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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Lecture-09
Analysis and Design of Two-way Slab System without Beams
(Flat Plate and Flat Slabs)By: Prof Dr. Qaisar Ali
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1
Civil Engineering Department
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Topics Addressed
Two Way Slabs
Behavior
Types
Analysis and Design Considerations
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 3
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Topics Addressed
Direct Design Method
Introduction
Limitations
Frame Analysis Steps for Flat Plates and Flat Slabs
Frame marking
Column and middle strips marking
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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
Two Way Slabs
Behavior
A slab having bending in both directions is called two-way slab (Long span/short span < 2).
25′ 25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 5
20′
20′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Behavior
Short direction moments in two-way slab.
Short
25′
25′
25′
25′
20′
20′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 6
Short Direction
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Behavior
Long direction moments in two-way slab.
25′
25′
25′
25′
20′
20′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 7
Long Direction
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Behavior: More Demand (Moment) in short directiondue to size of slab
∆central Strip = (5/384)wl4/EI
As these imaginary strips are part of monolithic slab, the deflection at anypoint, of the two orthogonal slab strips must be same:
∆a = ∆b
(5/384)w l 4/EI = (5/384)w l 4/EI
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
(5/384)wala4/EI = (5/384)wblb4/EI
wa/wb = lb4/la4 wa = wb (lb4/la4)
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
Two Way Slabs
Types
Wall Supported
Beam supported
Flat Plate
Flat slab
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Waffle Slab
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Analysis
Unlike beams and columns, slabs are two dimensionalmembers. Therefore their analysis except one-way slabsystems is relatively difficult.
Design
Once the analysis is done the design is carried out in the
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Once the analysis is done, the design is carried out in theusual manner. So no problem in design, problem is only inanalysis of slabs.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs
Approximate Analysis Methods of ACI
Slab System Applicable Analysis MethodsOne-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
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 11
q y
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
Direct Design Method
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 12
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
IntroductionIn DDM, frames rather than panels are analyzed as is done inanalysis of two way slabs with beams using ACI momentcoefficients.
Interior Frame
Exterior Frame
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 13
Interior Frame
Interior Frame
Exterior Frame
20′
20′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Introduction
25′
20′
25′ 25′ 25′
For complete analysis of slab system, frames areanalyzed in E-W and N-S directions.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
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E-W FramesN-S Frames
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Introduction
Though DDM is useful for analysis of slabs, speciallywithout beams, the method is applicable with somelimitations as discussed next.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 15
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Limitations (ACI 13.6.1)Uniformly distributed loading (L/D ≤ 2)Uniformly distributed loading (L/D ≤ 2)
ll11 ll11≥2≥2ll1 1 /3 /3 Three or more Three or more spans spans
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
ll22 Column offsetColumn offset≤ ≤ ll2 2 /10/10
Rectangular slab Rectangular slab panels (2 or less:1)panels (2 or less:1)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Limitations (ACI 13.6.1): Example
1515′′ 1515′′If ≥10If ≥10′′DDM APPLICABLE as 2/3 (15) = 10DDM APPLICABLE as 2/3 (15) = 10′′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
1515′′ 1515′′If <10If <10′′DDM NOT APPLICABLEDDM NOT APPLICABLE
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
Step No. 01: Slab is divided in frames (for E-W analysis,slab is divided into E-W frames and vice versa for N-Sanalysis).25′
20′
25′ 25′ 25′
Interior Frame
Exterior Frame
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 18
20′
20′
Interior Frame
Interior Frame
Exterior Frame
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
Step No. 01 (continued):
An interior frame
Interior Frame l
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 19
Interior Frame
l1
l2
20′
20′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
25′
20′
25′ 25′ 25′
Step No. 01 (continued):
Interior Frame lHalf width of panel
Marking an E-W Interior Frame
Panel Centerline
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
20
Interior Frame
l1
l2 on one side
Half width of panel on other side
Col Centerline
Panel Centerline
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
25′
20′
25′ 25′ 25′
Step No. 01 (continued):
Exterior Frame
l
Marking an E-W Exterior Frame
Note: For exterior framesl2 = Panel width/2 +h2/2
l2Half width of panel on one side
h2/2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
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l1
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
Step No. 02: A frame is divided further into strips known ascolumn and middle strips (Defined in ACI 13.2).
Column Strip: A column strip is a design strip with a width on eachside of a column centerline equal to 25 percent of l1 or l2, whicheveris less.
Middle Strip: Middle strips are design strips bounded by two column
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Middle Strip: Middle strips are design strips bounded by two columnstrips.
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25′
20′
25′ 25′ 25′
20′
20′
l2Column stripFull Middle strip
Half Middle strip
l2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
Step No. 02 (continued): Why a frame is divided into columnand middle strips?
Because the slab portion on the column centerline will offer moreresistance than the rest of the slab.
25′ 25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 23
20′
20′
20′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
Step No. 02 (continued):
25′
20′
25′ 25′ 25′
CS/2C S
M.S/2l2Half Column strip
a) Marking Column Strip
b) Middle Strip
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 24
20′
20′
CS/2 = Least of l1/4 or l2/4
CS/2 C.S
M.S/2
l1
ln
p
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Step No. 02 (continued): Frame and strips in 3D.
Direct Design Method
½ l t i idth l /4 (l ) /4 hi h i i i
(l2)B
½-Middle strip
½-Middle strip
Column strip
½ column strip width: l1/4 or (l2)B/4, whichever is minimum(l2)A
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
½ Middle strip
ln
½ column strip width: l1/4 or (l2)A/4, whichever is minimum
25
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis
25′
20′
25′ 25′ 25′
Step No. 02 (continued): For l1 = 25′ and l2 = 20′, CS and MSwidths are given as follows.
5′10′
5′l2Half Column strip
a) Marking Column Strip
b) Middle Strip
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
26
CS/2 = Least of l1/4 or l2/4
l2/4 = 20/4 = 5′
5′10′
5′
l1
ln
p
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 03: Calculate Static Moment (Mo) for interior span ofp ( o) pframe. 25′
20′
25′ 25′ 25′
MoMo =wu l2 ln
2
8l2
ln
Span of frame
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 04: Longitudinal Distribution of Static Moment (Mo).p g ( o)
M+
M − M −M − = 0.65Mo
M + = 0.35Mo
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
+ 0.35 o 20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.p p
M − = 0.65Mo
M + = 0.35Mo
0.60M +
0.75M − 0.75M −
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
+ 0.35 o 20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 03: Calculate Static Moment (Mo) for exterior span ofp ( o) pframe. 25′
20′
25′ 25′ 25′
MoMo =wu l2 ln
2
8l2
ln
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
30
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 04: Longitudinal distribution of static moment (Mo).p g ( o)
Mext − = 0.26Mo
M ext+ = 0.52Mo
Mext+
Mext− Mint−
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
ext+ 0.5 o
Mint- = 0.70Mo
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.p p
M ext+ = 0.52Mo
Mext − = 0.26Mo0.60Mext+
1.00Mext− 0.75Mint−
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
ext+ 0.5 o
Mint- = 0.70Mo
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.p p
M ext+ = 0.52Mo
Mint- = 0.70Mo
Mext − = 0.26Mo
25′
20′
25′ 25′ 25′
0.60Mext+
1.00Mext− 0.75Mint− 0.60M+
0.75M− 0.75M−
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M - = 0.65Mo
M + = 0.35Mo
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisExample 1: Analyze the flat slab shown below using DDM The slabExample 1: Analyze the flat slab shown below using DDM. The slabsupports a live load of 144 psf. All columns are 14″ square. Take fc′ =4 ksi and fy = 60 ksi.
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Step A: Sizes
ACI table 9 5 (c) is used for finding flat plate and flat slabACI table 9.5 (c) is used for finding flat plate and flat slabthickness.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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
Direct Design Method
Step A: Sizes
Exterior panel governs ThereforeExterior panel governs. Therefore,
hf = ln/30 = [{25 – (2 × 14/2)/12}/30] × 12 = 9.53″ (ACI minimum requirement)
Take hf = 10″
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 36
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Step B: Loads
Service Dead Load (D L) = γ l bhfService Dead Load (D.L) γslabhf
= 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
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
= 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
Direct Design Method
Frame AnalysisStep No. 01 : Marking E-W Interior Frame.
Interior Frame lHalf width of panel
Panel Centerline
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 38
Interior Frame
l1
l2 on one side
Half width of panel on other side
Col Centerline
Panel Centerline 20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 02 : Marking column and middle strips.
5′ 10′
5′l2Half Column strip
a) Marking Column Strip
b) Middle Strip
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 39
CS/2 = Least of l1/4 or l2/4
l2/4 = 20/4 = 5′
5′10′
5′
l1
ln
p
20′
20′
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 03: Static Moment (Mo) calculation.
25′
20′
25′ 25′ 25′
20′
Mo = wul2ln2/8
= 540 ft-kip
l2
Mo = 540 ft-k Mo = 540 ft-k
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′l1
ln =23.83′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 04: Longitudinal distribution of Static Moment (Mo).
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
25′
20′
25′ 25′ 25′
20′
Mext+
Mext− Mint-
M+
M − M −
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M − = 0.65Mo = 351M + = 0.35Mo = 189
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 04: Longitudinal distribution of Static Moment (Mo).
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
25′
20′
25′ 25′ 25′
20′
281
140 378
189
351 351
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M − = 0.65Mo = 351M + = 0.35Mo = 189
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
0.60M+
0.75M − 0.75M −0.60Mext+
1.00Mext − 0.75Mint−
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
M − = 0.65Mo = 351M + = 0.35Mo = 189
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.
0.60M+
0.75M − 0.75M −0.60Mext+
140 0.75Mint−
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M − = 0.65Mo = 351M + = 0.35Mo = 189
100 % of M ext- goes to column strip and remaining to middle strip
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
113
0.75M − 0.75M −168
140 0.75Mint −
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
M − = 0.65Mo = 351M + = 0.35Mo = 189
60 % of Mext+ & M + goes to column strip and remaining to middle strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame AnalysisStep No. 05: Lateral distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
113
263 263168
140 283
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
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
Direct Design Method
Frame Analysis (E-W Interior Frame)Step No. 05: Lateral distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
113
263 263168
140 283
Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378
M 0 65M 351
112/2 94/2 88/2 76/20 88/2
112/2 94/2 88/2 76/20 88/2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
M − = 0.65Mo = 351M + = 0.35Mo = 189
112/2 94/2 88/2 76/20 88/2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (E-W Interior Frame)Step No. 05: Lateral distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
263 263168
140
112/2 94/2 76/20 88/2
113
283
88/211.2 9.4 8.8 7.6
14.016.8
28.3 26.3
11.3
26.3
8.8 5′ half middle strip
5′ half middle strip
10′ column strip
Mu (per foot width)= M / strip width
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
5 half middle strip
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25
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (E-W Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.
25′
20′
25′ 25′ 25′
20′
Mext- = 0.26Mo = 74 Mext+ = 0.52Mo = 148 Mint- = 0.70Mo = 200
89Mo = 285.68 ft-kip
l2 =10.58′
60
0
74 150 140 140
59 50 46 40 46
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
M - = 0.65Mo = 186M + = 0.35Mo = 100
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (E-W Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.
15.94l2 =10.58′
10.75
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
Mo = 285.68 ft-kip
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M - = 0.65Mo = 186M + = 0.35Mo = 100
20′
20′
50
26
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (N-S Interior Frame)Step No. 05: Lateral distribution to column and middle strips.
Mext- = 0.26Mo = 110 Mext+ = 0.52Mo = 219 Mint- = 0.70Mo = 295
Mo = 421.5 ft-kip131 88/2
1100 0
22174/2
206 69/2
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M - = 0.65Mo = 274M + = 0.35Mo = 148
l2 =25′
88.859/2
206 69/2
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (N-S Interior Frame)Step No. 05: Lateral distribution to column and middle strips.
Mext- = 0.26Mo = 110 Mext+ = 0.52Mo = 219 Mint- = 0.70Mo = 295
Mo = 421.5 ft-kip13.1
11.00 0
22.14.9
20.6
8.8
4.6
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M - = 0.65Mo = 274M + = 0.35Mo = 148
l2 =25′
8.883.9
20.6 4.6
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (N-S Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.
Mext − = 0.26Mo = 58Mext+ = 0.52Mo = 114 Mint − = 0.70Mo = 154
Mo = 220.5 ft-kip69
58
115.5
107.3
0
45
38.5
35.75
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M − = 0.65Mo = 143M + = 0.35Mo = 77
l2 =13.08′
46.2
107.3
30.8
35.75
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (N-S Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.
Mext − = 0.26Mo = 58Mext+ = 0.50Mo = 110 Mint − = 0.70Mo = 154
Mo = 220.5 ft-kip12.32
10.4
20.69
19.2
0
6
5.13
4.76
25′
20′
25′ 25′ 25′
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
M − = 0.65Mo = 143M + = 0.35Mo = 77
l2 =13.08′
8.27 4.12
19.2 4.76
20′
20′
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (E-W Direction Moments)
25′ 25′ 25′ 25′
8.8
11.3
26.3 26.316.8
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′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
55
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Frame Analysis (N-S Direction moments)
25′ 25′ 25′ 25′
8.88
13.1
3.9
11.00 0
22.14.9
20.6
8.8
4.6
8.27
12.32
10.4
20.69
19.2
0
6
4.12
5.13
4.76
0
8.8
4.6
25
20′
25 25 25
20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20.6 4.619.2 4.76 4.6
20′
56
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Comparison with SAPEW direction moments from SAP
14.9 (15.94)
24 (13)
24 (26.8)
10.0 (10.75)
24 (25.1)
24 (25.1)
11.0(11.87)
16.0(16.8)
0(0)
20(14)
8(10)
28(28.3)
8(8)
10.5(11.3)
8(9.2)
28(26.3)
6.8(9.2)
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)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 57
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Comparison with SAPNS direction moments from SAP
4.5(0)
24 (10.4)
25(11)
4.5 (0)
25(11)
7(5.8)
2.6(5.13)
10(11.9)
20(20.69)
12(12.7)
22(22.1)
7(5.6)
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)
22(22.1)
22(20.6)
9(8.88)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 58
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design MethodExample 2
Analysis results of the slab shown below using DDM are presentednext. The slab supports a live load of 60 psf. Superimposed deadload is equal to 40 psf. All columns are 14″ square. Take fc′ = 3 ksiand fy = 40 ksi.
25′
20′
25′ 25′ 25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20′
20′
59
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 2Calculation summaryCalculation summary
Slab thickness hf = 10″
Factored load (wu) = 0.294 ksf
Column strip width = 5′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 60
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 2E-W Direction Moments (units: kip-ft)
33.9
88
204 204130
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′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
20
20′
61
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 2N-S Direction moments (units: kip-ft)
3 9
0 0
1714.9
33.9
26.5
35 8
53.2
44.3
89.5
83.1
0
17.7
11 9
14.9
13.9
0
28.5
22.8
25′
20′
25′ 25′ 25′
20′68 4
101
84.7
159
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
3.9
26.5
35.8 11.9
83.1 13.9
22.8 20
20′
68.4
159
62
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design MethodExample 3
Analysis results of the slab shown below using DDM are presentednext. The slab supports a live load of 60 psf. Superimposed deadload is equal to 40 psf. All columns are 12″ square. Take fc′ = 3 ksiand fy = 40 ksi.
20′
15′
20′ 20′ 20′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
15′
15′
63
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 3Calculation summaryCalculation summary
Slab thickness hf = 8″
Factored load (wu) = 0.264 ksf
Column strip width = 3.75′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 64
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 3E-W Direction Moments (units: kip-ft)
14.5
37.5
87.1 87.155.8
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′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
15
15′
65
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Example 3N-S Direction moments (units: kip-ft)
3 9
0 0
67.94.9
13.5
10.5
14 3
21.2
17.7
35.7
33.1
0
7.1
4 8
5.9
5.5
0
11.3
9.1
20′
15′
20′ 20′ 20′
15′27 2
40.4
33.6
63.1
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
3.9
10.5
14.3 4.8
33.1 5.5
9.1 15
15′
27.2
63.1
66
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Maximum spacing and minimum reinforcementMaximum spacing and minimum reinforcement requirement
Maximum spacing (ACI 13.3.2):
smax = 2 hf in each direction.
Minimum Reinforcement (ACI 7.12.2.1):
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Asmin = 0.0018 b hf for grade 60.
Asmin = 0.002 b hf for grade 40 and 50.
67
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs
At least 3/4” cover for fire or corrosion protection.
Slab
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
3/4″
Support
68
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs
In case of two way slabs supported on beams, short-direction barsare normally placed closer to the top or bottom surface of the slab,with the larger effective depth because of greater moment in shortdirection.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 69
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs
However in the case of flat plates/slabs, the long-direction negativeand positive bars, in both middle and column strips, are placedcloser to the top or bottom surface of the slab, respectively, with thelarger effective depth because of greater moment in long direction.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 70
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs
ACI 13.3.8.5 requires that all bottom bars within the column strip ineach direction be continuous or spliced with length equal to 1.0 ld , ormechanical or welded splices.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 71
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs
At least two of the column strip bars in each direction mustpass within the column core and must be anchored at exteriorsupports (ACI 13.3.8.5).
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 72
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Detailing of flexural reinforcement for columnsupported two-way slabs
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 73
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (Requirements of ACI Code)
Standard Bar Cut off Points (Practical (Recommendation):
For column and middle strips both
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 74
38
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Summary
Decide about sizes of slab and columns The slab depth canDecide about sizes of slab and columns. The slab depth canbe 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 anddimensions of all frames (exterior and interior)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
For a particular span of frame, find static moment (Mo =wul2ln2/8).
75
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Summary
Find longitudinal distribution of static moment:Find longitudinal distribution of static moment:
Exterior span (Mext - = 0.26Mo; M ext + = 0.52Mo; Mint - = 0.70Mo)
Interior span (Mint - = 0.65Mo; M int + = 0.35Mo)
Find lateral Distribution of each longitudinal moment:
100 % of Mext – goes to column strip
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
60 % of Mext + and Mint+ goes to column strip
75 % of Mint – goes to column strip
The remaining moments goes to middle strips
Design and apply reinforcement requirements (smax = 2hf)
76
39
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of Two Way Slab Systems for Shear
(Flat Plate and Flat Slabs)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 77
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
TopicsShear in Slabs Without Beams
Two-way shear (punch out shear)o ay s ea (pu c out s ea )
Shear strength of slab in punching shear
Various Design Options for Shear
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 78
40
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)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.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 79
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Two way shear (Punch out shear): Critical section
In shear design of beams, the critical section is taken at adistance “d” from the face of the support.
Beam
Shear crack
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
d
Beam
80
41
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Two way shear (Punch out shear): Critical section
In shear design of flat plates, the critical section is an areataken at a distance “d/2” from all face of the support.
Critical perimeterColumn
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Slab thickness (h)d/2
d/2d = h − cover
Tributary Area, At
Slab
81
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Two way shear (Punch out shear): Critical section
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 82
42
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Two way shear (Punch out shear): Critical perimeter, bo
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
bo = 2(c1+d)+2(c2+d)
83
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Two way shear (Punch out shear): Critical perimeter, bo
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
bo = 2(c1+d/2)+ (c2+d)
84
43
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Shear Strength of Slab in punching shear:
ΦVn = ΦVc + ΦVs
ΦVc is least of:
Φ4√ (fc′)bod
Φ(2 + 4/βc) √ (fc′)bod
Φ{(αsd/bo +2} √ (fc′)bod
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Φ{(αsd/bo 2} √ (fc )bod
βc = longer side of column/shorter side of column
αs = 40 for interior column, 30 for edge column, 20 for corner columns
85
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Shear Strength of Slab:
When ΦVc ≥ Vu (Φ = 0.75) O.K, Nothing required.
When ΦVc < Vu, then either increase ΦVc = Φ4√ (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)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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)
86
44
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Shear Strength of Slab:Shear Strength of Slab:
And/ or provide shear reinforcement (ΦVs) in the form of:
Integral beams
Bent Bars
Shear heads
Shear studs
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Shear studs
87
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Drop Panels (ACI 9.5.3.2 and 13.3.7.1):
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 88
45
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Column Capital:Column Capital:
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 89
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Shear in slab without beams
Minimum depth of slab in case of shear reinforcement to beprovided as integral beams or bent bars:
ACI 11.12.3 requires the slab effective depth d to be at least6 in., but not less than 16 times the diameter of the shearreinforcement.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
When bent bars and integral beams are to be used, ACI11.12.3.1 reduces ΦVc by 2
90
46
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.
Two Way Slabs (General)
p
Calculation of Punching shear demand (Vu):
Critical perimeter:d = h – 1 = 9″bo = 4(c+d)
= 4(14+9) = 92″
Tributary area (excluding area
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
y ( gof bo):At = (25×20) – (14+9)2/144
= 496.3 ft2
wu = 0.3804 kip/ft2
Vu = wu × At = 189 kip
91
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.
Two Way Slabs (General)
p
Calculation of Punching shear capacity (ΦVc):
√ (fc′)bod=√(4000)×92×9/1000=52 k
ΦVc is least of:
Φ4√ (fc′)bod = 156 k
Φ (2 + 4/β ) √ (f ′)b d = 234 k
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Φ (2 + 4/βc) √ (fc )bod = 234 k
Φ{(αsd/bo +2} √ (fc′)bod = 230 k
Therefore,
ΦVc = 156 k < Vu (190 k) , N.G
92
47
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 01): Drop panels
In drop panels, the slab thickness in the vicinity of the columns is increasedto increase the shear capacity (ΦVc = Φ4√ (fc′)bod) of concrete.
The increased thickness can be computed by equating Vu to ΦVc andsimplifying the resulting equation for “d” to calculate required “h”.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 93
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 01): Drop panels
25/6 = 4.25′
20/6 = 3.5′
Equate Vu to ΦVc:Vu = ΦVc
189 = 0.75 × 4 √ (fc′) × 92 × dd = 10.82″Therefore, h = d+1≈ 12″
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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′
94
48
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 02): Column Capitals
Occasionally, the top of the columns will be flared outward, as shown infigure. This is known as column capital.
This is done to provide a larger shear perimeter at the column and toreduce the clear span, ln, used in computing moments.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
ACI 6.4.6 requires that the capital concrete be placed at the same time asthe slab concrete. As a result, the floor forming becomes considerablymore complicated and expensive.
95
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 02): Column CapitalsEquate Vu to ΦVc:Vu = ΦVc
190 = 0.75 × 4 √ (fc′) × bo × 9bo = 111.26″
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Now,bo = 4 (c + d) 111.26 = 4(c + 9)Simplification gives,c = 18.8 ≈ 19″
96
49
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 02): Column Capitals
According to ACI code, θ < 45o
y = 2.5/ tanθ
c = 19″
2.5″
yθ
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Let θ = 30o, then y ≈ 4.35″
For θ = 20o, y ≈ 7″ 14″capital
column
y
97
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 03): Integral Beams
Vertical stirrups are used inconjunction with supplementaryhorizontal bars radiating outwardin two perpendicular directionsfrom the support to form what are
Vertical stirrupsFor 4 sides, total stirrup area is 4 times individual 2 legged stirrup area
lv
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
pptermed integral beams containedentirely within the slab thickness.
In such a way, critical perimeter isincreased
Horizontal barsIncreasedcriticalperimeter
98
50
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 03): Integral Beams
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
bo = 4R + 4c
99
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 03): Integral Beams
ΦVc = 156 kips
When integral beams are to be used, ACI 11.12.3 reduces ΦVc by 2.Therefore ΦVc = 156/2 = 78 kips
Using 3/8″ Φ, 2 legged (0.22 in2), 4 (side) = 4 × 0.22 = 0.88 in2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Spacing (s) = ΦAvfyd/ (Vu – ΦVc)
s = 0.75 × 0.88 × 60 × 9/ (190 – 78) = 3.18 ≈ 3″
Maximum spacing allowed d/2 = 6/2 = 3″ controls.
100
51
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 03): Integral BeamsFour #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)(lv – c1/2), therefore,
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
R = √ (2) x, and eqn. (1) becomes,
bo = 4√ (2) x + 4c1
bo = 4√ (2){(3/4)(lv – c1/2)} + 4c1
Or bo = 4.24lv – 2.12c1 + 4c1= 4.24lv + 1.88c1
Therefore lv ≈ 20″
101
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p
Design for shear (option 03): Integral Beams details.
lv = 20″ ≈ 24″ or 2′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 102
52
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
DDM Limitations:
For slabs with beams between supports on all sides (ACI 13.6.1.6):
Where,
0.2 ≤ α1l22/α2l1
2 ≤ 5.0
α = EcbIb / EcsIs
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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
103
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
DDM Limitations:
Explanation of Ib and Is:
α = EcbIb / EcsIs
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 104
53
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Example on α calculation
hf = 7″, hw = 18″, bw = 12″
Effective flange width
bw + 2hw = 48″, bw + 8hf = 68, 48″ governs
Ib = 33060 in4 OR,
20′20′
25′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
IT-section ≈ 2Irectangle section & IL-section ≈ 1.5Irectangle section
Ib = 2 × 12 × 243/12 = 27648 in4
Is = (10 + 10) × 12 × 73/12 = 6860 in4
α = Ib/Is = 33060/ 6860 = 4.82
105
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Longitudinal Distribution of Static Moments
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 106
54
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Longitudinal Distribution of Static Moments
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 107
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
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 toassign moments to column strip.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 108
55
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
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 toassign moments to column strip.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 109
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
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 toassign moments to column strip.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 110
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Middle Strip Moments
The remaining moments are assigned to middle strip in accordancewith ACI 13.6.6.
Beams between supports shall be proportioned to resist 85 percent ofcolumn strip moments if α1l2/l1 {Where l2 shall be taken as full span
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
column strip moments if α1l2/l1 {Where l2 shall be taken as full spanlength irrespective of frame location (exterior or interior)} is equal to orgreater than 1.0 (ACI 13.6.5.1).
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Graph A4
Lateral distribution of longitudinal moments can also be done usingGraph A.4 (Design of Concrete Structures, Nilson 13th Ed)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 112
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional
Direct Design Method
Requirements for Slab with Beams
Lateral Distribution of Longitudinal Moments
In graph A.4, l2 shall be
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
taken as full spanlength irrespective offrame location (exterioror interior).
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Example on graph A4:
Find the lateral distribution to column strip of positive andinterior negative moments using graph A4. Take
l2/l1 = 1.3
αl /l > 1
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
αl2/l1 > 1.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional
Direct Design Method
Requirements for Slab with Beams
Example on graph A4
l2/l1 = 1.3
αl /l > 1
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
αl2/l1 > 1
65 % of positive longitudinal moment will go to column strip
65 % of interior negative longitudinal moment will go to column strip
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor (βt)
In the presence of an exterior beam, all of the exterior negativefactored moment goes to the column strip, and none to the middlestrip, unless the beam torsional stiffness is high relative to the flexuralstiffness of the supported slab.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Torsional stiffness factor βt is the parameter accounting for this effect.βt reflects the relative restraint provided by the torsional resistance ofthe effective transverse edge beam.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor (βt)
For a considered frame, the transverse edge beam
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
provides restraint through its torsional resistance.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Torsional Stiffness Factor (βt)
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 118
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt:
Where walls are used as supports along column lines, they can beregarded as very stiff beams with an α1l2/l1 value greater than one.
Where the exterior support consists of a wall perpendicular to thedirection in which moments are being determined βt may be taken as
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
direction in which moments are being determined, βt may be taken aszero if the wall is of masonry without torsional resistance.
βt may be taken as 2.5 for a concrete wall with great torsionalresistance that is monolithic with the slab.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt:
βt can be calculated using the following formula:
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 120
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt:
Where,
Ecb = Modulus of elasticity of beam concrete;
Ecs = Modulus of elasticity of slab concrete
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
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 l1 and having width bounded by panel centerlines in l2 direction.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt:
C for βt determination can be calculated using the following formula.
y2y2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
x2x1
y1
x2
x1
y1122
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example): For determination of E-W frame exteriornegative moment distribution to column strip, find βt for beam marked. Takeslab depth = 7″ and Ecb = Ecs.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Exterior edge beam(12″ × 24″)
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example):
βt = EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
7″
hw ≤ 4hf = 17″
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
12″
24″
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example):
βt = EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
C = {1 – 0.63×12/24}{123 ×24/3} + {1 – 0.63×7/17}{73 ×17/3} = 10909 in4
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structuresx1 =12″
y1= 24″x2 = 7″
y2 = 17″
12
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example):
βt = EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
C = {1 – 0.63×12/17}{123 ×17/3} + {1 – 0.63×7/29}{73 ×29/3} = 8249 in4
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structuresx1 =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
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example):
βt = EcbC/(2EcsIs) = C/ (2Is)
Calculation of C:
Therefore, C = 10909 in4
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 127
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is)
Calculation of Is:
Is = bhf3/12 = (20 × 12) × 73/12 = 6860 in4
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
b
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional Requirements for Slab with Beams
Direct Design Method
Lateral Distribution of Longitudinal Moments
Determination of βt (Example):
βt = C/ (2Is)
= 10909/ (2 × 6860) = 0.80
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 129
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Additional
Direct Design Method
Requirements for Slab with Beams
Lateral Distribution ofLongitudinal Moments
Once βt is known,
βt = 0.8 90 % of exterior negative moment goes to column strip
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
exterior negative moment in column strip can be found. For,
l2/l1 = 1.3
αl2/l1 > 1 and βt = 0.8
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Minimum thickness for two way slab:
For 0 2 ≤ α ≤ 2:For 0.2 ≤ αm ≤ 2:
But not less than 5 in. fy in psi.
For αm > 2:
( )2.0536200,000
8.0
m
yn
−+
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=αβ
fl
h
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
But not less than 3.5 in. fy in psi.β936
200,0008.0 y
n
+
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
fl
h
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Two Way Slabs (General)Minimum thickness for two way Slab:
h = Minimum slab thickness without interior beams.
ln = length of clear span in direction that moments are beingdetermined, measured face-to-face of supports.
β = ratio of clear spans in long to short direction of two-wayslabs.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
αm = average value of α for all beams on edges of a panel.
For αm < 0.2, use the ACI table 9.5 (c).
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Direct Design Method
Special Reinforcement at exterior corner of SlabThe reinforcement at exterior ends of the slab shall be provided as per ACIThe reinforcement at exterior ends of the slab shall be provided as per ACI13.3.6 in top and bottom layers as shown.
The positive and negative reinforcement in any case, should be of a size andspacing equivalent to that required for the maximum positive moment (per footof width) in the panel.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 133
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
References
ACI 318-02
Design of Concrete Structures (Chapter 13) 13th Ed byDesign of Concrete Structures (Chapter 13), 13 Ed. by Nilson, Darwin and Dolan.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 134
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
The End
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 135