STEEL MEMBER DESIGN (EN :2005)

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Oliver Parsons
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1 GEODOMISI Ltd. - Dr. Costas Sachpazis Consulting Company for App'd by STEEL MEMBER DESIGN (EN :2005) In accordance with EN :2005 incorporating Corrigenda February 2006 and April details type; UKC 356x406x287 (Corus Advance) Steel grade; S275 From table 3.1: Nominal values of yield strength f y and ultimate tensile strength f u for hot rolled structural steel Nominal thickness of element; t = max(t f, t w) = 36.5 mm Nominal yield strength; f y = 275 N/mm 2 Nominal ultimate tensile strength; f u = 430 N/mm 2 Modulus of elasticity; E = N/mm 2 1

2 Consulting Company for App'd by Partial factors Resistance of cross-sections; γ M0 = 1.00 Resistance of members to instability; γ M1 = 1.00 Resistance of tensile members to fracture; γ M2 = 1.25 Lateral restraint Distance between major axis restraints; Distance between minor axis restraints; L y = 5000 mm L z = 5000 mm Effective length factors Effective length factor in major axis; K y = Effective length factor in minor axis; K z = Effective length factor for torsion; K LT = Classification of cross sections ε = [235 N/mm 2 / f y] = 0.92 Internal compression parts subject to bending and compression - Table 5.2 (sheet 1 of 3) Width of section; c = d = mm α = min([h / 2 + N Ed / (2 t w f y) - (t f+ r)] / c, 1) = c / t w = 13.9 ε <= 396 ε / (13 α - 1); Class 1 Outstand flanges - Table 5.2 (sheet 2 of 3) Width of section; c = (b - t w - 2 r) / 2 = 173 mm c / t f = 5.1 ε <= 9 ε; Class 1 is class 1 Check shear Height of web; Shear area factor; η = Design shear force parallel to z axis; h w = h - 2 t f = mm h w / t w < 72 ε / η Shear buckling resistance can be ignored V z,ed = 200 kn Shear area - cl 6.2.6(3); A v = max(a - 2 b t f + (t w + 2 r) t f, η h w t w) = 9378 mm 2 Design shear resistance - cl 6.2.6(2); Design shear force parallel to y axis; V c,z,rd = V pl,z,rd = A v (f y / [3]) / γ M0 = 1489 kn PASS - Design shear resistance exceeds design shear force V y,ed = 30 kn Shear area - cl 6.2.6(3); A v = max(2 b t f - (t w + 2 r) t f, A - (h w t w)) = mm 2 Design shear resistance - cl 6.2.6(2); V c,y,rd = V pl,y,rd = A v (f y / [3]) / γ M0 = 4656 kn PASS - Design shear resistance exceeds design shear force 2

3 Consulting Company for App'd by Check bending moment major (y-y) axis Design bending moment; M y,ed = 450 knm Design bending resistance moment - eq 6.13; M c,y,rd = M pl,y,rd = W pl.y f y / γ M0 = knm Slenderness ratio for lateral torsional buckling Correction factor - Table 6.6; k c = C 1 = 1 / k 2 c = 2.75 Curvature factor; g = [1 - (I z / I y)] = Poissons ratio; ν = 0.3 Shear modulus; G = E / [2 (1 + ν)] = N/mm 2 Unrestrained length; L = 1.00 L z = 5000 mm Elastic critical buckling moment; M cr = C 1 π 2 E I z / (L 2 g) [I w / I z + L 2 G I t / (π 2 E I z)] = knm Slenderness ratio for lateral torsional buckling; λ LT = [W pl.y f y / M cr] = Limiting slenderness ratio; λ LT,0 = 0.4 λ LT < λ LT,0 - Lateral torsional buckling can be ignored Design resistance for buckling Buckling curve - Table 6.5; b Imperfection factor - Table 6.3; α LT = 0.34 Correction factor for rolled sections; β = 0.75 LTB reduction determination factor; φ LT = 0.5 [1 + α LT ( λ LT - λ LT,0) + β λ 2 LT ] = LTB reduction factor - eq 6.57; χ LT = min(1 / [φ LT + (φ 2 LT - β λ 2 LT )], 1, 1 / λ 2 LT ) = Modification factor; f = min(1-0.5 (1 - k c) [1-2 ( λ LT - 0.8) 2 ], 1) = Modified LTB reduction factor - eq 6.58; χ LT,mod = min(χ LT / f, 1) = Design buckling resistance moment - eq 6.55; M b,rd = χ LT,mod W pl.y f y / γ M1 = knm PASS - Design buckling resistance moment exceeds design bending moment Check bending moment minor (z-z) axis Design bending moment; M z,ed = 125 knm Design bending resistance moment - eq 6.13; M c,z,rd = M pl,z,rd = W pl.z f y / γ M0 = 811 knm PASS - Design bending resistance moment exceeds design bending moment Check compression Design compression force; Design resistance of section - eq 6.10; Slenderness ratio for major (y-y) axis buckling Critical buckling length; Critical buckling force; N Ed = 4500 kn N c,rd = N pl,rd = A f y / γ M0 = kn L cr,y = L y K y = 3500 mm N cr,y = π 2 E SEC3 I y / L 2 cr,y = kn 3

4 Consulting Company for App'd by Slenderness ratio for buckling - eq 6.50; λ y = [A f y / N cr,y] = Design resistance for buckling b Imperfection factor - Table 6.1; α y = 0.34 Buckling reduction determination factor; φ y = 0.5 [1 + α y ( λ y - 0.2) + λ 2 y ] = Buckling reduction factor - eq 6.49; χ y = min(1 / [φ y + (φ 2 y - λ 2 y )], 1) = N b,y,rd = χ y A f y / γ M1 = kn Slenderness ratio for minor (z-z) axis buckling Critical buckling length; L cr,z = L z K z = 5000 mm Critical buckling force; N cr,z = π 2 E SEC3 I z / L 2 cr,z = kn Slenderness ratio for buckling - eq 6.50; λ z = [A f y / N cr,z] = Design resistance for buckling c Imperfection factor - Table 6.1; α z = 0.49 Buckling reduction determination factor; φ z = 0.5 [1 + α z ( λ z - 0.2) + λ 2 z ] = Buckling reduction factor - eq 6.49; χ z = min(1 / [φ z + (φ 2 z - λ 2 z )], 1) = N b,z,rd = χ z A f y / γ M1 = kn Check torsional and torsional-flexural buckling Torsional buckling length factor; K T = 1.00 Torsional buckling length; L cr,t = max(l y, L z) K T = 5000 mm Distance from shear centre to centroid in y axis; y 0 = 0.0 mm Distance from shear centre to centroid in z axis; z 0 = 0.0 mm Radius of gyration; i 0 = [i 2 y + i 2 z ] = mm Elastic critical torsional buckling force; N cr,t = 1 / i 2 0 [G I t + π 2 E SEC3 I w / L 2 cr,t ] = kn Torsion factor; β T = 1 - (y 0 / i 0) 2 = Elastic critical torsional-flexural buckling force N cr,tf = N cr,y / (2 β T) [1 + N cr,t / N cr,y - [(1 - N cr,t / N cr,y) (y 0 / i 0) 2 N cr,t / N cr,y]] = kn Elastic critical buckling force; N cr = min(n cr,t, N cr,tf) = kn Slenderness ratio for torsional buckling - eq 6.52; λ T = [A f y / N cr] = Design resistance for buckling c Imperfection factor - Table 6.1; α T = 0.49 Buckling reduction determination factor; φ T = 0.5 [1 + α T ( λ T - 0.2) + λ 2 T ] = Buckling reduction factor - eq 6.49; χ T = min(1 / [φ T + (φ 2 T - λ 2 T )], 1) = N b,t,rd = χ T A f y / γ M1 = kn 4

5 Consulting Company for App'd by Combined bending and axial force Normal force to plastic resistance force ratio; n = N Ed / N pl,rd = 0.45 Web area to gross area ratio; a w = min((a - 2 b t f) / A, 0.5) = 0.20 Design plastic moment resistance (y-y) - eq 6.13; M pl,y,rd = W pl.y f y / γ M0 = knm Reduced plastic mnt resistance (y-y)- eq 6.36; M N,y,Rd = M pl,y,rd min((1 - n) / (1-0.5 a w), 1) = knm Design plastic moment resistance (z-z) - eq 6.13; M pl,z,rd = W pl.z f y / γ M0 = knm Reduced plastic mnt resistance (z-z) - eq 6.38; M N,z,Rd = M pl,z,rd (1 - ((n-a w) / (1- a w)) 2 ) = knm Parameter introducing effect of biaxial bending; α_bi = 2.00 Parameter introducing effect of biaxial bending; β_bi = max(5 n, 1) = 2.24 Interaction formula eq (6.41); (M y,ed / M N,y,Rd) α_bi + (M z,ed / M N,z,Rd) β_bi = PASS - Reduced bending resistance moment exceeds design bending moment Check combined bending and compression Equivalent uniform moment factors - Table B.3; C my = C mz = C mlt = Interaction factors k ij for members susceptible to torsional deformations - Table B.2 Characteristic moment resistance; M y,rk = W pl.y f y = knm Characteristic moment resistance; M z,rk = W pl.z f y = 811 knm Characteristic resistance to normal force; N Rk = A f y = kn Interaction factors; k yy = C my [1 + min( λ y - 0.2, 0.8) N Ed / (χ y N Rk / γ M1)] = k zy = max(1, λ z) N Ed / ((C mlt ) χ z N Rk / γ M1) = k zz = C mz [1 + min(2 λ z - 0.6, 1.4) N Ed / (χ z N Rk / γ M1)] = k yz = 0.6 k zz = Interaction formulae - eq 6.61 & eq 6.62; N Ed / (χ y N Rk / γ M1) + k yy M y,ed / (χ LT M y,rk / γ M1) + k yz M z,ed / (M z,rk / γ M1) = N Ed / (χ z N Rk / γ M1) + k zy M y,ed / (χ LT M y,rk / γ M1) + k zz M z,ed / (M z,rk / γ M1) = PASS - Combined bending and compression checks are satisfied 5

Lateral Torsional Buckling (sections class 1-3) - Column Item COx

Page /7 Lateral Torsional Buckling (sections class -3) - according to EN 993--:005 (EC3) section and eqn. numbers refer to this code (Form EC3-LTB_06-04-8.mcd - adopted) Profile chosen Profile "HEA40"