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7/9/2020
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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
drqaisarali@uetpeshawar.edu.pk
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
Topics Addressed
Introduction
Analysis and Design of slabs
Strip Method of Analysis for One-way Slabs
Basic Design Steps
Example
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
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
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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.
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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 opposing beams.
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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.
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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
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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.
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Introduction
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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.
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Introduction
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 10
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
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 11
Two-way Joist
Introduction
Jamia Haqqania Mosque at Akora Khattak
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
12
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-320: Reinforced Concrete Design-I 13
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
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 14
Analysis and Design of
One way Slabs
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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 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
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 16
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
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 17
One way & Two way Slabs
Case 1: One way Slab Case 2: Two way Slab
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.
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One way Slab
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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.
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One way Slab
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 opposing 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.
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Analysis of One-way Slab
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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 is equal or greater than 2.
Note: Strip method of analysis is not applicable to flat plates etc.,
even if bending is in one direction.
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Analysis of One-way Slab
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
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 23
Basic Design Steps
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). Fy should be in psi.
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)
24
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
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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)
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 26
Basic Design Steps
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.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 27
Basic Design Steps
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)
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 28
Basic Design Steps
Design:
Bar Placement
In the case of beam the No. of bars = n = As/Ab
While in slab we have
n = As/Ab ..................................(a)
For a unit width b = 12″ with center to center
spacing “s” then the above equation will become
b/s = As/Ab b = 12″
12/s = As/Ab
s(in) = (Ab/As) × 12
s(in) = (Ab/As) × 12
b
los
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 29
Basic Design Steps
Design:
Flexural Reinforcement (ACI 7.6.1.1):
Minimum Main reinforcement Requirement:
Asmin = 0.0018 Ag (Ag = Grass Area of concrete = bhf)
Maximum Spacing Requirement (ACI 7.7.2.3):
Main Reinforcement
Least of 3hf or 18″
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 30
Basic Design Steps
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.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 31
Basic Design Steps
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.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 32
Basic Design Steps
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.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 33
Basic Design Steps
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”
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.
34
Slab
9″
9″
12”h
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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″
35
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 36
Example
Slab Design
Solution:
Step No. 02: Loads
Self-weight of slab = (7.5 / 12) x 0.150 = 0.09375 ksf
SDL = 0.120 ksf (SDL = Superimposed dead load)
LL = 0.100 ksf (LL = Live Load)
wu = 1.2 (self-weight + SDL) + 1.6 LL
wu = 1.2 (0.09375 + 0.120) + (1.6 x 0.100)
wu = 0.4165 ksf
For 1 foot strip width, Wu = 0.4165 ksf x 1ft = 0.4165 kips/ft
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 37
Example
Slab Design
Solution:
Step No. 03: Analysis
For unit strip width, (01 foot of slab):
Mu = Wu lc/c2 / 8 = 0.4165 x 122 / 8
= 7.497 ft-kips/ft
Mu = 7.497 x 12 = 89.96 in-kip/ft
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
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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”
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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/ft
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:
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/ft
After Trails, As = 0.27 in2/ft
40
Example
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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.
41
Example
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 42
Ab = Area of bar in in2,
As = Design steel area in in2/ft
Example
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
Using #4 bar, with area Ab= 0.2 in2
s = 12n
= 12AsAb
= (Ab /As) x 12 = (0.2/ 0.27) x 12 = 8.88″ say 8.5″
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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
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
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″
44
Example
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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
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
Step No. 05: Drafting
46
Example
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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
47
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.
48
Example
Main reinforcement
Shrinkage reinforcementSpacer
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I
A
A
49
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 50
Drafting in 3D
Section B-B
Main Reinforcement
#4 bar @ 8.5″ c/c
Shrinkage Reinforcement
#4 bar @ 14.5″ c/c
B
B
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 51
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 318-19)
52
References
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