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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 A s,
you aresolving for
A s /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 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
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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 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)
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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
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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 < 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!!!!
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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
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”)
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T&S Calculation
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Layout
Flexural Steel
Temperature & Shrinkage Steel
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Example Problem
Materials: f’ c = 3 ksi, f y = 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… 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
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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. 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
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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!!!
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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*Ln
2 /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 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
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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
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
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Final Design
Slab Thickness = 5” Longitudinal Steel = #4 @ 12” O.C. @ mid-depth
Transverse Steel = #3 @ 12” O.C.