Soil Compressibility & Settlement
Faculty of Engineering – Cairo University Third Year Civil Soil Mechanics Spring 2015
Soil Settlement
Settlements are vertical deformations of a soil mass under the effect of an applied stress causing vertical movement of the supported structure Settlements are one of the safety and performance criteria in geotechnical assessment of structures Excessive settlements can result in structural damage to a building frame, or loss of functionality.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
1 Soil Settlement
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Soil Settlement
Uniform settlements: equal settlements across the area of the structure Differential settlements: unequally foundation settlements in different areas of the structure. Differential settlement can result in severe structural damage while uniform settlements are of less consequences
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
2 Soil Settlement
Differential settlements
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Soil Settlement
Differential settlements
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
3 Define: Compressibility
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.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
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. e a v p e e-p curve
eo = initial void ratio eo po = initial effective stress e e 1 p e1 = final void ratio
p1 = final effective stress
e = eo –e1 p = p –p p p 1 o o 1 p
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
4 Settlement of Cohesive Soils
Coefficient of volume compressibility (mv): is the volume decrease of a unit volume of soil per unit increase of effective pressure during compression.
ΔV ΔeVs
Vo (1 eo )Vs 1 Δe a v mv Δp Δp (1 eo ) Δp (1 eo )
eoVs
Initial Condition Final Condition
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Settlement of Cohesive Soils
For a thin layer, p at mid depth is considered as an average stress within the layer. q ΔH V Δe H V (1 eo ) Δe Δe Δp ΔH S H H (1 e ) ΔP (1 e ) p o o H a S v ΔpH Clay (1 eo )
S mvΔpH e H where: eo S = settlement 1+e mv = coefficient of volume compressibility o H p = stresses at mid depth of layer due to 1 added loads H = total thickness of layer
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
5 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-layer is pi. q
S mviΔpiHi
p1 H1 Clay p2 H2
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Settlement of Cohesionless Soils
Sand may be considered as an elastic material with Young’s modulus (E) Young’s Modulus (E): is the slope for the linear portion of the stress-strain curve. q strain 1 stress E H 1 H p E p H S p Sand H E 1 S pH E E 100 (v. loose) -800 (v. dense) kg/cm2
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
6 Example
Compute the settlement due to compressibility of the sand and clay layers at points (a) and (b) of the building shown in the 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. Sand: = 1.70 t/m3 E = 500 kg/cm2
2 Clay: mv = 0.03 cm /kg
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Example
2 qnet =q–h = 20 – 1.70x3.0 = 14.9 t/m 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,
2 na = 191 a,sand =na/200 (qnet) = 191/200 x 14.9 = 14.2 t/m 2 nb =50 b,sand =nb/200 (qnet) = 50/200 x 14.9 = 3.7 t/m
a b
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
7 Example
2 qnet =q–h = 20 – 1.70x3.0 = 14.9 t/m For clay layer: z = 11.0 m = 1” = 2.5 cm L = 30 m 30x2.5/11 = 6.8 cm, B = 20 m 20x2.5/11 = 4.5 cm
2 na = 144 a,clay = n/200 (qnet) = 144/200 x 14.9 = 10.7 t/m 2 nb = 47.3 b,clay = n/200 (qnet) = 47.3/200 x 14.9 = 3.5 t/m
a b
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Example
Sa =Ssand +Sclay
Ssand =1/Ea,sandH 2 2 H = 10.0 m E = 500 kg/cm a,sand = 14.2 t/m
Ssand = 1/500 x 14.2/10 x 1000 = 2.84 cm
Sclay =mv a,clayH 2 2 H = 2.0 m mv =0.03cm/kg a,clay = 10.7 t/m
Sclay = 0.03 x 10.7/10 x 200 = 6.42 cm
Sa =Ssand +Sclay = 2.84 + 6.42 = 9.26 cm
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
8 Example
Sb =Ssand +Sclay
Ssand =1/Eb,sandH 2 2 H = 10.0 m E = 500 kg/cm b,sand =3.7t/m
Ssand = 1/500 x 3.7/10 x 1000 = 0.74 cm
Sclay =mv b,clayH 2 2 H = 2.0 m mv =0.03cm/kg b,clay =3.5t/m
Sclay = 0.03 x 3.5/10 x 200 = 2.1 cm
Sb =Ssand +Sclay = 0.74 + 2.1 = 2.84 cm
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Time of settlement compared to construction time P Loading Diagram
Construction period Time Time Settlement of Sand 1 S pH E
S Time
Settlement of Clay S mvΔpH
S
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
9 Causes of Settlement
Static Loads
Dynamic Loads
Groundwater table lowering
Loads from adjacent buildings
Capillary forces
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Theory of Consolidation
Consolidation: is the process of squeezing water out of saturated soil under the effect of loads.
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 the effect of loading is time dependent.
Therefore, parameters governing consolidation process include: soil properties, stress (p), time (t), drainage conditions.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
10 Mechanism of Consolidation Process
Analogy:
Confined saturated soil Piston for Valve ~ soil (solids and voids filled loading permeability with water)
Versus
Container filled with water that has a spring and a valve Water ~ pore-water
Remember: water is Spring ~ Container Solids incompressible
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Mechanism of Consolidation Process
For Sands: high permeability ~ valve is open, water gets out immediately (t = 0), excess pore water pressure u = 0
Stress = p
At t = 0 (immediately), u = 0 Load carried by spring
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
11 Mechanism of Consolidation Process For Clays: low permeability ~ valve is partially open, water gets out slowly (time-dependent), u decreases with t Stress = p Stress = p
At t = 0 (immediately), u = p At t = t1, 0 u < p Spring carry no load Load carried by water and spring
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Mechanism of Consolidation Process
For Clays:
Stress = p Stress = p
At t = t1, 0 u < p At t = long time, u = 0 Load carried by water and spring Load carried by spring
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
12 Mechanism of Consolidation Process
If we apply a certain load, and the valve is closed, then water (incompressible material) carries the entire added load. If the valve is opened, water flows out, and thus its volume is reduced. As volume is reduced, the spring starts to be compressed and carries part of the load. Meanwhile, excess pore water pressure decreases. The process continues until sufficient amount of water escapes from the valve, resulting in compression of the spring to carry the entire load. At this point, the excess pore water pressure decreases to zero, the spring carries the entire load, the system is in equilibrium.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Application on Soil
Stress distribution before application of load
GWT GS
Sand
Clay
Gravel
σt u σ'
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
13 Application on Soil
Short time (immediately) after application of q (t = 0)
q GWT GS
Sand q
Clay q
Gravel q q
σt u σ'
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Application on Soil
Long time after application of q (t = h)
q GWT GS
Sand q
Clay
Gravel q q
σt u σ'
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
14 Application on Soil
Total stress: σt Σγh q Pore water pressure:
. t=0 u γwhw q (clay) u γwhw (sand)
. t=h u γwhw (clay)
u γwhw (sand) At time t ??? Effective stress: . t=0 σ' Σγ'h (clay) σ' Σγ'h q (sand)
. t=h σ' Σγ'h q (clay) σ' Σγ'h q (sand)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Rate of Consolidation
q GWT GS
Sand t=0 q
Clay q
Gravel q q
σt u σ'
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
15 Rate of Consolidation
u
t = 0
t = t1
Clay t = t2
t = t3
t = ∞
q
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Assumptions of Theory of Consolidation
1. Clay is homogeneous, isotropic, and saturated
2. Water and clay particles are incompressible
3. Darcy’s law is valid
4. The clay layer is laterally confined
5. One-dimensional compression, and one-dimensional flow
6. Consolidation parameters from test applies to clay layer from which sample for test was taken
7. Soil properties are constant with time
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
16 Consolidation Test
Objectives: . Volume change-effective pressure relationship . Stress history of soil . Volume change – time – pore water dissipation relationship
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test
Dial gauge Compression Load cell
Loading frame
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17 Consolidation Test
Dial gauge
Compression cell Lever Arm
Load
Consolidation Test set-up
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test
Dial gauge
Compression cell Lever Arm
Load =
Po
Consolidation Test set-up
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
18 Consolidation Test
Loading Head Dial gauge Saturated Metal ring sample
Porous stone
Filter paper
Compression Cell/Oedometer Porous stone
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test
Loading Head Dial gauge Load (P ) Saturated 1 Metal ring sample
Porous stone
Filter paper
Compression Cell/Oedometer Porous stone
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
19 Consolidation Test
Equipment:
1. Compression cell / Oedometer
2. Loading frame to apply weights (Po)
3. Dial gage to measure soil compression
Note:
Load acting on the cell (P1) is magnified by the lever arm
ratio of the loading frame (LAR), where: P1 = LAR x Po
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test
Procedure:
1. Trim the sample into the metal ring and place filter paper on both sides of sample.
2. Measure initial conditions of sample: ho, eo, wo, Gs
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 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.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
20 Consolidation Test
Procedure:
6. Place the hanging load (Po) such that the pressure on the clay 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.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test
Loading scheme, for example:
0.25, 0.5, 1.0, 2.0, 4.0, 8.0, 16.0 (kg/cm2) Loading
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 elapsed time (t).
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
21 Consolidation Test Data Reduction
For each loading increment, plot h-log t:
Time (min) 0.1 1 10 100 1000 0
10
20
30
40 mm) -2 50
h (x 10 60
70
80
90
100
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test Data Reduction
At the end of each loading increment:
1. Calculate effective pressure (p) = Load x LAR/sample area
2. Measure h = sample compression during this loading increment e h 3. Calculate void ratio (e):
eo e h 1+eo h 1 e h o o o 1
e = eo – e
Plot e-p curve
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
22 Consolidation Test Data Reduction
2 Example: LAR = 3, sample area = 41.85 cm , ho = 2.54 cm, eo = 0.636
Load (kg) 0 7 14.5 29 58 116 232
Dial Reading x 10-2 (mm) = h 0 89 135 223 359 550 705
Stress (p) = Load x lever arm 7 x 3/41.85 0 1.04 2.08 4.16 8.32 16.63 ratio/area (kg/cm2) = 0.50
0.89/25.4 (1+0.636) e = h/h (1+e ) 0 0.087 0.1436 0.2312 0.3543 0.4541 o o = 0.0573
0.636-0.0573 e = e - e 0.636 0.5490 0.4924 0.4048 0.2817 0.1819 o = 0.5787
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test Data Reduction Plot e-p curve:
0.7
0.6
0.5
0.4 e
0.3
0.2
0.1
0 0 5 10 15 20 p (kg/cm2)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
23 Consolidation Test Data Reduction
From e-p curve, for a certain stress range, calculate av
and mv, where:
e e a v p e o e e1 a v p mv (1 eo )
p p o 1 p
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test Data Reduction Plot e-log(p) curve:
0.7
0.6
0.5
eo 0.4 e 0.3 e
e1 0.2 log(p)
0.1
0 p p 0.1 1o 101 100 p (kg/cm2)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
24 Consolidation Test Data Reduction
From e-log(p) curve, the slope of the linear portion is the
Compression Index (Cc):
e C c log( p)
Determine preconsolidation pressure (pc): is the largest effective pressure that has been applied on the soil in its geological history.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Consolidation Test Data Reduction Determination of preconsolidation pressure:
0.7 5 0.6 a b 3 1 0.5 4
0.4 2 e 0.3
0.2
0.1
0 0.1 1pc 10 100 p (kg/cm2)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
25 Determination of Preconsolidation Pressure
1. Determine point of maximum curvature (a).
2. At (a), draw tangent to e-log(p) curve.
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).
6. The preconsolidation pressure (pc) is the x-coordinate of the point (b).
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Types of clays w.r.t. preconsolidation
1. Normally consolidated clay: Clay that has never been loaded in the past by more than the
existing 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: Clay that has been loaded partially by the existing effective
overburden pressure (po) pc < po
Define: Overconsolidation Ratio (OCR): p OCR c po
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
26 Determination of Compression Index
For normally consolidated clays, Cc can be calculated as follows:
From e-log(p) curve:
e C c log( p)
From liquid limit (empirical):
Cc 0.009(wL 10)
where wL is in (%)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Definitions
0.7 Recompression curve 0.6
0.5 Virgin compression 0.4 curve e 0.3
0.2
Unloading/reloading curve 0.1
0 0.1 1pc 10 100 p (kg/cm2)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
27 Definitions
From e-log(p) curve, the recompression index (Cr) is the slope of the equivalent linear portion of the unloading/reloading curve.
0.7 e C 0.6 r log( p) Cr 0.5
Empirical relationship for C : 0.4 r e 0.3 Cc Cr 0.2 Cr (1/10 1/ 5)Cc 0.1 Cr (1/ 8)Cc 0 0.1 1 10 100 p (kg/cm2)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Settlement
Calculate settlement (S) of normally consolidated clay layer
of thickness (H) due to added stress (p)given Cc of clay: 0.7 e p 0.6 Cc eo log( p) 0.5
e0 e1 0.4 e Cc e Cc log( p ) log( p ) 0.3 1 0 e1 0.2 e Cc 0.1 p1 0 log( ) p ~p p p 0.1 1c o 101 100 0 p (kg/cmp(kg/cm2)2) p1 e Cc log( )...... 1 p0 e h S ...... 2 1 eo ho H
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
28 Settlement
From 1 and 2:
p1 S Cc log( ) (1 eo ) po H C p S c H log( 1 ) 1 eo po
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Settlement
For overconsolidated clay, (po and p1) < pc:
0.7 e C 0.6 o r e1 0.5
0.4 e e 0.3 p
0.2
0.1
0 p p p 0.1o 11 c 10 100 2 p(kg/cmp (kg/cm2) ) C p S r H log( 1 ) 1 eo po
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
29 Settlement
For overconsolidated clay, po < pc and and p1 > pc
0.7 e o C 0.6 r
0.5
0.4 e e e1 Cc 0.3
0.2 p
0.1
0 p p p 0.1o 1c 101 100 p (kg/cmp(kg/cm2)2) C p C p S r H log( c ) c H log( 1 ) 1 eo po 1 eo pc
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Rate of Consolidation
Short time after application of load “q”
q GWT GS
Sand p
Clay p
Gravel p p
σt u σ'
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
30 Rate of Consolidation Sand u u Sand t = 0 z t = 0 u ' 2H t = t1
Clay Clay t = t2 t = t t = t3 Isochrone t = ∞ t = ∞ Sand Hydrostatic pore p p Sand water pressure Excess pore (GWT) water pressure
At time “t” and depth “z”: p = u + ’ = increase in total stress due to added load u = excess pore water pressure due to added load ’ = increase in effective stress due to added load
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Rate of Consolidation u Sand z t = 0 u ' 2H Clay t = t t = ∞
p Sand
Degree of Consolidation (Utz) at time “t” and depth “z”: ' p u u U 1 tz p p p
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
31 Rate of Consolidation
Average degree of consolidation (Ut) for the clay layer at time “t”: u Sand H U 2 U dz z t = 0 t tz u ' 0
2H umax where: Clay ' p u U t = t tz p p t = ∞ p Sand Ut = (area of ’ diagram)/(area of p diagram)
Ut = (area of p diagram – area of u diagram)/(area of p diagram) where: area of Δp diagram Δp 2H 2 area of u diagram u 2H 3 max Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Rate of Consolidation
Settlement of clay layer of thickness 2H at time t=∞:
Sultimate = mvp(2H)
Settlement of clay layer of thickness 2H at time t=t:
St = mv’(2H)
St = mvp Ut)(2H) = Ut Sultimate
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.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
32 Differential equation of consolidation
Equation governing rate of consolidation
Consider a differential element of soil
Qin Area = dA Volume = dV Surcharge load dy x ground surface y dx z dz
Qout
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Differential equation of consolidation
(dV ) Q Q Q ( )dz compression and flow 1-D t out in z
Q kidA Darcy’s law is valid
h k dA z u k ( )dA u = excess p.w.p Qin z w Area = dA Volume = dV k u dy dA z w dx x dz y (dV ) k u ( dA)dz z t z w z Qout (dV ) k 2u 2 dV Equation (1) soil is homogeneous t w z
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
33 Differential equation of consolidation
dV e e v dV dV dV 1 e v 1 e dVv e dV (dV ) (dV ) v dV 1 1 t t s dV dV dV 1 e s 1 e dVs 1
e ( dV ) (dV ) (dV ) (edV ) e e dV v 1 e s dV t t t t t s t 1 e (dV ) e dV Equation (2) t t 1 e
From equations 1 and 2,
k 2u e dV 2 dV w z t 1 e
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Differential equation of consolidation
k(1 e) 2u e 2 w z t e e ' (chain rule) t ' t e ' e a : a t v t v '
e u u (load doesn’t vary with time) av ( ) av : 0 t t t t t u Sand k(1 e) 2u u t = 0 z u 2 av ' w z t 2H u u k(1 e) 2u max Clay t a z 2 v w t = t u 2u k(1 e) k t = ∞ c : cv v 2 a m t z v w v w p Sand
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
34 Differential equation of consolidation
Coefficient of Consolidation: cv k 2 cv cm /sec mv w
where: k = soil permeability
mv = coefficient of volume compressibility
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Solution of differential equation of consolidation
Boundary conditions: Example: a clay layer, of thickness 2H, with top and bottom boundaries freely draining Sand
t = t1 u(z=0,t>0) = 0 as time increases, u decreases 2H u(z=2H,t>0) = 0 t = t Clay z 2
u Sand Initial condition: For wide fill, applied instantly to saturated soil, the initial excess pore-water pressure u(z,t=0) is independent of depth and equals the applied stress (p)
u(z,t=0) = ui = p
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
35 Solution of differential equation of consolidation
Solution:
2ui 2 u(z,t) (sin MZ)exp(M Tv ) ……………… Equation 3 m0 M where: M (2m 1) 2
cvt Tv H 2 Tv = Time factor
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Solution of differential equation of consolidation
Equation (3) was solved chart: u/p = f(Tv, z)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
36 Solution of differential equation of consolidation
2ui 2 u(z,t) (sin MZ)exp(M Tv ) m0 M
Solution in terms of degree of consolidation (U):
u u u ui U (z,t) i 1 u ui ui z u ui-u 2 2 U (z,t) 1 (sin MZ)exp(M Tv ) 2H m0 M t = t Average degree of consolidation for entire layer: U = area / area z By integration: 2H
udz 2 2 0 Ut 1 exp( M Tv ) Ut 1 2 m0 M ui
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Solution of differential equation of consolidation
Equation (4) was solved Table and chart: Tv = f(U)
U (%) 0 10 20 30 40 50 60 70 80 90 95 99 100
0.0 0.008 0.031 0.071 0.126 0.197 0.287 0.403 0.567 0.848 1.129 1.781 ∞ Tv
0
10
20
30
40
50 (%)
U 60
70
80
90
100 0 0.2 0.4 0.6 0.8 1 1.2
Tv
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
37 Solution of differential equation of consolidation
Time Factor: Tv
cvt Tv 2 hd
where:
Tv = time factor corresponding to degree of consolidation occurring at time t cv = 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
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Solution of differential equation of consolidation
Drainage Path: hd
Sand Sand
2H Clay 2H Clay
Sand Impervious Rock hd = H hd = 2H
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
38 Determination of cv - logarithm-of-time method
cv is determined from the consolidation test data
There is a cv value that corresponds to each loading
increment, i.e. cv is stress dependent.
Two methods to calculate cv: Logarithm-of-time method
Square–root-of-time method
For the following test data corresponding to a certain
loading increment, find cv.
Time (min) 0 0.25 0.5 1 2.25 4 6.25 10 15 30 60 120 240 885
Dial Reading x 10-2 135 153 155 161 173 184 193 202 209 215 219 220 221 223 (mm)
135-135 153-135 = h x 10-2 (mm) 20 26 38 49 58 67 74 80 84 85 86 88 = 0 =18
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Determination of cv - logarithm-of-time method
Time (min) 0.1 1 10 100 1000 t = t t = 4t t50 = 3.6 0 1 1 min
10 U = 0%
20
30
40
mm) U = 50% -2 50
h (x 10 60
70
80 U = 100%
90
100
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
39 Determination of cv - logarithm-of-time method
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 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
3. 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)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Determination of cv - logarithm-of-time method
4. Determine (h) or (e) corresponding to 50% consolidation at mid distance between 0% consolidation and 100% consolidation.
5. Determine t50
6. Calculate cv:
cvt Tv 2 hd
where: T50 = 0.197, hd = ½ sample height (sample drains from both
sides), and t = t50 from consolidation curve
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
40 Determination of cv – square root-of-time method
Square root time (min)0.5 00.5 5 10 15 20 25 30 35 (t90) 0 a 10 U = 0%
20
30
40 mm) -2 50
h (x 10 h (x 60 70 c U = 90%
80
90 b x 0.15x 100
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Determination of cv – square root-of-time method
1. Plot the (h) or (e) on the vertical axis (arithmetic scale), and the square root of time on horizontal axis (arithmetic scale)
2. Determine (h) or (e) corresponding to 90% consolidation:
Draw a straight line through the initial linear portion of the curve. This line will intersect the vertical axis at a point (a) that corresponds to U = 0%
Measure distance (x) between the vertical axis and the drawn line
Measure additional distance (0.15x) point (b)
Draw a line connecting points (a) and (b) and extend it until it intersects the original curve at point (c) corresponds to U = 90%
0.5 3. Determine (t90)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
41 Determination of cv – square root-of-time method
4. Calculate cv:
cvt90 T90 2 hd
where: T90 = 0.848, hd = ½ sample height (sample drains from both
sides), and t = t90 from consolidation curve
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Types of Compression
Time (min) 0.1 1 10 100h = 0 1000 0 Instantaneous/Immediate Consolidation 10 U = 0% 20
30
40 Primary Consolidation mm) -2 50
h (x 10 60
70
80 U = 100%
90 Secondary Consolidation
100
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
42 Types of Compression
1. Instantaneous/Immediate Compression: due to compression of entrapped air and machine parts
2. Primary Compression: due to consolidation by escapage of water (relief of excess pore-water pressure) under the effect of pressure (loading)
3. Secondary compression: compression at constant effective stress (no excess pore-water pressure)
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Secondary Compression Index
Time (min) 0.1 1 10t100 100 1000 0
10
20
30
40 mm) -2 50
h (x 10 60
70 U = 100% 80 h 90 log(t)
100
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
43 Secondary Consolidation Settlement (Ss)
Secondary compression index (C): e C log(t) where: h e (1 e ) H o
Secondary settlement (Ss):
C t1 Ss H log( ) 1 eo to
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Example
A rectangular building 24m 16m is founded on a raft and has its (a) (c) (b) 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 points a, b Sand and c. ii.The time required for these points to settle5.0cm,andthetimerequiredto undergo 90% consolidation.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
44 Example 2 i. qnet =20t/m z=8.0m H=5.0m Using Influence Chart:
Points a & b c L (m) 24 12 B (m) 8 8 m 3 1.5 n 1 1
Iz 0.193 0.203 (t/m2) 0.193x20x2 = 7.72 0.203x20x4 = 16.24
S = mv H 0.004x7.72x5 = 0.154 0.004x16.24x5 = 0.325 m m
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Example
(b) ii. (a) (c)
2 cv = 0.006 cm /sec
Points a c b
U = St/Sf =5/15.4 = 0.325 5/32.5 = 0.154 =5/15.4 = 0.325 From Chart 0.083 0.019 0.083
(Tv)
cv t 0.006 t 0.006 t 0.006 t T 0.019 0.083 v 2 0.083 2 2 2 h D (250) (500) (500)
t (days) 10.0 9.2 40.0
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
45 Example
(a) (c) (b)
Points a c b
U = St/Sf 0.90 0.90 0.90 From ChartExample0.848 0.848 0.848 (Tv) ii.
2 c t cv = 0.006 cm0.006/sec t 0.006 t 0.006 t v 0.848 Tv 2 2 0.848 2 0.848 2 h D (250) (500) (500) t (days) 102.2 409.0 409.0
St (cm) 0.90 x 15.4 = 13.9 cm 0.90 x 32.5 = 29.3 cm 0.90 x 15.4 = 13.9 cm
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
Potential Sources of Error
Sample disturbance: due to improper handling and/or trimming. Disturbance affects e-logp curve, and thus preconsolidation pressure. Larger samples help reduce effect of trimming on sample disturbance.
Kontopoulos, Nikolaos S., 2012
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
46 Potential Sources of Error
Specimen not completely filling the ring: erroneous volume change due to compression of voids.
Clogged porous stones: porous stones should be cleaned after every test to remove embedded soil particles.
Friction between specimen and consolidation ring. Part of load applied to specimen can be lost by side friction. This effect can be reduced by lining the consolidation ring with Teflon.
Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)
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