Soil mechanics miscellaneous


  1. The per cent voids in mineral aggregate (VMA) and per cent air voids (Vv) in a compacted cylindrical bituminous mix specimen are 15 and 4.5 respectively. The per cent voids filled with bitumen (VFB) for this specimen is









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    NA

    Correct Option: C

    NA


  1. In an unconsolidated undrained triaxial test, it is observed that an increase in cell pressure from 150 kPa to 250 kPa leads to a pore pressure increase of 80 kPa. It is further observed that, an increase of 50 kPa in deviatoric stress results in an increase of 25 kPa in the pore pressure. The value of Skemptons’s pore pressure parameter B is:









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    Skempton’s pore pressure parameter, B

    B =
    ∆U
    ∆σ3

    =
    80
    =
    80
    = 0.8
    250 – 150100

    Correct Option: C

    Skempton’s pore pressure parameter, B

    B =
    ∆U
    ∆σ3

    =
    80
    =
    80
    = 0.8
    250 – 150100



  1. A given cohesion less soil has emax = 0.85 and emin = 0.50. In the field, the soil is compacted to a mass density of 1800 kg/ m3 at a water content of 8%. Take the mass density of water as 1000 kg/m3 and Gs as 2.7. The relative density (in%) of the soil is









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    P =
    Grw(1 + w)
    1 + e

    ⇒ 1800 =
    2.7 × 1000 × (1.08)
    1 + e

    ∴ 1 + e =
    2.7 × 1000 × 1.08
    1800

    ∴ e = 0.02
    ID =
    emax - e
    × 100
    emax - emin

    =
    0.85 - 0.62
    × 100
    0.85 - 0.5

    = 65.714%

    Correct Option: D

    P =
    Grw(1 + w)
    1 + e

    ⇒ 1800 =
    2.7 × 1000 × (1.08)
    1 + e

    ∴ 1 + e =
    2.7 × 1000 × 1.08
    1800

    ∴ e = 0.02
    ID =
    emax - e
    × 100
    emax - emin

    =
    0.85 - 0.62
    × 100
    0.85 - 0.5

    = 65.714%


  1. A fine-grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when water content is between 15% and 28%. The soil behaves fluid-like when the water content is more than 40%. The ‘Activity’ of the soil is









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    Activity =
    Ip
    C

    Ip = WL – Wp
    = 40 – 28 = 12
    ∴ Activity =
    12
    = 0.3
    100 - 60

    Correct Option: C

    Activity =
    Ip
    C

    Ip = WL – Wp
    = 40 – 28 = 12
    ∴ Activity =
    12
    = 0.3
    100 - 60



  1. The soil profile below a lake with water level at elevation = 0 m and lake bottom at elevation = –10 m is shown in the figure, where k is the permeability coefficient. A piezometer (stand pipe) installed in the sand layer shows a reading of +10 m elevation. Assume that the piezometric head is uniform in the sand layer. The quantity of water (in m3/s) flowing into the lake from the sand layer through the silt layer per unit area of the lake bed is









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    By Darcy’s law
    q = Ki A
    We want q per unit area

    ∴ q =
    KiA
    K.i
    A

    K = Co-efficient of permeability = 10–6 m/s
    i = hydraulic gradient = L/h
    L = Length of silt layer in direction of flow
    L = 20 m. (from figure)
    h = difference in water levels of the overhead and bottom of lake.
    = 20 m. (from figure)
    ∴ q = K.i = K ×
    L
    h

    q = 10–6 ×
    20
    20

    q = 1 × 10–6 m3 /sec

    Correct Option: C

    By Darcy’s law
    q = Ki A
    We want q per unit area

    ∴ q =
    KiA
    K.i
    A

    K = Co-efficient of permeability = 10–6 m/s
    i = hydraulic gradient = L/h
    L = Length of silt layer in direction of flow
    L = 20 m. (from figure)
    h = difference in water levels of the overhead and bottom of lake.
    = 20 m. (from figure)
    ∴ q = K.i = K ×
    L
    h

    q = 10–6 ×
    20
    20

    q = 1 × 10–6 m3 /sec