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Foundation Settlement By Kamal Tawfiq, Ph.D., P.E. Fall 2008. Foundation Settlement :. Total Foundation Settlement = Elastic Settlement + Consolidation Settlement S total = S e + S c. {. Foundation Type (Rigid; Flexible) - PowerPoint PPT Presentation
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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)
By: Kamal Tawfiq, Ph.D., P.E.
* 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)
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.
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
By: Kamal Tawfiq, Ph.D., P.E.