Design of One Way SlabsDesign of One Way Slabs

CE A433 – RC DesignCE A433 – RC Design

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

Spring 2007Spring 2007

DefinitionDefinition

A One Way A One Way Slab is Slab is simply a simply a very wide very wide beam that beam that spans spans between between supportssupports

Span, LWidth, b

Design for a 12” WidthDesign for a 12” Width

When you When you solve for solve for AAss, you are , you are solving for solving for AAss/ft width./ft width.

Beam ProfileBeam Profile

Design variables: Thickness (h) and Design variables: Thickness (h) and ReinforcingReinforcing

Solving for Thickness, hSolving for Thickness, h

Thickness may controlled by either:Thickness may controlled by either: ShearShear FlexureFlexure DeflectionDeflection

Thickness Based on ShearThickness Based on Shear

Shear stirrups are Shear stirrups are not possible in a slab not possible in a slab so all you have is so all you have is VVcc for strength.for strength. ACI 318-05 ACI 318-05

11.5.6.1(a) exempts 11.5.6.1(a) exempts slabs from the slabs from the requirement that requirement that shear reinforcement shear reinforcement is required where is required where ever Vever Vuu exceeds exceeds VVcc/2./2.

wc

u

wccu

bf

Vd

dbfVV

2

2

Thickness Based on FlexureThickness Based on Flexure

Use the three equations that were Use the three equations that were presented earlier in the semester for presented earlier in the semester for computing bdcomputing bd22 for singly reinforced for singly reinforced concrete beams, using b = 12”.concrete beams, using b = 12”. Largest beam size (based on ALargest beam size (based on Asminsmin as as

specified in the code)specified in the code) Smallest beam size (based on the steel Smallest beam size (based on the steel

strain being .005)strain being .005) Smallest beam size not likely to have Smallest beam size not likely to have

deflection problems (c ~ .375cdeflection problems (c ~ .375cbb))

Thickness Base on Thickness Base on DeflectionDeflection

We haven’t covered deflection We haven’t covered deflection calculations yet.calculations yet.

See ACI 318-05 9.5.2See ACI 318-05 9.5.2 You must comply with the requirements You must comply with the requirements

of ACI 318-05 Table 9.5(a) if you want to of ACI 318-05 Table 9.5(a) if you want to totally ignore deflectionstotally ignore deflections

Other ConsiderationsOther Considerations

For thinner multi-span slabs, it might be For thinner multi-span slabs, it might be useful to put the steel at mid depth so that useful to put the steel at mid depth so that it can act as both positive and negative it can act as both positive and negative reinforcing.reinforcing. Then h = d*2Then h = d*2

Cover requirements are a bit differentCover requirements are a bit different See ACI 318-05 7.7.1(c)See ACI 318-05 7.7.1(c)

You might need to make allowance for a You might need to make allowance for a wear surfacewear surface

Flexural SteelFlexural Steel

Consider as a rectangular singly reinforced Consider as a rectangular singly reinforced beam where b = 12”beam where b = 12” MMuu << AAssffyy(d-A(d-Assffyy/(1.7f’/(1.7f’ccb)) b)) Solve for A Solve for Ass

The resulting AThe resulting Ass is the req’d A is the req’d Ass PER FOOT PER FOOT OF WIDTH.OF WIDTH.

Also consider min AAlso consider min Ass requirement ACI 318- requirement ACI 318-05 10.5.105 10.5.1

All bars can provide this AAll bars can provide this Ass by selecting an by selecting an appropriate spacingappropriate spacing Spacing = ASpacing = Abb/(req’d A/(req’d Ass/ft width)/ft width) Watch units!!!!Watch units!!!!

Spacing LimitsSpacing Limits

ACI 318-05 7.6.5 has an upper limit ACI 318-05 7.6.5 has an upper limit on bar spacingon bar spacing S S << min(3h, 18”) min(3h, 18”)

The lower limit is as used in previous The lower limit is as used in previous beam problems..beam problems.. The clear distance between bars The clear distance between bars >>

max(1”, max aggregate size/.75)max(1”, max aggregate size/.75)

Typical CalculationTypical CalculationControlling 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

Temperature & Shrinkage Temperature & Shrinkage SteelSteel

ACI 318-05 7.12ACI 318-05 7.12 Req’d AReq’d Ass/ft width = (12”)h/ft width = (12”)h

= 0.0020 for f= 0.0020 for fyy < 60 ksi < 60 ksi = 0.0018 for f= 0.0018 for fyy = 60 ksi = 60 ksi

This steel is placed TRANSVERSE to This steel is placed TRANSVERSE to the flexural steel.the flexural steel.

ACI 318-05 7.12.2.2 ACI 318-05 7.12.2.2 Spacing Spacing << min(5h,18”) min(5h,18”)

T&S CalculationT&S Calculation

LayoutLayout

Flexural Steel

Temperature & Shrinkage

Steel

Example ProblemExample Problem

Materials: f’Materials: f’cc = 3 ksi, f = 3 ksi, fyy = 60 ksi = 60 ksi Imposed Loads: Live = 100 psf, Dead = Imposed Loads: Live = 100 psf, Dead =

25 psf25 psf

Finding “h”Finding “h”

At this point, we have enough At this point, we have enough information to determine h using ACI information to determine h using ACI 318-05 Table 9.5a:318-05 Table 9.5a: Cantilevers: h Cantilevers: h >> L/10 = 24”/10 = 2.4” L/10 = 24”/10 = 2.4” Main Spans: h Main Spans: h >> L/24 = 120”/28 = 4.29” L/24 = 120”/28 = 4.29”

We still need to check shear and We still need to check shear and flexure requirements… but need flexure requirements… but need more info!more info!

Determine LoadsDetermine Loads

Consider only a 1 ft width of beam (b = 12”)Consider only a 1 ft width of beam (b = 12”) wwLLLL = 100 psf = 100 plf/ft width = 100 psf = 100 plf/ft width wwDLDL = 25 psf + weight of slab = 25 psf + weight of slab

Make a guess at a slab thickness or write the Make a guess at a slab thickness or write the equations of shear and moment in terms of slab equations of shear and moment in terms of slab thickness… Let’s try h = 6”… we will need to fix this thickness… Let’s try h = 6”… we will need to fix this later if it turns out to be greater.later if it turns out to be greater.

wwDLDL = 25 psf + (150 pcf)*.5 ft = 100 psf = 100 plf/ftw = 25 psf + (150 pcf)*.5 ft = 100 psf = 100 plf/ftw wwuu = 1.2(100 plf/ftw) + 1.6(100 plf/ftw) = 280 plf/ftw = 1.2(100 plf/ftw) + 1.6(100 plf/ftw) = 280 plf/ftw

An Almost Arbitrary An Almost Arbitrary DecisionDecision

We will place the steel at mid-depth of the We will place the steel at mid-depth of the slab so that it handles both positive and slab so that it handles both positive and negative momentsnegative moments This means that we only need to design for the This means that we only need to design for the

worst case moment (positive or negative) worst case moment (positive or negative) along the span.along the span.

As a result, d = h/2As a result, d = h/2 This is a good choice for a short relatively thin This is a good choice for a short relatively thin

(less than 8”) slab.(less than 8”) slab. This makes things pretty simple. Only have to This makes things pretty simple. Only have to

design one set of flexural steel!design one set of flexural steel!

Determine Maximum ShearsDetermine Maximum Shears

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

The cantilevers are exempt from 8.3 since The cantilevers are exempt from 8.3 since they are statically determinant (i.e. don’t they are statically determinant (i.e. don’t meet the criteria to use 8.3)meet the criteria to use 8.3) VVuu = w = wuu*L*Lnn = (280 plf/ftw)*(1.5 ft) = 420 lb/ftw = (280 plf/ftw)*(1.5 ft) = 420 lb/ftw

The two center spans are the sameThe two center spans are the same VVuu = w = wuu*L*Lnn/2= (280 plf/ftw)*(9 ft)/2 = 1260 /2= (280 plf/ftw)*(9 ft)/2 = 1260

lb/ftwlb/ftw

Determine Req’d h Based on Determine Req’d h Based on ShearShear

For our choice:For our choice: d = h/2 > Vd = h/2 > Vuu/[/[2sqrt(f’c)bw)]2sqrt(f’c)bw)] d d >> (1260 lb/ftw)/[.75(2)sqrt(3000)(12”)] (1260 lb/ftw)/[.75(2)sqrt(3000)(12”)] d d >> 1.28 in 1.28 in h h >> 2.56 in 2.56 in

Deflection criteria (Table 9.5a) still Deflection criteria (Table 9.5a) still controls!!!controls!!!

Determine Maximum Determine Maximum MomentsMoments

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

Max positive MMax positive Muu = w = wuu*L*Lnn22/16 = 1,418 ft-lb/ftw/16 = 1,418 ft-lb/ftw

Max negative MMax negative Muu = w = wuu*L*Lnn22/11 = 2,062 ft-lb/ftw/11 = 2,062 ft-lb/ftw

Cantilevers are statically Cantilevers are statically determinate: Ldeterminate: Lnn = 1.5 ft. = 1.5 ft. MMuu = w = wuu*L*Lnn

22/2 = 315 ft-lb/ftw/2 = 315 ft-lb/ftw Design for MDesign for Muu = 2,062 ft-lb/ftw = 2,062 ft-lb/ftw

Select “h” Based on FlexureSelect “h” Based on Flexure

Can use the equations derived for choosing Can use the equations derived for choosing the size of rectangular singly reinforced the size of rectangular singly reinforced beams earlier in the semester. beams earlier in the semester. Use b = 12” and solve for d.Use b = 12” and solve for d.

Try solving the equations for both max and Try solving the equations for both max and min size to bracket the possibilities.min size to bracket the possibilities. Max size (based on min reinforcing): h = 6.41 inMax size (based on min reinforcing): h = 6.41 in Min size (based on stl strain = 0.005): h = 3.40 Min size (based on stl strain = 0.005): h = 3.40

inin

Now… Make a Choice!Now… Make a Choice!

I choose to use h = 5”… it is in the range I choose to use h = 5”… it is in the range for flexure and meets Table 9.5a deflection for flexure and meets Table 9.5a deflection criteria and Shear Strength criteriacriteria and Shear Strength criteria

Other choices that meet the limits Other choices that meet the limits computed are also validcomputed are also valid

No real need to go back and fix the “h” No real need to go back and fix the “h” that our load estimate since they are close that our load estimate since they are close and the assumption was conservative, but and the assumption was conservative, but can do it to refine the design if we want to.can do it to refine the design if we want to.

Determine the Flexural Determine the Flexural SteelSteel

Solve the flexural design inequality for ASolve the flexural design inequality for Ass:: MMuu << AAssffyy(d-A(d-Assffyy/(1.7*f’/(1.7*f’ccb)) b)) AAss >> 0.199 in 0.199 in22/ftw/ftw Watch those units!!!Watch those units!!!

Also check to make sure that the minimum Also check to make sure that the minimum AAss is met is met AAss >> min(200,3sqrt(f’ min(200,3sqrt(f’cc))*b))*bwwd/fd/fyy = 0.100 in = 0.100 in22/ftw/ftw

The larger value controls The larger value controls Use AUse Ass >> 0.199 in 0.199 in22/ftw/ftw

Select the Flexural SteelSelect the Flexural Steel

Use #4 @ 12” O.C.Use #4 @ 12” O.C.

Consider T&S SteelConsider T&S Steel

For our case, For our case, = 0.0018 = 0.0018 Req’d AReq’d Ass >> 0.0018(12”)(5”) = 0.108 in 0.0018(12”)(5”) = 0.108 in22/ftw/ftw Max allowed spacing = min(18”,5h) = 18”Max allowed spacing = min(18”,5h) = 18” Compute some spacing and choose a bar:Compute some spacing and choose a bar:

For #3 bar: For #3 bar: s s << 0.11 in 0.11 in22 / (0.108 in / (0.108 in22/ftw) = 1.02 ft = 12.2 in/ftw) = 1.02 ft = 12.2 in

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

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

Final DesignFinal Design

Slab Thickness = 5”Slab Thickness = 5” Longitudinal Steel = #4 @ 12” O.C. @ mid-depthLongitudinal Steel = #4 @ 12” O.C. @ mid-depth Transverse Steel = #3 @ 12” O.C.Transverse Steel = #3 @ 12” O.C.