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Shallow Foundation MUHAMMAD AZRIL HEZMI AHMAD SAFUAN AB RASHID KAMARUDIN AHMAD

Shallow Foundation · and friction between the base of the footing and the soil and is also applicable to shallow and surface foundations. Strip footing Square footing Rectangular

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  • Shallow FoundationMUHAMMAD AZRIL HEZMI

    AHMAD SAFUAN AB RASHID

    KAMARUDIN AHMAD

  • Shallow Foundation�Shallow foundation is a foundation whose depth below the surface, z, is equal to or is less than its least dimension, B. z ≤B.

    � Type of shallow foundation:

    a) Pad footing

    b) Strip/Continuous footing

    c) Raft/ Mat foundation

  • Type of shallow foundation

    a) Pad footing

    Generally an individual foundation designed to carry a single column load although there are occasions when a pad foundation supports two or more columns.

  • Type of shallow foundation

    b) Strip / Continuous footing

    Often termed a continuous footing this foundation has a length significantly greater than its width. It is generally used to support a series of columns or a wall.

  • Type of shallow foundation

    c) Raft / Mat foundation

    This is a generic term for all types of foundations that cover large areas. A raft foundation is also called as a mat foundation.

  • Factors in the design

    � Adequate depth

    � Limiting settlement

    � Safe against shear failure

  • Adequate depth

    � The depth of footing must be sufficient to prevent any changes in surface conditions, horizontal loads and strong overturning moments.

    � To prevent frost action and volume change effect, the depth of footing should more than 1.2 m and 1.5 m respectively.

  • Limiting settlement� Guidelines to limiting settlement by Skempton and MacDonald, 1956:

    � Sand - Maximum total settlement 40 mm

    � Clay - Maximum total settlement 60 mm

  • Safe against shear failure� Shear failure occurs when the soil divides into separate blocks or zones

    which move fully or partially and tangentially with respect to each other, along slip surfaces.

    � Conventionally, the factor of safety to use in design against shear failure is more than 3.0.

  • Modes of Shear Failure� General shear failure

    � This occurs when a clearly defined slip surface forms under the footing and develops outward towards one or both sides and eventually to the ground surface.

  • Modes of Shear Failure� Local shear failure

    � Significant vertical movement may take place before any noticeable development of shear plane occurs.

  • Modes of Shear Failure� Punching shear failure

    � This is a downward movement of the foundation caused by soil shear failure only occurring along the boundaries of the wedge of soil immediately below the foundation.

  • Bearing capacity terms

    �Ultimate bearing capacity, qu or qf� The value of the contact pressure between the foundation and the soil

    which will produce shear failure in the soil.

    �Net ultimate bearing capacity, qnett� Net loading at which ground fails in shear.

    � qnett = qu - γD

    D

    B

    q

  • Bearing capacity terms

    �Safe bearing capacity, qsafe , qa or qall� Net ultimate bearing capacity divided with factor of safety (FS) plus the

    term γD.

    qsafe = qnett/FS + γD

    �Allowable bearing capacity� The maximum allowable net loading intensity on the soil allowing for both

    shear and settlement effects.

  • Bearing capacity analysis

    �Prandtl’s analysis� Prandtl (1921) proposed the following general equation for qu:

    qu = c cot Φ[ exp (∏tan Φ) tan 2 (450 + ∏/2) – 1]

    � For a surface footing with Φ= 0, Prandtl obtained

    qu = 5.14c

  • Bearing capacity analysis

    �Terzaghi’s analysis� Terzaghi (1943) produced a formula for qu which allows for the effects of cohesion

    and friction between the base of the footing and the soil and is also applicable to shallow and surface foundations.

    � Strip footing

    � Square footing

    � Rectangular footing

    � Circular footing

    � The coefficients Nc, Nq and Ng depend upon the soil’s angle of shearing resistance and can be obtained from next figure.

    γqcu γBN5.0γzNcNq ++=

    γqcu γBN4.0γzNcN3.1q ++=

    −++

    +=L

    B2.01γBN5.0γzN

    L

    B3.01cNq

    γqcu

    γqcu γBN3.0γzNcN3.1q ++=

  • Bearing capacity analysis

  • Bearing capacity analysis� A rectangular foundation, 2m x 4m is to be founded at a depth of 1 m

    below the surface of soft clay with unit weight, γ = 20 kN/m3. If the soil parameters are c = 24 kN/m2 and Φ = 00, determine the ultimate bearing capacity of the foundation.

  • Bearing capacity analysis� From Terzaghi’s chart,

    Nc = 5.7, Nq = 1 and Nγ = 0

    qu = [(24)(5.7) {(1)+(0.3)(2/4)} ] + [(20)(1)(1)] + [0]

    = 177.3 kN/m2

  • Bearing capacity analysis� A strip footing is 2 m wide and is founded at depth of 3 m in a soil of

    unit weight 19.3 kN/m3 with c = 10 kN/m2 and Φ = 250. Using factor of safety = 3.0, determine the values of safe bearing capacity of the foundation.

  • Bearing capacity analysis� From Terzaghi’s chart,

    � Nc = 25, Nq = 13 and Nγ = 10

    � qu = [(10)(25)] + [(19.3)(3)(13)] + [(0.5)(19.3)(2)(10)] = 1200 kN/m2

    � qnett = qu - γD

    = 1200 – [(19.3) (3)]

    = 1142.1 kN/m2

    � qsafe = qnett/FS + γD

    = (1142.1)/(3) + (19.3)(3)

    = 439 kN/m2

  • Bearing capacity analysis� A square foundation is 1.5 m x 1.5 m in plan. The soil supporting the

    foundation has a c = 15.2 kN/m2 and Φ= 200. The unit weight of soil, γ = 17.8 kN/m3. Determine the design load on the foundation with a factor of safety 4. Assume the depth of the foundation is 1 m.

  • Bearing capacity analysisFrom Terzaghi chart,

    Nc = 17.7, Nq = 7.4 and Nγ = 5

    qu = [(1.3)(15.2)(17.7)] + [(17.8)(1)(7.4)] + [(0.4)(17.8)(1.5)(5)]

    = 535 kN/m2

    qnett = qu –γD

    = 535 – [(17.8) (1)]

    = 517.2 kN/m2

    qsafe = qnett/FS + γD

    = (517.2) / (4) + (17.8)(1)

    = 147.1 kN/m2

  • Bearing capacity analysis

    Design load,

    Q = qsafe x base area

    = [(147)] [(1.5) (1.5)]

    = 330.75 kN

  • Bearing capacity analysis� Determine the breadth of a strip footing required to carry an inclusive

    load of 550 kN per metre run at a depth of 1.5 m in a fine soil with c = 90 kN/m2 and unit weight of soil, γ= 19 kN/m3. Assume the factor of safety is 3.

  • Bearing capacity analysisqsafe = Q / base area

    = 550 / [(B) (1)]

    = 550 / B

    From Terzaghi chart,

    Nc = 5.7, Nq = 1 and Nγ = 0

    qu = [(90)(5.7)] + [(19)(1.5)(1)] + [0]

    = 541.5 kN/m2

    qnett = qu - γD

    = 541.5 – [(19) (1.5)]

    = 513 kN/m2

    qsafe = qnett/FS + γD

    = (513) / (3) + (19) (1.5)

    = 199.5 kN/m2

  • Bearing capacity analysisTherefore,

    550 / B = 200

    B = 550 / 200

    B = 2. 75 m

    Use breadth of a strip footing = 2.8 m

  • Bearing capacity analysis

    � Skempton’s analysis

    � Skempton (1951) using Terzaghi’s analysis by showing that for a cohesive soil(Φ= 0), the value of Nc can be estimated from Skemptonchart or can be obtained from the formula

    � with a limiting value of Nc = 7.5(1+0.2B/L).

    +

    +=B

    z2.01

    L

    B2.015N c

  • Bearing capacity analysis

    Nc Vs depth (after Skempton, 1951)

  • Bearing capacity analysis

    � A rectangular foundation, 2m x 4m is to be founded at a depth of 1 m below the surface of soft clay with unit weight, γ= 20 kN/m3. If the soil parameters are c = 24 kN/m2 and Φ= 00, determine the ultimate bearing capacity of the foundation.

  • Bearing capacity analysis� From Skempton chart,

    Nq = 1 and Nγ = 0

    = 5 [(1) + (0.2) (2)/(4)] [(1) + (0.2) (1)/(2)]

    = 6.05

    qu = [(24)(6.05) {(1)+(0.3)(2/4)} ] + [(20)(1)(1)] + [0]

    = 186.9 kN/m2

  • Meyerhof’s analysis

  • Meyerhof’s analysis

    � Meyerhof (1963) proposed the following general equation for qu:

    where;Nq = tan

    2 (450 + Φ/2) exp (∏ tan Φ)

    Nc = (Nq – 1) cot Φ

    Nγ = (Nq – 1) tan 1.4 Φ

    Dc = 1 + 0.4(z/B)

    Dq = 1 + 2 tan Φ (1 – sin Φ)2 (z/B)

    Dg = 1

    Ic = Iq = (1 – α/900)2

    Ig = (1 – α/Φ)2

    γγγγ DISγBN5.0DISγzNDIScNq qqqqccccu ++=

  • Meyerhof’s analysis

  • Meyerhof’s analysis� A rectangular foundation, 2m x 4m is to be founded at a depth of 1 m

    below the surface of soft clay with unit weight, γ = 20 kN/m3. If the soil parameters are c = 24 kN/m2 and Φ = 00, determine the ultimate bearing capacity of the foundation. (F.S. = 3)

  • Meyerhof’s analysisNc = 5.14, Nq = 1 and Nγ = 0

    = (1) + [(2/4) (1 / 5.14)] = 1.1

    = (1) + [(2/4) (tan 00) = 1

    = (1) – [(0.4) (2/4)] = 0.8

    c

    q

    cN

    N.

    L

    B1S +=

    φtanL

    B1Sq +=

    L

    B4.01S −=γ

  • Meyerhof’s analysisDc = 1 + 0.4(z/B)

    = (1) + [(0.4) (1/2)] = 1.2

    Dq = 1 + 2 tan Φ (1 – sin Φ)2 (z/B)

    = (1) + [(2) (tan 00) (1 – sin 00)2 (1/2)] = 1

    Dγ = 1

  • Meyerhof’s analysisDue to no inclination load, therefore α= 00

    Ic = Iq = (1 – α/900)2

    = (1 – 0/90)2

    =1

    Iγ = (1 – α/∏)2

    = (1 – 0/0)2

    = 1

    = [(24)(5.14)(1.1)(1)(1.2)] + [(20)(1)(1)(1)(1)(1)] + [0]

    = 182.8 kN/m2

    γγγγ DISγBN5.0DISγzNDIScNq qqqqccccu ++=

  • Meyerhof’s analysisqnett = qu - γD

    = 182.8 – [(20) (1)]

    = 162.8 kN/m2

    qsafe = qnett/FS + γD

    = (162.8) / (3) + (20) (1)

    = 74.3 kN/m2

  • Hansen’s analysis� Hansen (1970) using Meyerhof’s analysis by suggested different value

    of Nγ that can be obtained from the formula

    � Nγ = 1.5 (Nq – 1) tan 1.1Φ

    � which can be applies to a strip footing only.

  • Hansen’s analysis� A strip footing is 2 m wide and is founded at depth of 3 m in a soil of

    unit weight 19.3 kN/m3 with c = 10 kN/m2 and Φ= 250. Using factor of safety = 3.0, determine the values of safe bearing capacity of the foundation.

  • Hansen’s analysis

    From bearing capacity table,

    Nc = 20.72 and Nq = 10.66

    Nγ = 1.5 (Nq – 1) tan 1.1Φ

    = 1.5 (10.66 – 1) [tan 1.1 (25)]

    = 7.54

    = (1) + [(2/ ) (10.66 / 20.72)]

    = 1

    = 1

    = 1

    c

    q

    cN

    N.

    L

    B1S +=

    φtanL

    B1Sq +=

    L

    B4.01S −=γ

  • Hansen’s analysisDc = 1 + 0.4(z/B)

    = (1) + [(0.4) (3/2)]

    = 1.6

    Dq = 1 + 2 tan f (1 – sinΦ)2 (z/B)

    = (1) + [(2) (tan 250) (1 – sin 250)2 (3/2)]

    = 1.8

    Dγ = 1

  • Hansen’s analysis

    Due to no inclination load, therefore α = 00

    Ic = Iq = (1 – α/900)2

    = (1 – 0/90)2

    =1

    Iγ = (1 – α/Φ)2

    = (1 – 0/0)2

    = 1

    = [(10)(20.72)(1)(1)(1.6)] + [(19.3)(3)(10.66)(1)(1)(1.8))] + [(0.5)(19.3)(2)(7.54)(1)(1)(1)]

    = 1588 kN/m2

    γγγγ DISγBN5.0DISγzNDIScNq qqqqccccu ++=

  • Hansen’s analysis

    qnett = qu - γD

    = 1588 – [(19.3) (3)]

    = 1530 kN/m2

    qsafe = qnett/FS + γD

    = (1530) / (3) + (19.3) (3)

    = 567.9 kN/m2

  • Effect of ground water on bearing capacity� If the water table is at or above the footing’s base, the soil’s submerged

    unit weight, γ’ (unit weight of soil, γsat – unit weight of water, γw) should be used.

    � If the water table is below the footing’s base, the water table assumed to have no effect and the soil’s full unit weight, γ should be used.

  • Effect of ground water on bearing capacityA strip footing of breadth 2.5 m is to be founded at a depth of 2 m in a well drained compact sand having the unit weight, γsat = 20 kN/m

    3 with c’ = 0 kN/m2 and Φ’ = 350. Assuming that the water table may rise to the surface and adopting a factor of safety of 3, determine the safe bearing capacity using Terzaghi’s analysis.

  • Effect of ground water on bearing capacity

    Nc = 57.8, Nq = 41.4 and Nγ = 42.4

    γ’ = γsat – γw

    qu = [(0)] + [(20 – 9.81)(2)(41.4)] + [(0.5)(20 – 9.81)(2.5)(42.4)]

    = 1384 kN/m2

    qnett = qu - γ’D

    = 1384 – [(20 – 9.81) (2)]

    = 1363.6 kN/m2

    qsafe = qnett/FS + γ’D

    = (1363.6)/(3) + (20 – 9.81)(2)

    = 474.9 kN/m2

    γqcu BN'γ5.0zN'γcNq ++=

  • Effect of eccentric loading on bearing capacity� If the foundation is subjected to eccentric load, the effective length, L’

    and effective width, B’ should be used in the calculation.

    � L’ = L – 2e; B’ = B – 2e

    B

    2e B'

    e

    cL

  • Effect of eccentric loading on bearing capacity� A continuous footing is 1.8 m wide and is founded at a depth of 0.75 m

    in a clay soil of unit weight, γ = 20 kN/m3 with c = 30 kN/m2 and Φ= 00. The foundation is to carry a vertical line load which will act at a distance of 0.5 m from the centre. Determine the value of safe bearing capacity using Meyerhof’s analysis by taking factor of safety = 3.

  • Effect of eccentric loading on bearing capacity

    From bearing capacity table ,

    Nc = 5.14, Nq = 1 and Nγ = 0

    Eccentricity of line load,

    e = 0.5 m

    Effective width of footing,

    B’ = B – 2e

    = 1.8 – [(2) (0.5)]

    = 0.8 m

    = (1) + [(0.8/ ) (10.66 / 20.72)] = 1

    = 1

    = 1

    c

    q

    cN

    N.

    L

    B1S +=

    φtanL

    B1Sq +=

    L

    B4.01S −=γ

  • Effect of eccentric loading on bearing capacityDc = 1 + 0.4(z/B)

    = (1) + [(0.4) (0.75/0.8)]

    = 1.17

    Dq = 1 + 2 tan f (1 – sinΦ)2 (z/B)

    = (1) + [(2) (tan 00) (1 – sin 00)2 (1/0.8)]

    = 1

    Dγ = 1

  • Effect of eccentric loading on bearing capacityDue to no inclination load, therefore α = 00

    Ic = Iq = (1 – α/900)2

    = (1 – 0/90)2

    =1

    Iγ = (1 – α/Φ)2

    = (1 – 0/0)2

    = 1

    = [(30)(5.14)(1)(1)(1.17)] + [(20)(0.75)(1)(1)(1)] + [0]

    = 195.4 kN/m2

    γγγγ DISγBN5.0DISγzNDIScNq qqqqccccu ++=

  • Effect of eccentric loading on bearing capacityqnett = qu - γD

    = 195.4 – [(20) (0.75)]

    = 180.4 kN/m2

    qsafe = qnett/FS + γD

    = (165.4) / (3) + (20) (0.75)

    = 75 kN/m2

  • Effect of eccentric loading on bearing capacity

    � A foundation is 3 m wide and 9 m long is to be founded at a depth of 1.5 m in a deep deposit of dense sand. The angle of shearing resistance of the sand is 350 and its unit weight is 19 kN/m3. The foundation is to carry a vertical line load which will act at a distance of 0.5 m from the centre. Determine the safe bearing capacity using Meyerhof’s analysis if it subjected to a vertical line load of 220 kN per meter run, together with a horizontal line load of 50 kN per meter run acting at the base of the foundation. Use factor of safety 3.

  • Effect of eccentric loading on bearing capacity

    From bearing capacity table,

    Nc = 46.12, Nq = 33.3 and Nγ = 45.23

    Eccentricity of line load, e = 0.3 m

    Effective width of footing,

    B’ = B – 2e

    = 3 – [(2) (0.3)]

    = 2.4 m

    = (1) + [(2.4/9) (33.3 / 46.12)] = 1.19

    = (1) + [(2.4/9) (tan 350) = 1.19

    = (1) – [(0.4) (2.4/9)] = 0.89

    c

    q

    cN

    N.

    L

    B1S +=

    φtanL

    B1Sq +=

    L

    B4.01S −=γ

  • Effect of eccentric loading on bearing capacityDc = 1 + 0.4(z/B)

    = (1) + [(0.4) (1.5/2.4)]

    = 1.25

    Dq = 1 + 2 tan Φ (1 – sin Φ)2 (z/B)

    = (1) + [(2) (tan 350) (1 – sin 350)2 (1.5/2.4)]

    = 1.16

    Dγ = 1

  • Effect of eccentric loading on bearing capacityThe foundation is effectively acted upon by an inclined load whose angle to a vertical load, therefore

    α = tan -1 (horizontal load / vertical load)

    α = tan -1 (50 / 220)

    α = 12.80

    Ic = Iq = (1 –α/900)2

    = (1 – 12.8/90)2

    =0.74

    Ig = (1 – α/Φ)2

    = (1 – 12.8/35)2

    = 0.4

    = [0] + [(19)(1.5)(33.3)(1.19)(0.74)(1.16)] + [(0.5)(19)(2.4)(45.23)(0.89)(0.4)(1)]

    = 1336.6 kN/m2

    γγγγ DISγBN5.0DISγzNDIScNq qqqqccccu ++=

  • Effect of eccentric loading on bearing capacity

    qnett = qu - γD

    = 1336.6– [(19) (1.5)]

    = 1308kN/m2

    qsafe = qnett/FS + γD

    = (1308) / (3) + (19) (1.5)

    = 464.5 kN/m2

  • SettlementTotal settlement : 50 mm for sand; 75 mm for clay.

    Differential settlement : 12 mm while the angular distortion 1/500.

    Two types of settlement:

    Immediate settlement : as soon as the load is applied.

    Consolidation settlement : process of pore water pressure dissipation from soil.

  • SettlementElastic settlement (immediate settlement) (Harr, 1966)

    where:

    B = width of footing

    qo = nett pressure from footing, kN/m2

    Es = modulus of elasticity of soil, kN/m2

    = possion’s ratio of soil

    = factor depend on footing flexibility

  • SettlementConsolidation settlement (primary settlement) (normally consolidation)

    where

    mv = coefficient of volume compressibility

    = stress increment

    H = soil thickness

  • SettlementConsolidation settlement (primary settlement) (normally consolidation)

    where

    Cc = compression index

    Ho = soil thickness

    eo = initial void ratio

    = effective vertical stress

    = stress increment

  • c

    o

    o

    c

    o

    c

    o

    rc

    σ'

    Δσσ'

    e

    HC

    σ'

    σ'

    e

    HCS

    ++

    ++

    = log1

    log1o

    o

    o

    rc

    σ'

    Δσσ'

    e

    HCS

    ++

    = log1

    Settlement of Over-consolidated Soil