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  • SabahShawkat CabinetofStructuralEngineering 20173.2 Reinforced Concrete Slabs

    Slabs are divided into suspended slabs. Suspended slabs may be divided into two

    groups:

    (1) slabs supported on edges of beams and walls

    (2) slabs supported directly on columns without beams and known as flat slabs. Supported

    slabs may be one-way slabs (slabs supported on two sides and with main reinforcement in one

    direction only) and two-way slabs (slabs supported on four sides and reinforced in two

    directions). In one-way slabs the main reinforcement is provided along the shorter span. In order

    to distribute the load, a distribution steel is necessary and it is placed on the longer side. One-

    way slabs generally consist of a series of shallow beams of unit width and depth equal to the

    slab thickness, placed side by side. Such simple slabs can be supported on brick walls and can

    be supported on reinforced concrete beams in which case laced bars are used to connect slabs

    to beams.

    Figure 3.2-1: One way slab, two-way slab, ribbed slab, flat slab, solid flat slab with drop

    panel, waffle slab

    In R.C. Building construction, every floor generally has a beam/slab arrangement and

    consists of fixed or continuous one-way slabs supported by main and secondary beams.

  • SabahShawkat CabinetofStructuralEngineering 2017

    Figure 3.2-2: Solid flat slab, solid flat slab with drop panels

    The usual arrangement of a slab and beam floor consists of slabs supported on cross-

    beams or secondary beams parallel to the longer side and with main reinforcement parallel to

    the shorter side. The secondary beams in turn are supported on main beams or girders extending

    from column to column. Part of the reinforcement in the continuous is bent up over the support,

    or straight bars with bond lengths are placed over the support to give negative bending

    moments.

    Figure 3.2-3: Types of the reinforced concrete slab systems

  • SabahShawkat CabinetofStructuralEngineering 20173.2.1 Flat Slabs

    Flat plate is defined as a two-way slab of uniform thickness supported by any

    combination of columns, without any beams, drop panels, and column capitals. Flat plates are

    most economical for spans from 4,5 to 7,5m, for relatively light loads, as experienced in

    apartments or similar building.

    -A flat slab is a reinforced concrete slab supported directly on and built monolithically with the

    columns, the flat slab is divided into middle strips and column strips. The size of each strip is

    defined using specific rules. The slab may be in uniform thickness supported on simple

    columns. These flat slabs may be designed as continuous frames. However, they are normally

    designed using an empirical method governed by specified coefficients for bending moments

    and other requirements which include the following:

    1. There should be not less than three rectangular bays in both longitudinal and transverse

    directions.

    2. The length of the adjacent bays should not vary by more than 10 %.

    Figure 3.2.1-1: Post punching behaviour of slab- critical section

    The general layout of the reinforcement is based on the both bending moments (in spans) and

    bending moments in addition to direct loads (on columns).

  • SabahShawkat CabinetofStructuralEngineering 2017

    Figure 3.2.1-2: Combined punching shear and transfer of moments

    Figure 3.2.1-3

  • SabahShawkat CabinetofStructuralEngineering 2017

    Figure 3.2.1-4

  • SabahShawkat CabinetofStructuralEngineering 2017

    3.2.1-1 Analysis and Design of Flat Plate

    To obtain the load effects on the elements of the floor system and its supporting

    members using an elastic analysis, the structure may be considered as a series of equivalent

    plane frames, each consisting of vertical members columns, horizontal members - slab.

    Such plane frames must be taken both longitudinally (in x-direction) and transversely (in y

    direction) in the building, to assure load transfer in both directions.

    For gravity load effects, these equivalent plane frames can be further simplified into

    continuous beams or partial frames consisting of each floor may be analysed separately together

    with the columns immediately above and below, the columns being assumed fixed at their far

    ends. Such a procedure is described in the Equivalent Frame Method. When frame geometry

    and loadings meet certain limitations, the positive and negative factored moments at critical

    sections of the slab may be calculated using moment coefficients, termed Direct Design

    Method. These two methods differ essentially in the manner of determining the longitudinal

    distribution of bending moments in the horizontal member between the negative and positive

    moment sections. However, the procedure for the lateral distribution of the moments is the same

    for both design methods.

    Figure 3.2.1.1-1: Steel shear heads, steel plats joined by welding

  • SabahShawkat CabinetofStructuralEngineering 2017Since the outer portions of horizontal members (slab) are less stiff than the part along the

    support lines, the lateral distribution of the moment along the width of the member is not

    Uniform. The procedure generally adopted is to divide the slab into column strips (along the

    column lines) and middle strips and then apportion the moment between these strips and the

    distribution of the moment within the width of each strip being assumed uniform.

    Figure 3.2.1.1-2: Moments and frames

  • SabahShawkat CabinetofStructuralEngineering 2017

    Figure: 3.2.1.1-3

    Example: 3.2-1 Design and calculation of Flat Plate

    Geometric Shapes

    Slab thickness

    The geometry of the building floor plans:

    Construction height of object:

    Dimensions columns:

    The peripheral dimensions of the beam:

    Figure: 3.2.1-1

    Load calculation

    Load per area

    Reinforced concrete slab thickness of 300 mm

    hd 300 mm

    l1 7.7 m lk 2.3 m l2 3.6 m ly 7.7 m

    kv 2.850 m

    bs 400 mm hs bs

    ho 0.5 m bo 0.30 m

    qdo hd 25kN

    m3 1.35 qdo 10.125

    kN

    m2

  • SabahShawkat CabinetofStructuralEngineering 2017floor layer:

    Live load (apartments):

    Total load on 1 m 2 of slab:

    Force load Peripheral masonry thickness of 400 mm YTONG:

    Total load acting on the console:

    Investigation replacement frame in the X-axis Frame 1:

    Calculation model

    Figure: 3.2.1-2

    load calculation

    q1d 3kN

    m2 1.4 q 1d 4.2

    kN

    m 2

    vd 2.0kN

    m2 1.5 vd 3

    kN

    m2

    qd qdo q1d vd qd 17.325kN

    m2

    F1 10kN

    m3 kv ly 400 mm 1.35 F1 118.503kN

    F1d F1 F1d 118.503kN

  • SabahShawkat CabinetofStructuralEngineering 2017Load width in a direction perpendicular to the x:

    Load in the x-direction:

    Calculation of internal forces

    Moment on a console:

    Moment of inertia:

    Transverse replacement frame:

    Central girders replacement of frame:

    column:

    Bending stiffness:

    Transverse replacement frame:

    Central girders replacement frame:

    Column

    zsx ly

    qdx qd zsx qdx 133.403kNm

    Mk F1d lk qdxlk

    2

    2

    Mk 625.407kN m

    Iply hd

    312

    Ip 0.017m4

    Ist Ip

    Is112

    bs hs3 Is 5.208 10 3 m

    4

    KpIpl1

    1000kN

    rad m2 Kp 2.25 kN

    mrad

    KstIstl2

    1000kN

    rad m2 Kst 4.813 kN

    mrad

    KsIskv

    1000kN

    rad m2 Ks 1.827 kN

    mrad

  • SabahShawkat CabinetofStructuralEngineering 2017

    Figure: 3.2.1-3

    Primary moments in node 9:

    Primary moments in node 10:

    Equilibrium conditions:

    Node 9

    Node 10:

    9 1 10 1 M910 1 M108 1 M109 1 M1012 1M1010' 1 M10'10 1 M911 1 M97 1

    M97o 0 kN m M910o112

    qdx l12 M911o 0 kN m

    M109o112

    qdx l12 M108o 0 kN m M1012o 0 kN m

    M1010112

    qdx l22 M10'10o M1010

    GivenM97 kN m M97o Ks 3 9 rad M911 kN m M911o Ks 2 9 rad M910 kN m M910o Kp 2 9 rad 10 rad M108 kN m M108o Ks 3 10 rad M109 kN m M109o Kp 2 10 rad 9 rad M1012 kN m M1012o Ks 2 10 rad M1010' kN m M1010 Kst 10 rad M10'10 kN m M10'10o Kst 10 rad

    Mk M97 kN m M910 kN m M911 kN m 0 kN m

  • SabahShawkat CabinetofStructuralEngineering 2017

    The calculated moments of individual members of equilibrium conditions:

    The computation of shear forces in the individual members:

    Maximum moment between 9-10 Mmax

    Maximum moment between 10-10 Mstr

    M109 kN m M1012 kN m M1010' kN m M108 kN m 0 kN mv Find M97 M910 M911 M109 M1012 M1010' M108 9 10 M10'10

    M910 v 1.0( ) kN m M910 691.408 kN mM10'10 v 9 0( ) kN m M10'10 282.659kN mM1010' v 5.0( ) kN m M1010' 282.659 kN mM911 v 2.0( ) kN m M911 26.401 kN mM109 v 3.0( ) kN m M109 545.787kN m

    V910o qdxl12

    V109o V910o

    V910 V910oM910 M109

    l1

    V910 532.511kN

    V109 V109oM910 M109

    l1

    V109 494.688 kN

    V1010 qdxl14

    V1010' V1010V1010' 256.8 kN

    v

    0

    01

    2

    3

    4

    5

    6

    7

    8

    9

    39.601-691.408

    26.401

    545.787

    -105.251

    -282.659

    -157.877

    7.223

    -28.797

    282.659

    a V910l1

    V910 V109 a 3.992m

    Mmax V910 a M910 qdx a2

    2 Mmax 371.423kN m

    Mstr V1010'l14 M1010' qdx

    l22