1. The document provides examples of calculating consolidation parameters such as void ratio, coefficient of consolidation, and primary consolidation settlement from given soil testing data.
2. Parameters like initial void ratio, applied pressure, and thickness of soil layers are used to determine the change in stress and void ratio to then calculate settlement.
3. Several methods are presented to calculate the average effective stress and stress change at different points to then determine the consolidation settlement under different boundary conditions, stress histories, and soil properties.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Roof Truss Design (By Hamza Waheed UET Lahore )Hamza Waheed
This presentation defines, describes and presents the most effective and easy way to design a roof truss with all the necessary steps and calculations based on Allowable Stress Design. Soft-wares like MD Solids, Truss Analysis have been used. It is most convenient way to design a roof truss which is being the most important structural components of All types of steel bridges.
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
Democratizing Fuzzing at Scale by Abhishek Aryaabh.arya
Presented at NUS: Fuzzing and Software Security Summer School 2024
This keynote talks about the democratization of fuzzing at scale, highlighting the collaboration between open source communities, academia, and industry to advance the field of fuzzing. It delves into the history of fuzzing, the development of scalable fuzzing platforms, and the empowerment of community-driven research. The talk will further discuss recent advancements leveraging AI/ML and offer insights into the future evolution of the fuzzing landscape.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
Event Management System Vb Net Project Report.pdfKamal Acharya
In present era, the scopes of information technology growing with a very fast .We do not see any are untouched from this industry. The scope of information technology has become wider includes: Business and industry. Household Business, Communication, Education, Entertainment, Science, Medicine, Engineering, Distance Learning, Weather Forecasting. Carrier Searching and so on.
My project named “Event Management System” is software that store and maintained all events coordinated in college. It also helpful to print related reports. My project will help to record the events coordinated by faculties with their Name, Event subject, date & details in an efficient & effective ways.
In my system we have to make a system by which a user can record all events coordinated by a particular faculty. In our proposed system some more featured are added which differs it from the existing system such as security.
1. Example 1:
Given d = 5 cm, Ho = 1.9 cm, Ws = 58 g, Gs = 2.7, constant, c = 1*10-4
cm
σ'
(kg/cm2
) 0 0.25 0.5 1.0 2.0 4.0
R 100 199 256 358 520 635
Determine: 1) void ratio for loading 2) compute parameters of consolidation
Solution:
cm094.1
1*2.7(2.5)*π
58
γAG
W
H
*
2
ws
s
s ===
806.0094.19.1HHH sov =−=−=
737.0
1.094
0.806
H
H
e
s
v
o
===
009.0
094.1
10*1*)100199(
H
H
e
4
s
1
1 =
−
=
∆
=∆
−
728.0009.0737.0eee 1o1 =−=∆−=
014.0
094.1
10*1*)100256(
H
H
e
4
s
2
2 =
−
=
∆
=∆
−
723.0014.0737.0eee 2o2 =−=∆−=
And so on
σ'
(kg/cm2
) 0 0.25 0.5 1.0 2.0 4.0
R 100 199 256 358 520 635
ΔH, cm 0 0.0099 0.0156 0.0258 0.042 0.0535
e 0.737 0.728 0.723 0.713 0.699 0.688
From figure (1)
014.0
12
713.0699.0
Δσ'
Δe
av =
−
−
−=−= cm2
/kg
0081.0
737.01
014.0
e1
a
m
o
v
v =
+
=
+
= cm2
/kg
From figure (2)
047.0
1
2log
713.0699.0
Δlogσ'
Δe
cc =
−
−=−= and
σ'c = 0.7 kg/cm2
Example 2:
The following compression readings were obtained in an oedometer test on a specimen
of saturated clay:
σ'
(kN/m2
) 0 54 107 214 429 858 1716 3432 0
Dial Guage after
24 hrs, mm
5.00
4.74
7
4.49
3
4.108
3.44
9
2.608 1.676 0.737 1.480
The initial thickness of the specimen was 19.0 mm and at the end of the test the water
content was 19.8% and the specific gravity was 2.73. Plot e-logσ curve and determine
2. the coefficient of compressibility between 100-200 and 1000-1500 kN/m2
. What is the
value of cc for the latter increment?
Solution:
Void ratio at the end of the test = ef = ωGs= 0.198*2.73 = 0.541
Void ratio at the start of the test = eo = ef + ∆e
Now,
ee1
e
e1
e
H
H
foo
∆++
∆
=
+
∆
=
∆
350.0e
e0.5411
e
0.19
48.15
=∆⇒
∆++
∆
=
−
eo = ef + ∆e = 0.541 + 0.35 = 0.891
In general the relationship between ∆e and ∆H is given by:
891.01
e
19.0
H
e1
e
H
H
oo
+
∆
=
∆
⇒
+
∆
=
∆
H0995.0e ∆=∆⇒ , and can be used to obtain the void ratio at the end of each
increment period.
σ'
(kN/m2
) 0 54 107 214 429 858 1716 3432 0
Dial Guage
after 24 hrs,
mm
5.00
4.74
7
4.49
3
4.108
3.44
9
2.608 1.676 0.737 1.480
∆H, mm 0
0.25
3
0.50
7
0.892
1.55
1
2.392 3.324 4.263 3.520
∆e 0
0.02
5
0.05
0
0.089
0.15
4
0.238 0.331 0.424 0.350
e
0.89
1
0.86
6
0.84
1
0.802
0.73
7
0.653 0.560 0.467 0.541
e1 = eo – ∆e = 0.891 – 0.025 =0.866
0.891 – 0.05 = 0.841
The e-logσ curve using these values is shown in figure below. Using Casagrande’s
construction the value of the preconsolidation pressure is 325 kN/m3
.
Between the pressure range
100 and 200 kN/m2
e100= 0.845 e200= 0.808, therefore
/kNm10*7.3
100200
808.0845.0
Δσ'
Δe
a 24
v
−
=
−
−
−=−=
/kNm10*005.2
845.01
10*7.3
e1
a
m 24
4
v
v
−
−
=
+
=
+
=
Between the pressure range
1000 and 1500 kN/m2
e1000= 0.632 e1500= 0.577, therefore
/kNm10*1.1
10001500
577.0632.0
Δσ'
Δe
a 24
v
−
=
−
−
−=−=
3. /kNm10*74.6
632.01
10*1.1
e1
a
m 25
4
v
v
−
−
=
+
=
+
=
and
312.0
100
1500log
577.0632.0
Δlogσ'
Δe
cc =
−
−=−=
Example 3:
Compute the expected settlement of a building constructed on a soil layer of 2.5m
thick its e = 0.72, the pressure increased from 1.5 kg/cm2
to 3.8 kg/cm2
and caused
reduction in void ratio by 15%.
Solution:
e1 = 0.72(1 – 0.15) = 0.612
method 1
cm15.7m157.05.2*
72.01
0.6120.72
H.
e1
Δe
S
o
c
==
+
−
=
+
=
method 2
047.0
5.18.3
612.072.0
Δσ'
Δe
av =
−
−
==
027.0
72.01
047.0
e1
a
m
o
v
v =
+
=
+
=
cm15.5m0.1551.5)(3.8*2.5*0.027σHmS **vc
===′∆= −
Example 4:
For the system shown below, determine the primary consolidation settlement if cs =
0.05 for the following:
1) σ'c = 100 kPa 2) σ'c = 150 kPa 3) σ'c = 175 kPa 4) σ'c = 250 kPa
Solution:
1) σ'c = 100 kPa
σ'o at midpoint of the clay layer
2
o
kN/m150.85
9.7*19.81)5(19.718.7*5'σ
=
++= −
1663.0
85.150
100
σ
σ
OCR '
o
'
c
<=== therefore, under
]
'Δσσ'
σ
log
σ'
σΔσ
[log
e1
Hc
S
avo
'
o
o
'
o
o
c
c
−
+
′+
+
=
425.25)10085.150(
2
1
)σσ(
2
1'Δσ '
c
'
oav
=−== − kPa
4. 162mm0.162m]
25.425150.85
150.85
log
150.85
50150.85
[log
1.11
2*0.83
Sc
==+
+
+
=
−
2) σ'c = 150 kPa
1
85.150
150
σ
σ
OCR '
o
'
c
≈== normal
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm98m098.0
150.85
50150.85
log
1.11
2*0.83
Sc
==
+
+
=
3) σ'c = 175 kPa
16.1
85.150
175
σ
σ
OCR '
o
'
c ===
kPa200.855085.150σΔσ'
o =+=′+
σ'c = 175 kPa
Because kPa200.85σΔσ'
o =′+ > σ'c = 175 kPa
c
'
o
o
c
o
'
c
o
s
c σ'
σΔσ
log
e1
Hc
σ'
σ
log
e1
Hc
S
′+
+
+
+
=
mm50.4m.05040
175
50150.85
log
1.11
2*0.83
150.85
175
log
1.11
2*0.05
Sc
==
+
+
+
+
=
4) σ'c = 250 kPa
Because kPa200.85σΔσ'
o =′+ < σ'c = 250 kPa
o
'
o
o
s
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm6m0.006
150.85
50150.85
log
1.11
2*0.05
Sc
==
+
+
=
Example 5:
A soil profile shown in figure below, if a uniform distributed load,Δσ , is applied at
the ground surface, what is the settlement of the clay layer caused by primary
consolidation if:
1. The clay is normally consolidated
2. The preconsolidation pressure (σ'c) = 190 kN/m2
3. σ'c = 170 kN/m2
Use cs=1/6 cc
Solution:
1. The clay is normally consolidated
the average effective stress at the middle
5. of the clay layer is
2
o
kN/m79.14
9.81)-2(199.81)-4(1814*2'σ
=
++=
cc = 0.009(LL-10) = 0.009 (40 – 10) = 0.27
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm132m321.0
79.14
10079.14
log
0.81
4*0.27
Sc
==
+
+
=
2. The preconsolidation pressure (σ'c) = 190 kN/m2
2'
o kN/m179.14σΔσ =′+
σ'c = 190 kN/m2
Because 2'
o kN/m179.14σΔσ =′+ < σ'c = 190 kN/m2
o
'
o
o
s
c σ'
σΔσ
log
e1
Hc
S
′+
+
= , cs = 1/6 cc = 1/6 * 0.27 = 0.045
mm36m0.036
79.14
179.14
log
0.81
4*0.045
Sc
==
+
=
3. σ'c = 170 kN/m2
Because '
o
'
c
'
o σσσΔσ >>′+
c
'
o
o
c
o
'
c
o
s
c σ'
σΔσ
log
e1
Hc
σ'
σ
log
e1
Hc
S
′+
+
+
+
=
mm46.9m.04690
170
179.14
log
0.81
4*0.27
179.14
170
log
0.81
4*0.045
Sc
==
+
+
+
=
Example 6:
For the foundation shown, estimate the consolidation settlement.
Solution:
The average effective stress at the middle of the clay layer is
2
o
kN/m52.833
9.81)-(16
2
2.5
9.81)0.5(17.516.5*2.5'σ
=
++= −
Using approximate method to calculate the 'Δσ at
mid-point of clay layer
2
v kN/m445.13
)25.32)(25.31(
1*2*150
z)z)(L(B
qBL
Δσ =
++
=
++
=
Or using m,n method
15.0
25.3
5.0
z
L
m === , and 31.0
25.3
1
z
B
n === f(m,n) = 0.02
2
kN/m124*150*02.0'Δσ ==
1.0
0.5
2.5
Sand
B*L=1m*2m
G.W.T
?=16.5kN/m3
?sat=17.5kN/m
3
Sand
Clay(N.C.)
q=150kN/m
1.5
?sat=16kN/m3
cc=0.32
cs=0.09
eo=0.8
2
6. 1.5
1.5
2.5
Sand
B*L G.W.T
?=15kN/m
3
?sat=18kN/m
3
Sand
Clay(N.C.)
Gs=2.7,LL=38%
?=35%
Q
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm43.8m3804.0
52.833
13.44552.833
log
0.81
2.5*0.32
Sc
==
+
+
=
Example 7:
In the previous example, if the foundation is circular in shape with diameter of 2.5m,
estimate the consolidation settlement.
Solution:
Using approximate method to calculate the 'Δσ at mid point of clay layer
2
2
2
2
2
kN/m355.28
)25.35.2(
5.2*150
Z)(D
qD
σΔ =
+
=
+
=′
or using
2.6
1.25
3.25
R
z
== , and 0
1.25
0
R
x
== from figure 0.18
q
Δσv
≈
2
kN/m27150*18.0σΔ ==′
mm82.9m8290.0
52.833
28.35552.833
log
0.81
2.5*0.32
Sc
==
+
+
=
Example 8:
Refer to figure shown below, given that B=1.5m, L=2.5m, and Q=120 kN. Calculate
the primary consolidation settlement of the foundation.
Solution:
0.945
1
2.7*0.35
S
sωG
eo
===
3
w
s
sat
clay kN/m18.49.81*
0.9451
0.9452.7
γ
e1
eG
γ =
+
+
=
+
+
=
2
o
kN/m5.5234
9.81)-(18.4
2
2.5
9.81)-1.5(1815*1.5'σ
=
++=
'Δσ at the base of the foundation = 2
32kN/m
2.5*1.5
120
=
To determine the net consolidation pressure at mid height of clay layer under the
center of the foundation, the foundation's base must be divided into four equal
0.75*1.25 rectangular areas.
z = 1.5+2.5/2 = 2.75m from the foundation's base
45.0
75.2
25.1
z
L
m === , and 27.0
75.2
75.0
z
B
n === f(m,n) = 0.047
2
kN/m016.64*32*047.0'Δσ ==
2'
o kN/m51.5396.01645.523σΔσ =+=′+
cc = 0.009(LL-10) = 0.009 (38 – 10) = 0.252
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
7. mm4.17m0174.0
45.523
51.539
log
0.9451
2.5*0.252
Sc
==
+
=
Example 9:
A building is supporting on a raft 45m x 30m, the net foundation pressure being 125
kN/m2
. The soil profile is shown in figure below. The value of mv = 0.35 m2
/MN.
Determine the final settlement under the center of the raft due to consolidation of the
clay.
Solution: at mid point of clay layer,
z = 24 – 3.5 + 2 = 23.5m
96.0
5.23
5.22
z
L
m === , and 64.0
5.23
15
z
B
n === f(m,n) = 0.14
2
kN/m704*125*14.0'Δσ ==
mm98m0.09870*4*10*0.35σHmS 3-
**vc
===′∆=
Example 10:
A stratum of normally loaded clay of 7m thick is located at a depth 12m below
ground level. The (ωn) of the clay is 43% and its LL is 48%. The Gs of the solid particles
is 2.76. the water table is located at a depth of 5m below the ground surface. The soil is
sand above the clay stratum. The submerged unit weight of sand is 11 kN/m3
and the
same weighs 18 kN/m3
above water table. The average increase in pressure at the center
of the clay stratum is 120 kN/m2
due to the weight of a building that will be constructed
on the sand above the clay stratum. Estimate the expected settlement of the structure.
Solution:
1) Determination of e and γsub for clay
s
w
W
W
ω = , kN076.2781.9*76.2*1γGVW wsss ===
8. kN643.11076.27*43.0ωWW sw ===
kN719.38076.27643.11WWW swt =+=+=
187.1
1
76.2*43.0
S
ωG
e s
o
===
3
o
t
t kN/m704.17
187.11
719.38
e1
W
γ =
+
=
+
= , equal to γsat or wsat γ
e1
esG
γ
+
+
=
3
sub kN/m7.8949.8117.704γ =−=
2) Determination of overburden pressure '
oσ
2'
o kN/m194.6293.5*7.8947*1118*5σ =+++=
3) Compression index
cc = 0.009(LL-10) = 0.009(48-10) = 0.342
4) Excess pressure ∆σ' = 120 kN/m2
5) Total Settlement
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm228m228.0
194.629
120194.629
log
1.1871
7*0.342
Sc
==
+
+
=
Example 11:
Soil investigation at a site gave the following information. Top soil up to a depth of
10.6m is fine sand, and below this lies soft clay layer of 7.6m thick. The water table is
at 4.6m below the ground surface. The submerged unit weight of sand is 10.4 kN/m3
,
and wet unit weight above water table is 17.6 kN/m3
. The water content of normally
consolidated clay is 40%, its LL is 45% and Gs is 2.78. The proposed construction will
transmit a net pressure of 120 kN/m2
. Find the average settlement of the clay layer.
Solution:
cc = 0.009(LL-10) = 0.009(45-10) = 0.315
112.1
1
78.2*4.0
S
ωG
e s
o
===
3
wsat kN/m078.189.81*
1.1121
1.1122.78γ
e1
esG
γ =
+
+=
+
+
=
3
sub
kN/m8.2689.8118.078γ =−=
Effective overburden pressure at the mid height of the clay layer is
2'
o kN/m174.7783.8*8.26810.4*617.6*4.6σ =+++=
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm257m257.0
174.778
120174.778
log
1.1121
7.6*0.315
Sc
==
+
+
=
Example 12:
Two points on a curve for normally consolidated clay have the following
coordinates: Point 1: e1 = 0.7, σ1 = 1 kg/cm2
9. Point 2: e2 = 0.6, σ2 = 3 kg/cm2
If the average overburden pressure on a 6m thick clay layer is 1.5 kg/cm2
, how much
settlement the clay layer experience due to additional load intensity of 1.6 kg/cm2
.
Solution:
21.0
13log
6.07.0
σ
σ
log
ee
c
1
2
21
c =
−
=
−
=
We need the initial void ratio eo at an overburden pressure of 1.5 kg/cm2
.
5.1
3
log21.0)6.0e(21.0
σ
σ
log
ee
c o
o
2
2o
c =−⇒=
−
=
0.6630.0630.6eo =+=⇒
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm239m239.0
1.5
1.61.5
log
0.6631
6*0.21
Sc
==
+
+
=
Example 13:
A footing has a size of 3m by 1.5m and it causes a pressure increment of 200 kN/m2
at its base. Determine the consolidation settlement at the middle of the clay layer.
Assume 2:1 pressure distribution.
Solution:
2
'
o
kN/m1.885
5.19*1.58.19*0.516*2.5σ
=
+++=
2
v
kN/m692.27
)5.33)(5.35.1(
3*5.1*200
z)z)(L(B
qBL
Δσ
=
++
=
++
=
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm93m093.0
51.88
27.69251.88
log
0.81
3*0.3
Sc
==
+
+
=
10. Point 2: e2 = 0.6, σ2 = 3 kg/cm2
If the average overburden pressure on a 6m thick clay layer is 1.5 kg/cm2
, how much
settlement the clay layer experience due to additional load intensity of 1.6 kg/cm2
.
Solution:
21.0
13log
6.07.0
σ
σ
log
ee
c
1
2
21
c =
−
=
−
=
We need the initial void ratio eo at an overburden pressure of 1.5 kg/cm2
.
5.1
3
log21.0)6.0e(21.0
σ
σ
log
ee
c o
o
2
2o
c =−⇒=
−
=
0.6630.0630.6eo =+=⇒
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm239m239.0
1.5
1.61.5
log
0.6631
6*0.21
Sc
==
+
+
=
Example 13:
A footing has a size of 3m by 1.5m and it causes a pressure increment of 200 kN/m2
at its base. Determine the consolidation settlement at the middle of the clay layer.
Assume 2:1 pressure distribution.
Solution:
2
'
o
kN/m1.885
5.19*1.58.19*0.516*2.5σ
=
+++=
2
v
kN/m692.27
)5.33)(5.35.1(
3*5.1*200
z)z)(L(B
qBL
Δσ
=
++
=
++
=
o
'
o
o
c
c σ'
σΔσ
log
e1
Hc
S
′+
+
=
mm93m093.0
51.88
27.69251.88
log
0.81
3*0.3
Sc
==
+
+
=