Question

8.33. A site consists of uniform medium sand with 5% fine content (D5o = 1.4 mm). The thickness of this sand layer is 12 m from the ground surface. The minimal and maximum void ratio values are 0.45 and 0.98, respectively. The groundwater table is at 1.5 m from the ground surface. The unit weights of the sand above and below the groundwater table are 18 kN/m² and 19 kN/mº, respectively. The SPT N60 value at a depth of 6.0 m is 5. This site is located in a seismically active area. The potential earthquake magnitude can be 7.0. The peak ground acceleration can reach as high as 0.3g. Compaction grouting is proposed to densify this possible liquefiable soil (please verify whether the sand is liquefiable first) to prevent potential liquefaction. Assume the average diameter of grout bulbs can reach 0.9 m and the heave of the ground is limited to be less than 25 mm during the installation. Determine the spacing of compaction grouting in a square pattern.

Answer 1

Peak ground acceleration                      =                      0.3g

Actual Depth of water table                  =                      1.5 m

Assumed Depth of water table during earthquake                   =                      0 m                 

Magnitude of earthquake                      =                      7.0      

Depth of Soil layer                                =                      6 m

Field SPT N                                          =                      5

Percentage of fines                               =                      5 %

i) Cyclic Stress Ratio induced by Design Earthquakes (CSR)

Cyclic stress ratio caused by earthquake, CSR =

0.65 (maxrac Ovo/06)

Where,

a_max      =          maximum acceleration during earthquake

z          =          depth below ground surface = 6.0 m

r_d         =          stress reduction factor = 1.0 - 0.00765z, for z ≤ 9.15 m    =             0.9541

g          =          acceleration due to gravity

σvo'        =          Total vertical overburden stress

=          18.0 x 1.5+ 4.5 x 19 = 112.5 kN/m²

σvo'        =          Effective vertical overburden stress considering GWT at   ground surface

=          (18.0-10) x 1.5 +(19.0 -10) x 4.5 = 52.50 kN/m²

σvo'        =          Effective vertical overburden stress considering actual GWT

=          (18.0) x 1.5 +(19.0 -10) x 4.5 = 67.5 kN/m²

Therefore, cyclic stress ratio during earthquake, CSR =   0.65*0.3*0.9541*112.552.50 = 0.399

ii) Cyclic Resistance Ratio (CRR_7.5)

CRR_7.5 = 134-N160+N160135+5010N160+452 -1200

σvo'        =          Total vertical overburden stress

=          18.0 x 1.5+ 4.5 x 19 = 112.5 kN/m²

σvo'        =          Effective vertical overburden stress considering GWT at   ground surface

=          (18.0-10) x 1.5 +(19.0 -10) x 4.5 = 52.50 kN/m²

σvo'        =          Effective vertical overburden stress considering actual GWT

=          (18.0) x 1.5 +(19.0 -10) x 4.5 = 67.5 kN/m²

Therefore, cyclic stress ratio during earthquake, CSR =   0.65*0.3*0.9541*112.552.50 = 0.399

ii) Cyclic Resistance Ratio (CRR_7.5)

CRR_7.5 = 134-N160+N160135+5010N160+452 -1200

      =          Total vertical overburden stress

=          18.0 x 1.5+ 4.5 x 19 = 112.5 kN/m²

       =          Effective vertical overburden stress considering GWT at   ground surface

=          (18.0-10) x 1.5 +(19.0 -10) x 4.5 = 52.50 kN/m²

       =          Effective vertical overburden stress considering actual GWT

=          (18.0) x 1.5 +(19.0 -10) x 4.5 = 67.5 kN/m²

Therefore, cyclic stress ratio during earthquake, CSR =   = 0.399

0.65 * 0.3 * 0.9541* 112.5 52.50

ii) Cyclic Resistance Ratio (CRR_7.5)

CRR_7.5 =

34-(N) - + *135 + *(10(N.) s. +45) 200

(N₁)₆₀              =          N₆₀ x C_N

Where,

C_N                                =          Overburden correction factor

                                    =       
100 VDC
=
100 67.5
=1.21 <1.7 so C_N = 1.21

(N₁)₆₀                         =          5 x 1.21 = 6.08

At 6m depth,   Percentage of fines, FC = 5 %

SPT correction for fines, (N₁)_60cs = α + β (N₁)₆₀      

α                =

10, exp [1.76 - (229)], for 5% < FC <35% for FC < 5% for 5% <FC < 35% for FC > 35%

= 0

β          =           

1, FC15 for FC < 5% , for 5% <FC <35% 1.2, for FC > 35%

            =          1

(N₁)_60cs                                    =          0 + 1 x 6.08    = 6.08 ~ 6

CRR_7.5                =           = 0.080

50 34-6 + 135+ (106+45) 200

Magnitude Scaling Factor, MSF        =

                                                            =

                                                      =          1.19

Factor of safety against liquefaction,

FoS                  =          (CRR _7.5 /CSR) x MSF

                        =          (0.08/0.399) x 1.19

                        =          0.24 <1

The soil is showing liquefaction Potential which necessitates the ground improvement.

Based on the above calculation, the target SPT is 24 based on trial and error.

Design :

Dia. of bulbs		=	0.9	m
Area of bulbs, As		=	0.636	m2
Field SPT N value, Ni		=	5
Target SPT value after improvement, Nt		=	24
Fines Content, FC		=	5	%
Max void ratio, emax		=	0.980
Min void ratio, emin		=	0.450
Effective Overburden Pressure, σvo'		=	67.500	kPa
ΔNf		=	0.000
Relative Density, Dr (Before Densification)		=	40.045	%
Initial Void ratio, e1		=	0.768
Relative Density, Dr1 (After Densification)		=	87.735	%
Target Void ratio , et		=	0.515
Effective compaction ratio ,Rc		=	0.728
Area replacement ratio, as		=	0.196
Spacing for square pattern , S		=	1.80

  
Fines content, Fc (%) ΔΝ, 0 to 5 5 to 10 10 to 20 higher than 20 1.2 (Fc-5) 6 + 0.2 (FC - 10) 8 + 0.1 (Fc -20)