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49254 Advanced Soil Mechanics And Foundation Design

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49254 Advanced Soil Mechanics And Foundation Design

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49254 Advanced Soil Mechanics And Foundation Design

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Course Code: 49254
University: University Of Technology Sydney is not sponsored or endorsed by this college or university

Country: Australia

List and discuss the factors affecting shear strengths of cohesive soils under static and dynamic loads.

P1? = 200
M = 0.87
Γ= 2.09
λ= 0.11 
Required: values of q, p’, v at failure
vo = N – λ Ln (P1) = 2.16 – 0.11 Ln (200) = 1.577
For undrained test (specific volume remains constant)
vo = vf = 1.577
vf = Γ – λ Ln (p’)
   1.577 = 2.09 – 0.11 Ln (p’)
   p’ = 106.02
We know that
  qf = M p’ = 0.87 x 106.02 = 92.24 
Provided: Diameter of soil specimen, D = 100 mm
The height of soil specimen, H = 200 mm
Axial effective stress = 500
Radial effective stress = 150
Axial displacements = 0.9 mm
Radial displacements = -0.06 mm
Required: Deviatoric stress, volumetric and shear strains, shear modulus, bulk modulus, and elastic modulus 

Calculate the deviatoric stress

σd? = σa? – σr?
      = 500-150
      = 350
Thus, the deviatoric stress is 350

Calculate the shear strain

εs  =  
Substituting the values we get;
εs =
The negative sign means that the strain is compressive in nature
Thus, the shear strain is 0.0012.
Calculate the initial volume’
Vi = 2 x 200
    = 1570796.327 mm3
Calculate the final volume
Vf = 2 ( H + Δa)
Substitute the values
Vf = 2 (200 + 0.9)
= 1574080.307 mm3
Calculate the volumetric strain
εs = fi)
Substitute the values
εs = -3
Hence, the volumetric strain is -3

Calculate shear modulus

G =σr?/εs
Thus, the shear modulus is
Calculate the bulk modulus
K = σd? / εv 
   = 350 / (2.09 x 10-3)
   = 167464.11
Thus, the bulk modulus is 167464.11
Calculate the strain
Εa = Δa/H =  10-3
Calculate the elastic modulus
E = σa? / εa = -3 = 111111.11 
Given that sample A soil is ixotropically consolidated, thus, p’ = Po = 400
Initial specific volume, V1 = 2.052 
Γ= 3.15
λ= 0.19

For undrained test Uf = U1 
Uf = 2.052 = Γ – λ Ln (p’)
2.052 = 3.15 – 0.19 Ln (p’)
p’ = 323.41
Pore pressure developed in sample = (Po1 –Pf1) + 1/3 (qf1)
                                                                = (400-323.4) + 1/3(307.24) = 185.68
Isotopically consolidated to 863  and allowed to swell P11 = P1 = 40  
Vo = 2.052
U1 = (P11 –P11) – 1/3 (qf1) = (823.41 -40) – 1/3 (307.24)
      = 180.99 
P1 = 350
Γ=2.76, and
Values of q, p’, and v at failure
vo = N – λ Ln (P1) = 2.88 – 0.16 Ln (350) = 1.943
vo = vf = Γ – λ Ln (p’)
1.943 = 2.76 – 0.16 Ln (p’)
p’ = 165.05
q = M p’ = 0.88 x 165.05 = 145.24  

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