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    Design of Flat Slabs

    Dr. Ayman Hussein Hosny

    Professor of RC Structures

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    Lecture Layout

    Types Analysis Concrete dimensions

    Slab thickness

    Empirical analysis

    Frame Analysis Drop thickness

    Columns

    Finite element

    Check unchin Column heads Beams (if any)

    Examples

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    Types of Flat Slabs

    Regular flat slabs

    Slab

    co umn

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    Flat slab with Drop Parts

    Slab

    drop

    column

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    Flat slab with Drop Parts

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    Failures of Flat Slab

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    Punching failure

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    Punching failure

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    Flat slab with Column Heads

    Flat slab

    Column head or

    column capital

    column

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    Flat slab with Column Heads

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    Flat Slab with Drop Part and

    o umn ea

    Flat slab

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    REAL BUILDING

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    Concrete Dimensions

    Minimum thickness (ts)= 150 mm

    Thickness depends on L

    11 22

    Without drop

    ts Lext/32 t L /36

    With drop

    s ext

    ts Lint/40

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    Example 1:

    Without drops

    ts is bigger of

    150 mm

    6000/32=187.5 mm

    = .

    Take ts = 200 mm

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    Example 2:

    ts

    is bigger of

    6000/36=166.6 mm

    = mm

    Take ts = 180 mm

    Note slight effect of

    dro anel here

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    When to use drop?

    If ts is large

    If heavy live load (> 5-8 kN/m2)

    In general, drop parts are not common in

    residential buildings The are common in factories and buildin s

    with false ceilings

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    Drop panels

    Drop panels below

    s a

    ona pane s

    above slab

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    d

    td

    ts

    /4

    , d s

    Drop panel width (X) min

    XLmin/2

    = m n

    Example

    t =80mm X=2750 mm

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    Columns

    Bmin=bigger of

    300 mm

    h/15 Spacing/20

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    Column Head

    Mostly used to

    safeguard against

    unchin max= 45 degrees

    max= min

    What if >45 degrees?

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    Marginal Beams

    Might be used

    Carry walls

    Resist lateral loads

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    Analysis of Flat Slabs

    Empirical method 3rd civil

    Frame analysis 3rd civil

    Yield line analysis

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    Empirical method conditions

    Columns lay on straight lines (or maximum

    eccen r c y=

    Number of spans at least three in each direction erence etween two a acent spans w t n

    Maximum difference between largest and smallest

    span n e same rec on w n

    Inner spans larger or equal outer spans

    ax mum span m n mum span .

    Constant slab thickness, ts

    ve oa < ea oa

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    Example

    Columns lay on straight lines (or

    =

    Number of spans at least three

    in each direction

    spans within 10%

    Maximum difference between

    arges an sma es span n e

    same direction within 20%

    Inner spans larger or equal outer

    spans

    Maximum span/minimum

    span1.30

    Constant slab thickness, ts

    Live load< 2* Dead load

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    Analysis Procedure

    Obtain loads, Ws Consider wall loads

    Get straining actions in long and short

    rec ons Divide into strips

    Design of sections

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    SOLUTION

    Divide slab into

    column and field

    strips in bothdirections

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    In case of drop panels

    Calculation in Long

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    Calculation in Long

    In long direction, L1

    Mo=WsL2(L1-2/3D)2/8

    o

    length L1 having a width of

    2 Distribute Mo between

    column and field strips as

    er the followin table

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    Design

    Compute moment per meter

    Choose rft (same regulations as solid

    5-10 bars/meter

    n mum ame er= mm

    Preferable to have constant number of bars

    per meter

    Calculation in Short

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    Calculation in Short

    In short direction, L2

    Mo=WsL1(L2-2/3D)2/8

    o 2having a width of L1

    Distribute Mo between column and fieldstrips as per same table

    Note that larger moment results in longer-

    N t

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    Notes

    Table ratios are based on column strip

    width equals field strip width

    ,field strip moment by the following ratio, R

    =rea w o e s r p a . . o . .

    Reduce column strip moments accordingly

    E l

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    Example

    F.S.=4.25 m

    C.S.=2.75 m

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    D t il f RFT

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    Details of RFT

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    Effect of transferred moment on slab

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    Effect of transferred moment on slab

    Moment transferred to column partially by flexure and

    par a y y ors on

    fflexure =

    1

    qtorison

    12

    1

    bf

    +

    2

    s flexure . s

    Punching

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    Punching

    q up

    =

    o

    up

    o=15.1=

    bo=2A+B bo

    =A+B30.1= 50.1=

    Punching check

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    Punching check

    =qq

    d columnexterior.......3

    .......

    =

    0

    ]20.0[80.0b

    qc

    cucup +=

    columncorner.......2=

    ]50.0[316.0 fbaq cucup +=

    fcu=

    2

    .c

    cup The least

    shall be.cup considered

    EXAMPLE

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    EXAMPLE

    Data

    Walls=3 kN/m2

    = 2Flooring=2 kN/m2

    cu= a

    Example

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    Example

    LLWallsFCtw css +++=

    2/143322524.0 mkNxws =+++=

    210

    /215.114

    mmd

    mkNxwus

    =

    ==

    257]5.35.3[21 kNxxQ ==

    0

    25700050.1

    9104552

    x

    mmxb ==

    .210910

    mmx

    q ==

    ]200[800 fd

    q cu+=EXAMPLE

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    ]20.0[80.0qcu +=EXAMPLE

    2

    0

    302102x

    c

    ==

    .5.1

    .910

    .

    a

    cup

    ..b

    qc

    cup =

    2/11.2

    5.1]

    35050.0[316.0 mmNq

    cup =+=

    316.0qc

    cucup =

    cupqq >

    2/41.1

    5.1

    30316.0 mmNqcup ==

    USE

    COLUMN2

    /6.1 mmNqcup =

    HEAD

    EXAMPLE

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    EXAMPLE

    ]20.0[80.0 fd

    q cucu +=

    2/14.125700050.1 mmNxq ==

    2

    0

    /64.130

    20.02102

    80.0 mmx

    c

    =+=

    50.0316.0

    5.11610

    fa cu+=

    2/11.2

    3080550.0316.0 mmN

    c

    =+=

    316.0

    5.1805

    fcu

    cup

    =

    230

    c

    ==