Design of the one-way slab in Rcd 2 course.pdf
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About This Presentation
this presentation is about the civil engineering point of view. in the design of the one way slab is discuss very deeply and easily explained.
Slide Content
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Lecture 02
Design of One-way Slab System
(Part - I)
By:
Prof. Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
[email protected]
www.drqaisarali.com
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 1
Lecture 02
Design of One-way Slab System
(Part - I)
By:
Prof. Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
[email protected]
www.drqaisarali.com
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 1
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Lecture Contents
+ General
e Design of Concrete Floor Systems
e Design of One-way Slab System
+ General Design Procedure
+ Design of 90' X 60' Hall (Example)
e Case Studies
e References
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Lecture Contents
+ General
e Design of Concrete Floor Systems
e Design of One-way Slab System
+ General Design Procedure
+ Design of 90' X 60' Hall (Example)
e Case Studies
e References
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Learning Outcomes
Q At the end of this lecture, students will be able to;
> Classify Concrete Floor Systems.
> Choose different framing systems for structures.
Analyze and Design one-way slabs for flexure by ACI
Approximate coefficients method.
v
> Recall basic design steps of shear and flexure.
> Understand the effect of beams’ spacing on slab moments.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Learning Outcomes
Q At the end of this lecture, students will be able to;
> Classify Concrete Floor Systems.
> Choose different framing systems for structures.
Analyze and Design one-way slabs for flexure by ACI
Approximate coefficients method.
v
> Recall basic design steps of shear and flexure.
> Understand the effect of beams’ spacing on slab moments.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
General
Q Concrete Floor Systems
e In reinforced concrete construction, Concrete Floor Systems are
broad and flat plates, usually horizontal, with top and bottom
surfaces parallel or nearly so.
e They are used to provide flat and useful surfaces. They may be
supported by:
- Reinforced concrete beams
- Masonry or reinforced concrete walls
- Structural steel members
+ Columns
- Ground.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General
Q Concrete Floor Systems
e In reinforced concrete construction, Concrete Floor Systems are
broad and flat plates, usually horizontal, with top and bottom
surfaces parallel or nearly so.
e They are used to provide flat and useful surfaces. They may be
supported by:
- Reinforced concrete beams
- Masonry or reinforced concrete walls
- Structural steel members
+ Columns
- Ground.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
General
Q Classification of Concrete Floor Systems
1. Flat Plate
+ Concrete floors directly supported on columns (with no beams)
are called flat plates.
e Punching shear is a typical problem in flat plates.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 5
General
Q Classification of Concrete Floor Systems
1. Flat Plate
+ Concrete floors directly supported on columns (with no beams)
are called flat plates.
e Punching shear is a typical problem in flat plates.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 5
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
General
Q Classification of Concrete Floor Systems
2. Flat Slab
e A flat plate that has a column capital or a
drop panel is referred to as a flat slab.
e The drop panel and column capital are
intended to reduce the punching shear
problem.
* Drop panel is a projection of slab at the vicinity of column.
= Column capital is a widening of the top of a support column
where it meets the soffit of a slab.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General
Q Classification of Concrete Floor Systems
2. Flat Slab
e A flat plate that has a column capital or a
drop panel is referred to as a flat slab.
e The drop panel and column capital are
intended to reduce the punching shear
problem.
* Drop panel is a projection of slab at the vicinity of column.
= Column capital is a widening of the top of a support column
where it meets the soffit of a slab.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General
Q Classification of Concrete Floor Systems
3. One-way Joist System
e Joist construction consists of a monolithic combination of
regularly spaced ribs with clear spacing of 3’ and a top slab
arranged to span in one direction or two orthogonal directions.
Rib
Peshawar University Auditorium
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 7
General
Q Classification of Concrete Floor Systems
3. One-way Joist System
e Joist construction consists of a monolithic combination of
regularly spaced ribs with clear spacing of 3’ and a top slab
arranged to span in one direction or two orthogonal directions.
Rib
Peshawar University Auditorium
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 7
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General
Q Classification of Concrete Floor Systems
4. Two-way Joist System
e 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.
e Like one-way joist system, a two-way system will be called as
two-way joist system if clear spacing between ribs (dome width)
does not exceed 30 inches.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General
Q Classification of Concrete Floor Systems
4. Two-way Joist System
e 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.
e Like one-way joist system, a two-way system will be called as
two-way joist system if clear spacing between ribs (dome width)
does not exceed 30 inches.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General
Q Classification of Concrete Floor Systems
4. Two-way Joist System — x
Q Classification of Concrete Floor Systems
4. Two-way Joist System — x
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General
Q Classification of Concrete Floor Systems
5. Beam-supported Slab Systems
e Slabs may be supported on two opposite sides only, or on all
four sides.
e Based on the behavior of bending, there are two types of such
slabs:
a. One-way slabs (bending in one direction)
b. Two-way slabs (bending in two direction)
e Both types will be discussed in detail later.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General
Q Classification of Concrete Floor Systems
5. Beam-supported Slab Systems
e Slabs may be supported on two opposite sides only, or on all
four sides.
e Based on the behavior of bending, there are two types of such
slabs:
a. One-way slabs (bending in one direction)
b. Two-way slabs (bending in two direction)
e Both types will be discussed in detail later.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of Concrete Floor Systems
Q Analysis
e Unlike beams and columns, Concrete Floors Systems are two
dimensional members. Therefore, their analysis except one-way
slab systems is relatively difficult.
Q Design
e Once the analysis is done, the design is carried out in the usual
manner. Hence there is no problem in design, problem is only in
analysis of slabs.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of Concrete Floor Systems
Q Analysis
e Unlike beams and columns, Concrete Floors Systems are two
dimensional members. Therefore, their analysis except one-way
slab systems is relatively difficult.
Q Design
e Once the analysis is done, the design is carried out in the usual
manner. Hence there is no problem in design, problem is only in
analysis of slabs.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of Concrete Floor Systems
Q Methods of Concrete Floor Analysis
1. Analysis using computer software (FEA)
= SAP 2000
» ETABS
« SAFE and so on.
2. ACI Approximate Method of Analysis
= Strip Method
For One-way Slabs
= ACI Approximate Coefficients Method
= Moment Coefficient Method
For Two-way Slabs
= Direct Design Method (DDM)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of Concrete Floor Systems
Q Methods of Concrete Floor Analysis
1. Analysis using computer software (FEA)
= SAP 2000
» ETABS
« SAFE and so on.
2. ACI Approximate Method of Analysis
= Strip Method
For One-way Slabs
= ACI Approximate Coefficients Method
= Moment Coefficient Method
For Two-way Slabs
= Direct Design Method (DDM)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q Definition (case I)
e Slab supported on two opposing sides only is called one-way
slab.
e In such case, the bending occurs perpendicular to the supports as
shown in the following figures.
Bending behavior of one-way slab
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q Definition (case I)
e Slab supported on two opposing sides only is called one-way
slab.
e In such case, the bending occurs perpendicular to the supports as
shown in the following figures.
Bending behavior of one-way slab
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q Definition (case Il)
e When the ratio of long to short span in a slab supported on all
sides is equal to or greater than 2 the slab is called one-way
slab.
e Because the major bending moment is always along the short
direction, these slabs are solely designed for the short direction.
Slab supported on all sides, but ß = lp/la = 2
Animation
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q Definition (case Il)
e When the ratio of long to short span in a slab supported on all
sides is equal to or greater than 2 the slab is called one-way
slab.
e Because the major bending moment is always along the short
direction, these slabs are solely designed for the short direction.
Slab supported on all sides, but ß = lp/la = 2
Animation
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q Strip Method of Analysis
e 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 I,
as shown in figure.
e This method is called as strip method of analysis.
Main
reinforcement
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 15
Design of One-way Slab System
Q Strip Method of Analysis
e 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 I,
as shown in figure.
e This method is called as strip method of analysis.
Main
reinforcement
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 15
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of One-way Slab System
Q Limitations of Strip Method of Analysis
e Applicable for one-way slabs only.
e Requires the slab to be supported on stiff beams or walls.
e It is not applicable to flat plates etc., even if bending is in one
direction.
Prof. Dr. Qaisar Ali
CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q Limitations of Strip Method of Analysis
e Applicable for one-way slabs only.
e Requires the slab to be supported on stiff beams or walls.
e It is not applicable to flat plates etc., even if bending is in one
direction.
Prof. Dr. Qaisar Ali
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of One-way Slab System
Q ACI Approximate Coefficients Method
e According to ACI 6.5.1, as an alternate to frame analysis, ACI
approximate moments shall be permitted for analysis of one-way
slabs with certain restrictions.
e This method is discussed in following slides.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q ACI Approximate Coefficients Method
e According to ACI 6.5.1, as an alternate to frame analysis, ACI
approximate moments shall be permitted for analysis of one-way
slabs with certain restrictions.
e This method is discussed in following slides.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q ACI Approximate Moment Coefficients
Posith
Moment X wut,
Spandrel support | 4/24 ne] [mn m] Im m] [ano DEA
Column support | 1/16
+ 1/12 for spans having I, < 10 ft)
+ 1/9 for 2 spans (for any I, )
Note: (i) For simply supported members, M = w,/?/8, where / = center to center distance.
(ii) To calculate negative moments, £, shall be the average of the adjacent clear span lengths.
einforced Concrete Design — Il
Prof. Dr. Qaisar Ali
Design of One-way Slab System
Q ACI Approximate Moment Coefficients
Posith
Moment X wut,
Spandrel support | 4/24 ne] [mn m] Im m] [ano DEA
Column support | 1/16
+ 1/12 for spans having I, < 10 ft)
+ 1/9 for 2 spans (for any I, )
Note: (i) For simply supported members, M = w,/?/8, where / = center to center distance.
(ii) To calculate negative moments, £, shall be the average of the adjacent clear span lengths.
einforced Concrete Design — Il
Prof. Dr. Qaisar Ali
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q ACI Approximate Shear Coefficients
1.15 145 145 145 145 1
‘Shear coefficient
x Waln/2
a
Note:
The above co-efficient is used for shear in beam only as slab has no significant shear.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q ACI Approximate Shear Coefficients
1.15 145 145 145 145 1
‘Shear coefficient
x Waln/2
a
Note:
The above co-efficient is used for shear in beam only as slab has no significant shear.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Q Limitations ACI Approximate Coefficients Method
4
E
Prof. Dr. Qaisar Ali
Design of One-way Slab System
Applicable to One-way Slabs and Beams only.
Loads must be uniformly distributed; not applicable for point loads.
The ratio of service live load to service dead load shall not exceed 3
(L/D < 3).
Suitable for two or more spans.
Members must be prismatic.
CE 416: Reinforced Concrete Design - Il
Q Limitations ACI Approximate Coefficients Method
4
E
Prof. Dr. Qaisar Ali
Design of One-way Slab System
Applicable to One-way Slabs and Beams only.
Loads must be uniformly distributed; not applicable for point loads.
The ratio of service live load to service dead load shall not exceed 3
(L/D < 3).
Suitable for two or more spans.
Members must be prismatic.
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of One-way Slab System
Q Limitations ACI Approximate Coefficients Method
6. The longer of two adjacent spans does not exceed the shorter by
more than 20 percent.
Uniformly distributed load (L/D < 3) L and D are unfactored loads
ae
ee
ee
ne
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of One-way Slab System
Q Limitations ACI Approximate Coefficients Method
6. The longer of two adjacent spans does not exceed the shorter by
more than 20 percent.
Uniformly distributed load (L/D < 3) L and D are unfactored loads
ae
ee
ee
ne
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Q Steps in the Design of a Typical RC Structure
a
2.
Prof. Dr. Qaisar Ali
General Design Procedure
Selection of Structural Configuration
Selection of sizes
Calculation of Loads
Analysis (Calculating Ultimate load effects/stresses)
Determination of Steel Area
Detailing of reinforcement
Drafting of structural members
CE 416: Reinforced Concrete Design - Il
Q Steps in the Design of a Typical RC Structure
a
2.
Prof. Dr. Qaisar Ali
General Design Procedure
Selection of Structural Configuration
Selection of sizes
Calculation of Loads
Analysis (Calculating Ultimate load effects/stresses)
Determination of Steel Area
Detailing of reinforcement
Drafting of structural members
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General Design Procedure
1. Selection of Structural Configuration
e The very first and most important step in Reinforced Concrete design
is to choose an appropriate structural configuration for the given
structure.
e Structural configuration means reasonable arrangement of beams,
girders and columns.
e Because several alternatives exist for a single structure, selecting a
suitable and cost-effective configuration is quite challenging.
It is primarily determined by the designer's prior experience.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General Design Procedure
1. Selection of Structural Configuration
e The very first and most important step in Reinforced Concrete design
is to choose an appropriate structural configuration for the given
structure.
e Structural configuration means reasonable arrangement of beams,
girders and columns.
e Because several alternatives exist for a single structure, selecting a
suitable and cost-effective configuration is quite challenging.
It is primarily determined by the designer's prior experience.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
General Design Procedure
2. Selection of Sizes
e ACI tables 7.3.1.1 & 9.3.1.1 give the minimum thickness for one-
way slabs and beams, respectively.
Solid one-way slal
Beams or ribbed one-
way slabs
+ L= Span length (Center to center length in case of interior spans and clear projection in case of
cantilevers) (Section 2.2).
+ [1] For f, other than 60,000 psi, the expressions in the table shall be multiplied by (0.4 + f, /100,000)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 24
General Design Procedure
2. Selection of Sizes
e ACI tables 7.3.1.1 & 9.3.1.1 give the minimum thickness for one-
way slabs and beams, respectively.
Solid one-way slal
Beams or ribbed one-
way slabs
+ L= Span length (Center to center length in case of interior spans and clear projection in case of
cantilevers) (Section 2.2).
+ [1] For f, other than 60,000 psi, the expressions in the table shall be multiplied by (0.4 + f, /100,000)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 24
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General Design Procedure
3. Calculation of Loads
e According to ACI 5.2, 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.
e 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.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General Design Procedure
3. Calculation of Loads
e According to ACI 5.2, 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.
e 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.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
General Design Procedure
4. Analysis
e The analysis is carried out for ultimate load including self weight
obtained from size of the slab and the applied dead and live loads.
e The maximum demand values are used for the purpose of design.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General Design Procedure
4. Analysis
e The analysis is carried out for ultimate load including self weight
obtained from size of the slab and the applied dead and live loads.
e The maximum demand values are used for the purpose of design.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General Design Procedure
5. Determination of Steel Area
e Once the load effects (Bending moment, Shear force, Axial force
and Torsion ) are calculated based on the anticipated loads, the
next step is to calculate the required area of steel to withstand
these load effects.
e The steel area can be determined by equating design capacity of a
structural member with demand.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General Design Procedure
5. Determination of Steel Area
e Once the load effects (Bending moment, Shear force, Axial force
and Torsion ) are calculated based on the anticipated loads, the
next step is to calculate the required area of steel to withstand
these load effects.
e The steel area can be determined by equating design capacity of a
structural member with demand.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
General Design Procedure
6. Detailing
e The calculated steel area is detailed as per the Code Provisions.
e The detailing step involves development lengths, steel curtailment,
splices, symmetrical arrangement of rebars, spacing requirements
etc.
7. Drafting
e Drafting is the final stage in the design which is the proper sketch
of dimensions and reinforcement detailing of a structural member
in different views.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
General Design Procedure
6. Detailing
e The calculated steel area is detailed as per the Code Provisions.
e The detailing step involves development lengths, steel curtailment,
splices, symmetrical arrangement of rebars, spacing requirements
etc.
7. Drafting
e Drafting is the final stage in the design which is the proper sketch
of dimensions and reinforcement detailing of a structural member
in different views.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Q Example 2.1
Prof. Dr. Qaisar Ali
Design of 90' x 60’ Hall
The slab of a Hall having interior dimensions 90' x 60’ with story height
of 20' is subjected to a uniform service live load of 40 psf. Additionally,
the slab includes superimposed dead load in the form of 3-inch mud
layer and 2-inch tile layer. The allowable bearing capacity of soil at the
depth of 5 ft. from finished floor level is 2.204 ksf. For all structural
members, consider concrete compressive strength, f/ = 3 ksi and
steel yield strength, f, = 60 ksi
Design the Hall with various structural configurations.
CE 416: Reinforced Concrete Design - Il
Q Example 2.1
Prof. Dr. Qaisar Ali
Design of 90' x 60’ Hall
The slab of a Hall having interior dimensions 90' x 60’ with story height
of 20' is subjected to a uniform service live load of 40 psf. Additionally,
the slab includes superimposed dead load in the form of 3-inch mud
layer and 2-inch tile layer. The allowable bearing capacity of soil at the
depth of 5 ft. from finished floor level is 2.204 ksf. For all structural
members, consider concrete compressive strength, f/ = 3 ksi and
steel yield strength, f, = 60 ksi
Design the Hall with various structural configurations.
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Given Data
Dimensions of Hall : 90' x 60' (interior)
Story height, h = 20’
N Brick
‘MASONRY
[ WALL
Loads:
> SDL: 3" Mud layer and 2" Tile layer
> Live load: 60 psf
fe =3 ksi
fy = 60 ksi
Q Required Data
Design the Hall with various structural configurations.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Given Data
Dimensions of Hall : 90' x 60' (interior)
Story height, h = 20’
N Brick
‘MASONRY
[ WALL
Loads:
> SDL: 3" Mud layer and 2" Tile layer
> Live load: 60 psf
fe =3 ksi
fy = 60 ksi
Q Required Data
Design the Hall with various structural configurations.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution
e Several structural configurations are possible for the given hall, some
of which are shown below
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution
e Several structural configurations are possible for the given hall, some
of which are shown below
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution
e Several structural configurations are possible for the given hall, some
of which are shown below
and many more...
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution
e Several structural configurations are possible for the given hall, some
of which are shown below
and many more...
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution
e However, in this lecture we will explore the following design options:
1. Option 1: Hall Design Without Interior Columns
a) Beam-slab system supported on walls.
b) Beam-slab system supported on exterior columns.
2. Option 2: Hall Design With Only Two Interior Columns
a) Beam-slab system supported on girder and walls.
b) Beam-slab system supported on girder and exterior columns.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution
e However, in this lecture we will explore the following design options:
1. Option 1: Hall Design Without Interior Columns
a) Beam-slab system supported on walls.
b) Beam-slab system supported on exterior columns.
2. Option 2: Hall Design With Only Two Interior Columns
a) Beam-slab system supported on girder and walls.
b) Beam-slab system supported on girder and exterior columns.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90’ X 60’ Hall
Option 1a
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90’ X 60’ Hall
Option 1a
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
> Step 1: Selection of Structural Configuration
Beams are supported
on walls
Centerline of beams
60" Brick masonry wall
All beams are spaced
at 10" cle.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
> Step 1: Selection of Structural Configuration
Beams are supported
on walls
Centerline of beams
60" Brick masonry wall
All beams are spaced
at 10" cle.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1a]
> Step 1: Selection of Structural Configuration
EA Endspan a Interior span -
<— 10 —k— 10 —
Assumed wall thickness = 18 in K— 9.25 —>|18"| 85 —|
‘Assumed beam width = 18 in
Section A-A
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 36
Design of 90’ x 60’ Hall
Q Solution [option 1a]
> Step 1: Selection of Structural Configuration
EA Endspan a Interior span -
<— 10 —k— 10 —
Assumed wall thickness = 18 in K— 9.25 —>|18"| 85 —|
‘Assumed beam width = 18 in
Section A-A
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 36
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
e Step 2: Selection of Sizes
l r
hina = 24 (for one end continuous slab) En . ren .
10.75 <— 101 —#— 10 —
= on = 0.448' or 5.4” <— 925 —|18"| 85 —
Section A-A
l
hmin2 = 25 (for both end continuous slab)
= 0.357’ or 4.3”
28 . Ñ
hmin = larger of (Mmin1, min2) = 5.4” . Finally take h = 6”
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
e Step 2: Selection of Sizes
l r
hina = 24 (for one end continuous slab) En . ren .
10.75 <— 101 —#— 10 —
= on = 0.448' or 5.4” <— 925 —|18"| 85 —
Section A-A
l
hmin2 = 25 (for both end continuous slab)
= 0.357’ or 4.3”
28 . Ñ
hmin = larger of (Mmin1, min2) = 5.4” . Finally take h = 6”
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 3: Calculation of Loads
Thickness
Concrete
J 0.120 (8/12) x 0.12 = 0.03
| 2 | 0.120 (2/12) x 0.12= 0.02
Total dead load 0.125 ksf
e Factored Load, Wy sap = 1.2D + 1.6L
= 1.2 x 0.125 + 1.6 x 0.04 = 0.214 ksf
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
38
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 3: Calculation of Loads
Thickness
Concrete
J 0.120 (8/12) x 0.12 = 0.03
| 2 | 0.120 (2/12) x 0.12= 0.02
Total dead load 0.125 ksf
e Factored Load, Wy sap = 1.2D + 1.6L
= 1.2 x 0.125 + 1.6 x 0.04 = 0.214 ksf
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
38
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Q Solution [option 1a]
% Slab Design
.
Prof. Dr. Qaisar Ali
Design of 90’ x 60’ Hall
Before proceeding to the analysis step, there is an important notion
that needs clarification.
The ACI Coefficients give reasonably accurate results only when the
supports (beams or walls), exhibit sufficient stiffness.
However, it must be kept in mind that; “Beams of any depth cannot
act as support to slabs; only beams with Specific depth can do so.”
The subsequent slides elaborate the above statement.
CE 416: Reinforced Concrete Design - Il
Q Solution [option 1a]
% Slab Design
.
Prof. Dr. Qaisar Ali
Design of 90’ x 60’ Hall
Before proceeding to the analysis step, there is an important notion
that needs clarification.
The ACI Coefficients give reasonably accurate results only when the
supports (beams or walls), exhibit sufficient stiffness.
However, it must be kept in mind that; “Beams of any depth cannot
act as support to slabs; only beams with Specific depth can do so.”
The subsequent slides elaborate the above statement.
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Relative Stiffness of Slab and Beam-Slab Behavior.
i Slab
JL
Beam acting like pseudo
Wall (true) support support for slab
x Less stiff beams
More stiff beams.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Relative Stiffness of Slab and Beam-Slab Behavior.
i Slab
JL
Beam acting like pseudo
Wall (true) support support for slab
x Less stiff beams
More stiff beams.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Behavior of flexible (Less stiff) beams.
= LI
Before Bending
After Bending
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Behavior of flexible (Less stiff) beams.
= LI
Before Bending
After Bending
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Behavior of stiff beams.
After Bending
Animation on relative Stiffness
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Behavior of stiff beams.
After Bending
Animation on relative Stiffness
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Slab Design
e The question that arises is how to determine the necessary beam
depth to effectively support and minimize deflection in slabs.
e The required beam depth is determined by plotting the depth of the
beam against the negative moments in the slab, as illustrated below.
128 in-klin
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 43
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Slab Design
e The question that arises is how to determine the necessary beam
depth to effectively support and minimize deflection in slabs.
e The required beam depth is determined by plotting the depth of the
beam against the negative moments in the slab, as illustrated below.
128 in-klin
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 43
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Concluding Remarks
= From the plot, it is cleared that ACI minimum depth equations for beams
may not always offer adequate stiffness under all circumstances. These
equations are specifically applicable to beams with normal load conditions
and typical spans.
= A beam depth of 60 inches is considered adequate to provide the
necessary stiffness for the slab.
Note: The determination of the necessary beam depth to act as support for
the slab depends on serviceability requirements, which is beyond the scope
of this course.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
+ Concluding Remarks
= From the plot, it is cleared that ACI minimum depth equations for beams
may not always offer adequate stiffness under all circumstances. These
equations are specifically applicable to beams with normal load conditions
and typical spans.
= A beam depth of 60 inches is considered adequate to provide the
necessary stiffness for the slab.
Note: The determination of the necessary beam depth to act as support for
the slab depends on serviceability requirements, which is beyond the scope
of this course.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 4: Analysis
+ When using 60" deep beams, the ACI coefficient method can be applied.
The results obtained are displayed in the figure.
Wu=0214kst
o m 1112 1112 1116 1112
— ale
—=115 le = 9.25- > 15 ke = 85: 15
1997 e
BuD
Cp)
o
1688 1686" 1546 | 1546
[1] For calculating negative moments, t, shall be the average of the adjacent clear span lengths (ACI:6.5.2)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 4: Analysis
+ When using 60" deep beams, the ACI coefficient method can be applied.
The results obtained are displayed in the figure.
Wu=0214kst
o m 1112 1112 1116 1112
— ale
—=115 le = 9.25- > 15 ke = 85: 15
1997 e
BuD
Cp)
o
1688 1686" 1546 | 1546
[1] For calculating negative moments, t, shall be the average of the adjacent clear span lengths (ACI:6.5.2)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
+ Calculate moment capacity provided by minimum reinforcement
Asmin = 0.0018bh = 0.0018(12)(6) = 0.129 in?/ft
Asminfy 0.129 x 60
ce =
= — = = 0.2! in.
Dasfb, D85x3x12 02>3in
0.253
OM, = OAsminfy (d- 3) = 0.9 x 0.129 x 60 (s = =>)
= 33.95 in.k/ft
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
+ Calculate moment capacity provided by minimum reinforcement
Asmin = 0.0018bh = 0.0018(12)(6) = 0.129 in?/ft
Asminfy 0.129 x 60
ce =
= — = = 0.2! in.
Dasfb, D85x3x12 02>3in
0.253
OM, = OAsminfy (d- 3) = 0.9 x 0.129 x 60 (s = =>)
= 33.95 in.k/ft
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
+ @M,, calculated from A, min is greater than all moments calculated in
Step No 4. Hence Minimum reinforcement governs.
Ag = 0.129 in?/ft
Using #3 bar with A, = 0.11in?
124, _ 12(0.11)
s= =
Au lim.
+ Provide #3 bars @ 10.2 in. c/c for both positive and negative steel.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
+ @M,, calculated from A, min is greater than all moments calculated in
Step No 4. Hence Minimum reinforcement governs.
Ag = 0.129 in?/ft
Using #3 bar with A, = 0.11in?
124, _ 12(0.11)
s= =
Au lim.
+ Provide #3 bars @ 10.2 in. c/c for both positive and negative steel.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
Smax = 3h or 18" whichever is greater
Smax = 3(6)or 18” = 18” > OK!
Finally, provide #3 bars @ 9 in. c/c
e Shrinkage and Temperature reinforcement
Agar = Amin = 0.129 in? /ft (#3@9" c/c)
Smax = 5(6)or 18" = 18" > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 5: Determination of Flexural Steel Area
Smax = 3h or 18" whichever is greater
Smax = 3(6)or 18” = 18” > OK!
Finally, provide #3 bars @ 9 in. c/c
e Shrinkage and Temperature reinforcement
Agar = Amin = 0.129 in? /ft (#3@9" c/c)
Smax = 5(6)or 18" = 18" > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 6: Detailing
e Positive reinforcement bars are placed in the direction of flexure
stresses and placed at the bottom to maximize the “d”.
+ 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.
e At the discontinuous end, the ACI code recommends to provide
reinforcement equal to 1/3 times the positive reinforcement provided
at the mid span.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 6: Detailing
e Positive reinforcement bars are placed in the direction of flexure
stresses and placed at the bottom to maximize the “d”.
+ 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.
e At the discontinuous end, the ACI code recommends to provide
reinforcement equal to 1/3 times the positive reinforcement provided
at the mid span.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 7: Drafting
Temperature Reinforcement
Main Reinforcement
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Slab Design
> Step 7: Drafting
Temperature Reinforcement
Main Reinforcement
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
exec ars wesc mero #esac
(Supporing bars) (Supporting
2:10" with hook
SLAB SECTION A-A
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 51
exec ars wesc mero #esac
(Supporing bars) (Supporting
2:10" with hook
SLAB SECTION A-A
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 51
Design of 90' x 60’ Hall
Q Solution
e Slab reinforcement arrangement at Site
Er” >
[= Spacer
AL
Le Fe £
Q Solution
e Slab reinforcement arrangement at Site
Er” >
[= Spacer
AL
Le Fe £
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 2: Selection of Sizes
e Assume 18” width of beam. The minimum depth of beam for a
simply supported case is given by
60+2(2)
16
1
Amin = 16
| = 3.84"
Amin = 46.08"
Take h = 60” (the reason is already explained earlier)
Effective depth, d = 60-3 = 57"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 2: Selection of Sizes
e Assume 18” width of beam. The minimum depth of beam for a
simply supported case is given by
60+2(2)
16
1
Amin = 16
| = 3.84"
Amin = 46.08"
Take h = 60” (the reason is already explained earlier)
Effective depth, d = 60-3 = 57"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 2: Selection of Sizes
Effective width of T- beam by is minimum of:
* by + 16hy = 18 + 16(6) = 114"
+ Dy + Sy =18+8.5 x 12 = 120"
ln 60 ,
+ bte = 1840 x 12 = 198
=]
NN
Therefore, by = 114"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 2: Selection of Sizes
Effective width of T- beam by is minimum of:
* by + 16hy = 18 + 16(6) = 114"
+ Dy + Sy =18+8.5 x 12 = 120"
ln 60 ,
+ bte = 1840 x 12 = 198
=]
NN
Therefore, by = 114"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 3: Calculation of Loads
+ The factored load on the beam is the sum of the slab load and its
own weight.
Tributary area, Ay
_ (Wastab X At)
Wupeam = Span length
+ 1.25W
Which can be simplified to
Wu,beam = Wusiab X by + 1.25W --- (2.1)
|<— span length —>|
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 3: Calculation of Loads
+ The factored load on the beam is the sum of the slab load and its
own weight.
Tributary area, Ay
_ (Wastab X At)
Wupeam = Span length
+ 1.25W
Which can be simplified to
Wu,beam = Wusiab X by + 1.25W --- (2.1)
|<— span length —>|
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 3: Calculation of Loads
bulk hy)
Ta “re
sw
_ 18(60 — 6)
~ 144
Tributary area, Ay
x 0.150 = 1.0125k/ft
Now,
Wapeam = 0.214 x 10 + 1.2(1.0125)
|<— span length —>|
Wupeam = 3.355k/ft
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 3: Calculation of Loads
bulk hy)
Ta “re
sw
_ 18(60 — 6)
~ 144
Tributary area, Ay
x 0.150 = 1.0125k/ft
Now,
Wapeam = 0.214 x 10 + 1.2(1.0125)
|<— span length —>|
Wupeam = 3.355k/ft
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 4: Analysis
+ Maximum shear force and bending moment is given by
Y, = 87.23 kip 3.355 kiprtt
en pe
A
1 i | SED
È -i H i
F H H
& H 19034.17 in-kips !
h BMD
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 4: Analysis
+ Maximum shear force and bending moment is given by
Y, = 87.23 kip 3.355 kiprtt
en pe
A
1 i | SED
È -i H i
F H H
& H 19034.17 in-kips !
h BMD
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 5: Determination of flexural steel area
Using direct method, we have
2.614M, 2.614 x 19034
= de go seed = ee "
a=d | FD 57 fr CRITE 129< hf
The beam will be designed as rectangular.
M, 19034
oca) 09x 60 (57-222)
As = 6.25 in? (8 #8 bars)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 5: Determination of flexural steel area
Using direct method, we have
2.614M, 2.614 x 19034
= de go seed = ee "
a=d | FD 57 fr CRITE 129< hf
The beam will be designed as rectangular.
M, 19034
oca) 09x 60 (57-222)
As = 6.25 in? (8 #8 bars)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 6: Check for flexural steel area
e Minimum reinforcement limit
À, ea 40 x 18x 57 = 3.42 in?
a = = 3.42 in
I = pe 60000
e Maximum reinforcement limit
_ febyd _3 18X57 _ ae
Asmax = 393 = 273 = 13:8 in’
Asmin < As < Asmax > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 6: Check for flexural steel area
e Minimum reinforcement limit
À, ea 40 x 18x 57 = 3.42 in?
a = = 3.42 in
I = pe 60000
e Maximum reinforcement limit
_ febyd _3 18X57 _ ae
Asmax = 393 = 273 = 13:8 in’
Asmin < As < Asmax > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
2x 0.75V3000 x 18 x 57
OV, = 20) fibyd = Tan
= 84.29 kip
Calculate design spacing sg using
_ DA fyd
TT
Using 2 legged #3 stirrups, A, = 2A, = 2(0.11) = 0.22 in?
_ 0.75 x 0.22 x 60 x 57
= = 191.94 in.
Sa 87.23 — 84.29 132m
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
2x 0.75V3000 x 18 x 57
OV, = 20) fibyd = Tan
= 84.29 kip
Calculate design spacing sg using
_ DA fyd
TT
Using 2 legged #3 stirrups, A, = 2A, = 2(0.11) = 0.22 in?
_ 0.75 x 0.22 x 60 x 57
= = 191.94 in.
Sa 87.23 — 84.29 132m
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
Calculate maximum spacing Smax
Avfy _ 0.22 x 60,000
A a
50b, 50x18 u
Avfy ___022% 60000 _ ¿7 gu
Smax = Least of + 075/f/b, 0.75V3000x18 Smax = 14.7" < sq
d_57 ,
yay =285
24"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
Calculate maximum spacing Smax
Avfy _ 0.22 x 60,000
A a
50b, 50x18 u
Avfy ___022% 60000 _ ¿7 gu
Smax = Least of + 075/f/b, 0.75V3000x18 Smax = 14.7" < sq
d_57 ,
yay =285
24"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
1. Check for Depth of Beam:
0.75 x 0.22 x 60 x 57
OÙ < 08/f'b,d OU = Da
= 38.39 kip
08y/f.'byd = 40V. = 4 x 84.29 = 337.16 kip > OV, > OK!
2. Check for Maximum spacing of stirrups:
OÙ <04/f'b,d , 0%, = 38.39 kip
04,/f.'byd = 20V, = 2 x 84.29 = 168.58 kip > OV, > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
1. Check for Depth of Beam:
0.75 x 0.22 x 60 x 57
OÙ < 08/f'b,d OU = Da
= 38.39 kip
08y/f.'byd = 40V. = 4 x 84.29 = 337.16 kip > OV, > OK!
2. Check for Maximum spacing of stirrups:
OÙ <04/f'b,d , 0%, = 38.39 kip
04,/f.'byd = 20V, = 2 x 84.29 = 168.58 kip > OV, > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
Ve = 103.17 kip
Vua = 87.23 kip
ee nr OV, = 84.29 kip
Ve
O E = 42.14 kip
OV./2 > Va
Ss
SI
v
=
Theoretically no reinforcement is needed —>1
30.75’ >
Finally provide #3 @ 12 c/c throughout.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 7: Determination of Shear reinforcement
Ve = 103.17 kip
Vua = 87.23 kip
ee nr OV, = 84.29 kip
Ve
O E = 42.14 kip
OV./2 > Va
Ss
SI
v
=
Theoretically no reinforcement is needed —>1
30.75’ >
Finally provide #3 @ 12 c/c throughout.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
For relatively deep beams, some reinforcement should be placed
near the vertical faces of the tension zone
(ACI R9.7.2.3). This is termed as skin reinforcement.
‘Skin reinforcement required
in tree regions.
Small strain, minimal
web cracking chance.
Large strain, significant web
‘Strain diagram cracking risk.
Shallow beam
Strain diagram
Relatively deep beam
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 64
For relatively deep beams, some reinforcement should be placed
near the vertical faces of the tension zone
(ACI R9.7.2.3). This is termed as skin reinforcement.
‘Skin reinforcement required
in tree regions.
Small strain, minimal
web cracking chance.
Large strain, significant web
‘Strain diagram cracking risk.
Shallow beam
Strain diagram
Relatively deep beam
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 64
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e For beams with depth exceeding
36 inches (h >36”), longitudinal
skin reinforcement shall be
uniformly distributed on both side
faces of the beam for a distance
h/2 from the tension face (ACI
9.7.2.3).
Skin Reinforcement for beams with h > 36"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
65
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e For beams with depth exceeding
36 inches (h >36”), longitudinal
skin reinforcement shall be
uniformly distributed on both side
faces of the beam for a distance
h/2 from the tension face (ACI
9.7.2.3).
Skin Reinforcement for beams with h > 36"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
65
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
= Amount of skin reinforcement
e Bar sizes No. 3 to No.5, with a minimum area of 0.1 in? per foot of depth
(0.1h), are typically provided (ACI R9.7.2.3).
= Spacing of skin reinforcement
+ Spacing shall not exceed Snax given by
40,000
40,000
) — 2.56.07 12( 7 ) whichever is lesser
's
Smax = ss
Is
Where; f, = 2/3f, and C, is clear cover.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
= Amount of skin reinforcement
e Bar sizes No. 3 to No.5, with a minimum area of 0.1 in? per foot of depth
(0.1h), are typically provided (ACI R9.7.2.3).
= Spacing of skin reinforcement
+ Spacing shall not exceed Snax given by
40,000
40,000
) — 2.56.07 12( 7 ) whichever is lesser
's
Smax = ss
Is
Where; f, = 2/3f, and C, is clear cover.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
Askin = 0.1h = 0.1 X 5 = 0.5 in? (where his in feet, not inches)
Using #3 bar with A, = 0.11 in?
No. of bars = 0.5/0.11 = 4.55 = 5 (for the purpose of symmetry, take 6 bars.)
e Hence, 6 - #3 bars will be provided on each face of beam. The range
up to which skin reinforcement is provided is given by
h/2 = 60/2 = 30” (from tension face)
Prof. Dr. Qaisar Ali
CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
Askin = 0.1h = 0.1 X 5 = 0.5 in? (where his in feet, not inches)
Using #3 bar with A, = 0.11 in?
No. of bars = 0.5/0.11 = 4.55 = 5 (for the purpose of symmetry, take 6 bars.)
e Hence, 6 - #3 bars will be provided on each face of beam. The range
up to which skin reinforcement is provided is given by
h/2 = 60/2 = 30” (from tension face)
Prof. Dr. Qaisar Ali
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e Spacing between adjacent skin rebars
72-#6 bars
3
y=30- (1543414141) = 2533"
n = no. of bars on one face = 3
Now, (343) -#8 bars
(equally spaced)
L, Spacer (17)
(4+4)-#8 bars
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 68
Design of 90’ x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e Spacing between adjacent skin rebars
72-#6 bars
3
y=30- (1543414141) = 2533"
n = no. of bars on one face = 3
Now, (343) -#8 bars
(equally spaced)
L, Spacer (17)
(4+4)-#8 bars
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 68
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e Maximum spacing for skin reinforcement
40,000 40,000
Someta 15( T, ) —2.5C, or 12( T, ) whichever is lesser
Ë 's
an 2.5(15)=1125" and 12 un = 2"
2/3(60,000) A 2/3(60,000)| —
As S = 8.4" < Sax = 11.25” , therefore provided spacing is OK.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1a]
% Beam Design
> Step 8: Determination of Skin Reinforcement
e Maximum spacing for skin reinforcement
40,000 40,000
Someta 15( T, ) —2.5C, or 12( T, ) whichever is lesser
Ë 's
an 2.5(15)=1125" and 12 un = 2"
2/3(60,000) A 2/3(60,000)| —
As S = 8.4" < Sax = 11.25” , therefore provided spacing is OK.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
6:43 bars
(skin reinforcement)
ELEVATION DETALS
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 70
6:43 bars
(skin reinforcement)
ELEVATION DETALS
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 70
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
#3@12'clc #3@12"cle
(8+3) -#3 bars
(equally spaced)
(8+3) - #3 bars
(equally spaced)
y Spacer (1)
222721
+
N (4+4) - #8 bars
E18" ——
Section A-A Section B-B
einforced Concrete Design — Il 71
Prof. Dr. Qaisar Ali
#3@12'clc #3@12"cle
(8+3) -#3 bars
(equally spaced)
(8+3) - #3 bars
(equally spaced)
y Spacer (1)
222721
+
N (4+4) - #8 bars
E18" ——
Section A-A Section B-B
einforced Concrete Design — Il 71
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90’ X 60’ Hall
Option 1b
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90’ X 60’ Hall
Option 1b
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of Structural Configuration
Beams are supported
‘on columns
Centerline of beams
Brick masonry wall
18" thick
All beams are spaced
at 10" cle.
90"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of Structural Configuration
Beams are supported
‘on columns
Centerline of beams
Brick masonry wall
18" thick
All beams are spaced
at 10" cle.
90"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of Structural Configuration
so Endspan lJ interior span Ly
aan ai Column 10° >< 10 À
fall width = 18 in En Bu N
‘Assumed beam width = 18 in So 8s
Section A-A
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of Structural Configuration
so Endspan lJ interior span Ly
aan ai Column 10° >< 10 À
fall width = 18 in En Bu N
‘Assumed beam width = 18 in So 8s
Section A-A
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
e Analysis
Wu =0.214kst
I CO Gana as! IT] 4
se
— x nr — Beam Beam
~— Colum
115 ~ L= 925" > 15 = 1=85- 15 ke
BMD ZL
(ke) ><
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
e Analysis
Wu =0.214kst
I CO Gana as! IT] 4
se
— x nr — Beam Beam
~— Colum
115 ~ L= 925" > 15 = 1=85- 15 ke
BMD ZL
(ke) ><
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
+ Moment capacity provided by minimum reinforcement 0M, =33.95
in.kip (calculated in option 1a) is greater than all the moments
shown on previous slide, therefore Minimum reinforcement governs.
e Hence Provide;
= Main reinforcement : #3 @ 9" c/c (both positive & negative)
= Shrinkage reinforcement : #3 @ 9” c/c
= Supporting bars : #3 @ 18" c/c
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
+ Moment capacity provided by minimum reinforcement 0M, =33.95
in.kip (calculated in option 1a) is greater than all the moments
shown on previous slide, therefore Minimum reinforcement governs.
e Hence Provide;
= Main reinforcement : #3 @ 9" c/c (both positive & negative)
= Shrinkage reinforcement : #3 @ 9” c/c
= Supporting bars : #3 @ 18" c/c
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
+ Drafting
Temperature Reinforcement
Main Reinforcement
#3@9" ce
Een Ent loos Baal pose aoe proc |
|
90" |
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Slab Design
+ Drafting
Temperature Reinforcement
Main Reinforcement
#3@9" ce
Een Ent loos Baal pose aoe proc |
|
90" |
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
exec ars wesc mero #esac
(Supporing bars) (Supporting
2:10" with hook
SLAB SECTION A-A
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 78
exec ars wesc mero #esac
(Supporing bars) (Supporting
2:10" with hook
SLAB SECTION A-A
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 78
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
e A 2D frame can be detached from a 3D system in the following
manner:
W, = 3.355 kit
leje= 61.5"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
e A 2D frame can be detached from a 3D system in the following
manner:
W, = 3.355 kit
leje= 61.5"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1b]
< Frame Analysis
+ Numerous methods are available for frame analysis. here, we will use the
Simplified Flexibility Method.
3.355 x 61.5 5
= 7 = 103.17 kip
(18 x 60°)/12
=> = 37.04
Be (18 x 18)/12 ay
3 Simplified Fle: ity Method:
3.355(61.5) x 12
M, = = 1405.20 in. ki A simplified flexibility method, used for
uneg = T2(61.5) + 8(20)(37.04) P | analyzing a single bay, nonswey,
symmetrical pin-supported frame under
3.355(61.5)? x 12 uniformly distributed load.
Mu pos = ————— - 1405.20 = 17628. 97 in. kip|
= 8 Monos Mu? Ely
nee = Ta em. Rr = Ee
a 1405.20 5.86 ui 12L + 8HK, El.
TE A
pos Manag
Prof. Dr. Qaisar Ali
einforced Concrete Design — II
80
Design of 90’ x 60’ Hall
Q Solution [option 1b]
< Frame Analysis
+ Numerous methods are available for frame analysis. here, we will use the
Simplified Flexibility Method.
3.355 x 61.5 5
= 7 = 103.17 kip
(18 x 60°)/12
=> = 37.04
Be (18 x 18)/12 ay
3 Simplified Fle: ity Method:
3.355(61.5) x 12
M, = = 1405.20 in. ki A simplified flexibility method, used for
uneg = T2(61.5) + 8(20)(37.04) P | analyzing a single bay, nonswey,
symmetrical pin-supported frame under
3.355(61.5)? x 12 uniformly distributed load.
Mu pos = ————— - 1405.20 = 17628. 97 in. kip|
= 8 Monos Mu? Ely
nee = Ta em. Rr = Ee
a 1405.20 5.86 ui 12L + 8HK, El.
TE A
pos Manag
Prof. Dr. Qaisar Ali
einforced Concrete Design — II
80
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
M, = 1405.20 in-kip M, = 1405.20 in-kip
W, = 3.355 Kt
V, = 5.86 kip
P, = 103.17 kip P, = 103.17 kip
Prof. Dr. Qaisar Ali
teinforced Concrete Design — II
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
M, = 1405.20 in-kip M, = 1405.20 in-kip
W, = 3.355 Kt
V, = 5.86 kip
P, = 103.17 kip P, = 103.17 kip
Prof. Dr. Qaisar Ali
teinforced Concrete Design — II
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
W, = 3.355 kift
ana
103.17 kip 103.17 kip
1405.20 in.kip
103.17 kip ons
an 103.17 kip
1405.20 in.kip 1405.20 in.kip L
5.86 kip 4— T TM | > 5.86 kip
20
TE 586 kip
20°
5.86 kip +
103.17 kip
103.17 kip
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
W, = 3.355 kift
ana
103.17 kip 103.17 kip
1405.20 in.kip
103.17 kip ons
an 103.17 kip
1405.20 in.kip 1405.20 in.kip L
5.86 kip 4— T TM | > 5.86 kip
20
TE 586 kip
20°
5.86 kip +
103.17 kip
103.17 kip
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
W, = 3.355 k/ft
1405.20 in.kip < RRRERRERERERER! a
¡A mel
61.5" >|
103.17
SFD (kip)
103.17
17628.87
BMD (in.kip)
1405.30 1405.30
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
W, = 3.355 k/ft
1405.20 in.kip < RRRERRERERERER! a
¡A mel
61.5" >|
103.17
SFD (kip)
103.17
17628.87
BMD (in.kip)
1405.30 1405.30
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
103.17 kip
1405.20 in.kip
0 10317 586 0 140530 9
y Y
5.86 kip
20
t 5.86 kip © 0 o
AFD (kip) SFD (kip) BMD (in.kip)
103.17 kip
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Frame Analysis
103.17 kip
1405.20 in.kip
0 10317 586 0 140530 9
y Y
5.86 kip
20
t 5.86 kip © 0 o
AFD (kip) SFD (kip) BMD (in.kip)
103.17 kip
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
+ Frame Analysis
103.17 kip 1762887 in.kip
1405.20 1405.20
inkip inkip
103.17k |
5.86 kip 5.86 kip Y
Combined Shear Force Diagram Combined Bending Moment Diagram
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
+ Frame Analysis
103.17 kip 1762887 in.kip
1405.20 1405.20
inkip inkip
103.17k |
5.86 kip 5.86 kip Y
Combined Shear Force Diagram Combined Bending Moment Diagram
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1b]
% Beam Design
= Flexural Design
e The flexural design summary is tabulated below.
Required A,
(in?)
5.79
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90’ x 60’ Hall
Q Solution [option 1b]
% Beam Design
= Flexural Design
e The flexural design summary is tabulated below.
Required A,
(in?)
5.79
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Beam Design
= Shear Design
e The shear design for the beam will remain the same as before.
% = 103.17 kip
_ Vua = 87.23 kip
30.75"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Beam Design
= Shear Design
e The shear design for the beam will remain the same as before.
% = 103.17 kip
_ Vua = 87.23 kip
30.75"
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
6:43 bars
(skin reinforcement)
ELEVATION DETALS
Prof. Dr. Qaisar Ali einforced Concrete Design — II 88
6:43 bars
(skin reinforcement)
ELEVATION DETALS
Prof. Dr. Qaisar Ali einforced Concrete Design — II 88
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
#3@12'clc #3@12"cle
(8+3) -#3 bars
(equally spaced)
(8+3) - #3 bars
(equally spaced)
y Spacer (1)
222721
+
N (4+4) - #8 bars
E18" ——
Section A-A Section B-B
einforced Concrete Design — II 89
Prof. Dr. Qaisar Ali
#3@12'clc #3@12"cle
(8+3) -#3 bars
(equally spaced)
(8+3) - #3 bars
(equally spaced)
y Spacer (1)
222721
+
N (4+4) - #8 bars
E18" ——
Section A-A Section B-B
einforced Concrete Design — II 89
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90’ x 60’ Hall
Q Solution [option 1b]
% Column Design
e Longitudinal Reinforcement of column can be determined using
Design Aids as follows:
1. Select appropriate dimensions:
b=h=18"
2. Calculate ratio y
Assuming d' = 2.5in
18 — 2(2.5)
Yo. 0
= 0.72 = 0.70
18
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 90
Design of 90’ x 60’ Hall
Q Solution [option 1b]
% Column Design
e Longitudinal Reinforcement of column can be determined using
Design Aids as follows:
1. Select appropriate dimensions:
b=h=18"
2. Calculate ratio y
Assuming d' = 2.5in
18 — 2(2.5)
Yo. 0
= 0.72 = 0.70
18
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 90
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Column Design
3. Calculate K, and R,, factor
x =e 103.17
n= Ofibh 0.65 x 3 x 18 x 18
Ka = 0.16
e Mi 1407
M Ofibh? 0.65 x 3 x 18 x 182
Ra = 0.12
For y = 0.70, £ = 3 ksi and fy = 60 ksi, The relevant Design Aid is
DA-2 (from Appendix)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Column Design
3. Calculate K, and R,, factor
x =e 103.17
n= Ofibh 0.65 x 3 x 18 x 18
Ka = 0.16
e Mi 1407
M Ofibh? 0.65 x 3 x 18 x 182
Ra = 0.12
For y = 0.70, £ = 3 ksi and fy = 60 ksi, The relevant Design Aid is
DA-2 (from Appendix)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Read py from the graph
Pg = 0.01
Calculate Area of steel
Ag = 0.01 x 182 = 3.24 in?
Using #6 bar
No. of bars = 324, 8
0.44
045 050
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
92
Read py from the graph
Pg = 0.01
Calculate Area of steel
Ag = 0.01 x 182 = 3.24 in?
Using #6 bar
No. of bars = 324, 8
0.44
045 050
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
92
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Column Design
= Spacing of Ties
e Using #3 tie bars for #6 main bars, spacing for Tie bars according
to ACI 25.7.2.1 is minimum of
a. 16 dia. of main bar =16 x 6/8 =12" c/c
b. 48 x dia. of tie bar = 48 x (3/8) =18" c/c
c. Least column dimension =18" c/c
e Finally use #3, tie bars @ 9" c/c
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Column Design
= Spacing of Ties
e Using #3 tie bars for #6 main bars, spacing for Tie bars according
to ACI 25.7.2.1 is minimum of
a. 16 dia. of main bar =16 x 6/8 =12" c/c
b. 48 x dia. of tie bar = 48 x (3/8) =18" c/c
c. Least column dimension =18" c/c
e Finally use #3, tie bars @ 9" c/c
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
+ Column Design (Drafting) Y MÍ
—— Beam
<— #3@9." cle
18" E La
<— 8- #6 bars
So ng
Section A-A h Lap Splice: B
x
<— #3@9" cle
18"
<— 8- #6 bars |— #3@9" cle
oa 8- #6 bars —}
x >
18" |
Section B-B
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
+ Column Design (Drafting) Y MÍ
—— Beam
<— #3@9." cle
18" E La
<— 8- #6 bars
So ng
Section A-A h Lap Splice: B
x
<— #3@9" cle
18"
<— 8- #6 bars |— #3@9" cle
oa 8- #6 bars —}
x >
18" |
Section B-B
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Footing Design
= Given Data
+ Column size = 18" x 18"
+ fel =3ksi
+ fy =60ksi
+ a = 2.204 ksf
+ Reaction at the support should be used in design of footing.
Factored load on column (reaction at support) = 103.17 kips
Service load on column (reaction at support) = 81.87 kips
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
% Footing Design
= Given Data
+ Column size = 18" x 18"
+ fel =3ksi
+ fy =60ksi
+ a = 2.204 ksf
+ Reaction at the support should be used in design of footing.
Factored load on column (reaction at support) = 103.17 kips
Service load on column (reaction at support) = 81.87 kips
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of sizes
" Thickness
Assume thickness of footing “h” as 15 in. Taking size of bar as #4;
day = h= y =h-= (Cc+dp)
z = |.
putting C, = 3" and d, = 4/8" | A a
davg) | ar *
davg = 15 — (3 + 4/8) = 11.5" $ a
3 À 4 3
2
~ #4
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of sizes
" Thickness
Assume thickness of footing “h” as 15 in. Taking size of bar as #4;
day = h= y =h-= (Cc+dp)
z = |.
putting C, = 3" and d, = 4/8" | A a
davg) | ar *
davg = 15 — (3 + 4/8) = 11.5" $ a
3 À 4 3
2
~ #4
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of sizes BEL
= Required bearing area
service load
de
req =
de = 4a — Yriu(Z — h) — ye(h)
Ge = 2.204 — 0.165 — 1.25) — 0.15(1.25) = 1.63 ksf
81.87
Area = Taq = 50.23 ft?
for square footing;
B = 50.23 = 7.09 ft. Take B = 7'
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 97
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 1: Selection of sizes BEL
= Required bearing area
service load
de
req =
de = 4a — Yriu(Z — h) — ye(h)
Ge = 2.204 — 0.165 — 1.25) — 0.15(1.25) = 1.63 ksf
81.87
Area = Taq = 50.23 ft?
for square footing;
B = 50.23 = 7.09 ft. Take B = 7'
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 97
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 2: Calculation of loads (bearing pressure)
Factored load
=
pvd
By substituting values;
_ 103.17
er
Qu = 2.11ksf
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 98
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 2: Calculation of loads (bearing pressure)
Factored load
=
pvd
By substituting values;
_ 103.17
er
Qu = 2.11ksf
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 98
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90’ x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Two-way shear (punching shear)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
99
Design of 90’ x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Two-way shear (punching shear)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
99
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Two-way shear (punching shear) 3 Solution [option 1h)
Veo = Qu [B? — (c+ davg) |
On putting values
Vip = 2.11[7? — (1.5 + 11.5/12)?]
Vip = 90.64 kips
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Two-way shear (punching shear) 3 Solution [option 1h)
Veo = Qu [B? — (c+ davg) |
On putting values
Vip = 2.11[7? — (1.5 + 11.5/12)?]
Vip = 90.64 kips
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Bending moment
_ QuBk?
M, 2
Here;
2.11(7)(2.75)2 Critical section
Lo a
M, = 55.849 ft. kip or 670.19 in. kip
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 101
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 3: Analysis
= Bending moment
_ QuBk?
M, 2
Here;
2.11(7)(2.75)2 Critical section
Lo a
M, = 55.849 ft. kip or 670.19 in. kip
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 101
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90’ x 60’ Hall
Q Solution [option 1b]
> Step 4: Applying Shear Check
Dey = 40 bodavg
bo = A(c + davg)
bo = 4(18 + 11.5) = 118"
Now,
On = 4(0.75)V3000(118)(11.5)/1000
Dey = 222.98 kip
Op = 222.98 kip > Vi,» = 90.64 kip > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 102
Design of 90’ x 60’ Hall
Q Solution [option 1b]
> Step 4: Applying Shear Check
Dey = 40 bodavg
bo = A(c + davg)
bo = 4(18 + 11.5) = 118"
Now,
On = 4(0.75)V3000(118)(11.5)/1000
Dey = 222.98 kip
Op = 222.98 kip > Vi,» = 90.64 kip > OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 102
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 5: Determination of Flexural Reinforcement
2.614(670.19)
ae Bo an
a=115- [1.5 - y =031
" 670.19 PR
is = A 030 = 1. in
0.9 x 60 (11.5 -=5=)
> Step 6: Check for Minimum Flexural Reinforcement
Asmin = 0.0018Bh = 0.0018 x (7 x 12) x 15 = 2.27 in?
As = 1.09 < Asmin = 2.27 > Ag min governs!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 5: Determination of Flexural Reinforcement
2.614(670.19)
ae Bo an
a=115- [1.5 - y =031
" 670.19 PR
is = A 030 = 1. in
0.9 x 60 (11.5 -=5=)
> Step 6: Check for Minimum Flexural Reinforcement
Asmin = 0.0018Bh = 0.0018 x (7 x 12) x 15 = 2.27 in?
As = 1.09 < Asmin = 2.27 > Ag min governs!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 7: Detailing of Reinforcement
Using #4 bar with A, = 0.20 in?
227
ne or
n
_B-20 65x12,
rem 7°
Check for maximum bar spacing
Smax = Least of 3h = 3 X 15 = 45”or 18”
Provided spacing is OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
@
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 7: Detailing of Reinforcement
Using #4 bar with A, = 0.20 in?
227
ne or
n
_B-20 65x12,
rem 7°
Check for maximum bar spacing
Smax = Least of 3h = 3 X 15 = 45”or 18”
Provided spacing is OK!
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
@
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 7: Detailing of Reinforcement
For crack control, the maximum spacing between the adjacent bars
shall not exceed Sing, (Table 24.3.2).
40,000 40,000
Smax = Least of 15 —2.5C and 12
fs fs
2 2
fs = 3 fy = 3 (60,000) = 40000 psi
= Least of 15 000 2.5(3) and 12 = 7.52
u A / Mu Lee 40,000)
Provided spacing of 7” is OK. Finally provide #4@ 6” c/c.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Design of 90' x 60’ Hall
Q Solution [option 1b]
> Step 7: Detailing of Reinforcement
For crack control, the maximum spacing between the adjacent bars
shall not exceed Sing, (Table 24.3.2).
40,000 40,000
Smax = Least of 15 —2.5C and 12
fs fs
2 2
fs = 3 fy = 3 (60,000) = 40000 psi
= Least of 15 000 2.5(3) and 12 = 7.52
u A / Mu Lee 40,000)
Provided spacing of 7” is OK. Finally provide #4@ 6” c/c.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Column 18°x18"
3" clear cover
ais (each side)
{equaly spaced) “+
FOOTING PLAN
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 106
Column 18°x18"
3" clear cover
ais (each side)
{equaly spaced) “+
FOOTING PLAN
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 106
Site Pictures
Q Actual pictures of a hall of almost the same size in
Peshawar University
TR, ERA
NS A
AO AÑ \
AMA
Q Actual pictures of a hall of almost the same size in
Peshawar University
TR, ERA
NS A
AO AÑ \
AMA
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©:
Site Pictures
Q Actual pictures of a hall of almost the same size in
Peshawar University
Curtailed and Skin Reinforcement
Hall after completion
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 108
Site Pictures
Q Actual pictures of a hall of almost the same size in
Peshawar University
Curtailed and Skin Reinforcement
Hall after completion
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 108
Updated: Oct 26,2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Case Studies
e In the subsequent slides two case studies are carried out to
investigate the variation of moments in beams, moments and
slab thickness due to change in spacing between the beams.
A
90" >|
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Case Studies
e In the subsequent slides two case studies are carried out to
investigate the variation of moments in beams, moments and
slab thickness due to change in spacing between the beams.
A
90" >|
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Variation of Slab Thickness and Moments Vs Spacing Between the
Beams.
5 74 8.4 94 10.4
6.4
sed 6.5)
Max moment in slab % kip)
Prof. Dr. Qaisar Ali
teinforced Concrete Design — II 110
Variation of Slab Thickness and Moments Vs Spacing Between the
Beams.
5 74 8.4 94 10.4
6.4
sed 6.5)
Max moment in slab % kip)
Prof. Dr. Qaisar Ali
teinforced Concrete Design — II 110
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Case Studies
Q Case Study 01
Variation of Slab Thickness Vs Spacing Between the Variation of Slab Moments Vs Spacing Between the
120.00
100.00
80.00
60.00
2 4 16 18
Beams’ spacing (ft.)
2 4 16
Beams’ spacing (ft.)
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 411
Case Studies
Q Case Study 01
Variation of Slab Thickness Vs Spacing Between the Variation of Slab Moments Vs Spacing Between the
120.00
100.00
80.00
60.00
2 4 16 18
Beams’ spacing (ft.)
2 4 16
Beams’ spacing (ft.)
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 411
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
3.
M, (#ve)
einforced Concrete Design — II
(Im.
E M, (ve)
112
Prof. Dr. Qaisar Ali
3.
M, (#ve)
einforced Concrete Design — II
(Im.
E M, (ve)
112
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Variation of Beam Moments Vs Spacing Between Beams
25000.00
20000.00
15000.00
10000.00
5000.00
Moment at Mid span
0.00
6 8 10
Beams' spacing(ft.)
einforced Concrete Design — II 113
Prof. Dr. Qaisar Ali
Variation of Beam Moments Vs Spacing Between Beams
25000.00
20000.00
15000.00
10000.00
5000.00
Moment at Mid span
0.00
6 8 10
Beams' spacing(ft.)
einforced Concrete Design — II 113
Prof. Dr. Qaisar Ali
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Case Studies
Q Conclusions
1. The thickness of slab increases linearly with an increase in spacing
between beams.
2. The moments in a slab increase quadratically with an increase in the
spacing between the beams.
3. The moments in beams increase linearly with an increase in the
spacing between the beams.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Case Studies
Q Conclusions
1. The thickness of slab increases linearly with an increase in spacing
between beams.
2. The moments in a slab increase quadratically with an increase in the
spacing between the beams.
3. The moments in beams increase linearly with an increase in the
spacing between the beams.
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
References
+ Design of Concrete Structures 14th / 15th edition by Nilson, Darwin and
Dolan.
+ Building Code Requirements for Structural Concrete (ACI 318-19)
=
Dvd hie Chain Dam Antes HL Nita
a)
..
2272725
Jae o
*
Design of
Concrete Structures
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 115
References
+ Design of Concrete Structures 14th / 15th edition by Nilson, Darwin and
Dolan.
+ Building Code Requirements for Structural Concrete (ACI 318-19)
=
Dvd hie Chain Dam Antes HL Nita
a)
..
2272725
Jae o
*
Design of
Concrete Structures
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 115
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan ©
Prof. Dr. Qaisar Ali
The End of Part - I
Option 2 of the example will be covered
in Part 2 of the lecture.
CE 416: Reinforced Concrete Design - Il
Prof. Dr. Qaisar Ali
The End of Part - I
Option 2 of the example will be covered
in Part 2 of the lecture.
CE 416: Reinforced Concrete Design - Il
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Table 4.3-1 Minimum Uniformly Distributed Live Loads, L,, and Minimum Concentrated Live Loads
Apartments (See Residential)
Access floor systems,
Ollice use
Computer use
Armories and drill rooms
Assembly arcas
Fixed seats (fastened to floors)
Lobbies
Movable seats
Platforms (assembl
Stage floors
viewing stands, grandstands, and
ble:
‘Stadiums and arenas with fixed seats
(fastened to the floor)
Other assembly areas
Baleonies and decks
Catwalks for maintenance access
Corridors
First floor
Other floors
Prof. Dr. Qaisar Ali
Unto, Ly pat (m?)
(2.40)
100 (4.79)
150 (7.18)
60 (2.87)
100 (4.79)
100 (4.79)
100 (4.79)
150 (7.18)
100 (4.79)
60 (2.87)
100 (4.99)
1.5 times the live load for the
area served. Not required to.
exceed 100 psf (4.79 N/m?)
40 (1.92)
100 (4.79)
Same as oceupancy served.
xcept as indicated
Live Load
(Gee. No)
Yes (4:
Yes (47.2)
No (47.5)
No (47.5)
No (4.75)
No.
No (47.5)
No (47.5)
No (47.5)
No (4.7.5)
No (47.5)
Yes (472)
Yes (472)
Yes (472)
Mukipe Story Live
Permited? (See. No)
Yes (472)
Yes (4.7.2)
No (47.5)
No (475)
No (47.5)
No (475)
No (47.5)
No (47.5)
No (47.5)
No 7.5)
No (47.5)
Yes (472)
s 472)
CE 416: Reinforced Concrete Design - Il
300 (1.33)
117
Table 4.3-1 Minimum Uniformly Distributed Live Loads, L,, and Minimum Concentrated Live Loads
Apartments (See Residential)
Access floor systems,
Ollice use
Computer use
Armories and drill rooms
Assembly arcas
Fixed seats (fastened to floors)
Lobbies
Movable seats
Platforms (assembl
Stage floors
viewing stands, grandstands, and
ble:
‘Stadiums and arenas with fixed seats
(fastened to the floor)
Other assembly areas
Baleonies and decks
Catwalks for maintenance access
Corridors
First floor
Other floors
Prof. Dr. Qaisar Ali
Unto, Ly pat (m?)
(2.40)
100 (4.79)
150 (7.18)
60 (2.87)
100 (4.79)
100 (4.79)
100 (4.79)
150 (7.18)
100 (4.79)
60 (2.87)
100 (4.99)
1.5 times the live load for the
area served. Not required to.
exceed 100 psf (4.79 N/m?)
40 (1.92)
100 (4.79)
Same as oceupancy served.
xcept as indicated
Live Load
(Gee. No)
Yes (4:
Yes (47.2)
No (47.5)
No (47.5)
No (4.75)
No.
No (47.5)
No (47.5)
No (47.5)
No (4.7.5)
No (47.5)
Yes (472)
Yes (472)
Yes (472)
Mukipe Story Live
Permited? (See. No)
Yes (472)
Yes (4.7.2)
No (47.5)
No (475)
No (47.5)
No (475)
No (47.5)
No (47.5)
No (47.5)
No 7.5)
No (47.5)
Yes (472)
s 472)
CE 416: Reinforced Concrete Design - Il
300 (1.33)
117
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Table 4.3-1 Minimum Uniformly Distributed Live Loads, L,, and Minimum Concentrated Live Loads
Live Load Mutipe-Story Live
‘Reduction Permits? Load Reduction
‘Occupancy or Use Uniform, L pst (Km) (Gee. No) Pernod? (See. No.)
ining rooms and restaurants 100 (4:79) No (4.7.5)
Dwellings (See Residential)
Elevator machine room grating (on area of
2 in, by 2 in, (50 mum by 50 mm))
ish light floor plate construction (on =
area of Lin. by 1 in. (25 mm by 25 mm))
Fire escapes 100 (4.79) Yes (472) Yes (472)
‘On single-family dwellings only 40 (1.92) Yes 472) Yes (47.2)
Fixed ladders = =
Garages (See Section 4.10)
Passenger vehicles 40 (1.92) 474) Yes (47.4)
"Trucks and buses See Sec. 4.102 =
Helipads (See Section 4.11)
Helicopter takeoff weight 3,000 Ib No ILD
(13.35 KN) or less
Helicopter takeoff weight more than No GIL)
3.000 Ib (13.35 kN)
Hospitals
Operating rooms, laboratorios Yes (47.2) Yes (47.2) 1.000 (4:45)
Patient rooms Yes (4.7.2) Yes (47.2) 1.000 (4.45)
Corridors above first floor 80 (3.83) Yes (47.2) Yes (47.2) 1.000 (4:45)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 118
Table 4.3-1 Minimum Uniformly Distributed Live Loads, L,, and Minimum Concentrated Live Loads
Live Load Mutipe-Story Live
‘Reduction Permits? Load Reduction
‘Occupancy or Use Uniform, L pst (Km) (Gee. No) Pernod? (See. No.)
ining rooms and restaurants 100 (4:79) No (4.7.5)
Dwellings (See Residential)
Elevator machine room grating (on area of
2 in, by 2 in, (50 mum by 50 mm))
ish light floor plate construction (on =
area of Lin. by 1 in. (25 mm by 25 mm))
Fire escapes 100 (4.79) Yes (472) Yes (472)
‘On single-family dwellings only 40 (1.92) Yes 472) Yes (47.2)
Fixed ladders = =
Garages (See Section 4.10)
Passenger vehicles 40 (1.92) 474) Yes (47.4)
"Trucks and buses See Sec. 4.102 =
Helipads (See Section 4.11)
Helicopter takeoff weight 3,000 Ib No ILD
(13.35 KN) or less
Helicopter takeoff weight more than No GIL)
3.000 Ib (13.35 kN)
Hospitals
Operating rooms, laboratorios Yes (47.2) Yes (47.2) 1.000 (4:45)
Patient rooms Yes (4.7.2) Yes (47.2) 1.000 (4.45)
Corridors above first floor 80 (3.83) Yes (47.2) Yes (47.2) 1.000 (4:45)
Prof. Dr. Qaisar Ali CE 416: Reinforced Concrete Design - Il 118
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 119
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 119
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
005 010 0.15 020 025 030 035 040 045 050
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 120
005 010 0.15 020 025 030 035 040 045 050
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 120
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 121
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 121
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 122
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 122
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 123
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 123
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 424
Prof. Dr. Qaisar Ali
einforced Concrete Design — Il 424
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 125
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 125
Updated: Oct 26, 2023 Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
0.05
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 126
0.05
Prof. Dr. Qaisar Ali
einforced Concrete Design — II 126
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