GROUND IMPROVEMENT WORKSHOP 11-12 JUNE 2010 PERTH, AUSTRALIA GROUND IMPROVEMENT IN EXTREME GROUND CONDITIONS Presented by Serge VARAKSIN CHAIRMAN OF T.C. Ground Improvement
EPEC Bang Bo: Works Procedure Gulf of Thailand Construction ti of a 2-km road embankment, 1.33m thick 2
Ground Conditions Layer 1: Weathered crust (OC clay) ~ 1-2m thick Layer 2: Soft Bangkok clay (NC clay) ~ 20m thick Layer 3: Firm / stiff clay (overlying alt. sand layers) > 25m 3
Engineering Properties Layer 1 C u /σ = 0.2 Layer 2 C u /σ = 0.3 Soft layer 1: 0 10m highly plastic very soft Bangkok clay (av. C r ~ 0.35; av. C v ~ 0.5 1 m 2 /y) Soft layer 2: 10 20m soft to medium firm clay 4
Potential Problems Final embankment height = 1.33m Fill to compensate settlement = 3m Total fill to be placed = 4.33m 2m Surcharge? Layer 1 (0 10m) C u ~ 4 10 kpa Contract t period of 12 months including consolidation and pavement works. 5
Criteria / Concerns LONG TERM CRITERIA Post construction performance ( t = 25 years ): Differential settlement < 1:750 Residual settlement < 40cm Stability : Factor of safety > 1.5 SHORT TERM CONCERNS Stability during construction (most critical) Handover date for turbine & other heavy structures (time constraint) Inadequate local source of fill materials and high price (reduce or eliminate surcharge?) 6
Preparation of Working Platform Encounter slip failures and lateral flows! 7
Design Scheme & Field Implementation Double drain grid design for layer 1 and layer 2 soft clay. Vacuum depressurization at 60 kpa for 7 months Primary Grid : 1 drain per m 2 ISOTROPIC CONSOLIDATION Secondary Grid :1 drain per 2m 2 8
Initial Conditions 9
Pre Engineering Tests 10
Horizontal Drains Connection 11
Installation of Geotechnical Instruments 12
Initial Depressurization Initial depressurization of 0.3 bar (30 kpa) achieved in 3-5 days. Then, proceed with embankment construction. 13
Advanced Stage of Consolidation 14
Settlement Results Settlement: Mean (40) = 2.63m Most critical was embankment remained STABLE! vacuum pumping δs/δt < 0.5mm/day 15
Post Treatment Strength After Before Design 16
Another Proven Record Before Menard Vacuum 1m fill : FS < 1 During Menard Vacuum 4m fill : FS > 1.5 17
1 st Successful Vacuum Project in Thailand Client: EPEC / Consultant: SEATEC / M.Contractor: ABB Alstom 18
1 st Successful Vacuum Project in Thailand Total area treated t by MV: 30,000 000 m2 Drains: Total length of drains: Equipment: Working time: 600,000 m (V+H drains) 4rigs+2vibro 3 months Vacuum: Pumping equipment: Pumping duration: Average total settlement: 15 pumps 7 months 2.65 m Total working time: 10 months 19
Soil Profile 21
Future Caisson Stability Analysis 22
Exhibited Design 23
As built conditions 24
Proposed solution Layer I 15% rock (φ = 45 ) + 85% clay (C u = 50 kpa) Layer II 25
Shear strength mobilised in exhibited design The upper layer I consists of compacted sand fill of 1.3m thick (ϕ = 35 o ; C = 0) and the lower layer II consists of natural undisturbed clay of 1.5m thick (ϕ = 0 o ; C = 250 kn/m 2 ). Assuming an overburden effective stress (σ ) of 275 kn/m 2, we have the following: Layer I: τ = C + σ tan φ = 275 * tan(35 o ) = 192kN / m 2 Layer II: τ = C + σ tan φ = 250kN / m 2 26
Shear strength mobilised for proposed solution The upper layer I consists of compacted rock mat of 1.3m thick with 30% of rock size 150mm to 200mm and 70% of rock size 200mm to 300mm (ϕ = 45 o ; C = 0) and the lower layer II consists of composite rock-clay soil layer of 1.5m thick with an area replacement ratio m = 15%. The rock inclusions with similar grading as above has ϕ r = 45 o (C r = 0) and the surrounding clay soil has C s = 50 kn/m 2 (ϕ s = 0 o ). Taking the same overburden effective stress (σ ) of 275 kn/m 2, we have the following: Case 1: Without considering consolidation effect with no gain in shear strength Layer I: ϕ = 45 o ; C = 0 Layer II: Computation for composite ϕ and C Using equation (7): C = C ( m) + C (1 m) = 50(1 0.15) 42.5kN / m r s = 2 Using equation (10): σ r σ s o tan φ = m tan φ r + (1 m) tan φs = 0.15(2.04) tan 45 + 0 σ σ tan φ = 0.306 ϕ = 17 o (for σ s /σ = 2.04 see calculation on page 8 below) Layer I: Layer II: o τ = C + σ tan φ = 275* tan(45 ) = 275kN / m τ = C + σ tan φ = 42.5 + 275 tan17 o 2 = 127kN / m 2 27
Total shear strength for exhibited design and proposed solution 28
Criteria retained for pounder design A - 35 tons submerged - 175X175 1.75 1.75 meter DR section B Design a stabilisation part that will rest on the side of print and guarantee penetration until stabilising part hits the crater side C Final design with final objective, -penetrating part 1.7 x 1.7 -length of penetration 1.7 m -stabilising part 2.4 x 2.4 m -ironing i i or compaction by turning over the pounder 29
View of pounder construction 33
Caisson construction yard 34
Pontoon PMT Testing 35
View of pounder ready to work 36
GPS Positionning system and grids 37
Quality control screen Liebherr 895 38
Depth of penetration versus number of blows Penetration Vs Blows 2 Penetratio on, m 1.5 Q2 P3 O4 Q4 Q3 P4 1 O6 Q6 Q5 P6 O5 P5 0.5 O3 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Blows 39
Danger of offshore pounding V max of pounder measured by radar 8 m/sec V effective reached after 6 m drop 6 m/sec Weight below water 35 tons V = 2gh 36 = 20 h Equivalent drop height 2 m Equivalent Energy: 70 Tm Compaction Depth According to Classical Formula D = 83m 8,3 If δ = 0,5 D = 4,15 m!!! 40
General Set up 43
Testing pontoon 44
View on staff system 45
Self bored slotted tube : first experiment of Menard RETROJET SYSTEM S 46
Actual staff systems 47
Staff dulling equipment 48
Sequences of self boring staff system 49
Actual method for drilling PMT in rock mound (100 mm 300 mm) 51
Empirical determination of friction angle by PMT (Yee & Varaksin method) 2.5 4 28 2 2 * = = C P C P m L m L l l φ φ φ φ 4 24 5 * 2 2 φ P = 7 40 4 * 2 φ P L 2 log log = m C P C L φ l φ = 4 40 * 2 2.12 φ P L = 4 2.5 * 2 P L 52
Reminder of technical solution SITE CONDITION SITE CONDITION Compacted sand fill Compacted ϕ = 35 ο sand fill ϕ = 35 ο undisturbed soil Variance 15% rock thickness (φ = 45 ) soften + soil 15% rock (φ = 45 ) + undisturbed soil Variance 85% 15% clay (C rock thickness u = (C50 u (φ = kpa) 50= soften kpa) 45 ) soil 85% + 15% clay rock (C u (φ = 50= kpa) 45 ) + 85% clay (C u = (C 50 u = kpa) 50 kpa) 85% clay (C u = 50 kpa) Natural undisturbed soil Natural undisturbed soil Natural undisturbed soil Natural undisturbed soil Natural undisturbed soil Natural undisturbed soil Layer I Layer II 53
Preliminary concept Original rock surface before compaction 1.3m φ=45 degree, Ar=15%, C=0kPa, Column Diameter=1.7m 15m 1.5m Cu=50kPa Cu=50kPa Cu=250kPa 54
After compaction actual results Original rock surface before compaction 0.2m 1.3m 1.3m φ=48 degree Ar=22% C=0kPa Column Dia=2.4m φ=40 degree φ=48 degree φ=40 degree Ar=22% Cu=50kPa C=0kPa Column Dia=2.4m Cu=50kPa φ=48 degree Ar=22% C=0kPa Column Dia=2.4m Cu=250kPa 55
Strength of layer I In summary, layer I and layer II shall have the following characteristics. o Layer I : ϕ = 49 o (harmonic mean value); C = 0, m=0.22 (22% of total area) For column Layer I: ϕ = 40 o ; C = 0, m=0.78 (78% of total t area) For in-between bt columns assuming lowest internal friction angle of 40 o. Layer I: Computation for composite ϕ and C C=0 tanφ comp = m tanφ c + (1 m) tanφ oc = 0.22 tan 49 o + 0.78 tan 40 φ comp = 42.2 56
Strength of layer II and total share strength Layer II: Computation for composite ϕ and C C = C ( m ) + C (1 m ) = 50(1 0.22) 39 kn / m r s = 2 σ r σ s o tan φ = m tanφr + (1 m) tanφs = 0.22(1.88) tan 49 + 0 σ σ tan φ = 0.48 ϕ = 25.4 o (for σ s /σ = 1.88) Hence, layer I and layer II shall have the shear strength as below: Layer I: τ = C + σ tanφ = concept =275kN/m 2 ) 275* tan(42.2 o ) = 249kN / m 2 (compared with proposed Layer II: τ = C + σ tanφ = 39 + 275tan 25.4 2 concept = 127kN/m 2 ) o = 170kN / m 2 (compared with proposed 57
Layout of CPT in sand area 59
Results of CPT in sand fill after DC Before Dynamic Compaction After Dynamic Compaction 60
GROUND IMPROVEMENT WORKSHOP 11-12 JUNE 2010 PERTH, AUSTRALIA