Detailing Members

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    Detailing of Members

    Beams

    Slabs, one-way and two-way spanning

    Flat slabs

    Columns

    Walls

    Deep beams

    Foundations: Pile caps, Column & wall footings

    Tying systems

    Detailing rules for particular situations

    Detailing comparisons EC2 v BS 8110

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    As,min = 0,26 (fctm/fyk)btd but 0,0013btd

    As,max = 0,04Ac

    Section at supports should be designed for a

    hogging moment 0,25 max. span moment

    Any design compression reinforcement () should beheld by transverse reinforcement with spacing 15

    BeamsEC2: Cl. 9.2

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    Tension reinforcement in a flanged beam at

    supports should be spread over the effective width

    (see 5.3.2.1)

    BeamsEC2: Cl. 9.2

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    Shear Design: Links

    Variable strut methodallows a shallower strut angle

    hence activating more links.

    As strut angle reduces concrete stress increases

    Angle = 45 V carried on 3 links Angle = 21.8 V carried on 6 links

    d

    V

    z

    x

    d

    x

    V

    z

    s

    EC2: Cl. 6.2.3

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    Where av 2dthe applied shear force, VEd, for a point load(eg, corbel, pile cap etc) may be reduced by a factor av/2d

    where 0.5 av 2d provided:

    dd

    av av

    The longitudinal reinforcement is fully anchored at the support. Only that shear reinforcement provided within the central 0.75av is

    included in the resistance.

    Short Shear Spans with Direct

    Strut ActionEC2: Cl. 6.2.3 (8)

    Note: see PD6687-1:2010 Cl 2.14 for more information.

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    Shear reinforcement

    Minimum shear reinforcement, w,min = (0,08(fck)/fyk

    Maximum longitudinal spacing, sl,max = 0,75d (1+cot)

    Maximum transverse spacing, st,max = 0,75d 600 mm

    EC2: Cl. 9.2.2

    For vertical links sl,max = 0,75d

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    (1) Sufficient reinforcement should be provided at all sections to resist the

    envelope of the acting tensile force, including the effect of inclined cracks

    in webs and flanges.

    (2) For members with shear reinforcement the additional tensile force, Ftd,

    should be calculated according to 6.2.3 (7). For members without shear

    reinforcement Ftd

    may be estimated by shifting the moment curve a

    distance al = d according to 6.2.2 (5). This "shift rule may also be used

    as an alternative for members with shear reinforcement, where:

    al = z (cot - cot )/2 = 0.5 z cot for vertical shear links

    z= lever arm, = angle of compression strut

    al = 1.125 d when cot = 2.5 and 0.45 d when cot = 1

    CurtailmentEC2: Cl. 9.2.1.3

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    For members without shear reinforcement this is satisfied with al = d

    a lFtd

    a l

    Envelope of (MEd /z+NEd)

    Acting tensile force

    Resisting tensile force

    lbd

    lbd

    lbd

    lbd

    lbd lbd

    lbd

    lbdFtd

    Shift rule

    Curtailment of reinforcementEC2: Cl. 9.2.1.3, Fig 9.2 Concise: 12.2.2

    For members with shear reinforcement: al = 0.5 zCot

    But it is always conservative to use al = 1.125d

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    lbd is required from the line of contact of the support.

    Simple support (indirect) Simple support (direct)

    As bottom steel at support 0.25As provided in the span

    Transverse pressure may only be taken into account with

    a direct support.

    Shear shift rule

    al

    Tensile Force Envelope

    Anchorage of Bottom

    Reinforcement at End SupportsEC2: Cl. 9.2.1.4

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    Simplified Detailing Rules for

    Beams

    How to.EC2

    Detailing section

    Concise: Cl 12.2.4

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    h /31

    h /21

    B

    A

    h /32h /22

    supporting beam with height h1

    supported beam with height h2 (h1 h2)

    The supporting reinforcement is in

    addition to that required for other

    reasons

    A

    B

    The supporting links may be placed in a zone beyond

    the intersection of beams

    Supporting Reinforcement at

    Indirect Supports

    Plan view

    EC2: Cl. 9.2.5Concise: Cl 12.2.8

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    Solid Slabs

    One-way and two-way spanning

    EC2: Cl 9.3

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    Curtailment as beams except for the Shift rule al = d

    may be used

    Flexural Reinforcement min and max areas as beam

    Secondary transverse steel not less than 20% main

    reinforcement

    Reinforcement at Free Edges

    Solid slabsEC2: Cl. 9.3

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    Where partial fixity exists, not taken into account in design: Internalsupports:As,top 0,25As of Mmax in adjacent spanEnd supports: As,top 0,15As of Mmax in adjacent span

    This top reinforcement should extend 0,2 adjacent span

    Solid slabsEC2: Cl. 9.3

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    Flat slabs

    EC2: Cl 9.4

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    What are flat slabs? Solid concrete floors of constant thickness

    They have flat soffits

    Flat Slabs - Introduction

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    Distribution of momentsEC2: Figure I.1 Concise Figure 5.11

    Particular rules for flat slabs

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    Distribution of moments

    EC2: Table I.1 Concise: Table 5.2

    Particular rules for flat slabs

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    Arrangement of reinforcement should reflect behaviour

    under working conditions.

    At internal columns top reinforcement of area 0.5Atshould be placed in a width = 0.25 panel width.

    At least two bottom bars should pass through internal

    columns in each orthogonal direction.

    Particular rules for flat slabs

    EC2: Cl. 9.4 Concise: 12.4.1

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    Design reinforcement at edge and corner reinforcement

    should be placed within the becz

    cy

    y

    be = cz + y

    A

    cz

    cyy

    A

    be = z + y/2

    z

    A

    Particular rules for flat slabsEC2: Figure 9.9 Concise Figure 5.12

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    Flat slabs Top steel example

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    8m

    Flat slabs Top steel example

    Equivalent frame method

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    TR 64 Figure 14

    Reduction in maximum hogging moment

    at columns

    Flat slabs Top steel example

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    Flat slabs Top steel example

    Flat slab example design information

    Area of reinforcement to resist full negative moment, At = 8000mm

    2

    Panel width = 8.0m

    Column strip width = 4.0m Middle strip width = 4.0m

    0.5 At in a width of 0.125 panel width either side of the column ie:

    4000mm2 in width of 2m

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    Distribution of top reinforcement

    Distance (m)

    600

    100

    200

    300

    400

    500

    0

    Bendingmom

    ent(kNm/m

    )

    0 1 2 3 4 5 6 7 8

    Centre column strip:4000 mm2 or 2000 mm2/m

    Outer column strip:0.75 x 8000 4000

    = 2000 mm2 or 1000 mm2/m

    Middle strip:0.25 x 8000 = 2000 mm2

    or 500 mm2/m

    Assume a total area of steel, At = 8000 mm2

    Distribution 75% Column strip & 25% Middle strip6000 mm2 2000 mm2

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    Punching shear

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    Punching shear does not use the Variable Strut inclination method

    and is similar to BS 8110 methods

    The basic control perimeter is set at 2dfrom the loaded area

    The shape of control perimeters have rounded corners

    bz

    by

    2d 2d 2d

    2du1

    u1 u1

    Punching Shear

    Cl. 6.4 Figure 8.3

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    kd

    Outer control

    perimeter

    Outer perimeter of shear

    reinforcement

    1.5d (2d if > 2d from

    column)

    0.75d

    0.5dA A

    Section A - A

    0.75d

    0.5d

    Outer control

    perimeter

    kd

    The outer control perimeter at

    which shear reinforcement is not

    required, should be calculatedfrom:

    uout,ef = VEd / (vRd,c d)

    The outermost perimeter of

    shear reinforcement should be

    placed at a distance not

    greater than kd( k = 1.5)

    within the outer control

    perimeter.

    Punching Shear Reinforcement (1)EC2: Cl. 6.4.5 Figures 12.5 & 12.6

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    1,5d

    2d

    d

    d

    > 2d

    1,5d

    uout

    uout,ef

    Where proprietary systems are used the control perimeter at which

    shear reinforcement is not required, uout or uout,ef(see Figure) should be

    calculated from the following expression:

    uout,ef = VEd / (vRd,c d)

    Punching Shear Reinforcement (2)EC2: Cl. 6.4.5 Figure 8.10

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    Columns

    EC2: Cl 9.5

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    h 4b

    min 12

    As,min = 0,10NEd/fyd but 0,002 Ac

    As,max= 0.04 Ac (0,08Ac at laps)

    Minimum number of bars in a circular column is 4.

    Where direction of longitudinal bars changes more than1:12 the spacing of transverse reinforcement should be

    calculated.

    Columns

    EC2: Cl. 9.5.2

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    scl,tmax = min {20 min; b ; 400mm}

    150mm

    150mm

    scl,tmax

    scl,tmax should be reduced by a factor 0,6:

    in sections within h above or below a beamor slab

    near lapped joints where > 14.

    A min of 3 bars is required in lap length

    scl,tmax = min {12 min; 0.6b ; 240mm}

    Columns

    EC2: Cl. 9.5.3

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    Column starter bars - foundations

    Column starter bars should be anchored, with a full compression

    lap length.

    Bends and hooks cannot be considered to contribute to the

    compression anchorage The compression anchorage length should therefore be taken as

    41 for concrete class C25/30 (smaller for higher concrete

    classes)

    The effect of cover cannot be considered (

    2) Whittle has recently reported on some testing that shows that it

    may be considered. Reduces anchorage length to 29 (class

    C25/30)

    Whittle, R.Are modern pad footings and pile caps too shallow?Concrete May

    2011 pp 53- 55, The Concrete Society

    MPA - The Concrete Centre 33

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    Walls

    EC2: Cl 9.6

    W ll

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    Walls

    As,vmin = 0,002Ac (half located at each face)

    As,vmax = 0.04Ac (0,08Ac at laps)

    svmax = 3 wall thickness or 400mm

    Vertical Reinforcement

    Horizontal Reinforcement

    As,hmin = 0,25 Vert. Rein. or 0,001Ac

    shmax = 400mm

    Transverse Reinforcement

    Where total vert. rein. exceeds 0,02Ac links required as

    for columns

    Where main rein. placed closest to wall links are required.

    (at least 4No. m2)

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    Foundations

    EC2: Cl 9.8

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    Tying systems

    EC2: Cl 9.10

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    Detailing rules for particularsituations

    PD 6687-1:2010 Annex B

    Replacement for EC2: Annex J

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    PD 6687-1:2010 Annex B

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    Detailing Comparisons

    d or 150 mm from main bar9.2.2 (8): 0.75 d 600 mm

    9.2.1.2 (3) or 15 from main bar

    st,max

    0.75d9.2.2 (6): 0.75 dsl,max

    0.4 b s/0.87 fyv9.2.2 (5): (0.08 b s fck)/fykAsw,min

    Links

    Table 3.28Table 7.3NSmax

    dg + 5 mm or 8.2 (2): dg + 5 mm or or 20mmsmin

    Spacing of Main Bar s

    0.04 bh9.2.1.1 (3): 0.04 bdAs,max

    0.002 bh--As,min

    Main Bar s i n Compr essi on

    0.04 bh9.2.1.1 (3): 0.04 bdAs,max

    0.0013 bh9.2.1.1 (1): 0.26fctm/fykbd

    0.0013 bd

    As,min

    ValuesClause / Val uesMain Bar s i n Tensi on

    BS 8110EC2Beams

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    Detailing Comparisons

    places of maximum moment:

    main: 2h 250 mm

    secondary: 3h 400 mm

    3dor 750 mmsecondary: 3.5h 450 mmSmax

    dg + 5 mm or 8.2 (2): dg + 5 mm or or 20mm

    9.3.1.1 (3): main 3h 400 mm

    smin

    Spacing of Bar s

    0.04 bh9.2.1.1 (3): 0.04 bdAs,max

    0.002 bh9.3.1.1 (2): 0.2As for single way

    slabs

    As,min

    Secondar y T r ansver se Bar s

    0.04 bh0.04 bdAs,max

    0.0013 bh9.2.1.1 (1): 0.26fctm/fykbd

    0.0013 bd

    As,min

    ValuesClause / Val uesMain Bar s i n Tensi on

    BS 8110EC2Slabs

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    Detailing Comparisons

    Columns

    150 mm from main bar9.5.3 (6): 150 mm from main bar

    129.5.3 (3): min (12min; 0.6 b;240 mm)Scl,tmax

    0.25 or 6 mm9.5.3 (1) 0.25 or 6 mmMin size

    Links

    0.06 bh9.5.2 (3): 0.04 bhAs,max

    0.004 bh9.5.2 (2): 0.10NEd/fyk 0.002bhAs,min

    Main Bar s i n Comp r ession

    1.5d9.4.3 (1):

    within 1st control perim.: 1.5d

    outside 1st control perim.: 2d

    St

    0.75d9.4.3 (1): 0.75dSr

    Spacing of Li nks

    Total = 0.4ud/0.87fyv9.4.3 (2): Link leg = 0.053 sr st

    (fck)/fyk

    Asw,min

    ValuesClause / Val uesLinks

    BS 8110EC2Punching Shear