Diseño de vigas y columnas de concreto ACI-318-05

"RECTBEAM" --- RECTANGULAR CONCRETE BEAM ANALYSIS/DESIGN

Program Description:

"RECTBEAM" is a spreadsheet program written in MS-Excel for the purpose of analysis/design of rectangular

beam or column sections. Specifically, the ultimate moment capacity, bar spacing for crack control, moments

of inertia for deflection, beam shear and torsion requirements, and member capacity for flexure (uniaxial and

biaxial) with axial load are calculated. There is also a worksheet which contains reinforcing bar data tables.This version is based on the ACI 318-05 Code.

This program is a workbook consisting of ten (10) worksheets, described as follows:

Worksheet Name DescriptionDoc This documentation sheet

Complete Analysis Beam flexure, shear, crack control, and inertia

Flexure Ultimate moment capacity of singly or doubly reinforced beams/sections

Crack Control Crack control - distribution of flexural reinforcing

Shear Beam or one-way type shear

Torsion Beam torsion and shear

Inertia Moments of inertia of singly or doubly reinforced beams/sections

Uniaxial Combined uniaxial flexure and axial load

Biaxial Combined biaxial flexure and axial load

Rebar Data Reinforcing bar data tables

Program Assumptions and Limitations:

1. This program follows the procedures and guidelines of the ACI 318-05 Building Code.

2. This program utilizes the following references:

a. "Design of Reinforced Concrete - ACI318-05 Code Edition", by Jack C. McCormac (7th Ed.)

b. "Notes of the ACI318-05 Building Code Requirements for Structural Concrete", by PCA

3. The "Complete Analysis" worksheet combines the analyses performed by four (4) of the individual

worksheets all into one. This includes member flexural moment capacity, as well as shear, crack control,

and inertia calculations. Thus, any items below pertaining to any of the similar individual worksheets

included in this one are also applicable here.

4. In the "Flexure", "Uniaxial", and "Biaxial" worksheets, when the calculated distance to the neutral axis, 'c',

is less than the distance to the reinforcement nearest the compression face, the program will ignore that

reinforcing and calculate the ultimate moment capacity based on an assumed singly-reinforced section.

5. In the "Uniaxial" and "Biaxial" worksheets, the CRSI "Universal Column Formulas" are used by this program

to determine Points #1 through #7 of the 10 point interaction curve. For the most part, these formulas yield

close, yet approximate results. However, these results should be accurate enough for most applications

and situations.

which was applicable up through the ACI 318-99 Code, they have been factored by (0.65/0.70) to account for

ACI 318-05 Code. This modification has been made to the equations applicable to Points #1 through #7.

7. In the "Uniaxial" and "Biaxial" worksheets, the CRSI "Universal Column Formulas", which are used by this

program, assume the use of the reinforcing yield strength, fy =60 ksi.

8. In the "Uniaxial" and "Biaxial" worksheets, this program assumes a "short", non-slender rectangular column

with symmetrically arranged and sized bars.

9. In the "Uniaxial" and "Biaxial" worksheets, for cases with axial load only (compression or tension) and no

moment(s) the program calculates total reinforcing area as follows:

Ast = (Ntb*Abt) + (Nsb*Abs) , where: Abt and Abs = area of one top/bottom and side bar respectively.

6. To account for the fact that the CRSI "Universal Column Formulas" originally utilized f =0.70 for compression,

the reduction in the factor f = 0.65 for compression beginning with ACI 318-02 Code and continuing with the

10. In the "Uniaxial" and "Biaxial" worksheets, for pure moment capacity with no axial load, the program assumes

bars in 2 outside faces parallel to axis of bending plus 50% of the total area of the side bars divided equally

by and added to the 2 outside faces, and program calculates reinforcing areas as follows:

for X-axis: As = A's = ((Ntb*Abt) + (0.50*Nsb*Abs))/2

for Y-axis: As = A's = ((Nsb*Asb+4*Atb) + (0.50*(Ntb-4)*Atb))/2

12. In the "Complete Analysis" and "Flexure" worksheets as well as the "Uniaxial" and "Biaxial" worksheets

doubly reinforced sections.

e = Mu*12/Pu, are determined from interpolation within the interaction curve for the applicable axis.

14. In the "Biaxial" worksheet, the biaxial capacity is determined by the following approximations:

a. For Pu >= 0.1*f'c*Ag, use Bresler Reciprocal Load equation:

b. For Pu < 0.1*f'c*Ag, use Bresler Load Contour interaction equation:

15. The "Rebar Data" worksheet contains tables of reinforcing bar data which include various bar properties,

reinforcing bar areas based on spacing, and various plain welded wire fabric properties.

16. This program contains numerous “comment boxes” which contain a wide variety of information including

explanations of input or output items, equations used, data tables, etc. (Note: presence of a “comment box”

is denoted by a “red triangle” in the upper right-hand corner of a cell. Merely move the mouse pointer to the

desired cell to view the contents of that particular "comment box".)

11. In the "Uniaxial" and "Biaxial" worksheets, for Point #8 (fPn = 0.1*f'c*Ag) on the interaction curve the

corresponding value of fMn is determined from interpolation between the moment values at Point #7

(balanced condition, f = 0.65) and Point #9 (pure flexure, f <= 0.90).

(for Point #9, pure flexure) the program first determines the strain in tension reinforcing (et), then the capacity

reduction factor (f),and finally ultimate moment capacity (fMn) based on actual input reinforcing.

a. For et >= 0.005 ("tension-controlled" section): f = 0.90.

b. For fy/Es < et < 0.005 ("transition" section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) < 0.90 (Es=29000 ksi)

c. For et <= fy/Es ("compression-controlled" section): f = 0.65

Note: the ACI 318-05 Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and

13. In the "Uniaxial" and "Biaxial" worksheets, design capacities, fPn and fMn, at design eccentricity,

1/fPn = 1/fPnx + 1/fPny - 1/fPo

Biaxial interaction stress ratio, S.R. = Pu/fPn <= 1

Biaxial interaction stress ratio, S.R. = (Mux/fMnx)^1.15 + (Muy/fMny)^1.15 <= 1

"RECTBEAM (318-05).xls" ProgramVersion 1.2

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RECTANGULAR CONCRETE BEAM/SECTION ANALYSISFlexure, Shear, Crack Control, and Inertia for Singly or Doubly Reinforced Sections

Per ACI 318-05 CodeJob Name: Subject:

Job Number: Originator: Checker: BeamSlab

Input Data: Exterior b=12'' Interior

Beam or Slab Section? BeamExterior or Interior Exposure? Exterior

Reinforcing Yield Strength, fy = 60 ksi

Concrete Comp. Strength, f 'c = 4 ksi h=18'' d=15''Beam Width, b = 12.000 in.

Depth to Tension Reinforcing, d = 15.000 in. As(min) =Total Beam Depth, h = 18.000 in. As=3 As(temp) =

Tension Reinforcing, As = 3.000 in.^2 Singly Reinforced Section As(max) =No. of Tension Bars in Beam, Nb = 4.000

Tension Reinf. Bar Spacing, s1 = 3.000 in. d' bClear Cover to Tension Reinf., Cc = 2.000 in.

Depth to Compression Reinf., d' = 0.000 in. A's f 's =Compression Reinforcing, A's = 0.000 in.^2 As2 =Working Stress Moment, Ma = 120.00 ft-kips h d As1 =Ultimate Design Moment, Mu = 170.00 ft-kips

Ultimate Design Shear, Vu = 20.00 kips >= 0.005, Tension-controlledTotal Stirrup Area, Av(stirrup) = 0.220 in.^2 As

Tie/Stirrup Spacing, s2 = 6.0000 in. Doubly Reinforced Section

Results:c =

Moment Capacity Check for Beam-Type Section: Crack Control (Distribution of Reinf.): a =0.85 Per ACI 318-05 Code:

c = 5.190 in. Es = 29000 ksi

a = 4.412 in. Ec = 3605 ksi Max. Tension Reinf. for Singly Reinforced Section per ACI 318-99 Code (for comparison only):0.02851 n = 8.040.01667 fs = 36.93 ksi f 'sb =0.00333 fs(used) = 36.93 ksi

As(min) = 0.600 in.^2 <= As = 3 in.^2, O.K. s1(max) = 11.24 in. >= s1 = 3 in., O.K.As(max) =N.A. (total for section)

As(temp) = N.A. in.^2 (total) Per ACI 318-95 Code (for reference only):0.02064 dc = 3.0000 in. Shear for Beam-Type Section:

As(max) = 3.716 in.^2 >= As = 3 in.^2, O.K. z = 139.61 k/in.

N.A. z(allow) = 145.00 k/in. >= z = 139.61 k/in.,

f 's = N.A. ksi O.K.

0.00567 >= 0.005, Tension-controlled

0.900172.72 ft-k >= Mu = 170 ft-k, O.K.

Av(prov) =Shear Capacity Check for Beam-Type Section: Moment of Inertia for Deflection: Av(min) =

17.08 kips fr = 0.474 ksi Av(req'd) =24.75 kips kd = 6.0125 in. s2(max) =41.83 kips >= Vu = 20 kips, O.K. 5832.00 in.^4

68.31 kips >= Vu-(phi)Vc = 2.92 kips, O.K. Mcr = 25.61 ft-k

Av(prov) = 0.220 in.^2 = Av(stirrup) 2818.77 in.^4

Av(req'd) = 0.026 in.^2 <= Av(prov) = 0.22 in.^2, O.K. 2848.08 in.^4 (for deflection) Crack Control per ACI 318-05 Code:Av(min) = 0.060 in.^2 <= Av(prov) = 0.22 in.^2, O.K. Es =

r(min) =r(temp) =

rb =r(max) (for f = 0.9) =

r(max) =

ec =e's =

As(max) (for f = 0.9) =

et =

f =fMn =

b1 = fMn =

rb = rb =r =

r(min) = r(max) =

r(temp) =

r(max) = f =

e's = fVc =

et = fVs(req'd) =f = fVn =

fMn = fVs(max) =

fVc =fVs =fVn = Ig =

fVs(max) =Icr =Ie =

B31

The factor, 'b1', shall be = 0.85 for concrete strengths (f'c) <= 4000 psi. For concrete strengths > 4,000 psi, 'b1' shal be reduced continuously at a rate of 0.05 for each 1000 psi of strength > 4000 psi, but 'b1' shall not be taken < 0.65.

B32

The value, 'c', is the distance from the compression face of the beam/section to the neutral axis, and is calculated as follows: For Singly Reinforced Beam/Section: c = (As*fy/(0.85*f'c*b))/b1 For Doubly Reinforced Beam/Section: Solving for "c" in following expression: As*fy = 0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es c = (-(A's*ec*Es-As*fy)+SQRT((A's*ec*Es-As*fy)^2-4*(0.85*f'c*b1*b)*(-A's*d'*ec*Es)))/(2(0.85*f'c*b1*b)) where: Es = 29000 ksi

H32

The Modulus of Elasticity for Steel, 'Es', is assumed = 29,000 ksi

B33

The value, 'a', is the depth of the assumed retangular compression stress block, and is calculated as follows: a = c*b1

H33

The Modulus of Elasticity for Concrete, 'Ec', is calculated as follows: Ec = 57*(f'c*1000)^(1/2) ksi Note: "normal" weight concrete is assumed.

B34

The Reinforcing Ratio, 'rb', is the reinforcing ratio producing balanced strain conditions and is calculated as follows: rb = 0.85*b1*f'c/fy*(87/(87+fy))

H34

The Modular Ratio, 'n', is calculated as follows: n = Es/Ec

B35

The Ratio of Tension Reinforcing provided, 'r ', is calculated as follows: r = As/(b*d)

H35

The working stress tension in the reinforcing, 'fs', is calculated as follows: fs = 12*Ma/(As*d*(1-(SQRT(2*As/(b*d)*n+(As/(b*d)*n)^2)-As/(b*d)*n)/3))

B36

The minimum required percentage of flexural reinforcing, 'r(min)', is calculated as follows: r(min) >= 3*(f'c)^(1/2)/fy >= 200/fy where: f'c and fy are in psi.

H36

The actual value of 'fs' used in the calculation of the required spacing of flexural tension reinforcing shall be the lesser of the calculated value of 'fs' based on the applied moment and 2/3*fy.

B37

The minimum required area of flexural reinforcing, 'As(min)', is calculated as follows: As(min) = r(min)*b*d

H37

The center-to-center spacing, 's', of the flexural tension reinforcing shall not exceed the following: s1(max) = 15*(40/fs)-2.5*Cc but not greater than 12*(40/fs).

B38

The minimum required percentage of temperature reinforcing, 'r(temp)', is calculated as follows: r(temp) = 0.0020 for fy = 40 and 50 ksi r(temp) = 0.0018 for fy = 60 ksi r(temp) = 0.0018*60/fy for fy > 60 ksi Note: minimum temperature reinforcing percentage is not used for beams, only slab sections.

B39

The minimum required area of temperature reinforcing, 'As(temp)', is calculated as follows: As(temp) = r(temp)*b*h Note: normally 1/2 of the entire percentage, r(temp), is used in each face of section.

B40

The Reinforcing Ratio, 'r(max)', is the maximum allowable reinforcing ratio and is calculated as follows: For Singly Reinforced Beam/Section: r(max) = 0.85*f'c*b1*(0.003/(0.003+et))/fy where: et = 0.004 Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for singly reinforced section. For Doubly Reinforced Beam/Section: r(max) = As(max)/(b*d) where: As(max) = (0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es)/fy c = ec*d/(ec+et) et = 0.004 Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for doubly reinforced section.

H40

The concrete cover, 'dc', is the distance from the tension face of the beam/section to the centerline of the tension reinforcing and is calculated as follows: dc = h-d

B41

The maximum allowable area of reinforcing, 'As(max)', is calculated as follows: For Singly Reinforced Beam/Section: As(max) = r(max)*b*d For Doubly Reinforced Beam/Section: As(max) = (0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es)/fy where: et = 0.004 c = ec*d/(ec+et) Note: Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for singly or doubly reinforced section.

H41

The factor limiting the distribution of flexural reinforcement, 'z', is calculated as follows: z = fs*(dc*2*dc*b/Nb)^(1/3)

B42

e's = calculated strain in compression reinforcing, and is determined as follows: e's = ec*(c-d')/c, where ec = 0.003 (assumed).

H42

The allowable factor limiting the distribution of flexural reinforcement, 'z(allow)', shall not exceed the following values: Interior Exposure: Exterior Exposure: Beams z = 175 z = 145 One-way slabs z = 156 z =129

B43

The actual stress in the compession reinforcing is: f 's = e's*Es = (1-d'/c)*87 where: Es = 29000 ksi for steel

B44

et = calculated strain in tension reinforcing, and is determined as follows: et = ec*(d-c)/c, where ec = 0.003 (assumed). If et >= 0.005, section is "tension-controlled". If fy/Es < et < 0.005, section is in "transition zone". If et <= fy/Es, section is "compression-controlled". Note: the ACI Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and doubly reinforced sections.

B45

f = capacity reduction factor for flexural tension, and is determined as follows for non-spiral reinforced sections: If et >= 0.005 (tension-controlled section): f = 0.90 If fy/Es < et < 0.005 (transition section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) < 0.90 If et <= fy/Es (compression-controlled section): f = 0.65

B46

The Ultimate Moment Capacity, 'fMn', of the beam/section is calculated as follows: For Singly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)) For Doubly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)+A's*f's*(d-d')) where: f = capacity reduction factor, 0.65 <= f <= 0.90 f's = e's*Es = (1-d'/c)*87 <= fy

B49

The ultimate shear strength provided by the concrete, 'fVc', is calculated as follows: For shear and flexure only (no axial load): fVc = 2*f*(f'c*1000)^(1/2)*b*d where: f = 0.75 Note: for members such as one-way slabs, footings, and walls, where minimum shear reinforcing is not required, if fVc >= Vu then the section is considered adequate. For beams, when Vu > fVc/2 , then a minimum area of shear reinforcement is required.

H49

The Modulus of Rupture of concrete, 'fr', is calculated as follows: fr = 7.5*(f'c*1000)^(1/2) ksi

B50

The ultimate shear strength provided by the shear reinforcing in beams, 'fVs', is calculated as follows: fVs = f*fy*d*Av/s2 where: f = 0.75 The required amount of shear strength to be provided by the shear reinforcing in beams is: fVs(req'd) = Vu-fVc >= 0. The required amount of shear strength to be provided must be <= fVs(max).

H50

The distance from the compression face of the beam/section to the neutral axis, 'kd', is calculated as follows: for singly reinforced beams/sections: kd = ((2*d*B+1)^(1/2)-1)/B where: Es = 29000 ksi (for reinforcing steel) Ec = 57*(f'c*1000)^(1/2) ksi (for normal weight concrete) n = Es/Ec B= b/(n*As) for doubly reinforced beams/sections: kd = ((2*d*B*(1+r*d'/d+(1+r)^2)^(1/2)-(1+r))/B where: B = b/(n*As) r = (n-1)*A's/(n*As)

H51

The Gross (uncracked) Moment of Inertia, 'Ig', is calculated as follows: Ig = b*h^3/12

B52

The maximum allowable ultimate shear strength to be provided by the shear reinforcing in beams, 'fVs(max)', is calculated as follows: fVs(max) <= 8*f*(f'c*1000)^(1/2)*b*d/1000 where: f = 0.75

H52

The Cracking Moment, 'Mcr', is calculated as follows: Mcr = fr*Ig/yt where: yt = h/2

H53

The Cracked Section Moment of Inertia, 'Icr', is calculated as follows: for singly reinforced: Icr = b*kd^3/3+n*As*(d-kd)^2 for doubly reinforced: Icr = b*kd^3/3+n*As*(d-kd)^2+(n-1)*A's*(kd-d')^2

B54

The required area of shear reinforcing, 'Av', perpendicular to axis of the beam is calculated as follows: Av = fVs*s2/(f*fy*d) >= Av(min) where: f = 0.75 fVs = max. of: (Vu - fVc) or 0 Note: Av = area of both legs of a closed tie or stirrup.

H54

The Effective Moment of Inertia, 'Ie', is calculated as follows: Ie = (Mcr/Ma)^3*Ig+(1-(Mcr/Ma)^3)*Icr <= Ig

B55

The minimum area of shear reinforcing, 'Av(min)', to be provided for beams is calculated as follows: Av(min) = 75*SQRT(f'c*1000)*b*s2/fy but not less than: 50*b*s2/(fy*1000) Note: Av(min) = area of both legs of a closed stirrup.


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s2(max) = 7.500 in. >= s2 = 6 in., O.K. Ec =n =

B56

The maximum allowable shear reinforcing (closed stirrup) spacing, 's2', shall not exceed d/2, nor 24". However, when fVs > 4*f*(f'c*1000)^(1/2)*b*d, then the maximum spacing given above shall be reduced by one-half. (Note: f = 0.75)


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RECTANGULAR CONCRETE BEAM/SECTION ANALYSISUltimate Moment Capacity of Singly or Doubly Reinforced Sections


Job Number: Originator: Checker: BeamSlab

Input Data: b=12'' a =

Beam or Slab Section? BeamReinforcing Yield Strength, fy = 60 ksi

Concrete Comp. Strength, f 'c = 4 ksi h=18'' d=15''Beam Width, b = 12.000 in.

Depth to Tension Reinforcing, d = 15.000 in.

Total Beam Depth, h = 18.000 in. As=3Ultimate Design Moment, Mu = 170.00 ft-kips Singly Reinforced SectionAs(temp) =

Tension Reinforcing, As = 3.000 in.^2 ###Depth to Compression Reinf., d' = 0.000 in. d' b

Compression Reinforcing, A's = 0.000 in.^2

A's

f 's =Results: h d

As1 =Stress Block Data: ###

As0.85 Doubly Reinforced Section

c = 5.190 in. c = (As*fy/(0.85*f'c*b))/Beta1a = 4.412 in.

Reinforcing Criteria: c =a =

0.016670.028510.00333 Max. Tension Reinf. for Singly Reinforced Section per ACI 318-99 Code (for comparison only):

As(min) = 0.600 <= As = 3 in.^2, O.K.N.A. f 'sb =

As(temp) = N.A. in.^2 (total)

0.02064 0.85*f'c*Beta1*(0.003/(0.003+0.004))/fy As(max) =As(max) = 3.716 in.^2

Ultimate Moment Capacity: >= As = 3 in.^2, O.K.

N.A.f 's = N.A. ksi

0.00567 >= 0.005, Tension-controlled0.900

172.72 ft-kips

>= Mu = 170 ft-kips, O.K.Comments:

r =r(min) =

r(temp) =

r(max) (for f = 0.9) =r(max) =

ec =e's =

As(max) (for f = 0.9) =

b1 = b1 = 1.05-0.05*f'c >= 0.65 f =fMn =

a = b1*c

r = r = As/(b*d) fMn =rb = rb = 0.85*b1*f'c/fy*(87/(87+fy)

r(min) = r(min) >= 3*SQRT(f'c)/fy >= 200/fyAs(min) = r(min)*b*d rb =

r(temp) = r(temp) = 0.0018*60/fy for fy >= 60, else 0.002-0.00002*(fy-50)As(temp) = r(temp)*b*h r(max) =

r(max) = r(max) =As(max) = r(max)*b*d for singly reinforced, or for doubly reinforced:As(max) = (0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es)/fy for c = ec*d/(ec+0.004)

e's = e's = ec*(c-d')/cf 's = e's*Es

et = et = ec*(d-c)/cf = f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) <= 0.90

fMn = for singly reinforced: fMn = f*(As*fy*(d-a/2))for doubly reinforced: fMn = f*(As1*fy(d-a/2)+A's*f 's*(d-d'))

B25

The factor, 'b1', shall be = 0.85 for concrete strengths (f'c) <= 4000 psi. For concrete strengths > 4,000 psi, 'b1' shal be reduced continuously at a rate of 0.05 for each 1000 psi of strength > 4000 psi, but 'b1' shall not be taken < 0.65.

B26

The value, 'c', is the distance from the compression face of the beam/section to the neutral axis, and is calculated as follows: For Singly Reinforced Beam/Section: c = (As*fy/(0.85*f'c*b))/b1 For Doubly Reinforced Beam/Section: Solving for "c" in following expression: As*fy = 0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es c = (-(A's*ec*Es-As*fy)+SQRT((A's*ec*Es-As*fy)^2-4*(0.85*f'c*b1*b)*(-A's*d'*ec*Es)))/(2(0.85*f'c*b1*b)) where: Es = 29000 ksi

B27

The value, 'a', is the depth of the assumed retangular compression stress block, and is calculated as follows: a = c*b1

B31

The Ratio of Tension Reinforcing provided, 'r ', is calculated as follows: r = As/(b*d)

B32

The Reinforcing Ratio, 'rb', is the reinforcing ratio producing balanced strain conditions and is calculated as follows: rb = 0.85*b1*f'c/fy*(87/(87+fy))

B33

The minimum required percentage of flexural reinforcing, 'r(min)', is calculated as follows: r(min) >= 3*(f'c)^(1/2)/fy >= 200/fy where: f'c and fy are in psi.

B34

The minimum required area of flexural reinforcing, 'As(min)', is calculated as follows: As(min) = r(min)*b*d

B35

The minimum required percentage of temperature reinforcing, 'r(temp)', is calculated as follows: r(temp) = 0.0020 for fy = 40 and 50 ksi r(temp) = 0.0018 for fy = 60 ksi r(temp) = 0.0018*60/fy for fy > 60 ksi Note: minimum temperature reinforcing percentage is not used for beams, only slab sections.

B36

The minimum required area of temperature reinforcing, 'As(temp)', is calculated as follows: As(temp) = r(temp)*b*h Note: normally 1/2 of the entire percentage, r(temp), is used in each face of section.

B37

The Reinforcing Ratio, 'r(max)', is the maximum allowable reinforcing ratio and is calculated as follows: For Singly Reinforced Beam/Section: r(max) = 0.85*f'c*b1*(0.003/(0.003+et))/fy where: et = 0.004 Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for singly reinforced section. For Doubly Reinforced Beam/Section: r(max) = As(max)/(b*d) where: As(max) = (0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es)/fy c = ec*d/(ec+et) et = 0.004 Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for doubly reinforced section.

B38

The maximum allowable area of reinforcing, 'As(max)', is calculated as follows: For Singly Reinforced Beam/Section: As(max) = r(max)*b*d For Doubly Reinforced Beam/Section: As(max) = (0.85*f'c*b1*c*b+A's*(c-d')/c*ec*Es)/fy where: et = 0.004 c = ec*d/(ec+et) Note: Amount of tension reinforcing is limited by minimum tension reinforcing strain of et = 0.004 for singly or doubly reinforced section.

B42

e's = calculated strain in compression reinforcing, and is determined as follows: e's = ec*(c-d')/c, where ec = 0.003 (assumed).

B43

The actual stress in the compession reinforcing is: f 's = e's*Es = (1-d'/c)*87 where: Es = 29000 ksi for steel

B44

et = calculated strain in tension reinforcing, and is determined as follows: et = ec*(d-c)/c, where ec = 0.003 (assumed). If et >= 0.005, section is "tension-controlled". If fy/Es < et < 0.005, section is in "transition zone". If et <= fy/Es, section is "compression-controlled". Note: the ACI Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and doubly reinforced sections.

B45

f = capacity reduction factor for tension, and is determined as follows for non-spiral reinforced sections: If et >= 0.005 (tension-controlled section): f = 0.90 If fy/Es < et < 0.005 (transition section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) < 0.90 If et <= fy/Es (compression-controlled section): f = 0.65

B46

The Ultimate Moment Capacity, 'fMn', of the beam/section is calculated as follows: For Singly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)) For Doubly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)+A's*f's*(d-d')) where: f = capacity reduction factor, 0.65 <= f <= 0.90 f's = e's*Es = (1-d'/c)*87 <= fy


6 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/SECTION ANALYSISBeam or One-Way Type Shear

Per ACI 318-05 CodeJob Name: Subject: Slab

Job Number: Originator: Checker:

Input Data:

Beam or Slab Section? BeamReinforcing Yield Strength, fy = 60 ksi

Concrete Comp. Strength, f 'c = 4 ksi. Av(prov) =Beam Width, b = 10.000 in. Av(min) =

Depth to Tension Reinforcing, d = 13.500 in. Av(req'd) =Total Beam Depth, h = 16.000 in.

Ultimate Design Shear, Vu = 20.00 kips

Ultimate Design Axial Load, Pu = 0.00 kips

Total Stirrup Area, Av(used) = 0.220 in.^2

Tie/Stirrup Spacing, s = 6.0000 in.

dResults:

Typical Critical Sections for Shear

For Beam:

12.81 kips

22.28 kips

35.08 kips >= Vu = 20 kips, O.K.

51.23 kips >= Vu-(phi)Vc = 7.19 kips, O.K.

Av(prov) = 0.220 in.^2 = Av(used)

Av(req'd) = 0.071 in.^2 <= Av(prov) = 0.22 in.^2, O.K.

Av(min) = 0.050 in.^2 <= Av(prov) = 0.22 in.^2, O.K.

s(max) = 6.750 in. >= s = 6 in., O.K.

Comments:

fVc =

fVs =fVs(req'd) =

fVn =fVs(max) =

d Vu Vu d d Vu

f =fVc =

Vu

Vu

fVc =fVs =

fVn = fVc+fVs =fVs(max) =

C31

The ultimate shear strength provided by the concrete, 'fVc', is calculated as follows: For shear and flexure only (no axial load): fVc = 2*f*(f'c*1000)^(1/2)*b*d For shear with axial compression: fVc = 2*f*(1+Pu/(2*Ag)*(f'c*1000)^(1/2)*b*d For shear with axial tension: fVc = 2*f*(1+Pu/(0.5*Ag)*(f'c*1000)^(1/2)*b*d where: f = 0.75 Ag = b*h Note: for members such as one-way slabs, footings, and walls, where minimum shear reinforcing is not required, if fVc >= Vu then the section is considered adequate. For beams, when Vu > fVc/2 , then a minimum area of shear reinforcement is required.

C32


C34


C36

The required area of shear reinforcing, 'Av', perpendicular to axis of the beam is calculated as follows: Av = fVs*s/(f*fy*d) >= Av(min) where: f = 0.75 fVs = max. of: (Vu - fVc) or 0 Note: Av = area of both legs of a closed stirrup.

C37

The minimum area of shear reinforcing, 'Av(min)', to be provided for beams is calculated as follows: Av(min) = 75*SQRT(f'c*1000)*b*s/fy but not less than: 50*b*s/(fy*1000) Note: Av(min) = area of both legs of a closed stirrup.

C38

The maximum allowable shear reinforcing (closed stirrup) spacing, 's', shall not exceed d/2, nor 24". However, when fVs > 4*f*(f'c*1000)^(1/2)*b*d, then the maximum spacing given above shall be reduced by one-half. (Note: f = 0.75)


7 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/SECTION ANALYSISCrack Control - Distribution of Flexural Reinforcing

Per ACI 318-05 CodeJob Name: Subject: Slab

Job Number: Originator: Checker: ExteriorInterior

Input Data: fs =fs(used) =

Beam or Slab Section? Beam b=10'' s(max) =Exterior or Interior Exposure? Exterior

Reinforcing Yield Strength, fy = 60 ksi

Concrete Comp. Strength, f 'c = 4 ksi dc =Beam Width, b = 10.000 in. h=16'' d=13.5''

Depth to Tension Reinforcing, d = 13.500 in. z(allow) =Total Beam Depth, h = 16.000 in.

Tension Reinforcing, As = 2.400 in.^2 As=2.4No. of Tension Bars in Beam, Nb = 4.000 dc=2.5''

Tension Reinf. Bar Spacing, s = 3.000 in. BeamClear Cover to Tension Reinf., Cc = 2.000 in.

Working Stress Moment, Ma = 75.00 ft-kips b

h d

As

dcOne-Way Slab

Results:

Per ACI 318-05 Code:

Es = 29000 ksi Es = modulus of elasticity for steelEc = 3605 ksi Ec = 57*SQRT(f'c*1000)

n = 8.04 n = Es/Ecfs = 32.18 ksi fs = 12*Ma/(As*d*(1-(SQRT(2*As/(b*d)*n+(As/(b*d)*n)^2)-As/(b*d)*n)/3))

fs(used) = 32.18 ksi fs(used) = minimum of: fs and 2/3*fys(max) = 13.64 in. s(max) = minimum of: 15*40/fs(used)-2.5*Cc and 12*40/fs(used)

>= s = 3 in., O.K.Per ACI 318-95 Code: (for reference only)

dc = 2.5000 in. dc = h-dz = 101.37 k/in. z = fs(used)*(dc*2*dc*b/Nb)^(1/3)

z(allow) = 145.00 z(allow) >= z = 101.37 k/in., O.K.

Note: The above calculation of the 'z' factor is done solely for comparison purposes to ACI 318-05 Code.

Comments:

2*dc

2*dc

B34


B35


B36


B37

The working stress tension in the reinforcing, 'fs', is calculated as follows: fs = 12*Ma/(As*d*(1-(SQRT(2*As/(b*d)*n+(As/(b*d)*n)^2)-As/(b*d)*n)/3))

B38

The actual value of 'fs' used in the calculation of the required spacing of flexural tension reinforcing shall be the lesser of the calculated value of 'fs' based on the applied moment and 2/3*fy.

B39

The center-to-center spacing, 's', of the flexural tension reinforcing shall not exceed the following: s(max) = 15*(40000/(fs*1000))-2.5*Cc but not greater than 12*(40/fs).

B43

The concrete cover, 'dc', is the distance from the tension face of the beam/section to the centerline of the tension reinforcing and is calculated as follows: dc = h-d

B44

The factor limiting the distribution of flexural reinforcement, 'z', is calculated as follows: z = fs*(dc*2*dc*b/Nb)^(1/3)

B45

The allowable factor limiting the distribution of flexural reinforcement, 'z(allow)', shall not exceed the following values: Interior Exposure: Exterior Exposure: Beams z = 175 z = 145 One-way slabs z = 156 z =129


8 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/SECTION ANALYSISMoment of Inertia of Singly or Doubly Reinforced Sections


Job Number: Originator: Checker: fr =Es =

Input Data: Ec =n =

Reinforcing Yield Strength, fy = 60 ksi b=10'' r =Concrete Comp. Strength, f 'c = 4 ksi B =

Beam/Section Width, b = 10.000 in. kd =Depth to Tension Reinforcing, d = 13.500 in. Ig =

Beam/Section Total Depth, h = 16.000 in. h=16'' d=13.5''Tension Reinforcing, As = 2.400 in.^2 Icr =

Depth to Compression Reinf., d' = 0.000 in. Ig/Icr =Compression Reinforcing, A's = 0.000 in.^2 As=2.4Working Stress Moment, Ma = 75.00 ft-kips Singly Reinforced Section

d' b

A's

h d

AsDoubly Reinforced Section

Results:

fr = 0.474 ksi fr = 7.5*SQRT(f'c*1000)/1000Es = 29000 ksi Es = modulus of elasticity for steelEc = 3605 ksi Ec = 57*SQRT(f'c*1000)

n = 8.04 n = Es/Ecr = N.A. r = (n-1)*A's/(n*As)

B = 0.5180 B = b/(n*As)kd = 5.5430 in. kd = (SQRT(2*d*B+1)-1)/B

3413.33 in.^4 Ig = b*h^3/12Mcr = 16.87 ft-k Mcr = fr*Ig/(h/2)/12

1790.06 in.^4 Icr = b*kd^3/3+n*As*(d-kd)^21.907 Ig/Icr = ratio of gross to cracked inertias

1808.52 in.^4 Ie = (Mcr/Ma)^3*Ig+(1-(Mcr/Ma)^3)*Icr <= Ig

Note:

Comments:

Ig =

Icr =Ig/Icr =

Ie =

Use effective moment of inertia, 'Ie', in deflection calculations.

B32

The Modulus of Rupture of concrete, 'fr', is calculated as follows: fr = 7.5*(f'c*1000)^(1/2) ksi

B33


B34


B35


B38

The distance from the compression face of the beam/section to the neutral axis, 'kd', is calculated as follows: for singly reinforced beams/sections: kd = ((2*d*B+1)^(1/2)-1)/B where: B= b/(n*As) for doubly reinforced beams/sections: kd = ((2*d*B*(1+r*d'/d+(1+r)^2)^(1/2)-(1+r))/B where: B = b/(n*As) r = (n-1)*A's/(n*As)

B39

The Gross (uncracked) Moment of Inertia, 'Ig', is calculated as follows: Ig = b*h^3/12

B40

The Cracking Moment, 'Mcr', is calculated as follows: Mcr = fr*Ig/yt where: yt = h/2

B41

The Cracked Section Moment of Inertia, 'Icr', is calculated as follows: for singly reinforced: Icr = b*kd^3/3+n*As*(d-kd)^2 for doubly reinforced: Icr = b*kd^3/3+n*As*(d-kd)^2+(n-1)*A's*(kd-d')^2

B43

The Effective Moment of Inertia, 'Ie', is calculated as follows: Ie = (Mcr/Ma)^3*Ig+(1-(Mcr/Ma)^3)*Icr <= Ig


9 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/SECTION ANALYSISBeam Torsion and Shear


Job Number: Originator: Checker:

Input Data:

Reinforcing Yield Strength, fy = 60 ksi b=16''Concrete Comp. Strength, f 'c = 4 ksi xo=12.5''

Beam Width, b = 16.000 in. Av(prov) =Depth to Tension Reinforcing, d = 23.500 in. Av(req'd) =

Total Beam Depth, h = 26.000 in. Av(min) =Ultimate Design Shear, Vu = 60.00 kips d=23.5''

Ultimate Design Torsion, Tu = 30.00 ft-kips 26''Ultimate Design Axial Load, Pu = 0.00 kips

Total Stirrup Area, Av+t(used) = 0.400 in.^2

Closed Stirrup Spacing, s = 7.0000 in. dt=1.75''Edge Distance to Tie/Stirrup, dt = 1.7500 in.

Beam Sectionyo =

Results: Acp =Ag =

For Shear: Pcp =Ph =

35.67 kips Aoh =60.43 kips Ao =96.10 kips >= Vu = 60 kips, O.K. Tcr =

142.68 kips >= Vu-(phi)Vc = 24.33 kips, O.K. Tu(limit) =Av(prov) = 0.400 in.^2 = Av+t(used) Tu(max) =Av(req'd) = 0.161 in.^2 <= Av(prov) = 0.4 in.^2, O.K. At(prov) =

Av(min) = 0.093 in.^2 <= Av(prov) = 0.4 in.^2, O.K. At(req'd) =s(max) = 11.750 in. >= s = 7 in., O.K. At(min) =

For Torsion:

Tu(limit) = 8.14 ft-kips < Tu = 30 kips, must consider torsion!

Tu(max) = 71.51 ft-kips >= Tu = 30 kips, O.K. Total (Av+t) =At(prov) = 0.119 in.^2 = (Av+t(used)-Av(req'd))/2 Total (Av+t)(min) =At(req'd) = 0.117 in.^2 <= At(prov) = 0.119 in.^2, O.K.

At(min) = 0.000 in.^2 <= At(prov) = 0.119 in.^2, O.K.

1.171 in.^2 >= Al(min) = 1.021 in.^2, thus use Al(req'd)

1.021 in.^2 < Al(req'd) = 1.171 in.^2, thus use Al(req'd)

For Combined Shear and Torsion:

Total (Av+t) = 0.395 in.^2 <= Av+t(prov) = 0.4 in.^2, O.K.

Total (Av+t)(min) = 0.093 in.^2 <= Av+t(prov) = 0.4 in.^2, O.K.

Comments:

f =fVc =

fVs =fVs(req'd) =

fVn =fVs(max) =

Al

h = yo At

As

f =

fVc =fVs =

fVn = fVc+fVs =fVs(max) =

Al(req'd) =

Al(min) =

Al(req'd) =

Al(min) =

C26

The ultimate shear strength provided by the concrete, 'fVc', is calculated as follows: For shear and flexure only (no axial load): fVc = 2*f*(f'c*1000)^(1/2)*b*d For shear with axial compression: fVc = 2*f*(1+Pu/(2*Ag)*(f'c*1000)^(1/2)*b*d For shear with axial tension: fVc = 2*f*(1+Pu/(0.5*Ag)*(f'c*1000)^(1/2)*b*d where: f = 0.75 Ag = b*h Note: for members such as one-way slabs, footings, and walls, where minimum shear reinforcing is not required, if fVc >= Vu then the section is considered adequate. For beams, when Vu > fVc/2 , then a minimum area of shear reinforcement is required.

C27


C29


C31

The required area of shear reinforcing, 'Av(req'd)', perpendicular to axis of the beam is calculated as follows: Av(req'd) = fVs*s/(f*fy*d) >= Av(min) where: f = 0.75 Note: Av(req'd) = area of two legs of a closed stirrup.

C32

The minimum area of shear reinforcing, 'Av(min)', to be provided for beams is calculated as follows: Av(min) = 75*SQRT(f'c*1000)*b*s/fy but not less than: 50*b*s/(fy*1000) Note: Av(min) = area of both legs of a closed stirrup.

C33

The maximum allowable shear reinforcing (closed stirrup) spacing, 's(max)', shall not exceed d/2, nor 24". However, when fVs > 4*f*(f'c*1000)^(1/2)*b*d, then the maximum spacing given above shall be reduced by one-half. (Note: f = 0.75)

C37

The effects of torsion may be neglected when the factored torsional moment, 'Tu', is less than 'Tu(limit)', which is calculated as follows: Tu(limit) = (f*SQRT(f'c*1000)*(Acp)^2/(Pcp)*SQRT(1+Pu/(4*Ag*SQRT(f'c*1000))))/12000 where: f = 0.75 Acp = b*h Pcp = 2*(b+h) xo = b-(2*dt) yo = h-(2*dt) Pu = (+) for compression, (-) for tension

C38

The maximum allowable torsion (with or without shear), 'Tu(max)', is calculated as follows: Tu(max) = SQRT((((fVc*1000/(b*d)+f*8*(SQRT(f'c*1000)))^2-(Vu*1000/(b*d))^2)*(1.7^2*Aoh^4/Ph^2)))/12000 where: f = 0.75 Acp = b*h Pcp = 2*(b+h) xo = b-(2*dt) yo = h-(2*dt) Ph = 2*(xo+yo) Aoh = xo*yo Ao = 0.85*Aoh

C40

The required area of transverse torsion reinforcing, 'At', perpendicular to axis of the beam is calculated as follows: At(req'd) = Tu*12*s/(f*2*Ao*fy) >= At(min) where: f = 0.75 Aoh = xo*yo Ao = 0.85*Aoh Note: At(req'd) = area of only one leg of a closed stirrup. At(prov) = (Av+t(used)-Av(req'd))/2

C41

The minimum area of transverse torsion reinforcing, 'At(min)', perpendicular to axis of the beam is calculated as follows: At(min) = (50*b*s/(fy*1000)-Av(req'd))/2 Note: At(min) = area of only one leg of a closed stirrup. At(prov) = (Av+t(used)-Av(req'd))/2

C42

The required area of longitudinal torsion reinforcing, 'Al(req'd)', parallel to axis of the beam is calculated as follows: Al = At(req'd)/s*Ph >= Al(min) where: xo = b-(2*dt) yo = h-(2*dt) Ph = 2*(xo+yo)

C43

The minimum area of longitudinal torsion reinforcing, 'Al(min)', parallel to axis of the beam is calculated as follows: Al(min) = 5*SQRT*(f'c*1000)*Acp/(fy*1000) - (max. of: 25*b*s/(fy*1000) or At(req'd)) /s*Ph where: Acp = b*h xo = b-(2*dt) yo = h-(2*dt) Ph = 2*(xo+yo)

C47

The required total area of transverse shear and torsion reinforcing, 'Av+t', perpendicular to axis of the beam is calculated as follows: Av+t = Av+2*At >= Av+t(min) Note: Av = area of two legs of a closed stirrup. At = area of only one leg of a closed stirrup.

C48

The required total area of transverse shear and torsion reinforcing, 'Av+t(min)', perpendicular to axis of the beam is calculated as follows: Av+t(min) = 50*b*s/(fy*1000) Note: Av = area of two legs of a closed stirrup. At = area of only one leg of a closed stirrup.


10 of 18 04/07/2023 20:24:08


11 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/COLUMN ANALYSISFor X-Axis Flexure with Axial Compression or Tension Load

Assuming "Short", Non-Slender Member with Symmetric Reinforcing (per ACI 318-05 Code)Job Name: Subject: Pu =

Job Number: Originator: Checker: Mux =Lx = b =

Input Data: Ly = h = Lx=18 dy =

Reinforcing Yield Strength, fy = 60 ksi. g =Concrete Comp. Strength, f 'c = 4 ksi

Total Member Width, Lx = 18.000 in. Ntb =Total Member Depth, Ly = 18.000 in. Abt =

Distance to Long. Reinforcing, d' = 2.500 in. Ly=18 Ntb=8Ultimate Design Axial Load, Pu = 200.00 kips Nsb=0 Abs = Ultimate Design Moment, Mux = 100.00 ft-kips

Total Top/Bot. Long. Bars, Ntb = 8Top/Bot. Longitudinal Bar Size = 8 d'=2.5 (typ.) Ast =

Total Side Long. Bars, Nsb = 0 Member Section Ag =Side Longitudinal Bar Size = 8

Results: As(min) =

X-axis Flexure and Axial Load Interaction Diagram Points As(max) =Location Comments X =Point #1 948.55 0.00 0.00 F1 =Point #2 758.84 0.00 0.00 F2 =Point #3 758.84 105.09 1.66 Min. eccentricity F3 =Point #4 640.36 170.72 3.20 F4 =Point #5 534.60 207.29 4.65 X-axis Flexure and Axial Load Interaction Diagram Points:Point #6 447.71 232.10 6.22 ###Point #7 303.20 261.59 10.35 ###Point #8 129.60 225.87 20.91 ###Point #9 0.00 199.20 Pure moment capacity ###Point #10 -341.28 0.00 0.00 Pure axial tension capacity ###

###Gross Reinforcing Ratio Provided: Poc =

0.01951 Pos =###

Member Uniaxial Capacity at Design Eccentricity: ###Interpolated Results from Above: P15c =

P15s =457.21 228.60 6.00 ###

###Effective Length Criteria for "Short" Column: P30c =

k*Lu <= 9.90 ft. (for k*Lu/r(min) <= 22) P30s =k*Lu <= 18.00 ft. (for k*Lu/r(min) <= 40) ###

###Pure Axial Compression Capacity without Reinforcing: Pbc =

572.83 kips Pbs =###

Tie Minimum Size and Maximum Spacing: ####3@16'' ###

b1 =

rtb =rss =

rg =r(min) =

r(max) =

fPnx (k) fMnx (ft-k) ey (in.)

Nom. max. compression = fPoAllowable fPn(max) = 0.8*fPo

0% rebar tension = 0 ksi




fPn = 0.1*f'c*Ag(Infinity)

rg =

fPnx (k) fMnx (ft-k) ey (in.)

fPn =

0 50 100 150 200 250 300

-600

-400

-200

0

200

400

600

800

1000

1200X-AXIS INTERACTION DIAGRAM

( - )fMnx ft k

()

fPn

xk

X

Y

E24

The values of eccentricity shown in this table are calculated as follows: ey = fMnx*12/fPnx where: fPnx and fMnx are the respective values associated with a specific point on the interaction diagram.

C25

The maximum nominal pure axial compression capacity (with no moment) is calculated from the CRSI Formula "A" as follows: fPn(1) = (0.65/0.7)*((0.595*(1-rg)*f'c+42*rg)*Ag) where: rg = Ast/Ag Ast = (Ntb+Nsb)*Ab Ab = area of one bar Ag = b*h Note: in above formula for fPn(1) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05. Note: this value of fPn is refered to as fPo in the ACI Code. The value of fPo can be also computed using the ACI Code equation as follows: fPo = f*(0.85*f'c*(Ag-Ast)+fy*Ast) where: f = 0.65 (per ACI 318-05)

C26

The maximum allowable axial compression capacity (with no moment) is calculated from CRSI Formula "B" as follows: fPn(2) = 0.80*fPn(1) = 0.80*fPo

C27

The maximum allowable axial compression capacity associated with minimum eccentricity is calculated from CRSI Formulas "B" as follows: fPn(3) = 0.80*fPn(1) = fPo

D27

The ultimate moment capacity associated with the maximum allowable axial compression capacity at minimum eccentricity is calculated from the CRSI Formulas "B" as follows: fMn(3) = 0.20*fPo*fMn(ot)/(fPo-fPn(ot)) where: fPo = fPn(1) fPn(ot) = fPn(4) (zero tension condition) fMn(ot) = fMn(4) (zero tension condition)

C28

The axial compression capacity associated with a 0% tension in reinforcing condition is calculated from the CRSI Formulas "C" as follows: For 0.50 <= g < 1.00, fPn(4) = (0.65/0.7)*(0.2975*f'c*(b1*(1+g)-rtb-(b1*(1+g)-1+g)*rss/g)*Ag + 21*(rtb+(2.7-1.4*g)*g*rss)*Ag) For g < 0.50, fPn(4) = (0.65/0.7)*(0.2975*f'c*b1*(1+g)*Ag + 0.35*rg*(174*g/(1+g)-0.85*f'c)*Ag) where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss Note: in above formulas for fPn(4) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

D28

The ultimate moment capacity associated with a 0% tension in reinforcing condition is calculated from the CRSI Formulas "C" as follows: For 0.50 <= g < 1.00, fMn(4) = (0.65/0.7)*((Poc)*(0.5-0.25*b1*(1+g))*h + 10.5*g*(rtb+X*F1*rss)*h*Ag)/12 Poc = 0.2975*f'c*(b1*(1+g)-rtb-(b1*(1+g)-1+g)*rss/g)*Ag For g < 0.5, fMn(4) = (0.65/0.7)*((Poc)*(0.5-0.25*b1*(1+g))*h + (Pos)*(rtb/rg)*0.5*g*h)/12 Poc =(0.2975*f'c*b1*(1+g)*Ag Pos = 0.35*rg*(174*g/(1+g)-0.85*f'c)*Ag where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss X = 0 for Nsb <= 6 , X = 0.33 for Nsb = 8 , X = 0.50 for Nsb = 10 , X = 1.0 for Nsb >= 12 F1 = 0 for Nsb<= 2 , F1 = 0.33 for Nsb =4 , F1 = 0.50 for Nsb = 6 , F1 = 0.28*g+0.19 <= 0.37 for Nsb >= 8 Note: in above formulas for fMn(4) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

C29

The axial compression capacity associated with a 25% tension (15 ksi) in reinforcing condition is calculated from the CRSI Formulas "D" as follows: For 0.55 <= g < 1.00, fPn(5) = (0.65/0.7)*(0.2975*f'c*(0.857*b1*(1+g)-rtb-(0.857*b1*(1+g)-1+g)*rss/g)*Ag + 15.75*(rtb+(0.5+g)*rss)*Ag) For g < 0.55, fPn(5) = (0.65/0.7)*(0.255*f'c*b1*(1+g)*Ag + 0.35*rg*((189*g-15)/(1+g)-0.85*f'c-15)*Ag) where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss Note: in above formulas for fPn(5) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

D29

The ultimate moment capacity associated with a 25% tension (15 ksi) in reinforcing condition is calculated from the CRSI Formulas "D" as follows: For 0.55 <= g < 1.00, fMn(5) = (0.65/0.7)*((P15c)*(0.5-0.214*b1*(1+g))*h + 13.125*(rtb+X*F2*rss)*g*h*Ag)/12 P15c = 0.2975*f'c*(0.857*b1*(1+g)-rtb-(0.857*b1*(1+g)-1+g)*rss/g)*Ag For g < 0.55, fMn(5) = (0.65/0.7)*((P15c)*(0.5-0.214285*b1*(1+g))*h + 0.175*((189*g-15)/(1+g)-0.85*f'c+15)*g*h*Ag*rtb)/12 P15c = 0.255*f'c*b1*(1+g)*Ag where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss X = 0 for Nsb <= 6 , X = 0.33 for Nsb = 8 , X = 0.50 for Nsb = 10 , X = 1.0 for Nsb >= 12 F2 = 0 for Nsb<= 2 , F2 = 0.33 for Nsb =4 , F2 = 0.50 for Nsb = 6 , F2 = 0.25*g+0.19 <= 0.37 for Nsb >= 8 Note: in above formulas for fMn(5) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

C30

The axial compression capacity associated with a 50% tension (30 ksi) in reinforcing condition is calculated from the CRSI Formulas "E" as follows: For 0.60 <= g < 1.00, fPn(6) = (0.65/0.7)*(0.2975*f'c*(0.75*b1*(1+g)-rtb-(0.75*b1*(1+g)-1+g)*rss/g)*Ag + 10.5*(rtb+1.8*g*rss)*Ag) For g < 0.60, fPn(6) = (0.65/0.7)*0.223*f'c*b1*(1+g)*Ag + 0.35*((204*g-30)/(1+g)-0.85*f'c-30)*(rtb+rss/3)*Ag where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss Note: in above formulas for fPn(6) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

D30

The ultimate moment capacity associated with a 50% tension (30 ksi) in reinforcing condition is calculated from the CRSI Formulas "E" as follows: For 0.60 <= g < 1.00, fMn(6) = (0.65/0.7)*((P30c)*(0.5-0.1859*b1*(1+g))*h + 15.75*(rtb+X*F3*rss)*g*h*Ag)/12 P30c = 0.2975*f'c*(0.75*b1*(1+g)-rtb-(0.75*b1*(1+g)-1+g)*rss/g)*Ag For g < 0.60, fMn(6) = (0.65/0.7)*((P30c)*(0.5-0.1859*b1*(1+g))*h + 0.175*((204*g-30)/(1+g)-0.85*f'c+30)*g*h*Ag*(rtb+X*rss/2))/12 P30c = 0.223*f'c*b1*(1+g)*Ag where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss X = 0 for Nsb <= 6 , X = 0.33 for Nsb = 8 , X = 0.50 for Nsb = 10 , X = 1.0 for Nsb >= 12 F3 = 0 for Nsb<= 2 , F3 = 0.33 for Nsb =4 , F3 = 0.50 for Nsb = 6 , F3 = 0.24*g+0.19 <= 0.37 for Nsb >= 8 Note: in above formulas for fMn(5) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05.

C31

The axial compression capacity associated with "balanced" failure condition at 100% tension (60 ksi) in reinforcing is calculated from the CRSI Formulas "F" as follows: For 0.667 <= g < 1.00, fPn(7) = (0.65/0.7)*(0.2975*f'c*(0.60*b1*(1+g)-rtb-(0.60*b1*(1+g)-1+g)*rss/g)*Ag + 7.5*(3.6*g-2.4)*rss*Ag) For g < 0.667, fPn(7) = (0.65/0.7)*(0.1785*f'c*b1*(1+g)*Ag + 0.35*((234*g-60)/(1+g)-0.85*f'c-60)*(rtb+X*rss/2)*Ag) where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss X = 0 for Nsb <= 6 , X = 0.33 for Nsb = 8 , X = 0.50 for Nsb = 10 , X = 1.0 for Nsb >= 12 Note: in above formulas for fPn(7) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05. Note: this program uses the "exact" solution for sections with symmetrical bars on only 2 faces as follows: fPn(bal) = f*(0.85*f'c*b*b1*c+A's*(fy-0.85*f'c)-As*fy) where: f = 0.65 (per ACI 318-05) c = 87*d/(fy+87) As = A's = Ast/2 d = h-d', d" = h/2-d'

D31

The ultimate moment capacity associated with "balanced" failure condition at 100% tension (60 ksi) in reinforcing is calculated from the CRSI Formulas "F" as follows: For 0.667 <= g < 1.00, fMn(7) = (0.65/0.7)*((Pbc)*(0.5-0.15*b1*(1+g))*h + 21*(rtb+X*F4*rss)*g*h*Ag)/12 Pbc = 0.2975*f'c*(0.60*b1*(1+g)-rtb-(0.60*b1*(1+g)-1+g)*rss/g)*Ag For g < 0.667, fMn(7) = (0.65/0.7)*((Pbc)*(0.5-0.148*b1*(1+g))*h + 0.175*((234*g-60)/(1+g)-0.85*f'c+60)*(rtb+X*rss/3)*g*h*Ag)/12 Pbc = 0.1785*f'c*b1*(1+g)*Ag where: b1 =0.85-0.5*(f'c-4) <= 0.65 g = (h-2*d')/h Ag = b*h rtb = Ntb*Ab/Ag rss = Nsb*Ab/Ag rg = rtb+rss X = 0 for Nsb <= 6 , X = 0.33 for Nsb = 8 , X = 0.50 for Nsb = 10 , X = 1.0 for Nsb >= 12 F4 = 0 for Nsb<= 2 , F4 = 0.33 for Nsb =4 , F4 = 0.50 for Nsb = 6 , F4 = 0.2*(g-0.6)+0.32 <= 0.37 for Nsb >= 8 Note: in above formulas for fMn(7) the ratio of 0.65/0.7 was applied to account for reduction in f factor for ACI 318-05. Note: this program uses the "exact" solution for sections with symmetrical bars on only 2 faces as follows: fMn(bal) = (f*(0.85*f'c*b*b1*c*(d-0.5*b1*c-d'')+A's*(fy-0.85*f'c)*(d-d'-d")+As*fy*d"))/12 where: f = 0.65 (per ACI 318-05) c = 87*d/(fy+87) As = A's = Ast/2 d = h-d', d" = h/2-d'

C32

The ultimate axial load corresponding to a value fPn = 0.1*f'c*Ag.

D32

The ultimate moment capacity corresponding to a value fPn = 0.1*f'c*Ag. It is determined by straight line interpolating a value between Point #7 and Point #9 as follows: fMn(8) = (fMn(9)-fMn(7))*(fPn(8)-fPn(7))/(fPn(9)-fPn(7))+fMn(7)

D33

The Ultimate Moment Capacity, 'fMn', of the beam/section for pure moment with no axial load is calculated as follows: For Singly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2))/12 For Doubly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)+A's*f's*(d-d'))/12 f's = e's*Es = (1-d'/c)*87 <= fy (Es = 29000 ksi for steel) e's = ec*(c-d')/c, where ec = 0.003 (assumed) where: f = capacity reduction factor, 0.65 <= f <= 0.90 et = ec*(d-c)/c, where ec = 0.003 (assumed) et >= 0.005 (tension-controlled section): f =0.90 fy/Es < et < 0.005 (transition section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) <= 0.90 et <= fy/Es (compression-controlled section): f = 0.65 Notes: 1. ACI Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and doubly reinforced sections. 2. When the calculated value of f's < 0 (in the case where the distance to the neutral axis, c, is less than d') this program neglects A's and the calculated moment capacity is based on a singly reinforced beam/section. 3. For pure moment capacity with no axial load, the program assumes bars in 2 outside faces parallel to axis of bending plus 50% of the total area of the side bars divided equally by and added to the 2 outside faces, and program calculates reinforcing areas as follows: for X-axis: As = A's = ((Ntb*Abt) + (0.50*Nsb*Abs))/2

C34

The maximum allowable axial tension capacity (with no moment) is calculated as follows: fPn(9) = f*Ast*fy where: f = 0.9 Ast = (Ntb+Nsb)*Ab Ab = area of one bar

C37

The actual gross reinforcing ratio provided, 'rg ', is calculated as follows: rg = rtb+rsb where: rtb = Ntb*Abt/Ag rsb = Nsb*Abs/Ag Ag = b*h Note: r(max) is defined as follows: r(max) = 0.08 for Pu >= 0.1*f'c*Ag r(max) = 0.75*rb+r ' *f'sb/fy

C41

This value of the ultimate axial load capacity, 'fPn', is evaluated at the design eccentricity based on 'Pu' and 'Mu'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

D41

This value of the ultimate moment capacity, 'fMn', is evaluated at the design eccentricity based on 'Pu' and 'Mu'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

E41

The value of the design eccentricity, 'e', shown in table is calculated as follows: e = Mu*12/Pu

C45

For a sway frame, compression member slenderness can be ignored (thus a "short" column) when: k*Lu/r <= 22 where: k = effective length factor (assumed > 1.0) Lu = unbraced length of column (in.) r = radius of gyration = 0.30*h (in.)

C46

For a braced frame, compression member slenderness can be ignored (thus a "short" column) when: k*Lu/r <= 34 - 12*(M1/M2) <= 40 where: k = effective length factor (assumed <= 1.0) Lu = unbraced length of column (in.) r = radius of gyration = 0.30*h (in.) M1 = smaller factored end moment = positive for single curvature = negative for double curvature M2 = larger factored end moment (always positive)

C49

The Pure Axial Compression Capacity (with no moment) for the member without reinforcing is calculated from ACI Eqn. 10-2 as follows: fPn = 0.80*f*(0.85*f'c*Ag) where: f = 0.65 (per ACI 318-05) Ag = b*h Note: this value of 'fPn' is calculated merely to demonstrate the unreinforced axial compression capacity, and thus to determine whether or not each longitudinal bar must be tied. If fPn >= Pu, then in theory ties are not really required, so every bar need not be tied as well. (Plotted as a "red diamond" on the interaction diagram.)

C52

The criteria for minimum tie bar size is as follows: Use: minimum #3 ties for <= #10 longitudinal bars minimum #4 ties for > #10 longitudinal bars The criteria for maximum tie bar spacing is as follows: Use smallest of: 48*(tie bar dia.) 16*(smaller of top/bottom or side long. bars) Lesser of Lx or Ly Note: refer to ACI 318-05 Code Section 7.10.5 for tie requirements in compression members, and Sections 11.5.5 and 11.5.6 for tie requirements for shear.


12 of 18 04/07/2023 20:24:08

###

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( - )fMnx ft k

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13 of 18 04/07/2023 20:24:08

RECTANGULAR CONCRETE BEAM/COLUMN ANALYSIS CALCULATIONS:For Biaxial Flexure with Axial Compression or Tension Load

Assuming "Short", Non-Slender Member with Symmetric Reinforcing (per ACI 318-05 Code) Universal Column Formulas from CRSI Manual - for X-axis:Job Name: Subject: Pu =

Job Number: Originator: Checker: Mux =Lx = b =

Input Data: Ly = h = Lx=18 dy =

Reinforcing Yield Strength, fy = 60 ksi. g =Concrete Comp. Strength, f 'c = 4 ksi

Total Member Width, Lx = 18.000 in. Ntb =Total Member Depth, Ly = 18.000 in. Abt =

Distance to Long. Reinforcing, d' = 2.500 in. Ly=18 Ntb=8 Nsb =Ultimate Design Axial Load, Pu = 200.00 kips Nsb=0 Abs = Ultimate Design Moment, Mux = 100.00 ft-kips

Ultimate Design Moment, Muy = 100.00 ft-kips

Total Top/Bot. Long. Bars, Ntb = 8 d'=2.5 (typ.) Ast =Top/Bot. Longitudinal Bar Size = 8 Ag =

Total Side Long. Bars, Nsb = 0 Member SectionSide Longitudinal Bar Size = 8

Results: As(min) =Gross reinforcing ratio provided:

0.01951 As(max) =X =

X-axis Flexure and Axial Load Interaction Diagram Points Y-axis Flexure and Axial Load Interaction Diagram Points F1 =Location Comments Location Comments F2 =Point #1 948.55 0.00 0.00 Point #1 948.55 0.00 0.00 F3 =Point #2 758.84 0.00 0.00 Point #2 758.84 0.00 0.00 F4 =Point #3 758.84 105.09 1.66 Min. eccentricity Point #3 758.84 87.47 1.38 Min. eccentricity X-axis Flexure and Axial Load Interaction Diagram Points:Point #4 640.36 170.72 3.20 Point #4 651.66 136.89 2.52 ###Point #5 534.60 207.29 4.65 Point #5 543.64 165.23 3.65 ###Point #6 447.71 232.10 6.22 Point #6 456.48 181.87 4.78 ###Point #7 303.20 261.59 10.35 Point #7 312.47 196.34 7.54 ###Point #8 129.60 225.87 20.91 Point #8 129.60 170.84 15.82 ###Point #9 0.00 199.20 Pure moment capacity Point #9 0.00 152.76 Pure moment capacity ###Point #10 -341.28 0.00 0.00 Pure axial tension capacity Point #10 -341.28 0.00 0.00 Pure axial tension capacity Poc =

Pos =Member Uniaxial Capacity at Design Eccentricity, ey: Member Uniaxial Capacity at Design Eccentricity, ex: ###

Interpolated Results from Above: Interpolated Results from Above: ###P15c =

457.21 228.60 6.00 376.52 188.26 6.00 P15s =###

Biaxial Capacity and Stress Ratio for Pu >= 0.1*f'c*Ag: Effective Length Criteria for "Short" Column: ###263.93 k*Lu <= 9.90 ft. (for k*Lu/r(min) <= 22) P30c =

S.R. = 0.758 k*Lu <= 18.00 ft. (for k*Lu/r(min) <= 40) P30s =###

Biaxial Stress Ratio for Pu < 0.1*f'c*Ag: Pure Axial Compression Capacity w/o Reinf.: Tie Min. Size & Max. Spac.: ###S.R. = N.A. 572.83 #3@16'' Pbc =

Pbs =

b1 =

rtb =rss =

rg =r(min) =

r(max) =rg =

fPnx (k) fMnx (ft-k) ey (in.) fPny (k) fMny (ft-k) ex (in.)

Nom. max. compression = fPo Nom. max. compression = fPoAllowable fPn(max) = 0.8*fPo Allowable fPn(max) = 0.8*fPo

0% rebar tension = 0 ksi 0% rebar tension = 0 ksi




fPn = 0.1*f'c*Ag fPn = 0.1*f'c*Ag(Infinity) (Infinity)

fPnx (k) fMnx (ft-k) ey (in.) fPny (k) fMny (ft-k) ex (in.)

fPn = kips fPn = 1/(1/fPnx + 1/fPny -1/fPo) <= 1.0

S.R. = Pu/fPn <= 1.0

S.R. = (Mux/fMnx)^1.15 + (Muy/fMny)^1.15 <= 1.0 fPn = kips fPn = 0.80*0.70*(0.85*f'c*Ag)

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1200Y-AXIS INTERACTION DIAGRAM

( - )fMny ft k

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fPn

yk

C23

The actual gross reinforcing ratio provided, 'rg ', is calculated as follows: rg = rtb+rsb where: rtb = Ntb*Abt/Ag rsb = Nsb*Abs/Ag Ag = b*h Note: r(max) is defined as follows: r(max) = 0.08 for Pu >= 0.1*f'c*Ag r(max) = 0.75*rb+r ' *f'sb/fy

E26

The values of eccentricity shown in this table are calculated as follows: ey = fMnx*12/fPnx where: fPnx and fMnx are the respective values associated with a specific point on the interaction diagram.

M26

The values of eccentricity shown in this table are calculated as follows: ex = fMny*12/fPny where: fPny and fMny are the respective values associated with a specific point on the interaction diagram.

C27


K27


C28


K28


C29


D29


K29


L29


C30


D30


K30


L30


C31


D31


K31


L31


C32


D32


K32


L32


C33


D33


K33


L33


C34


D34


K34


L34


D35

The Ultimate Moment Capacity, 'fMn', of the beam/section for pure moment with no axial load is calculated as follows: For Singly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2))/12 For Doubly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)+A's*f's*(d-d'))/12 f's = e's*Es = (1-d'/c)*87 <= fy (Es = 29000 ksi for steel) e's = ec*(c-d')/c, where ec = 0.003 (assumed) where: f = capacity reduction factor, 0.65 <= f <= 0.90 et = ec*(d-c)/c, where ec = 0.003 (assumed) et >= 0.005 (tension-controlled section): f =0.90 fy/Es < et < 0.005 (transition section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) <= 0.90 et <= fy/Es (compression-controlled section): f = 0.65 Notes: 1. ACI Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and doubly reinforced sections. 2. When the calculated value of f's < 0 (in the case where the distance to the neutral axis, c, is less than d') this program neglects A's and the calculated moment capacity is based on a singly reinforced beam/section. 3. For pure moment capacity with no axial load, the program assumes bars in 2 outside faces parallel to axis of bending plus 50% of the total area of the side bars divided equally by and added to the 2 outside faces, and program calculates reinforcing areas as follows: for X-axis: As = A's = ((Ntb*Abt) + (0.50*Nsb*Abs))/2 for Y-axis: As = A's =((Nsb*Abs+4*Abt)+0.50*(Ntb-4)*Abt)/2

L35

The Ultimate Moment Capacity, 'fMn', of the beam/section for pure moment with no axial load is calculated as follows: For Singly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2))/12 For Doubly Reinforced Beam/Section: fMn = f*(0.85*f'c*a*b*(d-a/2)+A's*f's*(d-d'))/12 f's = e's*Es = (1-d'/c)*87 <= fy (Es = 29000 ksi for steel) e's = ec*(c-d')/c, where ec = 0.003 (assumed) where: f = capacity reduction factor, 0.65 <= f <= 0.90 et = ec*(d-c)/c, where ec = 0.003 (assumed) et >= 0.005 (tension-controlled section): f =0.90 fy/Es < et < 0.005 (transition section): f = 0.65+0.25*(et-fy/Es)/(0.005-fy/Es) <= 0.90 et <= fy/Es (compression-controlled section): f = 0.65 Notes: 1. ACI Code requires that the strain in the tension reinforcing (et) >= 0.004 for both singly and doubly reinforced sections. 2. When the calculated value of f's < 0 (in the case where the distance to the neutral axis, c, is less than d') this program neglects A's and the calculated moment capacity is based on a singly reinforced beam/section. 3. For pure moment capacity with no axial load, the program assumes bars in 2 outside faces parallel to axis of bending plus 50% of the total area of the side bars divided equally by and added to the 2 outside faces, and program calculates reinforcing areas as follows: for X-axis: As = A's = ((Ntb*Abt) + (0.50*Nsb*Abs))/2 for Y-axis: As = A's =((Nsb*Abs+4*Abt)+0.50*(Ntb-4)*Abt)/2

C36


K36


C40

This value of the ultimate axial load capacity, 'fPnx', is evaluated at the design eccentricity based on 'Pu' and 'Mux'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

D40

This value of the ultimate moment capacity, 'fMnx', is evaluated at the design eccentricity based on 'Pu' and 'Mux'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

E40

The value of the design eccentricity, 'ey', shown in table is calculated as follows: ey = Mux*12/Pu

K40

This value of the ultimate axial load capacity, 'fPny', is evaluated at the design eccentricity based on 'Pu' and 'Muy'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

L40

This value of the ultimate moment capacity, 'fMny', is evaluated at the design eccentricity based on 'Pu' and 'Muy'. The value shown is interpolated (straight-line) from the interaction diagram points shown in table above.

M40

The value of the design eccentricity, 'ex', shown in table is calculated as follows: ex = Muy*12/Pu

C44

The biaxial capacity is determined by the following approximation for Pu >= 0.1*f'c*Ag, using the Bresler Reciprocal Load equation: 1/fPn = 1/fPnx + 1/fPny - 1/fPo solving for fPn, fPn = 1/(1/fPnx + 1/fPny - 1/fPo)

K44

For a sway frame, compression member slenderness can be ignored (thus a "short" column) when: k*Lu/r <= 22 where: k = effective length factor (assumed > 1.0) Lu = unbraced length of column (in.) r = radius of gyration = 0.30*h (in.)

C45

Biaxial interaction stress ratio, S.R. = Pu/fPn <= 1

K45

For a braced frame, compression member slenderness can be ignored (thus a "short" column) when: k*Lu/r <= 34 - 12*(M1/M2) <= 40 where: k = effective length factor (assumed <= 1.0) Lu = unbraced length of column (in.) r = radius of gyration = 0.30*h (in.) M1 = smaller factored end moment = positive for single curvature = negative for double curvature M2 = larger factored end moment (always positive)

C48

The biaxial stress ratio is determined by the following approximation for Pu < 0.1*f'c*Ag, using the Bresler Load Contour interaction equation: S.R. = (Mux/fMnx)^1.15 + (Muy/fMny)^1.15 <= 1

K48

The Pure Axial Compression Capacity (with no moment) for the member without reinforcing is calculated from ACI Eqn. 10-2 as follows: fPn = 0.80*f*(0.85*f'c*Ag) where: f = 0.65 (per ACI 318-05) Ag = b*h Note: this value of 'fPn' is calculated merely to demonstrate the unreinforced axial compression capacity, and thus to determine whether or not each longitudinal bar must be tied. If fPn >= Pu, then in theory ties are not really required, so every bar need not be tied as well. (Plotted as a "red diamond" on the interaction diagram.)

P48

The criteria for minimum tie bar size is as follows: Use: minimum #3 ties for <= #10 longitudinal bars minimum #4 ties for > #10 longitudinal bars The criteria for maximum tie bar spacing is as follows: Use smallest of: 48*(tie bar dia.) 16*(smaller of top/bottom or side long. bars) Lesser of Lx or Ly Note: refer to ACI 318-05 Code Section 7.10.5 for tie requirements in compression members, and Sections 11.5.5 and 11.5.6 for tie requirements for shear.

REINFORCING BAR DATA TABLES:

Reinforcing Bar PropertiesBar Size Diameter Area Perimeter Weight

(in.) (in.^2) (in.) (lbs./ft.)#3 0.375 0.11 1.178 0.376#4 0.500 0.20 1.571 0.668#5 0.625 0.31 1.963 1.043#6 0.750 0.44 2.356 1.502#7 0.875 0.60 2.749 2.044#8 1.000 0.79 3.142 2.670#9 1.128 1.00 3.544 3.400

#10 1.270 1.27 3.990 4.303#11 1.410 1.56 4.430 5.313#14 1.693 2.26 5.320 7.650#18 2.257 4.00 7.091 13.600

Typical specification: ASTM A615 Grade 60 Deformed Bars

Reinforcing Bar Area for Various Bar Spacings (in.^2/ft.)Spacing Bar Size

(in.) #3 #4 #5 #6 #7 #8 #9 #10 #113 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.24

3-1/2 0.38 0.69 1.06 1.51 2.06 2.71 3.43 4.35 5.354 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.68

4-1/2 0.29 0.53 0.83 1.17 1.60 2.11 2.67 3.39 4.165 0.26 0.48 0.74 1.06 1.44 1.90 2.40 3.05 3.74

5-1/2 0.24 0.44 0.68 0.96 1.31 1.72 2.18 2.77 3.406 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.12

6-1/2 0.20 0.37 0.57 0.81 1.11 1.46 1.85 2.34 2.887 0.19 0.34 0.53 0.75 1.03 1.35 1.71 2.18 2.67

7-1/2 0.18 0.32 0.50 0.70 0.96 1.26 1.60 2.03 2.508 0.17 0.30 0.46 0.66 0.90 1.19 1.50 1.91 2.34

8-1/2 0.16 0.28 0.44 0.62 0.85 1.12 1.41 1.79 2.209 0.15 0.27 0.41 0.59 0.80 1.05 1.33 1.69 2.08

9-1/2 0.14 0.25 0.39 0.56 0.76 1.00 1.26 1.60 1.9710 0.13 0.24 0.37 0.53 0.72 0.95 1.20 1.52 1.87

10-1/2 0.13 0.23 0.35 0.50 0.69 0.90 1.14 1.45 1.7811 0.12 0.22 0.34 0.48 0.65 0.86 1.09 1.39 1.70

11-1/2 0.115 0.21 0.32 0.46 0.63 0.82 1.04 1.33 1.6312 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56

Tension Development and Splice Lengths for f 'c=3,000 psi and fy=60 ksiDevelopment Class "B" Splice Standard 90 deg. Hook

Bar Size Top Bar Other Bar Top Bar Other Bar Embed. Leg Length Bend Dia.(in.) (in.) (in.) (in.) (in.) (in.) (in.)

#3 22 17 28 22 6 6 2-1/4#4 29 22 37 29 8 8 3#5 36 28 47 36 10 10 3-3/4#6 43 33 56 43 12 12 4-1/2#7 63 48 81 63 14 14 5-1/4#8 72 55 93 72 16 16 6#9 81 62 105 81 18 19 9-1/2

#10 91 70 118 91 20 22 10-3/4#11 101 78 131 101 22 24 12#14 121 93 --- --- 37 31 18-1/4#18 161 124 --- --- 50 41 24

Notes:1. Straight development and Class "B" splice lengths shown in above tables are based on uncoated bars assuming center-to-center bar spacing >= 3*db without ties or stirrups or >= 2*db with ties or stirrups, and bar clear cover >= 1.0*db. Normal weight concrete as well as no transverse reinforcing are both assumed.2. Standard 90 deg. hook embedment lengths are based on bar side cover >= 2.5" and bar end cover >= 2" without ties around hook.3. For special seismic considerations, refer to ACI 318-05 Code Chapter 21.



#3 19 15 24 19 6 6 2-1/4#4 25 19 32 25 7 8 3#5 31 24 40 31 9 10 3-3/4#6 37 29 48 37 10 12 4-1/2#7 54 42 70 54 12 14 5-1/4#8 62 48 80 62 14 16 6#9 70 54 91 70 15 19 9-1/2

#10 79 61 102 79 17 22 10-3/4#11 87 67 113 87 19 24 12#14 105 81 --- --- 32 31 18-1/4#18 139 107 --- --- 43 41 24




#3 17 13 22 17 6 6 2-1/4#4 22 17 29 22 6 8 3#5 28 22 36 28 8 10 3-3/4#6 33 26 43 33 9 12 4-1/2#7 49 37 63 49 11 14 5-1/4#8 55 43 72 55 12 16 6#9 63 48 81 63 14 19 9-1/2

#10 70 54 91 70 15 22 10-3/4#11 78 60 101 78 17 24 12#14 94 72 --- --- 29 31 18-1/4#18 125 96 --- --- 39 41 24


Tension Lap Splice ClassesFor Other than Columns For Columns

Area (Provided) / Area (Req'd) % of Bars Spliced Maximum Tension Stress % of Bars Spliced<= 50% > 50% in Reinforcing Bars <= 50 % > 50%

< 2 B B <= 0.5*fy A B>= 2 A B > 0.5*fy B B

Compression Development and Splice Lengths for fy=60 ksiBar Size Development Length (in.) Splice Length (in.)

f 'c=3000 f 'c=4000 f 'c=5000 f 'c=3000 f 'c=4000 f 'c=5000#3 9 8 8 12 12 12#4 11 10 9 15 15 15#5 14 12 12 19 19 19#6 17 15 14 23 23 23#7 19 17 16 27 27 27#8 22 19 18 30 30 30#9 25 22 21 34 34 34

#10 28 24 23 38 38 38#11 31 27 26 43 43 43#14 37 32 31 --- --- ---#18 50 43 41 --- --- ---

Notes:1. For development in columns with reinforcement enclosed with #4 ties spaced <= 4" on center, values above may be multiplied by 0.75, but shall not be < 8".2. For development in columns with reinforcement enclosed by spiral reinforcement >= 1/4" diameter and <= 4" pitch, values above may be multiplied by 0.83, but shall not be < 8".3. For splices in columns with reinforcement enclosed with #4 ties spaced <= 4" on center, values shown above may be multiplied by 0.83 for #3 thru #11 bars, but shall not be < 12".4. For splices in columns with reinforcement enclosed by spiral reinforcement >= 1/4" diameter and <= 4" pitch, values above may be multiplied by 0.75 for #3 thru #11 bars, but shall not be < 12".

Maximum Allowable Spacing of Column Ties (in.)Vertical Bar Size Tie Bar Size Tie Bar Size Tie Bar Size

#3 #4 #5#5 10 --- ---#6 12 --- ---#7 14 --- ---#8 16 16 ---#9 18 18 ---

#10 18 20 ---#11 --- 22 22#14 --- 24 27#18 --- 24 30

Notes:1. Maximum tie spacing should be <= least column dimension.2. For special seismic considerations, refer to ACI 318-05 Code Chapter 21.

Plain Welded Wire Reinforcement PropertiesWelded Wire Reinf. Wire Diameter Wire Area Reinf. Weight

Designation Each Way (in.) Each Way (in.^2/ft.) (psf)6x6 - W1.4xW1.4 0.135 0.028 0.216x6 - W2.0xW2.0 0.159 0.040 0.296x6 - W2.9xW2.9 0.192 0.058 0.426x6 - W4.0xW4.0 0.225 0.080 0.584x4 - W1.4xW1.4 0.135 0.042 0.314x4 - W2.0xW2.0 0.159 0.060 0.434x4 - W2.9xW2.9 0.192 0.087 0.624x4 - W4.0xW4.0 0.225 0.120 0.85

Notes:1. Welded wire reinforcement designations are some common stock styles assuming plain wire reinf. per ASTM Specification A185. (fy = 65,000 psi)2. First part of welded wire reinf. designation denotes the wire spacing each way.3. Second part of welded wire reinf. designation denotes the wire size as follows:

W1.4 ~= 10 gage , W2.0 ~= 8 gageW2.9 ~= 6 gage , W4.0 ~= 4 gage

Diseño de vigas y columnas de concreto ACI-318-05

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