5/16/2019
1
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Lecture 07
Design of Reinforced
Concrete Slabs
By: Prof. Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 2
Topics Addressed
Introduction
Analysis and Design of slabs
Strip Method of Analysis for One-way Slabs
Basic Design Steps
Example
5/16/2019
2
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 3
Objectives
At the end of this lecture, students will be able to
Classify slab systems
Analyze one-way slabs using Strip Method of Analysis
for flexure
Design one-way slab system for flexure
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Introduction
In reinforced concrete construction, slabs are used to provide flat,
useful surfaces.
A reinforced concrete slab is a broad, flat plate, usually horizontal,
with top and bottom surfaces parallel or nearly so.
It may be supported by reinforced concrete beams (and is usually
cast monolithically with such beams), by masonry or reinforced
concrete walls, by structural steel members, directly by columns, or
continuously by the ground.
4
5/16/2019
3
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Introduction
Beam Supported Slabs
Slabs may be supported on two opposite sides only, as shown in
Figure, in which case the structural action of the slab is essentially
one-way, the loads being carried by the slab in the direction
perpendicular to the supporting beams.
5
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Introduction
Beam Supported Slabs
Slabs may be supported by beams on all four sides, as shown in
figure, in which case the structural action essentially becomes
two-way.
6
5/16/2019
4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Flat Plate
Concrete slabs in some cases may be carried directly by
columns. Punching shear is a typical problem in flat plates.
7
Introduction
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Flat Slab
Flat slab construction is also beamless but incorporates a thickened slab
region in the vicinity of the column and often employs column capital.
Drop Panel: Thick part of slab in the vicinity of columns.
Column Capital: Column head of increased size.
Punching shear problem encounter in such kinds of slabs, can be reduced by
introducing drop panel and column capital.
8
Introduction
5/16/2019
5
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 9
Rib
One-way Joist
⚫ Joist construction consists of a monolithic combination of
regularly spaced ribs and a top slab arranged to span in one
direction or two orthogonal directions.
Introduction
Peshawar University Auditorium
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 10
Two-way Joist
Introduction
Jamia Haqqania Mosque at Akora Khattak
5/16/2019
6
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Analysis and Design of Slabs
11
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.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 12
Analysis Methods
⚫ Analysis using computer software (FEA)
⚫ SAFE, SAP 2000, ETABS etc.
⚫ ACI Approximate Method of Analysis
⚫ Strip Method for one-way slabs
⚫ Moment Coefficient Method for two way slabs
⚫ Direct Design Method for two way slabs
Analysis and Design of Slabs
5/16/2019
7
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 13
Analysis and Design of
One way Slabs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 14
Definition of One way Slab
Case 1 (Slab supported on two opposing sides): If a slab is supported
on two opposing sides, bending in the slab will be produced only along
the side perpendicular to the direction of supports. In this case the slab
will be called as one way slab.
One way Slabs
5/16/2019
8
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 15
Definition of One way Slab
Case 2 (Slab supported on all sides): If a slab is supported on all sides
and the ratio of long to short side is equal to or greater than 2, major
bending in the slab will be produced along the short direction and the
slab will be called as one way slab. If the ratio is less than 2, bending
will occur in both directions and the slab will be called as two way slab.
One-Way Slab Two-Way Slab
One way Slabs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 16
One way & Two way Slabs
Case 1: One way Slab Case 2: Two way Slab
5/16/2019
9
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Reason for more Demand (Moment) in short direction
⚫ Δcentral Strip = (5/384)wl4/EI
⚫ Consider two strips along the long and short direction as shown in the
figure. As these imaginary strips are part of monolithic slab, the
deflection at any point, of the two orthogonal slab strips must be same:
⚫ Δa = Δb
(5/384)wala4/EI = (5/384)wblb
4/EI
⚫ wa/wb = lb4/la
4→ wa = wb (lb
4/la4)
⚫ Thus, larger share of load (Demand) is
taken by the short direction.
17
One way Slabs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Reason for more Demand (Moment) in short direction
⚫ Ma = wa la2/k ……… (1)
⚫ Mb = wb lb2/k ……… (2)
⚫ Substitute wa = wb (lb4/la
4) in Equ. 1
⚫ Ma = (wb lb4/la
4) la2/k
⚫ Finally, Ma = Mb x (lb/la)2
⚫ Thus, Bending in short direction is
more than Bending in long direction.
18
One way Slabs
5/16/2019
10
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Strip method of analysis:
⚫ For the purpose of analysis and design, a unit strip of one way slab, cut
out at right angles to the supporting beams, may be considered as a
rectangular beam of unit width, with a depth h and a span la as shown.
⚫ The method is called as strip method of analysis.
19
Analysis of One-way Slabs
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Strip method of analysis:
⚫ The strip method of analysis for slabs having bending in one direction is
applicable only when: a rectangular
⚫ Slab is supported on two opposing sides on stiff beams or walls,
⚫ Slab is supported on all sides on stiff beams or walls with ratio of long
to short side greater than 2.
⚫ Note: Strip method of analysis is not applicable to flat plates etc.,
even if bending is in one direction.
20
Analysis of One-way Slabs
5/16/2019
11
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Basic Design Steps
Basic Steps for Design
⚫ Selection of Size
⚫ Calculation of Loads
⚫ Analysis
⚫ Design
⚫ Drafting
21
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Sizes: ACI Table 7.3.1.1 gives the minimum one way slab thickness.
l = Span length, defined on the next slide.
For fy other than 60,000 psi, the expressions in Table 7.3.1.1 shall be multiplied by
(0.4 + fy/100,000).
22
Basic Design Steps
5/16/2019
12
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Sizes (Definition of Span Length, l)
23
1) l = ln ; for integral supports such as beams and columns with ln ≤ 10′
2) l = Minimum of [(ln +hf) or c/c distance] ; for non-integral supports such as walls
with any distance & for integral supports (beams and columns) with ln > 10′
• l (span length) is used in calculating depth of members.
• ln (clear span) is used for determining moments using ACI coefficients.
• lc/c is (center to center distance) is used for analysis of simply supported beam.
Beam
Slab
Wall
lc/clc/c
ln ln
hf
Basic Design Steps
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Loads:
One way slabs are usually designed for gravity loading. As slabs
are two dimensional elements, loads are calculated per unit area .
Ultimate Load is calculated as follows:
Wu = 1.2wD + 1.6wL
Wu = load per unit area (small letter)
Wu = load per unit length (capital letter)
24
Basic Design Steps
5/16/2019
13
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Analysis:
⚫ The analysis is carried out for ultimate load including self weight
obtained from size of the slab and the applied dead and live
loads.
⚫ The maximum bending moment value is used for flexural design.
25
Basic Design Steps
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
⚫ Capacity Demand
⚫ Capacity or Design Strength = Strength Reduction Factor (f)
Nominal Strength
⚫ Demand = Load Factor Service Load Effects
⚫ Bar spacing (in inches) = Ab/As × 12
(Ab = Area of bar in in2, As = Design or required steel in in2/ft)
26
Basic Design Steps
5/16/2019
14
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
⚫ Flexural Reinforcement (ACI 7.6.1.1):
⚫ Minimum Main reinforcement Requirement:
⚫ For Grade 40, Asmin = 0.0020 Ag (Ag = Grass Area of concrete = bhf)
⚫ For Grade 60, Asmin = 0.0018 Ag
⚫ Maximum Spacing Requirement (ACI 7.7.2.3):
⚫ Main Reinforcement
⚫ Least of 3hf or 18”
27
Basic Design Steps
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
Shrinkage Reinforcement:
⚫ Concrete shrinks as it dries out.
⚫ It is advisable to minimize such shrinkage by using concrete with the
smallest possible amounts of water and cement compatible with other
requirements, such as strength and workability, and by thorough
moist-curing of sufficient duration.
⚫ However, no matter what precautions are taken, a certain amount of
shrinkage is usually unavoidable.
28
Basic Design Steps
5/16/2019
15
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
Shrinkage Reinforcement:
⚫ Usually, however, slabs and other members are joined rigidly to other
parts of the structure and cannot contract freely.
⚫ This results in tension stresses known as shrinkage stresses.
⚫ Since concrete is weak in tension, these temperature and shrinkage
stresses are likely to result in cracking.
29
Basic Design Steps
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
Shrinkage Reinforcement:
⚫ In one-way slabs, the reinforcement provided for resisting the
bending moments has the desired effect of reducing shrinkage
and distributing cracks.
⚫ However, as contraction takes place equally in all directions, it is
necessary to provide special reinforcement for shrinkage and
temperature contraction in the direction perpendicular to the
main reinforcement.
⚫ This added steel is known as temperature or shrinkage
reinforcement, or distribution steel.
30
Basic Design Steps
5/16/2019
16
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design:
⚫ Minimum reinforcement Requirement for shrinkage and
Temperature reinforcement:
⚫ Same as main reinforcement minimum requirement (ACI 7.6.1.1)
⚫ Reinforcement is placed perpendicular to main steel to control
shrinkage and temperature cracking.
⚫ Maximum Spacing Requirement (ACI 7.7.2.4):
⚫ Shrinkage Reinforcement
⚫ Least of 5hf or 18”
31
Basic Design Steps
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Example
Design the given 12 feet simply supported slab carrying a uniform
service dead load (excluding self weight) of 120 psf and a uniform
service live load of 100 psf. Concrete compressive strength (fc′) = 3 ksi
and steel yield strength (fy) = 60 ksi.
32
Slab
12′
11.25′
9″
9″
12”h
5/16/2019
17
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 01: Sizes
⚫ From ACI table 7.3.1.1
⚫ For 12′ length, hf,min = l/20
⚫ l = span length, minimum of (ln + hf) or lc/c
Take ln = 11.25′ and hf = 6″
ln + hf = 11.25 + 6/12 = 11.75′ or lc/c = 12′
Therefore l = 11.75′
⚫ hf,min= 11.75 x 12/20 = 7.05″ rounded to 7.5″
33
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 02: Loads
⚫ Self weight of slab = (7.5 / 12) x 150 = 93.75 psf
⚫ SDL = 120 psf (SDL = Superimposed dead load)
⚫ LL = 100 psf (LL = Live Load)
⚫ wu = 1.2 (self weight + SDL) + 1.6 LL
⚫ wu = 1.2 (93.75 + 120) + (1.6 x 100)
⚫ wu = 416.5 psf
⚫ For 1 foot strip width, Wu = 416.5 psf x 1ft = 416.5 lb/ft
34
Example
5/16/2019
18
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 03: Analysis
⚫ For unit strip width, (01 foot of slab):
⚫ Mu = Wu l2 / 8 = 416.5 x 122 / 8
= 7497 ft-lb
= 7.497 ft-kip
⚫ Mu = 7.497 x 12 = 89.96 in-kip
35
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
d
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ hf = 7.5″ ; d = 7.5 – 1 = 6.5″
⚫ As = Mu/ {Φfy (d – a/2)}
⚫ Calculate “As” by trial and success method
36
Example
ACI 20.6.1.3.1 Clear cover for slab is 0.75″.
d= hf – y
If #4 (dia 0.5″) bar is to be used
y = 0.75 + 0.5/2
y = 1″
hf
12”
5/16/2019
19
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ First Trial:
⚫ Assume a = 0.2hf = 0.2 x 7.5 = 1.5″
⚫ As = 89.96 / [0.9 × 60 × {6..5 – (1.5/2)}] = 0.29 in2
⚫ a = Asfy/ (0.85fc′bw)
= 0.29 × 60/ (0.85 × 3 × 12) = 0.57 inches
37
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ Second Trial:
⚫ As = 89.96 / [0.9 × 60 × {6..5 – (0.57/2)}] = 0.27 in2
⚫ a = Asfy/ (0.85fc′bw)
= 0.27 × 60/ (0.85 × 3 × 12) = 0.53 inches
⚫ After Trails, As = 0.27 in2/ft
38
Example
5/16/2019
20
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ Minimum reinforcement check:
⚫ Asmin = 0.0018Ag = 0.0018 bhf
⚫ Asmin = 0.0018 x 12 x 7.5
= 0.162 in2
⚫ As the design As = 0.27 in2 > 0.162 in2
⚫ Therefore As is ok.
39
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ Bar Placement:
⚫ No of bars = n = As/Ab
⚫ Bar spacing, s (in inches) = b/n ; where b = 12 inches
⚫ s = 12n
= 12AsAb
= (Ab /As) x 12 = (0.2/ 0.27) x 12 = 8.88″ say 8.5″
40
Ab = Area of bar in in2,
As = Design steel area in in2/ft
Example
5/16/2019
21
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Main Reinforcement:
⚫ Maximum Spacing Requirement
⚫ Least of 3hf or 18″,
⚫ 3hf = 3 x 7.5 = 22.5″
⚫ Provided spacing is OK
41
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Shrinkage/ Reinforcement:
⚫ Asmin = 0.0018Ag = 0.0018 bhf
⚫ Asmin = 0.0018 x 12 x 7.5 = 0.162 in2
⚫ Using #4 bar, with area Ab= 0.2 in2
⚫ Spacing = (0.20 / 0.162) x 12 = 14.81″ say 14.5″
42
Example
5/16/2019
22
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Solution:
⚫ Step No. 04: Design
⚫ Shrinkage/ Reinforcement:
⚫ Maximum Spacing Requirement
⚫ Least of 5hf or 18″, 5hf = 5 x 7.5 = 37.5″
⚫ Provided spacing is OK
43
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Step No. 05: Drafting
44
Example
5/16/2019
23
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
⚫ Step No. 05: Drafting
45
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Slab Design
Placement of reinforcement:
Main reinforcing bars are placed in the direction of flexure stresses and
placed at the bottom(at the required clear cover) to maximize the “d”,
effective depth.
46
Example
5/16/2019
24
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
A
A
47
Drafting in 3D
Section A-A
Shrinkage Reinforcement
#4 bar @ 14.5″ c/c
Main Reinforcement
#4 bar @ 8.5″ c/c
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 48
Drafting in 3D
Section B-B
Main Reinforcement
#4 bar @ 8.5″ c/c
Shrinkage Reinforcement
#4 bar @ 14.5″ c/c
B
B
5/16/2019
25
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 49
Practice Example
Class Activity: Design 10 feet simply supported slab to carry a uniform
service dead load (excluding self weight) of 40 psf and a uniform
service live load of 120 psf. Concrete compressive strength (fc′) = 3 ksi
and steel yield strength (fy) = 40 ksi.
Slab
10′
9.25′
9″
9″
12″h
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Design of Concrete Structures 14th / 15th edition by Nilson, Darwin
and Dolan.
Building Code Requirements for Structural Concrete (ACI 318-14)
50
References