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CE 366 SHALLOW FOUNDATIONS P.1) CANTILEVER FOOTING Question: Given: Q 1 = 1500 kN, Q 2 = 2500 kN, q all = 200 kN/m 2 Ignore the weight of footings and find dimensions B and B 2 of a cantilever footing for a uniform soil pressure distribution. Draw shear and bending moment distributions. B B 2 B Q 1 =1500 kN L=8m Q 2 =2500 kN 0.8 m 0.8 m B 1 =2.0m

CE 366 SHALLOW FOUNDATIONS...A rectangular combined footing which supports three columns is to be constructed on a sandy clay layer whose allowable bearing value (base pressure) is

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  • CE 366 – SHALLOW FOUNDATIONS

    P.1) CANTILEVER FOOTING

    Question:

    Given: Q1 = 1500 kN, Q2 = 2500 kN, qall = 200 kN/m2

    Ignore the weight of footings and find dimensions B and B2 of a cantilever footing for a uniform

    soil pressure distribution. Draw shear and bending moment distributions.

    B

    B2 B

    Q1=1500 kN

    L=8m

    Q2=2500 kN

    0.8 m 0.8 m

    B1=2.0m

  • Solution:

    B1=2m B

    B2

    Strap

    Locate ΣQ = Q1 + Q2

    B b=1500 × 8 / 4000=3

    For B1=2m

    L=8 m

    Q1 = 1500 kN Q1+Q2 =4000kN

    Q = 2500 kN

    0.4m

    a = 8.0 + 0.4 – 1.0 – 3.0

    a = 4.40m

    R1= (4000 × 3) / 7.4 = 1622kN

    R2 = 4000 – 1622 = 2378 kN

    Determine B2,

    qall = Q / (2 × B2)

    200 = 1620 / (2 × B2) B2 = 4 m

    1.0

    200 kN/m2

    Q1=1500 kN Q2-R2=122 kN

    0.6m

    R1=1622 kN

    122kN

    OR

    Without considering resultant (Q1 +Q2)

    Moment w.r.t Q2 or (R2);

    Q1x8- R1x7.4=0 R1=1622 kN

    From force equilibrium;

    ΣFvertical =0

    1500+25000-1622-R2 =0

    R2=2378 kN

    Determine

    qall = Q / (2

    200 = 1622 / (2 B2) B2 ≈ 4

    200 = 2378 / B2 B ≈ 3.45

    0.8 m 0.8 m

    FBD

    V(kN)

    M(kN.m)

    1500 kN

    900 kN.m

  • P.2) TRAPEZOIDAL FOOTING

    Question:

    Determine B1 and B2 of a trapezoidal footing for a uniform soil pressure of 300 kN/m2. (γconc =

    24kN/m3)

    M1=600kN-m

    Q1=3000kN

    M2=800kN-m

    Q2=2000kN M3=1200kN-m

    Q3=4000kN

    1

    2 6 6 2

    B1 B2

    Centroid of the base

    ΣQ

    x

    M1=600kN-m

    Q1=3000kN

    M2=800kN-m

    Q2=2000kN M3=1200kN-m

    Q3=4000kN

    1

    Solution:

    After finding the location of resultant force, you can decide whether the wider side of

    the trapezoid will be at the left or right side.

    2.0 m 6.0 m 6.0 m 2.0 m

    x Centroid of the base

    ΣQ=9000 kN

    B1 B2

  • General formulation for finding coordinates of a centroid;

    ̅ ̅̅̅̅ ̅̅̅̅

    ̅

    ̅̅̅̅ ̅̅̅̅

    ΣQ=9000kN

    Weight of footing= 16x24x(B1+B2)/2 = 192(B1+B2)

    Area of footing = 8(B1+B2)

    ΣFv=0 9000+192(B1+B2)=8(B1+B2)x300

    B1+B2 = 4.07m ………(1)

    ΣM=0 (moment about centroid of the base)

    3000(14-x)+600+2000(8-x)+800-4000(x-2)-1200+(wght of ftg)x0=(base pressure)x0

    x = 7.36m

    x=7.36=

    ………..(2)

    From (1) and (2) B1= 1.55m ; B2 = 2.52m

    Note: You can also take moment about the left or right side but keep in mind that weight and base

    pressure will also have moment about left or right side.

  • B =

    16

    2

    2

    6

    6

    b =

    6

    P.3) MAT FOUNDATION

    Question:

    A mat foundation rests on a sand deposit whose allowable bearing value is 150 kN/m2

    . Column

    loads are given in the figure. The thickness of the mat is 2.0 m (concrete = 24 kN/m3). Calculate

    base pressures assuming that the lines passing through the centroid of the mat and parallel to the

    sides are principal axes. Find the base pressure distribution beneath the base and check whether the

    mat foundation given is safe?

    l = 12 m

    1000 kN 2000 kN 3000 kN

    EXISTING

    STRUCTURE

    2500 kN 3500 kN

    4000 kN

    3500 kN

    2500 kN

    1000 kN 2000 kN 3000 kN 2000 kN 1000 kN

    2 m 6 m 6 m 6 m

    6 m 2 m

    L = 28 m

    2 m

    2 m

  • Solution:

    Area of foundation = 28 × 16 – 12 × 6 = 376 m2

    Total vertical load = Σ V = Column loads + Weight of mat=31000+(376) × 24 × 2 = 49048 kN

    * Center of gravity (CG) of mat:

    28 × 16 × 14 − 6 × 12 × (16 + 6) = 12.47

    376

    m from left

    28 × 16 × 8 − 6 × 12 × (3 + 10) = 7.04

    376

    m from bottom

    16 6

    • 3

    12.47 CG •

    10 7.04

    * Location of ΣQ :

    Take moment about the left side:

    = (1 / 49048) . [ 2 × (1000+2500+1000) + 8 × (2000+3500+2000) + 14 × (3000 +

    + 4000 + 3000) + 20 × (3500+2000) + 26 × (2500+1000) + 376 × 2 × 24 × 12.47 ]

    = 12.95 m from left

    Take moment about bottom side :

    = (1 / 49048) . [ 2 × (1000+2000+3000+2000+1000) + 8 × (2500 + 3500 + 4000 +

    + 3500 + 2500) + 14 × (1000+2000+3000) + 376 × 2 × 24 × 7.04 ]

    = 7.28 m from bottom

    Centroid of the

    16mx28m rectangular

    Centroid of the 12mx6m

    cut section due to

    existing structure

  • Eccentricity :

    e1 = 12.95 – 12.47 = 0.48 m

    e2 = 7.28 – 7.04 = 0.24 m

    1

    8.96

    2 • 2

    7.04

    12.47 15.53

    1

    M1

    28 m

    M2

    16 m

    M1 about 1-1 axis:

    M1 = ΣQ ⋅ e1 = 49048 ⋅ (0.48) = 23543 kN.m

    M2 about 2-2 axis:

    M2 = ΣQ ⋅ e2 = 49048 ⋅ (0.24) = 11772 kN.m

    ⎡ B ⋅ L3 2 ⎤ ⎡b ⋅ l 2 ⎤

    I1−1 = ⎢

    12 + B ⋅ L ⋅ (D

    1 ) ⎢ − ⎢

    12 + b ⋅ l ⋅ (d

    1 ) ⎢ =

    ⎣ ⎦ ⎣ ⎦

    ⎡ × 3

    ⎤ ⎡ 3 ⎤ 4

    = 16 28

    + 16 × 28 × (14 − 12.47) 2 ⎢ − ⎢ 6 ×12

    + 6 ×12 × (22 − 12.47) 2 ⎢ = 22915 m

    ⎢ 12 ⎦ ⎣ 12 ⎦

    ⎡ L ⋅ B 3 2 ⎤ ⎡ l ⋅ b 2 ⎤

    I 2−2 = ⎢

    12 + B ⋅ L ⋅ (D

    2 ) ⎢ − ⎢

    12 + b ⋅ l ⋅ (d

    2 ) ⎢ =

    ⎣ ⎦ ⎣ ⎦

    = ⎡ 28 ×163

    + 16 × 28 × (8 − 7.04) 2 ⎤ −

    ⎡12 × 6 + 12 × 6 × (13 − 7.04) 2 ⎢ = 7197 m

    4

    ⎢ 12

    ⎢ ⎢ ⎦ ⎣ ⎦

    e1=0.48m

    e2=0.24m

  • Note: In soil mechanics compression is taken as positive (+)

    q = ΣQ ±

    M 1 ⋅ y1 ± M 2 ⋅ y2

    Area I1−1 I 2−2

    q = 49048

    ± 23543 ⋅ y

    1 ± 11772 ⋅ y

    2 = 130.4 ± 1.03 y ± 1.64 y

    376 22915 7197 1 2

    q A

    = 130.4 ± 1.03 y1 ± 1.64 y

    2 = 130.4 + 1.03 ⋅ (3.53) + 1.64 ⋅ (2.96) = 138.9 kN/m

    2

    qB

    = 130.4 + 1.03 ⋅ (3.53) +1.64 ⋅ (8.96) = 148.7

    kN/m2

    qC

    = 130.4 −1.03 ⋅ (12.47) + 1.64 ⋅ (8.96) = 132.2

    kN/m2

    qD

    = 130.4 −1.03 ⋅ (12.47) −1.64 ⋅ (7.04) = 106

    qE = 134.9 kN/m2

    qF = 151.3 kN/m2

    kN/m2

    ΣQ = 49048 kN 1

    qC = 132.2kN/m2

    C B qB = 148.7 kN/m2

    qA = 138.9 kN/m2

    A e2 = 0.24

    m

    CG

    F qF = 151.3 kN/m2

    (MAX)

    2 2

    D

    qD = 106 kN/m

    2

    (MIN)

    e1 = 0.48 m

    E

    qE = 134.9 kN/m2

    1

    Since at all critical points stress values are almost < qall = 150 kN/m2

    given mat foundation is

    safe.

  • P.4) COMBINED FOOTING ANALYZED BY RIGID METHOD

    Question:

    A rectangular combined footing which supports three columns is to be constructed on a sandy clay

    layer whose allowable bearing value (base pressure) is 84 kN/m2.The thickness of the concrete

    footing is 0.85m. There is a 1.20m thick soil fill having same unit weight with sandy clay on the

    footing. Unit weight of the sandy clay and the concrete are 20 kN/m3 and 24 kN/m3 respectively.

    Analyze the footing by rigid method and plot shear and moment diagrams.

    800kN 600kN 500kN

    1.2

    0.85

    2m 6m 6m 2m

    B=?

    Solution:

    Neglecting column weights;

    ΣQnet=800+600+500+(24-20)x16xBx0.85 = 1900+54.4B

    e = ΣM

    ΣV

    = 800 x6 − 500 x6

    = 1900 + 54.4B

    1800

    1900 + 54.4B

    qmax =

    ΣV

    BxL (1 +

    6e ) = 84kN / m 2

    L

    84 = 1900 + 54.4B

    (1 + Bx16

    6x1800 ) (1900 + 54.4B) x16 ⇒ B = 2.0m

    Use B=2.0m qmax = 84 kN/m2

    ; qmin = 41.6 kN/m2

    Downward uniform pressure = 0.85 × (24-20) = 3.4 kN/m2

  • These are the diagrams related to forces and moments acting on the foundation. Explanations are

    at the next page;

    800kN 600kN 500kN

    2m 6m 6m 2m

    3.4

    84

    3.4

    41.6

    1.01

    a A b

    4.2

    B e

    7.7

    C g

    84– 3.4

    ≅ 80.5 c

    75.2

    d

    59.4 43.5 h

    f

    41.6 - 3.4 = 38.2

    311.4 319.2 336

    + j m

    _

    Shear

    (shear diagram is

    488.6

    -507

    280.8

    -444

    164 parabolic in fact,

    but we linearized

    it for simplicity)

    _ -99.4

    +

    Moment

    315 167

  • V = 80.5 + 75.2

    x 2 x 2 = 311.4 kN A 2

    area (abcd) B = 2

    VA’ = 311.4 – 800 = – 488.6 kN

    To find the centroid of a trapezoid on the horizontal axis;

    x x = L 2B1+ B2

    x: distance from shorter dimension

    B1 B2 3 B1+ B2

    L

    x (abcd) = 2 2x80.5 + 75.2

    =1.01m 3 80.5 + 75.2

    MA = 311.4 x 1.01 = 315 kN.m

    VBA = 80.5 + 59.4

    x 8 x 2 − 800 = 319.2 kN 2

    VBC = 319.2 – 600 = -280.8 kN

    x (aefc) = 8 2x80.5 + 59.4

    = 4.2 m 3 80.5 + 59.4

    M = ⎡80.5 + 59.4

    x 8 x 2⎤

    x 4.2 − 800 x 6 = − 99.4 kN.m

    B ⎢ ⎢ ⎣ ⎦

    VCB = 80.5 + 43.5

    x14 x 2 − 800 − 600 = 336 kN 2

    VC = 336 – 500 = -164 kN

    Check the end point; 80.5 + 38.2

    x16 x 2 − 800 − 600 − 500 = 0 OK 2

  • x (aghc) = 14 2x80.5 + 43.5

    = 7.7 m 3 80.5 + 43.5

    M = ⎡80.5 + 43.5

    x 14 x 2⎤

    x 7.7 − 800 x12 − 600 x 6 =167 kN.m

    C ⎢ ⎢ ⎣ ⎦

    Slope of V b/w x = 2 and x = 8 m Æ (488.6+319.2)/6=134.6 kN/m

    Equation of V b/w x = 2 and x = 8 m Æ V(x) = -488.6 + 134.6 (x) for V(x)=0 Æ x=3.6 m

    Base pressure @ x = 2 + 3.6 = 5.6 m Æ 65.6 kN/m2 Similarly @ x = 10.7 m V = 0

    Base pressure @ x =10.7 m Æ 52.2 kN/m2

    maximum points of the moment diagram;

    a A b j

    65.6

    B e m C g

    52.2 h

    f n

    c d k

    j and m are the points where shear forces are equal to zero (i.e. moment is max)

    the distance between point b and point j

    x (ajkc) = (2 + 3.6) 2x80.5 + 65.6

    = 2.9 m 3 80.5 + 65.6

    M = ⎡80.5 + 65.6

    x (2 + 3.6) x 2⎤

    x 2.9 − 800 x 3.6 = − 507 kN.m

    AB ⎢ ⎢ ⎣ ⎦

    x (amnc) = 5.73 m

    MBC = -444 kN.m