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Concrete Floor Systems
Analysis and Design of Slabs
Basic Design Steps
Example: Design of 90′ x 60′ Hall
References
At the end of this lecture, students will be able to
Recall basic design steps of shear and flexure
Classify Concrete Floor Systems
Choose different framing systems for structures.
Analyze and Design one-way slabs for flexure by
ACI Approximate method
Discuss effect of beam spacing on slab moments
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CONCRETE FLOOR
SYSTEMS
STRUCTURAL MEMBER
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Concrete Floor Systems
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Concrete Floor Systems
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Concrete Floor Systems
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Concrete Floor Systems
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Concrete Floor Systems
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Two possible arrangement of reinforcement
are shown a) straight bar b) bent bars are
used
Concrete Floor Systems
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Concrete Floor Systems
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Beam Supported Slabs
Beam Supported slab
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Beam supported slab (One way)
Beam supported slab
(One way)
Beam supported slab (Two way)
Concrete Floor Systems
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Punching shear is a typical
problem in flat plate
Flat Plate
Flat plate (Two way
slab)
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Flat Slab
Flat slab
(Two way
slab)
Drop Panel: Thick part of slab in the
vicinity of columns
Column Capital: Column head of
increased size
Punching shear can be reduced by
introducing drop panel
and column capital
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One way and two way joist (Beam used to support
floors or roofs) system (grid or waffle slab)
Joist: T-beams called joists are formed by creating
void spaces in what otherwise would be a solid
slab.
Peshawar University Auditorium
ribs
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One-way 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.
Peshawar University Auditorium
ribs
structural system
will be however
called as joist
system if the pan
width (clear spacing
between ribs) is less
than or equal to 30
inches (ACI 8.11.3).
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Two way joist system (grid or waffle slab)
A two-way joist system, or waffle slab, comprises evenly spaced concrete joists
spanning in both directions and a reinforced concrete slab cast integrally with
the joists.
joist
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Two way joist system (grid or waffle slab)
Solid slab
Two way joist
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One way and two way joist system (grid
or waffle slab)
One way joist slab system
Types of Slab
One way slab
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Two way Slab
Introduction
Types of Slab
One way slab
Two way Slab
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 RC beams, by masonry or reinforced walls, by structural
steel members, directly by columns or continuously by the ground
Flexural stresses are dominated in slab therefore slabs are called flexural
members
Shear forces are mostly controlled by thickness of slab
Torsional reinforcement are provided at the corner of slab (two way slab)
The reinforcement are provide parallel to surface of slab in both direction
depends on flexural behavior
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Torsional reinforcement are provided at the corner of slab (two way
slab)
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The slab which is supported only on two sides or supported on four sides with
longer to shorter side ratio equal to or greater than two i.e
aspect ratio M=L/S ≥ 2 (lb/la = ≥ 2)
In one way slab, loads travel along the shorter span of slab
Therefore main bars are provided along the deflection profile as shown in the
figure
In the other direction temperature/shrinkage/distribution reinforcement are
provided
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Temperature
reinforcement
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The slab which is supported on four sides, with longer to shorter side ratio
less than two i.e aspect ratio M=L/S < 2 (lb/la <2)
In two way slab, loads travel along the both span of slab
Therefore main bars are provided along both direction as shown in the
figure
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Design of one way slab
Flexural design of slab is same as design of rectangular beam with 12 inches
width (unit strip of 1 feet)
The minimum thickness of one ay slab is calculated as per ACI code, the table
is shown below
Simply supported (pin-pin)
One-end continuous (pin-fixed)
Both-end continuous (fixed-fixed)
Cantilever (fixed-free)
l/16
l/18.5
l/20
l/8
and Beam
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Design of slab
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.
Analysis Methods
Analysis using computer software (FEA)
SAP 2000, ETABS, SAFE 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
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Analysis and Design of One-Way Slab Systems
Strip Method of Analysis for One-way Slabs
For purposes 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 strip method of analysis and design of
slabs having bending in one direction is
applicable only when:
Slab is supported on only two sides on stiff
beams or walls,
Slab is supported on all sides on stiff beams
or walls with ratio of larger to smaller side
greater than 2.
Note: Not applicable to flat plates etc., even
if bending is primarily in one direction.
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Analysis and Design of One-Way Slab Systems
Strip Method of Analysis for One-way Slabs
Basic Steps for Structural Design
Step No. 01: Sizes:-Sizes of all structural and non structural elements are
decided.
Step No. 02: Loads:-Loads on structure are determined based on occupational
characteristics and functionality (refer Appendix C at the end of this lecture)
Step No. 03: Analysis:-Effect of loads are calculated on all structural elements
Step No. 04: Design:-Structural elements are designed for the respective load
effects following the code provisions.
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Basic Steps for Structural Design
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Basic Steps for Structural Design
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Basic Steps for Structural Design
Loads:
According to ACI 8.2 — Service loads shall be in accordance with the general
building code of which this code forms a part, with such live load reductions as
are permitted in the general building code.
BCP SP-2007 is General Building Code of Pakistan and it refers to ASCE 7-10
for minimum design loads for buildings and other structures.
One-way slabs are usually designed for gravity loading
(U = 1.2D + 1.6L).
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Basic Steps for Structural Design
Loads:
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Basic Steps for Structural Design
Analysis:
Chapter 8 of the ACI addresses provisions for the analysis of concrete
members.
According to ACI 8.3.3, as an alternate to frame analysis, ACI approximate
moments shall be permitted for analysis of one-way slabs with certain
restrictions, which are as follows:
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Uniaxial
Column Axial
Column
Biaxial
Column
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spandrel beam - when wall can not take weight of slab or floor , In such cases,
the beams are provided exterior walls at each floor level to support the wall load
and perhaps some roof load also , known as spandrel beam.
It is not necessary that under the beam always be wall, there can be column and
window but spandrel beam will provide only at exterior wall.
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interior support
Exterior support
Mid span
Mid span
interior support
Exterior support
Mid span
Mid span
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b
a
Shear for a slab
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Basic Steps for Structural Design
Design:
Capacity ≥ Demand
Capacity or Design Strength = Strength Reduction Factor (ϕ) x Nominal
Strength
Demand = Load Factor xService Load Effects
Bar spacing (in inches) = Ab/As × 12
(Ab = area of bar in in2, As = Design steel in in2/ft)
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Maximum spacing for main steel reinforcement in one way slab according to
ACI 7.6.5 is minimum of 3hf or 18’’
The shrinkage/ temperature reinforcement is calculated as per ACI code 7.12.2
as shown in below table
Maximum spacing for shrinkage steel in one way slab according to ACI 7.6.5 is
minimum of 5hf or 18’’
In slab Minimum Steel reinforcement as per ACI code = shrinkage/ temperature
reinforcement
Maximum steel reinforcement is same as for rectangular beam #3, #4 and #5
bars are commonly used in slabs
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Design of slab
The following Example illustrates the design of a one-way slab. It will be noted
that the code (7.7.1.c) COVER REQUIREMENT for reinforcement in slabs (#11 and
smaller bars) is ¾ in. clear, unless corrosion or fire protection requirements are
more severe
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Assume structural
configuration. Take time
to reach to a
reasonable arrangement
of beams, girders and
columns. It
depends on experience.
Several alternatives are
possible.
One way and Two way slab
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Assume structural
configuration. Take time to
reach to a reasonable
arrangement of beams,
girders and columns. It
depends on experience.
Several alternatives are
possible.
One way and Two way slab
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One way and Two way slab
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Wall width = 18 in, given
Assume beam width = 18 in
Slab
Beam
One way and Two way slab
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Two way Slab
100’x60’
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Two way Slab
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Two way Slab SLAB, BEAMS, COLUMN AND FOOTING
S1
S1 S1
S1
S2
S2
S2
S2
S3 S3
S4 S4
B1 B1
B2 B3 B2
B4
B5
B5
B4
C3
C1
C2
19’
20’
19’
20’
B2
DESIGN EXAMPLES
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Design slab and beams of a 90′ × 60′ Hall. The height of Hall is 20′. Concrete
compressive strength (fc′) = 3 ksi and steel yield strength (fy) = 40 ksi. Take 3″
mud layer and 2″ tile layer above slab. Take Live Load equal to 40 psf.
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Assume structural
configuration. Take time
to reach to a
reasonable arrangement
of beams, girders and
columns. It
depends on experience.
Several alternatives are
possible.
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Wall width = 18 in, given
Assume beam width = 18 in
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Mu = coefficient × wu × ln
2
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Or from table of areas of bars
In next slide
OR other way around bar spacing (in inches) = Ab/As × 12
Using # 3 bar (Ab = 0.11 and As = 0.144 so
Bar spacing (in inches) = Ab/As × 12 =9.166 c/c’’ #3@ 9.166 c/c’’
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Design of slab
A table of areas of bars in slabs such as Appendix A, Table A.6 is very useful in
such cases for selecting the specific bars to be used.
Temperature Steel
Primary Flexural
reinforcement
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Slab Design
Placement of positive reinforcement:
Positive reinforcing bars are placed in the direction of flexure stresses
and placed at the bottom (above the clear cover) to maximize the “d”,
effective depth.
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Slab Design
Placement of negative reinforcement:
Negative reinforcement bars are placed perpendicular to the direction of
support (beam in this case). At the support these rest on the reinforcement
of the beam.
beam
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Slab Design
Placement of negative reinforcement:
At the far end of bars, the chairs are provided to support the negative
reinforcement. As each bar will need a separate chair therefore to reduce the
number of chairs supporting bars are provided perpendicular to the direction of
negative reinforcement.
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Slab Design
Reinforcement at discontinuous support:
At the discontinuous end, the ACI code recommends to provide
reinforcement equal to 1/3 times the positive reinforcement provided at
the mid span.
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Slab Design
Step No 05: Drafting
Main reinforcement = #3 @ 9″ c/c (positive & negative)
Shrinkage reinforcement = #3 @ 9″ c/c
Supporting bars = #3 @ 18″ c/c
Chairs or supporting bars
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Beam Design
Step No 01: Sizes
Minimum thickness of beam (simply
supported) = hmin = l/16
l = clear span (ln) + depth of member (beam) ≤
c/c distance between supports
Let depth of beam = 5′
ln + depth of beam = 60′ + 5′ = 65′
c/c distance between beam supports = 60 + 2
× (9/12) = 61.5′
Therefore l = 61.5′
Depth (h) = (61.5/16)×(0.4+fy/100000)×12=
36.9″ (Minimum by ACI 9.3.1.1).
Take h = 5′ = 60″
d = h – 3 = 57″
bw = 18″ (assumed)
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Beam Design
Step No 02: Loads
Load on beam will be equal to
Factored load on beam from slab +
factored self weight of beam web
Factored load on slab = 0. 214 ksf
Load on beam from slab = 0. 214 ksf x 10
= 2.14 k/ft
Factored Self load of beam web =
=1.2 x (54 × 18/144) × 0.15 =1.215 k/ft
Total load on beam = 2.14 + 1.215 = 3.355 k/ft
10’
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Shear Force & Bending Moment Diagrams
Beam Design
Step No 03: Analysis
Vu = 87.23 kip
Mu = 19034 in-kip
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Beam Design
Step No 04: Design
Design for flexure
Step (a): According to ACI 6.3.2.1, beff for T-beam is minimum of:
16hf + bw = 16 × 6 + 18 =114″
(clear length of beam)/4 =(60′/4) ×12 + 18 =198″
clear spacing between beams + bw =8.5′ × 12 + 18 =120″
So beff = 114″
Step (b): Check if beam is to be designed as rectangular beam or T-beam.
Assume a = hf = 6″ and calculate As:
As =Mu/ {Φfy (d–a/2)} =19034/ {0.9 × 40 × (57–6/2)} = 9.79 in2
Re-calculate “a”:
a =Asfy/ (0.85fc′beff) =9.79 × 40/ (0.85 × 3 × 114) = 1.34″ < hf
Therefore design beam as rectangular beam.
After trials As = 9.38 in2 (Asmax = 20.83 in2 ;Asmin = 5.13 in2)
Therefore As = 9.38 in2 (12 #8 bars)
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Beam Design
(Asmax = 20.83
in2 ;Asmin = 5.13
in2)
Therefore As =
9.38 in2 (12 #8
bars)
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Beam Design
Step No 04: Design
Design for flexure
Skin Reinforcement : (ACI 9.7.2.3)
As the effective depth d of a beam is greater than 36 inches, longitudinal skin
reinforcement is required as per ACI 9.7.2.3.
Askin, = Main flexural reinforcement/2 = 9.60/2 = 4.8 in2
Range up to which skin reinforcement is provided:
d/2 = 56.625/2 = 28.3125″
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Beam Design
Step No 04: Design
Design for flexure
Skin Reinforcement
For #8 bar used in skin reinforcement,
ssk (skin reinforcement spacing)
is least of:
d/6 = 56.625/6 = 9.44″, 12″, or
1000Ab/(d – 30) = 1000×0.80/(56.625 – 30) = 30.05″
Therefore ssk = 9.44″ ≈ 9″
With this spacing, 3 bars on each face are required.
And for # 8 bar, the total area of skin reinforcement is:
Askin = 6 × 0.80 = 4.8 in2
skin reinforcement
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Beam Design
Step No 04: Design
Design for Shear
Vu = 87.23 kip
ΦVc = Φ2 √ fc′bwd = (0.75 × 2 × √3000 × 18 × 56.625)/1000 = 83.74 kip
ΦVc < Vu (Shear reinforcement is required)
sd = ΦAvfyd/(Vu – ΦVc)
Using #3, 2 legged stirrups with Av = 0.11 × 2 =0.22 in2}
sd = 0.75 × 0.22 × 40 × 56.625/(87.23 – 83.74) = 107″
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Beam Design
Step No 04: Design
Design for Shear
Maximum spacing and minimum reinforcement requirement as permitted by ACI
9.7.6.2.2 and 10.6.2.2 shall be minimum of:
Avfy/(50bw) =0.22 × 40000/(50 × 18) ≈ 9.5″
d/2 =56.625/2 =28.3″
24″
Avfy/ 0.75 √fc’ bw = 0.22 × 40000/ {(0.75 × √3000 × 18} = 11.90″
Therefore, smax = 9.5″
ΦVc /2 = 83.74/2 = 41.87 kips at a distance of 17.5 ft from face of the support.
Therefore no reinforcemnt is required in this zone, however, we will provide #3,
2-legged vertical stirrups at 12 in. c/c
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Beam Design
Step No 04: Design
Design for Shear
Other checks:
Check for depth of beam:
ΦVs ≤ Φ8 √fc’ bwd (ACI 22.5.1.2)
Φ8 √fc’ bwd = 0.75 × 8 × √3000 × 18 × 56.625/1000 = 334.96 k
ΦVs = (ΦAvfyd)/smax
= (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.3 k < 334.96 k, O.K.
So depth is O.K. If not, increase depth of beam.
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Beam Design
Step No 04: Design
Design for Shear
Other checks:
Check if “ ΦVs ≤ Φ4 √ fc′bwd ” (ACI 10.7.6.5.2):
If “ΦVs ≤ Φ4 √ fc′ bwd”, the maximum spacing (smax) is O.K.
Otherwise reduce spacing by one half.
Φ4√ fc′bwd = 0.75 × 4 × √3000 × 18 × 56.625/1000= 167.47 k
ΦVs = (ΦAvfyd)/sd
= (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.33 k < 167.47 k, O.K.
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Beam Design
Step No 05: Drafting
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In this design example, the beams were supported on walls.
This was done to simplify analysis.
For practical reasons, however, the beams must be supported
on columns and hence the structural analysis will be that of a
frame rather than simply supported beam.
In the subsequent slides, the analysis and design results for
beams supported on columns have been provided.
Design slab, beams, columns, and footing of a 90′ × 60′ Hall. The height of Hall
is 20′. Concrete compressive strength (fc′) = 3 ksi and steel yield strength (fy)
= 40 ksi. Take 3″ mud layer and 2″ tile layer above slab. Take Live Load equal
to 40 psf.
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Frame Analysis
3D model of the hall showing beams supported on columns.
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Frame Analysis
A 2D frame can be detached from a 3D system in the following manner
Column size = 18’’ × 18’’
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Frame Analysis
Various methods can be used for frame analysis. Using moment distribution
method, the following results can be Obtained:
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Frame Analysis
Slab Design
Slab design will remain the same as in case of beams supported on
walls.
Main reinforcement = #3 @ 9″ c/c (positive & negative)
Shrinkage reinforcement = #3 @ 9″ c/c
Supporting bars = #3 @ 18″ c/c
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Frame Analysis
Beam Design
Beam flexure design will be as follows:
Mu (+ve) = 17627 in-kips
Mu (-ve) = 1407 in-kips
As (+ve) = 8.68 in2
Use 6 #8 in 1st layer & 2 #8 + 4 #7 bars in 2nd layer)
As = (8)(0.80) + (4)(0.60) = 8.80 in2 (As,max = 0.0203bd = 20.83 in2 OK)
As (-ve) = 0.69 in2 (As,min = 0.005bd = 5.13 in2, so As,min governs)
Use 7 #8 bars (5 bars in 1st layer and 2 bars in 2nd layer)
As = (7)(0.80) = 5.60 in2
Beam shear design will be same as in previous case.
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Beam Design
(As,max = 0.0203bd =
20.83 in2 )
As (-ve) = 0.69 in2
(As,min = 0.005bd =
5.13 in2)
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Frame Analysis
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DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH
UNIAXIAL BENDING using DESIGN AIDS
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The ACI column interaction diagrams are used to design or analyze
columns for different situations. In order to correctly use these diagrams, it
is necessary to compute the value of γ (gamma), which is equal to the
distance from the center of the bars on one side of the column to the center
of the bars on the other side of the column divided by h,
Column cross sections for normalized
interaction curves in Appendix A,
DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH
UNIAXIAL BENDING using DESIGN AIDS
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Frame Analysis
Column Design: Using ACI Design Aids
Main Reinforcement Design
Size:
18 in. × 18 in.
Loads:
Pu = 103.17 kips
Mu = 1407 in-kips
Calculate the ratio γ , for 2.5 in. cover (c): g = (h – 2c) / h
= (18 – 5)/18 = 0.72
Calculate Kn, Kn = Pu/(Ø fc′Ag) = 103.17/(0.65 × 3 × 324) = 0.16
Calculate Rn, Rn = Mu/(Øfc′Agh) = 1407/(0.65 × 3 × 324 × 18) = 0.12
fc′ = 3 ksi, fy = 60 ksi
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Column Design
Main Reinforcement Design
For given material strength,
the column strength interaction
diagram gives the
following reinforcement ratio:
ρ = 0.01
Ast = 0.01 × 324 = 3.24 in2
Using 8 #6 bars
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Column Design
Tie Bars:
Using 3/8″ Φ (#3) tie bars for 3/4″ Φ (#6) main bars (ACI 9.7.6.4.2),
Spacing for Tie bars according to ACI 9.7.6.4.3 is minimum of:
16 × dia of main bar =16 × 6/8 =12″ c/c
48 × dia of tie bar = 48 × (3/8) =18″ c/c
Least column dimension =18″ c/c
Finally use #3, tie bars @ 12″ c/c
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Column Design
12″ c/c
12″ c/c
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Footing Design
Isolated column footing; square or rectangular
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Footing Design
Data Given:
Column size = 18″ × 18″
fc′ =3 ksi
fy = 40 ksi
qa = 2.204 k/ft2
Factored load on column = 103.17 kips (Reaction at the support)
Service load on column = 81.87 kips (Reaction at the support due to service
load)
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Footing Design
Sizes:
Assume h = 15 in.
davg = h – clear cover – one bar dia
= 15 – 3 – 1(for #8 bar) = 11 in.
Assume depth of the base of footing from ground level (z) = 5′
Weight of fill and concrete footing, W= γfill(z - h) + γch
=100 × (5 – 1.25) +150 × (1.25) = 562.5 psf = 0.5625 ksf
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Footing Design
Sizes:
Effective bearing capacity, qe = qa – W
= 2.204 – 0.5625 = 1.642 ksf
Bearing area, Areq = Service Load/ qe
= 81.87/1.642 = 49.86 ft2
Areq = B x B = 49.86 ft2 => B = 7 ft.
Critical Perimeter, bo = 4 x (c + davg)
= 4 × (18 + 11) =116 in
7 ft
7 ft
5.5 ft
5.5 ft
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Footing Design
Loads:
qu (bearing pressure for strength design of footing):
qu = factored load on column / Areq
= 103.17 / (7 × 7) = 2.105 ksf
Analysis:
Beam Shear
Beam shear failure at a distance “d” from face of column
27.014k
2.105*(7/2-1.5/2)-11/12)*7 quB
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Footing Design
Analysis:
Punching shear:
Punching shear failure at a
distance “d/2” from face of column
Vup = quB2 – qu(c + davg)2
Vup = (2.105 × 49) –2.105 × {(18+11)/12)}2
103.145-12.29
= 90.85 kip
qu(c + davg)2
quB2
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Footing Design
Analysis:
Flexural Analysis:
Flexural Failure at face of column
Mu = quBk2/2
k = (B – c)/2 = (7 x 12 –18)/2
= 33 in = 2.75´
Mu = 2.105 × 7 × 2.75 × 2.75/2
= 55.72 ft-k = 668.60 in-kip
quB
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Footing Design
Design:
Design for Punching Shear:
Vup = 90.85 kip
Punching shear capacity (ΦVcp)
= Φ4√fc’ bodavg
= 0.75 × 4 × 3000 × 116 × 11/1000
= 209.66 k >Vup, O.K
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Footing Design
Design:
Design for Flexure:
Mu = 668.60 kip-in
a = 0.2davg = 0.2 × 11 = 2.2″
As = Mu/ {Φfy(davg – a/2)} = 668.60/ {0.9 × 40 × (11 – 2.2/2)} = 1.87in2
a = Asfy/ (0.85fc′B) = 1.83 × 40/ (0.85 × 3 × 7 × 12) = 0.35″
After trials, As = 1.71 in2 (Asmin = 0.005Bdavg = 4.62 in2 so Asmin governs)
Now, the spacing can be calculated as follows:
Using #8 bars: No. of bars = 4.62/0.79 ≈ 6 bars.
Spacing = 6.5 × 12 /5 = 15 in. c/c
Hence 6 bars can be provided in the foundation if they are placed 15 in. c/c
(Max. spacing should not exceed 3h or 18 in.)
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Drafting
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Drafting
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IMPORTANT TABLES
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IMPORTANT TABLES
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IMPORTANT TABLES
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IMPORTANT TABLES
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Bar spacing (in inches) = Ab/As × 12
Using # 4 bar (Ab = 0.2 and As = 0.27 so
Bar spacing (in inches) = Ab/As × 12 =8.8 c/c’’ #4@ 8.8 c/c’’
#4@ 14 c/c’’
#4@ 24 c/c’’
#4@ 14 c/c’’
#4@ 17 c/c’’
#4@ 8.8 c/c’’
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Two possible arrangement of reinforcement
are shown a) straight bar b) bent bars are
used
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Two way Slab
The slab which is supported on four sides, with longer to shorter side ratio
less than two i.e aspect ratio M=L/S < 2 (lb/la <2)
In two way slab, loads travel along the both span of slab
Therefore main bars are provided along both direction as shown in the
figure
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Two way Slab
100’x60’
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Two way Slab
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Two way Slab SLAB BEAMS COLUMN AND FOOTING
S1
S1 S1
S1
S2
S2
S2
S2
S3 S3
S4 S4
B1 B1
B2 B3 B2
B4
B5
B5
B4
C3
C1
C2
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1 2 3 4 5 6 7 8 9
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1 2 3 4 5 6 7 8 9
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COEFFICIENTS FOR NEGATIVE
MOMENTS IN SLABs
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COEFFICIENTS FOR DEAD
LOAD POSITIVE ATIVE
MOMENTS IN SLABs
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COEFFICIENTS FOR LIVE
LOAD POSITIVE MOMENTS IN
SLABs
RCD
RATIO OF LOAD W IN la AND lb
DIRECTIONS FOR SHEAR IN
SLAB AND LOAD ON
SUPPORTS
RCD
RCD
RCD
la
Mb (-ve)
Ma (-ve)
Ma (-ve)
M
b
(+ve)
Ma (+ve)
lb
Mb (-ve)
RCD
Bar spacing (in inches) = Ab/As × 12
Using # 4 bar (Ab = 0.2 and As = 0.288 so
Bar spacing (in inches) = Ab/As × 12 =7 c/c’’ #4@ 7 c/c’’
#4@ 7 c/c’’
#4@ 5 c/c’’
RCD
#4@ 12 c/c’’
#3@ 6c/c’’
RCD
#3@ 4.5c/c’’
#3@ 12c/c’’
RCD
la
Mb (-ve)
Ma (-ve)
Ma (-ve)
M
b
(+ve)
Ma (+ve)
lb
Mb (-ve)
RCD
RCD
RCD
RCD
RCD
RCD
RCD

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Slab_design_RCD_II_Lec_1_beam_column_arrangment.pptx

  • 1. Concrete Floor Systems Analysis and Design of Slabs Basic Design Steps Example: Design of 90′ x 60′ Hall References At the end of this lecture, students will be able to Recall basic design steps of shear and flexure Classify Concrete Floor Systems Choose different framing systems for structures. Analyze and Design one-way slabs for flexure by ACI Approximate method Discuss effect of beam spacing on slab moments RCD
  • 8. RCD Two possible arrangement of reinforcement are shown a) straight bar b) bent bars are used
  • 11. Beam Supported slab RCD Beam supported slab (One way) Beam supported slab (One way) Beam supported slab (Two way)
  • 12. Concrete Floor Systems RCD Punching shear is a typical problem in flat plate Flat Plate Flat plate (Two way slab)
  • 13. RCD Flat Slab Flat slab (Two way slab) Drop Panel: Thick part of slab in the vicinity of columns Column Capital: Column head of increased size Punching shear can be reduced by introducing drop panel and column capital
  • 14. RCD One way and two way joist (Beam used to support floors or roofs) system (grid or waffle slab) Joist: T-beams called joists are formed by creating void spaces in what otherwise would be a solid slab. Peshawar University Auditorium ribs
  • 15. RCD One-way 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. Peshawar University Auditorium ribs structural system will be however called as joist system if the pan width (clear spacing between ribs) is less than or equal to 30 inches (ACI 8.11.3).
  • 16. RCD Two way joist system (grid or waffle slab) A two-way joist system, or waffle slab, comprises evenly spaced concrete joists spanning in both directions and a reinforced concrete slab cast integrally with the joists. joist
  • 17. RCD Two way joist system (grid or waffle slab) Solid slab
  • 18. Two way joist RCD One way and two way joist system (grid or waffle slab) One way joist slab system
  • 19. Types of Slab One way slab RCD Two way Slab
  • 20. Introduction Types of Slab One way slab Two way Slab 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 RC beams, by masonry or reinforced walls, by structural steel members, directly by columns or continuously by the ground Flexural stresses are dominated in slab therefore slabs are called flexural members Shear forces are mostly controlled by thickness of slab Torsional reinforcement are provided at the corner of slab (two way slab) The reinforcement are provide parallel to surface of slab in both direction depends on flexural behavior RCD
  • 21. Torsional reinforcement are provided at the corner of slab (two way slab) RCD
  • 22. The slab which is supported only on two sides or supported on four sides with longer to shorter side ratio equal to or greater than two i.e aspect ratio M=L/S ≥ 2 (lb/la = ≥ 2) In one way slab, loads travel along the shorter span of slab Therefore main bars are provided along the deflection profile as shown in the figure In the other direction temperature/shrinkage/distribution reinforcement are provided RCD Temperature reinforcement
  • 23. RCD The slab which is supported on four sides, with longer to shorter side ratio less than two i.e aspect ratio M=L/S < 2 (lb/la <2) In two way slab, loads travel along the both span of slab Therefore main bars are provided along both direction as shown in the figure
  • 24. RCD Design of one way slab Flexural design of slab is same as design of rectangular beam with 12 inches width (unit strip of 1 feet) The minimum thickness of one ay slab is calculated as per ACI code, the table is shown below Simply supported (pin-pin) One-end continuous (pin-fixed) Both-end continuous (fixed-fixed) Cantilever (fixed-free) l/16 l/18.5 l/20 l/8 and Beam
  • 25. RCD Design of slab 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. Analysis Methods Analysis using computer software (FEA) SAP 2000, ETABS, SAFE 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
  • 26. RCD Analysis and Design of One-Way Slab Systems Strip Method of Analysis for One-way Slabs For purposes 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 strip method of analysis and design of slabs having bending in one direction is applicable only when: Slab is supported on only two sides on stiff beams or walls, Slab is supported on all sides on stiff beams or walls with ratio of larger to smaller side greater than 2. Note: Not applicable to flat plates etc., even if bending is primarily in one direction.
  • 27. RCD Analysis and Design of One-Way Slab Systems Strip Method of Analysis for One-way Slabs Basic Steps for Structural Design Step No. 01: Sizes:-Sizes of all structural and non structural elements are decided. Step No. 02: Loads:-Loads on structure are determined based on occupational characteristics and functionality (refer Appendix C at the end of this lecture) Step No. 03: Analysis:-Effect of loads are calculated on all structural elements Step No. 04: Design:-Structural elements are designed for the respective load effects following the code provisions.
  • 28. RCD Basic Steps for Structural Design
  • 29. RCD Basic Steps for Structural Design
  • 30. RCD Basic Steps for Structural Design Loads: According to ACI 8.2 — Service loads shall be in accordance with the general building code of which this code forms a part, with such live load reductions as are permitted in the general building code. BCP SP-2007 is General Building Code of Pakistan and it refers to ASCE 7-10 for minimum design loads for buildings and other structures. One-way slabs are usually designed for gravity loading (U = 1.2D + 1.6L).
  • 31. RCD Basic Steps for Structural Design Loads:
  • 32. RCD Basic Steps for Structural Design Analysis: Chapter 8 of the ACI addresses provisions for the analysis of concrete members. According to ACI 8.3.3, as an alternate to frame analysis, ACI approximate moments shall be permitted for analysis of one-way slabs with certain restrictions, which are as follows:
  • 33. RCD
  • 34. RCD
  • 35. RCD
  • 37. RCD spandrel beam - when wall can not take weight of slab or floor , In such cases, the beams are provided exterior walls at each floor level to support the wall load and perhaps some roof load also , known as spandrel beam. It is not necessary that under the beam always be wall, there can be column and window but spandrel beam will provide only at exterior wall.
  • 38. RCD interior support Exterior support Mid span Mid span interior support Exterior support Mid span Mid span
  • 40. RCD Basic Steps for Structural Design Design: Capacity ≥ Demand Capacity or Design Strength = Strength Reduction Factor (ϕ) x Nominal Strength Demand = Load Factor xService Load Effects Bar spacing (in inches) = Ab/As × 12 (Ab = area of bar in in2, As = Design steel in in2/ft)
  • 41. RCD Maximum spacing for main steel reinforcement in one way slab according to ACI 7.6.5 is minimum of 3hf or 18’’ The shrinkage/ temperature reinforcement is calculated as per ACI code 7.12.2 as shown in below table Maximum spacing for shrinkage steel in one way slab according to ACI 7.6.5 is minimum of 5hf or 18’’ In slab Minimum Steel reinforcement as per ACI code = shrinkage/ temperature reinforcement Maximum steel reinforcement is same as for rectangular beam #3, #4 and #5 bars are commonly used in slabs
  • 42. RCD Design of slab The following Example illustrates the design of a one-way slab. It will be noted that the code (7.7.1.c) COVER REQUIREMENT for reinforcement in slabs (#11 and smaller bars) is ¾ in. clear, unless corrosion or fire protection requirements are more severe
  • 43. RCD Assume structural configuration. Take time to reach to a reasonable arrangement of beams, girders and columns. It depends on experience. Several alternatives are possible. One way and Two way slab
  • 44. RCD Assume structural configuration. Take time to reach to a reasonable arrangement of beams, girders and columns. It depends on experience. Several alternatives are possible. One way and Two way slab
  • 45. RCD One way and Two way slab
  • 46. RCD Wall width = 18 in, given Assume beam width = 18 in Slab Beam One way and Two way slab
  • 49. RCD Two way Slab SLAB, BEAMS, COLUMN AND FOOTING S1 S1 S1 S1 S2 S2 S2 S2 S3 S3 S4 S4 B1 B1 B2 B3 B2 B4 B5 B5 B4 C3 C1 C2 19’ 20’ 19’ 20’ B2
  • 51. Design slab and beams of a 90′ × 60′ Hall. The height of Hall is 20′. Concrete compressive strength (fc′) = 3 ksi and steel yield strength (fy) = 40 ksi. Take 3″ mud layer and 2″ tile layer above slab. Take Live Load equal to 40 psf. RCD
  • 52. RCD Assume structural configuration. Take time to reach to a reasonable arrangement of beams, girders and columns. It depends on experience. Several alternatives are possible.
  • 53. RCD Wall width = 18 in, given Assume beam width = 18 in
  • 54. RCD
  • 55. RCD
  • 56. RCD Mu = coefficient × wu × ln 2
  • 57. RCD Or from table of areas of bars In next slide OR other way around bar spacing (in inches) = Ab/As × 12 Using # 3 bar (Ab = 0.11 and As = 0.144 so Bar spacing (in inches) = Ab/As × 12 =9.166 c/c’’ #3@ 9.166 c/c’’
  • 58. RCD Design of slab A table of areas of bars in slabs such as Appendix A, Table A.6 is very useful in such cases for selecting the specific bars to be used. Temperature Steel Primary Flexural reinforcement
  • 59. RCD
  • 60. RCD
  • 61. RCD Slab Design Placement of positive reinforcement: Positive reinforcing bars are placed in the direction of flexure stresses and placed at the bottom (above the clear cover) to maximize the “d”, effective depth.
  • 62. RCD Slab Design Placement of negative reinforcement: Negative reinforcement bars are placed perpendicular to the direction of support (beam in this case). At the support these rest on the reinforcement of the beam. beam
  • 63. RCD Slab Design Placement of negative reinforcement: At the far end of bars, the chairs are provided to support the negative reinforcement. As each bar will need a separate chair therefore to reduce the number of chairs supporting bars are provided perpendicular to the direction of negative reinforcement.
  • 64. RCD Slab Design Reinforcement at discontinuous support: At the discontinuous end, the ACI code recommends to provide reinforcement equal to 1/3 times the positive reinforcement provided at the mid span.
  • 65. RCD Slab Design Step No 05: Drafting Main reinforcement = #3 @ 9″ c/c (positive & negative) Shrinkage reinforcement = #3 @ 9″ c/c Supporting bars = #3 @ 18″ c/c Chairs or supporting bars
  • 66. RCD Beam Design Step No 01: Sizes Minimum thickness of beam (simply supported) = hmin = l/16 l = clear span (ln) + depth of member (beam) ≤ c/c distance between supports Let depth of beam = 5′ ln + depth of beam = 60′ + 5′ = 65′ c/c distance between beam supports = 60 + 2 × (9/12) = 61.5′ Therefore l = 61.5′ Depth (h) = (61.5/16)×(0.4+fy/100000)×12= 36.9″ (Minimum by ACI 9.3.1.1). Take h = 5′ = 60″ d = h – 3 = 57″ bw = 18″ (assumed)
  • 67. RCD Beam Design Step No 02: Loads Load on beam will be equal to Factored load on beam from slab + factored self weight of beam web Factored load on slab = 0. 214 ksf Load on beam from slab = 0. 214 ksf x 10 = 2.14 k/ft Factored Self load of beam web = =1.2 x (54 × 18/144) × 0.15 =1.215 k/ft Total load on beam = 2.14 + 1.215 = 3.355 k/ft 10’
  • 68. RCD Shear Force & Bending Moment Diagrams Beam Design Step No 03: Analysis Vu = 87.23 kip Mu = 19034 in-kip
  • 69. RCD Beam Design Step No 04: Design Design for flexure Step (a): According to ACI 6.3.2.1, beff for T-beam is minimum of: 16hf + bw = 16 × 6 + 18 =114″ (clear length of beam)/4 =(60′/4) ×12 + 18 =198″ clear spacing between beams + bw =8.5′ × 12 + 18 =120″ So beff = 114″ Step (b): Check if beam is to be designed as rectangular beam or T-beam. Assume a = hf = 6″ and calculate As: As =Mu/ {Φfy (d–a/2)} =19034/ {0.9 × 40 × (57–6/2)} = 9.79 in2 Re-calculate “a”: a =Asfy/ (0.85fc′beff) =9.79 × 40/ (0.85 × 3 × 114) = 1.34″ < hf Therefore design beam as rectangular beam. After trials As = 9.38 in2 (Asmax = 20.83 in2 ;Asmin = 5.13 in2) Therefore As = 9.38 in2 (12 #8 bars)
  • 70. RCD Beam Design (Asmax = 20.83 in2 ;Asmin = 5.13 in2) Therefore As = 9.38 in2 (12 #8 bars)
  • 71. RCD Beam Design Step No 04: Design Design for flexure Skin Reinforcement : (ACI 9.7.2.3) As the effective depth d of a beam is greater than 36 inches, longitudinal skin reinforcement is required as per ACI 9.7.2.3. Askin, = Main flexural reinforcement/2 = 9.60/2 = 4.8 in2 Range up to which skin reinforcement is provided: d/2 = 56.625/2 = 28.3125″
  • 72. RCD Beam Design Step No 04: Design Design for flexure Skin Reinforcement For #8 bar used in skin reinforcement, ssk (skin reinforcement spacing) is least of: d/6 = 56.625/6 = 9.44″, 12″, or 1000Ab/(d – 30) = 1000×0.80/(56.625 – 30) = 30.05″ Therefore ssk = 9.44″ ≈ 9″ With this spacing, 3 bars on each face are required. And for # 8 bar, the total area of skin reinforcement is: Askin = 6 × 0.80 = 4.8 in2 skin reinforcement
  • 73. RCD Beam Design Step No 04: Design Design for Shear Vu = 87.23 kip ΦVc = Φ2 √ fc′bwd = (0.75 × 2 × √3000 × 18 × 56.625)/1000 = 83.74 kip ΦVc < Vu (Shear reinforcement is required) sd = ΦAvfyd/(Vu – ΦVc) Using #3, 2 legged stirrups with Av = 0.11 × 2 =0.22 in2} sd = 0.75 × 0.22 × 40 × 56.625/(87.23 – 83.74) = 107″
  • 74. RCD Beam Design Step No 04: Design Design for Shear Maximum spacing and minimum reinforcement requirement as permitted by ACI 9.7.6.2.2 and 10.6.2.2 shall be minimum of: Avfy/(50bw) =0.22 × 40000/(50 × 18) ≈ 9.5″ d/2 =56.625/2 =28.3″ 24″ Avfy/ 0.75 √fc’ bw = 0.22 × 40000/ {(0.75 × √3000 × 18} = 11.90″ Therefore, smax = 9.5″ ΦVc /2 = 83.74/2 = 41.87 kips at a distance of 17.5 ft from face of the support. Therefore no reinforcemnt is required in this zone, however, we will provide #3, 2-legged vertical stirrups at 12 in. c/c
  • 75. RCD Beam Design Step No 04: Design Design for Shear Other checks: Check for depth of beam: ΦVs ≤ Φ8 √fc’ bwd (ACI 22.5.1.2) Φ8 √fc’ bwd = 0.75 × 8 × √3000 × 18 × 56.625/1000 = 334.96 k ΦVs = (ΦAvfyd)/smax = (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.3 k < 334.96 k, O.K. So depth is O.K. If not, increase depth of beam.
  • 76. RCD Beam Design Step No 04: Design Design for Shear Other checks: Check if “ ΦVs ≤ Φ4 √ fc′bwd ” (ACI 10.7.6.5.2): If “ΦVs ≤ Φ4 √ fc′ bwd”, the maximum spacing (smax) is O.K. Otherwise reduce spacing by one half. Φ4√ fc′bwd = 0.75 × 4 × √3000 × 18 × 56.625/1000= 167.47 k ΦVs = (ΦAvfyd)/sd = (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.33 k < 167.47 k, O.K.
  • 77. RCD Beam Design Step No 05: Drafting
  • 78. RCD In this design example, the beams were supported on walls. This was done to simplify analysis. For practical reasons, however, the beams must be supported on columns and hence the structural analysis will be that of a frame rather than simply supported beam. In the subsequent slides, the analysis and design results for beams supported on columns have been provided.
  • 79. Design slab, beams, columns, and footing of a 90′ × 60′ Hall. The height of Hall is 20′. Concrete compressive strength (fc′) = 3 ksi and steel yield strength (fy) = 40 ksi. Take 3″ mud layer and 2″ tile layer above slab. Take Live Load equal to 40 psf. RCD
  • 80. RCD Frame Analysis 3D model of the hall showing beams supported on columns.
  • 81. RCD Frame Analysis A 2D frame can be detached from a 3D system in the following manner Column size = 18’’ × 18’’
  • 82. RCD Frame Analysis Various methods can be used for frame analysis. Using moment distribution method, the following results can be Obtained:
  • 83. RCD Frame Analysis Slab Design Slab design will remain the same as in case of beams supported on walls. Main reinforcement = #3 @ 9″ c/c (positive & negative) Shrinkage reinforcement = #3 @ 9″ c/c Supporting bars = #3 @ 18″ c/c
  • 84. RCD Frame Analysis Beam Design Beam flexure design will be as follows: Mu (+ve) = 17627 in-kips Mu (-ve) = 1407 in-kips As (+ve) = 8.68 in2 Use 6 #8 in 1st layer & 2 #8 + 4 #7 bars in 2nd layer) As = (8)(0.80) + (4)(0.60) = 8.80 in2 (As,max = 0.0203bd = 20.83 in2 OK) As (-ve) = 0.69 in2 (As,min = 0.005bd = 5.13 in2, so As,min governs) Use 7 #8 bars (5 bars in 1st layer and 2 bars in 2nd layer) As = (7)(0.80) = 5.60 in2 Beam shear design will be same as in previous case.
  • 85. RCD Beam Design (As,max = 0.0203bd = 20.83 in2 ) As (-ve) = 0.69 in2 (As,min = 0.005bd = 5.13 in2)
  • 87. RCD DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH UNIAXIAL BENDING using DESIGN AIDS
  • 88. RCD The ACI column interaction diagrams are used to design or analyze columns for different situations. In order to correctly use these diagrams, it is necessary to compute the value of γ (gamma), which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by h, Column cross sections for normalized interaction curves in Appendix A, DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH UNIAXIAL BENDING using DESIGN AIDS
  • 89. RCD Frame Analysis Column Design: Using ACI Design Aids Main Reinforcement Design Size: 18 in. × 18 in. Loads: Pu = 103.17 kips Mu = 1407 in-kips Calculate the ratio γ , for 2.5 in. cover (c): g = (h – 2c) / h = (18 – 5)/18 = 0.72 Calculate Kn, Kn = Pu/(Ø fc′Ag) = 103.17/(0.65 × 3 × 324) = 0.16 Calculate Rn, Rn = Mu/(Øfc′Agh) = 1407/(0.65 × 3 × 324 × 18) = 0.12 fc′ = 3 ksi, fy = 60 ksi
  • 90. RCD Column Design Main Reinforcement Design For given material strength, the column strength interaction diagram gives the following reinforcement ratio: ρ = 0.01 Ast = 0.01 × 324 = 3.24 in2 Using 8 #6 bars
  • 91. RCD Column Design Tie Bars: Using 3/8″ Φ (#3) tie bars for 3/4″ Φ (#6) main bars (ACI 9.7.6.4.2), Spacing for Tie bars according to ACI 9.7.6.4.3 is minimum of: 16 × dia of main bar =16 × 6/8 =12″ c/c 48 × dia of tie bar = 48 × (3/8) =18″ c/c Least column dimension =18″ c/c Finally use #3, tie bars @ 12″ c/c
  • 93. RCD Footing Design Isolated column footing; square or rectangular
  • 94. RCD Footing Design Data Given: Column size = 18″ × 18″ fc′ =3 ksi fy = 40 ksi qa = 2.204 k/ft2 Factored load on column = 103.17 kips (Reaction at the support) Service load on column = 81.87 kips (Reaction at the support due to service load)
  • 95. RCD Footing Design Sizes: Assume h = 15 in. davg = h – clear cover – one bar dia = 15 – 3 – 1(for #8 bar) = 11 in. Assume depth of the base of footing from ground level (z) = 5′ Weight of fill and concrete footing, W= γfill(z - h) + γch =100 × (5 – 1.25) +150 × (1.25) = 562.5 psf = 0.5625 ksf
  • 96. RCD Footing Design Sizes: Effective bearing capacity, qe = qa – W = 2.204 – 0.5625 = 1.642 ksf Bearing area, Areq = Service Load/ qe = 81.87/1.642 = 49.86 ft2 Areq = B x B = 49.86 ft2 => B = 7 ft. Critical Perimeter, bo = 4 x (c + davg) = 4 × (18 + 11) =116 in 7 ft 7 ft 5.5 ft 5.5 ft
  • 97. RCD Footing Design Loads: qu (bearing pressure for strength design of footing): qu = factored load on column / Areq = 103.17 / (7 × 7) = 2.105 ksf Analysis: Beam Shear Beam shear failure at a distance “d” from face of column 27.014k 2.105*(7/2-1.5/2)-11/12)*7 quB
  • 98. RCD Footing Design Analysis: Punching shear: Punching shear failure at a distance “d/2” from face of column Vup = quB2 – qu(c + davg)2 Vup = (2.105 × 49) –2.105 × {(18+11)/12)}2 103.145-12.29 = 90.85 kip qu(c + davg)2 quB2
  • 99. RCD Footing Design Analysis: Flexural Analysis: Flexural Failure at face of column Mu = quBk2/2 k = (B – c)/2 = (7 x 12 –18)/2 = 33 in = 2.75´ Mu = 2.105 × 7 × 2.75 × 2.75/2 = 55.72 ft-k = 668.60 in-kip quB
  • 100. RCD Footing Design Design: Design for Punching Shear: Vup = 90.85 kip Punching shear capacity (ΦVcp) = Φ4√fc’ bodavg = 0.75 × 4 × 3000 × 116 × 11/1000 = 209.66 k >Vup, O.K
  • 101. RCD Footing Design Design: Design for Flexure: Mu = 668.60 kip-in a = 0.2davg = 0.2 × 11 = 2.2″ As = Mu/ {Φfy(davg – a/2)} = 668.60/ {0.9 × 40 × (11 – 2.2/2)} = 1.87in2 a = Asfy/ (0.85fc′B) = 1.83 × 40/ (0.85 × 3 × 7 × 12) = 0.35″ After trials, As = 1.71 in2 (Asmin = 0.005Bdavg = 4.62 in2 so Asmin governs) Now, the spacing can be calculated as follows: Using #8 bars: No. of bars = 4.62/0.79 ≈ 6 bars. Spacing = 6.5 × 12 /5 = 15 in. c/c Hence 6 bars can be provided in the foundation if they are placed 15 in. c/c (Max. spacing should not exceed 3h or 18 in.)
  • 103.
  • 109. RCD
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  • 111. RCD
  • 112. RCD Bar spacing (in inches) = Ab/As × 12 Using # 4 bar (Ab = 0.2 and As = 0.27 so Bar spacing (in inches) = Ab/As × 12 =8.8 c/c’’ #4@ 8.8 c/c’’ #4@ 14 c/c’’ #4@ 24 c/c’’ #4@ 14 c/c’’ #4@ 17 c/c’’ #4@ 8.8 c/c’’
  • 113. RCD
  • 114. RCD Two possible arrangement of reinforcement are shown a) straight bar b) bent bars are used
  • 115. RCD
  • 116. RCD Two way Slab The slab which is supported on four sides, with longer to shorter side ratio less than two i.e aspect ratio M=L/S < 2 (lb/la <2) In two way slab, loads travel along the both span of slab Therefore main bars are provided along both direction as shown in the figure
  • 119. RCD Two way Slab SLAB BEAMS COLUMN AND FOOTING S1 S1 S1 S1 S2 S2 S2 S2 S3 S3 S4 S4 B1 B1 B2 B3 B2 B4 B5 B5 B4 C3 C1 C2
  • 120. RCD
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  • 127. RCD 1 2 3 4 5 6 7 8 9
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  • 129. 1 2 3 4 5 6 7 8 9 RCD
  • 131. RCD COEFFICIENTS FOR DEAD LOAD POSITIVE ATIVE MOMENTS IN SLABs
  • 132. RCD COEFFICIENTS FOR LIVE LOAD POSITIVE MOMENTS IN SLABs
  • 133. RCD RATIO OF LOAD W IN la AND lb DIRECTIONS FOR SHEAR IN SLAB AND LOAD ON SUPPORTS
  • 134. RCD
  • 135. RCD
  • 136. RCD la Mb (-ve) Ma (-ve) Ma (-ve) M b (+ve) Ma (+ve) lb Mb (-ve)
  • 137. RCD Bar spacing (in inches) = Ab/As × 12 Using # 4 bar (Ab = 0.2 and As = 0.288 so Bar spacing (in inches) = Ab/As × 12 =7 c/c’’ #4@ 7 c/c’’ #4@ 7 c/c’’ #4@ 5 c/c’’
  • 140. RCD la Mb (-ve) Ma (-ve) Ma (-ve) M b (+ve) Ma (+ve) lb Mb (-ve)
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