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CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
SOLUTION OF HOMEWORK 3
Question 1 (25%)
A 4m x 4m footing will be constructed on
a site underlain by the given soil profile.
Both the Standard Penetration Test and
Cone Penetration Test results are available
for the site. According to the CPT results,
the average cone tip resistances for the first
and second sand layers are 4 MPa and 5
MPa, respectively. The SPT is repeated at
several depths, the depths and the SPT-N
numbers are given in the table below. The
unit weight of the sand is 20 kN/m3 and no
ground water is observed at the site.
Test depth from the surface
3
5
7
10
SPT-N60
6
7
12
15
a) Calculate the settlement of the footing using Burland and Burbidge (1985) Method.
Ignore all of the correction factors. Use the average SPT-N value for the depth of
interest. The SPT-N values given in the table are N60 (without overburden
correction).
b) Calculate the settlement of the footing 1 year after the construction using
Schmertmann’s Method. Show all calculations necessary to fill out the table and use
the results from the table for the rest of your calculations.
Layer
No
Iz
Es
(Unit=………)
Δz
(Unit=………)
Iz xΔz / Es
1
2
c) What would be the settlement value you use for design? Please explain your reason.
1 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Solution 1
a)
.
1.71
.
6
210 ∗ 4
12 15
4
1.71
0.068
10 .
∗ 0.068 37.68
.
7
10
≅ 3.8
b)
Layer
1
4000
2
5000
Note that;
 for square footing;
0
0.1
B/2
0.5
2B
0.0
where; B : width of footing
 2.0
S C
 :
1
C
C
C
I
z
E
C C q
1
0.5

q
0.2 log
0.1
3 ∗ 20
1 0.5
0.86
210
1
1 0.2 log
1.2
0.1
2 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Layer
No
1
Depth
(m)
3.0-5.0

(m)
2.0
2
5.0-11.0
6.0
4000
Es
(kPa)
8000
0.3
7.5 ∗ 10
5000
10000
0.25
1.5 ∗ 10
Iz

22.5 ∗ 10
S C C q
I
z
E
0.86 ∗ 1.2 ∗ 210 ∗ 22.5 ∗ 10
0.0487m ≅ 4.9cm
c)
OPTION 1:
As the designer of this foundation, to be on the safe side (and to be prepared for the
worst case scenario), consider the highest settlement value calculated in part (a) using
SPT and part (b) using CPT data. = 49 mm
OPTION 2:
In order to give equal value to both methods, average of the immediate settlement values
calculated in part (a) using SPT and part (b) using CPT data will be considered = 43 mm
Note that allowable settlement for raft foundations in sand is 65 mm, and for single
foundations in sand 40 mm"
3 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Question 2 (15%)
For the cantilever pile wall presented in the figure given below, estimate the depth of
penetration, d, after penalizing the passive resistance by a factor of safety 1.5
Solution 2
SAND
K
tan
45
K
P
z
z
z
tan
1
K
1
0.25
γz
0m → P
5m → P
9m → P
∅
2
20
20
5 ∗ 18
q K
45
37
2
0.25
4.02
2c K
20 0.25
5kPa
5 ∗ 18 0.25
4 ∗ 19
27.5kPa
10 0.25
36.5kPa
4 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
CLAY
∅
P
z
z
9
γz
20
9m → P
dm → P
20
z
γz
0m → P
dm → P
q K
5 ∗ 18
5 ∗ 18
P
z
K
0→K
18d
1
2c
K
4 ∗ 19 1
4 ∗ 19
q K
2 ∗ 90
18d 1
2c
2 ∗ 90
2 ∗ 90
6kPa
2 ∗ 90
6
18dkPa
K
180kPa
180
18dkPa
Apply FS=1.5 to passive resistance
z
z
0m → P
dm → P
180
1.5
180 18d
1.5
120kPa
120
12dkPa
5 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Force (kN/m)
Force (kN/m)
H1
H1
H2
H2
H3
H3
H4
5*5=25
(27.5-5)*5/2=56.25
27.5*4
(36.5-27.5)*4/2=18
6*d
18d*d/2=9d2
(4)*10*(4)/2=80
Passive
120*d
12d*d/2=6d2
H5
H5
M
0→d
Moment arm to Moment (kN.m/m)
point A
6.5+d
162.5+25d
5.67+d
318.9375+56.25d
2+d
220+110d
1.33+d
23.94+18d
d/2
3d2
d/3
3d3
(4/3+d)
106.67+80d
d/2
d/3
60d2
2d3
8m
Question 3 (10%)
Section of an anchored sheet-pile wall is shown in in the figure below.
a) Using a factor of safety of 2.5 against the passive resistance, calculate the required depth
of penetration d for the sheet-pile.
b) Assuming the depth of penetration is d=6 m, calculate the required horizontal force, H,
at point A. Do not change the given penetration depth in your calculations.
6 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
c) Assume the required horizontal force at point A is 120 kN/m. If that lateral support is
to be provided by anchors placed at every 3 m intervals and inclined 20o from the
horizontal, calculate the axial force T in each anchor.
Solution 3
a)
K
tan
45
K
∅
2
1
K
tan
1
0.26
45
36
2
0.26
3.85
γz q K
2c K
0kPa
z 0m → P
8 ∗ 19 0.26 39.5kPa
z 8m → P
8 ∗ 19 10d 0.26 39.5 2.6dkPa
z 8 dm → P
γz q K
P
0kPa
z 0m → P
10 d 3.85 38.5dkPa
z dm → P
8
8
d
2d
39.5 ∗ ∗ 6
39.5 ∗ d ∗ 6
2.6 ∗ d ∗ d/2 ∗
M
2
3
2
3
d 2d
6
38.5 ∗ d ∗
2 3
0
2.5
d 6.7m
P
6
"Since water pressures on both sides of the wall are equal, their moments are not
considered"
7 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
a)
M
39.5 ∗
8
8
6
∗ 6
39.5 ∗ 6 ∗
2.6 ∗ 6 ∗ 6/2 ∗ 6/3
2
3
2
38.5 ∗ 6 ∗ 6/2 6/3
H ∗ 12 0
2.5
H 135kN
"A factor of safety of 2.5 will be applied to passive pressure"
b)
120 ∗
3
cos 20
383kN
Question 4 (25%)
There are two videos in this question, the links of which are given below:
Video 1:http://www.youtube.com/playlist?p=PLB468342B03660375
Video 2:
https://www.facebook.com/507784339262303/videos/970703199637079/?type=1&theater
The first video was taken from a residence construction in Ankara. In this project, a deep
excavation was necessary, where the lateral stability of soil was ensured using piles. In the link,
there are total of 14 video pieces uploaded on YouTube website, which were recorded during
the construction stages of one of these piles.
In the second video link, pile construction simulation is given (You can also watch them on the
Video Wall of CE department.)
Please answer the following questions to test your knowledge about piles. (In Video 1, as you
watch them, the questions will appear on the screen).
1. In video 1, what kind of pile (driven or cast-in-situ), do you think, was constructed on
the site? Answer the same question for Video 2.
2. For Video 1, describe the first stage of pile construction using your own words?
3. Why do you think these plastic parts are attached to the steel bars? A similar component
is also shown somewhere in Video2, what are those components called in Video 2?
4. Why do you think installation of a case is necessary for constructing the pile?
5. What is the difference between Video 1 and Video 2 in terms of installing temporary
casing? What is the limit for inclination of casing (therefore the pile) in Video 2?
6. As you hear in Video 1, there are two steel reinforcements with lengths 12 m and 11
m, the vertical overlap distance of these reinforcements is 1.5 m. What would be the
length of longitudinal steel bars that will stay outside the pile?
7. In Video 1, the workers are trying to keep the steel bars on the air. Why do you think
they do that?
8 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
8. In Video 1, where do you think water flooding out of the pile casing come from? In
Video, at what time of total of 4:55 minutes this happens?
Since the idea of this question is to take you to the construction site and show you the details
of constructing a pile, you are expected to answer any 6 out of these 8 questions. Only one
sentence is OK for properly answering a question.
Solution 4
1. Cast in-situ pile is constructed.
2. The longitudinal and spiral reinforcements are bonded according to prescribed
alignment/spacing/diameter.
3. The spacers are installed to maintain the spacing between the casing/edge of the pile
and the longitudinal reinforcements. – In the second video, these are called U-bars.
4. Case installation is necessary, mostly for cohesion-less soils to prevent cave-in
problem (i.e., loose soil particles drops into the borehole), in other words
stabilization of the borehole during the pile construction up to pouring concrete.
5. In Video 1, the pile casing in installed through rotating the casing into the soil using
the pile machine. In Video 2 however, it is directly installed into soil. No details are
given. The limits of inclination angle is 1/100.
6. 12 + 11 – 1.5 = 21.5 m is the total length of the combined steel bar. The length of
the borehole/pile is 21 m (from the video sound records). In addition, the steel bars
are installed on air to be able to fill concrete at the bottom of the pile. Therefore, at
least 0.5 m is supposed to be stay outside of the bore hole.
7. There may be several reasons: (1) to prevent some dirty/ particles stuck on the
tendons which may seriously lead to corrosion and decrease the capacity of the
reinforcement, (2) to satisfy perfect bonding between the reinforcement and
concrete. (iii) to prevent steel bars entering into the ground.
8. Water comes from the original ground water table which fills the hole after
excavating the borehole. In Video 2, this happens in between 3:47 – 3:50.
9 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Question 5 (25%)
N-J-N Engineering Inc. consulting company hires you for their project in Bangkok, Thailand.
The company provides you the following video. Watch the video at the link below:
https://www.youtube.com/watch?v=PP8tLAwv1kE
a) At a time when the pile driving machine was broken, the workers, joyfully, drove the
pile into ground. Assume that each Thai men weighs 75 kg and jumps from 3 cm drop
height. Using a Thailand-modified ENR pile driving formula, calculate the allowable
load capacity of the pile (kN). Qallowable = 3 · Wr · H / (s+0.5·c) .... where c is a constant
25 mm, and s is the “average set”, assume 10 mm/drop.
b) Assume that the soft Bangkok clay typically has a constant undrained shear strength of
20 kPa. Assume that the pile is a 30-cm diameter, hollow, concrete pile with inner
diameter of 20 cm with a length of 5 m. The soil will be plugged inside the hollow area
and will move together with the pile, therefore the pile can be considered as a closeended pile. Calculate the ultimate pile capacity, Qult, using the static formulas.
c) A narrow excavation with a 5-m-width, in front of these piles will be made in saturated
Bangkok clay with undrained shear strength of 20 kPa, and a unit weight of 18 kN/m3.
There is no hard-layer underneath, it is all soft Bangkok clay. Calculate the factor of
safety against base stability.
10 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Solution 5
a) There are 6 Thai men and each weighs 75 kg, total weight of “drop hammer”, Wr= 6×75
= 450 kg = 4414 N. Drop height is 0.03 m.
Qallowable = 3 × 4414 × 0.03 / (0.01 + 0.5 × 0.025) = 17656 N = 18 kN
b) Undrained shear strength of Bangkok clay = cu = 20 kPa.
Pile diameter D = 0.03 m, and pile length L = 5 m
Qult = Skin friction + Tip resistance = Unit skin friction × clyndrical surface area of the
pile + unit tip resistance × tip area of the pile.
Qult = (α× cu × π × D × L) + (9 × cu × π × D2/4)
Qult = (1.0× 20 × π × 0.3 × 5) + (9 × 20 × π × 0.32/4) = 107 kN
For cu = 20 kPa α = 1.0 according to the graph on page 215 of CE366 Lecture Notes.
c) If there is no hard layer underneath the excavation, the width of the failure zone should
be taken as B1 = 0.707 B, where B is the width of the excavation.
B1 = 0.707 × 5 = 3.5 m
Check the depth of tension cracks: σha = 0,
σha = σv · KA – 2 · c · √KA = 0
Since ϕ = 0, KA = 1.0
The depth of tension crack, z is:
z = 2 · c / γ = 2 · 20 / 18 = 2.2 m
11 CE 366 Foundation Engineering
Solution of Homework 3
Middle East Technical University
Department of Civil Engineering
Force (kN/m)
Moment arm about point
b (m)
B1/2 = 3.5 / 2 = 1.75
B1 = 3.5
Moment about point
b (kN.m/m)
-551.25
385
Weight = γ· H · B1 = 18 · 5 · 3.5 = 315
S1 = (2 · π · B1/4) · cu
= (2 · π · 3.5 / 4) · 20 = 110
S2 = (H - z) · cu = (5 - 2.2) · 20 = 56
B1 = 3.5
196
Passive resistance*
245
B1/2 = 3.5 / 2 = 1.75
= (2 · cu ) · B1 = (2 · 20) · 3.5 = 140
* when calculating the passive resistance, the triangular part of the passive lateral earth pressure
distribution is ignored (σhp = σv · KP + 2 · c · √KP = γ· z + 2 · cu ). This is, in a way, similar to
applying a reduction (FS) to passive resistance.
385
196 245
551.25
1.498
12