Analytical Method in Reinforced Concrete

Ramon V. Jarquio

Universal Publishers Boca Raton, Florida

USA 2004

Analytical Method in Reinforced Concrete

Copyright 2004 Ramon V. Jarquio All rights reserved.

Universal-Publishers.com Boca Raton, Florida

USA 2004

ISBN: 1-58112-534-8

Rev (a) Feb 2005

i

Foreword

For more than five decades, analysis for the ultimate strength of reinforced concrete section employed the equivalent rectangular stress block, finite-element methods and the interaction formula to predict the ultimate strength of reinforced concrete section subjected to axial and biaxial bending loads. The analytical method described in this book uses the classical strength of materials approach, basic calculus and the fundamental requirement of equilibrium conditions defined by F = 0 and M =0 in any structural analysis. For concrete, it uses the parabolic nature of the rising part of concrete stress/strain curve and the useable concrete strain set by codes of practice. For steel, it utilizes the linear property of steel stress/strain curve. These two material properties are linked to a common deformation as they resist external loads. This deformation is assumed linear with respect to the neutral axis defined by the compressive depth, c of the concrete section. One major difference between the graphical method and the analytical method is the introduction of the capacity axis, which can vary from 0 to pi /2 from the horizontal axis and defines the ultimate strength of a concrete section from uniaxial to any position of biaxial bending condition. This axis effectively eliminates the concept of interaction formula for biaxial bending in the graphical method of analysis. It also allows the determination of the centroid of internal forces, which can correspond to a specific external load, which is lacking in the graphical method. Use of the analytical method has also disclosed the diagonal of a rectangular column as the axis for maximum capacity to resist biaxial bending and for circular column section the capacity axis can lie between any two steel bars for equilibrium of internal and external forces. The concrete forces are determined by integration of the stress volumes at every position of the concrete compressive depth, using the true parabola in reference [13]. The value of the concrete forces is not affected by the useable concrete strain assigned by

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codes. The steel forces, however, is governed by the useable concrete strain, chosen in the analysis and the limiting condition for maximum yield stress of steel. In this book, limiting condition for yield stress of steel bar to maintain linear reduction from extreme tensile stress towards the neutral axis is taken from reference [5]. The reader may use other limiting shape of the steel strain such as the CRSI (Concrete Reinforcing Steel Institute) method in which the straight-line shape of the strain diagram is pivoted at the position of useable concrete strain equal to 0.003. The CRSI method will yield higher bar forces. The tabulation for bar forces can be easily prepared from the listed formulas for bar force. Moreover, the basic formulations are listed which the reader can input by drawing the strain diagrams for other values of useable concrete strain, for instance using the Canadian value of 0.0035 or any desired value beyond the concrete strain at ultimate concrete compressive stress. The example column capacity curves and tables from the Excel software program were the result of using the ACI useable concrete strain of 0.003. To have a rational comparison of results between graphical and analytical method, the column capacity axis concept was also employed for the first time to the ACI 318.10.2.6 rectangular stress block to determine the capacity of the column section. Derived formulas for rectangular stress block is listed in the Appendix. This analytical method and the comparison to graphical method is unique and never been shown before in any work on the subject. The analysis is limited to standard shapes of reinforced concrete such as the circular and rectangular section. The methodology, however, is applicable not only to concrete material but also to timber and steel whose stress/strain properties can be expressed in an equation form for analysis. The availability of personal computers will facilitate the efficient solution using the analytical approach illustrated in this book as compared to graphical methods of solution. For clarity, numbering of equations in each chapter starts from one and numbering for CFT columns is done separately for circular and rectangular CFT columns. In this book equations for shear

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and torsion are not included since these are adequately covered in many reinforced concrete textbook. Chapter 4 is included to illustrate the use of the analytical method of computing the area of spread footing under compression, the footing capacity curve, design of a rectangular footing and footing settlement using the standard soil settlement formula. To convert to SI units use the following conversion:

1 in. = 25.4 mm. 1kip = 4.448 kN

1 ksi = 6.895 Mpa 1 lb-ft. = 1.356 N-m

1 kN-m = 0.738 ft.- kip

The reader can plug into the equations any system of units such as the English, Metric or SI units of measures. Manual calculations involved in the solutions for reinforced concrete circular and rectangular columns using parabolic and standard method are shown in the Appendix. A computer program for ultimate strength of reinforced concrete circular and rectangular columns in Excel 97 is available for a price. If interested contact me at my address below. Ramon V. Jarquio, P. E. CM - SEI (ASCE), M. ACI E-mail: Parabola5@aol.com

Acknowledgements I am very grateful to my brother, Victor V. Jarquio, P.E. for verifying the veracity of the column equations using his own method for my son Lawrence M. Jarquio for his assistance in preparing the manuscript according to the publishers requirements.

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TABLE OF CONTENTS

Page

Chapter 1 - BEAMS 1 - 26

True parabolic stress method 1 - 8

Singly reinforced concrete beam 8 - 13

Doubly reinforced concrete beam 13 - 17

T-beam analysis 18 - 20

Beam with Biaxial Bending 20 - 26

Chapter 2 - COLUMN ANALYSIS 27 - 89

Circular column 27 - 49

Rectangular column 49 - 80

Column Capacity Axis 81 - 87

Chapter 3 - CFT COLUMN ANALYSIS 90 - 125

CFT Circular column 91 - 104

CFT Rectangular column 104 - 125

Chapter 4 - FOOTING FOUNDATION 126 - 172

Rectangular footing 126 - 138

Circular footing 138 - 140

Triangular footing 140 - 147

Footing Capacity Curve 147 - 161

Footing Design 162 - 167

Wall Footing Surcharge Load 167 - 169

Footing Settlement 170 - 172

APPENDIX 175 - 245

Rebar development 175 - 179

Concrete Forces 180 - 181

v

Equivalent Rectangle 192 - 194

Standard Rectangular Method Formulas 196 - 199

Manual Calculations 199 - 235

Soil Wedge 236 - 239

Footing Derivations 240 - 245

1

CHAPTER 1

BEAM ANALYSIS

1.0 INTRODUCTION Ultimate strength design in reinforced concrete uses the so-called parabolic-rectangular stress block method, commonly known as the Whitney equivalent rectangular stress block. This current method is crude and inefficient because the equivalent rectangle was referred to the simple parabola. In contrast, the true parabolic stress method presented herein utilizes the basic parabola, which closely fits the stress-strain curves of concrete material. This parabola is defined by the value of f c , the depth of compression, c and the useable value of linear strain, ec . The maximum area derived from these limiting parameters is the measure of the ultimate strength of the concrete section under consideration. This true parabolic stress method permitted under ACI 318.10.2.6 facilitates the rational, realistic and systematic calculations for the ultimate strength in reinforced concrete as compared to the imaginary rectangular stress block method. Worked out examples are presented to show the application of this method using manual calculations. 1.1 DEVELOPMENT OF THE THEORY It is an established fact that the rising part of the stress/strain data for concrete material closely approximates a parabola. Earlier investigators decided that to apply the basic parabola to the analysis for ultimate strength would be a cumbersome, if not a difficult procedure since the derived property of a parabola required for the proper manipulation is not available. Hence, Whitney employed the so-called parabolic-rectangular stress block method of approximation of the concrete compressive forces. The fact that it is called parabolic-rectangular clearly indicates the

2

inability of previous investigators to operate the basic parabola in the solution for ultimate strength of reinforced concrete members. The true parabolic stress method presented herein also precludes the averaging of concrete compressive stress which is assumed at 0.85 fc even near the neutral axis in the standard rectangular stress block method. The true parabolic stress method gives accurate values of the compressive stresses in the reinforcing steel bars. In Figure A4 of the Appendix is shown an overlay of the basic parabola to the curve of stress/strain of concrete material. The four basic assumptions common to the true parabolic stress method and the standard parabolic-rectangular method are as follows:

1. The rising part of the concrete stress/strain curve is parabolic.

2. The strain due to bending on a given section is linear with respect to the neutral axis.

3. The useable concrete strain is limited at a point beyond the maximum concrete stress at a strain value, which is assumed to be 0.003 by the ACI and 0.0035 by the Canadian method.

4. The area between the parabola and the concrete stress/strain curve is neglected.

Also, in Figure A2 of the Appendix is shown the derivation of the property of the parabolic curve, which will be utilized in the analysis for ultimate strength. With this property of the parabola and the application of algebra, trigonometry, analytic geometry and calculus, the solution for ultimate strength in reinforced concrete becomes feasible and systematic. 1.2 DERIVATION The analytical method in reinforced concrete is based on applying basic mathematics on the two important assumptions in predicting the ultimate strength of reinforced concrete section. One assumption is the fact that the rising part of the plot of stress/strain data for concrete material is almost parabolic or second-degree

3

curve. The other assumption is that the strain of the reinforced concrete section due to bending is linear with respect to the neutral axis. To express these two assumptions into an equation form was not discovered by previous investigators and therefore resorted to the graphical method of analysis started by Whitney. This graphical method with the interaction formula for biaxial external loads and finite-element method is currently what is being employed as standard tools for predicting the ultimate strength of reinforced concrete sections. This paper will show how the analytical method in reinforced concrete can be implemented, thereby eliminating the graphical method of analysis extensively being used today. Previous investigators failed to solve the equation of the true parabola y = f(x) to represent the concrete compressive stress, fc and the concrete compressive depth, c for concrete forces because the property of the parabolic curve to use in the analysis was not in print and has to be derived using analytic geometry. This property of the parabola is the key to implement the analytical method. Once the equation of the true parabola is known the calculation for concrete forces in reinforced concrete section is just basic calculus. The analytical method will be compared to the ACI rectangular stress block method of analysis by solving the true equivalent rectangular stress block consistent with the required equilibrium conditions, F = 0 and M = 0, which is fundamental in any structural analysis. The presentation will be limited to the calculations for concrete forces only. The calculations for steel bar forces is relatively easy since it is based on the linear property of the strain diagram and only geometric proportion is involved in the operation. Hence, it is not shown here but other works of the author may be consulted to find out how the bar forces should be calculated and added to the concrete forces for the ultimate strength of the reinforced concrete section. Derivations From analytic geometry we get the equation of the parabola with its axis parallel to the y- axis and opening below as

4

(x - a)2 = - 2p (y - b) (1)

In Figure 1, when a = L/2 and b = m, equation (1) becomes (x - L/2)2 = - 2p (y - m) (2)

Since the curve passes through the origin (0, 0)

(0 - L/2)2 = - 2p (0 - m) and 2p = (L/2)2/m (3)

Put equation (3) in equation (2) to obtain (x - L/2)2 = - (L/2)2 (y - m)/m (4)

Figure 1 - Property of a Parabola

Y (L/2, m)

m

y

XO

x L - x

L

5

Simplifying equation (4) will yield the relationship

y/x(L - x) = m/(L/2)2 (5) Equation (5) is the property of the parabola, which is the key to the analytical method for the calculation of concrete forces for ultimate strength analysis of reinforced concrete section Figure 2 shows the schematic overlay of the parabola with the plot of the stress/strain of concrete material from data of compressive tests on concrete cylinders. The concrete force is the product of concrete stress and the area of concrete section under it. Figure 3 will be used to derive the equation of the concrete stress in the analysis for the calculation of the concrete forces. The basic parabola for concrete force is defined by the maximum useable concrete strain allowed and the maximum area bounded by the curve OMN and the X-axis. From the property of a parabola expressed in equation (5) y/x(2 xm - x) = f c /(xm)2 or

y = f c x (2xm - x)/ (xm)2 (6)

and from the strain diagram,

xm / em = c/ ec or

xm = c em / ec (7)

Substitute equation (7) in (6) y = (ec f c / c2 em2) [ 2 c em x - ec x2 ] (8) The area under this curve is actually the concrete force on a unit width and by calculus is given by

A = 0

c

( ec f c / c2 em2 ) [ 2 c em x - ec x2 ] dx (9)

6

Figure 2 - Overlay of Basic Parabola to Concrete Stress/Strain

Integrating and evaluating limits

A = (c f c ec / 3 em2 )( 3 em - ec ) (10)

Differentiate equation (10) with respect to em , equate to zero

Basic parabola closely matchesdata in this region

Basic parabola divergesin this region

Y

Neglect area between curves

Stress/Strain curve from concrete

cylinders

y = f(x)

fcStress fc

ec

c X

O

Strain

7

Figure 3 - Basic parabola for Ultimate Strength Analysis and solve for the value of em, which will yield the maximum area under the curve.

A = ( c fc ec / 3) [ em2 (3) - ( 3 em - ec ) ( 2 em ) ] / em4 = 0 or em = (2/3) ec (11) Equation (11) confirms that when ec = 0.003, the strain at the maximum compressive stress, fc occurs at ec = 0.002 as indicated in the ACI 318 R - 83 Commentary on Building Code Requirements for Reinforced Concrete, Chapter 10. It also

Y

M

N

fc fc

XO em

parabolic stress

ecstraight-line strain

xm

neutral axis

c

8

justifies assumption one that the rising part of the concrete stress strain curve is parabolic. From equation (7) xm = (2/3) c (12) Equation (12) indicates that the maximum compressive stress f c passes through the centroid of the triangular strain diagram. Substitute equation (12) in equation (6) to obtain the equation of the basic parabola to use in the analysis for ultimate strength in reinforced concrete, i.e., y = (3 f c / 4 c2 ) ( 4 c x - 3 x2 ) (13)

Equation (13) is the basic parabola to use in the analysis for the ultimate strength of reinforced concrete. At the assumed point of failure, N fc = c( 2xm - c) f c / (xm)2 (14)

Put equation (12) in equation (14) and simplify

f c = 0.75 f c (15)

In Figure 2 using ACI value of ec = 0.003 and the depth of concrete in compression denoted by c, the total compressive force on this section per unit width is given by the area under the curve OMN and the X - axis, i.e.,

A = 0

c

(3 f c / 4 c2 ) ( 4 c x - 3x2 ) dx (16)

Integrating equation (16) and evaluating limits

A = 0.75 c f c (17)

The centroid or point of application of this force can be calculated

from the expression

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A x = 0

c

(3 f c / 4 c2 ) (4 c x2 - 3 x3 ) dx (18)

Integrating and evaluating limits

A x = 0.4375 c2 f c (19)

Therefore,

x = 0.583 c (20)

c - x = 0.417 c (21)

1.3 SINGLY REINFORCED CONCRETE BEAM In reinforced concrete beams, the maximum compressive force is developed from the edge to a depth equal to c. Figure 4 shows balanced condition of the compressive and tensile forces. The expressions for these forces are as follows: C = 0.75 c f c b (22) T = As fy = pbdfy (23) F = 0: C = T or 0.75 c f c b = pbdfy or p = 0.75c f c / dfy (24) Using linear distribution of strains with respect to the neutral axis, c = ec d/(ec + es) (25) When ec = 0.003 and fy is expressed in ksi equation (25) is simplified to

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c = 87 d/(87 + fy) (25a)

When ec = 0.003 and f is expressed in Mpa equation (25) is simplified to

c = 600 d/(600 + fy) (25b) Substitute equation (25) in equation (24) to solve for p at balanced condition, i.e. When ec = 0.003 and fc and fy are expressed in ksi pb = 65 f c / fy ( 87 + fy ) (26a) When ec = 0.003 and fc fy are expressed in Mpa pb = 450 fc /fy (600 + fy) (26b)

b

0.003 0.75fc0.417c fc

C d

c 2c/3

n. a.As d - 0.417c M

Tfy /29000

Strain Stress

Figure 4 - Singly Reinforced Concrete Beam

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To insure ductile failure, pmax = 0.75 pb as mandated by the ACI code. Therefore, for design use pmax = 49 f c /fy(87 + fy) (27a) when f c and fy are expressed in ksi. pmax = 338 fc /fy (600 + fy) (27b) For other values and units for fc and fy the reader should be able to derive the appropriate equations to use. In the following sections the English system of units will be employed. Minimum Reinforcement In Figure 4, assume plain concrete so that c = d/2 and for fc

substitute the value of the modulus of rupture equal to 7.5 ( fc)1/2. Hence, from equation (24) pmin = 0.75(d/2)(237{fc}1/2 )/1000fy or pmin = 0.09{fc}1/2 /fy (28) Moment Capacity Summing moments about the compressive force, C M = Asfy (d-0.417c) = pbdfy (d-0.417c) (29) Substitute equation (23), (24) and (27) in equation (29) to obtain M = Kbd2 (30) in which, K = 49(51 + fy) f c /(87 + fy)2 (31) Example Problem 1: Design a rectangular beam to resist a bending moment of 3500 in-kips, using f c = 5 ksi and fy = 60 ksi.

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Solution: Step 1: From equation (31), K = 49(51 + 60)(5)/(87 + 60)2 = 1.258 Step 2: From equation (30), bd2 = M/K = 3500/1.258 = 2782 Assume b = 10. Therefore, d = 2782 10/ = 16.68 Assume concrete cover to center of bars is 2.50 . Then, T = 16.68 + 2.50 = 19.18 say T = 20 Step 3: From equation (29) with d = 20-2.50 =17.58 As = 3500/60{17.50-0.417(87)(17.50)/(87 + 60)} = 4.43 sq. in. Use 8 - # 7 bars in two layers (As = 4.81 sq. in.) Step 4: Check As limitations.

From equation (28), pmin = 0.09 5 / 60 = 0.0034

As min = 0.0034(10)(17.50) = 0.58 sq. in. < 4.81 sq. in. OK

From equation (27), pmax = 5(49)/60(87 + 60) = 0.0283

As max = 0.0283(10)(17.50) = 4.96 sq. in. > 4.81 sq. in. OK Step 5: Check spacing of rebars and revise thickness, if necessary. In this case, T = 17.50 + 3.50 to comply with the ACI requirement for spacing of rebars. Hence, use T = 21 inches. Example Problem 2: Determine the ultimate strength capacity of the beam designed in Example 1.

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Solution: Step 1: Determine the maximum concrete force given As = 4.81 sq. in., fy = 60 ksi, fc = 5 ksi and d = 21 - 3.50 = 17.50 in.

c = 17.50 (87/140) = 10.36

Maximum concrete force = 0.75(10.36)(5)(10) = 389 kips

Step 2: Determine the rebar force

Maximum As to develop this = 389/60 = 6.47 sq. in. Available rebar force = 4.81 (60) = 289 kips. The corresponding depth of compressive concrete section to balance this force is given by 0.75 c (5)(10) = 289 or c = 289/37.5 = 7.70 Step 3: Determine the resisting moment of the beam section Mc = concrete moment at centerline = 289(10.50 - 0.417x7.70)

= 2104 kips and

Ms = rebar moment at centerline = 289(10.50 - 3.50) = 2020 kips

M = Mc + Ms = 2104 + 2020 = 4124 kips > 3500 kips OK Note: The ACI requirement to use 75% of maximum concrete force in the design of tension rebars will also allow the compressive reinforcement in a beam section to have a strain within the yield strength of the steel reinforcement. This can be observed from the strain diagram as the value of c is reduced from 10.36 to 7.70 and tensile rebar strain maintained at fy/29000. 1.4 DOUBLY REINFORCED CONCRETE BEAM In Figure 5, C1 = Asfy (32)

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C2 = 0.75c f c b (33) T = Asfy (34) From the strain diagram, es = ec (c-d)/c (35) From the property of the parabola, fc = (c-d){(c/3) + d} f c /(2c/3)2 (36) When ec = 0.003 and es = fy/29000 fy = (0.003/c){(c-d)(29000)} - fc (37) Put equation (36) in equation (37) and simplify fy = (1/c2){(87-0.75 f c)c2 - (87d + 1.50df c)c + 2.25d2fc fy max (38) Put equation (38) in equation (32) to obtain C1 = (As/c2){(87-0.75fc)c2 - (87d + 1.50dfc)c + 2.25d2fc} (39) F = 0: C1 + C2 = T (40) Put equation (33), (34) and (39) in equation (40) and simplify

c3 - (1.33c2/bfc){Asfy - As (87-0.75fc)} - (1.33cAs/bfc)(87d + 1.50d fc) + 3d2As/b (41) This is a cubical equation in the variable c which can be solved as follows: Let P = -(1.33/bfc){Asfy - As (87-0.75fc)} (42)

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Q = - (1.33As/bfc)(87d + 1.50d fc) (43) R = 3d2As/b (44) Equation (41) is then written as c3 + Pc2 + Qc + R = 0 (45) Transform equation (45) to x3 + ax + b = 0 (46) by setting c = x - P/3 (47) and a = (3Q - P2)/3 (48) b = (2P3 - 9PQ + 27R)/27 (49) Solution to equation (46) is given by x = A + B (50) in which,

bfc

c

0.003 d 0.75fc fces

As c - d C2 2c/3d d - d

n. a.As d - 0.417c M

fy /29000 T

Strain Stress

Figure 4 - Doubly Reinforced Concrete Beam

16

A = 3 (-b/2) + (b2/4) + (a3/27) (51) B = 3 (-b/2) - (b2/4) + (a3/27) (52) Use above formulas when (b2/4) + (a3/27) > 0 When (b2/4) + (a3/27) < 0, use the following trigonometric formulas: cos(3) = (3b)/2a ( / )a 3 (53) x = 2 ( / )a 3 . cos (54) Moment Capacity: M = 0 M = C1 (d-d) + C2(d-0.417c) (55) Example Problem 3: Determine the moment capacity of the beam in Example 1 if it is provided with 4-#7 bars as compression reinforcement with d = 2.00 and As = 7.22 sq. inches. Solution: Step 1: Calculate values of P, Q, and R from equation (42), (43) and (44) with As = 7.22 sq. in. and As = 2.41 sq. in.

P = (-1.33/10x5){7.22(60) - 2.41(87-0.75x5)} = - 6.19

Q = (-1.33x2.41/10x5){87x2 + 1.50x2x5} = - 12.12

R = 3(2)2(2.41)/10 = 2.89

Step 2: Calculate values of a and b using eq. (48) and (49). a = {3(-12.12) - (- 6.19)2}/3 = - 24.89

b = {2(- 6.19)3 - 9(- 6.19)(-12.12) + 27(2.89)}/27 = - 39.69

17

Step 3: (b2/4) + (a3/27) = 393.75 - 571.10 < 0. Use equation (53) and (54) to obtain

cos(3) = 3(- 39.69)/2(- 24.89) (24.89/3)1/2 = 0.83042

3 = 33.86

= 11.29

cos = 0.98066

x = 2 (24.89/3)1/2 (0.98066) = 5.65 Step 4: Determine the location of the neutral axis. From eq.(47)

c = 5.65 + 6.19/3 = 7.71 Step 5: Calculate the compressive forces, C1 and C2.

C1 = {(87-0.75x5)(7.71)2 - (87x2 + 1.50x2x5)(7.71) +

2.25(2)2x5}{2.41/(7.71)2}

fy = 59.49 < 60 OK

C1 = 144 kips

C2 = 0.75(7.71)(5)(10) = 289 kips

T = 7.71 x 60 = 433 kips

Step 6: Solve for the moment capacity using equation (55)

M = 144 (17.50 - 2.00) + 289 (17.50 - 0.417x7.71)

= 6360 in-kips

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1.5 T-BEAM ANALYSIS Case 1: When c t use beam formulas Case 2: When c > t, i.e., d > [(87 + fy/87]t, the following analysis is applicable. In Figure 6, concrete forces are determined as follows:

CF = [0.75fc/c2] c t

c

[4cx - 3x2] dx CF = [0.75fc/c2][ct2 + c2t - t3] (56)

CFxF = [0.75fc bF/c2] c t

c

[4cx2 - 3x3]dx CF x F = [0.75fc bF/12c2][12c3t + 6c2t2 - 20ct3 + 9t4] (57) CW = [0.75fc bW/c2]

c t

c

[4cx - 3x2] dx CW = [0.75fc bW/c2][c3 - c2t - ct2 + t3] (58)

CW x W = [0.75fc bw / c2 ] tc

0[4cx2 - 3x3 ] dx

CW x W = [0.75fc bW /12c2][7c4 - 12c3t - 6c2t2 + 20ct3 - 9t4] (59) Bar Forces, T = Asfy and concrete forces, CF & CW must satisfy equilibrium conditions, i.e., F = 0 and M = 0. F = 0: T = CF + CW (60) As = [0.75fc/c2fy][(bF - bW)(ct2 + c2t - t3) + bWc3] (61) M = 0: M = T(d - c) + CF x F + CW x W (62) M = Asfy(d - c) + [0.75fc/12c2][(bF - bW)(12c3t + 6c2t2 - 20ct3 +