Direct Design Method and Design Diagram. For Reinforced Concrete Columns and Shear walls

Transcription

1 Direct Design Method and Design Diagram For Reinforced Concrete Columns and Shear walls BY MAJID HOUSHIAR B.S., Isfahan University of Technology, 1988 M.S., University of Illinois at Chicago, Chicago, 2016 THESIS Submitted as partial fulfillment of the requirements for the degree of Master of Science in Civil Engineering in the Graduate College of the University of Illinois at Chicago, 2016 Chicago, Illinois Defense Committee: Mustafa Mahamid, Advisor Didem Ozevin, Chair Sheng-wei Chi

2 ACKNOWLEDGMENTS I would like to express the deepest appreciation to my advisor Dr. Mustafa Mahamid, for all his advising and contributions to the development of this thesis. Without his patience, flexibility, guidance, and faith in me during my research, this thesis would not have been completed. The good advice, support and most importantly his friendship during my research time, has been invaluable on both academic and personal level, for which I am extremely grateful. ii

3 TABLE OF CONTENTS 1) INTRODUCTION RESEARCH MOTIVATION AND BACKGROUND Literature Review and Accepted Column Design Procedure Previous Researches on RC Column Design Accepted Column Design Procedure Columns Under Combined Load and Uniaxial Moments Interaction diagram Column Capacity Tables Columns under combined Biaxial Load and Moments Approximate methods Load contours D Interaction Diagram PROBLEM STATEMENT RESEARCH OBJECTIVES AND SCOPE METHODOLOGY ) COLUMNS DESIGN THEORY AND BACKGROUND DESIGN OF COLUMNS SUBJECT TO AXIAL LOAD AND BENDING BACKGROUND THEORY AND ASSUMPTION ACI PROVISIONS INTERACTION DIAGRAMS A SAMPLE INTERACTION DIAGRAM PROGRAM iii

4 3) DIRECT DESIGN METHOD DIRECT DESIGN METHOD METHOD DEVELOPMENT MATHCAD CALCULATION SHEET NON-LINEAR SYSTEM OF EQUATIONS Numerical Solution Newton s method Newton s method pros and cons SPECIAL CASES, REAL AND IMAGINARY SOLUTIONS Acceptable Negative value for Ab Non-Acceptable Negative value for Ab Guidelines for Solving Non-Linear System of equations C-LANGUAGE PROGRAM FOR RECTANGULAR SECTIONS Vision and Scope Theory and Assumptions Program Interface and Output Result Interpretation Result Validation C-LANGUAGE PROGRAM FOR NON-RECTANGULAR COLUMNS & SHEAR WALLS Vision and Scope Theory and Assumptions Program Interface and Output Result Validation iv

5 4) DESIGN DIAGRAM DESIGN DIAGRAM METHOD DEVELOPMENT Theory and Assumption Design Diagram Development Design Diagram Characteristics SAMPLE MODEL Design Diagram for a Circular Column Design Diagram for an L-Shaped Shear Wall Design Diagram for an U-Shaped Shear Wall ) SUMMARY AND CONCLUSION SUMMARY CONCLUSION REFERENCES APPENDIX (A) METHOD VALIDATION v

6 Table of Figures Fig. 1. Sample interaction diagram... 5 Fig. 2. ACI rectangular column interaction diagrams... 6 Fig. 3. Load Contour Diagram Fig. 4. Biaxial Interaction Surface Fig. 5. L-Shaped Column Biaxial Interaction Surface Fig. 6. Cross section of concrete column Fig. 7. Interaction diagrams for section in Fig Fig. 8. Strain and Stress Distribution for Section Subject to Flexure and Axial Load Fig. 9. Strain Distribution and interaction diagram Fig D-Interaction Diagram by MathCad Fig. 11. A Generic Column Cross Section Fig. 12. A Rectangular Section Parameters Fig. 13. Acceptable Negative Reinforcement Area Fig. 14. Non-Acceptable Negative Reinforcement Area Fig. 15. Column Direct Design Program Fig. 16. Buttons in Column Direct Design Fig. 17. Design Procedure in Column Direct Design Fig. 18. Sample RC Section Fig x22in Section, Biaxial Load and Moments Fig x22in Section, Direct Design Method Fig x22in Section, Interaction Diagram by spcolumn Fig. 22. Column Section Covered by Direct Design Program vi

7 Fig. 23. Creating Shapes by Solid and Opening Triangles Fig. 24. Calculating triangle properties Fig. 25. Compression Zone Properties Calculation Fig. 26. The Advanced Direct Design Program Fig. 27. U-Shaped shear wall, Interaction Diagram by spcolumn Fig. 28. Design diagram flowchart Fig. 29. Required Area of Bars by Direct Design Method Fig. 30. Design diagram for a square section Fig. 31. Design Diagram created by Direct-Design program Fig. 32. Corresponding section to selected points in Fig Fig. 33. Model 1, Circular section Fig. 34. Design diagram for Model Fig. 35. Mode 2, L-Shaped shear wall Fig. 36. Design result for point 3 of Table Fig. 37. Design diagram for Model Fig. 38. Model 3, U-Shaped shear wall Fig. 39. Design diagram for Model vii

8 List of Tables Table 1. Column Capacity Table... 7 Table 2. ACI provisions for Column Design Table 3. Compression Zone Calculation Table 4. Design results for section shown in Fig Table 5. Design results for circular section in Fig Table 6. Design results for L-Shaped section in Fig Table 7. Design results for U-Shaped section in Fig viii

9 List of Symbols a = depth of compression zone at nominal flexural strength b = width of section c = neutral axis depth at nominal flexural strength Ab di E c Fsi fc fy Pu Mux Muy Xsi Ysi Pnc = area of each bar. = distance of i th bar from extreme fiber in compress ion = modulus of elasticity of concrete = force at i th bar = compressive strength of concrete = yield stress of reinforcing steel = applied axial load = applied moment about x axis = applied moment about y axis = X coordinate of the i th bar = Y coordinate of the i th bar = concrete nominal factored axial strength Mnxc = concrete nominal factored flexural strength about x-axis Mnyc = concrete nominal factored flexural strength about y-axis Pns = reinforcement nominal factored axial strength Mnxs = reinforcement nominal factored flexural strength about x-axis Mnys = reinforcement nominal factored flexural strength about y-axis Pn = nominal factored axial strength Mnx = nominal factored flexural strength about x-axis Mny = nominal factored flexural strength about y-axis θ = angle between neutral axis and x-axis. cu si = ultimate concrete compressive strain = strain in i th bar = strength reduction factor ix

10 SUMMARY Design of reinforced concrete columns and shear walls is an iterative process. Whether done manually or computer aided, the capacity of an assumed section is checked using interaction diagrams and the procedure continues until a satisfactory section is found. This research introduces two new concepts for design of reinforced concrete columns, Direct Design method and Design Diagram. Direct design method is an analytical approach by which the required area of reinforcement is determined directly without using an interaction diagram. In this approach, location of the neutral axis and required area of reinforcement is found directly by solving a nonlinear system of equations. Direct design method provides an optimum solution for a reinforced concrete section; the capacity of the section is exactly equal to the applied load and moments. For each column or shear wall, there are many optimum sections with different sizes and bar arrangements. A design diagram shows all possible optimum sections for a column or shear wall. This study provides an algorithm and a computer program for making design diagrams. Having a design diagram, designers do not need to go through a trial-and-error procedure to find an acceptable section; they can simply pick up an optimum section that best fits their requirements from the design diagram. x

11 1) INTRODUCTION 1.1 Research Motivation and Background A column is a vertical structural member supporting axial compressive loads, with or without moments. The cross-sectional dimensions of a column are generally considerably less than its height. Columns support vertical loads from the floors and roof or other parts of structure and transmit these loads to the foundations. When a structural element is only subjected to bending (like beams or slabs), it is possible to develop relatively simple equations for investigating that element. Column, usually are subjected to combined axial load and biaxial bending. Although it is possible to derive a family of equations to evaluate the strength of columns subjected to combined bending and axial loads for regular and simple column sections, these equations are tedious to use. Deriving these equations for irregular sections or shear wall is almost impossible. For this reason, most of the column design methods are depend on interaction diagrams and tables. Design of reinforced concrete column is a trial-and-error procedure. Usually, loads to which the column will be subjected are normally known, and it is desirable to find the section dimension, bar arrangements and the steel areas. In normal practice, a section is assumed and its capacity is checked using interaction diagrams and the procedure continue until a satisfactory section in found. The accepted approach for design of RC columns is time consuming and there is no guarantee that the found section is an optimum section. This research develops a design procedure in which designers can simply pick up an optimum section from a design diagram without going through a trail-and-error procedure. 1

12 1.1.1 Literature Review and Accepted Column Design Procedure This section reviews the history of previous researches on RC column design. Then various methods of reinforced concrete column design are discussed in brief Previous Researches on RC Column Design The design of reinforced concrete (RC) columns have been investigated extensively by numerous researchers (1999): - Anderson and Lee Chu and Pabarcius Bresler Furlong Aas-Jacobsen Fleming and Werner Parme et al Ramamurthy Gurfinkel Hsu and Mirza Heindahm and Bianchini Ansari Ross and Yen Hsu 1986, Gouwens 1975; developed simplified design aids for rectangular columns that are included in many textbooks. - Marin 1979; presented design aids For L-shaped columns 2

13 - Ramamurthy and Khan 1983; suggested two methods of design (the failure surface and the equivalent rectangular column) - Hsu 1985; presented theoretical and experimental results. - Hsu 1987, and Dundar 1990; reported Channel and box shaped columns - Hsu 1989; reported T-shaped columns With the availability of powerful and inexpensive microcomputers, the analysis of RC columns and composite sections under biaxial bending and axial load became amenable. - Dinsmore (1982) developed a computer program for practical design according to the American Concrete Institute code. - Brondum-Nielsen (1982, 1983,1987a,b) and Yen (1991) presented methods for the analysis of arbitrary cross sections. - Kawakami et al. (1985) developed a computer algorithm of RC members under biaxial bending including the effects of both prestressed and regular reinforcements. - Barzegar and Erasito (1995) developed a computer algorithm for an arbitrary cross section using the Whitney s rectangular stress block and integral methods. The most complete algorithm for the analysis. - Rodriguez and Dario (1999) developed a computer algorithm for biaxial interaction diagrams for short RC column of any cross section. Above researchers have developed many procedures and methods for design of RC column, but only a few of them have been widely accepted in engineering practice. Following chapter will cover the widely accepted RC column design procedures. 3

14 1.1.3 Accepted Column Design Procedure Accepted column design procedure could be classified into two main group: - Column under combined load and Uniaxial moment. - Column under combined load and Biaxial moment Columns Under Combined Load and Uniaxial Moments The strength of columns subjected to combined load and uniaxial bending could be evaluated by using interaction diagrams or tables. Typical diagrams for axial loads and bending moments about the principal axes of section are available for manual calculations. There are also many commercial computer programs that develop interaction diagrams or tables for given cross sections. The following sections cover creating and using of interaction diagram and tables for design of RC columns Interaction diagram An Interaction diagram can be generated by plotting the design axial load strength ϕpn against the corresponding design moment strength ϕmn; this diagram defines the usable strength of a section at different eccentricities of the load. Fig. 1 shows a sample interaction diagram for a rectangular RC column section. Interaction diagrams are useful for studying the strengths of columns with varying proportions of loads and moments. Any combination of loading that falls inside the curve is satisfactory, whereas any combination falling outside the curve represents failure. 4

15 Fig. 1. Sample interaction diagram If individual column interaction diagrams were prepared as described in the preceding sections, it would be necessary to have a diagram for each different column cross section, for each different set of concrete and steel grades, and for each different bar arrangement. The result would be an astronomical number of diagrams. The number can be tremendously reduced, however, if the diagrams are plotted with ordinates of: K n = P n f c A g (instead of Pn) R n = P n e f c A g h (instead of Mn) The resulting normalized interaction diagrams can be used for cross sections with widely varying dimensions. The ACI has prepared normalized interaction curves in this manner for the 5

16 different cross section and bar arrangement situations shown in Fig. 2 and for different grades of steel and concrete. In order to correctly use these diagrams, it is necessary to compute the value of γ (gamma), which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by h, the depth of the column (both values being taken in the direction of bending). Usually the value of γ obtained falls in between a pair of curves, and interpolation of the curve readings will have to be made. Fig. 2. ACI rectangular column interaction diagrams 6

17 Column Capacity Tables The column load capacity tables provide capacities for bending about both major and minor axes. These tables give the factored usable capacity for usual range of sizes of square, rectangular, and round columns. The appropriate table is entered with values of the factored load and moment, and the column dimensions and reinforcement are obtained. Table 1shows a sample column capacity table. Table 1. Column Capacity Table 7

18 Columns under combined Biaxial Load and Moments Many columns are subjected to biaxial bending, that is, bending about both axes. Corner columns in Buildings, or columns in building subjected to lateral loads like wind and earthquake are the most common cases. Bridge piers are almost always subject to biaxial bending. For shapes other than circular ones, it is necessary to consider the three-dimensional interaction effects Approximate methods During the past few decades, several approximate methods have been introduced for the design of columns with biaxial moments. One of the approximate methods that is useful in analysis and that can be handled with pocket calculators includes the use of the so-called reciprocal interaction equation, which was developed by Professor Boris Bresler of the University of California at Berkeley. This equation, follows: 1 P n = 1 P ox + 1 P oy 1 P 0 Rearranging variables yields: where P n = 1 P ox P 1 oy P 0 Pox = Maximum uniaxial load strength of the column with a moment of Mnx = Pn ey Poy = Maximum uniaxial load strength of the column with a moment of Mny = Pn ex Po = Maximum axial load strength with no applied moments 8

19 This equation is simple in form and the variables are easily determined. Axial load strengths Po, Pox, and Poy are determined using any of the methods presented above for uniaxial bending with axial load. Experimental results have shown the above equation to be reasonably accurate when flexure does not govern design. The equation should only be used when: P n 0.1 f c A g There are many other approximate methods for designing RC column under combined load and biaxial bending. The other well-known methods are: - Bresler Reciprocal Load Method - Bresler Load Contour Method - PCA Load Contour Method Load contours In this method, the failure surface is approximated by a family of curves corresponding to constant values of Pn. These curves, may be regarded as load contours. Fig. 3 shows a load contour diagram created by Joaquin Marin for an L-Shaped column. 9

20 Fig. 3. Load Contour Diagram D Interaction Diagram A uniaxial interaction diagram defines the load-moment strength along a single plane of a section under an axial load P and a uniaxial moment M. The biaxial bending resistance of an axially loaded column can be presented schematically as a surface formed by a series of uniaxial 10

21 interaction curves drawn radially from the P axis. The difficulty associated with the determination of the strength of reinforced columns subject to combined axial load and biaxial bending is primarily an arithmetic one. The bending resistance of an axially loaded column about a particular skewed axis is determined through iterations involving lengthy calculations. Fig. 4 shows a biaxial interaction surface. Fig. 4. Biaxial Interaction Surface Any combination of loading that falls inside the interaction surface is satisfactory, whereas any combination falling outside the interaction surface represents failure. Most of the modern column design software has the ability of creating biaxial (3D) interaction surface. Fig. 5 shows a biaxial interaction surface created by spcolumn program for an L-Shaped column. 11

22 Fig. 5. L-Shaped Column Biaxial Interaction Surface 1.2 Problem Statement Design of reinforced concrete columns and shear walls is a trial-and-error procedure. If the factored loads and moments are known and it is necessary to select a cross section to resist them, the procedure is referred to as design or proportioning. A design problem is solved by guessing a section, analyzing whether it will be satisfactory, revising the section, and reanalyzing it. The analysis portion of the problem for column section design is mostly carried out via interaction diagrams. The traditional design procedure is time consuming and does not necessarily lead to an optimum section. 12

23 405mm (16in) For example, assume that it is required to select a cross section for a short square reinforced concrete column subjected to the following factored load and moment: Pu = 2180 kn (490 kip) : Applied axial load. Mux = 190 kn-m (140 k-ft) : Applied moment about x-axis y f y = 420 MPa (60 ksi) f ' = 35 MPa (5.0 ksi) c x A b 405mm (16in) 65mm (2.5in) Fig. 6. Cross section of concrete column Fig. 6 shows an assumed section for the above column. Size and reinforcement arrangement of the section are known. It is required to investigate the assumed section with different bar areas (Ab) to find whether any of them would satisfy the applied loads. Fig. 7 shows a series of interaction diagrams, created for different bar areas (Ab) for the column section shown in Fig

24 Moment, ØMn (k-ft) Axial load, ØPn (kn) Axial load, ØPn (kip) Moment, ØMn (kn-m) 2 2 A b = mm (-0.40 in ) 2 2 A b = 0 mm (0 in ) 2 2 A b = 250 mm (0.40 in ) 2 2 A b = 500 mm (0.80 in ) 2 2 A b = 750 mm (1.20 in ) Fig. 7. Interaction diagrams for section in Fig. 6 Point 1 in Fig. 7 that represents the above applied loads is located on the plot for Ab=500 mm 2 (0.8in 2 ). It shows that the capacity of the section with Ab =500 mm 2 is equal to the applied loads. Other sections with Ab >500 mm 2 are over designed and sections with Ab <500 mm 2 are not adequate. The calculated bar area is an acceptable value otherwise the designer should revise the section dimension or bar arrangement and repeat the procedure. For a short rectangular reinforced concrete column subjected to the above factored load and moment, a 405 mm square section with four bars with Ab =500 mm 2 (#8 bar) is an optimum section. 14

25 It is possible to find other optimum sections for the above column with different dimensions, bar arrangements, and bar areas. For example, a 355 mm (14in) square section with 12 #9 bar is also another optimum solution. However, finding all optimum solutions for a column or shear wall using traditional methods is time consuming and impractical. This research proposes a general and efficient procedure by which all possible optimum solutions for a column or shear wall are found and presented on a design diagram. 1.3 Research Objectives and scope An efficient and straightforward method for designing reinforced concrete columns and shear walls is presented in this research. This method creates a design diagram for a column or shear wall based on the applied load and moments. The provided design diagram shows all possible optimum sections for the column. Design diagram is a practical tool by which designers do not need to go through a trial-and-error procedure for designing columns. Design of column would be much faster, easier, and efficient by using this method, since the designers have all optimum solutions in one diagram. 1.4 Methodology 1- The Direct Design method is presenting for designing RC columns without using interaction diagrams. 2- The method finds the required area of reinforcement by solving a non-linear system of equations. 3- A procedure has developed for solving the non-linear system of equations that always can find the solution. 15

26 4- A C-language program has developed for designing RC columns by implementing direct design method. 5- The new concept of Design Diagram has presented. A design diagram shows all possible optimum sections for a column or shear wall. 6- A procedure has presented for making design diagrams by implementing direct design method. 7- A C-language program has developed for making design diagrams. 16

27 2) COLUMNS DESIGN THEORY AND BACKGROUND This chapter covers the theory and background of column section strength evaluation based on the ACI standard. In addition, this chapter gives a short explanation of interaction diagrams and provide a sample calculation sheet for making simple uniaxial interaction diagram. 2.1 Design of Columns Subject to Axial Load and Bending Many columns are subjected to biaxial bending, that is, bending about both axes. Corner columns in buildings where beams and girders frame into the columns from both directions are the most common cases, but there are others, such as where columns are cast monolithically as part of frames in both directions or where columns are supporting heavy spandrel beams or lateral load like earthquake. Bridge piers are almost always subject to biaxial bending. This chapter covers the background theory and assumptions for designing RC columns under combined bending and axial load. 2.2 Background theory and Assumption The strength of a member or cross-section subject to combined bending and axial load, Mn and Pn, must satisfy the two following conditions: 1- static equilibrium and. 2- compatibility of strains. Equilibrium between the compressive and tensile forces includes the axial load Pn acting on the cross section. The general condition of the stress and the strain in concrete and steel at nominal strength of a member under combined flexure and axial compression is shown in Fig. 8. The tensile 17

28 or compressive force developed in the reinforcement is determined from the strain condition at the location of the reinforcement. Fig. 8. Strain and Stress Distribution for Section Subject to Flexure and Axial Load For a known strain condition, the corresponding load-moment strength, Pn and Mn, can be computed directly. Design or investigation of a short compression member (without slenderness effect) is based primarily on the strength of its cross-section. Strength of a cross-section under combined flexure and axial load must satisfy both force equilibrium and strain compatibility. The combined nominal axial load and moment strength (Pn, Mn) is then multiplied by the appropriate strength reduction 18

29 factor ϕ to obtain the design strength (ϕpn, ϕmn) of the section. The design strength must be equal to or greater than the required strength: (ϕpn, ϕmn) (Pu, Mu) 2.3 ACI Provisions The formulation developed in this research is according to ACI 318 Code. It is assumed that maximum allowable strain on concrete is εcu=0.003 and the strains in reinforcement and concrete are directly proportional to their distance from the neutral axis. Also, concrete stress of 0.85f c is uniformly distributed over the compression zone bounded by the cross section and a straight line parallel to the neutral axis at a distance a = β1 c from the fiber of maximum compressive strain, where c is distance from the neutral axis to the fiber of maximum compressive strain. Reinforcement is assumed elastic-perfectly plastic. However, it is possible to consider any stressstrain diagram for both the concrete and the reinforcement. The allowable area of longitudinal reinforcement for non-composite compression members is considered not less than 0.01Ag or more than 0.08Ag. Table 2 shows all ACI provisions related to design of RC columns. 19

30 Table 2. ACI provisions for Column Design ACI Design of Columns: Notation Es ksi modulus of elasticity of reinforcement fy ksi specified yield strength for reinforcement f'c ksi specified compressive strength of concrete fs ksi stress in the i-th layer of surface reinforcement Fs kips the force developed in reinforcement, kips A g in. 2 gross area of concrete section A st in. 2 total area of longitudinal reinforcement ϕ strength reduction factor ε s strain of reinforcement ε y yield strain of reinforcement εcu εt εty Maximum strain at the extreme concrete compression fiber net tensile strain in extreme layer of longitudinal tension reinforcement net tensile strain used to define compression-controlled section Required strength ϕs n ϕp n P u, ϕm n M u Interaction between load effects shall be considered: Load combinations U 1.4D 1.2D + 1.6L + 0.5(Lr or S or R) 1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W) 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R) 1.2D + 1.0E + 1.0L + 0.2S 0.9D + 1.0W 0.9D + 1.0E Matrials Es Es=29'000 ksi fy Minimum Maximum Special seismic systems None 60 ksi Others None 80 ksi f'c Minimum Maximum General 2.5 ksi None Special seismic systems (Normal) 3.0 ksi None Special seismic systems (Lightweight ) 3.0 ksi 5.0 ksi 20

31 Strength reduction factor ϕ Spirals Other Compression-controlled (ε t ε ty) Transition (ε ty < ε t< 0.005) (ε t - ε ty)/( ε ty) (ε t - ε ty)/( ε ty) Tension-controlled (ε t 0.005) Strain Strain in bars and concrete assumed proportional to the distance from neutral axis εcu ε cu = εty ε ty = f y / E s ε ty = (Grade 60) Stress & Strength Tensile strength of concrete shall be neglected in flexural and axial strength calculations fs f s = E s ε s (ε s < ε y) R Fs F s = A s f s f s = f y (ε s ε y) Compression zone Uniformly distributed concrete stress and its boundaries c, shall be measured perpendicular to the neutral axis β β 1 = f' c / Axial strength limitation P n,max = 0.80 [ 0.85 f' c (A g - A st) + f y A st ] (Ties) Compressive strength P n,max = 0.85 [ 0.85 f' c (A g - A st) + f y A st ] (Spirals) P nt,max = f y A st Tensile strength Reinforcement Limitation A g A st 0.08 A g 2.4 Interaction Diagrams Although it is possible to derive a family of equations to evaluate the strength of columns subjected to combined bending and axial loads for regular and simple column sections, these equations are very complicated and tedious to use. Deriving these equations for irregular sections or shear wall is almost impossible. For this reason, most of the column design methods are depend on interaction diagrams and tables. 21

32 ØPn An Interaction diagram can be generated by plotting the design axial load strength ϕpn against the corresponding design moment strength ϕmn; this diagram defines the usable strength of a section at different eccentricities of the load. Interaction diagrams for columns are generally computed by assuming a series of strain distributions, each corresponding to a particular point on the interaction diagram, and computing the corresponding values of P and M. Once enough such points have been computed, the results are plotted as an interaction diagram. Fig. 9 shows a sample interaction diagram and some of the assumed strain distributions used to create the interaction diagram. The corresponding point for each assumed strain distribution is shown on the interaction diagram. Interaction diagram Strain distribution Neutral Axis cu ØMn Fig. 9. Strain Distribution and interaction diagram 2.5 A Sample Interaction Diagram Program A MathCAD calculation sheet for creating uniaxial interaction diagrams is provided in this section. The calculation sheet calculates and draws the interaction diagram for a square section with four reinforcements. The program could be modified simply to cover other rectangular 22

33 sections with different bar arrangments. The calculation sheet shows the practical procedure for making interaction diagrams. Concrete Contribution: Reinforcement Coordinates: Reinforcement Strain, Stress, Forces: :Distance of each bar from extreme fiber in compression :Strain in each bar 23

34 :Gama for each bar :Stress in each bar :Force of each bar Reinforcement Contribution: Reinforcement & Concrete Contribution: Drawing Interaction Diagram: 24

35 Fig D-Interaction Diagram by MathCad 25

36 3) DIRECT DESIGN METHOD 3.1 Direct Design Method Direct design method is an analytical approach by which the required area of reinforcement is determined directly without using an interaction diagram. In this approach, location of the neutral axis and required area of reinforcement is found directly by solving a non-linear system of equations. Direct design method provides an optimum solution for a reinforced concrete section; the capacity of the section is exactly equal to the applied load and moments. This chapter covers the development of this method, for a generic column sections. 3.2 Method Development Consider the general cross section in Fig. 11. The coordinate system is referred to the centroid of the concrete section. The location of neutral axis is defined by variables c, θ. Where θ is the angle of neutral axis with the x-axis, and c is distance from the neutral axis to the fiber of maximum compressive strain y c a i d i X c Ac Yc x A b N.A. Fig. 11. A Generic Column Cross Section 26

37 Contribution of concrete Assume following functions calculate the contribution of concrete to the nominal axial strength, nominal flexural strengths about the x-axis and the y-axis respectively. Pnc (c, θ) Mnxc (c, θ) Mnyc (c, θ) : Concrete nominal axial strength : Concrete nominal flexural strengths about the x-axis : Concrete nominal flexural strengths about the y-axis The contribution of concrete to the nominal strength of cross section is: Pnc (c, θ) = 0.85f c * Ac (c, θ) (1) Mnxc (c, θ) = 0.85f c * Ac (c, θ) * Yc (c, θ) (2) Mnyc (c, θ) = 0.85f c * Ac (c, θ) * Xc (c, θ) (3) Where: Ac (c, θ) : Area of compression zone Xc (c, θ) : X coordinate of the centroid of Ac (c, θ) Yc (c, θ) : Y coordinate of the centroid of Ac (c, θ) Contribution of reinforcement From the maximum allowable strain in concrete εcu= 0.003, the strain compatibility condition and the location of neutral axis, c the strain εsi at the ith bar is: εsi =0.003(1- di /c) (4) Where di is distance of i th bar from extreme fiber in compression in the direction perpendicular to neutral axis. 27

38 Stress in each bar is determined from the stress-strain diagram of the reinforcement. The contribution of reinforcement to the nominal strength of cross section is: Pns (c, θ, Ab) = Σ Fsi (c, θ, Ab) (5) Mnxs (c, θ, Ab) = Σ Fsi (c, θ, Ab) * Ysi (6) Mnys (c, θ, Ab) = Σ Fsi (c, θ, Ab) * Xsi (7) Where: Fsi (c, θ, Ab) : force at the i th bar Xsi Ysi : X coordinate of the i th bar : Y coordinate of the i th bar The strength reduction factor ϕ (c, θ) is defined by the tensile strain in the extreme bar in tension at nominal strength. The nominal factored strength of section is: Pn (c, θ, Ab ) = (c, θ)* [Pnc (c, θ) + Pns (c, θ, Ab)] (8) Mnx(c, θ, Ab) = (c, θ)* [Mnxc (c, θ) + Mnxs (c, θ, Ab)] (9) Mny(c, θ, Ab) = (c, θ)* [Mnyc (c, θ) + Mnys (c, θ, Ab)] (10) The optimum design for a column is when the factored strengths ϕpn, ϕmnx and ϕmny in Eq. (8) - (10) are equal with the externally applied load and moments Pu, Mux and Muy respectively. Pn (c, θ, Ab) = Pu (11) Mnx(c, θ, Ab) = Mux (12) Mny(c, θ, Ab) = Muy (13) 28

39 By re ordering the above relations, the final system of equations is: P (c, θ, Ab) = Pn (c, θ, Ab) Pu = 0 (14) Mx (c, θ, Ab) = Mnx(c, θ, Ab) Mux = 0 (15) Mx (c, θ, Ab) = Mny(c, θ, Ab) Muy = 0 (16) Solving the system of Eq. (14) to (16), determines the three unknown variables C, θ, Ab. The location of N.A. is defined by C and θ, and Ab is the required area of each bar. 3.3 MathCAD calculation Sheet A sample MathCAD calculation sheet is provided in this section. This program shows the practical use of the direct design method developed in previous section. The program is written for a rectangular column section with four bars and could be used for sections with more bars or different shapes with a little modification. To compute the nominal strength of a section, it is necessary to compute the area and the moment of area of the compression zone. Presented program calculate above values by integrating over the section using the width function provided by Function (17). The function calculates the width of compression zone at any distance of y from origin. be(c,, y) = [β1 c -(h/2 - y)* cos(θ) ]* sin(θ) -1 (17) Function (18), calculates the distance of each bar from extreme fiber in compression in the direction perpendicular to the neutral axis in terms of θ and coordinate of each bar referred to origin of coordinate system. See Fig. 12. di(x, y, θ) = (b/2 - x)* sin(θ)+ (h/2-y) * cos(θ) (18) 29

40 Fig. 12. A Rectangular Section Parameters The provided program calculates C and θ, and Ab for the following applied loads: Pu = 500 kips (2225 KN) : Applied axial load. Mux = 1200 in-kips (135 KN-m) Muy = 700 in-kips (80 KN-m) : Applied moment about x-axis. : Applied moment about y-axis. SAMPLE MATHCAD PROGRAM SECTION PARAMETERS 30

41 CONCRETE CONTRIBUTION COORDINATES OF BARS STRAINS, STRESSES AND FORCES IN BARS 31

42 REINFORCEMENT CONTRIBUTION STRENGTH REDUCTION FACTOR For longitudinal bar enclosed by ties, NOMINAL STRENGTH OF SECTION APPLIED AXIAL LOAD AND MOMENTS NOMINAL STRENGTH OF SECTION 32

43 SOLVE BLOCK INITIAL GUESS FOR UNKWON VARIABLES For calculation of forces in bars, stress in bars is reduced by (γ * 0.85 * fc) where γ is γ = 1, for bars located in compression zone γ = 0, for bars located in tensile zone γ = 0~1, for bars located on neutral axis γ factor consider the effect of concrete replaced by reinforcement. 33

44 RESULTS The equations in first three lines of the solve block are the same equations (14)-(16), developed in previous section. To solve this system of equations it is necessary to have an initial approximation for the unknown variables. In this example, the required area of each bar for the section in Fig. 12 under the applied load and moments is Ab =1.2183in 2 (785 mm 2 ). In other words, for Ab = in 2 the factored strength of the section is exactly equal to the applied load and moments. It means that, the point corresponds to the applied load and moments is exactly on the failure surface of the section with this area of reinforcement. The functions ϕpn(), ϕmnx() and ϕmny() can be used also to develop a 3D failure surfaces. 3.4 Non-Linear System of equations Equations (14) to (16) make up a nonlinear system of equations. There is no closed form solution for this nonlinear system of equations. Therefore, the solution depends on numerical iteration techniques, like Newton s method Numerical Solution The main idea for solving a non linear system of equations is starting with an initial guess which is reasonably close to the true solution, then the function is approximated by its tangent plane (which can be computed using the tools of calculus), and one computes the intercepts of this 34

45 tangent plane. These intercepts will typically be a better approximation to the solution than the original guess, and the method can be iterated Newton s method Newton s method can be easily adapted to deal with systems of nonlinear equations, reducing the nonlinear problem to an infinite sequence of linear problems, as mentioned earlier. Eq. (17) shows Newton s method for nonlinear systems, and it is generally expected to give quadratic convergence. X (k+1) = X (k) J(X (k) ) 1 F(X (k) ) k>=0 (17) Newton s method could be written in term of equations (14) to (16) as follow (for simplicity Ab has replaced by A): c k+1 [ θ k+1 ] = [ A k+1 c k θ k P(c k, θ k, A k ) ] J(X (k) ) 1 [ Mx(c k, θ k, A k )] (18) A k My(c k, θ k, A k ) Where J(X (k) ) = c k P(c k, θ k, A k ) c k Mx(c k, θ k, A k ) [ My(c c k, θ k, A k ) k θ k P(c k, θ k, A k ) Mx(c θ k, θ k, A k ) k My(c θ k, θ k, A k ) k A k P(c k, θ k, A k ) A k Mx(c k, θ k, A k ) A k My(c k, θ k, A k ) ] (19) And Ck, θk, Ak Ck+1, θk+1, Ak+1 : value for unknown variables at K th iteration : value for unknown variables at (K+1) th iteration 35

46 The following steps are required for solving the non-linear system of Equations (14) to (16) by implementing Newton s method: 1- Assume initial value for unknown variables C, θ, Ab. 2- Calculate the Jacobean matrix J using equation (19). 3- Use equation (18) to find the new values for unknown variables C, θ, Ab. 4- Repeat the process until the calculated section capacity is close enough to the applied load and moments Newton s method pros and cons Solving a nonlinear system of equations by Newton s method is an iterative process and should start from an initial assumption for the unknown variables. Newton s method is accurate and converges to the solution very fast; however, the calculations may not converge if the initial assumption is not close to the solution. The weak point of Newton s method is that finding an initial assumption close to the solution is not always easy and practical. In addition, there is no guarantee that the provided initial assumption would converge to the solution. 3.5 Special cases, Real and Imaginary Solutions As mentioned earlier, in accepted approach for designing RC column the capacity of an assumed section is checked using interaction diagrams. In practice, interaction diagrams are generated based on real and reasonable sections. In direct design method, the appropriate section is found by solving a non-linear system of equations. In many cases, a negative (imaginary) value is found for unknown variables C, θ, Ab. 36

47 1- C is distance from the neutral axis to the fiber of maximum compressive strain. A negative value for C is not acceptable. Whenever the solution contains a negative value for C, the iteration should start from another initial assumption. 2- θ is the angle of neutral axis with the x-axis. A negative value for θ is acceptable. 3- Ab is the required area of reinforcement. Determining the validity of a negative value for Ab is complicated. The following sections will cover the different scenario for negative values of Ab Acceptable Negative value for Ab In some situations, the theoretical capacity of the column section without contribution of reinforcement is greater than the applied load and moments. In these cases, solving the non-linear system of equations will lead to a negative value for Ab. In fact, the negative value for Ab indicates that a smaller section is required to satisfy the applied load and moments. For example, consider the square column section shown in Fig. 13. The column is subjected to the following load and moment: Pu = 400 kips : Applied axial load. Mux = 75 k-ft : Applied moment about x-axis. Fig. 13(a) shows the stress diagram of concrete (without considering reinforcement bars). The neutral axis depth is C=11.3in. As it is shown in Fig. 13(a), the corresponding capacity of the column section without steel contribution is: ϕpn = 400 ϕmn x= 115 kip k-ft It means that the capacity of concrete without steel contribution is more than the applied load. In Fig. 13(b) the neutral axis depth is C=13.08in, Ab= -0.91in 2. the capacity of the section shown 37

48 405mm (16in) in Fig. 13(b) is exactly equal to applied load and moment. In other word, Negative Ab reduces the section capacity and makes it equal to the applied loads. For calculation of factored load and moments the strength reduction factor is considered ϕ=0.65. ( a ) ( b ) f y = 420 MPa (60 ksi) f ' = 35 MPa (5.0 ksi) c y 0.85fc' 0.85fc' 71.0 kips x A b N.A. N.A. 8.3 kips 405mm (16in) 65mm (2.5in) (Only Concrete) P = 400 kips M = 115 k-ft (Concrete & Bars) A = in 2 b P = 400 kips M = 75 k-ft Fig. 13. Acceptable Negative Reinforcement Area In general, whenever the capacity of a column section without contribution of steel is greater that the capacity of the section with negative reinforcement, the negative calculated value for bars is acceptable. In this case, there is no need to start a new iteration to find another solution Non-Acceptable Negative value for Ab In some cases, there are two set of solutions for the non-linear system of equations. One set of answer with positive Ab and one set with negative Ab. For example, consider the square column section shown in Fig. 14. The column is subjected to the following load and moment: Pu = 400 kips : Applied axial load. Mux = 129 k-ft : Applied moment about x-axis. 38

49 405mm (16in) ( a ) ( b ) f y = 420 MPa (60 ksi) f ' = 35 MPa (5.0 ksi) c y 0.85fc' kips 0.85fc' 32.4 kips A b x N.A kips N.A. 7.9 kips kips 10.7 kips 405mm (16in) 65mm (2.5in) A = in 2 b P = 400 kips M = 129 k-ft A = 0.18 in 2 b P = 400 kips M = 129 k-ft Fig. 14. Non-Acceptable Negative Reinforcement Area In Fig. 14 (a) C=6.3in, Ab=-2.18in 2, ϕ=0.76 and in Fig. 14 (b) C=10.8in, Ab=0.18in 2, ϕ=0.65. Both stress distributions have the same capacity. In this case, the negative Ab is not acceptable and the solution shown in Fig. 14 (b) is the acceptable solution Guidelines for Solving Non-Linear System of equations As mentioned earlier, Newton s method is accurate and converges to the solution very fast. However, the calculations may not converge if the initial assumption is not close to the solution. Therefore, any algorithm for solving nonlinear system of equations (14) to (16) should be able to: 1- Start with an initial assumption for unknown variables based on the column section properties and applied loads 2- provide an improved initial assumption whenever the calculation does not converge based on the previous initial assumption 3- Detect the cases that the found solution is not acceptable 39

50 4- Detect the cases that there is no reasonable solution for the given section (section is very big or very small). The following guidelines could help to consider a reasonable range for initial assumption: 1- For doubly symmetric sections 0 θ 90 and θ can be initially approximated as atan(mux/ Muy). 2- C > 0, and for large values of C there would be no significant changes in nominal strengths of section. 3- The bar area could be initially approximated as Ab = 650 mm 2 (1.0in 2 ) 4- The acceptable range of Ab is 200 mm 2 (0.31in 2 ) Ab 2580 mm 2 (4in 2 ) for bars No. 5 to No. 18 respectively (nine bars sizes), and the total area of reinforcement is limited from 1% to 8% of the gross section of concrete. 3.6 C-Language Program for Rectangular Sections A C-Language program based on direct design method has developed for designing reinforced concrete columns. The program solves the system of Eq. (14) to (16) by implementing Newton s method Vision and Scope The following Items show the vision and scope of the program: 1- Designs rectangular section with any arrangement of reinforcement. 2- User can define material properties and applied loads. 3- User can define and easily change section dimension and bar arrangements. 4- Program finds the required area of reinforcement for section and applied loads provided by user. 40

51 5- Program calculates the Design Ratio for section and applied loads provided by user. 6- Program can automatically find an optimum section for any set of applied load. 7- Program can optimize section parameter locally. 8- Program shows the defined section graphically. 9- Program provides printable output result Theory and Assumptions The program has an internal algorithm for guessing an initial assumption for unknown variables based on the column section properties and applied loads. This algorithm can provide an improved initial assumption whenever the calculation does not converge based on the previous initial assumption. Following advantages of the program make it appropriate for design engineering practice: 1- Does not require initial assumption provided by user. 2- Can always find the solution if there is a reasonable solution. 3- Can detect the cases that there is no reasonable solution for the given section (section is very big or very small). The program uses ACI provision for designing columns. The most difficult part in calculating the capacity of a concrete section is finding the area and the centroid of the compression zone. In general, in a rectangular section, the neutral axis could have four different cases. Table 3, shows the required relations for calculating the area and the moment of area of compression zone in above cases. 41

52 Table 3. Compression Zone Calculation 42

53 3.6.3 Program Interface and Output Fig. 15 shows the final Column Direct Design program. The program shows the efficiency and simplicity of direct design method. As user is typing a value or parameter, the design results are shown. Any of these on-fly calculations are equal to making and testing several 3D interaction diagrams and the results of this method are more accurate. Fig. 15. Column Direct Design Program Different parts of the program interface are as follow: - In Material - Load box, user defines material properties and applied loads. - In Section box, user defines section dimensions, bar arrangements, and bar diameters. - The message box, shows the required bar area and equivalent available bar size. - The design Ratio and reinforcement percentage are shown in the box below the shape.0 43

54 Fig. 16 shows the functionality of the button used in the program. This buttons help user to find a desired optimum section quickly. Fig. 16. Buttons in Column Direct Design Fig. 17 explains the design ration concept and shows the design procedure in brief. Fig. 17. Design Procedure in Column Direct Design 44

55 CALCULATION SUMMARY INPUT DATA: fy = 60 ksi fc = 5 ksi Es = ksi Cov = 2.5 in As_Req = in² NEUTRAL AXIS LOCATION: c = in Ɵ = 0 degree CONCRETE COMPRESSION ZONE: β1 = 0.8 a = in Ac = in² Xbar = 0 in Ybar = in Pn_C = 0.85 * fc * Ac = kips Mnx_C = 0.85 * fc * Ac * Ybar = in-kips Mny_C = 0.85 * fc * Ac * Xbar = 0 in-kips REINFORCEMENT: bar Xs Ys ds εs fs Fs Fs*Ys Fs*Ys Σ Pn_s = ΣFs = kips Mnx_s = Σ(Fs*Ys) = in-kips Mny_s = Σ(Fs*Xs) = 0 in-kips FACTORED NOMINAL STRENGTH: Ø = 0.65 ØPn = Ø ( Pn_c + Pn_s ) = 400 kips ØMnx = Ø ( Mnx_c + Mnx_s ) = 1548 in-kips ØMny = Ø ( Mny_c + Mny_s ) = 0 in-kips ØPn = 400 kips ØMnx = 129 ft-kips ØMny = 0 ft-kips 45

56 405mm (16in) Result Interpretation Table 4 shows the design results for the section in Fig. 6 for four different set of applied loads and moments. The last column in the table displays the computed required areas for each reinforcing bar. y f y = 420 MPa (60 ksi) f ' = 35 MPa (5.0 ksi) c x A b 405mm (16in) 65mm (2.5in) Fig. 18. Sample RC Section For load case 1, the computed area of reinforcement is an acceptable value. It shows that the section dimensions and bars arrangement are appropriate. Use #8 bars, Ab = 510 mm 2 (0.79 in 2 ). In load case 2, the calculated area of bars is negative. Theoretically, the concrete section without contribution of reinforcement can carry any load case that falls into this region. In this case, the designer shall provide the minimum required reinforcement, or reduce the size of the section. In load case 3, the calculated area of bars is so large that it exceeds #18 bars, Ab = 2580 mm 2 (4.0 in 2 ) the largest available bar sizes. Therefore, a larger cross section is required for the concrete strength and bar grade assumed. 46

57 560mm (22in) In load case 4, the concrete section is subjected to axial load and biaxial moments. The computed area of reinforcement is an acceptable value. Use #8 bars, Ab = 510 mm 2 (0.79 in 2 ). Table 4. Design results for section shown in Fig. 18 Input Result Load Pu Mu x Mu y A b case kn (kip) kn-m (k-ft) kn-m (k-ft) mm 2 (in 2 ) (490) 190 (140) (0.78) (400) 102 (75) (-0.91) (750) 339 (250) (4.13) (500) 135 (100) 102 (75) 468 (0.73) Result Validation The RC column section shown in Fig. 19 is used for testing the validity of the program and direct design method. The column is subjected to the following load and moment: Pu = 640 kips : Applied axial load. Mux = 250 k-ft : Applied moment about x-axis. Muy = 100 k-ft : Applied moment about y-axis. y f y = 420 MPa (60 ksi) f ' = 35 MPa (5.0 ksi) c x A b 405mm (16in) 50mm (2.0in) Fig x22in Section, Biaxial Load and Moments 47

58 Fig x22in Section, Direct Design Method Designing the RC column section shown in Fig. 19 by Direct design method shows that the required bar area is Ab=0.787in 2, see Fig. 20. For testing validation of the result, the column capacity is checked by spcolumn program (spcolumn, STRUCTUREPOINT, formerly the Engineering Software Group of the Portland Cement Association PCA). Fig. 21 shows the interaction diagram for RC column section shown in Fig. 19 developed by spcolumn program. As it is shown in Fig. 21 the point corresponding to the applied load is on the interaction diagram. It means that the capacity of the RC column section is exactly equal to the applied load and moments. In other words, the required area of bars found by direct design method is a correct value, and the Column Direct Design program found this value without trial-anderror and without using interaction diagram. 48

59 Following page shows spcolum output files: STRUCTUREPOINT - spcolumn v4.81 (TM) Page 2 Licensed to: StructurePoint. License ID: /14/ in.col 10:08 AM General Information: ==================== File Name: 16-22in.col Project: Column: 16 x 22 Engineer: M.H. Code: ACI Units : English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Material Properties: ==================== f'c = 5 ksi fy = 60 ksi Ec = ksi Es = ksi Ultimate strain = in/in Beta1 = 0.8 Section: ======== Rectangular: Width = 16 in Gross section area, Ag = 352 in^2 Ix = in^4 rx = in Xo = 0 in Depth = 22 in Iy = in^4 ry = in Yo = 0 in Reinforcement: ============== Bar Set: ASTM A615 Size Diam (in) Area (in^2) Size Diam (in) Area (in^2) Size Diam (in) Area (in^2) # # # # # # # # # # # Confinement: Tied; #3 ties with #10 bars, #4 with larger bars. phi(a) = 0.8, phi(b) = 0.9, phi(c) = 0.65 Layout: Rectangular Pattern: Sides Different (Cover to longitudinal reinforcement) Total steel area: As = 7.90 in^2 at rho = 2.24% Minimum clear spacing = 4.50 in Top Bottom Left Right Bars 3 # 8 3 # 8 2 # 8 2 # 8 Cover(in) Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in *** End of output *** 49

60 P (kip) 1400 (Pmax) y x 1 16 x 22 in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 0 M (21.8 ) (k-ft) 350 Date: 08/14/16 Time: 10:23:23 (Pmin) -600 spcolumn v4.81. Licensed to: StructurePoint. License ID: File: E:\UIC\Column_Paper\Thesis\Writing\16-22in.col Project: DirectDesign Method Column: 16 x 22 Engineer: M. H. f'c = 5 ksi fy = 60 ksi Ag = 352 in^2 10 #8 bars Ec = 4031 ksi Es = ksi As = 7.90 in^2 rho = 2.24% fc = 4.25 ksi Xo = 0.00 in Ix = in^4 e_u = in/in Yo = 0.00 in Iy = in^4 Beta1 = 0.8 Min clear spacing = 4.50 in Clear cover = 2.00 in Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) = 0.65 Fig x22in Section, Interaction Diagram by spcolumn 50

61 3.7 C-Language Program for Non-Rectangular Columns & Shear walls The Column Direct Design program introduced earlier was developed for designing the doubly symmetric rectangular sections. The main purpose of this program was investigating the following basic concept and issues in direct design method: 1- Studying the efficiency and usability of the method. 2- Finding an optimum method for solving the non-linear system of equations. 3- Studying the special cases like, negative bar area. 4- Testing the validity of the method. The direct design method is not limited to rectangular sections with doubly symmetric bar arrangement. This method could be used to calculate the required area of reinforcement for any column or shear wall section with any arbitrary bar arrangement. In addition, designing a section using the direct design method would provide the following information: 1- Location of neutral axis. 2- Shape and location of compression zone. 3- Stress in each reinforcement bar. Traditional design method usually cannot provide above information. A new more advanced C-Language program has been developed based on the above direct design method abilities for designing reinforced concrete columns. The program can design a wide variety of column sections and shear walls. In addition, the program shows the location of neutral axis, shape and location of compression zone, and stress in each reinforcement bar. 51

62 3.7.1 Vision and Scope The following Items show the vision and scope of the new Direct Design program: 1- Designs wide verity of column sections and shear walls. 2- User can define material properties and applied loads. 3- User can easily define and change the section dimension and bar arrangements. 4- Program finds the required area of reinforcement for the section and applied loads provided by user. 5- Program shows the defined section graphically. 6- Program can rotates the sections. 7- Shows neutral axis location, defined by its angle and the distance from fiber in maximum compression. 8- Shows the location and shape of the compression zone for each set of applied loads. 9- Shows the stress in each reinforcement bar. 10- Program has the flexibility of adding any new shape to its shape library. Fig. 22 shows all column section and shear walls covered by the new Direct Design program. The dimension signs on each shapes show the variable parameter in each shape. 52

63 Y X Y X Y X Square Rectangle Box Y X Y X Y X U C I Y Y X X Y X Wall L T Y X Y X Y X Circle Pipe Dumbbell Fig. 22. Column Section Covered by Direct Design Program 53

64 3.7.2 Theory and Assumptions The new Direct Design program uses the same theory and assumptions as the old program explained earlier, and is based on ACI provision for designing columns. As mentioned earlier, the most difficult part in calculating the capacity of a concrete section is finding the area and the centroid of the compression zone. The new program calculates the centroid and area of the compression zone by dividing the shape into triangles. In the first step, the solid part of the shape will be divided into some triangles. If the shape has some opening, they will be divided into triangles in next step (see Fig. 23) = Solid Opening Final Fig. 23. Creating Shapes by Solid and Opening Triangles The properties of the compression zone will be calculated by adding or subtracting the triangular shapes properties using the related equations. In general, each triangle could have for different position in regard to the compression zone boundary. When compression zone boundary line cuts a triangle, it would make a smaller triangle that it properties could be calculated in the same way (see Fig. 24). 54

65 1 a x,y c 3 Compression zone Boundary N.A. b 2 Fig. 24. Calculating triangle properties Finally, the properties of compression zone would be calculated using following relations (see Fig. 25). a = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 : first triangle side lenght b = (x 2 x 3 ) 2 + (y 2 y 3 ) 2 : second triangle side lenght c = (x 3 x 1 ) 2 + (y 3 y 1 ) 2 : third triangle side lenght s = (a + b + c)/2 : half of triangle perimeter Area = s(s a)(s b)(s c) x i = (x 1 + x 2 + x 3 )/3 y i = (y 1 + y 2 + y 3 )/3 : triangle area : centroid x coordinate of i th triangle : centroid y coordinate of i th triangle X = n (x m i=1 i A i ) solid j=1(x i A i ) opening n m i=1(a i ) solid j=1(a i ) opening : centroid x coordinate of shape Y = n (y m i=1 i A i ) solid j=1(y i A i ) opening n (A i ) solid m i=1 j=1(a i ) opening : centroid y coordinate of shape 55

66 N.A. N.A. 5 3 Solid Opening Fig. 25. Compression Zone Properties Calculation Program Interface and Output Fig. 26 shows the New Direct Design program. The program can design a verity of column sections and shear walls. For each set of applied load and moments, the program calculates the required area of bars. It also shows location and angle of neutral axis, compression zone boundaries, and stress in each bar. - In Material I Load box, user defines material properties and applied loads. - In Pattern box, user defines section pattern and dimensions, bar arrangements, and bar diameters. - The Result box, shows the required bar area and equivalent available bar size, and neutral axis parameters. - The Shape box, shows the section shape, bar arrangements, compression zone, and neutral axis. 56

67 Fig. 26. The Advanced Direct Design Program 57

68 3.7.4 Result Validation The shear wall section shown in Fig. 26 is used for testing the validity of the program and direct design method. The shear wall is subjected to the following load and moment: Pu = 5100 kips : Applied axial load. Mux = k-ft : Applied moment about x-axis. Designing the shear wall section shown in Fig. 26 by Direct design method shows that the required bar area is Ab=0.419n 2. For testing validation of the result, the shear wall capacity is checked by spcolumn program. Fig. 27 shows the interaction diagram for RC column section shown in Fig. 26 developed by spcolumn program. As it is shown in Fig. 27 the point corresponding to the applied load is on the interaction diagram. It means that the capacity of the shear wall section is exactly equal to the applied load and moments. In other words, the required area of bars found by direct design method is a correct value, and the Column Direct Design program found this value without trial-anderror and without using interaction diagram. 58

69 Following page shows spcolum output files: STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM) Page 2 Licensed to: StructurePoint. License ID: /14/16 C:\Users\mhoushiar\Desktop\U-Shaped.col General Information: ==================== File Name: C:\Users\mhoushiar\Desktop\U-Shaped.col Project: Validity Test Column: U-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Material Properties: ==================== Concrete: Standard Steel: Standard f'c = 4 ksi fy = 60 ksi Ec = 3605 ksi Es = ksi fc = 3.4 ksi Eps_yt = in/in Eps_u = in/in Beta1 = 0.85 Section: ======== Exterior Points No. X (in) Y (in) No. X (in) Y (in) No. X (in) Y (in) Gross section area, Ag = 3120 in^2 Ix = e+006 in^4 rx = in Xo = e-007 in Iy = e+006 in^4 ry = in Yo = -0 in Reinforcement: ============== Bar Set: ASTM A615 Size Diam (in) Area (in^2) Size Diam (in) Area (in^2) Size Diam (in) Area (in^2) # # # # # # # # # # # Confinement: Tied; #3 ties with #10 bars, #4 with larger bars. phi(a) = 0.8, phi(b) = 0.9, phi(c) = 0.65 Pattern: Irregular Total steel area: As = in^2 at rho = 1.21% Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in *** End of output *** 59

70 P (kip) y x (Pmax) 66 x 152 in 1 Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 Date: 08/15/16 0 (Pmin) M (0 ) (k-ft) Time: 07:22: STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: C:\Users\mhoushiar\Desktop\U-Shaped.col Project: Validity Test Column: U-Shaped Engineer: M.H. f'c = 4 ksi fy = 60 ksi Ag = 3120 in^2 88 #6 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 1.21% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = e+006 in^4 e_u = in/in Yo = in Iy = e+006 in^4 Beta1 = 0.85 Min clear spacing = 5.03 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) = 0.65 Fig. 27. U-Shaped shear wall, Interaction Diagram by spcolumn 60

71 4) DESIGN DIAGRAM 4.1 Design Diagram Direct design method provides an optimum solution for a reinforced concrete column or shear wall sections. This method considers section size and bar arrangement as the given parameters, and finds the required area of each bar. For each column or shear wall, many optimum sections could be found with different sizes and bar arrangements. A design diagram shows all possible practical optimum sections for a column or shear wall. 4.2 Method Development This section covers the theory and assumption of making a design diagram. Then explains how a design diagram is made and finally shows the advantages of using design diagrams Theory and Assumption Creating a design diagram is highly depending on the direct design method. For making a design diagram, it is required to calculate the required area of bars for several section dimensions and bar arrangements. It is not practical to make a design diagram using traditional design method. In this research design diagrams are developed based on ACI provision for designing columns. However, creating a design diagram is not limited to any structural standard Design Diagram Development A design diagram is created by investigating the required area of reinforcement for different section sizes and bar arrangements. The procedure starts from minimum acceptable dimension and minimum acceptable number of bars for the section. The number of bars increases step-by-step and the required area of reinforcement for each step is calculated using direct design method. By increasing the number of bars, the required area of each bar decreases and the process continues 61

72 until the calculated area for each bar is smaller than minimum available bar size. By repeating the same approach for different section dimensions, a matrix of required bar areas for different section sizes and bar arrangements is created. In the next step, then design diagram is made by interpolating the points in the matrix. The lines corresponding to the bar numbers are calculated by interpolation between the saved areas for each point in the matrix, and the reinforcement percentage lines are calculated with a similar approach by interpolating between reinforcement percentages. The procedure is shown by the flowchart in Fig. 28. To better clarifying the development of a design diagram, assume a RC column square section is subjected to the following load and moments: Pu = 2180 kn (490 kip) Mux = 190 kn-m (140 k-ft) Muy = 0 Cover to the center of bars is considered 65mm (2.5in) and the material properties of the column are: f c = 35 fy = 420 MPa (5.0 ksi) MPa (60 ksi) Fig. 29 shows required area of each bar calculated by direct design method. The required areas are calculated for several square sections, from 305mm (12in) to 430mm (17in), and from two bars to eight bars at each side for each section size. Fig. 30 shows the created design diagram by interpolating the required bar areas shown in Fig

73 Start Determine applied load and moments Assume material properties Determine section shape (rectangle, circle,...) start with: Min. section dimension start with: Min. Number of bars. Find: A, b Save: A, b increase Number of bars. No A < #5 b Yes increase Section dimension using Min. number of bars No Yes Interpolate results Draw design diagram Stop Fig. 28. Design diagram flowchart 63

74 8% 4% 3% 2% 1% Section dimension (in) Number of bars (each side) Section dimension (mm) Fig. 29. Required Area of Bars by Direct Design Method Section dimension (in) #8 #7 #6 7 6 #9 Number of bars (each side) 5 # Section dimension (mm) Fig. 30. Design diagram for a square section 64

75 Fig. 31 shows the Direct-Design program output for a RC column square section based on the above material properties and applied load and moment. The design diagram shown in Fig. 30 has created based on these results. Fig. 31. Design Diagram created by Direct-Design program 65

76 4.2.3 Design Diagram Characteristics Any point on a design diagram represents a section, with its capacity equal to applied loads. For example, any of the points 1, 2 and 3 in the design diagram shown in Fig. 30 represents an optimum section. Point 1 represents a 405mm (16in) square section, which has 2 #8 bars at each side. Point 2 shows a 380mm (15in) square section which has 3 #9 bars at each side (no bar is available between #8 and #9 so, designer should choose the bigger size). Point 3 shows a 355mm (14in) square section, which has 4 #9 bars at each side. Fig. 32 shows corresponding cross sections for points 1-3 in Fig. 30. Point - 1 Point - 2 Point bar each side 3 bar each side 4 bar each side #8 #9 #9 405mm (16in) 380mm (15in) 355mm (14in) Fig. 32. Corresponding section to selected points in Fig. 30 In addition, design diagrams show the acceptable range for section dimension and bar arrangement. For example, Fig. 30 shows that for the above applied loads and assumed material properties, it is not possible to have a square section smaller than 330mm (13in), or a 350mm (13.8in) square section, should have at least 4 #9 bar at each side. 66

77 4.3 Sample Model In this section three design diagram are created for the following sections: 1- Circular column section 2- L-Shaped shear wall 3- U-Shaped shear wall Design Diagram for a Circular Column Create the design diagram for the circular section shown in Fig. 33. Assume that f c = 28 MPa (4.0 ksi) fy = 420 MPa (60 ksi) bars clear cover is 50 mm (2.0 in), and section is subjected to the following load and moments. Pu = 900 kn (202 kip) Mux = 400 kn-m (295 k-ft) Muy = 200 kn-m (147 k-ft) f y = 420 MPa (60 ksi) f ' = 28 MPa (4.0 ksi) c ( n ) No. of bars D Fig. 33. Model 1, Circular section Solution 67

78 For creating the design diagram according to the flowchart given in Fig. 28 it is required to investigate the required area of reinforcement for different section dimensions and bar arrangements. Table 5 shows the design results for some sample points. The input data are section diameter (D), and number of bars in section (n). The outputs are the angle of neutral axis with x- axis, distance from the neutral axis to the fiber of maximum compressive strain (C), and required area of each bar (Ab). Table 5. Design results for circular section in Fig. 33 Input Results D n Angle C A b Points mm (in) - deg. mm (in) mm 2 (in 2 ) (23.6) (8.0) 505 (0.78) (23.6) (8.0) 355 (0.55) (21.6) (9.1) 705 (1.09) By repeating the sample calculation shown in Table 5 for other section diameters, and bar arrangements a matrix for output results would be prepared. The design diagram shown in Fig. 34 is created by interpolating the above results. Points 1-3 in Fig. 34 represent the corresponding points in Table 5. 68

79 4% 3% 2% 1% Number of bars Section dimension (in) #10 #9 #8 #7 # Section dimension (mm) Fig. 34. Design diagram for Model 1 69

80 4.3.2 Design Diagram for an L-Shaped Shear Wall Create the design diagram for the L-Shaped shear wall shown in Fig. 35. Assume that: f c = 28 MPa (4.0 ksi) fy = 420 MPa (60 ksi) Bars clear cover is 50 mm (2.0 in), and section is subjected to the following load and moments: Pu = 4,500 kn (1010 kip) Mux = Muy = 250 kn-m (-7375 k-ft) kn-m (185 k-ft) t Wall thickness f y = 420 MPa (60 ksi) f ' = 28 MPa (4.0 ksi) c 2500 mm (98 in) s Bar spacing t 1500 mm (59 in) Fig. 35. Mode 2, L-Shaped shear wall Solution 70

81 Table 6 shows design results for some sample points. The input data are wall thickness (t), and bar spacing (s). The output results are similar to Model 1. Table 6. Design results for L-Shaped section in Fig. 35 Input Results t s Angle C Ab Points mm (in) mm (in) deg. mm (in) mm 2 (in 2 ) (13.8) 100 (4.0) (30.8) 330 (0.51) (13.8) 150 (6.0) (31.3) 510 (0.79) (11.8) 150 (6.0) (35.3) 720 (1.12) Fig. 36 shows the required bar area, neutral axis, compression zone, and bar stresses for point 3 of Table 6 calculated by the direct design method. Fig. 37 shows the design diagram for the L- Shaped shear wall. Points 1-3 in Fig. 37 represent the corresponding points in Table f y f y f y -f y c = 895 mm (35.3 in) A b= 720 mm 2 (1.12 in ) 2 Fig. 36. Design result for point 3 of Table 6 71

82 3% 2% 1% Bar spacing (mm) Bar spacing (in) Wall thickness (in) #10 #9 300 #8 250 # # Wall thickness (mm) Fig. 37. Design diagram for Model 2 72

83 4.3.3 Design Diagram for an U-Shaped Shear Wall Determine the design diagram for the U-Shaped shear wall shown in Fig. 38. Assume that: f c = 28 MPa (4.0 ksi) fy = 420 MPa (60 ksi) Bars clear cover is 50 mm (2.0 in), and section is subjected to the following load and moments: Pu = kn (4500 kip) Mux = -13,560 Muy = 0 kn-m ( k-ft) kn-m (0 k-ft) t Wall thickness f y = 420 MPa (60 ksi) f ' = 28 MPa (4.0 ksi) c 2900 mm (114.2 in) s Bar spacing t 1000 mm (39.4 in) Solution Fig. 38. Model 3, U-Shaped shear wall 73

84 Bar spacing (mm) Bar spacing (in) 4% 3% 2% 1% Table 7 shows design results for some sample points. The input data are wall thickness (t), and bar spacing (s). The output results are similar to Model 2. Fig. 39 shows the design diagram for the shear wall. Points 1-3 in Fig. 39 represent the corresponding points in Table 7. Table 7. Design results for U-Shaped section in Fig. 38 Input Results t s Angle c Ab Points mm (in) mm (in) deg. mm (in) mm 2 (in 2 ) (11.0) 150 (6.0) (113.5) 670 (1.04) (12.2) 150 (6.0) (112.0) 515 (0.80) (12.2) 200 (7.9) (111.7) 705 (1.09) 8 9 Wall thickness (in) #10 #9 #8 #7 # Wall thickness (mm) Fig. 39. Design diagram for Model 3 5) SUMMARY AND CONCLUSION 74

85 5.1 Summary In this research two new concepts for design of reinforcement concrete columns and shear walls has presented. 1- Direct Design Method 2- Design Diagram For each concept, assumptions, background theory, and method development has explained in detail. Two C-Language program has developed for design of RC sections based on the above concepts: 1- Column Direct Design, for design of rectangular RC column sections. 2- Direct-Design, for design of a variety of RC columns and shear walls. The program can also create design diagram for columns and shear walls. The design diagram for several models has been created at the end of research to show practical use of above concepts and their simplicity and efficiency. 5.2 Conclusion Direct design method is an analytical method by which the required area of reinforcement for short reinforced concrete columns or shear walls is directly calculated without using interaction diagrams. Some other advantages of this method are: It could be used for any column or shear wall section with any arbitrary bar arrangement. It directly provides an optimum section without going through a trial-and-error procedure. It is not limited to any particular stress-strain diagram for concrete and reinforcement. The numerical solution of this method is very efficient, accurate, and fast in computer calculations. 75

86 Efficiency of direct design method provides a practical way for making design diagrams. For each column or shear wall, many optimum sections could be found with different size and bar arrangement; capacity of any of these sections is equal to applied load and moments. The advantages of design diagram are: Shows all possible optimum column or shear wall cross sections in one diagram. Eliminates trial-and-error procedure from column design. Shows the acceptable limitation for section dimension and bar arrangement. Designers can pick up an optimum section that best fits to their requirements from the design diagram. 76

87 REFERENCES spcolumn v5.50, Formerly pcacolumn, PCACOL, and IrrCOL. STRUCTUREPOINT, formerly the Engineering Software Group of the Portland Cement Association (PCA) CRSI, Design Handbook, 10 th Edition, 2008, 600 pages Concrete Reinforcing Steel Institute (CRSI), Schaumburg, IL ACI , Building Code Requirements for Structural Concrete and Commentary & PCA Notes on , PCA and ACI joint publication Wight, K., and Macgregor, G., Reinforced Concrete Mechanics and Design, 6 th edition, Pearson Education, Inc., Upper Saddle River, New Jersey Marin, Joaquin, Design Aids for L-Shaped Reinforced Concrete Columns, ACI Journal/ November 1979, Title No pp , American Concrete Institute McCormac, J., and Brown, R., Design of Reinforced Concrete, 10 th edition, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ Rodriguez J. A. and Dario Aristizabal-Ochoa J., 1999, Biaxial Interaction Diagram for Short RC Columns of any Cross Section, Journal of Structural Engineering Furlong R. W., Cheng-Tzu Thomas Hsu, and S. Ali Mirza, 2004, Analysis and Design of Concrete Columns for Biaxial Bending Overview, ACI Structural Journal ACI : Building Code Requirements for Structural Concrete and Commentary. American Concrete Institute. ACI committee 318, 2014, 520 pages, ISBN: Burden, R., and Faires, D., Numerical Analysis, 9 th Edition, Brooks/Cole 20 Channel Center Street Boston, MA02210 USA 77

88 Chapra S., and Canale R., Numerical Methods for Engineers, 6 th edition, The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY Gautschi, W., Numerical Analysis, Second Edition, Springer ScienceCBusiness Media, LLC, 233 Spring Street, New York, NY 10013, USA Jeffrey A., 2002, Advanced Engineering Mathematics, Harcourt/Academic Press 200 Wheeler Road, Burlington, Massachusetts 01803, USA 78

89 APPENDIX (A) Method Validation For showing the validation of direct design method all the model presented in Error! Reference source not found. have been design with the spcolumn program (spcolumn, STRUCTUREPOINT, formerly the Engineering Software Group of the Portland Cement Association PCA). For each model the outputs of Direct-Design program and outputs of spcolumn are given in the following pages. In all presented models the load case corresponding point is on the interaction diagram created by spcolumn. It shows that the selected section is an optimum solution and this section is found by direct design method without trial-and-error. Following shows the list of the model in this appendix: 1) Point 1 in Table 5. Design results for circular section in Fig. 33 2) Point 2 in Table 5. Design results for circular section in Fig. 33 3) Point 3 in Table 5. Design results for circular section in Fig. 33 4) Point 1 in Table 6. Design results for L-Shaped section in Fig. 35 5) Point 2 in Table 6. Design results for L-Shaped section in Fig. 35 6) Point 3 in Table 6. Design results for L-Shaped section in Fig. 35 7) Point 1 in Table 7. Design results for U-Shaped section in Fig. 38 8) Point 2 in Table 7. Design results for U-Shaped section in Fig. 38 9) Point 3 in Table 7. Design results for U-Shaped section in Fig

90 Point 1 in Table 5. Design results for circular section in Fig. 33 STRUCTUREPOINT - spcolumn v5.50 Column: Circular-1 Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

91 P (kip) 2000 y x (Pmax) (Pmax) 23.6 x 23.6 in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 Date: 08/19/ M (26 ) (k-ft) 500 Time: 07:23:51 (Pmin) -400 (Pmin) STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\Circular-1.col Project: Column: Circular-1 Engineer: M.H. f'c = 4 ksi fy = 60 ksi Ag = in^2 8 bars Ec = 3605 ksi Es = ksi As = 6.24 in^2 rho = 1.43% fc = 3.4 ksi e_yt = in/in Xo = in Ix = in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 6.12 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

92 Point 2 in Table 5. Design results for circular section in Fig. 33 STRUCTUREPOINT - spcolumn v5.50 Column: Circular-1 Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

93 P (kip) 2000 y x (Pmax) (Pmax) 23.6 x 23.6 in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 Date: 08/19/ M (26.5 ) (k-ft) 450 Time: 07:25:35 (Pmin) -400 (Pmin) STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\Circular-2.col Project: Column: Circular-1 Engineer: M.H. f'c = 4 ksi fy = 60 ksi Ag = in^2 11 bars Ec = 3605 ksi Es = ksi As = 6.05 in^2 rho = 1.39% fc = 3.4 ksi e_yt = in/in Xo = in Ix = in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 4.45 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

94 Point 3 in Table 5. Design results for circular section in Fig. 33 STRUCTUREPOINT - spcolumn v5.50 Column: Circular-1 Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

95 P (kip) 2000 y x (Pmax) (Pmax) 21.6 x 21.6 in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A M (26 ) (k-ft) 500 Date: 08/19/16 Time: 07:27:25 (Pmin) -800 (Pmin) STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\Circular-3.col Project: Column: Circular-1 Engineer: M.H. f'c = 4 ksi fy = 60 ksi Ag = in^2 10 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 2.99% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = in^4 e_u = in/in Yo = 0.00 in Iy = in^4 Beta1 = 0.85 Min clear spacing = 3.89 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

96 Point 1 in Table 6. Design results for L-Shaped section in Fig. 35 STRUCTUREPOINT - spcolumn v5.50 Column: L-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

97 P (kip) y x (Pmax) (Pmax) 65.9 x in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A M (179 ) (k-ft) Date: 08/23/16 Time: 07:24:20 (Pmin) (Pmin) STRUCTUREPOINT - spcolumn v5.50 Beta - Jul (TM). Licensed to: StructurePoint. License ID: A05D-26C53 File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\L-Shaped-1.col Project: Column: Engineer: f'c = 4 ksi fy = 60 ksi Ag = in^2 81 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 1.91% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = e+006 in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 2.92 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

98 Point 2 in Table 6. Design results for L-Shaped section in Fig. 35 STRUCTUREPOINT - spcolumn v5.50 Column: L-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

99 P (kip) y x (Pmax) (Pmax) 65.9 x in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 Date: 08/23/ M (179 ) (k-ft) Time: 07:22:15 (Pmin) (Pmin) STRUCTUREPOINT - spcolumn v5.50 Beta - Jul (TM). Licensed to: StructurePoint. License ID: A05D-26C53 File: d:\users\mhoushiar\documents\visual studio 2012\projects\coldirect\examples\l-shaped-2.col Project: Column: Engineer: f'c = 4 ksi fy = 60 ksi Ag = in^2 55 #8 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 2.01% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = e+006 in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 4.54 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

100 Point 3 in Table 6. Design results for L-Shaped section in Fig. 35 STRUCTUREPOINT - spcolumn v5.50 Column: L-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

101 P (kip) y x (Pmax) (Pmax) 64.9 x in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A615 Date: 08/19/ M (179 ) (k-ft) Time: 07:38:50 (Pmin) (Pmin) STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\L-Shaped-3.col Project: Column: Engineer: f'c = 4 ksi fy = 60 ksi Ag = in^2 54 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 3.26% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = e+006 in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 4.56 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

102 Point 1 in Table 7. Design results for U-Shaped section in Fig. 38 STRUCTUREPOINT - spcolumn v5.50 Column: U-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

103 P (kip) y x (Pmax) (Pmax) x in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural M (0 ) (k-ft) Bars: ASTM A615 Date: 08/19/16 (Pmin) (Pmin) Time: 07:40: STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\U-Shaped-1.col Project: Column: Engineer: f'c = 4 ksi fy = 60 ksi Ag = 2123 in^2 68 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 3.33% fc = 3.4 ksi e_yt = in/in Xo = 0.00 in Ix = e+006 in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 4.50 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

104 Point 2 in Table 7. Design results for U-Shaped section in Fig. 38 STRUCTUREPOINT - spcolumn v5.50 Column: U-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in

105 P (kip) y x (Pmax) (Pmax) x in Code: ACI Units: English Run axis: Biaxial Run option: Investigation Slenderness: Not considered Column type: Structural Bars: ASTM A M (180 ) (k-ft) Date: 08/19/16 Time: 07:40:59 (Pmin) (Pmin) STRUCTUREPOINT - spcolumn v5.50 Debug - Aug (TM). Licensed to: MH, SP. License ID: File: D:\Users\mhoushiar\Documents\Visual Studio 2012\Projects\ColDirect\Examples\U-Shaped-2.col Project: Column: Engineer: f'c = 4 ksi fy = 60 ksi Ag = in^2 67 #8 bars Ec = 3605 ksi Es = ksi As = in^2 rho = 2.28% fc = 3.4 ksi e_yt = in/in Xo = in Ix = e+006 in^4 e_u = in/in Yo = in Iy = in^4 Beta1 = 0.85 Min clear spacing = 4.52 in Clear cover = N/A Confinement: Tied phi(a) = 0.8, phi(b) = 0.9, phi(c) =

106 Point 3 in Table 7. Design results for U-Shaped section in Fig. 38 STRUCTUREPOINT - spcolumn v5.50 Column: U-Shaped Engineer: M.H. Code: ACI Units: English Run Option: Investigation Slenderness: Not considered Run Axis: Biaxial Column Type: Structural Factored Loads and Moments with Corresponding Capacities: ========================================================= Pu Mux Muy PhiMnx PhiMny PhiMn/Mu NA depth Dt depth eps_t Phi No. kip k-ft k-ft k-ft k-ft in in