CFD CSA A23.3 04
User Manual: CFD-CSA-A23.3-04
Open the PDF directly: View PDF .
Page Count: 83
Download | |
Open PDF In Browser | View PDF |
Concrete Frame Design Manual CSA A23.3-04 Concrete Frame Design Manual CSA A23.3-04 For ETABS® 2016 ISO ETA122815M24 Rev. 0 Proudly developed in the United States of America December 2015 Copyright Copyright Computers & Structures, Inc., 1978-2015 All rights reserved. The CSI Logo®, SAP2000®, ETABS®, and SAFE® are registered trademarks of Computers & Structures, Inc. Watch & LearnTM is a trademark of Computers & Structures, Inc. The computer programs SAP2000® and ETABS® and all associated documentation are proprietary and copyrighted products. Worldwide rights of ownership rest with Computers & Structures, Inc. Unlicensed use of these programs or reproduction of documentation in any form, without prior written authorization from Computers & Structures, Inc., is explicitly prohibited. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior explicit written permission of the publisher. Further information and copies of this documentation may be obtained from: Computers & Structures, Inc. www.csiamerica.com info@csiamerica.com (for general information) support@csiamerica.com (for technical support) DISCLAIMER CONSIDERABLE TIME, EFFORT AND EXPENSE HAVE GONE INTO THE DEVELOPMENT AND TESTING OF THIS SOFTWARE. HOWEVER, THE USER ACCEPTS AND UNDERSTANDS THAT NO WARRANTY IS EXPRESSED OR IMPLIED BY THE DEVELOPERS OR THE DISTRIBUTORS ON THE ACCURACY OR THE RELIABILITY OF THIS PRODUCT. THIS PRODUCT IS A PRACTICAL AND POWERFUL TOOL FOR STRUCTURAL DESIGN. HOWEVER, THE USER MUST EXPLICITLY UNDERSTAND THE BASIC ASSUMPTIONS OF THE SOFTWARE MODELING, ANALYSIS, AND DESIGN ALGORITHMS AND COMPENSATE FOR THE ASPECTS THAT ARE NOT ADDRESSED. THE INFORMATION PRODUCED BY THE SOFTWARE MUST BE CHECKED BY A QUALIFIED AND EXPERIENCED ENGINEER. THE ENGINEER MUST INDEPENDENTLY VERIFY THE RESULTS AND TAKE PROFESSIONAL RESPONSIBILITY FOR THE INFORMATION THAT IS USED. Contents Chapter 1 Chapter 2 Introduction 1.1 Organization 1-2 1.2 Recommended Reading/Practice 1-3 Design Prerequisites 2.1 Design Load Combinations 2-1 2.2 Design and Check Stations 2-3 2.3 Identifying Beams and Columns 2-3 2.4 Design of Beams 2-3 2.5 Design of Columns 2-4 2.6 Design of Joints 2-5 2.7 P-Delta Effects 2-6 i Concrete Frame Design CSA A23.3-04 Chapter 3 2.8 Element Unsupported Length 2-6 2.9 Choice of Input Units 2-7 Design Process 3.1 Notation 3-1 3.2 Design Load Combinations 3-4 3.3 Limits on Material Strength 3-5 3.4 Strength Resistance Factors 3-6 3.5 Column Design 3-7 3.5.1 3.5.2 3.5.3 3.5.4 3.6 3.7 ii Generation of Biaxial Interaction Surface Calculate Column Capacity Ratio Required Reinforcing Area Design Column Shear Reinforcement 3-7 3-10 3-16 3-16 Beam Design 3-26 3.6.1 Design Beam Flexural Reinforcement 3.6.2 Design Beam Shear Reinforcement 3-26 3-36 Joint Design 3-45 3.7.1 3.7.2 3.7.3 3.7.4 3-45 3-47 3-49 3-49 Determine the Panel Zone Shear Force Determine the Effective Area of Joint Check Panel Zone Shear Stress Beam-Column Flexural Capacity Ratios Appendix A Second Order P-Delta Effects Appendix B Member Unsupported Lengths and Computation of K-Factors Appendix C Concrete Frame Design Preferences Contents Appendix D Concrete Frame Overwrites Appendix E Error Messages and Warnings References iii Chapter 1 Introduction The design of concrete frames is seamlessly integrated within the program. Initiation of the design process, along with control of various design parameters, is accomplished using the Design menu. It should be noted that two design processes are available in CSiBridge: superstructure design (on the Design/Rating tab) and design of the individual elements comprising the structure (the Advanced > Frame Design commands). This manual addresses the second design process. Automated design at the object level is available for any one of a number of user-selected design codes, as long as the structures have first been modeled and analyzed by the program. Model and analysis data, such as material properties and member forces, are recovered directly from the model database, and no additional user input is required if the design defaults are acceptable. The design is based on a set of user-specified loading combinations. However, the program provides default load combinations for each design code supported in the program. If the default load combinations are acceptable, no definition of additional load combinations is required. In the design of columns, the program calculates the required longitudinal and shear reinforcement. However, the user may specify the longitudinal steel, in 1-1 Concrete Frame Design CSA A23.3-04 which case a column capacity ratio is reported. The column capacity ratio gives an indication of the stress condition with respect to the capacity of the column. The biaxial column capacity check is based on the generation of consistent three-dimensional interaction surfaces. It does not use any empirical formulations that extrapolate uniaxial interaction curves to approximate biaxial action. Interaction surfaces are generated for user-specified column reinforcing configurations. The column configurations may be rectangular, square or circular, with similar reinforcing patterns. The calculation of moment magnification factors, unsupported lengths and strength reduction factors is automated in the algorithm. Every beam member is designed for flexure and shear at output stations along the beam span. All beam-column joints are investigated for existing shear conditions. For Ductile and Moderately Ductile moment resisting frames, the shear design of the columns, beams and joints is based on the probable moment capacities of the members. Also, the program will produce ratios of the beam moment capacities with respect to the column moment capacities, to investigate weak beam/strong column aspects, including the effects of axial force. Output data can be presented graphically on the model, in tables for both input and output data, or on the calculation sheet prepared for each member. For each presentation method, the output is in a format that allows the engineer to quickly study the stress conditions that exist in the structure and, in the event the member reinforcing is not adequate, aids the engineer in taking appropriate remedial measures, including altering the design member without rerunning the entire analysis. 1.1 Organization This manual is designed to help you quickly become productive with the concrete frame design options of CSA A23.3-04. Chapter 2 provides detailed descriptions of the Deign Prerequisites used for CSA A23.3-04. Chapter 3 provides detailed descriptions of the code-specific process used for CSA 1-2 Organization Chapter 1 - Introduction A23.3-04. The appendices provide details on certain topics referenced in this manual. 1.2 Recommended Reading/Practice It is strongly recommended that you read this manual and review any applicable “Watch & Learn” Series™ tutorials, which are found on our web site, http://www.csiamerica.com, before attempting to design a concrete frame. Additional information can be found in the on-line Help facility available from within the program’s main menu. Recommended Reading/Practice 1-3 Chapter 2 Design Prerequisites This chapter provides an overview of the basic assumptions, design preconditions, and some of the design parameters that affect the design of concrete frames. In writing this manual it has been assumed that the user has an engineering background in the general area of structural reinforced concrete design and familiarity with CSA A23.3-04 codes. 2.1 Design Load Combinations The design load combinations are used for determining the various combinations of the load cases for which the structure needs to be designed/checked. The load combination factors to be used vary with the selected design code. The load combination factors are applied to the forces and moments obtained from the associated load cases and are then summed to obtain the factored design forces and moments for the load combination. For multi-valued load combinations involving response spectrum, time history, moving loads and multi-valued combinations (of type enveloping, square-root of the sum of the squares or absolute) where any correspondence between interacting quantities is lost, the program automatically produces multiple sub combinations using maxima/minima permutations of interacting quantities. 2-1 Concrete Frame Design CSA A23.3-04 Separate combinations with negative factors for response spectrum cases are not required because the program automatically takes the minima to be the negative of the maxima for response spectrum cases and the above described permutations generate the required sub combinations. When a design combination involves only a single multi-valued case of time history or moving load, further options are available. The program has an option to request that time history combinations produce sub combinations for each time step of the time history. Also an option is available to request that moving load combinations produce sub combinations using maxima and minima of each design quantity but with corresponding values of interacting quantities. For normal loading conditions involving static dead load, live load, wind load, and earthquake load, or dynamic response spectrum earthquake load, the program has built-in default loading combinations for each design code. These are based on the code recommendations and are documented for each code in the corresponding manuals. For other loading conditions involving moving load, time history, pattern live loads, separate consideration of roof live load, snow load, and so on, the user must define design loading combinations either in lieu of or in addition to the default design loading combinations. The default load combinations assume all load cases declared as dead load to be additive. Similarly, all cases declared as live load are assumed additive. However, each load case declared as wind or earthquake, or response spectrum cases, is assumed to be non additive with each other and produces multiple lateral load combinations. Also wind and static earthquake cases produce separate loading combinations with the sense (positive or negative) reversed. If these conditions are not correct, the user must provide the appropriate design combinations. The default load combinations are included in design if the user requests them to be included or if no other user-defined combination is available for concrete design. If any default combination is included in design, all default combinations will automatically be updated by the program any time the design code is changed or if static or response spectrum load cases are modified. Live load reduction factors can be applied to the member forces of the live load case on an element-by-element basis to reduce the contribution of the live load to the factored loading. 2-2 Design Load Combinations Chapter 2 - Design Prerequisites The user is cautioned that if moving load or time history results are not requested to be recovered in the analysis for some or all of the frame members, the effects of those loads will be assumed to be zero in any combination that includes them. 2.2 Design and Check Stations For each load combination, each element is designed or checked at a number of locations along the length of the element. The locations are based on equally spaced segments along the clear length of the element. The number of segments in an element is requested by the user before the analysis is made. The user can refine the design along the length of an element by requesting more segments. When using the CSA A23.3-04 design code, requirements for joint design at the beam to column connections are evaluated at the top most station of each column. The program also performs a joint shear analysis at the same station to determine if special considerations are required in any of the joint panel zones. The ratio of the beam flexural capacities with respect to the column flexural capacities considering axial force effect associated with the weak-beam/strongcolumn aspect of any beam/column intersection are reported. 2.3 Identifying Beams and Columns In the program, all beams and columns are represented as frame elements. But design of beams and columns requires separate treatment. Identification for a concrete element is accomplished by specifying the frame section assigned to the element to be of type beam or column. If any brace element exists in the frame, the brace element also would be identified as a beam or a column element, depending on the section assigned to the brace element. 2.4 Design of Beams In the design of concrete beams, in general, the program calculates and reports the required areas of steel for flexure and shear based on the beam moments, shears, load combination factors, and other criteria, which are described in detail in the code-specific chapters. The reinforcement requirements are calculated at a user-defined number of stations along the beam span. Design and Check Stations 2-3 Concrete Frame Design CSA A23.3-04 All the beams are only designed for major direction flexure and shear. Effects due to any axial forces, torsion and minor direction bending that may exist in the beams must be investigated independently by the user. In designing the flexural reinforcement for the major moment at a particular section of a particular beam, the steps involve the determination of the maximum factored moments and the determination of the reinforcing steel. The beam section is designed for the maximum positive and maximum negative factored moment envelopes obtained from all of the load combinations. Negative beam moments produce top steel. In such cases, the beam is always designed as a Rectangular section. Positive beam moments produce bottom steel. In such cases, the beam may be designed as a Rectangular beam or a T beam. For the design of flexural reinforcement, the beam is first designed as a singly reinforced beam. If the beam section is not adequate, the required compression reinforcement is calculated. In designing the shear reinforcement for a particular beam for a particular set of loading combinations at a particular station due to the beam major shear, the steps involve the determination of the factored shear force, the determination of the shear force that can be resisted by concrete, and the determination of the reinforcement steel required to carry the balance. Special considerations for seismic design are incorporated into the program for CSA A23.3-04. 2.5 Design of Columns In the design of the columns, the program calculates the required longitudinal steel, or if the longitudinal steel is specified, the column stress condition is reported in terms of a column capacity ratio, which is a factor that gives an indication of the stress condition of the column with respect to the capacity of the column. The design procedure for the reinforced concrete columns of the structure involves the following steps: 2-4 Generate axial force-biaxial moment interaction surfaces for all of the different concrete section types of the model. Check the capacity of each column for the factored axial force and bending moments obtained from each loading combination at each end of the Design of Columns Chapter 2 - Design Prerequisites column. This step is also used to calculate the required reinforcement (if none was specified) that will produce a capacity ratio of 1.0. The generation of the interaction surface is based on the assumed strain and stress distributions and some other simplifying assumptions. These stress and strain distributions and the assumptions are documented in Chapter 3. The shear reinforcement design procedure for columns is very similar to that for beams, except that the effect of the axial force on the concrete shear capacity must be considered. For certain special seismic cases, the design of columns for shear is based on the capacity shear. The capacity shear force in a particular direction is calculated from the moment capacities of the column associated with the factored axial force acting on the column. For each load combination, the factored axial load is calculated, using the load cases and the corresponding load combination factors. Then, the moment capacity of the column in a particular direction under the influence of the axial force is calculated, using the uniaxial interaction diagram in the corresponding direction, as documented in Chapter 3. 2.6 Design of Joints To ensure that the beam-column joint of special moment resisting frames possesses adequate shear strength, the program performs a rational analysis of the beam-column panel zone to determine the shear forces that are generated in the joint. The program then checks this against design shear strength. Only joints having a column below the joint are designed. The material properties of the joint are assumed to be the same as those of the column below the joint. The joint analysis is done in the major and the minor directions of the column. The joint design procedure involves the following steps: Determine the panel zone design shear force, Determine the effective area of the joint, and Check panel zone shear stress. The joint design details are documented in Chapter 3. Design of Joints 2-5 Concrete Frame Design CSA A23.3-04 2.7 P-Delta Effects The program design process requires that the analysis results include P-delta effects. The P-delta effects are considered differently for “braced” or “non-sway” and “unbraced” or “sway” components of moments in columns or frames. For the braced moments in columns, the effect of P-delta is limited to “individual member stability.” For unbraced components, “lateral drift effects” should be considered in addition to individual member stability effect. The program assumes that “braced” or “nonsway” moments are contributed from the “dead” or “live” loads. Whereas, “unbraced” or “sway” moments are contributed from all other types of loads. For the individual member stability effects, the moments are magnified with moment magnification factors, as documented in Chapter 3 of this manual. For lateral drift effects, the program assumes that the P-delta analysis is performed and that the amplification is already included in the results. The moments and forces obtained from P-delta analysis are further amplified for individual column stability effect if required by the governing code, as in the CSA A23.3-04 codes. Users should be aware that the default analysis option in the program is that P-delta effects are not included. The user can include P-delta analysis and set the maximum number of iterations for the analysis. The default number of iteration for P-delta analysis is 1. Further details on P-delta analysis are provided in Appendix A of this design manual. 2.8 Element Unsupported Lengths To account for column slenderness effects, the column unsupported lengths are required. The two unsupported lengths are l33 and l22. These are the lengths between support points of the element in the corresponding directions. The length l33 corresponds to instability about the 3-3 axis (major axis), and l22 corresponds to instability about the 2-2 axis (minor axis). Normally, the unsupported element length is equal to the length of the element, i.e., the distance between END-I and END-J of the element. The program, however, allows users to assign several elements to be treated as a single 2-6 P-Delta Effects Chapter 2 - Design Prerequisites member for design. This can be done differently for major and minor bending as documented in Appendix B of this design manual. The user has options to specify the unsupported lengths of the elements on an element-by-element basis. 2.9 Choice of Input Units English as well as SI and MKS metric units can be used for input. But the codes are based on a specific system of units. All equations and descriptions presented in the subsequent chapters correspond to that specific system of units unless otherwise noted. For example, the CSA A23.3-04 code is published in Millimeter-Newton-Second units. By default, all equations and descriptions presented in the chapter “Design for CSA A23.3-04” correspond to MillimeterNewton-Second units. However, any system of units can be used to define and design the structure in the program. It should be noted that two design processes are available in CSiBridge: superstructure design (on the Design/Rating tab) and design of the individual elements comprising the structure (the Advanced > Frame Design commands). This manual addresses the second design process. Choice of Input Units 2-7 Chapter 3 Design Process This chapter provides a detailed description of the code-specific algorithms the program uses to design concrete frames when the Canadian code CSAA23.3-04 (CSA 2004) is selected. For simplicity, all equations and descriptions presented in this chapter correspond to Newton-Millimeter-Second units unless otherwise noted. The program provides options to design or check Conventional, Moderately Ductile (moderate seismic risk areas), and Ductile (high seismic risk areas) Moment Resisting frames as required for seismic design. The details of the design criteria used for the different framing systems are described in this chapter. 3.1 Notation The various notations used in this chapter are described herein: Acv Area of concrete used to determine shear stress, sq-mm Acore Area of concrete core, sq-mm Ag Gross area of concrete, sq-mm As Area of tension reinforcement, sq-mm A′s Area of compression reinforcement, sq-mm 3-1 Concrete Frame Design CSA A23.3-04 3-2 As(required) Area of steel required for tension reinforcement, sq-mm Ast Total area of column longitudinal reinforcement, sq-mm Av Area of shear reinforcement, sq-mm a Depth of compression block, mm ab Depth of compression block at balanced condition, mm b Width of member, mm bf Effective width of flange (T-beam section), mm bw Width of web (T-beam section), mm Cm Coefficient, dependent upon column curvature, used to calculate moment magnification factor c Depth to neutral axis, mm cb Depth to neutral axis at balanced conditions, mm d Distance from compression face to tension reinforcement, mm d′ Concrete cover to center of reinforcing, mm ds Thickness of slab (T-beam section), mm Ec Modulus of elasticity of concrete, MPa Es Modulus of elasticity of reinforcement, assumed as 200,000 MPa f′ c Specified compressive strength of concrete, MPa fy Specified yield strength of flexural reinforcement, MPa fyh Specified yield strength of shear reinforcement, MPa h Dimension of beam or column, mm Ig Moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement, mm4 Ist Moment of inertia of reinforcement about centroidal axis of member cross-section, mm4 k Effective length factor L Clear unsupported length, mm Notation Chapter 3 - Design Process M1 Smaller factored end moment in a column, N-mm M2 Larger factored end moment in a column, N-mm Mc Factored moment to be used in design, N-mm Mns Non-sway component of factored end moment, N-mm Ms Sway component of factored end moment, N-mm Mf Factored moment at section, N-mm Mfx Factored moment at section about X-axis, N-mm Mfy Factored moment at section about Y-axis, N-mm Nf Factored axial load at section (tension positive), N Pb Axial load capacity at balanced strain conditions, N Pc Critical buckling strength of column, N Pr,max Maximum axial load strength allowed, N Po Axial load capacity at zero eccentricity, N Pf Factored axial load at section (compression positive), N Vc Shear resisted by concrete, N VD+L Shear force from span loading, N Vp Shear force computed from probable moment capacity, N Vf Factored shear force at a section, N Vs Shear force at a section resisted by steel, N α Reinforcing steel overstrength factor α1 Factor for obtaining average compressive stress in concrete block β Factor accounting for shear resistance of cracked concrete β1 Factor for obtaining depth of compression block in concrete βd Absolute value of the ratio of the maximum factored axial dead load moment to the maximum factored total load moment θ Angle of inclination of diagonal compressive stresses with the longitudinal axis of beam or column Notation 3-3 Concrete Frame Design CSA A23.3-04 3.2 δb Moment magnification factor for non-sway moments δs Moment magnification factor for sway moments εc Strain in concrete εs Strain in reinforcing steel φc Strength reduction factor for concrete φs Strength reduction factor for steel φm Strength reduction factor for member λ Shear strength factor Design Load Combinations The design load combinations are the various combinations of the prescribed load cases for which the structure is to be checked. The program creates a number of default design load combinations for a concrete frame design. Users can add their own design load combinations as well as modify or delete the program default design load combinations. An unlimited number of design load combinations can be specified. To define a design load combination, simply specify one or more load cases, each with its own scale factor. The scale factors are applied to the forces and moments from the load cases to form the factored design forces and moments for each design load combination. There is one exception to the preceding. For spectral analysis modal combinations, any correspondence between the signs of the moments and axial loads is lost. The program uses eight design load combinations for each such loading combination specified, reversing the sign of axial loads and moments in major and minor directions. As an example, if a structure is subjected to dead load, DL, and live load, LL, only, the CSA A23.3-04 design check may need only one design load combination, namely, 1.25 DL +1.5 LL. However, if the structure is subjected to wind, earthquake or other loads, numerous additional design load combinations may be required. The program allows live load reduction factors to be applied to the member forces of the reducible live load case on a member-by-member basis to reduce the contribution of the live load to the factored responses. 3-4 Design Load Combinations Chapter 3 - Design Process The design load combinations are the various combinations of the prescribed load cases for which the structure is to be checked. For this code, if a structure is subjected to dead load (DL), live load (LL), wind (WL), and earthquake (EL) loads, and considering that wind and earthquake forces are reversible, the following load combinations should be considered (CSA 4.1.3.2): 1.4 DL 1.25 DL + 1.50 LL (CSA 8.3.2, Table C.1, Case 1) (CSA 8.3.2, Table C.1, Case 2) 1.25 DL ± 1.40 WL 0.90 DL ± 1.40 WL 1.25 DL + 0.5 LL ± 1.40 WL (CSA 8.3.2, Table C.1, Case 4) (CSA 8.3.2, Table C.1, Case 4) (CSA 8.3.2, Table C.1, Case 4) 1.00 DL ± 1.00 EL 1.00 DL + 0.50 LL ± 1.00 EL (CSA 8.3.2, Table C.1, Case 5) (CSA 8.3.2, Table C.1, Case 5) These are also the default design load combinations in the program whenever the CSA A23.3-04 code is used. In generating the preceding default loading combinations, the importance factor is taken as 1. The user should use other appropriate design load combinations if roof live load is separately treated, or if other types of loads are present. PLL is the live load multiplied by the Pattern Live Load Factor. The Pattern Live Load Factor can be specified in the Preferences. When using the CSA A23.3-04 code, the program design assumes that a P-delta analysis has been performed. 3.3 Limits on Material Strength The upper and lower limits of f c′ should be 80 MPa and 20 MPa respectively, for all framing types (CSA 8.6.1.1, 21.2.6.1). f c′ ≤ 80 MPa (CSA 8.6.1.1, 21.2.6.1) f c′ ≥ 20 MPa (CSA 8.6.1.1) The upper limit of f y should be 500 MPa for all frames (CSA 8.5.1, 21.2.7.1) Limits on Material Strength 3-5 Concrete Frame Design CSA A23.3-04 When the compression strength of concrete used in design is beyond the given limits or when the yield strength of steel used in design exceeds the given limits, the code does not cover such cases. The code allows use of f c′ and fy beyond the given limits, provided special care is taken regarding the detailing and ductility (CSA 8.6.1.2, 21.2.7.1). The program does not enforce any of these limits for column P-M-M interaction check or design and flexure design of beam. The specified strengths are used for design. The user is responsible for using the proper strength values while defining the materials. For shear design, special limits are applicable (CSA 11.3.4, 11.3.6.4, 21.2.6.3) and enforced in the program as described herein. The code allows the use of reinforcement with fy less than 400 MPa in members resisting earthquake induced forces without any restriction. The program also allows the use of fy greater than 400 MPa. However, if fy is between 400 MPa and 500 MPa, the code requires that increased strain be taken into account because higher-yield-strength steel, in general, reduces ductility (CSA 21.2.7.1). The program also considers the increased strain through strain-compatibility relations during P-M-M interaction capacity calculations. 3.4 Strength Resistance Factors The strength resistance factor , ϕ, is material dependent and is defined as ϕc = 0.65for concrete and (CSA 8.4.2 ) 0.85for steel. ϕs = (CSA 8.4.3a) In some special cases, a member resistance factor, ϕm , is used as an additional reduction factor in addition to ϕc and ϕs (CSA 8.4.1). In connection with buckling resistance, ϕm is taken as 0.75 (CSA 10.15.3.1). The values of ϕc and ϕs given herein are the default values. These values can be modified in the Preferences. For structural concrete manufactured in pre-qualified manufacturing plants, ϕc can be taken as 0.7 (CSA 8.4.2, 16.1.3). 3-6 Strength Resistance Factors Chapter 3 - Design Process 3.5 Column Design The user may define the geometry of the reinforcing bar configuration of each concrete column section. If the area of reinforcing is provided by the user, the program checks the column capacity. However, if the area of reinforcing is not provided by the user, the program calculates the amount of reinforcing required for the column. The design procedure for the reinforced concrete columns of the structure involves the following steps: Generate axial force/biaxial moment interaction surfaces for all of the different concrete section types of the model. A typical biaxial interaction surface is shown in Figure 3-1. When the steel is undefined, the program generates the interaction surfaces for the range of allowable reinforcement: 1 to 8 percent for Conventional, Moderately Ductile Resisting frames (CSA 10.9.1 and CSA 10.9.2) and 1 to 6 percent for Ductile Moment Resisting frames (CSA 21.4.3.1). Calculate the capacity ratio or the required reinforcing area for the factored axial force and biaxial (or uniaxial) bending moments obtained from each loading combination at each station of the column. The target capacity ratio is taken as one when calculating the required reinforcing area. Design the column shear reinforcement. The following three subsections describe in detail the algorithms associated with this process. 3.5.1 Generation of Biaxial Interaction Surfaces The column capacity interaction volume is numerically described by a series of discrete points that are generated on the three-dimensional interaction failure surface. In addition to axial compression and biaxial bending, the formulation allows for axial tension and biaxial bending considerations. A typical interaction surface is shown in Figure 3-1. The coordinates of these points are determined by rotating a plane of linear strain in three dimensions on the section of the column. See Figure 3-2. The linear strain diagram limits the maximum concrete strain, εc , at the extremity of the section to 0.0035 (CSA 10.1.3). Column Design 3-7 Concrete Frame Design CSA A23.3-04 Figure 3-1 A typical column interaction surface The formulation is based consistently on the general principles of ultimate strength design (CSA 10.1), and allows for any doubly symmetric rectangular, square, or circular column section. The stress in the steel is given by the product of the steel strain and the steel modulus of elasticity, ε s Es , and is limited to the yield stress of the steel, f y (CSA 8.5.3.2). The area associated with each reinforcing bar is assumed to be placed at the actual location of the center of the bar, and the algorithm does not assume any further simplifications in the manner in which the area of steel is distributed over the cross-section of the column (such as an equivalent steel tube or cylinder), as shown in Figure 3-3. 3-8 Column Design Chapter 3 - Design Process Figure 3-2 Idealized strain distribution for generation of interaction surface The concrete compression stress block is assumed to be rectangular, with a stress value of α1 f c′ (CSA 10.1.7). See Figure 3-3. The depth of the stress block is β1c, where α= 0.85 − 0.0015 f c′ , 1 α1 ≥ 0.67 (CSA 10.1.7) β = 0.97 − 0.0025 f c′ , 1 β1 ≥ 0.67. (CSA 10.1.7) Column Design 3-9 Concrete Frame Design CSA A23.3-04 Figure 3-3 Idealization of stress and strain distribution in a column section The interaction algorithm provides a correction to account for the concrete area that is displaced by the reinforcement in the compression zone. The effects of the strength reduction factors are included in the generation of the interaction surfaces. The maximum compressive axial load is limited to Pr,max , where the maximum factored axial load resistance is given by ( ) (CSA 10.10.4) ( ) (CSA 10.10.4) Pr,max = 0.80 ϕc α1 f c′ Ag − Ast + ϕs f y Ast (tied column) Pr= 0.80 ϕc α1 f c′ Ag − Ast + ϕs f y Ast (spiral column). ,max 3.5.2 Calculate Column Capacity Ratio The column capacity is checked for each loading combination at each check station of each column. In checking a particular column for a particular loading combination at a particular location, the program uses the following steps: Determine the factored moments and forces from the load cases and the specified load combination factors to give Pf , M fx , and M fy . 3 - 10 Column Design Chapter 3 - Design Process Determine the moment magnification factors for the column moments. Apply the moment magnification factors to the factored loads obtained in the first step. Determine whether the point, defined by the resulting axial load and biaxial moment set, lies within the interaction volume. The following three sections describe in detail the algorithms associated with this process. 3.5.2.1 Determine Factored Moments and Forces The factored loads for a particular load combination are obtained by applying the corresponding load factors to all the load conditions, giving Pf, Mfx, and Mfy. The computed moments are further amplified by using “Moment Magnification Factors” to allow for “Lateral Drift Effect” and “Member Stability Effect.” 3.5.2.2 Determine Moment Magnification Factors The moment magnification factors are applied in two stages. First the moments are separated into their “sway” and “non-sway” components. The sway components are amplified for lateral drift effect (CSA 10.16). Although according to the code, this amplification is significant for “unbraced” frames only and may be avoided for “braced” frames, the program treats all frames uniformly to amplify non-sway components of moments. These amplified moments are further amplified for individual member stability effect. According to the code, the individual member stability effect is only significant if: kL > r kL > r 25 − Pf 10 M1 M2 for non-sway frames, and (CSA 10.15.2) , for sway frames. (CSA 10.16.4) fc′ Ag 35 Pf , fc′ Ag However, the program considers individual member stability effect for all compression column elements. Column Design 3 - 11 Concrete Frame Design CSA A23.3-04 3.5.2.2.1 Lateral Drift Effect For all frames, the moment magnification factor for lateral drift effect is applied only to the “sway” moment in the program. = M M ns + δ s M s (CSA 10.16.2) The moment obtained from analysis is separated into two components: the sway (Ms) and the non-sway (Mns) components. The non-sway or braced components, which are identified by “ns” subscripts are predominantly caused by gravity load. The sway components are identified by “s” subscripts. The sway moments are predominantly caused by lateral loads and are related to the cause of sidesway. The moment magnification factors in the major and minor directions can, in general, be different. The moment magnification factors for moments causing sidesway, δsx and δsy can be taken as 1.0 if a P-delta analysis is carried out (CSA 10.16.3.1). The program assumes that a P-delta analysis has been performed and, therefore, moment magnification factors for moments causing sidesway are taken as unity. For the P-delta analysis the load should correspond to a load combination of (1.25 DL + 1.40 LL)/ϕm (CSA 10.16.5), where ϕm is the strength reduction factor for stability and is equal to 0.75 (CSA 10.15.3). See also White and Hajjar (1991). The user is reminded of the special analysis requirements, especially those related to the value of EI used in analysis (CSA 10.14.1). The program analysis uses the EI of the gross cross-sectional area in conjunction with a multiplication factor. The multiplication factors are defined on a section-by-section basis, so that the effect of axial force and cracking can be modeled in a first order analysis. If the program assumptions are not satisfactory for a particular member, the user can explicitly specify values of δsx and δsy . 3.5.2.2.2 Member Stability Effects All compression members are designed using the factored axial load, Pf , from the analysis and a magnified factored moment, Mc. The magnified moment is computed as, 3 - 12 Column Design Chapter 3 - Design Process Mc = δb M 2 ≥ M 2 , (CSA 10.15.3.1) where M2 is the column maximum end moment obtained from elastic analysis after considering minimum eccentricity and lateral drift effect, and Mc is the maximum moment occurring at the end or at an interior point within the span of the column. The moment magnification factor, δb, for moments not causing sidesway associated with the major or minor direction of the column is given by = δb 1− Cm Pf ≥ 1.0, where (CSA 10.15.3.1) ϕm Pc 0.75, ϕm = Pc = π2 EI ( kL )2 (CSA 10.15.3.1) , (CSA 10.15.3.1) k is conservatively taken as 1, however the user can overwrite the value, EI is associated with a particular column direction given by EI = 0.4 Ec I g 1 + βd and , 0.6 0.4 Cm =+ Ma ≥ 0.4, Mb (CSA 10.15.3.1) (CSA 10.15.3.2) Ma and Mb are the moments at the ends of the column, and Mb is numerically larger than Ma . Ma / Mb is positive for single curvature bending and negative for double curvature bending. The preceding expression of Cm is valid if there is no transverse load applied between the supports. If transverse load is present on the span, or the length is overwritten, or for any other case, Cm = 1 (CSA 10.15.3.3). Cm can be overwritten by the user on an element- by-element basis. The pre-magnified factored moments are increased, if required, to obtain minimum eccentricities such that M /Pf is at least (15 + 0.03h) mm about each axis separately, where h is the dimension of the column in mm units in the corresponding direction (CSA 10.15.3.1). Column Design 3 - 13 Concrete Frame Design CSA A23.3-04 M ≥ Pf (15 + 0.03h ) (CSA 10.15.3.1) The moment magnification factor, δb, must be a positive number and greater than one. Therefore Pf m must be less than ϕmPc. If Pf is found to be greater than or equal to ϕmPc , a failure condition is declared. δb is taken as 1 for tension members. The preceding calculations use the unsupported length of the column. The two unsupported lengths are l22 and l33 corresponding to instability in the minor and major directions of the element, respectively. These are the lengths between the support points of the element in the corresponding directions. If the program assumptions are not satisfactory for a particular member, the user can explicitly specify values of δs and δb. 3.5.2.3 Determine Capacity Ratio As a measure of the stress condition of the column, a capacity ratio is calculated. The capacity ratio is basically a factor that gives an indication of the stress condition of the column with respect to the capacity of the column. Before entering the interaction diagram to check the column capacity, the moment magnification factors are applied to the factored loads to obtain Pf , Mfx , and Mfy. The point (Pf , Mfx , Mfy) is then placed in the interaction space, shown as point L in Figure 3-4. If the point lies within the interaction volume, the column capacity is adequate; however, if the point lies outside the interaction volume, the column is overstressed. This capacity ratio is achieved by plotting the point L and determining the location of point C. The point C is defined as the point where the line OL (if extended outwards) will intersect the failure surface. This point is determined by three-dimensional linear interpolation between the points that define the failure OL surface. See Figure 3-4. The capacity ratio, CR, is given by the ratio . OC If OL = OC (or CR = 1) the point lies on the interaction surface and the column is stressed to capacity. 3 - 14 Column Design Chapter 3 - Design Process Figure 3-4 Geometric representation of column capacity ratio If OL < OC (or CR < 1) the point lies within the interaction volume and the column capacity is adequate. If OL > OC (or CR > 1) the point lies outside the interaction volume and the column is overstressed. The maximum of all the values of CR calculated from each load combination is reported for each check station of the column, along with the controlling Pf , Mfx, and Mfy set and associated load combination number. If the reinforcing area is not defined, the program computes the reinforcement that will give an interaction ratio of unity. Column Design 3 - 15 Concrete Frame Design CSA A23.3-04 3.5.3 Required Reinforcing Area If the reinforcing area is not defined, the program computes the reinforcement that will give a column capacity ratio equal to the Utilization Factor Limit, which is set to 0.95 by default. 3.5.4 Design Column Shear Reinforcement The shear reinforcement is designed for each loading combination in the major and minor directions of the column. In designing the shear reinforcing for a particular column for a particular loading combination due to shear forces in a particular direction, the program uses the following steps: Determine the factored forces acting on the section, Mf, Pf, and Vf. Note that Mf and Pf are needed for the calculation of vc . Determine the shear stress, vc , that can be resisted by concrete alone. Calculate the reinforcement steel required to carry the balance. For Ductile and Moderately Ductile moment resisting concrete frames, the shear design of the columns is also based on the probable and nominal moment capacities, respectively, of the members in addition to the factored moments (CSA 21.4.5.1(b), 21.7.2.3(b)). Effects of the axial forces on the column moment capacities are included in the formulation. The following three sections describe in detail the algorithms associated with this process. 3.5.4.1 Determine Section Forces In the design of the column shear reinforcement of a Conventional moment resisting concrete frame, the forces for a particular load combination, namely, the column axial force, Pf , the column moment, Mf, and the column shear force, Vf , in a particular direction are obtained by factoring the program load cases with the corresponding load combination factors. In the shear design of Ductile moment resisting frames (seismic design), the following are checked in addition to the requirement for Conventional moment 3 - 16 Column Design Chapter 3 - Design Process resisting frames. In the design of Ductile moment resisting concrete frames, the design shear force, Vf , in a particular direction is also calculated from the probable moment capacities of the column associated with the factored axial force acting on the column (CSA 21.4.5.1(b)). For each load combination, the factored axial load, Pf , is calculated. Then, the positive and negative moment capacities, M +f and M −j , of the column in a particular direction under the influence of the axial force Pf are calculated using the uniaxial interaction diagram in the corresponding direction. The design shear force, Vf , is then given by V= Vp + VD + L f (CSA 21.4.5.1(b)) where, VP is the shear force obtained by applying the calculated probable ultimate moment capacities at the two ends of the column acting in two opposite directions. Therefore, VP is the maximum of VP1 and VP 2 , where VP1 = M I− + M J+ , and L VP 2 = M I+ + M J− , where L M I+ , M I− = Positive and negative moment capacities at end I of the column using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , M J+ , M J− = Positive and negative moment capacities at end J of the column using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , and L = Clear span of column. For Ductile moment resisting frames, α is taken as 1.25 (CSA 2.2, 21.4.5.1). VD + L is the contribution of shear force from the in-span distribution of gravity loads. For most of the columns, it is zero. Column Design 3 - 17 Concrete Frame Design CSA A23.3-04 The Vp is enforced not to be greater than the factored shear force component from specified gravity load increased by Rd Ro (CSA 21.4.5.1). In the design of Moderately Ductile moment resisting frames (seismic), the shear capacity of the column is also checked for the nominal shear due to the nominal ( ϕs =ϕc =1.0 ) moment capacities and the factored gravity load (CSA 21.7.2.3), in addition to the design checks required for Conventional moment resisting frames. The procedure for calculating nominal moment capacity is the same as that for computing the probable moment capacity for Ductile moment resisting frames, except that α is taken equal to 1 (CSA 2.2, 21.7.2.3a) rather than 1.25. The design shear strength is enforced not to be greater than the factored shear force with the seismic load increased by Rd Ro (CSA 21.7.2.3b). See also Table 3-1 for details. 3.5.4.2 Determine Concrete Shear Capacity Given the design force set M f , N f , and V f , the shear capacity provided by the concrete alone, Vc , is calculated as follows: Vc = ϕc λβ f c′bw d , (CSA 11.3.4) ϕc is the resistance factor for concrete. By default, it is taken as 0.65 (CSA8.4.2). For concrete produced in a pre-qualified manufacturing plants, its value can be taken as 0.70 (CSA 16.1.3). This value can be overwritten in the Preferences. λ 3 - 18 is the strength reduction factor to account for low density concrete (CSA 2.2). For normal density concrete, its value is 1 (CSA 8.6.5), which is the program default value. For concrete using lower density aggregate, the user can change the value of λ in the material properties. The recommended values for λ is as follows (CSA 8.6.5). Column Design Chapter 3 - Design Process 1.00, 0.85, λ = 0.75, for normal density concrete, for semi-low-density concrete in which all of the fine aggregate is natural sand, for semi-low-density concrete in which none of the fine aggregate is natural sand. β is the factor for accounting for the shear resistance of cracked concrete (CSA 2.2). Its value is normally between 0.1 and 0.4. It is determine according to section 11.3.6 of the Code, which is described in the sections that follow. bw is the width of the cross-section resisting the shear perpendicular to the shear force direction. For columns with rectangular cross-sections, bw is taken as the width of the section perpendicular to the shear direction. For columns with circular cross-sections, bw is taken as the average width appropriate for shear direction. d v is the effective shear depth. It is taken as the greater of 0.9d or 0.72h, dv = max {0.9d ,0.72h} (CSA 2.2) where d is the distance from the extreme compression fiber to the centroid of tension reinforcement, and h is the overall depth of the cross-section in the direction of shear force. The value of the β factor is preferably determine using the Simplified method (CSA 11.3.6.3) if applicable. When the conditions of Simplified method do not apply, the General method (CSA 11.3.6.4) is used. For designing concrete column sections for shear force, the special value of β does not apply (CSA 11.3.6.4) When the axial force is compressive, the specified yield strength of the longitudinal reinforcing f y does not exceed 400 MPa, and the specified concrete strength f c′ does not exceed 60 MPa, β is determined in accordance with the Simplified method, as follows (CSA 11.6.3.3): When the section contains at least the minimum transverse reinforcement, β is taken as 0.18 (CSA 11.6.3.3a). Column Design 3 - 19 Concrete Frame Design CSA A23.3-04 β = 0.18 (CSA 11.6.3.a) When the section contains no transverse reinforcement, β is determined based on the specified maximum nominal size of coarse aggregate, ag . For maximum-sized coarse aggregate not less than 20 mm, β is taken as: 230 β= 1000 + d v (CSA 11.3.6.3 b) where d v is the effective shear depth expressed in millimeter, as described in preceding sections. For maximum sized coarse aggregate less than 20 mm, β is taken as: 230 β= 1000 + sze where, S ze = 35 S z ≥ 0.85S z . 15 + ag (CSA 11.3.6.3 (c)) (CSA 11.3.6.3.c) In the preceding expression, the crack spacing parameter, S ze , shall be taken as the minimum of d v and the maximum distance between layers of distributed longitudinal reinforcement. However, S ze is conservatively taken as equal to d v . In summary, for simplified cases, β can be expressed as follows: if minimum transverse reinforcement is provided, 0.18, 230 β , if no transverse reinforcement provided, and ag ≥ 20mm, 1000+d v 230 , if no transverse reinforcement provided, and ag < 20mm. 1000+Sze 3 - 20 Column Design Chapter 3 - Design Process When the axial force is tensile, or the specified yield strength of the longitudinal reinforcing f y > 400 MPa, or the specified concrete strength f c′ > 60 MPa, β is determined in accordance with the General method, as follows (CSA 11.3.6.1, 11.3.6.4): = β 0.40 1300 • (1 + 1500ε x ) (1000 + S ze ) (CSA 11.3.6.4) In the preceding expression, the equivalent crack spacing parameter, S ze , is taken as equal to 300 mm if minimum transverse reinforcement is provided (CSA 11.3.6.4). Otherwise it is determines as stated in simplified method. if minimum transverse 300, Sze = reinforcement is provided, 35 Sz ≥ 0.85Sz , otherwise. 15 + ag The longitudinal strain, ε x , at mid-depth of the cross-section is computed from the following equation: M f d v + V f + 0.5 N f εx = s ( Es As ) (CSA 11.3.6.4) In evaluating the ε x , the following conditions apply: V f and M f are taken as positive quantities (CSA 11.3.6.4a) As is taken as the total area of longitudinal reinforcement in the column section. For the column section check option, the program uses the sum of user-defined reinforcement in the section. For the column section design option, the longitudinal reinforcement area is taken as the envelope of reinforcement required for all design load combinations. Actual provided reinforcement might be slightly higher than this quantity. The reinforcement should be developed to achieve full strength (CSA 11.3.6.3 b). Column Design 3 - 21 Concrete Frame Design CSA A23.3-04 If the value of ε x calculated from the preceding equation is negative, it is recalculated as follows: Mf = εx + V f + 0.5 N f dv ≥ − 0.0002 2 ( Es As + Ec Act ) (CSA 11.3.6.4 d) For sections closer than dv from the face of the support, εx is calculated based on Mf, Vf, and Nf at a section at a distance dv from the face of the support (CSA 11.3.6.4 d). If the axial tension is large enough to crack the flexural compression face of the section, the value of ε x is increased by a factor of 2 (CSA 11.3.6.4 e). The program uses a linear elastic stress distribution to check this condition. An upper limit on ε x is imposed as follows: ex ≤ 0.003 (CSA 11.3.6.4 f) ε x is positive for tensile action. N f is positive for tensile action. For Ductile moment resisting frames, an upper limit on β of 0.10 is enforced (CSA 21.4.5.2) as follows: β ≤ 0.10 (CSA 21.4.5.2) In both the Simplified method and the General method, the shear strength of the section due to concrete, vc , depends on whether the minimum transverse reinforcement is provided. To check this condition, the program performs the design in two passes. In the first pass, it is assumes that no transverse shear reinforcement is needed. When the program determines that shear reinforcement is need, the program performs the second pass with the assumption that at least minimum shear reinforcement is provided. 3.5.4.3 Determine Required Shear Reinforcement The minimum possible factored shear resistance, Vr,max is computed as follows: 3 - 22 Column Design Chapter 3 - Design Process Vr ,max = 0.25ϕc f c′bw d . (CSA 11.3.3) The shear reinforcement per unit spacing, Av s , is computed as follows: If V f ≤ Vc , Av =0 s (CSA 11.3.5.1) else if Vc < V f ≤ Vr ,max ( ) V f − Vc tan θ Av = s ϕs f yh d v (CSA 11.3.3, 11.3.5.1) else if V f > Vr ,max a failure condition is declared. (CSA 11.3.3) A minimum area of shear reinforcement is provided in the following regions (CSA 11.2.8.1): (a) in regions of flexural members where the factored shear force V f exceeds Vc , (b) in regions of beams and columns with an overall depth greater than 750 mm, and (c) in regions of beams and columns where the factored torsion T f exceeds 0.25 Tcr . Where the minimum shear reinforcement is required by section CSA 11.2.8.1, or by calculations, the minimum area of shear reinforcement per unit spacing is taken as: f′ Av ≥ c bw s fy (CSA 11.2.8.2) Column Design 3 - 23 Concrete Frame Design CSA A23.3-04 In the preceding, the term θ is used. Here θ is the angle of inclination of diagonal compressive stresses to the longitudinal axis of the member. The θ value is normally between 22 to 44 degrees. It is determined according to Section 11.3.6 of the Code. Similar to the β factor, which was described previously, the value of θ is preferably determined using the Simplified method (CSA 11.3.6.3), whenever applicable. The program uses the General method when the conditions for the Simplified methods are not satisfied (CSA 11.3.6.4). For designing concrete column sections for shear forces, the special value of θ does not apply (CSA 11.3.6.2). If the axial force is compressive, the specified yield strength of the longitudinal reinforcing fy does not exceed 400 MPa, and the specified concrete strength f c′ does not exceed 60 MPa, θ is taken to be 35 degrees (CSA 11.3.6.3). θ =35o for Pf ≤ 0 or f y ≤ 400MPa or f c′ ≤ 60MPa (CSA11.3.6.4) If the axial force is tensile, the specified yield strength of the longitudinal reinforcing fy > 400 MPa, and the specified concrete strength f c′ > 60 MPa, θ is determined using the General method as follows (CSA 11.3.6.4). θ= 29 + 7000ε x for Pf < 0 or f y > 400 MPa or f c′ > 60 MPa (CSA11.3.6.4) where ε x is the longitudinal strain at the mid-depth of the cross-section for the factored load. The calculation procedure has been described in the preceding sections. The maximum of all of the calculated Av s values obtained from each load combination is reported for the major and minor directions of the column, along with the controlling shear force and associated load combination number. For all columns and at any station, the minimum area of transverse circular hoop reinforcement is imposed as follows: 3 - 24 Column Design Chapter 3 - Design Process Ag f′ h Av ≥ 0.45 − 1 c core s Acore f yh 4 (CSA 10.9.4) In potential plastic hinge locations of Ductile moment resisting frames as described later, the minimum area of circular hoops and transverse stirrups is imposed as follows: Av Pf fc′ hcore ≥ s Po f yh 4 (Hoops) (CSA 21.4.4.2) Ag f′ h Av ≥ 0.45 − 1 c core s Acore f yh 4 (Hoops) (CSA 21.4.4.2, 10.9.4) Av f′ ≥ 0.09 c hcore s f yh (Stirrups) (CSA 21.4.4.2) Ag Av hcore ≥ 0.2 K n K p s Acore (Stirrups) (CSA 21.4.4.2) where K n = Pf n1 = and K p Po ( n1 − 2 ) Transverse reinforcement shall be provided over a length, lo , from the face of each joint and on both sides of any section where flexural yielding can occur as a result of inelastic displacement of the frame. The length, lo , is determined as follows: (a) where Pf ≤ 0.5φc f c′ Ag , lo shall not be less than 1.5 times the largest member cross-section dimension or 1/6 of the clear span of the member; and (b) where Pf > 0.5φc f c′ Ag , lo shall not be less than 2.0 times the largest member cross-section dimension or 1/6 of the clear span of the member (CSA 21.4.4.5). Column Design 3 - 25 Concrete Frame Design CSA A23.3-04 The column shear reinforcement requirements reported by the program are based on the preceding considerations. Any other minimum stirrup requirements to satisfy spacing or volumetric requirements must be investigated independently of the program by the user. 3.6 Beam Design In the design of concrete beams, the program calculates and reports the required areas of steel for flexure and shear based on the beam moments, shears, load combination factors and other criteria described in the subsections that follow. The reinforcement requirements are calculated at a user-defined number of check stations along the beam span. All of the beams are designed for major direction flexure and shear only. Effects due to any axial forces, minor direction bending, and torsion that may exist in the beams must be investigated independently by the user. The beam design procedure involves the following steps: Design beam flexural reinforcement Design beam shear reinforcement 3.6.1 Design Beam Flexural Reinforcement The beam top and bottom flexural steel is designed at a user-defined number of design stations along the beam span. In designing the flexural-reinforcement for the major moment of a particular beam for a particular section, the program uses the following steps: Determine the maximum factored moments Determine the reinforcing steel 3.6.1.1 Determine Factored Moments In the design of flexural reinforcement of Ductile (CSA 21.5), Moderately Ductile (CSA 21.7), or Conventional (CSA 21.8) moment resisting concrete frame beams, the factored moments for each load combination at a particular 3 - 26 Beam Design Chapter 3 - Design Process beam station are obtained by factoring the corresponding moments for different load cases with the corresponding load factors. The beam section is then designed for the maximum positive M +f and maximum negative M −f factored moments obtained from all of the load combinations. Positive beam moments produce bottom steel. In such cases, the beam may be designed as a Rectangular or T-beam. Negative beam moments produce top steel. In such cases, the beam is always designed as a Rectangular section. 3.6.1.2 Determine Required Flexural Reinforcement In the flexural reinforcement design process, the program calculates both the tension and compression reinforcement. Compression reinforcement is added when the applied design moment exceeds the maximum moment capacity of a singly reinforced section. The user has the option of avoiding the compression reinforcement by increasing the effective depth, the width, or the grade of concrete. The design procedure is based on the simplified rectangular stress block, as shown in Figure 3-5 (CSA 10.1.7). Figure 3-5 Rectangular beam design Beam Design 3 - 27 Concrete Frame Design CSA A23.3-04 Furthermore, it is assumed that the compression carried by concrete is less than that which can be carried at the balanced condition (CSA 10.1.4). When the applied moment exceeds the moment capacity at the balanced condition, the area of compression reinforcement is calculated on the assumption that the additional moment will be carried by compression and additional tension reinforcement. The design procedure used by the program for both rectangular and flanged sections (L and T-beams) is summarized in the sections that follow. It is assumed that the design ultimate axial force in a beam is negligible; hence, all of the beams are designed for major direction flexure and shear only. 3.6.1.2.1 Design for Flexure of a Rectangular Beam In designing for a factored negative or positive moment, M f , (i.e., designing top or bottom steel) the depth of the compression block is given by a , as shown in Figure 3-5, where, a =− d d2 − 2| M f | α1 f c′ϕc b , (CSA 10.1) where the value of ϕc is 0.65 (CSA 8.4.2) in the preceding and following equations. Also , α1 , β1 , and cb are calculated as follows: α1 =− 0.85 0.0015 f c′ ≥ 0.67, (CSA 10.1.7) β= 0.97 − 0.0025 f c′ ≥ 0.67, and 1 (CSA 10.1.7) cb = 700 d. 700 + f y (CSA 10.5.2) The balanced depth of the compression block is given by ab = β1cb . (CSA 10.1.4, 10.1.7) If a ≤ ab (CSA 10.5.2), the area of tensile steel reinforcement is then given by 3 - 28 Beam Design Chapter 3 - Design Process As = Mf a ϕs f y d − 2 . This steel is to be placed at the bottom if M f is positive, or at the top if M f is negative. If a > ab (CSA 10.5.2), compression reinforcement is required and is calculated as follows: The factored compressive force developed in the concrete alone is given by C = ϕc α ac f c′bab , and (CSA 10.1.7) the factored moment resisted by the concrete and bottom steel is a M = C d − b . fc 2 The moment resisted by compression steel and tensile steel is M= M f − M fc . fs So the required compression steel is given by As′ = M fs ( ϕ f ′ − ϕ α f ′ ) ( d − d ′) s s c , where 1 c c − d′ = f s′ 0.0035 Es ≤ fy. c (CSA 10.1.2 , 10.1.3) The required tensile steel for balancing the compression in concrete is As1 = M fc a fy d − b 2 ϕs , and the tensile steel for balancing the compression in steel is Beam Design 3 - 29 Concrete Frame Design CSA A23.3-04 As 2 = M fs f y ( d − d ′ ) ϕs . Therefore, the total tensile reinforcement A = As1 + As 2 , and total coms pression reinforcement is As′ . As is to be placed at the bottom, and As′ is to be placed at the top if M f is positive, and vice versa. 3.6.1.2.2 Design for Flexure of a T-Beam Flanged Beam Under Negative Moment In designing for a factored negative moment, M f , (i.e., designing top steel), the contribution of the flange to the strength of the beam is ignored. The design procedure is therefore identical to the one used for Rectangular beams, except that in the corresponding equations, b is replaced by b. Flanged Beam Under Positive Moment If M f > 0, the depth of the compression block is given by (see Figure 3-6). a =− d d2 − 2M f α1 f c′ϕc b f . (CSA 10.1) where the value of ϕc is 0.65 (CSA 8.4.2) in the preceding and following equations. See Figure 3-6. Also α1 , β1 , and cb are calculated as follows: α= 0.85 − 0.0015 f c′ ≥ 0.67, 1 (CSA 10.1.7) β= 0.97 − 0.0025 f c′ ≥ 0.67, and 1 (CSA 10.1.7) cb = 700 d. 700 + f y (CSA 10.5.2) The depth of the compression block under balanced condition is given by ab = β1cb . 3 - 30 Beam Design (CSA 10.1.4, 10.1.7) Chapter 3 - Design Process If a ≤ d s , the subsequent calculations for As are exactly the same as previously performed for the rectangular section design. However, in this case, the width of the beam is taken as b f . Whether compression reinforcement is required depends on if a > ab . Figure 3-6 T-beam design If a > d s , calculation for As is performed in two parts. The first part is for balancing the compressive force from the flange, C f , and the second part is for balancing the compressive force from the web, Cw . As shown in, ( ) Cf = α1 f c′ b f − bw d s . Therefore, As1 = C f ϕc f y ϕs (CSA 10.1.7) and the portion of M f that is resisted by the flange is d M ff = C f d − s 2 ϕc . Therefore, the balance of the moment, M f to be carried by the web is given by Beam Design 3 - 31 Concrete Frame Design CSA A23.3-04 M= M f − M ff . fw The web is a Rectangular section of dimensions bw and d , for which the depth of the compression block is recalculated as a1 =− d d2 − 2 M fw α1 f c′ϕc bw . (CSA 10.1) If a1 ≤ ab (CSA 10.5.2), the area of tensile steel reinforcement is then given by As 2 = M fw a ϕs f y d − 1 2 , and = A As1 + As 2 . s This steel is to be placed at the bottom of the T-beam. If a1 > ab (CSA 10.5.2), compression reinforcement is required and is calculated as follows: The factored compressive force in the concrete web alone is given by C =ϕc α1 f c′bab , and (CSA 10.1.7) the factored moment resisted by the concrete web and tensile steel is a M = Cd − b fc 2 . The moment resisted by compression steel and tensile steel is M = M fw − M fc . fs Therefore, the compression steel is computed as 3 - 32 Beam Design Chapter 3 - Design Process As′ = ( M fs ) ϕs f s′ − ϕc α1 f c′ ( d − d ′ ) , where c − d′ f s′ 0.0035 Es = ≤ fy. c (CSA 10.1.2, 10.1.3) The tensile steel for balancing compression in web concrete is As 2 = M fc a fy d − b 2 ϕs , and the tensile steel for balancing compression in steel is As 3 = M fs f y ( d − d ′ ) ϕs . Total tensile reinforcement As = As1 + As 2 + As 3 , and total compression reinforcement is As′ . As is to be placed at the bottom and As′ is to be placed at the top. 3.6.1.3 Minimum and Maximum Tensile Reinforcement The minimum flexural tensile steel required for a beam section is given by the minimum of the two limits: As ≥ As = 0.2 f c′ fy bw h, or 4 As (required) . 3 (CSA 10.5.1.2) (CSA 10.5.1.3) In addition, the minimum flexural tensile steel provided in a T section with flange under tension is given by the limit: As ≥ 0.004 ( b − bw ) d s . (CSA 10.5.3.1) Beam Design 3 - 33 Concrete Frame Design CSA A23.3-04 An upper limit of 0.04 times the gross web area on both the tension reinforcement and the compression reinforcement is arbitrarily imposed as follows: 0.04bd Rectangular beam As ≤ 0.04bw d T beam 0.04bd Rectangular beam As′ ≤ 0.04bw d T beam 3.6.1.4 Special Consideration for Seismic Design For Ductile moment resisting concrete frames (seismic design), the beam design satisfies the following additional conditions (see also Table 3-2 for a comprehensive listing): The minimum longitudinal reinforcement shall be provided at both the top and the bottom. Any of the top and bottom reinforcement shall not be less than As ( min ) . As ( min ) ≥ 1.4 bw d fy (CSA 21.3.2.1) The beam flexural steel is limited to a maximum given by As ≤ 0.025bw d . (CSA 21.3.2.1) At any end (support) of the beam, the beam positive moment capacity (i.e., associated with the bottom steel) would not be less than one-half of the beam negative moment capacity (i.e., associated with the top steel) at that end (CSA 21.3.2.2). Neither the negative moment capacity nor the positive moment capacity at any of the sections within the beam would be less than one-fourth of the maximum of positive or negative moment capacities of any of the beam end (support) stations (CSA 21.3.2.2). For Conventional moment resisting concrete frames (seismic design), the beam design would satisfy the following conditions: 3 - 34 Beam Design Chapter 3 - Design Process At any support of the beam, the beam positive moment capacity would not be less than one-third of the beam negative moment capacity at that end (CSA 21.7.2.1.1). Neither the negative moment capacity nor the positive moment capacity at any of the sections within the beam would be less than one-fifth of the maximum of positive or negative moment capacities of any of the beam end (support) stations (CSA 21.7.2.1.1). Table 3-2: Design Criteria Type of Check/ Design Strength of Materials Conventional Moment Resisting Frames (Non-Seismic) Moderately Ductile Moment Resisting Frames (Seismic) Ductile Moment Resisting Frames (Seismic) fc′ ≤ 80 MPa fc′ ≤ 80 MPa fc′ ≤ 80 MPa fc′ ≤ 500 MPa fc′ ≤ 500 MPa fc′ ≤ 500 MPa fc′ ≤ 8 MPa fc′ ≤ 8 MPa fc′ ≤ 8 MPa Column Check (interaction) NLDa Combinations NLDa Combinations NLDa Combinations Column Design (Interaction) NLDa Combinations 1% < ρ < 8% NLDa Combinations 1% < ρ < 8% NLDa Combinations 1% < ρ < 6% α = 1.0 NLDa Combinations Modified NLDa Combinations or Nominal Capacity Shear (Vp) ϕ = 1.0 and α = 1.0 Column Shears Minimum hoop Beam Design Flexure Beam Min. Moment Override Check NLDa Combinations ρ ≤ 0.04 ρ ≥ 0.2 fc′ f y Minimum hoop reinforcement NLDa Combinations ρ ≤ 0.04 ρ ≥ 0.2 fc′ f y Minimum hoop reinforcement Special reinforcement in hinge ρ ≥ 0.2 1 − M uEND 3 1 M u+SPAN ≥ max M u+ , M u− 5 END 1 M u−SPAN ≥ max M u+ , M u− 5 END { { ϕ = 1.0 and α = 1.25 Strength based on β ≤ 0.1, θ ≥ 45° at potential plastic hinge region NLDa Combinations ρ ≤ 0.04 M u+END ≥ No Requirement NLDa Combinations and Probable Capacity Shear (Vp) } } fc′ f y M u+END ≥ 1 − M uEND 2 1 M u+SPAN ≥ max M u+ , M u− 4 END 1 M u−SPAN ≥ max M u+ , M u− 4 END { { Beam Design } } 3 - 35 Concrete Frame Design CSA A23.3-04 Table 3-2: Design Criteria Type of Check/ Design Conventional Moment Resisting Frames (Non-Seismic) Moderately Ductile Moment Resisting Frames (Seismic) Ductile Moment Resisting Frames (Seismic) NLDa Combinations NLDa Combinations or Nominal Capacity Shear (Vp) with α = 1.0 and ϕ = 1.0 plus VD+L, or modified load combination with earthquake load amplified by RdRo. NLDa Combinations and Probable Capacity Shear (Vp) with α = 1.0 and ϕ = 1.0 plus VD+L, or modified load combination with earthquake load amplified by RdRo; strength based on β = 0 and β = 45 degree in potential plastic hinge. Beam Design Shear 3.6.2 Design Beam Shear Reinforcement The shear reinforcement is designed for each load combination at a user-defined number of stations along the beam span. In designing the shear reinforcement for a particular beam for a particular loading combination at a particular station due to the beam major shear, the program uses the following steps: Determine the factored forces acting on the section, M f and V f . Note that M f is needed for the calculation of vc . Determine the shear stress, vc , that can be resisted by concrete alone. Calculate the reinforcement steel required to carry the balance. For Ductile and Moderately ductile moment resisting frames, the shear design of the beams is also based on the probable and nominal moment capacities of the members, respectively (CSA 21.3.4.1, 21.7.2.3). The following three sections describe in detail the algorithms associated with these steps. 3.6.2.1 Determine Shear Force and Moment In the design of the beam shear reinforcement of an Conventional moment resisting concrete frame, the shear forces and moments for a particular load combination at a particular beam section are obtained by factoring the associated shear forces and moments with the corresponding load combination factors. 3 - 36 Beam Design Chapter 3 - Design Process In the design of Ductile moment resisting concrete frames (seismic design), the shear capacity of the beam is also checked for the probable shear due to the probable moment capacities and the factored gravity load, in addition to the design checks required for Conventional moment resisting frames. The procedure for calculating the design shear force in a beam from probable moment capacities is the same as that described for a column in the Column Design section. See also Table 3-2 and Figure 3-7 for more details. The design shear force V f is then given by V= V p + VD + L f (CSA 21.3.4.1) where, V p is the shear force obtained by applying the calculated probable ultimate moment capacities at the two ends of the beams acting in two opposite directions. Therefore, V p is the maximum of VP1 and VP 2 , where V= P1 (M − I + M J+ ) L , and V= P2 (M − I + M J+ ) L , where M I+ = Moment capacity at end I, with bottom steel in tension, using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , M J+ = Moment capacity at end J, with bottom steel in tension, using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , M I− = Moment capacity at end I, with top steel in tension, using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , M J− = Moment capacity at end J, with top steel in tension, using a steel yield stress value of αf y and no ϕ factors ( ϕs =ϕc =1.0 ) , and L = Clear span of beam. The overstrength factor α is always taken as 1.25 for Ductile moment resisting frames (CSA 2.2, 21.3.4.1). VD + L is the contribution of shear force from the in-span distribution of gravity loads. Beam Design 3 - 37 Concrete Frame Design CSA A23.3-04 In the design of Moderately Ductile moment resisting frames (seismic), the shear capacity of the beam is also checked for the nominal shear due to the nominal ( ϕs =ϕc =1.0 ) moment capacities and the factored gravity load (CSA 21.7.2.3(a), 2.2), in addition to the design checks required for Conventional moment resisting frames. The procedure for calculating nominal moment capacity is the same as that for computing the probable moment capacity for Ductile moment resisting frames, except that α is taken equal to 1 (CSA 21.7.2.3.(a), 2.2) rather than 1.25. The design shear strength is enforced not to be greater than the factored shear force with the seismic load amplified by a factor of Rd Ro (CSA 21.3.4.1, 21.7.2.3(b)). See also Table 3-2 for details. POINT OF INFLECTION Vu COLUMN ABOVE TOP OF BEAM COLUMN HEIGHT (H) PANEL ZONE M uL TL Vuh CR TR CL M uR COLUMN BELOW Vu POINT OF INFLECTION ELEVATION Figure 3-7 Column shear force Vu 3 - 38 Beam Design Chapter 3 - Design Process 3.6.2.2 Determine Concrete Shear Capacity Given the design force set M f and V f , the shear capacity provided by the concrete alone, Vc , is calculated as follows: Vc = ϕc λβ f c′ bw d , (CSA 11.3.4) ϕc is the resistance factor for concrete. By default it is taken as 0.65 (CSA8.4.2). For concrete produced in a pre-qualified manufacturing plants, its value can be taken as 0.70 (CSA 16.1.3). This value can be overwritten in the Preferences. λ is the strength reduction factor to account for low density concrete (CSA 2.2). For normal density concrete, its value is 1 (CSA 8.6.5), which is taken by the program as the default value. For concrete using lower density aggregate, the user can change the value of λ in the material property data. The recommended value for λ is as follows (CSA 8.6.5). 1.00, 0.85, λ = 0.75, for normal density concrete, for semi-low-density concrete in which all of the fine aggregate is natural sand, for semi-low-density concrete in which none of the fine aggregate is natural sand. β is the factor for accounting for the shear resistance of cracked concrete (CSA 2.2). Its value is normally between 0.1 and 0.4. It is determine according to the section 11.3.6 of the Code, which is described in the sections that follow. bw is the effective web width. For rectangular beam, it is the width of the beam. For T-beam, it is the width of the web of the beam. d v is the effective shear depth. It is taken as the greater of 0.9d or 0.72h, d v = max {0.9d ,0.72h} (CSA 2.2) Beam Design 3 - 39 Concrete Frame Design CSA A23.3-04 where d is the distance from the extreme compression fiber to the centroid of tension reinforcement, and h is the overall depth of the cross-section in the direction of shear force. The value of the β factor is preferably taken as the Special value (CSA 11.3.6.2) or it is determined using the Simplified method (CSA 11.3.6.3), if applicable. When the conditions of the Special value or Simplified method do not apply, the General method is used to determine the β factor (CSA 11.3.6.4). If the overall beam depth, h, is less than 250 mm or if the beam depth of a T-beam before the slab is not greater than one-half of the width of web or 350 mm, β is taken as 0.21 (CSA 11.3.6.2) When the specified yield strength of the longitudinal reinforcing fy does not exceed 400 MPa, and the specified concrete strength f c′ does not exceed 60 MPa, β is determined in accordance with the Simplified method, as follows (CSA 11.6.3.3): When the section contains at least the minimum transverse reinforcement, β is taken as 0.18 (CSA 11.6.3.3a). β = 0.18 (CSA 11.6.3.a) When the section contains no transverse reinforcement, β is determined based on the specified maximum nominal size of coarse aggregate, ag . For maximum size of coarse aggregate not less than 20 mm, β is taken as: 230 β= 1000 + d v (CSA 11.3.6.3 b) where is the effective shear depth expressed in millimeter, which is described in preceding sections. For maximum size of coarse aggregate less than 20 mm, β is taken as: 230 β= 1000 + sze 3 - 40 Beam Design (CSA 11.3.6.3 c) Chapter 3 - Design Process where, S ze = 35 S z ≥ 0.85S z 15 + ag (CSA 11.3.6.3.c) In the preceding expression, the crack spacing parameter, S ze , shall be taken as the minimum of d v and the maximum distance between layers of distributed longitudinal reinforcement. However, S ze is conservatively taken as equal to dv . In summary, for Simplified cases, β can be expressed as follows: if minimum transverse reinforcement is provided, 0.18, 230 β , if no transverse reinforcement provided, and ag ≥ 20mm, 1000 + d v 230 , if no transverse reinforcement provided, and ag < 20mm. 1000 + S ze When the specified yield strength of the longitudinal reinforcing fy > 400 MPa, or the specified concrete strength f c′ > 60 MPa, β is determined in accordance with the General method as follows (CSA 11.3.6.1, 11.3.6.4): = β 0.40 1300 • + ε + S ze ) 1 1500 1000 ( x) ( (CSA 11.3.6.4) In the preceding expression, the equivalent crack spacing parameter, S ze , is taken as equal to 300 mm if minimum transverse reinforcement is provided (CSA 11.3.6.4). Otherwise it is determine as stated in the Simplified method. if minimum transverse reinforcement is provided, 300 Sze = 35 Sz ≥ 0.85Sz otherwise. 15 + ag The longitudinal strain, ε x , at mid-depth of the cross-section is computed from the following equation: M f dv + V f εx = s ( Es As ) (CSA 11.3.6.4) Beam Design 3 - 41 Concrete Frame Design CSA A23.3-04 In evaluating the ε x , the following conditions apply: V f and M f are taken as positive quantities (CSA 11.3.6.4a) As is taken as the total area of longitudinal reinforcement in the beam section. It is taken as the envelope of reinforcement required for all design load combinations. Actual provided reinforcement might be slightly higher than this quantity. The rebar should be developed to achieve full strength (CSA 11.3.6.3 b). For sections closer than d v from the face of the support, ε x is calculated based on M f and V f at a section at a distance d v from the face of the support (CSA 11.3.6.4 d). An upper limit on ε x is imposed as follows: ex ≤ 0.003 (CSA 11.3.6.4f) ε x is positive for tensile action. For Ductile moment resisting frames, in the region of potential plastic hinge, the shear reinforcement is designed assuming β is equal to zero: β =0 (CSA 21.3.4.2) In both the Simplified and General methods, the shear strength of the section due to concrete, vc , depends on whether the minimum transverse reinforcement is provided. To check this condition, the program performs the design in two passes. In the first pass, it is assumes that no transverse shear reinforcement is needed. When the program determines that shear reinforcement is need, the program performs the second pass with the assumption that at least minimum shear reinforcement is provided. 3.6.2.3 Determine Required Shear Reinforcement The minimum possible factored shear resistance, Vr ,max is computed as follows: Vr ,max = 0.25ϕc f c′bw d . 3 - 42 Beam Design (CSA 11.3.3) Chapter 3 - Design Process The shear reinforcement per unit spacing, Av s , is computed as follows: If V f ≤ Vc Av =0 s (CSA 11.3.5.1) else if Vc < V f ≤ Vr ,max ( ) V f − Vc tan θ Av = s ϕs f yh d v (CSA 11.3.3, 11.3.5.1) else if V f > Vr ,max a failure condition is declared. (CSA 11.3.3) A minimum area of shear reinforcement is provided in the following regions (CSA 11.2.8.1): (a) in regions of flexural members where the factored shear force V f exceeds Vc , (b) in regions of beams and columns with an overall depth greater than 750 mm, and (c) in regions of beams and columns where the factored torsion T f exceeds 0.25 . Where the minimum shear reinforcement is required by section CSA 11.2.8.1, or by calculations, the minimum area of shear reinforcement per unit spacing is taken as Tcr . f′ Av ≥ 0.06 c bw s fy (CSA 11.2.8.2) In the preceding, the term θ is used. Here θ is the angle of inclination of diagonal compressive stresses to the longitudinal axis of the member. The θ value is Beam Design 3 - 43 Concrete Frame Design CSA A23.3-04 normally between 22 to 44 degree. It is determined according to section 11.3.6 in the Code. Similar to the β factor, which was described previously, the value of θ is preferably taken as the Special value (CSA 11.3.6.2) or it is determined using the Simplified method (CSA 11.3.6.3), whenever applicable. The program uses the General method when conditions for the Simplified method are not satisfied (CSA 11.3.6.4). If the overall beam depth, h, is less than 250 mm or if the depth of the T-beam below the slab is not greater than one-half of the width of web or 350 mm, θ is taken as 42 degree (CSA 11.3.6.2). If the specified yield strength of the longitudinal reinforcing fy does not exceed 400 MPa, and the specified concrete strength f c′ does not exceed 60 MPa, θ is taken to be 35 degree (CSA 11.3.6.3). θ =35o for Pf ≤ 0 or f y ≤ 400 MPa or 60 MPa (CSA11.3.6.4) If the axial force is tensile, the specified yield strength of the longitudinal reinforcing fy > 400 MPa, and the specified concrete strength f c′ > 60 MPa, θ is determined using the General method as follows (CSA 11.3.6.4), θ= 29 + 7000ε x for Pf < 0 or f y > 400 MPa or 60 MPa (CSA11.3.6.4) where ε x is the longitudinal strain at the mid-depth of the cross-section for the factored load. The calculation procedure has been described in preceding sections. The maximum of all of the calculated Av s values, obtained from each load combination, is reported for the major directions of the beam along with the controlling shear force and associated load combination number. The beam shear reinforcement requirements reported by the program are based purely on the considerations herein. Any other minimum stirrup requirements to satisfy spacing and volumetric requirements must be investigated independently of the program by the user. 3 - 44 Beam Design Chapter 3 - Design Process 3.7 Joint Design To ensure that the beam-column joint of Ductile and Moderately Ductile moment resisting frames possesses adequate shear strength, the program performs a rational analysis of the beam-column panel zone to determine the shear forces that are generated in the joint. The program then checks this against design shear strength. Only joints having a column below the joint are designed. The material properties of the joint are assumed to be the same as those of the column below the joint. The joint analysis is completed in the major and the minor directions of the column. The joint design procedure involves the following steps: Determine the panel zone design shear force, Vuh Determine the effective area of the joint Check panel zone shear stress The following three sections describe in detail the algorithms associated with these steps. 3.7.1 Determine the Panel Zone Shear Force For a particular column direction, major or minor, the free body stress condition of a typical beam-column intersection is shown in Figure 3-4. The force Vuh is the horizontal panel zone shear force that is to be calculated. The forces that act on the joint are Pf, Vf, M Lf and M Rf . The forces Pf and Vf are axial force and shear force, respectively, from the column framing into the top of the joint. The moments M Lf and M Rf are obtained from the beams framing into the joint. The joint shear force Vuh is calculated by resolving the moments into C and T forces. Noting that TL = CL and TR = CR , Vuh = TL + TR − Vu Joint Design 3 - 45 Concrete Frame Design CSA A23.3-04 The location of C or T forces is determined by the direction of the moment. The magnitude of C or T forces is conservatively determined using basic principles of ultimate strength theory (CSA 10.1). The moments and the forces from beams that frame into the joint in a direction that is not parallel to the major or minor direction of the column are resolved along the direction that is being investigated, thereby contributing force components to the analysis. Also, the program calculates the positive and negative moments, considering the fact that the concrete cover may be different for the direction of moment. In the design of Ductile moment resisting concrete frames, the evaluation of the design shear force is based on the maximum probable moment capacities (with reinforcing steel overstrength factor, α, and no ϕ factors) of the beams framing into the joint (CSA 21.5.1.2, 21.7.2.4.1). In the design of Moderately Ductile moment resisting frames, the evaluation of the design shear force is based on the nominal capacities (with no reinforcing steel overstrength factor, i.e., α =1, and no ϕ factors, i.e., ϕc =ϕs =1 of the beams framing into the joint (CSA 21.7.2.4.1). The C and T force are based on these moment capacities. The program calculates the column shear force, Vu , from the beam moment capacities as follows: Vu ,beam = M uL + M uR H It should be noted that the points of inflection shown on Figure 3-4 are taken as midway between actual lateral support points for the columns. If there is no column at the top of the joint, the shear force from the top of the column is taken as zero. The program also checks the capacity shear that can be developed in the column based on its positive and negative moment capacities for the presence of factored axial load, Pf . + top + M u− bot M u− top + M u+ bot M Vu ,col = min u , H H Vu is taken as the minimum of the two values Vu ,beam and Vu ,col . 3 - 46 Joint Design Chapter 3 - Design Process Vu = min {Vu ,beam ,Vu ,col } The effects of load reversals, as illustrated in Case 1 and Case 2 of Figure 3-8 are investigated, and the design is based on the maximum of the joint shears obtained from the two cases. 3.7.2 Determine the Effective Area of Joint The joint area that resists the shear forces is assumed always to be rectangular in plan view. The dimensions of the rectangle correspond to the major and minor dimensions of the column below the joint, except if the beam framing into the joint is very narrow. The effective width of the joint area to be used in the calculation is limited to two times the width of the beam. The area of the joint is assumed not to exceed the area of the column below. The joint area for joint shear along the major and minor directions is calculated separately (CSA 2.3). The effective joint shear area, AJ , is given by AJ = min {bcol hcol , 2bw hcol } (CSA 2.3) where, bcol is the width of column section perpendicular to direction of shear, hcol is the depth of column section in the direction of shear, and bw is the width of the beam or its web. For circular column, both bcol and hcol are taken equal to the overall diameter of the column. For a section designer section, bcol and hcol are based on equivalence. It should be noted that if the beam frames into the joint eccentrically, the preceding assumptions may not be conservative and the user should investigate the acceptability of the particular joint. Joint Design 3 - 47 Concrete Frame Design CSA A23.3-04 Figure 3-8 Beam-Column joint analysis 3 - 48 Joint Design Chapter 3 - Design Process 3.7.3 Check Panel Zone Shear Stress The panel zone shear force Vuh is compared to the factored shear resistance of the joint Vc (CSA 21.5.4.1, 21.7.2.4.2). Vc is calculated from the following equation: ′ 2.2ϕc f c AJ , for joints confined on all four sides, Vc 1.6ϕc f c′ AJ , for joints confined on three faces or on two opposite faces, = 1.3ϕc f c′ AJ , for all other joints. ϕc is the resistance factor for concrete. By default, it is taken as 0.65 (CSA 8.4.2). For concrete produced in a pre-qualified manufacturing plant, its value can be taken as 0.70 (CSA 16.1.3). Its value can be overwritten in the Preferences. f c′ is the specified compressive strength of concrete. It is assumed to be the same as that of the column below the joint. The f c′ should not be greater than 8 MPa, f c′ ≤ 8MPa AJ is the effective joint shear area appropriate for the direction. A beam that frames into a face of a column at the joint is considered in the program to provide confinement to the joint if at least three-quarters of the face of the joint is covered by the framing member (CSA 21.5.4.1, 21.7.2.4.2). 3.7.4 Beam-Column Flexural Capacity Ratios At a particular joint for a particular column direction, major or minor, the program will calculate the ratio of the sum of the beam moment capacities to the sum of the column moment capacities. For Ductile and Moderately Ductile moment resisting frames, the following CSA provision must be satisfied (CSA 21.4.2.2, 21.7.2.2.2). Joint Design 3 - 49 Concrete Frame Design CSA A23.3-04 ∑M ≥ ∑M c For Ductile moment resisting frames ∑M c (CSA 21.4.2.2, 21.7.2.2.2) b ∑M c and ∑M b are defined as follows: is the sum of the nominal flexural strength of columns, M nc , framing into the joint evaluated at the center of the joint. Individual column flexural strength is calculated for the associated factored axial forces, Pf . The nominal capacity is calculated with no overstrength factor, i.e., α =1, and no ϕ factors or ϕc =ϕs =1. ∑M b is the sum of the maximum probable flexural strength of beams, M pb , framing into the joint evaluated at the face of the joint. The probable capacity is calculated with an overstrength factor, i.e., α =1.25, and no ϕ factors or ϕc =ϕs =1. For Moderately Ductile moment resisting frames, ∑M c and ∑M b are de- fined as follows: ∑M c is the sum of the factored flexural strength of columns, ϕM nc , framing into the joint evaluated at the center of the joint. Individual column flexural strength is calculated for the associated factored axial forces, Pf . The nominal capacity is calculated with no overstrength factor, i.e., α =1, and with appropriate ϕc and ϕs factors. ∑M b is the sum of the nominal flexural strength of beams, M pb , framing into the joint evaluated at the face of the joint. The probable capacity is calculated with an overstrength factor, i.e., α =1, and no ϕ factors, i.e., ϕc =ϕs =1 . The moment capacities of beams that frame into the joint in a direction that is not parallel to the major or minor direction of the column are resolved along the direction that is being investigated and the resolved components are added to the summation. The column capacity summation includes the column above and the column below the joint. For each load combination, the factored axial force, Pf , in each 3 - 50 Joint Design Chapter 3 - Design Process of the columns is calculated from the load combinations. For each load combination, the moment capacity of each column under the influence of the corresponding factored axial load Pf is then determined separately for the major and minor directions of the column, using the uniaxial column interaction diagram; see Figure 3-8. The moment capacities of the two columns are added to give the capacity summation for the corresponding load combination. The maximum capacity summations obtained from all of the load combinations is used for the beam-column capacity ratio. The beam-column capacity ratio is determined for a beam-column joint only when the following conditions are met: The frame is a Ductile or Moderately Ductile moment resisting frame. When a column exists above the beam-column joint, it is a concrete column. All of the beams framing into the column are concrete beams. The connecting member design results are available. The load combination involves seismic load. The beam-column flexural capacity ratios are reported only for Ductile and Moderately Ductile moment resisting frames involving seismic design load combinations. If this ratio is greater than 1.0, a warning message is printed in the output file. Joint Design 3 - 51 APPENDICES Appendix A Second Order P-Delta Effects Typically, design codes require that second order P-delta effects be considered when designing concrete frames. They are the global lateral translation of the frame and the local deformation of members within the frame. Consider the frame object shown in Figure A-1, which is extracted from a story level of a larger structure. The overall global translation of this frame object is indicated by ∆. The local deformation of the member is shown as δ. The total second order P-delta effects on this frame object are those caused by both ∆ and δ. The program has an option to consider P-delta effects in the analysis. When P-delta effects are considered in the analysis, the program does a good job of capturing the effect due to the ∆ deformation shown in Figure A-1, but it does not typically capture the effect of the δ deformation (unless, in the model, the frame object is broken into multiple elements over its length). A-1 Design Manual Concrete Frame CSA A23.3-04 Figure A-1 The Total Second Order P-Delta Effects on a Frame Element Caused by Both ∆ and δ Consideration of the second order P-delta effects are generally achieved by computing the flexural design capacity using a formula similar to that shown in the following equation. MCAP = aMnt + bMlt MCAP = where, Flexural design capacity required Mnt = Required flexural capacity of the member assuming there is no joint translation of the frame (i.e., associated with the δ deformation in Figure A-1) Mlt = Required flexural capacity of the member as a result of lateral translation of the frame only (i.e., associated with the ∆ deformation in Figure A-1) a = Unitless factor multiplying Mnt b = Unitless factor multiplying Mlt (assumed equal to 1 by the program; see below) When the program performs concrete frame design, it assumes that the factor b is equal to 1 and calculates the factor a. That b = 1 assumes that P-delta effects have been considered in the analysis, as previously described. Thus, in general, when performing concrete frame design in this program, consider P-delta effects in the analysis before running the program. A-2 Appendix A Appendix B Member Unsupported Lengths and Computation of K-Factors The column unsupported lengths are required to account for column slenderness effects. The program automatically determines the unsupported length ratios, which are specified as a fraction of the frame object length. Those ratios times the frame object length gives the unbraced lengths for the members. Those ratios can also be overwritten by the user on a member-by-member basis, if desired, using the overwrite option. There are two unsupported lengths to consider. They are L33 and L22, as shown in Figure B-1. These are the lengths between support points of the member in the corresponding directions. The length L33 corresponds to instability about the 3-3 axis (major axis), and L22 corresponds to instability about the 2-2 axis (minor axis). B-1 Concrete Frame Design CSA A23.3-04 Figure B-1 Axis of bending and unsupported length In determining the values for L22 and L33 of the members, the program recognizes various aspects of the structure that have an effect on these lengths, such as member connectivity, diaphragm constraints, and support points. The program automatically locates the member support points and evaluates the corresponding unsupported length. It is possible for the unsupported length of a frame object to be evaluated by the program as greater than the corresponding member length. For example, assume a column has a beam framing into it in one direction, but not the other, at a floor level. In that case, the column is assumed to be supported in one direction only at that story level, and its unsupported length in the other direction will exceed the story height. B-2 Member Unsupported Lengths and Computation of K-Factors Appendix C Concrete Frame Design Preferences The concrete frame design preferences are general assignments that are applied to all of the concrete frame members. The design preferences should be reviewed and any changes from the default values made before performing a design. The following table lists the design preferences that are specific to using CSA A23.3-04; the preferences that are generic to all codes are not included in this table. Table C-1 Preferences Item Time History Design Possible Values Envelopes, Step-by-Step Default Value Description Envelopes Toggle for design load combinations that include a time history designed for the envelope of the time history, or designed step-by-step for the entire time history. If a single design load combination has more than one time history case in it, that design load combination is designed for the envelopes of the time histories, regardless of what is specified here. C-1 Concrete Frame Design CSA A23.3-04 Possible Values Item Default Value Description Number Interaction Curves Multiple of 4 ≥4 24 Number of equally spaced interaction curves used to create a full 360 deg interaction surface (this item should be a multiple of four). We recommend 24 for this item. Number Any odd value ≥5 11 Number of points used for defining a single curve in a concrete frame; should be odd Consider Minimum Eccentricity No, Yes Yes Phi (Steel) >0 0.85 Phi (Concrete) >0 0.6 The strength reduction factor for concrete Pattern Live Load Factor ≥0 0.75 The scale factor for performing pattern loading for live loads Utilization Factor Limit >0 0.95 Stress ratios that are less than or equal to this value are considered acceptable C-2 Appendix C Toggle to consider if minimum eccentricity is considered in design. Strength reduction factor for rebar Appendix D Concrete Frame Overwrites The concrete frame design overwrites are basic assignments that apply only to those elements to which they are assigned. Table D-1 lists concrete frame design overwrites for CSA A23.3-04. Default values are provided for all overwrite items. Thus, it is not necessary to specify or change any of the overwrites. However, at least review the default values to ensure they are acceptable. When changes are made to overwrite items, the program applies the changes only to the elements to which they are specifically assigned. Table D-1 Overwrites Item Current Design Section Possible Values Any defined concrete section Default Value Description Analysis section The design section for the selected frame objects. When this overwrite is applied, any previous auto select section assigned to the frame object is removed. D-1 Concrete Frame Design CSA A23.3-04 Item Possible Values Default Value Element Type Ductile Moderately Ductile, Conventional From Reference Frame type per moment frame definition given in CSA 2.2. The program default is Ductile, which the user can overwrites if needed. Calculated The reduced live load factor. A reducible live load is multiplied by this factor to obtain the reduced live load for the frame object. Specifying 0 means the value is program determined. Calculated Unbraced length factor for buckling about the frame object major axis. This item is specified as a fraction of the frame object length. Multiplying this factor times the frame object length gives the unbraced length for the object. Specifying 0 means the value is program determined. Calculated Unbraced length factor for buckling about the frame object minor axis. Multiplying this factor times the frame object length gives the unbraced length for the object. Specifying 0 means the value is program determined. This factor is also used in determining the length for lateral-torsional buckling. Calculated See CSA, 10.15.1. Effective length factor for buckling about the frame object major axis. This item is specified as a fraction of the frame object length. Calculated See CSA, 10.15.1. Effective length factor for buckling about the frame object minor axis. This item is specified as a fraction of the frame object length. Calculated See CSA 10.15.3.2. Factor relating actual moment diagram to an equivalent uniform moment diagram about the frame object major axis. Live Load Reduction Factor Unbraced Length Ratio (Major) Unbraced Length Ratio (Minor) Effective Length Factor (K Major) Effective Length Factor (K Minor) Moment Coefficient (Cm Major) D-2 Appendix D ≤0 ≥0 ≥0 >0 >0 >0 Description Appendix D - Concrete Frame Overwrites Item Moment Coefficient (Cm Minor) NonSway Moment Factor (Db major) NonSway Moment Factor (Db minor) Sway Moment Factor (Ds major) Sway Moment Factor (Ds minor) Possible Values >0 >0 >0 1 1 Default Value Description Calculated See CSA 10.15.3.2. Factor relating actual moment diagram to an equivalent uniform moment diagram about the frame object minor axis. Calculated See CSA 10.15.3.1. The magnified non-sway moment factor amplified for the effects of member curvature about the frame object major axis. Calculated See CSA 10.15.3.1. The magnified non-sway moment factor amplified for the effects of member curvature about the frame object minor axis. From Reference See CSA 10.16.3.1. The magnified sway moment factor shall be taken as the column end moments calculated using a second-order analysis about the frame object major axis. From Reference See CSA 10.16.3.1. The magnified sway moment factor shall be taken as the column end moments calculated using a second-order analysis about the frame object minor axis. Overwrites D-3 Appendix E Error Messages and Warnings Table E-1 provides a complete list of Concrete Errors messages and Warnings. Table E-1 Error Messages Error Number Description 1 Beam concrete compression failure 2 Reinforcing required exceeds maximum allowed 3 Shear stress exceeds maximum allowed 4 Column design moments cannot be calculated 5 Column factored axial load exceeds Euler Force 6 Required column concrete area exceeds maximum 7 Flexural capacity could not be calculated for shear design 8 Concrete column supports non-concrete beam/column E-1 Concrete Frame Design CSA A23.3-04 Error Number E-2 Description 9 k ∗ L/r > 115 , zeta_ 2 < 0 , eta < 1.0 (GB50010 7.3.10) 10 Column is overstressed for P-M-M 11 Axial compressive capacity for concrete exceeded (TBM 6.4.2) 12 Beam frames into column eccentrically (11.6.3) 13 Torsion exceeds maximum allowed 14 Reinforcing provided is below minimum required 15 Reinforcing provided exceeds maximum allowed 16 Tension reinforcing provided is below minimum required 17 k ∗ L/r > 30 (GB 7.3.10) 21 The column is not ductile. Beam/column capacity ratio is not needed. 22 The load is not seismic. Beam/column capacity ratio is not needed. 23 There is no beam on top of column. Beam/column capacity ratio is not needed. 24 At least one beam on top of column is not of concrete. Beam/column capacity ratio is not calculated. 25 The column on top is not of concrete. Beam/column capacity ratio is not calculated. 26 The station is not at the top of the column. Beam/column capacity ratio is not needed. 27 The column is not ductile. Joint shear ratio is not needed. Appendix E Appendix E - Error Messages and Warnings Error Number Description 28 The load is not seismic. Joint shear ratio is not needed. 29 There is no beam on top of column. Joint shear ratio is not needed. 30 At least one beam on top of column is not of concrete. Joint shear ratio is not calculated. 31 The column on top is not of concrete. Joint shear ratio is not needed. 32 The station is not at the top of the column. Joint shear ratio is not needed. 33 Beam/column capacity ratio exceeds limit. 34 Joint shear ratio exceeds limit. 35 Capacity ratio exceeds limit. 36 All beam s around the joint have not been designed. Beam/column capacity ratio is not calculated. 37 At least one beam around the joint have failed. Beam/column capacity ratio is not calculated. 38 The column above the joint have not been designed. Beam/column capacity ratio is not calculated. 39 The column above the joint have failed. Beam/column capacity ratio is not calculated. 40 All beams around the joint have not been designed. Joint shear ratio is not calculated. 41 At least one beam around the joint have failed. Joint shear ratio is not calculated. Output Details E- 3 Concrete Frame Design CSA A23.3-04 Error Number E-4 Description 42 The column above the joint has not been designed. Joint shear ratio is not calculated. 43 The column above the joint has failed. Joint shear ratio is not calculated. 45 Shear stress due to shear force and torsion together exceeds maximum allowed. Appendix E References CSA, 2004. A23.3-44, De sign of Concrete Structures, Canadian Standards Association, Rexdale, Ontario, Canada. CSI, 2008b. CSI Analysis Reference Manual, Computers and Structures, Inc., Berkeley, California. White, D. W. and J. F. Hajjar, 1991. “Application of Second-Order Elastic Analysis in LRFD: Research to Practice,” Engineering Journal, American Institute of Steel Construction, Inc., Vol. 28, No. 4. i
Source Exif Data:
File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.6 Linearized : No Author : CSL#003444 Comments : Company : Computers and Structures Create Date : 2015:09:15 06:27:49-07:00 Modify Date : 2016:02:10 16:21:14-08:00 Subject : Title : Tagged PDF : No XMP Toolkit : Adobe XMP Core 4.2.1-c043 52.372728, 2009/01/18-15:08:04 Metadata Date : 2016:02:10 16:21:14-08:00 Creator Tool : Acrobat PDFMaker 11 for Word Format : application/pdf Creator : CSL#003444 Document ID : uuid:57703d7b-a9e0-4a34-bad9-1fdf7b865a84 Instance ID : uuid:6ac86e93-555e-4c27-a185-b5f879010008 Producer : Adobe PDF Library 11.0 Page Layout : SinglePage Page Count : 83EXIF Metadata provided by EXIF.tools