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CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
1
HOMEWORK 2
Homework 2 is due on December 26, 2017, Tuesday at 16:59.
CE363 Homeworks are to be submitted to the βCE363 Homework boxβ in Soil
Mechanics Lab.
Soil Mechanics lab door is locked every day at 17:00 and homeworks cannot be
submitted under the door.
Homework solutions will be posted to the course website on December 26, 2017,
Tuesday at 17:00.
Unless otherwise stated, use Ξ³water = 10 kN/m3
Question 1 (a:15%, b:10%)
A very wide granular embankment that is 2-m-thick, having unit weight of 20 kN/m3, is placed
over the ground surface of a soft (normally consolidated) clay deposit that is 3-m thick. Below the
clay layer, there is impervious rock, and water table is at the ground surface of soft clay. Saturated
unit weight of clay is 18 kN/m3, it is fully saturated and its initial average moisture content is 23%,
Gs=2.67, compression index is 0.35 and coefficient of consolidation is 6 m2/year.
a) Plot time versus consolidation settlement of clay layer until average degree of
consolidation is 95%. Note that horizontal axis should be time (in months), and vertical
axis should be settlement (vertical axis should increase in downward direction).
b) After the fill is placed, at the time when the ground surface settlement is measured as 15
cm,
- what is the average degree of consolidation of the clay layer?
- what is the degree of consolidation at 2 m depth below the top of the clay?
- what is the total pore water pressure and excess pore water pressure at 2 m depth below
the top of clay?
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
2
EMBANKMENT
IMPERVIOUS ROCK
N.C SOFT CLAY
2 m
3 m
π₯π = 20 ππ π3β β 2 π = 40 πππ (ππ’π π‘π π£πππ¦ π€πππ ππππππππππ‘)
π0β² = 1.5 π β (18 ππ π3β β 10 ππ π3β ) = 12 πππ (ππππ‘πππππ¦ ππ‘ π‘βπ πππ β ππππ‘β ππ ππππ¦ πππ¦ππ)
ππβ² = 12 πππ + 40 πππ = 52 πππ (πππππ π π‘πππ π ππ‘ π‘βπ πππ β ππππ‘β ππ ππππ¦ πππ¦ππ)
Since π€ β πΊπ = ππ β π,
π0 = π€0 β πΊπ = 23% β 2.67 = 0.614 (ππ = 1, ππ’πππ¦ π ππ‘π’πππ‘ππ)
Given, πΆπ = 0.35 & the soil is normally consolidated,
ππ =πΆπ
1 + π0β π» β log (
ππβ²
π0β²) =
0.35
1 + 0.614β 3 β log (
52
12) = 0.414 π = 41.4 ππ
a) Consolidation settlement vs time graph is asked. Given,
ππ£ = 6 π2 π¦πππβ = 0.5 π2/ππππ‘β
ππ£ =ππ£ β π‘
π»2, π€βπππ π» (ππ π, π·) = 3 π (π ππππ π‘βπ πππ¦ππ ππ βπππ ππππ ππ. )
ππ£ πππ ππ πππππ‘ππ π€ππ‘β ππ£πππππ ππππππ ππ ππππ ππππππ‘πππ π€ππ‘β ππππππ€πππ πππππ’πππ :
πππ π < 60 % β ππ£ =π
4β π2
πππ π > 60% β ππ£ = β0.933 β log(1 β π) β 0.085
π»ππππ, π‘ β ππ£ β π β π
πΉππ πππ π‘ππππ,
π΄π‘ π‘ = 3 ππππ‘βπ , ππ£ =0.5 π2 ππππ‘ββ β 3 ππππ‘βπ
9 π2= 0.167
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
3
π = β4 β ππ£
π= β
4 β 0.167
3.14= 0.461 = 46.1 % (π < 60%)
π = 41.4 ππ β 46.1 % = 19.1 ππ
π΄π‘ π‘ = 12 ππππ‘βπ , ππ£ =0.5 π2 ππππ‘ββ β 12 ππππ‘βπ
9 π2= 0.667
π = 1 β 10β(ππ£+0.085)/0.933 = 1 β 10β0.667+0.085
0.933 = 0.844 = 84.4 % (π > 60%)
π = 41.4 ππ β 84.4 % = 34.9 ππ
Similarly,
t (months) Tv U Settlement (cm)
0 0.000 0.000 0.0
0.1 0.006 0.084 3.5
0.5 0.028 0.188 7.8
1 0.056 0.266 11.0
2 0.111 0.376 15.6
3 0.167 0.461 19.1
4 0.222 0.532 22.0
5 0.278 0.595 24.6
6 0.333 0.644 26.7
7 0.389 0.689 28.6
8 0.444 0.729 30.2
9 0.500 0.764 31.6
10 0.556 0.794 32.9
11 0.611 0.821 34.0
12 0.667 0.844 34.9
13 0.722 0.864 35.8
14 0.778 0.881 36.5
15 0.833 0.896 37.1
16 0.889 0.910 37.7
17 0.944 0.921 38.2
18 1.000 0.931 38.6
19 1.056 0.940 38.9
20 1.111 0.948 39.3
20.3 1.129 0.950 39.4
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
4
b) πβππ π = 15 ππ, π =15 ππ
41.4 ππ= 36.2 % πππππππ π πππππ ππ ππππππππ πππππ
ππ£ = π
4β (36.2 %)2 = 0.103
π·πππ‘β ππππ π‘ππ ππ π‘βπ ππππ¦ = 2 π, πππ₯πππ’π ππππππππ πππ π‘ππππ = 3 π
ππ§ = 0.14 , π πππππ ππ ππππππππ πππππ ππ‘ 2 π ππππ‘β πππππ€ π‘βπ π‘ππ ππ π‘βπ ππππ¦
0
5
10
15
20
25
30
35
40
45
0 2 4 6 8 10 12 14 16 18 20 22
Set
tlem
ent
(cm
)
Time (months)
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
5
ππ§ =π’ππ β π’π
π’ππ, π€βπππ π’ππ = ππππ‘πππ ππ₯πππ π ππππ ππππ π π’ππ, π’π = ππ₯πππ π ππππ ππππ π π’ππ
π’ππ = π₯π = 40 πππ
πΉππ ππ§ = 0.14 = 40 β π’π
40,
π’π = 34.4 πππ, ππ₯πππ π ππππ ππππ π π’ππ ππ‘ 2 π ππππ‘β πππππ€ π‘βπ π‘ππ ππ π‘βπ ππππ¦
π΅πππππ ππππππππππ‘ ππππ π‘ππ’ππ‘πππ, π’0 = 2 π β 10 ππ π3β = 20 πππ
πππ‘ππ ππππ π€ππ‘ππ ππππ π π’ππ π€βππ π = 15 ππ , π’π = 20 πππ + 34.4 πππ = 54.4 πππ
Question 2 (a: 14%; b: 5 %)
Following results are obtained in a series of CU triaxial tests performed on saturated samples of
Ankara clay.
a) Calculate the effective stress-based modified shear strength parameters, intercept a' and
slope angle ' , by filling up the following table and plotting the modified Mohr-Coulomb
failure envelopes on the chart provided below.
Test Number 1 2
Confining Pressure
(kPa) 40 80
Deviatoric Stress
(kPa) 120 170
Pore water pressure
at failure (kPa) 15 25
'3,f (kPa) 25 55
'1,f (kPa) 145 225
'1,f+'3,f)/2 (kPa) 85 140
'1,f-'3,f)/2 (kPa) 60 85
From the figure above: a' = 21.4 kPa & ' = 24.4o
sin(πβ²) = tan(πΌβ²) β sin(πβ²) = tan(24.4) β πβ² = 27Β°
πβ² =πβ²
cos(πβ²)=
21.4
cos(27)= 24 πππ
y = 0.4545x + 21.364
0
50
100
150
200
250
0 50 100 150 200 250
(1'-
3')
/2
(1'+3')/2
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
6
b) Vane shear test is used to determine the undrained shear strength of the same clay in Part
(a). If the clay is known to have a sensitivity ratio of 3 and undrained shear strength of
cu=35 kPa, estimate the peak and the residual torques. The shear vane used in the test has
a diameter of 75 mm and length of 150 mm.
)62
(32 DLD
cT u
πππππ = π β 35 β ((0.0752π₯0.15
2) + (
0.0753
6)) = 54.1 π. π
ππππ ππ‘ππ£ππ‘π¦ πππ‘ππ = πππππ
ππππ πππ’ππ= 3 β ππππ πππ’ππ = 18.0 π. π
Question 3 (4%)
A silty clay specimen was placed in a triaxial cell and sheared as part of an unconfined
compression test. If it is sheared at deviator stress of 50 kN/m2,
Draw the Mohr circle corresponding to failure state. Estimate the followings:
i. Confining stress, Ο3β¦β¦β¦0β¦β¦β¦. ...kPa........
ii. Major principal stress, Ο1β¦β¦50β¦β¦β¦β¦. .kPa..........
iii. Unconfined compressive strength, quβ¦β¦β¦50β¦β¦β¦. .kPa.
iv. Undrained shear strength, cuβ¦β¦25β¦β¦β¦β¦. ..kPa.........
0
10
20
30
40
50
60
0 10 20 30 40 50 60
t(k
Pa)
(kPa)
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
7
Question 4 (a: 7%, b: 3%)
The one dimensional consolidation test data for a soil sample is given below:
Effective stress (kPa) 25 50 100 150 200 250 300
void ratio 0.96 0.93 0.89 0.84 0.78 0.71 0.65
The pre-consolidation pressure of the soil sample is πππ ππ·π and initial void ratio is given as
eo=0.98.
a) Determine the coefficient of volume of compressibility, compression index, and reload
(recompression) index assuming that the initial effective stress on the sample is 155 kPa
and the final effective stress will be 245 kPa.
ππ =1
1+π0β (
βπ
βπ) =
1
1+0.84β (
0.84β0.71
250β150) = 7.06 β 10β4π2/ππ (e0 for Ο = 150 kPa)
πΆπ =(π0 β πππ)
log Ο π β log ππ=
βπ
log ππ/ππ=
(0.93 β 0.89)
log (10050
)= 0.133
The closest void ratio data for stress level of 155 kPa and 245 kPa are 150 kPa and 250 kPa in the
table given above. Therefore Cc is calculated based on these stress levels. Alternatively you may
determine void ratio values for 155 kPa and 245 kPa and calculate Cc.
0.50
0.60
0.70
0.80
0.90
1.00
10 60 110 160 210 260 310 360
e
Ο' (kPa)
e-Ο'
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
8
πΆπ =(πππ β π)
log π β log ππ=
βπ
log π/ππ=
(0.84 β 0.71)
log (250150
)= 0.586
b) Calculate the consolidation settlement of a 5 m-thick clay layer in part (a) if initial and final
effective stresses at mid-depth are ππ ππ·π πππ πππ ππ·π respectively.
ππ =πΆπ
(1 + ππ)β π» β log (
ππ
ππ) +
πΆπ
(1 + ππ)β π» β log (
ππ
ππ)
ππ =0.133
(1 + 0.98)β 5 β log (
115
50) +
0.586
(1 + 0.98)β 5 β log (
200
115)
ππ = 0.121 + 0.356 = 0.477 π = 47.7 ππ
Question 5 (a:4%, b:4%, c:4%)
A direct shear apparatus has ππ ππ sample diameter. Following data were recorded at failure
during the tests caried out over the samples of a clayey sand using this device.
Do not perform area correction
π΅ (ππ΅) π» (ππ΅) π (ππ·π) π (ππ·π)
Test 1 0.15 0.15
Test 2 0.45 0.36
a) Complete the table and plot the failure envelope of the tested soil. When plotting, use an
appropriate scaling for the axes and clearly indicate the stresses defining the envelope.
Ο =N
A=
150 N
3117.2 ππ2= 0.048 πππ = 48 πππ
Ο =T
A=
150 N
3117.2 ππ2= 0.048 πππ = 48 πππ
Ο =N
A=
450 N
3117.2 ππ2= 0.1444 πππ = 144.4 πππ
Ο =T
A=
360 N
3117.2 ππ2= 0.1155 πππ = 115.5 πππ
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
9
π΅ (ππ΅) π» (ππ΅) π (ππ·π) π (ππ·π)
Test 1 0.15 0.15 48 48
Test 2 0.45 0.36 144.4 115.5
b) Calculate the shear strength parameters of the clayey sand.
β β² = ππππ‘ππ (βπ¦
βπ₯) = ππππ‘ππ (
βπ¦
βπ₯) = ππππ‘ππ (
115.5 β 48
144.4 β 48) = ππππ‘ππ (
67.5
96.4)
β β 35 Β°
π = πβ² + πβ² β π‘ππβ β²
115.5 = πβ² + 144.4 β π‘ππ35
πβ² β 14.4 πππ
c) At what shear stress would this soil fail if the normal stress is ππ ππ·π.
π = πβ² + πβ² β π‘ππβ β²
π = 14.4 + 85 β π‘ππ35 = 73.9 πππ
0
20
40
60
80
100
120
140
0 20 40 60 80 100 120 140 160
Ο
(kP
a)
Ο (kPa)
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
10
Question 6 (a:6%, b:4%)
During a consolidated undrained triaxial test on a saturated clay sample, failure is reached at a
principle stress difference of 385 kPa. The consolidation (all-round) pressure was 250 kPa and the
pore pressure was measured at failure as 145 kPa.
a) Plot the total and effective stress Mohr circles corresponding to the stress state at failure.
Use an appropriate scale when drawing and clearly indicate the necessary data to define
the circles.
Given
βπ = 385 πππ
π3 = 250 πππ
π’ = 145 πππ
π3 = 250 πππ
π1 = π3 + βπ = 250 + 385 = 635 πππ
πΒ΄1 = π1 β π’ = 635 β 145 = 490 πππ
πΒ΄3 = π3 β π’ = 250 β 145 = 105 πππ
b) If the effective angle of internal shear of this clay is known to be 30Β°, calculate the effective
cohesion parameter.
π πππΒ΄ =
πΒ΄1 β πΒ΄3
2πΒ΄1 + πΒ΄3
2+
πΒ΄π‘πππΒ΄
π ππ30 =
490 β 1052
490 + 1052
+πΒ΄
π‘ππ30
Solving for πΒ΄ = 50.5 πππ
0
50
100
150
200
250
300
350
400
450
500
-200 -100 0 100 200 300 400 500 600 700 800 900 1000
t(k
Pa)
, ' (kPa)
TSCESC
105 kPa 250 kPa
u
635 kPa
cΒ΄=50.5 kPa
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
11
Question 7 (a:12%, b:4%, c:4%)
Consider the retaining wall shown in the figure below:
a) Using Rankineβs theory of lateral earth pressure calculate and plot the active and passive
earth pressure distributions at the back and in front of the wall respectively.
πΎππ πππβ1 =1 β π ππβ β²
1 + π ππβ β²=
1 β π ππ27
1 + π ππ27= 0.375
πΎππ πππβ2 =1 β π ππβ β²
1 + π ππβ β²=
1 β π ππ30
1 + π ππ30= 0.333
πΎππ πππβ2 =1 + π ππβ β²
1 β π ππβ β²=
1 + π ππ30
1 β π ππ30= 3
Calculate the active pressure (z=depth from ground surface)
πββ² = ππ£
β² β πΎπ β 2 β πβ² β βπΎπ
πββ² = ππ£
β² β πΎπ + 2 β πβ² β βπΎπ
π΄π‘ π§ = 0 π; πββ² = 18 β 0 β 0.375 β 2 β 10 β β0.375 = β12.25 πππ
π΄π‘ π§ = 4 π; πββ² = 18 β 4 β 0.375 β 2 β 10 β β0.375 = 14.75 πππ
π΄π‘ π§ = 4 π; πββ² = 18 β 4 β 0.333 β 2 β 0 β β0.333 = 24 πππ
π΄π‘ π§ = 7π; πββ² = ((18 β 4 + (21 β 10) β 3) β 0.333 β 2 β 0 β β0.333 = 35 πππ
4 m
3 m
Soil-1
cΒ΄=10 kPa
ΟΒ΄=27Β°
Ξ³=18 kN/m3
Soil-2
cΒ΄=0 kPa
ΟΒ΄=30Β°
Ξ³sat=21 kN/m3
CE 363 Soil Mechanics Middle East Technical University
Homework 2 Department of Civil Engineering
12
Calculate the passive pressure (z=depth from ground surface)
π΄π‘ π§ = 0 π; πββ² = (21 β 10) β 0 β 3 + 2 β 0 β β3 = 0 πππ
π΄π‘ π§ = 3 π; πββ² = (21 β 10) β 3 β 3 + 2 β 0 β β3 = 99 πππ
4 m
3 m
12.25 kPa
14.75 kPa
24 kPa
35 kPa 30 kPa30 kPa 99 kPa
b) Calculate the depth of tension crack at the back of the wall.
πββ² = ππ£
β² β πΎπ β 2 β πβ² β βπΎπ = 0
πββ² = 0 = 18 β π§ β 0.375 β 2 β 10 β β0.375
π§ =12.25
18 β 0.375= 1.81 π
c) Calculate the total passive resistance in front of the wall.
Total passive resistance (including water pressure)
πΉπ = (99 β 3 β1
2) = 148.5 ππ/π
πΉπ€ = 30 β 3 β1
2= 45 ππ/π
Active Pressure
Distribution Passive Pressure
Distribution
Hydrostatic Pressure
Distribution
Hydrostatic Pressure
Distribution
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