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EGB373-TUT 06-Week 07- Consolidation Settlement in soils-Answers-Colour.pdf

EGB373-TUT 06-Week 07- Consolidation Settlement in soils-Answers-Colour

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Oct 30, 2023

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1.000 1.050 1.100 1.150 1.200 1.250 1.300 10 100 1000 Log( Effective stress kPa) Void Ratio Point with minimum radius (maximum curvature = 1/R) A Horizontal line at A Tangent at A Bisector s c ’ = 100 kPa The pre-consolidation pressure, σ c ’= 100 kPa Q1

1.000 1.050 1.100 1.150 1.200 1.250 1.300 10 100 1000 Log( Effective stress kPa) Void Ratio Slope: Compression index, C c Compression index, C c = 0.68 Select two points on the line AB: A B (90, 1.150) e 1 = 1.150 s 1 ’ =90 kPa Q1 conti.. (150, 1.000) e 2 = 1.000 s 2 ’ =150 kPa 68 . 0 ) 90 / 150 log( 150 . 0 90 log 150 log ) 000 . 1 150 . 1 ( log log ) ( ' 1 ' 2 2 1         s s e e C c

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1.000 1.050 1.100 1.150 1.200 1.250 1.300 10 100 1000 Log( Effective stress kPa) Void Ratio Slope: Re-compression index, C r Re-compression index, C r = 0.15 Select two points on the line DE: D E (20, 1.250) e 1 = 1.250 s 1 ’ =20 kPa Q1 conti.. (1.200, 43) e 2 = 1.200 s 2 ’ =43 kPa 15 . 0 ) 20 / 43 log( 050 . 0 20 log 43 log ) 200 . 1 250 . 1 ( log log ) ( ' 1 ' 2 2 1         s s e e C r

Q2 e o =1.11 e c e f e s 0 ’=50 s c ’ =100 s f ’ =115 Log(σ ’ ) Δσ ’ =65kPa C r =0.08 C c =0.4 C r = 0.08 C c =0.4 Over consolidation ratio (OCR) = 2 kPa OCR c c c 100 50 2 ' 50 ' 2 ' ' 0        s s s s Pre-consolidation pressure (s c ’ ) = 100 e 0 = 1.11 s 0 ’ =50 kPa Ds ’ = 65 kPa s f ’  s 0 ’ + Ds ’ = 50+65 =115 kPa

Q2 Conti.. 086 . 1 50 100 log 08 . 0 11 . 1 50 100 log 08 . 0 ' ' log 0                            D  c c c o c r e e e e e C s s e o =1.11 e c e f e s 0 ’=50 s c ’ =100 s f ’ =115 Log(σ ’ ) Δσ ’ =65kPa C r =0.08 C c =0.4

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Q2 conti.. 062 . 1 100 115 log 4 . 0 086 . 1 100 115 log 4 . 0 ' ' log                            D  f f f c c f c e e e e e C s s Final Void Ratio e o =1.11 e c =1.086 e f e s 0 ’=50 s c ’ =100 s f ’ =115 Log(σ ’ ) Δσ ’ =65kPa C r =0.08 C c =0.4

Q3 H=3m 0.5m q=75kPa B=6m Δ 8 . 0 6 . 2 0   e G s g sat =? Initial effective stress at mid-depth s 0 ’=? If s 0 ’ and OCR (=4) are known, pre-consolidation pressure, s c ’, can be calculated Stress increment at mid-depth of clay layer due to embankment on the surface ( Ds ’) can be calculated C c (=0.21) and C r (=0.035) are given. Then, total primary consolidation settlement of the clay layer, S p can be calculated

Q3 Conti.. 8 . 0 6 . 2 0   e G s w r s e eS G g g + +  1 w r s e eS G   + +  1 H=3m 0.5m q o =75kPa B=6m Δ g sat =? Initial effective stress at mid-depth s 0 ’=? 1 , 8 . 0 , 6 . 2 0    r s S e G 3 / 50 . 18 8 . 9 8 . 0 1 1 8 . 0 6 . 2 m kN sat   +  +  g

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Q3 Conti.. H=3m 0.5m q o =75kPa B=6m Δ g sat =18.50 kN/m 3 Initial effective stress at mid-depth s 0 ’=? 2 depth - mid at 2 depth - mid at 2 depth - mid at / 95 . 17 8 . 9 75 . 27 ' / 8 . 9 8 . 9 0 . 1 / 75 . 27 5 . 18 5 . 1 m kN m kN u m kN          s s Before applying embankment load (at mid-depth of clay layer, z = 1.5 m) Initial effective stress at mid-depth s 0 ’=17. 95 kN/m 2

e o =0.8 e c =? e s 0 ’=17.95 s c ’ =? s f ’ = ? Log(σ ’ ) Δσ ’ = ? C r =0.035 C c =0.21 Q3 Conti.. 4 (OCR) ratio ion consolidat Over  2 0 / 8 . 71 95 . 17 4 ' 95 . 17 ' 4 ' ' m kN OCR c c c       s s s s

Stress increment at mid depth due to embankment loading (strip loading) ( ) 8 . 0 ) 9 . 126 sin( 216 . 2 9 . 126 2 tan 2 0 . 2 5 . 1 0 . 3 2 / 2 tan 0 . 1 ) 0 cos( ) 2 cos( , 2 2 0 0 1                +       rad Z           B=6 Z=1.5m 3.0m α β   z s D   ) 2 cos( sin      s + +  D q q = 75 kN/m 2 At the centre of the strip foundation   2 / 9 . 71 0 . 1 8 . 0 214 . 2 75 m kN x  +  D  s Q3 Conti… Stress increment at mid-depth of clay layer due to embankment on the surface , Ds ’ = 71.9 kN/m 2

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e o =0.8 e c =? e s 0 ’=17.95 s c ’ =71.8 s f ’ = 89.85 Log(σ ’ ) Δσ ’ = 71.9 C r =0.035 C c =0.21 4 (OCR) ratio ion consolidat Over  2 0 / 8 . 71 95 . 17 4 ' 95 . 17 ' 4 ' ' m kN OCR c c c       s s s s Stress increment at mid-depth of clay layer due to embankment on the surface , Ds ’ = 71.9 kN/m 2 779 . 0 021 . 0 8 . 0 95 . 17 8 . 71 log 035 . 0 log / /                     D  c c o o c r e e e e C s s Q3 Conti… e c =?

mm m S S x x S e H C e H C S p p p c o c C o c r p c o 69 069 . 0 034 . 0 035 . 0 8 . 71 99 . 89 log 779 . 0 1 0 . 3 21 . 0 95 . 17 8 . 71 log 8 . 0 1 0 . 3 035 . 0 log 1 log 1 / / / / / 0 / / /   +        + +       +          D + + +         +   D + s s s s s s s s e o =0.8 e c =0.779 e s 0 ’=17.95 s c ’ =71.8 s f ’ = 89.85 Log(σ ’ ) Δσ ’ = 71.9 C r =0.035 C c =0.21 Q3 Conti…

Q4 H e 1 e e S o 1 o c         +   Consider the mid depth of each layer and appropriate values of e are obtained from the given table Pressure [kN/m 2 27 54 107 214 429 214 107 54 Void ratio 1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 sand Layer - 1 4m 8m q = 4x21 = 84kPa 2m 2m 2m 2m Mid-depth =5m Layer - 2 Layer - 3 Layer - 4 g sat = 19 kN/m 3 g sat = 19 kN/m 3 Clay z Mid-depth =7m Mid-depth =11m Mid-depth =9m Table -1

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Q4 Conti… mm mm H e e e H e e S f o p 2 . 87 10 2 225 . 2 128 . 1 225 . 1 1 1 3 0 0 ) 1 (                    +           + D  2 0 / 46 ) 8 . 9 5 ( ) 19 1 ( ) 19 4 ( ' m kN     +   s Effective vertical stress before loading on the surface, s 0 ’ Consider mid-depth of layer -1 ( z =5 m) Corresponding void ration, e 0 , from the table-1: e 0 = 1.225 Effective vertical stress after loading on the surface, s f ’ 2 0 / 130 84 46 ' ' ' m kN f  +  D +  s s s Corresponding void ration, e f , from the table-1: e f = 1.128 Total primary consolidation settlement of layer-1, S p(1) Pressure [kN/m 2 27 54 107 214 429 214 107 54 Void ratio 1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 Similarly, the total primary consolidation settlement of each sub-layer can be calculated as shown in the table-2 (next slide) to get total primary consolidation settlement of the clay layer Table -1

Layer σ 0 / (kN/m 2 ) σ f / (kN/m 2 ) e 0 (Table-1) e f (Table-1) e 0 -e f S p (mm) 1 46.00 130.00 1.225 1.128 0.097 87.2 2 64.40 148.40 1.203 1.115 0.088 79.9 3 82.80 166.20 1.177 1.102 0.075 68.9 4 101.20 185.20 1.152 1.088 0.067 59.5 Total primary consolidation settlement 295.5 Q4 Conti… Table -2 H e e e H e e S f o p         +           + D  0 0 1 1

(46,1.025) (130,1.0096) e D e D 18 Q4 Conti… Consider heave in layer-1 Pressure [kN/m 2 27 54 107 214 429 214 107 54 Void ratio 1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 0154 . 0 025 . 1 0096 . 1 0       D f e e e H e e e H e e S f o p         +           + D  0 0 1 1 ) 130 ' ( 128 . 1 0 0 kPa e   s mm m H 2000 2   mm S p 47 . 14 2000 128 . 1 1 0154 . 0          +   Similarly, heave in other layers can be calculated as shown in the next slide

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Layer σ 0 / (kN/m 2 ) σ f / (kN/m 2 ) e 0 (Table-1) S s (mm) 1 130 46 1.128 -0.0154 -14.5 2 148.4 64.4 1.115 -0.0139 -13.1 3 168.8 82.8 1.102 -0.0118 -11.2 4 185.2 101.2 1.088 -0.0094 -9.0 Total heave -47.8 H e e e H e e S f o p         +           + D  0 0 1 1 Pressure [kN/m 2 27 54 107 214 429 214 107 54 Void ratio 1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 Q4 Conti… e D

Total settlement=S e +S p Where- S e -elastic settlement S p -consolidation settlement Q5

Elastic Settlement of clay layer mm m S E B q S e s e 29 029 . 0 40000 35 105 3 . 0 98 . 0 1 0          Q5 Conti… 32m H = 30m B= 35m q=105kN/m 2 d = 2 m E s =40 MN/m 2 chart) the (From 3 . 0 Foundation Circular 86 . 0 35 30 1     B H chart) the (From 98 . 0 06 . 0 35 2 0     B d

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Elastic Settlement of clay layer Q5 Conti… 98 . 0 0   06 . 0 35 2   B d 86 . 0 35 30   B H 3 . 0 1  

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Q5 Conti… (a) Considering single layer 32m H = 30m B= 35m q=105kN/m 2 d = 2 m E s =40 MN/m 2 c qI  D s 0 ) 2 / 35 ( 0 ) 2 / ( 86 . 0 ) 2 / 35 ( 15 ) 2 / (     B r B z Mid-depth of the clay layer, z = 15 m At the centre of the circular foundation, r=0 723 . 0 I table, the From c  2 0 / 9 . 75 723 . 0 105 ' m kN I q c     D s m v =0.14 m 2 /MN mm H m S v p 8 . 318 30000 ) 1000 /( 9 . 75 14 . 0 ) ' (     D  s

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Q5 Conti… (b) Dividing into 6 sub-layers B= 35m q=105kN/m 2 D f = 2m ΔH=5 m 0=>2.5m 0=>7.5m 0=>12.5m 0=>17.5m 0=>22.5m 0=>27.5m (1) (2) (3) (4) (5) (6) z c I q  D s 0 ) 2 / 35 ( 0 ) 2 / ( 14 . 0 ) 2 / 35 ( 5 . 2 ) 2 / (     B r B z 996 . 0 I table, the From c  Consider mid-depth of layer-1 at the centre z = 2.5 m, r =0 2 / 6 . 104 996 . 0 105 ' m kN qI c     D s mm H m S v p 2 . 73 5000 ) 1000 /( 6 . 104 14 . 0 ) ' ( ) 1 (     D  s

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Layer Z(m) Z/(B/2) I c (Table) Δσ ’ (kN/m 2 ) S (ad)p (mm) 1 2.5 0.14 0.996 104.58 73.2 2 7.5 0.43 0.938 98.49 68.9 3 12.5 0.71 0.806 84.63 59.2 4 17.5 1 0.646 67.83 47.5 5 22.5 1.29 0.509 53.12 37.2 6 27.5 1.57 0.405 42.52 27.8 S (oedo)p 315.8 m H MN m H m S v p 0 . 5 , / 14 . 0 m ) ' ( 2 v ) 1 (   D  s 2 / 105 ' m kN q I q c   D s Q5 Conti… (b) Dividing into 6 sub-layers

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Due to the thickness of clay layer relative to the size of foundation, there will be a significant lateral strain. Therefore, Skempton-Bjerrum method is appropriate. Q5 Conti… ) ( ) ( oedo p corrected p S S   Pore-pressure coefficient, A = 0.65 Circular foundation, H/B =30/35=0.86 From the chart,  = 0.75

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(b) Considering 6 sub-layers S p = μ S (oedo)p =0.75x315.8mm=236.9 mm Total settlement=S e +S p =27+236.9=263.9 mm Q5 Conti… (a) Considering single clay layer S p = μ S (oedo)p =0.75x318.8mm=239.1 mm Total settlement=S e +S p =27+239.1=266.1mm

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