Soil mechanics miscellaneous

Soil mechanics miscellaneous

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Soil Mechanics

Soil mechanics miscellaneous
  1. In a drained triaxial compression test, a saturated specimen of a cohesion less sand fails under a deviatoric stress of 3 kg/cm2 when the cell pressure is 1 kgf/cm2. The effective angle of shearing resistance of sand is about








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    C = 0 (Cohesion less sand)
    σ3 = 1 kg/cm2
    σa = 3 kg/cm2
    σ1 = σ3 + σd
    = 1 + 3 = 4 kg/cm2
    we know,
    σ1 = σ3 + tan2α + 2c tan α
    ⇒ 4 = 1 tan2 α + c
    ∴ tan2 α = 4
    ⇒ tan α = 2
    ∴ α = tan–12

    α = 45 +
    Φ
    		= tan–1 2
    2

    ∴ Φ = 36.87° ≈ 37°

    Correct Option: A

    C = 0 (Cohesion less sand)
    σ3 = 1 kg/cm2
    σa = 3 kg/cm2
    σ1 = σ3 + σd
    = 1 + 3 = 4 kg/cm2
    we know,
    σ1 = σ3 + tan2α + 2c tan α
    ⇒ 4 = 1 tan2 α + c
    ∴ tan2 α = 4
    ⇒ tan α = 2
    ∴ α = tan–12

    α = 45 +
    Φ
    		= tan–1 2
    2

    ∴ Φ = 36.87° ≈ 37°

  1. The void ratio and specific gravity of a soil are 0.65 and 2.72 respectively. The degree of saturation (in percent) corresponding to water content of 20% is








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    e =
    wG
    Sr

    ⇒ 0.65 =
    0.2 × 2.72
    Sr

    ∴ Sr = 0.837 = 83.7%

    Correct Option: C

    e =
    wG
    Sr

    ⇒ 0.65 =
    0.2 × 2.72
    Sr

    ∴ Sr = 0.837 = 83.7%

  1. The coefficients of permeability of a soil in horizontal and vertical directions are 3.46 and 1.5 m/day respectively. The base length of a concrete dam resting in this soil is 100 m. When the flow net is developed for this soil with 1:25 scale factor in the vertical direction, the reduced base length of the dam will be








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    kx = 3.46 m/d
    ky = 1.5 m/d

    = 65.84 m
    Apply scale factor of 1: 25

    65.84
    		= 2.63 m
    25

    Correct Option: A

    kx = 3.46 m/d
    ky = 1.5 m/d

    = 65.84 m
    Apply scale factor of 1: 25

    65.84
    		= 2.63 m
    25

  1. Data from a sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 micron IS sieve. The liquid limit and plastic limit of the finer fraction was found to be 45 and 33 percents respectively. The group symbol of the given soil as per IS:1498-1970 is








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    Since more than 50% n of sample pass through 75µ sieve, it is fine grained soil.
    wL = 45%
    Q WL > 35% and < 50%, it is intermediate compressible
    (I) Equation of A-line
    Ip = 0.73 (WL – 20)
    = 0.73 (45 – 20) = 18.25%
    ∴ Ip is below A line,
    ∴ it is silty soil (m)
    ∴ Group symbol is MI

    Correct Option: B

    Since more than 50% n of sample pass through 75µ sieve, it is fine grained soil.
    wL = 45%
    Q WL > 35% and < 50%, it is intermediate compressible
    (I) Equation of A-line
    Ip = 0.73 (WL – 20)
    = 0.73 (45 – 20) = 18.25%
    ∴ Ip is below A line,
    ∴ it is silty soil (m)
    ∴ Group symbol is MI

  1. The avoid ratios at the densest, loosest and the natural states of a sand deposit are 0.2, 0.6 and 0.4, respectively, the relative density of the deposit is








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    ID =
    emax - e
    		× 100
    emax - emin

    =
    0.6 - 0.4
    		× 100
    0.6 - 0.2

    =
    0.2
    		× 100 = 50%
    0.4

    Correct Option: C

    ID =
    emax - e
    		× 100
    emax - emin

    =
    0.6 - 0.4
    		× 100
    0.6 - 0.2

    =
    0.2
    		× 100 = 50%
    0.4