CEE 461: Reinforced Concrete Design, slide 7-1 Module 7: Flexural Analysis and Design of One-Way Slabs CEE 461: Reinforced Concrete Design, slide 7-2 What is a one-way slab? direction of slab span direction of beam span column reaction
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CEE 461: Reinforced Concrete Design, slide 7-1
Module 7:
Flexural Analysis and Design ofOne-Way Slabs
CEE 461: Reinforced Concrete Design, slide 7-2
What is a one-way slab?
direction of slab span
direction of beam span
column reaction
CEE 461: Reinforced Concrete Design, slide 7-3
What is a one-way slab?
girderor beam
beamor joist
slab spans in shortdirection betweensupporting beams
2L
L ratio aspect
a
b !=Lb
La
CEE 461: Reinforced Concrete Design, slide 7-4
Deflected Shape
deflected shapeis like a barrel
neglect any loadtransfer inlong direction
CEE 461: Reinforced Concrete Design, slide 7-5
Load Distribution
Consider uniform loadings appliedto series of parallel one-foot widestrips.
Gravity loads applied uniformlyover slab area (expressed in ksf)are treated as uniform loads onslab strips (expressed in k/ft).
CEE 461: Reinforced Concrete Design, slide 7-6
Trump Tower, Chicago: Cast-in-Place ConcreteFlat-Plate Construction
Two-Way Slab
CEE 461: Reinforced Concrete Design, slide 7-7
Slab Reinforcement
note: beamreinforcement notshown
column column
girder
longitudinal bottomreinforcement
span 1 span 2
longitudinal topreinforcement
transversetemperatureand shrinkagereinforcement
a
a
CEE 461: Reinforced Concrete Design, slide 7-8
Slab Reinforcement
Section “a” - “a”
d h
As bar
min. clearcover = ¾”(318-7.7.1)
temperature and shrinkagereinforcement (318-7.12) = 0.0018bh for Grade 60,smax= 5h < 18”
unit width of slab(1)
(1) typically 1 foot or 1 meters
"12AA bar s foot per s =
12d
A
sd
A foot per sbar s ==!
s(2)
(2) smax= 3h < 18” (318-7.6.5)
CEE 461: Reinforced Concrete Design, slide 7-9
Slab Gravity Loadingssample live loads per ASCE 7: Table 4.1(in psf)assembly areas 100bowling alleys 75corridors 100ballrooms 100hospital operating rooms 60library stack rooms 150heavy manufacturing 250offices 50dwellings 40hotel rooms 40roofs 20classrooms 40stadiums 60shops 100
12
hpcf)(150
h load dead concrete
!
= "
(in psf when h=inches)
CEE 461: Reinforced Concrete Design, slide 7-10
Slab Serviceability
cantilever
both-ends continuous
one-end continuous
simply supported
Minimum SlabThickness
End RestraintCondition
20
l
24
l
28
l
10
l
Table 9.5(a) : Minimum Thickness, h, for One-Way Slabs1
1If slab is supportingconstruction likely to bedamaged by large deflections,then need to calculatedeflections rather than usetable.
CEE 461: Reinforced Concrete Design, slide 7-11
Example: Slab Design
20’ 20’
Plan View
aa
Section “a” - “a”
Design a one-way slab to resist a live load of 60 psf.Use f’c = 4000 psi and Grade 60 reinforcement.
Given:
20’ 20’
I0 I0a
cb
w = 1.0 k/ft
M, kip -ft
38.7 k-ft
25.0 k-ft
20’ 20’
I0 I0aa
cbbb
w = 1.0 k/ft
M, kip -ft
38.7 k-ft
25.0 k-ft
N
CEE 461: Reinforced Concrete Design, slide 7-12
Example: Slab Designper Table 9.5a: one-end discontinuous "10
24
)12(20h
ft/in
min == use a 10” slab
20’ 20’
I0 I
0a
cb
w = 1.0 k/ft
M, kip-ft
38.7 k-ft
25.0 k-ft
38.7(0.150+0.096) -12.5(0.112)=8.12 k-ft
50(0.150+0.096)=12.3 k-ftMoment Envelope
1.2D+1.6L on both spans:
1.2D+1.6L on left span, 0.9D on right span:
20’ 20’
I0 I
0a
cb
w = 1.0 k/ft
M, kip-ft
38.7 k-ft
25.0 k-ft
38.7(0.150+0.096) -12.5(0.112)=8.12 k-ft
50(0.150+0.096)=12.3 k-ftMoment Envelope
20’ 20’
I0 I
0aa
cbbb
w = 1.0 k/ft
M, kip-ft
38.7 k-ft
25.0 k-ft
38.7(0.150+0.096) -12.5(0.112)=8.12 k-ft
50(0.150+0.096)=12.3 k-ftMoment Envelope
1.2D+1.6L on both spans:
1.2D+1.6L on left span, 0.9D on right span:
psf125)pcf150(12
"10w
D ==
ftk
uD 150.0)125.0(2.1w!
==
ftk
uD 112.0)125.0(9.0w !==
ftk
uL 096.0)060.0(6.1w !==
or
Loadings:
CEE 461: Reinforced Concrete Design, slide 7-13
Example: Slab Design
use #5@12” top in EW (As/ft = 0.31in2)
ft/in
c
y
y
us
2
32.0)95.0)(0.9)(60(9.0
12x3.12
f'
f 59.01df
MA ==
!!"
#$$%
&'
='
()
Top Reinforcement:
0029.00.9x12
31.0 top ==!
975.04.0
(60)0029.059.01
f'
f 59.01
c
y=!
"
#$%
&'=!!
"
#$$%
&'
(
ok
ft/ins
2
21.0 )97.0)(0.9)(60(9.0
12x12.8A ==+
Bottom Reinforcement:"6.17"12x
21.0
31.0 s ==
use #5@16” bottom in EW(As/ft = 0.23in2)
CEE 461: Reinforced Concrete Design, slide 7-14
Example: Slab Designshrinkage and temperature reinforcement = 0.0018(12”)(10”) = 0.216 in2/ft
use #5@18” bottom or top in NS(As/ft = 0.21in2)
#5@16”#5@18”
#5@12”#3@18”#5@18”
Final Slab Design
10”
20’
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