Sachpazis Steel Member Analysis & Design (EN1993-1!1!2005)

8/12/2019 Sachpazis Steel Member Analysis & Design (EN1993-1!1!2005)

1/6

GEODOMISI Ltd. - Dr. Costas Sachpazis Civil & Geotechnical Engineering Consulting Company for

Structural Engineering, Soil Mechanics, Rock Mechanics,Foundation Engineering & Retaining Str uctures.

Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -Mobile: (+30) 6936425722 & (+44) 7585939944,

[email protected]

Project: Steel Member Analysis & Design, In accordance with

EN1993-1-1:2005 incorporating Corrigenda February 2006and April 2009 and the recommended values.

Job Ref.

www.geodomisi.com

SectionCivil & Geotechnical Engineering Sheet no./rev. 1

Calc. by

Dr. C. Sachpazis Date

30/04/2014

Chk'd byDate App'd by Date

STEEL MEMBER DESIGN (EN1993-1-1:2005)In accordance with EN1993-1-1:2005 incorporating Corrigenda February 2006

and April 2009 and the recommended values

Section detailsSection type; UKC 305x305x240 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 , tw) = 37.7 mm

Nominal yield strength; f y = 275 N/mm 2 Nominal ultimate tensile strength; f u = 430 N/mm 2 Modulus of elasticity; E = 210000 N/mm 2

318.4

23 3 5 2

. 5

3 7

. 7

3 7

. 7


2/6



Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -Mobile: (+30) 6936425722 & (+44) 7585939944,

[email protected]



Job Ref.

www.geodomisi.com


Calc. by


30/04/2014


Partial factors - Section 6.1 Resistance of cross-sections; M0 = 1.00 Resistance of members to instability; M1 = 1.00 Resistance of tensile members to fracture; M2 = 1.25

Lateral restraintDistance between major axis restraints; L y = 4200 mmDistance between minor axis restraints; L z = 4200 mm

Effective length factorsEffective length factor in major axis; K y = 0.700

Effective length factor in minor axis; K z = 1.000

Effective length factor for torsion; K LT = 1.000

Classification of cross sections - Section 5.5 = [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 = 246.7 mm

= min([h / 2 + N Ed / (2 tw f y) - (t f + r)] / c, 1) =1.000

c / t w = 11.6


3/6



Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -Mobile: (+30) 6936425722 & (+44) 7585939944,

[email protected]



Job Ref.

www.geodomisi.com


Calc. by


30/04/2014


Shear area - cl 6.2.6(3); A v = max(2 b tf - (t w + 2 r) t f , A - (h w tw)) =24206 mm 2

Design shear resistance - cl 6.2.6(2); V c,y,Rd = V pl,y,Rd = A v (f y / [3]) / M0 = 3843.2 kNPASS - Design shear resistance exceeds design shear force

Check bending moment major (y-y) axis - Section 6.2.5Design bending moment; M y,Ed = 420 kNm

Design bending resistance moment - eq 6.13; M c,y,Rd = M pl,y,Rd = W pl.y f y / M0 = 1167.9 kNm

Slenderness ratio for lateral torsional bucklingCorrection factor - Table 6.6; k c = 0.603

C1 = 1 / k c2 = 2.75

Curvature factor; g = [1 - (I z / Iy)] = 0.827 Poissons ratio; = 0.3 Shear modulus; G = E / [2 (1 + )] = 80769 N/mm 2 Unrestrained length; L = 1.00 Lz = 4200 mmElastic critical buckling moment;

Mcr = C 1 2 E Iz / (L2 g) [Iw / Iz + L 2 G It / (2 E Iz)] = 20672.7 kNmSlenderness ratio for lateral torsional buckling; LT = [Wpl.y f y / Mcr ] = 0.238 Limiting slenderness ratio; LT,0 = 0.4

LT < LT,0 - Lateral torsional buckling can be ignored

Design resistance for buckling - Section 6.3.2.1Buckling curve - Table 6.5; b

Imperfection factor - Table 6.3; LT = 0.34 Correction factor for rolled sections; = 0.75 LTB reduction determination factor; LT = 0.5 [1 + LT ( LT - LT,0 ) + LT2] =0.494

LTB reduction factor - eq 6.57; LT = min(1 / [ LT + (LT2 - LT2)], 1, 1 / LT2) =

1.000 Modification factor; f = min(1 - 0.5 (1 - k c) [1 - 2 ( LT - 0.8) 2], 1) =0.927

Modified LTB reduction factor - eq 6.58; LT,mod = min( LT / f, 1) = 1.000 Design buckling resistance moment - eq 6.55; M b,Rd = LT,mod W pl.y f y / M1 = 1167.9 kNm

PASS - Design buckling resistance moment exceeds design bending moment

Check bending moment minor (z-z) axis - Section 6.2.5Design bending moment; M z,Ed = 110 kNm

Design bending resistance moment - eq 6.13; M c,z,Rd = M pl,z,Rd = W pl.z f y / M0 = 536.4 kNmPASS - Design bending resistance moment exceeds design bending moment


4/6



Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -Mobile: (+30) 6936425722 & (+44) 7585939944,

[email protected]



Job Ref.

www.geodomisi.com


Calc. by


30/04/2014


Biaxial bending - Section 6.2.9Plastic moment resistance (y-y); M N,y,Rd = M pl,y,Rd = 1167.9 kNmPlastic moment resistance (z-z); M N,z,Rd = M pl,z,Rd = 536.4 kNm

Normal force to plastic resistance force ratio; n = N Ed / N pl,Rd = 0.41

Parameter introducing effect of biaxial bending; _ bi = 2.00 Parameter introducing effect of biaxial bending; _ bi = max(5 n, 1) = 2.05

Interaction formula eq (6.41); (M y,Ed / MN,y,Rd ) _bi + (M z,Ed / MN,z,Rd ) _bi = 0.168 PASS - Biaxial bending check is satisfied

Check compression - Section 6.2.4Design compression force; N Ed = 3440 kN

Design resistance of section - eq 6.10; N c,Rd = N pl,Rd = A f y / M0 = 8409.2 kN

Slenderness ratio for major (y-y) axis bucklingCritical buckling length; L cr,y = L y Ky = 2940 mmCritical buckling force; N cr,y = 2 E SEC3 Iy / Lcr,y 2 = 153948.9 kNSlenderness ratio for buckling - eq 6.50; y = [A f y / N cr,y ] = 0.234

Design resistance for buckling - Section 6.3.1.1Buckling curve - Table 6.2; b

Imperfection factor - Table 6.1; y = 0.34 Buckling reduction determination factor; y = 0.5 [1 + y ( y - 0.2) + y2] = 0.533 Buckling reduction factor - eq 6.49; y = min(1 / [ y + (y2 - y2)], 1) = 0.988 Design buckling resistance - eq 6.47; N b,y,Rd = y A f y / M1 = 8308.5 kN

PASS - Design buckling resistance exceeds design compression force

Slenderness ratio for minor (z-z) axis bucklingCritical buckling length; L cr,z = L z Kz = 4200 mmCritical buckling force; N cr,z = 2 E SEC3 Iz / Lcr,z 2 = 23868.7 kNSlenderness ratio for buckling - eq 6.50;

z = [A f

y / N

cr,z] = 0.594

Design resistance for buckling - Section 6.3.1.1Buckling curve - Table 6.2; c

Imperfection factor - Table 6.1; z = 0.49 Buckling reduction determination factor; z = 0.5 [1 + z ( z - 0.2) + z2] = 0.773 Buckling reduction factor - eq 6.49; z = min(1 / [ z + (z2 - z2)], 1) = 0.789 Design buckling resistance - eq 6.47; N b,z,Rd = z A f y / M1 = 6636.5 kN

PASS - Design buckling resistance exceeds design compression force

Check torsional and torsional-flexural buckling - Section 6.3.1.4Torsional buckling length factor; K T = 1.00


5/6


6/6



Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -Mobile: (+30) 6936425722 & (+44) 7585939944,

[email protected]



Job Ref.

www.geodomisi.com


Calc. by


30/04/2014


Interaction factors k ij for members susceptible to torsional deformations -

Table B.2Characteristic moment resistance; M y,Rk = W pl.y f y = 1167.9 kNmCharacteristic moment resistance; M z,Rk = W pl.z f y = 536.4 kNmCharacteristic resistance to normal force; N Rk = A f y = 8409.2 kNInteraction factors; k yy = C my [1 + min( y - 0.2, 0.8) NEd / (y NRk /M1)] = 0.406

kzy = 1 - 0.1 max(1, z) NEd / ((C mLT - 0.25) z NRk / M1) = 0.654

kzz = C mz [1 + min(2 z - 0.6, 1.4) NEd / (z NRk / M1)] = 0.783

kyz = 0.6 kzz = 0.470 Interaction formulae - eq 6.61 & eq 6.62; N Ed / (y NRk / M1) + k yy My,Ed / (LT My,Rk / M1)

+ k yz Mz,Ed / (M z,Rk / M1) = 0.656 NEd / (z NRk / M1) + k zy My,Ed / (LT My,Rk / M1)+ k zz Mz,Ed / (M z,Rk / M1) = 0.914

PASS - Combined bending and compression checks are satisfied

Documents

Sachpazis Steel Member Analysis & Design (EN1993-1!1!2005)