Soil mechanics miscellaneous

Soil mechanics miscellaneous

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Soil Mechanics

Soil mechanics miscellaneous
  1. A layer of normally consolidated, saturated silty clay of 1 m thickness is subjected to one dimensional consolidation under a pressure increment of 20 kPa. The properties of the soil are: specific gravity = 2.7, natural moisture content = 45%, compression index = 0.45, and recompression index = 0.05. The initial average effective stress within the layer is 100 kPa. Assuming Terzaghi’s theory to be applicable, the primary consolidation settlement (rounded off to the nearest mm) is








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    G = 2.7, w = 45% = 0.45, H = 1 m
    Sr = 1 (standard)
    Cc = 0.45, Cr = 0.05, σ0 = 100 kPa ∆σ = 20 kPa.

    e0 =
    wG
    		=
    .45 × 2.7
    		= 1.215
    Sr1

    Consolidation settlement, Sf or ∆H
    = H.
    Cc
    		log10
    σf
    1 + e0σ0

    = H.
    Cc
    		log10
    σ0 + ∆σ
    1 + e0σ0

    = 1 ×
    .45
    		× log10
    100 + 20
    1 + 1.215100

    = .016 m = 16 mm

    Correct Option: D

    G = 2.7, w = 45% = 0.45, H = 1 m
    Sr = 1 (standard)
    Cc = 0.45, Cr = 0.05, σ0 = 100 kPa ∆σ = 20 kPa.

    e0 =
    wG
    		=
    .45 × 2.7
    		= 1.215
    Sr1

    Consolidation settlement, Sf or ∆H
    = H.
    Cc
    		log10
    σf
    1 + e0σ0

    = H.
    Cc
    		log10
    σ0 + ∆σ
    1 + e0σ0

    = 1 ×
    .45
    		× log10
    100 + 20
    1 + 1.215100

    = .016 m = 16 mm

  1. Identical surcharges are placed at ground surface at sites X and Y, with soil conditions shown alongside and water table at ground surface. The silty clay layres at X and Y are identical. The thin sand layer at Y is continouous and freedraining with a very large discharge capacity. If primary consolidation at X is estimated to complete in 36 months, what would be the corresponding time for compeltion of primary cosnolidation at Y?








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    Time factor,

    Tv =
    Cvt
    d2

    ∴ t ∝ d2
    At site X, d1 = 10 m and t1 = 36 months
    At site Y, d2 = 5 m and t2 =?
    t1
    		=
    d12
    t2d22

    ∴ t2 = 36 × 52
    ∴ t2 =
    36 × 52
    		= 9 months
    102

    Correct Option: C

    Time factor,

    Tv =
    Cvt
    d2

    ∴ t ∝ d2
    At site X, d1 = 10 m and t1 = 36 months
    At site Y, d2 = 5 m and t2 =?
    t1
    		=
    d12
    t2d22

    ∴ t2 = 36 × 52
    ∴ t2 =
    36 × 52
    		= 9 months
    102

  1. A saturated clay stratum draining both at the top and bottom undergoes 50 percent consolidation in 16 years under an applied load. If an additional drainage layer were present at the middle of the clay stratum, 50 percent consolidation would occur in








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    Time factor,

    Tv =
    Cv.t
    		 t ∝ d2
    d2

    d = H for single drainage d = H/2 for double drainage.
    I case.
    t1 = 16 years, d = H/2
    I case.
    t2 = ?
    d =
    (H/2)
    		=
    H
    24

    t1
    		=
    (H/2)2
    t2(H/4)2

    (t ∝ d2)

    Correct Option: B

    Time factor,

    Tv =
    Cv.t
    		 t ∝ d2
    d2

    d = H for single drainage d = H/2 for double drainage.
    I case.
    t1 = 16 years, d = H/2
    I case.
    t2 = ?
    d =
    (H/2)
    		=
    H
    24

    t1
    		=
    (H/2)2
    t2(H/4)2

    (t ∝ d2)

Direction: The average effective overburden pressure on 10 m thick homogeneous saturated clay layer is 150 K Pa. Consolidation test on undisturbed soil sample taken from the clay layer showed that the void ratio decreased from 0.6 to 0.5 by increasing the stress intensity from 100 kPa to 200 kPa. (G = 2.65)

  1. The total consolidation settlement of the clay layer due to the construction of a structure imposing an additional stress intensity of 200 kPa is








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    Cc =∆e
    log10
    σ1
    σ0

    ∴ ∆e = Cc × log10
    σ1
    σ0

    = 0.209 × log10
    200 + 150
    		= .0769
    150

    ∆e
    		=
    ∆H
    1 + e0H

    ∴ ∆H =
    ∆e.H
    		=
    .0769 × 10
    		= .49 ≈ 0.5 m
    1 + e01 + .563

    Correct Option: D

    Cc =∆e
    log10
    σ1
    σ0

    ∴ ∆e = Cc × log10
    σ1
    σ0

    = 0.209 × log10
    200 + 150
    		= .0769
    150

    ∆e
    		=
    ∆H
    1 + e0H

    ∴ ∆H =
    ∆e.H
    		=
    .0769 × 10
    		= .49 ≈ 0.5 m
    1 + e01 + .563

  1. A 4 m thick layer of normally consolidated clay has an average void ratio of 1.30. Its compression index is 0.6 and coefficient of consolidation is 1 m2 /yr. If the increase in vertical pressure due to foundation load on the clay layer is equal to the existing effective overburden pressure, the change in the thickness of the clay layer is_______mm








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    314
    Change in thicknesss, ΔH

    =
    Cc
    		H0log10
    σ0 + Δσ0
    1 + e0σ0

    CC = 0.6, e0 = 1.30, H0 = 4
    Δσ = σ0
    ∴ ΔH =
    0.6
    		 × 2 × log10
    σ0 + Δσ0
    1 + 1.30σ0

    =
    0.6
    		 × 2 log10 (2)
    2.3

    = 0.314 m
    = 314 mm

    Correct Option: C

    314
    Change in thicknesss, ΔH

    =
    Cc
    		H0log10
    σ0 + Δσ0
    1 + e0σ0

    CC = 0.6, e0 = 1.30, H0 = 4
    Δσ = σ0
    ∴ ΔH =
    0.6
    		 × 2 × log10
    σ0 + Δσ0
    1 + 1.30σ0

    =
    0.6
    		 × 2 log10 (2)
    2.3

    = 0.314 m
    = 314 mm