16 Reinforced Concrete Design Design of Slabs - .xyzlibvolume3.xyz/.../designofslabs/designofslabstutorial2.pdf · 16 Types of Slabs Load Paths and Framing Concepts One-way Slabs
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1616
� Types of Slabs
� Load Paths and Framing Concepts
� One-way Slabs
� Two-way Slabs
Mongkol JIRAVACHARADET
Reinforced Concrete DesignReinforced Concrete Design
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Design of Slabs
Types of Slab
One-way slab Two-way slabOne-way slab
Flat plate slab Flat slab Grid slab
Think we’ll need some additional framing members???
Load Path / Framing Possibilities
Ln = 4.4 m
Ln = 8.2 m
Ln = 3.2 m
Ln = 3.6 m
Plan
Framing Concepts
Let’s use a simple example
for our discussion…
Think about relating it to your
design project.
Column spacing 8 m c-c
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…
But we need to support the
load from these new beams,
so we will need additional
supporting members
Framing Concepts
We will need to shorten up
the span with additional beams
We again assume that we’ll
have major girders running in
one direction in our one-way
system.
Framing Concepts
Now let’s go back through with
a slightly different load path.
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
Assume walls are not there for structural support, but
consider that the may help you in construction (forming)
Identify which column lines are best suited to having
major framing members (i.e. girders)
ExampleExample
Condo Floor PlanCondo Floor Plan
1.0 m
LS
Main reinforcement
������������� ��������������
One-way Slab
Design of one-way slabs is like design of parallel 1m beams.
Design of One-way Slab (L > 2S)
S
L 1 m
������������� ��������� 1 ���
S
w
Minimum Thickness (ACI)
Simply
supported
One end
continuous
Both ends
continuousCantilever
L/20 L/24 L/28 L/10
*multiplied by 0.4 + fy/7,000 for steel other than SD40
ACI Design ProvisionACI Design Provision
Shrinkage and temperature reinforcement
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
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 4,0000.0014
yf
×≥
Effect of column width
b b
L
b/212
2wL
2
wL
A′A
b/2
B
w
Moment at A’:
( )
+−−=
−
+−=
8412
2
2/
2212
22
22
wbwLbwL
bwbwLwL
( )
+−−=
−−=
1261212
222wbwLbwLbLw
M
If A’ and B’ are fiexed against rotation,
B′
12
2wL
2
wL
Typical reinforcement in a oneTypical reinforcement in a one--way slabway slab
Exterior span
Bottom bars
Top bars at
exterior beams
Top bars at
exterior beams
Interior span
Temperature bars
(a) Straight top and bottom bars
Exterior span
Bottom bars
Bent bar Bent bars
Interior span
Temperature bars
(b) Alternate straight and bent bars
3 @
8 m
= 24 m
4 @ 12 m = 48 m
G1
A AS1 S2 S3
Example: Design one-way slab as shown below to carry the live
load 500-kg/m2 fc’ = 210 kg/cm2, fy = 2,400 kg/cm
2
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
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
ksc 6.1755.101009.0
100176222=
××
×==
bd
MR u
n φ
0077.085.0
211
85.0'
'
=
−−=
c
n
y
c
f
R
f
fρ < 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 OK
Select RB9@0.18 (As = 3.53 cm2/m)
Detailing of one-way slab
3
1L
4
1L
8
1L
L1
Temp. steel
4.0 �.
.13 �.
1.0 �. 1.3 �.
RB9@0.10
RB9@0.18 RB9@0.07
4.0 �.
.13 �.
1.0 �. 1.3 �.RB9@0.07
� ���������
RB9@0.18 RB9@0.14 ������#$
Design of Two-way Slab (L < 2S)
L
S
Min. Thickness:
t ≥ 9 cm ≥ Perimeter/180 = 2(L+S)/180
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
45o 45o
45o 45o
A
D C
B
S
L
Short span (BC):
Floor load = w kg/sq.m
Tributary area = S2/4 sq.m
Load on beam = wS/4 wS/3 kg/m
Long span (AB): Span ratio m = S/L
Tributary area = SL/2 - S2/4 = sq.m
Load on beam kg/m
−m
mS 2
4
2
−2
3
3
2mwS
Moment Coefficient Method
S/4
S/4
S/2
L/4 L/2 L/4
%&'�(��
%&'��
%&'��
%&'�(�� %&'��%&'��
-Ms
-Ms
+Ms
+ML
-ML-ML
Middle strip moment: MM = CwS2
Column strip moment: MC = 2MM/3
������������������ ������ ( C )
"��#�����
$%�����$%��&��
�)��)��*+ � m
���'�&(,���-('-/���) �� �
-/��0�)�) �� �
,���-'��12��(��3)��
1.0 0.9 0.8 0.7 0.6 0.5
0.033-
0.025
0.040-
0.030
0.048-
0.036
0.055-
0.041
0.063-
0.047
0.083-
0.062
0.033-
0.025
���4�%�% �5 �6��67&�,���-('-/���) �� �
-/��0�)�) �� �
,���-'��12��(��3)��
0.0410.0210.031
0.0480.0240.036
0.0550.0270.041
0.0620.0310.047
0.0690.0350.052
0.0850.0420.064
0.0410.0210.031
���4�%�% �5 �� �6��,���-('-/���) �� �
-/��0�)�) �� �
,���-'��12��(��3)��
0.0490.0250.037
0.0570.0280.043
0.0640.0320.048
0.0710.0360.054
0.0780.0390.059
0.0900.0450.068
0.0490.0250.037
������������������ ������ ( C )
"��#�����
$%�����$%��&��
�)��)��*+ � m
���4�%�% �5 ����6��,���-('-/���) �� �
-/��0�)�) �� �
,���-'��12��(��3)��
1.0 0.9 0.8 0.7 0.6 0.5
0.0580.0290.044
0.0660.0330.050
0.0740.0370.056
0.0820.0410.062
0.0900.0450.068
0.0980.0490.074
0.0580.0290.044
���4�%�% �5 ��756��,���-('-/���) �� �
-/��0�)�) �� �
,���-'��12��(��3)��
-0.0330.050
-0.0380.057
-0.0430.064
-0.0470.072
-0.0530.080
-0.0550.083
-0.0330.050
Bar detailing in slab
L1 L2
L1/7 L1/4
L1/3
L2/4
L2/3����������������
Bar detailing in beam
L1 L2
L1/8
L1/3 L2/3
L1/8
������������� ������
��� ����
��� ��
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/cm
2
3.80
4.00
4.805.00
Floor plan
0.10
0.50
0.20 0.20
Cross section
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
As,min = 0.0018(100)(10) = 1.8 cm2/m
Short span
Moment coeff. C
-M(������������) +M -M(���������)
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
2
2 2
86,60017.11 kg/cm
0.9 100 7.5
un
MR
bdφ= = =
× ×
0.85 21 1 0.0045
0.85
c n
y c
f R
f fρ
′= − − = ′
As = 0.0045(100)(7.5) = 3.36 4�.2 > As,min
(� �63��(7�1���8� 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
2
2 2
66,30017.44 kg/cm
0.9 100 7.5
un
MR
bdφ= = =
× ×
0.85 21 1 0.0046
0.85
c n
y c
f R
f fρ
′= − − = ′
As = 0.0046(100)(6.5) = 2.97 4�.2 > As,min
(� �63��(7�1���8� DB10@0.20 (As=3.90 4�.2)
����� 8�9�"����8:��;� � �� ��7�
%��9� 9(2�: Vu = wuS/4 = (846)(4.0)/4
= 846 ��./�.
�<�(8��8'%��9� � ��2� φVc = 0.85(0.53) (100)(7.5)
= 5234 ��./�. OK
240
0.50
0.10
0.20 4.80 0.20
DB10@0.20 � ���������DB10@0.40 ������#$
DB10@0.40 ������#$
���6��&��
0.70
1.20
1.20
1.60
0.50
0.10
0.20 3.80 0.20
DB10@0.20 � ���������DB10@0.40 ������#$
DB10@0.40 ������#$
���6�����
0.95
1.300.55
0.95
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