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Monday, January 22, 2018

Design Example of Steel Beams According to Eurocode 3


Introduction
This post deals with the design of simply supported I-beam section subjected to permanent and variable loads according to Eurocode 3. The design involves selecting the appropriate section that will satisfy limit state requirements.

Problem Statement
It is desired to select an appropriate section to satisfy ultimate and serviceability limit state requirements for a laterally restrained simply supported beam that is subjected to the following loads;
Permanent Load Gk = 38 kN/m
Variable Load Qk = 12 kN/m

The length of the beam = 7.5m

Solution
At ultimate limit state,
PEd = 1.35Gk + 1.5Qk
PEd = 1.35(38) + 1.5(12) = 69.3 kN/m



An advanced UK beam S275 is to be used for this design.
Fy = 275 N/mm2
γm0 = 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)


The required section is supposed to have a plastic modulus about the y-y axis that is greater than;
Wpl,y = My,Edγm0/Fy

 Wpl,y = (487.265 × 103 × 1.0)/275 = 1771.872 cm3

From steel tables, try section 457 x 191 x 82        Wpl,y = 1830 cm3

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Properties
 h = 460.0mm; b = 191.3mm; d = 407.6mm; tw = 9.9mm; tf; r = 10.2mm; A = 104 cm4; Iy = 37100 cm4; Iz = 1870 cm4; Wel,y = 1610 cm3; Wel,y = 1830 cm3

hw = h - 2tf = 428.0mm

 E (Modulus of elasticity) = 210000 N/mm(Clause 3.2.6(1))

Classification of section
 ε = √(235/Fy) = √(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)

Outstand flange: flange under uniform compression c = (b - tw - 2r)/2 = [191.3 - 9.9 - 2(10.2)]/2 = 80.5mm

c/tf = 80.5/16.0 = 5.03

The limiting value for class 1 is c/tf  ≤ 9ε = 9 × 0.92
5.03 < 8.28
Therefore, outstand flange in compression is class 1

Internal Compression Part (Web under pure bending)
c = d = 407.6mm
c/tw = 407.9/9.9 = 41.17

The limiting value for class 1 is c/tw ≤ 72ε = 72 × 0.92 = 66.24
41.17 < 66.24
Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.



Member Resistance Verification
Moment Resistance
For the structure under consideration, the maximum bending moment occurs where the shear force is zeo. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2))

MEd/Mc,Rd ≤  1.0 (clause 6.2.5(1))

Mc,Rd = Mpl,Rd = (Mpl,y × Fy)/γm0

Mc,Rd = Mpl,Rd = [(1830 × 275)/1.0] × 10-3 = 503 kNm

MEd/Mc,Rd = 487.265/503 = 0.9687 < 1.0 Ok

Shear Resistance (clause 6.6.2)
The basic design requirement is;

VEd/Vc,Rd ≤  1.0

Vc,Rd = Vpl,Rd = Av(F/ √3)/γm0 (for class 1 sections)
For rolled I-section with shear parallel to the web, the shear area is;

Av = A - 2btf + (tw + 2r)tf (for class 1 sections) but not less than ηhwtw

Av = (104 × 102 - (2 × 191.3 × 16) + [9.9 + 2(10.2)] × 16 = 4763 mm2
η = 1.0 (conservative)
ηhwt= (1.0 × 428 × 9.9) = 4237 mm2
4763 > 4237
Therefore, Av = 4763 mm2

The shear resistance is therefore;
Vc,Rd = Vpl,Rd = [4763 × (275 / √3)/1.0]  × 10-3 = 756 kN

VEd/Vc,Rd = 259.875/756 = 0.343 < 1.0 Ok

Shear Buckling
Shear buckling of the unstiffnened web will not need to be considered if;

hw/t≤  72ε/η

hw/t= 428.0/9.9 = 43
72ε/η  = (72 ×  0.92)/1.0  = 66

43 < 66 Therefore shear buckling need not be considered.

Serviceability limit state
Vertical deflections are computed based on variable loads. Permanent loads need not be considered.(BS EN 1993-1-1 NA 2.23)

Qk = 12 kN/m

w = 5ql4/384EI

w = (5 × 12 × 75004)/(384 × 210000 × 37100 × 104) = 6.345mm 
Span/360 = 7500/360 = 20.833mm (BS EN 1993-1-1 NA 2.23)

 6.345mm < 20.833mm. Therefore, deflection is satisfactory

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1 comment:

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