1. Steel Beam Design
Short Description
Steel Beam Design...
Description
STRUCTURA AL STEELWORK & TIMBER DESIGN N - ECS328 STEEL BEAM M DESIGN NOV13 – MAC14 M
C Clause
R Remarks Example of o Design fo or Laterally Restrained R Beam n is loaded with w uniform mly distributted loading g A steel beam section Gk = 8 kN//m and Qk = 6 kN/m. Beam B was ffully restrain ned and seat on 100mm m bearings at a each end. (Ignore sself-weight of steel beam)
E EN1990 1. DESIGN LOADING / ACTION Table A1.2(B) on, FEd = γG Gk + γQ Qk (eq 6.10) Combinattion of Actio = (1.35 x 8kN/m) + (1.5 x 6kN/m) = 19.8 kN/m Design Ben nding Mom ment, MEd (FEd x L2) / 8 = (19.8 kN/m k x 8m x 8m) / 8 = 158.4 kNm k
MEd = 158.4 kNm m
Design She ear Force, VEd (FEd x L) / 2
= (19.8 kN/m k x 8m) / 2 = 79.2 kN N
1
VEd = 79.2 kN
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
2. SELECTION OF SECTION Required Plastic Modulus , Wply = MEd / (fy / γMO ) 6.1 (1) Table 3.1
Partial factor for particular resistance, γM Resistance of cross-sections whatever the class, γMO = 1.0 Assume the nominal thk. of flange and web is less than 40mm and advance UKB S275 to be used, .: yield strength, fy = 275N/mm2 .: Required Plastic Modulus , Wply = MEd / (fy / γMO ) =158.4 x 106 / (275 / 1.0) =576cm3 From Table of Properties, select 356 x 171 x 51kg/m UKB provided 896cm3 of Plastic Modulus
356 x 171 x 51kg/m UKB
Section Properties for 356 x 171 x 51kg/m UKB Depth of Section, h Width of Section, b Thk. Flanges, tf Thk. of Web, tw Root Radius, r Depth between Fillet, d Ratio Local Buckling (flange), cf/tf Ratio of Local Buckling (web), cw/tw Second Moment Area, Iyy Second Moment Area, Izz Radius of Gyration, iy Radius of Gyration, iz Elastic Modulus, Wely Elastic Modulus, Welz Plastic Modulus, Wply Plastic Modulus, Wplz Area of Section, A
2
= 355 mm = 171.5 mm = 11.5 mm = 7.4 mm = 10.2 mm = 311.6 mm = 6.25 mm = 42.1mm = 14,100 cm4 = 968cm cm4 = 14.8 cm = 3.86 cm = 796 cm3 = 113 cm3 = 896 cm3 =174 cm3 =64.9 cm2
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
3. CLASIFICATION OF SECTION Table 3.1
Take steel grade as S275 and tf = 11.5 mm < 40 mm .: fy = 275 N/mm2
fy = 275 N/mm2
Web Table 5.2(1)
ε = 0.92 and cw/tw = 42.1 < 72 ε .: Web section was Class 1 Flange
Table 5.2(2)
ε = 0.92 and cf/tf = 6.25 < 9 ε .: Flange section was Class 1 Both section was Class 1
.: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR 6.2.6 6.2.6(1)
Shear The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17)
6.2.6(2) 6.2.6(3a)
Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 6400 – 2( 171.5)(11.5) + (7.5 + 2(10.2))11.5 = 2776.35 mm2 hw = h – 2tf hw = 355 – 2(11.5) = 332 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 332 x 7.5 = 2490 mm2 Since Av > nhwtw .: take Av = 2776.35 mm2
3
Av = 2776.35 mm2
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
.: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [ 2776.35 (275/√3)] / 1.0 = 440.8 kN
Vc,Rd =440.8 kN
VEd /Vc,Rd ≤ 1 (eq. 6.17) 79.2 / 440.8 = 0.18 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance 6.2.8(1)(2)(3)
Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 In this case, VEd < 0.5 Vc,Rd .: no reduction of moment resistance, Mc,Rd
6.2.5 6.2.5(1)
Bending The design value of the bending moment MEd at each crosssection shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12)
6.2.5(2)
For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (896x103 x 275)/ 1.0 = 246.4kNm
Mc,Rd = 246.4kNm
MEd /Mc,Rd =0.64 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
6.2.6(6)
5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN 1993-1-5, if: hw/tw > 72ε/n (eq. 6.22) hw/tw = 332 / 7.5 = 44.3 72ε/n = 72(0.92)/1.0 = 66.24 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling
BS EN 1993-15:2006 8(1)
6. RESISTANCE TO FLANGE INDUCED BUCKLING To prevent the compression flange buckling in the plane of the web, the following criterion should be met: hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw = 332 x 7.5 = 2490 mm2 Af = b x tf = 171.5 x 11.5 = 1972.25 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/275) (√(2490/1972.25)) = 257.41 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web
7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING) 5
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
6.2(1)
For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should be taken as: FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2)
6.4(1)
Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6 kF = 2 + 6 [(ss + c)/ hw]
Figure 6.1 (c) 6.5(3)
For type (c) in Figure 6.1, effective loaded length, ly should be taken as the smallest value obtained from the equation 6.11 or equation 6.12 : ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) ly = le + tf √m1 + m2 (eq. 6.12) le = kF E tw2/2fyw hw ≤ ss + c (eq.6.13)
6.5(1)
m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9)
kF = 2 + 6 [(ss + c)/ hw] (Figure 6.1 (c)) = 2 + 6 [(100 + 0)/ 332] 6
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
= 3.81 le = kF E tw2/2 fyw hw ≤ ss + c (eq.6.13)
kF = 3.81
le = kF E tw2/2 fyw hw = (3.81x 210 x103 x 7.52)/(2 x 275 x 332) = 246.47mm ss + c = 100 + 0 = 100mm Since le = kF E tw2/2 fyw hw ≥ ss + c .: take le = 100mm m1 = fyf bf / fyw tw (eq. 6.8) =(275 x 171.5) / (275 x 7.5) = 22.87
le = 100mm
m2 = 0.02 ( hw/tf)2 = 0.02 ( 332/11.5) 2 = 16.70
m1 = 22.87
ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) = 100 + 11.5 √ [(22.87/2) + (100/11.5)2 + 16.7] = 217.13mm
m2 = 16.70
ly = le + tf √m1 + m2 (eq. 6.12) = 100 + 11.5 √22.87 + 16.7 = 172.34mm Choose the lesser value from eq. 6.11 and eq. 6.12 .: ly = 172.34 ly = 172.34mm
Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (3.81) (210x103) ( 7.53 / 332) = 915, 024 N 7
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) = √ (( 172.34 x 7.5 x 275) / 915,024) = 0.62 Since λF = 0.62 > 0.5 .: value for m2 acceptable.
Fcr = 915, 024 N
λF = 0.62
XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.62 = 0.81 ≤ 1.0 .: take XF = 0.81 Leff = XF ly (eq. 6.2) = 0.81 x 172.34 = 139.6mm
XF = 0.81
FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (275 x 139.6 x 7.5)/1.0
Leff = 139.6mm
Since FRD > VEd .: section is adequate for resistance of web to transverse force
FRD = 287.93 kN
= 287.93 kN
8. DEFLECTION dactual = 5wL4 / 384 EI = (5x6x80004)/(384x210x103x141x106) = 10.8mm dallow = span/360 = 8000/360 = 22.22mm Since dallow > dactual .: section is satisfactory for deflection
Clause
Remarks Example of Design for Laterally Restrain Beam Support for a conveyor. Part of the support for a conveyor consists 8
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STRUCTURA AL STEELWORK & TIMBER DESIGN N - ECS328 STEEL BEAM M DESIGN NOV13 – MAC14 M
of a pair of o identical beams as shown in fiigure below w. Each be eam is connectted to a sta anchion at end A by a cleat and is supporrted on across beam at D by boltin ng through the conne ecting flang ges. Lateral resstraint is prrovided by y transverse e beams a at A, B and E connected d to rigid supports. s Th he loads sh hown are a at the ultimate limit state.
1. DESIGN LOADING / ACTION
MEd = 825 kNm
VEd = 375 kN
2. SELECTIO ON OF SECTTION Required Plastic P Mod dulus , Wply = MEd / (fy / γMO ) 9
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
6.1 (1)
Partial factor for particular resistance, γM Resistance of cross-sections whatever the class, γMO = 1.0
Table 3.1
Assume the nominal thk. of flange and web is less than 40mm and advance UKB S355 to be used, .: yield strength, fy = 355N/mm2 .: Required Plastic Modulus , Wply = MEd / (fy / γMO ) =825 x 106 / (355 / 1.0) =2324cm3 From Table of Properties, select 533 x 210 x 101 kg/m UKB provided 2610cm3 of Plastic Modulus
533 x 210 x 101 kg/m UKB
Section Properties for 533 x 210 x 101 kg/m UKB Depth of Section, h Width of Section, b Thk. Flanges, tf Thk. of Web, tw Root Radius, r Depth between Fillet, d Ratio Local Buckling (flange), cf/tf Ratio of Local Buckling (web), cw/tw Second Moment Area, Iyy Second Moment Area, Izz Radius of Gyration, iy Radius of Gyration, iz Elastic Modulus, Wely Elastic Modulus, Welz Plastic Modulus, Wply Plastic Modulus, Wplz Area of Section, A
= 536.7 mm = 210 mm = 17.4 mm = 10.8 mm = 12.7 mm = 476.5 mm = 4.99 mm = 44.1 mm = 61,500 cm4 = 2690 cm4 = 21.9 cm = 4.57 cm = 2290 cm3 = 256 cm3 = 2610 cm3 =399 cm3 =129 cm2
3. CLASIFICATION OF SECTION Table 3.1
Take steel grade as S355 and tf = 17.4 mm < 40 mm .: fy = 355 N/mm2 10
fy = 355 N/mm2 MAT
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
Web Table 5.2(1)
ε = 0.81 and cw/tw = 44.1 < 72 ε .: Web section was Class 1 Flange
Table 5.2(2)
ε = 0.81 and cf/tf = 4.99 < 9 ε .: Flange section was Class 1 Both section was Class 1
.: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR 6.2.6
Shear
6.2.6(1)
The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17)
6.2.6(2)
Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18)
6.2.6(3a)
Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 12,900 – 2( 210)(17.4) + (10.8 + 2(12.7))17.4 = 6221.8 mm2 hw = h – 2tf hw = 536.7 – 2(17.4) = 501.9 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 501.9 x 10.8 = 5420.52 mm2 Since Av > nhwtw .: take Av = 6221.8 mm2
Av = 6221.8 mm2
.: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [6221.8 (355/√3)] / 1.0 = 1275.22 kN
Vc,Rd =1275.22 kN
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
VEd /Vc,Rd ≤ 1 (eq. 6.17) 375 / 1275.22 = 0.29 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance 6.2.8(1)(2)(3)
Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 VEd = 375kN and 0.5 Vc,Rd = 0.5 x 1275.22 = 637.61kN In this case, VEd < 0.5 Vc,Rd .: no reduction of yield strength, fy for moment resistance, Mc,Rd
6.2.5
Bending
6.2.5(1)
The design value of the bending moment MEd at each crosssection shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12)
6.2.5(2)
For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (2.61 x 106 x 355) / 1.0 = 926.6kNm
Mc,Rd = 926.6kNm
MEd /Mc,Rd =0.89 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance
5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN 12
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
1993-1-5, if: 6.2.6(6)
hw/tw > 72ε/n (eq. 6.22) hw/tw = 501.9 / 10.8 = 46.47 72ε/n = 72(0.81)/1.0 = 58.32 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling
BS EN 1993-15:2006
6. RESISTANCE TO FLANGE INDUCED BUCKLING
8(1)
To prevent the compression flange buckling in the plane of the web, the following criterion should be met: hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw = 501.9 x 10.8 = 5,420.52 mm2 Af = b x tf = 210 x 17.4 = 3,654 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/355) (√(5,420.52/3,654)) = 216.15 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web
7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING) 6.2(1)
For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should be taken as: 13
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2) 6.4(1)
Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6
Figure 6.1 (c)
kF = 2 + 6 [(ss + c)/ hw]
6.5(2)
For type (a) and (b) in Figure 6.1, effective loaded length, ly should be taken as : ly = ss + 2tf ( 1 + √ m1+m2 ) (eq. 6.11) but ly ≤ distance between adjacent transverse stiffeners.
6.5(1)
m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9)
At D, two UB intersect which 533×210×101kg/m upper load carrying beam ABCDE and 610×229×140kg/m lower support beam at D. From Table of Properties, UB section for 610×229×140kg/m 14
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
Depth of Section, h Width of Section, b Thk. Flanges, tf Thk. of Web, tw Root Radius, r
617.2 mm 230.2 mm 22.1 mm 13.1 mm 12.7 mm
hw = (h−2tf) = 617.2 – (2×22.1) = 573 mm ss = 2tf + tw + (2 − √2)rb = (2 × 22.1) + 13.1 + (2 − √2)12.7 = 64.74 < hw = 573 mm
kF = 6 + 2 (hw/α)2 (Figure 6.1 (a)). Assume α is large value =6 m1 = fyf bf / fyw tw (eq. 6.8) =(355 x 210) / (355 x 10.8) = 19.4
kF = 6
m1 = 22.87
m2 = 0.02 ( = 0.02 (501.9/17.4) 2 = 16.64 hw/tf)2
m2 = 16.64
ly = ss + 2tf ( 1 + √ m1+m2 ) = 64.74 + 2(17.4) (1+√19.4 + 17.4) = 310.65 mm
Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (6) (210x103) ( 10.83 / 501.9) = 2,846.2 x 103 N
15
ly = 310.65 mm
Fcr = 2,846.2x103 N
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) = √ ((310.65 x 10.8 x 355) / 2,846.2 x 103) = 0.65 XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.65 = 0.77 ≤ 1.0 .: take XF = 0.77
λF = 0.65
XF = 0.77
Leff = XF ly (eq. 6.2) = 0.77 x 310.65 = 239.2 mm
Leff = 239.2 mm
FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (355 x 239.2 x 10.8)/1.0
= 917.1 kN
FRD = 917.1 kN
Since FRD > VEd at D .: section is adequate for resistance of web to transverse force
Additional notes on resistance of web to transverse force (web buckling) BS EN 1993-1-5distinguishes between two types of forces applied through a flange to the web: 16
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STRUCTURA AL STEELWORK & TIMBER DESIGN N - ECS328 STEEL BEAM M DESIGN NOV13 – MAC14 M
(a) Forces resisted by y shear in th he web for l oading typ pes (a) and (c) (b) Forces transferred d through th he web dire ectly to the e other flang ge for loading g type (b)
Figure 6.1: Buckling Co oefficient forr different typ pes of load a application (BS EN 1993-1-5) 1
For loading g types (a) and (c) the e web is like ely to fail ass a result off (i) crushing g of the web close to the t flange accompan nied by yielding off the flange e, the comb bined effec ct sometime es referred tto as web cru ushing (ii) Localize ed buckling g and crush hing of the w web benea ath the flange, the e combined d effect som metimes re ferred to ass web crippling.
For loading g type (b) the t web is likely to fail as a result of (i) web cru ushing (ii) Buckling g of the we eb over most of the de epth of the member
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STRUCTURA AL STEELWORK & TIMBER DESIGN N - ECS328 STEEL BEAM M DESIGN NOV13 – MAC14 M
Clause 6.3 3 (1) mentio oned that le ength of stifffness beariing ss should d ed load is e be taken as a the dista ance over which w applie effectively distributed d at slope of o 1:1 (BS EN N 1993-1-5, FFigure 6.2). However ss should nott be taken as a larger than hw
Simply sup pported late erally restra ained beam m The beam shown in figure f below w is fully latterally restra ained along g its length an nd has bea aring lengtths of 50 mm at the e un-stiffen ned 18 8
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STRUCTURA AL STEELWORK & TIMBER DESIGN N - ECS328 STEEL BEAM M DESIGN NOV13 – MAC14 M
supports and 75 mm m under the conc centrated load. Che eck resistance of 533 x 210 x 92 UKB U using SS275 steel grade for the loading off uniformly distributed d load (UDL) was gk = 15kN/m and d qk = 30kN/m and conce entrated loa ad Gk = 40kkN and Qk = 50kN
1. DESIGN LOADING / ACTION Combinattion of Actio on UDL, FEd = γG Gk + γQ Qk (eq 6.10) = (1.35 x 15 5kN/m) + (1 1.5 x 30kN/m m) = 62.5 kN/m m = γG Gk + γQ Qk (eq 6.10) = (1.35 x 40 0kN) + (1.5 x 50N) = 129 kN
Combinattion of Actio on PL, FEd
DESIGN BE ENDING MO OMENT, MEd (FEd1 x L2) / 8 = (62.5 kN N/m x 6.5m m x 6.5m) / 8 = 330.1 kNm (FEd2 x L) / 4 = (129 kN N x 6.5m) / 4 = 209.6 kNm k 30.1 kNm + 209.6 2 kNm ∑MEd = 33 = 53 39.7 kNm DESIGN SH HEAR FORCE, VEd (FEd1 x L) / 2 FEd2 / 2
= (62.5 kN/m k x 6.5m m) / 2 = 203.13 3 kN
= 129 kN / 2 = 64.5 kN
∑FEd = 267.63 kN ON OF SECTTION 2. SELECTIO Section Pro operties forr 533 x 210 x 92kg/m U KB
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
Depth of Section, h Width of Section, b Thk. Flanges, tf Thk. of Web, tw Root Radius, r Depth between Fillet, d Ratio Local Buckling (flange), cf/tf Ratio of Local Buckling (web), cw/tw Second Moment Area, Iyy Second Moment Area, Izz Radius of Gyration, iy Radius of Gyration, iz Elastic Modulus, Wely Elastic Modulus, Welz Plastic Modulus, Wply Plastic Modulus, Wplz Area of Section, A
= 533.1 mm = 209.3 mm = 15.6 mm = 10.1 mm = 12.7 mm = 476.5 mm = 5.57 mm = 47.2mm = 55,200 cm4 = 2390 cm4 = 21.7 cm = 4.51 cm = 2070 cm3 = 228 cm3 = 2360 cm3 =355 cm3 =117 cm2
3. CLASIFICATION OF SECTION Take steel grade as S275 and tf = 15.6 mm < 40 mm .: fy = 275 N/mm2 Web ε = 0.92 and cw/tw = 47.2 < 72 ε .: Web section was Class 1 Flange ε = 0.92 and cf/tf = 5.57 < 9 ε .: Flange section was Class 1 .: Both section was Class 1
4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear 20
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
The design value of the shear force VEd at each cross section shall satisfy: VEd /Vc,Rd ≤ 1 (eq. 6.17) Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) Shear area, Av for rolled I section Av = A – 2btf + (tw + 2r) tf but not less than nhwtw Av = 11,700 – 2(209.3)(15.6) + (10.1 + 2(12.7))15.6 = 5723.64 mm2 hw = h – 2tf hw = 533.1 – 2(15.6) = 501.9 mm n may be conservatively taken equal to 1.0. .: nhwtw = 1.0 x 501.9 x 10.1 = 5069.2 mm2 Since Av > nhwtw .: take Av = 5723.64 mm2 .: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [ 5723.64 (275/√3)] / 1.0 = 908.75 kN VEd /Vc,Rd ≤ 1 (eq. 6.17) 267.63 / 908.75= 0.29 Since VEd /Vc,Rd ≤ 1 .: Section was satisfactory for shear resistance
Bending & Shear If VEd > 0.5 Vc,Rd .: moment resistance, Mc,Rd should be taken as the design resistance of the cross-section, calculated using a reduced 21
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
yield strength, (1 – ρ) fy where ρ = ((2VEd/Vpl,Rd) - 1)2 In this case, VEd < 0.5 Vc,Rd .: no reduction of moment resistance, Mc,Rd Bending The design value of the bending moment MEd at each crosssection shall satisfy: MEd /Mc,Rd ≤ 1 (eq. 6.12) For Class 1 section, Mc,Rd = Mpl,Rd = (Wpl x fy)/ γMO (eq. 6.13) .: Mc,Rd = (2360x103 x 275)/ 1.0 = 649kNm MEd /Mc,Rd = 539.7 kNm/649 kNm = 0.83 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance 5. RESISTANCE TO SHEAR BUCKLING In addition the shear buckling resistance for webs without intermediate stiffeners should be according to section 5 of EN 1993-1-5, if: hw/tw > 72ε/n (eq. 6.22) hw/tw = 501.9/10.1 = 49.7 72ε/n = 72(0.92)/1.0 = 66.24 Since hw/tw < 72ε/n .: no further check on resistance for shear buckling
6. RESISTANCE TO FLANGE INDUCED BUCKLING To prevent the compression flange buckling in the plane of the web, the following criterion should be met: 22
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
hw/tw ≤ k (E/fy)( √(Aw/Afc)) (eq.8.1) Aw = hw x tw = 501.9x10.1 = 5069.2 mm2 Af = b x tf = 209.3 x15.6 = 3265.08 mm2 For Class 1, plastic rotation utilize, hence k = 0.3 k (E/fy)( √(Aw/Afc) = 0.3 (210x103/275) (√(5069.2/3265.08)) = 285.45 Since hw/tw ≤ k (E/fy)( √(Aw/Afc)) .: section is prevent the compression flange buckling in the plane of the web
7. RESISTANCE OF WEB TO TRANSVERSE FORCE (WEB BUCKLING) For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should 23
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
be taken as: FRD = (fyw x Leff x tw)/γm (eq. 6.1) Effective length for resistance to transverse forces, Leff Leff = XF ly (eq. 6.2) Reduction factor, XF XF = 0.5 / λF ≤ 1.0 (eq. 6.3) λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) For webs without longitudinal stiffeners kF should be obtained from Figure 6.1. For webs with longitudinal stiffeners kF may calculated from the equation 6.6 kF = 2 + 6 [(ss + c)/ hw] For type (c) in Figure 6.1, effective loaded length, ly should be taken as the smallest value obtained from the equation 6.11 or equation 6.12 : ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) ly = le + tf √m1 + m2 (eq. 6.12) le = kF E tw2/2fyw hw ≤ ss + c (eq.6.13) m1 = fyf bf / fyw tw (eq. 6.8) m2 = 0.02 ( hw/tf)2 if λF > 0.5 (eq. 6.9) m2 = 0 if λF ≤ 0.5 (eq. 6.9)
kF = 2 + 6 [(ss + c)/ hw] (Figure 6.1 (c)) = 2 + 6 [(50 + 0)/ 501.9] = 2.6 le = kF E tw2/2 fyw hw ≤ ss + c (eq.6.13) 24
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
le = kF E tw2/2 fyw hw = (2.6 x 210 x103 x 10.12)/(2 x 275 x 501.9) = 201.77 mm ss + c = 50 + 0 = 50mm Since le = kF E tw2/2 fyw hw ≥ ss + c .: take le = 50mm m1 = fyf bf / fyw tw (eq. 6.8) =(275 x 209.3) / (275 x 10.1) = 20.72 m2 = 0.02 ( hw/tf)2 = 0.02 (501.9/15.6) 2 = 20.7 ly = le + tf √ [(m1/2) + (le/tf)2 + m2] (eq. 6.11) = 50 + 15.6 √ [(20.72/2) + (50/15.6)2 + 20.7] = 150.28 mm ly = le + tf √m1 + m2 (eq. 6.12) = 50 + 15.6 √20.72 + 20.7 = 150.4 mm Choose the lesser value from eq. 6.11 and eq. 6.12 .: ly = 150.29
Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (2.6) (210x103) (10.13 / 501.9) = 1008.75 x 103 N λF = √ ( ly tw fyw / Fcr ) (eq. 6.4) 25
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STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
= √ ((150.29 x 10.1 x 275) / 1008.75 x 103) = 0.64 XF = 0.5 / λF ≤ 1.0 (eq. 6.3) = 0.5 / 0.64 = 0.78 ≤ 1.0 .: take XF = 0.78 Leff = XF ly (eq. 6.2) = 0.78 x 150.29 = 117.23 mm FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (275 x 117.23 x 10.1)/1.0
= 325.61 kN VEd / FRD = 267.63 / 325.61 = 0.82 Since FRD > VEd .: section is adequate for resistance of web to transverse force
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