One Way reinforced concrete slabs

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    Design of One Way Slabs

    CE A433 RC Design

    T. Bart Quimby, P.E., Ph.D.Spring 2007

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    Definition

    A One WaySlab is simply

    a very widebeam thatspans betweensupports

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    Design for a 12 Width

    When yousolve for As,

    you aresolving for

    As/ft width.

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    Beam Profile

    Design variables: Thickness (h) and Reinforcing

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    Solving for Thickness, h

    Thickness may controlled by either:

    Shear

    Flexure

    Deflection

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    Thickness Based on Shear

    Shear stirrups are notpossible in a slab soall you have is fV

    c

    forstrength. ACI 318-05

    11.5.6.1(a) exemptsslabs from the

    requirement that shearreinforcement isrequired where ever Vuexceeds fVc/2.

    wc

    u

    wccu

    bf

    Vd

    dbfVV

    2

    2

    f

    ff

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    Thickness Based on Flexure

    Use the three equations that werepresented earlier in the semester for

    computing bd2

    for singly reinforcedconcrete beams, using b = 12. Largest beam size (based on Asmin as specified

    in the code)

    Smallest beam size (based on the steel strainbeing .005)

    Smallest beam size not likely to havedeflection problems (c ~ .375cb)

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    Thickness Base on Deflection

    We havent covered deflection calculationsyet.

    See ACI 318-05 9.5.2You must comply with the requirements ofACI 318-05 Table 9.5(a) if you want to totally

    ignore deflections

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    Other Considerations

    For thinner multi-span slabs, it might be usefulto put the steel at mid depth so that it can act

    as both positive and negative reinforcing. Then h = d*2

    Cover requirements are a bit different

    See ACI 318-05 7.7.1(c)

    You might need to make allowance for a wearsurface

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    Flexural Steel

    Consider as a rectangular singly reinforced beamwhere b = 12 Mu < fAsfy(d-Asfy/(1.7fcb)) Solve for As

    The resulting Asis the reqd As PER FOOT OFWIDTH.

    Also consider min As requirement ACI 318-0510.5.1

    All bars can provide this As by selecting anappropriate spacing Spacing = Ab/(reqd As/ft width) Watch units!!!!

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    Spacing Limits

    ACI 318-05 7.6.5 has an upper limit onbar spacing

    S < min(3h, 18)

    The lower limit is as used in previousbeam problems..

    The clear distance between bars > max(1,max aggregate size/.75)

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    Typical Calculation

    Controlling Flexural Steel Requirement 0.294 in^2/ftw

    Bar Ab db max s Use s Act. As Act d pMn Mu/pMn c Stl Strain

    (in^2) (in) (in) (in) (in^2/ftw) (in) (ft-k/ftw) (in)

    #3 0.11 0.375 4.49 4.50 0.293 8.06 10.36 0.927 0.507 0.04466

    #4 0.20 0.500 8.16 8.50 0.282 8.00 9.90 0.970 0.488 0.04613

    #5 0.31 0.625 12.65 13.00 0.286 7.94 9.95 0.965 0.495 0.04510

    #6 0.44 0.750 17.95 18.00 0.293 7.88 10.11 0.950 0.507 0.04355

    #7 0.60 0.875 24.48 18.00 0.400 7.81 13.53 0.710 0.692 0.03087

    #8 0.79 1.000 32.23 18.00 0.527 7.75 17.45 0.550 0.911 0.02252

    #9 1.00 1.128 40.80 18.00 0.667 7.69 21.59 0.445 1.153 0.01699

    #10 1.27 1.270 51.82 18.00 0.847 7.62 26.64 0.361 1.465 0.01260

    #11 1.56 1.410 63.65 18.00 1.040 7.55 31.73 0.303 1.799 0.00958

    #14 2.25 1.693 91.81 18.00 1.500 7.40 42.53 0.226 2.595 0.00556

    #18 4.00 2.257 163.21 18.00 2.667 7.12 61.93 0.155 4.614 0.00163

    Note: Check development lengths

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    Temperature & Shrinkage Steel

    ACI 318-05 7.12

    Reqd As/ft width = (12)hr r = 0.0020 for fy < 60 ksi r = 0.0018 for fy = 60 ksi

    This steel is placed TRANSVERSE to the

    flexural steel.ACI 318-05 7.12.2.2

    Spacing < min(5h,18)

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    T&S Calculation

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    Layout

    Flexural Steel

    Temperature &Shrinkage Steel

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    Example Problem

    Materials: fc = 3 ksi, fy = 60 ksi

    Imposed Loads: Live = 100 psf, Dead = 25 psf

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    Finding h

    At this point, we have enough informationto determine h using ACI 318-05 Table

    9.5a: Cantilevers: h > L/10 = 24/10 = 2.4

    Main Spans: h > L/24 = 120/28 = 4.29

    We still need to check shear and flexurerequirements but need more info!

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    Determine Loads

    Consider only a 1 ft width of beam (b = 12)

    wLL = 100 psf = 100 plf/ft width

    wDL = 25 psf + weight of slab Make a guess at a slab thickness or write the equations

    of shear and moment in terms of slab thickness Letstry h = 6 we will need to fix this later if it turns out to

    be greater. wDL = 25 psf + (150 pcf)*.5 ft = 100 psf = 100 plf/ftw

    wu = 1.2(100 plf/ftw) + 1.6(100 plf/ftw) = 280 plf/ftw

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    An Almost Arbitrary Decision

    We will place the steel at mid-depth of the slabso that it handles both positive and negativemoments

    This means that we only need to design for the worstcase moment (positive or negative) along the span.

    As a result, d = h/2

    This is a good choice for a short relatively thin (less

    than 8) slab. This makes things pretty simple. Only have to design

    one set of flexural steel!

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    Determine Maximum Shears

    Use ACI 318-05 8.3 (the slab meets the criteria!)to compute internal forces (or you can do a fullelastic analysis)

    The cantilevers are exempt from 8.3 since theyare statically determinant (i.e. dont meet thecriteria to use 8.3) V

    u

    = wu

    *Ln

    = (280 plf/ftw)*(1.5 ft) = 420 lb/ftw

    The two center spans are the same Vu = wu*Ln/2= (280 plf/ftw)*(9 ft)/2 = 1260 lb/ftw

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    Determine Reqd h Based on Shear

    For our choice:

    d = h/2 > Vu/[f2sqrt(fc)bw)]

    d > (1260 lb/ftw)/[.75(2)sqrt(3000)(12)]

    d > 1.28 in

    h > 2.56 in

    Deflection criteria (Table 9.5a) stillcontrols!!!

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    Determine Maximum Moments

    Main spans: Ln = 9 ft Can use ACI 318-05 8.3:

    Max positive Mu = wu*Ln2

    /16 = 1,418 ft-lb/ftw Max negative Mu = wu*Ln

    2/11 = 2,062 ft-lb/ftw

    Cantilevers are statically determinate: Ln =1.5 ft. Mu = wu*Ln2/2 = 315 ft-lb/ftw

    Design for Mu = 2,062 ft-lb/ftw

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    Select h Based on Flexure

    Can use the equations derived for choosing thesize of rectangular singly reinforced beams

    earlier in the semester. Use b = 12 and solve for d.

    Try solving the equations for both max and minsize to bracket the possibilities.

    Max size (based on min reinforcing): h = 6.41 in

    Min size (based on stl strain = 0.005): h = 3.40 in

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    Now Make a Choice!

    I choose to use h = 5 it is in the range forflexure and meets Table 9.5a deflection criteria

    and Shear Strength criteria Other choices that meet the limits computed are

    also valid

    No real need to go back and fix the h that our

    load estimate since they are close and theassumption was conservative, but can do it torefine the design if we want to.

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    Determine the Flexural Steel

    Solve the flexural design inequality for As: Mu < fAsfy(d-Asfy/(1.7*fcb))

    As

    > 0.199 in2/ftw

    Watch those units!!!

    Also check to make sure that the minimum As ismet

    As > min(200,3sqrt(fc))*bwd/fy = 0.100 in

    2

    /ftw

    The larger value controls Use As > 0.199 in

    2/ftw

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    Select the Flexural Steel

    Use #4 @ 12 O.C.

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    Consider T&S Steel

    For our case, r = 0.0018

    Reqd As > 0.0018(12)(5) = 0.108 in2/ftw

    Max allowed spacing = min(18,5h) = 18 Compute some spacing and choose a bar:

    For #3 bar:

    s < 0.11 in2 / (0.108 in2/ftw) = 1.02 ft = 12.2 in

    For #4 bar: s < 22.2 in use 18 #3 is the better choice!

    Use #3 @ 12 O.C. for T&S steel

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    Final Design

    Slab Thickness = 5 Longitudinal Steel = #4 @ 12 O.C. @ mid-depth

    Transverse Steel = #3 @ 12 O.C.