49254 Advanced Soil Mechanics And Foundation Design

Answer:

Provided:

P₁^? = 200

M = 0.87

N=2.16

Γ= 2.09

λ= 0.11 

Required: values of q, p’, v at failure

Solution

v_o = N – λ Ln (P₁) = 2.16 – 0.11 Ln (200) = 1.577

For undrained test (specific volume remains constant)

v_o = v_f = 1.577

v_f = Γ – λ Ln (p’)

   1.577 = 2.09 – 0.11 Ln (p’)

   p’ = 106.02

We know that

  q_f = 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 

1. Calculate the deviatoric stress

σ_d? = σ_a? – σ_r?

      = 500-150

      = 350

Thus, the deviatoric stress is 350

1. Calculate the shear strain

ε_s  =  

Substituting the values we get;

ε_{s =}

The negative sign means that the strain is compressive in nature

Thus, the shear strain is 0.0012.

Calculate the initial volume’

V_i = ² x 200

    = 1570796.327 mm³

Calculate the final volume

V_f = ² ( H + Δa)

Substitute the values

V_f = ² (200 + 0.9)

= 1574080.307 mm³

Calculate the volumetric strain

ε_{s = fi})

Substitute the values

ε_{s =} ^-3

Hence, the volumetric strain is ^-3

1. Calculate shear modulus

G =σ_r^?/ε_s

=  

Thus, the shear modulus is

Calculate the bulk modulus

K = σ_d^? / ε_v 

   = 350 / (2.09 x 10^-3)

   = 167464.11

Thus, the bulk modulus is 167464.11

Calculate the strain

Ε_a = Δ_a/H =  10^-3

Calculate the elastic modulus

E = σ_a^? / ε_a = ^-3 = 111111.11 

Given that sample A soil is ixotropically consolidated, thus, p’ = P_o = 400

Initial specific volume, V₁ = 2.052 

 M=0.95

Γ= 3.15

λ= 0.19

For undrained test U_f = U₁ 

U_f = 2.052 = Γ – λ Ln (p’)

2.052 = 3.15 – 0.19 Ln (p’)

p’ = 323.41

Pore pressure developed in sample = (P_o¹ –P_f¹) + 1/3 (q_f¹)

                                                                = (400-323.4) + 1/3(307.24) = 185.68

Sample

Isotopically consolidated to 863  and allowed to swell P¹₁ ⁼ P₁ = 40  

V_o = 2.052

U₁ = (P₁¹ –P₁¹) – 1/3 (q_f¹) = (823.41 -40) – 1/3 (307.24)

      = 180.99 

P₁ = 350

M=0.88

N=2.88,

Γ=2.76, and

 λ=0.16 

Values of q, p’, and v at failure

Solution

v_o = N – λ Ln (P₁) = 2.88 – 0.16 Ln (350) = 1.943

v_o = v_f = 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