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    Footing

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    rudolf
    The document summarizes the design of a footing to support a column with an axial load of 1480 kN. Key calculations include: 1. Determining the footing depth (h) of 600mm and width (d) of 505mm based on soil bearing capacity and shear stress limits. 2. Sizing the footing width (B) at 2500mm and length (D) at 3160mm based on service load calculations. 3. Checking reinforcement requirements in the long and short directions of the footing and ensuring code compliance for moment capacity, crack control, and shear strength.

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    © All Rights Reserved
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    Footing

    Uploaded by

    rudolf
    3 views4 pages
    The document summarizes the design of a footing to support a column with an axial load of 1480 kN. Key calculations include: 1. Determining the footing depth (h) of 600mm and width (d) of 505mm based on soil bearing capacity and shear stress limits. 2. Sizing the footing width (B) at 2500mm and length (D) at 3160mm based on service load calculations. 3. Checking reinforcement requirements in the long and short directions of the footing and ensuring code compliance for moment capacity, crack control, and shear strength.

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    The document summarizes the design of a footing to support a column with an axial load of 1480 kN. Key calculations include: 1. Determining the footing depth (h) of 600mm and width (d) of 505mm based on soil bearing capacity and shear stress limits. 2. Sizing the footing width (B) at 2500mm and length (D) at 3160mm based on service load calculations. 3. Checking reinforcement requirements in the long and short directions of the footing and ensuring code compliance for moment capacity, crack control, and shear strength.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    3 views4 pages

    Footing

    Uploaded by

    rudolf
    The document summarizes the design of a footing to support a column with an axial load of 1480 kN. Key calculations include: 1. Determining the footing depth (h) of 600mm and width (d) of 505mm based on soil bearing capacity and shear stress limits. 2. Sizing the footing width (B) at 2500mm and length (D) at 3160mm based on service load calculations. 3. Checking reinforcement requirements in the long and short directions of the footing and ensuring code compliance for moment capacity, crack control, and shear strength.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 4

    Title FOOTING

    Done By Lecturer Group 1


    Eng. Nyereyemhuka

    REFERENCES CALCULATIONS OUTPUT

    FOOTING

    Live = 253.2 kN
    Dead =1227.55
    Total axial loads = 1480KN
    MXX = 33.32 KNm
    Soil bearing capacity = 200kN/m2

    Calculating footing depth (h and d) at ultimate limit state

    NU = 1480.75 KN

    Ultimate shear stress (Vu)


    = lesser 0.8 √ 35 ,5 N /mm2
    =4.77 N/mm2

    Perimeter of column u
    =( 450 × 2 ) +(300 × 2)
    =1500 mm

    Shear stress at the face of column

    N
    v=
    Column perimeter × d

    assuming v is VU/2

    V U 1480.75 ×10 3
    =
    2 1500 × d

    Making d the subject of formula


    3
    1480.75 ×10
    d=
    2.35× 1500

    d = 420 mm

    Assume blinding concrete cover = 75 mm


    Assume 20 mm diameter bars

    h=420+75+ 20
    = 515mm TAKING h=600mm

    d=600−75−20
    = 505mm

    Calculating the footing area (B and D) at serviceability limit state

    Ns = 1.0Gk + 1.0Qk + W footing (hγcA)


    = 1227.55 + 253.2 + ( 0.6 × 24 × A ¿
    = 1480.75 +14.4A

    Soil bearing pressure at service limit state = Ns/A

    1480.75+14.4 A
    200=
    A

    1480.75
    A=
    185.6

    = 7.97 m2

    Taking B = 2500 mm and D = 3160 mm

    1
    M 33.32
    e= =
    N 1480.75
    = 0.02

    D
    Since eccentricity e < there is no tension under the footing
    6

    Reinforcement in the long direction

    Calculating stress at the face of the column

    N 6M
    P1 = −¿ 2
    BD BD

    1480.75 6 × 33.32
    −
    2.5× 3.16 2.5 ×3.162

    = 179.43 KN

    N 6M
    P2 = +
    BD BD 2

    1480.75 6 ×33.32
    = +
    2.5× 3.16 2.5 × 3.162

    = 195.45 KN

    L+ hc
    Pface = P1 + (P2 - P1)
    3.16

    1.355+0.6
    Pface = 179.43 + (16.02)
    3.16

    = 189.34 KN/m2

    F1 = PfaceLB
    = 189.34 × 1.355× 2.5
    =631.23 kN

    F2 = 0.5(P2 - Pface) LB
    ¿ 0.5 ×6.11× 1.355 ×2.5
    =10.35 kN

    M = F1(L/2) + F2(2L/3)

    1.355 2× 1.355
    = 631.23 +10.35
    2 3

    = 438.91 kNm

    M
    K= 2
    f cu b d

    438.91
    = 2
    35× 2500 × 420

    = 0.028436 Since 0.028436 < 0.156 OK

    Z
    D
    =0 ,5+ 0.25−
    √
    0.028436
    0.9

    = 0.967 Which is > 0.95 so we take z= 0.95d

    Z = 0.95 × 420
    = 399mm

    M
    A st required =
    0 , 95 f y Z
    2
    6
    438.91× 10
    ¿
    0.95 ×460 × 399

    = 2517 mm2

    Provide 9T20 at spacing 277 mm centres


    A s provided=2830 mm2

    Reinforcement in the short direction

    N
    P=
    BD

    1480.75
    = = 187.44 KN
    2.5× 3.16

    F = P LD
    =187.44 × 1.1× 3.16
    = 651 kN

    M = F(L/2)

    1.1
    651 ×
    2
    =358.05 KN

    M
    K= 2
    f cu b d

    6
    358.05 ×10
    ¿ 2
    35× 2500 ×420

    = 0.0231 Since 0.0231 < 0.156 OK

    Z
    D
    =0 ,5+ 0.25−
    √0.0231
    0.9

    0.973 Which is > 0.95 so we take z= 0.95d

    Z = 0.95 × 420
    = 399mm

    M
    A st required =
    0 , 95 f y Z

    6
    358.05× 10
    0.95 ×460 × 399

    =2053 mm2

    Provide 7T20 at spacing at 357 mm centres


    A s provided=2200 mm2

    Direct shear check

    P1 = 179.43 KN
    P2 = 195.45 KN
    V [(P2 + Pd)/2]B(L-d)

    =452.21 KN

    Shear stress

    V = V/Bd = 0.43

    From table 3.8 BS8110

    100 A S 100 ×2830


    = =
    Bd 2500 ×420
    3
    = 0.27

    Take 400/d = 1
    1 35
    0.79 ×0.283 ×1 × 250.5
    Vc = [ ]
    1.25
    Vc = 0.489 N/mm2

    Since v < vc shear is OK

    Punching shear check

    Check punching shear at 1.5d

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