Concrete Beam/Column Analysis/Design

Doc"RECTBEAM" --- RECTANGULAR CONCRETE BEAM ANALYSIS/DESIGNProgram Description:"RECTBEAM" is a spreadsheet program written in MS-Excel for the purpose of analysis/design of rectangularbeam or column sections. Specifically, the required flexural reinforcing, ultimate moment capacity, bar spacingfor crack control, moments of inertia for deflection, beam shear and torsion requirements, and member capacityfor flexure (uniaxial and biaxial) with axial load are calculated. There is also a worksheet which containsreinforcing bar data tables.This program is a workbook consisting of eleven (11) worksheets, described as follows:Worksheet NameDescriptionDocThis documentation sheetComplete AnalysisBeam flexure, shear, crack control, and inertiaFlexure(As)Flexural reinforcing for singly or doubly reinforced beams/sectionsFlexure(Mn)Ultimate moment capacity of singly or doubly reinforced beams/sectionsCrack ControlCrack control - distribution of flexural reinforcingShearBeam or one-way type shearTorsionBeam torsion and shearInertiaMoments of inertia of singly or doubly reinforced beams/sectionsUniaxialCombined uniaxial flexure and axial loadBiaxialCombined biaxial flexure and axial loadRebar DataReinforcing bar data tablesProgram Assumptions and Limitations:1. This program follows the procedures and guidelines of the ACI 318-99 Building Code.2. The "Complete Analysis" worksheet combines the analyses performed by four (4) of the individualworksheets 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 worksheetsincluded in this one are also applicable here.3. In the "Flexure(As)" worksheet, the program will display a message if compression reinforcing is required,when the beam/section cannot handle the ultimate design moment with tension reinforcing only. Then adoubly-reinforced design is performed.4. In the "Flexure(As)" worksheet for a singly reinforced beam/section, when the required flexural reinforcing isless than the Code minimum, then the program will use the lesser value of either 4/3 times the required valueor the minimum value as the amount to actually use for design.5. In the "Flexure(Mn)", "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 thatreinforcing and calculate the ultimate moment capacity based on an assumed singly-reinforced section.6. In the "Uniaxial" and "Biaxial" worksheets, the CRSI "Universal Column Formulas" are used by this programto determine points #1 through #7 of the 10 point interaction curve.7. In the "Uniaxial" and "Biaxial" worksheets, the CRSI "Universal Column Formulas", which are used by thisprogram, 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 columnwith symmetrically arranged and sized bars.9. In the "Uniaxial" and "Biaxial" worksheets, for cases with axial load only (compression or tension) and nomoment(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.10. In the "Uniaxial" and "Biaxial" worksheets, for pure moment capacity with no axial load, the program assumesbars in 2 outside faces parallel to axis of bending plus 50% of the total area of the side bars divided equallyby 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))/2for Y-axis: As = A's = ((Nsb*Asb+4*Atb) + (0.50*(Ntb-4)*Atb))/211. In the "Uniaxial" and "Biaxial" worksheets, design capacities, fPn and fMn, at design eccentricity,e = Mu*12/Pu, are determined from interpolation within the interaction curve for each axis.12. In the "Uniaxial" and "Biaxial" worksheets, when the design eccentricity falls between the "balanced" point(Point #7) and point of pure flexure (Point #9) the program uses f = 0.7 at Point #7 and f = 0.9 at Point #9.However, it should be noted that the Code permits the value of 'f' to be increased linearly from a startingvalue of 0.70 at fPn = 0.1*f 'c*Ag (Point #8), up to the maximum value of 0.9 at Point #9, using:f = 0.90 - 2*Pu/(f 'c*Ag).13. 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:1/fPn = 1/fPnx + 1/fPny - 1/fPoBiaxial interaction stress ratio, S.R. = Pu/fPn = 200/fywhere: f'c and fy are in psi.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))The Reinforcing Ratio, 'r(max)', is the maximum allowable reinforcing ratio and is calculated as follows: For Singly Reinforced Beam/Section: r(max) = 0.75*rb For Doubly Reinforced Beam/Section: r(max) = 0.75*rb+r '*f'sb/fy where: r ' = A's/(b/d) f'sb = 87*(1-d'/d)*(87+fy)/87 = Av(min) where: f = 0.85 fVs = max. of: (Vu - fVc) or 0Note: Av = area of both legs of a closed tie or stirrup.The minimum area of shear reinforcing, 'Av(min)', to be provided for beams is calculated as follows: Av(min) = 50*b*s/(fy*1000)Note: Av(min) = area of both legs of a closed tie or stirrup.The maximum allowable shear reinforcing (tie/stirrup) spacing, 's', shall not exceed d/2, nor 24". However,when fVs > f*4*(f'c*1000)^(1/2)*b*d , then the maximum spacing given above shall be reduced by one-half. (Note: f = 0.85)The Modulus of Elasticity for Steel, 'Es', is assumed = 29,000 ksiThe Modulus of Elasticity for Concrete, 'Ec', is calculated as follows: Ec = 57*(f'c*1000)^(1/2) ksiNote: "normal" weight concrete is assumed.The Modular Ratio, 'n', is calculated as follows: n = Es/EcThe working stress tension in the reinforcing, 'fs', is calculated as follows: fs = 12*Ma/(n*As*d*(1-((2*As/b*d)+(n*As/(b*d))^2)^(1/2)-n*As/(b*d))/3))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 0.6*fy.The center-to-center spacing, 's', of the flexural tension reinforcing shall not exceed the following: s = 540/fs-2.5*Ccbut not greater than 12*(36/fs).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-dThe factor limiting the distribution of flexural reinforcement, 'z', is calculated as follows: z = fs*(dc*2*dc*b/Nb)^(1/2)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 = 145One-way slabs z = 156 z =129The Modulus of Rupture of concrete, 'fr', is calculated as follows: fr = 7.5*(f'c*1000)^(1/2) ksiThe Gross (uncracked) Moment of Inertia, 'Ig', is calculated as follows: Ig = b*h^3/12The Cracking Moment, 'Mcr', is calculated as follows: Mcr = fr*Ig/ytwhere: yt = h/2The 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')^2The Effective Moment of Inertia, 'Ie', is calculated as follows: Ie = (Mcr/Ma)^3*Ig+(1-(Mcr/Ma)^3)*Icr 60 ksiNote: minimum temperature reinforcing percentage is not used for beams, only slab sections.The minimum required area of temperature reinforcing, 'As(temp)', is calculated as follows: As(temp) = r(temp)/2*b*hNote: 1/2 of the entire percentage, r(temp), is used in each face of section.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)/Bwhere: 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))/Bwhere: B = b/(n*As) r = (n-1)*A's/(n*As)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: c = (Asmax*fy/(0.85*f'c*b))/b1The value, 'a', is the depth of the assumed retangular compression stress block, and is calculated as follows: a = c*b1The Ratio of Tension Reinforcing, 'r', is calculated as follows: For Singly Reinforced Beam/Section: r = As/(b*d) r = (f*fy-((f*fy)^2-4*(f*0.59*fy^2/f'c)*(12*Mu/b*d^2))^(1/2))/(2*(f*0.59*fy^2/f'c)) where: f = 0.9 For Doubly Reinforced Beam/Section: If A's yields, where f's = (1-d'/c)*87 >= fy , then r =(0.75*rb)+(A's/(b*d)) If A's does not yield, where f's = (1-d'/c)*87 < fy , then r = (0.75*rb)+(A's/(b*d))*(f's/fy)The area of required tension reinforcing, 'As', is calculated as follows: For Singly Reinforced Beam/Section: As = r*b*d For Doubly Reinforced Beam/Section: If A's yields, where f's = (1-d'/c)*87 >= fy , then As = (0.75*rb*b*d)+(Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')) If A's does not yield, where f's = (1-d'/c)*87 < fy , then As = (0.75*rb*b*d)+((Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')))*(f's/fy) where: f = 0.9For the case of a doubly reinforced beam/section only, the compression stress in the compression reinforcing, 'f's', is calculated as follows: If f's yields, where (1-d'/c)*87 >= fy , then f's = fy If f's does not yield, then f's = (1-d'/c)*87 < fyFor the case of a doubly reinforced beam/section only, the area of required compression reinforcing, 'A's', is calculated as follows: If A's yields, where f's = (1-d'/c)*87 >= fy , then A's = (Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')) If A's does not yield, where f's = (1-d'/c)*87 < fy , then As = ((Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')))*(f's/fy) where: f = 0.9The required area of flexural reinforcing to be used, 'As(use)', for singly reinforced beam-type sections is calculated as follows: If As < As(min) then As(use) = lesser of r(min)*b*d or (4/3)*r*b*d , otherwise As(use) = As.

The required area of flexural reinforcing to be used, 'As(use)', for singly reinforced slab-type sections is calculated as follows: If As < As(temp) then As(use) = r(temp)/2*b*h (per face) , otherwise As(use) = As.

Flexure(As)RECTANGULAR CONCRETE BEAM/SECTION DESIGNCALCULATIONS:Flexural Reinforcing for Singly or Doubly Reinforced SectionsPer ACI 318-99 CodeSingly Reinforced Section:Doubly Reinforced Section:Job Name:Subject:Job Number:Originator:Checker:Beamb1 =0.85b1 =0.85Slabr =0.01727rb =0.02851Input Data:As =2.332r(max) =0.02138(for tension only)As(max) =2.886in.^2 (for tension only)Beam or Slab Section?Beamb=10''a =5.093(for tension only)Reinforcing Yield Strength, fy =60ksifMn(max) =142.27ft-k (>= Mu)Concrete Comp. Strength, f 'c =4ksiA 's required?NoBeam Width, b =10.000in.c =4.841c =N.A.Depth to Tension Reinforcing, d =13.500in.h=16''d=13.5''a =4.115= b1*ca =N.A.= b1*cTotal Beam Depth, h =16.000in.f 's yields?N.A.Ultimate Design Moment, Mu =120.00ft-kipsf 's =N.A.ksiDepth to Compression Reinf., d' =0.000in.As=2.332A 's =N.A.in.^20Singly Reinforced SectionAs =N.A.in.^20.0000r(min) =0.00333r(min) =N.A.Results:0.0000r(temp) =N.A.r(temp) =N.A.d'brb =0.02851rb =N.A.Stress Block Data:f 'sb =N.A.ksiA'sr(max) =0.02138= 0.75*rbr(max) =N.A.= 0.75*rb+r' *f 'sb/fyReinforcing Bar Area for Various Bar Spacings (in.^2/ft.)b1 =0.85SpacingBar Sizec =4.841in.hdAs(min) =0.450in.^2As(min) =N.A.in.^2(in.)#3#4#5#6#7#8#9#10#11a =4.115in.As(temp) =N.A.in.^2/faceAs(temp) =N.A.in.^230.440.801.241.762.403.164.005.086.24As(max) =2.886in.^2As(max) =N.A.in.^23-1/20.380.691.061.512.062.713.434.355.35Reinforcing Criteria:As40.330.600.931.321.802.373.003.814.68Doubly Reinforced Section4-1/20.290.530.831.171.602.112.673.394.16rb =0.0285150.260.480.741.061.441.902.403.053.74r(min) =0.003330.005-1/20.240.440.680.961.311.722.182.773.40As(min) =0.450in.^260.220.400.620.881.201.582.002.543.12r(temp) =N.A.(total)6-1/20.200.370.570.811.111.461.852.342.88As(temp) =N.A.in.^2/face70.190.340.530.751.031.351.712.182.67r(max) =0.021387-1/20.180.320.500.700.961.261.602.032.50As(max) =2.886in.^280.170.300.470.660.901.191.501.912.348-1/20.160.280.440.620.851.121.411.792.20Computed Reinforcing:90.150.270.410.590.801.051.331.692.089-1/20.140.250.390.560.761.001.261.601.97r =0.01727100.130.240.370.530.720.951.201.521.87As =2.332in.^210-1/20.130.230.350.500.690.901.141.451.78(4/3)*As =3.109in.^2110.120.220.340.480.650.861.091.391.70f 's =N.A.ksi011-1/20.1150.210.320.460.630.821.041.331.63A's =N.A.in.^2120.110.200.310.440.600.791.001.271.56As(use) =2.332in.^2Comments:00

&R"RECTBEAM.xls" ProgramVersion 3.1&C&P of &N&R&D &TThe Ratio of Tension Reinforcing, 'r', is calculated as follows: For Singly Reinforced Beam/Section: r = As/(b*d) r = (f*fy-((f*fy)^2-4*(f*0.59*fy^2/f'c)*(12*Mu/b*d^2))^(1/2))/(2*(f*0.59*fy^2/f'c)) where: f = 0.9 For Doubly Reinforced Beam/Section: If A's yields, where f's = (1-d'/c)*87 >= fy , then r =(0.75*rb)+(A's/(b*d)) If A's does not yield, where f's = (1-d'/c)*87 < fy , then r = (0.75*rb)+(A's/(b*d))*(f's/fy)The factor, 'b1', shall be = 0.85 for concrete strengths (f'c) 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.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: c = (Asmax*fy/(0.85*f'c*b))/b1The value, 'a', is the depth of the assumed retangular compression stress block, and is calculated as follows: a = c*b1The 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))The Reinforcing Ratio, 'r(max)', is the maximum allowable reinforcing ratio and is calculated as follows: For Singly Reinforced Beam/Section: r(max) = 0.75*rb For Doubly Reinforced Beam/Section: r(max) = 0.75*rb+r '*f'sb/fy where: r ' = A's/(b/d) f'sb = 87*(1-d'/d)*(87+fy)/87 = 3*(f'c)^(1/2)/fy >= 200/fywhere: f'c and fy are in psi.The maximum allowable area of reinforcing, 'As(max)', is calculated as follows: For Singly Reinforced Beam/Section: As(max) = 0.75*rb*b*d For Doubly Reinforced Beam/Section: As(max) = (0.75*rb+r '*f'sb/fy)*b*dThe area of required tension reinforcing, 'As', is calculated as follows: For Singly Reinforced Beam/Section: As = r*b*d For Doubly Reinforced Beam/Section: If A's yields, where f's = (1-d'/c)*87 >= fy , then As = (0.75*rb*b*d)+(Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')) If A's does not yield, where f's = (1-d'/c)*87 < fy , then As = (0.75*rb*b*d)+((Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')))*(f's/fy) where: f = 0.9For the case of a doubly reinforced beam/section only, the area of required compression reinforcing, 'A's', is calculated as follows: If A's yields, where f's = (1-d'/c)*87 >= fy , then A's = (Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')) If A's does not yield, where f's = (1-d'/c)*87 < fy , then As = ((Mu/f-0.85*f'c*a*b*(d-a/2))/(fy*(d-d')))*(f's/fy) where: f = 0.9For the case of a doubly reinforced beam/section only, the compression stress in the compression reinforcing, 'f's', is calculated as follows: If f's yields, where (1-d'/c)*87 >= fy , then f's = fy If f's does not yield, then f's = (1-d'/c)*87 < fyThe minimum required area of flexural reinforcing, 'As(min)', is calculated as follows: As(min) = r(min)*b*dThe 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 ksiNote: minimum temperature reinforcing percentage is not used for beams, only slab sections.The minimum required area of temperature reinforcing, 'As(temp)', is calculated as follows: As(temp) = r(temp)/2*b*hNote: 1/2 of the entire percentage, r(temp), is used in each face of section.The required area of flexural reinforcing to be used, 'As(use)', for singly reinforced beam-type sections is calculated as follows: If As < As(min) then As(use) = lesser of r(min)*b*d or (4/3)*r*b*d , otherwise As(use) = As.

The required area of flexural reinforcing to be used, 'As(use)', for singly reinforced slab-type sections is calculated as follows: If As < As(temp) then As(use) = r(temp)/2*b*h (per face) , otherwise As(use) = As.

Flexure(Mn)RECTANGULAR CONCRETE BEAM/SECTION ANALYSISCALCULATIONS:Ultimate Moment Capacity of Singly or Doubly Reinforced SectionsPer ACI 318-99 CodeSingly Reinforced Section:Doubly Reinforced Section:Job Name:Subject:f 's yields?N.A.Job Number:Originator:Checker:Beamb1 =0.85b1 =N.A.Slabc =4.983c =N.A.Input Data:a =4.235= b1*ca =N.A.= b1*cr(prov) =0.01778= As/(b*d)r(prov) =N.A.= As/(b*d)Beam or Slab Section?Beamb=10''r(min) =0.00333r(min) =N.A.Reinforcing Yield Strength, fy =60ksir(temp) =N.A.r(temp) =N.A.Concrete Comp. Strength, f 'c =4ksirb =0.02851rb =N.A.Beam Width, b =10.000in.f 'sb =N.A.ksiDepth to Tension Reinforcing, d =13.500in.h=16''d=13.5''r(max) =0.02138= 0.75*rbr(max) =N.A.= 0.75*rb+r' *f 'sb/fyTotal Beam Depth, h =16.000in.f 's =N.A.ksiTension Reinforcing, As =2.400in.^2As(min) =0.450in.^2As(min) =N.A.in.^2Depth to Compression Reinf., d' =0.000in.As=2.4As(temp) =N.A.in.^2/faceAs(temp) =N.A.in.^2Compression Reinforcing, A's =0.000in.^2Singly Reinforced SectionAs(max) =2.886in.^2As(max) =N.A.in.^20fMn =122.93ft-kfMn =N.A.ft-k00.000d'b00.000Results:00.000A'sReinforcing Bar Area for Various Bar Spacings (in.^2/ft.)Stress Block Data:SpacingBar Sizehd(in.)#3#4#5#6#7#8#9#10#11b1 =0.8530.440.801.241.762.403.164.005.086.24c =4.983in.3-1/20.380.691.061.512.062.713.434.355.35a =4.235in.As40.330.600.931.321.802.373.003.814.68Doubly Reinforced Section4-1/20.290.530.831.171.602.112.673.394.16Reinforcing Criteria:50.260.480.741.061.441.902.403.053.740.005-1/20.240.440.680.961.311.722.182.773.40r =0.0177860.220.400.620.881.201.582.002.543.12rb =0.028516-1/20.200.370.570.811.111.461.852.342.88r(min) =0.0033370.190.340.530.751.031.351.712.182.67As(min) =0.450in.^2 = As = 2.4 in.^2, O.K.9-1/20.140.250.390.560.761.001.261.601.97100.130.240.370.530.720.951.201.521.87Ultimate Moment Capacity:10-1/20.1260.230.350.500.690.901.141.451.78110.120.220.340.480.650.861.091.391.70fMn =122.93ft-kips011-1/20.1150.210.320.460.630.821.041.331.63f 's =N.A.ksi0.00120.110.200.310.440.600.791.001.271.56Note:fMn should be >= MuComments:

&R"RECTBEAM.xls" ProgramVersion 3.1&C&P of &N&R&D &TThe Ratio of Tension Reinforcing provided, 'r', is calculated as follows: r = As/(b*d)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 = 0.9 f's = (1-d'/c)*87 = 200/fywhere: f'c and fy are in psi.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))The Reinforcing Ratio, 'r(max)', is the maximum allowable reinforcing ratio and is calculated as follows: For Singly Reinforced Beam/Section: r(max) = 0.75*rb For Doubly Reinforced Beam/Section: r(max) = 0.75*rb+r '*f'sb/fy where: r ' = A's/(b/d) f'sb = 87*(1-d'/d)*(87+fy)/87 60 ksiNote: minimum temperature reinforcing percentage is not used for beams, only slab sections.The minimum required area of temperature reinforcing, 'As(temp)', is calculated as follows: As(temp) = r(temp)/2*b*hNote: 1/2 of the entire percentage, r(temp), is used in each face of section.

ShearRECTANGULAR CONCRETE BEAM/SECTION ANALYSISCALCULATIONS:Beam or One-Way Type ShearPer ACI 318-99 CodeBeamFor Beam:Job Name:Subject:SlabfVc =14.51kipsJob Number:Originator:Checker:fVs =45.90kipsfVn = fVc+fVs =60.41kipsInput Data:fVs(max) =58.06kipsAv(prov) =0.800in.^2Beam or Slab Section?BeamAv(min) =0.050in.^2Reinforcing Yield Strength, fy =60ksiAv(req'd) =0.048in.^2Concrete Comp. Strength, f 'c =4ksi.s(max) =6.750in.Beam Width, b =10.000in.Depth to Tension Reinforcing, d =13.500in.For One-Way Slab-Type Section:Total Beam Depth, h =16.000in.d VuVu dd VufVc =14.51kipsUltimate Design Shear, Vu =20.00kipsUltimate Design Axial Load, Pu =0.00kipsTotal Stirrup Area, Av(stirrup) =0.400in.^2Tie/Stirrup Spacing, s =6.0000in.0.00000.00VudResults:VuFor Beam:Typical Critical Sections for ShearfVc =14.51kipsfVs =45.90kips0.00fVn = fVc+fVs =60.41kips >= Vu = 20 kips, O.K.fVs(max) =58.06kips >= Vs = 45.9 kips, O.K.Av(prov) =0.800in.^2 =Av(stirrup)*(12/s)Av(req'd) =0.048in.^2 f*4*(f'c*1000)^(1/2)*b*d , then the maximum spacing given above shall be reduced by one-half. (Note: f = 0.85)The minimum area of longitudinal torsion reinforcing, 'Al', parallel to axis of the beam is calculated as follows: Al(min) = 5*(f'c*1000)^(1/2)*Acp/(fy*1000)-At(req'd)/s*Ph where: Acp = b*h xo = b-(2*dt) yo = h-(2*dt) Ph = 2*(xo+yo)The required area of longitudinal torsion reinforcing, 'Al', 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)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.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 = 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.The ultimate shear strength provided by the concrete, 'fVc', is calculated as follows: For shear and flexure only (no axial load): fVc = 2*(f'c*1000)^(1/2)*b*d For shear with axial compression: fVc = 2*(1+Pu/(2*Ag)*(f'c*1000)^(1/2)*b*d For shear with axial tension: fVc = 2*(1+Pu/(0.5*Ag)*(f'c*1000)^(1/2)*b*d where: Ag = b*hNote: 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.

UniaxialRECTANGULAR CONCRETE BEAM/COLUMN ANALYSISCALCULATIONS:For X-Axis Flexure with Axial Compression or Tension LoadPlot Scale FactorsAssuming "Short", Non-Slender Member with Symmetric ReinforcingUniversal Column Formulas from CRSI Manual - for X-axis:X-axisY-axisJob Name:Example #1Subject:Pu =200.000kips150700Job Number:Originator:Checker:Mux =100.000ft-kipsLx = b =18.000in.Input Data:Ly = h =18.000in.Lx=18dy =15.000in.Calculate Pure Moment Capacity for All Bars Distributed Between 2 Faces:Reinforcing Yield Strength, fy =60ksi.g =0.667SBF =0.50Side bar factor0.75Concrete Comp. Strength, f 'c =3ksib1 =0.85As = A's =2.765in.^2Total Member Width, Lx =18.000in.Ntb =6f 's yields?NoTotal Member Depth, Ly =18.000in.Abt =0.79c =3.449in.Distance to Long. Reinforcing, d' =3.000in.Ly=18Ntb=6Nsb =2a =2.932= b1*c (in.)Ultimate Design Axial Load, Pu =200.00kipsNsb=2Abs =0.79r =0.01024Ultimate Design Moment, Mux =100.00ft-kipsrtb =0.014630rtb = Ntb*Abt/(b*h)r(min) =0.00333Total Top/Bot. Long. Bars, Ntb =6rss =0.004877rss = Nsb*Abs/(b*h)rb =0.02138Top/Bot. Longitudinal Bar Size =8d'=3 (typ.)Ast =6.32f 'sb =57.60ksiTotal Side Long. Bars, Nsb =2Member SectionAg =324.00in.^2r(max) =0.02587= 0.75*rb+r' *f 'sb/fy1Side Longitudinal Bar Size =80rg =0.019506rg = rtb+rssf 's =11.33ksi23r(min) =0.010000fMn =164.79ft-kips0fPn(max)4Results:As(min) =3.240in.^205r(max) =0.08000000.000e(min)fs = 0X-axis Flexure and Axial Load Interaction Diagram PointsAs(max) =25.920in.^200.000fs = 0.25*fy 6LocationfPnx (k)fMnx (ft-k)ey (in.)CommentsX =0.000fPnfs = 0.5*fyPoint #1832.500.000.00Nom. max. compression = fPoF1 =0.0007Point #2666.000.000.00Allowable fPn(max) = 0.8*fPoF2 =0.000fs = fyPoint #3666.0079.001.42Min. eccentricityF3 =0.0008Point #4541.75137.963.060% rebar tension = 0 ksiF4 =0.000Point #5438.99164.074.4825% rebar tension = 15 ksiX-axis Flexure and Axial Load Interaction Diagram Points:9fMnPoint #6371.15181.675.8750% rebar tension = 30 ksi1fPo(Mu=0) =832.50kipsPoint #7237.58202.4810.23100% rebar tension = 60 ksifMn(Mu=0) =0ft-kipse(1) =0in.Using ACI Code Eqn. for fPn(max):10Point #897.20162.1920.02fPn = 0.1*f'c*Ag2fPn(max) =666.00kipsfPn(max) =666.00kipsInteraction DiagramPoint #90.00164.79(Infinity)Pure moment capacityfMn(Mu=0) =0ft-kipse(2) =0in.Point #10-341.280.00-0.00Pure axial tension capacity3fPn(max) =666.00kipsfMn(e=0.1*h) =79.00ft-kipse(3) =1.423in.Gross Reinforcing Ratio Provided:fPn(ot)c =403.14rg =0.01951fPn(ot)s =138.6200.000004fPn(ot) =541.75kipsMember Uniaxial Capacity at Design Eccentricity:fMn(ot) =137.96ft-kipse(4) =3.056in.Interpolated Results from Above:fPn(15)c =344.98fPnx (k)fMnx (ft-k)ey (in.)fPn(15)s =94.01364.54182.276.005fPn(15) =438.99kips0fMn(15) =164.07ft-kipse(5) =4.485in.Effective Length Criteria for "Short" Column:fPn(30)c =301.47k*Lu = 0.1*f'c*Ag:Effective Length Criteria for "Short" Column:fMn(15) =359.09ft-kipse(5) =5.440in.fPn =507.56kips fPn = 1/(1/fPnx + 1/fPny -1/fPo) = 0.1*f'c*Ag r(max) = 0.75*rb+r ' *f'sb/fyThe criteria for minimum tie bar size is as follows: Use: minimum #3 ties for #10 longitudinal barsThe 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 LyNote: refer to ACI 318-99 Code Section 7.10.5 for tie requirements in compression members, and Sections 11.5.4 and 11.5.5 for tie requirements for shear.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 =0.7 Ag = b*hNote: 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 diagrams.)

Biaxial

Pn,MnPu,MuPo(max)e(min)fs=0fs=0.25*fyfs=0.5*fyfs=fyPu=0.1*f'c*AgPlot Scale FactorsAxial Cap. w/o Reinf.fMnx (ft-k)fPnx (k)X-AXIS INTERACTION DIAGRAM

Rebar DataREINFORCING BAR DATA TABLES:Reinforcing Bar PropertiesBar SizeDiameterAreaPerimeterWeight(in.)(in.^2)(in.)(lbs./ft.)#30.3750.111.1780.376#40.5000.201.5710.668#50.6250.311.9631.043#60.7500.442.3561.502#70.8750.602.7492.044#81.0000.793.1422.670#91.1281.003.5443.400#101.2701.273.9904.303#111.4101.564.4305.313#141.6932.265.3207.650#182.2574.007.09113.600Typical specification: ASTM A615 Grade 60 Deformed BarsReinforcing Bar Area for Various Bar Spacings (in.^2/ft.)Spacing:SpacingBar Size3(in.)#3#4#5#6#7#8#9#10#113.530.440.801.241.762.403.164.005.086.2443-1/20.380.691.061.512.062.713.434.355.354.540.330.600.931.321.802.373.003.814.6854-1/20.290.530.831.171.602.112.673.394.165.550.260.480.741.061.441.902.403.053.7465-1/20.240.440.680.961.311.722.182.773.406.560.220.400.620.881.201.582.002.543.1276-1/20.200.370.570.811.111.461.852.342.887.570.190.340.530.751.031.351.712.182.6787-1/20.180.320.500.700.961.261.602.032.508.580.170.300.470.660.901.191.501.912.3498-1/20.160.280.440.620.851.121.411.792.209.590.150.270.410.590.801.051.331.692.08109-1/20.140.250.390.560.761.001.261.601.9710.5100.130.240.370.530.720.951.201.521.871110-1/20.130.230.350.500.690.901.141.451.7811.5110.120.220.340.480.650.861.091.391.701211-1/20.1150.210.320.460.630.821.041.331.6312.5120.110.200.310.440.600.791.001.271.5613Tension Development and Splice Lengths for f 'c=3,000 psi and fy=60 ksiDevelopmentClass "B" SpliceStandard 90 deg. HookBar SizeTop BarOther BarTop BarOther BarEmbed.Leg LengthBend Dia.(in.)(in.)(in.)(in.)(in.)(in.)(in.)#322172822662-1/4#429223729883#53628473610103-3/4#64333564312124-1/2#76348816314145-1/4#87255937216166#981621058118199-1/2#10917011891202210-3/4#1110178131101222412#1412193------373118-1/4#18161124------504124Notes:1. Straight development and Class "B" splice lengths shown in above tables arebased on uncoated bars assuming center-to-center bar spacing >= 3*db withoutties 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 318-99 Code Chapter 21.Tension Development and Splice Lengths for f 'c=4,000 psi and fy=60 ksiDevelopmentClass "B" SpliceStandard 90 deg. HookBar SizeTop BarOther BarTop BarOther BarEmbed.Leg LengthBend Dia.(in.)(in.)(in.)(in.)(in.)(in.)(in.)#319152419662-1/4#425193225783#5312440319103-3/4#63729483710124-1/2#75442705412145-1/4#86248806214166#97054917015199-1/2#10796110279172210-3/4#11876711387192412#1410581------323118-1/4#18139107------434124Notes:1. Straight development and Class "B" splice lengths shown in above tables arebased on uncoated bars assuming center-to-center bar spacing >= 3*db withoutties 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 318-99 Code Chapter 21.Tension Development and Splice Lengths for f 'c=5,000 psi and fy=60 ksiDevelopmentClass "B" SpliceStandard 90 deg. HookBar SizeTop BarOther BarTop BarOther BarEmbed.Leg LengthBend Dia.(in.)(in.)(in.)(in.)(in.)(in.)(in.)#317132217662-1/4#422172922683#5282236288103-3/4#6332643339124-1/2#74937634911145-1/4#85543725512166#96348816314199-1/2#1070549170152210-3/4#11786010178172412#149472------293118-1/4#1812596------394124Notes:1. Straight development and Class "B" splice lengths shown in above tables arebased on uncoated bars assuming center-to-center bar spacing >= 3*db withoutties 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 318-99 Code Chapter 21.Tension Lap Splice ClassesFor Other than ColumnsFor ColumnsArea (Provided) / Area (Req'd)% of Bars SplicedMaximum Tension Stress% of Bars Spliced 50%in Reinforcing Bars 50%< 2BB= 2AB> 0.5*fyBBCompression Development and Splice Lengths for fy=60 ksiBar SizeDevelopment Length (in.)Splice Length (in.)f 'c=3000f 'c=4000f 'c=5000f 'c=3000f 'c=4000f 'c=5000#3988121212#411109151515#5141212191919#6171514232323#7191716272727#8221918303030#9252221343434#10282423383838#11312726434343#14373231---------#18504341---------Notes:1. For development in columns with reinforcement enclosed with #4 ties spaced= 1/4" diameter and