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Don’t forget to visit our companion site http://www.vulcanhammer.org

Use subject to the terms and conditions of the respective websites.

ENCE 3610ENCE 3610Soil MechanicsSoil Mechanics

Lecture 9Settlement and Consolidation of Soils

Overview

Soil is a nonhomogeneous porous material consisting of three phases Solids Fluid (normally water) Air

Soil deformation may occur by change in: Stress Water content Soil mass Temperature

When soil deformation takes place below a foundation, it results in settlement

Settlement of soils below foundations is the most common cause of foundation failure

Stages of Settlement Elastic Settlement Se:

settlement due to immediate elastic deformation before major particle rearrangement

Primary Consolidation Settlement Sc: settlement due to major particle rearrangement

Secondary Consolidation Settlement Ss: further particle rearrangement after primary settlement

Total Settlement St = Se + Sc + Ss

Other Types of Settlement Dynamic forces Expansive soil Collapsible soil

Types of Settlement

Settlement and Consolidation

Volume change in soils is not instantaneous: it takes place over time

How long it takes depends upon the type of soil and its permeability

Terms Settlement refers to the

distance the structure moves

Consolidation refers to the time it takes to move a certain distance

Soil Type Considerations Rate of consolidation

depends upon permeability of the soil

Cohesive soils are slow, use classic consolidation theory

Cohesionless soils are more rapid

Hough’s Method (ENCE 3610)

Schmertmann’s Method (ENCE 4610,) incorporates elasticity considerations

Elastic Settlement

Based on theory of elasticity

Load applied at a point or over an area on a semi-infinite half space

Can estimate both deflections and stresses

Theory of Boussinesq most commonly used; will discuss stresses later

Initial Settlement:Perloff's Method Variables

q = unit load on foundation (assuming uniform loading)

B = major dimension of foundation

I = combined influence factor

ν = Poisson’s Ratio Es = Modulus of

Elasticity of the soil

sc E

qBIS21

Variables for Perloff’s Method

Perloff Method Example

Solution (using method presented) Since method

determines settlement at corners, divide up foundations into quarters, determine settlement at corners and add using superposition

Perloff’s Method Example

Use 10’ x 10’ Foundation B = L = 10’ H = 10’ L/B = 1 H/B = 1 ν = 0.5 From table, I = 0.15

'9.020

5.0115.01044

1

2

2

sc E

qBIS

Overview of Primary Settlement and Consolidation Consolidation is the process

of gradual transfer of an applied load from the pore water to the soil structure as pore water is squeezed out of the voids.

The amount of water that escapes depends on the size of the load and compressibility of the soil.

The rate at which it escapes depends on the coefficient of permeability, thickness, and compressibility of the soil.

Two questions need to be answered by consolidation analysis: How much will the structure

settle? Generally referred to

as the “settlement” question

How long will it take for it to do so? Generally referred to

as the “consolidation” question

Principle of Consolidation

Consolidation Test

A test intended to replicate the process of primary (and secondary) settlement (distance) and time (consolidation) in the field

Necessary to run with undisturbed specimens, as disturbance changes the soil skeleton structure and thus the results

Apparatus

Consolidometer The consolidometer

has a rigid base, a consolidation ring, porous stones, a rigid loading plate, and a support for a dial indicator

Fixed or floating ring

Preparation of Specimen

Procedure

Consecutive Loads on Specimen

.5 ksf 1 ksf 2 ksf 4 ksf 8 ksf 16 ksf 32 ksf

Times to Record Data for Each Load

0.1 min 0.2 min 0.5 min 1.0 min 2.0 min 4.0 min 8.0 min 15 min

30 min

1 hr

2 hr

4 hr

8 hr

24 hr

Record the dial reading for each load increment at completion of primary consolidation for each load (usually 24 hours)

Remove load in decrements, recording dial indicator readings

Conduct a water content test on specimen after unloading is complete

Typical Results of Consolidation Tests Interpreted as a Stress-Strain Problem

The plot is semi-logarithmic with the ordinate (y-axis) logarithmic and the abscissa (x-axis) natural.

The stress-strain analogy is physically exact but does not quite correspond to theory of elasticity because the compression is not elastic (and not path independent.)

This is commonly used in the Netherlands, and is also the basis for Hough’s Method.

Typical Consolidation Results Expressed as a Function of Void Ratio

Frequently e0→

Note axes reversed

Used primarily for the consolidation of saturated clays

Primary Consolidation Settlement

Se→

Primary Compression Settlement

Strain can be defined as ε = Se/H

Primary Compression Settlement is thus more commonly written as

(normally consolidated soils) Ho = original height of

layer being compressed

eo = original void ratio

o

o

o

cc e

HCS

log1

Non-Laboratory Estimates of Compression Index

Note that “virgin” consolidation does not “kick in” until after an effective stress point is passed. The soil has been compressed to some degree by that effective stress before we add additional stress

Normally Consolidated Soils

These are used if full consolidation tests are either not possible, or for preliminary analysis

Sample Settlement Problem(Cohesive Soils) Given Two-Layer Soil Profile

Clean well-graded fine to coarse sand layer above clay layer and at surface N60 = 20 35' thick Unit weight of soil above

phreatic surface = 110 pcf Submerged weight of soil

below water table= 65 pcf Depth of water table = 15’

Clay stratum, normally consolidated LL= 45 25' thick Water Content = 40% Specific Gravity of solids

= 2.78

Given Building placed on

surface Square Foundation, 30’

x 30’ Uniform Load of 10 ksf

on entire foundation

Find Average settlement of

building due to primary settlement of clay layer

Sample Settlement Problem(Cohesive Soils)

Step 1: Compute Effective Stresses for Soil Profile

Sample Settlement Problem(Cohesive Soils) Step 2: Compute

Foundation Load for Center of Layer Depth of Center of

Layer = (15 + 20 + 25/2) = 47.5’

Step 3: Compute Compression Index and Void Ratio for Clay Layer Compression Index

Void Ratio

psf 1498

305.47

1

10000

1

2

2

=Δσ

+

=Δσ

B

z+

q=Δσ

v

v

v

Sample Settlement Problem(Cohesive Soils) Step 4: Compute Settlement Using Formula

for Cohesive Soils

"75.6

3607

14983607log

112.11

25315.012

log1

c

c

o

o

o

cc

S

S

e

HCS

Overconsolidation or Preconsolidation

☞ Overconsolidation complicates the analysis of primary settlement because the ground being compressed acts as if the previous overburden pressure is still being applied, irrespective of current overburden conditions

☞ Requires dividing the settlement analysis into two parts

☞ Sometimes referred to as preconsolidation

The ratio of the maximum overburden stress a soil has experienced to the present overburden stress

If OCR >1, the soil is overconsolidated

If OCR ~1, the soil is normally consolidated

If OCR <1, the soil is underconsolidated

cOCR

Effect on Primary Settlement

sC

Settlement in Overconsolidated Soils☞ Formula, σ’o + Δσ’ > σ’c (to the right of “F”)

☞ Formula, σ’o + Δσ’ < σ’c (to the left of “F”)

c

o

o

c

o

c

o

sc e

HC

e

HCS

log1

log1

o

o

o

sc e

HCS

log1

Overconsolidated Example☞ Given

☞ Soil with void ratio-log pressure as shown

☞ 6' thick clay layer under sand layer

☞ Two possible additional loadings:

☞ 5 tsf☞ 7.5 tsf

☞ Find☞ OCR☞ Settlement for each

case

Overconsolidated Example

☞ OCR = 6.6/0.3 = 22☞ 5 tsf additional load

☞ With overburden: 5 + 0.3 = 5.3 tsf

☞ 0.3 < 5.3 < 6.6, thus use

Spc = (6)(12)/(1+1.05)(0.078)log10(5.3/0.3)=3.42”

☞ 7.5 tsf additional load☞ With overburden: 7.5 +

0.3 = 7.8 tsf☞ 6.6 < 7.8, thus use

Sc = (6)(12)/(1+1.05)((0.078)log10(6.6/0.3)+(0.42)log10(7.8/6.6)) = 4.75”

o

o

o

sc e

HCS

log1

c

o

o

c

o

c

o

sc e

HC

e

HCS

log1

log1

Multiple Layers

☞ Multiple layers can be handled by computing the single layer consolidation for each and summing the displacements

☞ Take care to properly compute or assign the additional vertical stresses generated by the surface load for each layer (not all of the stresses will be the same)

☞ Take care when dealing with overconsolidated soils to properly understand the consolidation region for each layer

Secondary Compression of Cohesive Soils

Computation of Secondary Compression☞ Formula

☞ Cα = coefficient of secondary consolidation☞ tf = time at end of secondary consolidation compression☞ tp = time at end of time period of interest in primary

settlement computations☞ ep = void ratio at the beginning of secondary

compression

p

f

p

f

ps t

tHC

t

t

eH

CS loglog1

Example of Secondary Compression

☞ Given☞ As shown☞ Primary Settlement Data

☞Expected time for primary consolidation ~ 13 years

☞eo = 1 (at beginning of primary consolidation)

☞Normally consolidated clay, Cc = 0.21

☞ Desired life of structure = 110 years

☞ Cα = 0.02

☞ Find☞Secondary

Consolidation Settlement

Secondary Compression Example

☞ Primary Compression Calculation

☞ Compute void ratio ep at end of primary consolidation

☞ Secondary Compression Calculation

Total Compression = 6.35” + 1.84” = 8.19”

"35.675.

8.75.log

11121621.0

log1

o

o

o

cc e

HCS

"84.113

110log

93.011216

02.0log1

p

f

ps t

t

eH

CS

93.075.

8.75.log21.01log

o

ocoop Ceeee

Hough’s Method for Settlement of Shallow Foundations in Cohesionless Soils

• Recommended by FHWA (and thus AASHTO, state DOT’s etc.) because studies show that it is consistent and conservative

• Described in Soils and Foundations Reference Manual, pp. 7-15 to 7-19

• Requires a “layer by layer” analysis and thus should be done with a spreadsheet

Hough’s Method

1. Subdivide subsurface soil profile into approximately 3-m (10-ft) layers based on stratigraphy to a depth of about three times the footing width.

• Make sure layers are approximately homogeneous with relation to SPT blow count, unit weight and type of soil.

• Divide layers at phreatic surface (water table)2. Correct SPT blowcounts to an (N1)60 value, using methods previously discussed.3. Determine bearing capacity index (C′) using corrected SPT blowcounts, N′,

determined in Step 2.4. Calculate the effective vertical stress, σ’o, at the midpoint of each layer and the

average bearing capacity index for that layer.5. Calculate the increase in stress at the midpoint of each layer, Δσ, using 2:1

method (for this course.)6. Calculate the settlement in each layer, Ho, under the applied load using the

following formula:

7. Sum the settlements in each layer to determine the total settlement

o

oo

C

HS

log

Bearing Capacity Index Factors for Hough Method

Note max. value

kPa) Units,(SI 2100

ksf) Units,(U.S. 2 2

60601

voN

voN

N

C

C

NCN

Sample Settlement Problem(Cohesionless Soils, Hough’s Method) Given Two-Layer Soil Profile

Clean well-graded fine to coarse sand layer above clay layer and at surface N60 = 20 35' thick Unit weight of soil above

phreatic surface = 110 pcf Submerged weight of soil

below water table= 65 pcf Depth of water table = 15’

Clay stratum, normally consolidated LL= 45 25' thick Water Content = 40% Specific Gravity of solids

= 2.78

Given Building placed on

surface Square Foundation, 30’

x 30’ Uniform Load of 10 ksf

on entire foundation

Find Average settlement of

building due to settlement of sand layer

Sample Settlement Problem(Cohesionless Soils, Hough’s Method) Step 1: Compute Effective Stress Profile and

Foundation Stress Profile for Bearing Strata and Foundation

Sample Settlement Problem(Cohesionless Soils, Hough’s Method) Step 2: Correct SPT

Values for Overburden Pressure Layer 1:

(N1)60 = (1.6)(20) = 31 Layer 2: CN = 0.9,

(N1)60 = (0.9)(20) = 19

Step 3: Determine Values for C’

ksf) Units,(U.S. 2 1.56825.0

2

ksf) Units,(U.S. 2 2

60601

N

voN

N

C

C

NCN

Sample Settlement Problem(Cohesionless Soils, Hough’s Method) Step 4: Compute Settlements of Sand Layers

Layer 1:

Layer 2: S = 1.33”

"88.1825

6400825

90

'1512log

o

oo

C

HS

Time Rate of Consolidation

Overview of Time Rate Computations☞ Consolidation is a process

that takes place over time and is the result of two processes:

☞ Dissipation of excess pore water pressures created by surcharge or new foundation loads at the surface. This takes place when the excess water is drained from the soil

☞ Rearrangement of the soil particles (“soil skeleton”) to reflect new effective stress conditions created by foundation or surcharge

Assumptions Clay-water system is

homogeneous Saturation is complete Water in

incompressible Soil particles are

incompressible Flow is in one direction

only (in the direction of compression)

Darcy’s Law is valid

Consolidation ModelSign ConventionBetter Reversed

Coefficient of Compressibility

Coefficient of VolumeCompressibility

pmpe

an

ee

n

pe

a

vo

v

o

v

1

ng,Substituti

1

ipRelationshPorosity

tu

mtp

mzv

tzv

tdzdv

n

dzAtAvv

VV

n

vv

tb

T

w

ng,Substituti

u(z,t): Pressure vs. Distance and Time Variable

Continuity of Flow:

Darcy's Law

tu

mzu

k

zu

kzv

zu

kzh

kkiv

vw

w

w

2

2

2

2

1

Equating,

1

ating,Differenti

1

Consolidation Equation

☞ Rearranging,

☞ Boundary conditions are the conditions of drainage at the surfaces of the consolidating layer

☞ Initial conditions are the effective stress and pore water conditions

Coefficient of Consolidation

Key variable in primary consolidation calculations

Tends to be a constant for a given soil because the ratio of k to mv tends to remain constant

● Both of these variables tend to be directly proportional to the void ratio of the soil

2

2

2

2

zu

czu

mk

tu

vwv

vwv m

kc

Drainage Conditions

☞ Speed of the drainage depends upon the conditions at the boundaries of the drainage

☞ Drainage can be one or two ways☞ Theory based on one

way drainage☞ Distribution of initial pore

pressure can affect time rate of consolidation (see diagram at right)

Solution of Consolidation Equation

Solution Equation is a second

order, parabolic differential equation (similar to heat equation)

Solution is an infinite Fourier series with orthogonality and eigenfunctions

Solution depends upon both boundary and initial conditions

Consider the case of one way drainage

● Boundary Conditions At top of layer: p = 0

(free flow) At bottom of layer: ∂p/∂z = 0 (no flow)

● Initial Conditions p = p0 = (stress

induced by applied load, remains constant)

● Solution

2

22

412

1

1

212cos

1214 H

tci

i

i

o

v

ehz

iip

p

Time Factor for Consolidation

☞ Time Factor

☞ t = time of consolidation☞ Tv = time factor for vertical

drainage☞ Hdr = longest distance of

drainage to pervious surface☞ Usually depth of layer for

single drainage, half of layer depth for double drainage

☞ cv = coefficient of consolidation

2dr

vv H

tcT

vTi

i

i

o

ehz

iip

p 412

1

1 22

212cos

1214

Degree of Consolidation Degree of Consolidation

Expressed as a percentage Expresses ratio of both pore

pressures and settlement Allows relating settlement to

time (consolidation)

o

e

c

tc

u

u

S

SU 1

1

412

22

2

12

181

i

Ti v

ei

U

Time Factor vs. Average Degree of Consolidation

Assumes uniform initial pressure

Example of Time Factor Calculation☞ Given

☞ As shown☞ eo = 1.0

☞ Cc = 0.21

☞ cv = 0.03 ft3/day

☞ Find☞ Relationship of time

of consolidation with time factor

Solution of Time Factor CalculationDays Tv

10 0 0 020 0.01 10 0.6430 0.01 10 0.6440 0.02 15 0.9550 0.02 15 0.9560 0.03 20 1.2770 0.03 20 1.2780 0.04 25 1.5990 0.04 25 1.59

100 0.05 28 1.78200 0.09 32 2.03300 0.14 40 2.54400 0.19 48 3.05500 0.23 53 3.37600 0.28 60 3.81700 0.33 63 4800 0.38 68 4.32900 0.42 72 4.57

1000 0.47 75 4.76

Hdr 8Cv 0.03Spc 6.35

Percent Consolidation

Settlement, in.

days 2130803.0

22

ttHtc

Tdr

vv

Time Factor Calculation Solution

Coefficient of Consolidation

Logarithm of Time Method

Square Root of Time Method

Accelerating Consolidation Settlement

☞ Rate of consolidation settlement under natural conditions is frequently unacceptable for actual construction and structure use

☞ Previous Example: 75% settlement @ 1000 days = 2 years 8 months 26 days

☞ Accelerating settlement if sometimes necessary to complete construction and use structure

Concept of Consolidation Acceleration Provide a high-

permeability path for the water to escape, thus reducing the time for consolidation to take place

Type of acceleration Sand Drains Prefabricated Vertical

(Wick) Drains

Vertical Drain System

Sand Drain Installation

Wick Drains

☞ Geosynthetic used as a substitute to sand columns; arrayed in a similar manner to the sand drains

☞ Installed by being pushed or vibrated into the ground

Installation of Wick Drains

Questions