Consolidation data

Initial specimen height, ð»t(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 â€“ Soil Mechanics Lab

Ehsan Mehryaar, e-mail â€“ em355@njit.edu

Ghiwa Assaf, email â€“ ga338@njit.edu

1

Â§ Agenda

Class 6

Â§ Basic notions

Â§ Consolidation test (Experiment + Data + 1st part of table)

Â§ Consolidation Calculations (2nd part of the table)

Â§ Consolidation Write up (Report)

2

Decrease in void

ratio

Compaction

Basic Notions

â€¢

â€¢

â€¢

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

â€¢

â€¢

â€¢

â€¢

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

ðœ¹ð‘»ð’ð’•ð’‚ð’ = ðœ¹ð’† + ðœ¹ð’‘ + ðœ¹ð’”

Primary

consolidation

ðœ¹ð’‘

Elastic

ðœ¹ð’†

â€¢

Basic Notions

â€¢

â€¢

â€¢

â€¢

â€¢

â€¢

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

â€¢

â€¢

Due to gradual

dissipation of pore

pressure, induced by

external loading

Volume change: due to

expulsion of water

from the soil mass

Secondary

Consolidation

ðœ¹ð’”

â€¢

â€¢

âˆ†ð’– â‰ ðŸŽ

â€¢

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

âˆ†ð’– = ðŸŽ

â€¢

Important for organic

soils

4

Spring-cylinder

model

Basic Notions

Primary

consolidation

Secondary

consolidation

â€¢

Assumptions:

â€¢ Deformation of the compressive soil layer only occurs in one dimension

â€¢ 100% saturation for most settlement problems

â€¢

A soil will compress when loaded, because:

â€¢ The soil grains deform

â€¢ The soil grains relocate

â€¢ The water and air is squeezed from the voids

â€¢ 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

Â§ Standard: ASTM D-2435

Â§ One-dimension (1D) field consolidation can be simulated in laboratory

Â§ 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

ð‘‰ð‘’ð‘Ÿð‘¡ð‘–ð‘ð‘Žð‘™ ð·ð‘–ð‘Žð‘™ ð‘…ð‘’ð‘Žð‘‘ð‘–ð‘›ð‘” =

= ð¼ð‘›ð‘ ð‘–ð‘‘ð‘’ ð‘…ð‘–ð‘›ð‘”Ã— ð‘ ð‘šð‘Žð‘™ð‘™ð‘’ð‘ ð‘¡ ð‘‘ð‘–ð‘šð‘’ð‘›ð‘ ð‘–ð‘œð‘›Ã—#ð‘‘ð‘–ð‘£ð‘–ð‘ ð‘–ð‘œð‘›ð‘ ð‘œð‘“ ð‘–ð‘›ð‘ ð‘–ð‘‘ð‘’ ð‘Ÿð‘–ð‘›ð‘” Ã—#ð‘‘ð‘–ð‘£ð‘–ð‘ ð‘–ð‘œð‘›ð‘ ð‘œð‘“ ð‘œð‘¢ð‘¡ð‘ ð‘–ð‘‘ð‘’ ð‘Ÿð‘–ð‘›ð‘” ð‘ð‘’ð‘¡ð‘¤ð‘’ð‘’ð‘› 2 ð‘ð‘œð‘›ð‘ ð‘’ð‘ð‘¢ð‘¡ð‘–ð‘£ð‘’ ð‘›ð‘¢ð‘šð‘ð‘’ð‘Ÿð‘ +

+ ð‘‚ð‘¢ð‘ ð‘–ð‘‘ð‘’ ð‘…ð‘–ð‘›ð‘”Ã— ð‘ ð‘šð‘Žð‘™ð‘™ð‘’ð‘ ð‘¡ ð‘‘ð‘–ð‘šð‘’ð‘›ð‘ ð‘–ð‘œð‘› Ã—#ð‘‘ð‘–ð‘£ð‘–ð‘ ð‘–ð‘œð‘›ð‘ ð‘œð‘“ ð‘œð‘¢ð‘¡ð‘ ð‘–ð‘‘ð‘’ ð‘Ÿð‘–ð‘›ð‘” ð‘ð‘’ð‘¡ð‘¤ð‘’ð‘’ð‘› 2 ð‘ð‘œð‘›ð‘ ð‘’ð‘ð‘¢ð‘¡ð‘–ð‘£ð‘’ ð‘›ð‘¢ð‘šð‘ð‘’ð‘Ÿð‘

9

Sieve Analysis

#4

Dial reading vs ð‘¡ (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, ð»$ (&)

Height of solids, ð»(

ð‘Š( (ð‘‘ð‘Ÿð‘¦ ð‘¤ð‘’ð‘–ð‘”â„Žð‘¡)

ð»( = ðœ‹

Ã—ð· ) Ã—ðº( ðœŒ*

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

ð»%

mm

—

P

ð‘¡ð‘œð‘›/ð‘“ð‘¡ (

mm

âˆ†ð»

mm

0

A

—

ð»! (#)

mm

ð»! ())

—

—

I=B-A

—

0.25

B

—

P=ð»! ()) âˆ’ ð¼

—

—

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

—

—

—

—

—

—

ð‘’

—

—

—

—

—

—

—

—

7

Average

height during

consolidation

8

9

2nd PART

10

Fitting

time

11

Cv

ð»! (&%’)

mm

t90

sec

t50

sec

t90

ð‘šð‘š( /ð‘ ð‘’ð‘

t50

ð‘šð‘š( /ð‘ ð‘’ð‘

–ð»! ()) + ð‘ƒ

2

–ð‘ƒ+ð‘„

2

–ð‘„+ð‘…

2

–ð‘…+ð‘†

2

–ð‘†+ð‘‡

2

–ð‘ˆ+ð‘‡

2

–ð‘‰+ð‘ˆ

2

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–13

1

2

3

Pressure

Final

dial

reading

Change

in

specimen

height

Pressure

ð‘ƒ=

ð¹ð‘œð‘Ÿð‘ð‘’ ð¿ð‘œð‘Žð‘‘Ã—ðµð‘’ð‘Žð‘š ð‘ð‘œð‘’ð‘“ð‘“ð‘–ð‘ð‘–ð‘’ð‘›ð‘¡

=

ð´ð‘Ÿð‘’ð‘Ž

ð´ð‘Ÿð‘’ð‘Ž

Height of voids

ð»+ = ð»$ (,) âˆ’ ð»(

Initial specimen height, ð»$ (&)

Final void ratio

ð‘’=

ð»+

ð»(

1st PART

4

5

6

Final

specimen

height

Height Final

of

void

void ratio

ð»%

mm

—

P

ð‘¡ð‘œð‘›/ð‘“ð‘¡ (

mm

âˆ†ð»

mm

0

A

—

ð»! (#)

mm

ð»! ())

—

—

I=B-A

—

0.25

B

—

P=ð»! ()) âˆ’ ð¼

—

—

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

—

—

—

—

—

—

ð‘’

—

—

—

—

—

—

—

—

7

Average

height during

consolidation

8

9

2nd PART

10

Fitting

time

11

Cv

ð»! (&%’)

mm

t90

sec

t50

sec

t90

ð‘šð‘š( /ð‘ ð‘’ð‘

t50

L( /t

–ð»! ()) + ð‘ƒ

2

–ð‘ƒ+ð‘„

2

–ð‘„+ð‘…

2

–ð‘…+ð‘†

2

–ð‘†+ð‘‡

2

–ð‘ˆ+ð‘‡

2

–ð‘‰+ð‘ˆ

2

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–14

2nd Part

15

90% consolidation

50% consolidation

t90: Dial reading vs ð‘¡ (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

ð»%

mm

—

P

ð‘¡ð‘œð‘›/ð‘“ð‘¡ (

mm

âˆ†ð»

mm

0

A

—

ð»! (#)

mm

ð»! ())

—

—

I=B-A

—

0.25

B

—

P=ð»! ()) âˆ’ ð¼

—

—

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

—

—

—

—

—

—

ð‘’

—

—

—

—

—

—

—

—

7

Average

height during

consolidation

8

9

2nd PART

10

Fitting

time

11

Cv

ð»! (&%’)

mm

t90

sec

t50

sec

t90

ð‘šð‘š( /ð‘ ð‘’ð‘

t50

ð‘šð‘š( /ð‘ ð‘’ð‘

–ð»! ()) + ð‘ƒ

2

–ð‘ƒ+ð‘„

2

–ð‘„+ð‘…

2

–ð‘…+ð‘†

2

–ð‘†+ð‘‡

2

–ð‘ˆ+ð‘‡

2

–ð‘‰+ð‘ˆ

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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

17

90% consolidation

A- initial point of the curve in blue

Line of reference

t90: Dial reading vs ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

19

90% consolidation

A- initial point of the curve in blue

ðµ=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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

20

90% consolidation

A- initial point of the curve in blue

C = 3.45

ðµ=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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

21

90% consolidation

A- initial point of the curve in blue

C = 3.45

ðµ=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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

22

90% consolidation

ð‘Žð‘ð‘ð‘–ð‘ ð‘ ð‘Ž ð· =

ð‘¡*+ = 2

ð‘¡*+ = 4ð‘šð‘–ð‘›

A- initial point of the curve in blue

C = 3.45

ðµ=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 ð‘¡ (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 ð´ðµ through the early portion of the curve.

3.

Draw a line ð´ð¶ such that ð‘‚ð¶ = 1.15 ð‘‚ðµ.

4.

The abscissa of point ð·, which is the intersection of ð´ð¶ and the consolidation curve, gives

the square root of time for 90% consolidation ð‘¡*+ .

23

50% consolidation

ð‘¥

ð‘¥

Where it ð‘¡ð‘–ð‘šð‘’ = 0? â†’ ð‘‘ð‘¡- = 4ð‘¡,

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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

24

50% consolidation

ð‘¥

ð‘¥

ð‘¡, = 1

ð‘¡( = 4

ð‘¡, = 0.25

ð‘¡( = 1

ð‘¡- = 4ð‘¡,

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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

25

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘¥ = 25ð¸ âˆ’ 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

26

50% consolidation

ð‘¥

ð‘¥

Line DE

ð‘¡- = 4ð‘¡,

ð‘¥ = 25ð¸ âˆ’ 3 = 25Ã—10!”

ð‘¥ = 25ð¸ âˆ’ 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

27

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘! = 875ð¸ âˆ’ 3ð‘šð‘š

ð‘¥ = 25ð¸ âˆ’ 3

ð‘¥ = 25ð¸ âˆ’ 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

28

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘!

= 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

29

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘!

= 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

30

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘!

= 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 ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

ð‘‘”!!

= 960

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

ð´

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

31

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘!

= 875

ð¹

ð‘‘#!

= 917.5

1.

parabola on the natural scale. Select times ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

ð‘‘”!!

= 960

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

32

50% consolidation

ð‘¥

ð‘¥

ð‘¡- = 4ð‘¡,

ð‘‘!

= 875

ð‘¡#!

= 3ð‘šð‘–ð‘›

ð¹

ð‘‘#!

= 917.5

1.

parabola on the natural scale. Select times ð‘¡, and ð‘¡- on the curved portion such that ð‘¡- = 4ð‘¡,

ð‘‘”!!

= 960

Let the difference of specimen deformation during time (ð‘¡- âˆ’ ð‘¡, ) be equal to ð‘¥.

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 ð·ð¸ such that the vertical distance ðµð· is equal to ð‘¥. The deformation

corresponding to the line ð·ð¸ is ð‘‘0 (that is, deformation at 0% consolidation).

3.

Extend the straight-line portions of primary and secondary consolidations to intersect at ð´.

The ordinate of ð´ is represented by ð‘‘,++ that is, the deformation at the end of 100% primary

consolidation.

4.

The ordinate of point ð¹ on the consolidation curve represents the deformation at 50% primary

consolidation, and its abscissa represents the corresponding time (ð‘¡50).

33

90% consolidation

50% consolidation

t90: Dial reading vs ð‘¡ (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

ð»%

mm

—

P

ð‘¡ð‘œð‘›/ð‘“ð‘¡ (

mm

âˆ†ð»

mm

0

A

—

ð»! (#)

mm

ð»! ())

—

—

I=B-A

—

0.25

B

—

P=ð»! ()) âˆ’ ð¼

—

—

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

—

—

—

—

—

—

ð‘’

—

—

—

—

—

—

—

—

7

Average

height during

consolidation

8

9

2nd PART

10

Fitting

time

11

Cv

ð»! (&%’)

mm

t90

sec

t50

sec

t90

ð‘šð‘š( /ð‘ ð‘’ð‘

t50

ð‘šð‘š( /ð‘ ð‘’ð‘

–ð»! ()) + ð‘ƒ

2

–ð‘ƒ+ð‘„

2

–ð‘„+ð‘…

2

–ð‘…+ð‘†

2

–ð‘†+ð‘‡

2

–ð‘ˆ+ð‘‡

2

–ð‘‰+ð‘ˆ

2

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–34

We can use both ð‘¡.- and ð‘¡/- to measure ð‘+ . The equation for time

factor of consolidation is

ð‘‡+ =

ð‘+ ð‘¡

)

ð»01

ð»01 =

ð»$ 2+3

2

Where ð»01 is average longest drainage path

Consolidation

#17

ð»! &%’

Where for 50% consolidation ð‘‡+ = ð‘‡.- = 0.197. Therefore, we have

)

0.197ð»01

ð‘+ =

ð‘¡.-

For 90% consolidation ð‘‡+ = ð‘‡/- = 0.848

)

0.848ð»01

ð‘+ =

ð‘¡/-

35

Practical Engineering Method:

ð¶A ð‘œð‘Ÿ ð¶B

â€¢ not based on scientific facts

â€¢ rule of thumb

1

ð¶?

1

Consolidation

#17

ð¶A ð‘œð‘Ÿ ð¶B

1

ðœŽâ€²@

Pre-consolidation

pressure

Step 1 â€“ Draw the tangent to initial part of the graph.

Step 2 – Draw the tangent to second part of the graph.

Step 3 â€“ The abscissa of intersection of the two tangent is the pre-consolidation pressure

36

ð¶A ð‘œð‘Ÿ ð¶B

1

âˆ†ð‘’1

Recompression index or Swell index

0.8

Consolidation

#17

0.6

âˆ†ð‘’(

ð¶1 =

âˆ†ð‘’4

ð¶( =

,783 1.

,-1/

ð¶?

1

ð¶A ð‘œð‘Ÿ ð¶B

âˆ†61

ð¶1 =

âˆ†60

,-

783,-0.

0/

âˆ†61

7839:1.;7839:1/

1

Compression index

ð¶4 =

âˆ†6+

,783 +.

,-+/

ðœŽâ€²A,

ðœŽâ€²?,

ðœŽâ€²A-

ðœŽâ€²B,

ð‘ ð‘™ð‘œð‘ð‘’ =

ðœŽâ€²B-

ðœŽâ€²?-

ð‘ ð‘™ð‘œð‘ð‘’ =

ð‘Œ- âˆ’ ð‘Œ,

ð‘‹- âˆ’ ð‘‹,

ð‘Œ- âˆ’ ð‘Œ,

ð‘Œ- âˆ’ ð‘Œ,

=

log(ð‘‹- ) âˆ’ log(ð‘‹, ) ð‘™ð‘œð‘” ð‘‹ð‘‹,

37

Recompression index or Swell index

ð¶1 =

âˆ†61

,-

783,-1.

1/

ð¶( =

âˆ†60

,783 0.

,-0/

Compression index

ð¶4 =

âˆ†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

Â§ Das, B.M. 2002. Soil Mechanics Laboratory Manual. Oxford University Press.

Â§ Das, B.M., and K. Sobhan. 2018. Principles of Geotechnical Engineering. 9th ed. Cengage Learning

Â§ Ronaluna. 2012. â€œ1D Consolidation Testâ€. 2012. https://www.youtube.com/watch?v=3bvevFBNYw0

References

Â§

METUGeotech. 2012. â€œSoil Mechanics Laboratory Tests: Consolidation Test. 2012.

Â§ Kaliakin, V. N. (2017). Chapter 8â€”Example problems related to compressibility and settlement of

soils. Soil mechanics. Butterworth-Heinemann, London, 331-376

Â§ Holtz, R.D., W.D. Kovacs, and T.C Sheahan. 1981. An Introduction to Geotechnical Engineering. Second.

Pearson

40

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