Foundation Engineering. 02 Soil Compressibility
Foundation Engineering. 02 Soil Compressibility
Foundation Engineering. 02 Soil Compressibility
SOIL COMPRESSIBILITY
CE 543/511D: GEOTECHNICAL ENGINEERING 02 - FOUNDATION ENGINEERING
SOUTH
THE LEANING TOWER OF PISA
THE LEANING TOWER OF PISA
IN 2009
<0.26” PER YEAR
NORTH
THE KANSAI AIRPORT
THE KANSAI AIRPORT
THE KANSAI AIRPORT
3.2m
THE KANSAI AIRPORT
THE KANSAI AIRPORT
THE KANSAI AIRPORT
THE KANSAI AIRPORT
ELASTIC SETTLEMENT 𝑷
∆𝝈′
𝝈′ 𝒒=
𝑷
𝑨
𝜹𝑷
𝜹𝑬
𝜹𝑷
VOIDS
SOLIDS
VOIDS
SOLIDS
𝝈′
∆𝝈′
𝑷
SECONDARY CONSOLIDATION
SETTLEMENT 𝒒=
𝑷
𝑨
∆𝝈′ 𝑷
𝒒=
𝝈′ 𝑨
𝜹𝑷
𝜹𝑺 𝜹𝑬
𝜹𝑷
𝜹𝑺 VOIDS
SOLIDS
𝝈′
∆𝝈′
𝑷
SOIL SETTLEMENT 𝑷
𝒒=
𝑨
𝜹T = 𝜹𝐸 + 𝜹𝑃 + 𝜹𝑆
𝜹
ELASTIC SETTLEMENT: FAST
CONSOLIDATION SETTLEMENT:
SAND GRAVEL FAST
VOIDS
SILT CLAY SLOW
SOLIDS
SECONDARY COMPRESSION SETTLEMENT: SLOW
SOIL SETTLEMENT
FGS CGS
TIME FRAME PROCESS MAGNITUDE PROCESS MAGNITUDE
NEGLIGIBLE TO SMALL
DISTORTION
SHORT TERM DISTORTION NEGLIGIBLE TO SMALL
CONSOLIDATION SMALL TO MODERATE
CONSOLIDATION MODERATE TO LARGE SECONDARY
NEGLIGIBLE TO SMALL
LONG TERM SECONDARY COMPRESSION
SMALL TO LARGE
COMPRESSION
𝑷
ELASTIC SETTLEMENT 𝒒=
𝑷
𝑨
𝑞𝐵(1 − µ2 )
𝑆𝐸 = 𝐼𝑝
𝐸
𝜹𝑬
WHERE:
𝑞 - PRESSURE FROM THE FOOTING
𝐵 - BASE OF THR FOOTING
VOIDS
µ - POISSON’S RATIO OF THE SOIL
𝐸 - MODULUS OF ELASTICITY OF THE SOIL SOLIDS
𝐼𝑃 - INFLUENCE FACTOR
𝑷
PRIMARY CONSOLIDATION
SETTLEMENT
𝑷
𝒒=
𝑨
′ ′
𝜎𝑣0 + ∆𝜎𝑣′ = 𝜎𝑣𝑓 𝜹𝑷
CONSOLIDATION
𝜹𝑷 𝜹𝑷
∆𝑽𝒗 ∆𝑽𝒗
VOIDS VOIDS
𝑢0
𝑢𝑒
SOLIDS SOLIDS
PRESSURE
GAUGE
′ ′ ′ ′
𝜎𝑣0 + ∆𝜎𝑣′ = 𝜎𝑣𝑓 𝜎𝑣0 + ∆𝜎𝑣′ = 𝜎𝑣𝑓
CONSOLIDATION 𝝈′ CONSOLIDATION 𝝈′
∆𝝈′ ∆𝝈′
𝝈𝒗𝟎
∆𝑷 ∆𝝈𝒗
𝝈𝒐
𝑷
𝒕
𝝈𝒗𝟎
∆𝒖
𝒖𝟎
𝑢0 𝒕
𝝈𝒗𝟎
𝑢𝑒
𝝈′𝒗𝒇
∆𝝈′𝒗
𝝈′𝒗𝒐
𝒕
𝝈𝒗𝟎
∆𝝈𝒗
𝝈𝒐
𝒕
𝝈𝒗𝟎
FILL
𝜹𝑷
∆𝒖
𝒖𝟎
SOIL 𝑨 𝒕
𝝈𝒗𝟎
LAB 𝝈′𝒗𝒇
∆𝝈′𝒗
𝝈′𝒗𝒐
𝒕
OEDOMETER
DIAL GAGE 0
𝜎′ = 𝜎 − 𝑢 ∆𝐿
LOADING CAP 𝜀=
𝐿
𝛿𝑧
𝑃 ≈0
𝑷 𝜎′ = − 𝑢 ∆𝐻
𝐴 𝜀=
𝐻
𝑃
𝜎′ =
≈ 1” 𝐴
SOIL SAMPLE
𝜎1′ 𝜀1
CONFINING RING 𝜎2′ 𝜀2
3” − 4”
POROUS STONE 𝜎3′ 𝜀3
𝐻𝑜 ≈ 3/4” − 1”
𝜎4′ 𝜀4
https://youtu.be/mtuBbWXE0D8?t=536
0.0 0.0
RECOMPRESSION
CURVE
0.05 0.05
0.10 0.10
COMPRESSION
CURVE
STRAIN
STRAIN
0.15 0.15
REBOUND
0.20 0.20 CURVE
0.25 0.25
0 200 400 600 800 10 100 1000
3 2
4
STRAIN
𝑨
5
STRAIN
LOG OF EFFECTIVE STRESS LOG OF EFFECTIVE STRESS
𝝈′𝒄 𝝈′𝒄
0.0 0.0
0.05 0.05
0.10 0.10
STRAIN
STRAIN
0.15 0.15
0.20 0.20
0.25 0.25
10 100 1000 1000 10 100 1000 1000
0.05 1.60
0.10 1.20
VOID RATIO
STRAIN
0.15 0.80
0.20 0.40
0.25 0.00
10 100 1000 1000 10 100 1000 1000
𝐶𝐶 =
log 𝜎𝑏′ − log(𝜎𝑎′ )
𝑒𝑎 − 𝑒𝑏
𝐶𝐶 =
b
log 𝜎𝑏′ /𝜎𝑎′
𝑒𝑏
𝑒𝑐 − 𝑒𝑑
𝐶𝑆 =
LOG OF EFFECTIVE STRESS log 𝜎𝑑′ /𝜎𝑐′
𝜎𝑎′ 𝜎𝑏′
𝑒𝑎 − 𝑒𝑏
RECALL: 𝐶𝐶 = log 𝜎 ′ /𝜎 ′
COMPRESSION INDICES 𝑏 𝑎
𝑟𝑖𝑠𝑒 𝑑𝜀 ∆𝜀
c 𝑚= =
𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 =
d
𝑟𝑢𝑛 𝑑log 𝜎′ ∆log 𝜎′
∆𝑒𝑏 ∆𝑒𝑎 𝜀𝑏 − 𝜀𝑎
𝜀𝑏 − 𝜀𝑎 = − 𝐶. 𝑅 =
a 1+𝑒 1+𝑒 log 𝜎𝑏′ − log(𝜎𝑎′ )
𝜀𝑎
𝑒𝑏 − 𝑒 𝑒𝑎 − 𝑒 𝑒𝑎 − 𝑒𝑏
STRAIN
𝜀𝑏 − 𝜀𝑎 = −
1+𝑒 1+𝑒 𝐶. 𝑅 = 1 +′ 𝑒 ′
log 𝜎𝑏 /𝜎𝑎
𝑒𝑏 − 𝑒𝑎
𝜀𝑏 − 𝜀𝑎 = 𝐶𝑐
b
1+𝑒 𝐶. 𝑅 =
𝜀𝑏 1+𝑒
𝐶𝑆
LOG OF EFFECTIVE STRESS
𝑅𝑒𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 =
1+𝑒
𝜎𝑎′ 𝜎𝑏′
EMPIRICAL FORMULA FOR COMPRESSION INDEX, 𝑪𝑪 :
SKEMPTON (FOR REMOLDED CLAY) 𝐶𝐶 = 0.007 (𝐿𝐿 − 7)
(FOR UNDISTURBED CLAY) 𝐶𝐶 = 0.009 (𝐿𝐿 − 10)
1+𝑒
RENDON – HERRERO 𝐶𝐶 = 0.141 𝐺𝑆 1.2 ( 𝐺 )2.38
𝑆
𝑨 𝜎′ = 𝜎 − 𝑢
5
1
SOIL
5 4
3
STRAIN
6 𝑨 𝜎′ = 𝜎 − 𝑢
𝑂𝐶𝑀
0.15 1+𝑒
𝑂𝐶𝑅 = 1.0
0.20 OVERCONSOLIDATED
𝜎𝑐′ < 𝜎′
0.25
10 100 ′ 1000 1000 𝑂𝐶𝑀 > 0
𝝈′ 𝝈𝒄
EFFECTIVE STRESS 𝑂𝐶𝑅 > 1.0
0.0 𝐶𝑐 𝑎𝑛𝑑 𝐶𝑆 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
𝐶𝑆
𝑂𝐶𝑀 𝑎𝑛𝑑 𝑂𝐶𝑅 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
1+𝑒
0.05
0.10
𝐶𝑆
𝐶𝐶
1+𝑒 A
STRAIN
0.15 1+𝑒
0.20
B
0.25
10 100 1000 1000
EFFECTIVE STRESS
0.0 𝜎 ′ + ∆𝜎 ′ = 𝜎𝑓′
𝐶𝑆 𝐶𝑆 𝜎𝑐′
𝜀1 = log ′ CONSOLIDATION
1+𝑒 1+𝑒 𝜎
0.05
∆𝐿 ∆𝐻 𝑆𝑃
𝜀2 𝐶𝐶 𝜎𝑓′ RECALL: 𝜀=
𝐿
=
𝐻
=
𝐻
0.10 = log ′
1+𝑒 𝜎𝑐
𝐶𝐶 𝑆𝑃 = 𝜀𝐻
STRAIN
1+𝑒
0.15
OVERCONSOLIDATED SOIL
CASE II: 𝜎 ′ < 𝜎𝑐′ & (𝜎𝑓′ > 𝜎𝑐′ )
0.20
𝐶𝑆 𝜎𝑐′ 𝐶𝐶 𝜎𝑓′
𝑆𝑃 = [ log ′ + log ′ ]𝐻
1+𝑒 𝜎 1+𝑒 𝜎𝑐
0.25
10 100 1000 1000
𝝈′ 𝝈′𝒄 𝝈′𝒇 𝐶𝑆 𝐻 𝜎𝑐′ 𝐶𝐶 𝐻 𝜎𝑓′
EFFECTIVE STRESS 𝑆𝑃 = log ′ + log ′
1+𝑒 𝜎 1+𝑒 𝜎𝑐
0.0 𝜎 ′ + ∆𝜎 ′ = 𝜎𝑓′
𝐶𝑆 𝐶𝑆 𝜎𝑓′
𝜀 = log ′ CONSOLIDATION
1+𝑒 1+𝑒 𝜎
0.05
∆𝐿 ∆𝐻 𝑆𝑃
RECALL: 𝜀=
𝐿
=
𝐻
=
𝐻
0.10
𝐶𝐶 𝑆𝑃 = 𝜀𝐻
STRAIN
1+𝑒
0.15
OVERCONSOLIDATED SOIL
0.20 CASE I: 𝜎 ′ < 𝜎𝑐′ & (𝜎𝑓′ < 𝜎𝑐′ )
𝐶𝑆 𝜎𝑓′
0.25 𝑆𝑃 = [ log ′ ]𝐻
10 100 ′ ′ 1+𝑒 𝜎
𝝈′ 𝝈𝒇𝝈𝒄 1000 1000
1+𝑒
0.15
NORMALLY CONSOLIDATED SOIL
0.20 𝜎 ′ = 𝜎𝑐′
𝐶𝐶 𝜎𝑓′
𝑆𝑃 = [ log ′ ]𝐻
0.25 1+𝑒 𝜎
10 100 1000 1000
′
𝝈 = 𝝈′𝒄 𝝈′𝒇
𝐶𝐶 𝐻 𝜎𝑓′
EFFECTIVE STRESS 𝑆𝑃 = log ′
1+𝑒 𝜎
PROBLEM 01
A 2.1 LAYER OF CLAY IS BURIED BENEATH A 3M STRATUM OF VERY COMPACT GRANULAR SOIL.
COMPACT SAND UNDERLIES THE CLAY. THE LAYER OF GRANULAR SOIL IS COMPOSED OF
MATERIAL HAVING A UNIT WEIGHT OF 20.46 KN/M3. THE CLAY UNIT WEIGHT IS 16.52 KN/M3. A
LABORATORY COMPRESSION TEST ON A SAMPLE OF THE CLAY INDICATES A COMPRESSION
INDEX OF 0.40 AND A NATURAL VOID RATIO OF 1.30. A PLANNED BUILDING LOADING WILL CAUSE
A 26.38KPA STRESS INCREASE AT THE MIDDLE OF THE CLAY LAYER. THE PAST MAXIMUM
PRESSURE WAS 95.94KPA AND THE CS VALUE WAS 0.10
a. WHAT AMOUNT OF PRIMARY COMPRESSION OCCURS IN THE CLAY FOR THE INDICATED
CONDITIONS?
b. HOW MUCH PRIMARY COMPRESSION OF THE CLAY LAYER WOULD RESULT IF THE
GROUNDWATER TABLE WAS AT THE GROUND SURFACE (ALL OTHER CONDITION REMAINS)?
FOR EFFECTIVE STRESS:
𝜎′ = 𝜎 − 𝑢
𝜎 ′ = 20.46 3 + 16.52 1.05 − 0
𝛾 = 20.46𝑘𝑁/𝑚3
𝜎 ′ = 78.726𝑘𝑃𝑎
3.0𝑚 GRANULAR SOIL FOR FINAL EFFECTIVE STRESS:
𝜎𝑓′ = 𝜎 ′ + ∆𝜎′
𝐶𝑠 𝐻 𝜎𝑓′
𝜎𝑐′ = 95.94𝑘𝑃𝑎 𝐶𝑐 = 0.40 𝑆𝑝 = log ′
𝐶𝑠 = 0.10 1+𝑒 𝜎
∆𝜎′ = 26.38𝑘𝑃𝑎
0.10(2.1𝑚) 65.3755𝑘𝑃𝑎
OVERCONSOLIDATED CLAY: CASE I 𝑆𝑝 = log × 1000
1 + 1.30 38.9955𝑘𝑃𝑎
𝜎 ′ < 𝜎𝑐′ & 𝜎𝑐′ > 𝜎𝑓′
𝑆𝑝 = 20.4887𝑚𝑚
PROBLEM 02
THE SAND IN THE FIGURE HAS A HEIGHT OF 4.8M. THE GROUNDWATER TABLE IS 3.4M
BELOW THE GROUND SURFACE. THE UNIT WEIGHT OF SAND ABOVE THE WATER TABLE
IS 17.31 KN/M3 AND HAS A SATURATED UNIT WEIGHT OF 18.10 KN/M3 BELOW THE
WATER TABLE. THE SAND OVERLIES A CLAY LAYER 1.2M THICK HAVING A SATURATED
UNIT WEIGHT OF 16.5 KN/M3 AND A VOID RATIO OF 1.70. THE OVER CONSOLIDATION
RATIO IS 2.0. COMPRESSION INDICES ARE CS = 0.04 AND CC = 0.35.
IF A SQUARE FOOTING 3M X 3M RESTING ON THE SAND LAYER CARRIES A COLUMN
LOAD OF 3500 KN. THE BASE OF THE FOOTING IS 1.2M BELOW THE GROUND
SURFACE, COMPUTE THE SETTLEMENT DUE TO CONSOLIDATION OF THE CLAY LAYER.
FOR INITIAL EFFECTIVE STRESS:
1.2𝑚 3mx3m 𝛾 = 17.31𝑘𝑁/𝑚3 𝜎′ = 𝜎 − 𝑢
1 3.4𝑚 𝜎 ′ = 17.31 3.4 + 18.10 1.4 + 16.5 0.6 − 9.81(2)
4.8𝑚 𝜎 ′ = 74.474𝑘𝑃𝑎
2 GWT
4.2𝑚 𝛾𝑠𝑎𝑡 = 18.10𝑘𝑁/𝑚3 FOR PRECONSOLIDATION STRESS:
𝜎𝑐′
𝑂𝐶𝑅 = 2.0 =
𝜎′
𝛾𝑠𝑎𝑡 = 16.5𝑘𝑁/𝑚3 𝜎𝑐′ = 2.0(74.474)
1.2𝑚
2.1m 3m 2.1m 𝑒 = 1.70 𝜎𝑐′ = 148.948𝑘𝑃𝑎
FOR ADDITIONAL EFFECTIVE STRESS:
𝑂𝐶𝑅 = 2.0 𝐶𝑐 = 0.35
𝐶𝑠 = 0.04 ′
𝑃 𝑃 3500
OVERCONSOLIDATED CLAY: CASE I ∆𝜎 = = 2
= 2
= 67.5154𝑘𝑃𝑎
𝐴 𝐵+𝐻 3 + 4.2
𝐶𝑠 𝐻 𝜎𝑓′
FOR SETTLEMENT: 𝑆𝑝 = log ′ FOR FINAL EFFECTIVE STRESS:
1+𝑒 𝜎
0.04(1200𝑚𝑚) 141.9894𝑘𝑃𝑎 𝜎𝑓′ = 𝜎 ′ + ∆𝜎 ′ = 74.474 + 67.5154
𝑆𝑝 = log
1 + 1.70 74.474𝑘𝑃𝑎
𝜎𝑓′ = 141.9894𝑘𝑃𝑎
𝑆𝑝 = 4.9822𝑚𝑚
PROBLEM 03
TWO FOOTING “A” AND “B” REST IN A LAYER OF SAND 2.7M THICK. THE BOTTOM OF THE FOOTINGS ARE 0.90M
BELOW THE GROUND SURFACE. BENEATH THE SAND LAYER IS 1.8M CLAY LAYER. BENEATH THE CLAY LAYER IS A
HARD PAN. THE WATER TABLE IS AT A DEPTH OF 1.8M BELOW THE GROUND SURFACE. THE LAYER OF GRANULAR
SOIL IS COMPOSED OF MATERIAL HAVING A SATURATED UNIT WEIGHT OF 20.46 KN/M3 AND A UNIT WEIGHT OF
18.45 KN/M3. THE CLAY SATURATED UNIT WEIGHT IS 16.52 KN/M3. A LABORATORY COMPRESSION TEST
INDICATES THAT THE CLAY HAS A PRE-CONSOLIDATION PRESSURE OF 72KPA. COMPRESSION INDEX IS 0.30 AND
THE VALUE OF SWELL INDEX IS 0.05. VOID RATIO OF CLAY IS 1.50. DETERMINE:
A. THE STRESS INCREASE AT THE CENTER OF CLAY LAYER ASSUMING THAT FOOTING “A”, WITH A DIMENSION 2M
X 1.8M, DISTRIBUTES A PRESSURE AT AN ANGLE OF 2 VERTICAL TO 1 HORIZONTAL. THE LOAD OF THE FOOTING IS
3500kN.
B. THE SIZE OF THE SQUARE FOOTING “B” SO THAT THE SETTLEMENT OF THE CLAY LAYER IS THE SAME BENEATH
FOOTING “A” AND “B”.
C. DETERMINE THE SETTLEMENT BENEATH FOOTING “A”.
NOTE! NEGLECT THE EFFECT OF THE LOAD OF THE TWO FOOTING WITH EACH OTHER.
FOR INITIAL EFFECTIVE STRESS:
′ FOR STRESS IN INCREASE DUE TO FOOTING A:
𝜎 = 18.45 1.8 + 20.46 0.9 + 16.52 0.9 − 9.81(1.8)
′
𝑃 𝑃
′ ∆𝜎 = =
𝜎 = 48.834𝑘𝑃𝑎
𝐴 (𝐵 + 𝐻)(𝐿 + 𝐻)
FOR FINAL EFFECTIVE STRESS:
3500
𝜎𝑓′ = 48.834 + 165.4846 = 214.3186𝑘𝑃𝑎 ∆𝜎𝐴′ =
(2 + 2.7)(1.8 + 2.7)
A ∆𝜎𝐴′ = 165.4846𝑘𝑃𝑎
B
0.9𝑚 2mx1.8m 𝛾 = 18.45𝑘𝑁/𝑚3 FOR THE DIMENSION OF FOOTING B:
1.8𝑚
2.7𝑚 GRANULAR SOIL GWT ∆𝜎𝐴′ = ∆𝜎𝐵′
𝜎 ′ = 48.834𝑘𝑃𝑎 3500
∆𝜎𝐴′ =
(2 + 2.7)(1.8 + 2.7)
FOR FINAL EFFECTIVE STRESS:
𝜎𝑓′ = 48.834 + 165.4846 = 214.3186𝑘𝑃𝑎 ∆𝜎𝐴′ = 165.4846𝑘𝑃𝑎
𝑆𝑝𝐴 = 𝑆𝑝𝐵
𝐶𝑐 𝐻 𝜎 ′ + ∆𝜎𝐴′ 𝐶𝑐 𝐻 𝜎 ′ + ∆𝜎𝐵′
log = log
1+𝑒 𝜎𝑐′ 1+𝑒 𝜎𝑐′
𝜎 ′ + ∆𝜎𝐴′ 𝜎 ′ + ∆𝜎𝐵′
log ′ = log
𝜎𝑐 𝜎𝑐′
𝜎 ′ + ∆𝜎𝐴′ 𝜎 ′ + ∆𝜎𝐵′
′ =
𝜎𝑐 𝜎𝑐′
𝜎 ′ + ∆𝜎𝐴′ = 𝜎 ′ + ∆𝜎𝐵′
∆𝜎𝐴′ = ∆𝜎𝐵′
SECONDARY CONSOLIDATION SETTLEMENT
∆𝝈′
𝝈′
𝜹𝑷
𝜹𝑺 INITIAL COMPRESSION
𝜹𝑷
STRAIN
𝜹𝑺 VOIDS
PRIMARY CONSOLIDATION
SOLIDS
SECONDARY CONSOLIDATION
𝝈′
∆𝝈′ LOG OF TIME
SECONDARY CONSOLIDATION SETTLEMENT
SECONDARY COMPRESSION INDICES:
𝑑𝑒
𝐶𝛼 = INITIAL COMPRESSION
𝑑(log 𝑡)
𝑑𝜀
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 =
𝑑(log 𝑡)
STRAIN
𝐶𝛼
𝑆𝐶𝑅 = PRIMARY CONSOLIDATION
1 + 𝑒𝑝
𝑡 > 𝑡𝑝 𝑆𝑠 = 43.7468𝑚𝑚
𝑡 = 𝑡𝑝 + ∆𝑡
OVERCONSOLIDATED SOIL
𝑒
𝐶𝑠 𝜎c′
∆𝑒1 = 𝐶𝑠 log ′
CASE II: (𝜎𝑐′ >𝜎′) & (𝜎𝑓′ >𝜎𝑐′ )
𝜎
𝑒𝑝 = 𝑒 − (∆𝑒1 + ∆𝑒2 )
𝜎c′ 𝜎f′
𝑒𝑝 = 𝑒 − 𝐶𝑠 log ′ − 𝐶𝑐 log ′
𝜎 𝜎𝑐
𝜎f′
∆𝑒1 = 𝐶𝑐 log ′
VOID RATIO
𝐶𝑐 𝜎𝑐 OVERCONSOLIDATED SOIL
CASE I: (𝜎𝑐′ >𝜎′) & (𝜎𝑓′ <𝜎𝑐′ )
𝜎𝑓′
𝑒𝑝 𝑒𝑝 = 𝑒 − 𝐶𝑠 𝑙𝑜𝑔 ′
𝜎
𝑆𝑝 = 50.6066𝑚𝑚
𝑡𝑝 = 8𝑦𝑟𝑠 𝑡 = 20𝑦𝑟𝑠
SECONDARY COMPRESSION SETTLEMENT: 𝐶𝛼 = 0.008 𝑒𝑝 = 1.4749
𝐶𝛼 𝐻 𝑡
𝑆𝑐 = log 0.008(1800𝑚𝑚) 20𝑦𝑟𝑠
1 + 𝑒𝑝 𝑡𝑝 𝑆𝑐 = log
1 + 1.4749 8𝑦𝑟𝑠
𝑆𝑐 = 2.354𝑚𝑚
PROBLEM 06
THE FIGURE SHOWS A SOIL FORMATION OF 5M SAND AND 6M CLAY. THE
GROUNDWATER TABLE IS 4M BELOW THE GROUND SURFACE. IF A FILL WITH A
UNIT WEIGHT OF 17KN/M3 AND A HEIGHT 2M IS PLACED ON THE GROUND, FIND:
A. THE PRIMARY CONSOLIDATION SETTLEMENT OF THE CLAY LAYER IF IT IS
NORMALLY CONSOLIDATED WITH A COMPRESSION INDEX OF 0.38 AND A VOID
RATIO OF 1.52.
B. THE TOTAL CONSOLIDATION SETTLEMENT OF THE CLAY 5 YEARS AFTER
THE COMPLETION OF PRIMARY CONSOLIDATION SETTLEMENT. TIME FOR
COMPLETION OF PRIMARY SETTLEMENT IS 2 YEARS. SECONDARY COMPRESSION
INDEX IS 0.02
FILL 2𝑚
𝛾 = 17𝑘𝑁/𝑚3
𝛾 = 15𝑘𝑁/𝑚3
4𝑚
5𝑚 GRANULAR SOIL
GWT
𝛾𝑠𝑎𝑡 = 16𝑘𝑁/𝑚3
𝛾𝑠𝑎𝑡 = 18𝑘𝑁/𝑚3
6𝑚 𝑒 = 1.52
CLAY LAYER
𝐶𝑐 = 0.38
FOUNDATION ENGINEERING:
SOIL COMPRESSIBILITY
CE 543/511D: GEOTECHNICAL ENGINEERING 02 - FOUNDATION ENGINEERING