Foundation Settlement By Kamal Tawfiq, Ph.D., P.E. Fall 2008

Foundation Settlement

By

Kamal Tawfiq, Ph.D., P.E.

Fall 2008

Foundation Settlement:Total Foundation Settlement = Elastic Settlement + Consolidation Settlement

Stotal = Se + Sc Foundation Type (Rigid; Flexible)

Elastic Settlement or Immediate Settlement depends on

Settlement Location (Center or Corner){

Theory of Elasticity

Elastic Settlement

Time Depended Elastic Settlement (Schmertman & Hartman Method (1978){

By: Kamal Tawfiq, Ph.D., P.E.

Elastic settlement occurs in sandy, silty, and clayey soils.

Water

Water Table (W.T.)

VoidsSolids

Expulsion of the water

Consolidation Settlement (Time Dependent Settlement)


* Consolidation settlement occurs in cohesive soils due to the expulsion of the water from the voids.

* Because of the soil permeability the rate of settlement may varied from soil to another. * Also the variation in the rate of consolidation settlement depends on the boundary conditions.

SConsolidation = Sprimary + Ssecondary

Primary Consolidation Volume change is due to reduction in pore water pressure

Secondary Consolidation Volume change is due to the rearrangement of the soil particles (No pore water pressure change, Δu = 0, occurs after the primary consolidation)

Elastic Settlement

Se = Bqo (1 - μs) α2Es

2

(corner of the flexible

foundation)Se = Bqo (1 - μs) αEs

2

(center of the flexible

foundation)


Where α = 1/π [ ln ( √1 + m2 + m / √1 + m2 - m ) + m*ln ( √1 + m2 + 1 / √1 + m2 - 1 )

m = B/L

B = width of foundationL = length of foundation


3.0

2.5

2.0

1.5

1.0

L / B

3.02 843 5 6 7 1091

α,

αav

, α

rα

αavαr

For circular foundationα = 1

αav = 0.85αr = 0.88

Values of α, αav, and αr

Elastic Settlement Using the Strain Influence Factor: [Schmertman & Hartman Method (1978)]

Example:B x L

EsDf Iz

Depth, z

q = γ Df

ΔZ1

q

ΔZ2

ΔZ3

ΔZ4

Es3

Is3

Average IsAverage Es

The variation of the strain influence factor with depth below the foundation is shown in Figure 1. Note that, for square or circular foundations,

Iz = 0.1 at z = 0Iz = 0.5 at z = 0.5B

Iz = 0 at z = 2B

Similarly, for foundations with L/B ≥ 10

Iz = 0.2 at z = 0Iz = 0.5 at z = BIz = 0 at z = 4B

Se = C1 C2 ( q - q) ∑ (Iz / Es ) Δz

whereIs = strain influence factor

C1 = a correction factor for the depth of foundation embedment = 1 - 0.5 [q / (q - q)]

C2 = a correction factor to account for creep in soil = 1 + 0.2 log (time in years /0.1)

q = stress at the level of the foundationq = overburden pressure = γ Df

Elastic Parameters of Various Soils

Young’s Modulus, Es

Type of Soil MN/m2 Lb/in2 Poisson’s Ratio, s

Loose sand 10.35 - 24.15 1,500 - 3,500 0.20 - 0.40Medium dense sand 17.25 - 27.60 2,500 - 4,000 0.25 - 0.40Dense sand 34.50 - 55.20 5,000 - 8,000 0.30 - 0.45Silty sand 10.35 - 17.25 1,500 - 2,500 0.20 - 0.40Sand and Gravel 69.00 - 172.50 10,000-25,000 0.15 -0.35Soft clay 2.07 - 5.18 300 - 750Medium clay 5.18 - 10.35 750 - 1,500 0.20 - 0.50Stiff clay 10.35 - 24.15 1,500 - 3,500

Es (kN/m2) = 766NEs = 2qc

where N = standard penetration numberqc = static cone penetration resistance

Note: Any consistent set of units can be used

The Young’s modulus of normally consoliadated clays can be estimated as

Es = 250c to 500c

For overconsolidated clays

Es = 750c to 1000c

where c = undrained cohesion of clayey soil

Example:

2

12,000 20,000

Depth, z (m)

8

4

6

4,000

B x L = 3 m x 3m

Es (kN/m2)1.5 m

q = 160 kN/m2

Actual

0.2

6

2 1.5

4 Iz = 0.5

0.40

0.1 0.6

Averaged

Iz

= 17.8 kN/m3

Depth (m) Z (m) Es (kN/m2) Average Iz (Iz/Es). z (m3/kN)

0 - 1 1 8,000 0.233 0.291 x 10-4

1.0 - 1.5 0.5 10,000 0.433 0.217 x 10-4

1.5 - 4 2.5 10,000 0.361 0.903 x 10-4

4.0 - 6 2 16,000 0.111 0.139 x 10-4

= 1.550 x 10-4

C1 = 1 - 0.5 (q / q - q )

= 1 - 0.5 [ 17.68 x 1.5 / 160 - (17.8 x 1.5)]

C2 = 1 + 0.2 log (5/0.1) = 1.34

Hence

Sc = C1 . C2 (q -q) (Iz/Es) z

= (0.9)(1.34)[160-(17.8x1.5)](1.55x10-4)

= 249.2x10-4 m 24.9 mm

0

2B

Time = 5 years

Example:

2

12,000 20,000

Depth, z (m)

8

4

6

4,000

B x L = 3 m x 3m

Es (kN/m2)1.5 m

q = 160 kN/m2

Actual

0.2

6

2 1.5

4 Iz = 0.5

0.40

0.1 0.6

Averaged

Iz

= 17.8 kN/m3

Depth (m) Z (m) Es (kN/m2) Average Iz (Iz/Es). z (m3/kN)

0 - 1 1 8,000 0.233 0.291 x 10-4

1.0 - 1.5 0.5 10,000 0.433 0.217 x 10-4

1.5 - 4 2.5 10,000 0.361 0.903 x 10-4

4.0 - 6 2 16,000 0.111 0.139 x 10-4

= 1.550 x 10-4

C1 = 1 - 0.5 (q / q - q )

= 1 - 0.5 [ 17.68 x 1.5 / 160 - (17.8 x 1.5)]

C2 = 1 + 0.2 log (5/0.1) = 1.34

Hence

Sc = C1 . C2 (q -q) (Iz/Es) z

= (0.9)(1.34)[160-(17.8x1.5)](1.55x10-4)

= 249.2x10-4 m 24.9 mm

0

2B

Time = 5 years

Qdesign = Column Load

Uo

P P

OverburdenPressure

u =Excess Pore Water Pressure

u =Excess Pore Water Pressure

Po

Hdr = Hc /2

Stress Distribution2: 1 method

Hc = Layer Thickness


Rate of Consolidation

Settlement at any time = Stime

Stime = Sultimate * U% Sultimate= (Cc/1+eo) Hc . log [(Po + P)/Po]

U% = f (Tv) ....

Tv = f (cv) ......

Tv = (Hdr)

2

cv . t

Sand

Sand

Caly

HcHc/2

Stressed Zone

Consolidation Settlement

Consolidation Settlement (Primary Consolidation) = Sc = (Cc/1+eo) Hc . log [(Po + P)/Po]

Qdesign = Column Load

OverburdenPressure

Po


Sand

Sand

CalyB

2

1

2

1

StressDistribution

NormallyConsolidatedClay

21

p

p

Hsand

Hclay/2 Hclay

Cc

Log PPo

Void Ratio

Log P

P

Void Ratio

Po Po + P

Cc H log po + pSultimate = H =

Po 1 + eo ( )

Log P

P

Void Ratio

Po Po + P

21

p

Hsand

Hclay/2 Hclay

Loading Unloading

Po = sand . Hsand + ( clay - water ) . Hclay/2

NormallyConsolidated Soil

Clay

Sand

Sand

eo


Consolidation Settlement

Sand

Sand

Sand

Sand

Cs H PcSultimate = H =Po 1 + eo

( ) logCc H+

1 + eo log

Pc ( )Po + P2

Log P

P

Void Ratio

Po Po + P=Pc

21 p2

p2

Hsand

Hclay/2 Hclay

21

p

Hsand

Hclay/2 Hclay

Log P

Void Ratio

Po Pc

P2

Cs

Cs

Po + P2

The soil becomeoverconsolidated

soil

eo

Re loadingwith Heavy Load


Cs H PcSultimate = H =Po

( )1 + eo

log

Log P

P

Void Ratio

Po Pc

21 p2

p2

Hsand

Hclay/2 Hclay

21

p

Hsand

Hclay/2 Hclay

Log P

Void Ratio

Po Pc

P2

Cs

Po + P2The soil becomeoverconsolidated

soil

eo

Re loadingwith light Load


Log P

Void Ratio

OCR = Pc/Po

OCR = 1OCR > 1OCR > 4

Normally Consolidated

Heavily Over Consolidated

Over Consolidated

Pc

Determining The Preconsolidation Pressure (Pc)

13

2

4

5 6


Po

7

Cassagrande Graphical Method

Rate of Consolidation

Settlement at any time = Stime

Stime = Sultimate * U%

U% = f (Tv) ....

Tv =

Tv = Time Factor

t = time (month, day, or year)(Hdr)

2= Drainage PathHdr = H or H/2

Tv = f (cv) ......

(Hdr)2

cv . tCv = Coefficient of Consolidation

Cv is obtained from laboratory testing

Clay

Sand

Sand

Clay

Rock

Sand

Two way drainage Hdr = Hclay/2

One way drainage Hdr = Hclay

From Tables


Foundation Settlement By Kamal Tawfiq, Ph.D., P.E. Fall 2008

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