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    Given:: Soil 3 3 F Concrete

    Uploaded by

    aira
    The document provides design calculations for a rectangular footing with given dimensions and loadings. [1] The effective soil pressure and required size of the footing are calculated. [2] One-way and two-way shear capacities are checked against factored shear demands. [3] Steel reinforcement requirements are determined for the long and short directions of the footing. [4] Bearing capacity and development lengths of the reinforcement are checked. All design checks are satisfactory.

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    © All Rights Reserved
    Download as XLSX, PDF, TXT or read online from Scribd

    Given:: Soil 3 3 F Concrete

    Uploaded by

    aira
    12 views11 pages
    The document provides design calculations for a rectangular footing with given dimensions and loadings. [1] The effective soil pressure and required size of the footing are calculated. [2] One-way and two-way shear capacities are checked against factored shear demands. [3] Steel reinforcement requirements are determined for the long and short directions of the footing. [4] Bearing capacity and development lengths of the reinforcement are checked. All design checks are satisfactory.

    Original Description:

    Computation footing

    Original Title

    Foundation

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    © © All Rights Reserved

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    The document provides design calculations for a rectangular footing with given dimensions and loadings. [1] The effective soil pressure and required size of the footing are calculated. [2] One-way and two-way shear capacities are checked against factored shear demands. [3] Steel reinforcement requirements are determined for the long and short directions of the footing. [4] Bearing capacity and development lengths of the reinforcement are checked. All design checks are satisfactory.

    Copyright:

    © All Rights Reserved
    Download as XLSX, PDF, TXT or read online from Scribd
    Download as xlsx, pdf, or txt
    12 views11 pages

    Given:: Soil 3 3 F Concrete

    Uploaded by

    aira
    The document provides design calculations for a rectangular footing with given dimensions and loadings. [1] The effective soil pressure and required size of the footing are calculated. [2] One-way and two-way shear capacities are checked against factored shear demands. [3] Steel reinforcement requirements are determined for the long and short directions of the footing. [4] Bearing capacity and development lengths of the reinforcement are checked. All design checks are satisfactory.

    Copyright:

    © All Rights Reserved
    Download as XLSX, PDF, TXT or read online from Scribd
    Download as xlsx, pdf, or txt
    of 11

    RECTANGULAR FOOTING

    GIVEN:
    Footing (width)= 2140 mm
    Column= 450 mm
    DL= 825 KN
    LL= 668 KN
    Ø= 25 mm
    ƴsoil= 15.74 KN/m3
    ƴconcrete= 23.5 KN/m3
    Df= 1500 mm
    hconcrete= 600 mm
    f'c= 27.6 Mpa
    fy= 415 Mpa
    qa= 192 kPa
    d= 525 mm
    Ψe= 1.0
    Ψt = 1.0
    Ψs = 1.0
    λ= 1.0
    Ødowel 25.0 mm

    1. EFFECTIVE SOIL PRESSURE


    qe = qa-∑ƴh
    qe = qa - ɣchc - ɣshs
    hs = 900 mm
    qe = 163.734 kPa
    2. SIZE OF FOOTING
    A (𝐷𝐿+𝐿𝐿)/𝑞�
    =

    9.1184482148 m2
    A= 9.12 m2
    Length = 4.26
    Dimension = 4.26 m x 2.14 m

    qu = (1.2𝐷𝐿+1.6𝐿𝐿)/
    �𝑎𝑐𝑡𝑢𝑎�

    225.74561404 kPa
    qu = 225.750 kPa

    3. ONE WAY (WIDE BEAM) SHEAR


    Vu = qu*Ashaded
    666.6849 KN
    Vu = 666.685 KN

    (0.75)(𝟏/𝟔)(√(𝒇^′
    ØVc =
    𝒄))(bw)(d)
    737.79826702 KN
    ØVc = 737.798 KN

    Condition 𝑉𝑢 ≤ Ø𝑉𝑐
    SAFE

    4. TWO WAY (PUNCHING) SHEAR


    Vu = qu*Ashaded
    1844.2364063 KN
    Vu = 1844.236 KN
    (0.75)(𝟏/𝟑)(√(𝐟^′ �))
    ØVc = (b )(d)
    o
    2689.1712536 KN
    ØVc = 2689.172 KN

    Condition 𝐕𝐮 ≤ Ø𝐕�
    SAFE

    6. STEEL AREA REQUIREMENTS


    (Along Long Direction)
    Mu = (� 𝐮)(𝐤)(𝐛)
    (𝐤/𝟐)
    = 876.60006131 kN-m
    Mu = 876.6 kN-m

    Mu=ØbdR
    Mu=(0.9)bdR
    Ru = 1.6513022655 MPa
    Ru = 1.652 MPa

    0.85(𝑓^′ 𝑐)/𝑓� - [1 - √(1− 2𝑅𝑢/(0.85𝑓^′


    ρ= 𝑐))

    ρ = 0.0041317137

    (𝟏.�)/𝐟�
    ρmin= (𝟏.�)/𝐟�

    ρmin= 0.003373494

    ρmin=
    √(𝐟^′ �)/
    (�(𝐟�))

    ρmin= 0.0031648013
    use ρmin = 0.00337349

    use greater ρ = 0.00413171

    As = ρ bd
    As = 4641.98 mm2

    N = � �/� 𝐛

    9.4565647196 bars
    N= 10 bars

    SPACING

    STEEL REQUIREMENTS
    (Along Short Direction)
    (� 𝐮)(𝐤)(𝐛)
    (𝐤/𝟐)
    Mu = (� 𝐮)(𝐤)(𝐛)
    (𝐤/𝟐)
    343.33713619 kN-m
    Mu = 343.337 kN-m

    Mu=ØbdR
    Mu=(0.9)bdR
    0.3249001357 Mpa
    0.325 MPa

    0.85(𝑓^′ 𝑐)/𝑓� - [1 - √(1−


    ρ= 2𝑅𝑢/(0.85𝑓^′ 𝑐))

    ρ = 0.0007886335

    ρmin= 1.4/𝑓�

    0.003373494

    ρmin= (√𝑓′𝑐)/(4(𝑓�))

    ρmin= 0.0031648013

    use ρmin = 3.373E-03

    use gearter ρ = 0.00337349


    As = ρbd
    7544.8192771 mm2
    7544.82 mm2

    N = � �/ � �

    12.610718229 bars
    13 bars

    BAR DISTRIBUTION
    (𝑁 𝑐�𝑛𝑡�𝑟 𝑏𝑎𝑛�)/(𝑁 𝑖𝑛 𝑠ℎ𝑜𝑟𝑡 �𝑖𝑟�𝑐𝑡𝑖𝑜𝑛) = 2/(𝛽+1)

    (�𝑜𝑛𝑔 𝑠𝑖��
    β =𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔)/(𝑠ℎ𝑜𝑟𝑡 𝑠𝑖�� 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔)

    β = 1.99065421

    Ncenter band = 8.69375


    Ncenter band = 9 bars

    Nouter band = (𝑁𝑠ℎ𝑜𝑟𝑡 𝑠𝑖��−𝑁𝑐�𝑛𝑡�𝑟 𝑏𝑎𝑛�)/2

    Nouter band = 2 bars


    2 bars

    7. CHECK FOR BEARING CAPACITY


    N1= Ø(0.85)(f'c)(A1)
    = (0.65)(0.85)(f'c)(A1)
    N1 = 6862.05

    Pu = 1.2DL+1.6LL
    Pu = 2058.8 KN

    Condition N1 ≥� 𝐮
    SAFE! DOWEL IS NOT NECESSARY

    b)For Base of Footing


    N2= Ø(0.85)(f'c)(A√(�
    1
    ) 2/�1)

    but
    √(�2/
    �1) = 1.41118401

    √(�2/
    �1) =
    use 1.41118401
    N2= 4357.6268456 kN
    N2 = 4357.627 kN
    Condition � 𝟐≥�𝐮
    SAFE! DOWEL IS NOT NECESSARY

    if dowel is needed
    PEX = Pu - N1
    PEX = -4803.25 KN

    �𝑠�= 𝑃��/𝑓�

    -11574.096386

    N = � ��/� 𝐛

    N = -23.57855554
    -23
    use N= 4
    (minimum requirement for dowel)

    8.REQUIRED DEVELOPMENT LENGTH


    a) Along long direction

    1) For Tension
    Ld (actual) = 795 mm

    Ld = (𝐟� � � � � � � �𝐛)/
    (𝟏.𝟏 𝛌 √𝐟�
    (□(64&(� 𝐛+𝐊��)/�𝐛
    )))
    (�𝑏+𝐾𝑡𝑟)/�𝑏≤2.5
    (�𝑏+𝐾𝑡𝑟)/�𝑏
    = 2
    Use (�𝑏+𝐾𝑡𝑟)/�𝑏=
    2

    Ld= 897.65795416

    use Ld = 897.65795416

    but 𝐋� ≥𝟑𝟎𝟎𝐦𝐦
    so Ld is GOOD!

    2) For Compression`
    Ld = (0.24𝑓� �𝑏 )/(λ√(𝑓^′ 𝑐))

    Ld= 473.9633998

    Ld = 0.043(db)(fy)
    Ld = 446.125

    use Ld = 473.9633998

    but 𝐋� ≥𝟐𝟎𝟎𝐦𝐦
    so Ld is GOOD!

    b) Along Short Direction


    1) For Tension
    Ld (actual) = 797.6 mm

    (𝑓� Ψ� Ψ𝑠 Ld
    Ψ𝑡 �𝑏)/(1.1
    = λ √𝑓� (□(64&(𝑐𝑏+𝐾𝑡𝑟)/�𝑏)))

    (�𝑏+𝐾𝑡𝑟)/�𝑏≤2
    .5
    (�𝑏+𝐾𝑡𝑟)/�𝑏
    2

    Therefore
    Use (�𝑏+𝐾𝑡𝑟)/�𝑏= 2

    Ld= 897.65795416

    use Ld = 897.65795416

    but, 𝐋� ≥𝟑𝟎𝟎𝐦𝐦
    so Ld is GOOD!

    2) For Compression`
    Ld = (0.24𝑓� �𝑏 )/(λ√(𝑓^′
    𝑐))
    473.9633998

    Ld = 0.043(db)(fy)
    446.125

    use Ld = 473.9633998

    but𝐋� ≥𝟐𝟎𝟎𝐦𝐦
    so Ld is GOOD!

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