 
          1550
        
        
          Proceedings of the 18
        
        
          th
        
        
          International Conference on Soil Mechanics and Geotechnical Engineering, Paris 2013
        
        
          
            Proceedings of the 18
          
        
        
          
            th
          
        
        
          
            International Conference on Soil Mechanics and Geotechnical Engineering, Paris 2013
          
        
        
          utilization of recycled Bassanite has been verified. In the
        
        
          future, it is important to study the suppression technique for
        
        
          a variety of earthquakes, focusing attention on not only
        
        
          suppression of liquefaction, but also recycling costs.
        
        
          recycling of waste wood from earthquake damage and bamboo
        
        
          chips.
        
        
          4.  CONCLUSIONS
        
        
          (1) The liquefaction resistance is increased, due to the increase
        
        
          of cohesion from the added solidification material. For the
        
        
          added solidification material, the interparticles forces of
        
        
          sand due to this cohesion suppresses liquefaction. The
        
        
          magnitude of the suppression effect of liquefaction is
        
        
          influenced by the interparticles forces of the sand.
        
        
          5.  ACKNOWLEDGEMENTS
        
        
          The study presented in this paper was supported by Professor
        
        
          Timothy A. Newson (University of Western Ontario) and Dr.
        
        
          Takuro Fujikawa. This support is gratefully acknowledged.
        
        
          (2) The effect of shear deformation suppression also increases
        
        
          due to the development of apparent cohesion. For the case
        
        
          of short fiber and solidification material, liquefaction is
        
        
          suppressed by the cohesion of solidification material and
        
        
          shear deformation suppression.
        
        
          6.  REFERENCES
        
        
          (3) It is considered from the study results that, the utilization of
        
        
          discharged wood waste by the earthquake will have a
        
        
          similar effect as the effect of shear deformation suppression
        
        
          Yasuda S. 2011. The liquefaction damage of KANTO. Board of
        
        
          Tohoku-Pacific Ocean Earthquake disaster investigation report (in
        
        
          Japanese).
        
        
          Sasaki T. 2012. River bank and liquefaction. Public Works Research
        
        
          Institute,
        
        
        
          /
        
        
          of the short fibers. Additionally, the effectiveness of the
        
        
          Zen K. 1994. Remedial Measures for Reclaimed Land by Premixing
        
        
          Method.
        
        
          
            The Japanese Society of Soil Mechanics and Foundation
          
        
        
          
            Engineering
          
        
        
          , Vol.42, No.2, 37-42 (in Japanese).
        
        
          0
        
        
          0.2
        
        
          0.4
        
        
          0.6
        
        
          1
        
        
          0.8
        
        
          Sezaki M. 2011. Liquefaction and slope failures in the Great East Japan
        
        
          Earthquake.
        
        
          
            Text of road technology workshops
          
        
        
          (in Japanese).
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          0.20
        
        
          
        
        
          d
        
        
          =1.486(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (a) C=0%
        
        
          Figure 6. Effective stress path diagram
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%+F=1%
        
        
          
        
        
          d
        
        
          =1.489(g/cm
        
        
          3
        
        
          )
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          PT Line
        
        
          PT Line
        
        
          (b) C=0%+F=1%
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura
        
        
          C=1%+F=1%
        
        
          
        
        
          P'
        
        
          c
        
        
          
        
        
          
        
        
          d
        
        
          =1.481(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (c) C=1%+F=1%
        
        
          0 20 40 60 80 100 120
        
        
          C=0%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          (u/p
        
        
          c
        
        
          ')
        
        
          max
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          1
        
        
          10
        
        
          100
        
        
          1000
        
        
          C=0%
        
        
          C=2%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Number of cycles N (cycles)
        
        
          DA=5%
        
        
          Figure 7. The relationships between number of cycles and max pore
        
        
          pressure ratio
        
        
          0
        
        
          1
        
        
          2
        
        
          3
        
        
          4
        
        
          5
        
        
          0 20 40 60 80 100 12
        
        
          Figure 9. Liquefaction strength curve
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          N=20
        
        
          N=34
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Additive conditions
        
        
          C=0%
        
        
          C=2%
        
        
          C=1%
        
        
          +F=1%
        
        
          C=0%
        
        
          +F=1%
        
        
          0
        
        
          C=0%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          DA (%)
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          Figure 10. The relationships between additive conditions and
        
        
          liquefaction strength
        
        
          Figure 8. The relationships between number of cycles and double
        
        
          amplitude axial strain
        
        
          recycling of waste wood from earthquake damage and
        
        
          bamboo
        
        
          chips.
        
        
          4.  CONCLUSIONS
        
        
          (1) The liquefaction resistance is increased, due to the
        
        
          increase of cohesion from the added solidification
        
        
          material. For the added solidification material, the
        
        
          interparticles fo ces of sand due to this cohesion
        
        
          suppr ses liquefaction. The m gnitude of the
        
        
          suppression effect of liquefaction is influenced by the
        
        
          i terparticles forces of the sand.
        
        
          (2) The eff ct of shear deformation suppression lso
        
        
          incr as s due to the development of apparent
        
        
          cohesion. For the cas f short fiber and solidification
        
        
          material, liqu faction is suppressed by the cohesion of
        
        
          solidification material and shear deformation
        
        
          su pre sion.
        
        
          (3) It is considered from the study results that, the
        
        
          utilization of disc arged wood w ste by the
        
        
          earthquake will have a similar effec as the effect of
        
        
          shear de orm tion suppression
        
        
          of the short fibers. Additionally, the effectiveness of
        
        
          the
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          0.20
        
        
          
        
        
          d
        
        
          =1.486(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (a) C=0%
        
        
          Figure 6. Effective stress path diagram
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%+F=1
        
        
          
        
        
          d
        
        
          =1.489(g/cm
        
        
          3
        
        
          )
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          PT Line
        
        
          PT Line
        
        
          (b) C=0%+F=1%
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura
        
        
          C=1%+F=1%
        
        
          
        
        
          P'
        
        
          c
        
        
          
        
        
          
        
        
          d
        
        
          =1.481(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (c) C=1%+F=1%
        
        
          0
        
        
          0.2
        
        
          0.4
        
        
          0.6
        
        
          0.8
        
        
          1
        
        
          0 20 40 60 80 100 120
        
        
          C=0%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          (u/p
        
        
          c
        
        
          ')
        
        
          max
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          ' 0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          Figure 7. The relationships between number of cycles and max pore
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          1
        
        
          10
        
        
          100
        
        
          1000
        
        
          C=0%
        
        
          C=2%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Number of cycles N (cycles)
        
        
          DA=5%
        
        
          pressure ratio
        
        
          Figure 9. Liquefaction strength curve
        
        
          0
        
        
          1
        
        
          2
        
        
          3
        
        
          4
        
        
          5
        
        
          0 20 40 60 80 100
        
        
          C=0%
        
        
          C=0%+F
        
        
          C=1%+F
        
        
          DA (%)
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          N=20
        
        
          N=34
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Additive conditions
        
        
          C=0%
        
        
          C=2%
        
        
          C=1%
        
        
          +F=1%
        
        
          C=0%
        
        
          +F=1%
        
        
          Figure 8. The relationships between number of cycles and
        
        
          double amplitude axial strain
        
        
          Figure 10. The relationships between additive conditions and
        
        
          liquefaction strength
        
        
          recycling of waste wood from earthqu ke damage a d
        
        
          bamboo
        
        
          chips.
        
        
          4.  CONCLUSIONS
        
        
          (1) The liquefac ion resistance is increased, due to the
        
        
          increase of cohe io from the ad ed solidification
        
        
          mat rial. F r the adde li i i ti aterial, the
        
        
          interparticl s f rces of sand due to this cohesion
        
        
          suppr sses liquefaction. The magnitud of the
        
        
          suppression effect of liquefac on is influenced by the
        
        
          interparticles forc s of the sand.
        
        
          (2) The effect of shear def mation suppress on also
        
        
          inc e ses due t the development f appar nt
        
        
          cohesion. For the case of short fiber and solidificati
        
        
          mat rial, liquefacti n is suppresse by the cohesion of
        
        
          solidi icat on material and sh ar deformation
        
        
          uppression.
        
        
          (3) It is considered from the study results that, the
        
        
          utilization of discharged wood waste by the
        
        
          earthquake will have a similar ef ct as he effect of
        
        
          shear deformation suppression
        
        
          of the short fibers. Additionally, the effectiveness of
        
        
          the
        
        
          -150
        
        
          -1 0
        
        
          -50
        
        
          0
        
        
          50
        
        
          1 0
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          ffective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          0.20
        
        
          
        
        
          d
        
        
          =1.486(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (a) C=0%
        
        
          Figure 6. Effective stress path diagram
        
        
          -150
        
        
          -1 0
        
        
          -50
        
        
          0
        
        
          50
        
        
          100
        
        
          150
        
        
          0 20 4 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%+F=1%
        
        
          
        
        
          d
        
        
          =1.489(g/cm
        
        
          3
        
        
          )
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          PT Line
        
        
          PT Line
        
        
          (b) C=0%+F=1%
        
        
          -150
        
        
          -1 0
        
        
          -50
        
        
          0
        
        
          50
        
        
          1 0
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura
        
        
          C=1%+F=1%
        
        
          
        
        
          P'
        
        
          c
        
        
          
        
        
          
        
        
          d
        
        
          =1.481(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (c) C=1%+F=1%
        
        
          0
        
        
          .2
        
        
          0.4
        
        
          0.6
        
        
          0.8
        
        
          1
        
        
          0 20 40 60 80 100 120
        
        
          C=0%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          (u/p
        
        
          c
        
        
          ')
        
        
          max
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          Figure 7. The relationships between number of cycles and max pore
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          1
        
        
          10
        
        
          100
        
        
          1000
        
        
          C=0%
        
        
          C=2%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Number of cycles N (cycles)
        
        
          DA=5%
        
        
          pressure ratio
        
        
          Figure 9. Liquefaction strength curve
        
        
          0
        
        
          1
        
        
          2
        
        
          3
        
        
          4
        
        
          5
        
        
          0 20 40 60 80 100
        
        
          C=0%
        
        
          C=0%+F
        
        
          C=1%+F
        
        
          DA (%)
        
        
          Number of cycles N (cycles)
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          N=20
        
        
          N 34
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Additive conditions
        
        
          C=0%
        
        
          C=2%
        
        
          C=1%
        
        
          +F=1%
        
        
          C=0%
        
        
          +F=1%
        
        
          Figure 8. The relationships between number of cycles and
        
        
          double amplitude axial strain
        
        
          Figure 10. The relationships between additive conditions and
        
        
          liquefaction strength
        
        
          recycling of waste wood from earthquake damage and
        
        
          bamboo
        
        
          chips.
        
        
          4.  CONCLUSIONS
        
        
          (1) The liquefaction resistanc is increased, due to the
        
        
          increase of cohesion from the added solidification
        
        
          material. For the added solidification material, the
        
        
          interparticles f rces of sand due to this cohesion
        
        
          suppr sses liquefaction. Th m gnitude of the
        
        
          suppression e fect of liquefact on is influenced by the
        
        
          inter art c es f ces of the sand.
        
        
          (2) The ffec of shear deformation suppr ssion also
        
        
          incr ases due to t e develo m nt of apparent
        
        
          cohes on. For the case of sh t fib r and solid fication
        
        
          materi l, liquefact n is suppressed by the cohesion of
        
        
          solidification m teri l and
        
        
          hear deformation
        
        
          suppression.
        
        
          (3) It is considered from the study results that, the
        
        
          utilization of dis harged wood was e by the
        
        
          earthquake will have a similar effect as the ef ect of
        
        
          shear def rmation suppression
        
        
          of the short fibers. Additiona ly, h eff ctiveness of
        
        
          the
        
        
          -150
        
        
          -100
        
        
          -50
        
        
          0
        
        
          5
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          0.20
        
        
          
        
        
          d
        
        
          =1.486(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (a) C=0%
        
        
          Figure 6. ff ctive str ss path diagram
        
        
          -15
        
        
          -100
        
        
          -50
        
        
          0
        
        
          5
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura Sand
        
        
          C=0%+F=1%
        
        
          
        
        
          d
        
        
          =1.489(g/cm
        
        
          3
        
        
          )
        
        
          
        
        
          p'
        
        
          c
        
        
          
        
        
          PT Line
        
        
          PT Line
        
        
          (b) C= %+F=1%
        
        
          -15
        
        
          -100
        
        
          -50
        
        
          0
        
        
          5
        
        
          100
        
        
          150
        
        
          0 20 40 60 80 100 120
        
        
          Deviator stress q (kPa)
        
        
          Effective mean principal stress p' (kPa)
        
        
          Toyoura
        
        
          C=1%+F=1%
        
        
          
        
        
          P'
        
        
          c
        
        
          
        
        
          
        
        
          d
        
        
          =1.481(g/cm
        
        
          3
        
        
          )
        
        
          PT Line
        
        
          PT Line
        
        
          (c) C=1%+F=1%
        
        
          0
        
        
          0.2
        
        
          0.4
        
        
          0.6
        
        
          0.8
        
        
          1
        
        
          0 20 40 60 80 100 120
        
        
          C=0%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          (u/p
        
        
          c
        
        
          ')
        
        
          max
        
        
          Number of cycles N ( ycles)
        
        
          
        
        
          d
        
        
          /
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          Figure 7. The relationships between number of cycles and max pore
        
        
          0.1
        
        
          2
        
        
          3
        
        
          4
        
        
          0 5
        
        
          0.6
        
        
          1
        
        
          10
        
        
          100
        
        
          10 0
        
        
          C=0%
        
        
          C=2%
        
        
          C=0%+F=1%
        
        
          C=1%+F=1%
        
        
          Cyclic deviator
        
        
          stress ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Number of cycles N (cycles)
        
        
          DA=5%
        
        
          pressure ratio
        
        
          Figure 9. Liquefaction strength curve
        
        
          0
        
        
          1
        
        
          2
        
        
          3
        
        
          4
        
        
          5
        
        
          0 20 40 60 80 100
        
        
          0
        
        
          C=0%+F
        
        
          C=1%+F
        
        
          DA (%)
        
        
          Number of cycles N (cycl )
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.35
        
        
          
        
        
          d
        
        
          /2
        
        
          
        
        
          c
        
        
          '=0.20
        
        
          0.1
        
        
          0.2
        
        
          0.3
        
        
          .4
        
        
          0.5
        
        
          .6
        
        
          N=20
        
        
          N=34
        
        
          Cyclic deviator
        
        
          stre s ratio
        
        
          
        
        
          /p'
        
        
          c
        
        
          Additive conditions
        
        
          C=0%
        
        
          C=2%
        
        
          C=1%
        
        
          +F=1%
        
        
          C=0%
        
        
          +F=1%
        
        
          Figure 8. The relationships between number of cycles and
        
        
          double amplitude axial strain
        
        
          Figure 10. The relationships between additive conditions and
        
        
          liquefaction strength