6 soil compressibility_gb_dr. asmaa moddather
DESCRIPTION
Lecture about soil compressibility and consolidationTRANSCRIPT
PBW N302
Credit Hours
CEM/WEE/STE
Dr. Asmaa ModdatherSoil Mechanics and Foundations
Faculty of Engineering – Cairo University
FALL 2012
Definitions
• Compressibility: is the property through which
particles of soil are brought closer to each other, due
to escapage of air and/or water from voids under the
effect of an applied pressure.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Settlement of Cohesive Soils
• Coefficient of Compressibility (av): is the rate of change of
void ratio (e) with respect to the applied effective pressure
(p) during compression.
p
ea v
∆
∆=
e
e-p curve
e
eo = initial void ratio
po = initial effective stress
Dr. Asmaa Moddather – PBW N302 – Fall 2012
p
e∆
p∆
eo
e1
po p1
po = initial effective stress
e1 = final void ratio
p1 = final effective stress
∆e = eo – e1
∆p = p1 – po
• Coefficient of Volume Compressibility: (mv) is the volume decrease
of a unit volume of soil per unit increase of effective pressure during
Settlement of Cohesive Soils
of a unit volume of soil per unit increase of effective pressure during
compression
)e(1
a
Δp
Δe
)e(1
1
Δp
)Ve(1
ΔeV
Δp
V
Δv
mo
v
o
so
s
Ov
+=
+=
+==
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• For a thin layer , ∆p at mid depth is considered as an average
stress within the layer.q
Settlement of Cohesive Soils
stress within the layer.
)e(1
Δe
H
ΔH
o+=
H)e(1
Δp
ΔP
ΔeH
)e(1
ΔeSΔH
oo +=
+==
ΔpH)e(1
aS v
+=
H∆p
Clay
Dr. Asmaa Moddather – PBW N302 – Fall 2012
1
eo
1+eo H
∆e ∆H
ΔpH)e(1
So+
=
∆pHmS v=
mv = coefficient of volume compressibility
∆p = increase in effective stresses at mid depth of layer
H = thickness of layer
Settlement of Cohesive Soils
• For a thick layer the layer may be divided to number of sub-
layers, each of thickness Hi. The stress at mid-depth of each sub-
H1∆p1
q
iiH∆pmS vi∑=
layers, each of thickness Hi. The stress at mid-depth of each sub-
layer is ∆pi.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
H1
Clay
H2
∆p2
• Sand may be considered as elastic material with Young’s Modulus.
• Young’s Modulus (E): is the slope for the linear portion of the
Settlement of Cohesionless Soils
• Young’s Modulus (E): is the slope for the linear portion of the
stress-strain curve.
E
1
stress
strain=
E
1
Δp
H
ΔH
=
q
Dr. Asmaa Moddather – PBW N302 – Fall 2012
E
p
H
H ∆=
∆
E
p
H
S ∆=
pHE
1S ∆=
E
1mv =
2kg/cm dense) (v. 800-loose) (v. 100E =
EΔp
H∆p
Sand
Example
• Compute the settlement due to
compressibility of the sand layer and
consolidation of the clay layer at points (a)
and (b) of the building shown in Figure.
The building has a basement and is
founded on raft foundation. The stress
from the building at the foundation level
is as shown in the figure.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
is as shown in the figure.
Sand: γ = 1.70 t/m3 , E = 500 kg/cm2
Clay: mv = 0.03 cm2/kg.
� Due to distributed load:
qnet = q – γh = 20 – 1.70x3.0 = 14.9 t/m2
Example
For sand layer
z = 5.0 m = 1” = 2.5 cm
L = 30 m � 30x2.5/5.0 = 15.0 cm, B = 20 m � 20x2.5/5.0 = 10.0 cm,
na = 191 ∆σa,sand = n/200 (qnet) = 191/200 x 14.9 = 14.2 t/m2
nb = 50 ∆σb, sand = n/200 (qnet) = 50/200 x 14.9 = 3.7 t/m2
Dr. Asmaa Moddather – PBW N302 – Fall 2012
ab
� Due to distributed load:
qnet = q – γh = 20 – 1.70x3.0 = 14.9 t/m2
Example
For clay layer
z = 11.0 m = 1” = 2.5 cm
L = 30 m � 30x2.5/11.0 = 6.82 cm, B = 20 m � 20x2.5/11.0 = 4.55 cm,
na = 144 ∆σa,clay = n/200 (qnet) = 144/200 x 14.9 = 10.7 t/m2
nb = 47.3 ∆σb,clay = n/200 (qnet) = 47.3/200 x 14.9 = 3.5 t/m2
Dr. Asmaa Moddather – PBW N302 – Fall 2012
ab
Sa = Sa,sand + Sa,clay
Example
Sa,sand = 1/Esand ∆σa,sand H
H = 10.0 m E = 500 kg/cm2 ∆σa,sand = 14.2 t/m2
Sa,sand = 1/5000 x 14.2 x 10.0 = 0.0284 m = 2.84 cm
Sa,clay = mv ∆σa,clay H
H = 2.0 m m = 0.03 cm2/kg = 3.7 t/m2
Dr. Asmaa Moddather – PBW N302 – Fall 2012
H = 2.0 m mv = 0.03 cm2/kg ∆σa,clay = 3.7 t/m2
Sa,clay = 0.03/10 x 3.7 x 2.0 = 0.0222 m = 2.22 cm
Sa = Sa,sand + Sa,clay = 2.84 + 2.22 = 5.06 cm
Example
Sb = Sb,sand + Sb,clay
Sb,sand = 1/Esand ∆σb,sand H
H = 10.0 m E = 500 kg/cm2 ∆σb,sand = 10.7 t/m2
Sb,sand = 1/5000 x 10.7 x 10.0 = 0.0214 m = 2.14 cm
Sb,clay = mv ∆σb,clay H
H = 2.0 m m = 0.03 cm2/kg = 3.5 t/m2
Dr. Asmaa Moddather – PBW N302 – Fall 2012
H = 2.0 m mv = 0.03 cm2/kg ∆σb,clay = 3.5 t/m2
Sb,clay = 0.03/10 x 3.5 x 2.0 = 0.021 m = 2.10 cm
Sb = Sb,sand + Sb,clay = 2.14 + 2.10 = 4.24 cm
Settlement
H∆p
q
∆pHmS =
(Sand)
(Clay)
ΔpHE
1S =
Dr. Asmaa Moddather – PBW N302 – Fall 2012
∆pHmS v= (Clay)
Time of Settlement Compared to Construction Time
� Loading diagram:
Construction period
� Settlement of sandpH
E
1S ∆=
Time
Time
Dr. Asmaa Moddather – PBW N302 – Fall 2012
� Settlement of clay∆pHmS v=
Time
Causes of Settlement
• Static loads
•• Dynamic loads
• Groundwater lowering
• Capillary forces
•
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• Loads from adjacent structures
Theory of Consolidation
• Consolidation: is the process of squeezing water out of soil under the
effect loads, provided no lateral movement occurs.effect loads, provided no lateral movement occurs.
• In sandy soils, high permeability, drainage of water out of the soil
under the effect of loading happens immediately.
• In clay soils, low permeability, drainage of water out of the soil under
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• In clay soils, low permeability, drainage of water out of the soil under
the effect of loading is time dependent.
• Therefore, parameters governing consolidation process include: soil
properties, stress (p), time (t), drainage conditions.
Analogy:
Confined saturated soil (solids Valve ~ soil Piston for
Mechanism of Consolidation ProcessMechanism of Consolidation Process
and voids filled with water)
Versus
Container filled with water that
has a spring and a valve
Valve ~ soil permeability
Piston for loading
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Remember: water is incompressible
Spring ~ Solids
Water ~ pore-water
Container
• If we apply a certain load, and the valve is closed, then water
(incompressible material) carries the entire added load.
Mechanism of Consolidation ProcessMechanism of Consolidation Process
(incompressible material) carries the entire added load.
• If the valve is opened, water flows out, and thus its volume is
reduced.
• Since the total volume is reduced, the spring starts to be
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• Since the total volume is reduced, the spring starts to be
compressed and carries part of the load. Meanwhile, additional
pressure in the water decreases.
• The process continues until sufficient amount of water escapes
from the valve, which will result in the compression of the
Mechanism of Consolidation ProcessMechanism of Consolidation Process
from the valve, which will result in the compression of the
spring that would allow it to carry the entire load.
• At this point, the additional pressure in water decreases to zero,
the spring carries the entire load, the system is in equilibrium.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
For Sands: high permeability ~ valve is open, water gets out immediately
(t = 0), ∆u = 0
Mechanism of Consolidation ProcessMechanism of Consolidation Process
Stress = p
Dr. Asmaa Moddather – PBW N302 – Fall 2012
At t = 0 (immediately), ∆u = 0Load carried by spring
For Clays: low permeability ~ valve is partially open, water gets out
slowly (time-dependent), ∆u decreases with t
Stress = p
Mechanism of Consolidation ProcessMechanism of Consolidation Process
Stress = p
Stress = p
Dr. Asmaa Moddather – PBW N302 – Fall 2012
At t = 0 (immediately), ∆u = pSpring carry no load
At t = t1, 0 < ∆u < pLoad carried by water and spring
For Clays:
Mechanism of Consolidation ProcessMechanism of Consolidation Process
Stress = p
Stress = p
Dr. Asmaa Moddather – PBW N302 – Fall 2012
At t = long time, ∆u = 0Load carried by spring
At t = t2, 0 < ∆u < pLoad carried by water and spring
• Stress distribution before application of load
Application on SoilApplication on Soil
GSGWT
Sand
Clay
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Gravel
tσ 'σu
• Short time (immediately) after application of q (t = 0)
Application on SoilApplication on Soil
qGSGWT
Sand
Clay
q
q
q
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Gravel
tσ 'σu
q
q
• Long time after application of q (t = h)
Application on SoilApplication on Soil
qGSGWT
Sand
Clay
q
q
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Gravel
tσ 'σu
q
q
Application on Soil
• Total stress:
• Pore water pressure:
qΣγhσ t +=
• Pore water pressure:
o t = 0 � (clay)
� (sand)
o t = h � (clay)
� (sand)
• Effective stress:
qhγu ww +=
wwhγu =
wwhγu =
wwhγu = At time t ???
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• Effective stress:
o t = 0 � (clay)
� (sand)
o t = h � (clay)
� (sand)
qhΣγ'σ' +=
hΣγ'σ'=
qhΣγ'σ' +=
qhΣγ'σ' +=
Rate of Consolidation
qGSGWT
Sand
Clay
q
q
q
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Gravel
tσ 'σu
q
q
Rate of ConsolidationRate of Consolidation
Clay
u
t = 0
t = t1
t = t2
Dr. Asmaa Moddather – PBW N302 – Fall 2012
q
t = t3
t = t2
t = ∞
Assumptions of Theory of Consolidation
• Clay is homogeneous, isotropic, and saturated.
•• Water and clay particles are incompressible.
• Darcy’s law is valid.
• The clay layer is laterally confined.
• One-dimensional compression, and one-dimensional flow.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• One-dimensional compression, and one-dimensional flow.
•Consolidation parameters from test applies to clay layer from
which sample for test was taken.
• Soil properties are constant with time.
Consolidation Test
• Objectives:
o Volume change-effective pressure relationship
o Stress history of soil
o Volume change – time – pore water dissipation relationship
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Consolidation Test Load cell LoadDial gauge
Compression cell
Oedometer
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Consolidation Test
Load cell
Dial gauge
Compression cell
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Oedometer
Consolidation Test
Load cell
Dial gauge
Compression cell
Load
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Oedometer
Consolidation Test
Dial gaugeLoad Head
Dial gauge
Porous stone
Saturated sampleMetal ring
Filter paper
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Porous stone
Filter paper
Compression Cell/Oedometer
Consolidation Test
Dial gaugeLoad Head
Load (P )Dial gauge
Porous stone
Soil saturated sampleMetal ring
Filter paper
Load (P1)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Compression Cell/Oedometer Porous stone
Filter paper
Consolidation Test
• Equipments:
o Compression cell/Oedometero Compression cell/Oedometer
o Loading frame to apply weights (Po)
o Dial gauge to measure soil compression
Note:
Load acting on the cell (P1) is magnified by the lever arm ratio of the
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Load acting on the cell (P1) is magnified by the lever arm ratio of the
loading frame (LAR), where: P1 = LAR x P
• Procedure:
1. Measure initial conditions of sample: ho, eo, wo, Gs
Consolidation Test
o o o s
2. Trim the sample into the metal ring and place filter paper on both
sides of sample.
3. Place the ring into the compression cell between the two porous
stones.
4. A metal cap (loading head) is placed over the top porous stone, on
Dr. Asmaa Moddather – PBW N302 – Fall 2012
4. A metal cap (loading head) is placed over the top porous stone, on
which the load (P1) is applied.
5. Set-up the dial gage to zero reading. Fill the compression cell with
water until the top porous stone is covered with water.
• Procedure:
6. Place the hanging load (Po) such that the pressure on the clay
Consolidation Test
o
sample is equal to the first loading step.
7. Record readings of dial gage (compression of sample) with time
until compression stops (usually within 24 hours).
8. Repeat steps 6 and 7 for subsequent loading and unloading
increments.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
increments.
• Loading scheme, for example:
o 0.25, 0.5, 1.0, 2.0, 4.0, 8.0, 16.0 (kg/cm2) Loading
Consolidation Test
o 16.0, 4.0, 1.0, 0.25 (kg/cm2) Unloading
• Results: for each loading/unloading increment, we record:
dial gage (∆h) that measures sample compression versus
Dr. Asmaa Moddather – PBW N302 – Fall 2012
elapsed time (t).
Consolidation Test Data Reduction
• For each loading increment, plot ∆h-log t:
0.1 1 10 100 1000Time (min)
0
10
20
30
40
50
h (
x 1
0-2
mm
)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
60
70
80
90
100
∆∆ ∆∆h
(x
• At the end of each loading increment:
1. Calculate effective pressure (p) = Load x LAR/sample area
Consolidation Test Data Reduction
2. Measure ∆h = sample compression during this loading increment
3. Calculate void ratio (e):
oo h
∆h
e1
∆e=
+ eo
∆e ∆h
Dr. Asmaa Moddather – PBW N302 – Fall 2012
e = eo – ∆e
• Plot e-p curve
1
1+eo ho
232116582914.570Load (kg)
Example: LAR = 3, sample area = 41.85 cm2, ho = 25.4 mm, eo = 0.636
Consolidation Test Data Reduction
232116582914.570Load (kg)
705550359223135890Dial Reading x 10-2 (mm)= ∆h
16.638.324.162.081.047 x 3/41.85
= 0.500
Stress (p) = Load x lever arm ratio/area (kg/cm2)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0.45410.35430.23120.14360.0870.89/25.4 (1+0.636)= 0.0573
0∆e = ∆h/ho (1+eo)
0.18190.28170.40480.49240.54900.636-0.0573
= 0.57870.636e = eo - ∆e
• Plot e-p Curve
0.7
Consolidation Test Data Reduction
0.3
0.4
0.5
0.6
0.7
e
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0
0.1
0.2
0 2 4 6 8 10 12 14 16 18
p (kg/cm2)
• From e-p curve, for a certain stress range, calculate av and
mv, where:
Consolidation Test Data Reduction
mv, where:
p
ea v
∆
∆=
e
e∆
p∆
eo
e1
Dr. Asmaa Moddather – PBW N302 – Fall 2012
)e(1
am
o
vv
+=
ppo p1
• Plot e-log(p) curve:
0.7
Consolidation Test Data Reduction
0.3
0.4
0.5
0.6
e
e∆
eo
e1
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0
0.1
0.2
0.1 1 10 100
p (kg/cm2)
p)log(∆
po p1
e1
• From e-log(p) curve, the slope of the linear portion is the
Compression Index (Cc):
Consolidation Test Data Reduction
c
• Determine preconsolidation pressure (pc): is the largest
∆log(p)
∆eCc =
o1
1o
logplogp
ee
−
−= 402.0
log4.8log12.0
22.00.38=
−
−=
Dr. Asmaa Moddather – PBW N302 – Fall 2012
• Determine preconsolidation pressure (pc): is the largest
effective pressure that has been applied on the soil in its
geological history.
• Determination of preconsolidation pressure:
0.7
5
Consolidation Test Data Reduction
0.3
0.4
0.5
0.6
e
1
2
3
4
5
ba
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0
0.1
0.2
0.1 1 10 100
p (kg/cm2)
pc = 1.8 kg/cm2
1. Determine point of maximum curvature (a).
2. At (a), draw tangent to e-log(p) curve.
Determination of Preconsolidation Pressure
3. At (a), draw horizontal line.
4. Draw a bisector to the angle enclosed between the tangent and
horizontal lines.
5. Extend the linear portion of the curve until it intersects the bisector at
point (b).
Dr. Asmaa Moddather – PBW N302 – Fall 2012
point (b).
6. The preconsolidation pressure (pc) is the x-coordinate of the point (b).
1. Normally consolidated clay:
Clay that has never been loaded in the past by more than the existing
effective overburden pressure (p ) p ~ p
Types of Clay w.r.t. Preconsolidation
effective overburden pressure (po) pc ~ po
2. Overconsolidated (preconsolidated) clay:
Clay that has been loaded in the past by more than the existing effective
overburden pressure (po) pc > po
Causes: pre-existing structures, erosion of overburden
3. Underconsolidated clay:
Dr. Asmaa Moddather – PBW N302 – Fall 2012
3. Underconsolidated clay:
Clay that has been loaded partially by the existing effective overburden
pressure (po) pc < po
o
c
p
pOCR =Define: Overconsolidation Ratio (OCR):
� For normally consolidated clays, Cc can be calculated as follows:
� From e-log(p) curve:
Determination of Compression IndexDetermination of Compression Index
� From liquid limit:
∆log(p)
∆eCc =
Dr. Asmaa Moddather – PBW N302 – Fall 2012
where wL is in (%)
)100.009(wC Lc −≈
0.6
0.7
Recompression curve
DefinitionsDefinitions
0.2
0.3
0.4
0.5
e
Virgin compression curve
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0
0.1
0.1 1 10 100
p (kg/cm2)
Unloading/reloading curve
pc
� From e-log(p) curve, the recompression index (Cr) is the slope of
the equivalent linear portion of the unloading/reloading curve.
DefinitionsDefinitions
0.3
0.4
0.5
0.6
0.7
e
)log( p
eCr
∆
∆=
Cr
Cr
Cc
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0
0.1
0.2
0.1 1 10 100p (kg/cm2)
Cr
� Calculate settlement (S) of normally consolidated clay layer of
thickness (H) due to added stress (∆p) given Cc of clay:
∆e
SettlementSettlement
∆log(p)
∆eCc =
)log(p)log(p
eeC
01
10c
−
−=
)p
log(
∆eC
1c =
0.1
0.2
0.3
0.4
0.5
0.6
0.7
e
eo
e1
e Cc
∆p
Dr. Asmaa Moddather – PBW N302 – Fall 2012
)p
plog(
0
1
.......1..........).........p
plog(C∆e
0
1c=
......2..............................H
S
h
∆h
e1
∆e
oo
==+
0
0.1
0.1 1 10 100p (kg/cm2)
p1op~c p
� From 1 and 2:
)e(1S
)p
log(C 1 +=
SettlementSettlement
)e(1H
S)
p
plog(C o
o
1c +=
)p
plog(H
e1
CS
o
1
o
c
+=
Dr. Asmaa Moddather – PBW N302 – Fall 2012
� For overconsolidated clay, (po and p1) < pc:
0.7e
SettlementSettlement
0
0.1
0.2
0.3
0.4
0.5
0.6
e
eo
e1
Cr
∆p
Dr. Asmaa Moddather – PBW N302 – Fall 2012
)p
plog(H
e1
CS
o
1
o
r
+=
0
0.1 1 10 100p (kg/cm2)
p1 pcpo
� For overconsolidated clay, po < pc and and p1 > pc
0.7e
SettlementSettlement
0
0.1
0.2
0.3
0.4
0.5
0.6
e
eo
e1
Cr
∆p
Cc
Dr. Asmaa Moddather – PBW N302 – Fall 2012
)p
plog(H
e1
C)
p
plog(H
e1
CS
c
1
o
c
o
c
o
r
++
+=
0
0.1 1 10 100p (kg/cm2)
p1pcpo
Rate of ConsolidationRate of Consolidation
q
Short time after application of load “q”
GSGWT
Sand
Clay
q
q
q
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Gravel
tσ 'σu
q
q
Rate of consolidationRate of consolidation
u
t = 0
u
t = 0
Isochrone
Sand
zu σ'
Clay
t = 0
t = t1
t = t3
t = t2
t = ∞
t = t2
t = ∞
umax σ‘minClay
Sand
2H
hDu σ'
Dr. Asmaa Moddather – PBW N302 – Fall 2012
∆p ∆p Sand
At time “t” and depth “z”:
∆p = ∆u + ∆σ’ = increase in total stress due to added load∆u = excess pore water pressure due to added load∆σ’ = increase in effective stress due to added load
Hydrostatic pore water pressure
(GWT)
u
t = 0
Sand
z∆u ∆σ'
2H
Rate of consolidationRate of consolidation
Degree of Consolidation (Utz) at time “t” and depth “z”:
∆p
t = tt = ∞
Clay
Sand
2H
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Degree of Consolidation (Utz) at time “t” and depth “z”:
∆p
∆u1
∆p
∆u∆p
∆p
∆σU
'
tz −=−
==
� Average degree of consolidation (Ut) for the clay layer at time “t”:
Rate of ConsolidationRate of Consolidation
dzU2UDh
tzt ∫=
u
t = 0Sand
zdzU2U
0
tzt ∫=
diagram∆pofarea
diagramσ'ofareaU t
∆=
∆p
∆u∆p
∆p
∆σU
'
tz
−==
t = 0
t = t
t = ∞
umax σ‘minClay
2H
z
hDu σ'
Where
Dr. Asmaa Moddather – PBW N302 – Fall 2012
diagram∆pofarea
diagram∆pofarea
diagramuofarea-diagram∆pofareaU t
∆=
2H∆pdiagram∆pofarea =
2Hu3
2diagramuofarea max=∆
∆p
t = ∞
Sand
Where
� Settlement of clay layer of thickness 2H at time t=∞:
Sultimate = mv∆p (2H)
Rate of ConsolidationRate of Consolidation
� Settlement of clay layer of thickness 2H at time t=t:
St = mv∆σ’ (2H)
� Average degree of consolidation (U ) at time “t”:
Dr. Asmaa Moddather – PBW N302 – Fall 2012
� Average degree of consolidation (Ut) at time “t”:
Ut = St/Sultimate
The degree of consolidation at any time (t) is the percentage of
settlement that took place at this time w.r.t. to ultimate
consolidation settlement.
� Equation governing rate of consolidation
Differential Equation of ConsolidationDifferential Equation of Consolidation
� Consider a differential element of soil
Surcharge load
ground surface
Qin Area = dAVolume = dV
dx
dyx
z
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Qout
dzy z
kidAQ −=
)dzz
Q(QQ
t
(dV)inout
∂
∂=−=
∂
∂−compression and flow 1-D
Darcy’s law is valid
Differential Equation of ConsolidationDifferential Equation of Consolidation
kidAQ −=
dAz
hk
∂
∂−=
)dAγ
u(
zk
w∂
∂−=
dAz
u
γ
k
w ∂
∂−=
Darcy’s law is valid
u = excess p.w.pQin Area = dA
Volume = dV
dxdz
dy
xy
dA)dzz
u
γ
k(
zt
(dV)
w ∂
∂−
∂
∂=
∂
∂−
dVz
u
γ
k
t
(dV)2
2
w ∂
∂=
∂
∂Equation (1) soil is homogeneous
Qout
dz y
z
e1
e
dV
dVv
+= dV
e1
edVv
+=
dVv
de
Differential Equation of ConsolidationDifferential Equation of Consolidation
e1
dV
t
edV
t
e
t
)(edV
t
dV)e1
e(
t
)(dV
t
(dV)s
sv
+∂
∂=
∂
∂=
∂
∂=
∂
+∂
=∂
∂=
∂
∂
dVs
dV
1e1
1
dV
dVs
+= dV
e1
1dVs
+=
dVe(dV) ∂=
∂Equation (2)
e1tt +∂=
∂
From equations 1 and 2,
e1
dV
t
edV
z
u
γ
k2
2
w +∂
∂=
∂
∂
Equation (2)
t
e
z
u
γ
e)k(12
2
w ∂
∂=
∂
∂+
σ'ee ∂∂=
∂(chain rule)
Differential Equation of ConsolidationDifferential Equation of Consolidation
t
σ'
σ'
e
t
e
∂
∂
∂
∂=
∂
∂
t
ua)
t
u
t
σ(a
t
evv
∂
∂=
∂
∂−
∂
∂−=
∂
∂
t
σ'a
t
ev
∂
∂−=
∂
∂
(chain rule)
σ'
ea: v
∂
∂=
0t
σ: =
∂
∂ (load doesn’t vary with time)
t
ua
z
u
γ
e)k(1v2
2
∂
∂=
∂
∂+
tzγv2
w ∂∂
2
2
wv z
u
γa
e)k(1
t
u
∂
∂+=
∂
∂
2
2
vz
uc
t
u
∂
∂=
∂
∂
wvwv
vγm
k
γa
e)k(1c: =
+=
� Coefficient of Consolidation: cv
v
kc = cm2/sec
Differential Equation of ConsolidationDifferential Equation of Consolidation
where:k = soil permeabilitymv = coefficient of volume compressibility
wv
vγm
c = cm2/sec
Dr. Asmaa Moddather – PBW N302 – Fall 2012
� Boundary conditions:
Example: a clay layer, of thickness 2H, with top and bottom boundaries freely draining Sand
Differential Equation of ConsolidationDifferential Equation of Consolidation
� u(z=0,t) = 0
� u(z=2H,t) = 0
� u(z,t=0) = ui
Sand
Clay
u
t = t1
t = t2z
as time increases, u decreases2H
� Solution:
)TM((sinMZ)expM
2ut)u(z, v
2i −=∑∞
Solution of Differential Equation of ConsolidationSolution of Differential Equation of Consolidation
where:
Tv = Time factor
)TM((sinMZ)expM
t)u(z, v
0m
∑=
1)(2m2
πM +=
2
vv
H
tcT =
Tv = Time factor
� Solution in terms of degree of consolidation (U):
……………… Equation 3)TMexp(M
21U v
2
0m2
−−= ∑∞
=
Equation 3 was solved and the results were tabulated:
Tv = f(U)
Solution of Differential Equation of ConsolidationSolution of Differential Equation of Consolidation
2
d
vv
h
tcT =
where:T = time factor corresponding to degree of consolidation occurring at time t
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Tv = time factor corresponding to degree of consolidation occurring at time tcv = coefficient of consolidation (determined from consolidation test)hd = longest drainage path within clay layer
The above equation holds for consolidation of sample in the lab, and for clay layer consolidating in the field
0
10
20
3030
40
50
60
70
80
90
U (
%)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
90
100
0 0.2 0.4 0.6 0.8 1 1.2
Tv
10099959080706050403020100U (%)
∞1.7811.1290.8480.5670.4030.2870.1970.1260.0710.0310.0080.0Tv
� Drainage Path: hd
Sand Sand
Solution of Differential Equation of ConsolidationSolution of Differential Equation of Consolidation
2H Clay
Sand
2H Clay
Impervious Rockhd = H hd = 2H
Dr. Asmaa Moddather – PBW N302 – Fall 2012
Determination of cv
� cv is determined from the consolidation test data
� The is a c value that corresponds to each loading � The is a cv value that corresponds to each loading
increment, i.e. cv is stress dependent
� For the following test data corresponding to a certain
loading increment, find cv.
Dr. Asmaa Moddather – PBW N302 – Fall 2012
885240120603015106.2542.2510.50.250Time (min)
223221220219215209202193184173161155153135Dial Reading x 10-2 (mm)
888685848074675849382620153-135
=18135-135
= 0= ∆h x 10-2 (mm)
0.1 1 10 100 1000
Time (min)
t50 = 3.6 min0.25 1.00
� ∆h-log t Curve
Determination of cv
0
10
20
30
40
50
h (
x 1
0-2
mm
)
U = 0%
t50 = 3.6 min0.25 1.00
U = 50%
Dr. Asmaa Moddather – PBW N302 – Fall 2012
60
70
80
90
100
∆∆ ∆∆h
(x
U = 100%
1. Plot the (∆h) or (e) on the vertical axis (arithmetic scale), and the time (t) on
horizontal axis (logarithmic scale) S-shape
2. Determine (∆h) or (e) corresponding to 0% consolidation:
Determination of Determination of ccvv
2. Determine (∆h) or (e) corresponding to 0% consolidation:
� Consider (∆h1) corresponding to a small time (t1)
� Consider (∆h2) corresponding to (4t1)
� Determine the difference between (∆h1) and (∆h2) y
� Determine (∆h) corresponding to 0% consolidation at a distance
equivalent to y above (∆h1)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
1
3. Determine (∆h) or (e) corresponding to 100% consolidation:
� Extrapolate the linear portion at the middle of the consolidation curve
� Extrapolate the linear portion at the end of the consolidation curve
� The two lines intersect at a point corresponding to (∆h) or (e) at 100%
consolidation
1. Two points on the first portion of curve are selected for which the values
of “t” are in the ratio of 4:1.
2. The vertical distance between the two points is measured.
Determination of Determination of ccvv
2. The vertical distance between the two points is measured.
3. An equal distance set off above the first point fixes a line corresponding
to U = 0%.
4. The line corresponding to U = 100% passes through the intersection of
the two linear parts of the curve.
5. The line corresponding to U = 50% is located at the midway between
the two lines (U = 0% & U = 50%)
6. The intersection between the curve and the line (U = 50%) is
Dr. Asmaa Moddather – PBW N302 – Fall 2012
6. The intersection between the curve and the line (U = 50%) is
corresponding to t50.
2
D
vv
h
tcT =
t
hTc
2
Dvv =
50
2
Dv
t
0.197hc =
4. Determine (∆h) or (e) corresponding to 50% consolidation at mid
distance between 0% consolidation and 100% consolidation.
5. Determine t
Determination of Determination of ccvv
5. Determine t50
6. Calculate cv:
where: T = 0.197, h = ½ sample height (sample drains from both
2
d
vv
h
tcT =
Dr. Asmaa Moddather – PBW N302 – Fall 2012
where: T50 = 0.197, hd = ½ sample height (sample drains from both
sides), and t = t50 from consolidation curve
Types of Compression
0
0.1 1 10 100 1000
Time (min)Immediate consolidation
∆h = 00
10
20
30
40
50
h (
x 1
0-2
mm
)
U = 0%
Primary consolidation
Dr. Asmaa Moddather – PBW N302 – Fall 2012
60
70
80
90
100
∆∆ ∆∆h
(x
U = 100%
Secondary consolidation
1. Immediate: due to compression of entraped air and
machine parts.
Types of CompressionTypes of Compression
machine parts.
2. Primary: due to consolidation by escapage of water
(relief of excess pore-water pressure) under the effect of
pressure (loading)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
pressure (loading)
3. Secondary: compression at constant effective stress (no
excess pore-water pressure) [creep]
� A rectangular building 24 × 16 m is
shown in Figure and is founded on a
raft and has its foundation level at
ExampleExample
(a) (c) (b)
raft and has its foundation level at
2.0 m below ground surface. If the
building exerts a net stress of 2.0
kg/cm2 on soil at foundation
level, determine:
i. The ultimate settlement of
(a) (c)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
points a, b and c.
ii. The time required for these
points to settle 5.0 cm and the
time required undergo 90%
consolidation.
i. The ultimate settlement of points a, b and c.
qnet = 20 t/m2
Example
z = 8.0 m
H = 5.0 m
Using Influence Chart:
ca & bPoints
1224L (m)
88B (m)
Dr. Asmaa Moddather – PBW N302 – Fall 2012
88B (m)
1.53m
11n
0.2030.193Iz
0.203x20x4 = 16.240.193x20x2 = 7.72∆σ (t/m2)
0.004x16.24x5 = 0.325 m0.004x7.72x5 = 0.154 mS = mv ∆σ H
ii. The time required for these
points to settle 5.0 cm.
Example
(a) (c) (b)
cv = 0.006 cm2/sec
bcaPoints
=5/15.4 = 0.3255/32.5 = 0.154=5/15.4 = 0.325U = St/Sf
0.0830.0190.083From Chart
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0.0830.0190.083From Chart(Tv)
40.09.210.0t (days)
2
D
vv
h
tcT =
2(500)
t0.006019.0 =
2(250)
t0.006083.0 = 2(500)
t0.006083.0 =
ii. The time required for these
points to undergo 90%
Example
(a) (c) (b)
consolidation.
cv = 0.006 cm2/sec
bcaPoints
0.900.900.90U = St/Sf
0.8480.8480.848From Chart
Dr. Asmaa Moddather – PBW N302 – Fall 2012
0.8480.8480.848From Chart(Tv)
409.0409.0102.2t (days)
0.90 x 15.4 = 13.9 cm0.90 x 32.5 = 29.3 cm0.90 x 15.4 = 13.9 cmSt (cm)
2
D
vv
h
tcT =
2(500)
t0.006848.0 =
2(250)
t0.006848.0 =
2(500)
t0.006848.0 =