Lec06 Analysis and Design of T Beams (Reinforced Concrete Design I & Prof. Ab...Hossam Shafiq II
1) T-beams are commonly used structural elements that can take two forms: isolated precast T-beams or T-beams formed by the interaction of slabs and beams in buildings.
2) The analysis and design of T-beams considers the effective flange width provided by slab interaction or the dimensions of an isolated precast flange.
3) Two methods are used to analyze T-beams: assuming the stress block is in the flange and using rectangular beam theory, or using a decomposition method if the stress block extends into the web.
Prepared by madam rafia firdous. She is a lecturer and instructor in subject of Plain and Reinforcement concrete at University of South Asia LAHORE,PAKISTAN.
The document contains 10 examples involving calculation of earth pressures on retaining structures using Rankine's and Coulomb's theories. Example 1 calculates active earth pressure on a retaining wall with and without groundwater. Example 2 determines thrust on a wall with the water table rising. Example 3 finds active pressure, point of zero pressure and center of pressure for a cohesive soil. The remaining examples involve calculating earth pressures considering various soil properties and conditions.
The document provides derivations of design equations for reinforced concrete beams. It begins by deriving the equation for maximum moment capacity of a singly reinforced beam based on concrete strength as M=0.167*fck*b*d^2. It then derives equations for doubly reinforced beams where compression steel is also required. The document further derives equations for design of flanged beams depending on whether the neutral axis lies within the flange or web. It concludes by outlining design procedures for singly and doubly reinforced beams.
A group of 16 square piles extends 12 m into stiff clay soil, underlain by rock at 24 m depth. Pile dimensions are 0.3 m x 0.3 m. Undrained shear strength of clay increases linearly from 50 kPa at surface to 150 kPa at rock. Factor of safety for group capacity is 2.5. Determine group capacity and individual pile capacity.
The group capacity is calculated to be 1600 kN. The individual pile capacity is determined to be 100 kN. The factor of safety of 2.5 is then applied to determine the safe load capacity.
Three point loads and a uniform contact pressure on a circular foundation are used to calculate the vertical stress increase at various points below the foundations. The solutions involve determining shape factors from charts and formulas to calculate the stress contribution from each loading area. The stress increases are then summed to find the total vertical stress increase at the point of interest, which ranges from 0-186 kN/m^2 depending on the example.
Lec06 Analysis and Design of T Beams (Reinforced Concrete Design I & Prof. Ab...Hossam Shafiq II
1) T-beams are commonly used structural elements that can take two forms: isolated precast T-beams or T-beams formed by the interaction of slabs and beams in buildings.
2) The analysis and design of T-beams considers the effective flange width provided by slab interaction or the dimensions of an isolated precast flange.
3) Two methods are used to analyze T-beams: assuming the stress block is in the flange and using rectangular beam theory, or using a decomposition method if the stress block extends into the web.
Prepared by madam rafia firdous. She is a lecturer and instructor in subject of Plain and Reinforcement concrete at University of South Asia LAHORE,PAKISTAN.
The document contains 10 examples involving calculation of earth pressures on retaining structures using Rankine's and Coulomb's theories. Example 1 calculates active earth pressure on a retaining wall with and without groundwater. Example 2 determines thrust on a wall with the water table rising. Example 3 finds active pressure, point of zero pressure and center of pressure for a cohesive soil. The remaining examples involve calculating earth pressures considering various soil properties and conditions.
The document provides derivations of design equations for reinforced concrete beams. It begins by deriving the equation for maximum moment capacity of a singly reinforced beam based on concrete strength as M=0.167*fck*b*d^2. It then derives equations for doubly reinforced beams where compression steel is also required. The document further derives equations for design of flanged beams depending on whether the neutral axis lies within the flange or web. It concludes by outlining design procedures for singly and doubly reinforced beams.
A group of 16 square piles extends 12 m into stiff clay soil, underlain by rock at 24 m depth. Pile dimensions are 0.3 m x 0.3 m. Undrained shear strength of clay increases linearly from 50 kPa at surface to 150 kPa at rock. Factor of safety for group capacity is 2.5. Determine group capacity and individual pile capacity.
The group capacity is calculated to be 1600 kN. The individual pile capacity is determined to be 100 kN. The factor of safety of 2.5 is then applied to determine the safe load capacity.
Three point loads and a uniform contact pressure on a circular foundation are used to calculate the vertical stress increase at various points below the foundations. The solutions involve determining shape factors from charts and formulas to calculate the stress contribution from each loading area. The stress increases are then summed to find the total vertical stress increase at the point of interest, which ranges from 0-186 kN/m^2 depending on the example.
- Deep beams are defined as beams with a shear span to depth ratio of less than 2. They behave differently than ordinary beams due to two-dimensional loading and non-linear stress distributions.
- Deep beams transfer significant load through compression forces between the load and supports. Shear deformations are more prominent.
- Design of deep beams requires considering two-dimensional effects, non-linear stress distributions, and large shear deformations. Procedures include checking minimum thickness, designing for flexure and shear, and detailing reinforcement.
1. The nominal resisting moment of reinforced concrete beams with compression steel is calculated as the sum of two parts: the moment due to compression concrete and tensile steel, and the moment due to compression steel and tensile steel.
2. The strain in the compression steel is checked to determine if it has yielded, and then the compression stress is calculated.
3. The analysis procedure involves determining the neutral axis location, checking compression steel yield, and calculating section ductility and design moment strength.
The document provides details on the design of a reinforced concrete column footing to support a column load of 1100kN from a 400mm square column. It describes the design process which includes determining the footing size, calculating bending moment, reinforcement requirements, checking shear capacity and development length. The design example shows a 3.5m x 3.5m square footing with 12mm diameter bars at 100mm c/c is adequate to support the given load based on the specified material properties and design codes. Reinforcement and footing details are also provided.
This document provides an overview of the design of steel beams. It discusses various beam types and sections, loads on beams, design considerations for restrained and unrestrained beams. For restrained beams, it covers lateral restraint requirements, section classification, shear capacity, moment capacity under low and high shear, web bearing, buckling, and deflection checks. For unrestrained beams, it discusses lateral torsional buckling, moment and buckling resistance checks. Design procedures and equations for determining effective properties and capacities are also presented.
This document discusses the slope-deflection method for analyzing beams and frames. It provides the theory and equations of the slope-deflection method. Examples are included to demonstrate how to use the method to determine support reactions, member end moments, and draw bending moment and shear force diagrams.
This document discusses the design of beams. It defines different types of beams like floor beams, girders, lintels, purlins, and rafters. It describes how beams are classified based on their support conditions as simply supported, cantilever, fixed, or continuous beams. Commonly used beam sections include universal beams, compound beams, and composite beams. The document also covers plastic analysis of beams, classification of beam sections, and failure modes of beams.
This document provides an example of designing a rectangular reinforced concrete beam. It includes calculating the loads, bending moment, required tension reinforcement, checking shear capacity and deflection. For a simply supported beam with a uniformly distributed load, the document calculates the steel reinforcement area required using formulas and tables. It then checks that the beam satisfies requirements for shear capacity, minimum and maximum steel ratios, and deflection. The document also provides an example of designing a doubly reinforced beam.
1) The document discusses the analysis of flanged beam sections like T-beams and L-beams. It covers topics like effective flange width, positive and negative moment regions, and ACI code provisions for estimating effective flange width.
2) Examples are provided for analyzing a T-beam and an L-beam section. This includes calculating the effective flange width, checking steel strain, minimum reinforcement requirements, and computing nominal moments.
3) Reinforcement limitations for flange beams are also outlined, covering requirements for flanges in compression and tension.
The document provides information about calculating wind load on an industrial building located in Chennai, India. It gives the dimensions of the building as 15m x 30m with a frame span of 15m and column height of 6m. It outlines the process to calculate the design wind speed using factors for risk, terrain, and topography. It then calculates the design wind pressure and uses this to calculate the wind load on the walls and roof of the building, finding values of 28.8 kN for the walls and 38.7 kN for the roof.
This document is the sixth edition of the National Structural Code of the Philippines (NSCP) Volume I, which provides requirements for designing buildings, towers, and other vertical structures. It was published in 2010 by the Association of Structural Engineers of the Philippines. The code contains chapters on minimum design loads, materials, and other topics to guide structural design in compliance with the latest standards. The foreword expresses pride in the publication and updates to the code to regulate structural design for safety.
This document provides information on the structural design of a simply supported reinforced concrete beam. It includes:
- A list of students enrolled in an elementary structural design course.
- Equations and diagrams showing the forces and stresses in a reinforced concrete beam with a singly reinforced bottom section.
- Limits on the maximum depth of the neutral axis according to the grade of steel.
- Examples of analyzing the stresses and determining steel reinforcement for a given beam cross-section.
- A design example calculating the dimensions and steel reinforcement for a rectangular beam with a factored uniform load.
The document discusses the analysis of statically determinate structures, including defining idealized structures, explaining the principle of superposition and equations of equilibrium, classifying structures as determinate or indeterminate, and providing examples of determining reactions on beams and frames through applying the equations of equilibrium.
This document discusses different types of braced excavation systems used to support deep excavations, including soldier beams with lagging, sheet piles, and slurry trenches. It describes the design process for braced cuts, which involves analyzing stability, ground movements, and structural elements like sheet piles and struts. Methods for determining loads on structural elements using tributary area and equivalent beam approaches are presented. Factors affecting stability like heaving in soils are discussed. Design of structural components like struts, wales, and sheet piles is also covered.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Compression members are structural members subjected to axial compression or compressive forces. Their design is governed by strength and buckling capacity. Columns can fail due to local buckling, squashing, overall flexural buckling, or torsional buckling. Built-up columns use components like lacings, battens, and cover plates to help distribute stress more evenly and increase buckling resistance compared to a single member. Buckling occurs when a straight compression member becomes unstable and bends under a critical load.
The document discusses different types of soil settlement including immediate, primary, and secondary consolidation settlements. It provides formulas to calculate settlement, defines concepts like void ratio, compression index, coefficient of consolidation, and overconsolidation ratio. It also includes sample calculations for estimating primary consolidation settlement of a clay layer under a surcharge load based on laboratory consolidation test results and given soil properties.
The document discusses different methods of designing concrete structures, focusing on the limit state method. It describes the limit state method's goal of achieving an acceptable probability that a structure will not become unsuitable for its intended use during its lifetime. The document then discusses stress-strain curves for concrete and steel. It covers stress block parameters and equations for calculating the depth of the neutral axis and moment of resistance for singly reinforced concrete beams. The document concludes by providing examples of analyzing an existing beam section and designing a new beam section.
Stiffness method of structural analysisKaran Patel
This method is a powerful tool for analyzing indeterminate structures. One of its advantages over the flexibility method is that it is conducive to computer programming.
Stiffness method the unknowns are the joint displacements in the structure, which are automatically specified.
Lec04 Analysis of Rectangular RC Beams (Reinforced Concrete Design I & Prof. ...Hossam Shafiq II
This document discusses the ultimate flexural analysis of reinforced concrete beams according to building codes. It covers topics such as concrete stress-strain relationships, stress distributions at failure, nominal and design flexural strength, moments in beams, tension steel ratios, minimum steel requirements, ductile and brittle failure modes, and calculations for balanced and maximum steel ratios. Diagrams illustrate key concepts regarding stress blocks, strain distributions, and section types. Formulas are presented for determining balanced steel ratio, maximum steel ratio, and checking neutral axis depth.
Doubly reinforced beams have both tension and compression reinforcement, allowing for a shallower beam depth than a singly reinforced beam. There are two cases for the behavior of doubly reinforced beams at ultimate loading:
1) Case I occurs when both tension and compression steel yield. The neutral axis depth can be calculated and the moment capacities from compression steel, concrete, and tension steel determined.
2) Case II occurs when only the tension steel yields, and the compression steel does not yield. The strain in the compression steel must be calculated.
The document discusses the behavior of doubly reinforced beams under ultimate loading conditions for both cases when compression steel does and does not yield. It provides equations to calculate forces, strains, and moment
- Deep beams are defined as beams with a shear span to depth ratio of less than 2. They behave differently than ordinary beams due to two-dimensional loading and non-linear stress distributions.
- Deep beams transfer significant load through compression forces between the load and supports. Shear deformations are more prominent.
- Design of deep beams requires considering two-dimensional effects, non-linear stress distributions, and large shear deformations. Procedures include checking minimum thickness, designing for flexure and shear, and detailing reinforcement.
1. The nominal resisting moment of reinforced concrete beams with compression steel is calculated as the sum of two parts: the moment due to compression concrete and tensile steel, and the moment due to compression steel and tensile steel.
2. The strain in the compression steel is checked to determine if it has yielded, and then the compression stress is calculated.
3. The analysis procedure involves determining the neutral axis location, checking compression steel yield, and calculating section ductility and design moment strength.
The document provides details on the design of a reinforced concrete column footing to support a column load of 1100kN from a 400mm square column. It describes the design process which includes determining the footing size, calculating bending moment, reinforcement requirements, checking shear capacity and development length. The design example shows a 3.5m x 3.5m square footing with 12mm diameter bars at 100mm c/c is adequate to support the given load based on the specified material properties and design codes. Reinforcement and footing details are also provided.
This document provides an overview of the design of steel beams. It discusses various beam types and sections, loads on beams, design considerations for restrained and unrestrained beams. For restrained beams, it covers lateral restraint requirements, section classification, shear capacity, moment capacity under low and high shear, web bearing, buckling, and deflection checks. For unrestrained beams, it discusses lateral torsional buckling, moment and buckling resistance checks. Design procedures and equations for determining effective properties and capacities are also presented.
This document discusses the slope-deflection method for analyzing beams and frames. It provides the theory and equations of the slope-deflection method. Examples are included to demonstrate how to use the method to determine support reactions, member end moments, and draw bending moment and shear force diagrams.
This document discusses the design of beams. It defines different types of beams like floor beams, girders, lintels, purlins, and rafters. It describes how beams are classified based on their support conditions as simply supported, cantilever, fixed, or continuous beams. Commonly used beam sections include universal beams, compound beams, and composite beams. The document also covers plastic analysis of beams, classification of beam sections, and failure modes of beams.
This document provides an example of designing a rectangular reinforced concrete beam. It includes calculating the loads, bending moment, required tension reinforcement, checking shear capacity and deflection. For a simply supported beam with a uniformly distributed load, the document calculates the steel reinforcement area required using formulas and tables. It then checks that the beam satisfies requirements for shear capacity, minimum and maximum steel ratios, and deflection. The document also provides an example of designing a doubly reinforced beam.
1) The document discusses the analysis of flanged beam sections like T-beams and L-beams. It covers topics like effective flange width, positive and negative moment regions, and ACI code provisions for estimating effective flange width.
2) Examples are provided for analyzing a T-beam and an L-beam section. This includes calculating the effective flange width, checking steel strain, minimum reinforcement requirements, and computing nominal moments.
3) Reinforcement limitations for flange beams are also outlined, covering requirements for flanges in compression and tension.
The document provides information about calculating wind load on an industrial building located in Chennai, India. It gives the dimensions of the building as 15m x 30m with a frame span of 15m and column height of 6m. It outlines the process to calculate the design wind speed using factors for risk, terrain, and topography. It then calculates the design wind pressure and uses this to calculate the wind load on the walls and roof of the building, finding values of 28.8 kN for the walls and 38.7 kN for the roof.
This document is the sixth edition of the National Structural Code of the Philippines (NSCP) Volume I, which provides requirements for designing buildings, towers, and other vertical structures. It was published in 2010 by the Association of Structural Engineers of the Philippines. The code contains chapters on minimum design loads, materials, and other topics to guide structural design in compliance with the latest standards. The foreword expresses pride in the publication and updates to the code to regulate structural design for safety.
This document provides information on the structural design of a simply supported reinforced concrete beam. It includes:
- A list of students enrolled in an elementary structural design course.
- Equations and diagrams showing the forces and stresses in a reinforced concrete beam with a singly reinforced bottom section.
- Limits on the maximum depth of the neutral axis according to the grade of steel.
- Examples of analyzing the stresses and determining steel reinforcement for a given beam cross-section.
- A design example calculating the dimensions and steel reinforcement for a rectangular beam with a factored uniform load.
The document discusses the analysis of statically determinate structures, including defining idealized structures, explaining the principle of superposition and equations of equilibrium, classifying structures as determinate or indeterminate, and providing examples of determining reactions on beams and frames through applying the equations of equilibrium.
This document discusses different types of braced excavation systems used to support deep excavations, including soldier beams with lagging, sheet piles, and slurry trenches. It describes the design process for braced cuts, which involves analyzing stability, ground movements, and structural elements like sheet piles and struts. Methods for determining loads on structural elements using tributary area and equivalent beam approaches are presented. Factors affecting stability like heaving in soils are discussed. Design of structural components like struts, wales, and sheet piles is also covered.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Compression members are structural members subjected to axial compression or compressive forces. Their design is governed by strength and buckling capacity. Columns can fail due to local buckling, squashing, overall flexural buckling, or torsional buckling. Built-up columns use components like lacings, battens, and cover plates to help distribute stress more evenly and increase buckling resistance compared to a single member. Buckling occurs when a straight compression member becomes unstable and bends under a critical load.
The document discusses different types of soil settlement including immediate, primary, and secondary consolidation settlements. It provides formulas to calculate settlement, defines concepts like void ratio, compression index, coefficient of consolidation, and overconsolidation ratio. It also includes sample calculations for estimating primary consolidation settlement of a clay layer under a surcharge load based on laboratory consolidation test results and given soil properties.
The document discusses different methods of designing concrete structures, focusing on the limit state method. It describes the limit state method's goal of achieving an acceptable probability that a structure will not become unsuitable for its intended use during its lifetime. The document then discusses stress-strain curves for concrete and steel. It covers stress block parameters and equations for calculating the depth of the neutral axis and moment of resistance for singly reinforced concrete beams. The document concludes by providing examples of analyzing an existing beam section and designing a new beam section.
Stiffness method of structural analysisKaran Patel
This method is a powerful tool for analyzing indeterminate structures. One of its advantages over the flexibility method is that it is conducive to computer programming.
Stiffness method the unknowns are the joint displacements in the structure, which are automatically specified.
Lec04 Analysis of Rectangular RC Beams (Reinforced Concrete Design I & Prof. ...Hossam Shafiq II
This document discusses the ultimate flexural analysis of reinforced concrete beams according to building codes. It covers topics such as concrete stress-strain relationships, stress distributions at failure, nominal and design flexural strength, moments in beams, tension steel ratios, minimum steel requirements, ductile and brittle failure modes, and calculations for balanced and maximum steel ratios. Diagrams illustrate key concepts regarding stress blocks, strain distributions, and section types. Formulas are presented for determining balanced steel ratio, maximum steel ratio, and checking neutral axis depth.
Doubly reinforced beams have both tension and compression reinforcement, allowing for a shallower beam depth than a singly reinforced beam. There are two cases for the behavior of doubly reinforced beams at ultimate loading:
1) Case I occurs when both tension and compression steel yield. The neutral axis depth can be calculated and the moment capacities from compression steel, concrete, and tension steel determined.
2) Case II occurs when only the tension steel yields, and the compression steel does not yield. The strain in the compression steel must be calculated.
The document discusses the behavior of doubly reinforced beams under ultimate loading conditions for both cases when compression steel does and does not yield. It provides equations to calculate forces, strains, and moment
The document discusses the design of doubly reinforced beams using the limit state method. Doubly reinforced beams are needed when the required depth is restricted but the beam needs to resist a higher bending moment. This is done by providing compression reinforcement in addition to tension reinforcement. The minimum compression steel is 0.4% of the section area and maximum is 4%. The yield stress of compression steel is taken as 0.87 times the yield stress in tension. The analysis involves assuming a neutral axis depth and checking force equilibrium.
The document summarizes key concepts about pre-stressed concrete design. It discusses the working stress design (WSD) method, which assumes linear stress-strain behavior and uses allowable stress levels. The document outlines WSD assumptions and procedures for analyzing rectangular beams, including transformed section properties and determining steel ratio effects. It also describes the internal couple method and use of double reinforcement when maximum moment exceeds allowable.
Lec 11 12 -flexural analysis and design of beamsCivil Zone
This document provides information on the flexural analysis and design of reinforced concrete beams based on the ultimate strength design method. It discusses under-reinforced and over-reinforced failure modes. For under-reinforced beams, it describes the three stages of loading: uncracked stage, cracked stage, and steel yielding stage. Equations are derived for calculating the reinforcement ratio ρ. For over-reinforced beams, it discusses failure when the concrete reaches its strain limit before steel yields. The document also provides guidelines for determining if a section is under-reinforced or over-reinforced, and criteria for selection of the strength reduction factor Φ. In the end, it lists the data and outputs required for capacity analysis of a
This document provides information on the flexural analysis and design of reinforced concrete beams based on the ultimate strength design method. It discusses under-reinforced and over-reinforced failure modes. For under-reinforced beams, it describes the three stages of loading: initial uncracked stage, cracked stage, and steel yielding stage. Equations are derived for calculating the reinforcement ratio ρ. For over-reinforced beams, it discusses failure when the concrete reaches its strain limit before steel yields. The document also provides guidelines for determining if a section is under-reinforced or over-reinforced based on reinforcement ratio and steel strain limits. It concludes with an overview of analyzing the load carrying capacity of a singly reinforced rectangular beam using strength
This document provides information on analysis and design of reinforced concrete beams. It discusses key concepts such as modular ratio, neutral axis, stress diagrams, and types of reinforcement. It also defines under-reinforced, balanced, and over-reinforced beam sections. Several examples are provided to illustrate determination of neutral axis depth, moment of resistance, steel percentage, and stresses in concrete and steel reinforcement. Design aspects like maximum load capacity are also explained through examples.
This document provides definitions and design considerations for singly reinforced concrete beams. It defines key terms like overall depth, effective depth, clear cover, and neutral axis. It explains that a singly reinforced beam only has steel reinforcement in the tensile zone below the neutral axis. Beam design aims to select member dimensions and reinforcement amount to safely support loads over the structure's lifetime. Singly reinforced beams can be designed as balanced, under-reinforced, or over-reinforced sections depending on steel reinforcement ratio. Basic design rules cover effective span, depth, bearing capacity, deflection limits, and reinforcement requirements.
Sheryar Bismil
Student of Mirpur University of Science & Technology(MUST).
Student of Final Year Civil Engineering Department Main campus Mirpur.
Here we Gonna to learn about the basic to depth wise study of Plan Reinforced Concrete-i.
From basis terminology to wide information about the analysis and design of Concrete member like column,Beam,Slab,etc.
Sheryar Bismil
Student of Mirpur University of Science & Technology(MUST).
Student of Final Year Civil Engineering Department Main campus Mirpur.
Here we Gonna to learn about the basic to depth wise study of Plan Reinforced Concrete-i.
From basis terminology to wide information about the analysis and design of Concrete member like column,Beam,Slab,etc.
good for engineering students
to get deep knowledge about design of singly reinforced beam by working stress method.
see and learn about rcc structure....................................................
The document provides information on the design of singly reinforced concrete beams. It defines key terms like overall depth, effective depth, clear cover, neutral axis, and lever arm. It describes the types of beam sections as balanced, under-reinforced and over-reinforced. Under-reinforced beams are designed for economy and provide warning before failure, while over-reinforced beams fail suddenly from concrete overstress. The procedure for designing singly reinforced beams using the working stress method is outlined in steps involving calculating design constants, assuming beam dimensions, determining loads, finding steel area required, and checking for shear and deflection requirements.
T-Beam Design by USD method-10.01.03.102Sadia Mitu
This document defines and describes T-beams, which are concrete beams with a flange formed by a monolithically cast slab. It provides definitions of T-beams, explaining that the slab acts as a compression flange while the web below resists shear and separates bending forces. The document outlines the ultimate strength design method and effective flange width concept used in T-beam analysis and design. It then presents the design procedure for T-beams, discussing analysis of positive and negative bending moments as well as singly and doubly reinforced beams. Advantages and disadvantages of T-beams are listed at the end.
chapter 4 flexural design of beam 2021.pdfAshrafZaman33
This chapter discusses the flexural analysis and design of beams. It covers fundamental assumptions for bending and shear stresses in beams. It also discusses bending behavior of homogeneous and reinforced concrete beams. The chapter includes analysis of cracked and uncracked beam sections, and design for flexure including underreinforced, overreinforced and balanced conditions. It also covers design of doubly reinforced beams, T-beams and practical considerations like concrete cover and bar spacing.
This document contains a summary of key concepts related to the design of reinforced concrete structures. It begins with multiple choice questions testing knowledge of topics like modulus of rupture, bleeding of concrete, factors affecting concrete strength, and design philosophies. It then covers the design of various structural elements like beams, slabs, and shear reinforcement. Questions are included on the design of singly reinforced beams, doubly reinforced beams, flanged beams, shear design, bond and torsion. Key terms are also defined related to limit states and partial safety factors.
This document provides an overview of the design of rectangular reinforced concrete beams that are singly or doubly reinforced. It defines key assumptions in the design process including plane sections remaining plane after bending. It also covers evaluation of design parameters such as moment factors, strength reduction factors, and balanced reinforcement ratios. The design procedures for singly and doubly reinforced beams are described including checking crack width for singly reinforced beams. Figures are also provided to illustrate concepts such as stress distributions and the components of a doubly reinforced beam.
Design of rectangular & t beam using usdTipu Sultan
1) The document discusses the design of T-beams and rectangular reinforced concrete beams. It provides definitions of beams, T-beams, and their key components.
2) Methods for calculating the effective flange width of T-beams and analyzing the strengths of T-beam sections are presented. Design equations are given for singly and doubly reinforced beam design.
3) The design process described includes determining steel reinforcement areas for the flange and web of T-beams to resist nominal bending moments, based on the effective flange width and strength calculations.
The document discusses guidelines for detailing reinforcement in concrete structures. It begins by defining detailing as the preparation of working drawings showing the size and location of reinforcement. Good detailing ensures reinforcement and concrete interact efficiently. The document then discusses sources of tension in concrete structures from various loading conditions like bending, shear, and connections. It provides equations from AS3600-2009 for calculating minimum development lengths for reinforcing bars to develop their yield strength based on bar size, concrete strength, and transverse reinforcement. It also discusses lap splice requirements. In summary, the document provides best practice guidelines for detailing reinforcement to efficiently resist loads and control cracking in concrete structures.
This document discusses the behavior and design of doubly reinforced concrete beams. It describes that doubly reinforced beams have both tension and compression reinforcement to allow for shallower beam depths. There are two possible cases for doubly reinforced beams at the ultimate limit state: 1) both the tension and compression steel yield, or 2) only the tension steel yields, while the compression steel remains elastic. The document provides equations for analyzing each case to determine the forces in the steel and concrete and the beams' moment capacity.
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3. It provides examples of different types of bridges and explains the basic structural systems used in bridges, including simply supported, cantilever, and continuous beams as well as rigid frames.
Lec11 Continuous Beams and One Way Slabs(1) (Reinforced Concrete Design I & P...Hossam Shafiq II
The document discusses reinforced concrete continuity and analysis methods for continuous beams and one-way slabs. It describes how steel reinforcement must extend through members to provide structural continuity. The ACI/SBC coefficient method of analysis is summarized, which uses coefficient tables to determine maximum shear forces and bending moments for continuous beams and one-way slabs under various loading conditions in a simplified manner compared to elastic analysis. Requirements for applying the coefficient method include having multiple spans with ratios less than 1.2, prismatic member sections, and live loads less than 3 times dead loads.
Lec10 Bond and Development Length (Reinforced Concrete Design I & Prof. Abdel...Hossam Shafiq II
This document discusses bond and development length in reinforced concrete. It defines bond as the adhesion between concrete and steel reinforcement, which is necessary to develop their composite action. Bond is achieved through chemical adhesion, friction from deformed bar ribs, and bearing. Development length refers to the minimum embedment length of a reinforcement bar needed to develop its yield strength by bonding to the surrounding concrete. The development length depends on factors like bar size, concrete strength, bar location, and transverse reinforcement. It also provides equations from design codes to calculate the development length for tension bars, compression bars, bundled bars, and welded wire fabric. Hooked bars can be used when full development length is not available, and the document discusses requirements for standard hook geome
Lec09 Shear in RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)Hossam Shafiq II
This document discusses shear in reinforced concrete beams. It covers shear stress and failure modes, shear strength provided by concrete and steel stirrups, design according to code provisions, and critical shear sections. Key points include: transverse loads induce shear stress perpendicular to bending stresses; shear failure is brittle and must be designed to exceed flexural strength; nominal shear strength comes from concrete and steel stirrups according to code equations; design requires checking section adequacy and providing minimum steel area and maximum stirrup spacing. Critical shear sections for design are located a distance d from supports.
Lec05 Design of Rectangular Beams with Tension Steel only (Reinforced Concret...Hossam Shafiq II
The document discusses design considerations for rectangular reinforced concrete beams with tension steel only. It covers topics such as beam proportions, deflection control, selection of reinforcing bars, concrete cover, bar spacing, effective steel depth, minimum beam width, and number of bars. Beam proportions should have a depth to width ratio of 1.5-2 for normal spans and up to 4 for longer spans. Minimum concrete cover and bar spacings are specified to protect the steel. Effective steel depth is the distance from the extreme compression fiber to the steel centroid. Design assumptions must be checked against the final design.
Lec03 Flexural Behavior of RC Beams (Reinforced Concrete Design I & Prof. Abd...Hossam Shafiq II
The document discusses the behavior and analysis of reinforced concrete beams. It describes three stages that beams undergo as loading increases: 1) the uncracked concrete stage, 2) the cracked-elastic stage, and 3) the ultimate strength stage. It also discusses assumptions made in flexural theory, stress-strain curves for concrete and steel, and methods for calculating stresses in uncracked and cracked beams using the transformed area method. Key points covered include cracking moment, modular ratio, and the three-step transformed area method for cracked sections.
Better Builder Magazine brings together premium product manufactures and leading builders to create better differentiated homes and buildings that use less energy, save water and reduce our impact on the environment. The magazine is published four times a year.
This is an overview of my current metallic design and engineering knowledge base built up over my professional career and two MSc degrees : - MSc in Advanced Manufacturing Technology University of Portsmouth graduated 1st May 1998, and MSc in Aircraft Engineering Cranfield University graduated 8th June 2007.
Online train ticket booking system project.pdfKamal Acharya
Rail transport is one of the important modes of transport in India. Now a days we
see that there are railways that are present for the long as well as short distance
travelling which makes the life of the people easier. When compared to other
means of transport, a railway is the cheapest means of transport. The maintenance
of the railway database also plays a major role in the smooth running of this
system. The Online Train Ticket Management System will help in reserving the
tickets of the railways to travel from a particular source to the destination.
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...DharmaBanothu
The Network on Chip (NoC) has emerged as an effective
solution for intercommunication infrastructure within System on
Chip (SoC) designs, overcoming the limitations of traditional
methods that face significant bottlenecks. However, the complexity
of NoC design presents numerous challenges related to
performance metrics such as scalability, latency, power
consumption, and signal integrity. This project addresses the
issues within the router's memory unit and proposes an enhanced
memory structure. To achieve efficient data transfer, FIFO buffers
are implemented in distributed RAM and virtual channels for
FPGA-based NoC. The project introduces advanced FIFO-based
memory units within the NoC router, assessing their performance
in a Bi-directional NoC (Bi-NoC) configuration. The primary
objective is to reduce the router's workload while enhancing the
FIFO internal structure. To further improve data transfer speed,
a Bi-NoC with a self-configurable intercommunication channel is
suggested. Simulation and synthesis results demonstrate
guaranteed throughput, predictable latency, and equitable
network access, showing significant improvement over previous
designs
An In-Depth Exploration of Natural Language Processing: Evolution, Applicatio...DharmaBanothu
Natural language processing (NLP) has
recently garnered significant interest for the
computational representation and analysis of human
language. Its applications span multiple domains such
as machine translation, email spam detection,
information extraction, summarization, healthcare,
and question answering. This paper first delineates
four phases by examining various levels of NLP and
components of Natural Language Generation,
followed by a review of the history and progression of
NLP. Subsequently, we delve into the current state of
the art by presenting diverse NLP applications,
contemporary trends, and challenges. Finally, we
discuss some available datasets, models, and
evaluation metrics in NLP.
We have designed & manufacture the Lubi Valves LBF series type of Butterfly Valves for General Utility Water applications as well as for HVAC applications.
Lec07 Analysis and Design of Doubly Reinforced Beam (Reinforced Concrete Design I & Prof. Abdelhamid Charif)
1. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif
Analysis and Design of
Doubly Reinforced Beams
Definitions
• The steel that is occasionally used on the compression side of
beams is called compression steel.
• Beams with both tensile and compressive steel are also
referred to as doubly reinforced beams.
• Depth of compression steel is d’ = cover + ds + db’/2
2
d
b
sA
N.A.
'
d
'
sA
c
CE370 : Prof. Abdelhamid Charif
• Bending contribution of steel increases
with lever arm
• For tension steel a high value of depth d
is desirable and for compression steel a
low value of depth d’ is desirable.
2. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 2
Why use compression steel ?
• Compression steel is not normally required in beams because
high compressive strength of concrete decreases the need for
such reinforcement.
• Occasionally, however, space or aesthetic requirements limit
beams to such sizes that compression steel is needed in addition
to tensile steel.
• A beam designed with tension steel only but not satisfying
tension-control condition (εt < 0.005) must either have
compression reinforcement or increased dimensions.
• Compression steel may also be provided for practical reasons only
(to fix stirrups and prevent their movement during casting and
vibration)
3CE370 : Prof. Abdelhamid Charif
Why use compression steel ? (cont.)
• Compressive steel increases moment capacity of RC sections as
well as their ductility (higher ultimate curvature).
• Though expensive, compression steel makes beams tough and
ductile, enabling them to withstand large moments,
deformations, and stress reversals such as might occur during
earthquakes.
• Many codes for earthquake zones require that certain minimum
amounts of compression steel be included in flexural members.
• Compression steel reduces long-term deflections due to
shrinkage and plastic flow.
• Tests of doubly reinforced concrete beams have shown that
even if the compression concrete crushes, the beam may very
well not collapse if the compression steel is enclosed by stirrups.
4CE370 : Prof. Abdelhamid Charif
3. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 3
5
Effect of compression steel
on sustained load deflection
bd
A'
s
ratiosteelnCompressio'
CE370 : Prof. Abdelhamid Charif
6
Effect of compression steel on ductility
Limited effect on strength
CE370 : Prof. Abdelhamid Charif
Sometimes the neutral axis is close to the compression steel, the
strain and stress are therefore very small. Thus compression
steel adds little moment capacity to the beam. It can, however,
increase beam ductility and reduce long term deflections.
4. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 4
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif
Analysis of RC Beams with
Compression Steel
Analysis of RC beams
with compression steel
• In beams with compression steel (doubly reinforced),
the amount of tension steel is high (almost maximum).
There is normally no need to check minimum steel
• Flanges in T-beams provide extra compression
capacity. T-beams do not usually require compression
steel
• Analysis and design of doubly reinforced beams is
therefore limited to rectangular sections.
8CE370 : Prof. Abdelhamid Charif
5. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 5
Strain Compatibility in
Doubly Reinforced Beams
Tension steel in one layer or more.
Compression steel in one layer.
Tensile steel assumed yielding but in
analysis problems, this may be untrue
Compression steel may yield or not.
Strain expressions are derived from
strain compatibility of plane
sections (use of similar triangles).
9
d
b
sA
N.A.
003.0
t
'
s'd
'
sA
c
td
s
mind
min
• As , d , εs = Area, depth and strain at tension steel centroid
• dt , εt = Depth and strain at bottom tension layer (max. depth)
• dmin , εmin = Depth and strain at minimum depth tension layer
• A’s , d’ , ε’s = Area, depth and strain at compression steel
CE370 : Prof. Abdelhamid Charif
Strain Compatibility in
Doubly Reinforced Beams
10
Strain expressions are derived
using similar triangles :
ysy
ysss
s
s
t
t
s
f
E
f
c
dc
c
cd
c
cd
c
cd
'
''
'
'
min
min
if
if
:stresssteelnCompressio
'
003.0:nCompressio
003.0:iondepth tensMin.
003.0:tensionBottom
003.0:tensionCentroid
d
b
sA
N.A.
003.0
t
'
s'd
'
sA
c
td
s
mind
min
CE370 : Prof. Abdelhamid Charif
6. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 6
11
(a) Section
a
abfC cc
'
85.0
'
85.0 cf
ys fAT
(*)'''
sss fAC
d
b
sA
N.A.
003.0
'
s'd
'
sA
c
s
(b) Strains (c) Forces
Forces in Doubly Reinforced Beams
'''
''
'
85.0:concretedisplacedforaccountTo
concrete)displacedconsider(modify to:forcencompressioSteel
85.0:forcencompressioConcrete
assumed)yielding(tensile:forcetensionSteel
csss
sss
cc
ys
ffAC
fAC
abfC
fAT
(*) Compression steel
area displaces same
area of concrete
CE370 : Prof. Abdelhamid Charif
Displaced Concrete
• Compression steel bars displace an equal amount (area) of
compression concrete.
• The concrete compression force is therefore slightly
reduced
12
'''
''''''
''
85.0
85.0
:forceecompressivsteelthecorrectingbyaccountintoit
taketoconvenientmorethereforeisItarm.leversametheehasand
levelsteelncompressioat thelocatedisconcretedisplacedThis
85.0reductionForce
csss
scssssss
sc
ffAC
AfAfCAfC
Af
CE370 : Prof. Abdelhamid Charif
7. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 7
13
(a) Section
a
abfC cc
'
85.0
'
85.0 cf
ys fAT
''''
85.0 csss ffAC
d
b
sA
N.A.
003.0
'
s'd
'
sA
c
s
(b) Strains (c) Forces
Moment in Doubly Reinforced Beams
'''''
''''
85.0
2
85.0
steelileabout tensforcesncompressioofMomentmomentNominal
85.085.0:mequilibriuForce
ddffA
a
dabfM
ffAabffACCT
csscn
csscyssc
CE370 : Prof. Abdelhamid Charif
14
Nominal Moment in
Doubly Reinforced Beams
eq.(b)'''''
85.0
2
85.0momentNominal ddffA
a
dabfM csscn
eq.(a)
bf
ffAfA
a
ffAabffACCT
c
csys
csscyssc
s
'
'''
''''
85.0
85.0
:thusisdepthblockStress
85.085.0:mequilibriuForce
ysy
ysss
s
f
E
f
'
''
'
if
if
:stresssteelnCompressio
notoryieldedhassteelncompressiowhetherisdifficultymainThe
CE370 : Prof. Abdelhamid Charif
8. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 8
Analysis of Doubly Reinforced Beams
• Assume tension steel has yielded
• Assume compression steel has yielded
• Compute stress block depth a and neutral axis depth c
• Compute compression and tension steel strains
• If compression and tension steel have yielded, OK continue,
perform strain checks and compute nominal moment
• If compression steel has not yielded, express unknown stress
block depth and compression steel stress in terms of neutral
axis depth c and solve resulting quadratic equation. Perform
strain checks and compute nominal moment.
• If tension steel has not yielded, use adapted method
15CE370 : Prof. Abdelhamid Charif
Analysis of Doubly Reinforced Beams
(Compression steel has yielded)
16
c
cd
c
dc
a
c
bf
ffAfA
bf
ffAfA
a
fff
ss
c
cysys
c
csys
yss
s
003.0,003.0:strainsSteel
85.0
85.0
85.0
85.0
:eq.(a)fromdepthblockstressCompute
:yieldedhavesteeltensionandncompressioAssume
'
'
1
'
''
'
'''
'
''''
85.0
2
85.0
:eq.(b)hmoment witnominalcompute
:thenassumed,as,yieldedhavesteeltensionandncompressioIf
ddffA
a
dabfM cyscn
CE370 : Prof. Abdelhamid Charif
9. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 9
Analysis of Doubly Reinforced Beams
(Compression steel has not yielded) (1)
17
''''
''
''
'''
85.085.0
:equationmequilibriuforceBack to
600003.0200000
:yieldednothassteelncompressioIf
csscyssc
ysss
sssys
ffAabffACCT
f
c
dc
c
dc
Ef
Ef
085.060085.0
85.060085.0
''
'
'
1
'
'
'
'
1
'
ysscsc
cscys
fAAf
c
dc
Acbf
f
c
dc
AcbffA
Substituting the stress block depth (a = β1c) and the compression
steel stress in terms of neutral axis depth gives:
CE370 : Prof. Abdelhamid Charif
Analysis of Doubly Reinforced Beams
(Compression steel has not yielded) (2)
18CE370 : Prof. Abdelhamid Charif
Multiplying all terms by c and assembling leads to a quadratic
equation with respect to the neutral axis depth c :
eq.(b)'''''
2''
''''
1
'
'
1
'
'
'
1
'
''''2
85.0
2
85.0:OKIf
yieldncompressionoandyieldioncheck tensandstrainssteelCompute
1
4
1
2
:issolutionpositiveThe
85.0
600
85.0
with
0
ddffA
a
dabfM
RPP
dPRPP
c
b
A
R
bf
A
P
bf
fA
P
dPcRPPc
csscn
s
c
s
c
ys
10. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 10
19
Analysis Problem-1
MPa20andMPa420
shownbeamtheofcapacitymomentdesigntheDetermine
'
cy ff
All dimensions in mm
750
350
60'
d
202
324
624
66
684d
mm
d
dd
mm
d
dhd
b
s
b
s
60
2
20
1004
2
cover'
68466750
2
32
1004750
2
cover
'
2
2
'
2
2
3.628
4
20
2
0.3217
4
32
4
mmA
mmA
s
s
CE370 : Prof. Abdelhamid Charif
20
ysys ff ''
:yieldedhassteelncompressioAssume
mma
bf
ffAfA
a
c
cysys
527.184
3502085.0
2085.04203.6284200.3217
85.0
85.0
:depthblockstressCompute '
''
yieldingassumedOK0021.000217.0
09.217
6009.217
003.0003.0:strainsteelnCompressio
09.217
85.0
527.184
:depthN.A.
''
'
'
1
yss
s
c
dc
mm
a
c
Solution 1
control-OK tension005.000645.0
09.217
09.217684
003.0003.0:strainTension
tt
t
t
c
cd
CE370 : Prof. Abdelhamid Charif
11. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 11
21
mkNM
mkNmmNM
M
ddffA
a
dabfM
n
n
n
csscn
.9.7267.80790.0momentDesign
.7.807.107.807
606842085.04203.628
2
527.184
684350527.1842085.0
85.0
2
85.0
:eq.(b)usingmomentNominal
6
'''''
Solution 1 – Cont.
CE370 : Prof. Abdelhamid Charif
22
Analysis Problem-2
Analysis of a rectangular section 300 x 600 mm with six 20-mm bars
in two layers as tension steel and three 20-mm bars as compression
reinforcement. Net spacing between steel layers is 30 mm.
Stirrups have 10-mm diameter
MPafMPaf yc 42025'
2'
2
21
2min
12
1
48.942
96.1884
515
2
490
4905054020
54060
60101040'
mmA
mmA
mm
dd
d
mmdd
mmSdd
mmhdd
mmd
s
s
l
t
d
300
206
'd
203
td 600mind
CE370 : Prof. Abdelhamid Charif
12. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 12
23
ysys ff ''
:yieldedhassteelncompressioAssume
mma
bf
ffAfA
a
c
cysys
09.62
3002585.0
2585.042048.94242096.1884
85.0
85.0
:depthblockstressCompute '
''
yieldingNot0021.0000536.0
05.73
6005.73
003.0003.0:strainsteelnCompressio
05.73
85.0
09.62
:depthN.A.
''
'
'
1
yss
s
c
dc
mm
a
c
Solution 2
CE370 : Prof. Abdelhamid Charif
Solution 2 – Cont.
(Compression steel has not yielded)
24CE370 : Prof. Abdelhamid Charif
Neutral axis depth is the solution of a quadratic equation :
mmc
c
P
RP
b
A
R
bf
A
P
bf
fA
P
RPP
dPRPP
c
s
c
s
c
ys
046.105
1
696.335765.1041007.146
6035765.1044
1
2
696.335765.1041007.146
35765.104
85.03002585.0
48.942600
696.3
85.0300
48.942
1007.146
85.03002585.0
42096.1884
85.0
600
85.0
with
1
4
1
2
2
'
'
1
'
'
1
'
'
'
1
'
2''
''''
13. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 13
Solution 2 – Cont.
(Compression steel has not yielded old)
25
control-OK tensionandyieldOK005.0and
011.0
046.105
046.105490
003.0
003.0:depthminimumatstrainTension
)420(293.257
046.105
60046.105
600600
yieldingnotOK0021.000129.0
046.105
60046.105
003.0003.0:strainsteelnCompressio
mm105.046c
minmin
min
min
min
'
'
s
''
'
'
y
y
yss
s
c
cd
MPafMPa
c
dc
f
c
dc
CE370 : Prof. Abdelhamid Charif
Solution 2 – Cont.
(Compression steel has not yielded)
26
mkNM
mkNmmNM
M
mmcaMPaf
ddffA
a
dabfM
n
n
n
csscn
.06.33295.36890.0:momentDesign
.95.368.1095.368
605152585.0293.25748.942
2
289.89
515300289.892585.0
289.89046.10585.0and293.257:with
85.0
2
85.0:eq.(b)frommomentNominal
6
1
'
s
'''''
CE370 : Prof. Abdelhamid Charif
14. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 14
27
Observations about effect of compression steel :
• Three compression steel bars (50% of tension steel) have added
9.36 kN.m (2.9%) only to the beam flexural capacity, because of
the reduced stress (61 % of yield value).
• Little advantage in strength from compression steel
• Main advantage of compression steel is in increasing ductility
and reducing long term deflections
mkNM
mkNM
n
n
.06.33295.36890.0:momentDesign
.95.368:momentNominal
Discussion – Effect of compression steel
mkNM
mkNM
n
n
.70.322559.35890.0:momentDesign
.559.358:momentNominal
:steelncompressiowithoutbeamSame
CE370 : Prof. Abdelhamid Charif
28
Analysis Problem-3
Analysis of a rectangular section 250 x 600 mm with six 28-mm bars
in two layers as tension steel and two 20-mm bars as compression
reinforcement. Net spacing between steel layers is 28 mm.
Stirrups have 10-mm diameter
MPafMPaf yc 42018'
2'
2
21
12min
1
3.628202
5.3694286
508
2
4805653628
536141040
60101040'
mmA
mmA
mm
dd
d
mmSddd
mmhdd
mmd
s
s
l
t
CE370 : Prof. Abdelhamid Charif
d
250
286
'd
202
td 600mind
15. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 15
29
yielding)notlayersboth(yieldingnotsteelTension00103.0
05.399
05.399536
003.0003.0:strainsteelTension
yieldingncompressioOK0021.000255.0
05.399
6005.399
003.0003.0:strainsteelnCompressio
05.399
85.0
19.339
:depthN.A.
19.339
2501885.0
1885.04203.6284205.3694
85.0
85.0
:depthblockstressCompute
:yieldedhassteelncompressioAssume
1
1
''
'
'
1
'
''
''
yt
t
yss
s
c
cysys
ysys
c
cd
c
dc
mm
a
c
mma
bf
ffAfA
a
ff
Solution 3
CE370 : Prof. Abdelhamid Charif
Analysis of Doubly Reinforced Beams
Tension steel has not yielded
30CE370 : Prof. Abdelhamid Charif
As shown in the previous problem 3, tension steel does not yield
when the neutral axis depth is rather large, which usually leads to
yielding of the compression steel. We therefore assume initially that
compression steel has yielded.
d
b
1
2
s
s
A
A
'd
'
sA
1d2d
Lumping of tension layers cannot be
used since there is no yielding.
We treat the case of problem 3 with
two tension layers in a general way
using strain compatibility.
Both tension layers are not yielding.
2
21 25.1847283 mmAA ss
16. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 16
CE370 : Prof. Abdelhamid Charif 31
yss
ssss
ss
ff
c
dc
c
cd
f
c
cd
Ef
c
cd
c
cd
''
2
2
1
11
2
2
1
1
'
003.0:stressstrain /nCompressio
600600:stressesTension
003.0003.0:strainsTension
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Cont.
a
abfC cc
'
85.0
'
85.0 cf
111 ss fAT
'''
85.0 cyss ffAC
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s
1d
222 ss fAT
CE370 : Prof. Abdelhamid Charif 32
c
cd
A
c
cd
AffAcbf
TTCC
h
dT
h
dTd
h
C
ah
CM
sscysc
sc
scn
2
2
1
1
''
1
'
21
2211
600600)85.0(85.0
:mequilibriuForce
22
'
222
:centroidsectionaboutcomputedmomentNominal
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Cont.
a
abfC cc
'
85.0
'
85.0 cf
111 ss fAT
'''
85.0 cyss ffAC
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s
1d
222 ss fAT
17. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 17
CE370 : Prof. Abdelhamid Charif 33
1
)(
)(4
1
2
:issolutionpositiveThe
85.0
)85.0(
85.0
600
85.0
600
0)(
:asexpressed-rebecanhichequation wQuadratic
0)(600)(600)85.0(85.0
600600)85.0(85.0
2
21
221121
1
'
''
1
'
2
2
1
'
1
1
221121
2
221121
''2
1
'
2
2
1
1
''
1
'
QPP
dPdPQPP
c
bf
ffA
Q
bf
A
P
bf
A
P
dPdPcQPPc
dAdAcAAffAcbf
c
cd
A
c
cd
AffAcbf
c
cys
c
s
c
s
sssscysc
sscysc
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Cont.
CE370 : Prof. Abdelhamid Charif 34
mmc
c
QPP
dPdPQPP
c
bf
ffA
Q
bf
A
bf
A
PP
c
cys
c
s
c
s
54.320
1
)21.789.3409.340(
)4809.3405369.340(4
1
2
21.789.3409.340
1
)(
)(4
1
2
:issolutionpositiveThe
21.78
85.02501885.0
)1885.0420(3.628
85.0
)85.0(
9.340
85.02501885.0
25.1847600
85.0
600
85.0
600
:3problemfornapplicatioNumerical
2
2
21
221121
1
'
''
1
'
2
1
'
1
21
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Solution 3 - Cont.
18. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 18
CE370 : Prof. Abdelhamid Charif 35
kN
c
cd
ATkN
c
cd
AT
kNffACkNcbfC
MPaffMPaEfMPaEf
c
dc
c
cd
c
cd
mmc
ss
cysscc
ysssssss
ys
ys
ys
37.55160001.745600
27.254)85.0(16.104285.0
420,48.298,3.403
OK00244.0
54.320
6054.320
003.0
'
003.0
OK0014924.0
54.320
54.320480
003.0003.0
OK0020165.0
54.320
54.320536
003.0003.0
:forcesandstressesstrains,variousCompute54.320
2
22
1
11
''
1
'
'
2211
'
2
2
1
1
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Solution 3 - Cont.
CE370 : Prof. Abdelhamid Charif 36
mkNM
mkNM
h
dT
h
dTd
h
C
ah
CM
n
n
scn
.40.32978.50665.0
yielding)(No65.0
.78.506:obtainon wesubstitutiAfter
22
'
222
:momentNominal 2211
a
16.1042cC
'
85.0 cf
01.7451 T
27.254sC
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s
1d
37.5512 T
Analysis of Doubly Reinforced Beams
Tension steel has not yielded – Solution 3 - Cont.
19. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 19
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif
Design of doubly reinforced beams
according to ACI and SBC 304
RC design of beams with compression steel
according to ACI and SBC 304
• If a section, designed with tension steel only, is not tension-
controlled, compression steel is then required.
• The solution is to set tension steel strain equal to minimum
value of 0.005 and then design accordingly.
• The neutral axis depth and strains become therefore all known.
• With many tension layers, satisfying tension-control at the
bottom layer will in general satisfy yield condition at the
tension steel layers.
• If however it is suspected that this might not be true, then the
neutral axis depth will be chosen to satisfy both conditions.
38CE370 : Prof. Abdelhamid Charif
20. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 20
39
y
yy
d
cc
10003
3
:depthminimumsteelatYield
min
min
t
t
tt d
d
cc 375.0
8
3
005.0
:layerbottomatcontrol-Tension
y
t dd
c
10003
3
,
8
3
Min
:conditionsbothSatisfying
min
RC design of beams with compression steel
according to ACI and SBC 304
d
b
sA
N.A.
003.0
005.0t
'
s
'
sA
c
td
s
mind
y min
CE370 : Prof. Abdelhamid Charif
40
RC design of beams with compression steel
Design steps
cβa
dd
c
y
t
1
min
knownthusisblockstressecompressivconcreteofDepth
steeltensilemaximumProviding
10003
3
,
8
3
Min
conditionsstrainsteelnsionsatisfy tetoaxisneutralSet the2/
layersofnumberexpectedtoaccordingdepthsvariousEstimate1/
continuerequired,steelnCompressioIf
onlysteelh tensionDesign wit
required,notsteelnCompressioIf
2
85.0
:steelncompressiohoutmoment witnominalCompute3/
0
0
'
0
un
un
cn
MM
MM
a
dabfM
CE370 : Prof. Abdelhamid Charif
21. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 21
41
RC design of beams with compression steel
Design steps – Cont.
'''
'
'
'''''
85.0
2
85.0
:areasteelncompressioRequired
85.0
2
85.0
settingbyareasteelncompressiorequiredDetermine5/
ddff
a
dabf
M
A
ddffA
a
dabf
M
M
MM
cs
c
u
s
cssc
u
n
un
MPaEf
ff
c
dc
ssssys
ysys
s
''''
''
'
200000If
If
'
003.0
:stressandstrainsteelnCompressio4/
CE370 : Prof. Abdelhamid Charif
42
RC design of beams with compression steel
Design steps – Cont.
y
cssc
s
csscyssc
f
ffAabf
A
ffAabffACCT
''''
''''
85.085.0
:areasteeltensionRequired
85.085.0
mequilibriuforceusingareasteeltensionrequiredDetermine6/
CE370 : Prof. Abdelhamid Charif
Although the tension control condition is enforced from the start,
the actually provided steel areas (when choosing bar numbers) may
violate this condition and strain checks are therefore required.
The provided tension steel area must not be excessive as compared
to the required value. This topic will be elaborated later.
22. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 22
43
Design Problem-1
85.0MPa30andMPa420
kN.m510andkN.m753
1
'
cy
LD
ff
MM
mmd
mmdd
mmdd
mmhd
t
70'
67030
73030
700100
min
CE370 : Prof. Abdelhamid Charif
800
375
'd
'
sA
sA
d
Expecting two tension steel layers, the
various steel depths are estimated as :
Design the shown 375 x 800 mm rectangular beam subjected to
the following dead and live moments :
kN.mM
MMM
u
LDu
0.1392
)510(7.1)375(4.17.14.1:momentUltimate
44
Solution 1
mmca
mmc
dd
c
y
t
6875.23275.27385.0
75.27312.394,75.273Min
0021.010003
6703
,
8
7303
Min
10003
3
,
8
3
Min
conditionssteelnsionsatisfy tetoaxisneutralSet the
1
min
requiredsteelnCompressio).1392(
.8.11687.12989.0
.7.1298.107.1298
2
6875.232
7003756875.2323085.0
2
85.0
:steelncompressiohoutmoment witNominal
0
0
6
0
'
0
mkNMM
mkNM
mkNmNM
a
dabfM
un
n
n
cn
CE370 : Prof. Abdelhamid Charif
23. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 23
45
Solution 1 – Cont.
MPaff
c
dc
ysys
s
420)0021.0(
00223.0
75.273
7075.273
003.0
'
003.0
:stressandstrainsteelnCompressio
''
'
)4.1140223(8.997
707003085.0420
2
6875.232
7003756875.2323085.0
90.0
101392
85.0
2
85.0
:areasteelncompressioRequired
22'
6
'
'''
'
'
mmmmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s
CE370 : Prof. Abdelhamid Charif
Solution 1 – Cont.
46
)0.6434328(0.6235
420
3085.04208.9973756875.2323085.0
85.085.0
:areasteeltensionRequired
22
''''
mmmmA
A
f
ffAabf
A
s
s
y
cssc
s
CE370 : Prof. Abdelhamid Charif
4.1140223/8.997
:)(steelncompressioActual/Required
0.6434328/0.6235
:)(steeltensionActual/Required
'
2
2
s
s
A
mm
A
mm
800
375
'd
223
328
d
24. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 24
Using 32 mm for bar spacing (not less than bar diameter), and
10 mm stirrups, the maximum number of 32-mm bars in one layer is:
4
67.4
3232
0423210632375
cover26
max
n
n
Sd
ddSb
n
bb
bsb
47
Two layers are therefore required as assumed (4+4).
For layer spacing, we use 30 mm.
Solution 1 – Cont.
800
375
'd
223
328
d
CE370 : Prof. Abdelhamid Charif 48
Solution 1 – Cont.
375
'd
223
328
d
4.1140223/8.997:)(steelncompressioActual/Required
0.6434328/0.6235:)(steeltensionActual/Required
'2
2
s
s
Amm
Amm
mm
d
dd
mm
dd
d
mmdSddd
mmdd
d
dhdd
b
s
bl
t
b
st
61111040
2
cover'
703
2
6723230734
734161040800
2
cover
'
21
1min2
1
1
The difference between the required and provided values of steel
areas may lead to violation of the tension control condition. Strain
checks are therefore required using actual steel areas and depths .
25. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 25
CE370 : Prof. Abdelhamid Charif 49
Solution 1 – Cont.
375
'd
223
328
d
mmd
mm
dd
d
mmdd
mmdd t
61'
703
2
672
734
21
min2
1
The actual tension steel depths are slightly greater than was
initially safely assumed.
The actual compression steel depth is less than was assumed and
this also is on the safe side.
The steel strain checks are however still required because of the
excess in the provided steel areas.
50
ysys ff ''
:yieldedhassteelncompressiothatbeforefoundwasasAssume
yieldingassumedOK0021.000234.0
11.277
6111.277
003.0003.0:strainsteelnCompressio
11.277
85.0
54.235
:depthN.A.
54.235
3753085.0
3085.04204.11404200.6434
85.0
85.0
:depthblockstressactualCompute
''
'
'
1
'
''
yss
s
c
cysys
c
dc
mm
a
c
mma
bf
ffAfA
a
CE370 : Prof. Abdelhamid Charif
Solution 1 – Cont.
26. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 26
51
Solution 1 – Cont.
CE370 : Prof. Abdelhamid Charif
control-tensionOKNOT005.000495.0
00495.0
11.277
11.277734
003.0003.0
:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK0043.0
11.277
11.277672
003.0003.0
:depthminimumatcheckYield
min
min
t
t
t
y
c
cd
c
cd
The section turns out to be not tension controlled because in the
provided steel areas, there was more excess in the tension steel:
2'
2
6.1428.9974.1140:nCompressio
0.19900.62350.6434:Tension
:Required-ProvidedareasteelExcess
mmA
mmA
s
s
CE370 : Prof. Abdelhamid Charif 52
Solution 1 – Cont.
375
'd
282
328
d
2'
2
'2
2
7.2338.9975.1231:nCompressio
0.19900.62350.6434:Tension
5.1231282/8.997:)(steelncompressioActual/Required
0.6434328/0.6235:)(steeltensionActual/Required
mmA
mmA
Amm
Amm
s
s
s
s
mm
d
dd
mmddd
mmddmmdd
b
s
t
64141040
2
cover'
:depthsteelncempressioNew
7032/)(
672734
:beforeasaredepthssteelTension
'
21
min21
Let us now try another solution with more excess in compression
steel (keeping the same bar reinforcement in tension):
27. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 27
53
ysys ff ''
:yieldedhassteelncompressioagain thatAssume
yieldingassumedOK0021.00023.0
689.272
64689.272
003.0003.0:strainsteelnCompressio
689.272
85.0
786.231
:depthN.A.
786.231
3753085.0
3085.04205.12314200.6434
85.0
85.0
:depthblockstressactualCompute
''
'
'
1
'
''
yss
s
c
cysys
c
dc
mm
a
c
mma
bf
ffAfA
a
CE370 : Prof. Abdelhamid Charif
Solution 1 – Cont.
54
Solution 1 – Cont.
CE370 : Prof. Abdelhamid Charif
control-tensionOK005.000508.0
689.272
689.272734
003.0
003.0:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK00439.0
689.272
689.272672
003.0003.0
:depthminimumatcheckYield
min
min
t
t
t
y
c
cd
c
cd
OKmkNMM
mkNmmNM
M
ddffA
a
dabfM
un
n
n
csscn
.0.13923.148974.16549.0momentDesign
.74.1654.1074.1654
647033085.04205.1231
2
786.231
703375786.2313085.0
85.0
2
85.0:(optional)checkMoment
6
'''''
28. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 28
Over Design and Tension Control Condition
• Recall the maximum tensile steel ratio which corresponds to a
tensile strain of 0.005 (Figure a).
• To maintain force equilibrium, any further increase in the
tensile steel area would require a longer compression block
and thus a shift downwards for the neutral axis (Figure b).
• This reduces the steel strain below the limit of 0.005 and
violates the tension control condition of SBC code.
CE370 : Prof. Abdelhamid Charif 55
d
areasteelMaximum:(a)
max,sA
003.0
005.0t
maxc
d
steelmax.thanMore:(b)
max,sA
003.0
005.0t
maxc
Over Design and Tension Control Condition
• In the design of beams with tension steel only, the risk of over
design, generating non-tension-controlled beams, is avoided by
imposing a steel strain check using the actual provided steel
area and depth (equivalent to satisfying maximum steel limit).
• The over design risk is higher in beams requiring compression
steel, as the design is based on the limit condition.
• Excess in provided tensile steel area will also shift the neutral
axis downwards and violate the tension control condition.
CE370 : Prof. Abdelhamid Charif 56
d
steelRequired:(a)
sA
003.0
005.0t
maxc
d
in tensionexcessmorewithsteelProvided:(b)
psA ,
003.0
005.0t
maxc'
sA '
, psA
29. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 29
CE370 : Prof. Abdelhamid Charif 57
Over Design and Tension Control Condition
• Excess in tensile steel must be balanced by corresponding
excess in compression steel.
• Thus in order to avoid violation of the tension control condition,
the required compression steel area must be re-determined
using the actual tension steel area (after setting bar diameter
and bar number for tension steel).
• The new compression steel area is determined using force
equilibrium and using the actual provided tension steel area:
''
'
,'
2,
''''
,
85.0
85.0
:isareasteelncompressiorequiredNew
85.085.0
cs
cyps
s
csscypssc
ff
abffA
A
ffAabffACCT
Design Summary (if compression steel required
1. Assume initial steel depths according to expected layers (do not
overestimate tension depth / underestimate compression depth)
2. Set neutral axis depth to satisfy tension steel strain conditions
and deduce compression steel strain and stress.
3. Compute required compression steel using force equilibrium.
4. Deduce required tension steel using moment equilibrium.
5. Select bar diameter, find required bar number for tension steel,
and then determine the actual provided area.
6. Deduce the new required compression steel using force
equilibrium and the actual provided tension steel area.
7. Select bar diameter, find required bar number for compression
steel, and then determine the actual provided area
8. Determine final values of steel depths, and perform moment
check if required.
CE370 : Prof. Abdelhamid Charif 58
30. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 30
59
y
cssc
s
cs
c
u
s
ssssys
ysys
s
y
t
f
ffAabf
A
ddff
a
dabf
M
A
MPaEf
ff
c
dc
cβa
dd
c
''''
'''
'
'
''''
''
'
1
min
85.085.0
:areasteeltensionRequired/4
85.0
2
85.0
:areasteelncompressioRequired3/
200000If
If
'
003.0and
10003
3
,
8
3
Min
conditionsstrainsteelnsionsatisfy tetoaxisneutralSet the2/
layersofnumberexpectedtoaccordingdepthsvariousEstimate1/
CE370 : Prof. Abdelhamid Charif
RC design steps for beams with compression steel
60
RC design steps for beams with compression steel – Cont.
design-rethen),(unsafestillifand
checkmomentperformunsafeifdepths,steelfinalDeduce/8
4
withSelect/7
85.0
85.0
:areasteelncompressiorequiredNew6/
4
withSelect/5
'''
,
2'
'
'
'
2,''
''
'
,'
2,
,
2
b
b
un
bbps
b
b
b
s
bb
cs
cyps
s
bbpsb
b
s
b
MM
ANA
d
A
A
A
Nd
ff
abffA
A
ANA
d
A
A
A
Nd
CE370 : Prof. Abdelhamid Charif
• With this strategy, the tension control condition is
always satisfied.
31. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 31
61
Re - Solve Design Problem-1
85.0MPa30andMPa420 1 '
cy ff
mmd
mmdd
mmdd
mmhd
t
70'
67030
73030
700100
min
CE370 : Prof. Abdelhamid Charif
800
375
'd
'
sA
sA
d
No need to show again that compression
steel is required. We use the same initial
steel depths:
Design the shown 375 x 800 mm rectangular beam subjected to
an ultimate moment of 1392 kN.m:
62
Re-Solution 1
mmca
mmc
dd
c
y
t
6875.23275.27385.0
75.27312.394,75.273Min
0021.010003
6703
,
8
7303
Min
10003
3
,
8
3
Min
conditionssteelnsionsatisfy tetoaxisneutralSet the
1
min
CE370 : Prof. Abdelhamid Charif
MPaff
c
dc
ysys
s
420)0021.0(
00223.0
75.273
7075.273
003.0
'
003.0
:stressandstrainsteelnCompressio
''
'
32. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 32
63
Re-Solution 1 – Cont.
2'
6
'
'''
'
'
8.997
707003085.0420
2
6875.232
7003756875.2323085.0
90.0
101392
85.0
2
85.0
:areasteelncompressioRequired
mmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s
CE370 : Prof. Abdelhamid Charif
2
,
2
''''
0.6434Φ3288:barsΦ32Using
0.6235
420
3085.04208.9973756875.2323085.0
85.085.0
:areasteeltensionRequired
mmAN
mmA
f
ffAabf
A
psb
s
y
cssc
s
Re-Solution 1 – Cont.
64CE370 : Prof. Abdelhamid Charif
2'
,
2'
2,
'
2,
''
'
,'
2,
5.1231282Use65.1200
3085.0420
3756875.2323085.04200.6434
85.0
85.0
:areasteelncompressiorequiredNew
mmAmmA
A
ff
abffA
A
pss
s
cs
cyps
s
375
'd
282
328
d
The required compression steel area
jumped from 997.8 to 1200.65 mm2
because of the provided tension steel.
This time, we find directly the second
solution which was checked before.
33. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 33
65
Design Problem-2
0.85MPa30andMPa420 1 '
cy ff
CE370 : Prof. Abdelhamid Charif
600
375
'd
'
sA
sA
d
Design the shown 375 x 600 mm rectangular beam for an ultimate
moment of 780 kN.m and with the following data:
mmd
mmdd
mmdd
mmhd
t
60'
47030
53030
500100
min
Expecting two tension steel layers, the
various steel depths are estimated as :
66
Solution 2
mmca
mmc
dd
c
y
t
9375.16875.19885.0
75.19847.276,75.198Min
0021.010003
4703
,
8
5303
Min
10003
3
,
8
3
Min
1
min
requiredsteelnCompressio).780(
.15.60428.67190.0
.28.671.1028.671
2
9375.168
5003759375.1683085.0
2
85.0
0
0
6
0
'
0
mkNMM
mkNM
mkNmNM
a
dabfM
un
n
n
cn
CE370 : Prof. Abdelhamid Charif
Set the N.A. depth to satisfy tension steel conditions:
Nominal moment without compression steel :
34. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 34
67
Solution 2 – Cont.
MPaEf
c
dc
sss
ys
s
8.418002094.0200000
yieldedNot)0021.0(
002094.0
75.198
6075.198
003.0
'
003.0
:stressandstrainsteelnCompressio
''
'
'
2'
6
'
'''
'
'
09.1129
605003085.08.418
2
9375.168
5003759375.1683085.0
90.0
10780
85.0
2
85.0
:areasteelncompressioRequired
mmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s
CE370 : Prof. Abdelhamid Charif
Solution 2 – Cont.
68
2'
,
2'
2,
''
'
,'
2,
2
,
2
''''
64.1256204Use95.1152
3085.08.418
3759375.1683085.04200.4926
85.0
85.0
:areasteelncompressiorequiredNew
4)(4layersOK two0.4926288:bars28Using
66.4903
420
3085.08.41809.11293759375.1683085.0
85.085.0
:areasteeltensionRequired
mmAmmA
ff
abffA
A
mmA
mmA
A
f
ffAabf
A
pss
cs
cyps
s
ps
s
s
y
cssc
s
CE370 : Prof. Abdelhamid Charif
35. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 35
CE370 : Prof. Abdelhamid Charif 69
mm
d
dd
mm
dd
d
mmdSddd
mmdd
d
dhdd
b
s
bl
t
b
st
60101040
2
cover'
507
2
4782830536
536141040600
2
cover
'
21
1min2
1
1
375
'd
204
288
d600
Solution 2 – Cont.
Actual steel depths are :
The actual compression steel depth is unchanged whereas the
tension steel depths are slightly greater than was initially safely
assumed. No moment check is therefore required.
CE370 : Prof. Abdelhamid Charif 70
Solution 2 – Strain Check – Optional only
Assume as was found that compression steel has not yielded. The
neutral axis depth is thus the solution of a quadratic equation :
mmcammc
c
P
RP
b
A
R
bf
A
P
bf
fA
P
RPP
dPRPP
c
s
c
s
c
ys
20.165357.19485.0357.194
1
9423.37594.925384.254
607594.924
1
2
9423.37594.925384.254
7594.92
85.03753085.0
6.1256600
9423.3
85.0375
6.1256
5384.254
85.03753085.0
4200.4926
85.0
600
85.0
with
1
4
1
2
1
2
'
'
1
'
'
1
'
'
'
1
'
2''
''''
36. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 36
71
)420(8.414002074.0200000
yieldingnotOK0021.0002074.0
357.194
60357.194
003.0003.0:strainsteelnCompressio
''
s
''
'
'
MPafMPaEf
c
dc
yss
yss
s
CE370 : Prof. Abdelhamid Charif
Solution 2 – Cont.
controlTensionOK005.000527.0
357.194
357.194536
003.0003.0
:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK0044.0
357.194
357.194478
003.0003.0
:depthminimumatcheckYield
min
min
min
t
t
t
y
c
cd
c
cd
72
surprise)(No.0.780
.196.800105.88990.0:momentDesign
.105.889.10105.889
605073085.08.4146.1256
2
2.165
5073752.1653085.0
85.0
2
85.0
6
'''''
OKmkNMM
mkNM
mkNmmNM
M
ddffA
a
dabfM
un
n
n
n
csscn
CE370 : Prof. Abdelhamid Charif
Solution 2 – Cont.
Moment check (not required)
Since the tensile steel strain conditions are satisfied and the
actual steel depths are safer than assumed, the moment check is
not required. We still perform it.
37. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 37
Design of doubly reinforced beams
Final observations
• It is important in the final design stage to comply with the design
results. Any excess in the tensile steel area may result in
violation of tension-control condition.
• Excess in tensile steel must be balanced by corresponding excess
in compression steel.
• Excess of steel may increase moment capacity but may change
the type of beam behavior and failure
• Over-design is not only uneconomical, it may also affect
structural safety
• The design method described previously avoids systematically
over design leading to violation of tension control condition.
73CE370 : Prof. Abdelhamid Charif
CE370 : Prof. Abdelhamid Charif 74
Effect of tension steel ratio
These moment curvature
relationships, produced by
RC-TOOL software, show
the ductility of under
reinforced beams.
Despite their improved
strength, over reinforced
beams have little energy
absorption capacity, which
is the main factor in seismic
design.
38. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 38
CE370 : Prof. Abdelhamid Charif 75
Effect of compression steel ratio
These moment curvature
relationships, produced by
RC-TOOL software, show
that adequate compression
steel can recover the
ductility of a brittle beam.
A measure of ductility
(energy absorption
capacity) is given by the
area under the curve.
Maximum compression steel ?
• SBC and ACI codes do not fix any maximum limit to a beam total
steel reinforcement.
• The maximum tension steel ratio is satisfied through the tension
control condition.
• If the tension control check is not satisfied, then either the section
is increased or compression steel must be provided.
• There is a risk that the second option will be abusively used
resulting in excessive compression steel.
• Should not there be a limit on the compression steel ?
• The deficiency in the section may be such that the required
compression steel may be too large and even exceed tension steel.
• This strange situation is more likely in shallow beams which are
used for architectural reasons (popular in KSA).
CE370 : Prof. Abdelhamid Charif 76
39. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 39
CE370 : Prof. Abdelhamid Charif 77
Design Problem 3
Design of a shallow beam (depth less than width) 1000 x 320 mm.
In order to hide the beam, its thickness was limited to that of the
slab (320 mm).
This “hiding” architectural constraint is mechanically obsolete and
must be restrained by the structural community.
We design this beam for four different values of the ultimate
moment (300, 400, 500, 600 kN.m).
1000
320
MPa420MPa20 y
'
c ff
For simplification, we consider
the same steel depths in all
cases based on one layer in both
tension and compression steel.
d = h – 60 = 260 mm, d’ = 60 mm
78
Solution 3
mmca
mm
d
c
875.825.9785.0
5.97
8
2603
8
3
layer)tension(oneconditionssteelnsionsatisfy tetoaxisneutralSet the
1
cases.allinrequiredsteelnCompressio
.1.2779.30790.0
.9.307.109.307
2
875.82
2601000875.822085.0
2
85.0
:steelncompressiohoutmoment witNominal
0
0
6
0
'
0
un
n
n
cn
MM
mkNM
mkNmNM
a
dabfM
CE370 : Prof. Abdelhamid Charif
40. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 40
79
Solution 3 – Cont.
ys
y
s
sssys
s
ff
f
f
MPaEf
c
dc
of%5555.0
8.230001154.0200000
001154.0
5.97
605.97
003.0
'
003.0
:stressandstrainsteelnCompressio
'
'
'''
'
y
cssc
s
cs
c
u
s
f
ffAabf
A
ddff
a
dabf
M
A
''''
'''
'
'
85.085.0
:areasteeltensionRequired
85.0
2
85.0
:areasteelncompressioRequired
CE370 : Prof. Abdelhamid Charif
Solution 3 – Cont.
80CE370 : Prof. Abdelhamid Charif
300 594.2 3656.9 0.16 0.013
400 3193.1 4979.7 0.64 0.026
500 5792.0 6302.4 0.92 0.038
600 8390.8 6525.2 1.29 0.047
mkN
Mu
. bh
AA ss
'
2
'
mm
As
2
mm
As
s
s
A
A'
The following table presents the various results of required
tension and compression steel areas for the four ultimate
moment values. Comparison ratios are also given.
41. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 41
Maximum compression steel ?
• We can observe the rapid increase in the compression
steel area, until it exceeds the tension steel value.
• To avoid such nonsense situations, a limit must be set.
• Despite the “silence” of SBC and ACI codes on this
issue, it is recommended to limit compression steel
amount to a fraction of tension steel area (say 20 to
40%) or to fix a limit on the total steel ratio.
CE370 : Prof. Abdelhamid Charif 81
Analysis of a general multi-layer RC section
using strain compatibility method - 1
• Many problems were previously solved using strain compatibility.
• A general RC section with many steel layers is now considered
with partial yielding. The most strained layers are yielding
whereas the remaining layers are in the elastic stage.
• This allows tuning to any possible case.
• The steel area, depth and strain at each layer are :
CE370 : Prof. Abdelhamid Charif 82
• Asi , di , εsi = At
tension layer i
• εt = At bottom layer
(tension control, Ф)
• A’
sj , d’
j , ε’
sj = At
compression layer j
0.003
di
a =
β1c
Depths Strains Forces
c
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
'''
85.0 csjsj ffA
sisi fA
42. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 42
CE370 : Prof. Abdelhamid Charif 83
Analysis of a general multi-layer RC section
using strain compatibility method - 2
• Steel stresses are :
c
dc
Ef
ff
c
cd
Ef
ff
j
sjssjysj
ysjysj
i
sissiysi
ysiysi
'
'''
''
600:If
:If
600:If
:If
0.003
di
a =
β1c
Depths Strains Forces
c
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
'''
85.0 csjsj ffA
sisi fA
c
dc
c
cd
j
sj
i
si
'
'
003.0
003.0
• Steel strains are :
''''
85.085.0 csjsjcsisisjcsi ffAabffACCT
• Force equilibrium:
CE370 : Prof. Abdelhamid Charif 84
Analysis of a general multi-layer RC section
using strain compatibility method - 3
Starting from an initial guess, steel layers are split in yielding (y) and
elastic (e) parts, and using a = β1c, equilibrium equation becomes:
'''''
1
'
85.0
600
85.0
600
sjcejsejsyjyceiseisyiy AfcdA
c
AfcbfcdA
c
Af
''''
85.085.0 csjsjcsisi ffAabffA
Multiplying by c and grouping leads to a general quadratic equation :
''
0
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
0
''2
0If:caseSpecial
85.0
600
85.085.0
600
85.0
with0
ejejeiei
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyjejejeieio
dPdPcP
b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPPdPdPcPc
43. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 43
CE370 : Prof. Abdelhamid Charif 85
Analysis of a general multi-layer RC section
using strain compatibility method - 4
The positive solution of this quadratic equation is :
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
0
''2
85.0
600
85.085.0
600
85.0
with0
b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPPdPdPcPc
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyjejejeieio
1
4
1
2 2
0
''
0
P
dPdPP
c
ejejeiei
Depending on the layers and their yielding state, the term P0 may
be positive or negative. The expression of the solution is sign
dependent. The absolute value and (±) are required for generality.
(+) sign is used if the term P0 is negative and (-) is used otherwise.
CE370 : Prof. Abdelhamid Charif 86
Analysis of a general multi-layer RC section
using strain compatibility method - 5
All strains are deduced
and if the assumed
yielding is violated in
any layer, a new
iteration must be
carried out using the
last yielding state.
22
85.0
22
85.0 ''''' h
dfAd
h
ffA
ah
abfM isisijcsjsjcn
Bottom tensile layer strain εt is used to check tension control and
obtain the strength reduction factor leading to design moment ФMn
The nominal moment is obtained about the section centroid as :
0.003
di
a =
β1c
Depths Strains Forces
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
'''
85.0 csjsj ffA
sisi fA
h/2
44. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 44
CE370 : Prof. Abdelhamid Charif 87
Example of multi layer general section
Strain compatibility method - 1
Analyze the 600 x 1500 shown
section with the following
material and layer spacing data :
mmS
.εMPaf
.βMPaf
l
yy
'
c
30
00260520
85022 1
mm600
1000
204
2883
375
1500
208
162
162
There are two skin layers with two Φ16
bars each. Top skin layer (d =375mm)
is considered part of the compression
steel and the second layer (d =1000mm)
is part of tension steel.
CE370 : Prof. Abdelhamid Charif 88
mmd
mmdSdd
mm
d
dd
mmd
mmdSdd
mmdSdd
mmd
d
dhdd
bl
b
s
bl
bl
b
st
375
110203060
60101040
2
cover
1000
1320581378
1378581436
14361410401500
2
cover
'
3
''
1
'
2
'
'
1
4
23
12
1
1
Example of multi layer general section
Strain compatibility method - 2
mm600
1000
204
2883
375
1500
208
162
162
There are four tension layers and
three compression layers and the
depths are :
45. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 45
CE370 : Prof. Abdelhamid Charif 89
2
2
'
3
2
2
'
1
2
2
'
1
2
2
4
2
2
321
1.402
4
16
2
6.1256
4
20
4
2.2513
4
20
8
1.402
4
16
2
0.4926
4
28
8
mmA
mmA
mmA
mmA
mmAAA
s
s
s
s
sss
Example of multi layer general section
Strain compatibility method - 3
mm600
1000
204
2883
375
1500
208
162
162
The various areas are :
CE370 : Prof. Abdelhamid Charif 90
Example of multi layer general section
Strain compatibility method - 4
We start by assuming that all steel has yielded except the skin
layers which are closer to the neutral axis. All layers in the yield
parts (y) except tension layer 4 and compression layer 3 which are
in the elastic parts (e).
The neutral axis depth is given by :
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
02
0
''
0
85.0
600
85.085.0
600
85.0
with1
4
1
2
b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPP
P
dPdPP
c
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyj
ejejeiei
46. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 46
CE370 : Prof. Abdelhamid Charif 91
The various parameters are :
1802.8
85.0600
1.4026.12562.2513
2973.25
85.06002285.0
1.402600
85.0
600
5155.68
85.06002285.0
5206.1256
85.0
0309.137
85.06002285.0
5202.2513
85.0
2973.25
85.06002285.0
1.402600
85.0
600
5876.268
85.06002285.0
5200.4926
85.0
1
'
3
'
2
'
1
1
'
'
11
'
'
3'
3
1
'
'
2'
2
1
'
'
1'
1
11
'
4
4
1
'
1
321
b
AAA
b
A
R
bf
A
P
bf
fA
P
bf
fA
P
bf
A
P
bf
fA
PPP
ssssj
c
s
e
c
ys
y
c
ys
y
c
s
e
c
ys
yyy
Example of multi layer general section
Strain compatibility method - 5
CE370 : Prof. Abdelhamid Charif 92
mmc
dPdP
c
P
dPdPP
cP
P
P
RPPPPPPPRPPPPP
RPPP
PPPP
eeee
ejejeiei
eyyyeyyeiyiejyj
eyy
eyyy
415.614
1
802.557
37510002973.254
1
2
802.557
1
802.557
4
1
2
802.557
1
4
1
2
0
802.557
1802.82973.255876.26832973.255155.680309.137
1802.82973.255155.680309.137
2973.255876.268
22
'
3
'
344
2
0
''
0
0
0
0
'
4321
'
3
'
2
'
1
'''
0
''
3
'
2
'
1
4321
Example of multi layer general section
Strain compatibility method - 6
47. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 47
CE370 : Prof. Abdelhamid Charif 93
Example of multi layer general section
Strain compatibility method - 7
OK:00117.0
415.614
375415.614
003.0
OKNOT:00246.0
415.614
110415.614
003.0
OK:0027.0
415.614
60415.614
003.0
OK:001883.0
415.614
415.6141000
003.0
OK:00345.0
415.614
415.6141320
003.0
OK:0037.0
415.614
415.6141378
003.0
OK:0040.0
415.614
415.6141436
003.0
003.0003.0
'
3
'
2
'
1
4
3
2
1
'
'
ys
ys
ys
ys
ys
ys
ys
j
sj
i
si
c
dc
c
cd
c = 614.415 mm
Steel layer strains are :
The initial assumed
yielding state is violated
at the compressive layer 2
A second round of
calculations is therefore
required using the last
yielding state.
CE370 : Prof. Abdelhamid Charif 94
All tension parameters and some of the compression
parameters remain unchanged. The new values are :
mmc
dPdPdP
c
P
RPPPPPPPRPPPPP
bf
A
P
b
A
RPP
PPPP
eeeeee
eyyyeeyeiyiejyj
c
s
e
sj
ey
eyyy
656.6171
261.547
3752973.251100563.7910002973.254
1
2
261.547
1
261.547
4
1
2
261.547
261.5471802.82973.255876.26832973.250563.790309.137
0563.79
85.06002285.0
6.1256600
85.0
600
1802.82973.250309.137
2973.255876.268
2
2
'
3
'
3
'
2
'
244
0
'
4321
'
3
'
2
'
1
'''
0
1
'
'
2'
2
1
'
''
3
'
1
4321
Example of multi layer general section
Strain compatibility method - 8
48. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 48
CE370 : Prof. Abdelhamid Charif 95
Example of multi layer general section
Strain compatibility method - 9
OK:0011786.0
656.617
375656.617
003.0
OK:0024657.0
656.617
110656.617
003.0
OK:00271.0
656.617
60656.617
003.0
OK:001857.0
656.617
656.6171000
003.0
OK:00341.0
656.617
656.6171320
003.0
OK:00369.0
656.617
656.6171378
003.0
OK:003975.0
656.617
656.6171436
003.0
003.0003.0
'
3
'
2
'
1
4
3
2
1
'
'
ys
ys
ys
ys
ys
ys
ys
j
sj
i
si
c
dc
c
cd
c = 617.656 mm
Steel layer strains are :
The assumed yielding
state is now confirmed in
all layers.
The stresses are :
ssssys
ysys
Ef
MPaff
200000
520
CE370 : Prof. Abdelhamid Charif 96
Compute the various forces to check equilibrium :
OKkNT
NfAT
NfATTT
NffAC
NffAC
NffAC
NabfC
MPafmmca
sss
yssss
csss
csss
csss
cc
c
9.7833C:checkmequilibriuForce
0.1493404.3711.402
0.25615205204926
7.87255)7.187.235(1.40285.0
3.596181)7.1814.493(6.125685.0
2.1259867)7.18520(2.251385.0
2.589061260001.5252285.085.0
7.1885.001.525656.61785.0
444
1321
''
3
'
32
''
2
'
22
''
1
'
11
'
'
1
Example of multi layer general section
Strain compatibility method - 10
49. 03-Mar-13
CE 370 : Prof. Abdelhamid Charif 49
CE370 : Prof. Abdelhamid Charif 97
The nominal moment is obtained about the section centroid as :
2
1500
10004.3711.402
2
1500
13205204926
2
1500
13785204926
2
1500
14365204926
375
2
1500
)7.187.235(1.402110
2
1500
)7.1814.493(6.1256
60
2
1500
)7.18520(2.2513
2
01.525
2
1500
60001.5252585.0
22
85.0
22
85.0
2222
'''''
'
n
isisijcsjsjcn
isijsjcn
M
h
dfAd
h
ffA
ah
abfM
h
dTd
h
C
ah
CM
Example of multi layer general section
Strain compatibility method - 11
CE370 : Prof. Abdelhamid Charif 98
The nominal moment is :
The bottom tensile layer strain εt is used to check tension control
and obtain the strength reduction factor.
Example of multi layer general section
Strain compatibility method - 11
mkNMn
y
yt
st
.45.715347.90187932.0
7932.0
0026.0005.0
0026.0003975.0
0.250.65
005.0
0.250.65
ionIn transit003975.01
mkNmmNMn .47.9018.1047.9018 6