1. steel beam design
DESCRIPTION
Steel Beam DesignTRANSCRIPT
C
ETab
STRUCTURASTEEL BEAMNOV13 – M
Clause
EN1990 le A1.2(B)
AL STEELWORK &M DESIGN MAC14
Example o A steel beGk = 8 kN/on 100mmbeam)
1. DESIGN Combinat = ( = 1 Design Ben (FEd x L2) / 8 Design She (FEd x L) / 2
& TIMBER DESIGN
of Design fo
am section/m and Qk
m bearings a
LOADING /
tion of Actio1.35 x 8kN/9.8 kN/m
nding Mom
8 = (19.8 k = 158.4 k
ear Force, V
2 = (19.8 k = 79.2 kN
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or Laterally R
n is loaded w= 6 kN/m. B
at each en
/ ACTION
on, FEd = γG
m) + (1.5 x
ment, MEd
kN/m x 8m xkNm
VEd
kN/m x 8m) N
Restrained
with uniformBeam was fd. (Ignore s
G Gk + γQ Qk
6kN/m)
x 8m) / 8
/ 2
Beam
mly distributfully restrainself-weight
k (eq 6.10)
ted loadingned and se of steel
MAT
R
g at
MEd = VEd =
Remarks
= 158.4 kNm
= 79.2 kN
m
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.1 (1)
Table 3.1
2. SELECTION OF SECTION Required Plastic Modulus , Wply = MEd / (fy / γMO ) 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 Section Properties for 356 x 171 x 51kg/m UKB Depth of Section, h = 355 mm Width of Section, b = 171.5 mm Thk. Flanges, tf = 11.5 mm Thk. of Web, tw = 7.4 mm Root Radius, r = 10.2 mm Depth between Fillet, d = 311.6 mm Ratio Local Buckling (flange), cf/tf = 6.25 mm Ratio of Local Buckling (web), cw/tw = 42.1mm Second Moment Area, Iyy = 14,100 cm4 Second Moment Area, Izz = 968cm cm4 Radius of Gyration, iy = 14.8 cm Radius of Gyration, iz = 3.86 cm Elastic Modulus, Wely = 796 cm3 Elastic Modulus, Welz = 113 cm3 Plastic Modulus, Wply = 896 cm3 Plastic Modulus, Wplz =174 cm3 Area of Section, A =64.9 cm2
356 x 171 x 51kg/m UKB
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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Table 3.1
Table 5.2(1)
Table 5.2(2)
6.2.6
6.2.6(1)
6.2.6(2)
6.2.6(3a)
3. CLASIFICATION OF SECTION Take steel grade as S275 and tf = 11.5 mm < 40 mm .: fy = 275 N/mm2 Web ε = 0.92 and cw/tw = 42.1 < 72 ε .: Web section was Class 1 Flange ε = 0.92 and cf/tf = 6.25 < 9 ε .: Flange section was Class 1 .: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear 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 = 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
fy = 275 N/mm2 Both section was Class 1 Av = 2776.35 mm2
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.2.8(1)(2)(3)
6.2.5
6.2.5(1)
6.2.5(2)
.: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [ 2776.35 (275/√3)] / 1.0 = 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 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 Bending The design value of the bending moment MEd at each cross-section 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 = (896x103 x 275)/ 1.0 = 246.4kNm MEd /Mc,Rd =0.64 Since MEd /Mc,Rd < 1 .: section was satisfactory for moment resistance
Vc,Rd =440.8 kN Mc,Rd = 246.4kNm
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.2.6(6)
BS EN 1993-1-5:2006
8(1)
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 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)
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.2(1)
6.4(1)
Figure 6.1 (c)
6.5(3)
6.5(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) 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 [(100 + 0)/ 332]
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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= 3.81 le = kF E tw2/2 fyw hw ≤ ss + c (eq.6.13) 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 m2 = 0.02 ( hw/tf)2 = 0.02 ( 332/11.5) 2 = 16.70 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 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 Fcr = 0.9 kF E (tw3 / hw) (eq. 6.5) = 0.9 (3.81) (210x103) ( 7.53 / 332) = 915, 024 N
kF = 3.81 le = 100mm m1 = 22.87 m2 = 16.70 ly = 172.34mm
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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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. 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 FRD = (fyw x Leff x tw)/γm (eq. 6.1) = (275 x 139.6 x 7.5)/1.0 = 287.93 kN Since FRD > VEd .: section is adequate for resistance of web to transverse force 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
Fcr = 915, 024 N λF = 0.62 XF = 0.81 Leff = 139.6mm FRD = 287.93 kN
Clause Remarks
Example of Design for Laterally Restrain Beam Support for a conveyor. Part of the support for a conveyor consists
STRUCTURASTEEL BEAMNOV13 – M
AL STEELWORK &M DESIGN MAC14
of a pair ois connecton across Lateral resconnectedlimit state.
1. DESIGN
2. SELECTIO Required P
& TIMBER DESIGN
of identical ted to a sta beam at Dstraint is prd to rigid s
LOADING /
ON OF SECT
Plastic Mod
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beams as anchion at D by boltinrovided bysupports. Th
/ ACTION
TION
dulus , Wply =
9
shown in fi end A by
ng through y transversehe loads sh
= MEd / (fy /
igure belowa cleat an the connee beams ahown are a
γMO )
w. Each bed is suppor
ecting flangat A, B anat the ultim
MAT
eam rted ges. d E ate
MEd = VEd =
= 825 kNm
= 375 kN
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.1 (1) Table 3.1 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 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 Section Properties for 533 x 210 x 101 kg/m UKB Depth of Section, h = 536.7 mm Width of Section, b = 210 mm Thk. Flanges, tf = 17.4 mm Thk. of Web, tw = 10.8 mm Root Radius, r = 12.7 mm Depth between Fillet, d = 476.5 mm Ratio Local Buckling (flange), cf/tf = 4.99 mm Ratio of Local Buckling (web), cw/tw = 44.1 mm Second Moment Area, Iyy = 61,500 cm4 Second Moment Area, Izz = 2690 cm4 Radius of Gyration, iy = 21.9 cm Radius of Gyration, iz = 4.57 cm Elastic Modulus, Wely = 2290 cm3 Elastic Modulus, Welz = 256 cm3 Plastic Modulus, Wply = 2610 cm3 Plastic Modulus, Wplz =399 cm3 Area of Section, A =129 cm2
3. CLASIFICATION OF SECTION Take steel grade as S355 and tf = 17.4 mm < 40 mm .: fy = 355 N/mm2
533 x 210 x 101 kg/m UKB fy = 355 N/mm2
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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Table 5.2(1) Table 5.2(2) 6.2.6 6.2.6(1) 6.2.6(2) 6.2.6(3a)
Web ε = 0.81 and cw/tw = 44.1 < 72 ε .: Web section was Class 1 Flange ε = 0.81 and cf/tf = 4.99 < 9 ε .: Flange section was Class 1 .: Both section was Class 1 4.RESISTANCE OF CROSS- SECTION SUBJECT TO BENDING AND SHEAR Shear 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 = 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 .: Vc,Rd = Vpl, Rd = [ Av (fy /√3) ] / γMO (eq 6.18) = [6221.8 (355/√3)] / 1.0 = 1275.22 kN
Both section was Class 1 Av = 6221.8 mm2 Vc,Rd =1275.22 kN
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.2.8(1)(2)(3) 6.2.5 6.2.5(1) 6.2.5(2)
VEd /Vc,Rd ≤ 1 (eq. 6.17) 375 / 1275.22 = 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 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 Bending The design value of the bending moment MEd at each cross-section 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 = (2.61 x 106 x 355) / 1.0 = 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
Mc,Rd = 926.6kNm
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.2.6(6) BS EN 1993-1-5:2006 8(1) 6.2(1)
1993-1-5, if: 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 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
= 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) For un-stiffened or stiffened webs the design resistance to local buckling under transverse forces should be taken as:
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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6.4(1) Figure 6.1 (c) 6.5(2) 6.5(1)
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 (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. 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
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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Depth of Section, h 617.2 mm Width of Section, b 230.2 mm Thk. Flanges, tf 22.1 mm Thk. of Web, tw 13.1 mm Root Radius, r 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 m2 = 0.02 ( hw/tf)2 = 0.02 (501.9/17.4) 2 = 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
kF = 6 m1 = 22.87 m2 = 16.64 ly = 310.65 mm Fcr = 2,846.2x103 N
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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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 Leff = XF ly (eq. 6.2) = 0.77 x 310.65 = 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 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:
λF = 0.65 XF = 0.77 Leff = 239.2 mm FRD = 917.1 kN
STRUCTURASTEEL BEAMNOV13 – M
AL STEELWORK &M DESIGN MAC14
(a) Forces (b) Forces for loading
Figure 6.1:
For loading(i) crushingyielding ofas web cru(ii) Localizeflange, thecrippling.
For loading(i) web cru(ii) Buckling
& TIMBER DESIGN
resisted by transferredg type (b)
Buckling Co
g types (a) g of the wef the flangeushing ed bucklinge combined
g type (b) tushing g of the we
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17
y shear in thd through th
oefficient forEN 1
and (c) theb close to t
e, the comb
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the web is l
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STRUCTURASTEEL BEAMNOV13 – M
AL STEELWORK &M DESIGN MAC14
Clause 6.3be taken adistributedshould not
Simply sup The beamlength an
& TIMBER DESIGN
3 (1) mentioas the dista
d at slope ot be taken a
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shown in fnd has bea
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oned that leance over wof 1:1 (BS ENas larger th
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STRUCTURASTEEL BEAMNOV13 – M
AL STEELWORK &M DESIGN MAC14
supports resistance loading of= 30kN/m
1. DESIGN Combinat Combinat
DESIGN BE (FEd1 x L2) / (FEd2 x L) / 4 ∑MEd = 33 = 53 DESIGN SH (FEd1 x L) / 2 FEd2 / 2 = = ∑FEd = 2672. SELECTIO Section Pro
& TIMBER DESIGN
and 75 m of 533 x f uniformly dand conce
LOADING /
tion of Actio
tion of Actio
ENDING MO
8 = (62.5 kN = 330.1 k
4 = (129 kN = 209.6 k
30.1 kNm + 239.7 kNm
HEAR FORC
2 = (62.5 k = 203.13
129 kN / 2 64.5 kN
.63 kN ON OF SECT
operties for
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mm under 210 x 92 Udistributed entrated loa
/ ACTION
on UDL, FEd
on PL, FEd
OMENT, MEd
N/m x 6.5mNm
N x 6.5m) / kNm
209.6 kNm
E, VEd
kN/m x 6.5m3 kN
TION
r 533 x 210 x
9
the concUKB using Sload (UDL) ad Gk = 40k
= γG Gk + = (1.35 x 15 = 62.5 kN/m
= γG Gk + = (1.35 x 40 = 129 kN
m x 6.5m) / 8
4
m) / 2
x 92kg/m U
centrated S275 steel was gk = 1
kN and Qk =
γQ Qk (eq 5kN/m) + (1m
γQ Qk (eq 0kN) + (1.5 x
8
KB
load. Chegrade for 5kN/m and
= 50kN
6.10)
1.5 x 30kN/m
6.10)
x 50N)
MAT
eck the
d qk
m)
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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Depth of Section, h = 533.1 mm Width of Section, b = 209.3 mm Thk. Flanges, tf = 15.6 mm Thk. of Web, tw = 10.1 mm Root Radius, r = 12.7 mm Depth between Fillet, d = 476.5 mm Ratio Local Buckling (flange), cf/tf = 5.57 mm Ratio of Local Buckling (web), cw/tw = 47.2mm Second Moment Area, Iyy = 55,200 cm4 Second Moment Area, Izz = 2390 cm4 Radius of Gyration, iy = 21.7 cm Radius of Gyration, iz = 4.51 cm Elastic Modulus, Wely = 2070 cm3 Elastic Modulus, Welz = 228 cm3 Plastic Modulus, Wply = 2360 cm3 Plastic Modulus, Wplz =355 cm3 Area of Section, A =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
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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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
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
22 MAT
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 cross-section 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:
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
23 MAT
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
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
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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)
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
25 MAT
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)
STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328 STEEL BEAM DESIGN NOV13 – MAC14
26 MAT
= √ ((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