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.

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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

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

An advanced UK beam S275 is to be used for this design.
F_y = 275 N/mm²
γ_m0 = 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;
W_pl,y = M_y,Edγ_m0/F_y

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

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

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Properties
h = 460.0mm; b = 191.3mm; d = 407.6mm; t_w = 9.9mm; t_f; r = 10.2mm; A = 104 cm⁴; I_y = 37100 cm⁴; I_z = 1870 cm⁴; W_el,y = 1610 cm³; W_el,y = 1830 cm³

h_w = h – 2t_f = 428.0mm

E (Modulus of elasticity) = 210000 N/mm²  (Clause 3.2.6(1))

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

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

c/t_f = 80.5/16.0 = 5.03

The limiting value for class 1 is c/t_f  ≤ 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/t_w = 407.9/9.9 = 41.17

The limiting value for class 1 is c/t_w ≤ 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))

M_Ed/M_c,Rd ≤  1.0 (clause 6.2.5(1))

M_c,Rd = M_pl,Rd = (M_pl,y × F_y)/γ_m0

M_c,Rd = M_pl,Rd = [(1830 × 275)/1.0] × 10^-3 = 503 kNm

M_Ed/M_c,Rd = 487.265/503 = 0.9687 < 1.0 Ok

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

V_Ed/V_c,Rd ≤  1.0

V_c,Rd = V_pl,Rd = A_v(F_y / √3)/γ_m0 (for class 1 sections)
For rolled I-section with shear parallel to the web, the shear area is;

A_v = A – 2bt_f + (t_w + 2r)t_f (for class 1 sections) but not less than ηh_wt_w

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

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

V_Ed/V_c,Rd = 259.875/756 = 0.343 < 1.0 Ok

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

h_w/t_w ≤  72ε/η

h_w/t_w = 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)

Q_k = 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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