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    CIVL3112 - Structural Concrete Design1

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    Aditya Mandapati
    1) A square footing 3.4m wide is designed to support a 500mm x 400mm column carrying a service load of 4400kN. 2) The thickness of the footing is determined to be 960mm to resist the factored shear force of 5700kN from the column. 3) Punching shear is checked and found to be satisfactory with a factored shear of 5924kN resisting the applied factored shear of 4848kN. 4) Longitudinal reinforcement is determined to be 34 Y20 bars based on the minimum steel requirement of 10.412mm^2 governing over the moment demand of 1885kNm.

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    CIVL3112 - Structural Concrete Design1

    Uploaded by

    Aditya Mandapati
    22 views2 pages
    1) A square footing 3.4m wide is designed to support a 500mm x 400mm column carrying a service load of 4400kN. 2) The thickness of the footing is determined to be 960mm to resist the factored shear force of 5700kN from the column. 3) Punching shear is checked and found to be satisfactory with a factored shear of 5924kN resisting the applied factored shear of 4848kN. 4) Longitudinal reinforcement is determined to be 34 Y20 bars based on the minimum steel requirement of 10.412mm^2 governing over the moment demand of 1885kNm.

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    CIVL3112_ Structural Concrete Design1

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    1) A square footing 3.4m wide is designed to support a 500mm x 400mm column carrying a service load of 4400kN. 2) The thickness of the footing is determined to be 960mm to resist the factored shear force of 5700kN from the column. 3) Punching shear is checked and found to be satisfactory with a factored shear of 5924kN resisting the applied factored shear of 4848kN. 4) Longitudinal reinforcement is determined to be 34 Y20 bars based on the minimum steel requirement of 10.412mm^2 governing over the moment demand of 1885kNm.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    22 views2 pages

    CIVL3112 - Structural Concrete Design1

    Uploaded by

    Aditya Mandapati
    1) A square footing 3.4m wide is designed to support a 500mm x 400mm column carrying a service load of 4400kN. 2) The thickness of the footing is determined to be 960mm to resist the factored shear force of 5700kN from the column. 3) Punching shear is checked and found to be satisfactory with a factored shear of 5924kN resisting the applied factored shear of 4848kN. 4) Longitudinal reinforcement is determined to be 34 Y20 bars based on the minimum steel requirement of 10.412mm^2 governing over the moment demand of 1885kNm.

    Copyright:

    © All Rights Reserved
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    of 2

    CIVL4403 Structural Concrete Design 1 CIVL4403 Structural Concrete Design 2

    Design of Isolated Column Footing Design of Isolated Column Footing

    Self-study 6: Design of Isolated Column Footing Area of footing


    From the range of bearing pressure given, and allowable for the self-weight of the footing, the available bearing
    pressure is, say, 400kPa.
    A column 500mm by 400mm with ten Y24 bars carries a service dead load of 3000kN and service live load of Total service load= 3000 + 1400 = 4400kN,
    1400kN. A square footing is required for the column. The allowable bearing pressure for the foundation is Hence the required area= 4000⁄400 = 11.0m2 ; choose a footing 3.4m square.
    estimated to be 400 to 450 kPa. The concrete strength is fc′ = 30MPa. Thickness of footing
    For the ultimate strength design of the footing we consider the factored column load, P ∗ , and the corresponding
    upward pressure, q u
    P ∗ = 1.20 × 3000 + 1.5 × 1400 = 5700kN’
    q u = 5700⁄3.4^2 = 493kPa = 493 × 10−3 MPa
    To determine the required effective depth, we consider the beam shear on a critical section at distance d from
    the 500mm face of the column (Figure 1(b)).
    V ∗ = 493 × 10−3 × 3400(1450 − d) = 1720(1450 − d)N
    The shear strength of the section depends on the amount of the tensile steel used for bending. We assume a
    minimum quantity of:
    As per Clause 16.3.1 pmin = 0.19 (D/d)2 f’ct,f / fsy , but since depth is unknown use following expression instead
    pminn = 1.4⁄fsy = 1.4⁄400 = 0.0035 (conservative)
    Having Vuc from the following equation:
    1
    𝐴
    𝑠𝑡 ′ 3
    𝑉𝑢𝑐 = 𝛽1 𝐵𝑑 ( 𝐵𝑑 𝑓𝑐 ) (Clause 8.2.7.1)
    1⁄
    Vuc = 1.1 × 3400 × d(0.0035 × 30) 3 = 1764d N
    Equating V ∗ to ϕVuc′ we get d = 873mm.
    With a cover of 75mm and 20 mm diameter reinforcing bars:
    D = 873 + 10 + 75 = 958mm; Hence we take D = 960mm, d = 875mm, and dom = 865mm.
    Check actual pmin = 0.19 x (960/875)2 x 3.29 / 400 = 0.0019 << 0.0035 considered OK
    We must now check for punching shear. The shear perimeter (Figure 1 (a)) is a rectangular (400 + dom ) ×
    Figure 1. Details of Tut.20
    (500 + dom ) i.e. 1265 × 1365mm. The shear acting is: (Clause 9.2.3)
    V ∗ = 493 × 10−3 (34002 − 1265 × 1365) = 4848 kN
    Having equation:
    2
    𝑓𝑐𝑣 = 0.17 (1 + 𝛽 ) √𝑓𝑐′ < 0.34√𝑓𝑐′
    ℎ

    fcv = 0.34√30 = 1.86MPa


    CIVL4403 Structural Concrete Design 3
    Design of Isolated Column Footing

    u = 2(1265 + 1365) = 5260mm


    Also:
    𝑉𝑢𝑜 = 𝑢𝑑𝑜𝑚 𝑓𝑐𝑣
    ϕVuo = 0.7 × 5260 × 965 × 1.86 = 5924 kN;
    Which is greater than V ∗ and so the design of is satisfactory for punching shear.
    Tensile reinforcement for bending
    The critical section for bending is at the longer face (500mm face) of the column, where the length of the
    cantilevered portion is 0.5(3400 − 400) = 1500mm. The design ultimate moment is hence:
    M ∗ = 0.493 × 3400 × 15002 ⁄2 = 1885 × 106 Nmm = 1885kNm
    An estimate of the steel area required is now made:
    Ast = 1885 × 106 ⁄(0.8 × 0.9 × 875 × 400) = 7481mm2
    However, the minimum steel requirement is:
    Astmin = 0.0035 × 3400 × 875 = 10.412mm2 which governs. Use 34Y20 bars.
    (It is possible to use the Amin given in Cl 16.3.1 at this stage, however in this case the shear capacities will need
    to be re-checked using the reduced reinforcement area as they were calculated using a p of 0.0035)
    Tensile reinforcement in transverse direction:
    In the other direction, the effective depth is marginally less, by the diameter of a reinforcing bar. However, the
    moment is also slightly less because of the smaller column size and the minimum steel area applies. We again
    use 34 Y20 bars.

    Figure 2. Plan view of reinforcement

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