New Jersey Institute of Technology Soil Mech Consolidation Lab Report

  

Consolidation data
Initial specimen height, 𝐻t(i) (in): 0.872
Diameter, in: 2.572
Dry weight of sample, g: 103
Specific Gravity Solids: 2.7,
where 1 ton/ft2= 2000lb/ft2
Correction coefficient of consolidation device (Beam ratio): 11
Consolidation Dial Gauge: Figure 1
Order of Loads applied: 2kg, 8.5kg, 16.5kg, 10.5kg and 20.5kg
The class will start at 6.05am
Class 6
341 A – Soil Mechanics Lab
Ehsan Mehryaar, e-mail – em355@njit.edu
Ghiwa Assaf, email – ga338@njit.edu
1
§ Agenda
Class 6
§ Basic notions
§ Consolidation test (Experiment + Data + 1st part of table)
§ Consolidation Calculations (2nd part of the table)
§ Consolidation Write up (Report)
2
Decrease in void
ratio
Compaction
Basic Notions
•
•
•
Compressive mechanical energy to
densify the soil
Soil solids are rearranged in a
denser arrangement, by removal of
air-filled porosity
Improve engineering
characteristics
Consolidation
•
•
•
•
Static loads applied to saturated
soils
Over a period of time the increased
stresses are transferred to the soil
skeleton
Usually to undisturbed soil
deposits that has appreciable
amount of fines
Removal of water-filled porosity
3
Soil Settlement
𝜹𝑻𝒐𝒕𝒂𝒍 = 𝜹𝒆 + 𝜹𝒑 + 𝜹𝒔
Primary
consolidation
𝜹𝒑
Elastic
𝜹𝒆
•
Basic Notions
•
•
•
•
•
•
Occurs immediately
after a load in applied
No change in moisture
content
Elastic deformation
Use of Elastic theory
Important for granular
soils
Not important for
saturated clays
(permeability is too
low)
Effects are usually
neglected
Different categories of
soil settlement cause
by loads
•
•
Due to gradual
dissipation of pore
pressure, induced by
external loading
Volume change: due to
expulsion of water
from the soil mass
Secondary
Consolidation
𝜹𝒔
•
•
∆𝒖 ≠ 𝟎
•
Important for inorganic
clays and saturated
fine-grained soils with
low coefficient of
permeability
Occurs at constant
effective stress
Volume change: due to
adjustment of soil
fabric, which is the
rearrangement of
particles
∆𝒖 = 𝟎
•
Important for organic
soils
4
Spring-cylinder
model
Basic Notions
Primary
consolidation
Secondary
consolidation
•
Assumptions:
• Deformation of the compressive soil layer only occurs in one dimension
• 100% saturation for most settlement problems
•
A soil will compress when loaded, because:
• The soil grains deform
• The soil grains relocate
• The water and air is squeezed from the voids
• As pore fluid is squeezed out, the soil grains will rearrange themselves into
a more stable and denser configuration
5
Stage I: Elastic Settlement:
Initial compression due to preloading
Stage II: Primary consolidation
Excess pore water pressure gradually
transferred into effective stress due to
expulsion of pore water
Basic Notions
Stage III: Secondary consolidation
Occurs after complete dissipation of
the excess pore water pressure,
caused by plastic readjustment of soil
fabric
6
§ Standard: ASTM D-2435
§ One-dimension (1D) field consolidation can be simulated in laboratory
§ The experiment is performed in a consolidometer (also referred to as
oedometer)
Consolidation
#17
7
Sample
https://mediaspace.njit.edu/playlist/dedicated/177
447701/1_mo91yrnt/1_jvn50t84 (the loading of
the sample is manual)
https://www.youtube.com/watch?v=3bvevFBNY
w0 (the loading of the sample is automatic)
Experiment
Consolidation
#17
8
!
Consolidation
#17
Dial Gauge:
Inner and Outer Diameters
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐷𝑖𝑎𝑙 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 =
= 𝐼𝑛𝑠𝑖𝑑𝑒 𝑅𝑖𝑛𝑔× 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛×#𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑖𝑛𝑠𝑖𝑑𝑒 𝑟𝑖𝑛𝑔 ×#𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑖𝑛𝑔 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 +
+ 𝑂𝑢𝑠𝑖𝑑𝑒 𝑅𝑖𝑛𝑔× 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 ×#𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑖𝑛𝑔 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
9
Sieve Analysis
#4
Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in
reverse order and in regular scale)
10
Initial specimen height, 𝐻$ (&)
Height of solids, 𝐻(
𝑊( (𝑑𝑟𝑦 𝑤𝑒𝑖𝑔ℎ𝑡)
𝐻( = 𝜋
×𝐷 ) ×𝐺( 𝜌*
4
Consolidation
#17
11
1st Part
12
Consolidation
#17
1
2
3
Pressure
Final
dial
reading
Change
in
specimen
height
1st PART
4
5
6
Final
specimen
height
Height Final
of
void
void ratio
𝐻%
mm
—
P
𝑡𝑜𝑛/𝑓𝑡 (
mm
∆𝐻
mm
0
A
—
𝐻! (#)
mm
𝐻! ())
—
—
I=B-A
—
0.25
B
—
P=𝐻! ()) − 𝐼
—
—
J=C-B
—
0.5
C
—
Q=P-J
—
—
K=D-C
—
1
D
—
R=Q-K
—
—
L=E-D
—
2
E
—
S=R-L
—
—
M=F-E
—
4
F
—
T=S-M
—
—
N=G-F
—
8
G
—
U=T-N
—
—
O=H-G
—
16
H
—
V=U-O
—
—
—
—
—
—
𝑒
—
—
—
—
—
—
—
—
7
Average
height during
consolidation
8
9
2nd PART
10
Fitting
time
11
Cv
𝐻! (&%’)
mm
t90
sec
t50
sec
t90
𝑚𝑚( /𝑠𝑒𝑐
t50
𝑚𝑚( /𝑠𝑒𝑐
–𝐻! ()) + 𝑃
2
–𝑃+𝑄
2
–𝑄+𝑅
2
–𝑅+𝑆
2
–𝑆+𝑇
2
–𝑈+𝑇
2
–𝑉+𝑈
2
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
–13
1
2
3
Pressure
Final
dial
reading
Change
in
specimen
height
Pressure
𝑃=
𝐹𝑜𝑟𝑐𝑒 𝐿𝑜𝑎𝑑×𝐵𝑒𝑎𝑚 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
=
𝐴𝑟𝑒𝑎
𝐴𝑟𝑒𝑎
Height of voids
𝐻+ = 𝐻$ (,) − 𝐻(
Initial specimen height, 𝐻$ (&)
Final void ratio
𝑒=
𝐻+
𝐻(
1st PART
4
5
6
Final
specimen
height
Height Final
of
void
void ratio
𝐻%
mm
—
P
𝑡𝑜𝑛/𝑓𝑡 (
mm
∆𝐻
mm
0
A
—
𝐻! (#)
mm
𝐻! ())
—
—
I=B-A
—
0.25
B
—
P=𝐻! ()) − 𝐼
—
—
J=C-B
—
0.5
C
—
Q=P-J
—
—
K=D-C
—
1
D
—
R=Q-K
—
—
L=E-D
—
2
E
—
S=R-L
—
—
M=F-E
—
4
F
—
T=S-M
—
—
N=G-F
—
8
G
—
U=T-N
—
—
O=H-G
—
16
H
—
V=U-O
—
—
—
—
—
—
𝑒
—
—
—
—
—
—
—
—
7
Average
height during
consolidation
8
9
2nd PART
10
Fitting
time
11
Cv
𝐻! (&%’)
mm
t90
sec
t50
sec
t90
𝑚𝑚( /𝑠𝑒𝑐
t50
L( /t
–𝐻! ()) + 𝑃
2
–𝑃+𝑄
2
–𝑄+𝑅
2
–𝑅+𝑆
2
–𝑆+𝑇
2
–𝑈+𝑇
2
–𝑉+𝑈
2
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
–14
2nd Part
15
90% consolidation
50% consolidation
t90: Dial reading vs 𝑡 (the XX axis corresponds the
square root of time, the YY axis should be in reverse
order; and both axis must be in regular scale)
t50: Dial reading vs t (the XX axis corresponds to the
time and must be in log scale; the YY axis should be
in reverse order and in regular scale)
1
2
3
Pressure
Final
dial
reading
Change
in
specimen
height
1st PART
4
5
6
Final
specimen
height
Height Final
of
void
void ratio
𝐻%
mm
—
P
𝑡𝑜𝑛/𝑓𝑡 (
mm
∆𝐻
mm
0
A
—
𝐻! (#)
mm
𝐻! ())
—
—
I=B-A
—
0.25
B
—
P=𝐻! ()) − 𝐼
—
—
J=C-B
—
0.5
C
—
Q=P-J
—
—
K=D-C
—
1
D
—
R=Q-K
—
—
L=E-D
—
2
E
—
S=R-L
—
—
M=F-E
—
4
F
—
T=S-M
—
—
N=G-F
—
8
G
—
U=T-N
—
—
O=H-G
—
16
H
—
V=U-O
—
—
—
—
—
—
𝑒
—
—
—
—
—
—
—
—
7
Average
height during
consolidation
8
9
2nd PART
10
Fitting
time
11
Cv
𝐻! (&%’)
mm
t90
sec
t50
sec
t90
𝑚𝑚( /𝑠𝑒𝑐
t50
𝑚𝑚( /𝑠𝑒𝑐
–𝐻! ()) + 𝑃
2
–𝑃+𝑄
2
–𝑄+𝑅
2
–𝑅+𝑆
2
–𝑆+𝑇
2
–𝑈+𝑇
2
–𝑉+𝑈
2
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
–16
90% consolidation
A- initial point of the curve in blue
1.
Choose point A. In this case use the initial point of the curve in blue, where it intersects the
YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
17
90% consolidation
A- initial point of the curve in blue
Line of reference
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
A- initial point of the curve in blue
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
18
90% consolidation
A- initial point of the curve in blue
A
B
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
19
90% consolidation
A- initial point of the curve in blue
𝐵=3
A
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
20
90% consolidation
A- initial point of the curve in blue
C = 3.45
𝐵=3
A
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
21
90% consolidation
A- initial point of the curve in blue
C = 3.45
𝐵=3
A
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
22
90% consolidation
𝑎𝑏𝑐𝑖𝑠𝑠𝑎 𝐷 =
𝑡*+ = 2
𝑡*+ = 4𝑚𝑖𝑛
A- initial point of the curve in blue
C = 3.45
𝐵=3
A
1.
Choose point A and a line of reference. In this case use the initial point of the curve in blue,
where it intersects the YY axis.
t90: Dial reading vs 𝑡 (the XX axis corresponds the square
root of time, the YY axis should be in reverse order; and
both axis must be in regular scale)
2.
Draw a line 𝐴𝐵 through the early portion of the curve.
3.
Draw a line 𝐴𝐶 such that 𝑂𝐶 = 1.15 𝑂𝐵.
4.
The abscissa of point 𝐷, which is the intersection of 𝐴𝐶 and the consolidation curve, gives
the square root of time for 90% consolidation 𝑡*+ .
23
50% consolidation
𝑥
𝑥
Where it 𝑡𝑖𝑚𝑒 = 0? → 𝑑𝑡- = 4𝑡,
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
24
50% consolidation
𝑥
𝑥
𝑡, = 1
𝑡( = 4
𝑡, = 0.25
𝑡( = 1
𝑡- = 4𝑡,
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
25
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑥 = 25𝐸 − 3
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
26
50% consolidation
𝑥
𝑥
Line DE
𝑡- = 4𝑡,
𝑥 = 25𝐸 − 3 = 25×10!”
𝑥 = 25𝐸 − 3
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
27
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑! = 875𝐸 − 3𝑚𝑚
𝑥 = 25𝐸 − 3
𝑥 = 25𝐸 − 3
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
28
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑!
= 875
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
29
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑!
= 875
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
30
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑!
= 875
1.
The initial curved portion of the plot of deformation versus log t is approximated to be a
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
𝑑”!!
= 960
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
𝐴
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
31
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑!
= 875
𝐹
𝑑#!
= 917.5
1.
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
𝑑”!!
= 960
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
The initial curved portion of the plot of deformation versus log t is approximated to be a
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
32
50% consolidation
𝑥
𝑥
𝑡- = 4𝑡,
𝑑!
= 875
𝑡#!
= 3𝑚𝑖𝑛
𝐹
𝑑#!
= 917.5
1.
parabola on the natural scale. Select times 𝑡, and 𝑡- on the curved portion such that 𝑡- = 4𝑡,
𝑑”!!
= 960
Let the difference of specimen deformation during time (𝑡- − 𝑡, ) be equal to 𝑥.
2.
t50: Dial reading vs t (the XX axis corresponds to the time
and must be in log scale; the YY axis should be in reverse
order and in regular scale)
The initial curved portion of the plot of deformation versus log t is approximated to be a
Draw a horizontal line 𝐷𝐸 such that the vertical distance 𝐵𝐷 is equal to 𝑥. The deformation
corresponding to the line 𝐷𝐸 is 𝑑0 (that is, deformation at 0% consolidation).
3.
Extend the straight-line portions of primary and secondary consolidations to intersect at 𝐴.
The ordinate of 𝐴 is represented by 𝑑,++ that is, the deformation at the end of 100% primary
consolidation.
4.
The ordinate of point 𝐹 on the consolidation curve represents the deformation at 50% primary
consolidation, and its abscissa represents the corresponding time (𝑡50).
33
90% consolidation
50% consolidation
t90: Dial reading vs 𝑡 (the XX axis corresponds the
square root of time, the YY axis should be in reverse
order; and both axis must be in regular scale)
t50: Dial reading vs t (the XX axis corresponds to the
time and must be in log scale; the YY axis should be
in reverse order and in regular scale)
1
2
3
Pressure
Final
dial
reading
Change
in
specimen
height
1st PART
4
5
6
Final
specimen
height
Height Final
of
void
void ratio
𝐻%
mm
—
P
𝑡𝑜𝑛/𝑓𝑡 (
mm
∆𝐻
mm
0
A
—
𝐻! (#)
mm
𝐻! ())
—
—
I=B-A
—
0.25
B
—
P=𝐻! ()) − 𝐼
—
—
J=C-B
—
0.5
C
—
Q=P-J
—
—
K=D-C
—
1
D
—
R=Q-K
—
—
L=E-D
—
2
E
—
S=R-L
—
—
M=F-E
—
4
F
—
T=S-M
—
—
N=G-F
—
8
G
—
U=T-N
—
—
O=H-G
—
16
H
—
V=U-O
—
—
—
—
—
—
𝑒
—
—
—
—
—
—
—
—
7
Average
height during
consolidation
8
9
2nd PART
10
Fitting
time
11
Cv
𝐻! (&%’)
mm
t90
sec
t50
sec
t90
𝑚𝑚( /𝑠𝑒𝑐
t50
𝑚𝑚( /𝑠𝑒𝑐
–𝐻! ()) + 𝑃
2
–𝑃+𝑄
2
–𝑄+𝑅
2
–𝑅+𝑆
2
–𝑆+𝑇
2
–𝑈+𝑇
2
–𝑉+𝑈
2
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
–34
We can use both 𝑡.- and 𝑡/- to measure 𝑐+ . The equation for time
factor of consolidation is
𝑇+ =
𝑐+ 𝑡
)
𝐻01
𝐻01 =
𝐻$ 2+3
2
Where 𝐻01 is average longest drainage path
Consolidation
#17
𝐻! &%’
Where for 50% consolidation 𝑇+ = 𝑇.- = 0.197. Therefore, we have
)
0.197𝐻01
𝑐+ =
𝑡.-
For 90% consolidation 𝑇+ = 𝑇/- = 0.848
)
0.848𝐻01
𝑐+ =
𝑡/-
35
Practical Engineering Method:
𝐶A 𝑜𝑟 𝐶B
• not based on scientific facts
• rule of thumb
1
𝐶?
1
Consolidation
#17
𝐶A 𝑜𝑟 𝐶B
1
𝜎′@
Pre-consolidation
pressure
Step 1 – Draw the tangent to initial part of the graph.
Step 2 – Draw the tangent to second part of the graph.
Step 3 – The abscissa of intersection of the two tangent is the pre-consolidation pressure
36
𝐶A 𝑜𝑟 𝐶B
1
∆𝑒1
Recompression index or Swell index
0.8
Consolidation
#17
0.6
∆𝑒(
𝐶1 =
∆𝑒4
𝐶( =
,783 1.
,-1/
𝐶?
1
𝐶A 𝑜𝑟 𝐶B
∆61
𝐶1 =
∆60
,-
783,-0.
0/
∆61
7839:1.;7839:1/
1
Compression index
𝐶4 =
∆6+
,783 +.
,-+/
𝜎′A,
𝜎′?,
𝜎′A-
𝜎′B,
𝑠𝑙𝑜𝑝𝑒 =
𝜎′B-
𝜎′?-
𝑠𝑙𝑜𝑝𝑒 =
𝑌- − 𝑌,
𝑋- − 𝑋,
𝑌- − 𝑌,
𝑌- − 𝑌,
=
log(𝑋- ) − log(𝑋, ) 𝑙𝑜𝑔 𝑋𝑋,
37
Recompression index or Swell index
𝐶1 =
∆61
,-
783,-1.
1/
𝐶( =
∆60
,783 0.
,-0/
Compression index
𝐶4 =
∆6+
,-
783,-+.
+/
Consolidation
#17
38
Report
Sample Calculations
10%
Results including graphs and tables
20%
Discussion
20%
Summary and Conclusions
10%
References
2%
Quality of Presentation, graphs, tables etc.
8%
Please check outline
39
§ Das, B.M. 2002. Soil Mechanics Laboratory Manual. Oxford University Press.
§ Das, B.M., and K. Sobhan. 2018. Principles of Geotechnical Engineering. 9th ed. Cengage Learning
§ Ronaluna. 2012. “1D Consolidation Test”. 2012. https://www.youtube.com/watch?v=3bvevFBNYw0
References
§
METUGeotech. 2012. “Soil Mechanics Laboratory Tests: Consolidation Test. 2012.

§ Kaliakin, V. N. (2017). Chapter 8—Example problems related to compressibility and settlement of
soils. Soil mechanics. Butterworth-Heinemann, London, 331-376
§ Holtz, R.D., W.D. Kovacs, and T.C Sheahan. 1981. An Introduction to Geotechnical Engineering. Second.
Pearson
40
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