British Standard vs Eurocode 3 – Steel Stee l Building Design A comparison between BS 5950 and Eurocode 3. BS 5950 Terminology Force Capacity Mc Design strength p y Dead load Live load Wind load Symbol Elastic Modulus Plastic Modulus Radius of Gyration Torsion constant Warping constant Changes in load factor 1.4Gk + 1.6Qk
Classification: ε = (
275
p y
Eurocode 3 Action Resistance Mc,Rd Yield strength f y Permanent load Variable load Another variable load Z S r J H
Wel W pl i It Iw 1.35Gk + 1.5 Qk
) 0.5
Moment Resistance Class 1 and 2: M c = pyS
Tu Trung Nguyen – MSc. Structural Engineering
Classification: ε = (
235
f y
Class 1 and 2: M c,Rd =
) 0.5
f yW pl γ M 1
British Standard vs Eurocode 3 – Steel Stee l Building Design Class 3 semi-compact: Mc = pyZ or Mc = pySeff
Class 3: M c,Rd =
Class 4: slender Mc = pyZeff
Class 4: M c,Rd =
Low shear Fv < 60% Pv Shear Resistance Pv = 0.6pyA
γ M 1 f yW eff ,min γ M 1
Low Shear VEd < 50% V pl,Rd
Av VEd =
Shear Area Av = tD Shear bucking d/t > 70 ε Deflection - Serviceability LS Imposed load only
f yW el ,min
f y 3
γ M 1
Shear Area Av = A – 2bt f – (tw + 2r)tf ≈ 1.04tD = htw Shear bucking hw/tw > 72 ε Permanent action, δ 1 Variable action, δ 2 Pre-camber, δ 0
Span/360 – brittle Span/200 - general
δ max < L/250 δ 2 < L/350 – brittle
δ 3 < L/300 – general Compression Members Pc = Ag pc from table 23, 24 BS 5950 pc is a function of λ
N b,Rd =
χβ a Af y γ M 1
λ , non χ : reduction factor depends on λ , dimensional slenderness 1 ≤1 λ = 2 Φ + Φ − λ 2
Φ = 0.5[1 + a(λ − 0.2) + λ ] λ = λ =
Ncr =
β a Af y N cr π 2 EI eff 2
=
λ λ 1
E f y
; Ieff = 0.5 ho2 A + 2 μ I
L cr from table 6.8
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; λ 1 = π
British Standard vs Eurocode 3 – Steel Stee l Building Design
LTB Mx < M b/mLT and Mx < Mcx M b = p b x modulus p b from λ LT from table 16
λ LT = uνλ β w (see below) λ =
L E
1.0 for UB, UC same approach as for compression χ LT β aW pl , y f y M b,Rd = γ M 1 λ LT =
r y
1
Φ LT + Φ
2 LT
− λ LT
≤1
r y = the radius of gyration about the minor axis Class 1 and 2: β w = 1.0, M b = p bSx
Φ LT = 0.5[1 + a LT (λ LT − 0.2) + λ LT ]
Class 3 semi-compact: Mc = pyZ => β w = Zx/Sx
λ LT =
Mc = pySeff => β w = Sx,eff /Sx Class 4 slender cross-sections: β w = Zx,eff /Sx mLT from Table 18 Equal flange: ν =
1
[1 + 0.05(λ / x) 2 ]0.25 u = 0.9 , x = D/T equal flange, I and H
2
Mcr = C 1
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M cr
=
λ λ 1
π 2 EI Z I w L2
; λ 1 = π
I z
+
E f y
L2 GI t π 2 EI z
C1 results from bending diagram below aLT = 0.34 for rolled UC section aLT = 0.49 for rolled UB section Approx:
•
Quick determine λ LT , using λ LT = uνλ β w where λ =
W pl . y f y
Mcr =
0.9 Ah
⎛ L ⎞ ⎜⎜ ⎟⎟ ⎝ i z ⎠
2
1+
1 Lt f 2 ( ) 20 i z h
L E , conservative: u = 0.9, v=1, β w =1 r y
British Standard vs Eurocode 3 – Steel Building Design
Eq. 5.14 EC3 cl.5.3.3(1,2,3) NEd = MEd/h Deflection:
am L
0.5(1 +
1
) 500 m m: number of members to be restrained
e0 =
; am =
1. Load factor yM0 = 1,00,
yM1 = 1,00,
yM2 = 1,25
2. Tension: N Ed N t , Rd
≤ 1 ; Nt,Rd = min (
Af y 0.9 A Net f u ) ; y M 0 y M 2
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EC3:2005 cl.6.2.3 (1,2)
British Standard vs Eurocode 3 – Steel Stee l Building Design Where: f u = 1.5f y (cl.3.2.2 (1)) Anet = A – (number of bolts)x(Diameter of bolts)
cl.6.2.2.2
3.Compression N Ed N c , Rd
≤1
where
Nc,Rd =
Af y y M 0
for class 1,2 and 3
cl.6.2.4
4. Bending moment M Ed M c , Rd
≤1
cl.6.2.5
Class 1, 2
Class 3
Mc,Rd = M pl , Rd =
W pl f y
Mc,Rd = M el , Rd =
y M 0
Shear
V pl , Rd
W el , min f y y M 0
Mc,Rd =
cl.6.2.6
Av f y V Ed
Class 4
≤1
where
V pl,Rd =
3
y M 0
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W el ,eff f y y M 0
British Standard vs Eurocode 3 – Steel Stee l Building Design
5.Bending and Shear Cl.6.2.8 where ρ = (
N = (1 − ρ ) f y [W pl , y − My,V,Rd =
ρ Aw 4t w
y M 0
2V Ed
V pl , Rd
− 1) 2
2
] f y
≤ M c , Rd Where Aw =hwtw
6.Bending and axial force Class 1, and 2
cl.6.2.9.1
MEd < M N,Rd + For only rectangular section N M N,Rd = M pl,Rd [1- ( Ed ) 2 ] N pl , Rd + For doubly symmetrical I and H section or other flange section to resistance about y-y NEd < 0.25N pl,Rd
Tu Trung Nguyen – MSc. Structural Engineering
British Standard vs Eurocode 3 – Steel Stee l Building Design NEd <
0.5hwt w f y y M 0
Tu Trung Nguyen – MSc. Structural Engineering
British Standard vs Eurocode 3 – Steel Stee l Building Design
7.Guide for design curved beam to Eurocode 3 Determine out of plane bending stress: Direct stress σ 1 =
M W el
+
N A
R: radius of the section, 1 B = (b − t − 2r ) 2 Reduce design strength σ 2 2 ⎡ 2 ⎤ f yd ) − 3τ 2 ⎥ yd = ⎢ f y − 3( 2 ⎣ ⎦
Out-of-plane bending stress: σ 2 =
3σ 1 B 2
RT T: Thickness of flange. B is outstand of the flange
0.5
+
σ 2
2 - Check axial force ( Please Please see Section 3 ) - Check Bending capacity ( please see Section 4 ) - Check section capacity under axial load and bending moment at Section 6 ( more obviously ) Briefly with class 1 and 2: I and H section MEd < M N,z,Rd n=
N Ed N pl , Rd
a = (A-2btf )/A < 0.5
+n
M N,Rd = M pl,z,Rd
+n>a
M N,z,Rd = M pl,z,Rd [1+ (
n−a
1− n ) 2 ] = M pl,z,Rd ( ) 1− a 1 − 0.5a
- Check member buckling ( cl.6.3.3)
Tu Trung Nguyen – MSc. Structural Engineering
British Standard vs Eurocode 3 – Steel Stee l Building Design
Reference: - Eurocode 3:1993-1-1:2005: General rules for steel building design - Bristish Standards 5950-1:2000: Code of practice for design rolled and welded sections - Charles King and David Brown (2001) Design for curved beam, SCI
Tu Trung Nguyen – MSc. Structural Engineering
