One Way reinforced concrete slabs

7/29/2019 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



Definition

A One WaySlab is simply

a very widebeam thatspans betweensupports



Design for a 12” Width

When yousolve for A s,

you aresolving for

A s /ft width.



Beam Profile

Design variables: Thickness (h) and Reinforcing



Solving for Thickness, h

Thickness may controlled by either:

Shear

Flexure

Deflection



Thickness Based on Shear

Shear stirrups are notpossible in a slab soall you have is f V

c

forstrength. ACI 318-05

11.5.6.1(a) exemptsslabs from the

requirement that shearreinforcement isrequired where ever Vu exceeds f Vc /2.

wc

u

wccu

b f

V d

d b f V V

2

2

f

f f



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 A smin as specified

in the code)

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

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



Thickness Base on Deflection

We haven’t covered deflection calculationsyet.

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

ignore deflections



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



Flexural Steel

Consider as a rectangular singly reinforced beamwhere b = 12” Mu < f A sf y(d-A sf y /(1.7f’ cb)) Solve for A s

The resulting A s is the req’d A s PER FOOT OFWIDTH.

Also consider min A s requirement ACI 318-0510.5.1

All bars can provide this A s by selecting anappropriate spacing Spacing = A b /(req’d A s /ft width) Watch units!!!!



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)



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



Temperature & Shrinkage Steel

ACI 318-05 7.12

Req’d A s /ft width = (12”)hr r = 0.0020 for f y < 60 ksi r = 0.0018 for f y = 60 ksi

This steel is placed TRANSVERSE to the

flexural steel. ACI 318-05 7.12.2.2

Spacing < min(5h,18”)



T&S Calculation



Layout

Flexural Steel

Temperature & Shrinkage Steel



Example Problem

Materials: f’ c = 3 ksi, f y = 60 ksi

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



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!



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… Let’stry 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



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!



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. don’t 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



Determine Req’d h Based on Shear

For our choice:

d = h/2 > Vu /[f2sqrt(f’c)bw)]

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

d > 1.28 in

h > 2.56 in

Deflection criteria (Table 9.5a) stillcontrols!!!



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*Ln

2 /2 = 315 ft-lb/ftw

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



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



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.



Determine the Flexural Steel

Solve the flexural design inequality for A s: Mu < f A sf y(d-A sf y /(1.7*f’ cb))

A s

> 0.199 in2 /ftw

Watch those units!!!

Also check to make sure that the minimum A s ismet

A s > min(200,3sqrt(f’ c))*bwd/f y = 0.100 in

2

/ftw

The larger value controls Use A s > 0.199 in2 /ftw



Select the Flexural Steel

Use #4 @ 12” O.C.



Consider T&S Steel

For our case, r = 0.0018

Req’d A s > 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



Final Design

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

Transverse Steel = #3 @ 12” O.C.

One Way reinforced concrete slabs

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