03-EC7 and Its Impact on Existing Design (Dr ChianSC)

EUROCODE 7 AND ITS IMPACT ON EXISTING DESIGN AND TESTING OF PILESBCA-GeoSS EC7 Seminar24 April 2015

Dr Darren Chian Siau ChenNational University of Singapore

1

ObjectivesWorries of a Geotechnical Engineer in Singapore How do I design according to EC7? How do I know if I have designed correctly? How can I take advantage of EC7 to my favour?

IntroductionChange in Design Approach Working state design:

Analyse expected working condition, then apply factor of safety

Limit state design Analyse various unexpected conditions, where loadings

are increase and resistances are reduced using partial factors.

3

4Prior to Eurocode 7 =

+

Working Load Fwl

Permanent + Variation Loads

Fd = G Gk + Q QkEurocode 7

Rs

, = + &, = / + /

, ,, ,

Rb

Rs

Rb

Site Investigation for Piles Recommended minimum depth of

investigation below the base of the deepest pile, za, as shown in figure on right:

where bg is the smaller width of the pile group on plan and DF is the base diameter of the largest pile.

The depth za may be reduced to 2m if the foundation is built on competent strata with distinct (i.e. known) geology. With indistinct geology, at least one borehole should go to at least 5m. If bedrock is encountered, it becomes the reference level for za.

Greater depths of investigation may be needed for very large or highly complex projects or where unfavourable geological conditions are encountered.

5

6Eurocode 7 Geotechnical Framework

Design Methods/ConsiderationsDesign of piles shall be based on either: Empirical or analytical calculations Static load tests/dynamic load tests Ground tests

7

Design of Piles in CompressionPile Groups: Bearing resistance failure of the piles failing individually

and acting as a block shall both be checked. Generally a pile block can be analysed as a single large

diameter pile. The lower of these two to be taken as the design bearing

resistance (Rd).

8

General Procedure to Design for EC79

Calculate design action

Compute resistance from soil parameters

Derive resistance from load test / ground test

Convert resistance to characteristic values using model factor or correlation factors

Compute characteristic shaft and base resistance

Determine required penetration depth of pile where total characteristic resistance design action

Permanent actions, Gk Fixed, doesnt change much.

Variable actions, Qk Change at different times.

Favourable Action that will make fdn less susceptible to ULS condition.

Unfavourable Action that will make fdn more susceptible to ULS condition

Design values of actions, FdFd = G Gk + Q Qk

where G and Q are partial factor

11

ULS Design Value of Actions

12

Action Partial Factors

Action SymbolSet

A1 A2

PermanentUnfavourable

G1.35 1.0

Favourable 1.0 1.0

VariableUnfavourable

Q1.5 1.3

Favourable 0 0

Which Sets to Use? Two combinations to be checked:

Combination 1: Increase loading Combination 2: Reducing resistance

In Combination 2, set M1 is used for calculating resistances of piles or anchors and set M2 for calculating unfavourable actions on piles owing e.g. to negative skin friction or transverse loading.

13

Combination 1:A1 + M1 + R1

Combination 2:A2 + (M1 or M2) + R4

where + implies: to be combined with.

Design of Piles in CompressionDesign Axial Compression Load (Vd): Determined, using the load partial factors for Combination

1 and 2, from the characteristic applied loads and the self weight of the pile and includes any downdrag, heave or transverse loading on the pile, as appropriate.

Common practice to assume that the weight of the pile cancels the weight of the overburden at the foundation base. This is permitted by EC7, provided the two cancel approximately.

The implication of this is to reduce the effect of the pile weight in the design or omitting the weight of pile completely.

14

Design of Piles in CompressionPile weight may not cancel the weight of the overburden if: downdrag is significant (as it reduces the effective stress

at the base of a pile); the soil is light, such as a pile through peat; or the pile extends above the surface of the ground.

15

16

Example:Action from superstructure: - Permanent unfavourable action = 1000 kN- Variable unfavourable action = 500 kNSelf-weight of pile = 200 kN

Total permanent unfavourable action = 1000 + 200 = 1200 kN

Total variable unfavourable load = 500 kN

Set A1:Design Action, Vd = 1.35Gk + 1.5Qk

= 1.35 (1200) + 1.5 (500) = 2370 kN

Set A2:Design Action, Vd = 1.0Gk + 1.3Qk

= 1.0 (1200) + 1.3 (500) = 1850 kN

19

Characteristic Values of Ground Properties A cautious estimate of the value affecting the occurrence

of the limit state

Orr and Farrell, 1999

20

Characteristic Value (Xk) Statistical methods:

Students method (author is actually Gossett)

where t = Student valuen = no. of test resultsV = coefficient of variation [std deviation() / mean value(Xm)]

21

Characteristic Value (Xk) Statistical methods:

Schneiders method

where kn = factor (=0.5 for one half a std deviation below mean)

22

Characteristic Value (Xk)Example: The characteristic value of the undrained shear strength, cu

is to be determined from 10 triaxial tests: 20 kPa, 25 kPa, 23 kPa, 27 kPa, 35 kPa, 15 kPa, 18 kPa, 32 kPa, 25 kPaand 26 kPa.

23

Characteristic Value (Xk)Students method:

cu,ave = 24.6 kPa, = 6.059 kPa, V = 0.246

t with 10 test results, 95% confidence = 2.228

cuk = 24.6 - 6.059 x 2.228 / sqrt(10) = 24.1 kPa

Scheniders method:

cuk = 24.6 - 0.5 x 6.059 = 21.6 kPa

24

Characteristic Value (Xk)Say, lets take

cu = 1.0 (Set M1) cu = 1.4 (Set M2)

Therefore,

For M1: cu,d = 21.6 / 1.0 = 21.6 kPa

For M2: cu,d = 21.6 / 1.4 = 15.4 kPa

Characteristic Value (Xk)25

Compute Resistance from Soil ParametersShaft Resistance Per Unit Area: Undrained: qs,cal = cuk Drained: qs,cal = Ks 'v0 tan = 'v0

Base Resistance Per Unit Area: Undrained: qb,cal = Nc cuk , where Nc = 9 Drained: qb,cal = q Nq

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Note: qb 10 MPa (bored pile)

18 MPa (driven pile)

Derive Resistance from TestsPile Load Test: Select pile type, design load tests, carry out load tests

Determine the measured pile resistance, Rcmfrom the load tests

Ground Tests: Select pile type, design load tests, carry out insitutests (e.g. CPT, SPT, etc)

Determine the pile shaft and base resistances, Rsand Rb from correlations

30

Convert Resistance to Characteristic ValueModel Factor: Obtain characteristic base resistance per unit area, qbk

and shaft resistance per unit area, qsk by dividing the calculated values by a model factor (MF):

qbk = qb,cal/MF and qsk = qs,cal/MFwhere:MF = 1.4 (without pile load test), or

= 1.2 (with pile load test)

32

Convert Resistance to Characteristic ValueCorrelation Factor:Pile Load Test Obtain characteristic resistance, Rcm by dividing the

measured values by correlation factors ():Rck = min(Rcm,mean/1, Rcm,min/2)

Ground Test Obtain characteristic base resistance per unit area, qbk

and shaft resistance per unit area, qsk by dividing the derived values by correlation factors ():

qbk = min(qb,mean/3, qb,min/4)qsk = min(qs,mean/3, qs,min/4)

33

Convert Resistance to Characteristic Value The factor used to determine Rck from the Rcm values,

accounts for the uncertainty in determining bearing resistance for the working piles from the results of a limited number of load tests on similar piles installed in similar ground at a finite number of locations on the site.

The uncertainty depends on the number of test results available, so reduces as the number of test results increases.

Rck is the minimum value obtained applying the values to the mean and average of the load test results.

Ultimate load test and working load test generally 2.5 to 3 times and 1.5 to 2 times working load respectively, subject to structural capacity of the pile not being exceeded.

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Convert Resistance to Characteristic Value35

Number of pile load tests 1 2 3 4 5Factor 1 on mean Rcm 1.55 1.47 1.42 1.38 1.35Factor 2 on lowest Rcm 1.55 1.35 1.23 1.15 1.08

Static Pile Load Test

Number of tested piles 2 5 10 15 20Factor 5 on mean Rcm 1.94

(1.65)1.85 (1.57)

1.83 (1.56)

1.82 (1.55)

1.81 (1.54)

Factor 6 on lowest Rcm 1.90 (1.62)

1.76 (1.50)

1.70 (1.45)

1.67 (1.42)

1.66 (1.41)

Dynamic Impact Test

Note: For structures having sufficient stiffness and strength to transfer loads from

weak to strong piles, the above factors may be divided by 1.1.

Note: Bracket values are values when using dynamic impact tests with signal

matching. Other notes available in NA to SS EN 1997-1:2010.

Convert Resistance to Characteristic Value

36

Number of profiles of tests 1 2 3 4 5 7 10Factor 3 on mean value 1.55 1.47 1.42 1.38 1.36 1.33 1.30Factor 4 on lowest value 1.55 1.39 1.33 1.29 1.26 1.20 1.15

Ground Test

Note: For structures having sufficient stiffness and strength to transfer loads from

weak to strong piles, the above factors may be divided by 1.1.

Compute Characteristic Resistances For pile load test, separate Rck into base and shaft

components, Rbk and Rsk, where possible.

Calculate the characteristic shaft and base resistances from:

Rbk = Ab qbk and Rsk = qsk Aswhere:Ab = the nominal plan area of the base of the pileAs = the nominal surface area of the pile

Obtain the design resistances for each resistance set:Set R1: Rd = Rbk + Rsk or Rd = RckSet R4: Rd = Rbk/b + Rsk/s or Rd = Rck/t

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39

Resistance Partial Factors

Resistance Symbol

Set

R1

R4 without explicit

verification of SLS

R4 with explicit

verification of SLS

Base b 1.0 1.7 1.5Shaft (compression) s 1.0 1.5 1.3Total/combined (compression) t 1.0 1.7 1.5

Shaft in tension s,t 1.0 2.0 1.7

Driven Piles

40


Resistance Symbol

Set

R1

R4 without explicit

verification of SLS

R4 with explicit

verification of SLS



Bored Piles

41


Resistance Symbol

Set

R1

R4 without explicit

verification of SLS

R4 with explicit

verification of SLS



Continuous Flight Auger (CFA) Piles

Design Methods/ConsiderationsR4 with or without explicit verification of SLS? If serviceability is verified by load tests carried out on

more than 1% of the constructed piles to loads not less than 1.5 times the representative load or if settlement at serviceability limit state is of no concern.

Resistance is verified by a maintained load test taken to the calculated, unfactored ultimate resistance. Use the lower factors from R4 with explicit verification of

SLS.

42

Example 1: Compression

1. Calculate design action: Assume: ignore self weight of pile

Set A1: G = 1.35, Q = 1.5 Design action = 1.35x1000+1.5x300

= 1800kN

Set A2: G = 1.0, Q = 1.3 Design action = 1.0x1000+1.3x300

= 1390kN

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0m

4m

Sandk = 32k = 19kN/m3

Gk = 1000kNQk = 300 kN

ClayShaft: cuk = 80kPaBase: cuk = 100kPa

Pile base level?

0.6m diameter bored pile

Example 1

2. Compute resistance: Assume: Pile load test carried out

with verification of SLS

Apply model factor, MF=1.2. Set R1: s = 1.0, b = 1.0

For sand layer,Design shaft resistance, Rs,d1= [K

tan x As,1] / MF / s

= [0.7tan x (DL1)] / 1.2 / 1.0

= [0.7x(18x4/2)x(tan32)x(x0.6x4)] / 1.2= 98.94kN

45

0m

4m


Gk = 1000kNQk = 300 kN


Pile base level?


Example 12. Compute resistance: For clay layer,

Design shaft resistance, Rs,d2= [ cu x As,2] / MF / s= [0.49x80x(x0.6xL2)] / 1.2 / 1.0= 61.58 L2 kN

Design base resistance, Rb,d= [9 cu x Ab] / MF / b= [9x100x(x0.62/4)] / 1.2 / 1.0= 212.06 kN

Set R1: Total resistance= 98.94 + 61.58 L2 + 212.06= 311 + 61.58 L2 kN

46

0m

4m


Gk = 1000kNQk = 300 kN


Pile base level?


Example 1

2. Compute resistance: Assume: Pile load test carried out

with verification of SLS

Apply model factor, MF=1.2. Set R4: s = 1.4, b = 1.7

For sand layer,Design shaft resistance, Rs,d1= [K

tan x As,1] / MF / s

= [0.7tan x (DL1)] / 1.2 / 1.4

= [0.7x(18x4/2)x(tan32)x(x0.6x4)] / 1.68= 70.67kN

47

0m

4m


Gk = 1000kNQk = 300 kN


Pile base level?


Example 12. Compute resistance: For clay layer,

Design shaft resistance, Rs,d2= [ cu x As,2] / MF / s= [0.49x80x(x0.6xL2)] / 1.2 / 1.4= 43.99 L2 kN

Design base resistance, Rb,d= [9 cu x Ab] / MF / b= [9x100x(x0.62/4)] / 1.2 / 1.7= 124.74 kN

Set R4: Total resistance= 70.67 + 43.99 L2 + 124.74= 195.41 + 43.99 L2 kN

48

0m

4m


Gk = 1000kNQk = 300 kN


Pile base level?


Example 1

3. Determine penetration depth:

Set A1 vs Set R1:1800kN = 311 + 61.58 L2 kN1800 311 = 61.58 L2L2 = 24.18m

Set A2 vs Set R4:1390kN = 195.41 + 43.99 L2 kN1390 195.41 = 43.99 L2L2 = 27.16m

Hence, total penetration depth = 27.16 + 4 = 31.16m

49

0m

4m


Gk = 1000kNQk = 300 kN


Pile base level?





50







51



where + implies: to be combined with. Question:

Wouldnt the use of M2 in Combination 2 reduce the negative skin friction? Unconservative approach??

52

Material Partial Factors

Soil Parameter SymbolSet

M1 M2

Angle of shearing resistance1 ' 1.0 1.25

Effective cohesion c' 1.0 1.25

Undrained shear strength cu 1.0 1.4

Unconfined strength qu 1.0 1.4

1 Factor is applied to tan '

Example 2: Downdrag

53

Soft claycuk = 20 kPa

Stiff claycuk = 50 kPa

Cuk.neg = 20 kPa

Gk = 300 kN

Simpson and Driscoll, 1998

Surcharge placed at ground level which is sufficient to mobilise limiting negative skin friction between the pile and the soft clay.

Characteristic applied load, = V#= 300kN.

Characteristic downdrag force, = D#=DL*q,-= 0.3 5 20

= 94.2 kN.

Characteristic shaft resistance, = R#= 0.3 50 L2= 47.1L2 kN.

54

Example 2: Downdrag

Combination 1(F) - downdrag force taken as action Partial factors for actions :

Vertical load, V: 3

= 1.35 Downdrag, D:

3= 1.35

Partial factors for resistances:Shaft resistance, R:

= 1.0

Total design vertical load = F = V + D= V#

3+ D#

3

= 3001.35 + 94.21.35= 532.2 kN.

55

Example 2: Downdrag

Note: Downdrag is classified as a 'permanent' action because its variation is always in the same direction (monotonic) until the action attains a certain limit value (EC1, 1.5.3.3).

Combination 1(F) - downdrag force taken as action Design shaft resistance = R

= R#/

= 47.1 L2 / 1.0. Require R F.

Hence 47.1 L2 / 1.0 532.2 kN.L2 11.29 m.

Design force for concrete shaft = F = 532.2 kN.

56

Example 2: Downdrag

Combination 1(S) - settlement taken as action Partial factors for actions:

Vertical load, V: 3

= 1.35 Any partial factor applied to settlement would have no

effect in this case.


= 1.0

Partial factor for unfavourable soil strength transmitting effect of settlement to pile = 1.0.

Hence design downdrag force = 94.2 kN.

57

Example 2: Downdrag

Combination 1(S) - settlement taken as action Total design vertical load = F = V + D

= V# 3

+ 94.2

= 3001.35 + 94.2= 499.2 kN.

Design shaft resistance = R= R#/

= 47.1 L2 / 1.0.

58

Example 2: Downdrag

Combination 1(S) - settlement taken as action Require R F.

Hence 47.1 L2 / 1.0 499.2 kN.L2 10.6 m.


It is necessary to satisfy both Combinations 1 and 2, but the choice of force or displacement as the action is open to the designer.

59

Example 2: Downdrag

Combination 2(F) - downdrag force taken as action Partial factors for actions:

Vertical load, V: 3

= 1.0 Downdrag, D:

3= 1.0

Partial factors tor resistances:Shaft resistance, R:

= 1.3 (R4 with explicit verification of SLS)


3+ D#

3

= 3001.0 + 94.21.0= 394.2 kN.

60

Example 2: Downdrag

Combination 2(F) - downdrag force taken as action Design shaft resistance = R

= R#/

= 47.1 L2 / 1.3. Require R F.

Hence 47.1 L2 / 1.3 394.2 kN.L2 10.88 m.


61

Example 2: Downdrag

Combination 2(S) - settlement taken as action Partial factors for actions :

Vertical load, V: 3

= 1.0 Settlement:

3= 1.0


= 1.3

When settlement is taken to be the action, downdrag effect is transferred to the pile using the soil strength, which therefore acts in an unfavourable manner.

62

Example 2: Downdrag

Combination 2(S) - settlement taken as action EC7, 2.4.2 (11) says that a partial factor less than 1.0 must

be applied in such cases. It could be taken to be 1 / 89

from Material Table (or 1 /

from Resistance Table). This would give 1 / 1.4 = 0.714 (or 1 / 1.3 = 0.769).

Assumption that the displacement would mobilise all the available shaft adhesion in the soft clay. The characteristic value of this (D#) is 94.2 kN, as shown earlier.

63

Example 2: Downdrag

Combination 2(S) - settlement taken as actionD = 94.2 /

,

= 94.2 / 0.714= 131.9 kN


3+ D#

3

= 3001.0 + 131.9= 431.9 kN.

64

Example 2: Downdrag

Combination 2(S) - settlement taken as action Design shaft resistance = R

= R#/

= 47.1 L2 / 1.3. Require R F.

Hence 47.1 L2 / 1.3 431.9 kN.L2 11.92 m.


65

Example 2: Downdrag

66

Calculation :;(m)




67




Answer:Partial factor M2 is applied onto the soil parameter (e.g. cu) when downdrag effect is transferred to the pile using the soil strength, Downdrag is multiplied

with the M2 partial factor. Caution: Engineer should not double factor for the

same resistance.

Design of Piles in Compression

68



Number of tested piles 2 5 10 15 20Factor 5 on mean Rcm 1.94 1.85 1.83 1.82 1.81Factor 6 on lowest Rcm 1.90 1.76 1.70 1.67 1.66

Dynamic Impact Test


Ground Test


69




Dynamic Impact Test


Ground Test Question: Is it worthwhile to do more static pile load tests?


70




Dynamic Impact Test


Ground Test


71




Dynamic Impact Test


Ground TestAnswer:

Reduction of factor from 1.55 to 1.35 Increase in characteristic bearing resistance by

15%.Beneficial when you have many piles to install

within the zone of similar soil profile.


72




(1.65)1.85 (1.57)

1.83 (1.56)

1.82 (1.55)

1.81 (1.54)


1.76 (1.50)

1.70 (1.45)

1.67 (1.42)

1.66 (1.41)

Dynamic Impact Test

Question: Is it worthwhile to do more dynamic impact

tests?




73




(1.65)1.85 (1.57)

1.83 (1.56)

1.82 (1.55)

1.81 (1.54)


1.76 (1.50)

1.70 (1.45)

1.67 (1.42)

1.66 (1.41)

Dynamic Impact Test




74




(1.65)1.85 (1.57)

1.83 (1.56)

1.82 (1.55)

1.81 (1.54)


1.76 (1.50)

1.70 (1.45)

1.67 (1.42)

1.66 (1.41)

Dynamic Impact Test



Answer:Requires a fair number of dynamic test with signal

matching analysis in order to reduce value of correlation factor.

Note: Take lower of the two categories of factors.Beneficial when you have many piles to install

within the zone of similar soil profile.

Design of Piles in CompressionModel Factor vs Correlation Factor

Model Factor, MF = 1.4 (without pile load test), or = 1.2 (with pile load test)

Correlation Factor,

75





Correlation Factor,

76



Question: Which gives a more aggressive design?



Correlation Factor,

77





Correlation Factor,

78



Answer: Model factor is often more aggressive as the factor is

generally lower.

79


80


Answer: Implied FoS from EC7 lower than BS. Shorter piles are

expected? Not necessary. Application of characteristic value onto soil parameters (i.e. cautious estimate) may bring the design of pile using EC7

closer to BS.

Structural CapacityEurocode 2:

Partial factors for concrete and steel for ULS:

The recommended values of c and s in the serviceability limit state (SLS) is 1.0.

The partial factor for concrete c given above should be multiplied by a factor, kf, for calculation of design resistance of cast in place piles without permanent casing. The recommended value is 1.1.

81

Design Situations c for concrete s for reinforcing steelPersistent & Transient 1.5 1.15Accidential 1.2 1.0

Structural CapacityEurocode 2:Bored piles Arrangement of reinforcements to allow free flow of concrete. Min. diameter for longitudinal bars not be less than 16 mm. At least 6 longitudinal bars. Clear distance between bars should not exceed 200 mm

measured along the periphery of the pile. Permit ease of flow of concrete.

82

Structural CapacityEurocode 2:Bored piles

83

Cast-in-place bored pile cross-section, Ac

Min area of longitudinal reinforcement, As,bpmin

Ac 0.5 m2 As 0.5% Ac0.5 m2 < Ac 1.0 m2 As 25 cm2 (0.25 - 0.5% Ac) Ac > 1.0 m2 As 0.25% Ac

Note: Prior recommendation was As 0.5% Ac .

Structural CapacityEurocode 2: Concrete

fcd = fck / c= 0.85 fck / 1.5 [driven]= 0.85 fck / (1.5 x 1.1) [bored-reinforced]= 0.60 fck / (1.5 x 1.1) [bored-unreinforced)]

Steelfyd = fyk / s

= fyk / 1.15 Structural capacity of pile > Design Action/Load

Q = fcd Ac + fyd As > Fd = G Gk + Q Qk= 0.515 fck Ac + 0.87 fyk As [bored-reinforced]= 0.364 fck Ac [bored-unreinforced]

84

Note:Factor of 0.85 for flexure and axial loading (NA to EC2 3.1.6)Factor of 0.6 due to less ductile properties of plain concrete (NA to EC2 12.3.1)

Structural CapacityEurocode 2:

Concretefcd = fck / c

= 0.85 fck / 1.5 [driven]= 0.85 fck / (1.5 x 1.1) [bored-reinforced]= 0.60 fck / (1.5 x 1.1) [bored-unreinforced)]

Steelfyd = fyk / s

= fyk / 1.15 Structural capacity of pile > Design Action/Load

Q = fcd Ac + fyd As > Fd = G Gk + Q Qk= 0.515 fck Ac + 0.87 fyk As [bored-reinforced]= 0.364 fck Ac [bored-unreinforced]

85

BS 8004 and SS CP4:Q = (0.4 fcu Ac + 0.75 fy As)/FoS [bored-reinforced]

SS CP65:Q = 0.25 fcu Ac [bored-unreinforced]

Note: fcu fck & fy fyk

Structural CapacityEurocode 2: Table 3.1 Strength characteristics for concrete

86

Structural CapacityEurocode 2: Table 3.1 Strength characteristics for concrete

87

Cylinder: fcm = fck + 8 (MPa)Cylinder fck 0.8 fck,cube

Note: Strength of a cylinder is approx. 0.8 that of a cube. Research has shown that the ratio of strength of a cylinder

vs cube increases (>0.8) as the strength of concrete increases.

Structural CapacityHence, Bored - reinforced section:

Q = fcd Ac + fyd As= 0.515 fck Ac + 0.87 fyk As 0.412 fck,cube Ac + 0.87 fyk As

Bored - unreinforced section:Q = fcd Ac

= 0.364 fck Ac 0.291 fck,cube Ac

Note: In CP4 and CP65,Q = (0.4 fcu Ac + 0.75 fy As)/FoS [bored-reinforced]Q = 0.25 fcu Ac [bored-unreinforced]

88

THANK YOU

Dr Darren Chian Siau ChenNational University of [email protected]

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03-EC7 and Its Impact on Existing Design (Dr ChianSC)

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