Design of Rolled sections beams by Limit State Method as per IS 800 : 200Planning for Systematic design of Rolled beams : Step 1 : Geometrical properties of the cross-section of the Beam. Step 2 : Classfications of cross- sections : (CLAUSE 3.7) i.e : The local buckling can be avoided before the limit state is achieved by limiting the width to thickness ratio of each element of a cross-section subjected to compression due to axial force, moment or shear as per Table 2. a) Plastic sections b) Compact sections c) Semi-compact sections d) Slender sections Step 3 : Reisistance to shear buckling shall be verified as specified in Clause 8.4.2.1 i.e : d / tw Step 4 : > 67ε Check for maximum effective slenderness ratios (Table 15 : Effective length for simply supported beams) Step 5 : Section 8 : Design of members subjected to bending The factored design moment,M at any section, in a beam due to external actions shall satisfy Case 1 : Md M ≤ M Md = Factored design moment = Design bending strength of the section Laterally supported beam : A beam may be assumed to be adequately laterally supported if restraint member is capable of resisting a lateral force not less than 2.5 percent of the maximum force in the compression flange of the member. Case 2 : Laterally unsupported beam Case 1 : Laterally supported beam : when , 0.6Vd V < V Vd = Factored design shear force = Design shear strength of the section Md = βbZpfy / ϒmo To avoid irreversible deformation under serviceability loads, 1.2 Ze fy / ϒmo shall be less than incase of simply supported beams. otherwise, when Step 6 : V > 0.6Vd Md = Mdv Mdv = Design bending strength under hogh shear as def Holes in the tension zone The effect of holes in the tension and compression flange, in the design bending strength need not be considered if it satisfies the criteria given in : Clause 8.2.1.4 Step 7 : Shear lag effects The simple theory of bending is based on the assumption that plane sections remain plane after bending.In reality, shear strains cause the section to warp.The higher stresses are produced near the junction of a web and lower stresses at points away from the web.This phenomenon is known as shear lag.It results in a non-uniform stress distribution across the width of the flange. The shear lag effects in flanges may be disregarded provided it validate Clause 8.2.1.5 Case 2 : Laterally unsupported beam (CLAUSE 8.2.2) Resistance to lateral torsional buckling need not to be checked seprately in the following cases : a) Bending is about the minor axis of the section - As you are designing the section for moment about the minor axis only. b) Section is hollow or solid bars - These sections has high moment of resistance about both the axis. c) In case of major axis bending, λLT (as defined in laterally unsupported beam design) is less than 0.4 λLT λLT As the value of = sqrt (fy / fcr,b) , is less than 0.4, it means you are overdesiging your section fo particular force and moment and then, there is no need to chec for lateral torsional buckling moment. The design bending strength of laterally unsupported beam as governed by lateral torsional buckling is given by: Md βbZpfbd = Step 8 : Check for deflection ( Table 6 of IS 800 defines the deflection limits) Step 9 : Web buckling and Web crippling (As per 10.11 of Design of steel structures by N.Subramanian) Web crippling strength of the web also called as the web bearing capacity at supports. Member subjected to combined forces : (Section 9 of IS 800 2007) Case 1 : Combined shear and bending : when , < V Vd = Factored design shear force = Design shear strength of the section Md = > 0.6Vd otherwise, when V 0.6Vd V βbZpfy / ϒmo Mfd) ≤ 1. N / Nd + My / Mdy + Mz / Mdz ii) Semi-compact sections In the absence of high shear force.3) Under combined axial force and bending moment.2 Ze fy / ϒmo b) Semi-compact section : Mdv Case 2 : = Ze fy / ϒmo Combined Axial force and Bending moment : (Clause 9. . if the maximum longitudinal stress under combined axial force and bending. section strength as governed by material failure and member strength The Indian code (IS 800 : 2007) provisions are based on the Eurocode provisions and the code requires the following two checks to be performed : a) Local capacity check b) Overall buckling check a) Section strength : (Local capacity check) i) Palstic & compact sections Members subjected to combined axial force (compression & tension) and bending moment. the following should be satisfied : (My / Mndy)α1 + ((Mz / Mndyz)α2 Conservatively.β(Md .Md = Mdv = Mdv Design bending strength under hogh shear as defined in 9. semi-compact section design is satisfactory under combined axial force and bending.2 a) Plastic or Compact section : Mdv = Md . the platform beams shall be designed as a laterally unsupported beam. i) Bending and axial tension Meff = (M-Ψ T Zec / A) ≤ Md ii) Bending and axial compression P / Pdy + Ky Cmy My / Mdy + KLT Mz / Mdz ≤ P / Pdy + 0. the member will as a full lateral restraint to the compression flange of the main beam.Laterall bending of main beam will take palce between this unsupported length Ly. there exists an unsupported length Ly . As there is no slab over these beams (except roof paltform). .So.fx satisfies the following criteria : fx ≤ fy / ϒmo For cross-section without holes. 2) When a member is capable of resisting a lateral force not less than 2. of certain dimensions between two connecting beams which will act as lateral restraint to the main beam which has to be designed.6 Ky Cmy My/Mdy+ KZ Mz / Mdz ≤ Important Points : 1) For design purpose.5percent of the maximum force in the compression flange of the member. the above criteria reduces to. N / Nd + My / Mdy + Mz / Mdz b) Overall member strength : (Overall buckling check) Members subjected to combined axial force and bending moment shall be checked for overall buckling failure as given in this section. the beam will act as a laterally unsupported Beam. hod as per IS 800 : 2007 : ved by limiting ernal actions shall n moment strength of the section pported if restraint member percent of the maximum . 1. ate Clause 8.2.n shear force rength of the section Md case of simply supported strength under hogh shear as defined in 9.This phenomenon cross the width of the flange.2 sign bending strength need sections remain plane after r stresses are produced near b.5 ately in the following cases : . there is no need to check it rned by lateral torsional buckling uctures by N. ction 9 of IS 800 2007) .4 ou are overdesiging your section for a and then.g the section for moment about the resistance about both the axis. ed beam design) is less than 0.Subramanian) pacity at supports. .ar as defined in 9. the ≤ 1 ≤ 1 satisfactory under combined mbined axial force and bending.2 2 Ze fy / ϒmo as governed by de provisions and the code ) and bending moment. ≤ 1 hall be checked for overall 1 1 ally unsupported beam.5percent of the maximum force ateral restraint to the compression . xists an unsupported length act as lateral restraint to the will take palce between this 2. G.71 = 1.5d .9 .3 m Ly = 2.d/tw when c < 0.509 m c = 1. of the compression section from the neutral axis about y-y axis 46.Design of Plate girders as per IS 800 : 2007 (Limit state method) : Beam Lateral status of the beam Mz = Vz = = 11739 kN-m L/C = 3 300 kN = 44.G. (Zpyy) = 3.14E+06 mm3 rzz = 839.06 mm Lz = 24.c/tw Case 2: when 0.5d 345 εf 105 ≤ 351.d/tw when Governing case : c < 1.21E+10 mm Iyy = 4.68 mm C. = 8 v.5 m c/d εflange = 0.d/tw Case 1: when 3d ≥ c ≥ d .43 mm ryy = 76.3 m 2.74d ≤ c < d Depth of neutral axis from 1090 mm Case 3: = 200 mm neutral axis about z-z axis 760.28E+08 mm4 Zzz = 4.63E+07 mm Plastic modulus of section about 3 y-y axis.0E+05 Mpa = 7. Fa = = 2-flange = (400mm x 1-web = (2100mm x y 400mm 2180mm z tw = 20mm top flange about z-z axis = 400mm y b Check for minimum web thickness : (As per clause c Condition : When only transverse stiffene .78E+07 mm3 Zyy = 2.02 εweb = 1.act E At Unsupported My Web shall be considere .31 mm fyf = 240 Mpa fyw = = 240 Mpa 2.51 m KLz = KLy = KLz /rzz = 28.95 KLy / ryy = 33 c > 3d Governing case : C.6.74d 270 εw 105 ≤ 275.d/tw Case 1: Case 2: .d/tw = ≤ c < 1.41E+06 mm 24.1.d/tw = ≤ c < 0. (Zpzz) = 5.02 = h. of the compression section from the .4 Web satisfies the Serviceability crite Check for compressiom flange buckling requireme clause 8.74d Case 4: when Depth of neutral axis about y-y axis = Section used = Compact section Compact section Buckling class about y-y axis = Plastic section Section classification = Buckling class about z-z axis Vh 4 Flange classification = Web classification = Member no.act = Axial force.2(b) : Condition : When transverse stiffeners Checked to prevent the comp.08 mm Plastic modulus of section about 3 z-z axis. flange from buckling into when c ≥ 1.4E+04 mm² Izz = 5.5d . 54)) = τb = Nominal shear strength.35)= 105. Vdy = Therefore. µ cr .93 web with stiffeners Shear Buckling design methods : Simple post-critical method : a) Poissons ratio.1) d/tw = 105 67εw = 68. λLT (2EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5[1 + αLT + ( 1 / {ФLT+ [ Ф LT -λ LT] } ≤ sqrt(1.2) d / tw Kv = = 67εw Sqrt(Kv/5.1. βb = 1 = βb . fcr.00 14.b 2 0.5 (1.41 .4. minor axis .47 kN Section is safe in shear Section is in low shear Shear area abt.1 E)/(LLT/ry) 2 2 1.01E+11N-mm 0.37 or λLT = 3012 mm = 0.54 Mpa Shear carrying capacity = t w =sqrt(240/(sqrt(3)x239. ФLT = 0.92 kN = 4030. Zp .5 1 .40E+03N-mm = 0.56 Mpa Av x τb 42000sqmm Vn = Vcr = 5819. Vn / ϒmo Avz = 0.5 ≤ 0. Vdz = Check for shear (at support) therefore.37 .76 138. Vn / ϒmo Avy = 32000sqmm Vn = Vcr = 4433.2 Zefy/Mcr) = otherwise.Member is within Slenderless limit Compression flange buckling requirement is Check for shear buckling before yielding : (As per clause 8.49 = 1. Vn =Vcr = Shear area abt. sqrt( βb Z pfy / Mcr) ≤ = 2 2 0.e λw = k v 12 1 2 2 0.84 kN Section is safe in shear Section is in low shear Classification of the section based on shear capacity : About major axis z-z = Section is in low shear About minor axis y-y = Section is in low shear Check for design capacity of the section : Simply supported beam Md where.3 Factored maximum shear force = E d 2 = 239.2.61 112. fbd fbd λLT = = Mcr = fcr.b Mcr = = λLT = χLT fy / ϒmo χLT .34 Shear buckling analysis is required Check for resistance to shear buckling : (As per clause 8. LLT [1+ (1/20){(LLT/ry)/(hf/tf)} ] .52 kN = 5290. αLT . major axis . 37 Member may be designed as laterally suppo . λLT = 0.Therefore. Vn = Vp = Plastic shear resistance under pure shear. = Actual deflection in the transverse direction parallel to y-y axis.0kNm 560.act. h = Overall depth of the section. = Actual deflection in the lateral direction parallel to z-z axis. Avz = Shear area about z-z axis. tw = Thickness of web.00kNm 44.For laterally supported beam : Mdzz = ≤ 12281.per.8kNm = 0. tf = Thickness of flange. Vh = Maximum shear stress in the lateral direction paralel to z-z axis. Izz = Moment of inertia of the section about z-z axis. E = Youngs modulus of steel. ϒmo = Partial safety factor. KLy = Effective length about y-y axis. fy = Yield strength of steel. . Avy = Shear area about y-y axis.77 mm Safe in Deflection Symbols: Mz = Factored moment about major axis z-z. M = v.52 kN-m Mdzz > Mz Mz/Mdzz Safe in bending = 11739/12282 0. Lz = Span of the beam about z-z axis.act = L / 325 max.52 kN-m Mdzz = 12281.38 kN-m ≤ Mdyy > My Section is safe Check for Deflection : Unfactored moment abt z-z axis. KLz = Effective length about z-z axis. Iyy = Moment of inertia of the section about y-y axis. Zzz = Elastic modulus of the section about z-z axis. bf = Width of flange.= = z 7826. Ly = Unsupported length of the compression flange. ryy = Radius of gyration about y-y axis.31 mm 74. d = Depth of web. δv. δh. Zyy = Elastic modulus of the section about y-y axis.956 Mdyy Mdyy 12524.act.956 < 1 = = 744. Vz = Maximum shear stress in the transverse direction paralel to y-y axis. At = Total cross-section area. rzz = Radius of gyration about z-z axis. My = Factored moment about minor axis y-y. nz = ratio of actual applied axial force to the design axial strength for buckling about th & z axis respectively. . Mfd = Plastic design strength of the area of the cross-section excluding the shear area. Mdvzz = Design moment capacity of the section under high shear about z-z axis. α1 . Mndz = Design reduced flexural strength under combined axial force and the respective uniaxial moment acting alone. fcd = Design compreesive stress of axially loaded compression members. Ae = Effective sectional area χ = Stress reduction factor for different buckling class. Cmz = Equivalent uniform momemt factor.Md = Design moment of the whole section disregarding high shear force effect. CmLT = Equivalent uniform momemt factor lateral torsional buckling. α2 = Constants Cmy . Mndy . Mdzz = Design moment capacity of the section disregarding high shear force effect about Mdyy = Design moment capacity of the section disregarding high shear force effect about Mdv = Design moment capacity of the section under high shear. Pdy . ny . Mdy = Design moment capacity of the section about y-y. Pdz = Design strength under axial compression as governed bu buckling about minor(y) major(z) xais respectively. Mdvyy = Design moment capacity of the section under high shear about y-y axis.slenderness ratio and yield stre α = Imperfection factor KL/r = Effective slenderness ratio Ag = Gross area of cross-section Mdz = Design moment capacity of the section about z-z. . 1. ≤ 345 εf2 ≤ 345 εf .) : Beam B1 0 kN-m 0 kN 0.00 mm 0 kN ed = 40mm) 20mm) For E250 STEEL 40mm 2100mm =d z 40mm=tf =b s per clause 8.1(b) se stiffeners are provided 200 εw ≤ 200 εw ≤ ≤ 270 εw be considered as unstiffened bility criteria requirement : (As per stiffeners are provided uckling into the web.6. ϒmo = 1.0.1 [1 + αLT + ( λLT .2) + λLT 2] sqrt( fy/fcr.hf 3N-mm = 2140mm .b) .uirement is satisfied 1554 kN 5290 kN Hence safe . ally supported beam . 4kNm Safe in bending .560. effect about z-z. respective out minor(y) & kling about the y . and yield stress.ffect. shear area. effect about y-y. . 75 ≤ 345 . of the compression section from the .74d ≤ c < d Depth of neutral axis from 248.G.57E+08 mm Iyy = 7. Fa = = Section used = 2-flange = (300mm x 1-web = (465mm y x 300mm 497mm z tw = 12mm top flange about z-z axis = 300mm y b Check for minimum web thickness : (As per clause c Condition : When only transverse stiffene .74d Case 4: when Depth of neutral axis about y-y axis = Plastic section Plastic section Buckling class about y-y axis = Compact section Section classification = Buckling class about z-z axis Vh 4 Flange classification = Web classification = Member no.G.89 mm fyf = 250 Mpa fyw = = 250 Mpa 2. of the compression section from the neutral axis about y-y axis 48.75 KLy / ryy = 39.9 mm Lz = 8.83 mm C.5E+04 mm² Izz = 6.65E+06 mm3 Zyy = 4.c/tw Case 2: when 0.act = Axial force.7 m KLz = KLy = KLz /rzz = 40.48 m Ly = 2.5 mm Case 3: = 150 mm neutral axis about z-z axis 194.48 m 2. flange from buckling into when c ≥ 1.96E+06 mm Plastic modulus of section about 3 y-y axis.d/tw = ≤ 38.9 = 1 εweb = 1 = h.75 ≤ Web shall be considere 3d ≥ c ≥ d 200 εw 200 Web satisfies the Serviceability crite Check for compressiom flange buckling requireme clause 8.5d 345 εf2 38.19 c > 3d Governing case : C.6.d/tw Case 1: Case 2: . (Zpyy) = 7.2(b) : Condition : When transverse stiffeners Checked to prevent the comp.11 mm ryy = 68.Design of Plate girders as per IS 800 : 2007 (Limit state method) : Beam Lateral status of the beam Mz = Vz = = 381 kN-m L/C = 3 77 kN = 13.7 m c = 1.5d .53 mm Plastic modulus of section about 3 z-z axis.d/tw = ≤ c ≥ 1.37E+05 mm 8.35 m c/d εflange = 2.0E+05 Mpa = 1.d/tw when c < 0. = 54 v. (Zpzz) = 2.act E At Unsupported My .5d .21E+07 mm4 Zzz = 2.80E+05 mm3 rzz = 208.1.d/tw Case 1: when 3d ≥ c ≥ d .d/tw when Governing case : c < 1. 04E+03N-mm = 0. fbd fbd λLT = = Mcr = fcr.45E+09N-mm λLT = 0.79 kN-m > 0.2.46 Therefore.73 131.5 ≤ 0. Zp .Member is within Slenderless limit Compression flange buckling requirement is Check for shear buckling before yielding : (As per clause 8. χLT 0. fcr.Vy Design strength.Vz = 77.Vdz (abt.40kN Vdz = 732.2 Zefy/Mcr) = otherwise. fbd = 381.49 = 1. Vdy (about minor axis) ϒmo = 1.per.00kN Design strength.652 < 1 Section is safe 197.1 Vn = Vp = Avzfyw/sqrt(3) = (d.1.act = Lz / 325 max. αLT . λLT = .46 For laterally unsupported beam : ФLT = . major axis)= Vn / ϒmo ϒmo = 1.18kN Factored design shear force.09 mm Safe in Deflection = . M = v.1 Vn = Vp = Avy fyw/sqrt(3) = Vp = Vdy = = Section is safe in shear Section is in low shear Section is in low shear Classification of the section based on shear capacity : About major axis z-z = Section is in low shear About minor axis y-y = Section is in low shear Check for design capacity of the section : Simply supported beam Md where.87 .tw. ФLT = 0.48 or λLT = = 381/585 0.= = Shear carrying capacity 254. βb = 1 = βb . LLT [1+ (1/20){(LLT/ry)/(hf/tf)} ] .1 E)/(LLT/ry) 2 2 3.75 67εw = 67 Shear buckling analysis is not required Factored design shear force.652 3240 mm = 0.00kNm 13. λLT = 0.b 2 0.49 Member shall be designed as laterally unsupp = 0.00kNm Mdyy = Mdyy > My Safe in bending Mdzz > Mz Mz/Mdzz 2 2 0.fyw)/sqrt(3) Vp = 805.03 kN-m Check for shear (at support) Factored maximum shear force = Check for Deflection : Unfactored moment abt z-z axis.5 1 .89 mm 26.5 (1.b Mcr = = χLT fy / ϒmo χLT .5[1 + αLT + ( 1 / {ФLT+ [ Ф LT -λ LT] } ≤ sqrt(1.67 Mdzz = 584.1) d/tw = 38. sqrt( βb Z pfy / Mcr) ≤ = (2EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0. . ≤ 345 εf2 ≤ 345 εf .6.1(b) se stiffeners are provided 200 εw ≤ 200 εw ≤ ≤ 270 εw be considered as unstiffened ≥ d bility criteria requirement : (As per stiffeners are provided uckling into the web.) : Beam B3 0 kN-m 0 kN 0.1.00 mm 0 kN ed = 16mm) 12mm) For E250 STEEL 16mm 465mm =d z 16mm=tf =b s per clause 8. 64kN 1259.0. ϒmo = 1.b) .tf.1 [1 + αLT + ( λLT .2) + λLT 2] sqrt( fy/fcr.uirement is satisfied = xis) 0.00kN Vn / ϒmo = (2b.hf = 481mm 3N-mm ally unsupported beam Mpa > 0kNm Safe in Bending 201 kN 732 kN Hence safe .67kN Section is safe in shear .fyw)/sqrt(3) 1385. . 6 94.300 465 16 12 200 457 20 16 497 unit wt L 125.2 157 125.38 2 526.38 2 631.38 8.5168 1 367.38 8.0691 8.264 1 481.6 nos 8.0053 . 269 0.651 8.683356 1007.5859 0.835 .998.
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