16
Reinforced Concrete Design
Design of Slabs
Types of Slabs
Load Paths and Framing Concepts
One-way Slabs
Two-way Slabs
Mongkol JIRAVACHARADET
SURANAREE
UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING
SCHOOL OF CIVIL ENGINEERING
Types of Slab
One-way slab
Flat plate slab
One-way slab
Flat slab
Two-way slab
Grid slab
Load Path / Framing Possibilities
Ln = 3.2 m
Ln = 4.4 m
Ln = 3.6 m
Ln = 8.2 m
Think we’ll need some additional framing members???
Framing Concepts
Let’s use a simple example
for our discussion…
Column spacing 8 m c-c
Think about relating it to your
design project.
Plan
Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system
Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system
If we span between girders
with our slab, then we have
a load path, but if the spans
are too long…
Framing Concepts
We will need to shorten up
the span with additional beams
But we need to support the
load from these new beams,
so we will need additional
supporting members
Framing Concepts
Now let’s go back through with
a slightly different load path.
We again assume that we’ll
have major girders running in
one direction in our one-way
system.
This time, let’s think about
shortening up the slab span by
running beams into our girders.
Our one-way slab will transfer our load to the beams.
Two Load Path Options
Framing Concepts - Considerations
For your structure:
Look for a “natural” load path
Identify which column lines are best suited to having
major framing members (i.e. girders)
Assume walls are not there for structural support, but
consider that the may help you in construction (forming)
Example
Condo Floor Plan
One-way Slab
L
S
m
0
.
1
Main reinforcement
Design of one-way slabs is like design of parallel 1m beams.
Design of One-way Slab (L > 2S)
1 w
1m
L
S
S
Minimum Thickness (ACI)
Simply
supported
One end
continuous
Both ends
continuous
Cantilever
L/20
L/24
L/28
L/10
*multiplied
by 0.4 + fy/7,000 for steel other than SD40
ACI Design Provision
Shrinkage and temperature reinforcement
For structural slabs only; not intended for soil-supported slabs on grade
Ratio of reinforcement As to gross concrete area Ag : As/Ag
RB24 (fy = 2,400 ksc) . . . . . . . . . . . . . . . . . 0.0025
DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . . . . 0.0020
DB40 (fy = 4,000 ksc) . . . . . . . . . . . . . . . . . 0.0018
DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . . . . 0.0018 × 4,000 ≥ 0.0014
fy
Spacing ≤ 5 t ≤ 45 cm
Main Steel (short direction):
As ≥ ∅ 6 mm
Max. Spacing ≤ 3 t ≤ 45 cm
Min. Spacing ≥ f main steel ≥ 4/3 max agg. ≥ 2.5 cm
Effect of column width
b
A
b
A′
B′
B
L
Moment at A’:
w
2
wL2
12
wL
12
b/2
wL
2
b/2
wL
2
wL2 wL  b  w(b / 2 )
=−
+
 −
12
2 2
2
 wL2 wLb wb 2 

= − 
−
+
4
8 
 12
If A’ and B’ are fiexed against rotation,
2
 wL2 wLb wb 2 
w(L − b )

M =−
= − 
−
+
12
6
12 
 12
2
Typical reinforcement in a one-way slab
Top bars at
exterior beams
Bottom bars
Top bars at
exterior beams
Temperature bars
Exterior span
Interior span
(a) Straight top and bottom bars
Bent bar
Bottom bars
Bent bars
Temperature bars
Exterior span
(b) Alternate straight and bent bars
Interior span
Example: Design one-way slab as shown below to carry the live
G1
A
S1 S2 S3
A
3 @ 8 m = 24 m
load 500-kg/m2 fc’ = 210 kg/cm2, fy = 2,400 kg/cm2
0.4 + 2400/7000 = 0.74
min h = 400(0.74)/24 = 12.3 cm
USE h = 13 cm
DL = 0.13×2400 = 312kg/m2
4 @ 12 m = 48 m
wu = 1.4(312) + 1.7(500) = 1,286.8 kg/m2
clear span = 4 - 0.3 = 3.7 m
Mu = (1,286.8)(3.7)2/10 = 1,762 kg-m
ρmax = 0.75ρb = 0.75(0.0454) = 0.0341
USE RB9 with 2 cm covering: d = 13-2-0.45 = 10.55 cm
Mu
1762 × 100
Rn =
=
= 17.6 ksc
2
2
0.9 × 100 × 10.55
φ bd
0.85 f c'
ρ=
fy

2 Rn
1 − 1 −
'

0
.
85
f
c


 = 0.0077 < ρmax


OK
As = ρbd = 0.0077(100)(10.55) = 8.16 cm2/m
Select RB9@0.07 (As = 9.28 cm2/m)
Temp. steel = 0.0025(100)(13) = 3.25 < 9.28 cm2/m
Select RB9@0.18 (As = 3.53 cm2/m)
OK
L1
4
L1
3
Detailing of one-way slab
Temp. steel
L1
8
L1
RB9@0.18
RB9@0.07
.13 .
RB9@0.10
1.0 .
1.3 .
4.0 .
RB9@0.18
RB9@0.14 #$
.13 .
1.0 .
RB9@0.07
1.3 .
4.0 .
Design of Two-way Slab (L < 2S)
Min. Thickness:
t ≥ 9 cm ≥ Perimeter/180 = 2(L+S)/180
S
L
Reinforcement Steel:
As ≥ φ 6 mm ≥ Temp. steel
Max. Spacing ≤ 3 t ≤ 45 cm
Min. Spacing ≥ φ main steel ≥ 4/3 max agg. ≥ 2.5 cm
Load transfer from two-way Slab
D
C
45o
45o
Short span (BC):
Floor load = w kg/sq.m
S
Tributary area = S2/4 sq.m
45o
45o
A
B
Load on beam = wS/4
wS/3 kg/m
L
Long span (AB): Span ratio m = S/L
2
S
Tributary area = SL/2 - S2/4 =
4
Load on beam
wS  3 − m 2

3  2
 2 − m  sq.m


 m 
 kg/m


Moment Coefficient Method
%&'
%&'(
%&'
-Ms
%&'(
+Ms
+ML
-ML
%&'
%&'
S/4
-ML
S/4
-Ms
L/4
L/2
S/2
L/4
Middle strip moment: MM = CwS2
Column strip moment: MC = 2MM/3
( C )
$%
"#
1.0
'&(
,
-('-/)
-/0))
,
-'12(3)
4%%
56
67&
,
-('-/)
-/0))
,
-'12(3)
4%%
56
,
-('-/)
-/0))
,
-'12(3)
))*+ m
0.9 0.8 0.7
$%&
0.6
0.5
0.033 0.040 0.048 0.055 0.063 0.083 0.033
0.025 0.030 0.036 0.041 0.047 0.062 0.025
0.041 0.048 0.055 0.062 0.069 0.085 0.041
0.021 0.024 0.027 0.031 0.035 0.042 0.021
0.031 0.036 0.041 0.047 0.052 0.064 0.031
0.049 0.057 0.064 0.071 0.078 0.090 0.049
0.025 0.028 0.032 0.036 0.039 0.045 0.025
0.037 0.043 0.048 0.054 0.059 0.068 0.037
( C )
$%
"#
1.0
4%%
56
,
-('-/)
-/0))
,
-'12(3)
4%%
5756
,
-('-/)
-/0))
,
-'12(3)
))*+ m
0.9 0.8 0.7
$%&
0.6
0.5
0.058 0.066 0.074 0.082 0.090 0.098 0.058
0.029 0.033 0.037 0.041 0.045 0.049 0.029
0.044 0.050 0.056 0.062 0.068 0.074 0.044
0.033 0.038 0.043 0.047 0.053 0.055 0.033
0.050 0.057 0.064 0.072 0.080 0.083 0.050
Bar detailing in slab
L1/7
L1/3
L1/4
L1
L2/3
L2/4
L2
Bar detailing in beam
L1/3
L1/8
L2/3
L1/8
L1
L2
L/5
L/5
L = Example: Design two-way slab as shown below to carry the live
load 300-kg/m2 fc’ = 240 kg/cm2, fy = 2,400 kg/cm2
4.00
0.10
3.80
0.50
0.20
0.20
Cross section
5.00
4.80
Min h = 2(400+500)/180 = 10 cm
DL = 0.10(2,400) = 240 kg/m2
wu = 1.4(240)+1.7(300) = 846 kg/m2
m = 4.00/5.00 = 0.8
Floor plan
As,min = 0.0018(100)(10) = 1.8 cm2/m
Short span
-M()
+M
-M()
Moment coeff. C
0.032
0.048
0.064
Max. M = C w S 2 = 0.064 × 846 × 4.02 = 866 kg-m/1 m width
d = 10 - 2(covering) - 0.5(half of DB10) = 7.5 cm
Rn =
Mu
86, 600
2
=
=
17.11
kg/cm
φ bd 2 0.9 ×100 × 7.52
0.85 f c′ 
2 Rn 
ρ=
 1 − 1 −
 = 0.0045
fy 
0.85 f c′ 
As = 0.0045(100)(7.5) = 3.36 4.2 > As,min
(63
(718 DB10@0.20 (As=3.90 4.2)
Long span
Moment coeff. C
-M()
+M
-M()
0.025
0.037
0.049
Max. M = C w S 2 = 0.049 × 846 × 4.02 = 663 kg-m/1 m width
d = 10 - 2(covering) - 1.5(half of DB10) = 6.5 cm
Rn =
Mu
66,300
2
=
=
17.44
kg/cm
φ bd 2 0.9 ×100 × 7.52
0.85 f c′ 
2 Rn 
ρ=
 1 − 1 −
 = 0.0046
fy 
0.85 f c′ 
As = 0.0046(100)(6.5) = 2.97 4.2 > As,min
(63
(718 DB10@0.20 (As=3.90 4.2)
89"8:
;7
%
9
9(2:
Vu
= wuS/4 = (846)(4.0)/4
= 846 ./.
<(88'%
92 φVc = 0.85(0.53) 240 (100)(7.5)
= 5234 ./.
OK
DB10@0.40 #$
DB10@0.20 DB10@0.40 #$
0.10
0.50
0.95
1.30
0.55
0.95
3.80
0.20
6
0.20
DB10@0.40 #$
DB10@0.20 DB10@0.40 #$
0.10
0.50
0.70
1.20
0.20
1.20
1.60
4.80
6&
0.20