lecture 09 analysis and design of flat plate slabs_2011
TRANSCRIPT
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Lecture-09
Analysis and Design of Two-way Slab
System without Beams
(Flat Plate and Flat Slabs)
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Topics Addressed
Two Way Slabs
Behavior
Types
Analysis and Design Considerations
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Topics Addressed
Direct Design Method
Introduction
Limitations
Frame Analysis Steps for Flat Plates and Flat Slabs
Frame marking
Column and middle strips marking
Static moment calculation
Longitudinal distribution of static moment
Lateral distribution of longitudinal moment
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Behavior
A slab having bending in both directions is called two-way
slab (Long span/short span < 2).
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Behavior
Short direction moments in two-way slab.
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Short
Direction
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Behavior
Long direction moments in two-way slab.
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Long
Direction
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Behavior:More Demand (Moment) in short direction
due to size of slab
central Strip= (5/384)wl4/EI
As these imaginary strips are part of monolithic slab, the deflection at any
point, of the two orthogonal slab strips must be same:
a = b(5/384)wala
4/EI = (5/384)wblb4/EI
wa/wb= lb4/la
4 wa= wb(lb4/la
4)
Thus, larger share of load (demand) is taken by the shorter direction.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Types
Wall Supported
Beam supported
Flat Plate
Flat slab
Waffle Slab
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Analysis
Unlike beams and columns, slabs are two dimensional
members. Therefore their analysis except one-way slab
systems is relatively difficult.
Design
Once the analysis is done, the design is carried out in the
usual manner. So no problem in design, problem is only in
analysis of slabs.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Approximate Analysis Methods of ACI
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Slab System Applicable Analysis Methods
One-Way Slab Strip Method for one-way slabs
Two-way slabs supported on stiff
beams and walls
Moment Coefficient Method,
Direct Design Method,
Equivalent Frame Analysis Method
Two-way slabs with shallow
beams or without beams
Direct Design Method,
Equivalent Frame Analysis Method
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Introduction
In DDM, frames rather than panels are analyzed as is done in
analysis of two way slabs with beams using ACI moment
coefficients.
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Interior Frame
Exterior Frame
Interior Frame
Exterior Frame
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Introduction
For complete analysis of slab system, frames are
analyzed in E-W and N-S directions.
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E-W FramesN-S Frames
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Introduction
Though DDM is useful for analysis of slabs, specially
without beams, the method is applicable with some
limitations as discussed next.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Limitations (ACI 13.6.1)
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Uniformly distributed loading (L/D 2)
2
1 121/3Three or more spans
Column offset 2/10
Rectangular slab
panels (2 or less:1)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Limitations (ACI 13.6.1): Example
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15 15If 10
DDM APPLICABLE as 2/3 (15) = 10
15 15If
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 01 (continued):
An interior frame
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Interior Frame
l1
l2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis
Step No. 01 (continued):
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Interior Frame
l1
l2Half width of panel
on one side
Half width of panel
on other side
Marking an E-W Interior Frame
Col Centerline
Panel Centerline
Panel Centerline
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis
Step No. 01 (continued):
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Exterior Frame
l1
Marking an E-W Exterior Frame
Note: For exterior frames
l2= Panel width/2 +h2/2
l2Half width of panel
on one side
h2/2
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 02: A frame is divided further into strips known as
column and middle strips (Defined in ACI 13.2).
Column Strip:A column strip is a design strip with a width on each
side of a column centerline equal to 25 percent of l1or l2, whichever
is less.
Middle Strip: Middle strips are design strips bounded by two column
strips.
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l2Column strip
Full Middle strip
Half Middle strip
l2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 02 (continued): Why a frame is divided into column
and middle strips?
Because the slab portion on the column centerline will offer more
resistance than the rest of the slab.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 02 (continued):
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CS/2 = Least of l1/4 or l2/4
CS/2
CS/2C.S
M.S/2
M.S/2
l2
l1
ln
Half Column strip
a) Marking Column Strip
b) Middle Strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No. 02 (continued): Frame and strips in 3D.
Direct Design Method
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-Middle strip
-Middle strip
n
Column strip
column strip width: l1/4 or (l2)A/4, whichever is minimum
column strip width: l1/4 or (l2)B/4, whichever is minimum
(l2)A
(l2)B
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis
Step No. 02 (continued): For l1 = 25 and l2 = 20, CS and MS
widths are given as follows.
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CS/2 = Least of l1/4 or l2/4
l2/4 = 20/4 = 5
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10
5
5
l2
l1
ln
Half Column strip
a) Marking Column Strip
b) Middle Strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 03: Calculate Static Moment (Mo) for interior span of
frame.
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25 25 25
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MoMo=wu 2 n
2
8l2
ln
Span of frame
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 04: Longitudinal Distribution of Static Moment (Mo).
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M+
M M
M = 0.65Mo
M + = 0.35Mo
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 05: Lateral Distribution to column and middle strips.
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M = 0.65Mo
M + = 0.35Mo
0.60M +
0.75M 0.75M
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 03: Calculate Static Moment (Mo) for exterior span of
frame.
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25 25 25
20
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MoMo=wu 2 n
2
8l2
ln
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 04: Longitudinal distribution of static moment (Mo).
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Mext = 0.26Mo
M ext+ = 0.52Mo
Mint- = 0.70Mo
Mext+
Mext Mint
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 05: Lateral Distribution to column and middle strips.
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M ext+ = 0.52Mo
Mint- = 0.70Mo
Mext = 0.26Mo0.60Mext+
1.00Mext 0.75Mint
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 05: Lateral Distribution to column and middle strips.
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M ext+ = 0.52Mo
Mint- = 0.70Mo
M - = 0.65Mo
M + = 0.35Mo
Mext = 0.26Mo
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25 25 25
20
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0.60Mext+
1.00Mext 0.75Mint 0.60M+
0.75M 0.75M
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Example 1:Analyze the flat slab shown below using DDM. The slab
supports a live load of 144 psf. All columns are 14square. Take fc=
4 ksi and fy= 60 ksi.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Step A: Sizes
ACI table 9.5 (c) is used for finding flat plate and flat slab
thickness.
hmin= 5 inches (slabs without drop panels)
hmin= 4 inches (slabs with drop panels)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Step A: Sizes
Exterior panel governs. Therefore,
hf= ln/30 = [{25(2 14/2)/12}/30] 12 = 9.53 (ACI minimum
requirement)
Take hf= 10
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Step B: Loads
Service Dead Load (D.L) = slabhf
= 0.15 (10/12) = 0.125 ksf
Superimposed Dead Load (SDL) = Nil
Service Live Load (L.L) = 144 psf or 0.144 ksf
Factored Load (wu) = 1.2D.L + 1.6L.L
= 1.2 0.125 + 1.6 0.144 = 0.3804 ksf
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 01 : Marking E-W Interior Frame.
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Interior Frame
l1
l2Half width of panel
on one side
Half width of panel
on other side
Col Centerline
Panel Centerline
Panel Centerline
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 02 : Marking column and middle strips.
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CS/2 = Least of l1/4 or l2/4
l2/4 = 20/4 = 5
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10
5
5
l2
l1
ln
Half Column strip
a) Marking Column Strip
b) Middle Strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis
Step No. 03: Static Moment (Mo) calculation.
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Mo = wul2ln2/8
= 540 ft-kip
l2
l1
ln =23.83
Mo= 540 ft-k Mo= 540 ft-k
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis Step No. 04: Longitudinal distribution of Static Moment (Mo).
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Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
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Mext+
Mext Mint-
M+
M M
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis
Step No. 04: Longitudinal distribution of Static Moment (Mo).
42
Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
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20
281
140 378
189
351 351
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis Step No. 05: Lateral Distribution to column and middle strips.
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Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
0.60M+
0.75M 0.75M
0.60Mext+
1.00Mext 0.75Mint
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
0.60M+
Direct Design Method
Frame Analysis
Step No. 05: Lateral Distribution to column and middle strips.
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0.75M 0.75M
0.60Mext+
140 0.75Mint
Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
100 % of Mext-goes to
column strip and
remaining to middle strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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20
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Direct Design Method
Frame Analysis Step No. 05: Lateral Distribution to column and middle strips.
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113
0.75M 0.75M
168
140 0.75Mint
Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
60 % of Mext+& M+goes
to column strip and
remaining to middle strip
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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25 25 25
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Direct Design Method
Frame Analysis
Step No. 05: Lateral distribution to column and middle strips.
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113
263 263
168
140 283
Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
75 % of Mint-goes to
column strip and
remaining to middle strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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25 25 25
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20
Direct Design Method
Frame Analysis (E-W Interior Frame) Step No. 05: Lateral distribution to column and middle strips.
47
113
263 263
168
140 283
Mext = 0.26Mo = 140
Mext+ = 0.52Mo = 281
Mint = 0.70Mo = 378
M = 0.65Mo = 351M+ = 0.35Mo = 189
112/2 94/2 88/2 76/20 88/2
112/2 94/2 88/2 76/20 88/2
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
25
20
25 25 25
20
20
Direct Design Method
Frame Analysis (E-W Interior Frame)
Step No. 05: Lateral distribution to column and middle strips.
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263 263
168
140
112/2 94/2 76/20 88/2
113
283
88/211.2 9.4 8.8 7.6
14.0
16.8
28.3 26.3
11.3
26.3
8.8 5 half middle strip
5 half middle strip10 column strip
Mu(per foot width)
= M / strip width
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
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Direct Design Method
Frame Analysis (E-W Exterior Frame) Step No. 05: Lateral distribution to column and middle strips.
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Mext- = 0.26Mo = 74
Mext+ = 0.52Mo = 148
Mint- = 0.70Mo = 200
M - = 0.65Mo = 186M+ = 0.35Mo = 100
89Mo = 285.68 ft-kip
l2 =10.58
60
0
74 150 140 140
59 50 46 40 46
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (E-W Exterior Frame)
Step No. 05: Lateral distribution to column and middle strips.
50
15.94
l2 =10.5810.75
0
13.26 26.8 25.1 25.1
11.87 10 9.2 8 9.2
Mext- = 0.26Mo = 74
Mext+ = 0.52Mo = 148
Mint- = 0.70Mo = 200
M - = 0.65Mo = 186M+ = 0.35Mo = 100
Mo = 285.68 ft-kip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (N-S Interior Frame) Step No. 05: Lateral distribution to column and middle strips.
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Mext- = 0.26Mo = 110
Mext+ = 0.52Mo = 219
Mint- = 0.70Mo = 295
M - = 0.65Mo = 274M+ = 0.35Mo = 148
Mo = 421.5 ft-kip
l2 =25
88.8
131 88/2
59/2
1100 0
22174/2
206 69/2
206 69/2
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25 25 25
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (N-S Interior Frame)
Step No. 05: Lateral distribution to column and middle strips.
52
Mext- = 0.26Mo = 110
Mext+ = 0.52Mo = 219
Mint- = 0.70Mo = 295
M - = 0.65Mo = 274M+ = 0.35Mo = 148
Mo = 421.5 ft-kip
l2 =25
8.88
13.1
3.9
11.00 0
22.14.9
20.6
20.6
8.8
4.6
4.6
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (N-S Exterior Frame) Step No. 05: Lateral distribution to column and middle strips.
53
Mext = 0.26Mo = 58
Mext+ = 0.52Mo = 114
Mint = 0.70Mo = 154
M = 0.65Mo = 143M+ = 0.35Mo = 77
Mo = 220.5 ft-kip
l2 =13.08
46.2
69
58
115.5
107.3
107.3
0
45
30.8
38.5
35.75
35.75
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25 25 25
20
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (N-S Exterior Frame)
Step No. 05: Lateral distribution to column and middle strips.
54
Mext = 0.26Mo = 58
Mext+ = 0.50Mo = 110
Mint = 0.70Mo = 154
M = 0.65Mo = 143M+ = 0.35Mo = 77
Mo = 220.5 ft-kip
l2 =13.08
8.27
12.32
10.4
20.69
19.2
0
6
4.12
5.13
4.76
19.2 4.76
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (E-W Direction Moments)
55
8.8
11.3
26.3 26.3
16.8
14.0 28.3
9.4 7.60
11.2 8.80
15.94 10.75
0
13.26 26.8 25.1 25.1
11.87 10 9.2 8 9.2
25
20
25 25 25
20
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Frame Analysis (N-S Direction moments)
56
8.88
13.1
3.9
11.00 0
22.14.9
20.6
20.6
8.8
4.6
4.6
8.27
12.32
10.4
20.69
19.2
0
6
4.12
5.13
4.76
19.2 4.76
0
8.8
4.6
4.6
25
20
25 25 25
20
20
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Comparison with SAP
EW direction moments from SAP
57
11.0
(11.87)
14.9
(15.94)
16.0
(16.8)
0
(0)
24
(13)
20
(14)
8
(10)
24
(26.8)
28
(28.3)
8
(8)
10.0
(10.75)
10.5
(11.3)
8
(9.2)
24
(25.1)
28
(26.3)
6.8
(9.2)
24
(25.1)
28
(26.3)
16.0
(16.8)
20
(14)
28
(28)
10.5
(11.3)
28
(26.3)
28
(26.3)
12.5
(11.2)
0
(0)
9
(9.4)
8
(7.6)
9
(8.8)
7.7
(8.8)
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Comparison with SAP
NS direction moments from SAP
58
7
(5.8)
4.5
(0)
2.6
(5.13)
10
(11.9)
24
(10.4)
20
(20.69)
12
(12.7)
25
(11)
22
(22.1)
7
(5.6)
4.5
(0)
1.8
(5.13)
1.5
(4.76)
20
(19.2)
22
(20.6)
1
(4.6)
4.5
(4.12)
9
(8.27)
9
(8.88)
3.8
(4.12)
9
(12.7)
25
(11)
22
(22.1)
22
(20.6)
9
(8.88)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 2
Analysis results of the slab shown below using DDM are presented
next. The slab supports a live load of 60 psf. Superimposed dead
load is equal to 40 psf. All columns are 14square. Take fc= 3 ksi
and fy= 40 ksi.
59
25
20
25 25 25
20
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 2
Calculation summary
Slab thickness hf= 10
Factored load (wu) = 0.294 ksf
Column strip width = 5
60
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 2
E-W Direction Moments (units: kip-ft)
61
33.9
88
204 204
130
108.6 219
36.5 29.20
43.4 33.90
68 46.4
0
57 116 107 107
23 19.3 18 15.5 18
25
20
25 25 25
20
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 2
N-S Direction moments (units: kip-ft)
62
3.9
0 0
1714.9
33.9
26.5
26.5
35.8
53.2
44.3
89.5
83.1
0
17.7
11.9
14.9
13.9
83.1 13.9
0
28.5
22.8
25
20
25 25 25
20
20
68.4
101
84.7
159
159
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 3
Analysis results of the slab shown below using DDM are presented
next. The slab supports a live load of 60 psf. Superimposed dead
load is equal to 40 psf. All columns are 12square. Take fc= 3 ksi
and fy= 40 ksi.
63
20
15
20 20 20
15
15
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 3
Calculation summary
Slab thickness hf= 8
Factored load (wu) = 0.264 ksf
Column strip width = 3.75
64
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 3
E-W Direction Moments (units: kip-ft)
65
14.5
37.5
87.1 87.1
55.8
46.5 93.8
15.6 12.50
18.6 14.50
29.7 20
0
24.8 50 46.5 46.5
9.9 8.3 7.7 6.7 7.7
20
15
20 20 20
15
15
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Example 3
N-S Direction moments (units: kip-ft)
66
3.9
0 0
67.94.9
13.5
10.5
10.5
14.3
21.2
17.7
35.7
33.1
0
7.1
4.8
5.9
5.5
33.1 5.5
0
11.3
9.1
20
15
20 20 20
15
15
27.2
40.4
33.6
63.1
63.1
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code)
67
Maximum spacing and minimum reinforcement
requirement
Maximum spacing (ACI 13.3.2):
smax= 2 hf in each direction.
Minimum Reinforcement (ACI 7.12.2.1):
Asmin = 0.0018 b hffor grade 60.
Asmin = 0.002 b hf for grade 40 and 50.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code)
Detailing of flexural reinforcement for column
supported two-way slabs
At least 3/4cover for fire or corrosion protection.
68
3/4
Slab
Support
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code) Detailing of flexural reinforcement for column
supported two-way slabs
In case of two way slabs supported on beams, short-direction bars
are normally placed closer to the top or bottom surface of the slab,
with the larger effective depth because of greater moment in short
direction.
69
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code)
Detailing of flexural reinforcement for column
supported two-way slabs
However in the case of flat plates/slabs, the long-direction negative
and positive bars, in both middle and column strips, are placed
closer to the top or bottom surface of the slab, respectively, with the
larger effective depth because of greater moment in long direction.
70
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code) Detailing of flexural reinforcement for column
supported two-way slabs
ACI 13.3.8.5 requires that all bottom bars within the column strip in
each direction be continuous or spliced with length equal to 1.0 ld, or
mechanical or welded splices.
71
ld
Slab
Support
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code)
Detailing of flexural reinforcement for column
supported two-way slabs
At least two of the column strip bars in each direction must
pass within the column core and must be anchored at exterior
supports (ACI 13.3.8.5).
72
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code) Detailing of flexural reinforcement for column
supported two-way slabs
73
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
(Requirements of ACI Code) Standard Bar Cut off Points (Practical
Recommendation):
For column and middle strips both
74
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Summary
Decide about sizes of slab and columns. The slab depth can
be calculated from ACI table 9.5 (c).
Find Load on slab (wu= 1.2DL + 1.6LL)
On given column plan of building, decide about location and
dimensions of all frames (exterior and interior)
For a particular span of frame, find static moment (Mo =
wul2ln
2/8).
75
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Summary
Find longitudinal distribution of static moment:
Exterior span (Mext - = 0.26Mo; Mext += 0.52Mo; Mint -= 0.70Mo)
Interior span (Mint - = 0.65Mo; Mint += 0.35Mo)
Find lateral Distribution of each longitudinal moment:
100 % of Mextgoes to column strip
60 % of Mext +and Mint+goes to column strip
75 % of Mintgoes to column strip
The remaining moments goes to middle strips
Design and apply reinforcement requirements (smax= 2hf)
76
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Design of Two Way Slab Systems for
Shear
(Flat Plate and Flat Slabs)
77
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Topics
Shear in Slabs Without Beams
Two-way shear (punch out shear)
Shear strength of slab in punching shear
Various Design Options for Shear
Example
78
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
79
Shear in slab without beams
Two way shear (Punch out shear)
In addition to flexure, flat plates shall also be designed for two way shear
(punch out shear) stresses.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
80
Shear in slab without beams
Two way shear (Punch out shear): Critical section
In shear design of beams, the critical section is taken at a
distance dfrom the face of the support.
d
Beam
Shear
crack
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
81
Shear in slab without beams
Two way shear (Punch out shear): Critical section
In shear design of flat plates, the critical section is an area
taken at a distance d/2from all face of the support.
Slab thickness (h)
Critical perimeter
d/2d/2
d = h cover
Tributary Area, At
Column
Slab
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
82
Shear in slab without beams
Two way shear (Punch out shear): Critical section
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
85
Shear in slab without beams
Shear Strength of Slab in punching shear:
Vn= Vc+ Vs
Vcis least of:
4 (fc)bod
(2 + 4/c) (fc)bod
{(sd/bo+2} (fc)bod
c= longer side of column/shorter side of column
s= 40 for interior column, 30 for edge column, 20 for corner columns
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
86
Shear in slab without beams
Shear Strength of Slab:
When Vc Vu( = 0.75) O.K, Nothing required.
When Vc< Vu, then either increase Vc= 4 (fc)bod by:
Increasing d,depth of slab: This can be done by increasing the slab depth
as a whole or in the vicinity of column (Drop Panel)
Increasing bo, critical shear perimeter: This can be done by increasing
column size as a whole or by increasing size of column head (Column
capital)
Increasing fc (high Strength Concrete)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
87
Shear in slab without beams
Shear Strength of Slab:
And/ or provide shear reinforcement (Vs) in the form of:
Integral beams
Bent Bars
Shear heads
Shear studs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
88
Shear in slab without beams
Drop Panels (ACI 9.5.3.2 and 13.3.7.1):
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
89
Shear in slab without beams
Column Capital:
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
90
Shear in slab without beams
Minimum depth of slab in case of shear reinforcement to be
provided as integral beams or bent bars:
ACI 11.12.3 requires the slab effective depth d to be at least
6 in., but not less than 16 times the diameter of the shear
reinforcement.
When bent bars and integral beams are to be used, ACI
11.12.3.1 reduces Vcby 2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Calculation of Punching shear demand (Vu):
Two Way Slabs (General)
91
Critical perimeter:
d = h1 = 9bo= 4(c+d)
= 4(14+9) = 92
Tributary area (excluding area
of bo):
At= (2520)(14+9)2/144
= 496.3 ft2
wu= 0.3804 kip/ft2
Vu= wuAt= 189 kip
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Example: Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Calculation of Punching shear capacity (Vc):
Two Way Slabs (General)
92
(fc)bod=(4000)929/1000=52 k
Vcis least of:
4 (fc)bod = 156 k
(2 + 4/c) (fc)bod= 312 k
{(sd/bo+2} (fc)bod= 307 k
Therefore,
Vc= 156 k < Vu(190 k) , N.G
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
93
Example: Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 01): Drop panels
In drop panels, the slab thickness in the vicinity of the columns is increased
to increase the shear capacity (Vc= 4(fc)bod) of concrete.
The increased thickness can be computed by equating Vu to Vc and
simplifying the resulting equation for dto calculate required h.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
94
Example: Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 01): Drop panels
25/6 = 4.25
20/6 = 3.5
Equate Vuto Vc:
Vu= Vc
189 = 0.75 4 (fc) 92 d
d = 10.82
Therefore, h = d+1 12
This gives 2 drop panel.
According to ACI, minimum
thickness of drop panel = h/4 =
10/4 = 2.5, which governs.
Drop Panel dimensions:
25/6 4.25; 20/6 3.5
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
95
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 02): Column Capitals
Occasionally, the top of the columns will be flared outward, as shown in
figure. This is known as column capital.
This is done to provide a larger shear perimeter at the column and to
reduce the clear span,ln, used in computing moments.
ACI 6.4.6 requires that the capital concrete be placed at the same time as
the slab concrete. As a result, the floor forming becomes considerably
more complicated and expensive.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
96
Example: Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 02): Column Capitals
Equate Vuto Vc:
Vu= Vc
190 = 0.75 4 (fc) bo9
bo= 111.26
Now,
bo= 4 (c + d)
111.26 = 4(c + 9)
Simplification gives,
c = 18.8 19
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
97
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 02): Column Capitals
According to ACI code, < 45o
y = 2.5/ tan
Let = 30o, then y 4.35
For = 20o, y 7
c = 19
14
2.5
capital
column
y
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
98
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 03): Integral Beams
Vertical stirrups are used in
conjunction with supplementary
horizontal bars radiating outward
in two perpendicular directionsfrom the support to form what are
termed integral beams contained
entirely within the slab thickness.
In such a way, critical perimeter is
increased
Vertical stirrups
For 4 sides, total
stirrup area is 4
times individual 2
legged stirrup area
Horizontal bars
lv
Increased
critical
perimeter
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
99
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 03): Integral Beams
bo= 4R + 4c
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
100
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 03): Integral Beams
Vc= 156 kips
When integral beams are to be used, ACI 11.12.3 reduces Vc by 2.
Therefore Vc = 156/2 = 78 kips
Using 3/8,2 legged (0.22 in2), 4 (side) = 4 0.22 = 0.88 in2
Spacing (s) = Avfyd/ (VuVc)
s = 0.75 0.88 60 9/ (19078) = 3.18 3
Maximum spacing allowed d/2 = 6/2 = 3controls.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
101
Example:Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 03): Integral Beams
Four #5 bars are to be provided in each direction to hold the stirrups. We know minimum bo =
111.26
bo= 4R + 4c1........ (1)
R = (x2+ x2)
From figure, x = (3/4)(lvc1/2), therefore,
R = (2) x, and eqn. (1) becomes,
bo= 4(2) x + 4c1
bo= 4(2){(3/4)(lvc1/2)} + 4c1
Or bo= 4.24lv2.12c1+ 4c1= 4.24lv+ 1.88c1
Therefore lv20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
102
Example: Calculate the shear capacity of slab at 14column C1
of the 10flat plate shown.
Design for shear (option 03): Integral Beams details.
lv= 2024or 2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
DDM Limitations:
For slabs with beams between supports on all sides (ACI 13.6.1.6):
Where,
Ecb= Modulus of elasticity of beam concrete
Ecs= Modulus of elasticity of slab concrete
Ib= Moment of inertia of beam section
Is= Moment of inertia of slab section
Direct Design Method
103
0.2 a1l22/a2l1
25.0
a= EcbIb / EcsIs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
DDM Limitations:
Explanation of Iband Is:
Direct Design Method
104
= EcbIb/ EcsIs
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Example on calculation
hf= 7, hw= 18, bw= 12
Effective flange width
bw+ 2hw= 48, bw+ 8hf= 68, 48governs
Ib= 33060 in4 OR,
IT-section2Irectangle section& IL-section1.5Irectangle section
Ib= 2
12
243
/12 = 27648 in4
Is= (10 + 10) 12 73/12 = 6860 in4
= Ib/Is= 33060/ 6860 = 4.82
Direct Design Method
105
20
2025
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Longitudinal Distribution of Static Moments
Direct Design Method
106
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Longitudinal Distribution of Static Moments
Direct Design Method
107
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Column Strip Moments
ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to
assign moments to column strip.
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Column Strip Moments
ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to
assign moments to column strip.
Direct Design Method
109
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Column Strip Moments
ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to
assign moments to column strip.
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Middle Strip Moments
The remaining moments are assigned to middle strip in accordance
with ACI 13.6.6.
Beams between supports shall be proportioned to resist 85 percent of
column strip moments if 1l2/l1 {Where l2 shall be taken as full span
length irrespective of frame location (exterior or interior)} is equal to or
greater than 1.0 (ACI 13.6.5.1).
Direct Design Method
111
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Graph A4
Lateral distribution of longitudinal moments can also be done using
Graph A.4 (Design of Concrete Structures, Nilson 13thEd)
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional
Requirements for
Slab with Beams
Lateral Distribution of
Longitudinal Moments
In graph A.4, l2shall be
taken as full span
length irrespective of
frame location (exterior
or interior).
Direct Design Method
113
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Example on graph A4:
Find the lateral distribution to column strip of positive and
interior negative moments using graph A4. Take
l2/l1= 1.3
l2/l1> 1.
Direct Design Method
114
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional
Requirements for
Slab with Beams
Example on graph A4
l2/l1= 1.3
l2/l1> 1
Direct Design Method
115
65 % of positivelongitudinal moment
will go to column
strip
65 % of interiornegative longitudinal
moment will go to
column strip
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor(t)
In the presence of an exterior beam, all of the exterior negative
factored moment goes to the column strip, and none to the middle
strip, unless the beam torsional stiffness is high relative to the flexural
stiffness of the supported slab.
Torsional stiffness factor tis the parameter accounting for this effect.
t reflects the relative restraint provided by the torsional resistance of
the effective transverse edge beam.
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor (t)
Direct Design Method
117
For a considered frame, the
transverse edge beam
provides restraint through its
torsional resistance.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor (t)
Direct Design Method
118
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t:
Where walls are used as supports along column lines, they can be
regarded as very stiff beams with an 1l2/l1value greater than one.
Where the exterior support consists of a wall perpendicular to the
direction in which moments are being determined, t may be taken as
zero if the wall is of masonry without torsional resistance.
t may be taken as 2.5 for a concrete wall with great torsionalresistance that is monolithic with the slab.
Direct Design Method
119
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t:
tcan be calculated using the following formula:
Direct Design Method
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t:
Where,
Ecb= Modulus of elasticity of beam concrete;
Ecs= Modulus of elasticity of slab concrete
C = cross-sectional constant to define torsional properties
x = shorter overall dimension of rectangular part of cross section, in.
y = longer overall dimension of rectangular part of cross section, in.
Is= Moment of inertia of slab section spanning in direction l1and having
width bounded by panel centerlines in l2direction.
Direct Design Method
121
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t:
C for tdetermination can be calculated using the following formula.
Direct Design Method
122
x2x1
y1
y2
x2
x1
y1
y2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t (Example): For determination of E-W frame exterior
negative moment distribution to column strip, find tfor beam marked. Take
slab depth = 7and Ecb= Ecs.
Direct Design Method
123
Exterior edge beam
(12 24)
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example):
t= EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
Direct Design Method
124
12
24
7
hw 4hf= 17
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example):
t= EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
C = {10.6312/24}{12324/3} + {10.637/17}{7317/3} = 10909 in4
Direct Design Method
125x1=12
y1= 24x2= 7
y2= 17
12
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example):
t= EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
C = {10.6312/17}{12317/3} + {10.637/29}{7329/3} = 8249 in4
Direct Design Method
126x1=12
y1= 17
x2= 7
y2= 17 + 12 = 29
1
2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example):
t= EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
Therefore, C = 10909 in4
Direct Design Method
127
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example): t= EcbC/(2EcsIs) = C/ (2Is)
Calculation of Is:
Is= bhf3/12 = (20 12) 73/12 = 6860 in4
Direct Design Method
128
b
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
Determination of t(Example):
t= C/ (2Is)
= 10909/ (2 6860) = 0.80
Direct Design Method
129
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Additional
Requirements for
Slab with Beams
Lateral Distribution of
Longitudinal Moments
Once tis known,
exterior negative
moment in column strip
can be found. For,
l2/l1= 1.3
l2/l1> 1 and t= 0.8
Direct Design Method
130
t= 0.890 % of exterior negative moment goes to column strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
131
Minimum thickness for two way slab:
For 0.2 m 2:
But not less than 5 in. fyin psi.
For m> 2:
But not less than 3.5 in. fyin psi.
2.0536200,000
8.0
m
y
n
a
fl
h
936200,000
8.0 y
n
fl
h
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs (General)
132
Minimum thickness for two way Slab:
h = Minimum slab thickness without interior beams.
ln = length of clear span in direction that moments are being
determined, measured face-to-face of supports.
= ratio of clear spans in long to short direction of two-way
slabs.
m= average value of for all beams on edges of a panel.
For m< 0.2, use the ACI table 9.5 (c).
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Direct Design Method
Special Reinforcement at exterior corner of Slab
The reinforcement at exterior ends of the slab shall be provided as per ACI
13.3.6 in top and bottom layers as shown.
The positive and negative reinforcement in any case, should be of a size and
spacing equivalent to that required for the maximum positive moment (per foot
of width) in the panel.
133
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
References
ACI 318-02
Design of Concrete Structures (Chapter 13), 13thEd. by
Nilson, Darwin and Dolan.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011135
The End