British Standard vs Eurocode 3 – Steel Building Design PDF

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British Standard vs Eurocode 3 – Steel Building Design A comparison between BS 5950 and Eurocode 3. BS 5950 Eurocode 3 Terminology Force Action Capacity Mc Resistance Mc,Rd Design strength py Yield strength fy Dead load Permanent load Live load Variable load Wind load Another variable load Symbol El...

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British Standard vs Eurocode 3 – Steel Building Design Chung LipLeong

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British Standard vs Eurocode 3 – Steel Building Design A comparison between BS 5950 and Eurocode 3. BS 5950 Terminology Force Capacity Mc Design strength py 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 = (

Eurocode 3 Action Resistance Mc,Rd Yield strength fy Permanent load Variable load Another variable load Z S r J H

275 0.5 ) py

Moment Resistance Class 1 and 2: Mc = pyS

Tu Trung Nguyen – MSc. Structural Engineering

Wel Wpl i It Iw 1.35Gk + 1.5 Qk

Classification: ε = (

235 0.5 ) fy

Class 1 and 2: Mc,Rd =

f yW pl

γ M1

British Standard vs Eurocode 3 – Steel Building Design Class 3 semi-compact: Mc = pyZ or Mc = pySeff

Class 3: Mc,Rd =

Class 4: slender Mc = pyZeff

Class 4: Mc,Rd =

Low shear Fv < 60% Pv Shear Resistance Pv = 0.6pyA

Span/360 – brittle Span/200 - general

Compression Members Pc = Agpc from table 23, 24 BS 5950 pc is a function of λ

γ M1

f yWeff ,min

γ M1

Low Shear VEd < 50% Vpl,Rd Av

fy

γ M1

VEd = Shear Area Av = tD Shear bucking d/t > 70 ε Deflection - Serviceability LS Imposed load only

f yWel ,min

3

Shear Area Av = A – 2btf – (tw + 2r)tf ≈ 1.04tD = htw Shear bucking hw/tw > 72 ε

Permanent action, δ 1 Variable action, δ 2 Pre-camber, δ 0

δ max < L/250 δ 2 < L/350 – brittle δ 3 < L/300 – general

Nb,Rd =

χβ a Af y γ M1

χ : reduction factor depends on λ , non-

dimensional slenderness 1 ≤1 λ= Φ + Φ2 − λ

Φ = 0.5[1 + a(λ − 0.2) + λ ]

λ=

β a Af y N cr

π EI eff

2

=

λ ; λ =π λ1 1

E fy

; Ieff = 0.5 ho2 A + 2 μI L2 cr μ from table 6.8 2

Ncr =

Tu Trung Nguyen – MSc. Structural Engineering

British Standard vs Eurocode 3 – Steel Building Design

LTB Mx < Mb/mLT and Mx < Mcx Mb = pb x modulus pb from λLT from table 16

λLT = uνλ β w (see below) L λ= E ry ry = the radius of gyration about the minor axis Class 1 and 2: β w = 1.0, Mb = pbSx Class 3 semi-compact: Mc = pyZ => β w = Zx/Sx Mc = pySeff => β w = Sx,eff/Sx Class 4 slender cross-sections: β w = Zx,eff/Sx mLT from Table 18 1 Equal flange: ν = [1 + 0.05(λ / x) 2 ]0.25 u = 0.9 , x = D/T equal flange, I and H

Quick determine λLT , using λLT = uνλ β w where λ =

Tu Trung Nguyen – MSc. Structural Engineering

1.0 for UB, UC same approach as for compression χ LT β aW pl , y f y Mb,Rd =

λLT =

γ M1 1

Φ LT + Φ LT − λLT 2

≤1

Φ LT = 0.5[1 + aLT (λLT − 0.2) + λLT ] 2

λ LT =

W pl . y f y M cr

π 2 EI Z

=

λ ; λ =π λ1 1

E fy

I w L2 GI t + L2 I z π 2 EI z C1 results from bending diagram below aLT = 0.34 for rolled UC section aLT = 0.49 for rolled UB section

Mcr = C1

Approx:

•

Mcr =

0.9 Ah ⎛L⎞ ⎜⎜ ⎟⎟ ⎝ iz ⎠

2

1+

1 Lt f 2 ( ) 20 iz h

LE , conservative: u = 0.9, v=1, β w =1 ry

British Standard vs Eurocode 3 – Steel Building Design

Eq. 5.14 EC3 cl.5.3.3(1,2,3) NEd = MEd/h Deflection: a L 1 e0 = m ; am = 0.5(1 + ) m 500 m: number of members to be restrained

1. Load factor yM0 = 1,00,

yM1 = 1,00,

yM2 = 1,25

2. Tension:

Af y 0.9 ANet f u N Ed ; ) ≤ 1 ; Nt,Rd = min ( yM 0 yM 2 N t , Rd

Tu Trung Nguyen – MSc. Structural Engineering

EC3:2005 cl.6.2.3 (1,2)

British Standard vs Eurocode 3 – Steel Building Design Where: fu = 1.5fy (cl.3.2.2 (1)) Anet = A – (number of bolts)x(Diameter of bolts)

cl.6.2.2.2

3.Compression N Ed ≤1 N c , Rd

where Nc,Rd =

Af y yM 0

for class 1,2 and 3

cl.6.2.4

4. Bending moment M Ed ≤1 M c , Rd

cl.6.2.5

Class 1, 2

Mc,Rd = M pl , Rd =

Class 3

W pl f y yM 0

Mc,Rd = M el ,Rd =

Wel , min f y yM 0

Mc,Rd = cl.6.2.6

Shear VEd ≤1 V pl , Rd

Class 4

Av f y

where Vpl,Rd =

3 yM 0

Tu Trung Nguyen – MSc. Structural Engineering

Wel ,eff f y yM 0

British Standard vs Eurocode 3 – Steel Building Design

5.Bending and Shear Cl.6.2.8

N = (1 − ρ ) f y [W pl , y −

My,V,Rd =

ρAw 2 4t w

yM 0

where ρ = ( ]fy

≤ M c , Rd

2VEd − 1) 2 V pl , Rd

Where Aw =hwtw

6.Bending and axial force Class 1, and 2

cl.6.2.9.1

MEd < MN,Rd + For only rectangular section N MN,Rd = Mpl,Rd [1- ( Ed ) 2 ] N pl , Rd + For doubly symmetrical I and H section or other flange section to resistance about y-y NEd < 0.25Npl,Rd

Tu Trung Nguyen – MSc. Structural Engineering

British Standard vs Eurocode 3 – Steel Building Design NEd <

0.5hwt w f y yM 0

Tu Trung Nguyen – MSc. Structural Engineering

British Standard vs Eurocode 3 – Steel Building Design

7.Guide for design curved beam to Eurocode 3 Determine out of plane bending stress: 3σ B 2 M N Out-of-plane bending stress: σ 2 = 1 Direct stress σ 1 = + Wel A RT R: radius of the section, T: Thickness of flange. B is outstand of the flange 1 B = (b − t − 2r ) 2 Reduce design strength

σ σ ⎡ 2 ⎤ fyd = ⎢ f y − 3( 2 ) 2 − 3τ 2 ⎥ + 2 2 2 ⎣ ⎦ - Check axial force (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 < MN,z,Rd 0.5

N Ed N pl , Rd +na

MN,z,Rd = Mpl,z,Rd [1+ (

n=

a = (A-2btf)/A < 0.5

n−a 2 1− n ) ] = Mpl,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 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...