , the settlement of the fill layer
is
Continuing:
∆
= =
+
=
+
( . )
=( )
+ .
= .
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Step 4:
After the effective vertical stress passes the past maximum consolidation stress
and reaches the final effective vertical stress, the settlement of the fill layer is
∆
= = =
+ +
( . )
=( )
+ .
= .
Step 5:
Therefore, the total settlement of the fill layer is
= + = . + . = .
334
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24) An embankment constructed with a clay soil

has a height = , and a slope angle
= °. The soil has a cohesion = ,
unit weight = / , and friction angle
= °. A potential failure plane is an arc with its
center at the edge of the embankment, O, as
shown in the figure. Determine the factor of safety against slope rotational
instability, FS, for this plane.
A) . B) . C) . D) .
335
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The Answer is D
Step 1:
The radius of the potential failure plane is
( )
= = ≅ .
( ) ( °)
O
 x 
H WM
TFF
Step 2:
Let be the angle that bisects the top slope angle, then = °− =
°− °= °. The weight of the sliding soil block is
° °
= = = = ( . )
° ° °
≅ . N/
Step 3:
The centroid of a circular sector is on the line that bisects the sector at the distance
( ) ( . ) ( °)
from the center = = ° ≅ .
°
Step 4:
Use the arc center O as the point of rotation, the mobilized moment is
= ( )= . ( . ) ( °)
= ∙ /
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Step 5:
Since the soil has a zero friction angle, the resistance moment along the potential
failure plane is
= = = ( )
× °
=( ) ( . ) ( . )
°
= ∙ /
Step 6:
The factor of safety against slope instability, FS is
∙ /
= = ≅ .
∙ /
337
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25) An embankment constructed with

a sandy clay soil has a height =  WM
, and a slope angle = °. The H
soil has a cohesion = , unit
s
weight = / , and friction Ls
angle = °. The factor of safety
against slope instability, FS, for a
plane that has an angle from the horizontal surface, is 3. Determine the angle
.
A) ° B) ° C) ° D) °
338
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The Answer is A
Step 1:
The top width of the sliding wedge is
( ) ( )
= − = −
( ) ( ) ( ) ( °)
Step 2:
The weight of the sliding soil wedge is
( ) ( )
= = = ( . ) − ( )
( ) ( °)
= −
( ) ( °)
Step 3:
Since the length of the slip surface is = / ( ), the shear resistance along
the slip plane is
= + ( ) ( )= / ( )+ ( ) ( )
( )( )
=
( )
+ − ( ) ( °)
( ) ( °)
Step 4:
The mobilized shear force along the slip plane is
= ( )= − ( )
( ) ( °)
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Step 5:
( )( )
+ − ( ) ( °)
( ) ( ) ( °)
=
− ( )
( ) ( °)
=
By trial and error, it is obtained that ≅ °
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26) An uniform desposit soil is shown in the following diagram. Information of

soil in each layer is given. Calculate the effective stress for a soil element at point
A.
Gs=2.6, S=53%, w = 30%,

H1=4m = /
H2=2m Gs=2.6, S=100%, w = 40%,

= / A
A) 90kPa B) 80kPa C) 85kPa D) 75kPa
341
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The Answer is C
Step 1:From the diagram, we can see that there are two layers of soil. For each
layer, we need to calculate the unit weight. Equation about unit weight is given
as,
+
=
+
The voids ratio is unknown. Therefore, we need to calculate voids ratio of each
layer first.
Step 2:Equation of void ratio of soil is shown as,
To determine the value of void ratio, we need to find volume of voids, , and
volume of solids, . To determine the voids ratio, we can deduce its equation by,
= = = = = =
Therefore, for each layer, the voids ratio can be calculated by,
%
= =( / ) = .
( . / )( %)
%
= =( / ) = .
( . / )( %)
Step 3:For the layer above groundwater level, the unit weight is equal to,
+ . + ( %)( . )
= = ( . / )= . /
+ + .
For the layer below groundwater level, the unit weight is equal to,
+ . +( %)( . )
= = ( . / )= . /
+ + .
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Step 4:The total stress at point A can be then calculated as,

= + =( . / )( )+( . / )( )= .
The pore water pressure is given as,
= =( . / )( )= .
Effective pressure at point A can be finally calculated by,

= − = . − . =
~85kPa
343
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soil in each layer is given. Calculate the dry unit weight of this soil if the effective
stress at point A is 85kPa.
Gs=2.6, S=53%, w = 30%,

H1=4m
H2=2m Gs=2.6, S=100%, w = 40%,

A
344
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The Answer is B
Step 1:From the diagram, we can see that there are two layers of soil. For each
layer, we need to calculate the unit weight. Equation about unit weight is given
as,
+
=
+
The voids ratio is unknown. Therefore, we need to calculate voids ratio of each
layer first.
Step 2:Equation of void ratio of soil can be shown as,
To determine the value of void ratio, we need to find volume of voids, , and
volume of solids, . To determine the voids ratio, we can deduce its equation by,
= = = = = =
Therefore, for each layer, the voids ratio can be calculated by,
%
= = =
( . / )( %)
%
= = =
( . / )( %)
Step 3:For the layer above groundwater level, the unit weight is equal to,
+ . +( %)( )
= = ( . / )
+ +
For the layer below groundwater level, the unit weight is equal to,
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+ . +( %)( )
= = ( . / )
+ +

= +
= =( . / )( )= .

= − = + − . =
. +( %)( )
= ( . N/ )( )
+
. +( %)( )
+ ( . / )( )− .
+
= /
~14kN/m3
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soil in each layer is given. Assume that the effective stress at point A is 100kPa,
determine the value of H2.
H1=3m e = 0.6, S = 2%, Gs = 2.6
H2 e = 0.55, S = 16%, Gs = 2.6
H3=2m
= / A
A) 2.00m B) 2.50m C) 3.00m D) 3.50m
347
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The Answer is A
Step 1:From the diagram, we can see that there are three layers of soil. For the
first layer above groundwater level, the unit weight is equal to,
+ . + ( %)( . )
= = ( . / )= /
+ + .
For the second layer above groundwater level, the unit weight is equal to,
+ . +( %)( . )
= = ( . / )= /
+ + .
For the third layer below groundwater level, the unit weight is equal to,
= /

= + +
=( / )( )+( / ) +( / )( )
=( + )
= =( . / )( )= .
= − =( + ) − . =
Solve the equation above,
= ~2m
348
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H1
e = 0.6, S = 2%, Gs = 2.6
H2=2m e = 0.55, S = 16%, Gs = 2.6
H3=2m
= / A
A) 2.00m B) 2.50m C) 3.00m D) 3.50m
349
____________________________________________________________
The Answer is B
+ . + ( %)( . )
= = ( . / )= /
+ + .
+ . +( %)( . )
= = ( . / )= /
+ + .
= /

= + +
=( / ) +( / )( )
+( / )( )=( + )
= ( )=( . / )( )= .
= − =( + ) − . =
= . ~2.5m
350
____________________________________________________________

H1=3m e = 0.6, S = 2%, Gs = 2.6
H2=2m e = 0.55, S = 16%, Gs = 2.6
H3
γ = 18 / A
A) 3m B) 4m C) 5m D) 6m
351
____________________________________________________________
The Answer is A
+ . + ( %)( . )
= = ( . / )= /
+ + .
+ . +( %)( . )
= = ( . / )= /
+ + .
= /

= + +
=( / )( )+( / ) +( / )( )
=( + )
= =( . / )( )= .
= − =( + ) − . =
= ~2m
352
____________________________________________________________

H1=3m γ = 16 /
H2 γ = 17 /
H3=2m
γ = 18 / A
A) 2.00m B) 2.50m C) 3.00m D) 3.50m
353
____________________________________________________________
The Answer is A
= /
= /
= /

= + +
=( / )( )+( / ) +( / )( )
=( + )
= =( . / )( )= .
= − =( + H ) − . =
= ~2m
354
____________________________________________________________

H1 γ = 16 /
H2=3m γ = 18 /
H3=2m
γ = 18 / A
A) 2.3m B) 2.5m C) 3.5m D) 3.3m
355
____________________________________________________________
The Answer is D
= /
= /
= /

= + +
=( / ) +( / )( )+( / )( )
=( + )
= ( + )=( . / )( + )= P
= − =( + ) − =
= . ~3.3m
356
____________________________________________________________

determine the value of H1, H2, and H3 if the ratio of them is 2:2:1.
H1 γ = 16 /
H2 γ = 18 /
H3 A
γ = 18 /
A) 4.2m;4.2m; 2.1m B) 3.8m;3.8m; 1.9m

C) 3.6m;3.6m; 1.8m D) 4.0m;4.0m; 2.0m
357
____________________________________________________________
The Answer is C
= /
= /
= /

= + +
=( / ) +( / ) +( / )
=( )
= ( + )=( . / )( + )=( . )
= − =( ) −( . ) =( . ) =
= .
Therefore,
= = . ~3.6m; 3.6m; 1.8m
358
____________________________________________________________

soil in each layer is given. Determine the effective pressure at point A.
H1=3m Water
Gs =2.7, e=0.8
H2=5m
A
359
____________________________________________________________
The Answer is B
Step 1:From the diagram, we can see that there is only one layer of soil
submerged under water. For this layer, the unit weight is equal to,
+ . + .
= = ( . / )= /
+ + .

= + =( . / )( )+( k / )( )= .
= ( + )=( . / )( + )= .

= − = . − . =
~46kPa
360
____________________________________________________________

soil in each layer is given. Assume that the maximum effective pressure at point
A is 50kPa, determine the value of H2.
H1=3m Water
Gs =2.7, e=0.8
H2
A
A) 5.8m B) 4.6m C) 5.4m D) 6.3
361
____________________________________________________________
The Answer is C
+ . + .
= = ( . / )= /
+ + .

= + =( . / )( ) + (1 / )
=( . + )
= ( + )=( . / )( + )=( . + . )

= − =( . + ) −( . + . ) =
= . ~5.4m
362
____________________________________________________________

A is 46kPa, determine the value of H1.
H1 Water
Gs =2.7, e=0.8
H2=5m
A) 5.8m B) 4.6m C) 5.4m D) Any valu
363
____________________________________________________________
The Answer is D
+ . + .
= = ( . / )= /
+ + .

= + =( . / )( )+( / )( )= .
= ( + )=( . / )( + )= .

= − = . − . =
Therefore, the value of H1 will not influence the effective pressure at point A. The
answer D is correct.
364
____________________________________________________________

A is 200kPa, determine the value of surcharge.
H1=4m γ = 16kN/m
H2=5m γ = 18kN/m
365
____________________________________________________________
The Answer is B
Step 1:From the diagram, we can see that there are two layers of soil. For the first
layer, the unit weight is equal to,
= /
For the second layer, the unit weight is equal to,
= /
Step 2:The effective stress at point A can be then calculated as,

= + + = +( / )( )+( / )( )
=( + )
According to the problem, the effective pressure at point A is equal to

200kPa, the value of surcharge can be determined by,
=( + ) =
= ~46kPa
366
____________________________________________________________

soil in each layer is given. Assume that value of surcharge is 50kPa. Determine
the effective prssure at point A.
H1=4m γ = 16kN/m
H2=5m γ = 18kN/m
A) 100kPa B) 250kPa C) 200kPa D) 150Kp
367
____________________________________________________________
The Answer is C
= /
= /

= + + = +( / )( )+( / )( )=
~200kPa
368
____________________________________________________________

soil in each layer is given. At first, there is no surcharge at the top of soil. If H2 is
reduced as 0.5H2 and in order to keep constant effective pressure at point,
surcharge is added at top. Determine the value of this surcharge.
H1=4m γ = 16kN/m
H2=5m γ = 18kN/m
369
____________________________________________________________
The Answer is A
= /
= /
Step 2:Before changing the value of H2, the effective stress at point A can be then
calculated as,
= + =( / )( )+( / )( )=
After reducing the value of H2, the effective stress at point A can be calculated as,
= + + = +( / )( )+( / )( )
=
Solve equation above, =
~45kPa
370
____________________________________________________________

soil in each layer is given. Assume that the surcharge is 30kPa acting at the top of
soil as shown in the diagram. If the effective pressure at point A is 180kPa,
determine the effective density of second layer, .
H1=4m γ = 16kN/m
H2=5m γ
A) 16.8kN/m3 B) 18.4kN/m3
C) 17.2kN/m3 D) 19.6kN/m3
371
____________________________________________________________
The Answer is C
= /
For the second layer, the unit weight is unknown,

= + + = +( / )( )+ ( )=
Solve the equation above, we can find the effective density of second layers
as,
= . /
~17.2kN/m3
372
____________________________________________________________

soil in each layer is given. Assume that the surcharge is 45kPa acting at the top of
soil as shown in the diagram. If the effective pressure at point A is 200kPa,
determine the effective density of first layer, .
H1=6m γ
H2=4m γ = 18kN/m
A) 15.6kN/m3 B) 12.6kN/m3
C) 14.4kN/m3 D) 13.8kN/m3
373
____________________________________________________________
The Answer is D
layer, the unit weight is unknown,

= /

= + + = + ( )+( / )( )=
Solve the equation above, we can find the effective density of second layers
as,
= . /
~13.8kN/m3
374
____________________________________________________________
42) Information of a soil is shown in the following diagram. Assume that the
surcharge is 40kPa acting at the top of soil as shown in the diagram. Determine
the effective pressure at point A.
H1=7m G = 2.7, e = 1.0, w = 30%
375
____________________________________________________________
The Answer is A
Step 1:From the diagram, we can see that there is only one layer of soil. For this
+
=
+
( %)( . )
, = = = %, therefore,
.
+ . +( %)( . )
= = ( . / )= /
+ + .

= + = +( / )( )=
~160kPa
376
____________________________________________________________
surcharge is 40kPa acting at the top of soil as shown in the diagram. If the
effective pressure at point A is 150kPa. Determine the water content of this soil.
H1=6m G = 2.7, e = 1.0
A) 15.6% B) 14.2% C) 13.3% D) 12.1%
377
____________________________________________________________
The Answer is C
Step 1:From the diagram, we can see that there is only one layer of soil. The
effective pressure at point A can be calculated by,
= + = + ( )=
Solving this equation, we can know the value of density of this soil as,
= /
Step 2: According to equations of the density of this soil can be described as,
+
=
+
Therefore,
+ . +
= = ( . / )= /
+ +
= %
According to the following equation water content of soil is,
( %)( . )
= = = . %
.
~13.3%
378
____________________________________________________________
value of surcharge is unkown, which acting at the top of soil as shown in the
diagram. If the effective pressure at point A is 170kPa, determine the value of this
surcharge.
H1=8m G = 2.7, e = 1.0, w=20%
379
____________________________________________________________
The Answer is B
+
=
+
( %)( . )
.
+ . +( %)( . )
= = ( . / )= . /
+ + .

= + = +( . / )( )=
Solving this equation we got,
= .
~43kPa
380
____________________________________________________________
value of surcharge is 25kPa, which acting at the top of soil as shown in the
diagram. If the effective pressure at point A is 140kPa, determine the value of H1.
H1 G = 2.6, e = 1.0, w=10%
A) 8.2m B) 7.4m C) 6.6m D) 5.8m
381
____________________________________________________________
The Answer is A
+
=
+
( %)( . )
.
+ . +( %)( . )
= = ( . / )= /
+ + .

= + = +( / ) =
Solving this equation we got,
= .
~8.2m
382
____________________________________________________________
value of surcharge is 35kPa, which acting at the top of soil as shown in the
diagram. If the effective pressure at point A is 160kPa, determine the value of
specific gravity of this soil.
H1=8m e = 1.0, S=36%
A) 2.5 B) 2.6 C) 2.7 D) 2.8
383
____________________________________________________________
The Answer is D
Step 1:From the diagram, we can see that there is only one layer of soil. The
effective pressure at point A can be calculated by,
= + = + ( )=
Solving this equation, we can know the value of density of this soil as,
= . /
Step 2: According to equations of the density of this soil can be described as,
+
=
+
Therefore,
+ +( %)( )
= = ( . / )= . /
+ +
= . ~2.8
384
____________________________________________________________
47) Assume that there are two soils shown in the following diagram. For the left
one, there is no surcharge. For the right one, a surcharge acting on the top of soil
but the value is unknown. Assume that effective pressure at point A is the same as
that at point B. Determine the value of this surcharge.
H2=6m = /
H1=8m γ = 18kN/m
B
A
385
____________________________________________________________
The Answer is C
Step 1:From the diagram, we can see that there is only one layer of soil on both
right and left soil. The effective pressure at point A can be calculated by,
= =( / )( )=
The effective pressure at point B can be calculated by,
= + = +( / )( )=( + )
Step 2: According to the problem, effective pressure at point A is the same as that
at point B, so we got
4 =( + )
= ~25kPa
386
____________________________________________________________
and the value is 30kPa. Assume that effective pressure at point A is the same as
that at point B. Determine the value of H2.
H2 = /
H1=8m γ = 18kN/m
B
A
A) 4.3m B) 5.8m C) 6.7m D) 7.4m
387
____________________________________________________________
The Answer is C
= =( / )( )=
= + = +( / ) =( + )
=( + )
= . ~6.7m
388
____________________________________________________________
that at point B. Effective pressure at point A is 90kPa. Determine the ratio
between H1 and H2.
H2 = /
H1 γ = 18kN/m
B
A
A) 5:3 B) 3:5 C) 4:3 D) 3:4
389
____________________________________________________________
The Answer is A
= =( / ) =( )
= + = +( / ) =( + )
( ) =( + ) =
=
=
Therefore, the ratio between H1 and H2 is 5:3~5:3
390
____________________________________________________________
that at point B. Determine the densiy of the left soil, .
H2=2m = /
H1=4m γ
B
A
391
____________________________________________________________
The Answer is B
= = ( )=( )
= + = +( / )( )=
( ) =
= . /
~19kN/m3
392
____________________________________________________________
and the value is 40kPa. Determine the effective pressure ratio between A and B.
H2=4m = /
H1=4m = 18 /
B
A
A) 1:1 B) 1:3 C) 2:3 D) 4:3
393
____________________________________________________________
The Answer is C
= =( / )( )=
= + = +( / )( )=
Step 2: Therefore, the effective pressure ratio between A and B is,

: = : = :
~2:3
394
____________________________________________________________
52) There is a frictionless wall shown in the following figure. The information is
provded in this figure. Assume that q=20kN/m. Calculate the value of H so that
the moment at point A is zero.
= 28 /
q = 50 /
G = 2.6 = 0.9 P1 P2 H
A) 4.2m B) 4.8m C) 3.8m D) 3.2m
395
____________________________________________________________
The Answer is B
Step 1:From the diagram, we can see that there are two horizontal forces acting
towards the wall. We need to calculate these forces first. From the problem, the
friction angle of this soil can be found by,
/
= = =
/
Rankine active earth pressure coefficient can be calculated by,
= − = − =
Step 2: The saturated unit weight can be calculated by,

( + ) ( . + . )( . / )
= = = /
+ + .
Step 3:The lateral forces due to soil and pore water can be calculated respectively
as,
= ( − ) = ( / − . / )
=( . )
= = ( . / ) =( . )
Step 4:In order to make the moment at point A zero we got,
= + =( . ) +( . ) = ( / )
396
____________________________________________________________
= .
397
____________________________________________________________
provded in this figure. Assume that H1 is 5m and H2 is 3m. Calculate the value of
resultant force towards this wall.
G = 2.7 = 0.8
= G = 2.6 = 0.9
= 30
P1 P2 H1
P4 P3
H2
A) 12kN B) 14kN C) 16kN D) 18
398
____________________________________________________________
The Answer is A
Step 1:From the diagram, we can see that there are two parts of soil acting
towards wall. We need to calculate forces from both parts first. From the problem,
Rankine active earth pressure coefficient for right part can be calculated by,
= − = − =
Rankine passive earth pressure coefficient for left part can be calculated by,
= + = + =

( + ) ( . + . )( . / )
= = = /
+ + .
( + ) ( . + . )( . / )
= = = /
+ + .
Step 3:The lateral forces for right part can be calculated by,
= ( − ) = ( / − . / )( ) =
= = ( . / )( ) = .
The lateral forces for the left part can be calculated by,
= ( − ) = ( )( / − . / )( )
= .
= = ( . / )( ) = .
Step 4: The resultant force for this wall can be calculated by,
= + − − = . + . − − .
= .
399
____________________________________________________________
~12kN
400
____________________________________________________________
provded in this figure. Assume that H1 is 3m. H is 6m. F = 40kN. If moment at
point A in the left figure is the same as that in the right figure, calculate the value
of H2.
F
= 30
H1
1
H
2 H2
3 4
A) 3m B) 4m C) 5m D) 6m
401
____________________________________________________________
The Answer is A
Step 1:Rankine active earth pressure coefficient can be calculated by,
= − = − =

( + ) ( . + . )( . / )
= = = /
+ + .
Step 3: Lateral force for Area 1,
= = ( / )( ) = .
= + = ( )+ = +
Lateral force for Area 2,
= = ( / )( ) =
= = .
= ( − ) = ( / − . / )
= = ( . / )( )
402
____________________________________________________________
Step 4:The moment at point A can be calculated by,

= + + +
=( . )( + )+( )( . ) + 1.
+ . =( )( )

= ~3m
403
____________________________________________________________
55) Two figures show two frictionless wall as below. Information about each wall
is given. Assume that H1=4m and H2=5m, H3=1m, S=20kN/m. Assume that
moment at point A for left figure and right figure is the same. Determine the
density of soil in the right figure, .
= 16 /
= 30 = H2
1 H1 3
2
H3
A A
A) 16kN/m3 B) 18kN/m3 C) 20kN/m3 D) 2
404
____________________________________________________________
The Answer is A
= − = − =
= = ( / )( )= .
= = ( )=
= = ( / )( ) = .
= = ( )= .
= − = ( ) − ( . / )( )
= . − .
= = ( )= .
Step 2:The moment at point A is the same. We got,

+ = =( . )( )+( . )( . )
=( . − . )( . )
= . /
~16kN/m3
405
____________________________________________________________
56) A retaining wall is given in the following diagram. It is known H1=5m. q =

10kN/m. Determine the value of H2 when resultant force is equal to zero.
H1 q G = 2.6 = 0.9
= 30
H2
1 2
A) 2.8m B) 3.6m C) 4.2m D) 3.8
406
____________________________________________________________
The Answer is A
= − = − =

( + ) ( . + . )( . / )
= = = /
+ + .
= ( − ) = ( / − . / )( )
= . ( )
= = ( . / )( ) = . ( )
Step 3:The resultant force is equal to zero,

= + − = . ( ) + . ( ) −( / )( )=
= . ~2.8m
407
____________________________________________________________
moment at point A.
= /
M
= = 18 /
= 30
P1 P2 H1
P4 P3
H2
A) 80kN-m B) 70kN-m
C) 75kN-m D) 65kN-m
408
____________________________________________________________
The Answer is A
Step 1:From the diagram, we can see that there are two parts of soil acting towards
wall. We need to calculate forces from both parts first. From the problem, Rankine
active earth pressure coefficient for right part can be calculated by,
= − = − =
= + = + =
= ( − ) = ( / − . / )( ) =
= = ( . / )( ) = .
= = =
= ( − ) = ( )( / − . / )( ) = .
= = ( . / )( ) = .
= = =
Step 3: The moment at point A can be calculated by,

= + − −
=( . )( )+( . )( )−( )
−( . ) =− −
~80kN-m
409
____________________________________________________________
provded in this figure. Assume that H1 is 5m. Calculate the value of H2 so that
resultant force towards this wall is equal to zero.
= /
= = 18 /
= 30
P1 P2 H1
P4 P3
H2
A) 2.8m B) 3.0m C) 2.6m D) 2.
410
____________________________________________________________
The Answer is A
= − = − =
= + = + =
= ( − ) = ( / − . / )( ) =
= = ( . / )( ) = .
= ( − ) = ( )( / − . / ) = .
= = ( . / ) = .
Step 3: The resultant force for this wall can be calculated by,
= + − − = . + . − − . =
= .
~2.8m
411
____________________________________________________________
so that moment at point A is equal to zero.
M
= = 18 /
= 30
P1 P2 H1
P4 P3
H2
A) 22kN/m3 B) 20kN/m3
C) 24kN/m3 D) 26kN/m3
412
____________________________________________________________
The Answer is D
= − = − =
= + = + =
= ( − ) = ( / − . / )( ) =
= = ( . / )( ) = .
= = =
= ( − ) = ( )( − . / )( )
= = ( . / )( ) = .
= = =
Step 3: The moment at point A can be calculated by,

= + − −
= ( )( − . / )( ) ( )+( . )( )
−( ) −( . ) =

= . / ~26kN/m3
413
____________________________________________________________
provded in this figure. Assume that H1 is 3m and H2 is 3m. H is 6m. If moment at
of F.
F
= 17 /
= 30 H1
1
H
2 H2
3 4
A = / A
A) 35kN B) 45kN C) 40kN D) 30kN
414
____________________________________________________________
The Answer is C
= − = − =
= γ = ( / )( ) = .
= + = ( )+( )=
= = ( / )( ) =
= = ( )= .
= ( − ) = ( / − . / )( ) = . k
= = ( )=
= = ( . / )( ) = .
= = ( )=

= + + +
=( . )( )+( )( . )+( . )( )
+( . )( )= ( )
= . ~40kN
415
____________________________________________________________
of H2.
F
= 17 /
= 30 H1
1
H
2 H2
3 4
A = / A
A) 3m B) 4m C) 5m D) 6m
416
____________________________________________________________
The Answer is A
= − = − =
= = ( / )( ) = .
= + = ( )+ = +
= = ( / )( ) =
= = .
1
= ( − ) = ( / − . / )
= = ( . / )( )

= + + +
=( . )( + )+( )( . )+ .
+ . =( )( )

= ~3m
417
____________________________________________________________
of H1.
F
= 17 /
= 30 H1
1
H
2 H2
3 4
A = / A
A) 3m B) 4m C) 5m D) 6m
418
____________________________________________________________
The Answer is A
= − = − =
= = ( / )
= + = ( )+
= = ( / )( ) =
= = ( )= .
= ( − ) = ( / − . / )( ) = .
= = ( )=
= = ( . / )( ) = .
= = ( )=

= + + +
=( . ) ( )+ +( )( . )+( .8 )( )
+( . )( )=( )( )
= ~3m
419
____________________________________________________________
provded in this figure. Assume that H1 is 3m and H2 is 3m. H is 6m. F = 40kN. If
moment at point A in the left figure is the same as that in the right figure,
calculate the value of .
F
= 17 /
H1
1
H
2 H2
3 4
A = / A
A) 200 B) 350 C) 250 D) 300
420
____________________________________________________________
The Answer is D
= = ( / )( ) = .
= + = ( )+ =

= = ( / )( ) =
= = ( )= .
= (γ − ) = ( / − . / )( ) = .
= = ( )=
= = ( . / )( ) = .
= = ( )=

= + + +
=( . )( )+( )( . m) + ( . )( )
+( . )( )=( )( )
Step 3: Rankine active earth pressure coefficient can be found by,
= − =
421
____________________________________________________________
= ~300
422
____________________________________________________________
F
= 30 H1
1
H
2 H2
3 4
A = / A
423
____________________________________________________________
The Answer is B
= − = − =
= = ( ) = .
= + = ( )+ =
=K = ( ) =
= = ( )= .
= ( − ) = ( / − . / )( ) = .
= = ( )=
= = ( . / )( ) = .
= = ( )=

= + + +
=( . )( )+( )( . )+( . )( )
+( . )( )=( )( )
= / ~17kN/m3
424
____________________________________________________________
F
= 17 /
= 30 H1
1
H
2 H2
3 4
A A
425
____________________________________________________________
The Answer is C
= − = − =
= = ( / )( ) = .
= + = ( )+ =
= = ( / )( ) =
= = ( )= .
= ( − ) = ( − . / )( )
= = ( )=
= = ( .8 / )( ) = .
= = ( )=

= + + +
=( . )( )+( )( . )+ . ( − . ) ( )
+( . )( )=( )( )
= / ~19kN/m3
426
____________________________________________________________
is given. Assume that H1=4m and H2=5m. S=20kN/m. Assume that moment at
point A for left figure and right figure is the same. Determine the density of soil in
the right figure, .
= 16 /
= 30 = H2
1 H1 3
2
A A
A) 15kN/m3 B) 17kN/m3 C) 19kN/m3 D) 13kN/m
427
____________________________________________________________
The Answer is A
= − = − =
= = ( / )( )= .
= = ( )=
= = ( / )( ) = .
= = ( )= .
= = ( ) = .
= = ( ) = .7

+ = =( . )( )+( . )( . )
=( . )( . )
= . / ~15kN/m3
428
____________________________________________________________
is given. Assume that H1=4m and H2=5m. S=20kN/m. Assume that moment at
point A for left figure and right figure is the same and the moment is equal to
100kN-m. Determine the friction angle, .
S
= /
= 16 /
1 H2
H1 3
2
A A
A) 220 B) 440 C) 330 D) 550
429
____________________________________________________________
The Answer is B
Step 1:Lateral force for Area 1,

= = ( / )( )=
= = ( )=
= = ( / )( ) =
= = ( )= .
= = ( / )( ) = .
= = ( )= .

+ = =( )( )+( )( . )
=( . )( . )= −
= .
Step 3: Rankine active earth pressure coefficient can be found by,
= − = .

=
~330
430
____________________________________________________________
is given. Assume that H2=5m. S=20kN/m. Assume that moment at point A for left
figure and right figure is the same. Determine the value of H1.
S
= /
= 16 /
= 30 = H2
1 H1 3
2
A A
A) 3m B) 4m C) 5m D) 2m
431
____________________________________________________________
The Answer is B
= − = − =
= = ( / ) = .
= = (16 / )( ) = 2.7
1
=
3
1 1 1
= K = (15 / )(5 ) = 62.5
2 2 3
1 1
= = (5 ) = 1.7
3 3

1 1
+ = = (6.7 ) + 2.7 = (62.5 )(1.7 )
2 3
H =4
~4m
432
____________________________________________________________
69) A point A in a sandy soil is 5 m below the ground. The underground water
level is 2 m below the ground. It is known that the in situ total unit weight of the
soil above the water level is 14.6 kN/m3, and the saturated unit weight of soil
below the water level is 17.8 kN/m3. Determine the effective vertical soil pressure
at point A.
 (in situ)
H1=2 m
W. L.
H2=3 m
sat (below WL)
A) 23.5 kPa B) 35.2 kPa C) 53.2 kPa D) 72.4 kPa
433
____________________________________________________________
The Answer is C
Step 1:
u. =  − u.
Step 2:
The total stress is calculated by = γH + γ H = (14.6 kN/m )(2 m) +
(17.8 kN/m )(3 m) = 82.6 kPa
Step 3:
The pore water pressure u is calculated byu = γ H = (9.8 kN/m )(3 m) =
29.4 kPa
Step 4:
u =  − u = 82.6 kPa − 29.4 kPa = 53.2 kPa
434
____________________________________________________________
70) Deformations of soils are a function of
A) Effective stresses B) Total stresses

C) Pore water pressure D) All of the above
435
____________________________________________________________
The Answer is A
Based on theprincipal of effective stress, effective stress is equal to total stress
minus pore water pressure. It is the most important principle in soil mechanics.
Deformations of soils are a function of effective stresses not total stresses.
(A) is correct.
(B) is incorrect. Deformations of soils are not a function of total stresses.
(C) is incorrect.Pore water pressure is isotropic and only can cause volumetric
changes of soil solids, which is nearly incompressible. Pore water pressure does
not cause displacement of soil solids, or deformation of soils.
(D) is incorrect,since options (B) and (C) are incorrect.
436
____________________________________________________________
71) Apoint A is 5 m below the ground. The underground water level is 2 m below
the ground. There are two deposits of sandy soil at the site. The first deposit of
soil has a depth of 3 m, and a total unit weight (above the water level)of 13.2
kN/m3, a saturated unit weight of 16.5 kN/m3. The second deposit of soil has a
saturated unit weight of 17.6 kN/m3. Determine the effective vertical soil pressure
at point A.
1
H1=2 m
W. L. Sandy soil 1
H2=1m 1sat
H3=2 m 2sat
Sandy soil 2
A
437
____________________________________________________________
The Answer is B
u. =  − u.
Step 1:
The total stress  is calculated by = γ H + γ H +γ H = (13.2 kN/
m )(2 m) + (16.5 kN/m )(1 m) + (17.6 kN/m )(2 m) = 78.1 kPa
Step 2:
The pore water pressure u is calculated byu = γ H = (9.8 kN/m )(1 m +
2 m) = 29.4 kPa
Step 3:
u =  − u = 78.1 kPa − 29.4 kPa = 48.7 kPa
438
____________________________________________________________
73) The groundwater level falls at a site due to massive pumping. During such a
process, which of the following consequences may occur for the soil above the
pumping point?
A) Effective vertical stress in the soil increases, and ground level rises
B) Effective vertical stress in the soil decreases, and ground level falls
C) Effective vertical stress in the soil does not change, and ground level falls
D) Effective vertical stress in the soil increases, and ground level falls
439
____________________________________________________________
The Answer is D
Pumping of groundwater leads to water flow through soil, which exerts a
frictional drag on the soil particles resulting in head losses. Downward seepage
increases the resultant effective stress, while upward seepage decreases the
resultant effective stress. For the scenario of the problem, water in the soil above
the pumping point seeps downward, which increases the effective vertical stress.
With the effective vertical stress increasing, the vertical strain of the soil also
increases, which leads to a reduction of the ground level.
(A) is incorrect.The ground level should fall.
(B) is incorrect.The effective vertical stress in the soil should increase.
(C) is incorrect.The effective vertical stress in the soil should increase.
(D) is correct.
440
____________________________________________________________
74) Apoint A in a sandy soil is 8 m below the ground. The underground water
level is 3 m below the ground. It is known that the specific gavity of soil solids is
2.7, the water content of soil is 20% above the water level, and 35% below the
water content. Determine the effective vertical soil pressure at point A.
Gs=2.7, S=0.5
H1=3 m =20%
W. L.
=35%
H2=5 m
441
____________________________________________________________
The Answer is D
Step 1:
It is known that water content ω = × 100, specific gravity of soil solids
G = , void ratio e = , degree of saturation S = × 100, and water
volume V = . Therefore, Se = × 100 × = × 100, and G ω =
× × 100 = × 100. Therefore, Se = G ω.The void ratio of the soil
above the water level is

G ω 2.7 × 20%
e= = = 1.08
S 0.5
Step 2:
The unit weight of soil above the water level is
G + Se 2.7 + 0.5 × 1.08 kN kN
γ = γ = × 9.8 ≅ 15.27
1+e 1 + 1.08 m m
Step 3:
The void ratio of the soil below the water level is
G ω 2.7 × 35%
e= = = 0.945
S 1.0
The unit weight of soil below the water level is
G +e 2.7 + 0.945 kN kN
γ = γ = × 9.8 ≅ 18.37
1+e 1 + 0.945 m m
Step 4:
u. =  − u.
The total stress  is calculated by = γ H + γ H = (15.27 kN/m )(3 m) +
(18.37 kN/m )(5 m) = 137.66 kPa
442
____________________________________________________________
Step 5:
The pore water pressure u is calculated byu = γ H = (9.8 kN/m )(5 m) =
49 kPa
Step 6:
u =  − u = 137.66 kPa − 49 kPa = 88.66 kPa
443
____________________________________________________________
75. A 3 m high smooth retaining wall extends from the top of bedrock to the
ground surface. The soil behind of the retaining wall is homogeneous and
cohesionless, has an in-situ total unit weight of 16.8 kN/m3, and angle of
internal friction of 30. Based on the Rankine theory, what is the total active
resultant lateral earth force per unit length of retaining wall?
Retaining wall
H=3 m  (in situ)=16.8 kN/m3

=30
A) 25.2 kN/m B) 35.2 kN/m

C) 50.4kN/m D) 75.6kN/m
444
____________________________________________________________
The Answer is A
P136 NCEES Ref Manual edition 8 V2
Step 1:
°
= °− = °− = .
Step 2:
The active lateral earth pressure distribution is linear. The active lateral earth
pressure at any depth, h, below the ground surface is calculated by = =
.
Step 3:
The total active lateral earth force per unit length of retaining wall is
= = ( . )( . / )( ) = . /
445
____________________________________________________________
cohesionless, with an in-situ total unit weight of 16.8 kN/m3, and angle of
internal friction of 30. A lateral forace F is applied on the opposite side of the
wall. Based on the Rankine theory, what is the total passive resultant lateral
earth force per unit length of retaining wall?
Retaining wall
H1=1.5 m
F
 (in situ)=16.8 kN/m3
=30
H2=1.5 m
A) 25.2 kN/m B) 35.2 kN/m C) 50.4 kN/m D) 226.8 kN/m
446
____________________________________________________________
The Answer is A
Step 1:
Based on the Rankine theory, the Rankine passive earth pressure coefficient
°
= °+ = °+ =
Step 2:
The passive lateral earth pressure distribution is linear. The passive lateral earth
pressure at any depth, h, below the ground surface is calculated by = =
.
Step 3:
= = ( )( . / )( ) = . /
447
____________________________________________________________
77. A5 m high smooth retaining wall extends from the top of bedrock to the
cohesionless, has an in-situ total unit weight of 16.8 kN/m3, and angle of
internal friction of 30. Groundwater level is 2 m below the surface of the soil,
as shown in the figure below. The saturated unit weight of the soil is 18.2
kN/m3. Based on the Rankine theory, what is the total active resultant lateral
earth force per unit length of retaining wall?
Retaining wall
H1=2 m  (in situ)=16.8 kN/m3
=30
H3=3 m sat=18.2 kN/m3
448
____________________________________________________________
The Answer is C
Step 1:
°
= °− = °− = .
Step 2:
The total vertical stress  at the groundwater level is calculated by =
=( . / )( )= . . Since the pore water pressure at the
groundwater level is 0, the effective vertical stress = − =
. − = . .
Step 3:
At the bottom level of the retaining wall, the total vertical stress  is calculated
by
 = + =( . / )( )+( . / )( )=
. .
The pore water pressure is calculated by = =
( . / )( )= . .
The effective vertical stress = − = . −
. = . .
The distribution of effective vertical stress along depth, , is then
shown in the plot below:
Step 4:
The active lateral earth pressure at any depth, h, below the ground surface is
calculated by = .
449
____________________________________________________________
Step 5:
= ( + ) + ( + )
=( . ) ( + . )( )
+ ( . + . )( ) = . /
450
____________________________________________________________
ground surface. The wall is connnected to the bedrock by a frictionless hinge. The
soil behind of the retaining wall is homogeneous and cohesionless, has an in-situ
total unit weight of 16.8 kN/m3, and angle of internal friction of 30. A lateral
forced is applied at the opposite side of the wall, at a height of H1=1.5 m from the
top of bedrock. Based on Rankine theory, what is the minimum force per unit
length of retaining wall required to resist the overturning moment?
Retaining wall
H=3 m  (in situ)=16.8 kN/m3

=30
H1
451
____________________________________________________________
The Answer is A
Step 1:
°
= °− = °− = .
Step 2:
The active lateral earth pressure distribution is linear. The active lateral
earth pressure at any depth, h, below the ground surface is calculated by
= = .
Step 3:
= = ( . )( . / )( ) = . /
Based on Page 51 of the NCEES reference manual, the resultant, , acts

at a height above the bedrock of
= = .
Step 4:
H=3 m F
H1
1/3 H
452
____________________________________________________________
Summing moments on the retaining wall about the hinge gives
. ( )
= = = . /
.
453
____________________________________________________________
cohesionless, has an in-situ total unit weight of 15.4 kN/m3 and angle of internal
friction of 35. Groundwater level is 3 m below the surface of the soil, as shown
in the figure below. The saturated unit weight of the soil is 17.2 kN/m3. A
resisting force F is applied on the opposite side of the wall at the top of the wall so
that the soil behind the wall reaches the passive earth pressure condition. Based
on the Rankine theory, what is the total passive resultant lateral earth force per
unit length of retaining wall?
Retaining wall
F
H1=3 m  (in situ)=15.4 kN/m3
=35
H2=4 m sat=17.2 kN/m3
A) 252 kN/m B) 352 kN/m C) 574 kN/m D) 1156 kN/m
454
____________________________________________________________
The Answer is D
Step 1:
Based on the Rankine theory, the Rankinepassive earth pressure coefficient
°
= °+ = °+ = . .
Step 2:
The total vertical stress  at the groundwater level is calculated by =
=( . / )( )= . . Since the pore water pressure at the
groundwater level is 0, the effective vertical stress = − =
. − = . .
Step 3:
At the bottom level of the retaining wall, the total vertical stress  is
 = + =( . / )( )+( . /
)( )= .
The pore water pressure is calculated
by = =( . / )( )=
. .
The effective vertical stress = −
= − . = . .
The distribution of effective vertical stress
along depth, , is then shown in theplot:
455
____________________________________________________________
Step 4:
The passive lateral earth pressure at any depth, h, below the ground
surface is calculated by = .
Step 5:
The total passive lateral earth force per unit length of retaining wall is
= ( + ) + ( + )
=( . ) ( + . )( )
+ ( . + . )( ) = /
456
____________________________________________________________
80. A continuous footing three feet wide is founded 4 ft below the ground
surface in a clay soil for which = / , = / , and
= °. There is no groundwater at the site. Determine the ultimate bearing
capacity according to Terzaghi’s theory.
A) , B) ,
C) , D) ,
457
____________________________________________________________
The Answer is C
Step 1:
According to Terzaghi’s theory, the ultimate bearing capacity, , for a
concentrically loaded continuous footing of width B can be expressed as:
= + + .
in which c is the cohesion of the soil below the base of the footing; is the
vertical effective stress at the elevation of the footing base; B is the footing width;
is the unit weight of the soil below the footing base; , , and are bearing
capacity factors, which can be found in the NCEES reference manual.
Step 2:
From the given information, it is known that = = / , =
/ , = , = , and = °. From the figure on Page 138 of
the NCEES reference manual, it can be found that = . , = . , and
= . .
458
____________________________________________________________
Step 3:
= + + .
= ( . )+ ( )( . )
+ . ( ) ( . )= ,
459
____________________________________________________________
81. A shallow foundation is to be constructed below the ground surface in a

uniform cohesionless sand. It is found that the bearing capacity ratio for cohesion
of soil below the foundation, , is 50. What is the bearing capacity ratio for the
vertical effective stress at the elevation of the foundation base, ?
A) B) C) D)
460
____________________________________________________________
The Answer is A
Step 1:
From the figure on Page 138 of the NCEES reference manual, it can be found that
when = , the friction angle is 36.
Step2:
For a friction angle of 36, the bearing capacity ratio for the vertical effective
stress at the elevation of the foundation base, , is 38.
461
____________________________________________________________
82. A continuous footing is founded 5 ft below the ground surface in a clay for
which = / , = / , and = °. There is no
groundwater at the site. If the factor of safety FS is to be at least 3, recommend an
allowable bearing capacity according to Terzaghi’s theory.
A) , B) ,
C) , D) ,
462
____________________________________________________________
The Answer is B
Step 1:
According to Terzaghi’s theory, the ultimate bearing capacity, , for a
= + + .
Step 2:
From the given information, it is known that = = / , =
/ , = , and = °. From the figure on Page 138 of the
NCEES reference manual, it can be found that = , = , and = .
Step 3:
= + + . = ( )+ ( )( ) +
= ,
463
____________________________________________________________
Step 4:
For a factor of safety FS=3, the allowable bearing capacity , is
,
= = =
464
____________________________________________________________
83. A continuous footing 3 ft wide is founded 5 ft below the ground surface in

a uniform cohesionless sand for which = / , = , and = °.
There is no groundwater at the site. If the factor of safety FS is to be at least 3,
recommend the allowable bearing force per lineal foot of foundation according to
Terzaghi’s theory.
A) . B) . C) . D) .
465
____________________________________________________________
The Answer is D
Step 1:
Based on Terzaghi’s theory, the ultimate bearing capacity, , for a
= + + .
Step 2:
From the given information, it is known that = = / , = , =
, = , and = °. From the figure on Page 138 of the NCEES
reference manual, it can be found that = , and = .
466
____________________________________________________________
Step 3:
= + + .
= + ( )( )+ . ( ) ( )
= ,
Step 4:
For a factor of safety FS=3, the allowable bearing capacity , is
,
= = =
Step 5:
For a lineal foot of foundation, the base area is
= ( )=( )( )=
For a factor of safety FS=3, the allowable bearing force per lineal foot of
foundation is
,
= = ( )= , =
/
= .
467
____________________________________________________________
84. A total force of 2000 kN is to be supported by a square footing, which

directly rests on a sand ground. The sand is uniform, cohesionless, having a unit
weight = / , and friction angle = °. If the factor of safety F is 2,
determine the minimum width (B) of the footing according to Terzaghi’s theory.
(Based on Terzaghi’s theory, the ultimate bearing capacity, , for a square
footing of width B is calculated by = . + + . , in
which c is the cohesion of the soil below the base of the footing; is the
capacity factors.)
A) . B) . C) . D) .
468
____________________________________________________________
The Answer is C
Step 1:
Since the sand is cohesionless and the footing rests directly on the ground, c=0,
and H=0. Therefore, = . + + . = . .
Step 2:
For = °, from the figure on Page 138 of the NCEES reference manual, it can
be found that = .
Step 3:
The pressure on the footing should be less than the allowable bearing capacity of
the sand.
.
≤ =
( )( )
≥ = ≅ .
. . ( / )( )
469
____________________________________________________________
85. Which of the following foundation types are deep foundation?

I) Spread footings II) Piles III) Wall footings
IV) Mats V) Raft foundation
A) II B) II and IV C) III and V D) I, II, and V
470
____________________________________________________________
The Answer is A
P 135 C2 NCEES Ref Manual edition 8 V2
When the term deep foundations is used, it invariably means pile foundations. A
pile is a long structural member installed in the ground to transfer loads to soils at
some significant depths.
(A) is correct.
(B) is incorrect.Mats are shallow foundations.
(C) is incorrect.Both wall footings and mats are shallow foundations.
(D) is incorrect. Spread footings and raft foundations are both shallow
foundations. Raft foundations are the same as mat foundations.
471
____________________________________________________________
86. Through which is almost all the structural load on a friction pile
transferred to the soil?
A) The bottom end of the pile

B) Skin friction along the length of the pile
C) Both A and B
D) None of the above
472
____________________________________________________________
The Answer is B
Friction pile is one that transfers almost all the structural load to the soil by skin
friction along a substantial length of the pile.
(A) is incorrect.A pile that transfers almost all the structural load to the soil at the
bottom end of the pile is named an end bearing or point bearing pile.
(B) is correct.
(C) is incorrect, since A is incorrect.
(D) is incorrect, since B is correct.
473
____________________________________________________________
87. A strip wall footing of width 4 ft is embedded 3 ft below the ground surface in
a clayed sand for which = / , = / , and = °.
The groundwater level 30 ft below the footing base. Use a factor of safety
of 2 and assume the wall footing is sufficiently long, determine the allowable
bearing capacity of the wall footing.
A) , B) ,
C) , D) ,
474
____________________________________________________________
The Answer is B
Step 1:
The ultimate bearing capacity, , for a strip footing of width B is:
= + + .
in which c is the cohesion of the soil; is the effective unit weight of soil; is
the depth of footing below ground surface; B is the footing width; , , and
are bearing capacity factors for cohesion, depth, and unit weight, respectively.
Step 2:
From the given information, it is known that = / , = /
, = , and = °. From the figure on Page 138 of the NCEES
reference manual, it can be found that = . , = . , and = . .
Step 3:
= + + .
= ( . )+ ( )( . )
+ . ( )( )( . ) = ,
475
____________________________________________________________
Step 4:
For a factor of safety = , the allowable bearing capacity , is
,
= = = ,
476
____________________________________________________________
88. What type of shallow foundation should be used when the allowable soil
pressure is low or where an array of columns and/or walls are so close that
individual footings would overlap or nearly touch each other?
A) Strip footing B) Spread footing

C) Pile D) Mat foundation
477
____________________________________________________________
The Answer is D
when the allowable soil pressure is low or where an array of columns and/or walls
are so close that individual footings would overlap or nearly touch each other, a
mat or raft foundation is required.
(A) is incorrect.A strip footing, also known as continuous footing, is used for a
load-bearing wall, or for a row of columns which are closely spaced. It cannot be
used for an array of columns.
(B) is incorrect.A spread footing (or isolated or pad) footing is provided to

support an individual column. A spread footing is typically circular, square or
rectangular slab of uniform thickness.
(C) is incorrect.Piles are deep foundations, not shallow foundations.
(D) is correct. A mat or raft foundation is a large slab supporting a number of

columns and walls under the entire structure or a large part of the structure.
478
____________________________________________________________
89. A 9-inch diameter concrete pile is driven 50 ft into insensitive clay, which
has an in-situ unit weight of / , and an undrained shear strength
= / . The groundwater table is at the ground surface. Assume the
entire pile length is effective and the adhesive stress = . , determine the
bearing capacity of the pile due to friction.
A) 60 kips B) 83 kips C) 160 kips D) 500 kips
479
____________________________________________________________
The Answer is B
Step 1:
The bearing capacity of the pile due to friction is provided by skin friction,
, over the embedded length of the pile, which is the product of the
adhesive stress, , and the surface area of the shaft.
Step 2:
= . = . =
Step 3:
= = = ( )≅
= ≅ .
480
____________________________________________________________
90. Two figures show two frictionless wall as below. Information about each wall
is given. Assume that H1=4m and H2=5m. Assume that moment at point A for
left figure and right figure is the same. Determine the value of S.
S
= /
= 16 /
= 30 = H2
1 H1 3
2
A A
A) 20kN/m B) 30kN/m C) 25kN/m D) 35kN/m
481
____________________________________________________________
The Answer is A

30 1
K = 45 − = 45 − =
2 2 3

1
=K S = S(4 ) = 1.3
3
1 1
= = (4 ) = 2
2 2
1 1 1
= K = (16 / )(4 ) = 42.7
2 2 3
1 1
= = (4 ) = 1.3
3 3
1 1 1
= K = (15 / )(5 ) = 62.5
2 2 3
1 1
= = (5 ) = 1.7
3 3

+ = = (1.3 )(2 ) + (42.7 )(1.3 ) = (62.5 )(1.7 )
= 19.5 /
~20kN/m
482
____________________________________________________________
91. A retaining wall is given in the following diagram. It is known that H1=5m
and H2=3m. Determine the value of q in order to make resultant force of retaining
wall zero.
H1 q γ = 18 /
= 30
H2
1 2
A) 15kN/m B) 13kN/m C) 11kN/m D) 17kN/m
483
____________________________________________________________
The Answer is C

30 1
K = 45 − = 45 − =
2 2 3

1 1 1
= K ( − ) = (18 / − 9.8 / )(3 ) = 12.3
2 2 3
1 1
= = (9.8 / )(3 ) = 44.1
2 2
Step 3:The resultant force is zero. We got,

P + P = qH = 12.3kN + 44.1kN = q(5m)
q = 11.3kN/m
~11kN/m
484
____________________________________________________________
92. A retaining wall is given in the following diagram. It is known H2=3m. q =

10kN/m. Determine the value of H1 in order to make resultant force of
retaining wall zero.
H1 q γ = 18 /
= 30
H2
1 2
A) 6.3m B) 5.6m C) 5.2m D) 4.8m
485
____________________________________________________________
The Answer is B

30 1
K = 45 − = 45 − =
2 2 3

1 1 1
= K ( − ) = (18 / − 9.8 / )(3 ) = 12.3
2 2 3
1 1
= = (9.8 / )(3 ) = 44.1
2 2
Step 3:The resultant force is zero. We got,

+ = = 12.3 + 44.1 = (10 / )
= 5.6
~5.6m
486
____________________________________________________________
94. A retaining wall is given in the following diagram. It is known H1=5m and
H2=3m. q = 10kN/m. Calculate the moment at point A.
H1 q γ = 18 /
= 30
H2
1 2
A) 60kN-m B) 50kN-m C) 80kN-m D) 70kN-m
487
____________________________________________________________
The Answer is D

30 1
K = 45 − = 45 − =
2 2 3

1 1 1
= K ( − ) = (18 / − 9.8 / )(3 ) = 12.3
2 2 3
1 1
= = (9.8 / )(3 ) = 44.1
2 2
1 1
z =z = = (3 ) = 1
3 3
Step 3:The moment at point A can be obtained by,

M = qH − P z − P z = (10kN/m)(5m) − (12.3kN)(1m) −
(44.1kN)(1m) = 68.6kN − m~70kN-m
488
____________________________________________________________
95. A retaining wall is given in the following diagram. It is known H1=5m. q =

10kN/m. Determine the value of H2 when resultant force is equal to zero.
H1 q γ = 18 /
= 30
H2
1 2
A) 2.8m B) 3.6m C) 4.2m D) 3.8m
489
____________________________________________________________
The Answer is A

30 1
K = 45 − = 45 − =
2 2 3

1 1 1
= K ( − ) = (18 / − 9.8 / )( ) = 1.37( )
2 2 3
1 1
= = (9.8 / )( ) = 4.9( )
2 2
Step 3:The resultant force is equal to zero,

F = P + P − qH = 1.37(H ) + 4.9(H ) − (10kN/m)(5m) = 0
H = 2.8m~2.8m
490
____________________________________________________________
96. A 5m soil is assumed to be in the range of virgin compression. CC is given as

0.6. If the initial effective pressure p0 is equal to 300kN/m2 and ∆p is equal to
200kN/m2. In order to know the initial voids ratio, 100cm3 initial soil is selected
and we find that there is 45cm3 voids. Determine the settlement of this soil.
A) 32cm B) 37cm C) 43cm D) 48cm
491
____________________________________________________________
The Answer is B
Step 1:Settlement for soil in the range of virgin compression can be calculated
based on the equation shown below,
H p + ∆p
∆H = C log
1+e p
From this problem, we can know everything except for the initial voids ratio. We
need to calculate it first.
Step 2:The initial voids ratio can be calculated based on,

V V 45cm
e= = = = 0.8
V V−V 100cm − 45cm
Therefore, we can know the settlement of this soil by,
5m 300kN/m + 200kN/m
∆H = (0.6)log = 37cm
1 + 0.8 300kN/m
~37cm
492
____________________________________________________________
97. A 3m soil is assumed to be in the range of virginn compression. CC is given

as 0.6. If the initial effective pressure p0 is equal to 300kN/m2 and ∆p is equal to
200kN/m2. In order to know the initial voids ratio, 100cm3 initial soil is selected
and we find that there is 75cm3 soil solids. Determine the settlement of this soil.
A) 34cm B) 30cm C) 26cm D) 38cm
493
____________________________________________________________
The Answer is B
H p + ∆p
∆H = C log
1+e p

V V−V 100cm − 75cm
e= = = = 0.33
V V 75cm
/ /
∆H = (0.6)log = 30cm~30cm
. /
494
____________________________________________________________
98. A 3m soil is assumed to be in the range of virgin compression. CC is given as

0.5. If the initial effective pressure p0 is equal to 300kN/m2 and ∆p is equal to
200kN/m2. It is known that it has 30cm settlement. What is the porosity of this
soil for the initial state?
A) 0.3 B) 0.2 C) 0.1 D) 0.4
495
____________________________________________________________
The Answer is C
H p + ∆p
∆H = C log
1+e p
can calculate it by,
H p + ∆p 3m 300kN/m + 200kN/m
C log = (0.5)log = 30cm
1+e p 1+e 300kN/m
e = 0.11
Step 2:The initial porosity can be calculated based on,

.
n= = = 0.1~0.1
.
496
____________________________________________________________
99. A 3m soil is assumed to be in the range of virgin compression. If the initial

effective pressure p0 is equal to 300kN/m2 and ∆p is equal to 200kN/m2. It is
known that it has 30cm settlement. If the initial porosity is 0.5, then what is the
value of CC?
A) 0.9 B) 0.8 C) 0.7 D) 0.6
497
____________________________________________________________
The Answer is A
H p + ∆p
∆H = C log
1+e p
From this problem, we can know everything except for the CC and voids ratio. We
can calculate voids ratio by,
n 0.5
e= = =1
1 − n 1 − 0.5
Step 2:Based on the equation, we have the following as,

H p + ∆p 3m 300kN/m + 200kN/m
∆H = C log = C log
1+e p 1+1 300kN/m
= 30cm
C = 0.9~0.9
498
____________________________________________________________
100. A 3m soil is assumed to be in the range of virgin compression. If the initial

effective pressure p0 is equal to 300kN/m2. It is known that it has 30cm settlement
when adding pressure ∆p and CC is 0.7. If the initial porosity is 0.5, then what is
the value of ∆p?
499
____________________________________________________________
The Answer is D
H p + ∆p
∆H = C log
1+e p
From this problem, we can know everything except for the ∆pand voids ratio. We
can calculate voids ratio by,
n 0.5
e= = =1
1 − n 1 − 0.5

H p + ∆p 3m 300kN/m + ∆p
∆H = C log = (0.7)log = 30cm
1+e p 1+1 300kN/m
∆p = 279kN/m ~280kN/m2
500
____________________________________________________________
101. A 5m soil is given. CR is given as 0.1 and Cc is 0.6. If the initial effective
pressure p0 is equal to 300kN/m2 , ∆p is equal to 200kN/m2, and pc is 400kN/m2,
determine the settlement of this soil when the initial porosity of soil is 0.44.
A) 25cm B) 15cm C) 20cm D) 30cm
501
____________________________________________________________
The Answer is C
Step 1:Settlement for soil can be calculated based on the equation shown below,
H p p + ∆p
∆H = C log + C log
1+e p p
Step 2:The initial porosity is given as 0.44. Therefore, voids ratio can be
calculated by,
n 0.44
e= = = 0.8
1 − n 1 − 0.44
5m 400kN/m 500kN/m
∆H = (0.1)log + (0.6)log = 19.6cm
1 + 0.8 300kN/m 400kN/m
~20cm
502
____________________________________________________________
102. A 5m soil is given. CR is given as 0.1 and CC is 0.6. If the initial effective
pressure p0 is equal to 300kN/m2 , ∆p is equal to 200kN/m2, and pc is 400kN/m2.
In order to know the initial voids ratio, 100cm3 initial soil is selected and we find
that there is 45cm3 voids. Determine the settlement of this soil.
A) 35cm B) 30cm C) 25cm D) 20cm
503
____________________________________________________________
The Answer is D
H p p + ∆p
1+e p p

V V 45cm
e= = = = 0.8
V V−V 100cm − 45cm
5m 400kN/m 500kN/m
∆H = (0.1)log + (0.6)log = 19.6cm
1 + 0.8 300kN/m 400kN/m
~20cm
504
____________________________________________________________
103. A 3m soil is given. CR is given as 0.1 and CC is 0.6. If the initial effective
pressure p0 is equal to 300kN/m2 , ∆p is equal to 200kN/m2, and pc is 400kN/m2.
In order to know the initial voids ratio, 100cm3 initial soil is selected and we find
that there is 75cm3 soil solids. Determine the settlement of this soil.
A) 18cm B) 16cm C) 20cm D) 14cm
505
____________________________________________________________
The Answer is B
H p p + ∆p
1+e p p

V V−V 100cm − 75cm
e= = = = 0.33
V V 75cm
3m 400kN/m 500kN/m
∆H = (0.1)log + (0.6)log = 15.9cm
1 + 0.33 300kN/m 400kN/m
~16cm
506
____________________________________________________________
104. A 3m soil is givin. CR is given as 0.1 and Cc is 0.6. If the initial effective
pressure p0 is equal to 300kN/m2, ∆p is equal to 200kN/m2, and pc is 400kN/m2. It
is known that it has 11cm settlement. What is the porosity of this soil for the
initial state?
A) 0.25 B) 0.5 C) 0.75 D) 1
507
____________________________________________________________
The Answer is B
H p p + ∆p
1+e p p
can calculate it by,
H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 500kN/m
= (0.1)log + (0.6)log = 11cm
1+e 300kN/m 400kN/m
e=1
Step 2:The initial porosity can be calculated based on,

e 1
n= = = 0.5
1+e 1+1
508
____________________________________________________________
105. A 3m soil is givin. CR is given as 0.1 and Cc is unknown. If the initial

effective pressure p0 is equal to 300kN/m2, ∆p is equal to 200kN/m2, and pc is
400kN/m2. It is known that it has 11cm settlement. If the initial porosity is 0.5,
then what is the value of CC?
A) 0.5 B) 0.8 C) 0.6 D) 0.7
509
____________________________________________________________
The Answer is C
H p p + ∆p
1+e p p
From this problem, we can know everything except for the initial voids ratio and
CC. We can calculate voids ratio by,
n 0.5
e= = =1
1 − n 1 − 0.5

H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 500kN/m
= (0.1)log + C log = 11cm
1+1 300kN/m 400kN/m
C = 0.6~0.6
510
____________________________________________________________
106. A 3m soil is givin. CR is unknown and Cc is 0.6. If the initial effective

is known that it has 11cm settlement. If the initial porosity is 0.5, then what is the
value of CR?
A) 0.2 B) 0.1 C) 0.3 D) 0.4
511
____________________________________________________________
The Answer is B
H p p + ∆p
1+e p p
CR. We can calculate voids ratio by,
n 0.5
e= = =1
1 − n 1 − 0.5

H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 500kN/m
= C log + (0.6)log = 11cm
1+1 300kN/m 400kN/m
C = 0.1~0.1
512
____________________________________________________________
pressure p0 is equal to 300kN/m2 and pc is 400kN/m2. It is known that it has 11cm
settlement when adding pressure ∆p. If the initial porosity is 0.5, then what is the
value of ∆p?
513
____________________________________________________________
The Answer is B
H p p + ∆p
1+e p p
∆p. We can calculate voids ratio by,
n 0.5
e= = =1
1 − n 1 − 0.5

H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 300kN/m + ∆p
= (0.1)log + (0.6)log
1+1 300kN/m 400kN/m
= 11cm
∆p = 200kN/m ~200kN/m2
514
____________________________________________________________
pressure p0 is equal to 300kN/m2. It is known that it has 11cm settlement when
adding pressure ∆p. If the voids occupy 50% of total volume at initial state, then
what is the value of ∆p?
515
____________________________________________________________
The Answer is A
H p p + ∆p
1+e p p
∆p. We can calculate voids ratio by,
V 50%V
e= = =1
V 50%V

H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 300kN/m + ∆p
= (0.1)log + (0.6)log
1+1 300kN/m 400kN/m
= 11cm
∆p = 200kN/m ~200kN/m2
516
____________________________________________________________
109. A 3m soil is givin. It is known that the initial porosity is 0.5 and after
ultimate consolidation settlement it become to 0.45. Determine the ultiate
consolidation settlement in this soil layer.
A) 0.56m B) 0.48m C) 0.33m D) 0.50m
517
____________________________________________________________
The Answer is C
Step1:Ultimate consolidation settlement in soil layer can be determined by the

following equation,
e −e
S =ε H= H
1+e
From this problem, we can know the thickness of this soil layer as 3m. We need to
determine the voids ratio before and after ultimate consolidation settlement. They
can be calculated based on,
n 0.5
e = = =1
1−n 1 − 0.5
n 0.45
e = = = 0.8
1−n 1 − 0.45

.
S = H= (3m) = 0.33m~0.33m
.
518
____________________________________________________________
110. A 3m soil is givin. It is known that the initial voids ratio is 1 and after
ultimate consolidation settlement it is unknown. Determine the final porosity of
this soil if the ultimate consolidation settlement is 0.5m.
A) 0.5 B) 0.4 C) 0.3 D) 0.2
519
____________________________________________________________
The Answer is B

following equation,
e −e
S =ε H= H
1+e
determine the voids ratio after ultimate consolidation settlement first. It can be
obtained by,
e −e 1−e
H= (3m) = 0.5m
1+e 1 + 0.8
e = 0.7
Step 2:We can calculate the final porosity of this soil by,
.
n= = = 0.41~0.4
.
520
____________________________________________________________
111. A 3m soil is givin. It is known that the final voids ratio is 0.7 after ultimate
consolidation settlement but the initial voids ratio is unknown. Determine the
initial porosity of this soil if the ultimate consolidation settlement is 0.5m.
A) 0.5 B) 0.4 C) 0.3 D) 0.2
521
____________________________________________________________
The Answer is A

following equation,
−
= =
1+
determine the initial voids ratio first. It can be obtained by,
− − 0.7
= (3 ) = 0.5
1+ 1+
=1
Step 2:We can calculate the initial porosity of this soil by,
= = = 0.5~0.5
522
____________________________________________________________
112. A soil layer is givin. The thickness of this soil layer is unknown. It is known
that the final voids ratio is 0.7 after ultimate consolidation settlement and the
initial porosity is 0.5. Determine the thickness of this soil layer if the ultimate
consolidation settlement is 0.5m.
A) 2m B) 3m C) 4m D) 5m
523
____________________________________________________________
The Answer is B

following equation,
−
= =
1+
From this problem, we can know the initial porosity of this soil. We need to
determine the initial voids ratio first. It can be obtained by,
0.5
= = =1
1− 1 − 0.5
Step 2:We can calculate the thickness of this soil layer by,
− 1 − 0.7
= = = 0.5
1+ 1 + 0.7
= 3 ~3m
524
____________________________________________________________
113. A 3m soil is givin. It is known that the initial porosity is 0.5 and after
ultimate consolidation settlement it become to 0.45. Determine the approximate
settlement in this soil layer if the average degree of consolidation is 70%.
A) 0.34m B) 0.23m C) 0.18m D) 0.36m
525
____________________________________________________________
The Answer is B
Step1:To calculate the approximate settlement, we need to calculate the ultimate

consolidation settlement in soil layer first. It can be determined by the following
equation,
−
= =
1+
From this problem, we can know the thickness of this soil layer as 3m. We need
to determine the voids ratio before and after ultimate consolidation settlement.
They can be calculated based on,
0.5
= = =1
1− 1 − 0.5
0.45
= = = 0.8
1− 1 − 0.45
Step 2:The approximate settlement can be then calculated by,

− 1 − 0.8
= = = (70%) (3 ) = 0.23
1+ 1 + 0.8
~0.23m
526
____________________________________________________________
114. A 3m soil is givin. It is known that the initial voids ratio is 1 and after
ultimate consolidation settlement it is unknown. Determine the final porosity of
this soil if the approximate settlement is 0.2m and average degree of consolidation
is 50%.
A) 0.34 B) 0.26 C) 0.38 D) 0.42
527
____________________________________________________________
The Answer is D
Step 1:Approximate settlement can be calculated based on the equation as shown

below,
−
= =
1+
The final voids ratio based on this equation can be determined by,
− 1−
= (50%) (3 ) = 0.2
1+ 1+1
= 0.73
Step 2:The final porosity can be then calculated by,

.
= = = 0.42~0.42
.
528
____________________________________________________________
115. A 3m soil is givin. It is known that the final voids ratio is 0.3 after ultimate
consolidation settlement but the initial voids ratio is unknown. Determine the
initial porosity of this soil if the approximate settlement is 0.2m and average
degree of consolidation is 50%.
A) 0.5 B) 0.4 C) 0.3 D) 0.2
529
____________________________________________________________
The Answer is A

below,
−
= =
1+
The final voids ratio based on this equation can be determined by,
− − 0.73
= (50%) (3 ) = 0.2
1+ 1+
=1
Step 2:The initial porosity can be then calculated by,

= = = 0.5~0.5
530
____________________________________________________________
116. It is known that initial effective consolidation stress, p0 is equal to 200kN/m2

and final effective consolidation stress, p0 + ∆p is equa to 500kN/m2. The past
maximum consolidation stress, pc is 600kN/m2. For this soil in initial state, we
know that it has 100pcf dry density, 40% water content and 80% degree of
saturation. If the thickness of soil is 3m, and CR = 0.1, CC = 0.6, calculation the
settlement of this soil.
A) 44mm B) 55mm C) 66mm D) 33m
531
____________________________________________________________
The Answer is C
Step 1:According to this problem, we know that,

p = 200kN/m
p + ∆p = 500kN/m
p = 600kN/m > p + ∆p and p
Therefore, we can pick the equation to calculation the settlement as,
H p + ∆p
∆H = C log
1+e p
Step 2:Equation of void ratio of soil is,

V
e=
V
To determine the water content from voids ratio, we need to determine the voids
ratio equation by,
V W wW
V γ S γ S W w w
e= = S = = = =γ
V V V V V γ S γ S
Step 3:To get the value of water content, we can substitute all known value into
the equation above and we have,
w 40%
e=γ = (100pcf) = 0.8
γ S (62.5pccf)(80%)
Step 4: Based on the equation above, we can calculate the settlement as,
∆ /
∆H = C log = (0.1)log = 0.066m = 66mm~66mm
. /
532
____________________________________________________________
117. It is known that initial effective consolidation stress, p0 is equal to 200kN/m2

and final effective consolidation stress, p0 + ∆p is equa to 500kN/m2. The past
maximum consolidation stress, pc is 600kN/m2. For this soil in initial state, we
know that it has 40% water content and 80% degree of saturation. If the thickness
of soil is 3m, and CR = 0.1, CC = 0.6, calculation the dry density of this soil with
66mm settlement.
A) 100pcf B) 110pcf C) 120pcf D) 130p
533
____________________________________________________________
The Answer is A
Step 1:According to this problem, we know that,

p = 200kN/m
p + ∆p = 500kN/m
p = 600kN/m > p + ∆p and p
Therefore, we can pick the equation to calculation the settlement as,
H p + ∆p
∆H = C log
1+e p

V
e=
V
ratio equation by,
V W wW
V S γ S γ S W w w
e= = = = = =γ
V V V V V γ S γ S
40%
= = =
(62.5 )(80%) 125
Step 4: Based on the equation above, we can calculate the dry density as,
+∆ 3 500 /
∆ = = (0.1) = 0.066
1+ 1+ 200 /
125
= 66
= 100 ~100pcf
534
____________________________________________________________
118. A 5m soil is given. CR is given as 0.1 and Cc is 0.6. If the initial effective
pressure p0 is equal to 300kN/m2 , ∆p is equal to 200kN/m2, and pc is 400kN/m2,
determine the settlement of this soil when the initial dry density of soil is 100pcf,
the water content is 36% and the degree of saturation is 54%.
A) 17cm B) 20cm C) 23cm D) 15cm
535
____________________________________________________________
The Answer is A
H p p + ∆p
1+e p p

V
e=
V
ratio equation by,
V W wW
V γ S γ S W w w
e= = S = = = =γ
V V V V V γ S γ S
36%
= = (100 ) = 1.1.
(62.5 )(54%)
Step 4: Based on the equation above, we can calculate the settlement as,
5 400 / 500 /
∆ = (0.1) + (0.6) = 16.8
1 + 1.1 300 / 400 /
~17cm
536
____________________________________________________________
119. A 3m soil is givin. CR is unknown and Cc is 0.6. If the initial effective

is known that it has 11cm settlement. If the initial dry density of soil is 110pcf, the
water content is 38% and the degree of saturation is 66%, then what is the value
of CR?
A) 0.1 B) 0.2 C) 0.3 D) 0.4
537
____________________________________________________________
The Answer is A
H p p + ∆p
1+e p p
CR. We need to calculate it first.

V
e=
V
ratio equation by,
V W wW
V γ S γ S W w w
e= = S = = = =γ
V V V V V γ S γ S
w 38%
e=γ = (110pcf) =1
γ S (62.5pccf)(66%)

H p p + ∆p
C log + C log
1+e p p
3m 400kN/m 500kN/m
= C log + (0.6)log = 11cm
1+1 300kN/m 400kN/m
C = 0.1~0.1
538
____________________________________________________________
pressure p0 is equal to 300kN/m2 and pc is 400kN/m2. It is known that it has 11cm
settlement when adding pressure ∆p. If the initial dry density of soil is 110pcf, the
water content is 38% and the degree of saturation is 66%, then what is the value
of ∆p?
539
____________________________________________________________
The Answer is B
H p p + ∆p
1+e p p
∆p. We need to calculate it first.

V
e=
V
ratio equation by,
V W wW
V γ S γ S W w w
e= = S = = = =γ
V V V V V γ S γ S
38%
= = (110 ) =1
(62.5 )(66%)

+∆
+
1+
3 400 / 300 / +∆
= (0.1) + (0.6)
1+1 300 / 400 /
= 11
∆ = 200 / ~200kN/m2
540
____________________________________________________________
121. A highway embankment constructed with a sandy clay soil has a height
H = 20 ft. The clay soil has a cohesion c = 600 lbf/ft , and friction angle
ϕ = 20°. It is determined that the embankment slope will fail along a plane that
has an angle α = 15° from the horizontal surface, as shown in the figure below.
The weight of the sliding soil wedge is 42,000 lbf per foot long of the
embankment. Determine the available shearing resistance along the slip plane per
per foot long of the embankment.
A) 1,870 B) 10,870 C) 61,130 D) 20,078
541
____________________________________________________________
The Answer is C
Step 1:
The mobilized shear force along the slip plane is T = cL + W cos(α )tan(ϕ).
Step 2:
From the given information, it is known W = 42,000 lbf/ft, α = 15°, ϕ = 20°,
c = 600 lbf/ft , and H = 20 ft. Since
L = H/sin(α )
T = cL + W cos(α )tan(ϕ) = cH/sin(α ) + W cos(α )tan(ϕ)
lbf
600 (20 ft) lbf
= ft + 42,000 cos(15°)tan(20°)
sin(15°) ft
lbf
≅ 61,130
ft
542
____________________________________________________________
122. A highway embankment

constructed with a clay soil has a WM

height H = 8 m, and a slope angle
H
θ = 25°. The clay soil has a cohesion
c = 60 kPa, unit weight γ = s
19 kN/m , and friction angle ϕ = 30°. Ls
Determine the factor of safety against
slope instability, FS, for a plane that
has an angle α = 15° from the horizontal surface.
A) 1.4 B) 5.2 C) 6.9 D) 9.6
543
____________________________________________________________
The Answer is D
Step 1:
( ) ( )
The top width of the sliding wedge is w = − = − =
( ) ( ) ( °) ( °)
12.70 m
Step 2:
1 kN
W = γA = γ wH = 19 (0.5)(12.70 m)(8 m) = 965.2 kN/m
2 m
Step 3:
Since the length of the slip surface is L = H/sin(α ), the shear resistance along
the slip plane is
T = cL + W cos(α )tan(ϕ) = cH/sin(α ) + W cos(α )tan(ϕ)
(60 kPa)(8 m)
= + (965.2 kN/m)cos(15°)tan(30°)
sin(15°)
≅ 2393 kN/m
Step 4:
T = W sin(α ) = (965.2 kN/m)sin(15°) = 250 kN/m
Step 5:
2393 kN
T m
FS = = ≅ 9.6
T kN
250
m
544
____________________________________________________________
123. A sandy soil is stockpiled on a ground. The soil is cohesionless with a unit
weight γ = 20 kN/m , and friction angle ϕ = 35°. What is the maximum slope
angle θ the sand cone can reach?
A) 10° B) 15° C) 20° D) 35°
545
____________________________________________________________
The Answer is D
When the slope angle θ of the sand cone reaches its maximum, the sand on the
slope surface is in equilibrium, T =T .
WM

Step 1:
The mobilized shear force along the slope surface is
T = W sin(θ)
The shearing resistance along the slope surface is
T = cL + W cos(α )tan(ϕ) = 0 ∙ L + W cos(θ)tan(ϕ)
Step 2:
When T = T , we have
W sin(θ) = W cos(θ)tan(ϕ)
( )
Therefore, tan(ϕ) = ( )
= tan(θ). So
θ = ϕ = 35°
546
____________________________________________________________
124. Which of the following slope failure types are impossible?
toe Slip plane
(I) (II)
toe Slip plane
toe Slip plane
(III) (IV)
toe Slip plane
A) I B) II C) III D) None
547
____________________________________________________________
The Answer is D
(A) is incorrect.Slope failure in (I) is a type of rotational failure (toe slide) that is
common in homogeneous fine-grained soils.
(B) is incorrect.Slope failure in (II) is a type of rotational failure (base failure)

that is common in homogeneous fine-grained soils. A soft soil layer resting on a
stiff layer of soil is prone to base failure.
(C) is incorrect.Slope failure in (III) is a type of rotational failure (slope slide) that
is common in homogeneous fine-grained soils.
(D) is correct. Slope failure in (IV) is a type of translational slide that is common
in coarse-grained soils.
548
____________________________________________________________
125. An embankment constructed with

a sandy clay soil has a height H = 10 m,  WM
and a slope angle θ. The soil has a
H
cohesion c = 10 kPa, unit weight
γ = 19 kN/m , and friction angle s
ϕ = 20°. The factor of safety against Ls
slope instability, FS, for a plane that has
an angle α = 25° from the horizontal
surface, is 2. Determine the slope angle θ.
A) 10° B) 15° C) 20° D) 31°
549
____________________________________________________________
The Answer is D
Step 1:
The top width of the sliding wedge is
H H (10 m) (10 m)
w= − = −
tan(α ) tan(θ) tan(25°) tan(θ)
Step 2:
1 kN (10 m) (10 m)
W = γA = γ wH = 19 (0.5) − (10 m)
2 m tan(25°) tan(θ)
kN 1 1
= 950 −
m tan(25°) tan(θ)
Step 3:
Since the length of the slip surface is L = ( )

= (
= 23.66 m, the shear
°)
resistance along the slip plane is
T = cL + W cos(α )tan(ϕ)
= (10 kPa)(23.66 m)
kN 1 1
+ 950 − cos(25°)tan(20°)
m tan(25°) tan(θ)
kN kN 1 1
= 236.6 + 313.38 −
m m tan(25°) tan(θ)
Step 4:
kN 1 1
T = W sin(α ) = 950 − sin(25°)
m tan(25°) tan(θ)
550
____________________________________________________________
Step 5:
kN kN 1 1
T 236.6 + 313.38 −
m m tan(25°) tan(θ)
FS = = =2
T kN 1 1
950 − sin(25°)
m tan(25°) tan(θ)
It is obtained that
1 1
− = 0.4833
tan(25°) tan(θ)
tan(θ) = 0.602θ ≅ 31°
551
____________________________________________________________
126. A highway is to be built in a region where soft soil exists in the ground,
which will cause excessive settlement of the highway embankment. Which of the
following methods may be used to reduce the settlement of the embankment
during the service life of the highway?
(I) Precompression of the soft soil
(II) Installation of vertical drains in the soil under a surcharge load
(III) Stabilization of the soil using admixtures
(IV) Injection of grout into the ground
A) I and II B) III C) II and IV D) All of the above
552
____________________________________________________________
The Answer is D
Soft soils are soils that have high porosity and water content, and under load, will
produce large settlements over a very long time span. Soft soils can be improved
by a variety of methods, such as precompression, vertical drains, in-situ
densification, grouting, stabilization using admixtures, and reinforcement with
geosynthetics.
(A) is incorrect.(III) and (IV) are also correct. Stabilization of soil using
admixtures can increase the strength and stiffness of the soil. Injection of
cementitious or chemical grout into the ground may also increase the strength and
stiffness of the soil.
(B) is incorrect.(I), (II) and (IV) are also correct. Precompression of the soft soil
using a surcharge will consolidate the soil to a level that the future consolidation
of the soil under highway loads is limited to an acceptable value.
C) is incorrect.(I) and (II) are also correct.
(D) is correct.
553
____________________________________________________________
127. A highway embankment is to be built on a deposit of clay with a depth of 10

m. The properties of the clay are shown in the figure below. Before the
embankment is constructed, a uniformly distributed surcharge load p = 200 kPa
is applied on the ground for an extended period of time. After the consolidation of
clay under the surcharge is completed, the embankment is built. Assume the
embankment pressure on the clay can be treated as a uniformly distributed
q = 100 kPa. Determine the settlement of embankment due to the primary
consolidation of the clay layer.
p=200 kPa
Ground Surface
H=10 m Clay, =19 kN/m3

e0=1.1, Cc=0.32, CR=0.14
Gravel
Rock
q=100 kPa
Ground Surface
H=10 m Clay, =19 kN/m3

e0=1.1, Cc=0.32, CR=0.14
Gravel
Rock
A) 0.120 m B) 0.208 m C) 0.342 m D) 0.652 m
554
____________________________________________________________
The Answer is B
The surcharge load pre-consolidates the clay soil. After the surcharge is removed,
the settlement of the ground recovers to some extent. When the embankment
pressure is applied, the clay is recompressed, following the reversed strain-
pressure path of settlement recovery, to the previously reached maximum
settlement, and then follow the normal consolidation curve to a final settlement.
Step 1:
In this problem, since the embankment pressure q is less than the preconsolidation
pressure p, the compression of soil under embankment pressure is in the range of
recompression. Therefore, the recompression index C should be used for
calculation.
After the surcharge load is removed and before the embankment is constructed,
the effective vertical stress at center of the clay layer is
1 1 kN
σ = Hγ = (10 m) 19 = 95 kPa
2 2 m
Step 3:
After the embankment is constructed, the effective vertical stress at center of the
clay layer is
σ = σ + q = 95 kPa + 100 kPa = 195 kPa
Step 4:
The settlement of the embankment is
σ
∆e C log
σ
S = Hε = H =H
1+e 1+e
195 kPa
(0.14)log
= (10 m) 95 kPa
1 + 1.1
= 0.208 m
555
____________________________________________________________
128. A field density test was performed on compacted fill in accordance with
ASTM D 1556 Standard Test Method for Density and Unit Weight of Soil in
Place by the Sand-Cone Method. The following is a summary of the test data:
Weight of wet soil extracted from the sand cone hole 6.438 lb
Weight of sand needed to fill the hole
(funnel correction already applied) 4.125 lb
The bulk density of the sand used in the sand cone apparatus 82.4 pcf
Water Content Test:

Empty cup 0.462lb
Cup plus wet soil 1.832lb
Cup plus dry soil 1.720lb
The laboratory maximum dry density performed on the same soil is equal to 130
pcf. Based on this data, the relative compaction of the fill is most nearly:
A) 99% B) 95% C) 91% D) 63
556
____________________________________________________________
The Answers is C
Volume of hole = 4.125 lb/82.4 pcf = 0.0501 cf

γ = 6.438 lb/0.0501 cf = 128.6 pcf
W = (1.832 − 1. 720)/(1.720 − 0.462) = 0.112/1.258 = 0.0890
γ = 128.6/(1 + 0.0890) = 118.0
Relative compaction = γ /γ = 118.0/130.0 = 0. 0.908 or 91%
557
____________________________________________________________
129. A temporary slope will be excavated to the dimensions shown in the figure.
Laboratory testing has yielded the geotechnical parameters shown in the chart.
The safety factor for the failure surface shown is most nearly:
A) 1.4 B) 1.6 C) 1.8 D) 2.0
CLAY
= °
=
=
∅=
POTENTIAL FALURE SURFACE
BEDROCK
558
____________________________________________________________
The Answer is C
D = 20
D/H = 2
β = 30 Stability number = 0.172 = =
×
C ≅ 413
C 750
FS = = = 1.82
C 413
559
____________________________________________________________
130. During a standard penetration test (SPT), unusually low blow counts are
encountered in a foundation expected to be medium-dense to dense sand. This is
an indication that the following condition may be present:
A) The sampler drive shoe is badly damaged or worn due to too many drivings to
refusal.
B) Cobbles are encountered.
C) The sampler drive shoe is plugged.
D) The groundwater in the borehole is much lower than in situ conditions
immediately outside the bore hole.
560
____________________________________________________________
The Answers is D
Reference: Cheney and Chassie, Soils and Foundations Workshop

Manual, 2nd ed., National Highway Institute, 1993, p. 23.
Groundwater coming up into the casing or hollow stem augers can
cause the test zone to become "quick." This loosening will reduce the
strength of the soil
561
____________________________________________________________
131. The excavation shown in the figure is required to repair a broken fiber optic
line. The excavation is to be open for 18 hours. A train will pass the excavation
area each hour. As the competent person on-site, you must determine which of the
following statements is most correct:
A) OSHA Soil Type A--Excavation is safe for entry.

B) OSHA Soil Type A--Excavation should be sloped at 3/4: 1 or flatter prior to
entry.
C) OSHA Soil Type B over A-- Excavation should be sloped at 3/4: 1in upper 3 ft
and 3/4: 1 in lower 9 ft prior to entry.
D) OSHA Soil Type B-- Excavation should be sloped at 1:1 or flatter prior to
entry.
( ) =
/ :
( ) = .
562
____________________________________________________________
The Answers is D
Reference: OSHA 1926, Subpart P, Appendix A and B, 2006, p. 334.

Per definition for Soil Type A Exception (ii), the soil cannot be classified
as "A" as it will be subject to vibration from the railroad. Therefore, per
definition of Soil Type B (iv), the soil should be classified as "B." Per
Table B-1, Type B soils should be sloped at 1:1 or flatter.
563
____________________________________________________________
132. A soil layer is givin. The thickness of this soil layer is unknown. It is known
that the final voids ratio is 0.73 after ultimate consolidation settlement and the
initial porosity is 0.5. Determine the thickness of this soil layer if the approximate
settlement is 0.2m and average degree of consolidation is 50%.
A) 2m B) 3m C) 4m D) 5m
564
____________________________________________________________
The Answer is B
Step 1:First we need to calculate the initial voids ratio. It can be obtained by,
.
= = =
− − .

below,
−
= =
+
Based on this equation, only the thickness of soil layer is unknown. Then we can
solve it by,
− − .
=( %) = .
+ +
= ~3m
565
____________________________________________________________
133. A retaining wall is shown in the following figure. The soil profile is also
shown in the figure. According to Rankine active state, calculate the moment the
soil produces toward retaining wall.
=4m = / =
Water
=4m = . / = /
Fig. 19
A) 372.4 kN-m B) 456.9 kN-m

C) 542.5 kN-m D) 435.1 kN-m
566
____________________________________________________________
The Answer is B
Step 1:Rankine active earth pressure coefficient,
= − = − =
Step 2:The lateral earth pressure distribution with depth is shown:
Active
2
5 3 4
Pore water Soil Pore water
Step 3:Notice that area 4 and area 5 are the same with different direction forces
toward retaining wall, there is no need to calculate them.
Lateral force fore Area 1,
= = ( )( / )( ) = .
Location to base, = + = ( )+ = .
= = ( / )( )( )= .
Location to base, = = ( )=
= ( − )
567
____________________________________________________________
= ( / − . / )( ) = .
Location to base, = = ( )= .
Step 4:Summing up moments at the base:

= + +
=( . )( . )+( . )( )+( . )( . )
= .
568
____________________________________________________________
134. There is a frictionless wall shown in the following figure. Assume that the
soil will slip along the dash line and the angle is 600. Calculate the active lateral
force per unit length.
= /
H=5 m
Fig. 20
A) 73.3 kN B) 83.3 kN C) 68.8 kN D) 85.6 kN
569
____________________________________________________________
The Answer is B
Step 1:Rankine active earth pressure coefficient can be given by,
= ( − )= ( − )=
Step 2:The vertical stress at the surface for soil,

=
The vertical stress at the base for soil,
= =( / )( )=
Step 3:The lateral stress at base for soil,
( ) = = = ( )= .
Step 4:The lateral force due to soil,
= ( ) = ( . )( )= .
570
____________________________________________________________
retaining wall push the soil to slip along the dash line and the angle is 300.
Calculate the passive lateral force per unit length.
A) 700 kN B) 820 kN C) 680 kN D) 750 kN
H=5 m
= /
Fig. 21
571
____________________________________________________________
The Answer is B
Step 1:Rankine passive earth pressure coefficient can be given by,

= ( − )= ( − )=

=
= =( / )( )=
( ) = = = ( )=
= ( ) = ( )( )=
572
____________________________________________________________
soil will slip along the dash line and the angle is 600. Calculate moment soil
produce to retaining wall.
= /
H=6 m
Fig. 22
A) 328 kN-m B) 264 kN-m
C) 196 kN-m D) 236 kN-m
573
____________________________________________________________
The Answer is B
Step 1:Rankine active earth pressure coefficient can be given by,
= ( − )= ( − )=

=
= =( / )( )=
( ) = = = ( )=
= ( ) = ( )( )=
Location to base, = = ( )=
Step 5:Summing up moments at the base:

= =( )( )= −
574

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Practice Problems PE Exam

Part 3 : Soil Mechanics

1) A continuous foundation is given in the following diagram. Determine the

A) 10.35m B) 13.5m C) 13.0m D) 12.5m

Solve this equation,

2. A continuous foundation is given in the following diagram. Determine the

A) 4.6m B) 3.85m C) 5.2m D) 6.2m

N =30, ɣ =22, =18

Solve this equation,

3. A continuous square foundation is given in the following diagram. Determine

N =58, Nɣ =70, N =45

Solve this equation,

4. A continuous foundation is given in the following diagram. Determine the

A) 19kN/m3 B) 18kN/m3 C) 16kN/m3 D) 16.5kN/m3

Solve this equation,

A) 19kN/m3 B) 18kN/m3 C) 10.6kN/m3 D) 17kN/m3

Solve this equation,

Step 3: Solve the gross allowable bearing capacity of this foundation,

Step 3: Solve the gross allowable bearing capacity of this foundation,

A) 4.0m B) 3.5m C) 5.3m D) 4.5m

Step 3: Solve the gross allowable bearing capacity of this foundation,

A) 370kN B) 520kN C) 450kN D) 320kN

Step 3: Solve the gross allowable bearing capacity of this foundation,

Solve this equation we got,

A) 150kN B) 130kN C) 170kN D) 110kN

Step 3: Solve the gross allowable bearing capacity of this foundation,

A) 250kN B) 230kN C) 270kN D) 210kN

Step 3: Solve the gross allowable bearing capacity of this foundation,

Solve this equation we got,

12)A continuous foundation is given in the following. It is known that the

Step 3: Net allowable bearing load can be calculated by,

13)A continuous foundation is given in the following. It is known that the

A) 7.6m B) 8.3m C) 9.4m D) 10.2m

Step 3: Net allowable bearing load can be calculated by,

Solve the equations above we got,

14)A continuous foundation is given in the following. It is known that the

Step 3: Net allowable bearing load can be calculated by,

Solve the equations above we got,

15)A continuous foundation is given in the following. It is known that the

Step 3: Net allowable bearing load can be calculated by,

Solve the equations above we got,

A) 0 kPa B)23 kPa C) 35 kPa D) 51 kPa

A) 23 kPa B) 35 kPa C) 51 kPa D) 88 kPa

sat =18 kN/m3

A) 5.2 kPa B) 10.1 kPa C)16.3 kPa D) 18.5 kPa

sat =20 kN/m3

15.9 kPa 58.8 kPa

the lateral force (per unit length of wall) on the wall, , is

A) 20 kips B) 33 kips C) 60 kips D) 72 kips

A) 0.020 m B)0.040 m C) 0.060 m D) 0.080 m

H2=10 m Clay, sat=18 kN/m3

A) 0.4 m B) 1.2 m C) 2.2 m D) 3.4 m

24) An embankment constructed with a clay soil

25) An embankment constructed with

26) An uniform desposit soil is shown in the following diagram. Information of

Gs=2.6, S=53%, w = 30%,

H2=2m Gs=2.6, S=100%, w = 40%,

A) 90kPa B) 80kPa C) 85kPa D) 75kPa